112. Same as Problem 111 for
f () = cos(9) + cos(10)
This one is hard to do by hand, but is interesting because you get to see an example of
amplitude modulation, AM. Compare Figure 13 on page 31.
113. Work out the Fourier coefcient integrals for the func
1
5
10
15
20
5
10
15
20
5
10
15
20
5
10
15
20
-1
1
-1
1
-1
1
-1
Figure 37: Top: What frequencies could be hidden in this function? Second: How about
now? The function has been modied a little to make some features stand out. Compare
to the last two. Third
but somehow we do not know what the coefcients are yet. Dot with , for example.
You nd the number a2 by doing a dot product:
v = a1 + a2 + a3 k = 0 + a2 + 0 = a2
by the orthogonality and the fact that is a unit vector. So we can try the same
thing with fu
20
15
10
5
0
-5
-10
-15
-20
0
10
20
30
40
50
Figure 38: A chaotic motion.
There are systems for which the study of linear equations does not prepare you.
Notice in Figure 38 that the function plotted there is more complicated than those
in Figures 20 (pag
106. We need to check carefully that when u is a solution to the Laplace equation, then
so is 6u, or cu for any constant c. How can that be checked? Start with the rst order
derivative
u
r
cu
u
= c ? Then how about the second derivatives, say (cu) ? Is th
At least that will be a solution to the Laplace equation (see Problem 108).
Then we have to choose the coefcients ck if possible to get the boundary
values. It will work to take c1 = 6 and c2 = 3. Answer:
u(r, ) = 6r3 cos(3) + 3r6 sin(6)
Look, you just th
18.1 Laplace leads to Fourier
We are solving the boundary value problem
1
1
urr + ur + 2 u = 0
r
r
(0 < r < 1, 0 2)
with a specied value at the boundary circle
u(1, ) = f ()
and so far you have found (you did Problem 105, no?) a list of solutions to the
L
Figure 36: A membrane stretched, or the graph of a steady temperature.
So we are solving the boundary value problem
1
1
urr + ur + 2 u = 0
r
r
(0 < r < 1, 0 2)
with a specied value at the boundary circle
u(1, ) = f ().
P RACTICE : What can we try? We have
17.1 Sound
Here we start from the Euler equations, and imagine a small disturbance superimposed over an ambient stillness.
P RACTICE : Is it true that the ambient stillness u = 0, = (constant) is a solution to
the Euler equations?
We then look for approxi
From the differential equation we see this: suppose you have an interval like
[x1 , x2 ] where J0 is 0 at each end, and nonzero between. At any local min or
max where J0 = 0, the DE then tells us that J0 = J0 there. The graph cant
be concave up at a local
P RACTICE : Why does
fR fL
RL
tend to fx as R and L tend to x? Assume f differentiable.
We notice that we have two unknown functions, u and . Probably one equation
isnt going to be enough, so we look for another equation. Turning to conservation
R
of mass
goes from 0 to 2. One cycle in every
q2
T
x
1
seconds. The frequency is
T x1
2 .
For larger roots x2 etc you get shorter cycles, higher pitched sounds.
P RACTICE : When you tighten up the drum tension T , do the frequencies go up or
down? Also the drumm
96. If you are not clear about the limit we found, try this. First tell what this limit is:
lim
h0
Then how about
lim
h0
f (x + h) f (x)
h
(x + h)g(x + h) xg(x)
?
h
Finally the one we need
lim
dr0
(r + dr)ur (r + dr, t) rur (r, t)
dr
16.1 A new Function f
(0,b)
T
T
(0,0)
(a,0)
T
Figure 33: A triangle of the drumhead. See Problem 93.
P ROBLEMS
93. Imagine the tension forces on a small triangle of material located anywhere in the
drumhead, as in Figure 33. Suppose a triangle with vertices (0, 0), (a, 0), and
rutt =
T
(rur )r or equivalently
utt =
T
1
urr + ur
r
We will nd some solutions to this having u(1, t) = 0 at the rim of the drum.
The solutions will be vibrations of our drum model, and hopefully will resemble
vibrations of a real drum.
1
coefcient
r
of
Figure 31: A wave traveling initially to the left is shown at six times, t increasing from top
to bottom in the picture. It is reected downward and reversed left-right when it bounces
off a wall at the left side. Does this t Problem 90 or Problem 91 bette
10
9
8
7
6
5
4
3
2
1
0
2
1
0
1
2
3
4
5
6
7
8
Figure 30: The wave u(x, t) = e(xct) , graphed for various times. Note that to keep
the expression x ct constant as t increases you have to increase x also, so this wave is
moving to the right. It could make a
1.2
1
0.8
1
0.6
0.8
0.4
0.6
0.2
0.4
0
0.2
0.2
0.4
0
0.5
1
1.5
2
2.5
3
0
0
3.5
0.5
1
1.5
2
2.5
3
3.5
Figure 27: Left:The two terms in our solution, plotted for t = 0 and t = .3 Notice that the
term involving sin(2x) has decayed a lot more than the sin(x) t
ux (0, t) = 0
u(l, t) = 60
u(x, 0) = 40
This is a bar of length l, insulated at the left end and held at 60 degrees at the right
end. The initial temperature is 40 degrees all along its length.
14.2 Product Solutions
There is another class of solutions to
74. Describe in whole sentences what the following problem is about.
ut =
ux (0, t) =
ux (l, t) =
u(x, 0) =
uxx
0
0
400
Determine whether there is a steady-state solution to this problem.
75. Describe what this problem is about.
ut
u(0, t)
u(l, t)
u(x, 0)
and its rate of change is
R
cut dV.
The rate of heat ow through the boundary surface S is
k u n dA.
S
By the divergence theorem this last integral is equal to
(k u) dV.
R
Thus we have
R
cut dV =
(k u) dV.
R
Subtract to get
R
(cut k(uxx + uyy + uzz ) dV
u
new
t
u
u
1
x
u
2
3
x
Figure 23: A stencil.
For the second space derivative uxx , remember it is the x derivative of ux . So we
can do this:
u3 u2
u
u2x 1
uxx (x, t) x
x
u3 2u2 + u1
=
.
(x)2
The heat equation becomes the statement
unew = u2 +
t
(u3 2u2
R
Figure 22: A region R inside the main block of material, with some heat energy owing
between R and the rest of the block. Energy conducts into R through its boundary surface
S. There is no heat source or sink, chemical reaction etc., inside R. You can p
2.5
2
1.5
1
0.5
0
-0.5
-1
-1.5
-2
-3.5
-3
-2.5
-2
-1.5
-1
-0.5
0
Figure 18: Graph of the cubic polynomial a3 + 5a2 + 6a + 1. The roots are approximately
3.2, 1.6, .3.
and this corresponds to the complexity of Figure 17. What we want instead is to
understa
2.5
2
1.5
1
0.5
0
-0.5
-1
-1.5
0
2
4
6
8
10
12
Figure 20: The graph in Problem 66. Remember 6 is approximately 2.
12
Boundary Value Problems
T ODAY: A change from initial conditions. Boundary values. We nd that we
dont know everything about y = y after al