Solving this PDE? My goal is to solve this partial differential equation in $u-v$ coordinates:
$$u\frac{\partial}{\partial u}\left(u\frac{\partial \phi}{\partial u}\right) + v\frac{\partial}{\partial v}\left(v\frac{\partial \phi}{\partial v}\right)=0$$
Some answers below (very helpful) change the coordinates, but I'd like to solve it without changing coordinates.

How do you solve it?

I would show a better attempt but I am really stuck and don't know how to proceed. If someone can give me a reference where this PDE has been solved I could understand the solution.
I tried looking online for the solution but haven't found anything yet.
 A: $$u\frac{\partial}{\partial u}\left(u\frac{\partial \phi}{\partial u}\right) + v\frac{\partial}{\partial v}\left(v\frac{\partial \phi}{\partial v}\right)=0$$
HINT : $\quad\begin{cases}
u=e^x\\
v=e^y
\end{cases}\quad\implies\quad 
\frac{\partial^2\phi}{\partial x^2}+\frac{\partial^2\phi}{\partial y^2}=0.$
A: To explain JJacquelin's hint a bit better, you can rewrite your PDE as
$$ \bigg(u\frac{\partial}{\partial u}\bigg)^2 \phi + \bigg(v\frac{\partial}{\partial v}\bigg)^2 \phi = 0. $$
This looks pretty similar to the wave equation since we have squares of fairly nice first order differential operators. Let's try a coordinate change. Let $u = s(x)$, then
$$\frac{\partial f}{\partial u} = \frac{\partial f}{\partial x} \frac{dx}{du} $$
We would like to end with $\displaystyle\frac{\partial}{\partial x}$ instead of $\displaystyle u\frac{\partial}{\partial u}$, so what that tells us is that $\displaystyle\frac{dx}{du}$ is nothing other than $\displaystyle\frac{1}{u}$. However we know that $\displaystyle \frac{du}{dx} = \bigg(\frac{dx}{du}\bigg)^{-1} = u$. From this, you get $u = e^x$ just as JJacquelin suggested. Same goes for $v$.
A: For a problem like this I would almost automatically go to "separation of variables".
Let $\phi(u,v)= F(u)G(v)$.  Then $u\frac{\partial \phi}{\partial u}= uG(v)\frac{dF}{du}$ so $u\frac{\partial}{\partial u}\left(u\frac{\partial \phi}{\partial u}\right)= u^2G(v)\frac{d^2F}{du^2}+ uG(v)\frac{dF}{du}$.
$v\frac{\partial \phi}{\partial v}= vF(u)\frac{dG}{dv}$ so $v\frac{\partial}{\partial v}\left(v\frac{\partial \phi}{\partial v}\right)= v^2F(u)\frac{d^2G}{dv^2}+ vF(u)\frac{dG}{dv}$
The equation becomes $u^2G(v)\frac{d^2F}{du^2}+ uG(v)\frac{dF}{du}+ v^2F(u)\frac{d^2G}{dv^2}+ vF(u)\frac{dG}{dv}= 0$.
Divide byFG: $\frac{u^2\frac{d^2F}{du^2}+ u\frac{dF}{du}}{F(u)}+ \frac{v^2\frac{d^2G}{dv^2}+ v\frac{dG}{dv}}{G(v)}= 0$ or
$\frac{u^2\frac{d^2F}{du^2}+ u\frac{dF}{du}}{F(u)}= -\frac{v^2\frac{d^2G}{dv^2}+ v\frac{dG}{dv}}{G(v)}$.
Since the left side depends on u only and the right side depends on v only, in order to be equal for all u and v they must be equal to the same constant.  Calling that constant "$\alpha$",
$\frac{u^2\frac{d^2F}{du^2}+ u\frac{dF}{du}}{F(u)}= \alpha$ and
$\frac{v^2\frac{d^2G}{dv^2}+ v\frac{dG}{dv}}{G(v)}= -\alpha$.
Then $u^2\frac{d^2F}{du^2}+ u\frac{dF}{du}- \alpha F= 0$ and
$\frac{v^2\frac{d^2G}{dv^2}+ v\frac{dG}{dv}}{G(v)}+ \alpha G= 0$
Those are ordinary, linear differential equations.  They are specifically "Euler type" equations which can be converted to "constant coefficients" by change of variables.  Or we can look for solutions of the form $F= u^n$ and $G= v^m$ for some n and m.
The first equation becomes $u^2(n(n-2)u^{n-2})+ u(nu^{n-1})-\alpha u^n= u^n(n^2+ n- \alpha)= 0$.  If $u^n$ is not identically 0 (the "trivial solution") we must have $n^2+ n- \alpha= 0$.
The second equation becomes $v^2(m(m-1)v^{m-2}+ u(mu^{m-1})+ \alpha v^m= v^m(m^2+ m+ \alpha)= 0$ so we must have $m^2+ m+ \alpha= 0$.
The solutions to those, of course, depend on $\alpha$ and we would get possible values for  $\alpha$ from the boundary conditions.
