Poisson Equation: 'Dirichlet' type problem on all of $\mathbb{R}^N$ I've seen a great deal about solving the Dirichlet problem for Poisson's equation
\begin{equation}
\Delta u = f \quad \mbox{in} \quad \Omega \subset \mathbb{R}^N\\
u = 0  \quad \mbox{on} \quad \partial\Omega
\end{equation}
when $\Omega$ is bounded, and quite a bit when $\Omega$ is unbounded but has nontrivial complement (e.g. when it's all of $\mathbb{R}^N$ except for the unit ball). My question concerns when, if at all, this is possible when $\Omega = \mathbb{R}^N$.
Obviously, the boundary of $\mathbb{R}^N$ is meaningless, but it seems reasonable to consider an analogous problem in the class of functions with the 'boundary condition' $u \to 0$ as $|x| \to \infty$.
However, we quickly run into problems; to understand what the problem might entail I picked $f = \exp(-x^2)$ and  $\Omega = \mathbb{R}$ as friendly looking example and found to my dismay that the general solution takes the form
\begin{equation}
u(x) = \frac{\sqrt{\pi}}{2}x\mbox{erf}(x)+\frac{1}{2}\exp(-x^2)+Ax+B
\end{equation}
where erf$(x) = \frac{2}{\sqrt{\pi}}\int_0^x\exp(-t^2)\mathrm{d}t$ is the error function and $A,B$ are constants. 
The $x$erf$(x)$ term assures that for any choice of $A,B$ that $u\to\infty$ as $|x| \to \infty$, at least in one of the directions: if we pick $A=\frac{\sqrt{\pi}}{2}$ the forward direction decays appropriately but not so in the backward direction, and vice versa. 
Of course, on a bounded domain we can always make a choice of constants such that the Dirichlet boundary condition is fulfilled, in particular on $[-L,L]$ the choice $A=0$, $B=-\frac{\sqrt{\pi}}{2}L\mbox{erf}(L)-\frac{1}{2}\exp(-L^2)$, but clearly as $L\to\infty$, $B$ also grows without bound. 
My question then, is how bad is the situation? 


*

*Do things improve for $N\geq 2$? The greater variety of harmonic functions might mean we can find one such that summing it with the solution derived via convolution with the Newton Kernel will allow us to satisfy the decay conditions but then again, maybe not. How can we understand this?

*Can we find success by restricting the class of functions? In 1D we can guarantee that the integral of a function decays iff the mass of the function is zero; is there an analogy for integrating twice?

*Assuming the answer to either of these is positive, can we then say anything about the integrability of the resulting $u$ over $\mathbb{R}^N$? In particular, I'd like to be able to talk about quantities such as $||\Delta^{-1/2} f||_{L^2 (\mathbb{R}^N)}$, but I'm not sure this will be possible without moving into some kind of weighted spaces if at all.
 A: This is studied in potential theory: the function $u$ is the Newtonian potential of $f$, 
$$u(x)=\int_{\mathbb{R}^n} K(x-y)f(y)\,dy$$
where $K(x)=c_n|x|^{2-n}$ for $n\ne 2$ and $K(x)=c_2\log|x|$ for $n=2$.
In dimensions $n\ge 3$ the kernel $K$ decays at infinity, so $u(x)\to 0$ as $|x|\to\infty$ in this case, provided $f$ is reasonable (integrable and vanishing at infinity should be enough).
In dimensions $n\le 2$ the kernel is unbounded. Intuitively: if you place a charge somewhere, the potential it generates grows without a bound because there is not enough room in low dimensions for the electric field to dissipate rapidly enough.  The only hope to get a solution vanishing at infinity is to have zero net charge: $\int f = 0$. 
In two dimensions the condition $\int f = 0$ indeed helps: with it we can write
$$u(x)=\int_{\mathbb{R}^2} (\log|x-y|-\log|x|)f(y)\,dy$$
and now the function in parentheses decays as $|x|\to\infty$. So, an integrable $f$ with  $\int f = 0$ and good behavior at infinity will work.
In one dimension it's worse. A solution of $$u''(x)=(\operatorname{sign}x )\chi_{[-1,1]} (x)$$
cannot have zero limit at both $\infty $ and $-\infty$.  You also need the first moment of $f$ to be zero, $\int x\,f(x)=0$. Indeed, the second derivative of any function vanishing at $\pm\infty$ has zero first moment: 
$$
\int xu''(x) = - \int u'(x) = u(\infty)-u(-\infty) = 0
$$
Integrability of solution
The integrability depends on subtle cancellations that are difficult to quantify in the physical space. Usually one deals with  quantities like $||\Delta^{-1/2} f||_{L^2 (\mathbb{R}^N)}$ by means of the Fourier transform. It is essentially the norm of $f$ in the negative-order Sobolev space $H^{-1}$. 
