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I am wondering if the solution of Laplace's equation (e.g. electrostatic potential $\Phi$ which satisfies $\nabla^2\Phi=0$) is square-integrable? I am confused, because in one dimension in the large distance limit, $\Phi(x)\sim\frac{1}{x}$. So $\int_{-\infty}^{\infty}\Phi^2\ dx<\infty$. However, if I go to the large distance limit in the three dimensions, $\int_{r=0}^{\infty}\int_ {\theta=0}^{\pi}\int_{\phi=0}^{2\pi}\Phi(r)^2\ d^3{\bf r}$. This seems to diverge, even though $\Phi\sim\frac{1}{r}$. It is true that the higher order multipole terms converge, but the monopole term still seems to dominate.

For me it looks odd that such a basic principle should exhibit different behaviour in one and three dimensions. Can anyone kindly explain if I am going wrong?

Regards, Kolahal

kolahalb
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  • Harmonic functions are far from unique; there are many of them and many of them aren’t square-integrable. An easy example is a nonzero constant, and a nontrivial example is a linear function. – Qiaochu Yuan Oct 23 '20 at 08:53
  • Hello Qiaochu, thank you for the clarification. So, there is not a general rule, I suppose. – kolahalb Oct 23 '20 at 08:57
  • Ah, in fact I’m being silly: the only harmonic function on $\mathbb{R}^n$ which is square-integrable is the zero function. The example you give is a harmonic function on $\mathbb{R}^n$ minus the origin though. – Qiaochu Yuan Oct 23 '20 at 09:36
  • What in two dimensions? Isn't it square-integrable? as $\Phi\sim\frac{1}{r}$, I thought that the integral $\int_{-\infty}^{\infty}\Phi^2\ dx\ dy$ converges. In 1D also, it seems to hold...Isn't that true? – kolahalb Oct 23 '20 at 10:12
  • That integral diverges in two dimensions. In polar coordinates you’re integrating over circles and it diverges logarithmically. In one dimension it still diverges but due to the contribution near the origin, at least if I haven’t gotten too mixed up. – Qiaochu Yuan Oct 23 '20 at 18:55

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