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Ok, lets say I am going to solve the following equation:

$u_t + (\frac{u^2}{2})_x = \epsilon u_{xx}$

which connects the end conditions $u_-=1$ and $u_+ = 0$. According to my understanding of the method, the travelling wave is given by the following equation:

$2 \epsilon\frac{dU}{dy} = U^2 + U $

Where $U(y) = u(x,t) $ and $y := x - t/2 $, the solution is:

$u(x,t) = U(x - t/2) = \frac{1}{1+\exp{\frac{x - t/2}{2 \epsilon}}} $

If you take the limit of $\epsilon$ to zero, then (if I am not mistaken):

$u(x, t) = U(x - t/2) = 1/2 $

(...) And here I got stuck, I mean, I know that the solution $u(x,t)$ for inviscid burgers' is one on the left hand side of the shock $y=t/2$ and zero otherwise; but following the aforementioned analysis, I don't find a connection between the limit I got and the actual solution of the problem. Can you help me with the logic of this problem?

PS.: I am just an enthusiast on PDEs, and my background is definentely no mathematics, so please try not to be too harsh on me :)

Kbzon
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  • A couple of things you can determine the shock time when you get infinite gradient of the solution. Why do you expect the solution be the same for $u$ when you take $\epsilon$ to zero? I will work on this when I get home. – Chinny84 Jan 14 '15 at 17:46
  • Also taking the limit of the solution is 1 not $\frac{1}{2}$ – Chinny84 Jan 14 '15 at 17:51

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You made a mistake taking the limit $\epsilon\to0$. Note that $$\lim_{\epsilon\to0}\exp(a/\epsilon) =\begin{cases} \infty \quad &a>0 \\ 0\quad & a<0 \\ 1 \quad & a=0\end{cases}$$ Hence, the limit of $\dfrac{1}{ 1+\exp((x-t/2)/\epsilon )}$ is $$u(x,t) =\begin{cases} 0 \quad &x>t/2 \\ 1\quad & x<t/2 \\ 1/2 \quad & x=t/2\end{cases}$$ in agreement with what you know about the inviscid equation.

  • Hahahaha, I knew I was mistaken with that limit. Anyhow, for a guy who hasn't taken more than algebra I feel happy of what I obtained. Plus, thank god I found you guys. I will be bothering you in the future. – Kbzon Jan 15 '15 at 09:15