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Solved Problems

Z a
a nπb 2 nπx
α = − sech g (x) sin dx
n
1
nπ a a 0 a
∞

X nπx nπy
u(x, y) = β sin cosh
n
a a
n=1
Z a
nπb 2 nπx
β = sech g (x) sin dx
2
n
a a 0 a
Now we solve the problem for v.
v xx + v yy = 0, 0 < x < a, 0 < y < b,

v(0, y) = f (y), v(a, y) = f (y),
2
1
v (x, 0) = 0, v(x, b) = 0
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,

00
2
0
Y = −λ Y, Y (0) = Y (b) = 0,
which has the solutions,

(2n − 1)π (2n − 1)πy
λ = , Y = cos , n ∈ N.
n
n
2b 2b
The equation for X(y) becomes,

2
(2n − 1)π
00
X = X .
n
n
2b
We choose solutions that satisfy the conditions, X(a) = 0 and X(0) = 0, respec-
tively.

(2n − 1)π(a − x) (2n − 1)πx
sinh , sinh
2b 2b
The solution for v(x, y) has the form,

∞
X (2n − 1)πy (2n − 1)π(a − x) (2n − 1)πx
v(x, y)= cos γ sinh +δ sinh .
n
n
2b 2b 2b
n=1
We determine the coefficients from the inhomogeneous boundary conditions.

∞
(2n − 1)πy (2n − 1)πa
X
v(0, y) = γ cos sinh = f (y)
n
1
2b 2b
n=1
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