Cauchy Problem PDE with tree variables I have the following pde
$$w_x -y w_y-w_z = 0 \,\,,\,\,\, w = w(x,y,z)$$
$$w(x,y,x)=x+y$$
Using System of characteristic ODEs :
$$\frac{dx}{x}=\frac{dy}{-y}=\frac{dz}{-1}$$
I get:
$$C_1=xy\,\,,\,\,\, C_2=\ln(x)+z$$
General solution of the PDE on implicit form:
$$w(x,y,z)= F(xy,\ln(x)+z)$$
 is an arbitrary function.
My question is how to continue from here in order to find $w(x,y,z)$ that satisfies the initial conditions.
 A: $$w_x -y w_y-w_z = 0 $$
$$\frac{dx}{1}=\frac{dy}{-y}=\frac{dz}{-1}$$
First characteristic equation from solving $\frac{dx}{1}=\frac{dy}{-y}$
$$ye^x=c_1$$
Second characteristic equation from solving $\frac{dx}{1}=\frac{dz}{-1}$
$$x+z=c_2$$
General solution :
$$w(x,y,z)=F(ye^x,x+z)$$
Condition : $w(x,y,x)=x+y=F(ye^x,2x)$
Let $\begin{cases}X=ye^x\\Y=2x\end{cases}\quad\implies\quad \begin{cases} x=\frac{Y}{2}\\y=Xe^{-Y/2}\end{cases}$
$$F(X,Y)=\frac{Y}{2}+Xe^{-Y/2}$$
So, the function $F(X,Y)$ is determined. We put it into the above general solution where $X=ye^x$ and $Y=x+z$.
$$w=\frac{x+z}{2}+ye^xe^{-(x+z)/2}$$
$$\boxed{w(x,y,z)=\frac{x+z}{2}+y\:e^{\frac{x-z}{2}}}$$
A: I did the work with the wrong solution, so I post it :)
From the general solution you got and the initial conditions, the solution can be expressed in a compact form too (with some issues I didn't explore further as the bi-valuate nature of the solution). 
$$w(x,y,x)= F(xy,\ln(x)+x)=x+y$$
or
$$w(x,y,x)= F(xy,\ln(xe^x))=x+y$$
Now we set two new variables $u=xy$ and $v=\ln(xe^x)$
We need isolate $x$ and $y$ as function of $u$ and $v$ in order to find the value of $x+y$. As said it is necessary to use the Lambert W function:
$e^v=xe^x$, then $x=W_k(e^v)$, $k=0,-1$ and $y=u/W_k(e^v)$
We now have $F(u,v)=W_k(e^v)+u/W_k(e^v)$.
So, $w(x,y,z)=F(xy,\ln(xe^z))=W_k(xe^z)+xy/W_k(xe^z)$
