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In this notes, example 5, page 5. It gives an example on how to solve a PDE where characteristics might cross using the Rankine-Hugoniot condition. The example is in an initial-value problem for Burgers' equation $u_t + u u_x = 0$, with piecewise linear initial data $$ u(x,0) = \phi(x) = \left\lbrace\begin{aligned} &1 &&\text{if}\quad x<0\\ &1-x &&\text{if}\quad 0<x<1\\ &0 &&\text{if}\quad 1<x \end{aligned}\right. $$ I have two questions about it:

1). In my opinion, Rankine-Hugoniot condition is just a necessary condition for the discontinuous points of a solution. How does it guarantee the thing we get is indeed a solution?

2).Using the Rankine-Hugoniot condition we can only calculate the speed of the curve of discontinuities. How do we locate its exact formula?(i.e. how to choose one-point on it). Can we choose any arbitrary point where the characteristics cross?

Arctic Char
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z.z
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1 Answers1

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If needed, here is a sketch of the characteristic curves in $x$-$t$ plane, which may help to see better what happens:

characteristics

The solution obtained by the method of characteristics up to the breaking time satisfies $u = \phi (x-ut)$ for $t<1$, i.e. $$ u(x,t) = \left\lbrace\begin{aligned} &1 & &\text{if}\quad x \leq t \\ &\tfrac{1-x}{1-t} & &\text{if}\quad t \leq x\leq 1\\ &0 & &\text{if}\quad x\geq 1 \end{aligned}\right. $$ At the breaking time, the shock speed $s$ is given by the Rankine-Hugoniot condition $s = \tfrac12 (1 + 0)$. Therefore, after the breaking time $t\geq 1$, $$ u(x,t) = \left\lbrace\begin{aligned} &1 & &\text{if}\quad x < 1 +\tfrac12 (t-1) \\ &0 & &\text{if}\quad x > 1 +\tfrac12 (t-1) \end{aligned}\right. $$

To your questions:

  1. The Rankine-Hugoniot condition $[\![\frac{1}{2}u^2]\!] = s [\![u]\!]$ is a condition satisfied by discontinuous weak solutions of the Burgers' equation. A weak solution which satisfies the Rankine-Hugoniot condition is not necessarily a solution to the proposed initial-value problem $u(x,0) = \phi(x)$. To be a solution, it must be in agreement with the characteristics equation until they cross, i.e. the fact that $u$ is constant along the curves $u = \phi(x - ut)$. Moreover, since weak solutions are not unique, it must satisfy an additional entropy condition. In the case of Burgers' equation, the Lax entropy condition amounts to the following statement: "if characteristics cross, then a shock-wave arises; but if they separate, a rarefaction wave occurs".
  2. The location of a discontinuity is deduced from the characteristics equation, whereas its speed is deduced from the Rankine-Hugoniot condition. In the present case, a shock occurs at the time $t=1$ (characteristics intersect). Its speed deduced from the Rankine-Hugoniot condition is $s=\frac{1}{2}$.
EditPiAf
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  • My apologies if I am commenting on an old post, but how did you obtain the bounds $x \leq t$, $t \leq x \leq 1$ and $x \geq 1$ for the solution when $t < 1$? – CBBAM May 22 '21 at 17:56
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    @CBBAM Using $u=\phi(x-ut)$ and the piecewise definition of $\phi$ (see also the figure). Consider for instance the first option $\phi(x-ut)=1$ for $x-ut<0$. We have $u=1$ and $x-t <0$ which gives us the first line, etc. – EditPiAf May 23 '21 at 12:31
  • Thanks! So if I understood correctly, in the scalar case where $u_L > u_R$ or $u_R > u_L$ then we use the theorem which states either we have a rarefaction or a shock and use the corresponding bounds. In this case where the wave is not constant and so there is not an obvious shock we use the method you just described, which is to calculate directly from the original bounds? – CBBAM May 23 '21 at 16:18
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    @CBBAM Yes, that's it! – EditPiAf May 23 '21 at 21:03