Let Ω be a bounded region in the plane, with smooth boundary ∂Ω. Green's second identity states that for any smooth functions u,v on Ω
∫Ω(u∇2v−v∇2u)dx dy=∮∂Ωu(n⋅∇v)−v(n⋅∇u)ds
where n is the outward pointing normal to ∂Ω. Using this identity with v replaced by
G0(x;x0)=2π1ln(∥x−x0∥)=4π1ln((x−x0)2+(y−y0)2)
and taking care of the singular point (x,y)=(x0,y0), show that if u solves the Poisson equation ∇2u=−ρ then
u(x)=−∫ΩG0(x;x0)ρ(x0)dx0 dy0+∮∂Ω(u(x0)n⋅∇G0(x;x0)−G0(x;x0)n⋅∇u(x0))ds
at any x=(x,y)∈Ω, where all derivatives are taken with respect to x0=(x0,y0).
In the case that Ω is the unit disc ∥x∥⩽1, use the method of images to show that the solution to Laplace's equation ∇2u=0 inside Ω, subject to the boundary condition
u(1,θ)=δ(θ−α),
is
u(r,θ)=2π11+r2−2rcos(θ−α)1−r2
where (r,θ) are polar coordinates in the disc and α is a constant.
[Hint: The image of a point x0∈Ω is the point y0=x0/∥x0∥2, and then
∥x−x0∥=∥x0∥∥x−y0∥
for all x∈∂Ω.]