Relativistic Poisson Problem I am doing a research project related to general relativity, and there is a PDE that occurs given by
$$
\theta\Delta \theta = \omega
$$
where $\omega\leq 0$ is smooth with compact support and $\theta$ has asymptotic boundary conditions $\theta\to 1$ on $\mathbb{R}^3.$ I am specifically looking at the limiting behavior of solutions when $\omega$ is collapsing to a line Dirac delta. I have tried finding cylindrically symmetric solutions for cylindrically symmetric $\omega$ by writing out the Laplacian in these coordinate systems, but I do not think these equations are solvable in general. I would like to get any recommendations for techniques that might be useful in approaching this problem.
Since creating this post I realized that for understanding limiting behavior of solutions with high mass density concentrated in a small region, it's probably good enough to assume $\theta\to 0$ asymptotically. Then scaling $\theta$ by a factor gives a solution to the problem where $\omega$ is scaled by the square of that factor.
 A: This is too long for a comment. In your original post you mentioned
the case where $\omega$ tended to a negative point mass. For that
I can give examples. Take $m$ any $C^2$ function with
bounded support in $[0,\infty)$,  and scale it if necessary so that
$$
 \int_0^\infty s^{3/2}\big(m'(s)\big)^2\,ds = \frac{1}{8\pi}.
$$
Set
$$
 \theta(x) = h^{1/2}m(h|x|^2),
$$
where $h$ is a positive parameter.
Then one calculates, writing $r$ for $|x|$,
$$ 
 \omega(x)
 =
 \theta\Delta\theta
 =
 h^{3/2}m(hr^2)\Big(
     6m'(hr^2)+4hr^2m''(hr^2)
                \Big)
$$
We show now that $\omega$ tends to $-\delta$ as $h$ tends to $\infty$.
Let $f$ be any continuous function. Use a change of variable
$x= h^{-1/2}y$ to calculate
$$
 \int_{R^3}\omega(x)f(x)\,d^3x
 =
 \int_{R^3}
 h^{3/2}m(|y|^2)\Big(
     6m'(|y|^2)+4|y|^2m''(|y|^2)
           \Big)
            f(h^{-1/2}y)\frac{d^3y}{h^{3/2}}.
$$
As $h$ tends to $\infty$, this goes to
$$
 \int_{R^3}
m(|y|^2)\Big(
     6m'(|y|^2)+4|y|^2m''(|y|^2)
           \Big)
            f(0)\,d^3y
 $$ $$ =
 f(0)\int_0^\infty
         \Big(
     m(s) \big(
       6m'(s)+4sm''(s)
          \big)
         \Big)4\pi s\frac{ds}{2s^{1/2}}
$$
I have written $|y|^2 = s$ and integrated using spherical coordinates.
Integrate the second derivative term
by parts and use the bounded support of $m$ to see
that this is equal to
$$
 -f(0)\int_0^\infty 8\pi s^{3/2}\big(m'(s)\big)^2\,ds.
$$
With our assumed rescaling of $m$ this is
$$
 = -f(0)
$$
as required.
I don't see how to do a line version of this.
