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650 CHAPTER 18 The Potential Equation

18.5.1 The Upper Half-Plane
We will solve the problem

2
∇ u(x, y) = 0for −∞ < x < ∞, y > 0
and

u(x,0) = f (x) for −∞ < x < ∞.
This is a Dirichlet problem because the horizontal axis is the boundary of the upper half-plane.
We seek a bounded solution.
Put u(x,t) = X(x)T (t), and obtain

X + λX = 0, T − λT = 0.

The eigenvalues are λ = ω with ω ≥ 0, and the eigenfunctions are
2
X ω (x) = a ω cos(ωx) + b ω sin(ωx).
2

The equation for Y is Y −ω Y =0 with constant multiples of e −ωy as bounded solutions because
y ≥ 0. For each ω ≥ 0, we have a solution
u ω (x, y) = (a ω cos(ωx) + b ω sin(ωx))e −ωy

of Laplace’s equation. To obtain a solution satisfying the boundary condition, use the superposi-
tion
∞

u(x, y) = (a ω cos(ωx) + b ω sin(ωx))e −ωy dω.
0
We need
∞

u(x,0) = f (x) = (a ω cos(ωx) + b ω sin(ωx))dω.
0
The coefficients are the Fourier integral coefficients of f on the real line:
1 ∞ 1 ∞
a ω = f (ξ)cos(ωξ)dξ and b ω = f (ξ)sin(ωξ)dξ.
π π
−∞ −∞
Insert these coefficients into the integral expression for u(x, y):
1 ∞ ∞ −ωy
u(x, y) = [ f (ξ)cos(ωξ)cos(ωx) + f (ξ)sin(ωξ)sin(ωx)]e dξ dω
π 0 −∞
1 ∞ ∞
= cos(ω(ξ − x))e −ωy dω f (ξ)dξ.
π −∞ 0
The inner integral can be evaluated explicitly:
−ωy ∞
∞ e
cos(ω(ξ − x))e −ωy dω = [−y cos(ω(ξ − x)) + (ξ − x)sin(ω(ξ − x))]
2
y + (ξ − x) 2
0 0
y
= .
2
y + (ξ − x) 2
The solution of the Dirichlet problem for the upper half-plane is therefore
y ∞ f (ξ)
u(x, y) = dξ. (18.4)
2
π −∞ y + (ξ − x) 2

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October 14, 2010 15:27 THM/NEIL Page-650 27410_18_ch18_p641-666