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In two dimensions, Poisson's equation has the fundamental solution,

$$G(\mathbf{r},\mathbf{r'}) = \frac{\log|\mathbf{r}-\mathbf{r'}|}{2\pi}. $$

I was trying to derive this using the Fourier transformed equation, and the process encountered an integral that was divergent. I was able to extract the correct function eventually, but the math was sketchy at best. I am hoping someone could look at my work and possibly justify it. Here goes.

First off, make the assumption that $G$ only depends on the difference $\mathbf{v}=\mathbf{r}-\mathbf{r'}$. Now, let's write $G$ as an inverse Fourier Transform and take the Laplacian,

$$\nabla^2G(\mathbf{v}) = \int\frac{d^2k}{(2\pi)^2}(-k^2)e^{i\mathbf{k} \cdot \mathbf{v}} \hat{G}(\mathbf{k}) = \delta(\mathbf{v}) $$

For this to be a delta function, we require that $\hat{G}(\mathbf{k}) = -1/k^2$. Now taking the inverse Fourier Transform of $G$...

\begin{align*} G(\mathbf{v}) &= -\int\frac{d^2k}{(2\pi)^2} \frac{e^{i\mathbf{k}\cdot\mathbf{v}}}{k^2} = -\int\limits_{0}^{\infty} \int\limits_{0}^{2\pi} \frac{dkd\theta}{(2\pi)^2} \frac{e^{i|\mathbf{k}||\mathbf{v}|\cos\theta}}{k}\\ &= - \int\limits_0^{\infty}\frac{dk}{2\pi}\frac{J_0(kv)}{k} \end{align*}

Here $J_0$ is a Bessel function of the first kind. This integral is divergent as far as I can tell, but let's continue onward and take a derivative with respect to $|\mathbf{v}|$.

\begin{align*} \frac{dG}{dv} &= \int\limits_0^{\infty}\frac{dk}{2\pi} J_1(kv)\\ &= \frac{1}{2\pi v} \end{align*}

Then integrating this and setting the constant to zero we get the desired result...

$$ G(\mathbf{v}) = \frac{\log v}{2\pi} $$

Clearly this was a lot of heuristics, but I am hoping someone could justify some of this with distributions etc... Could someone tell me what on earth I have done and why it worked?

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  • $\begingroup$ I don't know the answer to your question but I noticed that the same problem occurs when deriving the Retarded greens function for the wave equation: the inverse Fourier transform has two simple poles in frequency space which make the integral divergent. However then we continue and get the correct result in the end... In that case one basically defines a complex integral along a slightly shifted path that gives a finite value. The same trick (cheat) is used in Quantum Field theory to compute the feynman propagator. $\endgroup$ – Cyclone Dec 1 '17 at 17:54
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It suffices to show \begin{align} \int_{\mathbb{R}^2} G(\textbf{r}, \textbf{r}') \nabla^2f(\textbf{r}')\ d^2\textbf{r}' = f(\textbf{r}). \end{align}

Observe we have \begin{align} \int \log|\textbf{r}-\textbf{r}'|\nabla^2f(\textbf{r}')\ d^2\textbf{r}'&=\ \int_{|\textbf{r}-\textbf{r}'|\leq\ \epsilon} \log|\textbf{r}-\textbf{r}'|\nabla^2f(\textbf{r}')\ d^2\textbf{r}'+\int_{|\textbf{r}-\textbf{r}'|>\epsilon} \log|\textbf{r}-\textbf{r}'|\nabla^2f(\textbf{r}')\ d^2\textbf{r}'\\ &=:\ I_1+I_2. \end{align}

Let us first focus on $I_1$. Note that we have the following estimate \begin{align} \left| I_1\right| \leq&\ \|f\|_{C^2(B(0, 1))} \int_{|\textbf{r}-\textbf{r}'|\leq\ \epsilon} \left|\log|\textbf{r}-\textbf{r}'|\right|\ d^2\textbf{r}'\\ \leq &\ -C\int^\epsilon_0 r\log r\ dr \leq C\epsilon \end{align} which goes to $0$ as $\epsilon \rightarrow 0$. For the $I_2$ term, observe we have that \begin{align} I_2 &=\ \int_{|\textbf{r}-\textbf{r}'|=\epsilon} \log|\textbf{r}-\textbf{r}'|\frac{\partial f}{\partial n}(\textbf{r}')\ dS(\textbf{r}')-\int_{|\textbf{r}-\textbf{r}'|>\epsilon} \nabla\log|\textbf{r}-\textbf{r}'|\cdot \nabla f(\textbf{r}')\ d^2\textbf{r}'\\ &=\ \int_{|\textbf{r}-\textbf{r}'|=\epsilon} \log|\textbf{r}-\textbf{r}'|\frac{\partial f}{\partial n}(\textbf{r}')- \frac{\partial \log|\textbf{r}-\textbf{r}'|}{\partial n}f(\textbf{r}')\ dS(\textbf{r}')\\ &=:\ J_1+J_2. \end{align} For the $J_1$ term we have the estimate \begin{align} |J_1| \leq |\log \epsilon| \int_{|\textbf{r}-\textbf{r}'| = \epsilon} \left|\frac{\partial f}{\partial n}(\textbf{r}')\right|\ dS(\textbf{r}') \leq C|\epsilon \log \epsilon| \end{align} which also goes to $0$ as $\epsilon \rightarrow 0$.

Lastly, observe \begin{align} -\int_{|\textbf{r}'|=\epsilon} \frac{\partial \log|\textbf{r}'|}{\partial n} f(\textbf{r}-\textbf{r}')\ dS(\textbf{r}')= - \int_{|\textbf{r}'| = \epsilon} \frac{f(\textbf{r}-\textbf{r}')}{|\textbf{r}'|} dS(\textbf{r}')\rightarrow -2\pi f(\textbf{r}'). \end{align} as $\epsilon \rightarrow 0$.

Note: If we use Fourier transform for the Poisson equation $\Delta u = \delta$, we get \begin{align} -4\pi^2|\xi|^2 \hat u= 1 \end{align} which means \begin{align} \hat u = \frac{-1}{4\pi^2|\xi|^2}. \end{align} Taking the Fourier inverse of $\hat u$ yields the desired result.

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  • $\begingroup$ Very nice proof. How would you show the inverse Fourier transform in the last line gives the desired result? Probably not through my method right? Could you perhaps add a note on this? $\endgroup$ – ClassicStyle Feb 20 '17 at 6:33
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    $\begingroup$ This doesn't answer the question as to how one can use distributions to write the $2-D$ Green (or Green's) function for Laplace's equation as a Fourier transform. $\endgroup$ – Mark Viola Feb 20 '17 at 7:11

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