Local shock curve by an augmented Rankine–Hugoniot equation

On a neighborhood where [F3] supplies the $C^2$ right-eigenvector frame, let $R(u)$ have columns $r_j(u)$ and let ${\mathrm{\ell }}_i(u{)}^T$ be row $i$ of $R(u{)}^{-1}$. Then $R^{-1}A=\mathrm{diag}({\lambda }_j)R^{-1}$, so ${\mathrm{\ell }}_i^T{r}_j={\delta }_{ij}$ and ${\mathrm{\ell }}_i$ is a left eigenvector. For $v$ near $r_i(\bar{u})$ define $G(u_0,\sigma ,v,s)=({\int }_0^{1}Df(u_0+t\sigma v)vdt-sv,{\mathrm{\ell }}_i(u_0)\cdot v-1)$. Since $Df$ is $C^2$, this $G$ is $C^1$ in all variables and vanishes at $(\bar{u},0,r_i(\bar{u}),{\lambda }_i(\bar{u}))$.

The derivative of $G$ in $(v,s)$ at that point sends $(w,\alpha )$ to $((A(\bar{u})-{\lambda }_{i}I)w-\alpha r_i(\bar{u}),{\mathrm{\ell }}_i(\bar{u})\cdot w)$. If it vanishes, pairing the first component with ${\mathrm{\ell }}_i(\bar{u})$ gives $\alpha =0$; simplicity then makes $w$ a multiple of $r_i(\bar{u})$, and the last component forces $w=0$. Hence this square derivative is invertible.

Apply [F4] with parameter $(u_0,\sigma )$ and unknowns $(v,s)$. It gives unique $C^1$ functions $v_i(\sigma ,u_0),s_i(\sigma ,u_0)$ on a product neighborhood. Set $S_i(\sigma ;u_0)=u_0+\sigma v_i(\sigma ,u_0)$. For $\sigma \mathrm{\ne }0$, multiplying the first equation by $\sigma$ and using the fundamental theorem of calculus gives $f(S_i(\sigma ;u_0))-f(u_0)=s_i(\sigma ,u_0)(S_i(\sigma ;u_0)-u_0)$. At $\sigma =0$, the unique normalized eigenvector is $v_i(0,u_0)=r_i(u_0)$; hence ${\mathrm{\partial }}_{\sigma }{S}_i(0;u_0)=r_i(u_0)$.

At any fixed $u_0\in N_0$, differentiate the first component of $G=0$ at $\sigma =0$. It gives $(A(u_0)-{\lambda }_{i}I){\mathrm{\partial }}_{\sigma }{v}_i(0,u_0)+\frac{1}{2}{D}^{2}f(u_0)[r_i,r_i]-{\mathrm{\partial }}_{\sigma }{s}_i(0,u_0)r_i=0$. Pairing with ${\mathrm{\ell }}_i(u_0)$ yields ${\mathrm{\partial }}_{\sigma }{s}_i(0,u_0)=\frac{1}{2}{\mathrm{\ell }}_i(u_0)\cdot D^{2}f(u_0)[r_i,r_i]=\frac{1}{2}D{\lambda }_i(u_0)r_i(u_0)$; the last equality follows by differentiating $A{r}_i={\lambda }_i{r}_i$ in direction $r_i$ and pairing with ${\mathrm{\ell }}_i$.

Shrink the base neighborhood to an open ball $N_0$ whose closure $K$ lies inside the IFT parameter neighborhood, and choose $\eta >0$ so $K\times [-\eta ,\eta ]$ lies there. By continuity and the nonzero value of $d_i(\bar{u})$, shrink $K$ so $\mathrm{\mid }{d}_i(u_0)\mathrm{\mid }$ and all eigenvalue gaps have positive lower bounds on $K$. The $C^1$ functions $(u_0,\sigma )\mapsto {\lambda }_i(S_i(\sigma ;u_0))$ and $s_i(\sigma ,u_0)$ have uniformly continuous $\sigma$ derivatives on the compact product $K\times [-\eta ,\eta ]$ by [F5]. Their first-order expansions at $\sigma =0$ therefore have uniform remainders over $u_0\in K$: ${\lambda }_i(S_i(\sigma ;u_0))={\lambda }_i(u_0)+\sigma d_i(u_0)+o(\mathrm{\mid }\sigma \mathrm{\mid }),s_i(\sigma ,u_0)={\lambda }_i(u_0)+\frac{1}{2}\sigma d_i(u_0)+o(\mathrm{\mid }\sigma \mathrm{\mid })\mathrm{.}$ For $\sigma d_i(u_0)<0$ and sufficiently small uniform $\mathrm{\mid }\sigma \mathrm{\mid }$, this gives ${\lambda }_i(S_i(\sigma ;u_0))<s_i(\sigma ,u_0)<{\lambda }_i(u_0)$. The other-family inequalities in [F2] follow by continuity from the positive gaps, after shrinking $N_0$ and $\eta$.