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I am solving the following partial differential equation, with some initial conditions, given by: $$u_t + u\,u_x = 0,\,\, u(x, 0) = \exp(-x^2)$$ and given that the characteristic lines pass through some point $(\xi, 0)$ on the $x$ axis. I have solved it using the method of characteristics, and the initial coniditons, and obtained the solution $$u(x, t) = \exp\left((ut-x)^2\right)$$

Now I need to find the breaking time. I know that this is the time when $u_x$ has an infinite gradient. But it looks like when I differentiate the solution, there will be no denominator and it won't be possible to find $t_b$ and the corresponding $x_b$?

EditPiAf
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Victoria
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1 Answers1

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As you mentioned, this equation (Burgers equation), being a particular case of scalar conservation law, can be solved by means of characteristic lines. And the schok emerging time can be computed likewise.

Let me denote $f(x)=u(x,0)=e^{-x^2}$. Let us consider two characteristic lines emerging from $\xi_0,\xi_1\in\mathbb{R}$. They are given by $$x=f(\xi_0) t +\xi_0 \quad\mbox{ and }\quad x=f(\xi_1) t +\xi_1,$$ respectively. They will intersect when $$f(\xi_0) t +\xi_0 = x=f(\xi_1) t +\xi_1,$$ that is, when $$t=\frac{\xi_1-\xi_0}{f(\xi_0)-f(\xi_1)}$$ Note that $t>0$ whenever $f$ is decreasing in the interval $[\xi_0,\xi_1]$.

If we take the limit $\xi_1\to\xi_0$, we observe that the breaking time for $\xi_0$ converges to $$t=-\frac{1}{f'(\xi_0)}.$$ Therefore, the minimal time at which the shock emerges is $$t^*=-\frac{1}{\min_{x\in\mathbb{R}}\{f'(x)\}}.$$

AugSB
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  • Very helpful answer, +1 – Victoria Mar 26 '15 at 20:22
  • so this basically amount to finding that $\phi '(x) = -2x,e^{-x^2}$ and then observing that $\phi '(x)$ is minimised when $x = 0$? – Victoria Mar 26 '15 at 20:45
  • That is what you have to do. But be careful when computing the minimum: it is not located in $x=0$. – AugSB Mar 27 '15 at 08:39
  • hmm, do you need to differentiate again? But won't the result be the same? – Victoria Mar 28 '15 at 01:45
  • You have to compute the minimum of $f'$, so you do have to differetiate again. In fact, note that $f$ does no attain any minimum (it just tends to 0 as $x\to\pm\infty$). – AugSB Mar 28 '15 at 11:09