For $\gamma=5/3$, the shock data give $\rho_{\rm ps}=4\rho_0$ and $u_{\rm ps}=3V_s/4$. Since $V_s=R/(3t)$, the prescribed <homologous spherical flow> is
$$
u(r,t)=\frac{r}{4t}.
$$
Put $\xi=r/R(t)$ and $\rho=\rho_0D(\xi)$. The spherical <continuity equation> becomes
$$
-\frac{\xi}{3}D'+\frac14(3D+\xi D')=0,
$$
so $D\propto\xi^9$. Matching the post-shock density gives
$$
\rho(r,t)=4\rho_0\xi^9.
$$
The <material acceleration> is
$$
\frac{Du}{Dt}=-\frac{3r}{16t^2}.
$$
The radial <Euler equations for an inviscid fluid> thus give $dp/dr=3\rho r/(16t^2)$. Integrating from the centre to the shock,
$$
p_{\rm ps}-p(0,t)
=\frac{3}{16t^2}\int_0^R\rho r\,dr
=\frac{3\rho_0R^2}{44t^2}.
$$
Since $p_{\rm ps}=\rho_0R^2/(12t^2)$,
$$
\boxed{p(0,t)=\frac{\rho_0R^2}{66t^2}
=\frac{2}{11}p_{\rm ps}}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 314 / 4 / b / Solution

{title2=$D\mathbf u/Dt$}

The material acceleration is the rate of change of velocity following a continuum parcel:
$$
\frac{D\mathbf u}{Dt}
=\frac{\partial\mathbf u}{\partial t}
+(\mathbf u\mathbin\cdot\nabla)\mathbf u.
$$


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