= Solution
For the <isothermal equation of state>, $p=c_s^2\rho$, with constant <isothermal sound speed> $c_s$. A nonzero steady plane-parallel flow has constant mass flux $\rho u=j$. Choose the positive flow direction for convenience. The <continuity equation> then gives $\rho'/\rho=-u'/u$, and the steady <Euler momentum equation> becomes
$$
uu'=-c_s^2\frac{\rho'}{\rho}-\Phi'\quad\Longrightarrow\quad\left(u-\frac{c_s^2}{u}\right)u'=-\Phi'.
$$
Integrating and absorbing the reference <velocity> into the constant yields, for the <Mach number> $\mathcal M=u/c_s$,
$$
\boxed{\tfrac12\mathcal M^2-\ln\mathcal M+\Phi/c_s^2=C.}
$$
For the opposite flow direction use $\mathcal M=|u|/c_s$; the same relation holds.
A smooth <sonic point> has $u=c_s$, so its momentum equation requires $\Phi'=0$. Moreover $f(\mathcal M)=\mathcal M^2/2-\ln\mathcal M$ satisfies $f'(1)=0$, $f''(1)=2$ and has its minimum $f(1)=1/2$. Thus, if the flow reaches $\mathcal M=1$ at $x_*$,
$$
f(\mathcal M)-\tfrac12=\frac{\Phi(x_*)-\Phi(x)}{c_s^2}\geq0.
$$
A sonic crossing is therefore possible at a maximum of the potential, with
$$
\boxed{C=\tfrac12+\Phi_{\max}/c_s^2.}
$$
At a nondegenerate maximum, the expansions give $(\mathcal M-1)^2=-\Phi''(x_*)(x-x_*)^2/(2c_s^2)+\cdots$. Choosing either smooth sign gives a crossing slope $\mathcal M'(x_*)=\pm\sqrt{-\Phi''(x_*)/(2c_s^2)}$, joining a <subsonic flow> to a <supersonic flow>, or the reverse. This establishes the <plane-parallel isothermal sonic transition at a potential maximum>. For a globally defined crossing, the same maximum must also be high enough for the Bernoulli relation to remain real over the intended domain.
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