LECTURE 17: SOLVING THE LAPLACE EQUATIONS II
MINGFENG ZHAO
August 04, 2015
Solving the Laplace equation with Neumann boundary condition in a rectangle
Recall that Divergence Theorem: If ∆u = uxx + uyy = 0 in a domain D, then
Z
Z
Z
∂u
dS(x, y).
0=
∆u dxdy =
div ∇u =
∂n
D
D
∂D
Therefore, in order to have a solution to the problem:


∆u = uxx + uyy = 0,

(x, y) ∈ D,
∂u
(x, y) = f (x, y), (x, y) ∈ ∂D.
∂n
Then we must have an additional condition that:
Z
f (x, y) dS(x, y) = 0.

 BC :
∂D
Theorem 1. Let D be a “nice” bounded domain in R2 and f (x, y) be a “nice” function on ∂D, then the problem


∆u = uxx + uyy = 0, (x, y) ∈ D,


 BC :
∂u
(x, y) = f (x, y),
∂n
(x, y) ∈ ∂D.
has a solution if and only if
Z
f (x, y) dS(x, y) = 0.
∂D
Moreover, any two solutions differ by a constant.
For the Laplacian









 BC :
(1)








 AC :
equation in a rectangle D = [0, a] × [0, b], since ∂D consists of four line segments, then
∆u = uxx + uyy = 0,
uy (x, 0) = f1 (x),
(x, y) ∈ D,
uy (x, b) = f2 (x),
0 ≤ x ≤ a,
ux (0, y) = g1 (y), ux (a, y) = g2 (y), 0 ≤ y ≤ b,
Z
Z a
Z a
Z b
Z b
∂u
dS(x, y) = −
f1 (x) dx +
f2 (x) dx −
g1 (y) dy +
g2 (y) dy.
0=
0
0
0
0
∂D ∂n
1
2
MINGFENG ZHAO
To solve (1), we need to solve the following five problems:
(2)








 BC :
uy (x, 0) = f1 (x) − α1 =: f˜1 (x),







 AC :
ux (0, y) = 0, ux (a, y) = 0,
Z a
[f1 (x) − α1 ] dx = 0.
∆u = uxx + uyy = 0,
(x, y) ∈ D,
uy (x, b) = 0,
0 ≤ x ≤ a,
0 ≤ y ≤ b,
0
(3)








 BC :
uy (x, 0) = 0,







 AC :
ux (0, y) = 0, ux (a, y) = 0,
Z a
[f2 (x) − α2 ] dx = 0.
∆u = uxx + uyy = 0,
(x, y) ∈ D,
uy (x, b) = f2 (x) − α2 =: f˜2 (x),
0 ≤ x ≤ a,
0 ≤ y ≤ b,
0
(4)
(5)









 BC :
uy (x, 0) = 0,








 AC :
ux (0, y) = g1 (y) − β1 =: g̃1 (y),
Z b
[g1 (y) − β1 ] dy = 0.
(x, y) ∈ D,
uy (x, b) = 0,
0 ≤ x ≤ a,
ux (a, y) = 0,
0 ≤ y ≤ b,
0









 BC :
uy (x, 0) = 0,








 AC :
ux (0, y) = 0, ux (a, y) = g2 (y) − β2 =: g̃2 (y),
Z b
[g2 (y) − β2 ] dy = 0.








 BC :







 AC :
(6)
∆u = uxx + uyy = 0,
∆u = uxx + uyy = 0,
(x, y) ∈ D,
uy (x, b) = 0,
0 ≤ x ≤ a,
0 ≤ y ≤ b,
0
∆u = uxx + uyy = 0,
(x, y) ∈ D,
uy (x, 0) = α1 ,
uy (x, b) = α2 ,
0 ≤ x ≤ a,
ux (0, y) = β1 ,
ux (a, y) = β2 ,
0 ≤ y ≤ b,
0 = −aα1 + aα2 − bβ1 + bβ2 .
All α1 , α2 , β1 and β2 are constant. It’s easy to see that the solution to (1) is just the sum of solutions to (2), (3),
(4), (5) and (6).
• For (6), we can solve it directly: To solve (6), we are looking for a solution of the form u(x, y) = F (x) + G(y)
to (6), then
ux (x, 0) = F 0 (x),
uy (x, y) = G0 (y).
LECTURE 17: SOLVING THE LAPLACE EQUATIONS II
3
By the boundary condition in (6), we need
F 0 (0) = β1 ,
F 0 (a) = β2 ,
G0 (0) = α1 ,
and G0 (b) = α2 .
Then let’s try quadratic polynomials for F and G, that is,
F (x) =
β2 − β1 2
x + β1 x,
2a
and G(y) =
α2 − α1 2
y + α1 y.
2b
In this case, we have
∆u = F 00 (x) + G00 (y)
=
=
=
α2 − α1
β2 − β1
+
a
b
aα2 − aα1 + bβ2 − bβ1
ab
0,
By the additional condition in (6).
Therefore, the solution to (6) is given by:
u(x, y) =
β2 − β1 2
α2 − α1 2
x + β1 x +
y + α1 y .
2a
2b
• We are going to solve (2), (3), (4), and (5) by using the method of separation of variables: In the following, we
(7)
just solve (2). First, let’s look the problem:



∆u = uxx + uyy = 0, (x, y) ∈ D,



BC : uy (x, b) = 0, 0 ≤ x ≤ a,





ux (0, y) = 0, ux (a, y) = 0, 0 ≤ y ≤ b.
Let u(x, y) = X(x)Y (y) be a non-zero separated solution to (7), then X 00 (x)Y (y) + X(x)Y 00 (y) = 0, that is,
−
X 00 (x)
Y 00 (y)
=
= λ.
X(x)
Y (y)
It’s easy to see that λx = λy = 0, that is, λ is a constant. Since ux (0, y) = ux (a, y) = 0, then X 0 (0) =
X 0 (a) = 0. Then X(x) satisfies:
(8)

 X 00 + λX = 0,
 X 0 (0) = X 0 (a) = 0.
For the eigenvalue problem (7), we know that the eigenvalues are λ =
nπ S
functions are X(x) = C cos
x , with n ∈ N {0}.
a
nπ 2
a
, and the corresponding eigen-
4
MINGFENG ZHAO
For λ =
nπ 2
a
, since uy (x, b) = 0, then Y 0 (b) = 0. Then Y satisfies

2

 Y 00 − λY = Y 00 − nπ Y = 0,
a

 Y 0 (b) = 0.
(9)
Then the solution to (9) is given by:
h nπ
i
2π
nπ
Y (y) = C e a y + e a b e− a y .
Hence, for any n ≥ 0, we can find a non-zero solution to (7):
i
h nπ
2π
nπ
un (x, y) = C e a y + e a b e− a y .
Let
u(x, y) =
∞
X
h nπ
i
nπ 2π
nπ
an e a y + e a b e− a y cos
x
a
n=0
be the solution to (2). It’s easy to see that uy (x, b) = 0 for 0 ≤ x ≤ a, and ux (0, y) = ux (a, y) = 0 for 0 ≤ y ≤ b.
Since uy (x, 0) = f˜1 (x), then
f˜1 (x)
=
∞
X
an ·
n=0
=
∞
X
an ·
n=1
h nπ
a
−e
2π
a b
·
nπ nπ i
x
cos
a
a
i
nπ 2nπ
nπ h
1 − e a b cos
.
a
a
By the additional condition in (2), we know that
a
Z
f˜1 (x) dx = 0.
0
Then for all n ≥ 1, we have
an ·
i 2Z a
nπ 2nπ
nπ h
1−e a b =
f˜1 (x) cos
x dx.
a
a 0
a
That is, we have,
an =
Z
2
h
nπ 1 − e
2nπ
a b
i
0
a
nπ f˜1 (x) cos
x dx,
a
for all n ≥ 1.
Therefore, the solution to (2) is given by:
u(x, y) =
∞
X
h nπ
i
nπ 2π
nπ
x ,
an e a y + e a b e− a y cos
a
n=0
LECTURE 17: SOLVING THE LAPLACE EQUATIONS II
5
where a0 is any constant and
an =
Z
2
h
nπ 1 − e
2nπ
a b
i
a
0
nπ f˜1 (x) cos
x dx,
a
for all n ≥ 1.
Solving the Laplace equation with Robin boundary condition in a rectangle
Let D = [0, a] × [0, b], consider the following Robin boundary condition for the Laplace equation:



∆u = uxx + uyy = 0, (x, y) ∈ D,



(10)
BC : u(x, 0) = 0, u(x, b) = f (x), 0 ≤ x ≤ a,





ux (0, y) = 0, ux (a, y) = 0, 0 ≤ y ≤ b.
First, let’s look the problem:



∆u = uxx + uyy = 0, (x, y) ∈ D,



(11)
BC : u(x, 0) = 0, 0 ≤ x ≤ a,





ux (0, y) = 0, ux (a, y) = 0, 0 ≤ y ≤ b.
Let u(x, y) = X(x)Y (y) be a non-zero separated solution to (11), then X 00 (x)Y (y) + X(x)Y 00 (y) = 0, that is,
−
Y 00 (y)
X 00 (x)
=
= λ.
X(x)
Y (y)
It’s easy to see that λx = λy = 0, that is, λ is a constant. Since ux (0, y) = ux (a, y) = 0, then X 0 (0) = X 0 (a) = 0.
Then X(x) satisfies:

 X 00 + λX = 0,
 X 0 (0) = X 0 (a) = 0.
(12)
nπ 2
For the eigenvalue problem (11), we know that the eigenvalues are λ =
, and the corresponding eigenfunctions
a
nπ S
are X(x) = C cos
x , with n ∈ N {0}.
nπ 2 a
For λ =
, since u(x, 0) = 0, then Y (0) = 0. Then Y satisfies
a

2

 Y 00 − λY = Y 00 − nπ Y = 0,
a
(13)

 Y (0) = 0.
Then the solution to (13) is given by:
Y (y) =


 Cy,
if n = 0,

nπ  C e nπ
a y − e− a y ,
if n 6= 1.
6
MINGFENG ZHAO
Hence, for any n ≥ 0, we can find a non-zero solution to (11):
un (x, y) =


 Cy,
if n = 0,

nπ nπ
 C e nπ
a y − e− a y cos
x , if n =
6 1.
a
Let
u(x, y) = a0 y +
∞
X
nπ nπ
nπ x
an e a y − e− a y cos
a
n=1
be the solution to (10). It’s easy to see that u(x, 0) = 0 for 0 ≤ x ≤ a, and ux (0, y) = ux (a, y) = 0 for 0 ≤ y ≤ b. Since
u(x, b) = f (x), then
f (x)
=
a0 b +
∞
X
nπ nπ
nπ an e a b − e− a b cos
x .
a
n=1
Then we have
a0 b =
1 2
·
2 a
a
Z
f (x) dx, ,
0
nπ
nπ 2
and an · e a b − e− a b =
a
Z
a
f (x) cos
0
nπ x dx,
a
for all n ≥ 1.
which implies that
1
a0 =
ab
Z
a
f (x) dx,
0
2
and an = nπ b
nπ
a
e
− e− a b
Z
a
0
nπ x dx,
f˜1 (x) cos
a
for all n ≥ 1.
Therefore, the solution to (10) is given by:
u(x, y) = a0 y +
∞
X
nπ nπ
nπ x
an e a y − e− a y cos
a
n=1
where
a0 =
1
ab
Z
a
f (x) dx,
and an =
0
e
nπ
a b
2
nπ
− e− a b
Z
0
a
nπ f˜1 (x) cos
x dx,
a
Solving the Laplace equation in a wedge
Example 1. Consider the following problem:
(14)








 BC :








1
1
urr + ur + 2 uθθ = 0, 0 < r < a, 0 < θ < α,
r
r
u(r, 0) = u(r, α) = 0, 0 < r ≤ a,
u(a, θ) = f (θ),
0 < θ < α,
u(r, θ) is bounded as r → 0.
for all n ≥ 1.
LECTURE 17: SOLVING THE LAPLACE EQUATIONS II
7
First, let’s look the following problem:

1
1


urr + ur + 2 uθθ = 0, 0 < r < a, 0 < θ < α,


r
r

(15)
BC : u(r, 0) = u(r, α) = 0, 0 < r ≤ a,





u(r, θ) is bounded as r → 0.
Let u(r, θ) = R(r)Θ(θ) be a non-zero separated solution to (15), then
1
1
R00 (r)Θ(θ) + R0 (r)Θ(θ) + 2 R(r)Θ00 (θ) = 0.
r
r
That is, we have
R00 (r) + 1r R0 (r)
Θ00 (θ)
=−
= λ.
R(r)
Θ(θ)
It’s easy to see that λr = λθ = 0, that is, λ is a constant. Since u(r, 0) = u(r, α) = 0, then Θ(0) = Θ(α) = 0. So Θ(θ)
satisfies:


 Θ00 + λΘ = 0,
(16)

 Θ(0) = Θ(α) = 0.
nπ 2
For the eigenvalue problem (16), we know that the eigenvalues are λ =
, and the corresponding eigenfunctions
α
nπ are Θ(θ) = C sin
θ , with n ∈ N.
α
nπ nπ
nπ
1
For each n ≥ 1, when λ =
, since R00 (r) + R0 (r) − λR(r) = 0, then R(r) = C1 r α + C2 r− α . Since u(r, θ) is
α
r
nπ
bounded as r → 0, then C2 = 0, that is, R(r) = C1 r α .
In summary, for any n ≥ 1, we can find a non-zero solution to (15):
un (r, θ) = Cr
nπ
α
sin
nπ θ .
α
Let
u(r, θ) =
∞
X
an r
nπ
α
sin
n=1
nπ θ
α
be the solution to (14), it’s easy to see that u(r, 0) = u(r, α) = 0. Since u(a, θ) = f (θ), then
f (θ) =
∞
X
an a
nπ
α
sin
n=1
nπ θ .
α
Then we get
an a
nπ
α
2
=
α
α
Z
f (θ) sin
nπ 0
α
dθ,
for all n ≥ 1.
dθ,
for all n ≥ 1.
That is,
nπ
2a− α
an =
α
Z
α
f (θ) sin
0
nπ α
8
MINGFENG ZHAO
Therefore, the solution to (14) is:
u(r, θ) =
∞
X
an r
nπ
α
sin
n=1
nπ θ ,
α
where
nπ
2a− α
an =
α
Z
α
f (θ) sin
nπ 0
α
dθ,
for all n ≥ 1.
Example 2. Let u0 and u1 be two constants, consider the following problem:








 BC :








(17)
Let w(r, θ) =
1
1
urr + ur + 2 uθθ = 0, 0 < r < a, 0 < θ < α,
r
r
u(r, 0) = u0 , u(r, α) = u1 , 0 < r ≤ a,
u(a, θ) = f (θ),
0 < θ < α,
u(r, θ) is bounded as r → 0.
u1 − u0
θ + u0 , then wr = wrr = wθθ = 0, w(r, 0) = u0 and w(r, α) = u1 . Let v(r, θ) = u(r, θ) − w(r, θ),
α
then v satisfies











1
1
vrr + vr + 2 vθθ = 0, 0 < r < a, 0 < θ < α,
r
r
u1 − u0
θ − u0 ,
BC : v(r, 0) = u(r, α) = 0, v(a, θ) = f (θ) −
α
v(r, θ) is bounded as r → 0.
0 < θ < α,
By Example 1, then
v(r, θ) =
∞
X
n=1
an r
nπ
α
sin
nπ θ ,
α
where
nπ
2a− α
an =
α
Z
α
0
nπ u1 − u0
dθ,
f (θ) −
θ − u0 sin
α
α
for all n ≥ 1.
Therefore, the solution to (17) is given by
∞
nπ X
nπ
u1 − u0
u(r, θ) = w(r, θ) + v(r, θ) =
θ + u0 +
an r α sin
θ ,
α
α
n=1
where
nπ
2a− α
an =
α
Z
0
α
nπ u1 − u0
f (θ) −
θ − u0 sin
dθ,
α
α
for all n ≥ 1.
LECTURE 17: SOLVING THE LAPLACE EQUATIONS II
Example 3. Consider the following








 BC :
(18)








9
problem:
1
1
urr + ur + 2 uθθ = 0, 0 < r < a, 0 < θ < α,
r
r
uθ (r, 0) = uθ (r, α) = 0, 0 < r ≤ a,
u(a, θ) = f (θ),
0 < θ < α,
u(r, θ) is bounded as r → 0.
First, let’s look the following problem:

1
1


urr + ur + 2 uθθ = 0, 0 < r < a, 0 < θ < α,


r
r

(19)
BC
:
u
(r,
0)
=
u
(r,
α) = 0,
θ
θ





u(r, θ) is bounded as r → 0.
Let u(r, θ) = R(r)Θ(θ) be a non-zero separated solution to (19), then
1
1
R00 (r)Θ(θ) + R0 (r)Θ(θ) + 2 R(r)Θ00 (θ) = 0.
r
r
That is, we have
R00 (r) + 1r R0 (r)
Θ00 (θ)
=−
= λ.
R(r)
Θ(θ)
It’s easy to see that λr = λθ = 0, that is, λ is a constant. Since uθ (r, 0) = uθ (r, α) = 0, then Θ0 (0) = Θ0 (α) = 0. So
Θ(θ) satisfies:


 Θ00 + λΘ = 0,
(20)

 Θ0 (0) = Θ0 (α) = 0.
For the eigenvalue problem (20), we know that the eigenvalues are λ =
nπ S
θ , with n ∈ N {0}.
are Θ(θ) = C cos
α
For each n ≥ 0, when λ =
nπ α
nπ 2
α
, and the corresponding eigenfunctions


 C1 + C2 ln r,
1
, since R00 (r) + R0 (r) − λR(r) = 0, then R(r) =

r

Since u(r, θ) is bounded as r → 0, then C2 = 0, that is, R(r) = C1 r
nπ
α
.
In summary, for any n ≥ 0, we can find a non-zero solution to (19):
un (r, θ) = Cr
nπ
α
cos
nπ θ .
α
Let
u(r, θ) =
∞
X
n=0
an r
nπ
α
cos
nπ θ
α
C1 r
nπ
α
+ C2 r−
if n = 0,
.
nπ
α
, if n ≥ 1.
10
MINGFENG ZHAO
be the solution to (18), it’s easy to see that uθ (r, 0) = uθ (r, α) = 0. Since u(a, θ) = f (θ), then
f (θ) =
∞
X
an a
nπ
α
cos
n=1
nπ θ .
α
Then we get
a0 =
1 2
·
2 α
Z
Z
α
α
f (θ) dθ,
and an a
nπ
α
=
0
2
α
Z
α
f (θ) cos
nπ 0
α
dθ,
for all n ≥ 1.
That is,
1
a0 =
α
nπ
f (θ) dθ,
0
2a− α
and an =
α
Z
α
f (θ) cos
nπ α
0
dθ,
for all n ≥ 1.
dθ,
for all n ≥ 1.
Therefore, the solution to (18) is:
u(r, θ) =
∞
X
an r
nπ
α
cos
n=0
nπ θ ,
α
where
a0 =
1
α
Z
nπ
α
f (θ) dθ,
0
and an =
2a− α
α
Z
α
f (θ) cos
0
nπ α
Department of Mathematics, The University of British Columbia, Room 121, 1984 Mathematics Road, Vancouver, B.C.
Canada V6T 1Z2
E-mail address: mingfeng@math.ubc.ca