HOMEWORK 8 SOLUTIONS
Section 5.6
5.
Since this is a Dirichlet problem, we use sin series. Let
u(x, t) =
utt (x, t) =
∞
X
n=1
∞
X
an (t) sin nx
bn (t) sin nx
n=1
uxx (x, t) =
∞
X
cn (t) sin nx
n=1
Then
2
bn =
π
Z
π
utt sin nxdx = a00n (t)
0
Z
2 π
u sin nxdx
cn =
π 0 xx
π
Z π
2
2n
= ux sin nx −
u cos nxdx
π
π 0 x
0
π
Z
2n
2n2 π
= − u cos nx −
u sin nxdx
π
π 0
0
2n
=−
(u(π, t)(−1)n − u(0, t)) − n2 an (t)
π
Now with the boundary conditions u(0, t) = u(π, t) = 0, we have
cn (t) = −n2 an (t)
From the PDE
utt = c2 uxx + et sin 5x
we have that
∞
X
a00n (t) sin nx = c2
n=1
∞
X
−n2 an (t) sin nx + et sin 5x
n=1
This gives rises to the ODEs for an :
(
0
a00n (t) + c2 n2 an (t) =
et
1
if n 6= 5
if n = 5
Solving for an , we get

An cos (cnt) + Bn sin (cnt)
an (t) =
An cos (cnt) + Bn sin (cnt) +
if n 6= 5
1
et
1 + n2 c2
if n = 5
,
where A0n s and Bn0 s are constants. The initial condition u(x, 0) = 0 implies that

An = 0
if n 6= 5
1
An +
=0
if n = 5
1 + n2 c2
Solving these, we get

A n = 0
if n 6= 5
1
A n = −
1 + n 2 c2
if n = 5
The initial condition ut (0) = sin 3x implies that


cnBn = 0
if n 6= 5 and n 6= 3


cnBn = 1
if n = 3

1

cnBn +
=0
if n = 5
1 + n 2 c2
solving for this, we get


Bn = 0



1
Bn =
cn


1

Bn = −
cn + n3 c3
if n 6= 5 and n 6= 3
if n = 3
if n = 5
Thus, the solution can be written as
1
1
1
1
t
u(x, t) =
sin (3ct) sin (3x)−
cos (5ct) +
sin (5ct) −
e sin (5x)
3c
1 + 25c2
5c + 125c3
1 + 25c2
2
8.
Since it is a Dirichlet problem, we use sin series. Let
u(x, t) =
ut (x, t) =
∞
X
n=1
∞
X
an (t) sin
nπx
l
bn (t) sin
nπx
l
cn (t) sin
nπx
l
n=1
uxx (x, t) =
∞
X
n=1
Then
2
bn =
l
2
cn =
l
Z
Z
l
ut sin
0
nπx
dx = a0n (t)
l
l
nπx
dx
l
0
l
Z
2
nπx 2nπ l
nπx
= ux sin
− 2
ux cos
dx
l
l 0
l
l
0
l
Z
2nπ
nπx nπx
2n2 π 2 l
= − 2 u cos
u sin
dx
−
3
l
l 0
l
l
0
2nπ
n2 π 2
= − 2 (u(l, t)(−1)n − u(0, t)) − 2 an (t)
l
l
uxx sin
From the boundary conditions u(0, t) = 0 and u(l, t) = At, we get
cn = (−1)n+1
n2 π 2
2nπA
t
−
a (t)
l2
l2 n
From the PDE
ut = kuxx ,
we have
∞
X
∞
X
nπx
nπx
bn (t) sin
=k
cn (t) sin
l
l
n=1
n=1
This gives rise to an ODE about an :
a0n (t) = (−1)n+1
Let’s denote
λ=
kn2 π 2
l2
2knπA
kn2 π 2
t
−
an (t)
l2
l2
and β = (−1)n+1
3
2knπA
.
l2
Then using the integrating factor eλt , we get the solution
Z t
−λt
an (t) = βe
seλs ds + e−λt an (0)
0
Using integration by parts, this formula simplifies to
1
β
−λt
t − (1 − e ) + e−λt an (0)
an (t) =
λ
λ
With the initial condition u(x, 0) = 0, we have an (0) = 0, and hence
β
1
−λt
an (t) =
t − (1 − e )
λ
λ
Finally the solution is
β
u(x, t) =
λ
where
λ=
kn2 π 2
l2
nπx
1
−λt
t − (1 − e ) sin
λ
l
and
β = (−1)n+1
4
2knπA
.
l2
Section 6.1
5.
Laplace operator has rotation invariance; the domain is a disk, hence is also
rotational symmetric; the boundary condition is u(x, y) = 0 on the boundary r = a,
hence has rotation symmetry. Hence we may assume the solution u(x, y) depends
only on r, not depending on θ. In this case,
1
∆u = urr + ur
r
and the Poisson equation is now
1
urr + ur = 1
r
Multiply this ODE by integrating factor r, we get
(rur )r = r,
integrating on both sides gives us
rur =
hence
ur =
r2
,
2
r
2
Do another integration we get
u(r) − u(0) =
r2
4
u(r) = u(0) +
r2
.
4
Hence
2
With the boundary condition u(a) = 0, we have u(0) = − a4 , hence the solution is
u(r) =
r 2 − a2
.
4
5
11. (extra problem)
Assume u and v both satisfy the same Dirichlet problem
∆u = 0
u=g
in D
∆v = 0
on ∂D
v=g
in D
on ∂D
Then w = u − v satisfies the following Dirichlet problem
∆w = 0
in D
w=0
on ∂D
Now consider the energy E defined as
Z
E=
w∆wdx
D
On the one hand
E=0
because ∆w = 0; on the other hand,
Z
E=
w∆wdx
by integration by parts
D
Z
=wn · ∇w −
∇w · ∇wdx
since w=0 on ∂D
D
∂D
Z
=−
∇w · ∇wdx
ZD
|∇w|2 dx ≤ 0
=−
D
And the only possibility that E = 0 is when ∇w = 0 throughout D, which means
that w is a constant throughout D. Now since w = 0 on the boundary, the only
possible constant for w is 0. Hence u − v = 0 thoughout D, and uniqueness is
proved.
6
Section 6.2
1.
We take the advice in the hint, and guess that the solution takes the following
form:
u(x, y) = Ax2 + By 2 + Cx + Dy + Exy + G.
Then
uxx + uyy = 2A + 2B,
ux = 2Ax + C + Ey,
uy = 2By + D + Ex.
From the PDE, we have
2A + 2B = 0,
which implies that
A = −B.
From the boundary conditions, we have
C + Ey = −a,
D + Ex = b,
2Aa + C + Ey = 0
2Bb + D + Ex = 0
Solving for the coefficients, we have
E = 0,
C = −a,
D = b,
1
A= ,
2
B=−
1
2
Hence the solution is
u(x, y) =
x2 y 2
−
− ax + by + G
2
2
Notice that G can be any constant. This is consistent with the fact that the solutions to the Laplace equation with Neumann boundary conditions are not unique.
However, they are unique up to a constant.
7
4. (extra problem)
We split the problem into two parts
(
∆v = 0
in {0 < x < 1, 0 < y < 1}
(∗)
v(x, 0) = x, v(x, 1) = 0, vx (0, y) = 0,
vx (1, y) = 0
and
(
∆w = 0
in {0 < x < 1, 0 < y < 1}
()
w(x, 0) = 0, w(x, 1) = 0, wx (0, y) = 0,
wx (1, y) = y 2
Then by the linearity of the problem, u = v + w solves the original problem
(
∆u = 0
in {0 < x < 1, 0 < y < 1}
u(x, 0) = x, u(x, 1) = 0, ux (0, y) = 0, ux (1, y) = y 2
• Now let’s first look at (∗). Assume
v(x, y) = X(x)Y (y),
then we get
X 00 (x) + λX(x) = 0,
Y 00 (y) = λY (y)
We have zero Neumann boundary value in x direction, so we know that
X(x) = cos nπx,
λ = n2 π 2 .
Hence Y (y) = An sinh(nπy) + Bn cosh(nπy). The case n = 0 gives Y (y) =
A0 y + B0 . The condition that v(x, 1) = 0 implies that Y (1) = 0, that is,
An tanh(nπ) + Bn = 0, so
Bn = −An tanh(nπ);
while for n = 0, we have B0 = −A0 .
Now the function v(x, y) can be written as
v(x, y) = A0 y − A0 +
∞
X
An (sinh(nπy) − tanh(nπ) cosh(nπy)) cos nπx
n=1
Now with the fact that u(x, 0) = x, we have that
x = −A0 −
∞
X
An tanh(nπ) cos nπx
n=1
According to the Fourier cosine series for x given in page 109 Example 4, we
have
2(1 − (−1)n )
1
An = 2 2
A0 = − ,
2
m π tanh nπ
Hence,
∞
1 1 X 2(1 − (−1)n )
v(x, y) = − y+ +
[sinh(nπy) − tanh(nπ) cosh(nπy)] cos nπx
2 2 n=1 m2 π 2 tanh nπ
8
• Now let’s look at (). Assume
w(x, y) = X(x)Y (y),
then we get
X 00 (x) = λX(x),
Y 00 (y) + λY (y) = 0
We have zero Dirichlet boundary value in y direction, so we know that
Y (y) = sin nπy,
λ = n2 π 2 .
Hence X(x) = An sinh(nπx) + Bn cosh(nπx). The condition that wx (0, y) = 0
implies that X 0 (0) = 0, that is, An = 0. So the function w(x, y) can be written
as
∞
X
w(x, y) =
Bn cosh(nπx) sin nπy
n=1
2
The condition that wx (1, y) = y implies that
2
y =
∞
X
nπBn sinh(nπ) sin(nπy)
n=1
By exercise 2 of section 5.1, the sine series of y 2 is given by
∞
X
2
y =
an sin nπx,
n=1
with
an = (−1)n+1
2
4
+ ((−1)n − 1)
nπ
(nπ)3
So we must have
(−1)n+1
4
2
+ ((−1)n − 1)
= nπBn sinh(nπ)
nπ
(nπ)3
Solving for Bn we get
1
Bn =
sinh(nπ)
(−1)
n+1
4
2
+ ((−1)n − 1)
2
(nπ)
(nπ)4
and
w(x, y) =
∞
X
n=1
1
sinh(nπ)
(−1)
n+1
2
4
+ ((−1)n − 1)
2
(nπ)
(nπ)4
cosh(nπx) sin nπy
Finally the solution
u(x, y) = v(x, y) + w(x, y)
∞
1
1 X 2(1 − (−1)n )
=− y+ +
[sinh(nπy) − tanh(nπ) cosh(nπy)] cos nπx
2
2 n=1 m2 π 2 tanh nπ
∞
X
1
2
4
n+1
n
+
(−1)
+ ((−1) − 1)
cosh(nπx) sin nπy
2
4
sinh(nπ)
(nπ)
(nπ)
n=1
9