Partial Differential Equations, Fritz John, Page 18, Problem 6 Consider the partial differential equation:
$$\frac{\partial R(u)}{\partial y} + \frac{\partial S(u)}{\partial x}=0$$
where $u$ is a function of $x$ and $y$ and $S'(u)=uR'(u)$. We call a function $u(x,y)$ a weak solution of this equation if it satisfies:
$$\int\int(R(u)\phi_y+S(u)\phi_x)dxdy=0$$
for any function $\phi(x,y)$ of class $C_0^{\infty}$. Then, show that the following jump condition is satisfied:
$$\frac{d\xi}{dy}=\frac{S(u^+)-S(u^-)}{R(u^+)-R(u^-)}$$
where $x=\xi(y)$ is a curve separating the $2$ regions of the $xy$-plane where the solution $u$ is defined.
My instructor told me that I should use integration by parts, but I have no clue where to start. Also, looking at the nature of the equations, I suspect the use of Green's theorem. 
 A: Evans' PDE book (p. 137) provides the following explanation, assuming that the solution $u$ is a classical solution on each side of the jump curve, which we'll call $C$. Pick a test function $\phi$ with compact support $V$ that lies on either side of $C$, defining $V_{l}$ and $V_{r}$ as in the picture:

By the definition of $u$ as a solution,
\begin{align*}
  0=\iint_{\mathbb{R}^{2}} R(u)\phi_{y}+S(u)\phi_{x}dxdy
    = \iint_{V_{l}}R(u)\phi_{y}+S(u)\phi_{x}dxdy + \iint_{V_{r}}R(u)\phi_{y}+S(u)\phi_{x}dxdy.
\end{align*}
Green's Theorem tells us
\begin{align*}
  \iint_{V_{l}}R(u)\phi_{y}+S(u)\phi_{x}dxdy
    & = -\iint_{V_{l}}(R(u)_{y}+S(u)_{x})\phi dxdy + \int_{C} S(u^{-})\nu_{1}+R(u^{-})\nu_{2})\phi dxdy \\
    & = \int_{C}S(u^{-})\nu_{1}+R(u^{-})\nu_{2})\phi dxdy,
\end{align*}
where $(\nu_{1},\nu_{2})$ is the outward unit normal, the double integral term vanishes because $R(u)_{y}+S(u)_{x}=0$, and we only pick up boundary data at $C$ because $\phi=0$ on the rest of the boundary of $V_{l}$. Doing the same with $V_{r}$, adding them together, and using the definition of solution above, we get
\begin{align*}
  \int_{C}[(S(u^{-})-S(u^{+}))\nu_{1}+(R(u^{-})-R(u^{+}))\nu_{2}]\phi dxdy=0
\end{align*}
This is true for all test functions $\phi$, and so
\begin{align*}
  (S(u^{-})-S(u^{+}))\nu_{1}+(R(u^{-})-R(u^{+}))\nu_{2}=0
\end{align*}
on $C$. Returning to the parametrization of $C$ as $x=\xi(y)$, we can write $\nu$ as
$$\nu=\frac{(1,-\dot\xi)}{||(1,-\dot\xi)||}.$$
Plugging this into the above equation gives the desired result.
I didn't use the hypothesis relating $R$ and $S$, so there could well be a problem here.
