follows: $$\frac{V_A^2}{C_s^2}=\frac{2}{\gamma \beta_{\infty}}$$ where $\beta_{\infty}=B_{x\infty}^2/p_{\infty}$ is the free stream plasma $\beta$. This then gives: $$\label{Mach_eqn}
\frac{1}{\gamma-1}\frac{p}{\rho}+\frac{1}{2}M_{\infty}^2(v^2+w^2)+\frac{B_x^2}{\rho}\frac{2}{\gamma \beta_{\infty}}=\frac{1}{\gamma-1}+\frac{1}{2}M_{\infty}^2+\frac{2}{\gamma \beta_{\infty}}.$$

As we have assumed that the flow is irrotational, we can assume that the flow is constant entropy. For a perfect gas (in normalised form) this can be expressed as: $p/\rho^{\gamma}=1$. We can also use the identity that $B_x^2=p/\beta'$, where $\beta' \beta_{\infty}$ defines the plasma $\beta$ of system. Using these identities, equation \[Mach\_eqn\] becomes $$\rho^{\gamma-1}+\frac{\gamma-1}{2}M_{\infty}^2(v^2+w^2 -1)+\frac{\rho^{\gamma-1}}{\beta'}\frac{2(\gamma-1)}{\gamma \beta_{\infty}}=1+\frac{2(\gamma-1)}{\gamma \beta_{\infty}}$$ To tidy up the equations, we define: $$\begin{aligned}
D(y,z)=1+\frac{1}{\beta'}\frac{2(\gamma-1)}{\gamma \beta_{\infty}}\\
C=1+\frac{2(\gamma-1)}{\gamma \beta_{\infty}}\end{aligned}$$ giving: $$\rho^{\gamma-1}+\frac{\gamma-1}{2}\frac{1}{D(y,z)}M_{\infty}^2(v^2+w^2 -1)=\frac{C}{D(y,z)}$$

For simplicity, we will now make the assumption that the plasma $\beta$ is approximately uniform. In reality, we would have to solve for this using the induction equation, but for simplicity this is neglected in this analysis. This leads to the following equation: $$\label{reduced_en}
\rho^{\gamma-1}+\frac{\gamma-1}{2}\frac{1}{D}M_{\infty}^2(v^2+w^2 -1)=1.$$ Now it is possible to combine equation \[reduced\_en\] with equation \[mass\_cons\] to determine an equation for the velocity field.

If we follow the derivation by @VD1975, by transforming to cylindrical coordinates (where the length-scale is normalised by the radius of the cylinder) we can then calculate the velocity potential $\phi$ (with all $\partial / \partial x=0$): $$\label{firstMstar}
\phi=\left(r+\frac{1}{r} \right)\cos \theta - {M_*}^2\left[ \left(\frac{13}{12}\frac{1}{r}-\frac{1}{2}\frac{1}{r^3}+\frac{1}{12}\frac{1}{r^5} 