Fourier transform of $\log|x|$ in $\mathbb{R}^2$ and the solution to $\Delta u = \delta_0$

Question

In my analysis class, we are covering tempered distributions now. I was given this two-part question in Fourier transforms of distributions.

a. We are asked to compute the Fourier transform of $\log |x|$ as a tempered distribution in $\mathbb{R}^2$. Here, $|x|$ is the 2d Euclidean norm of the vector $x$.

b. We are asked to find the fundamental solution to the Laplace equation in $\mathbb{R}^2$, $\Delta u = \delta_0$, where $u$ is a distribution and understood in the sense of weak solutions.

Here, we take the Laplacian $\Delta u = -\frac{\partial^2 u}{\partial x^2}-\frac{\partial^2 u}{\partial y^2}$. We take the Schwartz space of functions $S(\mathbb{R}^2)$ and its continuous dual space, $S'(\mathbb{R}^2)$, the space of tempered distributions. For $T \in S'(\mathbb{R}^2)$ and $\phi \in S(\mathbb{R}^2)$, we use the notation $$\langle T,\phi \rangle = T(\phi)$$ and when $T(x)$ is a function, we define $$\langle T,\phi \rangle = \int_{\mathbb{R}^2} T(x)\phi(x)dx$$ For the Foruier transform on tempered distributions, we define $$ \langle \mathcal{F}T,\phi \rangle = \langle T,\mathcal{F}\phi \rangle. $$ To be honest, I have no idea how to compute the Fourier transform of $\log|x|$ in $\mathbb{R}^2$ and how to use it to find the fundamental solution in part b, I do think I need to move to the Fourier domain in b but other than that I am lost. I thank anyone who can help with parts A and B.

Answer 1

The most coherent/memorable approach I know is not the most direct: first, compute that $\Delta |x|^s=s(s+n-1)\cdot |x|^{s-2}$ on $\mathbb R^n$. One can also check that the residue of the meromorphic family of tempered distributions $s\to |x|^s$ at $s=-n$ is an explicit constant multiple of $\delta$. Thus, for $n\not=2$, the limit as $s\to -n$ shows that $\Delta |x|^{2-n}$ is a constant multiple of $\delta$.

The squared factor $s^2$ disrupts this, moving the residue to the degree-one term in the (distribution-valued) Laurent expansion, but differentiating in $s$ gives reduces the degree. Thus, $$ \Delta(\log|x|\cdot |x|^s) \;=\; \Delta {\partial\over \partial s}|x|^s \;=\; {\partial\over \partial s}\Delta |x|^s \;=\; {\partial\over \partial s} s^2|x|^{s-2} $$ Taking the limit as $s\to 0$ gives $\Delta \log|x|$ in the first term, and captures the residue of $|x|^s$ in the last. Namely, $\Delta \log|x|=2\pi\cdot \delta$.

Taking Fourier transform of this shows that $|x|^2\cdot \widehat{\log|x|}$ is a constant multiple of $1$. Since $1/|x|^2$ is not locally integrable in $\mathbb R^2$, we cannot say that the Fourier transform of $\log|x|$ is (a constant multiple of) $1/|x|^2$...