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?

  • $\begingroup$ 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, 2017 at 17:54

3 Answers 3


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.

  • $\begingroup$ 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? $\endgroup$ Feb 20, 2017 at 6:33
  • 2
    $\begingroup$ 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, 2017 at 7:11

Strictly speaking the Green function isn't Fourier transferrable as it is not L2 integrable. Any math that attempt to show directly the FT relation is necessarily flimsy. One remedy is to multiply an exponential function that decreases to 0 toward infinity but remain a constant 1 effectively within any finite region of interest. This way you should be able to justify otherwise flimsy math rigorously by carefully consider the FT integral toward infinity. However, the result implies that the Green function is the limit of a sequence of L2 functions whos FT converges to -1/k^2 pointwise but not in L2, as the L2 limit does not exist.

  • $\begingroup$ This is not an answer $\endgroup$ Jul 15, 2021 at 18:40
  • $\begingroup$ Log(R) is not L2, but Log(R)exp(-\epsilon R) where $\epsilon>0$ is. The rest is a standard calculus exercise, followed by taking $\epsilon$ down to 0. The limit exists pointwise both in spatial and Fourier domains, but not in L2. This is essentially saying that a unilateral Laplace's transform exists but not the Fourier transform, as the GF is too singular to live in the L2 space. What kind of answer are you looking for? $\endgroup$
    – Jin Sun
    Jul 15, 2021 at 19:05

Although I think @Jacky Chong's answer is sufficient, it doesn't quite "demystify" why the OP's formal manipulations work. I will try to sketch why exactly the OP's method works and leave the details up to the reader.

Indeed, suppose that there exists some tempered distribution $G$ (which is realized by some function also denoted as $G$) such that $-\Delta G=\delta$. Then we know that for any test function $f$, we have \begin{align} \int (-\Delta G)^\wedge(y) f(y)dy=\int\hat{\delta}(y)f(y)dy \end{align} Where the hat notation represents Fourier transform (in the distribution sense). By definition, we equivalently have \begin{align} \int \hat{G}(y) y^2 f(y)&=\int f(y)dy \end{align} Therefore, from the LHS, you can see that the condition $-\Delta G=\delta$ only requires $\hat{G}$ to act on functions that are zero at $y=0$. Its action on functions nonzero at $y=0$ can be arbitrarily chosen as long is the definition is consistent. Indeed, $G$ is not unique due to the fact that we can add any function $g$ which satisfies $-\Delta g =0$, i.e., replace $G\mapsto G+g$. Hence, we can possibly use the "principal valued function", which I define as \begin{align} \int\text{PV}\left(\frac{1}{y^2}\right) h(y)dy =\int_{|y| \le 1} \frac{h(y)-h(0)}{y^2} +\int_{|y|> 1} \frac{h(y)}{y^2} \end{align} For any test function $h(y)$. It's then clear that $\hat{G} = \text{PV} (1/y^2)$ is a well-defined tempered distribution in 2-dimensions and also satisfies $-\Delta G=\delta$. With this proper definition, you can repeat the OP's method and find that \begin{equation} \left( \text{PV} \left(\frac{1}{y^2}\right)\right)^\vee=-\frac{1}{2\pi} \log|x|+C \end{equation} Where $\vee$ denote inverse Fourier transform and $C$ is some constant.


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