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First of all, I saw this post(Entropy Solution of the Burger's Equation). Is it correct? What I mean is that, is the answer of this post entropy solution? I know this is correct, but I feel suspicious.

The reason why I feel suspicious is related to that problem;

"Construct entropy solution of $u_{t} + (\frac{u^{4}}{4})_{x} = 0$ with $u(x,0) =1$ if $x<0$, $u(x,0)=0$ otherwise."

First of all, I think I can use Lax-Olieniek formula since $F:=\frac{u^{4}}{4}$ is uniformly convex, so it ensure $G= (F')^{-1}$ is entropy solution. (Note that $x(s,t) = \frac{1}{4}t$). Then, some calculation denotes $G = u^{-3}$ and $u = 1$ if $x<t$, $u=0$ otherwise. So $G = u$. Is it right? I think I derive all the stuff properly,but I'm not sure it is entropy solution or not. Could you verify my construction?

EditPiAf
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user124697
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1 Answers1

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The unique weak solution which satisfies the Lax-Oleinik entropy condition is of the form $$ u(x,t)=\left\{ \begin{array}{lll} u_l & \text{if} & x<st, \\ u_r & \text{if} & x>st, \end{array} \right. $$ where $s$ is given by the Rankine-Hugoniot condition: $$ s=\frac{f(u_l)-f(u_r)}{u_l-u_r}. $$ In your case $$ u_l=1,\,\,\,u_r=0,\,\,\,s=\frac{1}{4}. $$

Yiorgos S. Smyrlis
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    Thank you for rapid comments. However, I don't understand how can $u(x,t)$ you suggested be derived. Actually, Since parametrizable representation of shock curve is $s(t) =\frac{1}{4}t$, you know, then implicit solution is $u(x-u^{3}t,0)$. Hence if $u=1$, then $x-t < 0$, which implies $x< t$, otherwise $u=0$ so $x-0 >0$, which implies $x>t$. Is it wrong? – user124697 Mar 06 '14 at 13:53