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$\newcommand{\x}{\mathbf{x}}$Let $\x$ denote a vector in $\mathbb{R}^3$, $|\x|$ its magintude and $\Delta=\frac{\partial^2}{\partial x 2}+\frac{\partial^2}{\partial y 2}+\frac{\partial^2}{\partial z 2}$ the usual Laplacian. What is $$\Delta\frac{1}{|\x|^2}$$ in the sense of distribution? That is, what is $$\iiint_{\mathbb{R}^3}\frac{1}{|\x|^2}\Delta\varphi(\x)d\x$$ for any test function $\varphi\in C^{\infty}_c(\mathbb{R}^3)$ (the subscript $c$ means compact support).

Attempt: I know how to do it for $\Delta\frac{1}{|\x|}$ (we get $\Delta\frac{1}{|\x|}=-4\pi\delta_0$, where $\delta_0$ is the delta distribution at $0$) and some others, and I am trying to reproduce the steps for this case but it does not seem to work. Let $\newcommand{\e}{\varepsilon}\e>0$ and consider the ball $B(0,\e)$ of radius $\e$ centered at $0$. Then, $$\iiint_{\mathbb{R}^3}\frac{1}{|\x|^2}\Delta\varphi(\x)d\x= \iiint_{\mathbb{R}^3\setminus B(0,\e)}\frac{1}{|\x|^2}\Delta\varphi(\x)d\x+\iiint_{B(0,\e)}\frac{1}{|\x|^2}\Delta\varphi(\x)d\x$$ Then, we can see that $$\left|\iiint_{B(0,\e)}\frac{1}{|\x|^2}\Delta\varphi(\x)d\x\right|\leq \iiint_{B(0,\e)}\frac{1}{|\x|^2}|\Delta\varphi(\x)|d\x \leq C\int_0^\e\frac{1}{\rho^2}\rho^2 d\rho\to0$$ as $\e\to 0$. Now, since $\varphi$ has compact support, there is a compact subset $\Omega\subseteq \mathbb{R}^3$ that includes $B(0,\e)$ such that $\varphi$ and all its derivatives vanish on $\Omega^c$. Thus, $$\iiint_{\mathbb{R}^3\setminus B(0,\e)}\frac{1}{|\x|^2}\Delta\varphi(\x)d\x = \iiint_{\Omega\setminus B(0,\e)}\frac{1}{|\x|^2}\Delta\varphi(\x)d\x$$ Now by Green's Second Identity, $$\iiint_{\mathbb{R}^3\setminus B(0,\e)}\left(\frac{1}{|\x|^2}\Delta\varphi(\x)-\varphi(\x)\Delta\frac{1}{|\x|^2}\right)d\x = \iint_{\partial\Omega}\left(\frac{1}{|\x|^2}\frac{\partial\varphi(\x)}{\partial\mathbf{n}}-\varphi(\x)\frac{\partial}{\partial\mathbf{n}}\frac{1}{|\x|^2}\right)dS+ \iint_{\partial B(0,\e)}\left(\frac{1}{|\x|^2}\frac{\partial\varphi(\x)}{\partial\mathbf{n}}-\varphi(\x)\frac{\partial}{\partial\mathbf{n}}\frac{1}{|\x|^2}\right)dS$$ Since $\varphi$ and all its derivatives vanish on $\partial \Omega$, we are left with $$\iiint_{\mathbb{R}^3\setminus B(0,\e)}\left(\frac{1}{|\x|^2}\Delta\varphi(\x)-\varphi(\x)\Delta\frac{1}{|\x|^2}\right)d\x= \iint_{\partial B(0,\e)}\left(\frac{1}{|\x|^2}\frac{\partial\varphi(\x)}{\partial\mathbf{n}}-\varphi(\x)\frac{\partial}{\partial\mathbf{n}}\frac{1}{|\x|^2}\right)dS$$ Now usually we can work with the right-hand side to find what it does as $\e\to 0$. But here I don't see how...

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Maybe try using a chain rule: $\Delta(|x|^{-2}) = \Delta(|x|^{-1})^2$? –  Neal Nov 14 '13 at 2:45

1 Answer 1

up vote 3 down vote accepted

Taking Fourier transform, we have

$$ \left( \Delta \frac{1}{|x|^{2}} \right)^{\wedge}(\xi) = -|\xi|^{2} \left( \frac{1}{|x|^{2}} \right)^{\wedge}(\xi). $$

Indeed, for any $\varphi \in \mathcal{S}(\Bbb{R}^{3})$,

\begin{align*} \langle (\Delta |x|^{-2})^{\wedge}, \varphi \rangle &= \langle \Delta |\xi|^{-2}, \hat{\varphi}(\xi) \rangle = \langle |\xi|^{-2}, \Delta \hat{\varphi}(\xi) \rangle = \langle |\xi|^{-2}, - (|x|^{2}\varphi)^{\wedge}(\xi) \rangle \\ &= - \langle (|x|^{-2})^{\wedge}(\xi), |\xi|^{2}\varphi(\xi) \rangle = - \langle |\xi|^{2} (|x|^{-2})^{\wedge}(\xi), \varphi(\xi) \rangle. \end{align*}

But we have

\begin{align*} \left( \frac{1}{|x|^{2}} \right)^{\wedge}(\xi) &= \int_{\Bbb{R}^{3}} \frac{e^{-i\xi\cdot x}}{|x|^{2}} \, dx = \int_{0}^{\infty} \int_{S^{2}} e^{-i\xi\cdot r \omega} d\sigma_{\omega} dr \\ &= \int_{0}^{\infty} \int_{0}^{2\pi} \int_{0}^{\pi} e^{-i|\xi| r \cos\phi} \sin\phi \, d\phi d\theta dr \\ &= 2\pi \int_{0}^{\infty} \int_{-1}^{t} e^{i|\xi| r t}\, dt dr \qquad (t = -\cos\phi) \\ &= \frac{4\pi}{|\xi|} \int_{0}^{\infty} \frac{\sin |\xi|r}{r} \, dt = \frac{2\pi^{2}}{|\xi|}. \end{align*}

So it follows that

$$ \left( \Delta \frac{1}{|x|^{2}} \right)^{\wedge}(\xi) = -|\xi|^{2} \left( \frac{1}{|x|^{2}} \right)^{\wedge}(\xi) = -2\pi^{2} |\xi| $$

and hence

$$ \Delta \frac{1}{|x|^{2}} = -2\pi^{2} \sqrt{-\Delta} \delta_{0}. $$

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Thank you very much for this answer. But what does $\sqrt{-\Delta}$ mean? –  Spenser Nov 14 '13 at 21:04
1  
@Spenser, It is a pseudo-differential operator defined by $$ \sqrt{-\Delta} u = \{ |\xi|\hat{u}(\xi) \}^{\vee}. $$ Since differential operators in the space-side becomes multiplication operators by polynomials, it makes sense to call it something similar to differential operator, hence pseudo-differential operator. It can also be defined by utilizing functional calculus. –  sos440 Nov 14 '13 at 21:26

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