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.

  • $\begingroup$ Aren't you forgetting some hypotheses, like a relation between $R$ and $S$? $\endgroup$ – Justpassingby Jan 16 '16 at 0:38
  • $\begingroup$ @Justpassingby I don't think so. The first PDE is all I have. $\endgroup$ – MathManiac Jan 16 '16 at 6:29
  • $\begingroup$ In the reference that you quote in the title, go back one or two pages and look at where $R$ and $S$ first show up. $\endgroup$ – Justpassingby Jan 16 '16 at 10:47
  • $\begingroup$ @Justpassingby Oh right. I'll update the problem. Sorry for the mistake! $\endgroup$ – MathManiac Jan 16 '16 at 20:06

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:

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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.

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  • $\begingroup$ Why does $\phi=0$ on rest of the boundary apart from $C$? $\endgroup$ – MathManiac Jan 17 '16 at 4:47
  • $\begingroup$ Because $V$ is the support of $\phi$. $\endgroup$ – Mark Perlman Jan 17 '16 at 9:04

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