Consider the burgers equation:
$$u_t + ( \frac{u^2}{2} )_x =0 $$
And the function
$$u(t,x) = \begin{cases} \frac{x}{t} & \hbox{if } |x| < \sqrt{t} \\ 0 & \hbox{if } |x| > \sqrt{t} \end{cases}$$
It has two discontinuity curves given by $\gamma_1 (s) : [0,+\infty[ \to (\sqrt{s},s)$ and $\gamma_2 (s) : [0,+\infty[ \to (-\sqrt{s},s)$ For any $s>0$ the trace of $u$ at the left of $\gamma_1(s)$ is $u(s,\gamma_1(s)^-) = \frac{1}{\sqrt{s}}$, the trace at the right is $u(s,\gamma_1(s)^+) = 0$. The characteristic speed of the state $\frac{1}{\sqrt{s}}$ is $\frac{1}{\sqrt{s}}$, the one of the state $0$ is $0$, the speed of the discontinuity is $\sigma(\gamma_1(s))=\frac{1}{2\sqrt{s}}$, thus the Oleinik conditions are respected: $$ 0 < \frac{1}{2\sqrt{s}} < \frac{1}{\sqrt{s}}$$ The Rankine-Hugoniot conditions are also respected :
$$\frac{f(u(s,\gamma_1(s)^-)) - f(u(s,\gamma_1(s)^+))}{u(s,\gamma_1(s)^-)- u(s,\gamma_1(s)^+)} = \frac{1}{2\sqrt{s}}=\sigma(\gamma_1(s))$$
Thus one might think that this is an entropy solution to the Burgers' equation, nonetheless, the entropy solution for this equation with initial data $u_0(x)=0$ is the constant $0$. Can someone tell me what is the problem ?
