Paper 4, Section II, 17C

Methods | Part IB, 2018

Let Ω\Omega be a bounded region in the plane, with smooth boundary ∂Ω\partial \Omega. Green's second identity states that for any smooth functions u,vu, v on Ω\Omega

∫Ω(u∇2v−v∇2u)dx dy=∮∂Ωu(n⋅∇v)−v(n⋅∇u)ds\int_{\Omega}\left(u \nabla^{2} v-v \nabla^{2} u\right) \mathrm{d} x \mathrm{~d} y=\oint_{\partial \Omega} u(\mathbf{n} \cdot \nabla v)-v(\mathbf{n} \cdot \nabla u) \mathrm{d} s

where n\mathbf{n} is the outward pointing normal to ∂Ω\partial \Omega. Using this identity with vv replaced by

G0(x;x0)=12πln⁡(∥x−x0∥)=14πln⁡((x−x0)2+(y−y0)2)G_{0}\left(\mathbf{x} ; \mathbf{x}_{0}\right)=\frac{1}{2 \pi} \ln \left(\left\|\mathbf{x}-\mathbf{x}_{0}\right\|\right)=\frac{1}{4 \pi} \ln \left(\left(x-x_{0}\right)^{2}+\left(y-y_{0}\right)^{2}\right)

and taking care of the singular point (x,y)=(x0,y0)(x, y)=\left(x_{0}, y_{0}\right), show that if uu solves the Poisson equation ∇2u=−ρ\nabla^{2} u=-\rho then

u(x)=−∫ΩG0(x;x0)ρ(x0)dx0 dy0+∮∂Ω(u(x0)n⋅∇G0(x;x0)−G0(x;x0)n⋅∇u(x0))ds\begin{aligned} u(\mathbf{x})=-\int_{\Omega} G_{0}\left(\mathbf{x} ; \mathbf{x}_{0}\right) \rho\left(\mathbf{x}_{0}\right) \mathrm{d} x_{0} \mathrm{~d} y_{0} \\ &+\oint_{\partial \Omega}\left(u\left(\mathbf{x}_{0}\right) \mathbf{n} \cdot \nabla G_{0}\left(\mathbf{x} ; \mathbf{x}_{0}\right)-G_{0}\left(\mathbf{x} ; \mathbf{x}_{0}\right) \mathbf{n} \cdot \nabla u\left(\mathbf{x}_{0}\right)\right) \mathrm{d} s \end{aligned}

at any x=(x,y)∈Ω\mathbf{x}=(x, y) \in \Omega, where all derivatives are taken with respect to x0=(x0,y0)\mathbf{x}_{0}=\left(x_{0}, y_{0}\right).

In the case that Ω\Omega is the unit disc ∥x∥⩽1\|\mathbf{x}\| \leqslant 1, use the method of images to show that the solution to Laplace's equation ∇2u=0\nabla^{2} u=0 inside Ω\Omega, subject to the boundary condition

u(1,θ)=δ(θ−α),u(1, \theta)=\delta(\theta-\alpha),

is

u(r,θ)=12π1−r21+r2−2rcos⁡(θ−α)u(r, \theta)=\frac{1}{2 \pi} \frac{1-r^{2}}{1+r^{2}-2 r \cos (\theta-\alpha)}

where (r,θ)(r, \theta) are polar coordinates in the disc and α\alpha is a constant.

[Hint: The image of a point x0∈Ω\mathbf{x}_{0} \in \Omega is the point y0=x0/∥x0∥2\mathbf{y}_{0}=\mathbf{x}_{0} /\left\|\mathbf{x}_{0}\right\|^{2}, and then

∥x−x0∥=∥x0∥∥x−y0∥\left\|\mathbf{x}-\mathbf{x}_{0}\right\|=\left\|\mathbf{x}_{0}\right\|\left\|\mathbf{x}-\mathbf{y}_{0}\right\|

for all x∈∂Ω.]\mathbf{x} \in \partial \Omega .]

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