Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2024 / Paper 1 / 1

For a continuously differentiable vector field $\mathbf{u}$ on a bounded region $V\subset {\mathbb{R}}^3$ with piecewise smooth closed boundary $S=\partial V$, the divergence theorem statesHere $\nabla \cdot \mathbf{u}$ is its divergence, $dV$ is the volume element, $dS$ is the surface area element, $\hat{\mathbf{n}}$ is the outward unit normal, and $d\mathbf{S}=\hat{\mathbf{n}}dS$. Thus a volume integral of the divergence equals the outward flux through the boundary.

Since $\mathbf{a}$ is constant, the product rule gives $\nabla \cdot (\mathbf{a}\psi )=\mathbf{a}\cdot \nabla \psi$. Applying the divergence theorem therefore givesThis holds for every constant vector $\mathbf{a}$, so each component of the difference vanishes. Consequently the gradient has the boundary representation

The divergence of a cross product and $\nabla \times \mathbf{a}=0$ give $\nabla \cdot (\mathbf{a}\times \mathbf{A})=-\mathbf{a}\cdot (\nabla \times \mathbf{A})$. On the boundary the scalar triple product gives $(\mathbf{a}\times \mathbf{A})\cdot \hat{\mathbf{n}}=-\mathbf{a}\cdot (\hat{\mathbf{n}}\times \mathbf{A})$. Thus the divergence theorem givesBecause $\mathbf{a}$ is arbitrary, the curl identity follows:

Use Einstein summation convention and write ${\partial }_j={\nabla }_j$. Since ${\partial }_j{r}_i={\delta }_{ij}$, where ${\delta }_{ij}$ is the Kronecker delta, the product rule givesThis proves $\boxed{P_i={\partial }_j(r_i{w}_j)-w_i}$. Now take $w_j={\partial }_j\psi$. Integrating this identity and applying the divergence theorem to the first term givesPart (b) changes the last volume integral of the gradient into a surface integral, yieldingHere $d{S}_i={\hat{n}}_{i}dS$; the Laplacian is ${\nabla }^2\psi ={\partial }_j{\partial }_j\psi$.

Apply the product rule to the tensor field, without assuming any symmetry of $\Sigma$:Integrating, the divergence theorem converts each of the two divergences into a boundary term:The repeated indices are summed according to Einstein summation convention, while $i$ labels the component of the resulting vector identity.

Choose the symmetric tensor fieldThe product rule and equality of mixed partial derivatives giveThis is the Euclidean scalar-gradient stress identity. Substitute it into part (e), and use ${\Sigma }_{ki}{\hat{n}}_k=({\partial }_n\psi ){\partial }_i\psi -\frac{1}{2}|\nabla \psi {|}^2{\hat{n}}_i$, where ${\partial }_n\psi =\hat{\mathbf{n}}\cdot \nabla \psi$. Reassembling the components gives the required pure surface integral:Sufficient smoothness of $\psi$ is understood for the displayed derivatives and applications of the divergence theorem.