Selected Solutions for Homework 4
Math 2280
Oct. 31, 2003
1. (Problem 3.5.2)
We want to solve the heat equation
@t u = @x2 u; u(0; t) = 0 = u(; t); u(x; 0) = 30 sin(x):
We know that the general form of the solution is
X
u(x; t) =
an e?n2 t sin(nx);
where an are the Fourier coecients of u(x; 0). However, the initial data is already in Fourier series, so we can
just red o these coecients: a1 = 30 and all the other coecients are zero. Thus
u(x; t) = 30e?t sin(x):
2. (Problem 3.5.10) We want to compute the Fourier coecients for the solution to the heat equation in the case
of inhomogeneous boundary data. Here we have
u(0; t) = T1 ; u(L; t) = T2 ; u(x; 0) = f (x):
The steady state temperature u0 is the linear function interpolating the two boundary data points, which is
u0 (x) = T1 + (T2 ?LT1)x :
Then the initial data for the modied problem is
f (x) ? u0 (x) = f (x) ? T1 + (T2 ?LT1)x :
This means that the Fourier coecients are given by
Z L
bn = L2 [f (x) ? T1 + (T2 ?LT1 )x ] sin((nx)=L)dx
0
Z L
Z L
= L2
f (x) sin((nx)=L)dx ? L2 [T1 + (T2 ?LT1 )x ] sin((nx)=L)dx
= L2
Z
0
L
0
f (x) sin((nx)=L)dx ? 2LT1
Z L
2
= L f (x) sin((nx)=L)dx
0
Z
L
0
sin((nx)=L)dx + 2(T2L?2 T1)
Z
L
0
0
x sin((nx)=L)dx
L
L Z L cos((nx)=L)dx]
Lx
L
2(
T
?
T
)
2
T
2
1
1
L
+ L n cos((nx)=L)j0 + L2 [? n cos((nx)=L) + n
= L2
= L2
L
Z
Z
0
0
L
T1 ((?1)n ? 1) +
f (x) sin((nx)=L)dx + 2n
f (x) sin((nx)=L)dx ? 2
T1 + (?1)n+1 T2
n
0
2(T2 ? T1 )(?1)n
n
:
3. (Problem 3.6.2)
We want to solve the heat equation
@t u = @x2 u; @x u(0; t) = 0 = @x u(; t); u(x; 0) = x:
The general solution is given by
where
u(x; t) = a0 +
X
u(x; 0) = a0 +
an e?n2 t cos(nx)
X
an cos(nx):
0
We have to nd the Fourier coecients for the function f (x) = x. First,
a0 = 1
Then for n 1,
2 xdx = 2x = 2 :
0
0
Z
Z +0 x cos(nx)dx = 2 [ nx sin(nx) ? n1 sin(nx)dx]
0
0
Z 2
2
2
= ? n sin(nx)dx = n2 cos(nx)j0 = n2 ((?1)n ? 1):
an = 2
Z
0
This last quantity is 0 for n even and ?4=(n2) for n odd. So we have
X
1 cos(nx)e?n2 t :
u(x; t) = 2 ? 4
2
n
n odd
4. (Problem 3.6.7)
We have a solution u to a heat equation with insulated ends, so
@t u = c2 @x2 u;
Let
@x u(0; t) = 0 = @x u(L; t):
U (t) = L1
Z
L
0
u(x; t)dx
denote the average temperature at time t. There are two ways to show that U is constant: one can use the
hint in the book or one can use the dierential equation.
First I'll present the solution which doesn't follow the hint in the book. We have
dU = 1 d Z L u(x; t)dx = 1 Z L @ u(x; t)dx
dt
L dt 0
L 0 t
2Z L
2
= cL
@x2 u(x; t)dx = cL @x u(x; t)jxx==oL = 0:
0
Here we have used the fact that u is bounded and L is nite to dierentiate underneath the integral sign. We
have also used the fundamental theorem of calculus and the boundary conditions on u.
There is another solution, where you use the fact that the general solution for u is given by
u(x; t) = a0 +
1
X
1
an e?n2 t cos((nx)=L):
Then evaluating U , we nd
L
Z
Z
L
U (t) = L1 u(x; t)dx = L1 [a0 + an e?n2 t cos((nx)=L)dx
0
0
Z L
2t Z L
?
n
?n2 t
X
L ) sin((nx)=L)L = a :
= L1
a0 dx + e L
cos((nx)=L)dx = a0 + e L (? n
0
0
X
0
0
One can justify bringing the sum out of the integral using the fact that the series an is in l2 , i.e.
to get an a priori bound. More precisely,
ju(x; t)j X
P
a2n < 1,
jan je?n2 t ;
which is bounded independent of t.
Of course, both solutions are correct. However, I nd the rst solution more satisfying because one does not
need to know anything about the general form of the solution. This is really just my personal taste.
5. (Problem 3.7.1)
We want to solve the wave equation
@t2u = 1 (@x2 u+@y2u); u(x; y; 0) = sin(3x) sin(y); @t u(x; y; 0) = 0; u(0; y; t) = u(1; y; t) = u(x; 0; t) = u(x; 1; t) = 0:
The general form of the solution is given by
u(x; y; t) =
X
m;n
p
p
sin(mx) sin(ny)[am;n cos( n2 + m2 t) + bm;n sin( n2 + m2 t)]:
Evaluating at t = 0, we see that the coecients are given by
u(x; y; 0) =
X
m;n
am;n sin(mx) sin(ny);
@t u(x; y; 0) =
X
m;n
p
bm;n m2 + n2 sin(mx) sin(ny):
Because @t u(x; y; 0) = 0, we see that the coecients bm;n are all zero. Also, the initial data u(x; y; 0) is already
in Fourier series, so a3;1 = 1 and the rest of the coecients are zero. Thus we have
p
u(x; y; t) = sin(3x) sin(y) cos( 10t):
6. (Problem 3.7.9)
We want to nd the nodla lines for the function
u(x; y; t) = sin(4x) sin(y) cos(t):
These are the lines which stay at the equilibrium position for all time; in other words, we are looking for the
(x; y) such that
u(x; y; t) = 0 for all t:
Thus we must
0 = sin(4x) sin(x);
0 x 1;
0 y 1:
This forces either sin(y) = 0, in which case y = 0; 1, or sin(4x) = 0, in which case x = 0; 1=4=; 1=2; 3=4; 1. So
the nodal lines are the boundary of the square, together with the lines x = 1=4, x = 1=2, and x = 3=4. Here is
a picture of them:
x = 1=2
x = 1=4
PSfrag replacements
x = 3=4