2.1.1 Using the formula,
1
u( x, t) = (e x+ct + e x
2
1
= (e x+ct + e x
2
Z
1 x+ct
)+
sin x dx
2c x ct
1
ct
) + cos( x ) cos(ct).
c
ct
2.1.2 Using the formula,
1
1
u( x, t) = [log(1 + ( x + ct)2 ) + log(1 + ( x ct)2 )] +
2
2c
1
= [log(1 + ( x + ct)2 )(1 +
Section 5.2
2.
Since is a rational number, we have = p/q for some integers p and q. Now
for any x, we must have
cos(x + 2q) = cos x,
and
cos(x + 2q) = cos(x + 2p) = cos(x).
Hence
cos x + cos(x)
is periodic with period 2q.
4.
(a) If (x) is an odd function,
HOMEWORK 6 SOLUTIONS
Section 4.1
2.
We have heat equation here:
ut = kuxx .
The initial condition is
u(x, 0) = 1,
and the boundary conditions are
u(0, t) = u(l, t) = 0.
This is a Dirichlet problem, and the solution according to formula (17) on page 87,
is
we see from part (b) of the exercise that these two should be the same.
(b) We set x = 0 and get the following identity:
1
l2 X
2l2
n0
0= +
( 1)n
cos (
)
2
6 n=1
(n)
l
1
l2 2l2 X ( 1)n
= + 2
6
n=1 n2
Hence we have
1
X
( 1)n+1
2
=
n2
12
n=1
11
Case 3,
=
2
< 0, For this case we have
c2 X 00 (x)
=
X(x)
2
.
Solving it we get
x
x
+ B sin
,
c
c
The boundary condition X(0) = 0 implies that A = 0, and the boundary
condition X(l) = 0 implies that
X(x) = A cos
l
= 0.
c
B sin
We dont want to get trivial
Section 5.1
2.
We have l = 1, and (x) = x2 .
(a) The coefficients of the sine series are given by
Z 1
An = 2
x2 sin (nx)dx
0
Z 1
2x2
4
1
=
cos(nx) |0 +
x cos (nx)dx
n
n 0
Z 1
2x2
4x
4
1
1
=
cos(nx) |0 +
sin(nx) |0
sin (nx)dx
n
(n)2
(n)2 0
2x2
4x
4
1
=
cos
(a) From the discussion of the above three cases, we see that the first case
and the third case < 0 can be combined as
=
n2 2 k
,
l
where
=0
n = 0, 1, 2, 3, . . .
And this forms the eigenvalues of our problem. Notice that here we have a k but
in the book
4.
We assume that
u(x, t) = T (t)X(x),
then
ut = T 0 (t)X(x),
and
uxx = T (t)X 00 (x).
Plug back into the PDE, we get
T 0 (t)X(x) = kT (t)X 00 (x).
Divide both sides by T (t)X(x) and set it to be equal to , we obtain
T 0 (t)
kX 00 (x)
=
=
T (t)
X(x)
This
HOMEWORK 5 SOLUTIONS
5
As before, lets let a = x c(t s) and b = x + c(t s). Then we have
t d
1
=
[f (a, s) + f (b, s)] ds
2c 0 dx
t
1
@f @a @f @s
@f @b @f @s
=
+
+
+
ds
2c 0
@a @x @s @x
@b @x @s @x
t
1
=
fa (a, s) + fb (b, s)ds
2c 0
t
1
=
fa (a, s) +
4.
Lets assume u(x, t) = X(x)T (t), then the PDE can be written as
T 00 (t)X(x) = c2 T (t)X 00 (x)
rX(x)T 0 (t),
move rX(x)T 0 (t) to the other side, we get
T 00 (t)X(x) + rX(x)T 0 (t) = c2 T (t)X 00 (x).
Hence we obtain an eigenvalue problem
c2 X 00 (x)
HOMEWORK 5 SOLUTIONS
Section 3.3
1
Well solve the inhomogeneous diusion equation on the half-line with Dirichlet boundary conditions using the reflection method. The setup of the problem is as follows:
u kuxx = f (x, t) 0 < x < , 0 < t <
u(0, t) = 0
u(x,
5.
(a) The coefficients of the cosine series are given by
2
B0 =
l
Z
l
0
l2
x2
dx =
2
3
and
Z
2 l x2
nx
Bn =
cos(
)dx
l 0 2
l
Z l
x2
nx l
2
nx
=
sin(
) |0
x sin (
)dx
n
l
n 0
l
Z l
x2
nx l
2xl
nx l
2l
nx
=
sin(
) |0 +
cos(
) |0
cos (
)dx
2
2
n
l
(n)
l
(n)
Section 4.2
1.
We assume that
u(x, t) = T (t)X(x),
then
ut = T 0 (t)X(x),
and
uxx = T (t)X 00 (x).
Plug back into the PDE, we get
T 0 (t)X(x) = kT (t)X 00 (x).
Divide both sides by T (t)X(x) and set it to be equal to , we obtain
T 0 (t)
kX 00 (x)
=
=
T (t
2
HOMEWORK 5 SOLUTIONS
W (x, t) =
=
0
+
t
0
=
+
0
0
=
0
=
0
= +
0
0
0
t
S(x y, t s)feven (y, s)dyds
S(x y, t) (y)dy
S(x y, t s)f (y, s)dyds +
t
S(x y, t s)f (y, s)dyds
t
0
0
0
S(x y, t s)f (y, s)dyds
0
S(x + y, t) (y)dy
0
[S(x y, t) + S(x + y, t)] (y)d
Case 3,
2
=
< 0, with
T 0 (t) =
2
> 0, we have
T (t),
X 00 (x) =
and
2
k
X(x).
Solving this, we get
T (t) = e
2t
,
p
p
X(x) = A cos ( x/ k) + B sin ( x/ k)
and
The boundary condition X(0) = 0 implies that A = 0; the other boundary
condition X 0 (l) = 0 i
From the first equation we get
A(e
p
l/ k
e
p
l/ k
p
l/ k
) = B(e
e
p
l/ k
),
hence A = B, plug this into the second equation, and take out the common
factors, we get
p
p
2B
p e l/ k (1 e 2 l/ k ) = 0,
k
this implies that B = 0 because 1 e 2
So we get X(x
For the second part of the integral, I2 , we have
Z 1
(x+y)2
1
I2 = p
e 4kt y dy
4kt Z0
1
x2 2xy y 2 4kty
1
4kt
=p
e
dy
4kt Z0
1
x2 2(x+2kt)y y 2
1
4kt
=p
e
dy
4kt Z0
1
x2 +(x+2kt)2 (x+2kt)2 2(x+2kt)y
1
4kt
=p
e
4kt Z0
1
x2 +(x+2kt)2 (x+2kt+y)2
1
4kt
=p
e
Section 3.2
1.
For the Neumann problem
utt
c2 uxx = 0,
u(x, 0) = (x),
ut (x, 0) = (x),
ux (0, t) = 0
with 0 < x < 1, we extend and to even functions h(x) = (x) for x
0,
h(x) = ( x) for x < 0, and g(x) = (x) for x
0, g(x) = ( x) for x < 0.
Consider the wav
Math 110A: Homework 3 solutions
2.2.1 From the energy conservation, we have that
1
E=
2
Z +
ru2t
+
Tu2x
1
dx =
2
Z +
ry2 + Tfx2 dx = 0.
By the vanishing theorem (A.1), it follows that ut ( x, t)2 = 0 and u2x ( x, t) = 0 for all
values of x and t. Therefor
Math 110A: Notes 21
1. Vanishing theorems
Theorem 1 Let f be a continuous function. If f ( x )
Rb
a f ( x ) dx = 0, then f ( x ) = 0 for all x 2 [ a, b ].
0 for all x 2 [ a, b] and
We wont ask you to prove this but you should know the intuition behind the
Math 110A: notes for ODE and other stuff
1. Operator
Definition
What is an operator?
Answer: In the context of this course, anything that sends functions to functions is an operator.
Examples: L =
u to xu.
x
sends a function u to u x . L = x (multiplica
2.4.6 Using the polar coordinate transformation dxdy = rdrdq, we can compute this directly.
Z
2 Z
Z
Z Z
2
2
2
2
x2
e
dx =
e x dx
e y dy =
e ( x +y ) dx dy
0
0
0
0
=
=
=
0
0
Z p/2
R
0
e
x2
dx =
p
2
0
p
.
4
r2
re
1
e
2
0
Z p/2
1
=
Therefore,
0
Z p/2 Z
Since u only depends on r, we can simplify this to
1
Du = urr + ur .
r
Therefore, the heat equation is ut = k (urr + 1r ur ).
1.3.7 In a similar manner to 1.3.6, we will transform into spherical coordinate ( x, y, z) !
(r sin q cos f, r sin q sin f, r cos
HOMEWORK 4 SOLUTIONS
Section 3.1
1.
According to formula (6) in page 59, we have
Z 1
Z 1
(x y)2
1
1
y
u(x, t) = p
e 4kt e dy p
e
4kt 0
4kt 0
(x+y)2
4kt
For the first part of the integral, I1 , we have
Z 1
(x y)2
1
I1 = p
e 4kt y dy
4kt Z0
1
x2 +2xy y 2 4k
on the half line 0 < x < 1, we first extend the initial condition (x) to an even
function defined on the whole real line, such that (x) = (x) when x 0 and
(x) = ( x) when x < 0. Now we consider a new PDE
ut
with
is
kuxx = 0;
u(x, 0) = (x),
1 < x < 1. Acco
f is any function of one variable. From the condition u(0, x ) = f ( 2x ) = sin x,
y
we can plug in x = y/2 to see that f (y) = sin( 2 ). After figuring out f , we can
finally find out what u is;
2x 3t
u(t, x ) = f (3t 2x ) = sin
2
1.2.2 Let v = uy , then
2.3.2 (a) Since the new maximum on the larger domain has to be larger than the previous one, this implies that if we increase T, then M(t) has to increase (or remain
the same) as well.
(b) Using the same idea as in the previous part, m( T ) has to be decr