Cauchy problem for $u_t + uu_x =0$ Discuss the solution of $u_t+ uu_x= 0$ with the following Cauchy data:
a) $u(x,0)= x ; 0\leq x\leq 1$
b)$u(x,0)=1/2; 0\leq x\leq 1$
Sketch the domains in the $x-t $ plane where the solutions are determined. Does the solution develop a singularity at any time? If yes, discuss the reasons for the break down of the solution.
My trials:
For a) by the methos of characteristic I got the solution u(x,t)=x/(1+t). So it has singularity at t=-1, is it possible? And then for b) methos of characteristics is not giving a solution. So if someone explain the solution of this kind problems it will be of great help. Thanks in advance.
 A: Suppose the parameters for characteristics are $\eta$ and $\xi$.
The characteristics are defined by
$$
\frac{\partial t}{\partial \eta} = 1, \frac{\partial x}{\partial \eta} = u, \frac{\partial u}{\partial \eta} = 0.
$$
Solving gives $t(\eta, \xi) = \eta + c_1(\xi)$, $u(\eta, \xi) = c_3(\xi)$, $x(\eta, \xi) = c_3(\xi)\eta + c_2(\xi)$.
Pick the parametrization of the initial solution curve: $t(0, \xi) = 0$ and $x(0, \xi) = \xi$. It follows that $c_1 = 0$, $c_2(\xi) = \xi$ and $t = \eta$. The initial condition determines $c_3$:
a) $u(0, \xi) = \xi$, so $c_3(\xi) = \xi$. It follows that $x = \xi(\eta + 1)$. Since the initial condition is only given for $\xi \in [0, 1]$, the domain in which the solution may exist is $(x, t) \in \{(\xi(\eta + 1), \eta) \mid \xi \in [0, 1], \eta \in \mathbb R\}$. Inverting the transformation $(\xi, \eta) \mapsto (x, t)$, we get $\xi = \frac{x}{t+ 1}$, so the singularity exists at $t = -1$. The solution is $u = \xi = \frac{x}{t+1}$.
b) $u(0, \xi) = \frac 12$, so $c_3(\xi) = \frac 12$. It follows that $x = \frac 12\eta + \xi$. The domain in which the solution may exist is $(x, t) \in \{(\frac 12\eta + \xi, \eta) \mid \xi \in [0, 1], \eta \in \mathbb R\}$. Inverting the transformation, we get $\xi = x - \frac 12 t$. The solution is $u = \frac 12$. There is no singularity. The solution is valid throughout the domain $\{ (x, t) \mid \frac t2 \le x \le \frac t2 + 1, t \in \mathbb R \}$.
