= Solution
Integrate each conservative <ideal magnetohydrodynamics> law through a thin pillbox around the stationary shock. Mass conservation gives
$$
[\rho u_x]=0.
$$
The MHD momentum-flux tensor is
$$
\rho\mathbf u\mathbf u
+\left(p+\frac{B^2}{2\mu_0}\right)I
-\frac{\mathbf B\mathbf B}{\mu_0}.
$$
Its normal and tangential components give
$$
\left[\rho u_x^2+\Pi-\frac{B_x^2}{\mu_0}\right]=0,
\qquad
\left[\rho u_xu_y-\frac{B_xB_y}{\mu_0}\right]=0,
$$
where $\Pi=p+B^2/(2\mu_0)$ is total pressure. The solenoidal condition gives $[B_x]=0$, while steady <Faraday's law> gives continuity of tangential electric field,
$$
[-E_z]=[u_xB_y-u_yB_x]=0.
$$
Finally, material energy flux plus the normal <Poynting vector> component gives
$$
\left[\rho u_x\left(\frac12u^2+h\right)
-\frac{E_zB_y}{\mu_0}\right]=0,
$$
where $h$ is <Enthalpy>[specific enthalpy] per unit mass.
Gravity is a bounded volume force, so its integral across a shock whose thickness tends to zero vanishes. Viscosity and resistivity inside the layer may produce entropy; consequently entropy flux need not be equal on both sides, although the second law requires nonnegative net entropy production.
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