Local Helmholtz chart and partial-Legendre derivative identities

Put $f(v,s)=e_s(v,s)$ and $a_0=e_{ss}(v_0,s_0)>0$. Since $U$ is open and $e_{ss},e_{sv}$ are continuous, choose $\delta ,\rho >0$ so the closed rectangle $R=[v_0-\delta ,v_0+\delta ]\times [s_0-\rho ,s_0+\rho ]$ is contained in $U$, $f_s=e_{ss}\ge a_0\mathrm{/}2=:a>0$ on $R$, and $\mathrm{\mid }{f}_v\mathrm{\mid }=\mathrm{\mid }{e}_{sv}\mathrm{\mid }\le B$ on $R$ for some finite $B$. By [F2] on the two vertical segments, $f(v_0,s_0-\rho )\le T_0-a\rho <T_0<T_0+a\rho \le f(v_0,s_0+\rho )$. Continuity of the endpoint values in $v$ therefore gives smaller open intervals $V$ about $v_0$ and $J$ about $T_0$, with $J\subset (0,\mathrm{\infty })$, such that $f(v,s_0-\rho )<T<f(v,s_0+\rho )$ for all $(v,T)\in V\times J$.

For fixed $v\in V$, [F2] and $f_s\ge a$ imply that $s\mapsto f(v,s)$ is strictly increasing on $[s_0-\rho ,s_0+\rho ]$. By [F3], for every $T\in J$ there is $s(v,T)\in (s_0-\rho ,s_0+\rho )$ with $f(v,s(v,T))=T$; strict increase gives uniqueness in that interval.

Take $(v,T),(v^{\mathrm{\prime }},T^{\mathrm{\prime }})\in V\times J$ and write $s=s(v,T)$, $s^{\mathrm{\prime }}=s(v^{\mathrm{\prime }},T^{\mathrm{\prime }})$. Applying [F2] in the $s$ variable and using $f_s\ge a$ gives $a\mathrm{\mid }{s}^{\mathrm{\prime }}-s\mathrm{\mid }\le \mathrm{\mid }f(v,s^{\mathrm{\prime }})-f(v,s)\mathrm{\mid }=\mathrm{\mid }f(v,s^{\mathrm{\prime }})-T\mathrm{\mid }$. Since $f(v^{\mathrm{\prime }},s^{\mathrm{\prime }})=T^{\mathrm{\prime }}$, [F2] in the $v$ variable and $\mathrm{\mid }{f}_v\mathrm{\mid }\le B$ give $\mathrm{\mid }f(v,s^{\mathrm{\prime }})-T\mathrm{\mid }\le B\mathrm{\mid }v-v^{\mathrm{\prime }}\mathrm{\mid }+\mathrm{\mid }T-T^{\mathrm{\prime }}\mathrm{\mid }$. Hence $s(v,T)$ is locally Lipschitz and, in particular, continuous.

At a fixed $(v,T)$, let $h,k$ be increments in $v,T$ and set $\mathrm{\Delta }s=s(v+h,T+k)-s(v,T)$. The bound of step 3.1 gives $\mathrm{\Delta }s=O(\mathrm{\mid }h\mathrm{\mid }+\mathrm{\mid }k\mathrm{\mid })$. By the definition of the total derivative of the $C^1$ function $f$ at $(v,s(v,T))$, $k=f_{v}h+f_s\mathrm{\Delta }s+o(\mathrm{\mid }h\mathrm{\mid }+\mathrm{\mid }\mathrm{\Delta }s\mathrm{\mid })$, so $\mathrm{\Delta }s=-\frac{f_v}{f_s}h+\frac{1}{f_s}k+o(\mathrm{\mid }h\mathrm{\mid }+\mathrm{\mid }k\mathrm{\mid })$. Thus $s_v=-e_{sv}\mathrm{/}{e}_{ss}$ and $s_T=1\mathrm{/}{e}_{ss}$. These derivatives are continuous because $e$ has continuous total derivatives through order two, $s$ is continuous, and $e_{ss}\ge a>0$; therefore the unique local branch is continuously differentiable.

By [F4] and the identity $e_s(v,s(v,T))=T$, differentiation of $a=e(v,s(v,T))-Ts(v,T)$ gives $a_v=e_v+(e_s-T)s_v=e_v=-p$ and $a_T=e_s{s}_T-s-T{s}_T=-s$.

Differentiating the first-derivative identities of step 5.1 and using step 4.1 yields $a_{TT}=-s_T=-1\mathrm{/}{e}_{ss}$ and $a_{vv}={\mathrm{\partial }}_v(e_v(v,s(v,T)))=e_{vv}+e_{vs}{s}_v=e_{vv}-e_{vs}{e}_{sv}\mathrm{/}{e}_{ss}$. These derivatives are continuous by the stated regularity, so $a$ is $C^2$. All derivatives are evaluated at the chart point $(v,s(v,T))$.