Homework 6
May 16, 2011
1
Recall from page 81, the definition of a periodic function. In particular, we have
Definition 0.1. We say that a function is periodic of period T 6= 0 if f (x + T ) = f (x) for all values
of x in the domain of f . The collection
Homework 1
1
May 16, 2011
1.1
1 a. 1
b. 2
c. 1
d. 4
e. 5
2e Verification is simple.
3b Verification checks out. Hint: dont forget your chain rule.
3c The answer to this non-homework problem will be in the full solutions.
4b Verification checks out; again,
Homework 10
1
May 16, 2011
6.3
1c Let f (x) = xec|x| ; find F[f (x)]. Though this is an application of a formula you will prove
in problem 9, for now you can integrate by parts in the definition easily enough to arrive at
1
4ic
F[f (x)] =
2
2 (c2 + 2)
Si
Homework 4
May 16, 2011
1
1.7
3 Solve
y 00 + y = 0
0
(0.1)
y (0) = 0
(0.2)
y() = 0
(0.3)
Hint: consider the characteristic equation and find general solutions for any different cases
that arise. Then, choose only the constants/cases which give nontrivial
Homework 5
May 16, 2011
1
Note again that having the answer in any and all of these problems is not terribly impressive
from a grading standpoint, which is why we provide solutions: it is knowing and doing the work in
between the problem and answer which
Homework 2
May 16, 2011
1
Recall, the definition of linearity for an operator L[u]:
Definition 0.1 (Linearity of PDE operator). A PDE in operator form, i.e. expressed as L[u] = f
like in class, is linear if for any two solutions u1 and u2 and any real num
Homework 12
1
May 16, 2011
Section 9.2
1
Problem 1
We have the wave equation in 3D,
utt c2 u = 0,
u = u(x, y, z, t)
Show that if we choose to separate only the time variable, by letting u(x, y, z, t) = (x, y, z)T (t),
we arrive at the Helmholtz equation i
Homework 8
1
May 16, 2011
Definition 1.1. Laplace Transform The Laplace Transform of a function f (x, t), denoted by
F (x, s) = L[f (x, t)], is defined to be
+
est f (x, t)dt
(1.1)
F (x, s) = L[f (x, t)] =
0
Though we wont be calculating this integral, f
Homework 7
1
May 16, 2011
4.4
1b Given the problem
utt = uxx ,
0 < x < L,
u(x, 0) = f (x)
ut (x, 0) = g(x),
u(0, t) = T
ux (L, t) = a,
t>0
(4.1)
0<x<L
(4.2)
t>0
(4.3)
Introduce a new function v(x, t) which will have homogeneous boundary conditions in the
MTH 3201 Midterm Test 1: Fall 2010: Solutions
Problem 1
Solve Laplaces equation uxx + uyy = 0 in the plane region cfw_0 < x < 2, 0 < y < 1 subject to
zero Neumann boundary conditions on the two edges y=0 and y=1, along with Dirichlet boundary
conditions
u
Homework 14
1
May 16, 2011
7.3
8 We start with the Bessel equation,
x2 y 00 + xy 0 + (x2 2 )y = 0
and apply the substitution u = xy, so y = u/ x. Then
1
1 3
y 0 = x 2 u0 x 2 u
2
and
1
3
3 5
y 00 = x 2 u00 x 2 u0 + x 2 u
4
Plugging these into the equation,
Homework 13
1
May 16, 2011
9.3
1. a. Calculate in cartesian and polar coordinates that xy = 0.
b.
(2x3 + 3x2 y y 2 ) = 12x + 6y 2
The calculation in polar coordinates, while tedious, yields the same result.
2. a.
y 2 (x2 + y 2 )4 = 14r2 sin2 + 2r2 cos2 =
Homework 3
1
May 16, 2011
See the end for a proof of something used in class (exercise 36)!
1.6
3,24 The PDE y 2 ux + x2 uy = 0, u = (x, y) separates into the ODEs
X 0 + x2 X = 0
Y 0 y 2 Y = 0
which we can integrate and recombine to arrive at the solution
Homework 15
1
May 16, 2011
7.2
2 Derive the Legendre Polynomials P3 , P4 , P5 .
P3 (x): We calculate 3 = 3 4 = 12, and have a5 = a7 = = 0. We are free to choose
a0 = a2 = = 0, so our polynomial solution to Legendres equation is
y(x) = a1 x + a3 x3
From th
Homework 11
1
May 16, 2011
6.4
1 We wish to solve
ut k 2 uxx = 0
(
T1
u(x, 0) =
T2
< x < +,
t>0
x<0
x>0
(6.1)
(6.2)
Luckily, we already know how to form the solution to problem (6.1), (6.2); we can plug in
there, and put this in terms of the error functi
Homework 9
May 16, 2011
1
6.2
1. Calculate the Fourier cosine and sine transforms of ecx . The easiest way that I know (and
also the way it is told to do in the exercise) is to calculate both at once, by writing
2 cs
2 cx
cx
cx
e
cos(x)dx + i
e
sin(x)dx
F