For $N$ equal <weak shocks> with individual <pressure> ratio $1+\delta$ and total ratio $P_s=(1+\delta)^N$, <entropy production in a perfect-gas shock> gives
$$
\frac{\Delta s}{c_v}
=\frac{\gamma^2-1}{12\gamma^2}\delta^2\log P_s\,[1+O(\delta)].
$$
At fixed $P_s>1$, this tends to zero as the compression is divided into increasingly many smaller <weak shocks>. A single finite <normal shock wave> instead gives the positive value $\log P_s-\gamma\log D(P_s)$. Gradual compression can consequently approach reversible <isentropic flow> while an abrupt compression generates finite <entropy>.

Entropy production in successive weak shocks

For a <perfect gas> with constant <specific heat capacity> $c_v$, the <first law of thermodynamics> gives
$$
ds=c_v\,d\log\left(\frac p{\rho^\gamma}\right).
$$
Using the <pressure-density Hugoniot relation for a perfect gas>, the <entropy production in a perfect-gas shock> is therefore
$$
\boxed{\frac{[s]}{c_v}=\log P-\gamma\log\left[\frac{(\gamma+1)P+\gamma-1}{(\gamma-1)P+\gamma+1}\right].}
$$
This is positive for $P>1$, as required by the <Second law of thermodynamics> for the physical compressive <normal shock wave>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 314 / 2 / c / Solution

Put $S(P)=[s]/c_v$. Differentiating the expression in part (c) gives
$$
S'(P)=\frac{(\gamma^2-1)(P-1)^2}{P[(\gamma+1)P+\gamma-1][(\gamma-1)P+\gamma+1]}.
$$
For $P=1+\delta$, the denominator is $4\gamma^2+O(\delta)$, so
$$
S'(1+\delta)=\frac{\gamma^2-1}{4\gamma^2}\delta^2+O(\delta^3).
$$
Integrating from the unshocked state, where $S(1)=0$, proves the cubic <entropy production in a perfect-gas shock>:
$$
\boxed{\frac{[s]}{c_v}=\frac{\gamma^2-1}{12\gamma^2}\delta^3+O(\delta^4).}
$$
The first-order and second-order terms vanish: a <weak shock> agrees with reversible compression through second order.

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 314 / 2 / d / Solution

For the <pressure-density Hugoniot relation for a perfect gas>, the change in <specific entropy> is $[s]/c_v=\log P-\gamma\log D(P)$, with $c_v$ the <specific heat capacity> at constant volume. Its derivative is
$$
\frac{d([s]/c_v)}{dP}
=\frac{(\gamma^2-1)(P-1)^2}{P[(\gamma+1)P+\gamma-1][(\gamma-1)P+\gamma+1]}.
$$
It is positive for a compressive <normal shock wave> with $P>1$. For a <weak shock> with $P=1+\delta$, integrating the leading term gives
$$
\frac{[s]}{c_v}=\frac{\gamma^2-1}{12\gamma^2}\delta^3+O(\delta^4).
$$
Thus the <entropy production> is cubic in the small <pressure> jump, even though the <mass density> and <temperature> changes already appear at first order.

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