I want to solve the Hopf equation (preferably with the method of characteristics) $$u_t + uu_{x}=0$$ with initial condition $$f(x) = \left\{ \begin{array}{ll} 1- |x| &, |x| \leq 1 \\ 0 &, |x| \geq 1\\ \end{array} \right. $$
Using the method of characteristics we can easily find $$ u=f(x-ut)$$ Thus we have $$u = \left\{ \begin{array}{ll} 1- x+ut &, x-ut \geq 0 \\ 0 +x - ut &, x-ut \leq 0\\ \end{array} \right. $$ but it doesn't seem like I can solve this to get $u$ as a function only of $x,t$ and not $u$ itself. Any ideas?
Additionaly, I would like to find the breaking time $t_b$ and breaking point $x_b$, which doesn't require analytically solving the problem, just finding the characteristics, which are: $$x=f(ξ)t +ξ $$ Now if $f$ where differentiable we would have $ξ_x =\frac{1}{f'(ξ)t+1}$ and thus the breaking time would be at $t$ such that $f'(ξ)t+1=0$, which gives $t_b=1$ if we ignore the $x \in (-1,0)$ part (since breaking wont happen where $f'(x) >0$). Is this correct? Also what is the breaking point, since $f'(ξ)t+1=0$ at $t=1$ for all $x \in (0,1)$?