Given the Burgers equation $$u_{y}+uu_x=0$$ with $y>0$. Suppose $u$ is continuous for all $y>0$ and $u_x$ has a jump discontinuity on the smooth curve $x=\xi(y)$ while $u$ is $C^1$ on either side of $\xi$.
Show that $u$ is a weak solution iff on either side of $\xi$ u satisfies the PDE pointwise, and $\frac{\mathrm{d}\xi}{\mathrm{d}y}(y)=u(\xi(y),y)$.
My approach: The "only if" part: Clearly $u$ satisfies the PDE in either side of $\xi$, the equation for $\xi$ may be derived by the Rankine-Hugeniot condition with the fact that $u$ is continuous: $$\frac{\mathrm{d}\xi}{\mathrm{d}y}=\frac{\frac{(u^+)^2}{2}-\frac{(u^-)2}{2}}{u^+-u^-}=\frac{u^{+}+u^-}{2}=u(\xi(y),y)$$, in which $u^+:=u(\xi(y)^+,y)$ and $u^-:=u(\xi(y)^-,y)$
Conversely, for the "if" part let's suppose above equation for $\xi$ is satisfied. We need to show $$\frac{\mathrm{d}}{\mathrm{d}y}\int_{a}^{b}u(x,y)\mathrm{d}x+\frac{u^2(b,y)}{2}-\frac{u^2(a,y)}{2}=0, ~~\forall a<b, y>0$$ There is really only one case that needs a proof: $a<\xi(y)<b$. Since $\frac{\mathrm{d}}{\mathrm{d}y}\int_{a}^{b}u(x,y)\mathrm{d}x=\frac{\mathrm{d}}{\mathrm{d}y}\int_{a}^{\xi(y)}u(x,y)\mathrm{d}x+\frac{\mathrm{d}}{\mathrm{d}y}\int_{\xi(y)}^{b}u(x,y)\mathrm{d}x=(u^--u^+)\xi'(y)+\int_{a}^{b}u_y\mathrm{d}x$
and both $\int_{a}^{b}u_y\mathrm{d}x$ and $\frac{u^2(b,y)}{2}-\frac{u^2(a,y)}{2}$ vanish as $a\to \xi(y)^-, b\to \xi(y)^+$, moreover, $=(u^--u^+)\xi'(y)=0$ due to $u^-=u^+$ (continuity of $u$), therefore the integration identity is valid in this case. Thus, $u$ is a weak solution.
My question: Please notice that in the "if" part I did not use the condition $\frac{\mathrm{d}\xi}{\mathrm{d}y}(y)=u(\xi(y),y)$ as there seems no need to use it since $u^--u^+=0$ implies $(u^--u^+)\xi'(y)=0$ for any $\xi$, is my proof correct?