A conservation law $u_t + \phi(u)_x = 0$ is considered.
For a flux $\phi(u)$ satisfying $\phi'' (u) > 0$, show that the entropy condition in the form: $u(x + a, t) − u(x, t) \leq \frac{aE}{t}$, for some $E > 0$ and all $x, t, a > 0$, implies the inequality, $\phi'(u^- ) > \gamma'(t) > \phi'(u^+ )$, where $u^-$ and $u^+$ are the values of $u$ behind and in front of the shock respectively, and $\gamma'(t)$ is the shock speed.
So we have $u(x+a,t)-u(x,t)=(\phi')^{-1}(\frac{x+a}{t})-(\phi')^{-1}(\frac{x}{t})$. I don't know how to proceed. Any help would be appreciated.
Let us consider that $u$ has a single discontinuity with jump $u^+-u^-$ at $x=\gamma(t)$. Since $u$ is nonsmooth, we go back to the integral definition of the conservation law $$ \frac{\text d}{\text d t} \int_{x_1}^{x_2} u\,\text d x = \phi (u|_{x=x_1})- \phi (u|_{x=x_2}) . $$ If $x_1<\gamma(t)<x_2$, the identity $$ \frac{\text d}{\text d t} \int_{x_1}^{x_2} u\,\text d x = \int_{x_1}^{\gamma(t)} u_t\,\text d x + \int_{\gamma(t)}^{x_2} u_t\,\text d x + \gamma'(t)\left(u^--u^+\right) $$ and the conservation law over $[x_1, \gamma (t)]$, $[\gamma (t), x_2]$ yield the Rankine-Hugoniot condition. Now, using the strict convexity (since $\phi'' > 0$) of the flux $\phi$ over $[\min\lbrace u^-, u^+\rbrace, \max\lbrace u^-, u^+\rbrace]$, we have the inequalities $$ \phi' (\min\lbrace u^-, u^+\rbrace) < \overbrace{\frac{\phi (u^+) - \phi (u^-)}{u^+-u^-}}^{\gamma'(t)} < \phi' (\max\lbrace u^-, u^+\rbrace) . $$ Taking the limit $\epsilon\to 0^+$ in the entropy condition $$ u|_{x=\gamma (t)+\epsilon} - u|_{x=\gamma (t)-\epsilon} \leq \frac {E\epsilon}{t} $$ imposes $u^+ < u^-$. Hence, the Lax entropy condition is obtained: $$ \phi' (u^-) > {\gamma'(t)} > \phi' (u^+) . $$