Strict convexity and entropy variables for the ideal-gas Euler flux

Set $\epsilon =E-\mathrm{\mid }m{\mathrm{\mid }}^2\mathrm{/}(2\rho )$ and $u=m\mathrm{/}\rho$. Rewriting the entropy gives $\eta =-c_v\rho \log ((\gamma -1)\epsilon )+c_v\gamma \rho \log \rho -s_0\rho$. For an increment $\delta U=(a,b,c)$, its first internal-energy variation is $\delta \epsilon =c-u\cdot b+\mathrm{\mid }u{\mathrm{\mid }}^{2}a\mathrm{/}2$, and its second directional variation is $D^2\epsilon [\delta U,\delta U]=-\mathrm{\mid }b-ua{\mathrm{\mid }}^2\mathrm{/}\rho$. The chain rule therefore yields $D^2\eta [\delta U,\delta U]=c_v(\frac{\gamma }{\rho }{a}^2-\frac{2}{\epsilon }a\delta \epsilon +\frac{\rho }{{\epsilon }^2}(\delta \epsilon {)}^2+\frac{1}{\epsilon }\mathrm{\mid }b-ua{\mathrm{\mid }}^2)$. The first three terms have matrix $\left(\begin{array}{cc}\gamma \mathrm{/}\rho & -1\mathrm{/}\epsilon \\ -1\mathrm{/}\epsilon & \rho \mathrm{/}{\epsilon }^2\end{array}\right)$, whose leading diagonal entry is positive and determinant is $(\gamma -1)\mathrm{/}{\epsilon }^2>0$. The last term is nonnegative. If the full quadratic form is zero, then $a=\delta \epsilon =0$ and $b=0$; the formula for $\delta \epsilon$ then gives $c=0$. Thus the Hessian is positive definite.

Differentiating the same expression for $\eta$ gives $\beta =c_v\rho \mathrm{/}\epsilon =1\mathrm{/}T$ and $v=(c_v\gamma -s-\beta \mathrm{\mid }u{\mathrm{\mid }}^2\mathrm{/}2,\beta u,-\beta )$. Positive definiteness from step 1.1 makes $D_{U}v=D_U^2\eta$ invertible, so the inverse function theorem gives a smooth local inverse near each state. Since $\mathcal{G}$ is convex by [F2], integrating the positive Hessian along the segment between any two distinct states also gives strict convexity of $\eta$ on $\mathcal{G}$.

Fix $j$ and an arbitrary variation $dU=(d\rho ,dm,dE)$, and write $d{u}_j=(d{m}_j-u_{j}d\rho )\mathrm{/}\rho$. Differentiating the flux in [F1] gives $d{F}_j=(u_{j}d\rho +\rho d{u}_j,u_{j}dm+md{u}_j+e_{j}dp,u_j(dE+dp)+(E+p)d{u}_j)$. With $v_m=\beta u$ and $v_E=-\beta$ from step 2.1, the $dp$ terms in $v^{T}d{F}_j-u_j{v}^{T}dU$ cancel, and the coefficient of $d{u}_j$ is $\rho v_{\rho }+\beta \rho \mathrm{\mid }u{\mathrm{\mid }}^2-\beta (E+p)=-\rho s=\eta$, using $E=\epsilon +\rho \mathrm{\mid }u{\mathrm{\mid }}^2\mathrm{/}2$, $p=(\gamma -1)\epsilon$, and $\beta \epsilon =c_v\rho$. Thus $v^{T}d{F}_j=u_j{v}^{T}dU+\eta d{u}_j=d{q}_j$ for every variation, proving ${\mathrm{\nabla }}_U{q}_j=v^{T}D{F}_j$ directly.

Direct substitution of $v$, $F_j$, and $q_j$ gives $v^T{F}_j-q_j=(\rho (c_v\gamma -s-\beta \mathrm{\mid }u{\mathrm{\mid }}^2\mathrm{/}2)+\beta \rho \mathrm{\mid }u{\mathrm{\mid }}^2-\beta E+\rho s)u_j$. Using $E=\epsilon +\rho \mathrm{\mid }u{\mathrm{\mid }}^2\mathrm{/}2$ and $\beta \epsilon =c_v\rho$, the coefficient simplifies to $c_v(\gamma -1)\rho =R\rho$. This proves ${\psi }_j=\rho R{u}_j$ and completes the stated entropy-pair and potential identities.