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I'm trying to show that the Green's function for the Laplace operator $-\nabla^2$ is badly behaved at infinity. I.e.

$$\int d^dx|G(x,y)|^2=\infty$$ for $d=1,2,3$. What happens when $d>4$?

I know the 1D Green's function is given by


but I'm not sure how to generalize this. Could someone push me in the right direction?

Update: for d=1, I have $\int |G(x,y)|^2 dx=\frac{1}{4}\int dx |x-y|^2=\infty$. This is trivial. How do I show this same thing for d=2, 3? And then what happens when $d=4$? I know $G(x, y)\propto|x-y|^{-(d-2)}$ for d>2.

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up vote 2 down vote accepted

It may be easier to start with vector Green's functions for $\nabla$ and then work up to Green's functions for the Laplacian.

$G_1$, the Green's function for $\nabla$, can be easy found through the Generalized Stokes theorem.

$$\oint G_1(y-x) \wedge d^{N-1}y = \int \delta(y-x) \, d^Ny = i_N$$

For $N=3$, we can pick a ball for the volume integral. This relates the values of $G_1$ on the boundary of the ball to a volume integral over that ball. We can conclude that $|G_1|$ should be constant over the surface of the ball and that its direction should be radially outward. Let the radius of the ball be $R$, and we conclude that

$$\oint G_1(y-x) \wedge d^{N-1}y = i_N S_N |G_1(R)|$$

where $S_N$ is the surface area, and $|G_1(R)|$ signifies the magnitude of $G_1$ for any argument with magnitude $R$ (a slight abuse of notation). The result is

$$i_N S_N |G_1(R)| = i_N$$

so $|G_1(R)| = 1/S_N$. In 3d, this would tell us that the magnitude of the Green's function is $1/4\pi R^2$, which is absolutely true. Only a couple steps remain to build the vector Green's function. We said the direction had to be radial, so that the result is

$$G_1(x-y) = \frac{1}{S_N(|x-y|)} \frac{x-y}{|x-y|}$$

where $S_N(R)$ is the "surface area" of a ball with radius $R$ in $N$ dimensions.

Now, to find $G_2$, the Green's function for the Laplacian, invoke radial symmetry to find that

$$G_2(x) = \int_\infty^{|x|} \frac{1}{S_N(r)} \, dr$$

(Referencing to infinity here is a choice, but an incredibly convenient one for making the math work out.) Quick check: $S_3(r) = 4\pi r^2$. The result in 3d is then

$$\int_\infty^{|x|} \frac{1}{4\pi r^2} \, dr = \left. -\frac{1}{4\pi r} \right|_\infty^{|x|} = -\frac{1}{4\pi |x|}$$

This is indeed the Green's function for the Laplacian in 3 dimensions.

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If $\Omega = \{ \vec{X} : \|\vec{X}\| < a \}$ (ball of radius $a$),

$$ G(\vec{X},\vec{Y}) = \begin{cases} \frac{1}{(2-n)\omega_n} \left(\|\vec{X}-\vec{Y}\|^{2-n} - \frac{1}{a^{2-n}}\|\vec{Y}\|^{2-n}\|\vec{X} - \vec{Z}\|^{2-n}\right) & n > 2\\ \frac{1}{2\pi}\log\left(\frac{\|\vec{X} - \vec{Y}\|}{\|\vec{X} - \vec{Z}\|} \frac{a}{\|\vec{Y}\|}\right) & n = 2 \end{cases} $$ where $\vec{Z} = \frac{a^2\vec{Y}}{\|\vec{Y}\|^2}$ and $\omega_n$ is the area of the unit sphere.

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I don't think I needed to actually compute the Green's function for higher dimensions. I just need to show that it is not in the space $L^2(R^d)$; i.e. the integral I wrote in my initial post. – Alex Nov 27 '12 at 5:45
@Alex Then I guess you'll have to use an analysis argument. The thing is -as you can see from its expression-, that the Green's function does decay well for $n>2$, and the divergent behavior should come from the volume element. At least that's my guess. – Pragabhava Nov 27 '12 at 5:54

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