Consider the transport equation on $[0,T] \times \mathbb{R}$: $$\partial_{t}u+ q(t,x)\partial_{x}u=0$$ where the initial data $u(0, \cdot) = u_{0}(\cdot)$ is Lipschitz continuous in space and time and $q(t,x) = sgn(x).$
Goal: Construct a continuous solution using the method of characteristics.
Attempt: The method of characteristics tells us that we need to solve the system $$\begin{aligned} &\frac{dt}{ds}=1,~ t(0) = 0, \\ &\frac{dx}{ds}=q,~ x(0) = \tau, \\ &\frac{du}{ds}=0,~ u(0)= u_{0}(\tau). \end{aligned}$$ Solving gives us that the set of characteristic lines is given by the set $$ \Gamma = \{ x-sgn(x)t = \tau ~| ~\tau \in \mathbb{R}\}. $$ So, wherever the characteristics are defined, the solution is given by $$u(t,x) = u_{0}(x - sgn(x)t). $$ From here, I am confused on how to arrive at the final solution which should be
$$u(x,t)= \begin{cases} u_{0}(x-tsgn(x)) & \text{ if } t \leq |x|, \\ h(t-|x|) & \text{ if } t \geq |x|. \end{cases}$$ where $h \in Lip[0,T]$ is such that $h(0) = u_{0}(0)$.
Sketching the characteristics is the confusing part because the characteristic lines are themselves discontinuous! We have a bunch of disjoint lines of gradient $\pm 1$ which I am unsure how to interpret. Is the solution really constant along these discontinuous lines? How can we arrive at the final solution?
I would appreciate any help.
Some more thoughts: The only continuous characteristic is given by $t = |x|$. I sketched a few of these characteristic lines (I took $\tau = 0, -1, 1, 2$) and saw that below $t=|x|$ we have some intersection, so I guess the solution is not defined there. One way to define the solution in this region $t < |x|$ in a continuous way is to join together with a continuous function $h$ which agrees with the value of $u_{0}$ along $t=|x|$. This would explain the final answer. But I am unsure if these ramblings are true because, well, the characteristics look super weird and I don't know if the method of characteristics is even valid in such a scenario.