Ideal-gas acoustic wave curves have the stated pressure response

For $0<p\le p_K$, put $ϵ=-1$ for the 1-family and $ϵ=1$ for the 3-family. For every density $\rho$ between ${\rho }_K$ and ${\rho }_{\ast }:={\rho }_K(p\mathrm{/}{p}_K{)}^{1\mathrm{/}\gamma }$, define $p(\rho )=p_K(\rho \mathrm{/}{\rho }_K{)}^{\gamma }$, $c(\rho )=c_K(\rho \mathrm{/}{\rho }_K{)}^{(\gamma -1)\mathrm{/}2}$, $u(\rho )=u_K+2ϵ(c(\rho )-c_K)\mathrm{/}(\gamma -1)$, and $E(\rho )=p(\rho )\mathrm{/}(\gamma -1)+\rho u(\rho {)}^2\mathrm{/}2$. This path stays in the admissible Euler domain because $\rho ,p(\rho )>0$. Differentiating gives $p^{\mathrm{\prime }}=c^2$, $u^{\mathrm{\prime }}=ϵc\mathrm{/}\rho$, and, in conserved coordinates, $d(\rho ,\rho u,E)\mathrm{/}d\rho =(1,u+ϵc,H+ϵuc)=r_{ϵ}$ by [F2]. Thus it is the acoustic integral curve, extended along this positive-density path; uniqueness of the local integral curve identifies it with the local rarefaction branch wherever that branch is defined. At $\rho ={\rho }_{\ast }$, this gives $c_{\ast }=c_K(p\mathrm{/}{p}_K{)}^{(\gamma -1)\mathrm{/}(2\gamma )}$ and $u_{\ast }-u_K=2ϵ(c_{\ast }-c_K)\mathrm{/}(\gamma -1)$, which equals $-f_K(p)$ for the 1-family and $f_K(p)$ for the 3-family by [F3].

For a compressive shock from side state $K$ to star state $\ast$ with trial pressure $p>p_K$ and speed $s$, set $w=u-s$, $v=1\mathrm{/}\rho$, $e=pv\mathrm{/}(\gamma -1)$, $j=\rho w$, and $E^{\mathrm{\prime }}=E-sm+\frac{1}{2}{s}^2\rho =\rho e+\frac{1}{2}\rho w^2$. Write $[g]=g_{\ast }-g_K$, $F_{\rho }=\rho u$, $F_m=\rho u^2+p$, $F_E=u(E+p)$, and $F_{E^{\mathrm{\prime }}}=F_E-s{F}_m+\frac{1}{2}{s}^2{F}_{\rho }$; for a right-going 3-shock, reversing the physical state order changes both sides of each RH equation by the same sign. Mass RH makes $j$ equal on both sides; momentum RH minus $s$ times mass RH makes $p+j^{2}v$ equal. Since $p\mathrm{\ne }{p}_K$, this identity forces $j\mathrm{\ne }0$. The linear combination of energy, momentum, and mass RH gives $[F_{E^{\mathrm{\prime }}}]=s[E^{\mathrm{\prime }}]$. Substitution gives $F_{E^{\mathrm{\prime }}}-s{E}^{\mathrm{\prime }}=u(E+p)-s(\rho u^2+p)+\frac{1}{2}{s}^2\rho u-s(E-sm+\frac{1}{2}{s}^2\rho )=(u-s)(E^{\mathrm{\prime }}+p)=w(E^{\mathrm{\prime }}+p)$, so $[w(E^{\mathrm{\prime }}+p)]=0$. Since $w=jv$, division by $j$ gives equality of $e+\frac{1}{2}{w}^2+pv$ on both sides. Using $w=jv$, the energy equality reads $e_{\ast }-e_K+\frac{1}{2}{j}^2(v_{\ast }^2-v_K^2)+p{v}_{\ast }-p_K{v}_K=0$. From $j^2(v_{\ast }-v_K)=-(p-p_K)$, the kinetic term is $-\frac{1}{2}(p-p_K)(v_{\ast }+v_K)$; combining it with the pressure terms gives $e_{\ast }-e_K+\frac{1}{2}(p+p_K)(v_{\ast }-v_K)=0$.

Let $p>p_K$ be the trial pressure, set $q=p\mathrm{/}{p}_K>1$, and use $\beta =(\gamma -1)\mathrm{/}(\gamma +1)$. Substituting $e=pv\mathrm{/}(\gamma -1)$ into step 1.2 and dividing by $p_K{v}_K$ gives $(qx-1)\mathrm{/}(\gamma -1)+(q+1)(x-1)\mathrm{/}2=0$, where $x=v_{\ast }\mathrm{/}{v}_K$. Solving this linear equation gives $x=(\beta q+1)\mathrm{/}(q+\beta )$ and ${\rho }_{\ast }\mathrm{/}{\rho }_K=1\mathrm{/}x=(q+\beta )\mathrm{/}(\beta q+1)$. Hence $1-x=(1-\beta )(q-1)\mathrm{/}(q+\beta )$, and using $q=p\mathrm{/}{p}_K$, $B_K=\beta p_K$, and $A_K=(1-\beta )\mathrm{/}{\rho }_K$ gives $v_K-v_{\ast }=(p-p_K)A_K\mathrm{/}(p+B_K)$. Define $V:=v_K\sqrt{(p-p_K)\mathrm{/}(v_K-v_{\ast })}>0$. Since $1-x=(v_K-v_{\ast })\mathrm{/}{v}_K$, $V(1-x)=\sqrt{(p-p_K)(v_K-v_{\ast })}=(p-p_K)\sqrt{A_K\mathrm{/}(p+B_K)}=f_K(p)$. For a left 1-shock choose $s=u_K-V$ and $j={\rho }_{K}V$; mass RH gives $u_{\ast }=u_K-V(1-x)$. For a right 3-shock oriented from the star state to $K$, choose $s=u_K+V$ and $j=-{\rho }_{K}V$; mass RH gives $u_{\ast }=u_K+V(1-x)$. These choices establish the response signs.

Put $q=p\mathrm{/}{p}_K$ and $x=v_{\ast }\mathrm{/}{v}_K=(\beta q+1)\mathrm{/}(q+\beta )$, so $0<x<1$. The relative shock speeds on the two sides have magnitudes $V=\mathrm{\mid }j\mathrm{\mid }{v}_K$ and $W=\mathrm{\mid }j\mathrm{\mid }{v}_{\ast }=xV$. Steps 1.2 and 2.1 give $V^2\mathrm{/}{c}_K^2=(q+\beta )\mathrm{/}(\gamma (1-\beta ))>1$ and $W^2\mathrm{/}{c}_{\ast }^2=(\beta q+1)\mathrm{/}(\gamma q(1-\beta ))<1$. The first inequality follows from $q+\beta >1+\beta =2\gamma \mathrm{/}(\gamma +1)$, and the second from $\beta q+1<\gamma q(1-\beta )$, equivalent to $q>1$. For a 1-shock, $s=u_{\ast }-W<u_{\ast }$ and $W<c_{\ast }$ give ${\lambda }_{-}(u_{\ast })<s$; $V>c_K$ gives $s<{\lambda }_{-}(u_K)$. For a 3-shock, $s=u_{\ast }+W>u_{\ast }$ and $W<c_{\ast }$ give $s<{\lambda }_{+}(u_{\ast })$; $V>c_K$ gives ${\lambda }_{+}(u_K)<s$. The other-family Lax inequalities follow from ${\lambda }_{-}<{\lambda }_0<{\lambda }_{+}$ and $s<u_{\ast }$ for a 1-shock or $s>u_{\ast }$ for a 3-shock.

Equality of the left and right star velocities is $u_L-f_L(p)=u_R+f_R(p)$, which rearranges to $\mathrm{\Phi }(p)=0$. At such a positive pressure the two Euler states have common $u_{\ast },p$ and differ only in density. From [F1], $F(U)-u_{\ast }U=(0,p,u_{\ast }p)$ is independent of density, so $F(U_R^{\ast })-F(U_L^{\ast })=u_{\ast }(U_R^{\ast }-U_L^{\ast })$. By [F2], $u_{\ast }$ is the middle characteristic speed on both sides, so this is a linearly-degenerate contact.