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646 CHAPTER 18 The Potential Equation

This can be rearranged to

1 π
u(r,θ) = f (ξ)dξ
2π −π
1 r π
∞ n
+ f (ξ)cos(n(ξ − θ))dξ. (18.2)
π R
n=1 −π
If we combine terms in equation (18.2), we obtain

1 π r
∞ n
u(r,θ) = 1 + 2 cos(n(ξ − θ)) f (ξ)dξ, (18.3)
2π −π R
n=1
which will be used to derive Poisson’s integral formula in the next section.

EXAMPLE 18.2
We will solve the Dirichlet problem for the disk with a radius of 4 about the origin if u(4,θ) =
2
f (θ) = θ . Using equation (18.1), the solution is
1 π
2
u(r,θ) = ξ dξ
2π −π
1 r π π
∞ n
2
2
+ ξ cos(nξ)dξ cos(nθ) + ξ sin(nξ)dξ sin(nθ)
π 4
n=1 −π −π
∞
1 4(−1) n n
r
2
= π + cos(nθ).
3 n 2 4
n=1
EXAMPLE 18.3
We will solve the Dirichlet problem
2
2
2
∇ u(x, y) = 0for x + y < 9
and
2
2
2
2
u(x, y) = x y for x + y = 9.
Convert this problem to polar coordinates. Let U(r,θ) = u(r cos(θ),r sin(θ)). On the boundary,
2
2
2
2
U(3,θ) = 9cos (θ)9sin (θ) = 81sin (θ)cos (θ) = f (θ).
The solution in polar coordinates is
1 π 2 2
U(r,θ) = 81cos (ξ)sin (ξ)dξ
2π −π
1 r π 2 2
∞ n
+ 81cos (ξ)sin (ξ)cos(nξ)dξ cos(nθ)
π 3 −π
n=1
π
2
2
+ 81cos (ξ)sin (ξ)sin(nξ)dξ sin(nθ) .
−π
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October 14, 2010 15:27 THM/NEIL Page-646 27410_18_ch18_p641-666