Characteristics method, an implicit solution I am solving the following PDE:
$$xy(u_x-u_y)=(x-y)u$$
using the characteristic curves method.
To obtain the characteristic curves I have to solve:
$$\dfrac{dx}{xy}=\dfrac{dy}{-xy}=\dfrac{du}{(x-y)u}$$
From $\dfrac{dx}{xy}=\dfrac{dy}{-xy}$ we have $x+y=c$, with $c$ a constant.Substituting in $\dfrac{dx}{xy}=\dfrac{du}{(x-y)u}$ this implies $\dfrac{du}{u}=\dfrac{(x-y)dx}{xy}=\dfrac{(2x+c)dx}{x(c-x)}$.
Integrating on both sides we have $\ln(u)=-\ln((c-x)x)+k_1$ this implies $u(x,y)=\dfrac{k}{xy}$, with $k$ constant.
But check out the solution and it says that the general solution is given by $\Phi(x+y,xyu)=0$ for any arbitrary function $\Phi$.
And then he says that in general if the solutions to the characteristic system are constant, let's say $f (x, y) = c$ and $g (x, y, u) = k$, with $k$ and $c$ constant, then the solution will be given by $\Phi(f,g)=0$, could someone explain to me why you say that please? and see if my solution is correct. Thanks
 A: In method of characteristics, we reduce the quasilinear PDE as an ODE along characteristic curves and hence solve it for points on a characteristic curve.  But at the end of the day, we need to go back and bundle these curves together to form our solution (at least locally).
What you have therefore shown is that, along any characteristic curve, the value of $x+y$ and $xyu$ are constant.  But they are allow to change as we move to another characteristic curve (indeed, each characteristic curve is given by $\{x+y=c\}\cap\{xyu=k\}$ locally (some $(c,k)\in\mathbb{R}^2$)).  Our solution surface $(x,y,u)$ is locally a 1-parameter family of characteristic curves $(c,k)$ but we don't actually know how $c,k$ are related to each other unless we are given initial data.  In other words, we have an arbitrary freedom of reasonable (see next paragraph) parametrisation of this family $(c,k)$ as $\Phi(c,k)=0$, then we have our general solution $\Phi(x+y,xyu)=0$.
Indeed you can check $\Phi(x+y,xyu(x,y))=0$ solves the given PDE locally:  differentiating with respect to $x,y$ gives
\begin{align*}
0&=\frac{\partial}{\partial x}\Phi(x+y,xyu)\\
&=\Phi_{,1}(x+y,xyu)+\Phi_{,2}(x+y,xyu)(yu+xyu_x)\\
0&=\frac{\partial}{\partial y}\Phi(x+y,xyu)\\
&=\Phi_{,1}(x+y,xyu)+\Phi_{,2}(x+y,xyu)(xu+xyu_y)
\end{align*}
If $\Phi_{,2}(x+y,xyu)\neq 0$ (which you must have to say anything about $u$ since the only way for $u$ to come in is through the second slot of $\Phi$), this gives the PDE
$$
yu+xyu_x = xu+xyu_y
$$
which is just a simple rearrangement of the given PDE.  In the general case, we want $\Phi$ to be regular parametrisation (i.e., not both $\Phi_{,1}$ and $\Phi_{,2}$ vanish simultaneously when $\Phi=0$).  Similarly for higher dimensions.
