I am trying to solve the following problem and have looked at similar problems with different initial conditions but I am still having trouble piecing everything together. Here is the question:
Consider the initial value problem
$u_t + uu_x=0, \quad x \in \mathbb{R},\quad t>0$
$u(x,0)=\begin{cases} 1-x^2 & |x|\leq1 \\ 0 & |x|>1 \end{cases}$
Sketch the characteristic diagram. At what time $t_b$ does the wave break? Find a formula for the solution.
This is what I have so far:
$\frac{dx}{dt} = u(x,t) = \phi(\xi) \quad$ where $\quad \phi(x)=u(x,0)$ and $\xi \in \mathbb{R}$.
Hence, for $|\xi| \leq 1 \quad \frac{dx}{dt} = 1-\xi^2 \implies x = t-t\xi^2 = t(1-\xi^2)$
and for $|\xi| > 1 \quad \frac{dx}{dt} = 0 \implies x = \xi$
Therefore I think I can draw the following (rough) characteristic curves.
I'm not sure if this is correct though so please let me know if this is wrong.
Also I tried to find the solution, that is:
Knowing that $x = t-t\xi^2$ in the $|\xi|\leq1$ case, we can show that $\xi = \sqrt{\frac{t-x}{t}}$ and hence $u(x,t)=\phi(\xi)=1-\left( \sqrt{\frac{t-x}{t}}\right)^2=\frac{x}{t}$.
And similarly, in the $|\xi|>1$ case, $u(x,t)=0$
So $u(x,t)=\begin{cases} \frac{x}{t} & |x|\leq1 \\ 0 & |x|>1 \end{cases}$
I then need to work out the breaking time which I have tried to do but don't have an answer.
If you can help I'd be very appreciative.