Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

$$G(x,y)=-\frac{|x-y|}{2}$$

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.

share|improve this question

2 Answers 2

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.

share|improve this answer

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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.