I am trying to solve the following:
$ u(x,y) u_x (x,y) + u_y(x,y)=0 \\ x\in\mathbb{R}, y>0\\ $
with the condition $u(0,y)=y, y>0$
Starting with the characteristics I find that for $u=z$ (that is the way I have been taught to solve such problems)
$ \frac{dx}{ds}=z $
$ \frac{dy}{ds}=1 $
$ \frac{dz}{ds}=0 $
In other cases where for example $\frac{dz}{ds}=1$, I would go on like this $\frac{dx}{z}=\frac{dy}{1}=\frac{dz}{1}$ and manipulate this system in order to get an expression for $x,y,z$ and extract two new equations $\phi$ and $\psi$.
Any ideas or suggestions would be highly appreciated
$$u\:u_x+u_y=0$$ System of characteristic ODEs : $$\frac{dx}{u}=\frac{dy}{1}=\frac{du}{0}$$ This implies $du=0$ and so a first characteristic equation : $$u=c_1$$ A second characteristic equation comes from $\frac{dx}{c_1}=\frac{dy}{1}\quad\implies\quad c_1y-x=c_2$ $$u\:y-x=c_2$$ The general solution expressed on implicit form $c_2=F(c_1)$ is : $$y\:u-x=F(u)$$ $F$ is an arbitrary function to be determined according to the boundary condition.
CONDITION : $u(0,y)=y\quad\implies\quad y\:y-0=F(y)$ $$F(y)=y^2$$ Now the function $F$ is known. We put it into the above general solution where the argument is $(u)$ thus $F(u)=u^2$ : $$y\:u-x=u^2$$ Solving for $u$ leads to : $$u=\frac12\left(y\pm\sqrt{y^2-4x} \right)$$ The condition $u(0,y)=y$ and $y>0$ determines the sign. The solution is : $$u(x,y)=\frac12\left(y+\sqrt{y^2-4x} \right)$$