Questions about a linear PDE Consider the following linear PDE:
$$u\frac{\partial}{\partial u}\left(u\frac{\partial f}{\partial u}\right) + v\frac{\partial}{\partial v}\left(v\frac{\partial f}{\partial v}\right)=\frac{\partial^2 f}{\partial u^2} + \frac{\partial^2 f}{\partial v^2}\tag{1}$$
where we have $f(u,v)$.

Has anyone studied this PDE in the literature?


Are there weak solutions to $(1)?$

I constructed this PDE by deriving the Laplacian in two different coordinate systems and then made the coordinate variables the same and equated the expressions, which is what you see on the LHS and RHS. I'm not sure what this process is called or if anyone has used it before.
 A: Separation of variables: let $f(u,v)=U(u)V(v)$. Substituting into the PDE and simplifying we get
$$
\frac{1}{U}\left[uU'+u^2 U''-U'' \right]=-\frac{1}{V}\left[vV'+v^2 V''-V'' \right]
$$
Introducing the arbitrary constant $k^2$
$$\tag{1}
uU'+(u^2-1)U''=k^2U 
$$
$$\tag{2}
vV'+(v^2-1)V''=-k^2V 
$$
Let us study (1). The results will carry over to (2) with the replacement $k^2 \to-k^2$. This ODE may be cast in Sturm–Liouville form
$$\tag{3}
\frac{d}{du}\left( \sqrt{1-u^2} U'\right) =-k^2 \frac{U}{\sqrt{1-u^2}}
$$
With weight $w(u)=1/\sqrt{1-u^2}$ and eigenvalue $k^2$. For a boundary value problem posed on $-1<u<1$, eq (3) is a regular Sturm–Liouville equation$^\dagger$. With the assistance of Mathematica, we find the two solutions
$$\tag{4}
U^1_k(u)=\sin \left[k \tan^{-1} \left( \frac{u}{\sqrt{1-u^2}}\right)\right] \\
U^2_k(u)=\cos\left[k \tan^{-1} \left( \frac{u}{\sqrt{1-u^2}}\right)\right] 
$$
The general solution to (1) is then
$$\tag{5}
U(u)=\sum\limits_k (A^1_k U^1_k(u) + A^2_k U^2_k(u))
$$
With coefficients $A^1$ and $A^2$ chosen to match boundary conditions. The sum is over all allowed values of $k$ consistent with the boundary data. For example, with the Dirichlet problem $U(0)=0$, $U(1)=0$, all $A^2$ vanish, and $k=2n$, for $n \in \mathbb{Z}$
We  can find the solutions to (2) with the replacement $k\to ik$ in (4)
$$\tag{6}
V^1_k(v)=\sinh \left[k \tan^{-1} \left( \frac{v}{\sqrt{1-v^2}}\right)\right] \\
V^2_k(v)=\cosh\left[k \tan^{-1} \left( \frac{v}{\sqrt{1-v^2}}\right)\right] 
$$
The general solution for $f$ is
$$\tag{7}
f(u,v)=\sum\limits_k (A^1_k U^1_k(u) + A^2_k U^2_k(u))(B^1_k V^1_k(v) + B^2_k V^2_k(v))
$$
In order to actually find coefficients once boundary data is set, you will need to use the orthogonality of the eigenfunctions with weight $w$. For the $U$ equation, these read
$$\tag{8}
\int\limits_{u_1}^{u_2}du\ (1-u^2)^{-1/2} U^i_k(u)U^j_{k'}(u)  =\delta^{ij}\delta_{k,k'}
$$
Where $i,j=1,2$, and $u_1<u<u_2$. An identical equation follows for the $V$. Continuing the example, if $V(0)=0$ and $V(1)=1$ we may apply (8) to find $A^1_n = \frac{\sin^2(n \pi /2)}{n \sinh(n \pi)}$. Thus the solution with these boundary conditions and $U(0)=0$, $U(1)=0$ is
$$\tag{9}
f(u,v)=\sum\limits_{n} \frac{ \sin^2(n \pi/2)}{n \sinh(n\pi)}\sin(2n \tan^{-1}(u/\sqrt{1-u^2}))\sinh(2n \tan^{-1}(v/\sqrt{1-v^2}))
$$
Here is a plot of the solution in (9)

$\dagger$ For $u>1$ the equivalent of (3) reads
$$
\frac{d}{du}\left( \sqrt{u^2-1} U'\right) =k^2 \frac{U}{\sqrt{u^2-1}}
$$
In this case, the eigenvalue is $-k^2$, the weight $w(u)=1/\sqrt{u^2-1}$, and in the solutions (4), all trig/ hyperbolic functions are to  be  exchanged.
