2
$\begingroup$

I've recently been studying for qualification exams for my master's program. I've run into a few problems that I'm stuck on and hope that I can get some help here.

We consider the hyperbolic conservation law $$u_t + ((1 - u)u)_x = 0.$$ This can model, for example, traffic density $u$, where the cars have speed $1 - u$, moving slower in heavy traffic. We consider the initial condition $$u(x, 0) = \begin{cases} 1/4 & x \leq 1/4 \\ x & 1/4 \leq x \leq 1 \\ 1 & 1 \leq x \end{cases}$$ Solve the equation using the method of characteristics. A shock will form. Resolve the motion of the shock.

I haven't learned anything about nonlinear PDEs/waves and don't know where to start with this problem. The only thing I can see is that the PDE can be transformed into the following: $$u_t + (1 - 2u)u_x = 0$$

Thanks in advance for any help!

$\endgroup$
1
$\begingroup$

A good resource is Evan's book on PDE; he has a section on the method of characteristics. The basic idea is as follows. Given any suitable PDE with boundary conditions, we may turn this PDE into a series of ODE as follows. We pick a point $(x_0, 0)$ in the boundary (so here, the "boundary" is time $t = 0$). It would really be great if there were some path $y(s)$ connecting $(x_0, 0)$ to an arbitrary point $(x, t)$ in the interior (i.e. for time $t > 0$), and then set up an ODE along this path.

It turns out that for this specific PDE, what happens is that along this path $y$, the solution $u$ is constant. What this means is the following. Taken any $(x, t) \in \mathbb{R} \times (0, \infty)$. Then if we can connected $(x, t)$ to $(x_0, 0)$ along one of these characteristic curves $y$, we know that $u(x, t) = u(x_0, 0)$, since $u$ is constant on these curves.

Again, the general theory is helpful to see why, but that is a whole section in Evans' text. I will try to give a very brief introduction here.

Let $G(q, z, y)$ be a real-valued function. Into the variable $q$ here, we will input the tuple $(u_x, u_t)$; so $q$ represents the data of the gradient of $u$. The variable $z$ here will represent $u(x(s))$, where $x : I \to \mathbb{R} \times [0, \infty)$ is our characteristic curve (which is at this point unknown). The variable $y$ is just $y = (x, t)$. This function $G$ is the PDE function we are solving; that is, we are solving the equation $G = 0$. The general form of the characteristic equations that concern us now are $$ \dot{z}(s) = q\cdot D_qG, \ \ \ \dot{x}(s) = D_qG. $$ Well for this problem at hand, $G(q, z, y) =(1 - 2z, 1) \cdot q$, as \begin{align*} G(q, z, y) &= u_t + (1 - 2u)u_x \\ &= (1 - 2u, 1) \cdot (u_x, u_t) \\ &= (1 - 2z, 1)\cdot q. \end{align*} Thus, $D_qG$ is just the vector $(1 - 2z, 1)$. So the second characteristic equation is $\dot{x} = (1 - 2z, 1)$. So our characteristic curve $x$ satisfies this ODE.

Now the magic happens with $\dot{z}$, as this reduces to $$ \dot{z}(s) = q\cdot D_qG = (u_x, u_t)\cdot (1 - 2z, 1) = 0; $$ this is zero, remember, because this is exactly the PDE you are solving for! Since $z(s) = u(x(s))$, what this is literally saying is that the solution $u$ is constant (zero derivative!) along the characteristic curve $x$.

It is now a matter of integrating $\dot{z}$ along these curves to find the solution $u$. For a very good illustration of the method of characteristics, try solving the transport equation with constant coefficients, also in Evans.

As for shocks, I'll just suggest Evans, since this answer is already very long.

$\endgroup$
1
$\begingroup$

Let us introduce the function $f : x\mapsto u(x,0)$. A good start with such initial-value problems for transport (hyperbolic) equations is to draw the characteristic curves in the $x$-$t$ plane, obtained through the method of characteristics. These curves have equation $x(t) = (1-2f(x_0))\, t + x_0$:

characteristics

The breaking time $t_b$ when such curves intersect corresponds to the generation of a shock wave: $$ t_b = \frac{-1}{\min \partial_x\left(1-2f(x)\right)} = \frac{1}{2} \, . $$ Until characteristics intersect $t<t_b$, the solution is given by the method of characteristics. It can be obtained by solving $u = f(x - (1-2u)t)$: $$ u(x,t) = \left\lbrace \begin{aligned} &1/4 &&\textstyle\text{if}\quad x < \frac{1}{4} + \frac{1}{2}t\\ &(t-x)/(2t-1) &&\textstyle\text{if}\quad \frac{1}{4} + \frac{1}{2}t < x < 1 - t\\ & 1 &&\textstyle\text{if}\quad 1 - t < x \end{aligned}\right. $$ One can observe that the characteristics cross at the time $t_b$ and the abscissa $x_b = 1/2$. After the formation of the shock, the shock speed is given by the Rankine-Hugoniot condition $$ s = \frac{1(1-1) - \frac{1}{4}(1-\frac{1}{4})}{1 - \frac{1}{4}} = -\frac{1}{4} \, . $$ The solution for $t>t_b$ is therefore $$ u(x,t) = \left\lbrace \begin{aligned} &1/4 &&\textstyle\text{if}\quad x < \frac{1}{2} - \frac{1}{4}t\\ &1 &&\textstyle\text{if}\quad \frac{1}{2} - \frac{1}{4}t < x \end{aligned}\right. $$

N.B. : this is the LWR traffic flow model.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.