Math 257/316 Assignment 9 Solutions

 uxx + uyy = 0 0 < x < 2 , 0 < y < 1
ux (0, y) = 0
ux (2, y) = 1
1. Find the solution to the boundary value problem:
.

u(x, 0) = 0
u(x, 1) = 0
Seeking product solutions u(x, y) = X(x)Y (y) leads as usual to
X 00
Y 00
=−
= λ.
X
Y
The choice of sign in front of λ is motivated by the fact that it is the ”Y problem”
Y 00 (y) = −λY (y), Y (0) = 0 = Y (1)
which has the 2 zero BCs. Solutions are, for n = 1, 2, 3, . . .,
Yn (y) = sin(nπy),
λn = n2 π 2 .
Then the ”X problem” is
X 00 (x) = n2 π 2 X(x), X 0 (0) = 0,
whose solution is any constant multiple of cosh(nπx). The general solution of the
PDE and the 3 zero BCs is a linear combination of these product solutions
u(x, t) =
∞
X
cn cosh(nπx) sin(nπy).
n=1
So ux =
P∞
n=1 nπcn sinh(nπx) sin(nπy),
1 = ux (2, y) =
∞
X
and the remaining BC is
nπcn sinh(2nπ) sin(nπy),
n=1
a Fourier sine series, and so
2
cn =
nπ sinh(2nπ)
Z
(
1
sin(nπx)dx =
0
1
4
n2 π 2 sinh(2nπ)
0
n odd
n even
,
and
u(x, y) =
∞
4 X
1
cosh((2k + 1)πx) sin((2k + 1)πy).
2
2
π
(2k + 1) sinh(2(2k + 1)π)
k=0
2. Find the solution to Laplace’s equation in the semi-infinite strip, {(x, y) | 0 < x <
2, y > 0}, with the following boundary conditions:
5πx
3πx
− 2 sin
, lim u(x, y) = 0.
u(0, y) = 0, ux (2, y) = 0 u(x, 0) = sin
y→∞
4
4
Separation of variables yields
X 00
Y 00
=−
= −λ.
X
Y
The ”X problem” is
X 00 (x) = −λX(x), X(0) = 0, X 0 (2) = 0.
√
√
√
So X(x) = A sin( √
λx) + B cos(
λx).
X(0)
=
0
=⇒
B
=
0,
so
X(x)
=
A
sin(
λx).
√
√
Then 0 = X 0 (2) = λA cos( λ2) implies 2 λ = π/2 + nπ, n = 0, 1, 2, 3, . . . .. That
2
is, λn = π16 (1 + 2n)2 , and Xn (x) = sin((1 + 2n)πx/4). The ”Y problem” is
Y 00 (y) = λY (y),
lim Y (y) = 0,
y→∞
whose solution (with λ = λn ) is any constant multiple of
√
Yn (y) = e−
So
u(x, t) =
∞
X
λn y
.
π
cn e− 4 (1+2n)y sin((1 + 2n)πx/4).
n=0
To satisfy the remaining BC
sin(3πx/4) − 2 sin(5πx/4) = u(x, 0) =
∞
X
cn sin((1 + 2n)πx/4)
n=0
we take c1 = 1, c2 = −2, and c0 = c3 = c4 = c5 = · · · = 0, so that
u(x, t) = e−3πy/4 sin(3πx/4) − 2e−5πy/4 sin(5πx/4).
2
3. Consider Laplace’s equation, uxx + uyy = 0, in the unit square, {(x, y) | 0 < x <
1, 0 < y < 1}, with the following Neumann boundary conditions:
1
ux (0, y) = 0, ux (1, y) = y − , uy (x, 0) = 0, uy (x, 1) = 0.
2
Determine whether or not this problem has a solution, and if it does, find the solution
(which would be unique only up to a constant!).
Since
1
Z
(y − 1/2)dy = 1/2 − 1/2 = 0,
0
the net flux through the boundary is zero. Hence there is a solution (and we can add
any constant to get the full family of solutions). The usual separation of variables
(taking into account the 3 zero BCs) leads to
u(x, y) = c0 +
∞
X
cn cosh(nπx) cos(nπy).
n=1
So ux =
P∞
n=1 cn nπ sinh(nπx) cos(nπy),
y − 1/2 = ux (1, y) =
and the non-zero BC is then
∞
X
cn nπ sinh(nπ) cos(nπy).
n=1
This is a Fourier cosine series, but without the constant term! So the only way it
can be satisfied is if the constantR term in the Fourier cosine series of y − 1/2 is zero.
1
Which, of course, it is: a0 = 2 0 (y − 1/2)dy = 0 (this is the condition we checked
above). The other coefficients are, for n = 1, 2, 3, . . .,
Z 1
2
cn =
(y − 1/2) cos(nπy)dy
nπ sinh(nπ) 0
Z 1
Z
2
1
1
1 1
1
=
y
sin(nπy)|0 −
sin(nπy)dy −
cos(nπy)dx
nπ sinh(nπ) nπ
nπ 0
2 0
(
4
− n3 π3 sinh(nπ)
n odd
2
= 3 3
cos(nπy)|10 =
.
n π sinh(nπ)
0
n even
So
u(x, t) = c0 −
4
π3
∞
X
n=1,n
1
odd
where c0 may take any value.
3
n3 sinh(nπ)
cosh(nπx) cos(nπy)
4. Find the solution of Laplace’s equation in a wedge of angle 0 < α < 2π with insulating
boundary conditions on the sides θ = 0 and θ = α:
1
1
0 < r < a, 0 < θ < α,
urr + ur + 2 uθθ = 0,
r
r
uθ (r, 0) = 0, uθ (r, α) = 0,
u(a, θ) = f (θ),
u(r, θ) bounded as r → 0.
Separation of variables u(r, θ) = R(r)Θ(θ) leads to
1
1
0 = R00 Θ + R0 Θ + 2 RΘ00
r
r
or
r2
R00
R0
Θ00
+r
=−
= λ.
R
R
Θ
The ”Θ problem”
Θ00 (θ) = −λΘ(θ), Θ0 (0) = 0 = Θ0 (α)
has (as usual) solutions
Θn (θ) = cos(nπθ/α),
λn = n2 π 2 /α2 ,
n = 0, 1, 2, 3, . . . .
Then the ”R problem” is the Euler equation
r2 R00 (r) + rR0 (r) − λn R(r) = 0
whose solutions are of the form R(r) = rµ with (indicial equation)
p
p
0 = µ2 − λn = (µ + λn )(µ − λn ).
√
The roots are µ = ± λn . So if n = 1, 2, 3, . . ., the general solution is
√
R(r) = Ar
λn
√
+ Br−
λn
.
√
To ensure R is bounded as r → 0, we set B = 0, and take R(r) = Ar
then λn = 0, and so (double root of the indicial equation)
λn .
If n = 0,
R(r) = A + B ln(r).
Again to ensure R is bounded as r → 0, we set B = 0, leaving just a constant function.
Thus putting everything together,
u(r, θ) = c0 +
∞
X
n=1
4
cn rnπ/α cos(nπθ/α).
It remains to satisfy the BC
f (θ) = u(a, θ) = c0 +
∞
X
cn anπ/α cos(nπθ/α).
n=1
This is a Fourier cosine series. So
c0 =
1
α
Z
Z
α
α
f (θ)dθ,
0
and for n = 1, 2, 3, . . . ,
2
cn = a−nπ/α
α
f (θ) cos(nπθ/α)dα.
0
5