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:
\(\ds \int_0^{2 \pi} \int_0^R r \ln r \, \d r \d \theta\) | \(=\) | \(\ds 2 \pi \int_0^R r \ln r \, \d r\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 4 R^2 \paren {\ln R^2 - 1}\) | ||||||||||||
\(\ds \) | \(<\) | \(\ds \infty\) |
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:
- $T_{\Delta \frac {\ln r} {2 \pi} } = \delta_{\tuple {0, 0} }$
$\blacksquare$
Sources
- 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): Chapter $\S 6.2$: A glimpse of distribution theory. Derivatives in the distributional sense