Jump discontinuity for Burgers equation $u_{y} +uu_{x} =0$ 
Let $u$ be a $C^{1}$-solution of $u_{y} +uu_{x} =0$ in each of two regions separated by a curve $x =\xi(y)$. Let $u$ be continuous, but $u_{x}$ have a discontinuity on the curve. Prove that $\frac{d\xi}{dy} = u$ and hence the curve is a characteristic.

I expressed $u_{y} +uu_{x}=0$ as $(u_{y}^{+} -u_{y}^{-} +u(u_{x}^{+}-u_{x}^{-})=0$ but am unsure where to go from there. 

The Rankine-Hugoniot condition reads
$$
\frac{\text d \xi}{\text d y} = \frac12(u^++u^-)
$$
where $u^\pm$ are the values of $u$ on each side of the curve $x=\xi(y)$. Since $u$ is continuous across the curve, we have $u^\pm = u|_{x=\xi}$, which ends the proof (recall that characteristics satisfy the Lagrange-Charpit equation $\text d x/u =\text d y /1$). See also this thread.
