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In Tao' blog:https://terrytao.wordpress.com/2009/04/19/245c-notes-3-distributions/#comment-551104 Consider the Poisson equation: $$\Delta K =\delta$$

The Exercise 37: Show that for $d=2$, a fundamental solution to the Poisson equation is given by the locally integrable function $K(x)= \frac{1}{2\pi} \log |x|$

We can get the solution for $d\geq 3$ by the transform of $$\widehat{|x|^{-\alpha}}(\xi)=C|\xi|^{-(d-\alpha)} (*)$$. But how can we use a similar method to get $K(x)$? Tao mentions that "requiring a logarithmic correction to the usual formula". Does it mean we need to take the log of (*)?

Hermi
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The usual formula gives $$ \Delta^{-1}(x) = \frac{|x|^{2-d}}{\gamma_{d}(2-d)} $$ as the solution to $\Delta\left[\Delta^{-1}(x)\right]=\delta(x)$ in $d$ dimensions, where $$ \gamma_{d} = \frac{2\pi^{\frac{d}{2}}}{\Gamma\left(\tfrac{d}{2}\right)} $$ is the $d$-dimensional solid angle. Consider now $d=2-\epsilon$ for small $\epsilon$. Expanding the above expression we find $$ \Delta^{-1}(x) = \frac{1}{\epsilon} + \frac{1}{2\pi}\,\log|x| + \cdots $$ where the dots denote either terms that vanish as $\epsilon\to0$ or are independent of $|x|$. Dropping the $x$-independent terms and sending $\epsilon\to0$ we find $$ \Delta^{-1}(x)=\frac{1}{2\pi}\,\log|x| $$ for $d=2$.

Indeed, consider $$ \frac{1}{2\pi}\int_{\mathbb R^2} \log|x|\ \Delta \chi (x)\,dx = \lim_{\epsilon\to0}\frac{1}{2\pi}\int_{\mathbb R^2\setminus B_\epsilon} \log|x|\ \Delta \chi (x)\,dx $$ for any smooth test function $\chi$, where $B_\epsilon$ denotes a disk centered at the origin with radius $\epsilon$. Integrating by parts twice we have: $$\begin{aligned} \lim_{\epsilon\to0} \frac{1}{2\pi}\int_{\mathbb R^2\setminus B_\epsilon} \log|x|\ \Delta \chi (x)\,dx &= 0 - \lim_{\epsilon\to0}\frac{1}{2\pi}\int_{\mathbb R^2\setminus B_\epsilon} \frac{1}{|x|^2} \langle x, \nabla\chi (x)\rangle\,dx\\ &= + \lim_{\epsilon\to0} \frac{1}{2\pi\epsilon}\int_{\partial B_\epsilon}\chi(x)dS(x) + 0 = \chi(0)\,. \end{aligned} $$ Therefore, $$ \Delta^{-1}(x) = \frac{1}{2\pi}\,\log|x|\,. $$

Brightsun
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    Yes, @BobOakley the Taylor series is $|x|^\epsilon=e^{\epsilon\log |x|}=1+\epsilon\log|x|+\cdots$. – Brightsun Apr 09 '20 at 16:37
  • @BobOakley I was having a look at your other question but it seems that it has been deleted. In general, in my experience it is hard to calculate the Fourier transforms of distributions directly using the definition. Usually it is more convenient to calculate it from functions and then take some limit that goes back to the distribution you're interested in... – Brightsun Apr 09 '20 at 16:40
  • @BobOakley I know, but I think it might be hard. I suggest you try to prove that $\int_{\mathbb R^2} e^{i \xi \cdot x}d\xi/|\xi|=A/|x|$. This should be essentially the derivative of the equation you want. – Brightsun Apr 09 '20 at 16:47