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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:

enter image description here

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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:

characteristics

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.

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    $\begingroup$ @EzioBosso Seems that the numerical value of the slopes/scales in the figure are incorrect by a factor 2, but the general shape of the sketch is correct. $\endgroup$ – EditPiAf Dec 6 '20 at 13:05

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