I am trying to solve
$u_{t}+uu_{x}=0, \quad x\in\mathbb{R},\quad t>0$
with the initial values
$u(x,0)= \begin{cases} 1-x^2, & |x| \leq 1 \\ 0, & |x| > 1 \end{cases}$
I also need to sketch the characteristic diagram and find the breaking time $t_b$.
I have tried searching on the internet and using my notes to find a solution, however, I always end up confused and without a solution.
This is what (little) I have done so far:
$\frac{dX(t)}{dt}=u(X(t),t) \quad$ -$\quad$ I think I am trying to get this in the form of an ODE so that I can solve simply.
This is where different sources go in different directions. Some then differentiate $u(X(t),t)$ and seem to get back to the original PDE and some go on and make it equal to the initial conditions and I cannot figure out what the correct method is, or what the underlying principles are.
Let $U(t) = u(X(t),t)$. Then $$ \dot{U}(t) = u_t(X(t),t) + \dot{X}(t)u_x(X(t),t) $$ Hence, if we set $\dot{X}(t) = u(X(t),t) = U(t)$, then $\dot{U}(t) = u_t(X(t),t) + u(X(t),t)u_x(X(t),t) = 0$. The set of equations are thus \begin{align*} \dot{X}(t) &= U(t) \\ \dot{U}(t) &= 0 \end{align*} Suppose we tried to solve the above equations with $X(t) = x$. The two equations imply that $U(t) = U(0)$ and $X(t) = X(0) + \int\limits_{0}^{t}{\dot{X}(t)\text{ d}t} = X(0) +\int\limits_{0}^{t}{U(t)\text{ d}t} = X(0) + U(0)t$. Thus, we want to find $X(0)$ such that $$ x = X(0) + u(X(0),0)t. \tag{1}$$ If $|x|>1$, then we can obviously choose $X(0) = x$, in which case $U(t) = U(0) = u(x,0) = 0$. If $|x|\le 1$, then we want to find $|X(0)|\le 1$ such that $$ x = X(0) + t(1-X(0)^2).$$ Now, for $t<\frac{1}{2}$, the mapping $y\mapsto y + t(1-y^2)$ has derivative $y\mapsto 1-2ty$, which is positive on $[-1,1]$, so $y\mapsto y+t(1-y^2)$ is invertible on $[-1,1]$ (and, by checking the endpoints, we see that it maps $[-1,1]$ onto $[-1,1]$). This means that for each $|x|\le 1$ there is a unique $|X(0)|\le 1$ such that $x = X(0) + t(1-X(0)^2)$. (The explicit solution is given by the quadratic formula, but I think this is the easiest way to show that we can solve the equation in the way we want.)
As such, the method of characteristics is valid for $t<\frac{1}{2}$, and it is not hard to show that (1) does not have a unique solution $|X(0)|\le 1$ for $|x|\le 1$ if $t>\frac{1}{2}$, i.e. the method of characteristics fails for $t>\frac{1}{2}$.