Paper 3, Section II, B

Vector Calculus | Part IA, 2019

(a) The function uu satisfies 2u=0\nabla^{2} u=0 in the volume VV and u=0u=0 on SS, the surface bounding VV.

Show that u=0u=0 everywhere in VV.

The function vv satisfies 2v=0\nabla^{2} v=0 in VV and vv is specified on SS. Show that for all functions ww such that w=vw=v on SS

VvwdV=Vv2dV\int_{V} \nabla v \cdot \nabla w d V=\int_{V}|\nabla v|^{2} d V

Hence show that

Vw2dV=V{v2+(wv)2}dVVv2dV\int_{V}|\boldsymbol{\nabla} w|^{2} d V=\int_{V}\left\{|\boldsymbol{\nabla} v|^{2}+|\boldsymbol{\nabla}(w-v)|^{2}\right\} d V \geqslant \int_{V}|\boldsymbol{\nabla} v|^{2} d V

(b) The function ϕ\phi satisfies 2ϕ=ρ(x)\nabla^{2} \phi=\rho(\mathbf{x}) in the spherical region x<a|\mathbf{x}|<a, with ϕ=0\phi=0 on x=a|\mathbf{x}|=a. The function ρ(x)\rho(\mathbf{x}) is spherically symmetric, i.e. ρ(x)=ρ(x)=ρ(r)\rho(\mathbf{x})=\rho(|\mathbf{x}|)=\rho(r).

Suppose that the equation and boundary conditions are satisfied by a spherically symmetric function Φ(r)\Phi(r). Show that

4πr2Φ(r)=4π0rs2ρ(s)ds4 \pi r^{2} \Phi^{\prime}(r)=4 \pi \int_{0}^{r} s^{2} \rho(s) d s

Hence find the function Φ(r)\Phi(r) when ρ(r)\rho(r) is given by ρ(r)={ρ0 if 0rb0 if b<ra\rho(r)=\left\{\begin{array}{ll}\rho_{0} & \text { if } 0 \leqslant r \leqslant b \\ 0 & \text { if } b<r \leqslant a\end{array}\right., with ρ0\rho_{0} constant.

Explain how the results obtained in part (a) of the question imply that Φ(r)\Phi(r) is the only solution of 2ϕ=ρ(r)\nabla^{2} \phi=\rho(r) which satisfies the specified boundary condition on x=a|\mathbf{x}|=a.

Use your solution and the results obtained in part (a) of the question to show that, for any function ww such that w=1w=1 on r=br=b and w=0w=0 on r=ar=a,

U(b,a)w2dV4πabab\int_{U(b, a)}|\nabla w|^{2} d V \geqslant \frac{4 \pi a b}{a-b}

where U(b,a)U(b, a) is the region b<r<ab<r<a.

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