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For a modified Burgers' equation $u_t +uu_x = -u^2$ with initial conditions $u(0,x) = 1,$ for $x<0$; $u(0,x) = 0$, for $x>0$. I can get the solution by using the method of characteristics. But in the end, I want to know whether the shock will happen in this case, if so, can we still use Rankine-Hugoniot to describe the interface between two states?

EditPiAf
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jean
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    The way I was taught, Rankine-Hugoniot is effectively a physical constraint on the problem: at the outset of a shock, the weak solution to an evolution PDE becomes non-unique, and Rankine-Hugoniot is a way of "selecting" the "physical" solution. So you should see whether the "physics" make sense in this setting. – Ian Apr 30 '17 at 21:52
  • @jean : Well, you get the solution using the method of characteristics. Editing what you get would enable to check if your preliminary result is correct. – JJacquelin May 01 '17 at 08:34
  • @Ian The Rankine-Hugoniot condition has nothing to do with physics, it simply comes from the notion of weak solution and does not "select" a weak solution. The entropy condition comes from physical considerations and selects a unique weak solution. The Rankine-Hugoniot condition can hold when entropy condition fails, for instance. – Jeff May 02 '17 at 03:40

1 Answers1

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  1. Hasn't the shock already happened? You are prescribing initial data that exhibits a discontinuity. What more do you need?

  2. I perfectly agree with Ian's comment in that the ultimate arbiter of whether the Rankine-Hugoniot conditions should be applied is in the physics. Furthermore, your equation is not obviously a conservation law so the interpretation would be a bit tricky. But at least mathematically there doesn't seem to be any particular difficulty in specifying the jump condition and the entropy condition as you would normally.

    For the jump condition: Your equation can be rewritten in integral form $$ \partial_t \int_a^b u~\mathrm{d}x + \frac12 u^2 \Big|_a^b + \int_a^b u^2 ~\mathrm{d}x = 0 $$ Supposing you have the shock front travelling along the line $y(t)$, with the solution being classical to the left and right of it. You have that $$ \partial_t \int_a^b u ~\mathrm{d}x = \int_a^y \partial_t u ~\mathrm{d}x + \int_y^b \partial_t u ~\mathrm{d}x + \dot{y} (u_l - u_r) $$ where $u_l = \lim_{x \to y^-} u(t,x)$ and $u_r = \lim_{x\to y^+} u(t,x)$. Now, since $\int_a^b u^2 = \int_a^y u^2 + \int_y^b u^2$ even when $u$ has a jump discontinuity, you arrive at the same jump condition that $$ \dot{y} = \frac12 (u_l + u_r) \tag{jump}$$ as the Burgers' equation.

    The entropy condition can also be imposed since the principal part of the equation is the same as Burgers.

    The only caveat in studying shocks for this particular equation is that along characteristic curves the solution satisfy $u' = - u^2$, and if the initial data is such that $u(0,x)$ is negative somewhere, one will have to deal with the situation where non-shock-type singularities occur.

Willie Wong
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