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18.5 Dirichlet Problem for Unbounded Regions 649

In this form, the real part is easily identified, yielding

∞ 2
n 1 −r
1 + 2 r cos(nζ) = .
1 +r − 2r cos(ζ)
2
n=1
Therefore, the solution of the Dirichlet problem for the unit disk is
1 π 1 −r 2
u(r,θ) = f (ξ)dξ.
2
2π −π 1 +r − 2r cos(ξ − θ)
This is Poisson’s integral formula. For a disk of radius R, a change of variables gives us
the solution
2
1 π R −r 2
u(r,θ) = f (ξ)dξ.
2
2
2π −π R +r − 2Rr cos(ξ − θ)

EXAMPLE 18.4
The solution of the problem of Example 18.2 also can be written

1 π 16 −r 2 2
u(r,θ) = ξ dξ
2
2π −π 16 +r − 8r cos(ξ − θ)
16 −r 2 π ξ 2
= dξ
2
2π −π 16 +r − 8r cos(ξ − θ)
for 0 ≤ r < 4,−π ≤ θ ≤ π. This integral solution may be more suitable than the infinite series
solution if we want to approximate values at specific points.

SECTION 18.4 PROBLEMS

−θ
In each of Problems 1 through 4, find an integral formula 4. R = 6, f (θ) = e ;(5.5,3π/5),(4,2π/7),(1,π),
for the solution of the Dirichlet problem. Use a numeri- (4,9π/4)
cal integration routine to approximate u(r,θ) at the given 5. Show that, for 0 ≤r < 1,
points.
1 π 1 −r 2
n
1. R = 1, f (θ) = θ;(1/2,π),(3/4,π/3),(0.2,π/4) r sin(nθ)= sin(nξ)dξ.
2π −π 1 +r − 2r cos(ξ − θ)
2
2. R = 4, f (θ) = sin(4θ);(1,π/6),(3,7π/2),(1,π/4),
(2.5,π/12)
3
n
3. R = 15, f (θ) = θ − θ;(4,π),(12,π/6),(8,π/4), Hint: Notice that r sin(nθ) is harmonic (in polar coor-
(7,π/3) dinates) and use Poisson’s formula.
18.5 Dirichlet Problem for Unbounded Regions

When D is unbounded (has points arbitrarily far from the origin), we may use a Fourier integral
or transform to solve a Dirichlet problem on D.

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October 14, 2010 15:27 THM/NEIL Page-649 27410_18_ch18_p641-666