Find the maximal $t_0>0$ such that the Cauchy problem: $$uu_x+u_t=0 \qquad{} u(x,0)=e^{-x^2/2}$$ exists in $\Bbb{R}\times[0,t)$; i.e. find the first time $t=t_0$ when the shock develops.
This is review for my PDE class, using Evans' textbook.
A shock develops when two characteristics intersect, so lets compute the characteristics:
$$\dot{x}=z$$ $$\dot{t}=1$$ $$\dot{z}=0$$
Note $x(0)=x_0$, $t(0)=0$ and $z(0)=e^{-x_0^2/2}$. We may solve the ODEs now:
$$z(s)=e^{-x_0^2/2}$$ $$t(s)=s$$ $$x(s)=se^{-x_0^2/2}+x_0$$
So our characteristic curves are $(se^{-x_0^2/2}+x_0,s)$
Now we need to find when they cross. So pick $x_1,x_2$ for $x_0$ and find intersection: $$se^{-x_1^2/2}+x_1=se^{-x_2^2/2}+x_2$$
We can solve this and get:
$$s=\frac{x_1-x_2}{e^{-x_2^2/2}-e^{-x_1^2/2}}$$
Now we want to minimize this, I assume. I took the derivatives and set them equal to $0$ but it gets really ugly.
Does anyone have any hints? I'd prefer hints to full answers.
Per hint, we want to find for which $s$ is the function $f_x(x)=se^{-x^2/2}+x$ injective. It is injective when $f_s'(x)>0$. So compute:
$$f_s'(x)=-sxe^{-x^2/2}+1>0$$
So:
$$1>sxe^{-x^2/2}$$
Since $s>0$ we can consider $x>0$ so we divide:
$$\frac{e^{x^2/2}}{x}>s$$
We find the minimum of $\frac{e^{x^2/2}}{x}$ which is $e^{1/2}$ at $x=1$. So we know that for $s<\sqrt{e}$ the function is injective. Thus our $t_0=\sqrt{e}$.