Breaking time of Burgers equation

Question

I am studying Burgers equation $$ u_t+uu_x=0, \qquad u(x,0) = u_0(x) , $$ and I am trying to prove the expression for the breaking time $t_c$. I obtain that $$ t_c=-\dfrac{1}{\min\limits_{x_1,x_2\in\mathbb{R}}\dfrac{1}{x_2-x_1}\int_{x_1}^{x_2}u_0'(x)dx}, $$ but I do not know why this is equal to $$ t_c=-\dfrac{1}{\min\limits_{x\in\mathbb{R}}u_0'(x)} $$ Thanks a lot!

Answer 1

This is slightly sloppy notation. You’re missing a condition that $x_1\ne x_2$; otherwise the denominator is not defined. Also, in the first result the minimum should be the infimum, since for general $u_0$ the infimum may not actually be attained.

If you correct these two inaccuracies, the result follows naturally: If you take $x_1=x_0-\epsilon$ and $x_2=x_0+\epsilon$, where $x_0=\operatorname{argmin}\limits_{x\in\mathbb{R}}u_0'(x)$, the quotient goes to $x_0=\min\limits_{x\in\mathbb{R}}u_0'(x)$ for $\epsilon\to0$, and on the other hand

$$ \frac1{x_2-x_1}\int_{x_1}^{x_2}u_0'(x)\mathrm dx\ge\frac1{x_2-x_1}\int_{x_1}^{x_2}\min\limits_{x\in\mathbb{R}}u_0'(x)\mathrm dx=\min\limits_{x\in\mathbb{R}}u_0'(x)\;. $$

Thus $\min\limits_{x\in\mathbb{R}}u_0'(x)$ is the infimum of the quotient.