Using the characteristics method, show that the Cauchy Problem for the
quasi-linear equation $$u_t + x u u_x = 0 \qquad u(0, x) = \phi(x) =
\frac \pi 2 – \arctan(x)$$ has two shock times, $t^*_\pm $, one in the
future and one in the past. Determine $t^*_\pm $.
The characteristic curves are the solutions to the ODE system
$$ \dot t = 1 \qquad t(0) = 0$$
$$ \dot x = ux \qquad x(0) = x_0$$
$$ \dot u = 0 \qquad u(0) = \phi (x_0)$$
This yields the solutions $$x(t) = \exp(t\phi(x_0))+x_0+1$$
If this expression can be inverted into a function $x_0(t, x)$ then there is no shock, but if that is the case then $x(t)$ must be monotone as function of $x_0$. Thus, define $F_t(x) = \exp(t\phi(x))+x+1$ : I need to find the values of $t$ for which $F' >0$ for all $x$. This reduces to:
$$ \frac s {1+x^2} \exp\left (s\left (\frac \pi 2-\arctan(x)\right )\right ) < 1$$
At which point I am stuck. I cannot solve this inequality, but also I have the nagging feeling that I made a mistake somewhere earlier. Did I? And what are the shock times?
Best Answer
As outlined by OP's self-answer, $$ x(t) = x_0\exp(t\phi(x_0)) \, . $$ Several methods can be used to find the shock times (see e.g. this post). Here we examine the non-ambiguous dependence to the initial data. One observes that $\text d x/\text d x_0$ vanishes at $t=t_S$ such that $$ t_S = \frac{-1}{x_0\phi'(x_0)} = x_0 + \frac{1}{x_0} \, . $$ Thus, for positive times, the classical solution breaks at the breaking time $$t^*_+ = \inf_{x_0>0} t_S = 2\, .$$ Similarly one gets $t^*_- = -2$ for negative times. This may be confirmed by a plot of the characteristics for several $x_0$.