If $\mathbf u$ is parallel to $\mathbf B$, write $\mathbf u=\kappa\mathbf B$. Then
$$
\mathbf u(\mathbf u\mathbin\cdot\mathbf B)=u^2\mathbf B,
$$
and the flux from part a reduces to
$$
\mathbf F_c=\mathbf B\left(\frac{u^2}{2}+h+\Phi\right)
=C_B\mathbf B,
$$
where $C_B$ is the gravitational <Bernoulli function>. If it is constant along field lines, $\mathbf B\cdot\nabla C_B=0$, then
$$
\nabla\cdot\mathbf F_c
=\mathbf B\cdot\nabla C_B+C_B\nabla\cdot\mathbf B=0.
$$
The source also vanishes because the flow is homentropic. Hence $\partial_th_c=0$, and every fixed volume satisfies
$$
\boxed{\frac d{dt}\int_V\mathbf u\cdot\mathbf B\,dV=0}.
$$
Thus the <cross-helicity> is conserved even for this time-dependent aligned flow.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 314 / 4 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-314/4/b/solution