gaiaecologicaq2

2022-07-08

Laplace's equation on the plane: how much boundary data must be specified to guarantee existence and uniqueness?
My question stemmed from a specific problem, so let's jump right in. I want to solve Laplace's equation in plane polars (r,θ) on the domain (r,θ)∈[1,∞)×[0,2π) subject to boundary conditions
u(1,θ)=f(θ),limr→∞u(r,θ)=0,
where f is a known function on [0,2π). It can be shown (e.g. via separation of variables) that the general solution for u that is bounded at infinity is given by
u(r,θ)=a0+∑n=1∞r−n(ancosnθ+bnsinnθ),
which we can match against the Fourier coefficients of f to find an and bn. However we see that the solution only strictly decays at infinity if a0=0. This is a restriction on the allowed boundary data f(θ) in order for the solution to exist.
So my question comes in two parts:
Is there a way to get around this? I've been reading a solution in which they write the constant a0 as a Fourier series expansion of a square wave defined on the domain [0,4π). However surely this throws up half-angles that cannot be matched with u.
Assuming the answer to the above question is no, can someone provide an argument as to why the constant seen at r=1 has to be the same as the constant at infinity? It seems to me that in the above formulation we have specified too much boundary data and therefore the solution doesn't exist. But why??Thanks in advance!

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Miguidi4y

Expert

Let u a (real) harmonic function on the exterior of a disc of radius k centered at the origin and continuous on the boundary circle.
By Green's theorem, one gets that there is a constant B (not depending on R) given by 2πB=∫2π0R∂u∂r(R,θ)dθ,R>k, which immediately implies that the harmonic function u(r,θ)−Blogr has a harmonic conjugate, so it is the real part of an analytic function on the disc exterior.
This implies that ∫2π0u(R,θ)dθ=A+2πBlogR,R>k
(if 2u=2Rf=f+f¯ the Laurent series of f gives that ∫2π0u(R,θ)dθ is constant in R since only the free term in the Laurent series has non zero integral on a circle and then apply the above to u−Blogr).
All the above is general stuff about harmonic function on an annulus or disc exterior; applied to the OP problem, we get that A=B=0 by taking R→∞ and using that limr→∞u(r,θ)=0; but now taking R→1 we get the neceesary condition that ∫2π0f(θ)dθ=0. The condition is also sufficient as noted in the OP.

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