# Homogeneous Quasi-linear PDE, Method of Characteristics.

I want to solve the PDE:

$$\frac{\partial u}{\partial t}+(1-2u)\frac{\partial u}{\partial x}=0$$

for $x \leq0$ and $t \geq 0$, with initial condition $u(0,x)=\frac{1}{2}$, and boundary condition;

$u(t,0)=\left\{\begin{matrix} \frac{t}{4}+\frac{1}{2}, 0\leq t\leq 2 & \\ 1, t \geq 2& \end{matrix}\right.$

by using the method of characteristics.

The method of characteristics gives us that $u$ is constant along characteristics which have equation defined by $\frac{dx}{dt}=1-2u$.

For a characteristic originating from the x axis at a point $(x,t)=(x_0,0)$, this tells us that on characteristics defined by $x(t)=x_0$, we have $u(x(t),t)=u(x(0),0)=\frac{1}{2}$. i.e. the characteristics originating on the x axis are vertical straight lines on which $u=\frac{1}{2}$.

For a characteristic originating from the t axis at a point $(x,t)=(0,\tau)$, we have 2 cases, depending on the value of $\tau$.

In both cases $\frac{du}{dt}=0$, on $\frac{dx}{dt}=1-2u$, so u is constant on the characteristics, so $u(x(\tau),\tau)=u(0,\tau)$ on characteristics defined by $x(t)=(1-2u(0,\tau))t+C$, where $C$ is constant. Applying our boundary condition this gives us:

$u=\frac{\tau}{4}+\frac{1}{2}$, on $x(t)=-\frac{\tau}{2}(t-\tau)$, for $0\leq\tau\leq2$, and:

$u=1$ on $x(t)=-(t-\tau)$, for $\tau\geq2$.

Up to here I'm fine, but when I draw up a plot of the characteristics in the t-x plane I get very confused. The x-axis characteristics are all vertical lines. The t-axis characteristics for $0\leq\tau\leq2$ are straight line originating from $(x,t)=(0,\tau)$ with slope that varies. As $\tau$, decreases from 2 to 0, the slope of these characteristics (in the t-x plane), goes from -1 and approaces $-\infty$, as $\tau$, approaches zero, meaning that each of these characteristics intersect.

As $u$ is constant on a characteristic, this is a contradiction, and so we expect a shock wave (implying a discontinuous solution).

Also for $\tau\geq2$, the characteristics are all parallel lines of slope -1, simply translated in the positive t direction (upwards in the t-x plane).

What is really confusing me is that the three types of characteristics are all intersecting, and that the $0\leq\tau\leq2$ characteristics are intersecting themselves infinitely often, and this is maling it hard for me to apply the shock criterion. At the origin, the characteristics firast intersect, so we expect a shock will immediately form. It will propagate with shock velocity given by: $\frac{dx_s}{dt}=\frac{(u(1-u))|_+-(u(1-u))|_-}{(u)|_+-(u)|_-}$

Where the subscripts denote the value of u just to the right or jsut to the left of the shock.

I really don't know how to continue from here, especially because of the three types of intersecting characteristics and the $0\leq\tau\leq2$ characteristics intersecting with themselves. I would really be grateful for any hints or ideas or help. I've tried a lot of things to resolve this issue, but none have gotten anywhere and it would take a long time to write them out. I think if I can just undbrstnad the shock condition/formation here, I can solve this problem.

## 1 Answer

An alternative way for solving thanks to the method of characteristics. So, you can compare with your result.

The next figure shows $u$ as a function of $x$ at various time. 