Green's Function for 2D Poisson Equation

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

The fundamental solution $G$ to the 2D Poisson equation satisfies the distributional equation $$ \frac{\partial^2 G}{\partial x^2} + \frac{\partial^2 G}{\partial y^2} = \delta . $$

Taking the Fourier transform gives $(i\xi)^2 \hat G + (i\eta)^2 \hat G = 1,$ so $$ \hat G(\xi, \eta) = -\frac{1}{\xi^2 + \eta^2} . $$

Now we take the inverse Fourier transform w.r.t. $\eta$: $$ \mathcal{F}_{\eta}^{-1} \{ -\frac{1}{\xi^2 + \eta^2} \} = -\frac{1}{2|\xi|} \mathcal{F}_{\eta}^{-1} \{ \frac{2|\xi|}{\xi^2 + \eta^2} \} = - \frac{1}{2|\xi|} e^{-|\xi||y|} . $$

Next we want to take the inverse Fourier transform of the previous result: $$ \mathcal{F}_{\xi}^{-1} \{ - \frac{1}{2|\xi|} e^{-|\xi||y|} \} $$ How do we do that? The solution is to start with a derivative of it: $$ \frac{d}{d|y|} \mathcal{F}_{\xi}^{-1} \{ - \frac{1}{2|\xi|} e^{-|\xi||y|} \} = -\frac{1}{2} \mathcal{F}_{\xi}^{-1} \{ e^{-|\xi||y|} \} = -\frac{1}{2} \frac{1}{2\pi}\frac{2|x|}{x^2+y^2} = -\frac{1}{4\pi}\frac{2|x|}{|x|^2+|y|^2} . $$ Taking the antiderivative then gives us $$ G(x,y) = -\frac{1}{4\pi} \ln(|x|^2+|y|^2) = -\frac{1}{2\pi} \ln\sqrt{x^2+y^2} . $$