Reading a bit about Lighthill-Whitham-Richards (LWR) Traffic PDEs, and I've realize I don't quite understand how to introduce shocks when a PDE has both an initial and boundary condition.
I propose a simple example:
Example. Define the following LWR with initial and boundary conditions
$$
\begin{align}
&u_t + 2uu_x = 0 \\
&u(x,0) = 1,\quad x > 0\\
&u(0,t) = 2,\quad 0 < t < 1\\
&u(0,t) = 1,\quad t > 1
\end{align}
$$
What I understand: We use the Method of Characteristics to solve, i.e.
$$\frac{dt}{1} = \frac{dx}{2u} = \frac{du}{0}.$$
From this it is easy to see that we get the implicit solution, $u(x,t) = F(x-tu)$. For the initial condition, we have characteristic curve, $x = tF(s) + s$. For $s > 0$, we get that $F(s) = 1$, and so
$$x = t + s \quad\Rightarrow\quad s = x – t \quad\Rightarrow\quad u(x,t) = 1,~ x > t.$$
Issue: So, now we look at the BP. Then using the same characteristic, $x = tF(s) + s$, we get
$$
\begin{align}
0 < t < 1 &\quad\Rightarrow\quad F(s) = 2 &\quad\Rightarrow\quad s = x – 2t &\quad\Rightarrow\quad u(x,t) = 2,~2t < x < 1 + 2t. \\
t > 1 &\quad\Rightarrow\quad F(s) = 1 &\quad\Rightarrow\quad s = x – t &\quad\Rightarrow\quad u(x,t) = 1,~x > 1 + t.
\end{align}
$$
I don't known how to visualize these in the $x-t$ plane (below line $x = t$), so that I can introduce shocks (two of them I think?). My picture so far would be:
Best Answer
One needs to be careful when applying the initial/boundary conditions for the Lagrange-Charpit system $$ \frac{dt}{1} = \frac{dx}{2u} = \frac{du}{0} , $$ which gives $u = C_1$ and $x = 2ut+C_2$. The previous equations yield $$u = C_1 = F(C_2) = F(x - 2u t) $$ or equivalently $$u = C_1 = G\big({-\tfrac{C_2}{2C_1}}\big) = G(t-x/(2u)) \, . $$ We may rewrite the equation of characteristic curves starting from the initial condition as $x = 2u t + x_0$, along which $u = F(x_0)$ is uniformly equal to the value $u(x_0,0)$. The equation of characteristic curves starting from the boundary condition reads $t = x/(2u) + t_0$ where $u = G(t_0)$ is uniformly equal to the value $u(0,t_0)$. Here is what the curves look like:
To determine the shock and rarefaction waves, you will need to use the conservative form obtained by substituting $2uu_x = (u^2)_x$. Note that the Lighthill-Witham-Richards traffic flow model may be recovered if we set $u = \frac12 - \rho$ where $0\leq \rho\leq 1$ is the car density, so that $$ u_t + 2u u_x = -[\rho_t + (1 - 2\rho) \rho_x] = 0 \, . $$ Hence, the proposed values of $u$ in the problem statement aren't representative of a physical configuration.