Can I characterize functions (in 2D), which will have compactly supported/support contained Poisson solution?

Question

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 $$ u_\text{rect}(x,y) = \begin{cases} 1, & \text{for }|x| < A, |y| <B,\\ 0. & \text{elsewhere}. \end{cases} $$ Or $u$'s, which describe thickness of the 3D ball projected to $x,y$ $$ u_\text{ball}(x,y) = \begin{cases} \sqrt{r^2 - y^2 - x^2}, &\text{for }x^2 + y^2 < r^2, \\ 0 & \text{elsewhere}. \end{cases} $$ These functions have the property that if I compute their Laplacian , even in the distribution 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.

Answer 1

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 $D_k$.