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For $1 < p < 2$, let $u_p$ be a unique solution to

$u_t + \left(\frac{1}{p}|u|^p\right)_x = 0$,

for initial condition

$u_0(x) = \begin{cases} 1 \quad x > 0, \\ 0 \quad x < 0. \end{cases}$

I now want to determine the solution $u(t, x) = \lim \limits_{p \to 1} u_p(t, x)$ of the limit problem

$u_t + (|u|)_x = 0$

with same same initial condition. My problem is that I don't really know how to get started with this or which ansatz to use. I tried using the Lax-Oleinik formula given by the Evans, but got no results.

EditPiAf
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Hamilton
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1 Answers1

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Given that the initial data is non-negative, let's assume that $u$ is non-negative. Therefore, we are left with $$ u_t + u_x = 0 ,\qquad u(x,0) = u_0(x) $$ where $u_0$ is the step function. The solution $u = u_0(x-t)$ is indeed non-negative, therefore this solution is the one obtained for $p=1$. Note that this solution is a contact discontinuity $$ u(x,t) = \left\lbrace \begin{aligned} 1 \qquad x> st,\\ 0 \qquad x< st. \end{aligned} \right. $$ with speed $s=1$.

EditPiAf
  • 20,898
  • Thank you for your answer!

    Do you perhaps also have an idea how to get that solution as a limit of the solutions u_p?

     – Hamilton Jun 19 '20 at 19:41
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    @Hamilton The solution is likely to be a discontinuity for other values of $p$ too. According to Rankine-Hugoniot, the shock speed is likely to be dependent on $p$ (see this post where the theory is presented). You may need to use the weak form of the PDE – EditPiAf Jun 22 '20 at 08:29