Prove Lax entropy condition for conservation law with convex flux 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.
 A: 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:
$$
\phi' (u^-) > {\gamma'(t)} > \phi' (u^+) .
$$
