Suppose all arguments are in $\mathbb{R}^3$, $G(x):\mathbb{R}^3\rightarrow \mathbb{C}$ is a smooth function vanishes at point set $x_j\in\Omega$$j=1\cdots,m$,with multiplicity as 1,
And the equation is
$G\Delta u + 2\nabla G\nabla u + Ghu = Gf$, $h$ and $f$ are some smooth functions.
I wonder if the solution exists.
Or could someone tell me where to find the references about this. The equation is only degenerate at finite points.
Thanks.
[UPDATE] I used another substitution, then the problem turned out to be:
Is there a function $\rho(x):\mathbb{R}^3\rightarrow \mathbb{C}$ s.t.
$$\Delta\rho +\dfrac{\nabla G}{G}\nabla\rho + ik(f(x) - \nabla\rho\cdot\nabla\rho) = 0$$
here $G(x)$ is the same as the previous one, vanishes at some points $x_j$.
So we can see, either we need $\nabla G\cdot\nabla\rho = 0$ at all $x_j$, or $\nabla\rho$ vanishes at $x_j$.
[UPDATE] It is proved that in this form, there is no solution, if $n(x)$ is arbitrary.
The original question is the existence of a function. I put an idea there. And if we treat the problem in this way by solving a nonlinear equation, assuming $n(x)$ is complex and can be extended to the whole domain, we can expand $\rho = \sum_{i=0}^{\infty} \rho_j(\xi,z)\overline{\xi}^j$, and suppose $\rho_j$ is harmonic.
Then we can get a recursion formula for each $\rho_j$, but unfortunately, the recursion cannot be solved if $n(x)$ is arbitrary.
I have posted a solution in that problem, in a very different way, but similar idea.