# Finding The Laplacian of $\frac{1}{|\vec x|^2}$ in the sense of distributions

I am currently trying to find the Laplacian of $u(x)=\frac{1}{|\vec x|^2}$ in $3$ Dimensions in the sense of distributions. Now, let us have a test function $\phi(x)$. We consider two cases, when $0\notin \text{supp}\>\phi$, and when $0\in \text{supp }\phi$. Now in first case, it is obvious that the Laplacian in the sense of distributions is just the pointwise Laplacian. Now I considered when $0\in\text{supp }\phi$. Then, let $B(0,\epsilon)$ be some ball centered at $0$ with radius $\epsilon$, and consider $\Omega$ the support of $\phi$. Then, consider $\Omega_\epsilon=\Omega\setminus B(0,\epsilon)$. Integrating over this using Green's Second Identity yielded:$$\iiint_{\Omega_{\epsilon}}\frac{1}{|x|^2}\Delta\phi dx = \iiint_{\Omega_{\epsilon}}\Delta\left(\frac{1}{|x|^2}\right)\phi dx+\iint_{\partial \Omega_{\epsilon}}\frac{1}{|x|^2} \frac{\partial\phi}{\partial\vec n} dS_{x}-\iint_{\partial \Omega_{\epsilon}}\frac{\partial}{\partial\vec n}\left(\frac{1}{|x|^2}\right)\phi dS_x\>\>\>\>(1)$$ Now, I focused on the last integral on the right hand side above: $$\iint_{\partial \Omega_{\epsilon}}\frac{\partial}{\partial\vec n}\left(\frac{1}{|x|^2}\right)\phi dS_x=8\pi\epsilon\left(\frac{1}{4\pi\epsilon^2}\iint_{\partial B(0,\epsilon)}\phi dS_x\right)$$ By the Mean Value Property, the value in the brackets above gave $\phi(0)$ and if we take $\epsilon\rightarrow 0$, we get the above tends to $0$. Now I looked at the second integral on the right hand side of $(1)$. $$\iint_{\partial B(0,\epsilon)}\frac{1}{|x|^2}\frac{\partial\phi}{\partial \vec n} dS_x = \frac{1}{\epsilon^2}\iint_{\partial B(0,\epsilon)}\frac{\partial\phi}{\partial\vec n}dS_x$$ Now since $\phi$ is bounded as it is compact, its derivatives hold the same characteristics, and thus: $$\frac{1}{\epsilon^2}\iint_{\partial B(0,\epsilon)}\frac{\partial\phi}{\partial\vec n}dS_x\leq C\cdot \frac{1}{\epsilon^2}\cdot 4\pi\epsilon^2$$ The above gives a constant. Thus, am I supposed to conclude that the distributional Laplacian when the support of $\phi$ includes $0$ is simply the pointwise Laplacian plus some arbitrary constant?

• In three dimensions? – md2perpe Dec 17 '17 at 21:27
• @md2perpe ah yes, forgot to mention the dimension. – Felicio Grande Dec 17 '17 at 21:30

No need for "two cases": the second case is the general case, since we are never going to use the assumption $0\in \operatorname{supp}\phi$.

### Incorrect

By the Mean Value Property, the value in the brackets above gave $ϕ(0)$

The function $\phi$ is not harmonic, so it does not satisfy the Mean Value Property. But it is true that $$\frac{1}{4\pi\epsilon^2}\iint_{\partial B(0,\epsilon)}\phi dS_x \to \phi(0)$$ because $\phi$ is continuous. However, there is more to say about this term...

### Incorrect

$$\iint_{\partial \Omega_{\epsilon}}\frac{\partial}{\partial\vec n}\left(\frac{1}{|x|^2}\right)\phi dS_x=8\pi\epsilon\left(\frac{1}{4\pi\epsilon^2}\iint_{\partial B(0,\epsilon)}\phi dS_x\right)$$ The normal derivative of $1/|x|^2$ on the sphere of radius $\epsilon$ is $2\epsilon^{-3}$, not $2\epsilon^{-1}$. As a result, this term tends to infinity, which is troublesome (but see below).

### Correct but incomplete

$$\frac{1}{\epsilon^2}\iint_{\partial B(0,\epsilon)}\frac{\partial\phi}{\partial\vec n}dS_x\leq C\cdot \frac{1}{\epsilon^2}\cdot 4\pi\epsilon^2$$ This term actually tends to $0$, because the divergence theorem implies $$\iint_{\partial B(0,\epsilon)}\frac{\partial\phi}{\partial\vec n}dS_x = \iiint_{B(0, \epsilon)} \Delta \phi(x)\,dx \le C \epsilon^3$$

### Conclusion

We need the limit as $\epsilon\to 0$ of $$\iiint_{\Omega_{\epsilon}}\Delta\left(\frac{1}{|x|^2}\right)\phi dx -2\epsilon^{-3}\iint_{\partial \Omega_{\epsilon}} \phi dS_x \tag{A}$$ The pointwise Laplacian should not be casually dismissed: it is not locally integrable, hence does not define a distribution. The formula (A) represents its regularization; the subtracted term tempers the singularity at $0$.

Using the divergence formula, one can compute $$\iiint_{\Omega_{\epsilon}}\Delta\left(\frac{1}{|x|^2}\right) dx = 2\epsilon^{-3} (4\pi \epsilon^2)$$ which allows us to rewrite (A) as $$\iiint_{\Omega_{\epsilon}}\Delta\left(\frac{1}{|x|^2}\right)(\phi-\phi(0))\,dx -2\epsilon^{-3}\iint_{\partial \Omega_{\epsilon}}( \phi-\phi(0)) dS_x \tag{B}$$ Here the second term tends to zero because $\phi(x) - \phi(0 ) = \nabla \phi(0) x + O(|x|^2)$, and the integral of $x$ over a sphere is zero. Final answer: the Laplacian is $$\phi\mapsto \lim_{\epsilon\to 0} \iiint_{\Omega_{\epsilon}}\Delta\left(\frac{1}{|x|^2}\right)(\phi-\phi(0))\,dx$$ where you can put the formula for pointwise $\Delta\left(\frac{1}{|x|^2}\right)$ (some constant times $|x|^{-4}$) to make it more explicit.

This is not "pointwise Laplacian plus a constant", this is pointwise Laplacian regularized.

• I cannot see that you show $\Delta \left( \frac{1}{|x|^2} \right) = \delta$ in the sense of distributions as you are not using derivatives of $\phi$. – md2perpe Dec 17 '17 at 21:23
• (a) I use $\Delta \phi$ and $\nabla \phi$; (b) the Laplacian of this function is not the Dirac delta. You may be thinking of $1/|x|$. – user357151 Dec 17 '17 at 21:27
• Yes, I was too fast when I wrote that formula. – md2perpe Dec 17 '17 at 21:36