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10.5 Poisson’s formula

A much more interesting case is the Dirichlet problem for a circle. The rotational invariance
of ∆ provides a hint that the circle is a natural shape for harmonic functions.
Let’s consider the problem

2
2
2
u xx + u yy = 0 for x + y < a , (10.24)
2
2
2
u = h(θ) for x + y = a . (10.25)
with radius a and any boundary data h(θ).
Our method, naturally, is to separate variables in polar coordinates: u = R(r)Θ(θ).
We can write
1 1 1 1
00
0
00
0 = u xx + u yy = u rr + u r + u θθ = R Θ + R Θ + RΘ .
r r 2 r r 2
2
Dividing by RΘ and multiplying by r , we find that
Θ” + λΘ = 0 (10.26)
0
2
00
r R + rR − λR = 0. (10.27)
These are ordinary differential equations, easily solved. What boundary conditions do
we associate with them?
For Θ(θ) we naturally require periodic BCs:

Θ(θ + 2π) = Θ(θ) for − ∞ < θ < ∞.

2
Thus λ = n and Θ(θ) = A cos nθ + B sin nθ (n = 1, 2, . . .) There is also the solution
λ = 0 with Θ(θ) = A.
The equation for R is also easy to solve because it is of the Euler type with solutions
2
α
of the form R(r) = r . Since λ = n it reduces to
α
2 α
α
α(α − 1)r + αr − n r = 0
n
whence α = ±n. Thus R(r) = Cr + Dr −n and we have the separated solutions
D
n
u = (Cr + )(A cos nθ + B sin nθ) (10.28)
r n
for n = 1, 2, 3, . . . In case n = 0, we need a second linearly independent solution of
(10.27) (besides R = constant). It is R = log r, as one learns in ODE courses. So we
also have the solutions
u = C + D log r. (10.29)

All of the solutions (10.28) and (10.29) we have found are harmonic functions in the disk
D, except that half of them are infinite at the origin (r = 0). But we have not yet used
any boundary condition at all in the r variable. The interval is 0 < r < a.At r = 0 some
of the solutions (r − n and log r) are infinite: We reject them. The requirement that

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