# Fractional Laplacian

Q(1) It is well known that if $s \rightarrow 1$ then the fractional Laplacian converges to the classical Laplacian but the form of the Laplacian still remains in the non-local form whereas it is known that the Laplacian is a local differential operator.

Why is this?.

(2) Also why do we choose the boundary condition as $u=0$ $\in \mathbb{R}^N\setminus\Omega$ and not just in $\partial\Omega$, if the domain $\Omega$ is bounded in $\mathbb{R}^N$.

• Note that for $s=1$, we have $\int_{\mathbb{R}^N}\frac{u(x)-u(y)}{|x-y|^{N+2}}dy$ which is non local even if the support of $u$ is compact. Nov 19, 2016 at 22:34
• To answer (1), the expression looks nonlocal, but turns out to be local. The reason, roughly speaking, is that the kernel is singular, i.e., its integral is infinite. So the finite non-local parts that appear to be in the formula are washed away and you only see the local behavior of $u$ near the singularity at $x=y$. To put it another way, it is a weighted average of $u(x)-u(y)$, but there is an infinite amount of weight near $x=y$ and a finite amount of "nonlocal" weight elsewhere.
– Jeff
Nov 20, 2016 at 3:13
• Thanks Jeff. Few questions. What do you mean by a "non-local" weight?. Secondly, isn't the argument same for $0<s<1$?. I am unable to see even heuristically that it is a loacl operator. Nov 20, 2016 at 22:52
• When $s<1$, the kernel is integrable (if you include the $u(x)-u(y)$), and is truly nonlocal. I will try to write a more detailed answer if I have time. The basic idea is that you are averaging with a non-integrable kernel, so your intuition about what is local and nonlocal is incorrect. An infinite amount of weight near $x=y$ can render the non-local part where $|x-y|>\varepsilon$ negligible.
– Jeff
Nov 21, 2016 at 2:44
• @Alexander note that $\int_{\mathbb{R}^N}\frac{u(x)-u(y)}{|x-y|^{N+2}}dy=\infty$ unless $u$ is constant. be carefull Dec 16, 2017 at 15:17

The expression you are referring to (I believe) is

$$\Delta^su(x) =c_{n,s} \int_{\mathbb{R}^n} \frac{u(y) - u(x)}{|x-y|^{n+2s}} \, dy,$$

which holds for $0 < s < 1$. Here, $c_{n,s}$ is a constant depending on $n$ and $s$. The expression does not hold for $s=1$; in fact it is not even integrable. Consider for example $u(x) = |x|^2$ so we have

$$\Delta^s u(0) = c_{n,s}\int_{\mathbb{R}^n} \frac{1}{|y|^{n + 2(s-1)}} \, dy.$$

When $s=1$, the integrand is $1/|y|^n$, which not integrable, and the principal value of the integral is infinity.

However, as you said, as $s\to 1^-$, $\Delta^su(x) \to \Delta u(x)$. What is happening is that $c_{n,s} \to 0^+$ as $s\to 1^-$, the integral is blowing up as $s\to 1^-$, and the two cancel each other out perfectly to give you $\Delta u(x)$.

One way you can view this is that when $s<1$ is very close to $1$, you are averaging with the kernel $c_{n,s}/|y|^{n+2(s-1)}$, which is very close to being not integrable at $y=0$. Hence, it puts an enormous amount of its weight near $y=0$, and the closer $s$ is to $1$, the more of its weight (proportionally) is concentrated at $y=0$. In the limit as $s\to 1^-$ it becomes local, though the integral expression is no longer valid. You could make the expression valid by writing

$$\Delta u(x) = \lim_{s\to 1^-} c_{n,s} \int_{\mathbb{R}^n} \frac{u(y) - u(x)}{|x-y|^{n+2s}}\, dy.$$

A shorter explanation: If you average with a kernel that has infinite mass, regions with finite mass are negligible.

• thanks for the argument, this is pretty clear. :) Nov 21, 2016 at 8:43
• oops!...hang on, but isn't the $c_{n,s}\rightarrow\frac{4}{\sqrt{\pi}}\Gamma(\frac{N+2}{2})$ as $s\rightarrow 1^{-}$ and not to $0$?. Sorry f there goes any calculation error. Nov 21, 2016 at 8:51
• The expression is $c_{n,s} = \frac{4^s\Gamma(n/2+s)}{\pi^{n/2}|\Gamma(-s)|}$. The Gamma function has a pole at $-1$, so the denominator tends to infinity as $s\to 1^-$.
– Jeff
Nov 21, 2016 at 17:09