Fundamental Solution to 2D Laplace's Equation

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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:

\(\ds \int_0^{2 \pi} \int_0^R r \ln r \, \d r \d \theta\) \(=\) \(\ds 2 \pi \int_0^R r \ln r \, \d r\)
\(\ds \) \(=\) \(\ds \frac 1 4 R^2 \paren {\ln R^2 - 1}\)
\(\ds \) \(<\) \(\ds \infty\)

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:

\(\ds \map {T_{\Delta \ln r} } \phi\) \(=\) \(\ds \map {T_{\ln r} } {\Delta \phi}\) Definition of Distributional Derivative
\(\ds \) \(=\) \(\ds \int_{\R^2} \ln r \Delta \phi \rd \mathbf x\)
\(\ds \) \(=\) \(\ds \int_{\norm {\mathbf x}_2 \mathop < R} \ln r \Delta \phi \rd \mathbf x\) Definition of Test Function
\(\ds \) \(=\) \(\ds \lim_{\epsilon \to 0} \int_{\epsilon \mathop < \norm {\mathbf x}_2 \mathop < R} \ln r \Delta \phi \rd \mathbf x\)

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 with respect to the boundary, while $\rd s$ is a small element of the boundary contour $\partial\Omega$.

For the first integral we have the following estimate:

\(\ds \size {\int_{\norm {\mathbf x}_2 \mathop = \epsilon} \ln r \dfrac {\partial \phi}{\partial n} \rd \map s {\mathbf x} }\) \(=\) \(\ds \ln \epsilon \size{ \int_{\norm {\mathbf x}_2 \mathop = \epsilon} \nabla \phi \cdot \map {\mathbf n} {\mathbf x} \rd \map s {\mathbf x} }\)
\(\ds \) \(\le\) \(\ds \ln \epsilon \int_{\norm {\mathbf x}_2 \mathop = \epsilon} \size {\nabla \phi \cdot \map {\mathbf n} {\mathbf x} } \rd \map s {\mathbf x}\)
\(\ds \) \(=\) \(\ds \ln \epsilon \int_{\norm {\mathbf x}_2 \mathop = \epsilon} \norm {\nabla \phi \cdot \map {\mathbf n} {\mathbf x} }_2 \rd \map s {\mathbf x}\)
\(\ds \) \(\le\) \(\ds \ln \epsilon \int_{\norm {\mathbf x}_2 \mathop = \epsilon} \sqrt{ \norm {\nabla \phi}_2 \norm {\map {\mathbf n} {\mathbf x} }_2} \rd \map s {\mathbf x}\) Cauchy-Schwarz inequality
\(\ds \) \(\le\) \(\ds \ln \epsilon \int_{\norm {\mathbf x}_2 \mathop = \epsilon} \sqrt{\norm {\nabla \phi}_2} \rd \map s {\mathbf x}\) $\norm {\map {\mathbf n} {\mathbf x} }_2 = 1$
\(\ds \) \(\le\) \(\ds \ln \epsilon \sup_{\norm {\mathbf x}_2 \mathop = \epsilon} \sqrt{\norm {\nabla \phi}_2} \int_{\norm {\mathbf x}_2 \mathop = \epsilon} \rd \map s {\mathbf x}\)
\(\ds \) \(\le\) \(\ds \sup_{\norm {\mathbf x}_2 \mathop = \epsilon} \sqrt{\norm {\nabla \phi}_2} 2 \epsilon \ln \epsilon\)

Furthermore:

\(\ds \lim_{\epsilon \mathop \to 0} 2 \pi \epsilon \ln \epsilon\) \(=\) \(\ds 2 \pi \lim_{\epsilon \mathop \to 0} \frac {\ln \epsilon}{\frac 1 \epsilon}\) Indeterminate Form: $0 \cdot \infty$
\(\ds \) \(=\) \(\ds 2 \pi \lim_{\epsilon \mathop \to 0} \frac {\frac 1 \epsilon}{- \frac 1 {\epsilon^2} }\) L'Hospital's Rule
\(\ds \) \(=\) \(\ds - 2 \pi \lim_{\epsilon \mathop \to 0} \epsilon\)
\(\ds \) \(=\) \(\ds 0\)

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:

\(\ds - \int_{\norm {\mathbf x}_2 \mathop = \epsilon} \dfrac {\partial \ln r}{\partial n} \map \phi {\mathbf x} \rd \map s {\mathbf x}\) \(=\) \(\ds \int_{\norm {\mathbf x}_2 \mathop = \epsilon} \dfrac {\partial \ln r}{\partial r} \phi \rd \map s {\mathbf x}\) $\map {\mathbf n} {\map S \epsilon} = - \hat {\mathbf r}$
\(\ds \) \(=\) \(\ds \int_{\norm {\mathbf x}_2 \mathop = \epsilon} \frac 1 r \map \phi {\mathbf x} \rd \map s {\mathbf x}\)
\(\ds \) \(=\) \(\ds \frac 1 \epsilon \int_{\norm {\mathbf x}_2 \mathop = \epsilon} \map \phi {\mathbf x} \rd \map s {\mathbf x}\)
\(\ds \) \(=\) \(\ds 2 \pi \paren {\frac 1 {2 \pi \epsilon} \int_{\norm {\mathbf x}_2 \mathop = \epsilon} \map \phi {\mathbf x} \rd \map s {\mathbf x} }\)

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:

\(\ds \size {\frac 1 {2 \pi \epsilon} \int_{\norm {\mathbf x}_2 \mathop = \epsilon} \map \phi {\mathbf x} \rd \map s {\mathbf x} - \map \phi {\mathbf 0} }\) \(=\) \(\ds \size {\frac 1 {2 \pi \epsilon} \int_{\norm {\mathbf x}_2 \mathop = \epsilon} \map \phi {\mathbf x} \rd \map s {\mathbf x} - \frac 1 {2 \pi \epsilon} \int_{\norm {\mathbf x}_2 \mathop = \epsilon} \map \phi {\mathbf 0} \rd \map s {\mathbf x} }\)
\(\ds \) \(=\) \(\ds \frac 1 {2 \pi \epsilon} \int_{\norm {\mathbf x}_2 \mathop = \epsilon} \size {\map \phi {\mathbf x} - \map \phi {\mathbf 0} } \rd \map s {\mathbf x}\)
\(\ds \) \(\le\) \(\ds \frac 1 {2 \pi \epsilon} \cdot \eta \int_{\norm {\mathbf x}_2 \mathop = \epsilon} \rd \map s {\mathbf x}\)
\(\ds \) \(=\) \(\ds \frac 1 {2 \pi \epsilon} \cdot \eta \cdot 2\pi \epsilon\)
\(\ds \) \(=\) \(\ds \eta\)

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:

\(\ds \lim_{\epsilon \to 0} \int_{\epsilon \mathop < \norm{\mathbf x}_2 \mathop < R} \ln r \Delta \phi \rd \mathbf x\) \(=\) \(\ds 2 \pi \lim_{\epsilon \to 0} \paren {\frac 1 {2 \pi \epsilon} \int_{\norm {\mathbf x}_2 \mathop = \epsilon} \map \phi {\mathbf x} \rd \map s {\mathbf x} }\)
\(\ds \) \(=\) \(\ds 2 \pi \map \phi {\mathbf 0}\)
\(\ds \) \(=\) \(\ds 2 \pi \map {\delta_{\tuple {0, 0} } } \phi\) Definition of Dirac Delta Distribution

So:

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

By linearity of the distribution:

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

$\blacksquare$


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