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What we usually call "fundamental solution of the Laplace operator" is the following function defined on $\mathbb{R}^n\setminus\{0\}$: $$\tag{1}\Phi(x)=\begin{cases} \frac{-1}{2\pi} \log r & n=2 \\ \frac{-1}{(2-n)n\alpha(n)} \frac{1}{r^{n-2}} & n \ge 3 \end{cases}$$ where $r=\lvert x \rvert$ is the radial coordinate. The main property of this function, which justifies its name, is that it is a distributional solution of the equation $$\tag{2}-\Delta \Phi = \delta, $$ but it certainly is not the unique solution of such equation. Indeed the solutions of (2) are exactly the distributions $T$ such that $T=\Phi + h$ for a harmonic (entire) function $h$. In this sense there are "a lot" of fundamental solutions of the Laplace operator.

Question. Why do we always choose the one given in equation (1)?

A first reason that comes to mind is that the only fundamental solutions having radial symmetry are of the form $$E=\Phi+ C, $$ for a constant $C$. But then, why do we choose to set $C=0$? Is this choice purely cosmetic? Especially in dimension $2$, where the fundamental solution is unbounded both at $0$ and at $\infty$, this seems to me to be the case.

Giuseppe Negro
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    Yes, I think it's cosmetic. I mean, when you have a standard function, it seems silly to go adding a constant to it. – Ray Yang Feb 11 '13 at 13:39

1 Answers1

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Forgive me if I sound a little rusty, but if I remember correctly, the $1/(2 \pi)$ factor is needed in the 2D Green's function of the Laplacian because there is a factor of $2 \pi$ obtained in integrating around a circle of radius $\epsilon$ (in applying Green's theorem to the fundamental equation) about the origin. In 3D, we see a factor of $1/(4 \pi)$ because we integrate over the surface of a sphere of radius $\epsilon$ about the origin.

Ron Gordon
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  • Thank you, but that's no problem. I edited the title to make clearer that the problem is with the additive constant. – Giuseppe Negro Jan 31 '13 at 23:12
  • My apologies for misunderstanding what you wanted. – Ron Gordon Jan 31 '13 at 23:16
  • No need to apologize, I must admit that I am being rather vague. To add concreteness, I'll review the derivation of the fundamental solution one finds in Evans' textbook. We look for a radial solution of $-\Delta u =0$, so we write $\Delta$ in polar coordinates and arrive at the equation $$\frac{-1}{r^{n-1}}\frac{\partial}{\partial r}\left( r^{n-1}\frac{\partial}{\partial r}\right) u=0, $$ from which we infer that $$u(r)=\begin{cases} C_0\log r + C_1&n=2 \ C_0\frac{1}{r^{n-2}}+ C_1& n\ge 3\end{cases}$$[...] – Giuseppe Negro Feb 01 '13 at 00:50
  • [...] Now, reasoning as you suggest, we determine the value of $C_0$. But the author also sets $C_1=0$, without further explanation. That's what I am looking for, some explanation on this seemingly arbitrary choice. – Giuseppe Negro Feb 01 '13 at 00:52
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    I decided to refresh myself and consulted my copy of Guenther & Lee, specifically, Prob. 7 of Sec. 8.3. Sigh...you nailed it. "Let $L = -1/(2 \pi)$, $M=0$ so $w(r)=(1/(2 \pi)) \log{(1/r)}$ is the fundamental solution." It treats it like a convention. At least in 3D, you could say that you were imposing the field to be zero at $\infty$. But here, the field is unbounded there, so what does it matter? Maybe you could say that $C_1$ could be considered as providing a scale for $r$, and the choice of $C_1$ chooses the distance scale. – Ron Gordon Feb 01 '13 at 01:08
  • Thank you very much for this comment, the fact that you agree with my question is clearing many doubts. As you say, the question has little significance on a physical ground, at least in the electrostatic interpretation, since adding a constant to an electrostatic potential does not alter the field it generates. I think that this convention is "useful cosmetics". Maybe, but I am not sure, this choice of constants makes so that the Fourier transform of the fundamental solution does not contain any $\delta$. – Giuseppe Negro Feb 01 '13 at 01:27