How can the following PDE be solved? I am working on this problem on PDE's : 
$f(x, y)\frac{{\partial}f}{{\partial}x} + \frac{{\partial}f}{{\partial}y} = 1$ with $f(u,u) = \frac{u}{2} , 0< u < 1$.
I tried to solve this PDE with the method of characteristics but since $f(x, y)$ is not given, I did not know how to continue.
Any ideas? 
 A: Remark: You can use existence theorems to conclude that a solution exists around the initial curve. Compare Theorem 1.1. in "Partial Differential Equations" by Prasad & Ravindran:

Let $x_0,y_0,u_0$ be $C^1$ in a closed interval and $a,b,c \in C^1$ in some domain of $(x,y,u)$-space containing the initial curve $\Gamma:\tau \mapsto (x_0(\tau),y_0(\tau),u_0(\tau))$. Further, let $$\begin{vmatrix} x_0' & y_0' \\ a & b \end{vmatrix}_\Gamma \neq 0.$$
  Then there exists a unique solution $u=u(x,y)$ of the quasilinear equation 
  $$a(x,y,u) u_x + b(x,y,u) u_y = c(x,y,u)$$
  in the neighborhood of the curve $\gamma : \tau \mapsto (x_0(\tau),y_0(\tau))$ and satisfying $u_0(\tau)=u(x_0(\tau),y_0(\tau))$.

In your case we have $\Gamma :\tau \mapsto (\tau,\tau,\tfrac{\tau}{2})$ hence
$$\begin{vmatrix} x_0' & y_0' \\ a & b\end{vmatrix}_\Gamma=\begin{vmatrix} 1 & 1 \\ \tfrac{\tau}{2} & 1 \end{vmatrix}=1-\tfrac{\tau}{2}$$
and for $\tau\neq 2$ we have unique solution of the PDE. By assumption we have $0<\tau<1$ so this is fine.

Computation: We have the characteristic ODEs $$x_s=f, y_s=1, f_s=1$$ with the initial curve $\gamma(\tau)=(\tau,\tau)=(x_0,y_0)$ and $f|_\gamma=\tau/2$.
We solve them and get $y=s+\tau$, $f=s+\frac{\tau}{2}$, $x'(s)=s+\frac{\tau}{2}$ i.e. 
\begin{align} x(s) &=\int s+ \frac{\tau}{2} ds+\tau=\frac{s^2}{2}+ \tau(\frac{s}{2}+1) \\
&=\frac{s^2}{2}+(y-s)(s/2+1)=s^2/2+ys/2+y-s^2/2-s \end{align}
hence $x-y=s(y/2-1)$ which yields
\begin{align} s&=2\frac{x-y}{y-2} \\ \tau&=y-2\frac{x-y}{y-2}=\frac{y^2-2x}{y-2} \end{align}
thus

$$f=2\frac{x-y}{y-2}+\frac{1}{2} \frac{y^2-2x}{y-2}=\frac{x-2y+y^2/2}{y-2}$$

Checking the initial condition
$$f(u,u)=\frac{-u+u^2/2}{u-2}=\frac{u(-1+u/2)}{u-2}=u/2$$
and $f \cdot f_x+f_y=1$ as it can easily be computed. The 'solution space' seems to be $\mathbb{R} \times \mathbb{R}_{y>2}$ or $\mathbb{R} \times \mathbb{R}_{y<2}$.
