For $N$ equal <weak shocks>, $(1+\delta)^N=P_s$, hence
$$
N=\frac{\log P_s}{\log(1+\delta)}=\frac{\log P_s}{\delta}[1+O(\delta)].
$$
Adding the individual increases in <specific entropy> gives the <entropy production in successive weak shocks>
$$
\frac{\Delta s_{\rm weak}}{c_v}
=\frac{\gamma^2-1}{12\gamma^2}\delta^2\log P_s[1+O(\delta)].
$$
By contrast, a single <normal shock wave> with the same total <pressure> ratio gives
$$
\frac{\Delta s_{\rm single}}{c_v}=\log P_s-\gamma\log D(P_s)
=\log P_s-\gamma\log\frac{\gamma+1}{\gamma-1}+O(P_s^{-1}).
$$
Thus \b[a single strong shock produces more entropy] than the chain of sufficiently small <weak shocks>. At fixed $P_s$, the chain approaches zero <entropy production> as $\delta\to0$, while the single-shock result stays positive. The gradual compression approaches reversible <isentropic flow>.

Solution

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

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>.

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 Incoming links: Entropy production in successive weak shocks