# Solving $u_y + (1-2u)\cdot u_x = 0$ using characteristic equations

I need to solve the following partial differential equation: $$F(x, y, u, p, q) = u_y + (1-2u)\cdot u_x = 0$$ with $$u(x, 0) = \left\{\begin{array}{cc} \frac{1}{4} & x < 0 \\ \frac{3}{4} & x>0 \end{array}\right.$$ Here we have $p = u_x$ and $q = u_y$. $u(x,y)$ is a continously differentiable function and $-\infty < x < \infty$ and $0 < y < \infty$.

I use the characteristic differential equations to get a solution. These are $$\left\{\begin{array}{ll} \frac{dx}{ds} = F_p = 1-2u \\ \frac{dy}{ds} = F_q = 1\\ \frac{du}{ds} = pF_p+qF_q = p(1-2u) + q\\ \frac{dp}{ds} = -F_x - pF_u = 2p^2\\ \frac{dq}{ds} = -F_y - qF_u = 2pq \end{array}\right.$$

The initial conditions are, if $s=0$, that $x = \lambda, y = 0,$ $$u = \left\{\begin{array}{cc} \frac{1}{4} & \lambda < 0\\ \frac{3}{4} & \lambda > 0\end{array}\right.$$ and, using that $$0 = \frac{du}{d\lambda} = p(\lambda)\frac{dx}{d\lambda} + q(\lambda)\frac{du}{d\lambda} = p(\lambda)\cdot 1 + 0$$ and $$0 = q + (1-2u)p = q + (1-2)\cdot 0 = q$$ we get $p=q=0$.

From this, it is easily seen that $y = s$. But from here, I'm stuck. The general solution of $\frac{dp}{ds} = 2p^2$ is $p(s) = \frac{1}{c_1 - 2s}$. It impossible for this function to sattisfy the initial condition, so does that mean that this partial differential equation doesn't have a solution?

Another thing is that the initial condition for $u(x, 0)$ isn't continous, so how can the solution of this problem be continously differentiable?

The PDE $$u_y + (1-2u)u_x = 0$$ is the Lighthill-Whitham-Richards traffic flow equation. The proposed initial-value problem may be interpreted as a one-way road with the car density $$u=1/4$$ before the abscissa $$x=0$$ (fast cars), and the car density $$u=3/4$$ after the abscissa $$x=0$$ (slow cars). In mathematical words, this is a Riemann problem, which cannot be solved by using the method of characteristics alone: we should consider weak solutions. The Lax entropy condition tells that the solution is a shock wave. Its speed is given by the Rankine-Hugoniot condition $$s = 1- \frac{1}{4} - \frac{3}{4} = 0$$. Therefore, the solution is a static shock wave u(x,y) = \left\lbrace \begin{aligned} &1/4 &&\text{if}\quad x<0\\ &3/4 &&\text{if}\quad x>0 \end{aligned}\right.

$\dfrac{dy}{dt}=1$ , letting $y(0)=0$ , we have $y=t$

$\dfrac{du}{dt}=0$ , letting $u(0)=u_0$ , we have $u=u_0$

$\dfrac{dx}{dt}=1-2u=1-2u_0$ , letting $x(0)=f(u_0)$ , we have $x=(1-2u_0)t+f(u_0)=(1-2u)y+f(u)$ , i.e. $u=F(x+(2u-1)y)$

$u(x,0)=\begin{cases}\dfrac{1}{4}&x<0\\\dfrac{3}{4}&x>0\end{cases}$ :

$F(x)=\begin{cases}\dfrac{1}{4}&x<0\\\dfrac{3}{4}&x>0\end{cases}$

$\therefore u=\begin{cases}\dfrac{1}{4}&x+(2u-1)y<0\\\dfrac{3}{4}&x+(2u-1)y>0\end{cases}=\begin{cases}\dfrac{1}{4}&x-\dfrac{y}{2}<0\\\dfrac{3}{4}&x+\dfrac{y}{2}>0\end{cases}$

Hence $u(x,y)=\begin{cases}\dfrac{1}{4}&x<\dfrac{y}{2}\\\dfrac{3}{4}&x>-\dfrac{y}{2}\\c&\text{otherwise}\end{cases}$

• I'm not sure I follow this. What is $f$ in $x(0) = f(u_0)$. It seems to me like it is an initial condition, but how exactly is it defined? Is it the inverse of $F$? If so, if $u = F(x + (2u-1)y)$, shouldn't $u(x,y)$ be a function of $x+(2u-1)y$ instead of $x$, like in your last expression for $u(x,y)$? – Pierre Jun 27 '15 at 23:04
• @ImPact Although the general solution is of the form $u=F(x+(2u-1)y)$ , the constant-type condition leads the function only possible to be the constant-type function. – doraemonpaul Jun 28 '15 at 0:42