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Solved Problems

The general solutions of this equation are

(
c + c z, for n = 0,
1
2
u (z) =
n
nz
c e +c e −nz for n 6= 0.
1
2
The bounded solutions are
 −nz
c e , for n > 0,

u (z) = c, for n = 0, = c e −|n|z .
n

c e , for n < 0,
 nz
We substitute the series into the initial condition at z = 0 to determine the multi-
plicative constants.

∞
X inθ
u(θ, 0) = u (0) e = f(θ)
n
n=−∞
Z 2π
1
u (0) = f(θ) e −inθ dθ ≡ f n
n
2π 0
Thus the solution is
∞
X
u(θ, z) = f e inθ −|n|z .
e
n
n=−∞
Note that
1 Z 2π
u(θ, z) → f = f(θ) dθ
0
2π 0
as z → +∞.

Exercise 205. Solve Laplace’s equation in a rectangle.

w xx + w yy = 0, 0 < x < a, 0 < y < b,
w(0, y) = f (y), w(a, y) = f (y),
2
1
w (x, 0) = g (x), w(x, b) = g (x)
2
y
1
Proceed by considering w = u + v where u and v are harmonic and satisfy

u(0, y) = u(a, y) = 0, u (x, 0) = g (x), u(x, b) = g (x),
2
y
1
v(0, y) = f (y), v(a, y) = f (y), v (x, 0) = v(x, b) = 0.
2
y
1
Solution: The decomposition of the problem is shown in Figure 10.1.
First we solve the problem for u.
u xx + u yy = 0, 0 < x < a, 0 < y < b,
u(0, y) = u(a, y) = 0,

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