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.
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.