Can I characterize functions (in 2D), which will have compactly supported/support contained Poisson solution I have the problem of solving Poisson equation in 2D
$$\Delta u = f$$
Let's say for a moment I want to solve it on $\mathbb{R}^2$, for $f(x,y), x\in \mathbb{R}, y\in \mathbb{R}$.
I know however that my solution $u$ must have compact support or more precisely support within the support of $f$ to be physically relevant for my application. But for many of $f$'s the $u$ will just run away so I believe there must be some characterization of such $f$'s that their $u$'s do not run away.
I am interested e.g.in $u$'s with jump discontinuities such as rectangle
\begin{equation}
u_{rect}(x,y) =  1, for |x| < A, |y| <B,  \quad u_{rect}(x,y) = 0 \text{elsewhere}.
\end{equation}
Or $u$'s, which describe thickness of the 3D ball projected to $x,y$
$$u_{ball}(x,y) = \sqrt{r^2  - y^2 - x^2}, for  x^2 + y^2 < r^2, \quad u_{rect}(x,y) = 0 \text{elsewhere}.$$
These functions has the property, that if I compute their Laplace operator, even in the distributive sense, the support of $f = \Delta u$ is not  much bigger then the support of $u$ itself. So I know there are such $f$'s, which I want.
Can I somehow approach the inverse problem and characterize such $f$'s so that the support of $u$ that solves $\Delta u = f$ will be always contained inside the support of $f$? So I need something as if the $u$ is nonzero at $x$, then $f(x)$ is nonzero or $x$ lies on a line segment between $x_1$ $x_2$ so that $f(x_1)$ and $f(x_2)$ are nonzero.
 A: Since we are in the plane I use complex notation.
The general solution of your equation is the sum of the
potential and an arbitrary harmonic function:
$$u(z)=\frac{1}{2\pi}\int\int\log|z-\zeta| f(\zeta)dm_\zeta+h(z)=P(z)+h(z),$$
where $dm$ is the area element, and $h$ is harmonic in the plane. Let $K$ be the support of $f$, and $D$ the exterior component of $C\backslash K$. We conclude from this formula that for $u$ to have compact support, we must have
$P(z)=-h(z)$ in a neighborhood of $\infty$. In other words,
$P$ must have a harmonic extension from a neighborhood of $\infty$ to the whole plane. For this $f$ has to satisfy
infinitely many conditions. For $z$ sufficiently large,
we have
$$\log|z-\zeta|=\log|z|-\sum_{k=1}^\infty\Re\frac{\zeta^k}{k z^k}.$$ Substituting this to our formula for $P$, and integrating, we obtain a series in the powers of $z$. The condition that $P$ has a harmonic extension is that
all terms of this series must vanish:
$$\int\int f(\zeta)d\mu_\zeta=0, \quad \mbox{and}\quad
\int\int \Re\left(\zeta^k\right)f(\zeta)d\mu_\zeta=0,\quad k=1,2,\ldots.$$
These are necessary and sufficient conditions to have a solution with compact support.
Remark. If $K$ separates the plane (the complement also has bounded components $D_j$, these conditions are not sufficient for support to be contained in the support of $f$.
You will have additional conditions for
$P(z)$ in $D_k$ to match the extension of $P(z)$ from $D$ to  $D_k$.
