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664 CHAPTER 18 The Potential Equation

which is a necessary condition for this problem to have a solution. Write the solution
∞
1 1 1 π
2
U(r,θ) = a 0 + r n cos(n(ξ − θ))cos(ξ)sin (ξ)dξ.
2 π n −π
n=1
Evaluate this integral:
π
2
cos(n(ξ − θ))cos(ξ)sin (ξ)dξ =
−π
⎧
⎪0 for n = 2,4,5,6,7,···
⎨
π cos(θ)/4 for n = 1
⎪
−π cos (θ) + 3π cos(θ)/4 for n = 3.
⎩ 2
The solution is
1 1 1 3 3 3
U(r,θ) = a 0 + r cos(θ) + r −cos (θ) + cos(θ)
2 4 3 4
1 1 1 1
3
3
3
= a 0 + r cos(θ) − r cos (θ) + r cos(θ).
2 4 3 4
2
2
To convert this solution to rectangular coordinates, let x =r cos(θ) and r = x + y to obtain
2
1 1 1 1
2
2
3
u(x, y) = a 0 + x − x + (x + y )
2 4 3 4
with a 0 as an arbitrary constant.
18.8.3 A Neumann Problem for the Upper Half-Plane
To illustrate a Neumann problem for an unbounded set, consider:
2
∇ u(x, y) = 0for −∞ < x < ∞, y > 0
and
∂u
(x,0) = f (x) for −∞ < x < ∞.
∂y
Again, notice that ∂u/∂y is the derivative of u normal to the horizontal axis, which is the
boundary of the upper half-plane.
Assume that
∞

f (x)dx = 0
−∞
as a necessary condition for a solution to exist.
We can solve this problem by separation of variables. However, there is an elegant device
for reducing this problem to one we have already solved. Let
∂u
v = .
∂y
Then
2
2
∂ 2 ∂u ∂ 2 ∂u ∂ ∂ u ∂ u
2
∇ v = + = + = 0.
∂x 2 ∂y ∂y 2 ∂y ∂y ∂x 2 ∂y 2

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October 14, 2010 15:27 THM/NEIL Page-664 27410_18_ch18_p641-666