I am studying conservation laws, i was doing the proof for the rankine hugoniot condition, in the middle of the proof they have used divergence theorem , but I do not understand how they have used that. There is no triple integral work, so how can they use divergence theorem ?
I have added the picture of the proof and highlited the part where the author has used divergence theorem.
Edit : now that I got the solution I don't understand this step. How to get this step
The trick is to realise $v_t$ and $v_x$ as the divergence of certain functions. Let $\nabla = (\partial_x, \partial_t)$ be the differential operator. Then \begin{align*} \iint_{\Omega^-} uv_t\, dxdt & = \iint_{\Omega^-}u\nabla\cdot(v,0)\, dxdt \\ & = \int_{\partial\Omega^-} u(v,0)\cdot\underline{n}\, ds - \iint_{\Omega^-} (\nabla u)\cdot(v,0)\, dxdt. \end{align*} Now, $\partial\Omega^-$ consists of the $x$-axis and the curve $x=\xi(t)$, but there is no contribution of the first integral over the $x$-axis since $v(x,0) = 0$ by assumption. With $\underline{n} = (\nu_1,\nu_2)$, we then have \begin{align*} \iint_{\Omega^-} uv_t\, dxdt & = \int_{x=\xi(t)} u^-(v,0)\cdot(\nu_1,\nu_2)\, ds - \iint_{\Omega^-} (u_t, u_x)\cdot(v,0)\, dxdt \\ & = \int_{x=\xi(t)} u^-v\nu_1\, ds - \iint_{\Omega^-} u_tv\, dxdt. \end{align*} You can apply the same argument to the other integral $$ \iint_{\Omega^-} f(u)v_x\, dxdt $$ and I will leave this as an exercise for you.