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In 2D, the Poisson equation

$$ \nabla^2 \Phi(\vec{x}) = \delta^{(2)}(\vec{x}) $$

admits the solution

$$ \Phi(\vec{x}) = \frac{1}{2\pi}\log(|\vec{x}|) + \text{constant}\,. \tag{1} $$

By using Fourier transforms, we know that the integral

$$ \Phi(\vec{x}) = \int \frac{d^2 \vec{k}}{(2\pi)^2} \frac{-e^{i \vec{k}\cdot \vec{x}}}{\vec{k}^2} \tag{2} $$

is a formal solution but it is divergent.

What is the consistent way to give a meaning to Eq.(2) such that we can use complex analysis to compute the solution Eq.(1)? I don't want to solve this equation by using polar coordinates. I want to use the analytic structure of the integrand in Eq.(2), together will complex analysis theorems, e.g. Cauchy theorems. So my question would be really: what is the path in the complex $\vec{k}$-plane that allows me to integrate Eq.(2)?

Denoting $\vec{k}=(k_1,k_2)$, we see that in the complex $k_1$-plane there is a pole at $k_1=\pm i k_2$. Is there a prescription to reduce the computation of Eq.(1) to computation of residues, for example?

apt45
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1 Answers1

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We see that

$$\mathscr{F}\{\Phi\}(\vec k)=\frac{1}{k^2}\tag1$$

for $|\vec k|\ne 0$. But $(1)$ is not true for all $\vec k\in \mathbb{R}^2$.

We can solve, instead, the problem

$$\nabla^2 \Phi(\vec x)-\varepsilon^2 \Phi(\vec x)=\delta^{(2)}(\vec x)\tag2$$

where $\varepsilon\in \mathbb{R}$. Now, taking the 2-D Fourier transform $(2)$, we can assert that

$$\mathscr{F}\{\Phi\}(\vec k)=-\frac{1}{k^2+\varepsilon^2}\tag3$$

Taking the inverse Fourier transform of $(3)$ and applying the residue theorem, we see that

$$\begin{align} \Phi(\vec x)&=-\frac1{(2\pi) ^2}\int_{-\infty}^\infty \int_{-\infty}^\infty \frac{e^{ik_xx}e^{ik_yy}}{k_x^2+k_y^2+\varepsilon^2}\,dk_x\,dk_y\\\\ &=-\frac1{(2\pi) ^2}\int_{-\infty}^\infty 2\pi i \left(\frac{e^{ik_xx}e^{-\sqrt{k_x^2+\varepsilon^2}\,|y|}}{2i\sqrt{k_x^2+\varepsilon^2}}\right) \,dk_x\\\\ &=-\frac1{2\pi}\int_0^\infty \frac{e^{-\sqrt{k_x^2+\varepsilon^2}\,|y|}}{\sqrt{k_x^2+\varepsilon^2}} \cos(k_x x)\,dk_x\tag4 \end{align}$$

In THIS ANSWER, I showed that the right-hand side of $(4)$ is equal to $K_0(\varepsilon \sqrt{x^2+y^2})$. Thus, we have from $(4)$ that

$$\Phi(\vec x)=-\frac1{2\pi}K_0(\varepsilon |\vec x|)\tag 5$$

Finally, using the small argument asymptotic expansion for the modified Bessel function of the second kind reveals that

$$\begin{align} \Phi(\vec x)&\sim -\frac1{2\pi}\left(-\gamma-\log\left(\frac{\varepsilon |\vec x|}{2}\right)\right)\\\\ &=\frac1{2\pi}\log(|\vec x|)+\text{constant} \end{align}$$

as was to be shown!

Mark Viola
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