I have an initial value problem:
$$ u_t - u^2u_x = 0 \quad \text{with} \quad u(x,0) = g(x) = \begin{cases} -\frac{1}{2}, & x\leq 0 \\ 1, & 0 < x <1 \\ \frac{1}{2}, & x \geq 1 \end{cases} $$
The characteristic system is
\begin{align*} \frac{\partial t}{\partial\tau} = 1, \hspace{30pt} \frac{\partial x}{\partial\tau} = -u^2, \hspace{30pt} \frac{\partial u}{\partial\tau} = 0 \end{align*}
with the initial curve given by \begin{align*} t(0) = 0, \hspace{30pt} x(0)=\xi, \hspace{30pt} u(0)=g(\xi) \end{align*} Note here that $u_{\tau}=0$ implies that the solution will be constant along the characteristics, and the characteristics will be straight lines. We have, \begin{align*} t = \tau, \hspace{30pt} u = \xi, \hspace{30pt} x = \xi-g(\xi)^2\tau \end{align*} Now, \begin{align*} x = \begin{cases} \xi - \frac{\tau}{4}, & \xi\leq 0 \\ \xi - \tau, & 0 < \xi <1 \\ \xi - \frac{\tau}{4}, & \xi \geq 1 \end{cases} \end{align*} Therefore the base characteristics are, \begin{align*} x &= \xi_0-\frac{\tau}{4} \hspace{10pt} \text{for } \xi_0 \leq 0 \\ x &= \xi_0-\tau \hspace{10pt} \text{for } 0\leq\xi_0 \geq 1 \\ x &= \xi_0-\frac{\tau}{4} \hspace{10pt} \text{for } \xi_0 \geq 1 \end{align*}
I can plot the base characteristics no problem, and I know that there are missing characteristics. My problem is finding them. I get, $$ u(x,t) = g(\xi) = \begin{cases} -\frac{1}{2}, & x\leq -\frac{t}{4} \\ 1, & -t < x <1-t \\ \frac{1}{2}, & x \geq 1-\frac{t}{4} \end{cases} $$
First of all, I can get my missing characteristics for $1-t<x<1-\frac{t}{4}$. Also my cases overlap, so I know $u(x,t)$ will be multivalued, but finding these missing characteristics is proving trouble some. Any ideas would be appreciated.
