Paper 3, Section I, B

Vector Calculus | Part IA, 2021

(a) Prove that

×(ψA)=ψ×A+ψ×A(A×B)=B×AA×B\begin{aligned} &\nabla \times(\psi \mathbf{A})=\psi \mathbf{\nabla} \times \mathbf{A}+\boldsymbol{\nabla} \psi \times \mathbf{A} \\ &\nabla \cdot(\mathbf{A} \times \mathbf{B})=\mathbf{B} \cdot \boldsymbol{\nabla} \times \mathbf{A}-\mathbf{A} \cdot \boldsymbol{\nabla} \times \mathbf{B} \end{aligned}

where A\mathbf{A} and B\mathbf{B} are differentiable vector fields and ψ\psi is a differentiable scalar field.

(b) Find the solution of 2u=16r2\nabla^{2} u=16 r^{2} on the two-dimensional domain D\mathcal{D} when

(i) D\mathcal{D} is the unit disc 0r10 \leqslant r \leqslant 1, and u=1u=1 on r=1r=1;

(ii) D\mathcal{D} is the annulus 1r21 \leqslant r \leqslant 2, and u=1u=1 on both r=1r=1 and r=2r=2.

[Hint: the Laplacian in plane polar coordinates is:

2u=1rr(rur)+1r22uθ2.]\left.\nabla^{2} u=\frac{1}{r} \frac{\partial}{\partial r}\left(r \frac{\partial u}{\partial r}\right)+\frac{1}{r^{2}} \frac{\partial^{2} u}{\partial \theta^{2}} . \quad\right]

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