A restricted nonlocal operator bridging together the Laplacian and the fractional Laplacian

Abstract

In this work we introduce volume constraint problems involving the nonlocal operator $(-\Delta )_{\delta }^{s}$, closely related to the fractional Laplacian $(-\Delta )^{s}$, and depending upon a parameter $\delta >0$ called horizon. We study the associated linear and spectral problems and the behavior of these volume constraint problems when $\delta \rightarrow 0^+$ and $\delta \rightarrow +\infty$. Through these limit processes on $(-\Delta )_{\delta }^{s}$ we derive spectral convergence to the local Laplacian and to the fractional Laplacian as $\delta \rightarrow 0^+$ and $\delta \rightarrow +\infty$ respectively, as well as we prove the convergence of solutions of these problems to solutions of a local Dirichlet problem involving $(-\Delta )$ as $\delta \rightarrow 0^+$ or to solutions of a nonlocal fractional Dirichlet problem involving $(-\Delta )^s$ as $\delta \rightarrow +\infty$.

Appendix A: Proof of secondary results

Proof of Lemma 2

It is immediate to see that $\langle \cdot ,\cdot \rangle _{\mathbb {H}_0^{\delta ,s}}$ is a scalar product on $\mathbb {H}_0^{\delta ,s}(\Omega )\times \mathbb {H}_0^{\delta ,s}(\Omega )$. Let us see that $\mathbb {H}_0^{\delta ,s}(\Omega )$ is complete with respect to the norm . Let $\{v_n\}_{n\in \mathbb {N}}$ be a Cauchy sequence in $\mathbb {H}_0^{\delta ,s}(\Omega )$. Hence, given $varepsilon>0$, and keeping in mind Proposition 1 and Lemma 1, there exist $c>0$ and $n_0=n_0(\varepsilon )\in \mathbb {N}$ such that, for $m,n\ge n_0$, we have

(A.1)

and consequently $\{v_n\}_{n\in \mathbb {N}}$ is also a Cauchy sequence in $L^2(\Omega )$, and since this is a complete space, there exists $v\in L^2(\Omega )$ such that $v_n\rightarrow v$ in $L^2(\Omega )$ as $n\rightarrow +\infty$. Arguing as in [32, Lemma 7], it is easily proved that $v\in \mathbb {H}_0^{\delta ,s}(\Omega )$ and that $\{v_n\}_{n\in \mathbb {N}}$ converges in $\mathbb {H}_0^{\delta ,s}(\Omega )$. Hence, the space $\mathbb {H}_0^{\delta ,s}(\Omega )$ is a complete space. $\square$

Proof of Lemma 3

Since the sequence $\{v_j\}_{j\in \mathbb {N}}$ is bounded in $\mathbb {H}_0^{\delta ,s}(\Omega )$, because of Proposition 1, it is bounded in $H^s(\Omega _{\delta })$ so that it is also bounded in $H^s(\Omega )$ and, hence, in $L^2(\Omega )$. Observe that, although $v=0$ on $\partial _{\delta }\Omega$, we have $\Vert \cdot \Vert _{H^s(\Omega )}\ne \Vert \cdot \Vert _{H^s(\Omega _{\delta })}$ since the latter norm takes into account the interaction between $\Omega$ and $\partial _{\delta }\Omega$. Then, by [15, Corollary 7.2], there exists $v\in L^p(\Omega )$ such that, up to a subsequence,

$$\begin{aligned} v_j\rightarrow v\quad \text {in}\ L^p(\Omega ),\ \text {as}\ j\rightarrow +\infty ,\quad \text {for any }p\in [2,2_s^*). \end{aligned}$$

As mentioned before, the proof of Proposition 2 can be done similarly as to the proof [33, Proposition 9] so that we sketch the main steps.

Let us consider the energy functional

and observe that $\big \langle \mathcal {J}'(u)\big |v\big \rangle =\frac{c_{N,s}}{2}\langle u,v\rangle _{\mathbb {H}_0^{\delta ,s}}=\big \langle \mathcal {J}'(v)\big | u\big \rangle$ for all $u,v\in \mathbb {H}_0^{\delta ,s}(\Omega )$.

Lemma 8

Let $\mathcal {E}\subseteq \mathbb {H}_0^{\delta ,s}(\Omega )$ be a weakly closed subspace and $\mathcal {E}_1=\{u\in \mathcal {E}:\Vert u\Vert _{L^2(\Omega )}=1\}$. Then, there exists $u^{\star }\in \mathcal {E}_1$ such that

$$\begin{aligned} \min \limits _{u\in \mathcal {E}_1}\mathcal {J}(u)=\mathcal {J}(u^{\star }). \end{aligned}$$

Moreover,

$$\begin{aligned} \begin{aligned} \langle \mathcal {J}'(u^{\star })|\varphi \rangle&=\frac{c_{N,s}}{2}\int _{\Omega _{\delta }}\int _{\Omega _{\delta }\cap B(x,\delta )}\frac{(u^{\star }(x)-u^{\star }(y))(\varphi (x)-\varphi (y)}{|x-y|^{N+2s}}dydx\\&=\lambda ^{\star }\int _{\Omega }u^{\star }(x)\varphi (x)dx, \end{aligned} \end{aligned}$$

for all $\varphi \in \mathcal {E}$ and

(A.2)

Proof

Let $\{u_j\}\subset \mathcal {E}_1$ be a minimizing sequence for $\mathcal {J}$, i.e.,

$$\begin{aligned} \mathcal {J}(u_j)\rightarrow \inf \limits _{u\in \mathcal {E}_1}\mathcal {J}(u)\ge 0. \end{aligned}$$

(A.3)

Hence, the sequence $\mathcal {J}(u_j)$ is bounded in $\mathbb {R}$ and by definition of $\mathcal {J}$, the sequence is also bounded. Moreover, because of Lemma 2 the space $\mathbb {H}_0^{\delta ,s}(\Omega )$ is a Hilbert space and, hence, it is also a reflexive space. Then, up to a subsequence (that we do nor relabel), $u_j$ converges weakly in $\mathbb {H}_0^{\delta ,s}(\Omega )$ to some $u^{\star }\in \mathcal {E}$ since $\mathcal {E}$ is weakly closed. Therefore, we conclude

$$\begin{aligned} \big \langle \mathcal {J}'(u_j)\big |\varphi \big \rangle \rightarrow \big \langle \mathcal {J}'(u^{\star })\big |\varphi \big \rangle \quad \text {for all}\ \varphi \in \mathbb {H}_0^{\delta ,s}(\Omega ). \end{aligned}$$

On the other hand, since is bounded, because of Lemma 3, we have

$$\begin{aligned} u_j\rightarrow u^{\star }\ \text {in}\ L^2(\Omega ), \end{aligned}$$

so that $\Vert u^{\star }\Vert _{L^2(\Omega )}=1$ and then $u^{\star }\in \mathcal {E}_1$. Observe that the functional $\mathcal {J}$ is continuous and convex in $\mathbb {H}_0^{\delta ,s}(\Omega )$, so that $\mathcal {J}$ is weakly lower semicontinuous in $\mathbb {H}_0^{\delta ,s}(\Omega )$. Then,

$$\begin{aligned} \liminf _{j\rightarrow +\infty }\mathcal {J}(u_j)\ge \mathcal {J}(u^{\star })\ge \inf \limits _{u\in \mathcal {E}_1} \mathcal {J}(u), \end{aligned}$$

and, by (A.3), we conclude $\mathcal {J}(u^{\star })=\min \limits _{u\in \mathcal {E}_1}\mathcal {J}(u)$. Next, we prove that $\lambda ^{\star }=\mathcal {J}(u^{\star })$. Let us set $\varepsilon \in (-1,1)$, $\varphi \in \mathcal {E}$ and $u_{\varepsilon }=\frac{u^{\star }+\varepsilon \varphi }{\Vert u^{\star }+\varepsilon \varphi \Vert _{L^2(\Omega )}}\in \mathcal {E}_1$. Then,

Because of the minimality of $u^{\star }$ we have $\left. \frac{d}{d\varepsilon }\mathcal {J}(u_{\varepsilon })\right| _{\varepsilon =0}=0$, from where we conclude . $\square$

Proof of Proposition 2-(1)

Taking $\mathcal {E}=\mathbb {H}_0^{\delta ,s}(\Omega )$ in Lemma 8 we find that minimum defining $\lambda _1^{\delta ,s}$ exists and $\lambda _1^{\delta ,s}$ is an eigenvalue of problem ($EP_{\delta }^{s}$), proving (2.7). Moreover, because of Lemma 8, the minimum defining $\lambda _1^{\delta ,s}$ is attained at some function $\varphi _1^{\delta ,s}\in \mathbb {H}_0^{\delta ,s}(\Omega )$ with $\Vert \varphi _1^{\delta ,s}\Vert _{L^2(\Omega )}=1$. Then, by (A.2), we conclude

proving (2.8). The proof of the non-negativity of the first eigenfunction follows using the triangle inequality and the minimality of $\varphi _1^{\delta ,s}$ as in [33, Proposition 9-b)]. To prove that $\lambda _1^{\delta ,s}$ is simple we can argue exactly as in [33, Proposition 9-c)] so we omit the details. $\square$

Proof of Proposition 2-(2)

Let us define $\lambda _k^{\delta ,s}$ by (2.12) and observe that, by Lemma 8 with

$$\begin{aligned} \mathcal {E}=\mathbb {P}_{k}^{\delta }=\{u\in \mathbb {H}_0^{\delta ,s}(\Omega ):\langle u,\varphi _j^{\delta ,s}\rangle _{\mathbb {H}_0^{\delta ,s}}=0,\ \forall j=1,\ldots ,k-1\}, \end{aligned}$$

which is weakly closed by Lemma 2; the minimium in (2.12) exists, it is attained at some function $\varphi _k^{\delta ,s}\in \mathbb {P}_{k}^{\delta }$ and

$$\begin{aligned} \frac{c_{N,s}}{2}\langle \varphi _{k}^{\delta ,s},\phi \rangle _{\mathbb {H}_0^{\delta ,s}}=\lambda _{k}^{\delta ,s}\int _{\Omega }\varphi _{k}^{\delta ,s}\phi \, dx,\qquad \text {for all }\phi \in \mathbb {P}_{k}^{\delta }. \end{aligned}$$

(A.4)

In order to show that $\lambda _{k}^{\delta ,s}$ is an eigenvalue associated to the eigenfunction $\varphi _{k}^{\delta ,s}$, it remains to prove that (A.4) holds for any $\phi \in \mathbb {H}_0^{\delta ,s}(\Omega )$. This can be done inductively as in [33, Proposition 9-d)]. Note that the initial induction step has been proved in Proposition 2-(1). Let us write

$$\begin{aligned} \mathbb {H}_0^{\delta ,s}(\Omega )=span\{\varphi _1^{\delta ,s},\ldots ,\varphi _k^{\delta ,s}\}\oplus \mathbb {P}_{k+1}^{\delta } \end{aligned}$$

and, given $\phi \in \mathbb {H}_0^{\delta ,s}(\Omega )$, write $\phi =\phi _1+\phi _2$ with $\phi _1=\sum _{i=1}^{k}c_i\varphi _i^{\delta ,s}$ and $\phi _2\in \mathbb {P}_{k+1}^{\delta }$.

Testing the equation for $\varphi _{k+1}^{\delta ,s}$ against $\phi _2=\phi -\phi _1$ and $\varphi _i^{\delta ,s}$ for $i=1,\ldots ,k$ it is easily proved that

$$\begin{aligned} \frac{c_{N,s}}{2}\langle \varphi _{k+1}^{\delta ,s},\phi \rangle _{\mathbb {H}_0^{\delta ,s}}-\lambda _{k+1}^{\delta ,s}\int _{\Omega }\varphi _{k+1}^{\delta ,s}\phi \, dx=0, \end{aligned}$$

so that (A.4) holds for any $\phi \in \mathbb {H}_0^{\delta ,s}(\Omega )$.

We continue by proving (2.10). Since $\displaystyle \mathbb {P}_{k+1}^{\delta }\subseteq \mathbb {P}_{k}^{\delta }\subseteq \mathbb {H}_0^{\delta ,s}(\Omega )$ we have $0<\lambda _1^{\delta ,s}\le \lambda _2^{\delta ,s}\le \ldots \le \lambda _{k}^{\delta ,s}\le \ldots$. Moreover, since $\varphi _2^{\delta ,s}\in \mathbb {P}_{2}^{\delta }$, by Proposition 2-(1) we get $\lambda _1\ne \lambda _2$ and (2.10) follows. To prove (2.11) let us assume that $\lambda _{k}^{\delta ,s}\rightarrow c<+\infty$ so that the sequence is bounded in $\mathbb {H}_0^{\delta ,s}(\Omega )$. Hence, because of Lemma 3 we conclude that, up to a subsequence, $\varphi _{k}^{\delta ,s}\rightarrow \overline{\varphi }\in \ L^2(\Omega )$ as $k\rightarrow +\infty$ and $\{\varphi _{k}^{\delta ,s}\}_{k\in \mathbb {N}}$ is a Cauchy sequence. On the other hand, letting $k,h\in \mathbb {N}$, $k>h$, since $\varphi _{k}^{\delta ,s}\in \mathbb {P}_{k}^{\delta }$ we have that $\langle \varphi _{k}^{\delta ,s},\varphi _{h}^{\delta ,s}\rangle _{\mathbb {H}_0^{\delta ,s}}=0$. Therefore $\{\varphi _{k}^{\delta ,s}\}_{k\in \mathbb {N}}$ is an orthogonal sequence in $\mathbb {H}_0^{\delta ,s}(\Omega )$. Moreover, taking in mind that $\varphi _{k}^{\delta ,s}$ are eigenfunctions, it follows that

$$\begin{aligned} 0=\frac{c_{N,s}}{2}\langle \varphi _{k}^{\delta ,s},\varphi _{h}^{\delta ,s}\rangle _{\mathbb {H}_0^{\delta ,s}}=\lambda _k^{\delta ,s}\int _{\Omega }\varphi _{k}^{\delta ,s}\varphi _{h}^{\delta ,s}dx, \end{aligned}$$

(A.5)

and we conclude that $\{\varphi _{k}^{\delta ,s}\}_{k\in \mathbb {N}}$ is also an orthogonal sequence in both $\mathbb {H}_0^{\delta ,s}(\Omega )$ and $L^2(\Omega )$. Then, $\Vert \varphi _k^{\delta ,s}-\varphi _j^{\delta ,s}\Vert _{L^2(\Omega )}^2=\Vert \varphi _k^{\delta ,s}\Vert _{L^2(\Omega )}^2+\Vert \varphi _j^{\delta ,s}\Vert _{L^2(\Omega )}^2=2$ and we get a contradiction with the fact that $\{\varphi _{k}^{\delta ,s}\}_{k\in \mathbb {N}}$ is a Cauchy sequence. Therefore, $\lambda _k\rightarrow +\infty$ as $k\rightarrow +\infty$.

To complete the proof of Proposition 2-(2), it remains to prove that the sequence of eigenvalues given by (2.12) exhust all the eigenvalues of problem ($EP_{\delta }^{s}$). This can be done exactly as in [33, Proposition 9-d)] so we omit the details. $\square$

Proof of Proposition 2-(3)

The orthogonality of the set $\{\varphi _k^{\delta ,s}\}_{k\in \mathbb {N}}$ in $\mathbb {H}_0^{\delta ,s}(\Omega )$ and $L^2(\Omega )$ follows by (A.5), so it remains to prove that it is a basis for both $\mathbb {H}_0^{\delta ,s}(\Omega )$ and $L^2(\Omega )$. This follows using a standard Fourier analysis technique exactly as in [33, Proposition 9-f)] so we omit the details for the sake of brevity. $\square$

Proof of Proposition 2-(4)

Let $n\in \mathbb {N},\ n\ge 1$ and $\lambda _k^{\delta ,s}>0$ be an eigenvalue satisfying (2.15), i.e.,

$$\begin{aligned} \lambda _{k-1}^{\delta ,s}<\lambda _{k}^{\delta ,s}=\ldots =\lambda _{k+n}^{\delta ,s}<\lambda _{k+n+1}^{\delta ,s}. \end{aligned}$$

Because of Proposition 2-(2), equation (2.12), every function belonging to$span\{\varphi _k^{\delta ,s},\ldots ,\varphi _{k+n}^{\delta ,s}\}$ is an eigenfunction corresponding to $\lambda _k^{\delta ,s}=\ldots =\lambda _{k+n}^{\delta ,s}$. It remains to prove that each eigenfunction $\psi \in \mathbb {H}_0^{\delta ,s}(\Omega )$, $\psi \ne 0$, associated to $\lambda _{k}^{\delta ,s}$ belongs to $span\{\varphi _k^{\delta ,s},\ldots ,\varphi _{k+n}^{\delta ,s}\}$. Let us write

$$\begin{aligned} \mathbb {H}_0^{\delta ,s}(\Omega )=span\{\varphi _k^{\delta ,s},\ldots ,\varphi _{k+n}^{\delta ,s}\}\oplus \left( span\{\varphi _k^{\delta ,s},\ldots ,\varphi _{k+n}^{\delta ,s}\}\right) ^{\bot } \end{aligned}$$

and, consequently, $\psi =\psi _1+\psi _2$ with the functions $\psi _1\in span\{\varphi _k^{\delta ,s},\ldots ,\varphi _{k+n}^{\delta ,s}\}$ and $\psi _2\in \left( span\{\varphi _k^{\delta ,s},\ldots ,\varphi _{k+n}^{\delta ,s}\}\right) ^{\bot }$. Using the orthogonality of the eigenfunctions and arguing as in [33, Proposition 9-g)] it can be easily proven that $\psi _2\equiv 0$ and, hence, $\psi =\psi _1\in span\{\varphi _k,\ldots ,\varphi _{k+h}\}$. $\square$

Up to minor changes indicated below, the proof of Lemma 5 is similar to the proof of [31, Proposition 4] so that we sketch the main steps of the proof.

Proof of Lemma 5

Let $\varphi \in \mathbb {H}_0^{\delta ,s}(\Omega )$, $\varphi \ne 0$ and $\lambda >0$ such that

$$\begin{aligned} \langle \varphi , \phi \rangle _{\mathbb {H}_0^{\delta ,s}}=\lambda \int _{\Omega }\varphi \,\phi \,dx\quad \text {for all } \phi \in \mathbb {H}_0^{\delta ,s}(\Omega ). \end{aligned}$$

Up to multiplication by a constant, we can assume that $\Vert \varphi \Vert _{L^2(\Omega )}^2=\varepsilon$ with $\varepsilon >0$ to be chosen. Next, for $k\in \mathbb {N}$, let

$$\begin{aligned} C_k:= 1-2^{-k},\ v_k:= \varphi -C_k,\ w_k:=v_k^+\ \text {and}\ U_k:= \Vert w_k\Vert _{L^2(\Omega )}^2, \end{aligned}$$

where $v^+:=\max \{v,0\}$. Notice that $w_k\in \mathbb {H}_0^{\delta ,s}(\Omega )$, since $\mathbb {H}_0^{\delta ,s}(\Omega )$ is a linear space and $\varphi (x)-C_k=-C_k\le 0$ a.e. $x\in \partial _{\delta }\Omega$ so that $w_k=(-C_k)^+=0$ on $\partial _{\delta }\Omega$. Moreover, $0\le w_k\le |\varphi |+|C_k|\le |\varphi |+1\in L^2(\Omega )$ because $\Omega$ is a bounded domain and $\lim \nolimits _{k\rightarrow +\infty }w_k=(\varphi -1)^+$. Hence, by the Dominated Convergence Theorem,

$$\begin{aligned} \lim \limits _{k\rightarrow +\infty } U_k=\int _{\Omega }\left[ (\varphi -1)^+\right] ^2dx. \end{aligned}$$

(A.6)

Proceeding in a similar way as in [31, Proposition 4] we can prove that

$$\begin{aligned} \int _{\Omega _{\delta }}\int _{\Omega _{\delta }\cap B(x,\delta )}\frac{|w_{k+1}(x)-w_{k+1}(y)|^2}{|x-y|^{N+2s}}dxdy\le \lambda 2^{k+1} U_k. \end{aligned}$$

(A.7)

In addition, since $\{w_{k+1}>0\}\subseteq \{w_k>2^{-(k+1)}\}$ for any $k\in \mathbb {N}$,

$$\begin{aligned} \begin{aligned} U_k&=\Vert w_k\Vert _{L^2(\Omega )}^2\ge \int _{\{w_k>2^{-(k+1)}\}}w_k^2(x)dx\ge 2^{-2(k+1)}|\{w_k>2^{-(k+1)}\}|\\&\ge 2^{-2(k+1)}|\{w_{k+1}>0\}|. \end{aligned} \end{aligned}$$

(A.8)

On the other hand, using Hölder inequality with exponents $p=\frac{2_s^*}{2}$ and $q=\frac{N}{2s}$, the fractional Sobolev inequality (cf.[15, Theorem 6.5]) and Proposition 1, we find,

$$\begin{aligned} U_{k+1}\le c\int _{\Omega _{\delta }}\int _{\Omega _{\delta }\cap B(x,\delta )}\frac{|w_{k+1}(x)-w_{k+1}(y)|^2}{|x-y|^{N+2s}}dxdy\left| \{w_{k+1}>0\}\right| ^{\frac{2s}{N}}, \end{aligned}$$

for a positive constant $c=c(N,s,\delta )$. As a consequence, by (A.7) and (A.8),

$$\begin{aligned} \begin{aligned} U_{k+1}&\le \left( c2^{k+1}\lambda U_k\right) \left( 2^{2(k+1)}U_k\right) ^{\frac{2s}{N}}=2^{1+\frac{4s}{N}}c\lambda \left( 2^{1+\frac{4s}{N}}\right) ^kU_k^{1+\frac{2s}{N}}\\&\le \left[ \left( 1+2^{1+\frac{4s}{N}}c\lambda \right) 2^{1+\frac{4s}{N}}\right] ^kU_k^{1+\frac{2s}{N}}=\theta ^k U_k^{\beta }, \end{aligned} \end{aligned}$$

with $\beta := 1+\frac{2s}{N}$ and $\theta =\theta (N,s,\delta ,\lambda )>1$. Next, let us choose $\varepsilon >0$ such that $\varepsilon ^{\beta -1}<\frac{1}{\theta ^{1/(\beta -1)}}$ and fix $\eta \in (\varepsilon ^{\beta -1},\frac{1}{\theta ^{1/(\beta -1)}})$. Let us note that, since $\theta >1$ and $\beta >1$, we have $\eta \in (0,1)$. Arguing inductively, it is easily proved that, for any $k\in \mathbb {N}$,

$$\begin{aligned} U_k\le \varepsilon \eta ^k. \end{aligned}$$

Then, by (A.6), we get $\displaystyle \int _{\Omega }\left[ (\varphi -1)^+\right] ^2dx=\lim \limits _{k\rightarrow +\infty }U_k=0$ since $\eta \in (0,1)$. Thus, $(\varphi -1)^+=0$ a.e. in $\Omega$ so that $\varphi \le 1$ a.e. in $\Omega$. By replacing $\varphi$ with $-\varphi$ it follows that $\Vert \varphi \Vert _{L^{\infty }(\Omega )}\le 1$. Taking in mind the scaling $\Vert \varphi \Vert _{L^2(\Omega }^2=\varepsilon$, we conclude

$$\begin{aligned} \Vert \varphi \Vert _{L^{\infty }(\Omega )}\le C\Vert \varphi \Vert _{L^2(\Omega )}, \end{aligned}$$

for a constant $C>0$ depending on $N,s,\delta$ and $\lambda$. $\square$