Fundamental Solution to 2D Laplace's Equation

Theorem
Let $\delta_{\tuple {0, 0}} \in \map {\DD'} {\R^2}$ be the Dirac delta distribution.

Let $\Delta = \dfrac {\partial^2} {\partial x^2} + \dfrac {\partial^2} {\partial y^2}$ be the Laplacian in $2$-dimensional Euclidean space.

Then in the distributional sense:


 * $\ds \map \Delta {\frac {\ln r} {2 \pi}} = \delta_{\tuple {0, 0}}$

where $r = \sqrt {x^2 + y^2}$.

Proof
Let $\norm {\,\cdot\,}_2$ be the 2-norm.

Local integrability
Let $u : \R^2 \to \R$ be a radial real function, say $\map u {\mathbf x} = \map f r$, where $r = \norm {\mathbf x}_2$.

By Laplacian in Polar Coordinates:


 * $\ds \Delta f = \map {f''} r + \frac {\map {f'} r} r$

Thus:


 * $\Delta \map \ln r = 0$

Furthermore, for all $R > 0$ we have:

Hence, $\ln r$ is a locally integrable function:


 * $\paren {\mathbf x \mapsto \ln r} \in \map {L^1_{loc}} {\R^2}$

Fundamental Solution
Let $\phi \in \map \DD {\R^2}$ be a test function with support on the open ball $\map {B_R} {\mathbf 0}$.

Then:

Let $\Omega := \set {\mathbf x \in \R^2 : \epsilon < \norm {\mathbf x}_2 < R}$ with the boundary $\partial \Omega = \map S {\epsilon} \cup \map S {R}$ where


 * $\map S \epsilon = \set {\mathbf x : \norm {\mathbf x}_2 = \epsilon}$


 * $\map S {R} = \set {\mathbf x : \norm {\mathbf x}_2 = R}$

By Green's Identities and the definition of test function:


 * $\ds \int_{\epsilon \mathop < \norm {\mathbf x}_2 \mathop < R} \ln r \Delta \phi \rd \mathbf x = \int_{\norm {\mathbf x}_2 \mathop = \epsilon} \ln r \dfrac {\partial \phi}{\partial n} \rd s - \int_{\norm {\mathbf x}_2 \mathop = \epsilon} \dfrac {\partial \ln r}{\partial n} \phi \rd s$

where:


 * $\dfrac {\partial \phi} {\partial n} = \nabla \phi \cdot \map {\mathbf n} {\mathbf x}$

and $\map n {\mathbf x}$ is the unit normal pointing outward the boundary, while $\rd s$ is a small element of the boundary contour $\partial\Omega$.

For the first integral we have the following estimate:

Furthermore:

Thus:


 * $\ds \lim_{\epsilon \to 0} \size {\int_{\norm {\mathbf x}_2 \mathop = \epsilon} \ln r \dfrac {\partial \phi}{\partial n} \rd \map s {\mathbf x} } \le 0$

For the second integral we have:

By definition, the test function $\phi$ is continuous.

Let $\eta > 0$.

Then there is $\epsilon_0 > 0$ such that $\epsilon_0 > \epsilon > 0$ and:


 * $\norm {\mathbf x}_2 < \epsilon_0 \implies \size {\map \phi {\mathbf x} - \map \phi {\mathbf 0} } < \eta$

Hence:

Since $\eta$ was arbitrary:


 * $\ds \forall \eta \in \R_{>0} : \exists \epsilon_0 > \epsilon : \forall \mathbf x \in \R^2 : \norm {\mathbf x}_2 < \epsilon_0 \implies \size {\frac 1 {2 \pi \epsilon} \int_{\norm {\mathbf x}_2 \mathop = \epsilon} \map \phi {\mathbf x} \rd \map s {\mathbf x} - \map \phi {\mathbf 0} } < \eta$

Thus:

So:


 * $\map {T_{\Delta \ln r}} \phi = 2 \pi \map {\delta_{\tuple {0, 0}}} \phi$

By linearity of the distribution:


 * $T_{\Delta \frac {\ln r}{2\pi}} = \delta_{\tuple{0,0}}$