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?
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dy}{dt}=1$ , 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 $u(x,y)=\begin{cases}\dfrac{1}{4}&x<\dfrac{y}{2}\\\dfrac{3}{4}&x>-\dfrac{y}{2}\\c&\text{otherwise}\end{cases}$