Paper 3, Section II, A

Vector Calculus | Part IA, 2007

The function ϕ(x,y,z)\phi(x, y, z) satisfies 2ϕ=0\nabla^{2} \phi=0 in VV and ϕ=0\phi=0 on SS, where VV is a region of R3\mathbb{R}^{3} which is bounded by the surface SS. Prove that ϕ=0\phi=0 everywhere in VV.

Deduce that there is at most one function ψ(x,y,z)\psi(x, y, z) satisfying 2ψ=ρ\nabla^{2} \psi=\rho in VV and ψ=f\psi=f on SS, where ρ(x,y,z)\rho(x, y, z) and f(x,y,z)f(x, y, z) are given functions.

Given that the function ψ=ψ(r)\psi=\psi(r) depends only on the radial coordinate r=xr=|\mathbf{x}|, use Cartesian coordinates to show that

ψ=1rdψdrx,2ψ=1rd2(rψ)dr2\nabla \psi=\frac{1}{r} \frac{d \psi}{d r} \mathbf{x}, \quad \nabla^{2} \psi=\frac{1}{r} \frac{d^{2}(r \psi)}{d r^{2}}

Find the general solution in this radial case for 2ψ=c\nabla^{2} \psi=c where cc is a constant.

Find solutions ψ(r)\psi(r) for a solid sphere of radius r=2r=2 with a central cavity of radius r=1r=1 in the following three regions:

(i) 0r10 \leqslant r \leqslant 1 where 2ψ=0\nabla^{2} \psi=0 and ψ(1)=1\psi(1)=1 and ψ\psi bounded as r0r \rightarrow 0;

(ii) 1r21 \leqslant r \leqslant 2 where 2ψ=1\nabla^{2} \psi=1 and ψ(1)=ψ(2)=1\psi(1)=\psi(2)=1;

(iii) r2r \geqslant 2 where 2ψ=0\nabla^{2} \psi=0 and ψ(2)=1\psi(2)=1 and ψ0\psi \rightarrow 0 as rr \rightarrow \infty.

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