Let $M$ be a closed Riemannian manifold, $g$ be a smooth function with compact support on $M\times\mathbb{R}$. Equip $M\times \mathbb{R}$ with the standard product metric.
My problem : Determine whether there exist a smooth function $u$ such that \begin{align} \begin{cases} \Delta u = g\\ \text{there exist $\delta>0$ with }\lVert e^{\delta\lvert t\rvert}u\rVert_{W^{2,2}}<\infty. \end{cases} \end{align} Here, $\Delta$ is the Laplace-Beltrami operator, and $t$ is the coordinate on $\mathbb{R}$, and $W^{2,2}$ means $W^{2,2}(M\times\mathbb{R})$.
My attempt : From cases in bounded domain with boundary value condition, I guess such $u$ exists and is unique. However, standard textbooks on elliptic PDEs does not treat asymptotic limiting conditions.
If I compose the fundamental solution $G$ to define $u:=\int_{M\times\mathbb{R}}G(x,y)g(y)dy$, I get $\Delta u=g$, with $u\in C^2$. From elliptic regularity, smoothness should follows (This should be local I guess). Once the vanishing of $u$ at infinity is developed, the exponential decay should follows, since harmonic functions on $M\times \mathbb{R}$ vanishes at infinity decays exponentially (by separation of variables and Sturm-Liouville theory); on the other hand, outside the support of $g$, $u$ satisfies $\Delta u=0$.
I think I have read some material saying that the Laplace operator between weighted Sobolev space $\Delta:W^{2,2}_{\delta}\to L^2_{\delta}$ is invertible, but I am not so certain. Some physics book develop the asymptotic vanishing using distribution theory and Fourier transform in $\mathbb{R}^n$. However in my case, the eigenfunctions of $\Delta$ are at worst of exponential growth (w.r.t. $t$), I don't think they can define a tempered distribution. (Schwartz space is only of polynomial decay.)
Any help is appreciated! Special cases for where $M$ is a surface or even $S^1$ is also appreciated!