# Paper 3, Section II, C

(a) For smooth scalar fields $u$ and $v$, derive the identity

$\nabla \cdot(u \nabla v-v \nabla u)=u \nabla^{2} v-v \nabla^{2} u$

and deduce that

\begin{aligned} \int_{\rho \leqslant|\mathbf{x}| \leqslant r}\left(v \nabla^{2} u-u \nabla^{2} v\right) d V=\int_{|\mathbf{x}|=r}\left(v \frac{\partial u}{\partial n}-u \frac{\partial v}{\partial n}\right) d S \\ &-\int_{|\mathbf{x}|=\rho}\left(v \frac{\partial u}{\partial n}-u \frac{\partial v}{\partial n}\right) d S \end{aligned}

Here $\nabla^{2}$ is the Laplacian, $\frac{\partial}{\partial n}=\mathbf{n} \cdot \nabla$ where $\mathbf{n}$ is the unit outward normal, and $d S$ is the scalar area element.

(b) Give the expression for $(\nabla \times \mathbf{V})_{i}$ in terms of $\epsilon_{i j k}$. Hence show that

$\nabla \times(\nabla \times \mathbf{V})=\nabla(\nabla \cdot \mathbf{V})-\nabla^{2} \mathbf{V}$

(c) Assume that if $\nabla^{2} \varphi=-\rho$, where $\varphi(\mathbf{x})=O\left(|\mathbf{x}|^{-1}\right)$ and $\nabla \varphi(\mathbf{x})=O\left(|\mathbf{x}|^{-2}\right)$ as $|\mathbf{x}| \rightarrow \infty$, then

$\varphi(\mathbf{x})=\int_{\mathbb{R}^{3}} \frac{\rho(\mathbf{y})}{4 \pi|\mathbf{x}-\mathbf{y}|} d V .$

The vector fields $\mathbf{B}$ and $\mathbf{J}$ satisfy

$\nabla \times \mathbf{B}=\mathbf{J}$

Show that $\nabla \cdot \mathbf{J}=0$. In the case that $\mathbf{B}=\nabla \times \mathbf{A}$, with $\nabla \cdot \mathbf{A}=0$, show that

$\mathbf{A}(\mathbf{x})=\int_{\mathbb{R}^{3}} \frac{\mathbf{J}(\mathbf{y})}{4 \pi|\mathbf{x}-\mathbf{y}|} d V$

and hence that

$\mathbf{B}(\mathbf{x})=\int_{\mathbb{R}^{3}} \frac{\mathbf{J}(\mathbf{y}) \times(\mathbf{x}-\mathbf{y})}{4 \pi|\mathbf{x}-\mathbf{y}|^{3}} d V$

Verify that $\mathbf{A}$ given by $(*)$ does indeed satisfy $\nabla \cdot \mathbf{A}=0$. [It may be useful to make a change of variables in the right hand side of $(*)$.]