= Solution
Since $Df/Dt=0$, the density ansatz has
$$
\frac1\rho\frac{D\rho}{Dt}=\frac{\dot\rho_0}{\rho_0}.
$$
The velocity divergence is
$$
\nabla\mathbin\cdot\mathbf u=A+B+C
=\frac d{dt}\log(abc).
$$
The <mass conservation> equation therefore gives
$$
\boxed{\rho_0\propto(abc)^{-1}}.
$$
Similarly,
$$
\frac1p\frac{Dp}{Dt}
=\frac{\dot\rho_0}{\rho_0}+\frac{\dot T}{T}
=-\gamma\nabla\mathbin\cdot\mathbf u,
$$
and hence
$$
\boxed{T\propto(abc)^{-(\gamma-1)}}.
$$
The $x$ component of the material acceleration is
$$
\frac{Du_x}{Dt}=(\dot A+A^2)x=\frac{\ddot a}{a}x.
$$
Because $d\widehat p/df=\widehat\rho$,
$$
-\frac1\rho\partial_xp
=-\frac1{\rho_0\widehat\rho}
\rho_0T\widehat\rho\left(-\frac{2x}{a^2}\right)
=\frac{2Tx}{a^2}.
$$
Combining this with $-\partial_x\Phi=-\Omega^2x$ gives
$$
\boxed{\ddot a+\Omega^2a=\frac{2T}{a}}.
$$
The $y$ and $z$ components give the analogous equations for $b$ and $c$. Finally, $\widehat p(0)=0$ makes $p=0$ on the material free surface, so both dynamic and <kinematic boundary condition> requirements are satisfied.
Solved by gpt-5.6-sol high.
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