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Solved Problems

w = g 2 (x) u = g 2 (x) v = 0

f 1 (y) f 2 (y) 0 0 f 1 (y) f 2 (y)
=
= ∅w = 0 = = = u ∅u = 0 u + = ∅v = 0 =
w w v v
w y = g 1 (x) w y = g 1 (x) v y = 0

Figure 10.1: Decomposition of the problem.

u (x, 0) = g (x), u(x, b) = g (x)
2
1
y
We substitute the separation of variables u(x, y) = X(x)Y (y) into Laplace’s equa-
tion.
X 00 Y 00 2
= − = −λ
X Y
We have the eigenvalue problem,

2
00
X = −λ X, X(0) = X(a) = 0,
which has the solutions,

nπ nπx
λ = , X = sin , n ∈ N.
n
n
a a
The equation for Y (y) becomes,

nπ
2
00
Y = Y ,
n
n
a
which has the solutions,
n o n nπy nπy o
e nπy/a , e −nπy/a or cosh , sinh .
a a

It will be convenient to choose solutions that satisfy the conditions, Y (b) = 0 and
0
Y (0) = 0, respectively.

nπ(b − y) nπy
sinh , cosh
a a

The solution for u(x, y) has the form,

∞

X nπx nπ(b − y) nπy
u(x, y) = sin α sinh + β cosh .
n
n
a a a
n=1
We determine the coefficients from the inhomogeneous boundary conditions. (Here
we see how our choice of solutions for Y (y) is convenient.)

∞
X nπ nπx nπb
u (x, 0) = − α sin cosh = g (x)
n
1
y
a a a
n=1
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