Take $B_{y2}=-B_{y1}\ne0$ and choose
$$
A_1=A_2=1.
$$
Then both tangential momentum fluxes vanish, so that jump condition is satisfied. The density relation from part (c) gives $\rho_2=\rho_1$, and mass conservation gives $u_{x2}=u_{x1}$. Alignment makes $u_{y2}=-u_{y1}$, so $u_2^2=u_1^2$.

Because $B_{y2}^2=B_{y1}^2$, normal momentum now gives
$$
\boxed{p_2=p_1}.
$$
The energy condition is then automatic. Thus density and pressure are continuous while the tangential magnetic field and velocity reverse direction. This is a $180^\circ$ <rotational discontinuity in magnetohydrodynamics>, carrying no compression or entropy jump.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 314 / 3 / e / Solution

A rotational discontinuity changes the tangential direction of the magnetic field and velocity while leaving density, pressure, and normal components continuous. The normal flow is Alfvénic.


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 Incoming links: Rotational discontinuity in magnetohydrodynamics