Vector Calculus | Part IA, 2006

In a region RR of R3\mathbb{R}^{3} bounded by a closed surface SS, suppose that ϕ1\phi_{1} and ϕ2\phi_{2} are both solutions of 2ϕ=0\nabla^{2} \phi=0, satisfying boundary conditions on SS given by ϕ=f\phi=f on SS, where ff is a given function. Prove that ϕ1=ϕ2\phi_{1}=\phi_{2}.

In R2\mathbb{R}^{2} show that

ϕ(x,y)=(a1coshλx+a2sinhλx)(b1cosλy+b2sinλy)\phi(x, y)=\left(a_{1} \cosh \lambda x+a_{2} \sinh \lambda x\right)\left(b_{1} \cos \lambda y+b_{2} \sin \lambda y\right)

is a solution of 2ϕ=0\nabla^{2} \phi=0, for any constants a1,a2,b1,b2a_{1}, a_{2}, b_{1}, b_{2} and λ\lambda. Hence, or otherwise, find a solution ϕ(x,y)\phi(x, y) in the region x0x \geqslant 0 and 0ya0 \leqslant y \leqslant a which satisfies:

ϕ(x,0)=0,ϕ(x,a)=0,x0ϕ(0,y)=sinnπya,ϕ(x,y)0 as x,0ya\begin{aligned} &\phi(x, 0)=0, \quad \phi(x, a)=0, \quad x \geqslant 0 \\ &\phi(0, y)=\sin \frac{n \pi y}{a}, \quad \phi(x, y) \rightarrow 0 \quad \text { as } \quad x \rightarrow \infty, \quad 0 \leqslant y \leqslant a \end{aligned}

where aa is a real constant and nn is an integer.

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