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Unformatted text preview: Notes on Diffy Qs
Differential Equations for Engineers by Jiˇ í Lebl
r
July 16, 2010 2 A
Typeset in LTEX. Copyright c 2008–2010 Jiˇí Lebl
r This work is licensed under the Creative Commons AttributionNoncommercialShare Alike 3.0
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During the writing of these notes, the author was in part supported by NSF grant DMS0900885.
See http://www.jirka.org/diffyqs/ for more information (including contact information). Contents
Introduction
0.1 Notes about these notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
0.2 Introduction to diﬀerential equations . . . . . . . . . . . . . . . . . . . . . . . . .
1 2 3 First order ODEs
1.1 Integrals as solutions . . . . . . . . . . .
1.2 Slope ﬁelds . . . . . . . . . . . . . . . .
1.3 Separable equations . . . . . . . . . . . .
1.4 Linear equations and the integrating factor
1.5 Substitution . . . . . . . . . . . . . . . .
1.6 Autonomous equations . . . . . . . . . .
1.7 Numerical methods: Euler’s method . . . .
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. Higher order linear ODEs
2.1 Second order linear ODEs . . . . . . . . . .
2.2 Constant coeﬃcient second order linear ODEs
2.3 Higher order linear ODEs . . . . . . . . . . .
2.4 Mechanical vibrations . . . . . . . . . . . . .
2.5 Nonhomogeneous equations . . . . . . . . .
2.6 Forced oscillations and resonance . . . . . . .
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. Systems of ODEs
3.1 Introduction to systems of ODEs . . . . . . . .
3.2 Matrices and linear systems . . . . . . . . . . .
3.3 Linear systems of ODEs . . . . . . . . . . . .
3.4 Eigenvalue method . . . . . . . . . . . . . . .
3.5 Two dimensional systems and their vector ﬁelds
3.6 Second order systems and applications . . . . .
3.7 Multiple eigenvalues . . . . . . . . . . . . . .
3.8 Matrix exponentials . . . . . . . . . . . . . . .
3.9 Nonhomogeneous systems . . . . . . . . . . .
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131 4
4 CONTENTS
Fourier series and PDEs
4.1 Boundary value problems . . . . . . . . . . . . . .
4.2 The trigonometric series . . . . . . . . . . . . . .
4.3 More on the Fourier series . . . . . . . . . . . . .
4.4 Sine and cosine series . . . . . . . . . . . . . . . .
4.5 Applications of Fourier series . . . . . . . . . . . .
4.6 PDEs, separation of variables, and the heat equation
4.7 One dimensional wave equation . . . . . . . . . .
4.8 D’Alembert solution of the wave equation . . . . .
4.9 Steady state temperature and the Laplacian . . . . . .
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. 143
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204 5 Eigenvalue problems
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5.1 SturmLiouville problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211
5.2 Application of eigenfunction series . . . . . . . . . . . . . . . . . . . . . . . . . . 219
5.3 Steady periodic solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222 6 The Laplace transform
229
6.1 The Laplace transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229
6.2 Transforms of derivatives and ODEs . . . . . . . . . . . . . . . . . . . . . . . . . 236
6.3 Convolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242 Further Reading 247 Index 249 Introduction
0.1 Notes about these notes This book originated from my class notes for teaching Math 286, diﬀerential equations, at the
University of Illinois at UrbanaChampaign in fall 2008 and spring 2009. It is a ﬁrst course on
diﬀerential equations for engineers. The standard book for the course is Edwards and Penney,
Diﬀerential Equations and Boundary Value Problems [EP], fourth edition. The structure of these
present notes, therefore, reﬂects the structure of [EP], at least as far as the material that is covered in
the course. Many examples and applications are taken more or less from this book, though they also
appear in many other sources, of course. Other books I have used as sources of information and
inspiration are E.L. Ince’s classic (and inexpensive) Ordinary Diﬀerential Equations [I], and also
my undergraduate textbooks, Stanley Farlow’s Diﬀerential Equations and Their Applications [F],
which is now available from Dover, and Berg and McGregor’s Elementary Partial Diﬀerential
Equations [BM]. See the Further Reading chapter at the end of the book.
I taught the course with the IODE software (http://www.math.uiuc.edu/iode/). IODE
is a free software package that is used either with Matlab (proprietary) or Octave (free software).
Projects and labs from the IODE website are referenced throughout the notes. They need not be
used for this course, but I think it is better to use them. The graphs in the notes were made with
the Genius software (see http://www.jirka.org/genius.html). I have used Genius in class to
show essentially these and similar graphs.
These notes are available from http://www.jirka.org/diffyqs/. Check there for any
A
possible updates or errata. The LTEX source is also available from the same site for possible
modiﬁcation and customization.
I would like to acknowledge Rick Laugesen. I have used his handwritten class notes the ﬁrst
time I taught the course. My organization of these present notes, and the choice of material covered,
is heavily inﬂuenced by his class notes. Many examples and computations are taken from his
notes. For spotting errors and other suggestions, I would also like to acknowledge (in no particular
order): John P. D’Angelo, Sean Raleigh, Jessica Robinson, Michael Angelini, Leonardo Gomes,
Jeﬀ Winegar, Ian Simon, Thomas Wicklund, Eliot Brenner, Sean Robinson, Jannett Susberry, Dana
AlQuadi, Cesar Alvarez, Cem Bagdatlioglu, Nathan Wong, Alison Shive, Shawn White, and
probably others I have forgotten. Finally I would like to acknowledge NSF grant DMS0900885.
The organization of these notes to some degree requires that they be done in order. Hence, later
5 6 INTRODUCTION chapters can be dropped. The dependence of the material covered is roughly given in the following
diagram:
Introduction
Chapter 1
Chapter 2
w ' Chapter 6 Chapter 3
w Chapter 4
Chapter 5
There are some references in chapters 4 and 5 to material from chapter 3 (some linear algebra),
but these references are not absolutely essential and can be skimmed over, so chapter 3 can safely be
dropped, while still covering chapters 4 and 5. The notes are done for two types of courses. Either
at 4 hours a week for a semester (Math 286 at UIUC):
Introduction, chapter 1 (plus the two IODE labs), chapter 2, chapter 3, chapter 4, chapter 5 (or 6).
Or a shorter version (Math 285 at UIUC) of the course at 3 hours a week for a semester:
Introduction, chapter 1 (plus the two IODE labs), chapter 2, chapter 4, (maybe chapter 5 or 6).
IODE need not be used for either version. If IODE is not used, some additional material may
need to be covered instead.
The lengths of the chapter on Laplace transform ( chapter 6) and the chapter on SturmLiouville
( chapter 5) are approximately the same and are interchangable. Laplace transform is not normally
covered at UIUC 285/286. I think it is essential that any notes on diﬀerential equations at least
mention the Laplace transform. 0.2. INTRODUCTION TO DIFFERENTIAL EQUATIONS 0.2 7 Introduction to diﬀerential equations Note: more than 1 lecture, §1.1 in [EP] 0.2.1 Diﬀerential equations The laws of physics are generally written down as diﬀerential equations. Therefore, all of science
and engineering use diﬀerential equations to some degree. Understanding diﬀerential equations is
essential to understanding almost anything you will study in your science and engineering classes.
You can think of mathematics as the language of science, and diﬀerential equations are one of
the most important parts of this language as far as science and engineering are concerned. As
an analogy, suppose that all your classes from now on were given in Swahili. Then it would be
important to ﬁrst learn Swahili, otherwise you will have a very tough time getting a good grade in
your other classes.
You have already seen many diﬀerential equations without perhaps knowing about it. And
you have even solved simple diﬀerential equations when you were taking calculus. Let us see an
example you may not have seen:
dx
+ x = 2 cos t.
(1)
dt
Here x is the dependent variable and t is the independent variable. Equation (1) is a basic example
of a diﬀerential equation. In fact it is an example of a ﬁrst order diﬀerential equation, since it
involves only the ﬁrst derivative of the dependent variable. This equation arises from Newton’s law
of cooling where the ambient temperature oscillates with time. 0.2.2 Solutions of diﬀerential equations Solving the diﬀerential equation means ﬁnding x in terms of t. That is, we want to ﬁnd a function
of t, which we will call x, such that when we plug x, t, and dx into (1), the equation holds. It is the
dt
same idea as it would be for a normal (algebraic) equation of just x and t. We claim that
x = x(t) = cos t + sin t
is a solution. How do we check? We simply plug x into equation (1)! First we need to compute
We ﬁnd that dx = − sin t + cos t. Now let us compute the left hand side of (1).
dt dx
.
dt dx
+ x = (− sin t + cos t) + (cos t + sin t) = 2 cos t.
dt
Yay! We got precisely the right hand side. But there is more! We claim x = cos t + sin t + e−t is also
a solution. Let us try,
dx
= − sin t + cos t − e−t .
dt 8 INTRODUCTION Again plugging into the left hand side of (1)
dx
+ x = (− sin t + cos t − e−t ) + (cos t + sin t + e−t ) = 2 cos t.
dt
And it works yet again!
So there can be many diﬀerent solutions. In fact, for this equation all solutions can be written in
the form
x = cos t + sin t + Ce−t
for some constant C . See Figure 1 for the graph of a few of these solutions. We will see how we
can ﬁnd these solutions a few lectures from now.
It turns out that solving diﬀerential equations
can be quite hard. There is no general method
that solves any given diﬀerential equation. We
will generally focus on how to get exact formulas
for solutions of certain diﬀerential equations, but
we will also spend a little bit of time on getting
approximate solutions.
For most of the course we will look at ordinary diﬀerential equations or ODEs, by which we
mean that there is only one independent variable
and derivatives are only with respect to this one
variable. If there are several independent variables, we will get partial diﬀerential equations or
PDEs. We will brieﬂy see these near the end of
Figure 1: Few solutions of dx + x = 2 cos t.
dt
the course.
Even for ODEs, which are very well understood, it is not a simple question of turning a crank to get answers. It is important to know when
it is easy to ﬁnd solutions and how to do so. Even if you leave much of the actual calculations
to computers in real life, you need to understand what they are doing. For example, it is often
necessary to simplify or transform your equations into something that a computer can actually
understand and solve. You may need to make certain assumptions and changes in your model to
achieve this.
To be a successful engineer or scientist, you will be required to solve problems in your job that
you have never seen before. It is important to learn problem solving techniques, so that you may
apply those techniques to new problems. A common mistake is to expect to learn some prescription
for solving all the problems you will encounter in your later career. This course is no exception.
0 1 2 3 4 5 3 3 2 2 1 1 0 0 1 1 0 0.2.3 1 2 3 4 5 Diﬀerential equations in practice So how do we use diﬀerential equations in science and engineering? First, we have some real
world problem that we wish to understand. We make some simplifying assumptions and create a 0.2. INTRODUCTION TO DIFFERENTIAL EQUATIONS 9 mathematical model. That is, we translate the real world situation into a set of diﬀerential equations.
Then we apply mathematics to get some sort of a mathematical solution. There is still something
left to do. We have to interpret the results. We have to ﬁgure out what the mathematical solution
says about the real world problem we started with.
Learning how to formulate the mathematical
Real world problem
model and how to interpret the results is what
your physics and engineering classes do. In this
abstract
interpret
course we will focus mostly on the mathematical
analysis. Sometimes we will work with simple real
solve
Mathematical
Mathematical
world examples, so that we have some intuition
model
solution
and motivation about what we are doing.
Let us look at an example of this process. One of the most basic diﬀerential equations is the
standard exponential growth model. Let P denote the population of some bacteria on a Petri dish.
Let us suppose that there is enough food and enough space. Then the rate of growth of bacteria will
be proportional to the population. I.e. a large population grows quicker. Let t denote time (say in
seconds). Our model will be
dP
= kP,
dt
for some positive constant k > 0.
Example 0.2.1: Suppose there are 100 bacteria at time 0 and 200 bacteria at time 10s. How many
bacteria will there be 1 minute from time 0 (in 60 seconds)?
First we have to solve the equation. We claim that a solution is given by
P(t) = Cekt ,
where C is a constant. Let us try.
dP
= Ckekt = kP.
dt
And it really is a solution.
OK, so what now? We do not know C and we do not know k. But we know something. We
know that P(0) = 100, and we also know that P(10) = 200. Let us plug these conditions in and see
what happens.
100 = P(0) = Cek0 = C,
200 = P(10) = 100 ek10 .
Therefore, 2 = e10k or ln 2
10 = k ≈ 0.069. So we know that
P(t) = 100 e(ln 2)t/10 ≈ 100 e0.069t . At one minute, t = 60, the population is P(60) = 6400. See Figure 2 on the next page. 10 INTRODUCTION Let us talk about the interpretation of the results. Does this mean that there must be exactly
6400 bacteria on the plate at 60s? No! We have made assumptions that might not be exactly true.
But if our assumptions are reasonable, then there will be approximately 6400 bacteria. Also note
that in real life P is a discrete quantity, not a real number. However, our model has no problem
saying that for example at 61 seconds, P(61) ≈ 6859.35.
Normally, the k in P = kP will be known, and
you will want to solve the equation for diﬀerent
initial conditions. What does that mean? Suppose
k = 1 for simplicity. Now suppose we want to
solve the equation dP = P subject to P(0) = 1000
dt
(the initial condition). Then the solution turns out
to be (exercise)
P(t) = 1000 et . 0 10 20 30 40 50 60 6000 6000 5000 5000 4000 4000 3000 3000 2000 2000 We will call P(t) = Cet the general solution,
as every solution of the equation can be written
in this form for some constant C . Then you will
need an initial condition to ﬁnd out what C is to
ﬁnd the particular solution we are looking for. Figure 2: Bacteria growth in the ﬁrst 60 secGenerally, when we say “particular solution,” we onds.
just mean some solution.
1000 1000 0 0 0 10 20 30 40 50 60 Let us get to what we will call the four fundamental equations. These appear very often and it
is useful to just memorize what their solutions are. These solutions are reasonably easy to guess
by recalling properties of exponentials, sines, and cosines. They are also simple to check, which
is something that you should always do. There is no need to wonder if you have remembered the
solution correctly.
First such equation is,
dy
= ky,
dx
for some constant k > 0. Here y is the dependent and x the independent variable. The general
solution for this equation is
y( x) = Cekx .
We have already seen that this is a solution above with diﬀerent variable names.
Next,
dy
= −ky,
dx
for some constant k > 0. The general solution for this equation is
y( x) = Ce−kx . 0.2. INTRODUCTION TO DIFFERENTIAL EQUATIONS 11 Exercise 0.2.1: Check that the y given is really a solution to the equation.
Next, take the second order diﬀerential equation
d2 y
= −k2 y,
dx2
for some constant k > 0. The general solution for this equation is
y( x) = C1 cos(kx) + C2 sin(kx).
Note that because we have a second order diﬀerential equation, we have two constants in our general
solution.
Exercise 0.2.2: Check that the y given is really a solution to the equation.
And ﬁnally, take the second order diﬀerential equation
d2 y
= k2 y,
2
dx
for some constant k > 0. The general solution for this equation is
y( x) = C1 ekx + C2 e−kx ,
or
y( x) = D1 cosh(kx) + D2 sinh(kx).
For those that do not know, cosh and sinh are deﬁned by
e x + e− x
,
2
e x − e− x
sinh x =
.
2 cosh x = These functions are sometimes easier to work with than exponentials. They have some nice familiar
d
properties such as cosh 0 = 1, sinh 0 = 0, and dx cosh x = sinh x (no that is not a typo) and
d
sinh x = cosh x.
dx
Exercise 0.2.3: Check that both forms of the y given are really solutions to the equation.
An interesting note about cosh: The graph of cosh is the exact shape a hanging chain will make
and it is called a catenary. Contrary to popular belief this is not a parabola. If you invert the graph
of cosh it is also the ideal arch for supporting its own weight. For example, the gateway arch in
Saint Louis is an inverted graph of cosh (if it were just a parabola it might fall down). This formula
is actually inscribed inside the arch:
y = −127.7 ft · cosh( x/127.7 ft) + 757.7 ft. 12 INTRODUCTION 0.2.4 Exercises Exercise 0.2.4: Show that x = e4t is a solution to x − 12 x + 48 x − 64 x = 0.
Exercise 0.2.5: Show that x = et is not a solution to x − 12 x + 48 x − 64 x = 0.
Exercise 0.2.6: Is y = sin t a solution to dy 2
dt = 1 − y2 ? Justify. Exercise 0.2.7: Let y + 2y − 8y = 0. Now try a solution of the form y = erx for some (unknown)
constant r. Is this a solution for some r? If so, ﬁnd all such r.
Exercise 0.2.8: Verify that x = Ce−2t is a solution to x = −2 x. Find C to solve for the initial
condition x(0) = 100.
Exercise 0.2.9: Verify that x = C1 e−t + C2 e2t is a solution to x − x − 2 x = 0. Find C1 and C2 to
solve for the initial conditions x(0) = 10 and x (0) = 0.
Exercise 0.2.10: Using properties of derivatives of functions that you know try to ﬁnd a solution to
( x )2 + x2 = 4.
Exercise 0.2.11: Solve:
a) dA
= −10A, A(0) = 5.
dt b) dH
= 3H, H (0) = 1.
dx c) dy
= 4y, y(0) = 0, y (0) = 1.
dx d) dx
= −9 x, x(0) = 1, x (0) = 0.
dy Chapter 1
First order ODEs
1.1 Integrals as solutions Note: 1 lecture (or less), §1.2 in [EP]
A ﬁrst order ODE is an equation of the form
dy
= f ( x, y),
dx
or just
y = f ( x, y).
In general, there is no simple formula or procedure one can follow to ﬁnd solutions. In the next few
lectures we will look at special cases where solutions are not diﬃcult to obtain. In this section, let
us assume that f is a function of x alone, that is, the equation is
y = f ( x). (1.1) We could just integrate (antidiﬀerentiate) both sides with respect to x.
y ( x) dx = f ( x) dx + C, that is
y( x) = f ( x) dx + C. This y( x) is actually the general solution. So to solve (1.1), we ﬁnd some antiderivative of f ( x) and
then we add an arbitrary constant to get the general solution.
Now is a good time to discuss a point about calculus notation and terminology. Calculus
textbooks muddy the waters by talking about integral as primarily the socalled indeﬁnite integral.
13 14 CHAPTER 1. FIRST ORDER ODES The indeﬁnite integral is really the antiderivative (in fact the whole oneparameter family of
antiderivatives). There really exists only one integral and that is the deﬁnite integral. The only
reason for the indeﬁnite integral notation is that we can always write an antiderivative as a (deﬁnite)
integral. That is, by the fundamental theorem of calculus we can always write f ( x) dx + C as
x f (t) dt + C.
x0 Hence the terminology to integrate when we may really mean to antidiﬀerentiate. Integration is
just one way to compute the antiderivative (and it is a way that always works, see the following
examples). Integration is deﬁned as the area under the graph, it only happens to also compute
antiderivatives. For sake of consistency, we will keep using the indeﬁnite integral notation when we
want an antiderivative, and you should always think of the deﬁnite integral.
Example 1.1.1: Find the general solution of y = 3 x2 .
Elementary calculus tells us that the general solution must be y = x3 + C . Let us check: y = 3 x2 .
We have gotten precisely our equation back.
Normally, we also have an initial condition such as y( x0 ) = y0 for some two numbers x0 and y0
( x0 is usually 0, but not always). We can then write the solution as a deﬁnite integral in a nice way.
Suppose our problem is y = f ( x), y( x0 ) = y0 . Then the solution is
x y( x ) = f ( s) ds + y0 . (1.2) x0 Let us check! We compute y = f ( x) (by fundamental theorem of calculus) and by Jupiter, y is a
x0
solution. Is it the one satisfying the initial condition? Well, y( x0 ) = x f ( x) dx + y0 = y0 . It is!
0
Do note that the deﬁnite integral and the indeﬁnite integral (antidiﬀerentiation) are completely
diﬀerent beasts. The deﬁnite integral always evaluates to a number. Therefore, (1.2) is a formula
we can plug into the calculator or a computer, and it will be happy to calculate speciﬁc values for us.
We will easily be able to plot the solution and work with it just like with any other function. It is not
so crucial to always ﬁnd a closed form for the antiderivative.
Example 1.1.2: Solve 2 y = e− x , y(0) = 1. By the preceding discussion, the solution must be
x y( x) = 2 e− s ds + 1.
0 Here is a good way to make fun of your friends taking second semester calculus. Tell them to ﬁnd
the closed form solution. Ha ha ha (bad math joke). It is not possible (in closed form). There is
absolutely nothing wrong with writing the solution as a deﬁnite integral. This particular integral is
in fact very important in statistics. 1.1. INTEGRALS AS SOLUTIONS 15 Using this method, we can also solve equations of the form
y = f (y).
Let us write the equation in Leibniz notation.
dy
= f (y).
dx
Now we use the inverse function theorem from calculus to switch the roles of x and y to obtain
dx
1
=
.
dy
f (y)
What we are doing seems like algebra with dx and dy. It is tempting to just do algebra with dx
and dy as if they were numbers. And in this case it does work. Be careful, however, as this sort of
handwaving calculation can lead to trouble, especially when more than one independent variable is
involved. At this point we can simply integrate
1
dy + C.
f (y) x(y) =
Finally, we try to solve for y. Example 1.1.3: Previously, we guessed y = ky (for some k > 0) has solution Cekx . We can actually
do it now. First we note that y = 0 is a solution. Henceforth, we assume y 0. We write
1
dx
=.
dy ky
We integrate to obtain
1
ln y + D,
k
where D is an arbitrary constant. Now we solve for y (actually for y).
x(y) = x = y = ekx−kD = e−kD ekx .
If we replace e−kD with an arbitrary constant C we can get rid of the absolute value bars (we can
do this as D was arbitrary). In this way, we also incorporate the solution y = 0. We get the same
general solution as we guessed before, y = Cekx .
Example 1.1.4: Find the general solution of y = y2 .
First we note that y = 0 is a solution. We can now assume that y
dx
1
= 2.
dy y 0. Write 16 CHAPTER 1. FIRST ORDER ODES We integrate to get
−1
+ C.
y
We solve for y = C1 x . So the general solution is
−
x= 1
or
y = 0.
C−x
Note the singularities of the solution. If for example C = 1, then the solution “blows up” as we
approach x = 1. Generally, it is hard to tell from just looking at the equation itself how the solution
is going to behave. The equation y = y2 is very nice and deﬁned everywhere, but the solution is
only deﬁned on some interval (−∞, C ) or (C, ∞).
y= Classical problems leading to diﬀerential equations solvable by integration are problems dealing
with velocity, acceleration and distance. You have surely seen these problems before in your
calculus class.
Example 1.1.5: Suppose a car drives at a speed et/2 meters per second, where t is time in seconds.
How far did the car get in 2 seconds (starting at t = 0)? How far in 10 seconds?
Let x denote the distance the car traveled. The equation is
x = et/2 .
We can just integrate this equation to get that
x(t) = 2et/2 + C.
We still need to ﬁgure out C . We know that when t = 0, then x = 0. That is, x(0) = 0. So
0 = x(0) = 2e0/2 + C = 2 + C.
Thus C = −2 and x(t) = et/2 − 2. Now we just plug in to get where the car is at 2 and at 10 seconds. We obtain
x(2) = 2e2/2 − 2 ≈ 3.44 meters, x(10) = 2e10/2 − 2 ≈ 294 meters. Example 1.1.6: Suppose that the car accelerates at a rate of t2 m/s2 . At time t = 0 the car is at the 1
meter mark and is traveling at 10 m/s. Where is the car at time t = 10.
Well this is actually a second order problem. If x is the distance traveled, then x is the velocity,
and x is the acceleration. The equation with initial conditions is
x = t2 , x(0) = 1, x (0) = 10. Well, what if we call x = v, and then we have the problem
v = t2 , v(0) = 10. Once we solve for v, we can then integrate and ﬁnd x.
Exercise 1.1.1: Solve for v and then solve for x. Then ﬁnd x(10) to answer the question. 1.1. INTEGRALS AS SOLUTIONS 1.1.1 17 Exercises Exercise 1.1.2: Solve dy
dx = x2 + x for y(1) = 3. Exercise 1.1.3: Solve dy
dx = sin(5 x) for y(0) = 2. Exercise 1.1.4: Solve dy
dx = 1
x 2 −1 for y(0) = 0. Exercise 1.1.5: Solve y = y3 for y(0) = 1.
Exercise 1.1.6 (challenging): Solve y = (y − 1)(y + 1) for y(0) = 3.
Exercise 1.1.7: Solve dy
dx = 1
y +1 for y(0) = 0. Exercise 1.1.8: Solve y = sin x for y(0) = 0.
Exercise 1.1.9: A spaceship is traveling at the speed 2t2 + 1 km/s (t is time in seconds). It is pointing
directly away from earth and at time t = 0 it is 1000 kilometers from earth. How far from earth is it
at one minute from time t = 0?
Exercise 1.1.10: Solve
integral. dx
dt = sin(t2 ) + t, x(0) = 20. It is OK to leave your answer as a deﬁnite 18 CHAPTER 1. FIRST ORDER ODES 1.2 Slope ﬁelds Note: 1 lecture, §1.3 in [EP]
At this point it may be good to ﬁrst try the Lab I and/or Project I from the IODE website:
http://www.math.uiuc.edu/iode/.
As we said, the general ﬁrst order equation we are studying looks like
y = f ( x, y).
In general, we cannot simply solve these kinds of equations explicitly. It would be good if we could
at least ﬁgure out the shape and behavior of the solutions, or if we could even ﬁnd approximate
solutions for any equation. 1.2.1 Slope ﬁelds As you have seen in IODE Lab I (if you did it), the equation y = f ( x, y) gives you a slope at each
point in the ( x, y)plane. We can plot the slope at lots of points as a short line through the point
( x, y) with the slope f ( x, y). See Figure 1.1.
3 2 1 0 1 2 3 3 2 1 0 1 2 3 3 3 3 3 2 2 2 2 1 1 1 1 0 0 0 0 1 1 1 1 2 2 2 2 3 3 3
3 2 1 0 1 2 Figure 1.1: Slope ﬁeld of y = xy. 3 3
3 2 1 0 1 2 3 Figure 1.2: Slope ﬁeld of y = xy with a graph
of solutions satisfying y(0) = 0.2, y(0) = 0, and
y(0) = −0.2. We call this picture the slope ﬁeld of the equation. If we are given a speciﬁc initial condition
y( x0 ) = y0 , we can look at the location ( x0 , y0 ) and follow the slopes. See Figure 1.2.
By looking at the slope ﬁeld we can get a lot of information about the behavior of solutions. For
example, in Figure 1.2 we can see what the solutions do when the initial conditions are y(0) > 0,
y(0) = 0 and y(0) < 0. Note that a small change in the initial condition causes quite diﬀerent 1.2. SLOPE FIELDS 19 behavior. On the other hand, plotting a few solutions of the equation y = −y, we see that no matter
what y(0) is, all solutions tend to zero as x tends to inﬁnity. See Figure 1.3.
3 2 1 0 1 2 3 3 3 2 2 1 1 0 0 1 1 2 2 3 3
3 2 1 0 1 2 3 Figure 1.3: Slope ﬁeld of y = −y with a graph of a few solutions. 1.2.2 Existence and uniqueness We wish to ask two fundamental questions about the problem
y = f ( x, y), y( x0 ) = y0 . (i) Does a solution exist?
(ii) Is the solution unique (if it exists)?
What do you think is the answer? The answer seems to be yes to both does it not? Well, pretty
much. But there are cases when the answer to either question can be no.
Since generally the equations we encounter in applications come from real life situations, it
seems logical that a solution always exists. It also has to be unique if we believe our universe is
deterministic. If the solution does not exist, or if it is not unique, we have probably not devised the
correct model. Hence, it is good to know when things go wrong and why.
Example 1.2.1: Attempt to solve:
1
y= ,
x y(0) = 0. Integrate to ﬁnd the general solution y = ln  x + C . Note that the solution does not exist at x = 0.
See Figure 1.4 on the next page. 20 CHAPTER 1. FIRST ORDER ODES 3 2 1 0 1 2 3 3 2 1 0 1 2 3 3 3 3 3 2 2 2 2 1 1 1 1 0 0 0 0 1 1 1 1 2 2 2 2 3 3 3
3 2 1 0 1 2 3 Figure 1.4: Slope ﬁeld of y = 1/x. 3
3 2 1 0 1 2 3 Figure 1.5: Slope ﬁeld of y = 2 y with two
solutions satisfying y(0) = 0. Example 1.2.2: Solve:
y = 2 y, y(0) = 0. See Figure 1.5. Note that y = 0 is a solution. But also the function x2 if x ≥ 0, y( x) = 2
− x if x < 0.
It is actually hard to tell from the slope ﬁeld that the solution will not be unique. Is there any
hope? Of course there is. It turns out that the following theorem is true. It is known as Picard’s
theorem∗ .
Theorem 1.2.1 (Picard’s theorem on existence and uniqueness). If f ( x, y) is continuous (as a
f
function of two variables) and ∂ y exists and is continuous near some ( x0 , y0 ), then a solution to
∂
y = f ( x, y), y( x0 ) = y0 , exists (at least for some small interval of x’s) and is unique.
Note that the problems y = 1/x, y(0) = 0 and y = 2 y, y(0) = 0 do not satisfy the hypothesis
of the theorem. Even if we can use the theorem, we ought to be careful about this existence business.
It is quite possible that the solution only exists for a short while.
Example 1.2.3: For some constant A, solve:
y = y2 ,
∗ y(0) = A. Named after the French mathematician Charles Émile Picard (1856 – 1941) 1.2. SLOPE FIELDS 21 We know how to solve this equation. First assume that A 0, so y is not equal to zero at least
for some x near 0. So x = 1/y2 , so x = −1/y + C , so y = C1 x . If y(0) = A, then C = 1/A so
−
y= 1
.
1/A − x If A = 0, then y = 0 is a solution.
For example, when A = 1 the solution “blows up” at x = 1. Hence, the solution does not exist
for all x even if the equation is nice everywhere. The equation y = y2 certainly looks nice.
For the most of this course we will be interested in equations where existence and uniqueness
holds, and in fact holds “globally” unlike for the equation y = y2 . 1.2.3 Exercises Exercise 1.2.1: Sketch direction ﬁeld for y = e x−y . How do the solutions behave as x grows? Can
you guess a particular solution by looking at the direction ﬁeld?
Exercise 1.2.2: Sketch direction ﬁeld for y = x2 .
Exercise 1.2.3: Sketch direction ﬁeld for y = y2 .
Exercise 1.2.4: Is it possible to solve the equation y = xy
cos x for y(0) = 1? Justify. √
Exercise 1.2.5: Is it possible to solve the equation y = y  x for y(0) = 0? Is the solution unique?
Justify. 22 1.3 CHAPTER 1. FIRST ORDER ODES Separable equations Note: 1 lecture, §1.4 in [EP]
When a diﬀerential equation is of the form y = f ( x), we can just integrate: y = f ( x) dx + C .
Unfortunately this method no longer works for the general form of the equation y = f ( x, y).
Integrating both sides yields
y= f ( x, y) dx + C. Notice dependence on y in the integral. 1.3.1 Separable equations Let us suppose that the equation is separable. That is, let us consider
y = f ( x)g(y),
for some functions f ( x) and g(y). Let us write the equation in the Leibniz notation
dy
= f ( x)g(y).
dx
Then we rewrite the equation as
dy
= f ( x) dx.
g(y)
Now both sides look like something we can integrate. We obtain
dy
=
g(y) f ( x) dx + C. If we can ﬁnd closed form expressions for these two integrals, we can, perhaps, solve for y.
Example 1.3.1: Take the equation
y = xy.
First note that y = 0 is a solution, so assume y 0 from now on. Write the equation as dy
=
y x dx + C. We compute the antiderivatives to get
ln y = x2
+ C.
2 dy
dx = xy, then 1.3. SEPARABLE EQUATIONS
Or 23
x2 x2 x2 y = e 2 +C = e 2 eC = De 2 , where D > 0 is some constant. Because y = 0 is a solution and because of the absolute value we
actually can write:
x2
y = De 2 ,
for any number D (including zero or negative).
We check:
x2
x2
y = Dxe 2 = x De 2 = xy.
Yay!
We should be a little bit more careful with this method. You may be worried that we were
integrating in two diﬀerent variables. We seemed to be doing a diﬀerent operation to each side. Let
us work this method out more rigorously.
dy
= f ( x)g(y)
dx
We rewrite the equation as follows. Note that y = y( x) is a function of x and so is dy
!
dx 1 dy
= f ( x)
g(y) dx
We integrate both sides with respect to x.
1 dy
dx =
g(y) dx f ( x) dx + C. We can use the change of variables formula.
1
dy =
g(y) f ( x) dx + C. And we are done. 1.3.2 Implicit solutions It is clear that we might sometimes get stuck even if we can do the integration. For example, take
the separable equation
xy
y= 2
.
y +1
We separate variables
y2 + 1
1
dy = y +
dy = x dx.
y
y 24 CHAPTER 1. FIRST ORDER ODES Now we integrate to get
x2
y2
+ ln y =
+ C.
2
2
Or maybe the easier looking expression (where D = 2C ):
y2 + 2 ln y = x2 + D.
It is not easy to ﬁnd the solution explicitly as it is hard to solve for y. We will, therefore, call this
solution an implicit solution. It is easy to check that implicit solutions still satisfy the diﬀerential
equation. In this case, we diﬀerentiate to get
y 2y + 2
= 2 x.
y It is simple to see that the diﬀerential equation holds. If you want to compute values for y you
might have to be tricky. For example, you can graph x as a function of y, and then ﬂip your paper.
Computers are also good at some of these tricks, but you have to be careful.
We note above that the equation also has a solution y = 0. In this case, it turns out that the
general solution is y2 + 2 ln y = x2 + C together with y = 0. These outlying solutions such as y = 0
are sometimes called singular solutions. 1.3.3 Examples Example 1.3.2: Solve x2 y = 1 − x2 + y2 − x2 y2 , y(1) = 0.
First factor the right hand side to obtain
x2 y = (1 − x2 )(1 + y2 ).
We separate variables, integrate and solve for y
1 − x2
y
=
,
1 + y2
x2
1
y
= 2 − 1,
2
1+y
x
−1
arctan(y) =
− x + C,
x
−1
y = tan
− x+C .
x
Now solve for the initial condition, 0 = tan(−2 + C ) to get C = 2 (or 2 + π, etc. . . ). The solution we
are seeking is, therefore,
−1
y = tan
−x+2 .
x 1.3. SEPARABLE EQUATIONS 25 Example 1.3.3: Suppose Bob made a cup of coﬀee, and the water was boiling (100 degrees Celsius)
at time t = 0 minutes. Suppose Bob likes to drink his coﬀee at 70 degrees. Let the ambient (room)
temperature be 26 degrees. Furthermore, suppose Bob measured the temperature of the coﬀee at 1
minute and found that it dropped to 95 degrees. When should Bob start drinking?
Let T be the temperature of coﬀee, let A be the ambient (room) temperature. Then for some k
the temperature of coﬀee is:
dT
= k ( A − T ).
dt
For our setup A = 26, T (0) = 100, T (1) = 95. We separate variables and integrate (C and D will
denote arbitrary constants)
1 dT
= −k ,
T − A dt
ln(T − A) = −kt + C, (note that T − A > 0)
T − A = D e−kt ,
T = D e−kt + A.
That is, T = 26 + D e−kt . We plug in the ﬁrst condition 100 = T (0) = 26 + D and hence D = 74.
Now we have T = 26 + 74 e−kt . We plug in 95 = T (1) = 26 + 74 e−k . Solving for k we get
−
k = − ln 957426 ≈ 0.07. Now we solve for the time t that gives us a temperature of 70 degrees. That
ln 70−26 74
is, we solve 70 = 26 + 74e−0.07t to get t = − 0.07 ≈ 7.43 minutes. So Bob can begin to drink the
coﬀee at about 7 and a half minutes from the time Bob made it. Probably about the amount of time
it took us to calculate how long it would take.
2 Example 1.3.4: Find the general solution to y = − xy (including singular solutions).
3
First note that y = 0 is a solution (a singular solution). So assume that y 0 and write
−3
y = x,
y2
3 x2
=
+ C,
y
2
3
6
y= 2
=2
.
x /2 + C
x + 2C 1.3.4 Exercises Exercise 1.3.1: Solve y = x/y.
Exercise 1.3.2: Solve y = x2 y.
Exercise 1.3.3: Solve dx
= ( x2 − 1) t, for x(0) = 0.
dt 26 CHAPTER 1. FIRST ORDER ODES Exercise 1.3.4: Solve dx
= x sin(t), for x(0) = 1.
dt Exercise 1.3.5: Solve dy
= xy + x + y + 1. Hint: Factor the right hand side.
dx Exercise 1.3.6: Solve xy = y + 2 x2 y, where y(1) = 1.
Exercise 1.3.7: Solve dy y2 + 1
=
, for y(0) = 1.
dx x2 + 1 Exercise 1.3.8: Find an implicit solution for dy x2 + 1
=
, for y(0) = 1.
dx y2 + 1 Exercise 1.3.9: Find explicit solution for y = xe−y , y(0) = 1.
Exercise 1.3.10: Find explicit solution for xy = e−y , for y(1) = 1.
Exercise 1.3.11: Find explicit solution for y = ye− x , y(0) = 1. It is alright to leave a deﬁnite
integral in your answer.
2 Exercise 1.3.12: Suppose a cup of coﬀee is at 100 degrees Celsius at time t = 0, it is at 70 degrees
at t = 10 minutes, and it is at 50 degrees at t = 20 minutes. Compute the ambient temperature. 1.4. LINEAR EQUATIONS AND THE INTEGRATING FACTOR 1.4 27 Linear equations and the integrating factor Note: 1 lecture, §1.5 in [EP]
One of the most important types of equations we will learn how to solve are the socalled linear
equations. In fact, the majority of the course will focus on linear equations. In this lecture we will
focus on the ﬁrst order linear equation. A ﬁrst order equation is linear if we can put it into the
following form:
y + p( x)y = f ( x).
(1.3)
The word “linear” here means linear in y. The dependence on x can be more complicated.
Solutions of linear equations have nice properties. For example, the solution exists wherever
p( x) and f ( x) are deﬁned, and has the same regularity (read: it is just as nice). But most importantly
for us right now, there is a method for solving linear ﬁrst order equations.
First we ﬁnd a function r( x) such that
r( x)y + r( x) p( x)y = d
r( x)y .
dx Then we can multiply both sides of (1.3) by r( x) to obtain
d
r( x)y = r( x) f ( x).
dx
Now we integrate both sides. The right hand side does not depend on y and the left hand side is
written as a derivative of a function. Afterwards, we can solve for y. The function r( x) is called the
integrating factor and the method is called the integrating factor method.
We are looking for a function r( x) such that if we diﬀerentiate it, we get the same function back
multiplied by p( x). That seems like a job for the exponential function! Let
r ( x) = e p( x)dx We compute:
y + p( x)y = f ( x),
e p( x)dx y +e
d
e
dx p( x)dx e p( x)y = e p( x)dx f ( x), y =e p( x)dx f ( x), p( x)dx p( x)dx y=
y = e− e p( x)dx p( x)dx f ( x) dx + C,
e p( x)dx f ( x) dx + C . Of course, to get a closed form formula for y we need to be able to ﬁnd a closed form formula
for the integrals appearing above. 28 CHAPTER 1. FIRST ORDER ODES Example 1.4.1: Solve
2 y + 2 xy = e x− x , y(0) = −1. First note that p( x) = 2 x and f ( x) = e x− x . The integrating factor is r( x) = e
multiply both sides of the equation by r( x) to get
2 2 2 2 p( x) dx = e x . We
2 2 e x y + 2 xe x y = e x− x e x ,
d x2
e y = ex .
dx
We integrate
2 e x y = e x + C,
2 2 y = e x− x + Ce− x .
Next, we solve for the initial condition −1 = y(0) = 1 + C , so C = −2. The solution is
2 2 y = e x− x − 2e− x .
Note that we do not care which antiderivative we take when computing e
add a constant of integration, but those constants will not matter in the end. p( x)dx . You can always Exercise 1.4.1: Try it! Add a constant of integration to the integral in the integrating factor and
show that the solution you get in the end is the same as what we got above.
An advice: Do not try to remember the formula itself, that is way too hard. It is easier to
remember the process and repeat it.
Since we cannot always evaluate the integrals in closed form, it is useful to know how to write
the solution in deﬁnite integral form. A deﬁnite integral is something that you can plug into a
computer or a calculator. Suppose we are given
y + p( x)y = f ( x), y( x0 ) = y0 . Look at the solution and write the integrals as deﬁnite integrals.
− y( x) = e x
x0 p( s) ds x t e x0 p( s) ds f (t) dt + y0 . (1.4) x0 You should be careful to properly use dummy variables here. If you now plug such a formula into a
computer or a calculator, it will be happy to give you numerical answers.
Exercise 1.4.2: Check that y( x0 ) = y0 in formula (1.4). 1.4. LINEAR EQUATIONS AND THE INTEGRATING FACTOR 29 Exercise 1.4.3: Write the solution of the following problem as a deﬁnite integral, but try to simplify
as far as you can. You will not be able to ﬁnd the solution in closed form.
2 y + y = ex −x , y(0) = 10. Example 1.4.2: The following is a simple application of linear equations and this type of a problem
is used often in real life. For example, linear equations are used in ﬁguring out the concentration of
chemicals in bodies of water.
A 100 liter tank contains 10 kilograms of salt dissolved in 60 liters of water. Solution of water
and salt (brine) with concentration of 0.1 kilograms per liter is ﬂowing in at the rate of 5 liters a
minute. The solution in the tank is well stirred and ﬂows out at a rate of 3 liters a minute. How
much salt is in the tank when the tank is full?
Let us come up with the equation. Let x denote the kilograms of salt in the tank, let t denote the
time in minutes. Then for a small change ∆t in time, the change in x (denoted ∆ x) is approximately
∆ x ≈ (rate in × concentration in)∆t − (rate out × concentration out)∆t.
Dividing through by ∆t and taking the limit ∆t → 0 we see that
dx
= (rate in × concentration in) − (rate out × concentration out).
dt
In our example, we have
rate in = 5,
concentration in = 0.1,
rate out = 3,
x
x
concentration out =
=
.
volume 60 + (5 − 3)t
Our equation is, therefore,
dx
x
= (5 × 0.1) − 3
.
dt
60 + 2t
Or in the form (1.3)
dx
3
+
x = 0.5.
dt 60 + 2t
Let us solve. The integrating factor is
r(t) = exp 3
3
dt = exp ln(60 + 2t) = (60 + 2t)3/2 .
60 + 2t
2 30 CHAPTER 1. FIRST ORDER ODES
We multiply both sides of the equation to get
(60 + 2t)3/2 dx
3
+ (60 + 2t)3/2
x = 0.5(60 + 2t)3/2 ,
dt
60 + 2t
d
(60 + 2t)3/2 x = 0.5(60 + 2t)3/2 ,
dt
(60 + 2t)3/2 x = 0.5(60 + 2t)3/2 dt + C,
(60 + 2t)3/2
dt + C (60 + 2t)−3/2 ,
2 x = (60 + 2t)−3/2
x = (60 + 2t)−3/2
x= 1
(60 + 2t)5/2 + C (60 + 2t)−3/2 ,
10 60 + 2t
+ C (60 + 2t)−3/2 .
10 We need to ﬁnd C . We know that at t = 0, x = 10. So
10 = x(0) =
or 60
+ C (60)−3/2 = 6 + C (60)−3/2 ,
10 C = 4(603/2 ) ≈ 1859.03. We are interested in x when the tank is full. So we note that the tank is full when 60 + 2t = 100,
or when t = 20. So
x(20) = 60 + 40
+ C (60 + 40)−3/2 ≈ 10 + 1859.03(100)−3/2 ≈ 11.86.
10 The concentration at the end is approximately 0.1186 kg/liter and we started with 1/6 or 0.167 kg/liter. 1.4.1 Exercises In the exercises, feel free to leave answer as a deﬁnite integral if a closed form solution cannot be
found. If you can ﬁnd a closed form solution, you should give that.
Exercise 1.4.4: Solve y + xy = x.
Exercise 1.4.5: Solve y + 6y = e x .
Exercise 1.4.6: Solve y + 3 x2 y = sin( x) e− x , with y(0) = 1.
3 Exercise 1.4.7: Solve y + cos( x)y = cos( x).
Exercise 1.4.8: Solve 1
x 2 +1 y + xy = 3, with y(0) = 0. 1.4. LINEAR EQUATIONS AND THE INTEGRATING FACTOR 31 Exercise 1.4.9: Suppose there are two lakes located on a stream. Clean water ﬂows into the ﬁrst
lake, then the water from the ﬁrst lake ﬂows into the second lake, and then water from the second
lake ﬂows further downstream. The in and out ﬂow from each lake is 500 liters per hour. The ﬁrst
lake contains 100 thousand liters of water and the second lake contains 200 thousand liters of water.
A truck with 500 kg of toxic substance crashes into the ﬁrst lake. Assume that the water is being
continually mixed perfectly by the stream. a) Find the concentration of toxic substance as a function
of time (in seconds) in both lakes. b) When will the concentration in the ﬁrst lake be below 0.001 kg
per liter. c) When will the concentration in the second lake be maximal.
Exercise 1.4.10: Newton’s law of cooling states that dx = −k( x − A) where x is the temperature,
dt
t is time, A is the ambient temperature, and k > 0 is a constant. Suppose that A = A0 cos(ωt) for
some constants A0 and ω. That is, the ambient temperature oscillates (for example night and day
temperatures). a) Find the general solution. b) In the long term, will the initial conditions make
much of a diﬀerence? Why or why not.
Exercise 1.4.11: Initially 5 grams of salt are dissolved in 20 liters of water. Brine with concentration
of salt 2 grams of salt per liter is added at a rate of 3 liters a minute. The tank is mixed well and is
drained at 3 liters a minute. How long does the process have to continue until there are 20 grams of
salt in the tank?
Exercise 1.4.12: Initially a tank contains 10 liters of pure water. Brine of unknown concentration of
salt is ﬂowing in at 1 liter per minute. The water is mixed well and drained at 1 liter per minute. In
20 minutes there are 15 grams of salt in the tank. What is the concentration of salt in the incoming
brine? 32 1.5 CHAPTER 1. FIRST ORDER ODES Substitution Note: 1 lecture, §1.6 in [EP]
Just like when solving integrals, one method to try is to change variables to end up with a
simpler equation to solve. 1.5.1 Substitution The equation
y = ( x − y + 1)2 .
is neither separable nor linear. What can we do? How about trying to change variables, so that in
the new variables the equation is simpler. We will use another variable v, which we will treat as a
function of x. Let us try
v = x − y + 1.
We need to ﬁgure out y in terms of v , v and x. We diﬀerentiate (in x) to obtain v = 1 − y . So
y = 1 − v . We plug this into the equation to get
1 − v = v2 .
In other words, v = 1 − v2 . Such an equation we know how to solve.
1
dv = dx.
1 − v2
So
1
v+1
= x + C,
ln
2
v−1
v+1
= e2 x+2C ,
v−1
or v+1 = De2 x for some constant D. Note that v = 1 and v = −1 are also solutions.
v −1
Now we need to “unsubstitute” to obtain
x−y+2
= De2 x ,
x−y
and also the two solutions x − y + 1 = 1 or y = x, and x − y + 1 = −1 or y = x + 2. We solve the ﬁrst
equation for y.
x − y + 2 = ( x − y)De2 x ,
x − y + 2 = Dxe2 x − yDe2 x ,
−y + yDe2 x = Dxe2 x − x − 2,
y (−1 + De2 x ) = Dxe2 x − x − 2,
y= Dxe2 x − x − 2
.
De2 x − 1 1.5. SUBSTITUTION 33 Note that D = 0 gives y = x + 2, but no value of D gives the solution y = x.
Substitution in diﬀerential equations is applied in much the same way that it is applied in
calculus. You guess. Several diﬀerent substitutions might work. There are some general things to
look for. We summarize a few of these in a table.
When you see Try substituting yy
y2 y
(cos y)y
(sin y)y
y ey y2
y3
sin y
cos y
ey Usually you try to substitute in the “most complicated” part of the equation with the hopes of
simplifying it. The above table is just a rule of thumb. You might have to modify your guesses. If a
substitution does not work (it does not make the equation any simpler), try a diﬀerent one. 1.5.2 Bernoulli equations There are some forms of equations where there is a general rule for substitution that always works.
One such example is the socalled Bernoulli equation† .
y + p( x)y = q( x)yn .
This equation looks a lot like a linear equation except for the yn . If n = 0 or n = 1, then the equation
is linear and we can solve it. Otherwise, a change of coordinates v = y1−n transforms the Bernoulli
equation into a linear equation. Note that n need not be an integer.
Example 1.5.1: Solve
xy + y( x + 1) + xy5 = 0, y(1) = 1. First, the equation is Bernoulli ( p( x) = ( x + 1)/ x and q( x) = −1). We substitute
v = y1−5 = y−4 , v = −4y−5 y . In other words, (−1/4) y5 v = y . So
xy + y( x + 1) + xy5 = 0,
− xy5
v + y( x + 1) + xy5 = 0,
4
−x
v + y−4 ( x + 1) + x = 0,
4
−x
v + v( x + 1) + x = 0,
4
† There are several things called Bernoulli equations, this is just one of them. The Bernoullis were a prominent Swiss
family of mathematicians. These particular equations are named for Jacob Bernoulli (1654 – 1705). 34 CHAPTER 1. FIRST ORDER ODES and ﬁnally 4( x + 1)
v = 4.
x
Now the equation is linear. We can use the integrating factor method. In particular, we will use
formula (1.4). Let us assume that x > 0 so  x = x. This assumption is OK, as our initial condition
is x = 1. Let us compute the integrating factor. Here p( s) from formula (1.4) is −4( s+1) .
s
v− x e1 x
p( s) ds = exp
1 − e x
1 p( s) ds e−4 x+4
−4( s + 1)
ds = e−4 x−4 ln( x)+4 = e−4 x+4 x−4 =
,
s
x4 4 x+4 ln( x)−4 =e = e4 x−4 x4 . We now plug in to (1.4)
v( x) = e− x
1 x t e 1 p( s) ds 4 dt + 1 p( s) ds
1
x = e4 x−4 x4 4
1 e−4t+4
dt + 1 .
t4 Note that the integral in this expression is not possible to ﬁnd in closed form. As we said before, it
is perfectly ﬁne to have a deﬁnite integral in our solution. Now “unsubstitute”
x y−4 = e4 x−4 x4 4
1 e − x +1 y=
x4 1.5.3 e−4t+4
dt + 1 ,
t4 x e−4t+4
t4
1 dt + 1 1/4 . Homogeneous equations Another type of equations we can solve by substitution are the socalled homogeneous equations.
Suppose that we can write the diﬀerential equation as
y =F y
.
x Here we try the substitutions
v= y
x and therefore y = v + xv . We note that the equation is transformed into
v + xv = F (v) or xv = F (v) − v or v
1
=.
F (v) − v x 1.5. SUBSTITUTION 35 Hence an implicit solution is
1
dv = ln  x + C.
F (v) − v
Example 1.5.2: Solve
x2 y = y2 + xy, y(1) = 1. We put the equation into the form y = (y/x) + y/x. Now we substitute v = y/x to get the separable
equation
xv = v2 + v − v = v2 ,
2 which has a solution
1
dv = ln  x + C,
v2
−1
= ln  x + C,
v
−1
v=
.
ln  x + C
We unsubstitute
−1
y
=
,
x ln  x + C
−x
y=
.
ln  x + C
We want y(1) = 1, so −1
−1
=
.
ln 1 + C
C
Thus C = −1 and the solution we are looking for is
1 = y(1) = y= 1.5.4
−x
.
ln  x − 1 Exercises Exercise 1.5.1: Solve y + y( x2 − 1) + xy6 = 0, with y(1) = 1.
Exercise 1.5.2: Solve 2yy + 1 = y2 + x, with y(0) = 1.
Exercise 1.5.3: Solve y + xy = y4 , with y(0) = 1.
Exercise 1.5.4: Solve yy + x = x2 + y2 . Exercise 1.5.5: Solve y = ( x + y − 1)2 .
Exercise 1.5.6: Solve y = x2 −y2
,
xy with y(1) = 2. 36 CHAPTER 1. FIRST ORDER ODES 1.6 Autonomous equations Note: 1 lecture, §2.2 in [EP]
Let us consider problems of the form
dx
= f ( x),
dt
where the derivative of solutions depends only on x (the dependent variable). These types of
equations are called autonomous equations. If we think of t as time, the naming comes from the
fact that the equation is independent of time.
Let us come back to the cooling coﬀee problem. Newton’s law of cooling says that
dx
= −k( x − A),
dt
where x is the temperature, t is time, k is some constant and A is the ambient temperature. See Figure 1.6 for an example with k = 0.3 and A = 5.
Note the solution x = A (in the example x = 5). We call these types of solutions the equilibrium
solutions. The points on the x axis where f ( x) = 0 are called critical points. The point x = A is
a critical point. In fact, each critical point corresponds to an equilibrium solution. Note also, by
looking at the graph, that the solution x = A is “stable” in that small perturbations in x do not lead
to substantially diﬀerent solutions as t grows. If we change the initial condition a little bit, then as
t → ∞ we get x → A. We call such critical points stable. In this simple example it turns out that all
solutions in fact go to A as t → ∞. If a critical point is not stable we would say it is unstable.
0 5 10 15 20 0 10 10 5 5 10 15 20 5 5.0 2.5 2.5 0.0 0.0 2.5 5 7.5 5.0 0 10.0 7.5 0 10.0 2.5 5 10 10
0 5 10 15 20 Figure 1.6: Slope ﬁeld and some solutions of
x = −0.3 ( x − 5). 5.0 5.0
0 5 10 15 20 Figure 1.7: Slope ﬁeld and some solutions of
x = 0.1 x (5 − x). 1.6. AUTONOMOUS EQUATIONS 37 Let us consider the logistic equation
dx
= kx( M − x),
dt
for some positive k and M . This equation is commonly used to model population if we know the
limiting population M , that is the maximum sustainable population. The logistic equation leads to
less catastrophic predictions on world population than x = kx. In the real world there is no such
thing as negative population, but we will still consider negative x for the purposes of the math.
See Figure 1.7 on the facing page for an example. Note two critical points, x = 0 and x = 5.
The critical point at x = 5 is stable. On the other hand the critical point at x = 0 is unstable.
It is not really necessary to ﬁnd the exact solutions to talk about the long term behavior of the
solutions. For example, from the above we can easily see that 5 if x(0) > 0, lim x(t) = 0
if x(0) = 0, t→∞ DNE or − ∞ if x(0) < 0. Where DNE means “does not exist.” From just looking at the slope ﬁeld we cannot quite decide
what happens if x(0) < 0. It could be that the solution does not exist for t all the way to ∞. Think
of the equation x = x2 , we have seen that it only exists for some ﬁnite period of time. Same can
happen here. In our example equation above it will actually turn out that the solution does not exist
for all time, but to see that we would have to solve the equation. In any case, the solution does go to
−∞, but it may get there rather quickly.
Often we are interested only in the long term behavior of the solution and we would be doing
unnecessary work if we solved the equation exactly. It is easier to just look at the phase diagram or
phase portrait, which is a simple way to visualize the behavior of autonomous equations. In this
case there is one dependent variable x. So draw the x axis, mark all the critical points and then draw
arrows in between. If f ( x) > 0 draw an up arrow and if it is negative draw a down arrow. x=5
x=0 Armed with the phase diagram, it is easy to sketch the solutions approximately.
Exercise 1.6.1: Try sketching a few solutions simply from looking at the phase diagram. Check
with the preceding graphs if you are getting the type of curves. 38 CHAPTER 1. FIRST ORDER ODES
Once we draw the phase diagram, we can easily classify critical points as stable or unstable. unstable stable Since any mathematical model we cook up will only be an approximation to the real world,
unstable points are generally bad news.
Let us think about the logistic equation with harvesting. Logistic equations are commonly used
for modeling population. Suppose an alien race really likes to eat humans. They keep a planet
with humans on it and harvest the humans at a rate of h million humans per year. Suppose x is the
number of humans in millions on the planet and t is time in years. Let M be the limiting population
when no harvesting is done. k > 0 is some constant depending on how fast humans multiply. Our
equation becomes
dx
= kx( M − x) − h.
dt
Multiply out and solve for critical points
dx
= −kx2 + kMx − h.
dt
Critical points A and B are
kM + (kM )2 − 4hk
kM − (kM )2 − 4hk
B=
.
2k
2k
Exercise 1.6.2: Draw the phase diagram for diﬀerent possibilities. Note that these possibilities are
A > B, or A = B, or A and B both complex (i.e. no real solutions). Hint: Fix some simple k and M
and then vary h.
A= For example, let M = 8 and k = 0.1. When h = 1, then A and B are distinct and positive.
The graph we will get is given in Figure 1.8 on the next page. As long as the population stays
above B, which is approximately 1.55 million, then the population will not die out. If ever the
population drops below B, humans will die out, and the fast food restaurant serving them will go
out of business.
When h = 1.6, then A = B. There is only one critical point that is unstable. When the population
is above 1.6 million it will tend towards this number. If it ever drops below 1.6 million, humans will
die out on the planet. This scenario is not one that we (as the human fast food proprietor) want to be
in. A small perturbation of the equilibrium state and we are out of business. There is no room for
error. See Figure 1.9 on the facing page
Finally if we are harvesting at 2 million humans per year, the population will always plummet
towards zero, no matter how well stocked the planet starts. See Figure 1.10 on the next page. 1.6. AUTONOMOUS EQUATIONS
0 5 10 39 15 20 0 5 10 15 20 10.0 10.0 10.0 10.0 7.5 7.5 7.5 7.5 5.0 5.0 5.0 5.0 2.5 2.5 2.5 2.5 0.0 0.0 0.0
0 5 10 15 20 Figure 1.8: Slope ﬁeld and some solutions of
x = 0.1 x (8 − x) − 1.
0 5 0.0
0 5 10 15 20 Figure 1.9: Slope ﬁeld and some solutions of
x = 0.1 x (8 − x) − 1.6.
10 15 20 10.0 10.0 7.5 7.5 5.0 5.0 2.5 2.5 0.0 0.0
0 5 10 15 20 Figure 1.10: Slope ﬁeld and some solutions of x = 0.1 x (8 − x) − 2. 1.6.1 Exercises Exercise 1.6.3: Let x = x2 . a) Draw the phase diagram, ﬁnd the critical points and mark them
stable or unstable. b) Sketch typical solutions of the equation. c) Find lim x(t) for the solution with
t→∞
the initial condition x(0) = −1.
Exercise 1.6.4: Let x = sin x. a) Draw the phase diagram for −4π ≤ x ≤ 4π. On this interval
mark the critical points stable or unstable. b) Sketch typical solutions of the equation. c) Find
limt→∞ x(t) for the solution with the initial condition x(0) = 1.
Exercise 1.6.5: Suppose f ( x) is positive for 0 < x < 1, it is zero when x = 0 and x = 1, and it is
negative for all other x. a) Draw the phase diagram for x = f ( x), ﬁnd the critical points and mark 40 CHAPTER 1. FIRST ORDER ODES them stable or unstable. b) Sketch typical solutions of the equation. c) Find lim x(t) for the solution
t→∞
with the initial condition x(0) = 0.5.
Exercise 1.6.6: Start with the logistic equation dx = kx( M − x). Suppose that we modify our
dt
harvesting. That is we will only harvest an amount proportional to current population. In other
words we harvest hx per unit of time for some h > 0 (Similar to earlier example with h replaced
with hx). a) Construct the diﬀerential equation. b) Show that if kM > h, then the equation is still
logistic. c) What happens when kM < h? 1.7. NUMERICAL METHODS: EULER’S METHOD 1.7 41 Numerical methods: Euler’s method Note: 1 lecture, §2.4 in [EP]
At this point it may be good to ﬁrst try the Lab II and/or Project II from the IODE website:
http://www.math.uiuc.edu/iode/.
As we said before, unless f ( x, y) is of a special form, it is generally very hard if not impossible
to get a nice formula for the solution of the problem
y = f ( x, y), y( x0 ) = y0 . What if we want to ﬁnd the value of the solution at some particular x? Or perhaps we want to
produce a graph of the solution to inspect the behavior. In this section we will learn about the basics
of numerical approximation of solutions.
The simplest method for approximating a solution is Euler’s method‡ . It works as follows: We
take x0 and compute the slope k = f ( x0 , y0 ). The slope is the change in y per unit change in x. We
follow the line for an interval of length h on the x axis. Hence if y = y0 at x0 , then we will say that
y1 (the approximate value of y at x1 = x0 + h) will be y1 = y0 + hk. Rinse, repeat! That is, compute
x2 and y2 using x1 and y1 . For an example of the ﬁrst two steps of the method see Figure 1.11.
1 0 1 2 3 1 0 1 2 3 3.0 3.0 3.0 3.0 2.5 2.5 2.5 2.5 2.0 2.0 2.0 2.0 1.5 1.5 1.5 1.5 1.0 1.0 1.0 1.0 0.5 0.5 0.5 0.5 0.0 0.0 0.0
1 0 1 2 3 0.0
1 0 1 Figure 1.11: First two steps of Euler’s method with h = 1 for the equation y =
conditions y(0) = 1. 2 y2
3 3 with initial More abstractly, for any i = 1, 2, 3, . . ., we compute
xi+1 = xi + h, yi+1 = yi + h f ( xi , yi ). The line segments we get are an approximate graph of the solution. Generally it is not exactly the
solution. See Figure 1.12 on the next page for the plot of the real solution and the approximation.
‡ Named after the Swiss mathematician Leonhard Paul Euler (1707 – 1783). Do note the correct pronunciation of
the name sounds more like “oiler.” 42 CHAPTER 1. FIRST ORDER ODES
1 0 1 2 3 3.0 3.0 2.5 2.5 2.0 2.0 1.5 1.5 1.0 1.0 0.5 0.5 0.0 0.0
1 0 1 2 3 Figure 1.12: Two steps of Euler’s method (step size 1) and the exact solution for the equation y =
with initial conditions y(0) = 1. y2
3 Let us see what happens with the equation y = y2/3, y(0) = 1. Let us try to approximate y(2)
using Euler’s method. In Figures 1.11 and 1.12 we have essentially graphically approximated
y(2) with step size 1. With step size 1 we have y(2) ≈ 1.926. The real answer is 3. So we are
approximately 1.074 oﬀ. Let us halve the step size. Computing y4 with h = 0.5, we ﬁnd that
y(2) ≈ 2.209, so error of about 0.791. Table 1.1 on the facing page gives the values computed for
various parameters.
Exercise 1.7.1: Solve this equation exactly and show that y(2) = 3.
The diﬀerence between the actual solution and the approximate solution we will call the error.
We will usually talk about just the size of the error and we do not care much about its sign. The
main point is, that we usually do not know the real solution, so we only have a vague understanding
of the error. If we knew the error exactly . . . what is the point of doing the approximation?
We notice that except for the ﬁrst few times, every time we halved the interval the error
approximately halved. This halving of the error is a general feature of Euler’s method as it is a ﬁrst
order method. In the IODE Project II you are asked to implement a second order method. A second
order method reduces the error to approximately one quarter every time we halve the interval.
Note that to get the error to be within 0.1 of the answer we had to already do 64 steps. To get
it to within 0.01 we would have to halve another three or four times, meaning doing 512 to 1024
steps. That is quite a bit to do by hand. The improved Euler method from IODE Project II should
quarter the error every time we halve the interval, so we would have to approximately do half as
many “halvings” to get the same error. This reduction can be a big deal. With 10 halvings (starting
at h = 1) we have 1024 steps, whereas with 5 halvings we only have to do 32 steps, assuming
that the error was comparable to start with. A computer may not care about this diﬀerence for a
problem this simple, but suppose each step would take a second to compute (the function may be 1.7. NUMERICAL METHODS: EULER’S METHOD 43 h Approximate y(2) Error Error
Previous error 1
0.5
0.25
0.125
0.0625
0.03125
0.015625
0.0078125 1.92593
2.20861
2.47250
2.68034
2.82040
2.90412
2.95035
2.97472 1.07407
0.79139
0.52751
0.31966
0.17960
0.09588
0.04965
0.02528 0.73681
0.66656
0.60599
0.56184
0.53385
0.51779
0.50913 Table 1.1: Euler’s method approximation of y(2) where of y = y2/3, y(0) = 1.
substantially more diﬃcult to compute than y2/3). Then the diﬀerence is 32 seconds versus about
17 minutes. Note: We are not being altogether fair, a second order method would probably double
the time to do each step. Even so, it is 1 minute versus 17 minutes. Next, suppose that we have to
repeat such a calculation for diﬀerent parameters a thousand times. You get the idea.
Note that we do not know the error! How do we know what is the right step size? Essentially
we keep halving the interval, and if we are lucky, we can estimate the error from a few of these
calculations and the assumption that the error goes down by a factor of one half each time (if we are
using standard Euler).
Exercise 1.7.2: In the table above, suppose you do not know the error. Take the approximate values
of the function in the last two lines, assume that the error goes down by a factor of 2. Can you
estimate the error in the last time from this? Does it (approximately) agree with the table? Now do
it for the ﬁrst two rows. Does this agree with the table?
Let us talk a little bit more about the example y = y2/3, y(0) = 1. Suppose that instead of the
value y(2) we wish to ﬁnd y(3). The results of this eﬀort are listed in Table 1.2 on the next page for
successive halvings of h. What is going on here? Well, you should solve the equation exactly and
you will notice that the solution does not exist at x = 3. In fact, the solution goes to inﬁnity when
you approach x = 3.
Another case when things can go bad is if the solution oscillates wildly near some point. Such
an example is given in IODE Project II. In this case, the solution may exist at all points, but even
a better approximation method than Euler would need an insanely small step size to compute the
solution with reasonable precision. And computers might not be able to handle such a small step
size anyway.
In real applications we would not use a simple method such as Euler’s. The simplest method
that would probably be used in a real application is the standard RungeKutta method (we will not
describe it here). That is a fourth order method, meaning that if we halve the interval, the error
generally goes down by a factor of 16. 44 CHAPTER 1. FIRST ORDER ODES
h Approximate y(3)
1
0.5
0.25
0.125
0.0625
0.03125
0.015625
0.0078125 3.16232
4.54329
6.86079
10.80321
17.59893
29.46004
50.40121
87.75769 Table 1.2: Attempts to use Euler’s to approximate y(3) where of y = y2/3, y(0) = 1.
Choosing the right method to use and the right step size can be very tricky. There are several
competing factors to consider.
• Computational time: Each step takes computer time. Even if the function f is simple to
compute, we do it many times over. Large step size means faster computation, but perhaps
not the right precision.
• Roundoﬀ errors: Computers only compute with a certain number of signiﬁcant digits. Errors
introduced by rounding numbers oﬀ during our computations become noticeable when the
step size becomes too small relative to the quantities we are working with. So reducing step
size may in fact make errors worse.
• Stability: Certain equations may be numerically unstable. Small errors lead to large errors
down the line. In the worst case the numerical computations might be giving us bogus
numbers that look like a correct answer. Just because the numbers have stabilized after
successive halving, does not mean that we must have the right answer. What may also happen
is that the numbers may never stabilize no matter how many times we halve the interval.
We have seen just the beginnings of the challenges that appear in real applications. Numerical
approximation of solutions to diﬀerential equations is an active research area for engineers and
mathematicians. For example, the general purpose method used for the ODE solver in Matlab and
Octave (as of this writing) is a method that appeared in the literature only in the 1980s. 1.7.1 Exercises dx
Exercise 1.7.3: Consider
= (2t − x)2 , x(0) = 2. Use Euler’s method with step size h = 0.5 to
dt
approximate x(1). 1.7. NUMERICAL METHODS: EULER’S METHOD 45 dx
Exercise 1.7.4: Consider
= t − x, x(0) = 1. a) Use Euler’s method with step sizes h =
dt
1, 1/2, 1/4, 1/8 to approximate x(1). b) Solve the equation exactly. c) Describe what happens to the
errors for each h you used. That is, ﬁnd the factor by which the error changed each time you halved
the interval.
Exercise 1.7.5: Approximate the value of e by looking at the initial value problem y = y with
y(0) = 1 and approximating y(1) using Euler’s method with a step size of 0.2. 46 CHAPTER 1. FIRST ORDER ODES Chapter 2
Higher order linear ODEs
2.1 Second order linear ODEs Note: less than 1 lecture, ﬁrst part of §3.1 in [EP]
Let us consider the general second order linear diﬀerential equation
A( x)y + B( x)y + C ( x)y = F ( x).
We usually divide through by A( x) to get
y + p( x)y + q( x)y = f ( x), (2.1) where p( x) = B( x)/A( x), q( x) = C ( x)/A( x), and f ( x) = F ( x)/A( x). The word linear means that the equation
contains no powers nor functions of y, y , and y .
In the special case when f ( x) = 0 we have a socalled homogeneous equation
y + p( x)y + q( x)y = 0. (2.2) We have already seen some second order linear homogeneous equations.
y + k2 y = 0
y −k y=0
2 Two solutions are: y1 = cos(kx), Two solutions are: y1 = e ,
kx y2 = sin(kx). y2 = e−kx . If we know two solutions of a linear homogeneous equation, we know a lot more of them.
Theorem 2.1.1 (Superposition). Suppose y1 and y2 are two solutions of the homogeneous equation
(2.2). Then
y( x) = C1 y1 ( x) + C2 y2 ( x),
also solves (2.2) for arbitrary constants C1 and C2 .
47 48 CHAPTER 2. HIGHER ORDER LINEAR ODES That is, we can add solutions together and multiply them by constants to obtain new and diﬀerent
solutions. We will prove this theorem because the proof is very enlightening and illustrates how
linear equations work.
Proof: Let y = C1 y1 + C2 y2 . Then
y + py + qy = (C1 y1 + C2 y2 ) + p(C1 y1 + C2 y2 ) + q(C1 y1 + C2 y2 )
= C1 y1 + C2 y2 + C1 py1 + C2 py2 + C1 qy1 + C2 qy2
= C1 (y1 + py1 + qy1 ) + C2 (y2 + py2 + qy2 )
= C1 · 0 + C2 · 0 = 0.
The proof becomes even simpler to state if we use the operator notation. An operator is an
object that eats functions and spits out functions (kind of like what a function is, but a function eats
numbers and spits out numbers). Deﬁne the operator L by
Ly = y + py + qy.
The diﬀerential equation now becomes Ly = 0. The operator (and the equation) L being linear
means that L(C1 y1 + C2 y2 ) = C1 Ly1 + C2 Ly2 . The proof above becomes
Ly = L(C1 y1 + C2 y2 ) = C1 Ly1 + C2 Ly2 = C1 · 0 + C2 · 0 = 0.
Two diﬀerent solutions to the second equation y − k2 y = 0 are y1 = cosh(kx) and y2 = sinh(kx).
x
−x
x
−x
Let us remind ourselves of the deﬁnition, cosh x = e +e and sinh x = e −e . Therefore, these are
2
2
solutions by superposition as they are linear combinations of the two exponential solutions.
The functions sinh and cosh are sometimes more convenient to use than the exponential. Let us
review some of their properties.
cosh 0 = 1
d
cosh x = sinh x
dx
cosh2 x − sinh2 x = 1 sinh 0 = 0
d
sinh x = cosh x
dx Exercise 2.1.1: Derive these properties using the deﬁnitions of sinh and cosh in terms of exponentials.
Linear equations have nice and simple answers to the existence and uniqueness question.
Theorem 2.1.2 (Existence and uniqueness). Suppose p, q, f are continuous functions and a, b0 , b1
are constants. The equation
y + p( x)y + q( x)y = f ( x),
has exactly one solution y( x) satisfying the initial conditions
y(a) = b0 , y (a) = b1 . 2.1. SECOND ORDER LINEAR ODES 49 For example, the equation y + k2 y = 0 with y(0) = b0 and y (0) = b1 has the solution
y( x) = b0 cos(kx) + b1
sin(kx).
k The equation y − k2 y = 0 with y(0) = b0 and y (0) = b1 has the solution
y( x) = b0 cosh(kx) + b1
sinh(kx).
k Using cosh and sinh in this solution allows us to solve for the initial conditions in a cleaner way
than if we have used the exponentials.
The initial conditions for a second order ODE consist of two equations. Common sense tells us
that if we have two arbitrary constants and two equations, then we should be able to solve for the
constants and ﬁnd a solution to the diﬀerential equation satisfying the initial conditions.
Question: Suppose we ﬁnd two diﬀerent solutions y1 and y2 to the homogeneous equation (2.2).
Can every solution be written (using superposition) in the form y = C1 y1 + C2 y2 ?
Answer is aﬃrmative! Provided that y1 and y2 are diﬀerent enough in the following sense. We
will say y1 and y2 are linearly independent if one is not a constant multiple of the other.
Theorem 2.1.3. Let p, q, f be continuous functions and take the homogeneous equation (2.2). Let
y1 and y2 be two linearly independent solutions to (2.2). Then every other solution is of the form
y = C1 y1 + C2 y2 .
That is, y = C1 y1 + C2 y2 is the general solution.
For example, we found the solutions y1 = sin x and y2 = cos x for the equation y + y = 0. It is
obvious that sine and cosine are not multiples of each other. If sin x = A cos x for some constant A,
we let x = 0 and this would imply A = 0 = sin x, which is preposterous. So y1 and y2 are linearly
independent. Hence
y = C1 cos x + C2 sin x
is the general solution to y + y = 0.
We will study the solution of nonhomogeneous equations in § 2.5. We will ﬁrst focus on ﬁnding
general solutions to homogeneous equations. 2.1.1 Exercises Exercise 2.1.2: Show that y = e x and y = e2 x are linearly independent.
Exercise 2.1.3: Take y + 5y = 10 x + 5. Find (guess!) a solution. 50 CHAPTER 2. HIGHER ORDER LINEAR ODES Exercise 2.1.4: Prove the superposition principle for nonhomogeneous equations. Suppose that y1
is a solution to Ly1 = f ( x) and y2 is a solution to Ly2 = g( x) (same linear operator L). Show that y
solves Ly = f ( x) + g( x).
Exercise 2.1.5: For the equation x2 y − xy = 0, ﬁnd two solutions, show that they are linearly
independent and ﬁnd the general solution. Hint: Try y = xr .
Note that equations of the form ax2 y + bxy + cy = 0 are called Euler’s equations or CauchyEuler equations. They are solved by trying y = xr and solving for r (we can assume that x ≥ 0 for
simplicity).
Exercise 2.1.6: Suppose that (b − a)2 − 4ac > 0. a) Find a formula for the general solution
of ax2 y + bxy + cy = 0. Hint: Try y = xr and ﬁnd a formula for r. b) What happens when
(b − a)2 − 4ac = 0 or (b − a)2 − 4ac < 0?
We will revisit the case when (b − a)2 − 4ac < 0 later.
Exercise 2.1.7: Same equation as in Exercise 2.1.6. Suppose (b − a)2 − 4ac = 0. Find a formula
for the general solution of ax2 y + bxy + cy = 0. Hint: Try y = xr ln x for the second solution.
If you have one solution to a second order linear homogeneous equation you can ﬁnd another
one. This is the reduction of order method.
Exercise 2.1.8: Suppose y1 is a solution to y + p( x)y + q( x)y = 0. Show that
y2 ( x) = y1 ( x) e− p( x) dx y1 ( x) 2 dx is also a solution.
If you wish to come up with the formula for reduction of order yourself, start by trying
y2 ( x) = y1 ( x)v( x). Then plug y2 into the equation, use the fact that y1 is a solution, substitute w = v ,
and you have a ﬁrst order linear equation in w. Solve for w and then for v. When solving for w, make
sure to include a constant of integration. Let us solve some famous equations using the method.
Exercise 2.1.9 (Chebychev’s equation of order 1): Take (1 − x2 )y − xy + y = 0. a) Show that
y = x is a solution. b) Use reduction of order to ﬁnd a second linearly independent solution. c)
Write down the general solution.
Exercise 2.1.10 (Hermite’s equation of order 2): Take y − 2 xy + 4y = 0. a) Show that y = 1 − 2 x2
is a solution. b) Use reduction of order to ﬁnd a second linearly independent solution. c) Write
down the general solution. 2.2. CONSTANT COEFFICIENT SECOND ORDER LINEAR ODES 2.2 51 Constant coeﬃcient second order linear ODEs Note: more than 1 lecture, second part of §3.1 in [EP]
Suppose we have the problem
y − 6y + 8y = 0, y(0) = −2, y (0) = 6. This is a second order linear homogeneous equation with constant coeﬃcients. Constant coeﬃcients
means that the functions in front of y , y , and y are constants, not depending on x.
To guess a solution, think of a function that you know stays essentially the same when we
diﬀerentiate it, so that we can take the function and its derivatives, add some multiples of these
together, and end up with zero.
Let us try a solution of the form y = erx . Then y = rerx and y = r2 erx . Plug in to get
y − 6y + 8y = 0,
r e − 6rerx + 8erx = 0,
2 rx r2 − 6r + 8 = 0
(r − 2)(r − 4) = 0. (divide through by erx ), Hence, if r = 2 or r = 4, then erx is a solution. So let y1 = e2 x and y2 = e4 x .
Exercise 2.2.1: Check that y1 and y2 are solutions.
The functions e2 x and e4 x are linearly independent. If they were not linearly independent we
could write e4 x = Ce2 x implying that e2 x = C , which is clearly not possible. Hence, we can write
the general solution as
y = C1 e2 x + C2 e4 x .
We need to solve for C1 and C2 . To apply the initial conditions we ﬁrst ﬁnd y = 2C1 e2 x + 4C2 e4 x .
We plug in x = 0 and solve.
−2 = y(0) = C1 + C2 ,
6 = y (0) = 2C1 + 4C2 .
Either apply some matrix algebra, or just solve these by high school math. For example, divide the
second equation by 2 to obtain 3 = C1 + 2C2 , and subtract the two equations to get 5 = C2 . Then
C1 = −7 as −2 = C1 + 5. Hence, the solution we are looking for is
y = −7e2 x + 5e4 x .
Let us generalize this example into a method. Suppose that we have an equation
ay + by + cy = 0, (2.3) 52 CHAPTER 2. HIGHER ORDER LINEAR ODES where a, b, c are constants. Try the solution y = erx to obtain
ar2 erx + brerx + cerx = 0,
ar2 + br + c = 0.
The equation ar2 + br + c = 0 is called the characteristic equation of the ODE. Solve for the r by
using the quadratic formula.
√
−b ± b2 − 4ac
.
r1 , r2 =
2a
Therefore, we have er1 x and er2 x as solutions. There is still a diﬃculty if r1 = r2 , but it is not hard to
overcome.
Theorem 2.2.1. Suppose that r1 and r2 are the roots of the characteristic equation.
(i) If r1 and r2 are distinct and real (when b2 − 4ac > 0), then (2.3) has the general solution
y = C1 er1 x + C2 er2 x .
(ii) If r1 = r2 (happens when b2 − 4ac = 0), then (2.3) has the general solution
y = (C1 + C2 x) er1 x .
For another example of the ﬁrst case, take the equation y − k2 y = 0. Here the characteristic
equation is r2 − k2 = 0 or (r − k)(r + k) = 0. Consequently, e−kx and ekx are the two linearly
independent solutions.
Example 2.2.1: Find the general solution of
y − 8y + 16y = 0.
The characteristic equation is r2 − 8r + 16 = (r − 4)2 = 0. The equation has a double root
r1 = r2 = 4. The general solution is, therefore,
y = (C1 + C2 x) e4 x = C1 e4 x + C2 xe4 x .
Exercise 2.2.2: Check that e4 x and xe4 x are linearly independent.
That e4 x solves the equation is clear. If xe4 x solves the equation, then we know we are done. Let
us compute y = e4 x + 4 xe4 x and y = 8e4 x + 16 xe4 x . Plug in
y − 8y + 16y = 8e4 x + 16 xe4 x − 8(e4 x + 4 xe4 x ) + 16 xe4 x = 0.
We should note that in practice, doubled root rarely happens. If coeﬃcients are picked truly
randomly we are very unlikely to get a doubled root.
Let us give a short “proof” for why the solution xerx works when the root is doubled. This case
rx
rx
is really a limiting case of when the two roots are distinct and very close. Note that e 2 2 −e11 is a
r −r
solution when the roots are distinct. When r1 goes to r2 in the limit this is like taking derivative of
erx using r as a variable. This limit is xerx , and hence this is a solution in the doubled root case. 2.2. CONSTANT COEFFICIENT SECOND ORDER LINEAR ODES 2.2.1 53 Complex numbers and Euler’s formula It may happen that a polynomial has some complex roots. For example, the equation r2 + 1 = 0
has no real roots, but it does have two complex roots. Here we review some properties of complex
numbers.
Complex numbers may seem a strange concept especially because of the terminology. There
is nothing imaginary or really complicated about complex numbers. A complex number is simply
a pair of real numbers, (a, b). We can think of a complex number as a point in the plane. We
add complex numbers in the straightforward way, (a, b) + (c, d) = (a + c, b + d). We deﬁne a
multiplication by
def
(a, b) × (c, d) = (ac − bd, ad + bc).
It turns out that with this multiplication rule, all the standard properties of arithmetic hold. Further,
and most importantly (0, 1) × (0, 1) = (−1, 0).
Generally we just write (a, b) as a + ib, and we treat i as if it were an unknown. We can just
do arithmetic with complex numbers just as we would do with polynomials. The property we just
mentioned becomes i2 = −1. So whenever we see i2 , we can replace it by −1. The numbers i and −i
are roots of r2 + 1 = 0.
Note that engineers often use the letter j instead of i for the square root of −1. We will use the
mathematicians’ convention and use i.
Exercise 2.2.3: Make sure you understand (that you can justify) the following identities:
• i2 = −1, i3 = −i, i4 = 1,
• 1
= −i,
i • (3 − 7i)(−2 − 9i) = · · · = −69 − 13i,
• (3 − 2i)(3 + 2i) = 32 − (2i)2 = 32 + 22 = 13,
• 1
3−2i = 1 3+2i
3−2i 3+2i = 3+2i
13 = 3
13 + 2
i.
13 We can also deﬁne the exponential ea+ib of a complex number. We can do this by just writing
down the Taylor series and plugging in the complex number. Because most properties of the
exponential can be proved by looking at the Taylor series, we note that many properties still hold
for the complex exponential. For example, e x+y = e x ey . This means that ea+ib = ea eib and hence if
we can compute eib easily, we can compute ea+ib . Here we will use the socalled Euler’s formula.
Theorem 2.2.2 (Euler’s formula).
eiθ = cos θ + i sin θ and e−iθ = cos θ − i sin θ. 54 CHAPTER 2. HIGHER ORDER LINEAR ODES Exercise 2.2.4: Using Euler’s formula, check the identities:
cos θ = eiθ + e−iθ
2 and sin θ = eiθ − e−iθ
.
2i
2 Exercise 2.2.5: Double angle identities: Start with ei(2θ) = eiθ . Use Euler on each side and
deduce:
cos(2θ) = cos2 θ − sin2 θ
and
sin(2θ) = 2 sin θ cos θ.
We also will need some notation. For a complex number a + ib we call a the real part and b the
imaginary part of the number. Often the following notation is used,
Re(a + ib) = a 2.2.2 and Im(a + ib) = b. Complex roots Now suppose that the equation ay + by + cy = 0 has a characteristic equation ar2 + br + c = 0 that
√
2
has complex roots. By quadratic formula the roots are −b± 2b −4ac . These are complex if b2 − 4ac < 0.
a
In this case we can see that the roots are
√
−b
4ac − b2
r1 , r2 =
±i
.
2a
2a
As you can see, you will always get a pair of roots of the form α ± iβ. In this case we can still write
the solution as
y = C1 e(α+iβ) x + C2 e(α−iβ) x .
However, the exponential is now complex valued. We would need to choose C1 and C2 to be
complex numbers to obtain a realvalued solution (which is what we are after). While there is
nothing particularly wrong with this approach, it can make calculations harder and it is generally
preferred to ﬁnd two realvalued solutions.
Here we can use Euler’s formula. First let
y1 = e(α+iβ) x and y2 = e(α−iβ) x . Then note that
y1 = eα x cos(β x) + ie x sin(β x),
y2 = eα x cos(β x) − ieα x sin(β x).
We note that linear combinations of solutions are also solutions. Hence,
y1 + y2
y3 =
= eα x cos(β x),
2
y1 − y2
y4 =
= eα x sin(β x),
2i
are also solutions. And furthermore they are realvalued. It is not hard to see that they are linearly
independent (not multiples of each other). Therefore, we have the following theorem. 2.2. CONSTANT COEFFICIENT SECOND ORDER LINEAR ODES 55 Theorem 2.2.3. Take the equation
ay + by + cy = 0.
If the characteristic equation has the roots α ± iβ (when b2 − 4ac < 0), then the general solution is
y = C1 eα x cos(β x) + C2 eα x sin(β x).
Example 2.2.2: Find the general solution of y + k2 y = 0, for a constant k > 0.
The characteristic equation is r2 + k2 = 0. Therefore, the roots are r = ±ik and by the theorem
we have the general solution
y = C1 cos(kx) + C2 sin(kx).
Example 2.2.3: Find the solution of y − 6y + 13y = 0, y(0) = 0, y (0) = 10.
The characteristic equation is r2 − 6r + 13 = 0. By completing the square we get (r − 3)2 + 22 = 0
and hence the roots are r = 3 ± 2i. By the theorem we have the general solution
y = C1 e3 x cos(2 x) + C2 e3 x sin(2 x).
To ﬁnd the solution satisfying the initial conditions, we ﬁrst plug in zero to get
0 = y(0) = C1 e0 cos 0 + C2 e0 sin 0 = C1 .
Hence C1 = 0 and y = C2 e3 x sin(2 x). We diﬀerentiate
y = 3C2 e3 x sin(2 x) + 2C2 e3 x cos(2 x).
We again plug in the initial condition and obtain 10 = y (0) = 2C2 , or C2 = 5. Hence the solution
we are seeking is
y = 5e3 x sin(2 x). 2.2.3 Exercises Exercise 2.2.6: Find the general solution of 2y + 2y − 4y = 0.
Exercise 2.2.7: Find the general solution of y + 9y − 10y = 0.
Exercise 2.2.8: Solve y − 8y + 16y = 0 for y(0) = 2, y (0) = 0.
Exercise 2.2.9: Solve y + 9y = 0 for y(0) = 1, y (0) = 1.
Exercise 2.2.10: Find the general solution of 2y + 50y = 0.
Exercise 2.2.11: Find the general solution of y + 6y + 13y = 0. 56 CHAPTER 2. HIGHER ORDER LINEAR ODES Exercise 2.2.12: Find the general solution of y = 0 using the methods of this section.
Exercise 2.2.13: The method of this section applies to equations of other orders than two. We will
see higher orders later. Try to solve the ﬁrst order equation 2y + 3y = 0 using the methods of this
section.
Exercise 2.2.14: Let us revisit Euler’s equations of Exercise 2.1.6 on page 50. Suppose now that
(b − a)2 − 4ac < 0. Find a formula for the general solution of ax2 y + bxy + cy = 0. Hint: Note
that xr = er ln x . 2.3. HIGHER ORDER LINEAR ODES 2.3 57 Higher order linear ODEs Note: somewhat more than 1 lecture, §3.2 and §3.3 in [EP]
After reading this lecture, it may be good to ﬁrst try Project III from the IODE website:
http://www.math.uiuc.edu/iode/.
Most equations that appear in applications tend to be second order. Higher order equations do
appear from time to time, but it is a general assumption of modern physics that the world is “second
order.”
The basic results about linear ODEs of higher order are essentially exactly the same as for
second order equations with 2 replaced by n. The important concept of linear independence is
somewhat more complicated when more than two functions are involved.
For higher order constant coeﬃcient ODEs, the methods are also slightly harder, but we will not
dwell on these. You can always use the methods for systems of linear equations from chapter 3 to
solve higher order constant coeﬃcient equations.
So let us start with a general homogeneous linear equation
y(n) + pn−1 ( x)y(n−1) + · · · + p1 ( x)y + p0 ( x)y = 0. (2.4) Theorem 2.3.1 (Superposition). Suppose y1 , y2 , . . . , yn are solutions of the homogeneous equation
(2.4). Then
y( x) = C1 y1 ( x) + C2 y2 ( x) + · · · + Cn yn ( x)
also solves (2.4) for arbitrary constants C1 , . . . , Cn .
We also have the existence and uniqueness theorem for nonhomogeneous linear equations.
Theorem 2.3.2 (Existence and uniqueness). Suppose p0 through pn−1 , and f are continuous functions and a, b0 , b1 , . . . , bn−1 are constants. The equation
y(n) + pn−1 ( x)y(n−1) + · · · + p1 ( x)y + p0 ( x)y = f ( x)
has exactly one solution y( x) satisfying the initial conditions
y(a) = b0 , 2.3.1 y (a) = b1 , ..., y(n−1) (a) = bn−1 . Linear independence When we had two functions y1 and y2 we said they were linearly independent if one was not the
multiple of the other. Same idea holds for n functions. In this case it is easier to state as follows.
The functions y1 , y2 , . . . , yn are linearly independent if
c1 y1 + c2 y2 + · · · + cn yn = 0,
has only the trivial solution c1 = c2 = · · · = cn = 0. If we can write the equation with a nonzero
constant, say c1 0, then we can solve for y1 as a linear combination of the others. If the functions
are not linearly independent, we say they are linearly dependent. 58 CHAPTER 2. HIGHER ORDER LINEAR ODES Example 2.3.1: Show e x , e2 x , e3 x are linearly independent.
Let us give several ways to show this fact. Many textbooks (including [EP] and [F]) introduce
Wronskians, but that is really not necessary here.
Let us write down
c1 e x + c2 e2 x + c3 e3 x = 0.
We use rules of exponentials and write z = e x . Then we have
c1 z + c2 z2 + c3 z3 = 0.
The left hand side is a third degree polynomial in z. It can either be identically zero, or it can have
at most 3 zeros. Therefore, it is identically zero and c1 = c2 = c3 = 0 and the functions are linearly
independent.
Let us try another way. As before we write
c1 e x + c2 e2 x + c3 e3 x = 0.
This equation has to hold for all x. What we could do is divide through by e3 x to get
c1 e−2 x + c2 e− x + c3 = 0.
As the equation is true for all x, let x → ∞. After taking the limit we see that c3 = 0. Hence our
equation becomes
c1 e x + c2 e2 x = 0.
Rinse, repeat!
How about yet another way. We again write
c1 e x + c2 e2 x + c3 e3 x = 0.
We can evaluate the equation and its derivatives at diﬀerent values of x to obtain equations for c1 ,
c2 , and c3 . Let us ﬁrst divide by e x for simplicity.
c1 + c2 e x + c3 e2 x = 0.
We set x = 0 to get the equation c1 + c2 + c3 = 0. Now diﬀerentiate both sides
c2 e x + 2c3 e2 x = 0.
We set x = 0 to get c2 + 2c3 = 0. We divide by e x again and diﬀerentiate to get 4c3 e2 x = 0. It is clear
that c3 is zero. Then c2 must be zero as c2 = −2c3 , and c1 must be zero because c1 + c2 + c3 = 0.
There is no one best way to do it. All of these methods are perfectly valid.
Example 2.3.2: On the other hand, the functions e x , e− x , and cosh x are linearly dependent. Simply
apply deﬁnition of the hyperbolic cosine:
cosh x = e x + e− x
.
2 2.3. HIGHER ORDER LINEAR ODES 2.3.2 59 Constant coeﬃcient higher order ODEs When we have a higher order constant coeﬃcient homogeneous linear equation. The song and
dance is exactly the same as it was for second order. We just need to ﬁnd more solutions. If the
equation is nth order we need to ﬁnd n linearly independent solutions. It is best seen by example.
Example 2.3.3: Find the general solution to
y − 3y − y + 3y = 0. (2.5) Try: y = erx . We plug in and get
r3 erx − 3r2 erx − rerx + 3erx = 0.
We divide through by erx . Then
r3 − 3r2 − r + 3 = 0.
The trick now is to ﬁnd the roots. There is a formula for the roots of degree 3 and 4 polynomials
but it is very complicated. There is no formula for higher degree polynomials. That does not mean
that the roots do not exist. There are always n roots for an nth degree polynomial. They might be
repeated and they might be complex. Computers are pretty good at ﬁnding roots approximately for
reasonable size polynomials.
Best place to start is to plot the polynomial and check where it is zero. Or you can try plugging
in. Sometimes it is a good idea to just start plugging in numbers r = −2, −1, 0, 1, 2, . . . and see if
you get a hit. There are some signs that you might have missed a root. For example, if you plug in
−2 into our polynomial you get −15. If you plug in 0 you get 3. That means there is a root between
−2 and 0 because the sign changed.
A good strategy at ﬁrst is to look for roots −1, 1, or 0, these are easy to see. Our polynomial
happens to have two roots r1 = −1 and r2 = 1. There must be 3 roots and the last root is reasonably
easy to ﬁnd. The constant term in a polynomial is the multiple of the negations of all the roots
because r3 − 3r2 − r + 3 = (r − r1 )(r − r2 )(r − r3 ). In our case we see that
3 = (−r1 )(−r2 )(−r3 ) = (1)(−1)(−r3 ) = r3 .
You should check that r3 = 3 really is a root. Hence we know that e− x , e x and e3 x are solutions
to (2.5). They are linearly independent as can easily be checked, and there are 3 of them, which
happens to be exactly the number we need. Hence the general solution is
y = C1 e− x + C2 e x + C3 e3 x .
Suppose we were given some initial conditions y(0) = 1, y (0) = 2, and y (0) = 3. Then
1 = y(0) = C1 + C2 + C3 ,
2 = y (0) = −C1 + C2 + 3C3 ,
3 = y (0) = C1 + C2 + 9C3 . 60 CHAPTER 2. HIGHER ORDER LINEAR ODES It is possible to ﬁnd the solution by high school algebra, but it would be a pain. The only sensible
way to solve a system of equations such as this is to use matrix algebra, see § 3.2. For now we note
that the solution is C1 = −1/4, C2 = 1 and C3 = 1/4. The speciﬁc solution to the ODE is
−1 − x
1
e + e x + e3 x .
4
4
Next, suppose that we have real roots, but they are repeated. Let us say we have a root r repeated
k times. In the spirit of the second order solution, and for the same reasons, we have the solutions
y= erx , xerx , x2 erx , . . . , xk−1 erx .
We take a linear combination of these solutions to ﬁnd the general solution.
Example 2.3.4: Solve
y(4) − 3y + 3y − y = 0.
We note that the characteristic equation is
r 4 − 3r 3 + 3r 2 − r = 0.
By inspection we note that r4 − 3r3 + 3r2 − r = r(r − 1)3 . Hence the roots given with multiplicity
are r = 0, 1, 1, 1. Thus the general solution is
y = (C1 + C2 x + C3 x2 ) e x + C4 . from r = 0 terms coming from r = 1 Similarly to the second order case we can handle complex roots. Complex roots always come in
pairs r = α ± iβ. Suppose we have two such complex roots, each repeated k times. The corresponding
solution is
(C0 + C1 x + · · · + Ck−1 xk−1 ) eα x cos(β x) + (D0 + D1 x + · · · + Dk−1 xk−1 ) eα x sin(β x).
where C0 , . . . , Ck−1 , D0 , . . . , Dk−1 are arbitrary constants.
Example 2.3.5: Solve
y(4) − 4y + 8y − 8y + 4y = 0.
The characteristic equation is
r4 − 4r3 + 8r2 − 8r + 4 = 0,
2 (r2 − 2r + 2) = 0,
2 (r − 1)2 + 1 = 0.
Hence the roots are 1 ± i, both with multiplicity 2. Hence the general solution to the ODE is
y = (C1 + C2 x) e x cos x + (C3 + C4 x) e x sin x.
The way we solved the characteristic equation above is really by guessing or by inspection. It is not
so easy in general. We could also have asked a computer or an advanced calculator for the roots. 2.3. HIGHER ORDER LINEAR ODES 2.3.3 61 Exercises Exercise 2.3.1: Find the general solution for y − y + y − y = 0.
Exercise 2.3.2: Find the general solution for y(4) − 5y + 6y = 0.
Exercise 2.3.3: Find the general solution for y + 2y + 2y = 0.
Exercise 2.3.4: Suppose that the characteristic equation for a diﬀerential equation is (r − 1)2 (r − 2)2 =
0. a) Find such a diﬀerential equation. b) Find its general solution.
Exercise 2.3.5: Suppose that a fourth order equation has a solution y = 2e4 x x cos x. a) Find such
an equation. b) Find the initial conditions that the given solution satisﬁes.
Exercise 2.3.6: Find the general solution for the equation of Exercise 2.3.5.
Exercise 2.3.7: Let f ( x) = e x − cos x, g( x) = e x + cos x, and h( x) = cos x. Are f ( x), g( x), and h( x)
linearly independent? If so, show it, if not, ﬁnd a linear combination that works.
Exercise 2.3.8: Let f ( x) = 0, g( x) = cos x, and h( x) = sin x. Are f ( x), g( x), and h( x) linearly
independent? If so, show it, if not, ﬁnd a linear combination that works.
Exercise 2.3.9: Are x, x2 , and x4 linearly independent? If so, show it, if not, ﬁnd a linear combination that works.
Exercise 2.3.10: Are e x , xe x , and x2 e x linearly independent? If so, show it, if not, nd a linear
combination that works. 62 CHAPTER 2. HIGHER ORDER LINEAR ODES 2.4 Mechanical vibrations Note: 2 lectures, §3.4 in [EP]
Let us look at some applications of linear second order constant coeﬃcient equations. 2.4.1 Some examples Our ﬁrst example is a mass on a spring. Suppose we have
a mass m > 0 (in kilograms) connected by a spring with spring
m
constant k > 0 (in newtons per meter) to a ﬁxed wall. There may be
some external force F (t) (in newtons) acting on the mass. Finally,
damping c
there is some friction measured by c ≥ 0 (in newtonseconds per
meter) as the mass slides along the ﬂoor (or perhaps there is a damper connected).
Let x be the displacement of the mass ( x = 0 is the rest position), with x growing to the right
(away from the wall). The force exerted by the spring is proportional to the compression of the
spring by Hooke’s law. Therefore, it is kx in the negative direction. Similarly the amount of force
exerted by friction is proportional to the velocity of the mass. By Newton’s second law we know
that force equals mass times acceleration and hence mx = F (t) − cx − kx or
k F (t) mx + cx + kx = F (t).
This is a linear second order constant coeﬃcient ODE. We set up some terminology about this
equation. We say the motion is
(i) forced, if F 0 (if F is not identically zero),
(ii) unforced or free, if F ≡ 0 (if F is identically zero),
(iii) damped, if c > 0, and
(iv) undamped, if c = 0.
This system appears in lots of applications even if it does not at ﬁrst seem like it. Many real
world scenarios can be simpliﬁed to a mass on a spring. For example, a bungee jump setup is
essentially a mass and spring system (you are the mass). It would be good if someone did the math
before you jump oﬀ the bridge, right? Let us give 2 other examples. E C
R L Here is an example for electrical engineers. Suppose that you have the
pictured RLC circuit. There is a resistor with a resistance of R ohms, an
inductor with an inductance of L henries, and a capacitor with a capacitance
of C farads. There is also an electric source (such as a battery) giving a
voltage of E (t) volts at time t (measured in seconds). Let Q(t) be the charge 2.4. MECHANICAL VIBRATIONS 63 in coulombs on the capacitor and I (t) be the current in the circuit. The relation between the two is
Q = I . By elementary principles we have that LI + RI + Q/C = E . If we diﬀerentiate we get
LI (t) + RI (t) + 1
I (t) = E (t).
C This is an nonhomogeneous second order constant coeﬃcient linear equation. Further, as L, R, and
C are all positive, this system behaves just like the mass and spring system. The position of the mass
is replaced by the current. Mass is replaced by the inductance, damping is replaced by resistance
and the spring constant is replaced by one over the capacitance. The change in voltage becomes the
forcing function. Hence for constant voltage this is an unforced motion.
Our next example is going to behave like a mass and spring system
only approximately. Suppose we have a mass m on a pendulum of length L.
We wish to ﬁnd an equation for the angle θ(t). Let g be the force of gravity.
Elementary physics mandates that the equation is of the form
θ+ L
θ g
sin θ = 0.
L This equation can be derived using Newton’s second law; force equals mass times acceleration.
The acceleration is Lθ and mass is m. So mLθ has to be equal to the tangential component of the
force given by the gravity. This is mg sin θ in the opposite direction. The m curiously cancels from
the equation.
Now we make our approximation. For small θ we have that approximately sin θ ≈ θ. This can
be seen by looking at the graph. In Figure 2.1 we can see that for approximately −0.5 < θ < 0.5 (in
radians) the graphs of sin θ and θ are almost the same.
1.0 0.5 0.0 0.5 1.0 1.0 1.0 0.5 0.5 0.0 0.0 0.5 0.5 1.0
1.0 1.0
0.5 0.0 0.5 1.0 Figure 2.1: The graphs of sin θ and θ (in radians). 64 CHAPTER 2. HIGHER ORDER LINEAR ODES Therefore, when the swings are small, θ is always small and we can model the behavior by the
simpler linear equation
g
θ + θ = 0.
L
Note that the errors that we get from the approximation build up so over a very long time, the
behavior might change more substantially. Also we will see that in a mass spring system, the
amplitude is independent of the period, this is not true for a pendulum. But for reasonably short
periods of time and small swings (for example if the length of the pendulum is very large), the
behavior is reasonably close.
In real world problems it is very often necessary to make these types of simpliﬁcations. Therefore, it is good to understand both the mathematics and the physics of the situation to see if the
simpliﬁcation is valid in the context of the questions we are trying to answer. 2.4.2 Free undamped motion In this section we will only consider free or unforced motion, as we cannot yet solve nonhomogeneous equations. Let us start with undamped motion where c = 0. We have the equation
mx + kx = 0.
If we divide by m and let ω0 = √ k/m, then we can write the equation as
x + ω2 x = 0.
0 The general solution to this equation is
x(t) = A cos(ω0 t) + B sin(ω0 t).
By a trigonometric identity, we have that for two diﬀerent constants C and γ, we have
A cos(ω0 t) + B sin(ω0 t) = C cos(ω0 t − γ).
√
It is not hard to compute that C = A2 + B2 and tan γ = B/A. Therefore, we let C and γ be our
arbitrary constants and write x(t) = C cos(ω0 t − γ).
Exercise 2.4.1: Justify the above identity and verify the equations for C and γ. Hint: Start with
cos(α − β) = cos(α) cos(β) + sin(α) sin(β) and multiply by C. Then think what should α and β be.
While it is generally easier to use the ﬁrst form with A and B to solve for the initial conditions,
the second form is much more natural. The constants C and γ have very nice interpretation. We
look at the form of the solution
x(t) = C cos(ω0 t − γ). 2.4. MECHANICAL VIBRATIONS 65 We can see that the amplitude is C , ω0 is the (angular) frequency, and γ is the socalled phase shift.
The phase shift just shifts the graph left or right. We call ω0 the natural (angular) frequency. This
entire setup is usually called simple harmonic motion.
Let us pause to explain the word angular before the word frequency. The units of ω0 are radians
per unit time, not cycles per unit time as is the usual measure of frequency. Because we know one
0
cycle is 2π radians, the usual frequency is given by ωπ . It is simply a matter of where we put the
2
constant 2π, and that is a matter of taste.
2
The period of the motion is one over the frequency (in cycles per unit time) and hence ωπ . That
0
is the amount of time it takes to complete one full oscillation.
Example 2.4.1: Suppose that m = 2 kg and k = 8 N/m. The whole mass and spring setup is sitting
on a truck that was traveling at 1 m/s. The truck crashes and hence stops. The mass was held in place
0.5 meters forward from the rest position. During the crash the mass gets loose. That is, the mass
is now moving forward at 1 m/s, while the other end of spring is held in place. The mass therefore
starts oscillating. What is the frequency of the resulting oscillation and what is the amplitude. The
units are the mks units (meterskilogramsseconds).
The setup means that the mass was at half a meter in the positive direction during the crash and
relative to the wall the spring is mounted to, the mass was moving forward (in the positive direction)
at 1 m/s. This gives us the initial conditions.
So the equation with initial conditions is
2 x + 8 x = 0,
x(0) = 0.5,
x (0) = 1.
√
√
We can directly compute ω0 = k/m = 4 = 2. Hence the angular frequency is 2. The usual
frequency in Hertz (cycles per second) is 2/2π = 1/π ≈ 0.318.
The general solution is
x(t) = A cos(2t) + B sin(2t).
Letting x(0) = 0.5 means A = 0.5. Then x (t) √ −2(0.5) sin(2t) + 2 B cos(2t). Letting x (0) = 1 we
=
√
get B = 0.5. Therefore, the amplitude is C = A2 + B2 = 0.5 ≈ 0.707. The solution is
x(t) = 0.5 cos(2t) + 0.5 sin(2t).
A plot of x(t) is shown in Figure 2.2 on the following page.
For the free undamped motion, if the solution is of the form
x(t) = A cos(ω0 t) + B sin(ω0 t),
this corresponds to the initial conditions x(0) = A and x (0) = ω0 B.
Therefore, it is easy to ﬁgure out A and B ﬁrst, and compute the amplitude and the phase shift
using A and B. In the example, we have already found C . Let us compute the phase shift. We know
that tan γ = B/A = 1. We take the arctangent of 1 and get approximately 0.785. As you may recall, 66 CHAPTER 2. HIGHER ORDER LINEAR ODES
1.0 0.0 2.5 5.0 7.5 10.0
1.0 0.5 0.5 0.0 0.0 0.5 0.5 1.0
0.0 2.5 5.0 7.5 1.0
10.0 Figure 2.2: Simple undamped oscillation. we still need to check if this γ is in the right quadrant. Since both A and B are positive, then γ
should be in the ﬁrst quadrant, and 0.785 radians really is in the ﬁrst quadrant.
Note: Many calculators and computer software do not only have the atan function for arctangent,
but also what is sometimes called atan2. This function takes two arguments, B and A, and returns
a γ in the correct quadrant for you. 2.4.3 Free damped motion Let us now focus on damped motion. Let us rewrite the equation
mx + cx + kx = 0,
as
x + 2 px + ω2 x = 0,
0
where
ω0 = k
,
m p= c
.
2m The characteristic equation is
r2 + 2 pr + ω2 = 0.
0
Using the quadratic formula we get that the roots are
r = −p ± p2 − ω2 .
0 The form of the solution depends on whether we get complex or real roots. We get real roots if and
only if the following number is nonnegative.
p2 − ω2 =
0 c
2m 2 − k
c2 − 4km
=
.
m
4m2 2.4. MECHANICAL VIBRATIONS 67 The sign of p2 − ω2 is the same as the sign of c2 − 4km. Thus we get real roots if and only if c2 − 4km
0
is nonnegative.
Overdamping
When c2 − 4km > 0, we say the system is overdamped. In this case, there are two distinct real roots r1
and r2 . Notice that both roots are negative. As
is negative.
The solution is
x(t) = C1 er1 t + C2 er2 t . p2 − ω2 is always less than p, then − p ±
0
0 25 50 75 p2 − ω2
0
100 1.5 1.5 Since r1 , r2 are negative, x(t) → 0 as t → ∞.
Thus the mass will tend towards the rest position
as time goes to inﬁnity. For a few sample plots
for diﬀerent initial conditions, see Figure 2.3.
Do note that no oscillation happens. In fact,
the graph will cross the x axis at most once. To see
why, we try to solve 0 = C1 er1 t + C2 er2 t . Therefore,
C1 er1 t = −C2 er2 t and using laws of exponents we
obtain
−C1
= e(r2 −r1 )t .
Figure 2.3: Overdamped motion for several difC2
ferent initial conditions.
This equation has at most one solution t ≥ 0. For
some initial conditions the graph will never cross the x axis, as is evident from the sample graphs.
1.0 1.0 0.5 0.5 0.0 0.0 0 25 50 75 100 Example 2.4.2: Suppose the mass is released from rest. That is x(0) = x0 and x (0) = 0. Then
x0
x(t) =
r1 er2 t − r2 er1 t .
r1 − r2
It is not hard to see that this satisﬁes the initial conditions.
Critical damping
When c2 − 4km = 0, we say the system is critically damped. In this case, there is one root of
multiplicity 2 and this root is − p. Therefore, our solution is
x(t) = C1 e− pt + C2 te− pt .
The behavior of a critically damped system is very similar to an overdamped system. After all a
critically damped system is in some sense a limit of overdamped systems. Since these equations are
really only an approximation to the real world, in reality we are never critically damped, it is a place
you can only reach in theory. You are always a little bit underdamped or a little bit overdamped. It
is better not to dwell on critical damping. 68 CHAPTER 2. HIGHER ORDER LINEAR ODES Underdamping
When c2 − 4km < 0, we say the system is
underdamped. In this case, the roots are complex. 0 5 10 15 20 25 30 = −p ± p2 − ω2
0
√ −1 ω2 − p2
0 1.0 0.5 0.5 0.0 0.0 0.5 r = −p ± 1.0 0.5 = − p ± iω1 ,
where ω1 = ω2 − p2 . Our solution is
0 x(t) = e− pt A cos(ω1 t) + B sin(ω1 t) ,
or
x(t) = Ce− pt cos(ω1 t − γ). 1.0 1.0
0 5 10 15 20 25 30 Figure 2.4: Underdamped motion with the envelope curves shown. An example plot is given in Figure 2.4. Note that
we still have that x(t) → 0 as t → ∞.
In the ﬁgure we also show the envelope curves Ce− pt and −Ce− pt . The solution is the oscillating
line between the two envelope curves. The envelope curves give the maximum amplitude of the
oscillation at any given point in time. For example if you are bungee jumping, you are really
interested in computing the envelope curve so that you do not hit the concrete with your head.
The phase shift γ just shifts the graph left or right but within the envelope curves (the envelope
curves do not change if γ changes).
Finally note that the angular pseudofrequency (we do not call it a frequency since the solution is
not really a periodic function) ω1 becomes lower when the damping c (and hence p) becomes larger.
This makes sense. When we change the damping just a little bit, we do not expect the behavior of
the solution to change dramatically. If we keep making c larger, then at some point the solution
should start looking like the solution for critical damping or overdamping, where no oscillation
happens.
On the other hand when c becomes smaller, ω1 approaches ω0 (ω1 is always smaller than ω0 ),
and the solution looks more and more like the steady periodic motion of the undamped case. The
envelope curves become ﬂatter and ﬂatter as p goes to 0. 2.4.4 Exercises Exercise 2.4.2: Consider a mass and spring system with a mass m = 2, spring constant k = 3, and
damping constant c = 1. a) Set up and ﬁnd the general solution of the system. b) Is the system
underdamped, overdamped or critically damped? c) If the system is not critically damped, ﬁnd a c
that makes the system critically damped. 2.4. MECHANICAL VIBRATIONS 69 Exercise 2.4.3: Do Exercise 2.4.2 for m = 3, k = 12, and c = 12.
Exercise 2.4.4: Using the mks units (meterskilogramsseconds), suppose you have a spring with
spring constant 4 N/m. You want to use it to weight items. Assume no friction. You place the mass
on the spring and put it in motion. a) You count and ﬁnd that the frequency is 0.8 Hz (cycles per
second). What is the mass? b) Find a formula for the mass m given the frequency ω in Hz.
Exercise 2.4.5: Suppose we add possible friction to Exercise 2.4.4. Further, suppose you do not
know the spring constant, but you have two reference weights 1 kg and 2 kg to calibrate your setup.
You put each in motion on your spring and measure the frequency. For the 1 kg weight you measured
1.1 Hz, for the 2 kg weight you measured 0.8 Hz. a) Find k (spring constant) and c (damping
constant). b) Find a formula for the mass in terms of the frequency in Hz. Note that there may be
more than one possible mass for a given frequency. c) For an unknown object you measured 0.2 Hz,
what is the mass of the object? Suppose that you know that the mass of the unknown object is more
than a kilogram.
Exercise 2.4.6: Suppose you wish to measure the friction a mass of 0.1 kg experiences as it slides
along a ﬂoor (you wish to ﬁnd c). You have a spring with spring constant k = 5 N/m. You take the
spring, you attach it to the mass and ﬁx it to a wall. Then you pull on the spring and let the mass go.
You ﬁnd that the mass oscillates with frequency 1 Hz. What is the friction? 70 2.5 CHAPTER 2. HIGHER ORDER LINEAR ODES Nonhomogeneous equations Note: 2 lectures, §3.5 in [EP] 2.5.1 Solving nonhomogeneous equations We have solved linear constant coeﬃcient homogeneous equations. What about nonhomogeneous
linear ODEs? For example, the equations for forced mechanical vibrations. That is, suppose we
have an equation such as
y + 5y + 6y = 2 x + 1.
(2.6)
We will generally write Ly = 2 x + 1 when the exact form of the operator is not important. We
solve (2.6) in the following manner. We ﬁnd the general solution yc to the associated homogeneous
equation
y + 5y + 6y = 0.
(2.7)
We call yc the complementary solution. We also ﬁnd a single particular solution y p to (2.6) in some
way and then
y = yc + y p
is the general solution to (2.6) (we will see why in a moment).
Note that y p can be any solution. Suppose you ﬁnd a diﬀerent particular solution y p . Write the
˜
diﬀerence as w = y p − y p . Then plug w into the left hand side of the equation to get
˜
w + 5w + 6w = (y p + 5y p + 6y p ) − (˜ p + 5˜ p + 6˜ p ) = (2 x + 1) − (2 x + 1) = 0.
y
y
y
In other words, w is a complementary solution. Using the operator notation the calculation becomes
simpler. As L is a linear operator and so we could just write
Lw = L(y p − y p ) = Ly p − Ly p = (2 x + 1) − (2 x + 1) = 0.
˜
˜
So w = y p − y p is a solution to (2.7). Any two solutions of (2.6) diﬀer by a solution to the
˜
homogeneous equation (2.7). The solution y = yc + y p includes all solutions to (2.6), since yc is the
general solution to the associated homogeneous equation.
Theorem 2.5.1. Let Ly = f ( x) be a linear ODE (not necessarily constant coeﬃcient). Let yc be
the general solution to the associated homogeneous equation Ly = 0 and let y p be any particular
solution to Ly = f ( x). Then the general solution to Ly = f ( x) is
y = yc + y p .
The moral of the story is that we can ﬁnd the particular solution in any old way. If we ﬁnd a
diﬀerent particular solution (by a diﬀerent method, or simply by guessing), then we still get the
same general solution. The formula may look diﬀerent, and the constants you will have to choose to
satisfy the initial conditions may be diﬀerent, but it is the same solution. 2.5. NONHOMOGENEOUS EQUATIONS 2.5.2 71 Undetermined coeﬃcients The trick is to somehow, in a smart way, guess one particular solution to (2.6). Note that 2 x + 1 is a
polynomial, and the left hand side of the equation will be a polynomial if we let y be a polynomial
of the same degree. Let us try
y = Ax + B.
We plug in to obtain
y + 5y + 6y = (Ax + B) + 5(Ax + B) + 6(Ax + B) = 0 + 5A + 6Ax + 6 B = 6Ax + (5A + 6 B).
So 6Ax + (5A + 6 B) = 2 x + 1. Therefore, A = 1/3 and B = −1/9. That means y p =
Solving the complementary problem (Exercise!) we get 1
3 x− 1
9 = 3 x −1
.
9 yc = C1 e−2 x + C2 e−3 x .
Hence the general solution to (2.6) is
y = C1 e−2 x + C2 e−3 x + 3x − 1
.
9 Now suppose we are further given some initial conditions. For example, y(0) = 0 and y (0) = 1/3.
First ﬁnd y = −2C1 e−2 x − 3C2 e−3 x + 1/3. Then
1
0 = y(0) = C1 + C2 − ,
9 1
1
= y (0) = −2C1 − 3C2 + .
3
3 We solve to get C1 = 1/3 and C2 = −2/9. The particular solution we want is
2
3 x − 1 3e−2 x − 2e−3 x + 3 x − 1
1
=
.
y( x) = e−2 x − e−3 x +
3
9
9
9
Exercise 2.5.1: Check that y really solves the equation (2.6) and the given initial conditions.
Note: A common mistake is to solve for constants using the initial conditions with yc and only
adding the particular solution y p after that. That will not work. You need to ﬁrst compute y = yc + y p
and only then solve for the constants using the initial conditions.
A right hand side consisting of exponentials, sines, and cosines can be handled similarly. For
example,
y + 2y + 2y = cos(2 x).
Let us simply ﬁnd some y p . We notice that we may have to also guess sin(2 x) since derivatives of
cosine are sines. We guess
y p = A cos(2 x) + B sin(2 x). 72 CHAPTER 2. HIGHER ORDER LINEAR ODES We plug y p into the equation and we get
−4A cos(2 x) − 4 B sin(2 x) − 4A sin(2 x) + 4 B cos(2 x) + 2A cos(2 x) + 2 B sin(2 x) = cos(2 x).
The left hand side must equal to right hand side. We group terms and we get that −4A + 4 B + 2A = 1
and −4 B − 4A + 2 B = 0. So −2A + 4 B = 1 and 2A + B = 0 and hence A = −1/10 and B = 1/5. So
y p = A cos(2 x) + B sin(2 x) = − cos(2 x) + 2 sin(2 x)
.
10 Similarly, if the right hand side contains exponentials we guess exponentials. For example, if
the equation is (where L is a linear constant coeﬃcient operator)
Ly = e3 x ,
we will guess y = Ae3 x . We note also that using the product rule for diﬀerentiation gives us a way to
combine these guesses. If we can guess a form for y such that Ly has all the terms needed to for the
right hand side, that is a good place to start. For example,
Ly = (1 + 3 x2 ) e− x cos(π x).
We will guess
y = (A + Bx + Cx2 ) e− x cos(π x) + (D + Ex + F x2 ) e− x sin(π x).
We will plug in and then hopefully get equations that we can solve for A, B, C, D, E , F . As you can
see this can make for a very long and tedious calculation very quickly. C’est la vie!
There is one hiccup in all this. It could be that our guess actually solves the associated
homogeneous equation. That is, suppose we have
y − 9y = e3 x .
We would love to guess y = Ae3 x , but if we plug this into the left hand side of the equation we get
y − 9y = 9Ae3 x − 9Ae3 x = 0 e3 x . There is no way we can choose A to make the left hand side be e3 x . The trick in this case is to
multiply our guess by x until we get rid of duplication with the complementary solution. That is
ﬁrst we compute yc (solution to Ly = 0)
yc = C1 e−3 x + C2 e3 x
and we note that the e3 x term is a duplicate with our desired guess. We modify our guess to
y = Axe3 x and notice there is no duplication anymore. Let us try. Note that y = Ae3 x + 3Axe3 x and
y = 6Ae3 x + 9Axe3 x . So
y − 9y = 6Ae3 x + 9Axe3 x − 9Axe3 x = 6Ae3 x . 2.5. NONHOMOGENEOUS EQUATIONS 73 So 6Ae3 x is supposed to equal e3 x . Hence, 6A = 1 and so A = 1/6. Thus we can now write the
general solution as
1
y = yc + y p = C1 e−3 x + C2 e3 x + xe3 x .
6
Now what about the case when multiplying by x does not get rid of duplication. For example,
y − 6y + 9y = e3 x .
Note that yc = C1 e3 x + C2 xe3 x . So guessing y = Axe3 x would not get us anywhere. In this case we
want to guess y = Ax2 e3 x . Basically, we want to multiply our guess by x until all duplication is gone.
But no more! Multiplying too many times will also make the process not work.
Finally what if the right hand side is several terms, such as
Ly = e2 x + cos x.
In this case we ﬁnd u that solves Lu = e2 x and v that solves Lv = cos x (do each term separately).
Then note that if y = u + v, then Ly = e2 x + cos x. This is because L is linear; we have Ly =
L(u + v) = Lu + Lv = e2 x + cos x. 2.5.3 Variation of parameters The method of undetermined coeﬃcients will work for many basic problems that crop up. But it
does not work all the time. It only works when the right hand side of the equation Ly = f ( x) has
only ﬁnitely many linearly independent derivatives, so that we can write a guess that consists of
them all. Some equations are a bit tougher. Consider
y + y = tan x.
Note that each new derivative of tan x looks completely diﬀerent and cannot be written as a linear
combination of the previous derivatives. We get sec2 x, 2 sec2 x tan x, etc. . . .
This equation calls for a diﬀerent method. We present the method of variation of parameters,
which will handle any equation of the form Ly = f ( x), provided we can solve certain integrals. For
simplicity, we will restrict ourselves to second order equations, but the method will work for higher
order equations just as well (the computations will be more tedious).
Let us try to solve the example
Ly = y + y = tan x.
First we ﬁnd the complementary solution Ly = 0. We get yc = C1 y1 + C2 y2 , where y1 = cos x and
y2 = sin x. Now to try to ﬁnd a solution to the nonhomogeneous equation we try
y p = y = u1 y1 + u2 y2 , 74 CHAPTER 2. HIGHER ORDER LINEAR ODES where u1 and u2 are functions and not constants. We are trying to satisfy Ly = tan x. That gives us
one condition on the functions u1 and u2 . Compute (note the product rule!)
y = (u1 y1 + u2 y2 ) + (u1 y1 + u2 y2 ).
We can still impose one more condition at our will to simplify computations (we have two unknown
functions, so we are allowed two conditions). We require that (u1 y1 + u2 y2 ) = 0. This makes
computing the second derivative easier.
y = u1 y1 + u2 y2 ,
y = (u1 y1 + u2 y2 ) + (u1 y1 + u2 y2 ).
Since y1 and y2 are solutions to y + y = 0, we know that y1 = −y1 and y2 = −y2 . (Note: If the
equation was instead y + ay + by = 0 we would have yi = −ayi − byi .) So
y = (u1 y1 + u2 y2 ) − (u1 y1 + u2 y2 ).
Now note that
y = (u1 y1 + u2 y2 ) − y,
and hence
y + y = Ly = u1 y1 + u2 y2 .
For y to satisfy Ly = f ( x) we must have f ( x) = u1 y1 + u2 y2 .
So what we need to solve are the two equations (conditions) we imposed on u1 and u2
u1 y1 + u2 y2 = 0,
u1 y1 + u2 y2 = f ( x).
We can now solve for u1 and u2 in terms of f ( x), y1 and y2 . You will always get these formulas for
any Ly = f ( x). There is a general formula for the solution you can just plug into, but it is better to
just repeat what we do below. In our case the two equations become
u1 cos( x) + u2 sin( x) = 0,
−u1 sin( x) + u2 cos( x) = tan( x).
Hence
u1 cos( x) sin( x) + u2 sin2 ( x) = 0,
−u1 sin( x) cos( x) + u2 cos2 ( x) = tan( x) cos( x) = sin( x). 2.5. NONHOMOGENEOUS EQUATIONS 75 And thus
u2 sin2 ( x) + cos2 ( x) = sin( x),
u2 = sin( x),
u1 = − sin2 ( x)
= − tan( x) sin( x).
cos( x) Now we need to integrate u1 and u2 to get u1 and u2 .
1
sin( x) − 1
ln
+ sin( x),
2
sin( x) + 1 u1 = u1 dx = − tan( x) sin( x) dx = u2 = u2 dx = sin( x) dx = − cos( x). So our particular solution is
y p = u1 y1 + u2 y2 = sin( x) − 1
1
cos( x) ln
+ cos( x) sin( x) − cos( x) sin( x) =
2
sin( x) + 1
1
sin( x) − 1
= cos( x) ln
.
2
sin( x) + 1 The general solution to y + y = tan x is, therefore,
y = C1 cos( x) + C2 sin( x) + 2.5.4 sin( x) − 1
1
cos( x) ln
.
2
sin( x) + 1 Exercises Exercise 2.5.2: Find a particular solution of y − y − 6y = e2 x .
Exercise 2.5.3: Find a particular solution of y − 4y + 4y = e2 x .
Exercise 2.5.4: Solve the initial value problem y + 9y = cos(3 x) + sin(3 x) for y(0) = 2, y (0) = 1.
Exercise 2.5.5: Setup the form of the particular solution but do not solve for the coeﬃcients for
y(4) − 2y + y = e x .
Exercise 2.5.6: Setup the form of the particular solution but do not solve for the coeﬃcients for
y(4) − 2y + y = e x + x + sin x.
Exercise 2.5.7: a) Using variation of parameters ﬁnd a particular solution of y − 2y + y = e x . b)
Find a particular solution using undetermined coeﬃcients. c) Are the two solutions you found the
same? What is going on?
Exercise 2.5.8: Find a particular solution of y − 2y + y = sin( x2 ). It is OK to leave the answer as
a deﬁnite integral. 76 2.6 CHAPTER 2. HIGHER ORDER LINEAR ODES Forced oscillations and resonance Note: 2 lectures, §3.6 in [EP]
k F (t)
m Let us return back to the mass on a spring example. We will
now consider the case of forced oscillations. That is, we will
consider the equation
mx + cx + kx = F (t) damping c for some nonzero F (t). The setup is again: m is mass, c is friction, k is the spring constant and F (t)
is an external force acting on the mass.
What we are interested in is some periodic forcing, such as noncentered rotating parts, or perhaps
even loud sounds or other sources of periodic force. Once we learn about Fourier series in chapter 4
we will see that we cover all periodic functions by simply considering F (t) = F0 cos(ωt) (or sine
instead of cosine, the calculations will be essentially the same). 2.6.1 Undamped forced motion and resonance First let us consider undamped (c = 0) motion for simplicity. We have the equation
mx + kx = F0 cos(ωt).
This equation has the complementary solution (solution to the associated homogeneous equation)
xc = C1 cos(ω0 t) + C2 sin(ω0 t),
√
where ω0 = k/m is the natural frequency (angular). It is the frequency at which the system “wants
to oscillate” without external interference.
Let us suppose that ω0 ω. Try the solution x p = A cos(ωt) and solve for A. Note that we need
not have sine in our trial solution as on the left hand side we will only get cosines anyway. If you
include a sine it is ﬁne; you will ﬁnd that its coeﬃcient will be zero (I could not ﬁnd a rhyme).
We solve using the method of undetermined coeﬃcients. We ﬁnd that
xp = F0
cos(ωt).
− ω2 ) m(ω2
0 We leave it as an exercise to do the algebra required.
The general solution is
x = C1 cos(ω0 t) + C2 sin(ω0 t) + F0
cos(ωt).
m(ω2 − ω2 )
0 2.6. FORCED OSCILLATIONS AND RESONANCE 77 or written another way
x = C cos(ω0 t − γ) + F0
cos(ωt).
− ω2 ) m(ω2
0 Hence it is a superposition of two cosine waves at diﬀerent frequencies.
Example 2.6.1: Take
0.5 x + 8 x = 10 cos(πt), x(0) = 0, x (0) = 0.
√
Let us compute. First we read oﬀ the parameters: ω = π, ω0 = 8/0.5 = 4, F0 = 10, m = 0.5.
The general solution is
x = C1 cos(4t) + C2 sin(4t) + 20
cos(πt).
16 − π2 Solve for C1 and C2 using the initial conditions. It is easy to see that C1 =
Hence
20
x=
cos(πt) − cos(4t) .
16 − π2
Notice the “beating” behavior in Figure 2.5.
First use the trigonometric identity
0 5 10 −20
16−π2 15 and C2 = 0. 20 10 A−B
A+B
sin
= cos B − cos A
2
2 5 5 0 0 5 2 sin 10 5 to get that
20
4−π
4+π
x=
2 sin
t sin
t.
2
16 − π
2
2
Notice that x is a high frequency wave modulated
by a low frequency wave.
10 10
0 5 10 15 20 Now suppose that ω0 = ω. Obviously, we
cannot try the solution A cos(ωt) and then use the Figure 2.5: Graph of 1620π2 cos(πt) − cos(4t) .
−
method of undetermined coeﬃcients. We notice
that cos(ωt) solves the associated homogeneous equation. Therefore, we need to try x p = At cos(ωt)+
Bt sin(ωt). This time we do need the sine term since the second derivative of t cos(ωt) does contain
sines. We write the equation
F0
x + ω2 x =
cos(ωt).
m
Plugging into the left hand side we get
2 Bω cos(ωt) − 2Aω sin(ωt) = F0
cos(ωt).
m 78 CHAPTER 2. HIGHER ORDER LINEAR ODES Hence A = 0 and B = F0
.
2mω Our particular solution is F0
2mω x = C1 cos(ωt) + C2 sin(ωt) + t sin(ωt) and our general solution is
F0
t sin(ωt).
2mω The important term is the last one (the particular solution we found). We can see that this term
t
F0
grows without bound as t → ∞. In fact it oscillates between 2F0ω and −mωt . The ﬁrst two terms only
m
2
2
2
oscillate between ± C1 + C2 , which becomes smaller and smaller in proportion to the oscillations
of the last term as t gets larger. In Figure 2.6 we see the graph with C1 = C2 = 0, F0 = 2, m = 1,
ω = π.
By forcing the system in just the right frequency we produce very wild oscillations. This
kind of behavior is called resonance or sometimes
pure resonance. Sometimes resonance is desired.
For example, remember when as a kid you could
start swinging by just moving back and forth on
the swing seat in the correct “frequency”? You
were trying to achieve resonance. The force of
each one of your moves was small, but after a
while it produced large swings.
On the other hand resonance can be destructive. In an earthquake some buildings collapse
while others may be relatively undamaged. This
1
is due to diﬀerent buildings having diﬀerent resoFigure 2.6: Graph of π t sin(πt).
nance frequencies. So ﬁguring out the resonance
frequency can be very important.
A common (but wrong) example of destructive force of resonance is the Tacoma Narrows bridge
failure. It turns out there was an altogether diﬀerent phenomenon at play there∗ .
0 5 10 15 20 5.0 5.0 2.5 2.5 0.0 0.0 2.5 2.5 5.0 5.0 0 2.6.2 5 10 15 20 Damped forced motion and practical resonance In real life things are not as simple as they were above. There is, of course, some damping. Our
equation becomes
mx + cx + kx = F0 cos(ωt),
(2.8)
for some c > 0. We have solved the homogeneous problem before. We let
p=
∗ c
2m ω0 = k
.
m K. Billah and R. Scanlan, Resonance, Tacoma Narrows Bridge Failure, and Undergraduate Physics Textbooks,
American Journal of Physics, 59(2), 1991, 118–124, http://www.ketchum.org/billah/BillahScanlan.pdf 2.6. FORCED OSCILLATIONS AND RESONANCE 79 We replace equation (2.8) with
x + 2 px + ω2 x =
0 F0
cos(ωt).
m We ﬁnd the roots of the characteristic equation of the associated homogeneous problem are r1 , r2 =
−p ± p2 − ω2 . The form of the general solution of the associated homogeneous equation depends
0 on the sign of p2 − ω2 , or equivalently on the sign of c2 − 4km, as we have seen before. That is
0 C1 er1 t + C2 er2 t if c2 > 4km, − pt xc = C1 e + C2 te− pt
if c2 = 4km, − pt
e C cos(ω t) + C sin(ω t) if c2 < 4km , 1
1
2
1
where ω1 = ω2 − p2 . In any case, we can see that xc (t) → 0 as t → ∞. Furthermore, there can
0
be no conﬂicts when trying to solve for the undetermined coeﬃcients by trying x p = A cos(ωt) +
B sin(ωt). Let us plug in and solve for A and B. We get (the tedious details are left to reader)
(ω2 − ω2 ) B − 2ω pA sin(ωt) + (ω2 − ω2 )A + 2ω pB cos(ωt) =
0
0 F0
cos(ωt).
m We get that
A=
B=
We also compute C = √ (ω2 − ω2 )F0
0
2 , 2 . m(2ω p)2 + m(ω2 − ω2 )
0
2ω pF0
m(2ω p)2 + m(ω2 − ω2 )
0 A2 + B2 to be
C= F0
m (2ω p) +
2 .
(ω2
0 − 2
ω2 ) Thus our particular solution is
xp = (ω2 − ω2 )F0
0
m(2ω p) +
2 m(ω2
0 − 2
ω2 ) cos(ωt) + 2ω pF0 Or in the other notation we have amplitude C and phase shift γ where (if ω
tan γ = 2 m(2ω p) + m(ω2 − ω2 )
0
2 2ω p
B
=2
.
A ω0 − ω2 sin(ωt)
ω0 ) 80 CHAPTER 2. HIGHER ORDER LINEAR ODES Hence we have
xp = F0 cos(ωt − γ).
2 m (2ω p)2 + (ω2 − ω2 )
0 F0
If ω = ω0 we see that A = 0, B = C = 2mω p , and γ = π/2.
The exact formula is not as important as the idea. You should not memorize the above formula,
you should remember the ideas involved. For diﬀerent forcing function F , you will get a diﬀerent
formula for x p . So there is no point in memorizing this speciﬁc formula. You can always recompute
it later or look it up if you really need it. For reasons we will explain in a moment, we will call xc the transient solution and denote it
by xtr . We will call the x p we found above the steady periodic solution and denote it by x sp . The
general solution to our problem is
x = xc + x p = xtr + x sp .
We note that xc = xtr goes to zero as t → ∞,
as all the terms involve an exponential with a
negative exponent. Hence for large t, the eﬀect
of xtr is negligible and we will essentially only
see x sp . Hence the name transient. Notice that
x sp involves no arbitrary constants, and the initial
conditions will only aﬀect xtr . This means that
the eﬀect of the initial conditions will be negligible after some period of time. Because of this
behavior, we might as well focus on the steady
periodic solution and ignore the transient solution. See Figure 2.7 for a graph of diﬀerent initial
conditions.
Notice that the speed at which xtr goes to zero
Figure 2.7: Solutions with diﬀerent initial condepends on p (and hence c). The bigger p is (the
ditions for parameters k = 1, m = 1, F0 = 1,
bigger c is), the “faster” xtr becomes negligible.
c = 0.7, and ω = 1.1.
So the smaller the damping, the longer the “transient region.” This agrees with the observation
that when c = 0, the initial conditions aﬀect the behavior for all time (i.e. an inﬁnite “transient
region”).
0 5 10 15 20 5.0 5.0 2.5 2.5 0.0 0.0 2.5 2.5 5.0 5.0 0 5 10 15 20 Let us describe what we mean by resonance when damping is present. Since there were no
conﬂicts when solving with undetermined coeﬃcient, there is no term that goes to inﬁnity. What we
will look at however is the maximum value of the amplitude of the steady periodic solution. Let C
be the amplitude of x sp . If we plot C as a function of ω (with all other parameters ﬁxed) we can ﬁnd
its maximum. We call the ω that achieves this maximum the practical resonance frequency. We call 2.6. FORCED OSCILLATIONS AND RESONANCE 81 the maximal amplitude C (ω) the practical resonance amplitude. Thus when damping is present we
talk of practical resonance rather than pure resonance. A sample plot for three diﬀerent values of c
is given in Figure 2.8. As you can see the practical resonance amplitude grows as damping gets
smaller, and any practical resonance can disappear when damping is large.
0.0 0.5 1.0 1.5 2.0 2.5 3.0 2.5 2.5 2.0 2.0 1.5 1.5 1.0 1.0 0.5 0.5 0.0 0.0
0.0 0.5 1.0 1.5 2.0 2.5 3.0 Figure 2.8: Graph of C (ω) showing practical resonance with parameters k = 1, m = 1, F0 = 1. The
top line is with c = 0.4, the middle line with c = 0.8, and the bottom line with c = 1.6.
To ﬁnd the maximum we need to ﬁnd the derivative C (ω). Computation shows
C (ω) = −4ω(2 p2 + ω2 − ω2 )F0
0
2 3/2 m (2ω p)2 + (ω2 − ω2 )
0 . This is zero either when ω = 0 or when 2 p2 + ω2 − ω2 = 0. In other words, C (ω) = 0 when
0
ω= ω2 − 2 p2
0 or ω = 0. It can be shown that if ω2 − 2 p2 is positive, then ω2 − 2 p2 is the practical resonance frequency
0
0
(that is the point where C (ω) is maximal, note that in this case C (ω) > 0 for small ω). If ω = 0 is
the maximum, then essentially there is no practical resonance since we assume that ω > 0 in our
system. In this case the amplitude gets larger as the forcing frequency gets smaller.
If practical resonance occurs, the frequency is smaller than ω0 . As damping c (and hence p)
becomes smaller, the closer the practical resonance frequency goes to ω0 . So when damping is very
small, ω0 is a good estimate of the resonance frequency. This behavior agrees with the observation
that when c = 0, then ω0 is the resonance frequency.
The behavior will be more complicated if the forcing function is not an exact cosine wave, but
for example a square wave. It will be good to come back to this section once you have learned about
the Fourier series. 82 2.6.3 CHAPTER 2. HIGHER ORDER LINEAR ODES Exercises Exercise 2.6.1: Derive a formula for x sp if the equation is mx + cx + kx = F0 sin(ωt). Assume
c > 0.
Exercise 2.6.2: Derive a formula for x sp if the equation is mx + cx + kx = F0 cos(ωt) + F1 cos(3ωt).
Assume c > 0.
Exercise 2.6.3: Take mx + cx + kx = F0 cos(ωt). Fix m > 0 and k > 0. Now think of the function
C (ω). For what values of c (solve in terms of m, k, and F0 ) will there be no practical resonance
(that is, for what values of c is there no maximum of C (ω) for ω > 0).
Exercise 2.6.4: Take mx + cx + kx = F0 cos(ωt). Fix c > 0 and k > 0. Now think of the function
C (ω). For what values of m (solve in terms of c, k, and F0 ) will there be no practical resonance
(that is, for what values of m is there no maximum of C (ω) for ω > 0).
Exercise 2.6.5: Suppose a water tower in an earthquake acts as a massspring system. Assume
that the container on top is full and the water does not move around. The container then acts as a
mass and the support acts as the spring, where the induced vibrations are horizontal. Suppose that
the container with water has a mass of m = 10, 000 kg. It takes a force of 1000 newtons to displace
the container 1 meter. For simplicity assume no friction. When the earthquake hits the water tower
is at rest (it is not moving).
Suppose that an earthquake induces an external force F (t) = mAω2 cos(ωt).
a) What is the natural frequency of the water tower.
b) If ω is not the natural frequency, ﬁnd a formula for the maximal amplitude of the resulting
oscillations of the water container (the maximal deviation from the rest position). The motion will
be a high frequency wave modulated by a low frequency wave, so simply ﬁnd the constant in front
of the sines.
c) Suppose A = 1 and an earthquake with frequency 0.5 cycles per second comes. What is the
amplitude of the oscillations. Suppose that if the water tower moves more than 1.5 meter, the tower
collapses. Will the tower collapse? Chapter 3
Systems of ODEs
3.1 Introduction to systems of ODEs Note: 1 lecture, §4.1 in [EP]
Often we do not have just one dependent variable and one equation. And as we will see, we
may end up with systems of several equations and several dependent variables even if we start with
a single equation.
If we have several dependent variables, suppose y1 , y2 , . . . , yn we can have a diﬀerential equation
involving all of them and their derivatives. For example, y1 = f (y1 , y2 , y1 , y2 , x). Usually, when we
have two dependent variables we would have two equations such as
y1 = f1 (y1 , y2 , y1 , y2 , x),
y2 = f2 (y1 , y2 , y1 , y2 , x),
for some functions f1 and f2 . We call the above a system of diﬀerential equations. More precisely, it
is a second order system. Sometimes a system is easy to solve by solving for one variable and then
for the second variable.
Example 3.1.1: Take the ﬁrst order system
y1 = y1 ,
y2 = y1 − y2 ,
with initial conditions of the form y1 (0) = 1, y2 (0) = 2.
We note that y1 = C1 e x is the general solution of the ﬁrst equation. We can then plug this y1 into
the second equation and get the equation y2 = C1 e x − y2 , which is a linear ﬁrst order equation that is
easily solved for y2 . By the method of integrating factor we get
e x y2 = C1 2 x
e + C2 ,
2
83 84
or y2 = CHAPTER 3. SYSTEMS OF ODES
C1 x
e
2 + C2 e− x . The general solution to the system is, therefore,
y1 = C1 e x ,
y2 = C1 x
e + C 2 e− x .
2 We can now solve for C1 and C2 given the initial conditions. We substitute x = 0 and ﬁnd that
C1 = 1 and C2 = 3/2.
Generally, we will not be so lucky to be able to solve like in the ﬁrst example, and we will have
to solve for all variables at once.
As an example application, let us think of mass and spring
systems again. Suppose we have one spring with constant k but
m2
m2
two masses m1 and m2 . We can think of the masses as carts, and
we will suppose that they ride along with no friction. Let x1 be the
displacement of the ﬁrst cart and x2 be the displacement of the second cart. That is, we put the two
carts somewhere with no tension on the spring, and we mark the position of the ﬁrst and second cart
and call those the zero position. That is, x1 = 0 is a diﬀerent position on the ﬂoor than the position
corresponding to x2 = 0. The force exerted by the spring on the ﬁrst cart is k( x2 − x1 ), since x2 − x1
is how far the string is stretched (or compressed) from the rest position. The force exerted on the
second cart is the opposite, thus the same thing with a negative sign. Newton’s second law states
that force equals mass times acceleration.
k m1 x1 = k( x2 − x1 ),
m2 x2 = −k( x2 − x1 ).
In this system we cannot solve for the x1 variable separately. That we must solve for both x1
and x2 at once is intuitively obvious, since where the ﬁrst cart goes depends exactly on where the
second cart goes and viceversa.
Before we talk about how to handle systems, let us note that in some sense we need only consider
ﬁrst order systems. Take an nth order diﬀerential equation
y(n) = F (y(n−1) , . . . , y , y, x).
Deﬁne new variables u1 , . . . , un and write the system
u1 = u2
u2 = u3
.
.
.
un−1 = un
un = F (un , un−1 , . . . , u2 , u1 , x). 3.1. INTRODUCTION TO SYSTEMS OF ODES 85 Now try to solve this system for u1 , u2 , . . . , un . Once you have solved for the u’s, you can discard u2
through un and let y = u1 . We note that this y solves the original equation.
A similar process can be followed for a system of higher order diﬀerential equations. For
example, a system of k diﬀerential equations in k unknowns, all of order n, can be transformed into
a ﬁrst order system of n × k equations and n × k unknowns.
Example 3.1.2: Sometimes we can use this idea in reverse as well. Let us take the system
x = 2y − x, y = x, where the independent variable is t. We wish to solve for the initial conditions x(0) = 1, y(0) = 0.
We ﬁrst notice that if we diﬀerentiate the ﬁrst equation once we get y = x and now we know
what x is in terms of x and y.
y = x = 2y − x = 2y − y .
So we now have an equation y + y − 2y = 0. We know how to solve this equation and we ﬁnd that
y = C1 e−2t + C2 et . Once we have y we can plug in to get x.
x = y = −2C1 e−2t + C2 et .
We solve for the initial conditions 1 = x(0) = −2C1 + C2 and 0 = y(0) = C1 + C2 . Hence, C1 = −C2
and 1 = 3C2 . So C1 = −1/3 and C2 = 1/3. Our solution is
2e−2t + et
x=
,
3 −e−2t + et
y=
.
3 Exercise 3.1.1: Plug in and check that this really is the solution.
It is useful to go back and forth between systems and higher order equations for other reasons.
For example, the ODE approximation methods are generally only given as solutions for ﬁrst order
systems. It is not very hard to adapt the code for the Euler method for a ﬁrst order equation to ﬁrst
order systems. We essentially just treat the dependent variable not as a number but as a vector. In
many mathematical computer languages there is almost no distinction in syntax.
In fact, this is what IODE was doing when you had it solve a second order equation numerically
in the IODE Project III if you have done that project.
The above example was what we will call a linear ﬁrst order system, as none of the dependent
variables appear in any functions or with any higher powers than one. It is also autonomous as the
equations do not depend on the independent variable t.
For autonomous systems we can easily draw the socalled direction ﬁeld or vector ﬁeld. That is,
a plot similar to a slope ﬁeld, but instead of giving a slope at each point, we give a direction (and a
magnitude). The previous example x = 2y − x, y = x says that at the point ( x, y) the direction in
which we should travel to satisfy the equations should be the direction of the vector (2y − x, x) with
the speed equal to the magnitude of this vector. So we draw the vector (2y − x, x) based at the point 86 CHAPTER 3. SYSTEMS OF ODES ( x, y) and we do this for many points on the xyplane. We may want to scale down the size of our
vectors to ﬁt many of them on the same direction ﬁeld. See Figure 3.1.
We can now draw a path of the solution in the plane. That is, suppose the solution is given by
x = f (t), y = g(t), then we can pick an interval of t (say 0 ≤ t ≤ 2 for our example) and plot all
the points f (t), g(t) for t in the selected range. The resulting picture is usually called the phase
portrait (or phase plane portrait). The particular curve obtained we call the trajectory or solution
curve. An example plot is given in Figure 3.2. In this ﬁgure the line starts at (1, 0) and travels along
the vector ﬁeld for a distance of 2 units of t. Since we solved this system precisely we can compute
x(2) and y(2). We get that x(2) ≈ 2.475 and y(2) ≈ 2.457. This point corresponds to the top right
end of the plotted solution curve in the ﬁgure.
1 0 1 2 3 1 0 1 2 3 3 3 3 3 2 2 2 2 1 1 1 1 0 0 0 0 1 1 1
1 0 1 2 3 Figure 3.1: The direction ﬁeld for x = 2y − x,
y = x. 1
1 0 1 2 3 Figure 3.2: The direction ﬁeld for x = 2y − x,
y = x with the trajectory of the solution starting
at (1, 0) for 0 ≤ t ≤ 2. Notice the similarity to the diagrams we drew for autonomous systems in one dimension. But
now note how much more complicated things become if we allow just one more dimension.
Also note that we can draw phase portraits and trajectories in the xyplane even if the system is
not autonomous. In this case however we cannot draw the direction ﬁeld, since the ﬁeld changes as
t changes. For each t we would get a diﬀerent direction ﬁeld. 3.1.1 Exercises Exercise 3.1.2: Find the general solution of x1 = x2 − x1 + t, x2 = x2 .
Exercise 3.1.3: Find the general solution of x1 = 3 x1 − x2 + et , x2 = x1 .
Exercise 3.1.4: Write ay + by + cy = f ( x) as a ﬁrst order system of ODEs.
Exercise 3.1.5: Write x + y2 y − x3 = sin(t), y + ( x + y )2 − x = 0 as a ﬁrst order system of ODEs. 3.2. MATRICES AND LINEAR SYSTEMS 3.2 87 Matrices and linear systems Note: 1 and a half lectures, ﬁrst part of §5.1 in [EP] 3.2.1 Matrices and vectors Before we can start talking about linear systems of ODEs, we will need to talk about matrices, so
let us review these brieﬂy. A matrix is an m × n array of numbers (m rows and n columns). For
example, we denote a 3 × 5 matrix as follows a11 a12 a13 a14 a15 A = a21 a22 a23 a24 a25 . a31 a32 a33 a34 a35
By a vector we will usually mean a column vector that is an n × 1 matrix. If we mean a row
vector we will explicitly say so (a row vector is a 1 × n matrix). We will usually denote matrices by
upper case letters and vectors by lower case letters with an arrow such as x or b. By 0 we will mean
the vector of all zeros.
It is easy to deﬁne some operations on matrices. Note that we will want 1 × 1 matrices to really
act like numbers, so our operations will have to be compatible with this viewpoint.
First, we can multiply by a scalar (a number). This means just multiplying each entry by the
same number. For example,
123
246
2
=
.
456
8 10 12
Matrix addition is also easy. We add matrices element by element. For example,
123
1 1 −1
23 2
+
=
.
456
02 4
4 7 10
If the sizes do not match, then addition is not deﬁned.
If we denote by 0 the matrix of with all zero entries, by c, d some scalars, and by A, B, C some
matrices, we have the following familiar rules.
A + 0 = A = 0 + A,
A + B = B + A,
(A + B) + C = A + ( B + C ),
c(A + B) = cA + cB,
(c + d)A = cA + dA. 88 CHAPTER 3. SYSTEMS OF ODES Another useful operation for matrices is the socalled transpose. This operation just swaps rows
and columns of a matrix. The transpose of A is denoted by AT . Example:
123
456 3.2.2 T 1 4 = 2 5 36 Matrix multiplication Let us now deﬁne matrix multiplication. First we deﬁne the socalled dot product (or inner product)
of two vectors. Usually this will be a row vector multiplied with a column vector of the same size.
For the dot product we multiply each pair of entries from the ﬁrst and the second vector and we sum
these products. The result is a single number. For example, a1 a2 b 1 a3 · b2 = a1 b1 + a2 b2 + a3 b3 . b3 And similarly for larger (or smaller) vectors.
Armed with the dot product we can deﬁne the product of matrices. First let us denote by rowi (A)
the ith row of A and by column j (A) the jth column of A. Now for an m × n matrix A and an n × p
matrix B we can deﬁne the product AB. We let AB be an m × p matrix whose i jth entry is
rowi (A) · column j ( B).
Do note how the sizes match up. Example: 1 0 −1 1 2 3 1 1 1 = 4 5 6 10 0
= 1·1+2·1+3·1
4·1+5·1+6·1 1·0+2·1+3·0
4·0+5·1+6·0 1 · (−1) + 2 · 1 + 3 · 0
621
=
4 · (−1) + 5 · 1 + 6 · 0
15 5 1 For multiplication we will want an analogue of a 1. This is the socalled identity matrix. The
identity matrix is a square matrix with 1s on the main diagonal and zeros everywhere else. It is
usually denoted by I . For each size we have a diﬀerent identity matrix and so sometimes we may
denote the size as a subscript. For example, the I3 would be the 3 × 3 identity matrix 1 0 0 I = I3 = 0 1 0 . 001 3.2. MATRICES AND LINEAR SYSTEMS 89 We have the following rules for matrix multiplication. Suppose that A, B, C are matrices of the
correct sizes so that the following make sense. Let α denote a scalar (number).
A( BC ) = (AB)C,
A( B + C ) = AB + AC,
( B + C )A = BA + CA,
α(AB) = (αA) B = A(α B),
IA = A = AI .
A few warnings are in order however.
(i) AB BA in general (it may be true by ﬂuke sometimes). That is, matrices do not commute. (ii) AB = AC does not necessarily imply B = C even if A is not 0.
(iii) AB = 0 does not necessarily mean that A = 0 or B = 0.
For the last two items to hold we would need to essentially “divide” by a matrix. This is where
matrix inverse comes in. Suppose that A is an n × n matrix and that there exists another n × n matrix
B such that
AB = I = BA.
Then we call B the inverse of A and we denote B by A−1 . If the inverse of A exists, then we call A
invertible. If A is not invertible we say A is singular or just say it is not invertible.
If A is invertible, then AB = AC does imply that B = C (this also means that the inverse is
unique). We just multiply both sides by A−1 to get A−1 AB = A−1 AC or IB = IC or B = C . It is also
−1
not hard to see that (A−1 ) = A. 3.2.3 The determinant We can now talk about determinants of square matrices. We deﬁne the determinant of a 1 × 1 matrix
as the value of its own entry. For a 2 × 2 matrix we deﬁne
det ab
cd = ad − bc. Before trying to compute determinant for larger matrices, let us ﬁrst note the meaning of the
determinant. Consider an n × n matrix as a mapping of Rn to Rn . For example, a 2 × 2 matrix A is a
mapping of the plane where x gets sent to A x. Then the determinant of A is the factor by which
the volume of objects gets changed. For example, if we take the unit square (square of sides 1) in
the plane, then A takes the square to a parallelogram of area det(A). The sign of det(A) denotes
changing of orientation (if the axes got ﬂipped). For example,
A= 11
.
−1 1 90 CHAPTER 3. SYSTEMS OF ODES Then det(A) = 1 + 1 = 2. Now let us see where the square with vertices (0, 0), (1, 0), (0, 1), and
(1, 1) gets sent. Obviously (0, 0) gets sent to (0, 0). Now
111
1
=
,
−1 1 0
−1 110
1
=
,
−1 1 1
1 111
2
=
.
−1 1 1
0 √
So it turns out that the image of the square is another square. This one has a side of length 2 and
is therefore of area 2.
If you think back to high school geometry, you may have seen a formula for computing the area
of a parallelogram with vertices (0, 0), (a, c), (b, d) and (a + b, c + d). And it is precisely
det ab
cd . The vertical lines here mean absolute value. The matrix
parallelogram. ab
cd carries the unit square to the given Now we can deﬁne the determinant for larger matrices. We deﬁne Ai j as the matrix A with the
ith row and the jth column deleted. To compute the determinant of a matrix, pick one row, say the ith
row and compute.
n (−1)i+ j ai j det(Ai j ). det(A) =
j =1 For example, for the ﬁrst row we get +a det(A ) if n is odd, 1n 1n
det(A) = a11 det(A11 ) − a12 det(A12 ) + a13 det(A13 ) − · · · −a det(A ) if n even. 1n
1n
We alternately add and subtract the determinants of the submatrices Ai j for a ﬁxed i and all j. For
example, for a 3 × 3 matrix, picking the ﬁrst row, we would get det(A) = a11 det(A11 ) − a12 det(A12 ) +
a13 det(A13 ). For example, 1 2 3 56
46
45 det 4 5 6 = 1 · det
− 2 · det
+ 3 · det 89
79
78 789
= 1(5 · 9 − 6 · 8) − 2(4 · 9 − 6 · 7) + 3(4 · 8 − 5 · 7) = 0.
The numbers (−1)i+ j det(Ai j ) are called cofactors of the matrix and this way of computing the
determinant is called the cofactor expansion. It is also possible to compute the determinant by
expanding along columns (picking a column instead of a row above).
Note that a common notation for the determinant is a pair of vertical lines.
ab
ab
= det
.
cd
cd 3.2. MATRICES AND LINEAR SYSTEMS 91 I personally ﬁnd this notation confusing since vertical lines for me usually mean a positive quantity,
while determinants can be negative. So I will not ever use this notation in these notes.
One of the most important properties of determinants (in the context of this course) is the
following theorem.
Theorem 3.2.1. An n × n matrix A is invertible if and only if det(A) 0. In fact, we have a formula for the inverse of a 2 × 2 matrix
ab
cd −1 = 1
d −b
.
ad − bc −c a Notice the determinant of the matrix in the denominator of the fraction. The formula only works if
the determinant is nonzero, otherwise we are dividing by zero. 3.2.4 Solving linear systems One application of matrices we will need is to solve systems of linear equations. This may be best
shown by example. Suppose that we have the following system of linear equations
2 x1 + 2 x2 + 2 x3 = 2,
x1 + x2 + 3 x3 = 5,
x1 + 4 x2 + x3 = 10.
Without changing the solution, we could swap equations in this system, we could multiply any
of the equations by a nonzero number, and we could add a multiple of one equation to another
equation. It turns out these operations always suﬃce to ﬁnd a solution.
It is easier to write the system as a matrix equation. Note that the system can be written as 2 2 2 x1 2 1 1 3 x2 = 5 . 1 4 1 x3
10
To solve the system we put the coeﬃcient matrix (the matrix on the left hand side of the equation)
together with the vector on the right and side and get the socalled augmented matrix 2 2 2 2 1 1 3 5 . 1 4 1 10
We then apply the following three elementary operations. 92 CHAPTER 3. SYSTEMS OF ODES (i) Swap two rows.
(ii) Multiply a row by a nonzero number.
(iii) Add a multiple of one row to another row.
We will keep doing these operations until we get into a state where it is easy to read oﬀ the answer or
until we get into a contradiction indicating no solution, for example if we come up with an equation
such as 0 = 1.
Let us work through the example. First multiply the ﬁrst row by 1/2 to obtain 1 1 1 1 1 1 3 5 . 1 4 1 10
Now subtract the ﬁrst row from the second and third row. 1 1 1 1 0 0 2 4 0309
Multiply the last row by 1/3 and the second row by 1/2. 1 1 1 1 0 0 1 2 0103
Swap rows 2 and 3. 1 1 1 1 0 1 0 3 0012
Subtract the last row from the ﬁrst, then subtract the second row from the ﬁrst. 1 0 0 −4 0 1 0 3 001 2
If we think about what equations this augmented matrix represents, we see that x1 = −4, x2 = 3,
and x3 = 2. We try this solution in the original system and, voilà, it works!
Exercise 3.2.1: Check that this solution works. 3.2. MATRICES AND LINEAR SYSTEMS 93 If we write this equation in matrix notation as
A x = b,
where A is the matrix
inverse, 222
113
141 and b is the vector 2
5
10 . The solution can be also computed with the x = A −1 A x = A −1 b .
One last note to make about linear systems of equations is that it is possible that the solution is
not unique (or that no solution exists). It is easy to tell if a solution does not exist. If during the
row reduction you come up with a row where all the entries except the last one are zero (the last
entry in a row corresponds to the right hand side of the equation) the system is inconsistent and
has no solution. For example if for a system of 3 equations and 3 unknowns you ﬁnd a row such as
[ 0 0 0 1 ] in the augmented matrix, you know the system is inconsistent.
You generally try to use row operations until the following conditions are satisﬁed. The ﬁrst
nonzero entry in each row is called the leading entry.
(i) There is only one leading entry in each column.
(ii) All the entries above and below a leading entry are zero.
(iii) All leading entries are 1.
Such a matrix is said to be in reduced row echelon form. The variables corresponding to columns
with no leading entries are said to be free variables. Free variables mean that we can pick those
variables to be anything we want and then solve for the rest of the unknowns.
Example 3.2.1: The following augmented matrix is in reduced row echelon form. 1 2 0 3 0 0 1 1 0000
If the variables are named x1 , x2 , and x3 , then x2 is the free variable and x1 = 3 − 2 x2 and x3 = 1.
On the other hand if during the row reduction process you come up with the matrix 1 2 13 3 0 0 1 1 , 00 0 3
there is no need to go further. The last row corresponds to the equation 0 x1 + 0 x2 + 0 x3 = 3, which
is preposterous. Hence, no solution exists. 94 3.2.5 CHAPTER 3. SYSTEMS OF ODES Computing the inverse If the coeﬃcient matrix is square and there exists a unique solution x to A x = b for any b, then A
is invertible. In fact by multiplying both sides by A−1 you can see that x = A−1 b. So it is useful to
compute the inverse, if you want to solve the equation for many diﬀerent right hand sides b.
The 2 × 2 inverse is basically given by a formula, but it is not hard to also compute inverses of
larger matrices. While we will not have too much occasion to compute inverses for larger matrices
than 2 × 2 by hand, let us touch on how to do it. Finding the inverse of A is actually just solving a
bunch of linear equations. If you can solve A xk = ek where ek is the vector with all zeros except a
1 at the kth position, then the inverse is the matrix with the columns xk for k = 1, . . . , n (exercise:
why?). Therefore, to ﬁnd the inverse we can write a larger n × 2n augmented matrix [ A I ], where I
is the identity. If you do row reduction and put the matrix in reduced row echelon form, then the
matrix will be of the form [ I A−1 ] if and only if A is invertible, so you can just read oﬀ the inverse
A−1 . 3.2.6 Exercises Exercise 3.2.2: Solve x= 12
34 5
6 by using matrix inverse. Exercise 3.2.3: Compute determinant of 9 −2 −6
−8 3 6
10 −2 −6 Exercise 3.2.4: Compute determinant of 1
4
6
8 2
0
0
0 3
5
7
10 1
0
0
1 .
. Hint: Expand along the proper row or column to make the calculations simpler.
123
111
010 Exercise 3.2.5: Compute inverse of . 123
456
78h not invertible? Is there only one such h? Are there several? Exercise 3.2.7: For which h is h11
0h0
11h not invertible? Find all such h. Exercise 3.2.8: Solve 9 −2 −6
−8 3 6
10 −2 −6 x= 1
2
3 Exercise 3.2.9: Solve 537
844
633 Exercise 3.2.6: For which h is
Inﬁnitely many. Exercise 3.2.10: Solve 3
3
0
2 2
3
2
3 x=
3
3
4
4 0
3
2
3 2
0
0 x= . .
2
0
4
1 . 3.3. LINEAR SYSTEMS OF ODES 3.3 95 Linear systems of ODEs Note: less than 1 lecture, second part of §5.1 in [EP]
First let us talk about matrix or vector valued functions. This is essentially just a matrix whose
entries depend on some variable. Let us say the independent variable is t. Then a vector valued
function x(t) is really something like x1 (t) x2 (t) . . x(t) = . . xn (t)
Similarly a matrix valued function is something such as a11 (t) a12 (t) · · · a1n (t) a21 (t) a22 (t) · · · a2n (t) A(t) = .
.
.
. .
.. .
.
.
. .
. an1 (t) an2 (t) · · · ann (t)
We can talk about the derivative A (t) or dA and this is just the matrix valued function whose i jth
dt
entry is ai j (t).
Rules of diﬀerentiation of matrix valued functions are similar to rules for normal functions. Let
A and B be matrix valued functions. Let c a scalar and C be a constant matrix. Then
(A + B)
(AB)
(cA)
(CA)
(AC ) =A +B
= A B + AB
= cA
= CA
=AC Do note the order in the last two expressions.
A ﬁrst order linear system of ODEs is a system that can be written as
x (t) = P(t) x(t) + f (t).
Where P is a matrix valued function, and x and f are vector valued functions. We will often suppress
the dependence on t and only write x = P x + f . A solution is of course a vector valued function x
satisfying the equation.
For example, the equations
x1 = 2tx1 + et x2 + t2 ,
x1
x2 =
− x2 + et ,
t 96 CHAPTER 3. SYSTEMS OF ODES can be written as
x= t2
2t e t
x+ t .
1/t −1
e We will mostly concentrate on equations that are not just linear, but are in fact constant coeﬃcient
equations. That is, the matrix P will be a constant and not depend on t.
When f = 0 (the zero vector), then we say the system is homogeneous. For homogeneous linear
systems we still have the principle of superposition, just like for single homogeneous equations.
Theorem 3.3.1 (Superposition). Let x = P x be a linear homogeneous system of ODEs. Suppose
that x1 , . . . , xn are n solutions of the equation, then
x = c1 x1 + c2 x2 + · · · + cn xn , (3.1) is also a solution. If furthermore this is a system of n equations (P is n × n), and x1 , . . . , xn are
linearly independent. Then every solution can be written as (3.1).
Linear independence for vector valued functions is essentially the same as for normal functions.
x1 , . . . , xn are linearly independent if and only if
c1 x1 + c2 x2 + · · · + cn xn = 0
has only the solution c1 = c2 = · · · = cn = 0.
The linear combination c1 x1 + c2 x2 + · · · + cn xn could always be written as
X (t) c,
where X (t) is the matrix with columns x1 , . . . , xn , and c is the column vector with entries c1 , . . . , cn .
X (t) is called the fundamental matrix, or fundamental matrix solution.
To solve nonhomogeneous ﬁrst order linear systems. We apply the same technique as we did
before.
Theorem 3.3.2. Let x = P x + f be a linear system of ODEs. Suppose x p is one particular solution.
Then every solution can be written as
x = xc + x p ,
where xc is a solution to the associated homogeneous equation ( x = P x).
So the procedure will be exactly the same. We ﬁnd a particular solution to the nonhomogeneous
equation, then we ﬁnd the general solution to the associated homogeneous equation and we add the
two.
Alright, suppose you have found the general solution x = P x + f . Now you are given an initial
condition of the form x(t0 ) = b for some constant vector b. Suppose that X (t) is the fundamental 3.3. LINEAR SYSTEMS OF ODES 97 matrix solution of the associated homogeneous equation (i.e. columns of X are solutions). The
general solution is written as
x(t) = X (t)c + x p (t).
Then we are seeking a vector c such that
b = x(t0 ) = X (t0 )c + x p (t0 ).
Or in other words we are solving the nonhomogeneous system of linear equations
X (t0 )c = b − x p (t0 )
for c.
Example 3.3.1: In § 3.1 we solved the following system
x1 = x1 ,
x2 = x1 − x2 .
with initial conditions x1 (0) = 1, x2 (0) = 2.
This is a homogeneous system, so f = 0. We write the system as
x= 10
x,
1 −1 x(0) = 1
.
2 We found the general solution was x1 = c1 et and x2 = c21 et + c2 e−t . Hence in matrix notation, the
fundamental matrix solution is
et 0
X (t) = 1 t −t .
ee
2
It is not hard to see that the columns of this matrix are linearly independent. To see this, just plug in
t = 0 and note that the two constant vectors are already linearly independent here.
Hence to solve the initial problem we solve the equation
X (0)c = b,
or in other words,
10
1
c=
.
1
1
2
2
After a single elementary row operation we ﬁnd that c =
x(t) = X (t)c =
This agrees with our previous solution. et
1t
e
2 01
=
e−t 3
2 1
3/2 . Hence our solution is 1t
e
2 et
.
3
+ 2 e−t 98 3.3.1 CHAPTER 3. SYSTEMS OF ODES Exercises Exercise 3.3.1: Write the system x1 = 2 x1 − 3tx2 + sin t, x2 = et x1 + 3 x2 + cos t as in the form
x = P(t) x + f (t).
1
Exercise 3.3.2: a) Verify that the system x = 1 3 x has the two solutions 1 e4t and −1 e−2t . b)
31
1
Write down the general solution. c) Write down the general solution in the form x1 =?, x2 =? (i.e.
write down a formula for each element of the solution). Exercise 3.3.3: Verify that 1
1 et and 1
−1 1 1 et are linearly independent. Hint: Just plug in t = 0. Exercise 3.3.4: Verify that 1 et and −1 et and
0
1
be a bit more tricky than in the previous exercise.
Exercise 3.3.5: Verify that t
t2 and t3
t4 1
−1
1 e2t are linearly independent. Hint: You must are linearly independent. 3.4. EIGENVALUE METHOD 3.4 99 Eigenvalue method Note: 2 lectures, §5.2 in [EP]
In this section we will learn how to solve linear homogeneous constant coeﬃcient systems
of ODEs by the eigenvalue method. Suppose we have a linear constant coeﬃcient homogeneous
system
x = P x,
where P is a constant square matrix. Suppose we try to adapt the method for the single constant
coeﬃcient equation by trying the function eλt . However, x is a vector. So we try x = veλt , where v is
an arbitrary constant vector. We plug this x into the equation to get
λveλt = Pveλt .
We divide by eλt and notice that we are looking for a scalar λ and a vector v that satisfy the equation
λv = Pv.
To solve this equation we need a little bit more linear algebra, which we now review. 3.4.1 Eigenvalues and eigenvectors of a matrix Let A be a constant square matrix. Suppose there is a scalar λ and a nonzero vector v such that
Av = λv.
We then call λ an eigenvalue of A and v is said to be a corresponding eigenvector.
Example 3.4.1: The matrix
because 21
01 has an eigenvalue of λ = 2 with a corresponding eigenvector 1
0 211
2
1
=
=2
.
010
0
0
If we rewrite the equation for an eigenvalue as
(A − λI )v = 0,
we notice that this equation has a nonzero solution v only if A − λI is not invertible. Were it
invertible, we could write (A − λI )−1 (A − λI )v = (A − λI )−1 0, which implies v = 0. Therefore, A
has the eigenvalue λ if and only if λ solves the equation
det(A − λI ) = 0.
This means that we will be able to ﬁnd an eigenvalue without ﬁnding the corresponding
eigenvector. The eigenvector will have to be found later, once λ is known. 100 CHAPTER 3. SYSTEMS OF ODES
211 Example 3.4.2: Find all eigenvalues of 1 2 0 .
002
We write 2 1 1
1 0 0
2 − λ
1
1 2−λ
0 =
det 1 2 0 − λ 0 1 0 = det 1 002
001
0
0
2−λ
= (2 − λ)2 (2 − λ)2 − 1 = −(λ − 1)(λ − 2)(λ − 3).
and so the eigenvalues are λ = 1, λ = 2, and λ = 3.
Note that for an n × n matrix, the polynomial we get by computing det(A − λI ) will be of degree
n, and hence we will in general have n eigenvalues.
To ﬁnd an eigenvector corresponding to λ, we write
(A − λI )v = 0,
and solve for a nontrivial (nonzero) vector v. If λ is an eigenvalue, this will always be possible.
211 Example 3.4.3: Find the eigenvector of 1 2 0 corresponding to the eigenvalue λ = 3.
002
We write 1 0 0 v1 −1 1 1 v1 2 1 1 (A − λI )v = 1 2 0 − 3 0 1 0 v2 = 1 −1 0 v2 = 0. 0 0 −1 v3
0 0 1 v3
002
It is easy to solve this system of linear equations. We write down the augmented matrix −1 1 1 0 1 −1 0 0 0 0 −1 0
and perform row operations (exercise: which ones?) 1 −1 0 0 0 1 000 until we get 0 0 . 0 The equations the entries of v have to satisfy are, therefore, v1 − v2 = 0, v3 = 0, and v2 is a free
variable. We can pick v2 to be arbitrary (but nonzero) and let v1 = v2 and of course v3 = 0. For
example, v = 1
1
0 . Let us verify that we really have an eigenvector corresponding to λ = 3: 2 1 1 1 3
1 1 2 0 1 = 3 = 3 1 . 0020
0
0 Yay! It worked. 3.4. EIGENVALUE METHOD 101 Exercise 3.4.1 (easy): Are the eigenvectors unique? Can you ﬁnd a diﬀerent eigenvector for λ = 3
in the example above? How are the two eigenvectors related?
Exercise 3.4.2: Note that when the matrix is 2 × 2 you do not need to write down the augmented
matrix and do row operations when computing eigenvectors (if you have computed the eigenvalues
correctly). Can you see why? Try it for the matrix 2 1 .
12 3.4.2 The eigenvalue method with distinct real eigenvalues OK. We have the system of equations
x = P x.
We ﬁnd the eigenvalues λ1 , λ2 , . . . , λn of the matrix P, and the corresponding eigenvectors v1 , v2 ,
. . . , vn . Now we notice that the functions v1 eλ1 t , v2 eλ2 t , . . . , vn eλn t are solutions of the system of
equations and hence x = c1 v1 eλ1 t + c2 v2 eλ2 t + · · · + cn vn eλn t is a solution.
Theorem 3.4.1. Take x = P x. If P is n × n and has n distinct real eigenvalues λ1 , λ2 , . . . , λn , then
there are n linearly independent corresponding eigenvectors v1 , v2 , . . . , vn , and the general solution
to this system of ODEs can be written as.
x = c1 v1 eλ1 t + c2 v2 eλ2 t + · · · + cn vn eλn t .
Example 3.4.4: Suppose we take the system 2 1 1 x = 1 2 0 x. 002
Find the general solution.
We have found the eigenvalues 1, 2, 3 earlier. We have found the eigenvector
1 eigenvalue 3. In similar fashion we ﬁnd the eigenvector −1 for the eigenvalue 1 and
0
eigenvalue 2 (exercise: check). Hence our general solution is t 3t
1 t 0 2t 1 3t e t+ e 3t x = −1 e + 0 e + 1 e = −e + e . 2t
0
1
0
e 1
1
0
0
1
−1 for the
for the Exercise 3.4.3: Check that this really solves the system.
Note: If you write a homogeneous linear constant coeﬃcient nth order equation as a ﬁrst order
system (as we did in § 3.1), then the eigenvalue equation
det(P − λI ) = 0.
is essentially the same as the characteristic equation we got in § 2.2 and § 2.3. 102 3.4.3 CHAPTER 3. SYSTEMS OF ODES Complex eigenvalues A matrix might very well have complex eigenvalues even if all the entries are real. For example,
suppose that we have the system
11
x=
x.
−1 1
Let us compute the eigenvalues of the matrix P =
det(P − λI ) = det 1−λ
1
−1 1 − λ 11
−1 1 . = (1 − λ)2 + 1 = λ2 − 2λ + 2 = 0. Thus λ = 1 ± i. The corresponding eigenvectors will also be complex. First take λ = 1 − i,
(P − (1 − i)I )v = 0,
i1
v = 0.
−1 i
It is obvious that the equations iv1 + v2 = 0 and −v1 + iv2 = 0 are multiples of each other. So we only
i
need to consider one of them. After picking v2 = 1, for example, we have an eigenvector v = 1 .
−i
In similar fashion we ﬁnd that 1 is an eigenvector corresponding to the eigenvalue 1 + i.
We could write the solution as
x = c1 i (1−i)t
−i (1+i)t
c ie(1−i)t − c2 ie(1+i)t
e
+ c2
e
= 1 (1−i)t
.
1
1
c1 e
+ c2 e(1+i)t 1 But then we would need to look for complex values c1 and c2 to solve any initial conditions. And
even then it is perhaps not completely clear that we get a real solution. We could use Euler’s formula
here and do the whole song and dance we did before, but we will not. We will do something a bit
smarter ﬁrst.
We claim that we did not have to look for the second eigenvector (nor for the second eigenvalue).
All complex eigenvalues come in pairs (because the matrix P is real).
¯
First a small side note. The real part of a complex number z can be computed as z+z , where the
2
bar above z means a + ib = a − ib. This operation is called the complex conjugate. Note that for a
real number a, a = a. Similarly we can bar whole vectors or matrices. If a matrix P is real, then
¯
P = P. We note that P x = P x = P x. Or
¯
(P − λI )v = (P − λI )v.
So if v is an eigenvector corresponding to eigenvalue a + ib, then v is an eigenvector corresponding
to an eigenvalue a − ib.
Now suppose that a + ib is a complex eigenvalue of P, v the corresponding eigenvector and
hence
x1 = ve(a+ib)t 3.4. EIGENVALUE METHOD 103 is a solution (complex valued) of x = P x. Then note that ea+ib = ea−ib and
x2 = x1 = ve(a−ib)t
is also a solution. The function
x3 = Re x1 = Re ve(a+ib)t = x1 + x1 x1 + x2
=
2
2 is also a solution. And it is realvalued! Similarly as Im z =
x4 = Im x1 = z−z
¯
2i is the imaginary part we ﬁnd that x1 − x2
.
2i is also a realvalued solution. It turns out that x3 and x4 are linearly independent.
Returning to our problem, we take
x1 = i (1−i)t
it
iet cos t + et sin t
e
=
e cos t − iet sin t = t
.
1
1
e cos t − iet sin t It is easy to see that
Re x1 = et sin t
,
et cos t Im x1 = et cos t
,
−et sin t are the solutions we seek.
Exercise 3.4.4: Check that these really are solutions.
The general solution is
x = c1 et sin t
et cos t
c et sin t + c2 et cos t
+ c2
= 1t
.
t
t
e cos t
−e sin t
c1 e cos t − c2 et sin t This solution is realvalued for real c1 and c2 . Now we can solve for any initial conditions that we
have.
The process follows. When you have complex eigenvalues, you notice that they always come
in pairs. You take one λ = a + ib from the pair, you ﬁnd a corresponding eigenvector v. You note
that Re ve(a+ib)t and Im ve(a+ib)t are also solutions to the equation, are realvalued and are linearly
independent. You go on to the next eigenvalue, which is either a real eigenvalue or another complex
eigenvalue pair. Hence, you will end up with n linearly independent solutions if you had n distinct
eigenvalues (real or complex).
You can now ﬁnd a realvalued general solution to any homogeneous system where the matrix
has distinct eigenvalues. When you have repeated eigenvalues, matters get a bit more complicated
and we will look at that situation in § 3.7. 104 3.4.4 CHAPTER 3. SYSTEMS OF ODES Exercises Exercise 3.4.5 (easy): Let A be an 3 × 3 matrix with an eigenvalue of 3 and a corresponding
eigenvector v = 1
−1
3 . Find Av. Exercise 3.4.6: a) Find the general solution of x1 = 2 x1 , x2 = 3 x2 using the eigenvalue method
(ﬁrst write the system in the form x = A x). b) Solve the system by solving each equation separately
and verify you get the same general solution.
Exercise 3.4.7: Find the general solution of x1 = 3 x1 + x2 , x2 = 2 x1 + 4 x2 using the eigenvalue
method.
Exercise 3.4.8: Find the general solution of x1 = x1 − 2 x2 , x2 = 2 x1 + x2 using the eigenvalue
method. Do not use complex exponentials in your solution.
Exercise 3.4.9: a) Compute eigenvalues and eigenvectors of A =
solution of x = A x.
Exercise 3.4.10: Compute eigenvalues and eigenvectors of −2 −1 −1
321
−3 −1 0 9 −2 −6
−8 3 6
10 −2 −6 . . b) Find the general 3.5. TWO DIMENSIONAL SYSTEMS AND THEIR VECTOR FIELDS 3.5 105 Two dimensional systems and their vector ﬁelds Note: 1 lecture, should really be in [EP] §5.2, but is in [EP] §6.2
Let us take a moment to talk about homogeneous systems in the plane. We want to think about
how the vector ﬁelds look and how this depends on the eigenvalues. So we have a 2 × 2 matrix P
and the system
x
x
=P
.
(3.2)
y
y
We will be able to visually tell how the vector ﬁeld looks once we ﬁnd the eigenvalues and
eigenvectors of the matrix.
Case 1. Suppose that the eigenvalues are real
and positive. Find the two eigenvectors and plot
them in the plane. For example, take the matrix
1 1 . The eigenvalues are 1 and 2 and the corre02
sponding eigenvectors are 1 and 1 . See Fig0
1
ure 3.3.
Now suppose that x and y are on the line determined by an eigenvector v for an eigenvalue λ.
x
That is, y = av for some scalar a. Then
x
x
=P
= P(av) = a(Pv) = aλv.
y
y 3 2 1 0 1 2 3 3 3 2 2 1 1 0 0 1 1 2 2 3 3
3 2 1 0 1 2 3 The derivative is a multiple of v and hence points
Figure 3.3: Eigenvectors of P.
along the line determined by v. As λ > 0, the
derivative points in the direction of v when a is
positive and in the opposite direction when a is
negative. Let us draw arrows on the lines to indicate the directions. See Figure 3.4 on the following
page.
We ﬁll in the rest of the arrows and we also draw a few solutions. See Figure 3.5 on the next
page. You will notice that the picture looks like a source with arrows coming out from the origin.
Hence we call this type of picture a source or sometimes an unstable node.
Case 2. Suppose both eigenvalues were negative. For example, take the negation of the matrix
in case 1, −1 −1 . The eigenvalues are −1 and −2 and the corresponding eigenvectors are the same,
0 −2
1 and 1 . The calculation and the picture are almost the same. The only diﬀerence is that the
0
1
eigenvalues are negative and hence all arrows are reversed. We get the picture in Figure 3.6 on the
following page. We call this kind of picture a sink or sometimes a stable node.
1
Case 3. Suppose one eigenvalue is positive and one is negative. For example the matrix 1 −2 .
0
1
The eigenvalues are 1 and −2 and the corresponding eigenvectors are the same, 1 and −3 . We
0 106
3 CHAPTER 3. SYSTEMS OF ODES
2 1 0 1 2 3 3 2 1 0 1 2 3 3 3 3 3 2 2 2 2 1 1 1 1 0 0 0 0 1 1 1 1 2 2 2 2 3 3 3
3 2 1 0 1 2 3 Figure 3.4: Eigenvectors of P with directions. 3 2 1 0 1 2 3
3 2 1 0 1 2 3 Figure 3.5: Example source vector ﬁeld with
eigenvectors and solutions. 3 3 2 1 0 1 2 3 3 3 3 3 2 2 2 2 1 1 1 1 0 0 0 0 1 1 1 1 2 2 2 2 3 3 3
3 2 1 0 1 2 3 Figure 3.6: Example sink vector ﬁeld with
eigenvectors and solutions. 3
3 2 1 0 1 2 3 Figure 3.7: Example saddle vector ﬁeld with
eigenvectors and solutions. reverse the arrows on one line (corresponding to the negative eigenvalue) and we obtain the picture
in Figure 3.7. We call this picture a saddle point.
The next three cases we will assume the eigenvalues are complex. In this case the eigenvectors
are also complex and we cannot just plot them on the plane.
Case 4. Suppose the eigenvalues are purely imaginary. That is, suppose the eigenvalues are ±ib.
01
1
For example, let P = −4 0 . The eigenvalues turn out to be ±2i and the eigenvectors are 2i and
1
1
−2i . We take the eigenvalue 2i and its eigenvector 2i and note that the real and imaginary parts 3.5. TWO DIMENSIONAL SYSTEMS AND THEIR VECTOR FIELDS 107 of vei2t are
Re 1 i2t
cos(2t)
e=
,
2i
−2 sin(2t) Im 1 i2t
sin(2t)
e=
.
2i
2 cos(2t) We can take any linear combination of them, and which one we take depends on the initial conditions.
For example, the real part is a parametric equation for an ellipse. Same with the imaginary part and
in fact any linear combination of them. It is not diﬃcult to see that this is what happens in general
when the eigenvalues are purely imaginary. So when the eigenvalues are purely imaginary, you get
ellipses for your solutions. This type of picture is sometimes called a center. See Figure 3.8.
3 2 1 0 1 2 3 3 2 1 0 1 2 3 3 3 3 3 2 2 2 2 1 1 1 1 0 0 0 0 1 1 1 1 2 2 2 2 3 3 3
3 2 1 0 1 2 Figure 3.8: Example center vector ﬁeld. 3 3
3 2 1 0 1 2 3 Figure 3.9: Example spiral source vector ﬁeld. Case 5. Now the complex eigenvalues have positive real part. That is, suppose the eigenvalues
1
are a ± ib for some a > 0. For example, let P = −4 1 . The eigenvalues turn out to be 1 ± 2i and
1
1
1
the eigenvectors are 2i and −1 i . We take 1 + 2i and its eigenvector 2i and ﬁnd the real and
2
(1+2i)t
imaginary of ve
are
Re 1 (1+2i)t
cos(2t)
e
= et
,
2i
−2 sin(2t) Im 1 (1+2i)t
sin(2t)
e
= et
.
2i
2 cos(2t) Now note the et in front of the solutions. This means that the solutions grow in magnitude while
spinning around the origin. Hence we get a spiral source. See Figure 3.9.
Case 6. Finally suppose the complex eigenvalues have negative real part. That is, suppose the
eigenvalues are −a ± ib for some a > 0. For example, let P = −1 −1 . The eigenvalues turn out to
4 −1 108 CHAPTER 3. SYSTEMS OF ODES be −1 ± 2i and the eigenvectors are −1 i and
2
the real and imaginary of ve(1+2i)t are 1
2i . We take −1 − 2i and its eigenvector Re and ﬁnd 1 (−1−2i)t
cos(2t)
e
= e−t
,
2i
2 sin(2t) Im 1
2i 1 (−1−2i)t
− sin(2t)
e
= e−t
.
2i
2 cos(2t) Now note the e−t in front of the solutions. This means that the solutions shrink in magnitude while
spinning around the origin. Hence we get a spiral sink. See Figure 3.10.
3 2 1 0 1 2 3 3 3 2 2 1 1 0 0 1 1 2 2 3 3
3 2 1 0 1 2 3 Figure 3.10: Example spiral sink vector ﬁeld. We summarize the behavior of linear homogeneous two dimensional systems in Table 3.1.
Eigenvalues Behavior real and both positive
real and both negative
real and opposite signs
purely imaginary
complex with positive real part
complex with negative real part source / unstable node
sink / stable node
saddle
center point / ellipses
spiral source
spiral sink Table 3.1: Summary of behavior of linear homogeneous two dimensional systems. 3.5. TWO DIMENSIONAL SYSTEMS AND THEIR VECTOR FIELDS 3.5.1 109 Exercises Exercise 3.5.1: Take the equation mx + cx + kx = 0, with m > 0, c ≥ 0, k > 0 for the massspring
system. a) Convert this to a system of ﬁrst order equations. b) Classify for what m, c, k do you get
which behavior. c) Can you explain from physical intuition why you do not get all the diﬀerent kinds
of behavior here?
Exercise 3.5.2: Can you ﬁnd what happens in the case when P = 1 1 . In this case the eigenvalue
01
is repeated and there is only one eigenvector. What picture does this look like?
Exercise 3.5.3: Can you ﬁnd what happens in the case when P =
the pictures we have drawn? 11
11 . Does this look like any of 110 3.6 CHAPTER 3. SYSTEMS OF ODES Second order systems and applications Note: more than 2 lectures, §5.3 in [EP] 3.6.1 Undamped mass spring systems While we did say that we will usually only look at ﬁrst order systems, it is sometimes more
convenient to study the system in the way it arises naturally. For example, suppose we have 3
masses connected by springs between two walls. We could pick any higher number, and the math
would be essentially the same, but for simplicity we pick 3 right now. Let us also assume no friction,
that is, the system is undamped. The masses are m1 , m2 , and m3 and the spring constants are k1 ,
k2 , k3 , and k4 . Let x1 be the displacement from rest position of the ﬁrst mass, and x2 and x3 the
displacement of the second and third mass. We will make, as usual, positive values go right (as x1
grows the ﬁrst mass is moving right). See Figure 3.11.
k1 k2
m1 k3
m2 k4
m3 Figure 3.11: System of masses and springs.
This simple system turns up in unexpected places. Note for example that our world really
consists of small particles of matter interacting together. When we try this system with many more
masses, we obtain a good approximation to how an elastic material will behave. By somehow taking
a limit of the number of masses going to inﬁnity, we obtain the continuous one dimensional wave
equation. But we digress.
Let us set up the equations for the three mass system. By Hooke’s law we have that the force
acting on the mass equals the spring compression times the spring constant. By Newton’s second
law we have that force is mass times acceleration. So if we sum the forces acting on each mass and
put the right sign in front of each term, depending on the direction in which it is acting, we end up
with the desired system of equations.
m1 x1 = −k1 x1 + k2 ( x2 − x1 )
m2 x2 = −k2 ( x2 − x1 ) + k3 ( x3 − x2 )
m3 x3 = −k3 ( x3 − x2 ) − k4 x3
We deﬁne the matrices m1 0 0 M = 0 m2 0 0 0 m3 and = −(k1 + k2 ) x1 + k2 x2 ,
= k2 x1 − (k2 + k3 ) x2 + k3 x3 ,
= k3 x2 − (k3 + k4 ) x3 . −(k1 + k2 ) k2
0 . k2
−(k2 + k3 )
k3
K= 0
k3
−(k3 + k4 ) 3.6. SECOND ORDER SYSTEMS AND APPLICATIONS 111 We write the equation simply as
M x = K x.
At this point we could introduce 3 new variables and write out a system of 6 equations. We claim
this simple setup is easier to handle as a second order system. We will call x the displacement
vector, M the mass matrix, and K the stiﬀness matrix.
Exercise 3.6.1: Repeat this setup for 4 masses (ﬁnd the matrices M and K ). Do it for 5 masses.
Can you ﬁnd a prescription to do it for n masses?
As with a single equation we will want to “divide by M .” This means computing the inverse of
M . The masses are all nonzero and M is a diagonal matrix, so comping the inverse is easy.
1 m1 0 0 0 1 0 . −1 M = m2 1
0 0 m3
This fact follows readily by how we multiply diagonal matrices. You should verify that MM −1 =
M −1 M = I as an exercise.
Let A = M −1 K . We look at the system x = M −1 K x, or
x = A x.
Many real world systems can be modeled by this equation. For simplicity, we will only talk about
the given massesandsprings problem. We try a solution of the form
x = veαt .
We compute that for this guess, x = α2 veαt . We plug our guess into the equation and get
α2 veαt = Aveαt .
We can divide by eαt to get that α2 v = Av. Hence if α2 is an eigenvalue of A and v is the
corresponding eigenvector, we have found a solution.
In our example, and in other common applications, it turns out that A has only real negative
eigenvalues (and possibly a zero eigenvalue). So we will study only this case. When an eigenvalue λ
is negative, it means that α2 = λ is negative. Hence there is some real number ω such that −ω2 = λ.
Then α = ±iω. The solution we guessed was
x = v cos(ωt) + i sin(ωt) .
By taking real and imaginary parts (note that v is real), we ﬁnd that v cos(ωt) and v sin(ωt) are
linearly independent solutions.
If an eigenvalue is zero, it turns out that both v and vt are solutions, where v is the corresponding
eigenvector. 112 CHAPTER 3. SYSTEMS OF ODES Exercise 3.6.2: Show that if A has a zero eigenvalue and v is the corresponding eigenvector, then
x = v(a + bt) is a solution of x = A x for arbitrary constants a and b.
Theorem 3.6.1. Let A be an n × n with n distinct real negative eigenvalues we denote by −ω2 >
1
−ω2 > · · · > −ω2 , and the corresponding eigenvectors by v1 , v2 , . . . , vn . If A is invertible (that is, if
n
2
ω1 > 0), then
n x(t) = vi ai cos(ωi t) + bi sin(ωi t) ,
i=1 is the general solution of
x = A x,
for some arbitrary constants ai and bi . If A has a zero eigenvalue, that is ω1 = 0, and all other
eigenvalues are distinct and negative, then the general solution becomes
n x(t) = v1 (a1 + b1 t) + vi ai cos(ωi t) + bi sin(ωi t) .
i =2 Note that we can use this solution and the setup from the introduction of this section even when
some of the masses and springs are missing. For example, when there are say 2 masses and only 2
springs, simply take only the equations for the two masses and set all the spring constants for the
springs that are missing to zero. 3.6.2 Examples Example 3.6.1: Suppose we have the system in Figure 3.12, with m1 = 2, m2 = 1, k1 = 4, and
k2 = 2.
k1 k2
m1 m2 Figure 3.12: System of masses and springs. The equations we write down are
20
−(4 + 2) 2
x=
x,
01
2
−2
or
x= −3 1
x.
2 −2 3.6. SECOND ORDER SYSTEMS AND APPLICATIONS 1
2 113 We ﬁnd the eigenvalues of A to be λ = −1, −4 (exercise). Now we ﬁnd the eigenvectors to be
1
and −1 respectively (exercise).
We check the theorem and note that ω1 = 1 and ω2 = 2. Hence the general solution is
x= 1
1
a1 cos(t) + b1 sin(t) +
a cos(2t) + b2 sin(2t) .
2
−1 2 The two terms in the solution represent the two socalled natural or normal modes of oscillation. And the two (angular) frequencies are the natural frequencies. The two modes are plotted
in Figure 3.13.
0.0 2.5 5.0 7.5 10.0 0.0 2.5 5.0 7.5 10.0 2 2 1.0 1.0 1 1 0.5 0.5 0 0 0.0 0.0 1 1 0.5 0.5 2 1.0 2
0.0 2.5 5.0 7.5 10.0 1.0
0.0 2.5 5.0 7.5 10.0 Figure 3.13: The two modes of the mass spring system. In the left plot the masses are moving in
unison and the right plot are masses moving in the opposite direction.
Let us write the solution as
x=
The ﬁrst term, 1
1
c cos(t − α1 ) +
c cos(2t − α2 ).
21
−1 2
1
c cos(t − α1 )
c1 cos(t − α1 ) = 1
,
2
2c1 cos(t − α1 ) corresponds to the mode where the masses move synchronously in the same direction.
On the other hand the second term,
1
c cos(2t − α2 )
c cos(2t − α2 ) = 2
,
−1 2
−c2 cos(2t − α2 )
corresponds to the mode where the masses move synchronously but in opposite directions.
The general solution is a combination of the two modes. That is, the initial conditions determine
the amplitude and phase shift of each mode. 114 CHAPTER 3. SYSTEMS OF ODES Example 3.6.2: We have two toy rail cars. Car 1 of mass 2 kg is traveling at 3 m/s towards the
second rail car of mass 1 kg. There is a bumper on the second rail car that engages at the moment
the cars hit (it connects to two cars) and does not let go. The bumper acts like a spring of spring
constant k = 2 N/m. The second car is 10 meters from a wall. See Figure 3.14.
k m1 m2
10 meters Figure 3.14: The crash of two rail cars. We want to ask several question. At what time after the cars link does impact with the wall
happen? What is the speed of car 2 when it hits the wall?
OK, let us ﬁrst set the system up. Let t = 0 be the time when the two cars link up. Let x1 be the
displacement of the ﬁrst car from the position at t = 0, and let x2 be the displacement of the second
car from its original location. Then the time when x2 (t) = 10 is exactly the time when impact with
wall occurs. For this t, x2 (t) is the speed at impact. This system acts just like the system of the
previous example but without k1 . Hence the equation is
20
−2 2
x=
x.
01
2 −2
or
x= −1 1
x.
2 −2 We compute the eigenvalues of A. It is not hard to see that the eigenvalues are 0 and −3
1
(exercise). Furthermore, the eigenvectors are 1 and −2 respectively (exercise). We note that
1
√
ω2 = 3 and we use the second part of the theorem to ﬁnd our general solution to be
√
√
1
1
(a1 + b1 t) +
a2 cos( 3 t) + b2 sin( 3 t)
1
−2
√
√
a1 + b1 t + a2 cos(√3 t) + b2 sin( √ t)
3
=
a1 + b1 t − 2a2 cos( 3 t) − 2b2 sin( 3 t) x= We now apply the initial conditions. First the cars start at position 0 so x1 (0) = 0 and x2 (0) = 0.
The ﬁrst car is traveling at 3 m/s, so x1 (0) = 3 and the second car starts at rest, so x2 (0) = 0. The ﬁrst
conditions says
a + a2
0 = x(0) = 1
.
a1 − 2a2 3.6. SECOND ORDER SYSTEMS AND APPLICATIONS 115 It is not hard to see that this implies that a1 = a2 = 0. We plug a1 and a2 and diﬀerentiate to get
√
√
b1 + √ b2 cos( √ t)
3
3
x (t) =
.
b1 − 2 3 b2 cos( 3 t)
So √
3
b1 + √ b2
3
= x (0) =
.
0
b1 − 2 3 b2 It is not hard to solve these two equations to ﬁnd b1 = 2 and b2 =
is (until the impact with the wall)
√ 1
2t + √ sin( 3 t) 3 x= 2t − √ sin(√3 t) . 2
3 1
√.
3 Hence the position of our cars Note how the presence of the zero eigenvalue resulted in a term containing t. This means that the
carts will be traveling in the positive direction as time grows, which is what we expect.
What we are really interested in is the second expression, the one for x2 . We have x2 (t) =
√
2
2t − √3 sin( 3 t). See Figure 3.15 for the plot of x2 versus time.
0 1 2 3 4 5 6 12.5 12.5 10.0 10.0 7.5 7.5 5.0 5.0 2.5 2.5 0.0 0.0
0 1 2 3 4 5 6 Figure 3.15: Position of the second car in time (ignoring the wall).
Just from the graph we can see that time of impact will be a little more than 5 seconds from
√
2
time zero. For this you have to solve the equation 10 = x2 (t) = 2t − √3 sin( 3 t). Using a computer
(or even a graphing calculator) we ﬁnd that timpact ≈ 5.22 seconds.
√
As for the speed we note that x2 = 2 − 2 cos( 3 t). At time of impact (5.22 seconds from t = 0)
we get that x2 (timpact ) ≈ 3.85.
√
The maximum speed is the maximum of 2 − 2 cos( 3 t), which is 4. We are traveling at almost
the maximum speed when we hit the wall. 116 CHAPTER 3. SYSTEMS OF ODES Now suppose that Bob is a tiny person sitting on car 2. Bob has a Martini in his hand and would
like to not spill it. Let us suppose Bob would not spill his Martini when the ﬁrst car links up with
car 2, but if car 2 hits the wall at any speed greater than zero, Bob will spill his drink. Suppose Bob
can move car 2 a few meters towards or away from the wall (he cannot go all the way to the wall,
nor can he get out of the way of the ﬁrst car). Is there a “safe” distance for him to be in? A distance
such that the impact with the wall is at zero speed?
The answer is yes. Looking at Figure 3.15 on the preceding page, we note the “plateau” between
t = 3 and t = 4. There is a point where the speed is zero. To ﬁnd it we need to solve x2 (t) = 0. This
√
2
4
is when cos( 3 t) = 1 or in other words when t = √π , √π , . . . and so on. We plug in the ﬁrst value to
3
3
2
4
obtain x2 √π = √π ≈ 7.26. So a “safe” distance is about 7 and a quarter meters from the wall.
3
3
Alternatively Bob could move away from the wall towards the incoming car 2 where another
8
safe distance is √π ≈ 14.51 and so on, using all the diﬀerent t such that x2 (t) = 0. Of course t = 0 is
3
always a solution here, corresponding to x2 = 0, but that means standing right at the wall. 3.6.3 Forced oscillations Finally we move to forced oscillations. Suppose that now our system is
x = A x + F cos(ωt). (3.3) That is, we are adding periodic forcing to the system in the direction of the vector F .
Just like before this system just requires us to ﬁnd one particular solution x p , add it to the general
solution of the associated homogeneous system xc and we will have the general solution to (3.3).
Let us suppose that ω is not one of the natural frequencies of x = A x, then we can guess
x p = c cos(ωt),
where c is an unknown constant vector. Note that we do not need to use sine since there are only
second derivatives. We solve for c to ﬁnd x p . This is really just the method of undetermined
coeﬃcients for systems. Let us diﬀerentiate x p twice to get
x p = −ω2 c cos(ωt).
Now plug into the equation
−ω2 c cos(ωt) = Ac cos(ωt) + F cos(ωt)
We can cancel out the cosine and rearrange the equation to obtain
(A + ω2 I )c = −F .
So −1 c = (A + ω2 I ) (−F ). 3.6. SECOND ORDER SYSTEMS AND APPLICATIONS 117 Of course this this is possible only if (A + ω2 I ) = (A − (−ω2 )I ) is invertible. That matrix is invertible
if and only if −ω2 is not an eigenvalue of A. That is true if and only if ω is not a natural frequency
of the system.
Example 3.6.3: Let us take the example in Figure 3.12 on page 112 with the same parameters as
before: m1 = 2, m2 = 1, k1 = 4, and k2 = 2. Now suppose that there is a force 2 cos(3t) acting on
the second cart.
The equation is
−3 1
0
x=
x+
cos(3t).
2 −2
2
We have solved the associated homogeneous equation before and found the complementary solution
to be
1
1
(a cos(t) + b1 sin(t)) +
(a cos(2t) + b2 sin(2t)) .
xc =
21
−1 2
We note that the natural frequencies were 1 and 2. Hence 3 is not a natural frequency, we can
try c cos(3t). We can invert (A + 32 I )
−3 1
+ 32 I
2 −2 −1 61
=
27 −1 = 7
40
−1
20 −1
40
3
20 . Hence,
−1 c = (A + ω I ) (−F ) =
2 7
40
−1
20 −1
40
3
20 0
=
−2 1
20
−3
10 . Combining with what we know the general solution of the associated homogeneous problem to
be we get that the general solution to x = A x + F cos(ωt) is
1
1
(a1 cos(t) + b1 sin(t)) +
(a2 cos(2t) + b2 sin(2t)) +
x = xc + x p =
2
−1 1
20
−3
10 cos(3t). The constants a1 , a2 , b1 , and b2 must then be solved for given any initial conditions.
If ω is a natural frequency of the system resonance occurs because you will have to try a
particular solution of the form
x p = c t sin(ωt) + d cos(ωt).
That is assuming that all eigenvalues of the coeﬃcient matrix are distinct. Note that the amplitude
of this solution grows without bound as t grows. 118 3.6.4 CHAPTER 3. SYSTEMS OF ODES Exercises Exercise 3.6.3: Find a particular solution to
x= 0
−3 1
x+
cos(2t).
2
2 −2 Exercise 3.6.4 (challenging): Let us take the example in Figure 3.12 on page 112 with the same
parameters as before: m1 = 2, k1 = 4, and k2 = 2, except for m2 , which is unknown. Suppose
that there is a force cos(5t) acting on the ﬁrst mass. Find an m2 such that there exists a particular
solution where the ﬁrst mass does not move.
Note: This idea is called dynamic damping. In practice there will be a small amount of damping
and so any transient solution will disappear and after long enough time, the ﬁrst mass will always
come to a stop.
Exercise 3.6.5: Let us take the Example 3.6.2 on page 114, but that at time of impact, cart 2 is
moving to the left at the speed of 3 m/s. a) Find the behavior of the system after linkup. b) Will the
second car hit the wall, or will it be moving away from the wall as time goes on. c) at what speed
would the ﬁrst car have to be traveling for the system to essentially stay in place after linkup.
Exercise 3.6.6: Let us take the example in Figure 3.12 on page 112 with parameters m1 = m2 = 1,
k1 = k2 = 1. Does there exist a set of initial conditions for which the ﬁrst cart moves but the second
cart does not? If so ﬁnd those conditions, if not argue why not. 3.7. MULTIPLE EIGENVALUES 3.7 119 Multiple eigenvalues Note: 1–2 lectures, §5.4 in [EP]
It may very well happen that a matrix has some “repeated” eigenvalues. That is, the characteristic
equation det(A − λI ) = 0 may have repeated roots. As we have said before, this is actually unlikely
to happen for a random matrix. If you take a small perturbation of A (you change the entries of A
slightly) you will get a matrix with distinct eigenvalues. As any system you will want to solve in
practice is an approximation to reality anyway, it is not indispensable to know how to solve these
corner cases. But it may happen on occasion that it is easier or desirable to solve such a system
directly. 3.7.1 Geometric multiplicity Take the diagonal matrix
A= 30
.
03 A has an eigenvalue 3 of multiplicity 2. We usually call the multiplicity of the eigenvalue in the
characteristic equation the algebraic multiplicity. In this case, there also exist 2 linearly independent
eigenvectors, 1 and 0 corresponding to the eigenvalue 3. This means that the socalled geometric
0
1
multiplicity of this eigenvalue is also 2.
In all the theorems where we required a matrix to have n distinct eigenvalues, we only really
needed to have n linearly independent eigenvectors. For example, let x = A x has the general
solution
1 3t
0 3t
x = c1
e + c2
e.
0
1
Let us restate the theorem about real eigenvalues. In the following theorem we will repeat eigenvalues according to (algebraic) multiplicity. So for A above we would say that it has eigenvalues 3 and
3.
Theorem 3.7.1. Take x = P x. If P is n × n and has n real eigenvalues (not necessarily distinct),
λ1 , . . . , λn , and if there are n linearly independent corresponding eigenvectors v1 , . . . , vn , and the
general solution to the ODE can be written as.
x = c1 v1 eλ1 t + c2 v2 eλ2 t + · · · + cn vn eλn t .
The geometric multiplicity of an eigenvalue of algebraic multiplicity n is equal to the number of
corresponding linearly independent eigenvectors we can ﬁnd. It is not hard to see that the geometric
multiplicity is always less than or equal to the algebraic multiplicity. We have, therefore, handled the
case when these two multiplicities are equal. If the geometric multiplicity is equal to the algebraic
multiplicity we say the eigenvalue is complete. 120 CHAPTER 3. SYSTEMS OF ODES In other words, the hypothesis of the theorem could be stated as saying that if all the eigenvalues
of P are complete, then there are n linearly independent eigenvectors and thus we have the given
general solution.
Note that if the geometric multiplicity of an eigenvalue is 2 or greater, then the set of linearly
independent eigenvectors is not unique up to multiples as it was before. For example, for the
1
diagonal matrix A above we could also pick eigenvectors 1 and −1 , or in fact any pair of two
1
linearly independent vectors. The number of eigenvalues is the number of free variables we obtain
when solving Av = λv. We then pick values for those free variables to obtain the eigenvectors. If
you pick diﬀerent values, you may get diﬀerent eigenvectors. 3.7.2 Defective eigenvalues If an n × n matrix has less than n linearly independent eigenvectors, it is said to be deﬁcient. Then
there is at least one eigenvalue with an algebraic multiplicity that is higher than its geometric
multiplicity. We call this eigenvalue defective and the diﬀerence between the two multiplicities we
call the defect.
Example 3.7.1: The matrix
31
03
has an eigenvalue 3 of algebraic multiplicity 2. Let us try to compute the eigenvectors.
0 1 v1
= 0.
0 0 v2
We must have that v2 = 0. Hence any eigenvector is of the form v01 . Any two such vectors are
linearly dependent, and hence the geometric multiplicity of the eigenvalue is 1. Therefore, the
defect is 1, and we can no longer apply the eigenvalue method directly to a system of ODEs with
such a coeﬃcient matrix.
The key observation we will use here is that if λ is an eigenvalue of A of algebraic multiplicity
m, then we will be able to ﬁnd m linearly independent vectors solving the equation (A − λI )m v = 0.
We will call these the generalized eigenvectors.
Let us continue with the example A = 3 1 and the equation x = A x. We have an eigenvalue
03
λ = 3 of (algebraic) multiplicity 2 and defect 1. We have found one eigenvector v1 = 1 . We have
0
the solution
x1 = v1 e3t .
In this case, let us try (in the spirit of repeated roots of the characteristic equation for a single
equation) another solution of the form
x2 = (v2 + v1 t) e3t . 3.7. MULTIPLE EIGENVALUES 121 We diﬀerentiate to get
x2 = v1 e3t + 3(v2 + v1 t) e3t = (3v2 + v1 ) e3t + 3v1 te3t .
x2 must equal A x2 , and A x2 = A(v2 + v1 t) e3t = Av2 e3t + Av1 te3t . By looking at the coeﬃcients of e3t and te3t we see 3v2 + v1 = Av2 and 3v1 = Av1 . This means that
(A − 3I )v1 = 0, (A − 3I )v2 = v1 . and If these two equations are satisﬁed, then x2 is a solution. We know the ﬁrst of these equations is
satisﬁed because v1 is an eigenvector. If we plug the second equation into the ﬁrst we obtain
(A − 3I )(A − 3I )v2 = 0, or (A − 3I )2 v2 = 0. If we can, therefore, ﬁnd a v2 that solves (A − 3I )2 v2 = 0, and such that (A − 3I )v2 = v1 , then we are
done. This is just a bunch of linear equations to solve and we are by now very good at that.
We notice that in this simple case (A − 3I )2 is just the zero matrix (exercise). Hence, any vector
v2 solves (A − 3I )2 v2 = 0. We just have to make sure that (A − 3I )v2 = v1 . Write
01a
1
=
.
00b
0
By inspection we see that letting a = 0 (a could be anything in fact) and b = 1 does the job. Hence
we can take v2 = 0 . Our general solution to x = A x is
1
x = c1 1 3t
0
1
c e3t + c2 te3t
e + c2
+
t e3t = 1
.
0
1
0
c2 e3t Let us check that we really do have the solution. First x1 = c1 3e3t + c2 e3t + 3c2 te3t = 3 x1 + x2 . Good.
Now x2 = 3c2 e3t = 3 x2 . Good.
Note that the system x = A x has a simpler solution since A is a triangular matrix. In particular,
the equation for x2 does not depend on x1 . Mind you, not every defective matrix is triangular.
Exercise 3.7.1: Solve x = 3 1 x by ﬁrst solving for x2 and then for x1 independently. Now check
03
that you got the same solution as we did above.
Let us describe the general algorithm. First for λ of multiplicity 2, defect 1. First ﬁnd an
eigenvector v1 of λ. Now ﬁnd a vector v2 such that
(A − 3I )2 v2 = 0,
(A − 3I )v2 = v1 . 122 CHAPTER 3. SYSTEMS OF ODES This gives us two linearly independent solutions
x1 = v1 eλt ,
x2 = v2 + v1 t eλt .
This machinery can also be generalized to larger matrices and higher defects. We will not go
over this method in detail, but let us just sketch the ideas. Suppose that A has a multiplicity m
eigenvalue λ. We ﬁnd vectors such that
(A − λI )k v = 0, but (A − λI )k−1 v 0. Such vectors are called generalized eigenvectors. For every eigenvector v1 we ﬁnd a chain of
generalized eigenvectors v2 through vk such that:
(A − λI )v1 = 0,
(A − λI )v2 = v1 ,
.
.
.
(A − λI )vk = vk−1 .
We form the linearly independent solutions
x1 = v1 eλt ,
x2 = (v2 + v1 t) eλt ,
.
.
.
xk = vk + vk−1 t + · · · + v2 t k −1
t k −2
+ v1
eλt .
(k − 2)!
(k − 1)! We proceed to ﬁnd chains until we form m linearly independent solutions (m is the multiplicity).
You may need to ﬁnd several chains for every eigenvalue. 3.7.3 Exercises Exercise 3.7.2: Let A = 5 −3
3 −1 . Solve x = A x. Exercise 3.7.3: Let A = 5 −4 4
030
−2 4 −1 . a) What are the eigenvalues? b) What is/are the defect(s) of the eigenvalue(s)? c) Solve x = A x.
Exercise 3.7.4: Let A = 210
020
002 . a) What are the eigenvalues? b) What is/are the defect(s) of the eigenvalue(s)? c) Solve x = A x in two diﬀerent ways and verify you get the same answer. 3.7. MULTIPLE EIGENVALUES
Exercise 3.7.5: Let A = 123 012
−1 −2 −2
−4 4 7 . a) What are the eigenvalues? b) What is/are the defect(s) of the 0 4 −2
−1 −4 1
0 0 −2 . a) What are the eigenvalues? b) What is/are the defect(s) of the 2 1 −1
−1 0 2
−1 −2 4 . a) What are the eigenvalues? b) What is/are the defect(s) of the eigenvalue(s)? c) Solve x = A x.
Exercise 3.7.6: Let A = eigenvalue(s)? c) Solve x = A x.
Exercise 3.7.7: Let A = eigenvalue(s)? c) Solve x = A x. Exercise 3.7.8: Suppose that A is a 2 × 2 matrix with a repeated eigenvalue λ. Suppose that there
are two linearly independent eigenvectors. Show that the matrix is diagonal, in particular A = λI. 124 3.8 CHAPTER 3. SYSTEMS OF ODES Matrix exponentials Note: 2 lectures, §5.5 in [EP] 3.8.1 Deﬁnition In this section we present a diﬀerent way of ﬁnding the fundamental matrix solution of a system.
Suppose that we have the constant coeﬃcient equation
x = P x,
as usual. Now suppose that this was one equation (P is a number or a 1 × 1 matrix). Then the
solution to this would be
x = ePt .
It turns out the same computation works for matrices when we deﬁne ePt properly. First let us write
down the Taylor series for eat for some number a.
eat = 1 + at + (at)2 (at)3 (at)4
+
+
+ ··· =
2
6
24 ∞
k =0 (at)k
.
k! Recall k! = 1 · 2 · 3 · · · k, and 0! = 1. We diﬀerentiate this series term by term
a + a2 t + a3 t2 a4 t3
(at)2 (at)3
+
+ · · · = a 1 + at +
+
+ · · · = aeat .
2
6
2
6 Maybe we can write try the same trick here. Suppose that for an n × n matrix A we deﬁne the matrix
exponential as
1
1
1
def
e A = I + A + A2 + A3 + · · · + Ak + · · ·
2
6
k!
Let us not worry about convergence. The series really does always converge. We usually write Pt
as tP by convention when P is a matrix. With this small change and by the exact same calculation
as above we have that
d tP
e = PetP .
dt
Now P and hence etP is an n × n matrix. What we are looking for is a vector. We note that in the
1 × 1 case we would at this point multiply by an arbitrary constant to get the general solution. In the
matrix case we multiply by a column vector c.
Theorem 3.8.1. Let P be an n × n matrix. Then the general solution to x = P x is
x = etP c,
where c is an arbitrary constant vector. In fact x(0) = c. 3.8. MATRIX EXPONENTIALS 125 Let us check. d tP
d
x=
e c = PetP c = P x.
dt
dt
Hence etP is the fundamental matrix solution of the homogeneous system. If we ﬁnd a way
to compute the matrix exponential, we will have another method of solving constant coeﬃcient
homogeneous systems. It also makes it easy to solve for initial conditions. To solve x = A x,
x(0) = b, we take the solution
x = etA b.
This equation follows because e0A = I , so x(0) = e0A b = b.
We mention a drawback of matrix exponentials. In general eA+B eA eB . The trouble is that
matrices do not commute, that is, in general AB BA. If you try to prove eA+B eA eB using the
Taylor series, you will see why the lack of commutativity becomes a problem. However, it is still
true that if AB = BA, that is, if A and B commute, then eA+B = eA eB . We will ﬁnd this fact useful.
Let us restate this as a theorem to make a point.
Theorem 3.8.2. If AB = BA, then eA+B = eA eB . Otherwise eA+B 3.8.2 eA eB in general. Simple cases In some instances it may work to just plug into the series deﬁnition. Suppose the matrix is diagonal.
a
For example, D = 0 0 . Then
b
ak 0
Dk =
,
0 bk
and
1
1
1 a2 0
1 a3 0
10
a0
ea 0
eD = I + D + D2 + D3 + · · · =
+
+
+
+ ··· =
.
01
0b
0 eb
2
6
2 0 b2
6 0 b3
So by this rationale we have that
eI = e0
0e and eaI = ea 0
.
0 ea This makes exponentials of certain other matrices easy to compute. Notice for example that
the matrix A = 5 −3 can be written as 2I + B where B = 3 −3 . Notice that 2I and B commute,
3 −3
3 −1
and that B2 = 0 0 . So Bk = 0 for all k ≥ 2. Therefore, eB = I + B. Suppose we actually want to
00
compute etA . 2tI and tB still commute (exercise: check this) and etB = I + tB, since (tB)2 = t2 B2 = 0.
We write
etA = e2tI +tB = e2tI etB = e2t 0
(I + tB) =
0 e2t
= e2t 0 1 + 3t −3t
(1 + 3t) e2t
−3te2t
=
.
0 e2t
3t
1 − 3t
3te2t
(1 − 3t) e2t 126 CHAPTER 3. SYSTEMS OF ODES So we have found the fundamental matrix solution for the system x = A x. Note that this matrix has
a repeated eigenvalue with a defect; there is only one eigenvector for the eigenvalue 2. So we have
found a perhaps easier way to handle this case. In fact, if a matrix A is 2 × 2 and has an eigenvalue
λ of multiplicity 2, then either it is diagonal, or A = λI + B where B2 = 0. This is a good exercise.
Exercise 3.8.1: Suppose that A is 2 × 2 and λ is the only eigenvalue. Then show that (A − λI )2 = 0.
Then we can write A = λI + B, where B2 = 0. Hint: First write down what does it mean for the
eigenvalue to be of multiplicity 2. You will get an equation for the entries. Now compute the square
of B.
Matrices B such that Bk = 0 for some k are called nilpotent. Computation of the matrix
exponential for nilpotent matrices is easy by just writing down the ﬁrst k terms of the Taylor series. 3.8.3 General matrices In general, the exponential is not as easy to compute as above. We cannot usually write any matrix
as a sum of commuting matrices where the exponential is simple for each one. But fear not, it is still
not too diﬃcult provided we can ﬁnd enough eigenvectors. First we need the following interesting
result about matrix exponentials. For any two square matrices A and B, we have
−1 eBAB = BeA B−1 .
This can be seen by writing down the Taylor series. First note that
2 ( BAB−1 ) = BAB−1 BAB−1 = BAIAB−1 = BA2 B−1 .
k And hence by the same reasoning ( BAB−1 ) = BAk B−1 . So now write down the Taylor series for
−1
eBAB
1
1
−1
2
3
eBAB = I + BAB−1 + ( BAB−1 ) + ( BAB−1 ) + · · ·
2
6
1 2 −1 1 3 −1
= BB−1 + BAB−1 + BA B + BA B + · · ·
2
6
12 13
= B I + A + A + A + · · · B−1
2
6
A −1
= Be B .
Now we will write a general matrix A as EDE −1 , where D is diagonal. This procedure is called
diagonalization. If we can do that, you can see that the computation of the exponential becomes
easy. Adding t into the mix we see that
etA = EetD E −1 . 3.8. MATRIX EXPONENTIALS 127 Now to do this we will need n linearly independent eigenvectors of A. Otherwise this method
does not work and we need to be trickier, but we will not get into such details in this course. We
let E be the matrix with the eigenvectors as columns. Let λ1 , . . . , λn be the eigenvalues and let v1 ,
. . . , vn be the eigenvectors, then E = [ v1 v2 · · · vn ]. Let D be the diagonal matrix with the
eigenvalues on the main diagonal. That is λ1 0 · · · 0 0 λ2 · · · 0 . . .
D=. . . .
.. . . . . 0 0 · · · λn
Now we write
AE = A[ v1 v2 · · · vn ]
= [ Av1 Av2 · · · Avn ]
= [ λ1 v1 λ2 v2 · · · λn vn ]
= [ v1 v2 · · · vn ]D
= ED.
Now the columns of E are linearly independent as these are the eigenvectors of A. Hence E is
invertible. Since AE = ED, we right multiply by E −1 and we get
A = EDE −1 .
This means that eA = EeD E −1 . With t is turns into λ1 t 0 ··· 0 e 0 eλ2 t · · · 0 −1 tA
tD −1
.
e = Ee E = E . .
. E .
.. .
.
.
. .
. λn t 0
0 ··· e (3.4) The formula (3.4), therefore, gives the formula for computing the fundamental matrix solution etA
for the system x = A x, in the case where we have n linearly independent eigenvectors.
Notice that this computation still works when the eigenvalues and eigenvectors are complex,
though then you will have to compute with complex numbers. Note that it is clear from the deﬁnition
that if A is real, then etA is real. So you will only need complex numbers in the computation and
you may need to apply Euler’s formula to simplify the result. If simpliﬁed properly the ﬁnal matrix
will not have any complex numbers in it.
Example 3.8.1: Compute the fundamental matrix solution using the matrix exponentials for the
system
x
12x
=
.
y
21y 128 CHAPTER 3. SYSTEMS OF ODES Then compute the particular solution for the initial conditions x(0) = 4 and y(0) = 2.
Let A be the coeﬃcient matrix 1 2 . We ﬁrst compute (exercise) that the eigenvalues are 3 and
21
1
−1 and the corresponding eigenvectors are 1 and −1 . Hence we write
1
−1 1 1 e3t 0 1 1
1 −1 0 e−t 1 −1
1 1 e3t 0 −1 −1 −1
=
1 −1 0 e−t 2 −1 1
−1 e3t e−t −1 −1
=
2 e3t −e−t −1 1 etA = = −1 −e3t − e−t −e3t + e−t
=
2 −e3t + e−t −e3t − e−t e 3 t +e − t
2
e 3 t −e − t
2 e3t −e−t
2
e3t +e−t
2 . The initial conditions are x(0) = 4 and y(0) = 2. Hence, by the property that e0A = I we ﬁnd that
the particular solution we are looking for is etA b where b is 4 . Then the particular solution we are
2
looking for is
x
=
y 3.8.4 e 3 t +e − t
2
e 3 t −e − t
2 e3t −e−t
2
e3t +e−t
2 4
2e3t + 2e−t + e3t − e−t
3e3t + e−t
=
=
.
2
2e3t − 2e−t + e3t + e−t
3e3t − e−t Fundamental matrix solutions We note that if you can compute the fundamental matrix solution in a diﬀerent way, you can use
this to ﬁnd the matrix exponential etA . The fundamental matrix solution of a system of ODEs is
not unique. The exponential is the fundamental matrix solution with the property that for t = 0
we get the identity matrix. So we must ﬁnd the right fundamental matrix solution. Let X be any
fundamental matrix solution to x = A x. Then we claim
etA = X (t) [X (0)]−1 .
Obviously if we plug t = 0 into X (t) [X (0)]−1 we get the identity. It is not hard to see that we can
multiply a fundamental matrix solution on the right by any constant invertible matrix and we still
get a fundamental matrix solution. All we are doing is changing what the arbitrary constants are in
the general solution x(t) = X (t)c. 3.8.5 Approximations If you think about it, the computation of any fundamental matrix solution X using the eigenvalue
method is just as diﬃcult as computation of etA . So perhaps we did not gain much by this new tool.
However, the Taylor series expansion actually gives us a very easy way to approximate solutions,
which the eigenvalue method did not. 3.8. MATRIX EXPONENTIALS 129 The simplest thing we can do is to just compute the series up to a certain number of terms. There
are better ways to approximate the exponential∗ . In many cases however, few terms of the Taylor
series give a reasonable approximation for the exponential and may suﬃce for the application. For
example, let us compute the ﬁrst 4 terms of the series for the matrix A = 1 2 .
21
5
12
t2 2 t3 3
22 2
3
+t
e ≈ I + tA + A + A = I + t
5 +t
2
6
21
22
tA 13
6
7
3 7
3
13
6
52
t
2
2 = 1+t+
+ 13 t3
2 t + 2 t2 + 7 t3
6
3
=
.
73
52
2t + 2t + 3 t
1 + t + 2 t + 13 t3
6
Just like the scalar version of the Taylor series approximation, the approximation will be better for
small t and worse for larger t. For larger t, we will generally have to compute more terms. Let us see
how we stack up against the real solution with t = 0.1. The approximate solution is approximately
(rounded to 8 decimal places)
e0.1 A ≈ I + 0.1 A + 0.12 2 0.13 3
1.12716667 0.22233333
A+
A=
.
0.22233333 1.12716667
2
6 And plugging t = 0.1 into the real solution (rounded to 8 decimal places) we get
e0.1 A = 1.12734811 0.22251069
.
0.22251069 1.12734811 This is not bad at all. Although if we take the same approximation for t = 1 we get (using the Taylor
series)
6.66666667 6.33333333
,
6.33333333 6.66666667
while the real value is (again rounded to 8 decimal places)
10.22670818 9.85882874
.
9.85882874 10.22670818
So the approximation is not very good once we get up to t = 1. To get a good approximation at
t = 1 (say up to 2 decimal places) we would need to go up to the 11th power (exercise). 3.8.6 Exercises Exercise 3.8.2: Find a fundamental matrix solution for the system x = 3 x + y, y = x + 3y.
Exercise 3.8.3: Find eAt for the matrix A =
∗ 23
02 . C. Moler and C.F. Van Loan, Nineteen Dubious Ways to Compute the Exponential of a Matrix, TwentyFive Years
Later, SIAM Review 45 (1), 2003, 3–49 130 CHAPTER 3. SYSTEMS OF ODES Exercise 3.8.4: Find a fundamental matrix solution for the system x1 = 7 x1 + 4 x2 + 12 x3 , x2 =
x1 + 2 x2 + x3 , x3 = −3 x1 − 2 x2 − 5 x3 . Then ﬁnd the solution that satisﬁes x =
Exercise 3.8.5: Compute the matrix exponential eA for A = 12
01 0
1
−2 . . Exercise 3.8.6: Suppose AB = BA (matrices commute). Show that eA+B = eA eB .
−1 Exercise 3.8.7: Use Exercise 3.8.6 to show that (eA )
invertible even if A is not. = e−A . In particular this means that eA is Exercise 3.8.8: Suppose A is a matrix with eigenvalues −1, 1, and corresponding eigenvectors 1 ,
1
0 . a) Find matrix A with these properties. b) Find the fundamental matrix solution to x = A x. c)
1
Solve the system in with initial conditions x(0) = 2 .
3
Exercise 3.8.9: Suppose that A is an n × n matrix with a repeated eigenvalue λ of multiplicity n.
Suppose that there are n linearly independent eigenvectors. Show that the matrix is diagonal, in
particular A = λI . Hint: Use diagonalization and the fact that the identity matrix commutes with
every other matrix. 3.9. NONHOMOGENEOUS SYSTEMS 3.9 131 Nonhomogeneous systems Note: 3 lectures (may have to skip a little), somewhat diﬀerent from §5.6 in [EP] 3.9.1 First order constant coeﬃcient Integrating factor
Let us ﬁrst focus on the nonhomogeneous ﬁrst order equation
x (t) = A x(t) + f (t),
where A is a constant matrix. The ﬁrst method we will look at is the integrating factor method. For
simplicity we rewrite the equation as
x (t) + P x(t) = f (t),
where P = −A. We multiply both sides of the equation by etP (being mindful that we are dealing
with matrices that may not commute) to obtain
etP x (t) + etP P x(t) = etP f (t).
We notice that PetP = etP P. This fact follows by writing down the series deﬁnition of etP ,
1
1
PetP = P I + I + tP + (tP)2 + · · · = P + tP2 + t2 P3 + · · · =
2
2
1
= I + I + tP + (tP)2 + · · · P = PetP .
2
We have already seen that d
dt etP = PetP . Hence,
d tP
e x(t) = etP f (t).
dt We can now integrate. That is, we integrate each component of the vector separately
etP x(t) =
−1 Recall from Exercise 3.8.7 that (etP ) etP f (t) dt + c. = e−tP . Therefore, we obtain x(t) = e−tP etP f (t) dt + e−tP c. 132 CHAPTER 3. SYSTEMS OF ODES Perhaps it is better understood as a deﬁnite integral. In this case it will be easy to also solve for
the initial conditions as well. Suppose we have the equation with initial conditions
x (t) + P x(t) = f (t), x(0) = b. The solution can then be written as
t x(t) = e e sP f ( s) ds + e−tP b. −tP (3.5) 0 Again, the integration means that each component of the vector e sP f ( s) is integrated separately. It is
not hard to see that (3.5) really does satisfy the initial condition x(0) = b.
0 x(0) = e−0P e sP f ( s) ds + e−0P b = I b = b.
0 Example 3.9.1: Suppose that we have the system
x1 + 5 x1 − 3 x2 = et ,
x2 + 3 x1 − x2 = 0,
with initial conditions x1 (0) = 1, x2 (0) = 0.
Let us write the system as
x+ 5 −3
et
x=
,
3 −1
0 x(0) = 1
.
0 We have previously computed etP for P = 5 −3 . We immediately also have e−tP , simply by negating
3 −1
t.
(1 + 3t) e2t
−3te2t
(1 − 3t) e−2t
3te−2t
etP =
,
e−tP =
.
3te2t
(1 − 3t) e2t
−3te−2t
(1 + 3t) e−2t
Instead of computing the whole formula at once. Let us do it in stages. First
t (1 + 3 s) e2 s
−3 se2 s
es
ds
2s
2s
3 se
(1 − 3 s) e
0 t t (1 + 3 s) e3 s
ds
3 se3 s e f ( s) ds =
sP 0 0 =
0 = te3t
(3t−1) e3t +1
3 . 3.9. NONHOMOGENEOUS SYSTEMS 133 Then
t x(t) = e e sP f ( s) ds + e−tP b −tP
0 =
=
= (1 − 3t) e−2t
3te−2t
−3te−2t
(1 + 3t) e−2t
te
t
−e
3 t
−e
3 + 1
3 −2t +t e −2t + te3t
(3t−1) e3t +1
3
−2t + 1
(1 − 3t) e−2t
3te−2t
−2t
−2t
−3te
(1 + 3t) e
0 (1 − 3t) e
−3te−2t (1 − 2t) e−2t
.
+ 1 − 2t e−2t
3 Phew!
Let us check that this really works.
x1 + 5 x1 − 3 x2 = (4te−2t − 4e−2t ) + 5(1 − 2t) e−2t + et − (1 − 6t) e−2t = et .
Similarly (exercise) x2 + 3 x1 − x2 = 0. The initial conditions are also satisﬁed as well (exercise).
For systems, the integrating factor method only works if P does not depend on t, that is, P is
constant. The problem is that in general
d
e
dt P(t) dt P(t) e P(t) dt , because matrices generally do not commute.
Eigenvector decomposition
For the next method, we note that the eigenvectors of a matrix give the directions in which the
matrix acts like a scalar. If we solve our system along these directions these solutions would be
simpler as we can treat the matrix as a scalar. We can put those solutions together to get the general
solution.
Take the equation
x (t) = A x(t) + f (t).
(3.6)
Assume that A has n linearly independent eigenvectors v1 , . . . , vn . Let us write
x(t) = v1 ξ1 (t) + v2 ξ2 (t) + · · · + vn ξn (t). (3.7) That is, we wish to write our solution as a linear combination of the eigenvectors of A. If we can
solve for the scalar functions ξ1 through ξn we have our solution x. Let us decompose f in terms of
the eigenvectors as well. Write
f (t) = v1 g1 (t) + v2 g2 (t) + · · · + vn gn (t). (3.8) 134 CHAPTER 3. SYSTEMS OF ODES That is, we wish to ﬁnd g1 through gn that satisfy (3.8). We note that since all the eigenvectors of A
are independent, the matrix E = [ v1 v2 · · · vn ] is invertible. We see that (3.8) can be written
as f = E g, where the components of g are the functions g1 through gn . Then g = E −1 f . Hence it is
always possible to ﬁnd g when there are n linearly independent eigenvectors.
We plug (3.7) into (3.6), and note that Avk = λk vk .
x = v1 ξ1 + v2 ξ2 + · · · + vn ξn
= A v1 ξ1 + v2 ξ2 + · · · + vn ξn + v1 g1 + v2 g2 + · · · + vn gn
= Av1 ξ1 + Av2 ξ2 + · · · + Avn ξn + v1 g1 + v2 g2 + · · · + vn gn
= v1 λ1 ξ1 + v2 λ2 ξ2 + · · · + vn λn ξn + v1 g1 + v2 g2 + · · · + vn gn
= v1 (λ1 ξ1 + g1 ) + v2 (λ2 ξ2 + g2 ) + · · · + vn (λn ξn + gn ).
If we identify the coeﬃcients of the vectors v1 through vn we get the equations
ξ1 = λ1 ξ1 + g1 ,
ξ2 = λ2 ξ2 + g2 ,
.
.
.
ξn = λn ξn + gn .
Each one of these equations is independent of the others. They are all linear ﬁrst order equations
and can easily be solved by the standard integrating factor method for single equations. That is, for
example for the kth equation we write
ξk (t) − λk ξk (t) = gk (t).
We use the integrating factor e−λk t to ﬁnd that
d
ξk (t) e−λk t = e−λk t gk (t).
dx
Now we integrate and solve for ξk to get
ξk (t) = eλk t e−λk t gk (t) dt + Ck eλk t . Note that if you are looking for just any particular solution, you could set Ck to be zero. If we leave
these constants in, we will get the general solution. Write x(t) = v1 ξ1 (t) + v2 ξ2 (t) + · · · + vn ξn (t),
and we are done.
Again, as always, it is perhaps better to write these integrals as deﬁnite integrals. Suppose that
we have an initial condition x(0) = b. We take c = E −1 b and note b = v1 a1 + · · · + vn an , just like
before. Then if we write
t ξk (t) = eλk t
0 e−λk s gk ( s) dt + ak eλk t , 3.9. NONHOMOGENEOUS SYSTEMS 135 we will actually get the particular solution x(t) = v1 ξ1 (t) + v2 ξ2 (t) + · · · + vn ξn (t) satisfying x(0) = b,
because ξk (0) = ak .
3/16
Example 3.9.2: Let A = 1 3 . Solve x = A x + f where f (t) = 22et for x(0) = −5/16 .
31
1
The eigenvalues of A are −2 and 4 and the corresponding eigenvectors are −1 and 1 respec1
tively. This calculation is left as an exercise. We write down the matrix E of the eigenvectors and
compute its inverse (using the inverse formula for 2 × 2 matrices)
t E= 11
,
−1 1 We are looking for a solution of the form x =
of the eigenvectors. That is we wish to write f = E −1 =
1
−1
2et
2t 1 1 −1
.
21 1 ξ1 + 1 ξ2 . We also wish to write f in terms
1
1
= −1 g1 + 1 g2 . Thus
1 1 1 −1 2et
g1
2et
et − t
= E −1
=
=t
.
2t
g2
e +t
2 1 1 2t
So g1 = et − t and g2 = et + t.
We further want to write x(0) in terms of the eigenvectors. That is, we wish to write x(0) =
3/16
1
1
−5/16 = −1 a1 + 1 a2 . Hence
a1
= E −1
a2 3
16
−5
16 1
4
−1
16 = . So a1 = 1/4 and a2 = −1/16. We plug our x into the equation and get that
1
1
1
1
1
1
ξ1 +
ξ2 = A
ξ1 + A
ξ2 +
g1 +
g
−1
1
−1
1
−1
12
= 1
1
1
1t
(−2ξ1 ) +
4ξ2 +
(et − t) +
(e − t).
−1
1
−1
1 We get the two equations
ξ1 = −2ξ1 + et − t,
ξ2 = 4ξ2 + et + t, 1
where ξ1 (0) = a1 = ,
4
−1
where ξ2 (0) = a2 =
.
16 We solve with integrating factor. Computation of the integral is left as an exercise to the student.
Note that you will need integration by parts.
ξ1 = e−2t e2t (et − t) dt + C1 e−2t = et t 1
− + + C1 e−2t .
324 136 CHAPTER 3. SYSTEMS OF ODES C1 is the constant of integration. As ξ1 (0) = 1/4, then 1/4 = 1/3 + 1/4 + C1 and hence C1 =
Similarly
et t
1
ξ2 = e4t e−4t (et + t) dt + C2 e4t = − − −
+ C2 e4t .
3 4 16 −1/3. As ξ2 (0) = 1/16 we have that −1/16 = −1/3 − 1/16 + C2 and hence C2 = 1/3. The solution is
1
x(t) =
−1
That is, x1 = e 4 t −e − 2 t
3 + et − e−2t 1 − 2t
1
+
+
1
3
4
3−12t
16 and x2 = e−2t +e4t +2et
3 + e4t − et 4t + 1
−
=
3
16 e4t −e−2t
+ 3−12t
3
16
−2t +e4t +2et
t−
e
+ 4165
3 . 4t−5
.
16 Exercise 3.9.1: Check that x1 and x2 solve the problem. Check both that they satisfy the diﬀerential
equation and that they satisfy the initial conditions.
Undetermined coeﬃcients
The method of undetermined coeﬃcients also works. The only diﬀerence here is that we will have
to take unknown vectors rather than just numbers. Same caveats apply to undetermined coeﬃcients
for systems as they do for single equations. This method does not always work. Furthermore if the
right hand side is complicated, we will have to solve for lots of variables. In this case we can think
of each element of an unknown vector as an unknown number. So in system of 3 equations if we
have say 4 unknown vectors (this would not be uncommon), then we already have 12 unknowns that
we need to solve for. The method can turn into a lot of tedious work. As this method is essentially
the same as it is for single equations, let us just do an example.
Example 3.9.3: Let A = −1 0 . Find a particular solution of x = A x + f where f (t) = et .
−2 1
Note that we can solve this system in an easier way (can you see how), but for the purposes of
the example, let us use the eigenvalue method plus undetermined coeﬃcients.
The eigenvalues of A are −1 and 1 and the corresponding eigenvectors are 1 and 0 respec1
1
tively. Hence our complementary solution is
t xc = α1 1 −t
0t
e + α2
e,
1
1 for some arbitrary constants α1 and α2 .
We would want to guess a particular solution of
x = aet + bt + c.
However, something of the form aet appears in the complementary solution. Because we do not yet
know if the vector a is a multiple of 0 we do not know if a conﬂict arises. It is possible that no
1 3.9. NONHOMOGENEOUS SYSTEMS 137 conﬂict arises, but to be safe we should also try btet . Here we ﬁnd the crux of the diﬀerence for
systems. You want to try both aet and btet in your solution, not just btet . Therefore, we try
x = aet + btet + ct + d.
Thus we have 8 unknowns. We write a = a1 , b =
a2
this into the equation. First let us compute x . b1
b2 ,c= c1
c2 , and d = d1
d2 , We have to plug x = a + b et + btet + c.
Now x must equal A x + f so
A x + f = Aaet + Abtet + Act + Ad + f =
= −a1
−b1
−c1
−d1
et
et +
t et +
t+
+
.
−2a1 + a2
−2b1 + b2
−2c1 + c2
−2d1 + d2
t Now we identify the coeﬃcients of et , tet , t and any constants.
a1 + b1 = −a1 + 1,
a2 + b2 = −2a1 + a2 ,
b1 = −b1 ,
b2 = −2b1 + b2 ,
0 = −c1 ,
0 = −2c1 + c2 + 1,
c1 = −d1 ,
c2 = −2d1 + d2 .
We could write this is an 8 × 9 augmented matrix and start row reduction, but it is easier to just do
this in an ad hoc manner. Immediately we see that b1 = 0, c1 = 0, d1 = 0. Plugging these back in
we get that c2 = −1 and d2 = −1. The remaining equations that tell us something are
a1 = −a1 + 1,
a2 + b2 = −2a1 + a2 .
So a1 = 1 and b2 = −1. a2 can be arbitrary and still satisfy the equation. We are looking for just a
2
single solution so presumably the simplest one is when a2 = 0. Therefore,
x = aet + btet + ct + d = 1
2 0 et + 1t
0
0
0
e
2
t et +
t+
=
.
t
−1
−1
−1
−te − t − 1 That is, x1 = 1 et , x2 = −tet − t − 1. You would add this to the complementary solution to get the
2
general solution of the problem. Notice also that both aet and btet really was needed. 138 CHAPTER 3. SYSTEMS OF ODES Exercise 3.9.2: Check that x1 and x2 solve the problem. Also try setting a2 = 1 and again check
these solutions. What is the diﬀerence between the two solutions we can obtain in this way?
As you can see, other than the handling of conﬂicts, undetermined coeﬃcients works exactly the
same as it did for single equations. However, the computations can get out of hand pretty quickly
for systems. The equation we had done was very simple. 3.9.2 First order variable coeﬃcient Just as for a single equation, there is the method of variation of parameters. In fact for constant
coeﬃcient systems, this is essentially the same thing as the integrating factor method we discussed
earlier. However this method will work for any linear system, even if it is not constant coeﬃcient,
provided you have somehow solved the associated homogeneous problem.
Suppose we have the equation
x = A(t) x + f (t).
(3.9)
Further, suppose that you have solved the associated homogeneous equation x = A(t) x and found
the fundamental matrix solution X (t). The general solution to the associated homogeneous equation
is X (t)c for a constant vector c. Just like for variation of parameters for single equation we try the
solution to the nonhomogeneous equation of the form
x p = X (t) u(t),
where u(t) is a vector valued function instead of a constant. Now substitute into (3.9) to obtain
x p (t) = X (t) u(t) + X (t) u (t) = A(t) X (t) u(t) + f (t).
But X is the fundamental matrix solution to the homogeneous problem so X (t) = A(t)X (t), and thus
X (t) u(t) + X (t) u (t) = X (t) u(t) + f (t).
Hence X (t) u (t) = f (t). If we compute [X (t)]−1 , then u (t) = [X (t)]−1 f (t). Now integrate to obtain
u and we have the particular solution x p = X (t) u(t). Let us write this as a formula
x p = X (t) [X (t)]−1 f (t) dt. Note that if A is constant and you let X (t) = etA , then [X (t)]−1 = e−tA and hence we get a solution
x p = etA e−tA f (t) dt, which is precisely what we got using the integrating factor method.
Example 3.9.4: Find a particular solution to
x= t2 1
t −1
t2
x+
(t + 1).
1t
1
+1 (3.10) 3.9. NONHOMOGENEOUS SYSTEMS 139 t
Here A = t21 1 1 −t1 is most deﬁnitely not constant. Perhaps by a lucky guess, we ﬁnd that
+
X = 1 −t solves X (t) = A(t)X (t). Once we know the complementary solution we can easily ﬁnd a
t1
solution to (3.10). First we ﬁnd [X (t)]−1 = t2 1
1t
.
+ 1 −t 1 Next we know a particular solution to (3.10) is
x p = X (t) [X (t)]−1 f (t) dt
1
1tt2
(t + 1) dt
2 + 1 −t 1 1
t
2t
dt
−t2 + 1 = 1 −t
t1 = 1 −t
t1 = 1 −t
t2
t 1 − 1 t3 + t
3 = 14
t
3
23
t+
3 t . Adding the complementary solution we have that the general solution to (3.10).
x= 1 −t c1
+
t 1 c2 14
t
3
23
t+
3 t = 1
c1 − c2 t + 3 t4
.
c2 + (c1 + 1) t + 2 t3
3 2
Exercise 3.9.3: Check that x1 = 1 t4 and x2 = 3 t3 + t really solve (3.10).
3 In the variation of parameters, just like in the integrating factor method we can obtain the general
solution by adding in constants of integration. That is, we will add X (t)c for a vector of arbitrary
constants. But that is precisely the complementary solution. 3.9.3 Second order constant coeﬃcients Undetermined coeﬃcients
We have already previously did a simple example of the method of undetermined coeﬃcients for
second order systems in § 3.6. This method is essentially the same as undetermined coeﬃcients for
ﬁrst order systems. There are some simpliﬁcations that you can make, as we did in § 3.6. Let the
equation be
x = A x + F (t),
where A is a constant matrix. If F (t) is of the form F0 cos(ωt), then you can try a solution of the
form
x p = c cos(ωt), 140 CHAPTER 3. SYSTEMS OF ODES and you do not need to introduce sines.
If the F is a sum of cosines, you note that we still have the superposition principle, so if
F (t) = F0 cos(ω0 t) + F1 cos(ω1 t), you could try a cos(ω0 t) for the problem x = A x + F0 cos(ω0 t),
and you would try b cos(ω1 t) for the problem x = A x + F0 cos(ω1 t). Then sum the solutions.
However, if there is duplication with the complementary solution, or the equation is of the form
x = A x + B x + F (t), then you need to do the same thing as you do for ﬁrst order systems.
Actually you will never go wrong with putting in more terms than needed into your guess. You
will just ﬁnd that the extra coeﬃcients will turn out to be zero. But it is useful to save some time
and eﬀort.
Eigenvector decomposition
If we have the system
x = A x + F (t),
we can do eigenvector decomposition, just like for ﬁrst order systems.
Let λ1 , . . . , λn be the eigenvalues and v1 , . . . , vn be the eigenvectors. Again form the matrix
E = [ v1 · · · vn ]. Write
x(t) = v1 ξ1 (t) + v2 ξ2 (t) + · · · + vn ξn (t).
Decompose F in terms of the eigenvectors
F (t) = v1 g1 (t) + v2 g2 (t) + · · · + vn gn (t).
And again g = E −1 F .
Now plug in and doing the same thing as before
x = v1 ξ1 + v2 ξ2 + · · · + vn ξn
= A v1 ξ1 + v2 ξ2 + · · · + vn ξn + v1 g1 + v2 g2 + · · · + vn gn
= Av1 ξ1 + Av2 ξ2 + · · · + Avn ξn + v1 g1 + v2 g2 + · · · + vn gn
= v1 λ1 ξ1 + v2 λ2 ξ2 + · · · + vn λn ξn + v1 g1 + v2 g2 + · · · + vn gn
= v1 (λ1 ξ1 + g1 ) + v2 (λ2 ξ2 + g2 ) + · · · + vn (λn ξn + gn ).
Identify the coeﬃcients of the eigenvectors to get the equations
ξ1 = λ1 ξ1 + g1 ,
ξ2 = λ2 ξ2 + g2 ,
.
.
.
ξn = λn ξn + gn .
Each one of these equations is independent of the others. Now solve each one of these using
the methods of chapter 2. Now write x(t) = v1 ξ1 (t) + · · · + vn ξn (t), and we are done; we have
a particular solution. If you have found the general solution for ξ1 through ξn , then again x(t) =
v1 ξ1 (t) + · · · + vn ξn (t) is the general solution. 3.9. NONHOMOGENEOUS SYSTEMS 141 Example 3.9.5: Let us do the example from § 3.6 using this method. The equation is
x= 0
−3 1
x+
cos(3t).
2
2 −2 The eigenvalues were −1 and −4, with eigenvectors
1
1
E −1 = 3 1 −1 . Therefore,
2 1
2 and 0
g1
11 1
=
= E −1 F (t) =
3 2 −1 2 cos(3t)
g2 1
−1 . Therefore E = 2
cos(3t)
3
−2
cos(3t)
3 11
2 −1 and . So after the whole song and dance of plugging in, the equations we get are
2
cos(3t),
3
2
ξ2 = −4 ξ2 − cos(3t).
3 ξ1 = −ξ1 + For each we can try the method of undetermined coeﬃcients and try C1 cos(3t) for the ﬁrst equation
and C2 cos(3t) for the second equation. We plug in to get
2
cos(3t),
3
2
−9C2 cos(3t) = −4C2 cos(3t) − cos(3t).
3
−9C1 cos(3t) = −C1 cos(3t) + Each of these equations we solve separately. We get −9C1 = −C1 + 2/3 and −9C2 = −4C2 − 2/3. And
hence C1 = −1/12 and C2 = 2/15. So our particular solution is
x= 1
2 −1
1
cos(3t) +
−1
12 2
cos(3t) =
15 1/20
−3/10 cos(3t). This solution matches what we got previously in § 3.6. 3.9.4 Exercises Exercise 3.9.4: Find a particular solution to x = x + 2y + 2t, y = 3 x + 2y − 4, a) using integrating
factor method, b) using eigenvector decomposition, c) using undetermined coeﬃcients.
Exercise 3.9.5: Find the general solution to x = 4 x + y − 1, y = x + 4y − et , a) using integrating
factor method, b) using eigenvector decomposition, c) using undetermined coeﬃcients.
Exercise 3.9.6: Find the general solution to x1 = −6 x1 + 3 x2 + cos(t), x2 = 2 x1 − 7 x2 + 3 cos(t), a)
using eigenvector decomposition, b) using undetermined coeﬃcients. 142 CHAPTER 3. SYSTEMS OF ODES Exercise 3.9.7: Find the general solution to x1 = −6 x1 + 3 x2 + cos(2t), x2 = 2 x1 − 7 x2 + 3 cos(2t),
a) using eigenvector decomposition, b) using undetermined coeﬃcients.
Exercise 3.9.8: Take the equation
x= 1
t 1 −1
1
t x+ t2
.
−t a) Check that
xc = c1 t sin t
t cos t
+ c2
−t cos t
t sin t is the complementary solution. b) Use variation of parameters to ﬁnd a particular solution. Chapter 4
Fourier series and PDEs
4.1 Boundary value problems Note: 2 lectures, similar to §3.8 in [EP] 4.1.1 Boundary value problems Before we tackle the Fourier series, we need to study the socalled boundary value problems (or
endpoint problems). For example, suppose we have
x + λ x = 0, x(a) = 0, x(b) = 0, for some constant λ, where x(t) is deﬁned for t in the interval [a, b]. Unlike before, when we
speciﬁed the value of the solution and its derivative at a single point, we now specify the value of
the solution at two diﬀerent points. Note that x = 0 is a solution to this equation, so existence of
solutions is not an issue here. Uniqueness of solutions is another issue. The general solution to
x + λ x = 0 will have two arbitrary constants present. It is, therefore, natural (but wrong) to believe
that requiring two conditions will guarantee a unique solution.
Example 4.1.1: Take λ = 1, a = 0, b = π. That is,
x + x = 0, x(0) = 0, x(π) = 0. Then x = sin t is another solution (besides x = 0) satisfying both boundary conditions. There are
more. Write down the general solution of the diﬀerential equation, which is x = A cos t + B sin t.
The condition x(0) = 0 forces A = 0. Letting x(π) = 0 does not give us any more information as
x = B sin t already satisﬁes both boundary conditions. Hence, there are inﬁnitely many solutions of
the form x = B sin t, where B is an arbitrary constant.
143 144 CHAPTER 4. FOURIER SERIES AND PDES Example 4.1.2: On the other hand, change to λ = 2.
x + 2 x = 0, x(0) = 0, x(π) = 0.
√
√
Then the general solution is x = A cos( 2 t) + B sin( 2 t). Letting x(0) = 0 still forces A = 0. We
√
√
apply the second condition to ﬁnd 0 = x(π) = B sin( 2 π). As sin( 2 π) 0 we obtain B = 0.
Therefore x = 0 is the unique solution to this problem.
What is going on? We will be interested in ﬁnding which constants λ allow a nonzero solution,
and we will be interested in ﬁnding those solutions. This problem is an analogue of ﬁnding
eigenvalues and eigenvectors of matrices. 4.1.2 Eigenvalue problems For basic Fourier series theory we will need the following three eigenvalue problems. We will
consider more general equations, but we will postpone this until chapter 5.
x + λ x = 0, x(a) = 0, x(b) = 0, (4.1) x + λ x = 0, x (a) = 0, x (b) = 0, (4.2) x (a) = x (b), (4.3) and
x + λ x = 0, x(a) = x(b), A number λ is called an eigenvalue of (4.1) (resp. (4.2) or (4.3)) if and only if there exists a nonzero
(not identically zero) solution to (4.1) (resp. (4.2) or (4.3)) given that speciﬁc λ. The nonzero
solution we found is called the corresponding eigenfunction.
Note the similarity to eigenvalues and eigenvectors of matrices. The similarity is not just
coincidental. If we think of the equations as diﬀerential operators, then we are doing the same exact
d2
thing. For example, let L = − dt2 . We are looking for nonzero functions f satisfying certain endpoint
conditions that solve (L − λ) f = 0. A lot of the formalism from linear algebra can still apply here,
though we will not pursue this line of reasoning too far.
Example 4.1.3: Let us ﬁnd the eigenvalues and eigenfunctions of
x + λ x = 0, x(0) = 0, x(π) = 0. For reasons that will be clear from the computations, we will have to handle the cases λ > 0,
λ = 0, λ < 0 separately. First suppose that λ > 0, then the general solution to x + λ x = 0 is
√
√
x = A cos( λ t) + B sin( λ t).
The condition x(0) = 0 implies immediately A = 0. Next
√
0 = x(π) = B sin( λ π). 4.1. BOUNDARY VALUE PROBLEMS 145 If B√ zero, then x is not a nonzero solution. So to get a nonzero solution we must have that
is
√
√
sin( λ π) = 0. Hence, λ π must be an integer multiple of π. In other words, λ = k for a
positive integer k. Hence the positive eigenvalues are k2 for all integers k ≥ 1. The corresponding
eigenfunctions can be taken as x = sin(kt). Just like for eigenvectors, we get all the multiples of an
eigenfunction, so we only need to pick one.
Now suppose that λ = 0. In this case the equation is x = 0 and the general solution is x = At + B.
The condition x(0) = 0 implies that B = 0, and x(π) = 0 implies that A = 0. This means that λ = 0
is not an eigenvalue.
Finally, suppose that λ < 0. In this case we have the general solution
√
√
x = A cosh( −λ t) + B sinh( −λ t).
Letting x(0) = 0 implies that A = 0 (recall cosh 0 = 1 and sinh 0 = 0). So our solution must be
√
x = B sinh( −λ t) and satisfy x(π) = 0. This is only possible if B is zero. Why? Because sinh ξ is
only zero for ξ = 0, you should plot sinh to see this. We can also see this from the deﬁnition of sinh.
t
−t
We get 0 = sinh t = e −e . Hence et = e−t , which implies t = −t and that is only true if t = 0. So
2
there are no negative eigenvalues.
In summary, the eigenvalues and corresponding eigenfunctions are
λk = k2 with an eigenfunction xk = sin(kt) for all integers k ≥ 1. Example 4.1.4: Let us compute the eigenvalues and eigenfunctions of
x + λ x = 0, x (0) = 0, x (π) = 0. Again we will have to handle the cases λ > 0,√ = 0, λ < 0 √
λ
separately. First suppose that λ > 0.
The general solution to x + λ x = 0 is x = A cos( λ t) + B sin( λ t). So
√
√
√
√
x = −A λ sin( λ t) + B λ cos( λ t).
The condition x (0) = 0 implies immediately B = 0. Next
√
√
0 = x (π) = −A λ sin( λ π).
√
√
Again A cannot be zero if λ is to be an eigenvalue, and sin( λ π) is only zero if λ = k for a positive
integer k. Hence the positive eigenvalues are again k2 for all integers k ≥ 1. And the corresponding
eigenfunctions can be taken as x = cos(kt).
Now suppose that λ = 0. In this case the equation is x = 0 and the general solution is x = At + B
so x = A. x (0) = 0 implies that A = 0. Obviously setting x (π) = 0 does not get us anything new.
This means that B could be anything (let us take it to be 1). So λ = 0 is an eigenvalue and x = 1 is a
corresponding eigenfunction.
√
√
Finally, let λ < 0. In this case we have the general solution x = A cosh( −λ t) + B sinh( −λ t)
and hence
√
√
√
√
x = A −λ sinh( −λ t) + B −λ cosh( −λ t). 146 CHAPTER 4. FOURIER SERIES AND PDES We have already seen (with roles of A and B switched) that for this to be zero at t = 0 and t = π it
implies that A = B = 0. Hence there are no negative eigenvalues.
In summary, the eigenvalues and corresponding eigenfunctions are
λk = k2 with an eigenfunction xk = cos(kt) for all integers k ≥ 1, and there is another eigenvalue
λ0 = 0 with an eigenfunction x0 = 1. The following problem is the one that leads to the general Fourier series.
Example 4.1.5: Let us compute the eigenvalues and eigenfunctions of
x + λ x = 0, x(−π) = x(π), x (−π) = x (π). You should notice that we have not speciﬁed the values or the derivatives at the endpoints, but rather
that they are the same at the beginning and at the end of the interval.
Let us skip λ < 0. The computations are the same and again we ﬁnd that there are no negative
eigenvalues.
For λ = 0, the general solution is x = At + B. The condition x(−π) = x(π) implies that A = 0
(Aπ + B = −Aπ + B implies A = 0). The second condition x (−π) = x (π) says nothing about B and
hence λ = 0 is an eigenvalue with a √
corresponding eigenfunction x = 1.
√
For λ > 0 we get that x = A cos( λ t) + B sin( λ t). Now
√
√
√
√
A cos(− λ π) + B sin(− λ π) = A cos( λ π) + B sin( λ π).
We remember that cos(−θ) = cos(θ) and sin(−θ) = − sin(θ). Therefore,
√
√
√
√
A cos( λ π) − B sin( λ π) = A cos( λ π) + B sin( λ π).
√
and hence either B = 0 or sin( λ π) = 0. Similarly (exercise) if we diﬀerentiate x and plug in the
√
second condition we ﬁnd that A = 0 or sin(√λ π) = 0. Therefore, unless we want A and B to both be
√
zero (which we do not) we must have sin( λ π) = 0. Hence, λ is an integer and the eigenvalues
are yet again λ = k2 for an integer k ≥ 1. In this case, however, x = A cos(kt) + B sin(kt) is an
eigenfunction for any A and any B. So we have two linearly independent eigenfunctions sin(kt) and
cos(kt). Remember that for a matrix we could also have had two eigenvectors corresponding to a
single eigenvalue if the eigenvalue was repeated.
In summary, the eigenvalues and corresponding eigenfunctions are
λk = k2
λ0 = 0 with the eigenfunctions
with an eigenfunction cos(kt)
x0 = 1. and sin(kt) for all integers k ≥ 1, 4.1. BOUNDARY VALUE PROBLEMS 4.1.3 147 Orthogonality of eigenfunctions Something that will be very useful in the next section is the orthogonality property of the eigenfunctions. This is an analogue of the following fact about eigenvectors of a matrix. A matrix is
called symmetric if A = AT . Eigenvectors for two distinct eigenvalues of a symmetric matrix are
orthogonal. That symmetry is required. We will not prove this fact here. The diﬀerential operators
we are dealing with act much like a symmetric matrix. We, therefore, get the following theorem.
Theorem 4.1.1. Suppose that x1 (t) and x2 (t) are two eigenfunctions of the problem (4.1), (4.2) or
(4.3) for two diﬀerent eigenvalues λ1 and λ2 . Then they are orthogonal in the sense that
b x1 (t) x2 (t) dt = 0.
a Note that the terminology comes from the fact that the integral is a type of inner product. We
will expand on this in the next section. The theorem has a very short, elegant, and illuminating
proof so let us give it here. First note that we have the following two equations.
x1 + λ1 x1 = 0 x2 + λ2 x2 = 0. and Multiply the ﬁrst by x2 and the second by x1 and subtract to get
(λ1 − λ2 ) x1 x2 = x2 x1 − x2 x1 .
Now integrate both sides of the equation.
b (λ1 − λ2 ) b x1 x2 dt =
a x2 x1 − x2 x1 dt
a
b =
a d
x x1 − x2 x1 dt
dt 2 = x2 x1 − x2 x1 b
t =a = 0. The last equality holds because of the boundary conditions. For example, if we consider (4.1) we
have x1 (a) = x1 (b) = x2 (a) = x2 (b) = 0 and so x2 x1 − x2 x1 is zero at both a and b. As λ1 λ2 , the
theorem follows.
Exercise 4.1.1 (easy): Finish the theorem (check the last equality in the proof) for the cases (4.2)
and (4.3).
We have seen previously that sin(nt) was an eigenfunction for the problem x + λ x = 0, x(0) = 0,
x(π) = 0. Hence we have the integral
π sin(mt) sin(nt) dt = 0,
0 when m n. 148 CHAPTER 4. FOURIER SERIES AND PDES Similarly π cos(mt) cos(nt) dt = 0, when m n. sin(mt) sin(nt) dt = 0, when m n, cos(mt) cos(nt) dt = 0, when m n, 0 And ﬁnally we also get
π
−π
π
−π and π
−π 4.1.4 cos(mt) sin(nt) dt = 0. Fredholm alternative We now touch on a very useful theorem in the theory of diﬀerential equations. The theorem holds
in a more general setting than we are going to state it, but for our purposes the following statement
is suﬃcient. We will give a slightly more general version in chapter 5.
Theorem 4.1.2 (Fredholm alternative∗ ). Exactly one of the following statements holds. Either
x + λ x = 0, x(a) = 0, x(b) = 0 (4.4) has a nonzero solution, or
x + λ x = f (t), x(a) = 0, x(b) = 0 (4.5) has a unique solution for every function f continuous on [a, b].
The theorem is also true for the other types of boundary conditions we considered. The theorem
means that if λ is not an eigenvalue, the nonhomogeneous equation (4.5) has a unique solution for
every right hand side. On the other hand if λ is an eigenvalue, then (4.5) need not have a solution
for every f , and furthermore, even if it happens to have a solution, the solution is not unique.
We also want to reinforce the idea here that linear diﬀerential operators have much in common
with matrices. So it is no surprise that there is a ﬁnite dimensional version of Fredholm alternative
for matrices as well. Let A be an n × n matrix. The Fredholm alternative then states that either
(A − λI ) x = 0 has a nontrivial solution, or (A − λI ) x = b has a solution for every b.
A lot of intuition from linear algebra can be applied for linear diﬀerential operators, but one
must be careful of course. For example, one obvious diﬀerence we have already seen is that in
general a diﬀerential operator will have inﬁnitely many eigenvalues, while a matrix has only ﬁnitely
many.
∗ Named after the Swedish mathematician Erik Ivar Fredholm (1866 – 1927). 4.1. BOUNDARY VALUE PROBLEMS 4.1.5 149 Application Let us consider a physical application of an endpoint problem. Suppose we have a tightly stretched
quickly spinning elastic string or rope of uniform linear density ρ. Let us put this problem into the
xyplane. The x axis represents the position on the string. The string rotates at angular velocity ω,
so we will assume that the whole xyplane rotates at angular velocity ω. We will assume that the
string stays in this xyplane and y will measure its deﬂection from the equilibrium position, y = 0,
on the x axis. Hence, we will ﬁnd a graph giving the shape of the string. We will idealize the string
to have no volume to just be a mathematical curve. If we take a small segment and we look at the
tension at the endpoints, we see that this force is tangential and we will assume that the magnitude
is the same at both end points. Hence the magnitude is constant everywhere and we will call its
magnitude T . If we assume that the deﬂection is small, then we can use Newton’s second law to get
an equation
T y + ρω2 y = 0.
Let L be the length of the string and the string is ﬁxed at the beginning and end points. Hence,
y(0) = 0 and y(L) = 0. See Figure 4.1.
y
y
0 L x Figure 4.1: Whirling string.
We rewrite the equation as y + ρω y = 0. The setup is similar to Example 4.1.3 on page 144,
T
except for the interval length being L instead of π. We are looking for eigenvalues of y + λy =
2
0, y(0) = 0, y(L) = 0 where λ = ρω . As before there are no nonpositive eigenvalues. With λ > 0, the
T
√
√
general solution to the equation is y = A cos( λ x) + B sin( λ x). The condition y(0) = 0 implies
√
√
that A = 0 as before. The condition y(L) = 0 implies that sin( λ L) = 0 and hence λ L = kπ for
some integer k > 0, so
ρω2
k2 π2
=λ= 2 .
T
L
What does this say about the shape of the string? It says that for all parameters ρ, ω, T not
2
22
satisfying the above equation, the string is in the equilibrium position, y = 0. When ρω = kLπ ,
2
T
then the string will “pop out” some distance B at the midpoint. We cannot compute B with the
information we have.
Let us assume that ρ and T are ﬁxed and we are changing ω. For most values of ω the string is
√
πT
in the equilibrium state. When the angular velocity ω hits a value ω = kL √ρ , then the string will pop
2 150 CHAPTER 4. FOURIER SERIES AND PDES out and will have the shape of a sin wave crossing the x axis k times. When ω changes again, the
string returns to the equilibrium position. You can see that the higher the angular velocity the more
times it crosses the x axis when it is popped out. 4.1.6 Exercises Hint for the following exercises: Note that when λ > 0, then cos
also solutions of the homogeneous equation. √
√
λ (t − a) and sin λ (t − a) are Exercise 4.1.2: Compute all eigenvalues and eigenfunctions of x + λ x = 0, x(a) = 0, x(b) = 0
(assume a < b).
Exercise 4.1.3: Compute all eigenvalues and eigenfunctions of x + λ x = 0, x (a) = 0, x (b) = 0
(assume a < b).
Exercise 4.1.4: Compute all eigenvalues and eigenfunctions of x + λ x = 0, x (a) = 0, x(b) = 0
(assume a < b).
Exercise 4.1.5: Compute all eigenvalues and eigenfunctions of x + λ x = 0, x(a) = x(b), x (a) =
x (b) (assume a < b).
Exercise 4.1.6: We have skipped the case of λ < 0 for the boundary value problem x + λ x =
0, x(−π) = x(π), x (−π) = x (π). Finish the calculation and show that there are no negative
eigenvalues. 4.2. THE TRIGONOMETRIC SERIES 4.2 151 The trigonometric series Note: 2 lectures, §9.1 in [EP] 4.2.1 Periodic functions and motivation As motivation for studying Fourier series, suppose we have the problem
x + ω2 x = f (t),
0 (4.6) for some periodic function f (t). We have already solved
x + ω2 x = F0 cos(ωt).
0 (4.7) One way to solve (4.6) is to decompose f (t) as a sum of cosines (and sines) and then solve many
problems of the form (4.7). We then use the principle of superposition, to sum up all the solutions
we got to get a solution to (4.6).
Before we proceed, let us talk a little bit more in detail about periodic functions. A function
is said to be periodic with period P if f (t) = f (t + P) for all t. For brevity we will say f (t) is
Pperiodic. Note that a Pperiodic function is also 2Pperiodic, 3Pperiodic and so on. For example,
cos(t) and sin(t) are 2πperiodic. So are cos(kt) and sin(kt) for all integers k. The constant functions
are an extreme example. They are periodic for any period (exercise).
Normally we will start with a function f (t) deﬁned on some interval [−L, L] and we will want
to extend periodically to make it a 2Lperiodic function. We do this extension by deﬁning a new
function F (t) such that for t in [−L, L], F (t) = f (t). For t in [L, 3L], we deﬁne F (t) = f (t − 2L), for
t in [−3L, −L], F (t) = f (t + 2L), and so on. We assumed that f (−L) = f (L). We could have also
started with f deﬁned only on the halfopen interval (−L, L] and then deﬁne f (−L) = f (L).
Example 4.2.1: Deﬁne f (t) = 1 − t2 on [−1, 1]. Now extend periodically to a 2periodic function.
See Figure 4.2 on the following page.
You should be careful to distinguish between f (t) and its extension. A common mistake is to
assume that a formula for f (t) holds for its extension. It can be confusing when the formula for f (t)
is periodic, but with perhaps a diﬀerent period.
Exercise 4.2.1: Deﬁne f (t) = cos t on [−π/2, π/2]. Now take the πperiodic extension and sketch its
graph. How does it compare to the graph of cos t. 4.2.2 Inner product and eigenvector decomposition Suppose we have a symmetric matrix, that is AT = A. We have said before that the eigenvectors of A
are then orthogonal. Here the word orthogonal means that if v and w are two distinct eigenvectors 152 CHAPTER 4. FOURIER SERIES AND PDES
3 2 1 0 1 2 3 1.5 1.5 1.0 1.0 0.5 0.5 0.0 0.0 0.5 0.5
3 2 1 0 1 2 3 Figure 4.2: Periodic extension of the function 1 − t2 . of A, then v, w = 0. In this case the inner product v, w is the dot product, which can be computed
as vT w.
To decompose a vector v in terms of mutually orthogonal vectors w1 and w2 we write
v = a1 w1 + a2 w2 .
Let us ﬁnd the formula for a1 and a2 . First let us compute
v, w1 = a1 w1 + a2 w2 , w1 = a1 w1 , w1 + a2 w2 , w1 = a1 w1 , w1 .
Therefore,
a1 = v, w1
.
w1 , w1 Similarly v, w2
.
w2 , w2
You probably remember this formula from vector calculus.
a2 = 1
Example 4.2.2: Write v = 2 as a linear combination of w1 = −1 and w2 = 1 .
3
1
First note that w1 and w2 are orthogonal as w1 , w2 = 1(1) + (−1)1 = 0. Then 2(1) + 3(−1)
−1
v, w1
=
=
,
1(1) + (−1)(−1)
2
w1 , w1
v, w2
2+3 5
=
a2 =
=.
1+1 2
w2 , w2 a1 = Hence −1 1
51
2
=
+
.
3
2 −1
21 4.2. THE TRIGONOMETRIC SERIES 4.2.3 153 The trigonometric series Instead of decomposing a vector in terms of the eigenvectors of a matrix, we will decompose a
function in terms of eigenfunctions of a certain eigenvalue problem. The eigenvalue problem we
will use for the Fourier series is
x + λ x = 0, x(−π) = x(π), x (−π) = x (π). We have previously computed that the eigenfunctions are 1, cos(kt), sin(kt). That is, we will want to
ﬁnd a representation of a 2πperiodic function f (t) as
a0
f (t) =
+
2 ∞ an cos(nt) + bn sin(nt).
n=1 This series is called the Fourier series† or the trigonometric series for f (t). We write the coeﬃcient
of the eigenfunction 1 as a20 for convenience. We could also think of 1 = cos(0t), so that we only
need to look at cos(kt) and sin(kt).
As for matrices we will want to ﬁnd a projection of f (t) onto the subspace generated by the
eigenfunctions. So we will want to deﬁne an inner product of functions. For example, to ﬁnd an we
want to compute f (t) , cos(nt) . We deﬁne the inner product as
def f (t) , g(t) = π f (t) g(t) dt.
−π With this deﬁnition of the inner product, we have seen in the previous section that the eigenfunctions
cos(kt) (including the constant eigenfunction), and sin(kt) are orthogonal in the sense that
cos(mt) , cos(nt) = 0
sin(mt) , sin(nt) = 0
sin(mt) , cos(nt) = 0 for m n,
for m n,
for all m and n. By elementary calculus for n = 1, 2, 3, . . . we have cos(nt) , cos(nt) = π and sin(nt) , sin(nt) =
π. For the constant we get that 1 , 1 = 2π. The coeﬃcients are given by
f (t) , cos(nt)
1
an =
=
cos(nt) , cos(nt)
π
f (t) , sin(nt)
1
bn =
=
sin(nt) , sin(nt)
π π f (t) cos(nt) dt,
−π
π f (t) sin(nt) dt.
−π Compare these expressions with the ﬁnitedimensional example. For a0 we get a similar formula
a0 = 2
† f (t) , 1
1
=
1, 1
π π f (t) dt.
−π Named after the French mathematician Jean Baptiste Joseph Fourier (1768 – 1830). 154 CHAPTER 4. FOURIER SERIES AND PDES Let us check the formulas using the orthogonality properties. Suppose for a moment that
a0
f (t) =
+
2 ∞ an cos(nt) + bn sin(nt).
n=1 Then for m ≥ 1 we have
a0
f (t) , cos(mt) =
+
2
= ∞ an cos(nt) + bn sin(nt) , cos(mt)
n=1 a0
1 , cos(mt) +
2 ∞ an cos(nt) , cos(mt) + bn sin(nt) , cos(mt)
n=1 = am cos(mt) , cos(mt) .
And hence am = f (t) , cos(mt)
cos(mt) , cos(mt) . Exercise 4.2.2: Carry out the calculation for a0 and bm .
Example 4.2.3: Take the function
f (t) = t
for t in (−π, π]. Extend f (t) periodically and write it as a Fourier series. This function is called the
sawtooth.
5.0 2.5 0.0 2.5 5.0 3 3 2 2 1 1 0 0 1 1 2 2 3 3
5.0 2.5 0.0 2.5 5.0 Figure 4.3: The graph of the sawtooth function.
The plot of the extended periodic function is given in Figure 4.3. Let us compute the coeﬃcients.
We start with a0 ,
1π
a0 =
t dt = 0.
π −π 4.2. THE TRIGONOMETRIC SERIES 155 We will often use the result from calculus that says that the integral of an odd function over a
symmetric interval is zero. Recall that an odd function is a function ϕ(t) such that ϕ(−t) = −ϕ(t).
For example the functions t, sin t, or (importantly for us) t cos(nt) are all odd functions. Thus
an = 1
π π
−π t cos(nt) dt = 0. Let us move to bn . Another useful fact from calculus is that the integral of an even function over a
symmetric interval is twice the integral of the same function over half the interval. Recall an even
function is a function ϕ(t) such that ϕ(−t) = ϕ(t). For example t sin(nt) is even.
bn =
=
=
=
=
We have used the fact that 1π
t sin(nt) dt
π −π
2π
t sin(nt) dt
π0
π
2 −t cos(nt)
1π
+
cos(nt) dt
π
n
n0
t =0
2 −π cos(nπ)
+0
π
n
−2 cos(nπ) 2 (−1)n+1
=
.
n
n 1 if n even, cos(nπ) = (−1) = −1 if n odd. n The series, therefore, is
∞
n=1 2 (−1)n+1
sin(nt).
n Let us write out the ﬁrst 3 harmonics of the series for f (t).
2 sin(t) − sin(2t) + 2
sin(3t) + · · ·
3 The plot of these ﬁrst three terms of the series, along with a plot of the ﬁrst 20 terms is given
in Figure 4.4 on the following page.
Example 4.2.4: Take the function 0 f (t) = π if −π < t ≤ 0,
if 0 < t ≤ π. 156 CHAPTER 4. FOURIER SERIES AND PDES
5.0 2.5 0.0 2.5 5.0 5.0 2.5 0.0 2.5 5.0 3 3 3 3 2 2 2 2 1 1 1 1 0 0 0 0 1 1 1 1 2 2 2 2 3 3 3 3 5.0 2.5 0.0 2.5 5.0 5.0 2.5 0.0 2.5 5.0 Figure 4.4: First 3 (left graph) and 20 (right graph) harmonics of the sawtooth function.
5.0 2.5 0.0 2.5 5.0 3 3 2 2 1 1 0 0 5.0 2.5 0.0 2.5 5.0 Figure 4.5: The graph of the square wave function. Extend f (t) periodically and write it as a Fourier series. This function or its variants appear often in
applications and the function is called the square wave.
The plot of the extended periodic function is given in Figure 4.5. Now we compute the
coeﬃcients. Let us start with a0
a0 =
Next,
an = 1
π 1
π π
−π f (t) dt = π
−π f (t) cos(nt) dt = 1
π
1
π π π dt = π.
0 π π cos(nt) dt = 0.
0 4.2. THE TRIGONOMETRIC SERIES 157 And ﬁnally
1π
f (t) sin(nt) dt
π −π
1π
=
π sin(nt) dt
π0
π
− cos(nt)
=
n
t =0 bn = 1 − cos(πn) 1 − (−1)n 2 =
=
= n
0 n
n if n is odd,
if n is even. The Fourier series is
π
+
2 ∞
n=1
n odd 2
π
sin(nt) = +
n
2 ∞ 2
sin (2k − 1) t .
2k − 1 k =1 Let us write out the ﬁrst 3 harmonics of the series for f (t).
2
π
+ 2 sin(t) + sin(3t) + · · ·
2
3
The plot of these ﬁrst three and also of the ﬁrst 20 terms of the series is given in Figure 4.6.
5.0 2.5 0.0 2.5 5.0 5.0 2.5 0.0 2.5 5.0 3 3 3 3 2 2 2 2 1 1 1 1 0 0 0 0 5.0 2.5 0.0 2.5 5.0 5.0 2.5 0.0 2.5 5.0 Figure 4.6: First 3 (left graph) and 20 (right graph) harmonics of the square wave function. We have so far skirted the issue of convergence. For example, if f (t) is the square wave function,
the equation
∞
π
2
f (t) = +
sin (2k − 1) t .
2 k=1 2k − 1 158 CHAPTER 4. FOURIER SERIES AND PDES is only an equality for such t where f (t) is continuous. That is, we do not get an equality for
t = −π, 0, π and all the other discontinuities of f (t). It is not hard to see that when t is an integer
multiple of π (which includes all the discontinuities), then
π
+
2 ∞
k =1 2
π
sin (2k − 1) t = .
2k − 1
2 We redeﬁne f (t) on [−π, π] as 0 f (t) = π π /2 if −π < t < 0,
if 0 < t < π,
if t = −π, t = 0, or t = π, and extend periodically. The series equals this extended f (t) everywhere, including the discontinuities. We will generally not worry about changing the function values at several (ﬁnitely many)
points.
We will say more about convergence in the next section. Let us however mention brieﬂy an
eﬀect of the discontinuity. Let us zoom in near the discontinuity in the square wave. Further, let
us plot the ﬁrst 100 harmonics, see Figure 4.7. You will notice that while the series is a very good
approximation away from the discontinuities, the error (the overshoot) near the discontinuity at
t = π does not seem to be getting any smaller. This behavior is known as the Gibbs phenomenon.
The region where the error is large does get smaller, however, the more terms in the series you take.
1.75 2.00 2.25 2.50 2.75 3.00 3.25 3.50 3.50 3.25 3.25 3.00 3.00 2.75 2.75 1.75 2.00 2.25 2.50 2.75 3.00 3.25 Figure 4.7: Gibbs phenomenon in action. We can think of a periodic function as a “signal” being a superposition of many signals of pure
frequency. For example, we could think of the square wave as a tone of certain base frequency. It
will be, in fact, a superposition of many diﬀerent pure tones of frequencies that are multiples of the 4.2. THE TRIGONOMETRIC SERIES 159 base frequency. On the other hand a simple sine wave is only the pure tone. The simplest way to
make sound using a computer is the square wave, and the sound will be a very diﬀerent from nice
pure tones. If you have played video games from the 1980s or so you have heard what square waves
sound like. 4.2.4 Exercises Exercise 4.2.3: Suppose f (t) is deﬁned on [−π, π] as sin(5t) + cos(3t). Extend periodically and
compute the Fourier series of f (t).
Exercise 4.2.4: Suppose f (t) is deﬁned on [−π, π] as t. Extend periodically and compute the
Fourier series of f (t).
Exercise 4.2.5: Suppose f (t) is deﬁned on [−π, π] as t3 . Extend periodically and compute the
Fourier series of f (t).
Exercise 4.2.6: Suppose f (t) is deﬁned on (−π, π] as −1 if −π < t ≤ 0, f (t) = 1 if 0 < t ≤ π.
Extend periodically and compute the Fourier series of f (t).
Exercise 4.2.7: Suppose f (t) is deﬁned on (−π, π] as t3 . Extend periodically and compute the
Fourier series of f (t).
Exercise 4.2.8: Suppose f (t) is deﬁned on [−π, π] as t2 . Extend periodically and compute the
Fourier series of f (t).
There is another form of the Fourier series using complex exponentials that is sometimes easier
to work with.
Exercise 4.2.9: Let
a0
f (t) =
+
2 ∞ an cos(nt) + bn sin(nt).
n=1 Use Euler’s formula eiθ = cos(θ) + i sin(θ) to show that there exist complex numbers cm such that
∞ f (t) = cm eimt .
m=−∞ Note that the sum now ranges over all the integers including negative ones. Do not worry about
convergence in this calculation. Hint: It may be better to start from the complex exponential form
and write the series as
∞
c0 + cm eimt + c−m e−imt .
m=1 160 4.3 CHAPTER 4. FOURIER SERIES AND PDES More on the Fourier series Note: 2 lectures, §9.2 – §9.3 in [EP]
Before reading the lecture, it may be good to ﬁrst try Project IV (Fourier series) from the
IODE website: http://www.math.uiuc.edu/iode/. After reading the lecture it may be good to
continue with Project V (Fourier series again). 4.3.1 2Lperiodic functions We have computed the Fourier series for a 2πperiodic function, but what about functions of diﬀerent
periods. Well, fear not, the computation is a simple case of change of variables. We can just rescale
the independent axis. Suppose that you have the 2Lperiodic function f (t) (L is called the half
π
period). Let s = L t, then the function
g( s) = f L
s
π is 2πperiodic. We want to also rescale all our sines and cosines. We want to write
f (t) = a0
+
2 ∞ an cos
n=1 nπ
nπ
t + bn sin
t.
L
L If we change variables to s we see that
a0
g( s) =
+
2 ∞ an cos(ns) + bn sin(ns).
n=1 We can compute an and bn as before. After we write down the integrals we change variables back to
t.
1π
1L
a0 =
g( s) ds =
f (t) dt,
π −π
L −L
1π
1L
nπ
an =
g( s) cos(ns) ds =
f (t) cos
t d t,
π −π
L −L
L
1π
1L
nπ
bn =
g( s) sin(ns) ds =
f (t) sin
t d t.
π −π
L −L
L
The two most common half periods that show up in examples are π and 1 because of the
simplicity. We should stress that we have done no new mathematics, we have only changed
variables. If you understand the Fourier series for 2πperiodic functions, you understand it for 2Lperiodic functions. All that we are doing is moving some constants around, but all the mathematics
is the same. 4.3. MORE ON THE FOURIER SERIES 161 Example 4.3.1: Let
f (t) = t for −1 < t ≤ 1, extended periodically. The plot of the periodic extension is given in Figure 4.8. Compute the Fourier
series of f (t).
2 1 0 1 2 1.00 1.00 0.75 0.75 0.50 0.50 0.25 0.25 0.00 0.00 2 1 0 1 2 Figure 4.8: Periodic extension of the function f (t).
We want to write f (t) =
is even and hence a0
2 + ∞
n=1 an cos(nπt) + bn sin(nπt). For n ≥ 1 we note that t cos(nπt) 1 an = f (t) cos(nπt) dt
−1
1 =2 t cos(nπt) dt
0 t
sin(nπt)
=2
nπ 1 1
t =0 −2 1
= 0 + 2 2 cos(nπt)
nπ 0
1
t =0 1
sin(nπt) dt
nπ 0 2 (−1)n − 1 =
= −4 2 π2
22
n
nπ Next we ﬁnd a0 if n is even,
if n is odd. 1 a0 = t dt = 1.
−1 You should be able to ﬁnd this integral by thinking about the integral as the area under the graph
without doing any computation at all. Finally we can ﬁnd bn . Here, we notice that t sin(nπt) is odd
and, therefore,
1 bn = f (t) sin(nπt) dt = 0.
−1 162 CHAPTER 4. FOURIER SERIES AND PDES Hence, the series is
1
+
2 ∞
n=1
n odd −4
cos(nπt).
n2 π2 Let us explicitly write down the ﬁrst few terms of the series up to the 3rd harmonic.
4
14
− 2 cos(πt) − 2 cos(3πt) − · · ·
2π
9π
The plot of these few terms and also a plot up to the 20th harmonic is given in Figure 4.9. You
should notice how close the graph is to the real function. You should also notice that there is no
“Gibbs phenomenon” present as there are no discontinuities.
2 1 0 1 2 2 1 0 1 2 1.00 1.00 1.00 1.00 0.75 0.75 0.75 0.75 0.50 0.50 0.50 0.50 0.25 0.25 0.25 0.25 0.00 0.00 0.00 0.00 2 1 0 1 2 2 1 0 1 2 Figure 4.9: Fourier series of f (t) up to the 3rd harmonic (left graph) and up to the 20th harmonic
(right graph). 4.3.2 Convergence We will need the one sided limits of functions. We will use the following notation
f (c−) = lim f (t),
t↑c and f (c+) = lim f (t).
t↓c If you are unfamiliar with this notation, limt↑c f (t) means we are taking a limit of f (t) as t approaches
c from below (i.e. t < c) and limt↓c f (t) means we are taking a limit of f (t) as t approaches c from
above (i.e. t > c). For example, for the square wave function 0 if −π < t ≤ 0, f (t) = (4.8)
π if 0 < t ≤ π, 4.3. MORE ON THE FOURIER SERIES 163 we have f (0−) = 0 and f (0+) = π.
Let f (t) be a function deﬁned on an interval [a, b]. Suppose that we ﬁnd ﬁnitely many points
a = t0 , t1 , t2 , . . . , tk = b in the interval, such that f (t) is continuous on the intervals (t0 , t1 ), (t1 , t2 ),
. . . , (tk−1 , tk ). Also suppose that all the one sided limits exist, that is, all of f (t0 +), f (t1 −), f (t1 +),
f (t2 −), f (t2 +), . . . , f (tk −) exist and are ﬁnite. Then we say f (t) is piecewise continuous.
If moreover, f (t) is diﬀerentiable at all but ﬁnitely many points, and f (t) is piecewise continuous,
then f (t) is said to be piecewise smooth.
Example 4.3.2: The square wave function (4.8) is piecewise smooth on [−π, π] or any other interval.
In such a case we simply say that the function is piecewise smooth.
Example 4.3.3: The function f (t) = t is piecewise smooth.
Example 4.3.4: The function f (t) = 1 is not piecewise smooth on [−1, 1] (or any other interval
t
containing zero). In fact, it is not even piecewise continuous.
√
Example 4.3.5: The function f (t) = 3 t is not piecewise smooth on [−1, 1] (or any other interval
containing zero). f (t) is continuous, but the derivative of f (t) is unbounded near zero and hence not
piecewise continuous.
Piecewise smooth functions have an easy answer on the convergence of the Fourier series.
Theorem 4.3.1. Suppose f (t) is a 2Lperiodic piecewise smooth function. Let
a0
+
2 ∞ an cos
n=1 nπ
nπ
t + bn sin
t
L
L be the Fourier series for f (t). Then the series converges for all t. If f (t) is continuous near t, then
f (t) =
Otherwise a0
+
2 ∞ an cos
n=1 f (t−) + f (t+) a0
=
+
2
2 nπ
nπ
t + bn sin
t.
L
L ∞ an cos
n=1 nπ
nπ
t + bn sin
t.
L
L If we happen to have that f (t) = f (t−)+ f (t+) at all the discontinuities, the Fourier series converges
2
to f (t) everywhere. We can always just redeﬁne f (t) by changing the value at each discontinuity
appropriately. Then we can write an equals sign between f (t) and the series without any worry. We
mentioned this fact brieﬂy at the end last section.
Note that the theorem does not say how fast the series converges. Think back the discussion of
the Gibbs phenomenon in last section. The closer you get to the discontinuity, the more terms you
need to take to get an accurate approximation to the function. 164 4.3.3 CHAPTER 4. FOURIER SERIES AND PDES Diﬀerentiation and integration of Fourier series Not only does Fourier series converge nicely, but it is easy to diﬀerentiate and integrate the series.
We can do this just by diﬀerentiating or integrating term by term.
Theorem 4.3.2. Suppose
a0
+
f (t) =
2 ∞ an cos
n=1 nπ
nπ
t + bn sin
t,
L
L is a piecewise smooth continuous function and the derivative f (t) is piecewise smooth. Then the
derivative can be obtained by diﬀerentiating term by term.
∞ f (t) =
n=1 nπ
nπ
−an nπ
bn nπ
sin
t+
cos
t.
L
L
L
L It is important that the function is continuous. It can have corners, but no jumps. Otherwise the
diﬀerentiated series will fail to converge. For an exercise, take the series obtained for the square
wave and try to diﬀerentiate the series. Similarly, we can also integrate a Fourier series.
Theorem 4.3.3. Suppose
f (t) = a0
+
2 ∞ an cos
n=1 nπ
nπ
t + bn sin
t,
L
L is a piecewise smooth function. Then the antiderivative is obtained by antidiﬀerentiating term by
term and so
∞
a0 t
an L
nπ
−bn L
nπ
F (t) =
+C +
sin
t+
cos
t.
2
nπ
L
nπ
L
n=1
where F (t) = f (t) and C is an arbitrary constant.
0
Note that the series for F (t) is no longer a Fourier series as it contains the a2 t term. The
antiderivative of a periodic function need no longer be periodic and so we should not expect a
Fourier series. 4.3.4 Rates of convergence and smoothness Let us do an example of a periodic function with one derivative everywhere.
Example 4.3.6: Take the function (t + 1) t f (t) = (1 − t) t if −1 < t ≤ 0,
if 0 < t ≤ 1, and extend to a 2periodic function. The plot is given in Figure 4.10 on the facing page.
Note that this function has one derivative everywhere, but it does not have a second derivative
derivative whenever t is an integer. 4.3. MORE ON THE FOURIER SERIES
2 165 1 0 1 2 0.50 0.50 0.25 0.25 0.00 0.00 0.25 0.25 0.50 0.50
2 1 0 1 2 Figure 4.10: Smooth 2periodic function. Exercise 4.3.1: Compute f (0+) and f (0−).
Let us compute the Fourier series coeﬃcients. The actual computation involves several integration by parts and is left to student.
1 a0 = 0 f (t) dt =
−1
1 1 (t + 1) t dt +
−1
0 f (t) cos(nπt) dt = an = (1 − t) t dt = 0,
0 −1
1 1 (t + 1) t cos(nπt) dt +
−1
0 (1 − t) t cos(nπt) dt = 0
0
1 bn = f (t) sin(nπt) dt =
(t + 1) t sin(nπt) dt +
−1 4(1 − (−1)n ) π38n3 if n is odd, =
=
0 π3 n3
if n is even.
−1 That is, the series is ∞
n=1
n odd 8
π3 n3 (1 − t) t sin(nπt) dt
0 sin(nπt). This series converges very fast. If you plot up to the third harmonic, that is the function
8
8
sin(πt) +
sin(3πt),
3
π
27π3
it is almost indistinguishable from the plot of f (t) in Figure 4.10. In fact, the coeﬃcient 278π3 is
already just 0.0096 (approximately). The reason for this behavior is the n3 term in the denominator.
The coeﬃcients bn in this case go to zero as fast as n13 goes to zero. 166 CHAPTER 4. FOURIER SERIES AND PDES It is a general fact that if you have one derivative, the Fourier coeﬃcients will go to zero
approximately like n13 . If you have only a continuous function, then the Fourier coeﬃcients will go
to zero as n12 . If you have discontinuities, then the Fourier coeﬃcients will go to zero approximately
as 1 . Therefore, we can tell a lot about the smoothness of a function by looking at its Fourier
n
coeﬃcients.
To justify this behavior take for example the function deﬁned by the Fourier series
∞ f (t) =
n=1 1
sin(nt).
n3 When we diﬀerentiate term by term we notice
∞ f (t) =
n=1 1
cos(nt).
n2 Therefore, the coeﬃcients now go down like n12 , which we said means that we have a continuous
function. The derivative of f (t) is deﬁned at most points, but there are points where f (t) is not
diﬀerentiable. It has corners, but no jumps. If we diﬀerentiate again (where we can) we ﬁnd that the
function f (t), now fails to be continuous (has jumps)
∞ f (t ) =
n=1 −1
sin(nt).
n This function is similar to the sawtooth. If we tried to dierentiate again we would obtain
∞ − cos(nt),
n =1 which does not converge!
Exercise 4.3.2: Use a computer to plot f (t), f (t) and f (t). That is, plot say the ﬁrst 5 harmonics
of the functions. At what points does f (t) have the discontinuities. 4.3.5 Exercises Exercise 4.3.3: Let 0 if −1 < t ≤ 0, f (t) = t if 0 < t ≤ 1, extended periodically. a) Compute the Fourier series for f (t). b) Write out the series explicitly up to
the 3rd harmonic. 4.3. MORE ON THE FOURIER SERIES
Exercise 4.3.4: Let 167 −t f (t) = 2
t if −1 < t ≤ 0,
if 0 < t ≤ 1, extended periodically. a) Compute the Fourier series for f (t). b) Write out the series explicitly up to
the 3rd harmonic.
Exercise 4.3.5: Let −t f (t) = 10
t 10 if −10 < t ≤ 0,
if
0 < t ≤ 10, extended periodically (period is 20). a) Compute the Fourier series for f (t). b) Write out the series
explicitly up to the 3rd harmonic.
Exercise 4.3.6: Let f (t) = ∞ 1 n13 cos(nt). Is f (t) continuous and diﬀerentiable everywhere? Find
n=
the derivative (if it exists everywhere) or justify why f (t) is not diﬀerentiable everywhere.
n 1)
Exercise 4.3.7: Let f (t) = ∞ 1 (−n sin(nt). Is f (t) diﬀerentiable everywhere? Find the derivative
n=
(if it exists everywhere) or justify why f (t) is not diﬀerentiable everywhere. Exercise 4.3.8: Let 0 if −2 < t ≤ 0, f (t) = t
if 0 < t ≤ 1, −t + 2 if 1 < t ≤ 2, extended periodically. a) Compute the Fourier series for f (t). b) Write out the series explicitly up to
the 3rd harmonic.
Exercise 4.3.9: Let
f (t) = et for −1 < t < 1 extended periodically. a) Compute the Fourier series for f (t). b) Write out the series explicitly up to
the 3rd harmonic. c) What does the series converge to at t = 1. 168 4.4 CHAPTER 4. FOURIER SERIES AND PDES Sine and cosine series Note: 2 lectures, §9.3 in [EP] 4.4.1 Odd and even periodic functions You may have noticed by now that an odd function has no cosine terms in the Fourier series and an
even function has no sine terms in the Fourier series. This observation is not a coincidence. Let us
look at even and odd periodic function in more detail.
Recall that a function f (t) is odd if f (−t) = − f (t). A function f (t) is even if f (−t) = f (t). For
example, cos(nt) is even and sin(nt) is odd. Similarly the function tk is even if k is even and odd
when k is odd.
Exercise 4.4.1: Take two functions f (t) and g(t) and deﬁne their product h(t) = f (t)g(t). a) Suppose
both are odd, is h(t) odd or even? b) Suppose one is even and one is odd, is h(t) odd or even? c)
Suppose both are even, is h(t) odd or even?
If f (t) and g(t) are both odd, then f (t) + g(t) is odd. Similarly for even functions. On the other
hand, if f (t) is odd and g(t) even, then we cannot say anything about the sum f (t) + g(t). In fact, the
Fourier series of any function is a sum of an odd (the sine terms) and an even (the cosine terms)
function.
In this section we are interested in odd and even periodic functions. We have previously deﬁned
the 2Lperiodic extension of a function deﬁned on the interval [−L, L]. Sometimes we are only
interested in the function on the range [0, L] and it would be convenient to have an odd (resp. even)
function. If the function is odd (resp. even), all the cosine (resp. sine) terms will disappear. What
we will do is take the odd (resp. even) extension of the function to [−L, L] and then we extend
periodically to a 2Lperiodic function.
Take a function f (t) deﬁned on [0, L]. On (−L, L] deﬁne the functions if 0 ≤ t ≤ L, def f (t)
Fodd (t) = − f (−t) if −L < t < 0, if 0 ≤ t ≤ L, def f (t)
Feven (t) = f (−t) if −L < t < 0. Extend Fodd (t) and Feven (t) to be 2Lperiodic. Then Fodd (t) is called the odd periodic extension of
f (t), and Feven (t) is called the even periodic extension of f (t).
Exercise 4.4.2: Check that Fodd (t) is odd and that Feven (t) is even.
Example 4.4.1: Take the function f (t) = t (1 − t) deﬁned on [0, 1]. Figure 4.11 on the facing page
shows the plots of the odd and even extensions of f (t). 4.4. SINE AND COSINE SERIES
2 1 0 169 1 2 2 1 0 1 2 0.3 0.3 0.3 0.3 0.2 0.2 0.2 0.2 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.1 0.1 0.1 0.1 0.2 0.2 0.2 0.2 0.3 0.3 0.3
2 1 0 1 2 0.3
2 1 0 1 2 Figure 4.11: Odd and even 2periodic extension of f (t) = t (1 − t), 0 ≤ t ≤ 1. 4.4.2 Sine and cosine series Let f (t) be an odd 2Lperiodic function. We write the Fourier series for f (t). We compute the
coeﬃcients an (including n = 0) and get
1
an =
L L f (t) cos
−L nπ
t d t = 0.
L That is, there are no cosine terms in the Fourier series of an odd function. The integral is zero
because f (t) cos (nπL t) is an odd function (product of an odd and an even function is odd) and the
integral of an odd function over a symmetric interval is always zero. Furthermore, the integral of an
even function over a symmetric interval [−L, L] is twice the integral of the function over the interval
π
[0, L]. The function f (t) sin nL t is the product of two odd functions and hence even.
bn = 1
L L f (t) sin
−L nπ
2
t dt =
L
L L f (t) sin
0 nπ
t d t.
L We can now write the Fourier series of f (t) as
∞ bn sin
n=1 nπ
t.
L Similarly, if f (t) is an even 2Lperiodic function. For the same exact reasons as above, we ﬁnd
that bn = 0 and
2L
nπ
an =
f (t) cos
t d t.
L0
L 170 CHAPTER 4. FOURIER SERIES AND PDES The formula still works for n = 0 in which case it becomes
L 2
a0 =
L
The Fourier series is then
a0
+
2 f (t) dt.
0 ∞ an cos
n=1 nπ
t.
L An interesting consequence is that the coeﬃcients of the Fourier series of an odd (or even)
function can be computed by just integrating over the half interval [0, L]. Therefore, we can compute
the Fourier series of the odd (or even) extension of a function by computing certain integrals over
the interval where the original function is deﬁned.
Theorem 4.4.1. Let f (t) be a piecewise smooth function deﬁned on [0, L]. Then the odd extension
of f (t) has the Fourier series
∞ Fodd (t) = bn sin
n=1 nπ
t,
L where
2
bn =
L L f (t) sin
0 nπ
t d t.
L The even extension of f (t) has the Fourier series
a0
Feven (t) =
+
2 ∞ an cos
n=1 nπ
t,
L where
an = 2
L L f (t) cos
0 nπ
t d t.
L π
π
The series ∞ 1 bn sin nL t is called the sine series of f (t) and the series a20 + ∞ 1 an cos nL t
n=
n=
is called the cosine series of f (t). It is often the case that we do not actually care what happens
outside of [0, L]. In this case, we can pick whichever series ﬁts our problem better.
It is not necessary to start with the full Fourier series to obtain the sine and cosine series. The
sine series is really the eigenfunction expansion of f (t) using the eigenfunctions of the eigenvalue
problem x + λ x = 0, x(0) = 0, x(L) = L. The cosine series is the eigenfunction expansion of f (t)
using the eigenfunctions of the eigenvalue problem x + λ x = 0, x (0) = 0, x (L) = L. We could
have, therefore, have gotten the same formulas by deﬁning the inner product
L f (t), g(t) = f (t)g(t) dt,
0 4.4. SINE AND COSINE SERIES 171 and following the procedure of § 4.2. This point of view is useful because many times we use a
speciﬁc series because our underlying question will lead to a certain eigenvalue problem. If the
eigenvalue value problem is not one of the three we covered so far, you can still do an eigenfunction expansion, generalizing the results of this chapter. We will deal with such a generalization
in chapter 5.
Example 4.4.2: Find the Fourier series of the even periodic extension of the function f (t) = t2 for
0 ≤ t ≤ π.
We want to write
∞
a0
an cos(nt),
f (t) =
+
2 n=1
where
a0 = 2
π π t2 dt =
0 2π2
,
3 and
π π
2 π2
4
2 21
an =
t cos(nt) dt =
t sin(nt) −
t sin(nt) dt
π0
πn
nπ 0
0
π
π
4
4(−1)n
4
cos(nt) dt =
.
= 2 t cos(nt) + 2
0
nπ
nπ 0
n2 Note that we have detected the “continuity” of the extension since the coeﬃcients decay as n12 . That
is, the even extension of t2 has no jump discontinuities. It will have corners, since the derivative
(which will be an odd function and a sine series) will have a series whose coeﬃcients decay only as
1
so the derivative will have jumps.
n
Explicitly, the ﬁrst few terms of the series are
π2
4
− 4 cos(t) + cos(2t) − cos(3t) + · · ·
3
9
Exercise 4.4.3: a) Compute the derivative of the even extension of f (t) above and verify it has jump
discontinuities. Use the actual deﬁnition of f (t), not its cosine series! b) Why is it that the derivative
of the even extension of f (t) is the odd extension of f (t). 4.4.3 Application We said that Fourier series ties in to the boundary value problems we studied earlier. Let us see this
connection in more detail.
Suppose we have the boundary value problem for 0 < t < L,
x (t) + λ x(t) = f (t), 172 CHAPTER 4. FOURIER SERIES AND PDES for the Dirichlet boundary conditions x(0) = 0, x(L) = 0. By using the Fredholm alternative
( Theorem 4.1.2 on page 148) we note that as long as λ is not an eigenvalue of the underlying
homogeneous problem, there will exist a unique solution. Note that the eigenfunctions of this
π
eigenvalue problem were the functions sin nL t . Therefore, to ﬁnd the solution, we ﬁrst ﬁnd the
Fourier sine series for f (t). We write x also as a sine series, but with unknown coeﬃcients. We
substitute the series for x into the equation and solve for the unknown coeﬃcients.
If we have the Neumann boundary conditions x (0) = 0, x (L) = 0, we do the same procedure
using the cosine series. These methods are best seen by examples.
Example 4.4.3: Take the boundary value problem for 0 < t < 1,
x (t) + 2 x(t) = f (t),
where f (t) = t on 0 < t < 1, and satisfying the Dirichlet boundary conditions x(0) = 0, x(1) = 0.
We write f (t) as a sine series
∞ f (t) = cn sin(nπt),
n=1 where 1 cn = 2 t sin(nπt) dt =
0 2 (−1)n+1
.
nπ We write x(t) as
∞ x(t) = bn sin(nπt).
n=1 We plug in to obtain
∞ x (t) + 2 x(t) = ∞ −bn n2 π2 sin(nπt) + 2
n=1
∞ = bn sin(nπt)
n=1 bn (2 − n2 π2 ) sin(nπt)
n=1
∞ = f (t) =
n=1 2 (−1)n+1
sin(nπt).
nπ Therefore,
bn (2 − n2 π2 ) =
or
bn = 2 (−1)n+1
nπ 2 (−1)n+1
.
nπ(2 − n2 π2 ) 4.4. SINE AND COSINE SERIES 173 We have thus obtained a Fourier series for the solution
∞ x(t) =
n=1 2 (−1)n+1
sin(nπt).
nπ (2 − n2 π2 ) Example 4.4.4: Similarly we handle the Neumann conditions. Take the boundary value problem
for 0 < t < 1,
x (t) + 2 x(t) = f (t),
where again f (t) = t on 0 < t < 1, but now satisfying the Neumann boundary conditions x (0) = 0,
x (1) = 0. We write f (t) as a cosine series
f (t) = c0
+
2 ∞ cn cos(nπt),
n=1 where 1 c0 = 2 t dt = 1,
0 and
1 cn = 2
0 −4 2 (−1)n − 1
22
t cos(nπt) dt =
= π n
0
2 n2 π if n odd,
if n even. We write x(t) as a cosine series
a0
+
x(t) =
2 ∞ an cos(nπt).
n=1 We plug in to obtain
∞ x (t) + 2 x(t) = ∞ −an n π cos(nπt) + a0 + 2 an cos(nπt) 22 n=1 n=1
∞ = a0 + an (2 − n2 π2 ) cos(nπt)
n=1 1
= f (t) = +
2 ∞
n=1
n odd −4
cos(nπt).
π2 n2 1
Therefore, a0 = 2 , an = 0 for n even (n ≥ 2) and for n odd we have an (2 − n2 π2 ) = −4
,
π2 n2 174 CHAPTER 4. FOURIER SERIES AND PDES or −4
.
− n2 π2 )
We have thus obtained a Fourier series for the solution
an = x(t) = 4.4.4 1
+
4 ∞
n=1
n odd n2 π2 (2 −4
cos(nπt).
n2 π2 (2 − n2 π2 ) Exercises Exercise 4.4.4: Take f (t) = (t − 1)2 deﬁned on 0 ≤ t ≤ 1. a) Sketch the plot of the even periodic
extension of f . b) Sketch the plot of the odd periodic extension of f .
Exercise 4.4.5: Find the Fourier series of both the odd and even periodic extension of the function
f (t) = (t − 1)2 for 0 ≤ t ≤ 1. Can you tell which extension is continuous from the Fourier series
coeﬃcients?
Exercise 4.4.6: Find the Fourier series of both the odd and even periodic extension of the function
f (t) = t for 0 ≤ t ≤ π.
Exercise 4.4.7: Find the Fourier series of the even periodic extension of the function f (t) = sin t
for 0 ≤ t ≤ π.
Exercise 4.4.8: Let
x (t) + 4 x(t) = f (t),
where f (t) = 1 on 0 < t < 1. a) Solve for the Dirichlet conditions x(0) = 0, x(1) = 0. b) Solve for
the Neumann conditions x (0) = 0, x (1) = 0.
Exercise 4.4.9: Let
x (t) + 9 x(t) = f (t),
for f (t) = sin(2πt) on 0 < t < 1. a) Solve for the Dirichlet conditions x(0) = 0, x(1) = 0. b) Solve
for the Neumann conditions x (0) = 0, x (1) = 0.
Exercise 4.4.10: Let
x (t) + 3 x(t) = f (t), x(0) = 0, x(1) = 0, where f (t) = ∞ 1 bn sin(nπt). Write the solution x(t) as a Fourier series, where the coeﬃcients are
n=
given in terms of bn .
Exercise 4.4.11: Let f (t) = t2 (2 − t) for 0 ≤ t ≤ 2. Let F (t) be the odd periodic extension. Compute
F (1), F (2), F (3), F (−1), F (9/2), F (101), F (103). Note: Do not compute using the sine series. 4.5. APPLICATIONS OF FOURIER SERIES 4.5 175 Applications of Fourier series Note: 2 lectures, §9.4 in [EP] 4.5.1 Periodically forced oscillation Let us return to the forced oscillations. We have a mass spring
system as before, where we have a mass m on a spring with spring
constant k, with damping c, and a force F (t) applied to the mass.
Suppose that the forcing function F (t) is 2Lperiodic for some
L > 0. We have already seen this problem in chapter 2 with a
simple F (t). The equation that governs this particular setup is
mx (t) + cx (t) + kx(t) = F (t). k F (t)
m damping c (4.9) We know that the general solution will consist of xc , which solves the associated homogeneous
equation mx + cx + kx = 0, and a particular solution of (4.9) we will call x p . For c > 0, the
complementary solution xc will decay as time goes on. Therefore, we are mostly interested in
particular solution x p that does not decay and is periodic with the same period as F (t). We call
this particular solution the steady periodic solution and we write it as x sp as before. The diﬀerence
in what we will do now is that we consider an arbitrary forcing function F (t) instead of a simple
cosine.
For simplicity, let us suppose that c = 0. The problem with c > 0 is very similar. The equation
mx + kx = 0
has the general solution
x(t) = A cos(ω0 t) + B sin(ω0 t),
k
where ω0 = m . Any solution to mx (t) + kx(t) = F (t) will be of the form A cos(ω0 t) + B sin(ω0 t) +
x sp . The steady periodic solution x sp has the same period as F (t).
In the spirit of the last section and the idea of undetermined coeﬃcients we will ﬁrst write c0
F (t) =
+
2 ∞ cn cos
n=1 nπ
nπ
t + dn sin
t.
L
L Then we write a proposed steady periodic solution x as
a0
x(t) =
+
2 ∞ an cos
n=1 nπ
nπ
t + bn sin
t,
L
L where an and bn are unknowns. We plug x into the diﬀerential equation and solve for an and bn in
terms of cn and dn . This process is perhaps best understood by example. 176 CHAPTER 4. FOURIER SERIES AND PDES Example 4.5.1: Suppose that k = 2, and m = 1. The units are the mks units (meterskilogramsseconds) again. There is a jetpack strapped to the mass, which ﬁres with a force of 1 newton for 1
second and then is oﬀ for 1 second, and so on. We want to ﬁnd the steady periodic solution.
The equation is, therefore,
x + 2 x = F (t),
where F (t) is the step function 0 F (t) = 1 if −1 < t < 0,
if 0 < t < 1, extended periodically. We write
F (t) = c0
+
2 ∞ cn cos(nπt) + dn sin(nπt).
n=1 We compute
1 cn = 1 F (t) cos(nπt) dt =
−1
1 c0 = for n ≥ 1, 1 F (t) dt =
−1
1 dn = cos(nπt) dt = 0
0 dt = 1,
0 F (t) sin(nπt) dt
−1
1 = sin(nπt) dt
0 − cos(nπt)
=
nπ
n = 1
t =0 2 πn if n odd,
if n even. 1 − (−1)
=
0 πn So
F (t) = 1
+
2 ∞
n=1
n odd 2
sin(nπt).
πn We want to try
a0
x(t) =
+
2 ∞ an cos(nπt) + bn sin(nπt).
n=1 Once we plug x into the diﬀerential equation x + 2 x = F (t), it is clear that an = 0 for n ≥ 1 as there
are no corresponding terms in the series for F (t). Similarly bn = 0 for n even. Hence we try
a0
x(t) =
+
2 ∞ bn sin(nπt).
n=1
n odd 4.5. APPLICATIONS OF FOURIER SERIES 177 We plug into the diﬀerential equation and obtain
∞ x + 2x = ∞ −bn n π sin(nπt) + a0 + 2 bn sin(nπt) 22 n=1
n odd n=1
n odd
∞ bn (2 − n2 π2 ) sin(nπt) = a0 +
n=1
n odd 1
= F (t) = +
2 ∞
n=1
n odd 2
sin(nπt).
πn 1
So a0 = 2 , bn = 0 for even n, and for odd n we get bn = 2
.
πn(2 − n2 π2 ) The steady periodic solution has the Fourier series
x sp (t) = 1
+
4 ∞
n=1
n odd 2
sin(nπt).
πn(2 − n2 π2 ) We know this is the steady periodic solution as it contains no terms of the complementary solution
and it is periodic with the same period as F (t) itself. See Figure 4.12 for the plot of this solution.
0.5 0.0 2.5 5.0 7.5 10.0
0.5 0.4 0.4 0.3 0.3 0.2 0.2 0.1 0.1 0.0
0.0 2.5 5.0 7.5 0.0
10.0 Figure 4.12: Plot of the steady periodic solution x sp of Example 4.5.1. 178 4.5.2 CHAPTER 4. FOURIER SERIES AND PDES Resonance Just like when the forcing function was a simple cosine, resonance could still happen. Let us assume
c = 0 and we will discuss only pure resonance. Again, take the equation
mx (t) + kx(t) = F (t).
When we expand F (t) and ﬁnd that some of its terms coincide with the complementary solution to
mx + kx = 0, we cannot use those terms in the guess. Just like before, they will disappear when
we plug into the left hand side and we will get a contradictory equation (such as 0 = 1). That is,
suppose
xc = A cos(ω0 t) + B sin(ω0 t),
where ω0 = Nπ
L for some positive integer N . In this case we have to modify our guess and try Nπ
Nπ
a0
x(t) =
+ t aN cos
t + bN sin
t+
2
L
L ∞ an cos
n=1
nN nπ
nπ
t + bn sin
t.
L
L In other words, we multiply the oﬀending term by t. From then on, we proceed as before.
Of course, the solution will not be a Fourier series (it will not even be periodic) since it contains these terms multiplied by t. Further, the terms t aN cos N π t + bN sin N π t will eventually
L
L
dominate and lead to wild oscillations. As before, this behavior is called pure resonance or just
resonance.
Note that there now may be inﬁnitely many resonance frequencies to hit. That is, as we change
the frequency of F (we change L), diﬀerent terms from the Fourier series of F may interfere with the
complementary solution and will cause resonance. However, we should note that since everything
is an approximation and in particular c is never actually zero but something very close to zero, only
the ﬁrst few resonance frequencies will matter.
Example 4.5.2: Find the steady periodic solution to the equation
2 x + 18π2 x = F (t),
where −1 F (t) = 1 if −1 < t < 0,
if 0 < t < 1, extended periodically. We note that
∞ F (t) =
n=1
n odd 4
sin(nπt).
πn Exercise 4.5.1: Compute the Fourier series of F to verify the above equation. 4.5. APPLICATIONS OF FOURIER SERIES 179 The solution must look like
x(t) = c1 cos(3πt) + c2 sin(3πt) + x p (t)
for some particular solution x p .
We note that if we just tried a Fourier series with sin(nπt) as usual, we would get duplication
when n = 3. Therefore, we pull out that term and multiply by t. We also have to add a cosine term
to get everything right. That is, we must try
∞ x p (t) = a3 t cos(3πt) + b3 t sin(3πt) + bn sin(nπt).
n=1
n odd
n3 Let us compute the second derivative.
x p (t) = −6a3 π sin(3πt) − 9π2 a3 t cos(3πt) + 6b3 π cos(3πt) − 9π2 b3 t sin(3πt)+
∞ (−n2 π2 bn ) sin(nπt). + n=1
n odd
n3 We now plug into the left hand side of the diﬀerential equation.
2 x p + 18π2 x = − 12a3 π sin(3πt) − 18π2 a3 t cos(3πt) + 12b3 π cos(3πt) − 18π2 b3 t sin(3πt)+
+ 18π2 a3 t cos(3πt) + 18π2 b3 t sin(3πt)+ ∞ + (−2n2 π2 bn + 18π2 bn ) sin(nπt).
n=1
n odd
n3 If we simplify we obtain
∞ 2 x p + 18π x = −12a3 π sin(3πt) + 12b3 π cos(3πt) +
2 (−2n2 π2 bn + 18π2 bn ) sin(nπt).
n=1
n odd
n3 This series has to equal to the series for F (t). We equate the coeﬃcients and solve for a3 and bn .
−1
4/(3π)
= 2,
−12π
9π
b3 = 0,
4
2
bn =
=3
nπ(18π2 − 2n2 π2 ) π n(9 − n2 ) a3 = for n odd and n 3. 180 CHAPTER 4. FOURIER SERIES AND PDES That is,
−1
x p (t) = 2 t cos(3πt) +
9π ∞
n=1
n odd
n3 2
sin(nπt).
− n2 ) π3 n(9 When c > 0, you will not have to worry about pure resonance. That is, there will never be
any conﬂicts and you do not need to multiply any terms by t. There is a corresponding concept of
practical resonance and it is very similar to the ideas we already explored in chapter 2. We will not
go into details here. 4.5.3 Exercises Exercise 4.5.2: Let F (t) = 1 + ∞ 1 n12 cos(nπt). Find the steady periodic solution to x + 2 x = F (t).
n=
2
Express your solution as a Fourier series.
Exercise 4.5.3: Let F (t) = ∞ 1 n13 sin(nπt). Find the steady periodic solution to x + x + x = F (t).
n=
Express your solution as a Fourier series.
Exercise 4.5.4: Let F (t) = ∞ 1 n12 cos(nπt). Find the steady periodic solution to x + 4 x = F (t).
n=
Express your solution as a Fourier series.
Exercise 4.5.5: Let F (t) = t for −1 < t < 1 and extended periodically. Find the steady periodic
solution to x + x = F (t). Express your solution as a Fourier series.
Exercise 4.5.6: Let F (t) = t for −1 < t < 1 and extended periodically. Find the steady periodic
solution to x + π2 x = F (t). Express your solution as a Fourier series. 4.6. PDES, SEPARATION OF VARIABLES, AND THE HEAT EQUATION 4.6 181 PDEs, separation of variables, and the heat equation Note: 2 lectures, §9.5 in [EP]
Let us recall that a partial diﬀerential equation or PDE is an equation containing the partial
derivatives with respect to several independent variables. Solving PDEs will be our main application
of Fourier series.
A PDE is said to be linear if the dependent variable and its derivatives appear at most to the
ﬁrst power and in no functions. We will only talk about linear PDEs. Together with a PDE, we
usually have speciﬁed some boundary conditions, where the value of the solution or its derivatives
is speciﬁed along the boundary of a region, and/or some initial conditions where the value of the
solution or its derivatives is speciﬁed for some initial time. Sometimes such conditions are mixed
together and we will refer to them simply as side conditions.
We will study three speciﬁc partial diﬀerential equations, each one representing a more general
class of equations. First, we will study the heat equation, which is an example of a parabolic PDE.
Next, we will study the wave equation, which is an example of a hyperbolic PDE. Finally, we will
study the Laplace equation, which is an example of an elliptic PDE. Each of our examples will
illustrate behavior that is typical for the whole class. 4.6.1 Heat on an insulated wire Let us ﬁrst study the heat equation. Suppose that we have a wire (or a thin metal rod) of length L
that is insulated except at the endpoints. Let x denote the position along the wire and let t denote
time. See Figure 4.13.
temperature u 0
insulation L x Figure 4.13: Insulated wire.
Let u( x, t) denote the temperature at point x at time t. The equation governing this setup is the
socalled onedimensional heat equation:
∂u
∂2 u
= k 2,
∂t
∂x
where k > 0 is a constant (the thermal conductivity of the material). That is, the change in heat at a
speciﬁc point is proportional to the second derivative of the heat along the wire. This makes sense; 182 CHAPTER 4. FOURIER SERIES AND PDES if at a ﬁxed t the graph of the heat distribution has a maximum (the graph is concave down), then
heat ﬂows away from the maximum. And viceversa.
We will generally use a more convenient notation for partial derivatives. We will write ut instead
2
of ∂u , and we will write u xx instead of ∂ xu . With this notation the heat equation becomes
∂t
∂2
ut = ku xx .
For the heat equation, we must also have some boundary conditions. We assume that the ends
of the wire are either exposed and touching some body of constant heat, or the ends are insulated.
For example, if the ends of the wire are kept at temperature 0, then we must have the conditions
u(0, t) = 0 and u( L , t ) = 0. If, on the other hand, the ends are also insulated we get the conditions
u x (0, t) = 0 and u x ( L , t ) = 0. In other words, heat is not ﬂowing in nor out of the wire at the ends. We always have two conditions
along the x axis as there are two derivatives in the x direction. These side conditions are called
homogeneous (that is, u or a derivative of u is set to zero).
Furthermore, suppose that we know the initial temperature distribution at time t = 0. That is,
u( x, 0) = f ( x),
for some known function f ( x). This initial condition is not a homogeneous side condition. 4.6.2 Separation of variables The heat equation is linear as u and its derivatives do not appear to any powers or in any functions.
Thus the principle of superposition still applies for the heat equation (without side conditions). If u1
and u2 are solutions and c1 , c2 are constants, then u = c1 u1 + c2 u2 is also a solution.
Exercise 4.6.1: Verify the principle of superposition for the heat equation.
Superposition also preserves some of the side conditions. In particular, if u1 and u2 are solutions
that satisfy u(0, t) = 0 and u(L, t) = 0, and c1 , c2 are constants, then u = c1 u1 + c2 u2 is still a solution
that satisﬁes u(0, t) = 0 and u(L, t) = 0. Similarly for the side conditions u x (0, t) = 0 and u x (L, t) = 0.
In general, superposition preserves all homogeneous side conditions.
The method of separation of variables is to try to ﬁnd solutions that are sums or products of
functions of one variable. For example, for the heat equation, we try to ﬁnd solutions of the form
u( x, t) = X ( x)T (t). 4.6. PDES, SEPARATION OF VARIABLES, AND THE HEAT EQUATION 183 That the desired solution we are looking for is of this form is too much to hope for. What is perfectly
reasonable to ask, however, is to ﬁnd enough “buildingblock” solutions of the form u( x, t) =
X ( x)T (t) using this procedure so that the desired solution to the PDE is somehow constructed from
these building blocks by the use of superposition.
Let us try to solve the heat equation
ut = ku xx with u(0, t) = 0, u(L, t) = 0, and u( x, 0) = f ( x). Let us guess u( x, t) = X ( x)T (t). We plug into the heat equation to obtain
X ( x)T (t) = kX ( x)T (t).
We rewrite as X ( x)
T (t)
=
.
kT (t)
X ( x)
This equation is supposed to hold for all x and all t. But the left hand side does not depend on x and
the right hand side does not depend on t. Therefore, each side must be a constant. Let us call this
constant −λ (the minus sign is for convenience later). We obtain the two equations
X ( x)
T (t)
= −λ =
.
kT (t)
X ( x)
Or in other words
X ( x ) + λ X ( x ) = 0,
T (t) + λkT (t) = 0.
The boundary condition u(0, t) = 0 implies X (0)T (t) = 0. We are looking for a nontrivial solution
and so we can assume that T (t) is not identically zero. Hence X (0) = 0. Similarly, u(L, t) = 0
implies X (L) = 0. We are looking for nontrivial solutions X of the eigenvalue problem X + λX = 0,
22
X (0) = 0, X (L) = 0. We have previously found that the only eigenvalues are λn = nLπ , for integers
2
nπ
n ≥ 1, where eigenfunctions are sin L x . Hence, let us pick the solutions
Xn ( x) = sin nπ
x.
L The corresponding T n must satisfy the equation
T n (t) + n2 π2
kT n (t) = 0.
L2 By the method of integrating factor, the solution of this problem is
T n (t) = e −n2 π2
kt
L2 . 184 CHAPTER 4. FOURIER SERIES AND PDES It will be useful to note that T n (0) = 1. Our buildingblock solutions are
un ( x, t) = Xn ( x)T n (t) = sin
We note that un ( x, 0) = sin nπ
L −n2 π2
nπ
x e L2 kt .
L x . Let us write f ( x) as the sine series
∞ f ( x) = bn sin
n=1 nπ
x.
L That is, we ﬁnd the Fourier series of the odd periodic extension of f ( x). We used the sine series as
it corresponds to the eigenvalue problem for X ( x) above. Finally, we use superposition to write the
solution as
∞
∞
−n2 π2
nπ
x e L2 kt .
u( x , t ) =
bn un ( x, t) =
bn sin
L
n=1
n=1
Why does this solution work? First note that it is a solution to the heat equation by superposition.
It satisﬁes u(0, t) = 0 and u(L, t) = 0, because x = 0 or x = L makes all the sines vanish. Finally,
plugging in t = 0, we notice that T n (0) = 1 and so
∞ u( x, 0) = ∞ bn un ( x, 0) =
n=1 bn sin
n=1 nπ
x = f ( x).
L Example 4.6.1: Suppose that we have an insulated wire of length 1, such that the ends of the wire
are embedded in ice (temperature 0). Let k = 0.003. Then suppose that initial heat distribution is
u( x, 0) = 50 x (1 − x). See Figure 4.14.
0.00 0.25 0.50 0.75 1.00 12.5 12.5 10.0 10.0 7.5 7.5 5.0 5.0 2.5 2.5 0.0
0.00 0.0
0.25 0.50 0.75 1.00 Figure 4.14: Initial distribution of temperature in the wire. 4.6. PDES, SEPARATION OF VARIABLES, AND THE HEAT EQUATION 185 We want to ﬁnd the temperature function u( x, t). Let us suppose we also want to ﬁnd when (at
what t) does the maximum temperature in the wire drop to one half of the initial maximum of 12.5.
We are solving the following PDE problem:
ut = 0.003 u xx ,
u(0, t) = u(1, t) = 0,
u( x, 0) = 50 x (1 − x) for 0 < x < 1. We write f ( x) = 50 x (1 − x) for 0 < x < 1 as a sine series. That is, f ( x) = 1 200 200 (−1)n 0 bn = 2
50 x (1 − x) sin(nπ x) dx = 3 3 −
= 400 3 n3 πn
π
0 π3 n3 0.00 0 ∞
n=1 bn sin(nπ x), where if n even,
if n odd. t 20
40 0.25
x 60
0.50 80 u(x,t) 100 0.75
1.00
12.5 12.5 10.0 10.0 7.5 7.5 5.0 5.0 2.5 2.5 0.0 0.0 11.700
10.400
9.100
7.800
6.500
5.200
3.900
2.600
1.300
0.000 0.25 0
20 0.50
40 x
0.75 60
t 80
100 1.00 Figure 4.15: Plot of the temperature of the wire at position x at time t.
The solution u( x, t), plotted in Figure 4.15 for 0 ≤ t ≤ 100, is given by the series:
∞ u( x , t ) =
n=1
n odd 400
22
sin(nπ x) e−n π 0.003 t .
3 n3
π 186 CHAPTER 4. FOURIER SERIES AND PDES Finally, let us answer the question about the maximum temperature. It is relatively easy to see
that the maximum temperature will always be at x = 0.5, in the middle of the wire. The plot of
u( x, t) conﬁrms this intuition.
If we plug in x = 0.5 we get
∞ u(0.5, t) =
n=1
n odd 400
22
sin(nπ 0.5) e−n π 0.003 t .
3 n3
π For n = 3 and higher (remember we are taking only odd n), the terms of the series are insigniﬁcant
compared to the ﬁrst term. The ﬁrst term in the series is already a very good approximation of the
function and hence
400 2
u(0.5, t) ≈ 3 e−π 0.003 t .
π
The approximation gets better and better as t gets larger as the other terms decay much faster. Let
us plot the function u(0.5, t), the temperature at the midpoint of the wire at time t, in Figure 4.16.
The ﬁgure also plots the approximation by the ﬁrst term.
0 25 50 75 100 12.5 12.5 10.0 10.0 7.5 7.5 5.0 5.0 2.5 2.5 0 25 50 75 100 Figure 4.16: Temperature at the midpoint of the wire (the bottom curve), and the approximation of
this temperature by using only the ﬁrst term in the series (top curve).
After t = 5 or so it would be hard to tell the diﬀerence between the ﬁrst term of the series for
u( x, t) and the real solution u( x, t). This behavior is a general feature of solving the heat equation. If
you are interested in behavior for large enough t, only the ﬁrst one or two terms may be necessary.
Let us get back to the question of when is the maximum temperature one half of the initial
maximum temperature. That is, when is the temperature at the midpoint 12.5/2 = 6.25. We notice on
the graph that if we use the approximation by the ﬁrst term we will be close enough. We solve
6.25 = 400 −π2 0.003 t
e
.
π3 4.6. PDES, SEPARATION OF VARIABLES, AND THE HEAT EQUATION 187 That is, 25
ln 6.400π
t=
≈ 24.5.
−π2 0.003
So the maximum temperature drops to half at about t = 24.5.
3 We mention an interesting behavior of the solution to the heat equation. The heat equation
“smoothes” out the function f ( x) as t grows. For a ﬁxed t, the solution is a Fourier series with
−n2 π2 coeﬃcients bn e L2 kt . If t > 0, then these coeﬃcients go to zero faster than any n1p for any power
p. In other words, the Fourier series has inﬁnitely many derivatives everywhere. Thus even if the
function f ( x) has jumps and corners, the solution u( x, t) as a function of x for a ﬁxed t > 0 is as
smooth as we want it. 4.6.3 Insulated ends Now suppose the ends of the wire are insulated. In this case, we are solving the equation
ut = ku xx with u x (0, t) = 0, u x ( L , t ) = 0, and u( x, 0) = f ( x). Yet again we try a solution of the form u( x, t) = X ( x)T (t). By the same procedure as before we plug
into the heat equation and arrive at the following two equations
X ( x ) + λ X ( x ) = 0,
T (t) + λkT (t) = 0.
At this point the story changes slightly. The boundary condition u x (0, t) = 0 implies X (0)T (t) = 0.
Hence X (0) = 0. Similarly, u x (L, t) = 0 implies X (L) = 0. We are looking for nontrivial solutions
X of the eigenvalue problem X + λX = 0, X (0) = 0, X (L) = 0. We have previously found that the
22
π
only eigenvalues are λn = nLπ , for integers n ≥ 0, where eigenfunctions are cos nL x (we include
2
the constant eigenfunction). Hence, let us pick solutions
Xn ( x) = cos nπ
x
L X0 ( x ) = 1. and The corresponding T n must satisfy the equation
T n (t) + n2 π2
kT n (t) = 0.
L2 For n ≥ 1, as before,
T n (t) = e −n2 π2
kt
L2 . For n = 0, we have T 0 (t) = 0 and hence T 0 (t) = 1. Our buildingblock solutions will be
un ( x, t) = Xn ( x)T n (t) = cos −n2 π2
nπ
x e L2 kt ,
L 188 CHAPTER 4. FOURIER SERIES AND PDES and
u0 ( x, t) = 1.
We note that un ( x, 0) = cos nπ
L x . Let us write f using the cosine series
f ( x) = a0
+
2 ∞ an cos
n=1 nπ
x.
L That is, we ﬁnd the Fourier series of the even periodic extension of f ( x).
We use superposition to write the solution as
a0
u( x , t ) =
+
2 ∞ a0
an un ( x , t ) =
+
2
n=1 ∞ an cos
n=1 −n2 π2
nπ
x e L2 kt .
L Example 4.6.2: Let us try the same equation as before, but for insulated ends. We are solving the
following PDE problem
ut = 0.003 u xx ,
u x (0, t) = u x (1, t) = 0,
u( x, 0) = 50 x (1 − x) for 0 < x < 1. For this problem, we must ﬁnd the cosine series of u( x, 0). For 0 < x < 1 we have
25
+
50 x (1 − x) =
3 ∞
n=2
n even −200
cos(nπ x).
π2 n2 The calculation is left to the reader. Hence, the solution to the PDE problem, plotted in Figure 4.17
on the next page, is given by the series
25
u( x, t) =
+
3 ∞
n=2
n even −200
22
cos(nπ x) e−n π 0.003 t .
2 n2
π Note in the graph that the temperature evens out across the wire. Eventually, all the terms except
the constant die out, and you will be left with a uniform temperature of 25 ≈ 8.33 along the entire
3
length of the wire. 4.6.4 Exercises Exercise 4.6.2: Suppose you have a wire of length 2, with k = 0.001 and an initial temperature
distribution of u( x, 0) = 50 x. Suppose that both the ends are embedded in ice (temperature 0). Find
the solution as a series. 4.6. PDES, SEPARATION OF VARIABLES, AND THE HEAT EQUATION 189 0.00 0
x 5 0.25 t
10 0.50 15
20 0.75 u(x,t) 25 1.00 30 12.5 12.5
10.0 10.0
7.5 7.5
5.0 11.700
10.400
9.100
7.800
6.500
5.200
3.900
2.600
1.300
0.000 5.0
2.5 2.5
0.0
0 0.0
0.00 5
0.25 10
15 0.50
20 t 0.75 25 x 30 1.00 Figure 4.17: Plot of the temperature of the insulated wire at position x at time t. Exercise 4.6.3: Find a series solution of
ut = u xx ,
u(0, t) = u(1, t) = 0,
u( x, 0) = 100
for 0 < x < 1.
Exercise 4.6.4: Find a series solution of
ut = u xx ,
u x (0, t) = u x (π, t) = 0,
u( x, 0) = 3 cos( x) + cos(3 x) for 0 < x < π. 190 CHAPTER 4. FOURIER SERIES AND PDES Exercise 4.6.5: Find a series solution of
1
u xx ,
3
u x (0, t) = u x (π, t) = 0,
10 x
u( x, 0) =
for 0 < x < π.
π ut = Exercise 4.6.6: Find a series solution of
ut = u xx ,
u(0, t) = 0, u(1, t) = 100,
u( x, 0) = sin(π x)
for 0 < x < 1.
Hint: Use the fact that u( x, t) = 100 x is a solution satisfying ut = u xx , u(0, t) = 0, u(1, t) = 100.
Then use superposition.
Exercise 4.6.7: Find the steady state temperature solution as a function of x alone, by letting t → ∞
in the solution from exercises 4.6.5 and 4.6.6. Verify that it satisﬁes the equation u xx = 0.
Exercise 4.6.8: Use separation variables to ﬁnd a nontrivial solution to u xx + uyy = 0, where
u( x, 0) = 0 and u(0, y) = 0. Hint: Try u( x, y) = X ( x)Y (y).
Exercise 4.6.9 (challenging): Suppose that one end of the wire is insulated (say at x = 0) and the
other end is kept at zero temperature. That is, ﬁnd a series solution of
ut = ku xx ,
u x (0, t) = u(L, t) = 0,
u( x, 0) = f ( x)
for 0 < x < L.
Express any coeﬃcients in the series by integrals of f ( x).
Exercise 4.6.10 (challenging): Suppose that the wire is circular and insulated, so there are no ends.
You can think of this as simply connecting the two ends and making sure the solution matches up at
the ends. That is, ﬁnd a series solution of
ut = ku xx ,
u(0, t) = u(L, t),
u( x, 0) = f ( x) u x (0, t) = u x (L, t),
for 0 < x < L. Express any coeﬃcients in the series by integrals of f ( x). 4.7. ONE DIMENSIONAL WAVE EQUATION 4.7 191 One dimensional wave equation Note: 1 lecture, §9.6 in [EP]
Suppose we have a string such as on a guitar of length L. Suppose we only consider vibrations
in one direction. That is let x denote the position along the string, let t denote time and let y denote
the displacement of the string from the rest position. See Figure 4.18.
y
y
0 L x Figure 4.18: Vibrating string. The equation that governs this setup is the socalled onedimensional wave equation:
ytt = a2 y xx ,
for some a > 0. We will assume that the ends of the string are ﬁxed and hence we get
y(0, t) = 0 and y(L, t) = 0. Note that we always have two conditions along the x axis as there are two derivatives in the x
direction.
There are also two derivatives along the t direction and hence we will need two further conditions
here. We will need to know the initial position and the initial velocity of the string.
y( x, 0) = f ( x) and yt ( x, 0) = g( x), for some known functions f ( x) and g( x).
As the equation is again linear, superposition works just as it did for the heat equation. And
again we will use separation of variables to ﬁnd enough buildingblock solutions to get the overall
solution. There is one change however. It will be easier to solve two separate problems and add
their solutions.
The two problems we will solve are
wtt = a2 w xx ,
w(0, t) = w(L, t) = 0,
w( x, 0) = 0
wt ( x, 0) = g( x) for 0 < x < L,
for 0 < x < L. (4.10) 192
and CHAPTER 4. FOURIER SERIES AND PDES
ztt = a2 z xx ,
z(0, t) = z(L, t) = 0,
z( x, 0) = f ( x)
zt ( x, 0) = 0 for 0 < x < L,
for 0 < x < L. (4.11) The principle of superposition will then imply that y = w + z solves the wave equation and
furthermore y( x, 0) = w( x, 0) + z( x, 0) = f ( x) and yt ( x, 0) = wt ( x, 0) + zt ( x, 0) = g( x). Hence, y is a
solution to
ytt = a2 y xx ,
y(0, t) = y(L, t) = 0,
(4.12)
y( x, 0) = f ( x)
for 0 < x < L,
yt ( x, 0) = g( x)
for 0 < x < L.
The reason for all this complexity is that superposition only works for homogeneous conditions
such as y(0, t) = y(L, t) = 0, y( x, 0) = 0, or yt ( x, 0) = 0. Therefore, we will be able to use the
idea of separation of variables to ﬁnd many buildingblock solutions solving all the homogeneous
conditions. We can then use them to construct a solution solving the remaining nonhomogeneous
condition.
Let us start with (4.10). We try a solution of the form w( x, t) = X ( x)T (t) again. We plug into the
wave equation to obtain
X ( x)T (t) = a2 X ( x)T (t).
Rewriting we get
T (t)
X ( x)
=
.
2 T (t)
a
X ( x)
Again, left hand side depends only on t and the right hand side depends only on x. Therefore, both
equal a constant, which we will denote by −λ.
X ( x)
T (t)
= −λ =
.
2 T (t)
a
X ( x)
We solve to get two ordinary diﬀerential equations
X ( x) + λX ( x) = 0,
T (t) + λa2 T (t) = 0.
The conditions 0 = w(0, t) = X (0)T (t) implies X (0) = 0 and w(L, t) = 0 implies that X (L) = 0.
22
Therefore, the only nontrivial solutions for the ﬁrst equation are when λ = λn = nLπ and they are
2
Xn ( x) = sin nπ
x.
L The general solution for T for this particular λn is
T n (t) = A cos nπa
nπa
t + B sin
t.
L
L 4.7. ONE DIMENSIONAL WAVE EQUATION 193 We also have the condition that w( x, 0) = 0 or X ( x)T (0) = 0. This implies that T (0) = 0, which in
turn forces A = 0. It will be convenient to pick B = nLa (you will see why in a moment) and hence
π
T n (t ) = nπa
L
sin
t.
nπa
L Our buildingblock solution will be
wn ( x, t) = nπ
nπa
L
sin
x sin
t.
nπa
L
L We diﬀerentiate in t, that is
nπ
nπa
x cos
t.
L
L (wn )t ( x, t) = sin
Hence, (wn )t ( x, 0) = sin nπ
x.
L We expand g( x) in terms of these sines as
∞ g( x) = bn sin
n=1 nπ
x.
L Using superposition we can just write down the solution to (4.10) as a series
∞ w( x , t ) = ∞ bn wn ( x, t) =
n=1 bn
n=1 nπ
nπa
L
sin
x sin
t.
nπa
L
L Exercise 4.7.1: Check that w( x, 0) = 0 and wt ( x, 0) = g( x).
Similarly we proceed to solve (4.11). We again try z( x, y) = X ( x)T (t). The procedure works
exactly the same at ﬁrst. We obtain
X ( x) + λX ( x) = 0,
T (t) + λa2 T (t) = 0.
and the conditions X (0) = 0, X (L) = 0. So again λ = λn =
Xn ( x) = sin n2 π2
L2 and nπ
x.
L This time the condition on T is T (0) = 0. Thus we get that B = 0 and we take
T n (t) = cos nπa
t.
L 194 CHAPTER 4. FOURIER SERIES AND PDES Our buildingblock solution will be
zn ( x, t) = sin nπ
nπa
x cos
t.
L
L We expand f ( x) in terms of these sines as
∞ f ( x) = cn sin
n=1 nπ
x.
L And we write down the solution to (4.11) as a series
∞ z( x, t) = ∞ cn zn ( x, t) =
n=1 cn sin
n=1 nπ
nπa
x cos
t.
L
L Exercise 4.7.2: Fill in the details in the derivation of the solution of (4.11). Check that the solution
satisﬁes all the side conditions.
Putting these two solutions together we will state the result as a theorem.
Theorem 4.7.1. Take the equation
ytt = a2 y xx ,
y(0, t) = y(L, t) = 0,
y( x, 0) = f ( x)
yt ( x, 0) = g( x)
where ∞ cn sin nπ
x.
L n=1
∞ g( x) =
n=1 (4.13) nπ
x.
L bn sin f ( x) =
and for 0 < x < L,
for 0 < x < L, Then the solution y( x, t) can be written as a sum of the solutions of (4.10) and (4.11). In other
words,
∞ y( x , t ) = bn
n=1
∞ = L
nπ
nπa
nπ
nπa
sin
x sin
t + cn sin
x cos
t
nπa
L
L
L
L sin
n=1 nπ
x
L bn L
nπa
nπa
sin
t + cn cos
t.
nπa
L
L 4.7. ONE DIMENSIONAL WAVE EQUATION 195 y
0.1
0 2 x Figure 4.19: Plucked string. Example 4.7.1: Let us try a simple example of a plucked string. Suppose that a string of length 2
is plucked in the middle such that it has the initial shape given in Figure 4.19. That is 0.1 x if 0 ≤ x ≤ 1, f ( x) = 0.1 (2 − x) if 1 < x ≤ 2. The string starts at rest (g( x) = 0). Suppose that a = 1 in the wave equation for simplicity.
We leave it to the reader to compute the sine series of f ( x). The series will be
∞ f ( x) =
n=1 Note that sin nπ
2 nπ
nπ
0.8
x.
sin
sin
n2 π2
2
2 is the sequence 1, 0, −1, 0, 1, 0, −1, . . . for n = 1, 2, 3, 4, . . .. Therefore,
f ( x) = 0.8
0.8
0 .8
π
3π
5π
x+
x − ···
sin x − 2 sin
sin
2
2
π
2
9π
2
25π
2 The solution y( x, t) is given by
∞ y( x, t) =
n=1
∞ =
m =1 = 0 .8
nπ
nπ
nπ
x cos
t
sin
sin
2 π2
n
2
2
2
0.8(−1)m+1
(2m − 1)π
(2m − 1)π
sin
x cos
t
22
2
2
(2m − 1) π π
3π
5π
0.8
π
0.8
3π
0 .8
5π
sin x cos t − 2 sin
x cos
t+
sin
x cos
t − ···
2
2
π
2
2
9π
2
2
25π
2
2 A plot for 0 < t < 3 is given in Figure 4.20 on the following page. Notice that unlike the heat
equation, the solution does not become “smoother,” the “sharp edges” remain. We will see the
reason for this behavior in the next section where we derive the solution to the wave equation in a
diﬀerent way.
Make sure you understand what the plot such as the one in the ﬁgure is telling you. For each
ﬁxed t, you can think of the function u( x, t) as just a function of x. This function gives you the shape
of the string at time t. 196 CHAPTER 4. FOURIER SERIES AND PDES
0.0 0
t 1
0.5 2 x
1.0 y(x,t) 3
0.10 1.5
2.0 0.00 0.00 0.05 0.05 0.10 y
0.05 0.05 y 0.10 0.110
0.088
0.066
0.044
0.022
0.000
0.022
0.044
0.066
0.088
0.110 0.0
0.5 0.10
0 1.0
x 1
1.5
t 2
3 2.0 Figure 4.20: Shape of the plucked string for 0 < t < 3. 4.7.1 Exercises Exercise 4.7.3: Solve
ytt = 9y xx ,
y(0, t) = y(1, t) = 0,
y( x, 0) = sin(3π x) + 1 sin(6π x)
4
yt ( x, 0) = 0 for 0 < x < 1,
for 0 < x < 1. ytt = 4y xx ,
y(0, t) = y(1, t) = 0,
y( x, 0) = sin(3π x) + 1 sin(6π x)
4
yt ( x, 0) = sin(9π x) for 0 < x < 1,
for 0 < x < 1. Exercise 4.7.4: Solve Exercise 4.7.5: Derive the solution for a general plucked string of length L, where we raise the
string some distance b at the midpoint and let go, and for any constant a (in the equation ytt = a2 y xx ). 4.7. ONE DIMENSIONAL WAVE EQUATION 197 Exercise 4.7.6: Suppose that a stringed musical instrument falls on the ﬂoor. Suppose that the
length of the string is 1 and a = 1. When the musical instrument hits the ground the string was in
rest position and hence y( x, 0) = 0. However, the string was moving at some velocity at impact
(t = 0), say yt ( x, 0) = −1. Find the solution y( x, t) for the shape of the string at time t.
Exercise 4.7.7 (challenging): Suppose that you have a vibrating string and that there is air resistance proportional to the velocity. That is, you have
ytt = a2 y xx − kyt ,
y(0, t) = y(1, t) = 0,
y( x, 0) = f ( x)
yt ( x, 0) = 0 for 0 < x < 1,
for 0 < x < 1. Suppose that 0 < k < 2πa. Derive a series solution to the problem. Any coeﬃcients in the series
should be expressed as integrals of f ( x). 198 CHAPTER 4. FOURIER SERIES AND PDES 4.8 D’Alembert solution of the wave equation Note: 1 lecture, diﬀerent from §9.6 in [EP]
We have solved the wave equation by using Fourier series. But it is often more convenient to
use the socalled d’Alembert solution to the wave equation‡ . This solution can be derived using
Fourier series as well, but it is really an awkward use of those concepts. It is much easier to derive
this solution by making a correct change of variables to get an equation that can be solved by simple
integration.
Suppose we have the wave equation
ytt = a2 y xx . (4.14) And we wish to solve the equation (4.14) given the conditions
y(0, t) = y(L, t) = 0 for all t,
y( x, 0) = f ( x)
0 < x < L,
yt ( x, 0) = g( x)
0 < x < L. 4.8.1 (4.15) Change of variables We will transform the equation into a simpler form where it can be solved by simple integration.
We change variables to ξ = x − at, η = x + at and we use the chain rule:
∂
∂ξ ∂
∂η ∂
∂
∂
=
+
=
+,
∂ x ∂ x ∂ξ ∂ x ∂η ∂ξ ∂η
∂
∂ξ ∂
∂η ∂
∂
∂
=
+
= −a + a .
∂t ∂t ∂ξ ∂t ∂η
∂ξ
∂η
We compute
∂2 y
∂
∂ ∂y ∂y
∂2 y
∂2 y
∂2 y
= 2 +2
,
y xx = 2 =
+
+
+
∂x
∂ξ ∂η ∂ξ ∂η
∂ξ
∂ξ∂η ∂η2
∂2 y
∂
∂
∂y
∂y
∂2 y
∂2 y
∂2 y
ytt = 2 = −a + a
−a + a
= a2 2 − 2a2
+ a2 2 .
∂t
∂ξ
∂η
∂ξ
∂η
∂ξ
∂ξ∂η
∂η
In the above computations, we have used the fact from calculus that
the wave equation,
∂2 y
0 = a2 y xx − ytt = 4a2
= 4a2 yξη .
∂ξ∂η
‡ ∂2 y
∂ξ∂η Named after the French mathematician Jean le Rond d’Alembert (1717 – 1783). = ∂2 y
.
∂η∂ξ Then we plug into 4.8. D’ALEMBERT SOLUTION OF THE WAVE EQUATION 199 Therefore, the wave equation (4.14) transforms into yξη = 0. It is easy to ﬁnd the general solution
to this equation by integrating twice. Let us integrate with respect to η ﬁrst§ and notice that the
constant of integration depends on ξ. We get yξ = C (ξ). Next, we integrate with respect to ξ and
notice that the constant of integration must depend on η. Thus, y = C (ξ) dξ + B(η). The solution
must, therefore, be of the following form for some functions A(ξ) and B(η):
y = A(ξ) + B(η) = A( x − at) + B( x + at). 4.8.2 The formula We know what any solution must look like, but we need to solve for the given side conditions. We
will just give the formula and see that it works. First let F ( x) denote the odd extension of f ( x), and
let G( x) denote the odd extension of g( x). Now deﬁne
A( x) = 1
1
F ( x) −
2
2a x G( s) ds, B( x) = 0 1
1
F ( x) +
2
2a x G( s) ds.
0 We claim this A( x) and B( x) give the solution. Explicitly, the solution is y( x, t) = A( x − at) + B( x + at)
or in other words:
x−at
1
1
1
1
F ( x − at) −
G( s) ds + F ( x + at) +
2
2a 0
2
2a
x+at
1
F ( x − at) + F ( x + at)
+
=
G( s) ds.
2
2a x−at x+at y( x, t) = G( s) ds
0 (4.16) Let us check that the d’Alembert formula really works.
1
1
y( x, 0) = F ( x) −
2
2a x
0 1
1
G( s) ds + F ( x) +
2
2a x G( s) ds = F ( x).
0 So far so good. Assume for simplicity F is diﬀerentiable. By the fundamental theorem of calculus
we have
−a
1
a
1
yt ( x, t) =
F ( x − at) + G( x − at) + F ( x + at) + G( x + at).
2
2
2
2
So
−a
1
a
1
yt ( x, 0) =
F ( x) + G( x) + F ( x) + G( x) = G( x).
2
2
2
2
Yay! We’re smoking now. OK, now the boundary conditions. Note that F ( x) and G( x) are odd.
x
Also 0 G( s) ds is an even function of x because G( x) is odd (to see this fact, do the substitution
§ We can just as well integrate with ξ ﬁrst, if we wish. 200 CHAPTER 4. FOURIER SERIES AND PDES s = −v). So
1
y(0, t) = F (−at) −
2
−1
=
F (at) −
2 1
2a
1
2a −at 1
1
G( s) ds + F (at) +
2
2a
0
at
1
1
G( s) ds + F (at) +
2
2a
0 at G( s) ds
0
at G( s) ds = 0.
0 Note that F ( x) and G( x) are 2L periodic. We compute
L−at
L+at
1
1
1
1
F (L − at) −
G( s) ds + F (L + at) +
G( s) ds
2
2a 0
2
2a 0
L
−at
1
1
1
= F (−L − at) −
G( s) ds −
G( s) ds +
2
2a 0
2a 0
L
at
1
1
1
+ F (L + at) +
G( s) ds +
G( s) ds
2
2a 0
2a 0
at
at
−1
1
1
1
=
F (L + at) −
G( s) ds + F (L + at) +
G( s) ds = 0.
2
2a 0
2
2a 0 y(L, t) = And voilà, it works.
Example 4.8.1: What the d’Alembert solution says is that the solution is a superposition of two
functions (waves) moving in the opposite direction at “speed” a. To get an idea of how it works, let
us do an example. Suppose that we have the simpler setup
ytt = y xx ,
y(0, t) = y(1, t) = 0,
y( x, 0) = f ( x),
yt ( x, 0) = 0.
Here f ( x) is an impulse of height 1 centered at x = 0.5: 0 if
0≤ 20 ( x − 0.45) if 0≤ f ( x) = 20 (0.55 − x) if 0.45 ≤ 0 if 0.55 ≤ x < 0.45,
x < 0.45,
x < 0.55,
x ≤ 1. The graph of this pulse is the top left plot in Figure 4.21 on the next page.
Let F ( x) be the odd periodic extension of f ( x). Then from (4.16) we know that the solution is
given as
F ( x − t) + F ( x + t)
y( x, t) =
.
2 4.8. D’ALEMBERT SOLUTION OF THE WAVE EQUATION 201 It is not hard to compute speciﬁc values of y( x, t). For example, to compute y(0.1, 0.6) we notice
x − t = −0.5 and x + t = 0.7. Now F (−0.5) = − f (0.5) = −20 (0.55 − 0.5) = −1 and F (0.7) =
f (0.7) = 0. Hence y(0.1, 0.6) = −12+0 = −0.5. As you can see the d’Alembert solution is much easier
to actually compute and to plot than the Fourier series solution. See Figure 4.21 for plots of the
solution y for several diﬀerent t.
0.00 0.25 0.50 0.75 1.00 0.00 0.25 0.50 0.75 1.00 1.0 1.0 1.0 1.0 0.5 0.5 0.5 0.5 0.0 0.0 0.0 0.0 0.5 0.5 0.5 0.5 1.0 1.0 1.0
0.00 0.25 0.50 0.75 1.00 0.00 0.25 0.50 0.75 1.00 1.0 0.00 0.25 0.50 0.75 1.00 0.00 0.25 0.50 0.75 1.00 1.0 1.0 1.0 1.0 0.5 0.5 0.5 0.5 0.0 0.0 0.0 0.0 0.5 0.5 0.5 0.5 1.0 1.0 1.0
0.00 0.25 0.50 0.75 1.00 0.00 1.0
0.25 0.50 0.75 1.00 Figure 4.21: Plot of the d’Alembert solution for t = 0, t = 0.2, t = 0.4, and t = 0.6. 4.8.3 Notes It is perhaps easier and more useful to memorize the procedure rather than the formula itself. The
important thing to remember is that a solution to the wave equation is a superposition of two waves
traveling in opposite directions. That is,
y( x, t) = A( x − at) + B( x + at). 202 CHAPTER 4. FOURIER SERIES AND PDES If you think about it, the exact formulas for A and B are not hard to guess once you realize what
kind of side conditions y( x, t) is supposed to satisfy. Let us give the formula again, but slightly
diﬀerently. Best approach is to do this in stages. When g( x) = 0 (and hence G( x) = 0) we have the
solution
F ( x − at) + F ( x + at)
.
2
On the other hand, when f ( x) = 0 (and hence F ( x) = 0), we let
x H ( x) = G( s) ds.
0 The solution in this case is
1
2a x+at G( s) ds = x−at −H ( x − at) + H ( x + at)
.
2a By superposition we get a solution for the general side conditions (4.15) (when neither f ( x) nor
g( x) are identically zero).
y( x, t) = F ( x − at) + F ( x + at) −H ( x − at) + H ( x + at)
+
.
2
2a (4.17) Do note the minus sign before the H .
Exercise 4.8.1: Check that the new formula (4.17) satisﬁes the side conditions (4.15).
Warning: Make sure you use the odd extensions F ( x) and G( x), when you have formulas for
f ( x) and g( x). The thing is, those formulas in general hold only for 0 < x < L, and are not usually
equal to F ( x) and G( x) for other x. 4.8.4 Exercises Exercise 4.8.2: Using the d’Alembert solution solve ytt = 4y xx , 0 < x < π, t > 0, y(0, t) = y(π, t) = 0,
y( x, 0) = sin x, and yt ( x, 0) = sin x. Hint: Note that sin x is the odd extension of y( x, 0) and yt ( x, 0).
Exercise 4.8.3: Using the d’Alembert solution solve ytt = 2y xx , 0 < x < 1, t > 0, y(0, t) = y(1, t) = 0,
y( x, 0) = sin5 (π x), and yt ( x, 0) = sin3 (π x).
Exercise 4.8.4: Take ytt = 4y xx , 0 < x < π, t > 0, y(0, t) = y(π, t) = 0, y( x, 0) = x(π − x), and
yt ( x, 0) = 0. a) Solve using the d’Alembert formula (Hint: You can use the sine series for y( x, 0).) b)
Find the solution as a function of x for a ﬁxed t = 0.5, t = 1, and t = 2. Do not use the sine series
here.
Exercise 4.8.5: Derive the d’Alembert solution for ytt = a2 y xx , 0 < x < π, t > 0, y(0, t) = y(π, t) = 0,
y( x, 0) = f ( x), and yt ( x, 0) = 0, using the Fourier series solution of the wave equation, by applying
an appropriate trigonometric identity. 4.8. D’ALEMBERT SOLUTION OF THE WAVE EQUATION 203 Exercise 4.8.6: The d’Alembert solution still works if there are no boundary conditions and the
initial condition is deﬁned on the whole real line. Suppose that ytt = y xx (for all x on the real line
and t ≥ 0), y( x, 0) = f ( x), and yt ( x, 0) = 0, where 0 if x < −1, x + 1 if −1 ≤ x < 0, f ( x) = − x + 1 if 0 ≤ x < 1, 0 if x > 1.
Solve using the d’Alembert solution. That is, write down a piecewise deﬁnition for the solution.
Then sketch the solution for t = 0, t = 1/2, t = 1, and t = 2. 204 CHAPTER 4. FOURIER SERIES AND PDES 4.9 Steady state temperature and the Laplacian Note: 1 lecture, §9.7 in [EP]
Suppose we have an insulated wire, a plate, or a 3dimensional object. We apply certain ﬁxed
temperatures on the ends of the wire, the edges of the plate or on all sides of the 3dimensional
object. We wish to ﬁnd out what is the steady state temperature distribution. That is, we wish to
know what will be the temperature after long enough period of time.
We are really looking for a solution to the heat equation that is not dependent on time. Let us
ﬁrst do this in one space variable. We are looking for a function u that satisﬁes
ut = ku xx ,
but such that ut = 0 for all x and t. Hence, we are looking for a function of x alone that satisﬁes
u xx = 0. It is easy to solve this equation by integration and we see that u = Ax + B for some
constants A and B.
Suppose we have an insulated wire, and we apply constant temperature T 1 at one end (say where
x = 0) and T 2 on the other end (at x = L where L is the length of the wire). Then our steady state
solution is
T2 − T1
u( x) =
x + T1.
L
This solution agrees with our common sense intuition with how the heat should be distributed in the
wire. So in one dimension, the steady state solutions are basically just straight lines.
Things are more complicated in two or more space dimensions. Let us restrict to two space
dimensions for simplicity. The heat equation in two variables is
ut = k(u xx + uyy ), (4.18) or more commonly written as ut = k∆u or ut = k 2 u. Here the ∆ and 2 symbols mean ∂∂x2 + ∂∂y2 .
We will use ∆ from now on. The reason for that notation is that you can deﬁne ∆ to be the right
thing for any number of space dimensions and then the heat equation is always ut = k∆u. The ∆ is
called the Laplacian.
OK, now that we have notation out of the way, let us see what does an equation for the steady
state solution look like. We are looking for a solution to (4.18) that does not depend on t. Hence we
are looking for a function u( x, y) such that
2 2 ∆u = u xx + uyy = 0.
This equation is called the Laplace equation¶ . Solutions to the Laplace equation are called harmonic
functions and have many nice properties and applications far beyond the steady state heat problem.
¶ Named after the French mathematician PierreSimon, marquis de Laplace (1749 – 1827). 4.9. STEADY STATE TEMPERATURE AND THE LAPLACIAN 205 Harmonic functions in two variables are no longer just linear (plane graphs). For example,
you can check that the functions x2 − y2 and xy are harmonic. However, if you remember your
multivariable calculus we note that if u xx is positive, u is concave up in the x direction, then uyy
must be negative and u must be concave down in the y direction. Therefore, a harmonic function can
never have any “hilltop” or “valley” on the graph. This observation is consistent with our intuitive
idea of steady state heat distribution.
Commonly the Laplace equation is part of a socalled Dirichlet problem . That is, we have some
region in the xyplane and we specify certain values along the boundaries of the region. We then try
to ﬁnd a solution u deﬁned on this region such that u agrees with the values we speciﬁed on the
boundary.
For simplicity, we will consider a rectangular region. Also for simplicity we will specify
boundary values to be zero at 3 of the four edges and only specify an arbitrary function at one
edge. As we still have the principle of superposition, you can use this simpler solution to derive the
general solution for arbitrary boundary values by solving 4 diﬀerent problems, one for each edge,
and adding those solutions together. This setup is left as an exercise.
We wish to solve the following problem. Let h and w be the height and width of our rectangle,
with one corner at the origin and lying in the ﬁrst quadrant.
(0, h)
∆u = 0,
u(0, y) = 0 for 0 < y < h,
u( x, h) = 0 for 0 < x < w,
u(w, y) = 0 for 0 < y < h,
u( x, 0) = f ( x) for 0 < x < w. (4.19)
(4.20)
(4.21)
(4.22)
(4.23) u=0 u=0 u=0 (0, 0) (w, h) u = f ( x) (w, 0) The method we will apply is separation of variables. Again, we will come up with enough
buildingblock solutions satisfying all the homogeneous boundary conditions (all conditions except
(4.23)). We notice that superposition still works for the equation and all the homogeneous conditions.
Therefore, we can use the Fourier series for f ( x) to solve the problem as before.
We try u( x, y) = X ( x)Y (y). We plug u into the equation to get
X Y + XY = 0.
We put the X s on one side and the Y s on the other to get
− X
Y
=
.
X
Y Named after the German mathematician Johann Peter Gustav Lejeune Dirichlet (1805 – 1859). 206 CHAPTER 4. FOURIER SERIES AND PDES The left hand side only depends on x and the right hand side only depends on y. Therefore, there is
X
some constant λ such that λ = −X = YY . And we get two equations
X + λ X = 0,
Y − λ Y = 0.
Furthermore, the homogeneous boundary conditions imply that X (0) = X (w) = 0 and Y (h) = 0.
Taking the equation for X we have already seen that we have a nontrivial solution if and only if
22
λ = λn = nwπ and the solution is a multiple of
2
nπ
x.
w
For these given λn , the general solution for Y (one for each n) is
Xn ( x) = sin nπ
nπ
y + Bn sinh
y.
(4.24)
w
w
We only have one condition on Yn and hence we can pick one of An or Bn to be something convenient.
It will be useful to have Yn (0) = 1, so we let An = 1. Setting Yn (h) = 0 and solving for Bn we get that
Yn (y) = An cosh Bn = − cosh
sinh nπh
w
nπh
w . After we plug the An and Bn we into (4.24) and simplify, we ﬁnd
Yn (y) = sinh
sinh nπ(h−y)
w
nπh
w . We deﬁne un ( x, y) = Xn ( x)Yn (y). And note that un satisﬁes (4.19)–(4.22).
Observe that
nπ
x.
un ( x, 0) = Xn ( x)Yn (0) = sin
w
Suppose
∞
nπ x
f ( x) =
bn sin
.
w
n=1
Then we get a solution of (4.19)–(4.23) of the following form. nπ(h−y)
∞
∞ nπ sinh w u( x, y) =
bn un ( x, y) =
bn sin
x w sinh nπh
n=1
n=1
w . As un satisﬁes (4.19)–(4.22) and any linear combination (ﬁnite or inﬁnite) of un must also satisfy
(4.19)–(4.22), we see that u must satisfy (4.19)–(4.22). By plugging in y = 0 it is easy to see that u
satisﬁes (4.23) as well. 4.9. STEADY STATE TEMPERATURE AND THE LAPLACIAN 207 Example 4.9.1: Suppose that we take w = h = π and we let f ( x) = π. We compute the sine series
for the function π (we will get the square wave). We ﬁnd that for 0 < x < π we have
∞ f ( x) =
n=1
n odd 4
sin(nx).
n Therefore the solution u( x, y), see Figure 4.22, to the corresponding Dirichlet problem is given as
∞ u( x, y) =
n=1
n odd 0.0 sinh n(π − y)
4
sin(nx)
.
n
sinh(nπ) 0.0
1.0 0.5
x y 0.5
1.5 1.0 2.0
2.5 1.5 3.0 u(x,y) 2.0
2.5 3.0 3.0
2.5
3.0
2.0
2.5
1.5
2.0
1.0
1.5 3.142
2.828
2.514
2.199
1.885
1.571
1.257
0.943
0.628
0.314
0.000 0.5
1.0
0.0
0.5 0.0
0.5 0.0 1.0
0.0 1.5 0.5
1.0 2.0
1.5 2.5 2.0
2.5
y 3.0 x 3.0 Figure 4.22: Steady state temperature of a square plate with three sides held at zero and one side
held at π. This scenario corresponds to the steady state temperature on a square plate of width π with 3
sides held at 0 degrees and one side held at π degrees. If we have arbitrary initial data on all sides, 208 CHAPTER 4. FOURIER SERIES AND PDES then we solve four problems, each using one piece of nonhomogeneous data. Then we use the
principle of superposition to add up all four solutions to have a solution to the original problem.
There is another way to visualize the solutions. Take a wire and bend it in just the right way so
that it corresponds to the graph of the temperature above the boundary of your region. Then dip the
wire in soapy water and let it form a soapy ﬁlm stretched between the edges of the wire. It turns out
that this soap ﬁlm is precisely the graph of the solution to the Laplace equation. Harmonic functions
come up frequently in problems when we are trying to minimize area of some surface or minimize
energy in some system. 4.9.1 Exercises Exercise 4.9.1: Let R be the region described by 0 < x < π and 0 < y < π. Solve the problem
∆ u = 0, u( x, 0) = sin x, u( x, π) = 0, u(0, y) = 0, u(π, y) = 0. Exercise 4.9.2: Let R be the region described by 0 < x < 1 and 0 < y < 1. Solve the problem
u xx + uyy = 0,
u( x, 0) = sin(π x) − sin(2π x),
u(0, y) = 0, u(1, y) = 0. u( x, 1) = 0, Exercise 4.9.3: Let R be the region described by 0 < x < 1 and 0 < y < 1. Solve the problem
u xx + uyy = 0,
u( x, 0) = u( x, 1) = u(0, y) = u(1, y) = C.
for some constant C. Hint: Guess, then check your intuition.
Exercise 4.9.4: Let R be the region described by 0 < x < π and 0 < y < π. Solve
∆u = 0, u( x, 0) = 0, u( x, π) = π, u(0, y) = y, u(π, y) = y. Hint: Try a solution of the form u( x, y) = X ( x) + Y (y) (diﬀerent separation of variables).
Exercise 4.9.5: Use the solution of Exercise 4.9.4 to solve
∆u = 0, u( x, 0) = sin x, u( x, π) = π, u(0, y) = y, u(π, y) = y. Hint: Use superposition.
Exercise 4.9.6: Let R be the region described by 0 < x < w and 0 < y < h. Solve the problem
u xx + uyy = 0,
u( x, 0) = 0, u( x, h) = f ( x),
u(0, y) = 0, u(w, y) = 0.
The solution should be in series form using the Fourier series coeﬃcients of f ( x). 4.9. STEADY STATE TEMPERATURE AND THE LAPLACIAN 209 Exercise 4.9.7: Let R be the region described by 0 < x < w and 0 < y < h. Solve the problem
u xx + uyy = 0,
u( x, 0) = 0, u( x, h) = 0,
u(0, y) = f (y), u(w, y) = 0.
The solution should be in series form using the Fourier series coeﬃcients of f (y).
Exercise 4.9.8: Let R be the region described by 0 < x < w and 0 < y < h. Solve the problem
u xx + uyy = 0,
u( x, 0) = 0, u( x, h) = 0,
u(0, y) = 0, u(w, y) = f (y).
The solution should be in series form using the Fourier series coeﬃcients of f (y).
Exercise 4.9.9: Let R be the region described by 0 < x < 1 and 0 < y < 1. Solve the problem
u xx + uyy = 0,
u( x, 0) = sin(9π x), u( x, 1) = sin(2π x),
u(0, y) = 0, u(1, y) = 0.
Hint: Use superposition.
Exercise 4.9.10: Let R be the region described by 0 < x < 1 and 0 < y < 1. Solve the problem
u xx + uyy = 0,
u( x, 0) = sin(π x),
u(0, y) = sin(πy),
Hint: Use superposition. u( x, 1) = sin(π x),
u(1, y) = sin(πy). 210 CHAPTER 4. FOURIER SERIES AND PDES Chapter 5
Eigenvalue problems
5.1 SturmLiouville problems Note: 2 lectures, §10.1 in [EP] 5.1.1 Boundary value problems We have encountered several diﬀerent eigenvalue problems such as:
X ( x) + λ X ( x) = 0
with diﬀerent boundary conditions
X (0) = 0
X (0) = 0
X (0) = 0
X (0) = 0 X ( L) = 0
X ( L) = 0
X ( L) = 0
X ( L) = 0 (Dirichlet) or,
(Neumann) or,
(Mixed) or,
(Mixed), . . . For example for the insulated wire, Dirichlet conditions correspond to applying a zero temperature
at the ends, Neumann means insulating the ends, etc. . . . Other types of endpoint conditions also
arise naturally, such as
hX (0) − X (0) = 0
hX (L) + X (L) = 0,
for some constant h.
These problems came up, for example, in the study of the heat equation ut = ku xx when we
were trying to solve the equation by the method of separation of variables. In the computation we
encountered a certain eigenvalue problem and found the eigenfunctions Xn ( x). We then found the
eigenfunction decomposition of the initial temperature f ( x) = u( x, 0) in terms of the eigenfunctions
∞ f ( x) = cn Xn ( x).
n=1 211 212 CHAPTER 5. EIGENVALUE PROBLEMS Once we had this decomposition and once we found suitable T n (t) such that T n (0) = 1, we noted
that a solution to the original problem could be written as
∞ cn T n (t)Xn ( x). u( x , t ) =
n=1 We will try to solve more general problems using this method. First, we will study second order
linear equations of the form
dy
d
p( x)
− q( x)y + λr( x)y = 0.
dx
dx (5.1) Essentially any second order linear equation of the form a( x)y + b( x)y + c( x)y + λd( x)y = 0 can
be written as (5.1) after multiplying by a proper factor.
Example 5.1.1 (Bessel):
x2 y + xy + λ x2 − n2 y = 0.
Multiply both sides by 1
x to obtain 12
d
dy
n2
n2
2
2
0=
x y + xy + λ x − n y xy + y + λ x −
y=
x
− y + λ xy.
x
x
dx dx
x
We can state the general SturmLiouville problem∗ . We seek nontrivial solutions to
dy
d
p( x)
− q( x)y + λr( x)y = 0,
dx
dx
α1 y(a) − α2 y (a) = 0,
β1 y(b) + β2 y (b) = 0. a < x < b,
(5.2) In particular, we seek λs that allow for nontrivial solutions. The λs for which there are nontrivial solutions are called the eigenvalues and the corresponding nontrivial solutions are called
eigenfunctions. Obviously α1 and α2 should not be both zero, same for β1 and β2 .
Theorem 5.1.1. Suppose p( x), p ( x), q( x) and r( x) are continuous on [a, b] and suppose p( x) > 0
and r( x) > 0 for all x in [a, b]. Then the SturmLiouville problem (5.2) has an increasing sequence
of eigenvalues
λ1 < λ2 < λ3 < · · ·
such that
lim λn = +∞ n→∞ and such that to each λn there is (up to a constant multiple) a single eigenfunction yn ( x).
Moreover, if q( x) ≥ 0 and α1 , α2 , β1 , β2 ≥ 0, then λn ≥ 0 for all n.
∗ Named after the French mathematicians Jacques Charles François Sturm (1803 – 1855) and Joseph Liouville (1809
– 1882). 5.1. STURMLIOUVILLE PROBLEMS 213 Note: Be careful about the signs. Also be careful about the inequalities for r and p, they must be
strict for all x! Problems satisfying the hypothesis of the theorem are called regular SturmLiouville
problems and we will only consider such problems here. That is, a regular problem is one where
p( x), p ( x), q( x) and r( x) are continuous, p( x) > 0, r( x) > 0, q( x) ≥ 0, and α1 , α2 , β1 , β2 ≥ 0.
When zero is an eigenvalue, we will usually start labeling the eigenvalues at 0 rather than 1 for
convenience.
Example 5.1.2: The problem y + λy, 0 < x < L, y(0) = 0, and y(L) = 0 is a regular SturmLiouville problem. p( x) = 1, q( x) = 0, r( x) = 1, and we have p( x) = 1 > 0 and r( x) = 1 > 0. The
22
π
eigenvalues are λn = nLπ and eigenfunctions are yn ( x) = sin( nL x). All eigenvalues are nonnegative
2
as predicted by the theorem.
Exercise 5.1.1: Find eigenvalues and eigenfunctions for
y + λ y = 0, y (0) = 0, y (1) = 0. Identify the p, q, r, α j , β j . Can you use the theorem to make the search for eigenvalues easier? (Hint:
Consider the condition −y (0) = 0)
Example 5.1.3: Find eigenvalues and eigenfunctions of the problem
y + λy = 0, 0 < x < 1,
hy(0) − y (0) = 0, y (1) = 0, h > 0. These equations give a regular SturmLiouville problem.
Exercise 5.1.2: Identify p, q, r, α j , β j in the example above.
First note that λ ≥ 0 by Theorem 5.1.1. Therefore, the general solution (without boundary
conditions) is
√
√
y( x) = A cos( λ x) + B sin( λ x)
if λ > 0,
y( x) = Ax + B
if λ = 0.
Let us see if λ = 0 is an eigenvalue: We must satisfy 0 = hB − A and A = 0, hence B = 0 (as
h > 0), therefore, 0 is not an eigenvalue (no eigenfunction).
Now let us try λ > 0. We plug in the boundary conditions.
√
0 = hA − λ B,
√
√
√
√
0 = −A λ sin( λ) + B λ cos( λ).
Note that if A = 0, then B = 0 and viceversa, hence both are nonzero. So B =
√
√
√
√
√
0 = −A λ sin( λ) + hA λ cos( λ). As A 0 we get
λ
√
√
√
0 = − λ sin( λ) + h cos( λ), hA
√,
λ and 214 CHAPTER 5. EIGENVALUE PROBLEMS or √
h
√ = tan λ.
λ Now use a computer to ﬁnd λn . There are tables available, though using a computer or a graphing
calculator will probably be far more convenient nowadays. Easiest method is to plot the functions
h/x and tan x and see for which x they intersect. There will be an inﬁnite number of intersections. So
√
√
denote by λ1 the ﬁrst intersection, by λ2 the second intersection, etc. . . . For example, when
h = 1, we get that λ1 ≈ 0.86, and λ2 ≈ 3.43. A plot for h = 1 is given in Figure 5.1. The appropriate
√
eigenfunction (let A = 1 for convenience, then B = h/ λ) is
h
yn ( x) = cos( λn x) + √ sin( λn x).
λn 0 2 4 6 4 4 2 2 0 0 2 2 4 4
0 2 4 Figure 5.1: Plot of 5.1.2 6 1
x and tan x. Orthogonality We have seen the notion of orthogonality before. For example, we have shown that sin(nx) are
orthogonal for distinct n on [0, π]. For general SturmLiouville problems we will need a more
general setup. Let r( x) be a weight function (any function, though generally we will assume it
is positive) on [a, b]. Then two functions f ( x), g( x) are said to be orthogonal with respect to the
weight function r( x) when
b f ( x) g( x) r( x) dx = 0.
a 5.1. STURMLIOUVILLE PROBLEMS 215 In this setting, we deﬁne the inner product as
b def f, g = f ( x) g( x) r( x) dx,
a and then say f and g are orthogonal whenever f , g = 0. The results and concepts are again
analogous to ﬁnite dimensional linear algebra.
The idea of the given inner product is that those x where r( x) is greater have more weight.
Nontrivial (nonconstant) r( x) arise naturally, for example from a change of variables. Hence, you
could think of a change of variables such that dξ = r( x) dx.
We have the following orthogonality property of eigenfunctions of a regular SturmLiouville
problem.
Theorem 5.1.2. Suppose we have a regular SturmLiouville problem
d
dy
p( x)
− q( x)y + λr( x)y = 0,
dx
dx
α1 y(a) − α2 y (a) = 0,
β1 y(b) + β2 y (b) = 0.
Let y j and yk be two distinct eigenfunctions for two distinct eigenvalues λ j and λk . Then
b y j ( x) yk ( x) r( x) dx = 0,
a that is, y j and yk are orthogonal with respect to the weight function r.
Proof is very similar to the analogous theorem from § 4.1. It can also be found in many books
including, for example, Edwards and Penney [EP]. 5.1.3 Fredholm alternative We also have the Fredholm alternative theorem we talked about before for all regular SturmLiouville problems. We state it here for completeness.
Theorem 5.1.3 (Fredholm alternative). Suppose that we have a regular SturmLiouville problem.
Then either
d
dy
p( x)
− q( x)y + λr( x)y = 0,
dx
dx
α1 y(a) − α2 y (a) = 0,
β1 y(b) + β2 y (b) = 0, 216 CHAPTER 5. EIGENVALUE PROBLEMS has a nonzero solution, or
dy
d
p( x )
− q( x)y + λr( x)y = f ( x),
dx
dx
α1 y(a) − α2 y (a) = 0,
β1 y(b) + β2 y (b) = 0,
has a unique solution for any f ( x) continuous on [a, b].
This theorem is used in much the same way as we did before in § 4.4. It is used when solving
more general nonhomogeneous boundary value problems. The theorem does not help us solve the
problem, but it tells us when a solution exists and when it exists if it is unique, so that we know
when to spend time looking for a solution. To solve the problem we decompose f ( x) and y( x) in
terms of the eigenfunctions of the homogeneous problem, and then solve for the coeﬃcients of the
series for y( x). 5.1.4 Eigenfunction series What we want to do with the eigenfunctions once we have them is to compute the eigenfunction
decomposition of an arbitrary function f ( x). That is, we wish to write
∞ f ( x) = cn yn ( x), (5.3) n=1 where yn ( x) the eigenfunctions. We wish to ﬁnd out if we can represent any function f ( x) in this way,
and if so, we wish to calculate cn (and of course we would want to know if the sum converges). OK,
so imagine we could write f ( x) as (5.3). We will assume convergence and the ability to integrate
the series term by term. Because of orthogonality we have
b f , ym = f ( x) ym ( x) r( x) dx
a
∞ = b yn ( x) ym ( x) r( x) dx cn
a n=1
b = cm ym ( x) ym ( x) r( x) dx = cm ym , ym .
a Hence,
f , ym
cm =
=
ym , ym b f ( x) ym ( x) r( x) dx a b
a 2 ym ( x) r( x) dx . (5.4) 5.1. STURMLIOUVILLE PROBLEMS 217 Note that ym are known up to a constant multiple, so we could have picked a scalar multiple of an
eigenfunction such that ym , ym = 1 (if we had an arbitrary eigenfunction ym , divide it by ym , ym ).
˜
˜˜
In the case that ym , ym = 1 we would have the simpler form cm = f , ym as we essentially did for
the Fourier series. The following theorem holds more generally, but the statement given is enough
for our purposes.
Theorem 5.1.4. Suppose f is a piecewise smooth continuous function on [a, b]. If y1 , y2 , . . . are the
eigenfunctions of a regular SturmLiouville problem, then there exist real constants c1 , c2 , . . . given
by (5.4) such that (5.3) converges and holds for a < x < b.
Example 5.1.4: Take the simple SturmLiouville problem
y + λ y = 0, 0<x< π
,
2 π
= 0.
2
The above is a regular problem and furthermore we actually know by Theorem 5.1.1 on page 212
that λ ≥ 0.
Suppose λ = 0, then the general solution is y( x) = Ax + B, we plug in the initial conditions to
get 0 = y(0) = B, and 0 = y ( π ) = A, hence λ = 0 is not an eigenvalue.
2
The general solution, therefore, is
√
√
y( x) = A cos( λ x) + B sin( λ x).
√
√
Plugging in the boundary conditions we get 0 = y(0) = A and 0 = y π = λ B cos λ π . B
2
2
√
√
cannot be zero and hence cos λ π = 0. This means that λ π must be an odd integral multiple of
2
2
√
π
, i.e. (2n − 1) π = λn π . Hence
2
2
2
λn = (2n − 1)2 .
y(0) = 0, y We can take B = 1. And hence our eigenfunctions are
yn ( x) = sin (2n − 1) x .
We ﬁnally compute π
2 π
.
4
0
So any piecewise smooth function on [0, π ] can be written as
2
sin (2n − 1) x 2 dx = ∞ f ( x) = cn sin (2n − 1) x ,
n=1 where
f , yn
cn =
=
yn , yn π
2 0 π
2 0 π
2 f ( x) sin (2n − 1) x d x 4
=
2
π
sin (2n − 1) x dx f ( x) sin (2n − 1) x d x. 0 Note that the series converges to an odd 2πperiodic (not πperiodic!) extension of f ( x). 218 CHAPTER 5. EIGENVALUE PROBLEMS Exercise 5.1.3 (challenging): In the above example, the function is deﬁned on 0 < x < π , yet the
2
series converges to an odd 2πperiodic extension of f ( x). Find out how is the extension deﬁned for
π
< x < π.
2 5.1.5 Exercises Exercise 5.1.4: Find eigenvalues and eigenfunctions of
y + λy = 0, y(0) − y (0) = 0, y(1) = 0. Exercise 5.1.5: Expand the function f ( x) = x on 0 ≤ x ≤ 1 using the eigenfunctions of the system
y + λy = 0, y (0) = 0, y(1) = 0. Exercise 5.1.6: Suppose that you had a SturmLiouville problem on the interval [0, 1] and came up
with yn ( x) = sin(γnx), where γ > 0 is some constant. Decompose f ( x) = x, 0 < x < 1 in terms of
these eigenfunctions.
Exercise 5.1.7: Find eigenvalues and eigenfunctions of
y(4) + λy = 0, y(0) = 0, y (0) = 0, y(1) = 0, y (1) = 0. This problem is not a SturmLiouville problem, but the idea is the same.
Exercise 5.1.8 (more challenging): Find eigenvalues and eigenfunctions for
dx
(e y ) + λe x y = 0,
dx y(0) = 0, y(1) = 0. Hint: First write the system as a constant coeﬃcient system to ﬁnd general solutions. Do note
that Theorem 5.1.1 on page 212 guarantees λ ≥ 0. 5.2. APPLICATION OF EIGENFUNCTION SERIES 5.2 219 Application of eigenfunction series Note: 1 lecture, §10.2 in [EP]
The eigenfunction series can arise even from higher order equations. Suppose we have an elastic
beam (say made of steel). We will study the transversal vibrations of the beam. That is, assume the
beam lies along the x axis and let y( x, t) measure the displacement of the point x on the beam at
time t. See Figure 5.2.
y y
x
Figure 5.2: Transversal vibrations of a beam.
The equation that governs this setup is
a4 ∂4 y ∂2 y
+
= 0,
∂ x4 ∂t2 for some constant a > 0.
Suppose the beam is of length 1 simply supported (hinged) at the ends. Suppose the beam is
displaced by some function f ( x) at time t = 0 and then let go (initial velocity is 0). Then y satisﬁes:
a4 y xxxx + ytt = 0 (0 < x < 1, t > 0),
y(0, t) = y xx (0, t) = 0,
y(1, t) = y xx (1, t) = 0,
y( x, 0) = f ( x), yt ( x, 0) = 0. (5.5) Again we try y( x, t) = X ( x)T (t) and plug in to get a4 X (4) T + XT = 0 or
X (4) −T
= 4 = λ.
X
aT
We note that we want T + λa4 T = 0. Let us assume that λ > 0. We can argue that we expect
vibration and not exponential growth nor decay in the t direction (there is no friction in our model
for instance). Similarly λ = 0 will not occur.
Exercise 5.2.1: Try to justify λ > 0 just from the equations. 220 CHAPTER 5. EIGENVALUE PROBLEMS Write ω4 = λ, so that we do not need to write the fourth root all the time. For X we get the
equation X (4) − ω4 X = 0. The general solution is
X ( x) = Aeω x + Be−ω x + C sin(ω x) + D cos(ω x).
Now 0 = X (0) = A + B + D, 0 = X (0) = ω2 (A + B − D). Hence, D = 0 and A + B = 0, or B = −A.
So we have
X ( x) = Aeω x − Ae−ω x + C sin(ω x).
Also 0 = X (1) = A(eω − e−ω ) + C sin ω, and 0 = X (1) = Aω2 (eω − e−ω ) − C ω2 sin ω. This means
that C sin ω = 0 and A(eω − e−ω ) = 2A sinh ω = 0. If ω > 0, then sinh ω 0 and so A = 0. This
means that C 0 otherwise λ is not an eigenvalue. Also ω must be an integer multiple of π. Hence
ω = nπ and n ≥ 1 (as ω > 0). We can take C = 1. So the eigenvalues are λn = n4 π4 and the
eigenfunctions are sin(nπ x).
Now T + n4 π4 a4 T = 0. The general solution is T (t) = A sin(n2 π2 a2 t) + B cos(n2 π2 a2 t). But
T (0) = 0 and hence we must have A = 0 and we can take B = 1 to make T (0) = 1 for convenience.
So our solutions are T n (t) = cos(n2 π2 a2 t).
As the eigenfunctions are just sines again, we can decompose the function f ( x) on 0 < x < 1
using the sine series. We ﬁnd numbers bn such that for 0 < x < 1 we have
∞ f ( x) = bn sin(nπ x).
n=1 Then the solution to (5.5) is
∞ y( x, t) = ∞ bn Xn ( x)T n (t) =
n=1 bn sin(nπ x) cos(n2 π2 a2 t).
n=1 The point is that Xn T n is a solution that satisﬁes all the homogeneous conditions (that is, all
conditions except the initial position). And since and T n (0) = 1, we have
∞ y( x, 0) = ∞ bn Xn ( x)T n (0) =
n=1 ∞ bn Xn ( x) =
n=1 bn sin(nπ x) = f ( x).
n=1 So y( x, t) solves (5.5).
Note that the natural (circular) frequency of the system is n2 π2 a2 . These frequencies are all
integer multiples of the fundamental frequency π2 a2 , so we will get a nice musical note. The exact
frequencies and their amplitude are what we call the timbre of the note.
The timbre of a beam is diﬀerent than for a vibrating string where we will get “more” of the
smaller frequencies since we will get all integer multiples, 1, 2, 3, 4, 5, . . . For a steel beam we will
get only the square multiples 1, 4, 9, 16, 25, . . . That is why when you hit a steel beam you hear a
very pure sound. The sound of a xylophone or vibraphone is, therefore, very diﬀerent from a guitar
or piano. 5.2. APPLICATION OF EIGENFUNCTION SERIES
Example 5.2.1: Let us assume that f ( x) =
by now) x( x−1)
.
10
∞ f ( x) =
n=1
n odd 221 On 0 < x < 1 we have (you know how to do this 4
sin(nπ x).
5π3 n3 Hence, the solution to (5.5) with the given initial position f ( x) is
∞ y( x , t ) =
n=1
n odd 5.2.1 4
sin(nπ x) cos(n2 π2 a2 t).
5π3 n3 Exercises Exercise 5.2.2: Suppose you have a beam of length 5 with free ends. Let y be the transverse
deviation of the beam at position x on the beam (0 < x < 5). You know that the constants are such
that this satisﬁes the equation ytt + 4y xxxx = 0. Suppose you know that the initial shape of the beam
is the graph of x(5 − x), and the initial velocity is uniformly equal to 2 (same for each x) in the
positive y direction. Set up the equation together with the boundary and initial conditions. Just set
up, do not solve.
Exercise 5.2.3: Suppose you have a beam of length 5 with one end free and one end ﬁxed (the
ﬁxed end is at x = 5). Let u be the longitudinal deviation of the beam at position x on the beam
(0 < x < 5). You know that the constants are such that this satisﬁes the equation utt = 4u xx . Suppose
x−
−
you know that the initial displacement of the beam is x505 , and the initial velocity is −(1005) in the
positive u direction. Set up the equation together with the boundary and initial conditions. Just set
up, do not solve.
Exercise 5.2.4: Suppose the beam is L units long, everything else kept the same as in (5.5). What
is the equation and the series solution.
Exercise 5.2.5: Suppose you have
a4 y xxxx + ytt = 0 (0 < x < 1, t > 0),
y(0, t) = y xx (0, t) = 0,
y(1, t) = y xx (1, t) = 0,
y( x, 0) = f ( x), yt ( x, 0) = g( x).
That is, you have also an initial velocity. Find a series solution. Hint: Use the same idea as we did
for the wave equation. 222 5.3 CHAPTER 5. EIGENVALUE PROBLEMS Steady periodic solutions Note: 1–2 lectures, §10.3 in [EP] 5.3.1 Forced vibrating string. Suppose that we have a guitar string of length L. We have studied the wave equation problem in this
case, where x was the position on the string, t was time and y was the displacement of the string.
See Figure 5.3.
y
y
0 L x Figure 5.3: Vibrating string. The problem is governed by the equations
ytt = a2 y xx ,
y(0, t) = 0,
y(L, t) = 0,
y( x, 0) = f ( x), yt ( x, 0) = g( x). (5.6) We saw previously that the solution is of the form
∞ y= An cos
n=1 nπa
nπ
nπa
t + Bn sin
t sin
x
L
L
L where An and Bn were determined by the initial conditions. The natural frequencies of the system
π
are the (circular) frequencies nLa for integers n ≥ 1.
But these are free vibrations. What if there is an external force acting on the string. Let us
assume say air vibrations (noise), for example a second string. Or perhaps a jet engine. For
simplicity, assume nice pure sound and assume the force is uniform at every position on the string.
Let us say F (t) = F0 cos(ωt) as force per unit mass. Then our wave equation becomes (remember
force is mass times acceleration)
ytt = a2 y xx + F0 cos(ωt).
with the same boundary conditions of course. (5.7) 5.3. STEADY PERIODIC SOLUTIONS 223 We will want to ﬁnd the solution here that satisﬁes the above equation and
y(0, t) = 0, y(L, t) = 0, y( x, 0) = 0, yt ( x, 0) = 0. (5.8) That is, the string is initially at rest. First we ﬁnd a particular solution y p of (5.7) that satisﬁes
y(0, t) = y(L, t) = 0. We deﬁne the functions f and g as
f ( x) = −y p ( x, 0), g( x) = − ∂y p
( x, 0).
∂t We then ﬁnd solution yc of (5.6). If we add the two solutions, we ﬁnd that y = yc + y p solves (5.7)
with the initial conditions.
Exercise 5.3.1: Check that y = yc + y p solves (5.7) and the side conditions (5.8).
So the big issue here is to ﬁnd the particular solution y p . We look at the equation and we make
an educated guess
y p ( x, t) = X ( x) cos(ωt).
We plug in to get
−ω2 X cos(ωt) = a2 X cos(ωt) + F0 cos(ωt)
or −ω2 X = a2 X + F0 after canceling the cosine. We know how to ﬁnd a general solution to
this equation (it is an nonhomogeneous constant coeﬃcient equation) and we get that the general
solution is
ω
ω
F0
X ( x) = A cos
x + B sin
x − 2.
a
a
ω
The endpoint conditions imply that X (0) = X (L) = 0, so
0 = X (0) = A −
or A = F0
ω2 F0
ω2 and F0
ωL
ωL
F0
cos
+ B sin
− 2.
2
ω
a
a
ω
ωL
Assuming that sin( a ) is not zero we can solve for B to get
0 = X ( L) = B=
Therefore,
F0
X ( x) = 2
ω −F0 cos
ω2 sin ωL
a
ωL
a −1 . cos ωL − 1 ω
ω
a cos x−
sin
x − 1 . ωL
a
a
sin a (5.9) 224 CHAPTER 5. EIGENVALUE PROBLEMS The particular solution y p we are looking for is
F0
y p ( x, t ) = 2
ω cos ωL − 1 ω
ω
a cos x−
sin
x − 1 cos(ωt). ωL
a
a
sin a Exercise 5.3.2: Check that y p works.
Now we get to the point that we skipped. Suppose that sin( ωL ) = 0. What this means is that ω is
a
equal to one of the natural frequencies of the system, i.e. a multiple of πLa . We notice that if ω is not
equal to a multiple of the base frequency, but is very close, then the coeﬃcient B in (5.9) seems
π
to become very large. But let us not jump to conclusions just yet. When ω = nLa for n even, then
ωL
cos( a ) = 1 and hence we really get that B = 0. So resonance occurs only when both cos( ωL ) = −1
a
π
and sin( ωL ) = 0. That is when ω = nLa for odd n.
a
We could again solve for the resonance solution if we wanted to, but it is, in the right sense, the
limit of the solutions as ω gets close to a resonance frequency. In real life, pure resonance never
occurs anyway.
The above calculation explains why a string will begin to vibrate if the identical string is plucked
close by. In the absence of friction this vibration would get louder and louder as time goes on. On
the other hand, you are unlikely to get large vibration if the forcing frequency is not close to a
resonance frequency even if you have a jet engine running close to the string. That is, the amplitude
will not keep increasing unless you tune to just the right frequency.
Similar resonance phenomena occur when you break a wine glass using human voice (yes this
is possible, but not easy† ) if you happen to hit just the right frequency. Remember a glass has much
purer sound, i.e. it is more like a vibraphone, so there are far fewer resonance frequencies to hit.
When the forcing function is more complicated, you decompose it in terms of the Fourier series
and apply the above result. You may also need to solve the above problem if the forcing function is
a sine rather than a cosine, but if you think about it, the solution is almost the same.
Example 5.3.1: Let us do the computation for speciﬁc values. Suppose F0 = 1 and ω = 1 and
L = 1 and a = 1. Then
y p ( x, t) = cos( x) − cos(1) − 1
sin( x) − 1 cos(t).
sin(1) Call B = cos(1)−1 for simplicity.
sin(1)
Then plug in t = 0 to get
f ( x) = −y p ( x, 0) = − cos x + B sin x + 1,
and after diﬀerentiating in t we see that g( x) = −
† ∂y p
( x, 0)
∂t = 0. Mythbusters, episode 31, Discovery Channel, originally aired may 18th 2005. 5.3. STEADY PERIODIC SOLUTIONS 225 Hence to ﬁnd yc we need to solve the problem
ytt = y xx ,
y(0, t) = 0, y(1, t) = 0,
y( x, 0) = − cos x + B sin x + 1,
yt ( x, 0) = 0.
Note that the formula that we use to deﬁne y( x, 0) is not odd, hence it is not a simple matter of
plugging in to apply the D’Alembert formula directly! You must deﬁne F to be the odd, 2periodic
extension of y( x, 0). Then our solution would look like
y( x, t) = F ( x + t) + F ( x − t)
cos(1) − 1
+ cos( x) −
sin( x) − 1 cos(t).
2
sin(1)
0
0.0 (5.10) t 1
2 0.2 3
4 x 5 0.5 y(x,t) 0.8
0.20
1.0
0.20 0.10 0.10 y y 0.00 0.00 0.10 0.10 0.240
0.148
0.099
0.049
0.000
0.049
0.099
0.148
0.197
0.254 0.20
0.0 0.20
0.2
0.5 0 x 1
2 0.8
3
t Figure 5.4: Plot of y( x, t) = 4 1.0
5 F ( x+t)+F ( x−t)
2 + cos( x) − cos(1)−1
sin(1) sin( x) − 1 cos(t). It is not hard to compute speciﬁc values for an odd extension of a function and hence (5.10) is a
wonderful solution to the problem. For example it is very easy to have a computer do it, unlike a
series solution. A plot is given in Figure 5.4. 226 5.3.2 CHAPTER 5. EIGENVALUE PROBLEMS Underground temperature oscillations Let u( x, t) be the temperature at a certain location at depth x underground at time t. See Figure 5.5. depth x Figure 5.5: Underground temperature.
The temperature u satisﬁes the heat equation ut = ku xx , where k is the diﬀusivity of the soil. We
know the temperature at the surface u(0, t) from weather records. Let us assume for simplicity that
u(0, t) = T 0 + A0 cos(ωt).
For some base temperature T 0 , then t = 0 is midsummer (could put negative sign above to make it
midwinter). A0 is picked properly to make this the typical variation for the year. That is, the hottest
temperature is T 0 + A0 and the coldest is T 0 − A0 . For simplicity, we will assume that T 0 = 0. ω is
picked depending on the units of t, such that when t = 1year, then ωt = 2π.
It seems reasonable that the temperature at depth x will also oscillate with the same frequency.
And this in fact will be the steady periodic solution, independent of the initial conditions. So we are
looking for a solution of the form
u( x, t) = V ( x) cos(ωt) + W ( x) sin(ωt).
for the problem
ut = ku xx , u(0, t) = A0 cos(ωt). (5.11) We will employ the complex exponential here to make calculations simpler. Suppose we have a
complex valued function
h( x, t) = X ( x) eiωt .
We will look for an h such that Re h = u. To ﬁnd an h, whose real part satisﬁes (5.11), we look for
an h such that
ht = kh xx ,
h(0, t) = A0 eiωt .
(5.12)
Exercise 5.3.3: Suppose h satisﬁes (5.12). Use Euler’s formula for the complex exponential to
check that u = Re h satisﬁes (5.11). 5.3. STEADY PERIODIC SOLUTIONS
Substitute h into (5.12). 227 iωXeiωt = kX eiωt . Hence,
kX − iωX = 0,
or
X − α 2 X = 0,
√
ω
√
where α = ± ik . Note that ± i = ± 1+2i so you could simplify to α = ±(1 + i)
general solution is
√ω
√ω
X ( x) = Ae−(1+i) 2k x + Be(1+i) 2k x . ω
.
2k Hence the We assume that an X ( x) that solves the problem must be bounded as x → ∞ since u( x, t) should
be bounded (we are not worrying about the earth core!). If you use Euler’s formula to expand the
complex exponentials, you will note that the second term will be unbounded (if B 0), while the
ﬁrst term is always bounded. Hence B = 0.
√ω
(1+i) 2k x
Exercise 5.3.4: Use Euler’s formula to show that e
will be unbounded as x → ∞, while
√ω
−(1+i) 2k x
e
will be bounded as x → ∞.
Furthermore, X (0) = A0 since h(0, t) = A0 eiωt . Thus A = A0 . This means that
√ω
√ω
√ω
√ω
h( x, t) = A0 e−(1+i) 2k x eiωt = A0 e−(1+i) 2k x+iωt = A0 e− 2k x ei(ωt− 2k x) .
We will need to get the real part of h, so we apply Euler’s formula to get
h( x, t) = A0 e − √ω
2k x cos ωt − Then ﬁnally
u( x, t) = Re h( x, t) = A0 e− ω
x + i sin ωt −
2k
√ω
2k x cos ωt − ω
x.
2k ω
x,
2k Yay!
ω
Notice the phase is diﬀerent at diﬀerent depths. At depth x the phase is delayed by x 2k .
For example in cgs units (centimetersgramsseconds) we have k = 0.005 (typical value for soil),
π
2π
ω
ω = seconds2in a year = 31,557,341 ≈ 1.99 × 10−7 . Then if we compute where the phase shift x 2k = π
we ﬁnd the depth in centimeters where the seasons are reversed. That is, we get the depth at which
summer is the coldest and winter is the warmest. We get approximately 700 centimeters, which is
approximately 23 feet below ground.
Be careful not to jump to conclusions. The temperature swings decay rapidly as you dig deeper.
√ω
The amplitude of the temperature swings is A0 e− 2k x . This decays very quickly as x grows. Let us
again take typical parameters as above. We also will assume that our surface temperature swing is 228 CHAPTER 5. EIGENVALUE PROBLEMS ±15◦ Celsius, that is, A0 = 15. Then the maximum temperature variation at 700 centimeters is only
±0.66◦ Celsius.
You need not dig very deep to get an eﬀective “refrigerator.” That is why wines are kept in a
cellar; you need consistent temperature. The temperature diﬀerential could also be used for energy.
A home could be heated or cooled by taking advantage of the above fact. Even without the earth
core you could heat a home in the winter and cool it in the summer. There is also the earth core, so
temperature presumably gets higher the deeper you dig. We did not take that into account above. 5.3.3 Exercises Exercise 5.3.5: Suppose that the forcing function for the vibrating string is F0 sin(ωt). Derive the
particular solution y p .
Exercise 5.3.6: Take the forced vibrating string. Suppose that L = 1, a = 1. Suppose that the
forcing function is the square wave that is 1 on the interval 0 < x < 1 and −1 on the interval
−1 < x < 0. Find the particular solution. Hint: You may want to use result of Exercise 5.3.5.
Exercise 5.3.7: The units are cgs (centimetersgramsseconds). For k = 0.005, ω = 1.991 × 10−7 ,
A0 = 20. Find the depth at which the temperature variation is half (±10 degrees) of what it is on the
surface.
Exercise 5.3.8: Derive the solution for underground temperature oscillation without assuming that
T 0 = 0. Chapter 6
The Laplace transform
6.1 The Laplace transform Note: 2 lectures, §10.1 in [EP] 6.1.1 The transform In this chapter we will discuss the Laplace transform∗ . The Laplace transform turns out to be a very
eﬃcient method to solve certain ODE problems. In particular, the transform can take a diﬀerential
equation and turn it into an algebraic equation. If the algebraic equation can be solved, applying
the inverse transform gives us our desired solution. The Laplace transform is also useful in the
analysis of certain systems such as electrical circuits, NMR spectroscopy, signal processing and
others. Finally, understanding the Laplace transform will also help with understanding the related
Fourier transform, which, however, requires more understanding of complex numbers. We will not
cover the Fourier transform.
The Laplace transform also gives a lot of insight into the nature of the equations we are dealing
with. It can be seen as converting between the time and the frequency domain. For example, take
the standard equation
mx (t) + cx (t) + kx(t) = f (t).
We can think of t as time and f (t) as incoming signal. The Laplace transform will convert the
equation from a diﬀerential equation in time to an algebraic (no derivatives) equation, where the
new independent variable s is the frequency.
We can think of the Laplace transform as a black box. It eats functions and spits out functions
in a new variable. We write L{ f (t)} = F ( s). It is common to write lower case letters for functions
in the time domain and upper case letters for functions in the frequency domain. We will use the
∗ Just like the Laplace equation and the Laplacian, the Laplace transform is also named after PierreSimon, marquis
de Laplace (1749 – 1827). 229 230 CHAPTER 6. THE LAPLACE TRANSFORM same letter to denote that one function is the Laplace transform of the other, for example F ( s) is the
Laplace transform of f (t). Let us deﬁne the transform.
∞ def L{ f (t)} = F ( s) = e− st f (t) dt.
0 We note that we are only considering t ≥ 0 in the transform. Of course, if we think of t as time there
is no problem, we are generally interested in ﬁnding out what will happen in the future (Laplace
transform is one place where it is safe to ignore the past). Let us compute the simplest transforms.
Example 6.1.1: Suppose f (t) = 1, then
∞ L{1} = e− st dt =
0 e− st
−s ∞
t =0 1
=.
s Of course, the limit only exists if s > 0. So L{1} is only deﬁned for s > 0.
Example 6.1.2: Suppose f (t) = e−at , then
∞ L{e } =
−at ∞
− st −at ee dt = 0 −( s+a)t e
0 e−( s+a)t
dt =
−( s + a) ∞ =
t =0 1
.
s+a Of course, the limit only exists if s + a > 0. So L{e−at } is only deﬁned for s + a > 0.
Example 6.1.3: Suppose f (t) = t, then using integration by parts
∞ L{t} = e− st t dt
0
∞ −te− st
1
=
+
s t =0 s
∞
1 e− st
=0+
s − s t =0
1
= 2.
s ∞ e− st dt
0 Again, the limit only exists if s > 0.
Example 6.1.4: A common function is the unit step function, which is sometimes called the
Heaviside function† . This function is generally given as 0 if t < 0, u(t) = 1 if t ≥ 0. † The function is named after the English mathematician, engineer, and physicist Oliver Heaviside (1850–1925).
Only by coincidence is the function “heavy” on “one side.” 6.1. THE LAPLACE TRANSFORM 231 Let us ﬁnd the Laplace transform of u(t − a), where a ≥ 0 is some constant. That is, the function
that is 0 for t < a and 1 for t ≥ a.
∞ L{u(t − a)} = ∞ e u(t − a) dt =
− st e 0 − st a e− st
dt =
−s ∞
t =a e−as
=
,
s where of course s > 0 (and a ≥ 0 as we said before).
By applying similar procedures we can compute the transforms of many elementary functions.
Many basic transforms are listed in Table 6.1.
f (t) L{ f (t)} C
t
t2
t3
tn
e−at
sin(ωt)
cos(ωt)
sinh(ωt)
cosh(ωt) C
s
1
s2
2
s3
6
s4
n!
sn+1
1
s +a
ω
s2 +ω2
s
s2 +ω2
ω
s2 −ω2
s
s2 −ω2
e−as
s u(t − a) Table 6.1: Some Laplace transforms (C , ω, and a are constants).
Exercise 6.1.1: Verify Table 6.1.
Since the transform is deﬁned by an integral. We can use the linearity properties of the integral.
For example, suppose C is a constant, then
∞ L{C f (t)} = ∞ e− stC f (t) dt = C
0 e− st f (t) dt = C L{ f (t)}.
0 So we can “pull out” a constant out of the transform. Similarly we have linearity. Since linearity is
very important we state it as a theorem.
Theorem 6.1.1 (Linearity of the Laplace transform). Suppose that A, B, and C are constants, then
L{A f (t) + Bg(t)} = AL{ f (t)} + BL{g(t)},
and in particular
L{C f (t)} = C L{ f (t)}. 232 CHAPTER 6. THE LAPLACE TRANSFORM Exercise 6.1.2: Verify the theorem. That is, show that L{A f (t) + Bg(t)} = AL{ f (t)} + BL{g(t)}.
These rules together with Table 6.1 on the previous page make it easy to already ﬁnd the Laplace
transform of a whole lot of functions already. It is a common mistake to think that Laplace transform
of a product is the product of the transforms. But in general
L{ f (t)g(t)} L{ f (t)}L{g(t)}. It must also be noted that not all functions have Laplace transform. For example, the function 1
t
2
does not have a Laplace transform as the integral diverges. Similarly tan t or et do not have Laplace
transforms. 6.1.2 Existence and uniqueness Let us consider in more detail when does the Laplace transform exist. First let us consider functions
of exponential order. f (t) is of exponential order as t goes to inﬁnity if
 f (t) ≤ Mect ,
for some constants M and c, for suﬃciently large t (say for all t > t0 for some t0 ). The simplest way
to check this condition is to try and compute
lim t→∞ f (t)
.
ect If the limit exists and is ﬁnite (usually zero), then f (t) is of exponential order.
Exercise 6.1.3: Use L’Hopital’s rule from calculus to show that a polynomial is of exponential
order. Hint: Note that a sum of two exponential order functions is also of exponential order. Then
show that tn is of exponential order for any n.
For an exponential order function we have existence and uniqueness of the Laplace transform.
Theorem 6.1.2 (Existence). Let f (t) be continuous and of exponential order for a certain constant
c. Then F ( s) = L{ f (t)} is deﬁned for all s > c.
You may have existence of the transform for other functions, that are not of exponential order,
but that will not relevant to us. Before dealing with uniqueness, let us also note that for exponential
order functions you also obtain that their Laplace transform decays at inﬁnity:
lim F ( s) = 0. s→∞ Theorem 6.1.3 (Uniqueness). Let f (t) and g(t) be continuous and of exponential order. Suppose
that there exists a constant C, such that F ( s) = G( s) for all s > C. Then f (t) = g(t) for all t ≥ 0. 6.1. THE LAPLACE TRANSFORM 233 Both theorems hold for piecewise continuous functions as well. Recall that piecewise continuous
means that the function is continuous except perhaps at a discrete set of points where it has jump
discontinuities like the Heaviside function. Uniqueness however does not “see” values at the
discontinuities. So you can only conclude that f (t) = g(t) outside of discontinuities. For example,
the unit step function is sometimes deﬁned using u(0) = 1 . This new step function, however, we
2
deﬁned has the exact same Laplace transform as the one we deﬁned earlier where u(0) = 1. 6.1.3 The inverse transform As we said, the Laplace transform will allow us to convert a diﬀerential equation into an algebraic
equation that we can solve. Once we do solve the algebraic equation in the frequency domain we
will want to get back to the time domain, as that is what we are really interested in. We, therefore,
need to also be able to get back. If we have a function F ( s), to be able to ﬁnd f (t) such that
L{ f (t)} = F ( s), we need to ﬁrst know if such a function is unique. It turns out we are in luck
by Theorem 6.1.3. So we can without fear make the following deﬁnition.
If F ( s) = L{ f (t)} for some function f (t). We deﬁne the inverse Laplace transform as
def L−1 {F ( s)} = f (t).
There is an integral formula for the inverse, but it is not as simple as the transform itself (requires
complex numbers). The best way to compute the inverse is to use the Table 6.1 on page 231.
1
Example 6.1.5: Take F ( s) = s+1 . Find the inverse Laplace transform.
We look at the table and we ﬁnd L−1 1
= e−t .
s+1 We note that because the Laplace transform is linear, the inverse Laplace transform is also linear.
That is,
L−1 {AF ( s) + BG( s)} = AL−1 {F ( s)} + BL−1 {G( s)}.
We can of course also just pull out constants. Let us demonstrate how linearity is used by the
following example.
s
Example 6.1.6: Take F ( s) = s s+++1 . Find the inverse Laplace transform.
3s
First we use the method of partial fractions to write F in a form where we can use Table 6.1 on
page 231. We factor the denominator as s( s2 + 1) and write
2 s2 + s + 1 A Bs + C
= +2
.
s3 + s
s
s +1
Hence A( s2 − 1) + s( Bs + C ) = s2 + s + 1. Therefore, A + B = 1, C = 1, A = 1. In other words,
F ( s) = s2 + s + 1 1
1
= +2
.
s3 + s
s s +1 234 CHAPTER 6. THE LAPLACE TRANSFORM By linearity of Laplace transform (and thus of its inverse) we get that
−1 L 1
1
s2 + s + 1
= L −1
+ L−1 2
= 1 + sin t.
3+s
s
s
s +1 A useful property is the socalled shifting property or the ﬁrst shifting property
L{e−at f (t)} = F ( s + a),
where F ( s) is the Laplace transform of f (t).
Exercise 6.1.4: Derive this property from the deﬁnition.
The shifting property can be used when the denominator is a more complicated quadratic that
may come up in the method of partial fractions. You always want to write such quadratics as
( s + a)2 + b by completing the square and then using the shifting property.
1
Example 6.1.7: Find L−1 s2 +4 s+8 .
First we complete the square to make the denominator ( s + 2)2 + 4. Next we ﬁnd L−1 s2 1
1
= sin(2t).
+4
2 Putting it all together with the shifting property we ﬁnd
L−1 s2 1
1
1
= L −1
= e−2t sin(2t).
2
+ 4s + 8
2
( s + 2) + 4 In general, we will want to be able to apply the Laplace transform to rational functions, that is
functions of the form
F ( s)
G ( s)
where F ( s) and G( s) are polynomials. Since normally (for functions that we are considering) the
Laplace transform goes to zero as s → ∞, it is not hard to see that the degree of F ( s) will always be
smaller than that of G( s). Such rational functions are called proper rational functions and we will
always be able to apply the method of partial fractions. Of course this means we will need to be
able to factor the denominator into linear and quadratic terms, which involves ﬁnding the roots of
the denominator. 6.1.4 Exercises Exercise 6.1.5: Find the Laplace transform of 3 + t5 + sin(πt).
Exercise 6.1.6: Find the Laplace transform of a + bt + ct2 for some constants a, b, and c. 6.1. THE LAPLACE TRANSFORM 235 Exercise 6.1.7: Find the Laplace transform of A cos(ωt) + B sin(ωt).
Exercise 6.1.8: Find the Laplace transform of cos2 (ωt).
Exercise 6.1.9: Find the inverse Laplace transform of 4
.
s2 −9 Exercise 6.1.10: Find the inverse Laplace transform of 2s
.
s 2 −1 Exercise 6.1.11: Find the inverse Laplace transform of 1
.
( s−1)2 ( s+1) t Exercise 6.1.12: Find the Laplace transform of f (t) = 0 if t ≥ 1,
if t < 1. 236 6.2 CHAPTER 6. THE LAPLACE TRANSFORM Transforms of derivatives and ODEs Note: 2 lectures, §7.2 –7.3 in [EP] 6.2.1 Transforms of derivatives Let us see how the Laplace transform is used for diﬀerential equations. First let us try to ﬁnd
the Laplace transform of a function that is a derivative. That is, suppose g(t) is a continuous
diﬀerentiable function of exponential order.
∞ L {g (t)} = e− st g (t) dt = e− st g(t)
0 ∞
t =0 ∞ (− s) e− st g(t) dt = −g(0) + sL{g(t)}. −
0 We can keep doing this procedure for higher derivatives. The results are listed in Table 6.2. The
procedure also works for piecewise smooth functions, that is functions that are piecewise continuous
with a piecewise continuous derivative. The fact that the function is of exponential order is used to
show that the limits appearing above exist. We will not worry much about this fact.
f (t ) L{ f (t)} = F ( s) g (t)
g (t)
g (t) sG( s) − g(0)
s2G( s) − sg(0) − g (0)
s3G( s) − s2 g(0) − sg (0) − g (0) Table 6.2: Laplace transforms of derivatives (G( s) = L{g(t)} as usual).
Exercise 6.2.1: Verify Table 6.2. 6.2.2 Solving ODEs with the Laplace transform If you notice, the Laplace transform turns diﬀerentiation essentially into multiplication by s. Let us
see how to apply this to diﬀerential equations.
Example 6.2.1: Take the equation
x (t) + x(t) = cos(2t), x(0) = 0, x (0) = 1. We will take the Laplace transform of both sides. By X ( s) we will, as usual, denote the Laplace
transform of x(t).
L{ x (t) + x(t)} = L{cos(2t)},
s
s2 X ( s) − sx(0) − x (0) + X ( s) = 2
.
s +4 6.2. TRANSFORMS OF DERIVATIVES AND ODES 237 We can plug in the initial conditions now (this will make computations more streamlined) to obtain
s2 X ( s) − 1 + X ( s) = s2 s
.
+4 We now solve for X ( s),
X ( s) = ( s2 s
1
+2
.
2 + 4)
+ 1)( s
s +1 We use partial fractions (exercise) to write
X ( s) = 1s
1s
1
−
+2
.
2+1
2+4
3s
3s
s +1 Now take the inverse Laplace transform to obtain
x(t) = 1
1
cos(t) − cos(2t) + sin(t).
3
3 The procedure is as follows. You take an ordinary diﬀerential equation in the time variable
t. You apply the Laplace transform to transform the equation into an algebraic (non diﬀerential)
equation in the frequency domain. All the x(t), x (t), x (t), and so on, will be converted to X ( s),
sX ( s) − x(0), s2 X ( s) − sx(0) − x (0), and so on. If the diﬀerential equation we started with was
constant coeﬃcient linear equation, it is generally pretty easy to solve for X ( s) and we will obtain
some expression for X ( s). Then taking the inverse transform if possible, we ﬁnd x(t).
It should be noted that since not every function has a Laplace transform, not every equation can
be solved in this manner. 6.2.3 Using the Heaviside function Before we move on to more general functions than those we could solve before, we want to consider
the Heaviside function. See Figure 6.1 on the following page for the graph. 0 if t < 0, u(t) = 1 if t ≥ 0. This function is useful for putting together functions, or cutting functions oﬀ. Most commonly
it is used as u(t − a) for some constant a. This just shifts the graph to the right by a. That is, it is a
function that is zero when t < a and 1 when t ≥ a. Suppose for example that f (t) is a “signal” and
you started receiving the signal sin t at time t. The function f (t) should then be deﬁned as 0 if t < π, f (t ) = sin t if t ≥ π. 238 CHAPTER 6. THE LAPLACE TRANSFORM
1.0 0.5 0.0 0.5 1.0 1.00 1.00 0.75 0.75 0.50 0.50 0.25 0.25 0.00 0.00 1.0 0.5 0.0 0.5 1.0 Figure 6.1: Plot of the Heaviside (unit step) function u(t). Using the Heaviside function, f (t) can be written as
f (t) = u(t − π) sin t.
Similarly the step function that is 1 on the interval [1, 2) and zero everywhere else can be written as
u(t − 1) − u(t − 2).
The Heaviside function is useful to deﬁne functions deﬁned piecewise. If you want the function t on
when t is in [0, 1] and the function −t + 2 when t is in [1, 2] and zero otherwise, you can use the
expression
t u(t) − u(t − 1) + (−t + 2) u(t − 1) − u(t − 2) .
Hence it is useful to know how the Heaviside function interacts with the Laplace transform. We
have already seen that
e−as
L{u(t − a)} =
.
s
This can be generalized into a shifting property or second shifting property.
L{ f (t − a) u(t − a)} = e−as L{ f (t)}. (6.1) Example 6.2.2: Suppose that the forcing function is not periodic. For example, suppose that we
had a mass spring system
x (t) + x(t) = f (t), x(0) = 0, x (0) = 0, where f (t) = 1 if 1 ≤ t < 3 and zero otherwise. We could imagine a mass and spring system where
a rocket was ﬁred for 2 seconds starting at t = 1. Or perhaps an RLC circuit, where the voltage was 6.2. TRANSFORMS OF DERIVATIVES AND ODES 239 being raised at a constant rate for 2 seconds starting at t = 1 and then held steady again starting at
t = 3.
We can write f (t) = u(t − 1) − u(t − 3). We transform the equation and we plug in the initial
conditions as before to obtain
e− s e−3 s
s2 X ( s) + X ( s) =
−
.
s
s
We solve for X ( s) to obtain
e− s
e−3 s
X ( s) =
−
.
s( s2 + 1) s( s2 + 1)
We leave it as an exercise to the reader to show that
1
= 1 − cos t.
+ 1) L −1
In other words L{1 − cos t} = 1
.
s( s2 +1) s( s2 So using (6.1) we ﬁnd L−1 e− s
= e− s L{1 − cos t} = 1 − cos(t − 1) u(t − 1).
s( s2 + 1) L −1 e−3 s
= e−3 s L{1 − cos t} = 1 − cos(t − 3) u(t − 3).
s( s2 + 1) Similarly Hence, the solution is
x(t) = 1 − cos(t − 1) u(t − 1) − 1 − cos(t − 2) u(t − 2).
The plot of this solution is given in Figure 6.2 on the next page. 6.2.4 Transforms of integrals A feature of Laplace transforms is that it is also able to easily deal with integral equations. That is,
equations in which integrals rather than derivatives of functions appear. The basic property, which
can be proved by applying the deﬁnition and again doing integration by parts, is the following.
t f (τ) dτ = L
0 1
F ( s).
s It is sometimes useful for computing the inverse transform to write
t f (τ) dτ = L−1
0 1
F ( s) .
s 240 CHAPTER 6. THE LAPLACE TRANSFORM
0 5 10 15 20 2 2 1 1 0 0 1 1 2 2
0 5 10 15 20 Figure 6.2: Plot of x(t). Example 6.2.3: To compute the inverse transform of
integration rule.
L −1 11
=
s s2 + 1 t 1
2+1
s L−1
0 1
s( s2 +1) we could proceed by applying this
t dτ = sin τ dτ = 1 − cos t.
0 Example 6.2.4: If an equation contains an integral of the unknown function the equation is called
an integral equation. For example, take the equation
t t2 = eτ x(τ) dτ. 0 If we apply the Laplace transform we obtain (where X ( s) = L{ x(t)})
1
2
1
= L{et x(t)} = X ( s − 1).
3
s
s
s
Thus
X ( s − 1) = 2
s2 or X ( s) = 2
.
( s + 1)2 We use the shifting property
x(t) = 2e−t t.
More complicated integral equations can also be solved using the convolution that we will learn
next. 6.2. TRANSFORMS OF DERIVATIVES AND ODES 6.2.5 241 Exercises Exercise 6.2.2: Using the Heaviside function write down the piecewise function that is 0 for t < 0,
t2 for t in [0, 1] and t for t > 1.
Exercise 6.2.3: Using the Laplace transform solve
mx + cx + kx = 0, x(0) = a, x (0) = b, where m > 0, c > 0, k > 0, and c2 − 4km > 0 (system is overdamped).
Exercise 6.2.4: Using the Laplace transform solve
mx + cx + kx = 0, x(0) = a, x (0) = b, where m > 0, c > 0, k > 0, and c2 − 4km < 0 (system is underdamped).
Exercise 6.2.5: Using the Laplace transform solve
mx + cx + kx = 0, x(0) = a, x (0) = b, where m > 0, c > 0, k > 0, and c2 = 4km (system is critically damped).
Exercise 6.2.6: Solve x + x = u(t − 1) for initial conditions x(0) = 0 and x (0) = 0.
Exercise 6.2.7: Show the diﬀerentiation of the transform property. Suppose L{ f (t)} = F ( s), then
show
L{−t f (t)} = F ( s).
Hint: Diﬀerentiate under the integral sign.
Exercise 6.2.8: Solve x + x = t3 u(t − 1) for initial conditions x(0) = 1 and x (0) = 0, x (0) = 0.
Exercise 6.2.9: Show the second shifting property: L{ f (t − a) u(t − a)} = e−as L{ f (t)}. 242 6.3 CHAPTER 6. THE LAPLACE TRANSFORM Convolution Note: 1 or 1.5 lectures, §7.2 in [EP] 6.3.1 The convolution We said that the Laplace transformation of a product is not the product of the transforms. All hope
is not lost however. We simply have to use a diﬀerent type of a “product.” Take two functions f (t)
and g(t) deﬁned for t ≥ 0. Deﬁne the convolution‡ of f (t) and g(t) as
t def ( f ∗ g)(t) = f (τ)g(t − τ) dτ. (6.2) 0 As you can see, the convolution of two functions of t is another function of t.
Example 6.3.1: Take f (t) = et and g(t) = t for t ≥ 0. Then
t ( f ∗ g)(t) = eτ (t − τ) dτ = et − t − 1. 0 To solve the integral we did one integration by parts.
Example 6.3.2: Take f (t) = sin(ωt) and g(t) = cos(ωt) for t ≥ 0. Then
t ( f ∗ g)(t) = sin(ωτ) cos ω(t − τ) dτ.
0 We will apply the identity
cos(θ) sin(ψ) = 1
sin(θ + ψ) − sin(θ − ψ) .
2 Hence,
t 1
sin(ωt) − sin(ωt − 2ωτ) dτ
02
t
1
1
=
τ sin(ωt) +
cos(2ωτ − ωt)
2
4ω
τ=0
1
= t sin(ωt).
2
Of course the formula only holds for t ≥ 0. We did assume that f and g are zero (or just not deﬁned)
for negative t.
( f ∗ g)(t) = ∞ For those that have seen convolution deﬁned before, you may have seen it deﬁned as ( f ∗ g)(t) = −∞ f (τ)g(t − τ) dτ.
This deﬁnition agrees with (6.2) if you deﬁne f (t) and g(t) to be zero for t < 0. When discussing the Laplace transform
the deﬁnition we gave is suﬃcient. Convolution does occur in many other applications, however, where you may have
to use the more general deﬁnition with inﬁnities.
‡ 6.3. CONVOLUTION 243 The convolution has many properties that make it behave like a product. Let c be a constant and
f , g, and h be functions then
f ∗ g = g ∗ f,
(c f ) ∗ g = f ∗ (cg) = c( f ∗ g),
( f ∗ g) ∗ h = f ∗ (g ∗ h).
The most interesting property for us, and the main result of this section is the following theorem.
Theorem 6.3.1. Let f (t) and g(t) be of exponential type, then
t L {( f ∗ g)(t)} = L f (τ)g(t − τ) dτ = L{ f (t)}L{g(t)}.
0 In other words, the Laplace transform of a convolution is the product of the Laplace transforms.
The simplest way to use this result is in reverse.
Example 6.3.3: Suppose we have the function of s deﬁned by
1
11
=
.
( s + 1) s2 s + 1 s2
We recognize the two entries of Table 6.2. That is
L −1 1
= e−t
s+1 L −1 11
=
s + 1 s2 and Therefore, t L−1 1
= t.
s2 τe−(t−τ) dτ = e−t + t − 1. 0 The calculation of the integral involved an integration by parts. 6.3.2 Solving ODEs The next example will demonstrate the full power of the convolution and Laplace transform. We
will be able to give a solution to the forced oscillation problem for any forcing function as a deﬁnite
integral.
Example 6.3.4: Find the solution to
x + ω2 x = f (t),
0
for an arbitrary function f (t). x(0) = 0, x (0) = 0, 244 CHAPTER 6. THE LAPLACE TRANSFORM We ﬁrst apply the Laplace transform to the equation. Denote the transform of x(t) by X ( s) and
the transform of f (t) by F ( s) as usual.
s2 X ( s) + ω2 X ( s) = F ( s),
0
or in other words
X ( s) = F ( s)
We know
L−1 1
sin(ω0 t)
.
=
2
ω0
+ ω0 s2 Therefore, t f (τ) x(t) =
0 or if we reverse the order t x(t) =
0 1
.
s2 + ω2
0 sin ω0 (t − τ)
dτ,
ω0 sin(ω0 t)
f (t − τ) dτ.
ω0 Let us notice one more thing with this example. We can now also notice how Laplace transform
handles resonance. Suppose that f (t) = cos(ω0 t). Then
t x(t) =
0 sin(ω0 τ)
1
cos ω0 (t − τ) dτ =
ω0
ω0 t cos(ω0 τ) sin ω0 (t − τ) dτ.
0 We have already computed the convolution of sine and cosine in Example 6.3.2. Hence
x(t) = 1
ω0 1
1
t sin(ω0 t) =
t sin(ω0 t).
2
2ω0 Note the t in front of the sine. This solution will, therefore, grow without bound as t gets large,
meaning we get resonance.
Using convolution you can also ﬁnd a solution as a deﬁnite integral for arbitrary forcing function
f (t) for any constant coeﬃcient equation. A deﬁnite integral is usually enough for most practical
purposes. It is usually not hard to numerically evaluate a deﬁnite integral. 6.3.3 Volterra integral equation A common integral equation is the Volterra integral equation§
t x(t) = f (t) + g(t − τ) x(τ) dτ,
0 § Named for the Italian mathematician Vito Volterra (1860 – 1940). 6.3. CONVOLUTION 245 where f (t) and g(t) are known functions and x(t) is an unknown we wish to solve for. To ﬁnd x(t),
we apply the Laplace transform to the equation to obtain
X ( s) = F ( s) + G( s)X ( s),
where X ( s), F ( s), and G( s) are the Laplace transforms of x(t), f (t), and g(t) respectively. We ﬁnd
F ( s)
.
1 − G( s) X ( s) = To ﬁnd x(t) we now need to ﬁnd the inverse Laplace transform of X ( s).
Example 6.3.5: Solve t x(t) = e−t + sinh(t − τ) x(τ) dτ.
0 We apply Laplace transform to obtain
X ( s) =
or
X ( s) = 1
s +1 1− 1
s2 −1 1
1
+2
X ( s),
s+1 s −1
= s−1
s
1
=2
−2
.
2−2
s
s −2 s −2 It is not hard to apply Table 6.1 on page 231 to ﬁnd
√
√
1
x(t) = cosh( 2 t) − √ sinh( 2 t).
2 6.3.4 Exercises Exercise 6.3.1: Let f (t) = t2 for t ≥ 0, and g(t) = u(t − 1). Compute f ∗ g.
Exercise 6.3.2: Let f (t) = t for t ≥ 0, and g(t) = sin t for t ≥ 0. Compute f ∗ g.
Exercise 6.3.3: Find the solution to
mx + cx + kx = f (t), x(0) = 0, x (0) = 0, for an arbitrary function f (t), where m > 0, c > 0, k > 0, and c2 − 4km > 0 (system is overdamped).
Write the solution as a deﬁnite integral.
Exercise 6.3.4: Find the solution to
mx + cx + kx = f (t), x(0) = 0, x (0) = 0, for an arbitrary function f (t), where m > 0, c > 0, k > 0, and c2 − 4km < 0 (system is underdamped).
Write the solution as a deﬁnite integral. 246 CHAPTER 6. THE LAPLACE TRANSFORM Exercise 6.3.5: Find the solution to
mx + cx + kx = f (t), x(0) = 0, x (0) = 0, for an arbitrary function f (t), where m > 0, c > 0, k > 0, and c2 = 4km (system is critically
damped). Write the solution as a deﬁnite integral.
Exercise 6.3.6: Solve t x(t) = e−t + cos(t − τ) x(τ) dτ.
0 Exercise 6.3.7: Solve t x(t) = cos t + cos(t − τ) x(τ) dτ.
0 Further Reading
[BM] Paul W. Berg and James L. McGregor, Elementary Partial Diﬀerential Equations, HoldenDay, San Francisco, CA, 1966.
[EP] C.H. Edwards and D.E. Penney, Diﬀerential Equations and Boundary Value Problems:
Computing and Modeling, 4th edition, Prentice Hall, 2008. [F] Stanley J. Farlow, An Introduction to Diﬀerential Equations and Their Applications, McGrawHill, Inc., Princeton, NJ, 1994. [I] E.L. Ince, Ordinary Diﬀerential Equations, Dover Publications, Inc., New York, NY, 1956. 247 248 FURTHER READING Index
acceleration, 16
addition of matrices, 87
algebraic multiplicity, 119
amplitude, 65
angular frequency, 65
antiderivative, 14
antidiﬀerentiate, 14
associated homogeneous equation, 70
atan2, 66
augmented matrix, 91
autonomous equation, 36
autonomous system, 85
beating, 77
Bernoulli equation, 33
boundary conditions for a PDE, 181
boundary value problem, 143
catenary, 11
CauchyEuler equation, 50
center, 107
cgs units, 227, 228
characteristic equation, 52
Chebychev’s equation of order 1, 50
cofactor, 90
cofactor expansion, 90
column vector, 87
commute, 89
complementary solution, 70
complete eigenvalue, 119
complex conjugate, 102
complex number, 53
complex roots, 54
constant coeﬃcient, 51, 96 convolution, 242
corresponding eigenfunction, 144
cosine series, 170
critical point, 36
critically damped, 67
d’Alembert solution to the wave equation, 198
damped, 66
damped motion, 62
defect, 120
defective eigenvalue, 120
deﬁcient matrix, 120
dependent variable, 7
determinant, 89
diagonal matrix, 111
matrix exponential of, 125
diagonalization, 126
diﬀerential equation, 7
direction ﬁeld, 85
Dirichlet boundary conditions, 172, 211
Dirichlet problem, 205
displacement vector, 111
distance, 16
dot product, 88, 152
dynamic damping, 118
eigenfunction, 144, 212
eigenfunction decomposition, 211, 216
eigenvalue, 99, 212
eigenvalue of a boundary value problem, 144
eigenvector, 99
eigenvector decomposition, 133, 140
ellipses (vector ﬁeld), 107
elliptic PDE, 181
249 250
endpoint problem, 143
envelope curves, 68
equilibrium solution, 36
Euler’s equation, 50
Euler’s equations, 56
Euler’s formula, 53
Euler’s method, 41
even function, 155, 168
even periodic extension, 168
existence and uniqueness, 20, 48, 57
exponential growth model, 9
exponential of a matrix, 124
exponential order, 232
extend periodically, 151
ﬁrst order diﬀerential equation, 7
ﬁrst order linear equation, 27
ﬁrst order linear system of ODEs, 95
ﬁrst order method, 42
ﬁrst shifting property, 234
forced motion, 62
systems, 116
Fourier series, 153
fourth order method, 43
Fredholm alternative
simple case, 148
SturmLiouville problems, 215
free motion, 62
free variable, 93
fundamental matrix, 96
fundamental matrix solution, 96, 125
general solution, 10
generalized eigenvectors, 120, 122
Genius software, 5
geometric multiplicity, 119
Gibbs phenomenon, 158
half period, 160
harmonic function, 204
harvesting, 38 INDEX
heat equation, 181
Heaviside function, 230
Hermite’s equation of order 2, 50
homogeneous equation, 34
homogeneous linear equation, 47
homogeneous side conditions, 182
homogeneous system, 96
Hooke’s law, 62, 110
hyperbolic PDE, 181
identity matrix, 88
imaginary part, 54
implicit solution, 24
inconsistent system, 93
indeﬁnite integral, 14
independent variable, 7
initial condition, 10
initial conditions for a PDE, 181
inner product, 88
inner product of functions, 153, 215
integral equation, 240, 244
integrate, 14
integrating factor, 27
integrating factor method, 27
systems, 131
inverse Laplace transform, 233
invertible matrix, 89
IODE
Lab I, 18
Lab II, 41
Project I, 18
Project II, 41
Project III, 57
Project IV, 160
Project V, 160
IODE software, 5
la vie, 72
Laplace equation, 181, 204
Laplace transform, 229
Laplacian, 204 INDEX
leading entry, 93
Leibniz notation, 15, 22
linear equation, 27, 47
linear ﬁrst order system, 85
linear operator, 48, 70
linear PDE, 181
linearity of the Laplace transform, 231
linearly dependent, 57
linearly independent, 49, 57
logistic equation, 37
with harvesting, 38
mass matrix, 111
mathematical model, 9
mathematical solution, 9
matrix, 87
matrix exponential, 124
matrix inverse, 89
matrix valued function, 95
method of partial fractions, 233
Mixed boundary conditions, 211
mks units, 65, 69, 176
multiplication of complex numbers, 53
multiplicity, 60
multiplicity of an eigenvalue, 119
natural (angular) frequency, 65
natural frequency, 76, 113
natural mode of oscillation, 113
Neumann boundary conditions, 172, 211
Newton’s law of cooling, 31, 36
Newton’s second law, 62, 63, 84, 110
nilpotent, 126
normal mode of oscillation, 113
odd function, 155, 168
odd periodic extension, 168
ODE, 8
onedimensional heat equation, 181
onedimensional wave equation, 191
operator, 48 251
ordinary diﬀerential equation, 8
orthogonal
functions, 147, 153
vectors, 151
with respect to a weight, 214
orthogonality, 147
overdamped, 67
parabolic PDE, 181
parallelogram, 90
partial diﬀerential equation, 8, 181
particular solution, 10, 70
PDE, 8, 181
period, 65
periodic, 151
phase diagram, 37
phase portrait, 37, 86
phase shift, 65
Picard’s theorem, 20
piecewise continuous, 163
piecewise smooth, 163
practical resonance, 81, 180
practical resonance amplitude, 81
practical resonance frequency, 80
product of matrices, 88
projection, 153
proper rational function, 234
pseudofrequency, 68
pure resonance, 78, 178
quadratic formula, 52
real part, 54
real world problem, 8
reduced row echelon form, 93
reduction of order method, 50
regular SturmLiouville problem, 213
repeated roots, 59
resonance, 78, 117, 178, 244
RLC circuit, 62
row vector, 87 252
saddle point, 106
sawtooth, 154
scalar, 87
scalar multiplication, 87
second order diﬀerential equation, 11
second order linear diﬀerential equation, 47
second order method, 42
second shifting property, 238
separable, 22
separation of variables, 182
shifting property, 234, 238
side conditions for a PDE, 181
simple harmonic motion, 65
sine series, 170
singular matrix, 89
singular solution, 24
sink, 105
slope ﬁeld, 18
solution, 7
solution curve, 86
source, 105
spiral sink, 108
spiral source, 107
square wave, 81, 156
stable critical point, 36
stable node, 105
steady periodic solution, 80, 175
steady state temperature, 190, 204
stiﬀness matrix, 111
SturmLiouville problem, 212
superposition, 47, 57, 96, 182
symmetric matrix, 147, 151
system of diﬀerential equations, 83
tedious, 72, 73, 79, 136
thermal conductivity, 181
three mass system, 110
timbre, 220
trajectory, 86
transient solution, 80
transpose, 88 INDEX
trigonometric series, 153
undamped, 64
undamped motion, 62
systems, 110
underdamped, 68
undetermined coeﬃcients, 71
for systems, 116
second order systems, 139
systems, 136
unforced motion, 62
unit step function, 230
unstable critical point, 36
unstable node, 105
variation of parameters, 73
systems, 138
vector, 87
vector ﬁeld, 85
vector valued function, 95
velocity, 16
Volterra integral equation, 244
wave equation, 181, 191, 198
weight function, 214 ...
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