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BY 4.0 license Open Access Published by De Gruyter December 31, 2020

Local versus nonlocal elliptic equations: short-long range field interactions

  • Daniele Cassani EMAIL logo , Luca Vilasi and Youjun Wang

Abstract

In this paper we study a class of one-parameter family of elliptic equations which combines local and nonlocal operators, namely the Laplacian and the fractional Laplacian. We analyze spectral properties, establish the validity of the maximum principle, prove existence, nonexistence, symmetry and regularity results for weak solutions. The asymptotic behavior of weak solutions as the coupling parameter vanishes (which turns the problem into a purely nonlocal one) or goes to infinity (reducing the problem to the classical semilinear Laplace equation) is also investigated.

MSC 2010: 35A15

1 Introduction and main results

Consider the following class of mixed local–nonlocal elliptic equations

(1.1) (Δ)suϵΔu=f(x,u) in Ω,u=0 in RNΩ,

where Ω is a bounded smooth domain of RN,N>2,s(0,1),ϵ>0 is a real parameter and

(Δ)su(x)=cN,sPVRNu(x)u(y)|xy|N+2sdy

is the fractional Laplacian operator. Here cN,s is a positive normalizing constant and PV stands for the Cauchy Principal Value.

Integro-differential equations of the form (1.1) arise naturally in the study of stochastic processes with jumps.

The generator of an N-dimensional Lévy process has the following general structure:

(1.2) Lu=i,jaijiju+jbjju+RN(u(x+y)u(x)yu(x))χB1(y)dν(y),

where ν is the Lévy measure and satisfies RNmin{1,|y|2}dν(y)<+,andχB1 is the usual characteristic function of the unit ball B1 of ℝN. The first term of (1.2) on the right-hand side corresponds to the diffusion, the second one to the drift, and the third one to the jump part.

In the particular case when ν = 0, namely when there are no jumps, 𝓛 turns into a classical and extensively studied second-order differential operator. On the other hand, the case when the process has no diffusion and no drift has attracted a lot of attention recently (we refer to the book [18] for the development of the existence theory for several nonlinear nonlocal problems; see also the survey [21] for related regularity properties of solutions). In particular, one of the most prominent operators belonging to this class is the fractional Laplacian (−Δ)s, which turns out to be the infinitesimal generator of isotropic s-stable Lévy processes and plays an important role in analysis and probability theory. The issue of the existence of solutions to fractional Laplacian problems, as well as the study of qualitative properties of such solutions, has been addressed by many authors and the literature devoted to the field grows up continuously (we mention, in particular, the following recent contributions [1, 4, 6, 7, 11, 17, 23, 25, 26] and the references therein). Finally, it is well-known, e.g. from [13, 19], that Markov processes with both diffusion and jump components are suitable to modeling many situations in finance and control theory, see also [8, 9, 10, 29].

For the above reasons, the study of the operator 𝓛 with diffusion, drift and jump components appears quite intriguing and, as far as we know, there are just a few contributions in this direction, mainly relying on a combination of probabilistic and analytic techniques. For instance in [12], under the assumption that the diffusion is uniformly elliptic and suitable conditions on ν, the author established a Harnack-type inequality and a regularity result for functions that are nonnegative in ℝN and harmonic in some domain. In [14, 15] the author derived interior Schauder estimates for both classical and weak solutions of the equation 𝓛 u = f.

Here we focus on the case 𝓛 = ϵ Δ –(–Δ)s, obtained from (1.2) by setting aij = ϵΔij, ϵ > 0, bj = 0 and ν given by the symmetric kernel |x - y|-N-2s. We study this problem from a purely analytical point of view. Precisely, the paper is divided into two parts. Firstly, for fixed ϵ > 0 (without loss of generality, we may assume ϵ = 1), we provide some basic results related to (1.1), including spectral properties, maximum principles, existence, nonexistence, symmetry and regularity of weak solutions. With these preliminary tools, in the second part we will address further issues related to the asymptotic profile, as ϵ → 0 as well as $\epsilon\to +\infty$, of ground states of the model problem

(1.3) (Δ)suϵΔu=|u|p2u in Ω,u>0 in Ω,u=0 in RNΩ.

If Ω is a bounded star-shaped domain, it was proved in [22] that (1.3) has no nontrivial bounded solutions if p2:=2NN2 . When ϵ = 0, problem (1.3) possesses a positive solution for 2<p<2s:=2NN2s,N>2s , and, if Ω is a bounded, C1,1 and star-shaped domain, do not exist positive bounded solution for p=2s and there are no nontrivial bounded solution for p>2s , see [23, 24]. It is natural to investigate the behavior of the ground state uϵ in (1.3), as ϵ → 0 (corresponding to the fractional case), which may converge, blow up or vanish, depending on the range of p.

On the other hand, in the case ϵ → + ∞, setting vϵ(x)=ϵ1p2uϵ(x) , the function vϵ satisfies

(1.4) ϵ1(Δ)svΔv=|v|p2v in Ω,v>0 in Ω,v=0 in RNΩ,

whose natural limit problem is

(1.5) Δv=|v|p2v in Ω,v>0 in Ω,v=0 in RNΩ.

By Sobolev embedding theorems and regularity theory, classical solutions to (1.5) do exist for 2 < p < 2* with N > 2, and thus one may suspect in this case that uϵ(x) blows up almost everywhere in Ω, as ϵ → + ∞.

We mention that the operator 𝓛 can be given a local realization by means of a Caffarelli-Silvestre-type extension [6]. However, if one proceeds as in [5] by directly extending uHs(Ω) to the cylinder CΩ := Ω × (0, +∞), i.e. considering CΩ as the domain of the extension w of u, with w = 0 on the lateral boundary L CΩ of CΩ, then one would end up with the fractional Laplacian operator defined by spectral decomposition, which on bounded domains does not coincide with (−Δ)s. The idea is then to start from the half-space R+N+1 and to require the trace of the extension to vanish a.e. outside Ω. This can be achieved as follows.

Define the space

Ys:=wHloc1(R+N+1):R+N+1y12s|w|2dxdy<+,

where R+N+1:=RN×(0,+) , equipped with the norm

wYs=R+N+1y12s|w|2dxdy1/2.

It is known from [6] that, for all uHs(RN) there exists a unique function E(u)∈ Ys, called the harmonic extension of u to R+N+1 , satisfying

(1.6) div(y12sE(u)(x,y))=0 in R+N+1,E(u)(x,0)=u(x) in RN,

and (–Δ)s u corresponds to the Dirichlet-to-Neumann map

(Δ)su(x)=ks1limy0y12sE(u)y(x,y),xRN,

for a suitable normalizing constant ks > 0. Next define the following subspace

Y0s:=wYs:w(x,0)=w|RN×{0}0 in RNΩ and w(x,0)H01(Ω)

whose elements have traces over ℝN which vanish outside Ω. We say that wY0s is a weak solution to

(1.7) div(y12sw(x,y))=0 in R+N+1w(x,0)=0 for all xRNΩ,ks1limy0y12swy(x,y)ϵΔw(x,0)=|w(x,0)|p2w(x,0) for all xΩ,

provided

(1.8) R+N+1y12swφdxdy+ksϵΩw(x,0)φ(x,0)dx=ksΩ|w(x,0)|p2w(x,0)φ(x,0)dx

for all φY0s . It is clear that if wY0s solves (1.7) in the weak sense, then its trace over ℝN weakly solves

(1.9) (Δ)suϵΔu=|u|p2u in Ω,u=0 in RNΩ.

Before presenting our main results, let us introduce the functional framework. Set Q := R N × R N C Ω × C Ω , with CΩ:=RNΩ . We denote by Xs(Ω) the linear space of Lebesgue measurable functions u:RNR such that the restriction of u to Ω belongs to H1(Ω) and

[u]s,Q2:=Q|u(x)u(y)|2|xy|N+2sdxdy<+.

A norm in Xs(Ω) is defined as follows

(1.10) uXs(Ω):=uH1(Ω)2+[u]s,Q212.

Let us also define the set

X0s:={uXs(Ω):u=0 a.e. in CΩ}.

For uX0s , we can write the norm (1.10) as

(1.11) uXs(Ω)=uH1(Ω)2+[u]s,RN212,

and, by Poincaré's inequality, this norm in X0s is also equivalent to

(1.12) us:=Ω|u|2dx+[u]s,RN212.

We cliam that (X0s,s) is an Hilbert space with scalar product

u,vs=Ωuvdx+RN×RN(u(x)u(y))(v(x)v(y))|xy|N+2sdxdy,u,vX0s.

For the proof, see Lemma 12.

Throughout this paper, we will use the symbol p,p[1,+] to denote the usual Lp-norm in ℝN, while the letters C, Ci will represent generic positive constants which may change from line to line.

Preliminary results

Let us begin with some spectral properties of the operator 𝓛 which, it is well known, are crucial for the solvability of linear and resonance nonlinear problems.

Theorem 1

Let s ∈ (0,1), N > 2, and consider the following eigenvalue problem:

(1.13) (Δ)suΔu=λu in Ω,u=0 in RNΩ.

Then:

  1. Problem (1.13) admits a positive eigenvalue λ1, characterized by

    (1.14) λ1=minuX0s{0}us2u22or,equivalently,λ1=minuX0s,u2=1us2,

    and a nonnegative eigenfunction e1X0s corresponding to λ1, satisfying e12=1andλ1=e1s2 ;

  2. λ1 is simple;

  3. the set of the eigenvalues of (1.13) consists of a sequence {λk},kN , with 0 < λ1 < λ2 ≤ … ≤ λk ≤ … and λk → + ∞ as k → + ∞. The eigenvalue λk+1 can be characterized by

    (1.15) λk+1=minuMk+1{0}us2u22,or,equivalently,λk+1=minuMk+1,u2=1us2,k=1,2,,

    where Mk+1={uX0s:u,ejs=0,j=1,2,,k} . Moreover, for any k ∈ ℕ, there exists ek+1Mk+1 that is an eigenfunction of 𝓛 corresponding to λk+1 and that verifies ek+12=1,λk+1=ek+1s2 ;

  4. the sequence {ek} of eigenfunctions corresponding to λk is an orthonormal basis of L2(Ω) and an orthogonal basis of X0s ;

  5. each eigenvalue λk has finite multiplicity; more precisely, if λk is such that λk1<λk==λk+h<λk+h+1 , for k > 1 and for some h ∈ ℕ, then the set of all the eigenfunctions corresponding to λk agrees with span{ek,…, ek+h}.

Theorem 1 can be viewed as the generalization of similar eigenvalue problems for the Laplacian and the fractional Laplacian operators, see the monograph [18, Chapter 3].

Next, we focus on maximum principles, both for classical and for weak supersolutions of the problem 𝓛u = 0, u|ℝN\Ω = 0. Define

Ls:=u:RNR s.t. u1+|x|N+2sL1RN.

Theorem 2

Assume that uLsC2(Ω) is lower semicontinuous on Ωˉ . If

(1.16) (Δ)suΔu0 in Ω,u0 in RNΩ,

then, u(x) ≥ 0 inN. Moreover, if u(x) = 0 at some point in Ω, then u(x) ≡ 0 inN.

Theorem 3

Assume that uH1(Ω), [u]s,ℝN < +∞, and u satisfies

(1.17) (Δ)suΔu0 in Ω,u0 in RNΩ.

Then u(x) ≥ 0 a.e. in Ω.

We have used the arguments of [7] to prove Theorem 2. Basically, if we have u(x0) = minΩ u(x) < 0 for some x0Ω, then (–Δ)su(x0) ≥ (–Δ)su(x0) – Δ u(x0) ≥ 0 and we reduce the problem to the fractional setting. On the other hand, by Stampacchia's theorem the proof of Theorem 3 relies only on local arguments.

In a classical nonlinear setting, we derive the following existence result for problem (1.1), where we set for simplicity ϵ = 1.

Theorem 4

Assume that f:Ω×RR satisfies the following conditions:

  1. f(x,t)C0(Ωˉ×R) and f(x,t) = 0 for t ≤ 0;

  2. there exist constants C1, C2 > 0 such that

    |f(x,t)|C1+C2|t|q,

    where 1 < q < 2* - 1;

  3. f(x,t) = o(|t|) as t → 0;

  4. there exist μ > 2 and R ≥ 0 such that for tR,

    0<μF(x,t)tf(x,t),

    where F(x,t):=0tf(x,ξ)dξ .

Then, the problem

(1.18) (Δ)suΔu=f(x,u) in Ω,u=0 in RNΩ

possesses a nonnegative solution uX0s .

Theorem 5 ([22],Theorem 1.3)

Assume that Ω ⊂ ℝN is a bounded star-shaped domain, fCloc0,1(Ωˉ×R) and

N22tf(x,t)NF(x,t)+xFx(x,t),forallxΩandtR.

If u is a bounded solution of (1.18), then u ≡ 0.

A consequence of Theorems 4 and 5 we have

Corollary 6

Assume that Ω ⊂ ℝN is a bounded domain, s ∈ (0,1) and p ∈ (2,2*). Then, the problem

(1.19) (Δ)suΔu=|u|p2u in Ω,u=0 in RNΩ

possesses a nonnegative bounded mountain pass solution uX0s . Moreover, if Ω is a bounded star-shaped domain and p ≥ 2*, then (1.19) does not admit nontrivial bounded solutions.

The mountain pass theorem underlies the proof of Theorem 4 and we further show that this mountain pass solution of (1.19) has actually minimal energy via Nehari's method.

Theorem 7

The mountain pass solution u of Corollary 6 is a ground state, i.e.

F(u)=infNF,

where

F(u):=12us21pΩ|u|pdx,uX0s,N:={uX0s{0}:F(u),u=0}.

The remaining results are related to symmetry properties and regularity of weak (classical) solutions.

Theorem 8

Assume that f: ℝ → ℝ is Lipschitz continuous and uLsC2(B(0,1)) is a positive solution of

(1.20) (Δ)suΔu=f(u) in B(0,1),u=0 in RNB(0,1).

Then u is radially symmetric, i.e. u(x) = v(|x|) for some decreasing function v:[0,1] → [0,+∞).

Theorem 9

Assume that uX0s is a weak solution of the problem

(1.21) (Δ)suΔu=f(x) in Ωu=0 in RNΩ.

Then the following facts hold:

  1. if fLq(Ω) with q>N2 , then there exists a constant C, depending only on N, usandfq , such that the positive weak solution of (1.21) satisfies

    uC;
  2. if fLq(Ω) with q=N2 , then there exists a constant α > 0 such that the positive weak solution of (1.21) satisfies

    Ωeαudx<+.

    In particular, uLq(Ω) for every q < +∞;

  3. if f is a positive function and fLq(Ω) with 2NN+2q<N2 , then there exists a constant C > 0 such that the positive weak solution of (1.21) satisfies

    uqCfq,q=qNN2q.

Note that the results in Theorem 9 are consistent with the ones known for the Laplace operator. As we will see in the proof, this is due to the choice of appropriate truncated functions which downplay the effects of the nonlocal term.

In the second part of this paper we aim at investigating the asymptotic behavior of ground states (weak solutions) of problem (1.3) when 2 < p < 2*, N > 2. For this purpose set

Sϵ:=inf{uϵ2:up=1}.

Note that Sϵ can be also written in the form

Sϵ=infuX0s{0}Pϵ(u),

where Pϵ(u)=uϵ2up2. Hereafter let us denote

uϵ:=ϵu22+[u]s,RN212.

The compactness of Sobolev embeddings yields the existence of non-negative minimizers wϵ of Sϵ. Then, by Lagrange's multiplier rule, there exists some λϵ > 0 such that

(1.22) (Δ)swϵϵΔwϵ=λϵ|wϵ|p2wϵ in Ω.

By multiplying both sides of (1.22) by wϵ and then integrating, we get λϵ = Sϵ. Moreover, by maximum principles, wϵ > 0.

The energy associated to (1.22) is

(1.23) J(u)=12uϵ21pSϵupp,uX0s.

Since wϵ is a solution of (1.22), we get wϵϵ2=Sϵwϵppandwϵp=1 , and thus

(1.24) J(wϵ)=12wϵϵ21pSϵwϵpp=p22pSϵ.

On the other hand, any nontrivial solution v of (1.22) necessarily satisfies

(1.25) J(v)=12vϵ21pSϵvpp=p22pSϵvpp.

We have also

(1.26) Sϵvϵ2vp2=Sϵvppvp2=Sϵvpp2,

which yields vp1 . This means that, by (1.25), v should satisfy

J(v)p22pSϵ.

This fact, together with (1.24), implies that wϵ is a ground state of equation (1.22) and so uϵ=Sϵ12pwϵ is a ground state of (1.3). Note that uϵ also satisfies Pϵ(uϵ)=Sϵ .

Let us stress the fact that we do not know whether the mountain-pass solution obtained from Theorem 4 and the minimal solution uϵ obtained above agree. Next we will focus on the minimal solution uϵ.

Before stating our main results, we introduce the following best constants

(1.27) S:=infuH0s{0}[u]s2up2 for 2<p<2s,

and

(1.28) S˜:=infuDs,2(RN){0}[u]s2u2s2 for p=2s,

where H0s is the linear space of Lebesgue measurable functions u:RNR such that u(x) = 0 a.e. in CΩ and [u]s := [u]{s,ℝN} < +∞, while Ds,2(ℝN) is the closure of C0(RN) with respect to the semi-norm [⋅]s,ℝN.

Main results

Theorem 10

The following facts hold:

  1. if 2<p<2s , then SϵS as ϵ → 0. Moreover, there exists a minimizer u0 of S and uϵu0inH0sasϵ0 ;

  2. if p=2s , then SϵS˜asϵ0 ;

  3. if 2s<p<2 , then Sϵ0asϵ0 . Moreover, there exist a minimizer uϵ such that uϵϵ0asϵ0 .

Theorem 11

If 2 < p < 2*, then Sϵ → 0 and uϵϵ+ as ϵ → + ∞.

Finally let us make a few comments on Theorems 10 and 11. We expect the minimizers uϵ obtained in (i) to be uniformly bounded with respect to ϵ and thus that the convergence should hold in the classical sense. In this context, the existence or nonexistence of positive solutions of the problem

(1.29) (Δ)suΔu=|u|p2u in RNorR+N

with 2<p<2s plays a very important role. However, the moving plane methods can not be applied directly here because one of the most important tools - the Kelvin transform - is not available. On the other hand, the classical minimizers uϵ obtained in (ii) blow up as ϵ → 0, up to a subsequence if necessary. Indeed, if not one has uϵu0 in classical sense for supp(u0) ⊂ Ω, and then u0 is a minimizer of S˜ , thus a contradiction. Obviously, for the case (iii), the minimizers uϵ vanishe a.e. in Ω. We stress that the constant

S:=infuDs,2(RN)D1,2(RN){0}u22+[u]s2up2 for 2s<p<2

plays a crucial role to get (iii), while to prove Theorem 11, we need the constant

Sˆ:=infuH01(Ω){0}u22up2 for 2<p<2.

The rest of this paper is organized as follows. We prove Theorems 19 in Section 2 whence Section 3 is devoted to prove Theorems 10 and 11.

2 Proofs of theorems 14 and 79

Let us prove first the following preliminary result

Lemma 12

(X0s,s) is a Hilbert space with scalar product

u,vs=Ωuvdx+RN×RN(u(x)u(y))(v(x)v(y))|xy|N+2sdxdy,u,vX0s.

Proof

It is straightforward to see that 〈 ⋅, ⋅〉s is a scalar product on X0s which induces the norm s . So, it suffices to prove that (X0s,s) is complete. Let {uk} be a Cauchy sequence in X0s , that is, for any ε > 0 , there exists k0 > 0 such that for k, jk0,

(2.1) ujuks<ε.

This implies that ujukH01(Ω)<ε . Since H01(Ω) is complete, there exists uH01(Ω) such that uku in H01(Ω) as k → +∞. Moreover, since uk = 0 a.e. in CΩ , we may define u = 0 a.e. in CΩ and then uku in H1(ℝN) as k → +∞. Thus, up to a subsequence, still denoted by $\{u_k\}$, uku a.e. in ℝN. Fatou's lemma and (2.1) yield

(2.2) [u]s,RN2limk+[uk]s,RN22limk+(ukuk0s2+uk0s2)<+,

and hence uX0s . Finally, by using again (2.1), we get

(2.3) ujuslim infk+ujuksε,

and uju in X0s as j → +∞, as desired.

Proof of Theorem 1. The proof is modeled on the one of the nonlocal operator LK studied in [18]. However, for reader's convenience, we give the detailed proof.

For simplicity set I(u):=us2 and M:={uX0s:u2=1} .

(1) Let {uj} be a minimizing sequence for I, that is,

limj+I(uj)=infuMI(u)0.

Then {I(uj)} is bounded, i.e. {uj} is bounded in X0s . Up to a subsequence, still denoted by {uj}, uju* in X0s for some uX0s and uju* in L2(Ω). Since ∥uj2 = 1, we get ∥u*2 = 1, that is, u* ∈ 𝓜. On the other hand,

(2.4) limj+I(uj)I(u)infuMI(u).

and this implies that I(u)=infuMI(u) .

Let t ∈ (-1,1), vX0s,Ct=u+tv2 and ut = (u* + tv)/Ct. Then ut ∈ 𝓜 and one has

(2.5) I(ut)=us2+2tu,vs+o(t)1+2tRNuvdx+o(t)=(us2+2tu,vs+o(t))12tRNuvdx+o(t)=I(u)+2tu,vsI(u)RNuvdx+o(t).

By the very definition of u* we get

(2.6) u,vs=I(u)RNuvdx.

Now, assume that λ1 is attained at some e1X0s with ∥e12 = 1. By (2.6), we get I(e1) = λ1. To prove that e1 ≥ 0, we first show that if e is an eigenfunction related to λ1 with ∥e2 = 1, then both e and |e| realize the minimum in (1.14). In fact, if x1{xRN:e(x)>0} and x2{xRN:e(x)<0} , then

||e(x1)||e(x2)||=|e(x1)+e(x2)|<e(x1)e(x2)=|e(x1)e(x2)|.

Thus, if both sets {xRN:e(x)>0}and{xRN:e(x)<0} have positive measure, we have I(|e|) < I(e). Note that |e|X0s and ∥|e|∥2 = ∥e2 = 1. Thus, since e1 is a minimizer, we get I(|e|) = I(e) = I(e1) = λ1. This also implies that e ≥ 0 or e ≤ 0 a.e. in Ω. By replacing e1 by |e1|, we may conclude that e1 ≥ 0.

(2) Suppose by contradiction that there exists another eigenfunction e˜1X0s,e˜1e1 , corresponding to λ1. Then, by (1) we may assume that e˜10 . Setting u1:=e˜1e˜12 and v1 := e1 - u1, we claim that v1 = 0. In fact, direct calculations show that v1 is also an eigenfunction related to λ1, and so from (1) we deduce v1 ≥ 0 or v1 ≤ 0 a.e. in Ω, that is, e12u12ore12u12 a.e. in Ω. However, since e122u122=0 , we get e12=u12 and hence v1 = 0 a.e. in Ω. This also implies that v1 = 0 a.e. in ℝN and therefore, if wX0s is an eigenfunction related to λ1, then w = ± ∥w2 e1 a.e. in ℝN.

(3) Since 𝓜k is a weakly closed subspace of X0s , arguing as in step (1), we deduce that λk+1 is attained at some ek+1X0s . On the other hand, being Mk+1Mk , we get 0 < λ1λ2 ≤ … ≤ λk ≤ … Now, we claim that λ1λ2. If not, e2 ∈ 𝓜2 would be also an eigenfunction related to λ1. By (2), e2 = ±∥e22e1 a.e. in ℝN. Thus, we would get 0=e1,e2s=±e22e1s2 , which yields e1 ≡ 0 a.e. in ℝN, a contradiction.

Analogously to the proof of (2.6), we have

(2.7) ek+1,ws=λk+1RNek+1wdx,wMk+1.

We claim that equality (2.7) holds for any wX0s , i.e., that ek+1 is an eigenfunction related to λk+1. Let us consider the decomposition X0s=span{e1,,ek}(span{e1,,ek})=span{e1,,ek}Mk+1 , and write any given vX0s as v = v1 + v2, with v1=j=1kcjej,cjR,andv2Mk+1 . Testing (2.7) with w = v2 we then obtain

ek+1,vsλk+1Ωek+1vdx=ek+1,v1sλk+1Ωek+1v1dx=j=1kcjek+1,ejsλk+1Ωek+1ejdx=0,

which proves the claim.

Next we show that λk → +∞ as k → +∞. Suppose, by contradiction, that λkC ∈ ℝ+ as k → +∞. Since eks2=λk , up a subsequence, ekje in L2(Ω) as kj → +∞. On the other hand, being 〈 eki, ekjs = 0 for kikj, we get ∫Ω eki ekjdx = 0 for kikj and thus

ekiekj22=eki22+ekj22=1,

which is a contradiction, as

ekiekj22ekie22+ekje220as ki,kj+.

Finally, let us show that any eigenvalue of (1.13) can be written in the form (1.15). Arguing again by contradiction, assume that there exists an eigenvalue λ ∉ {λk} and let e be an eigenfunction corresponding to λ with ∥e2 = 1. Then λ=es2e1s2=λ1 and this implies the existence of k ∈ ℕ such that λk < λ < λk+1. Thus, e ∉ 𝓜k+1 and so there exists j ∈ {1,2,⋯, k} such that 〈 e, ejs ≠ 0, a contradiction.

(4) Let us prove that {ek} is a basis of X0s by a standard Fourier analysis technique. Given uX0s , we define

uj=i=1ju,e˜ise˜i,

where e˜i=eieis . Letting vj = u - uj, we get

0vjs2=us2i=1ju,e˜is2,

that is,

ρj:=i=1ju,e˜is2us2.

Now, for l > j, we get

vlvjs2=i=j+1lu,e˜is2=ρlρj,

which implies that {vj} is a Cauchy sequence in ℝ, and thus there exists vX0s such that vjvinX0s as j → +∞. For ji we also have

vj,e˜is=u,e˜isv,e˜is=u,e˜isv,e˜s=0.

Passing to the limit, we get v,e˜is=0 for any i ∈ ℕ. This means that v* = 0 and therefore u_j = u - vju in X0sasj+.So,u=i=1u,e˜ise˜i , as we wanted.

In order to show that {ei} is a basis for L2(Ω), too, chosen vL2(Ω), we let vjC02(Ω) such that vvjL2(Ω)<1j . Since {ei} is a basis of X0s and X0sL2(Ω) , there exist kj ∈ ℕ and a function wj ∈ span {e1,…, ekj} such that vjwjs1j . Thus, by Sobolev embeddings,

vwj2vvj2+vjwj2vvj2+Cvjwjs1+Cj,

which yields the claim.

(5) X0s=span{ek,,ek+h}(span{ek,,ek+h}) . So, any eigenfunction uX0s{0} corresponding to λk admits the representation u = u1 + u2, with u1=i=kk+hcieispan{ek,,ek+h} for some ci ∈ ℝ, and u2 ∈ (span{ek,⋯,ek+h}). Obviously, 〈 u1, u2s = 0, which yields

(2.8) λku22=us2=u1s2+u2s2.

By definition, u1 is also an eigenfunction corresponding to λk. Thus,

λkΩu1u2dx=u1,u2s=0,

and

u22=u122+u222.

On the other hand, we have

(2.9) u1s2=i=kk+hci2eis2=λki=kk+hci2=λku122.

Recalling that u and u1 are eigenfunctions corresponding to λk, we deduce that u2 enjoys the same property and thus, u2(span{e1,,ek+h})=Mk+h+1 .

We claim that u2 ≡ 0. If not, we would have

(2.10) λk<λk+h+1=minuMk+h+1{0}us2u22u2s2u222.

Collecting (2.8), (2.9) and (2.10), we would obtain

λku22>λku122+λku222=λku22,

clearly a contradiction. Therefore, u = u1 ∈ span{ek,⋯,ek+h} and this completes the proof.  □

Proof of Theorem 2. Inspired by [7], assume that u(x) < 0 in Ω. Owing to the lower semi-continuity of u on Ωˉ , there exists some x0Ωˉ such that

u(x0)=minΩˉu(x)<0.

Moreover, since u ≥ 0 in ℝNΩ, we deduce that x0 is an interior point of Ω. Taking into account the fact that Δ u(x0) ≥ 0, we get

(Δ)su(x0)Δu(x0)cN,sPVRNu(x0)u(y)|x0y|N+2sdycN,sPVRNΩu(x0)u(y)|x0y|N+2sdy<0,

which contradicts the assumptions. Moreover, if u(x0) = 0 at some point x0Ω, then

(2.11) 0(Δ)su(x0)Δu(x0)cN,sPVRNu(y)|x0y|N+2sdy,

which forces u(x) ≡ 0 in ℝN.  □

As a consequence of the previous result one has

Corollary 13

Assume that uLsC2(Ω) is upper semi-continuous on Ωˉ . If

(2.12) (Δ)suΔu0inΩ,u0inRNΩ.

Then, u(x) ≤ 0 in Ω. Moreover, if u(x) = 0 at some point in Ω, then u(x) ≡ 0 in ℝN.

Proof of Theorem 3. Let G:RR be the truncation defined by G(t) = 0 for t ≤ 0 and G(t) = t for t > 0. Note that G is clearly a Lipschitz function and so, if vH1(Ω), we get G(v) ∈ H1(Ω) thanks to Stampacchia’s theorem. Moreover, [G(v)]s < +∞.

Choosing v = −u, then vH1(Ω), [v]s,RN<+ , and

(Δ)svΔv0inΩ,v0inRNΩ.

Using G(v) as a test function, we get

(2.13) ΩvG(v)dx+RN×RN(v(x)v(y))(G(v(x))G(v(y)))|xy|N+2sdxdy0,

and since (v(x) − v(y))(G(v (x))−G(v (y))) ≥ 0, we get

(2.14) ΩvG(v)dx0.

Now, by Stampacchia’s theorem (cf. [3]), the proof is standard. In fact, by (2.14) we get G′(v)∣∇ v2 = 0 a.e. in Ω and hence ∇ G(u) = 0 a.e. in Ω. Since G(v)H01(Ω) , we deduce G(v) = 0 a.e. in Ω and, as a consequence, v ≤ 0 and u = 0 a.e. in Ω.  □

Similarly, one has

Corollary 14

Assume that uH1(Ω), [u]s,RN<+ , and u satisfies

(Δ)suΔu0inΩ,u0inRNΩ.

Then, u(x) ≤ 0 a.e. in Ω.

The proof of the existence result established in Theorem 4 requires some preliminary steps. Since we focus on nonnegative solutions, we consider the truncated energy functional associated with problem (1.8), namely

(2.15) F(u):=12us2ΩF(x,u+)dx,uX0s,

where u+≔ max{u, 0}. Thanks to (f2), (f3) and Sobolev inequalities, F is well-defined, Fréchet differentiable and

(2.16) F(u),v=u,vsΩf(x,u+)vdx,u,vX0s.

In order to exploit the mountain pass theorem, let us first prove that F exhibits the suitable geometry.

Lemma 15

There exist ρ, δ>0 such that, for any uX0s satisfyingus = ρ, one has F(u)δ .

Proof. For any uX0s , by (f2), (f3) and Sobolev embeddings, we get

F(u)14us2CΩ|u|q+1dx14Cusq1us2.

By the definition of q, we can choose ρ>0 sufficiently small such that 14Cusq1>0 and the result follows by setting δ=14ρ2Cρq+1 .  □

Lemma 16

There exists eX0s satisfyinges > ρ and F(e)<0 .

Proof. Fix vX0s , with ∣vs = 1 and v > 0 a.e. in ℝN,, and let u = tv for t > 0. Then, by (f4) we obtain

F(u)12t21μtμΩ|v+|μdxC,ast+.

The assertion follows by taking e = tv, with t sufficiently large.  □

A key-ingredient is the Palais-Smale condition, (PS) for short.

Lemma 17

The functional F satisfies (PS).

Proof

Let cR , and let {uk} be a sequence in X0s such that

(2.17) limk+F(uk)=c

and

(2.18) limk+F(uk),v=0,vX0s.

By (f4), we have

(2.19) μF(uk)F(uk),ukμ22uks2C(R),

where C(R) > 0, and hence {uk} is bounded in X0s . Then, up to a subsequence still denoted by {uk}, there exists uX0s such that uku in X0s , uku in Lm(Ω), 1 < m<2*, and uku a.e. in Ω as k → +∞ (note that, since uk = 0 a.e. in CΩ , we may define u = 0 a.e. in CΩ) . By (2.18) and the boundedness of the sequence {uku}, we get

(2.20) limk+F(uk),(uku)=limk+uk,ukusΩf(x,uk+)(uku)dx=0.

On the other hand, by (f2) and the compactness of the embeddings X0sL1(RN) , X0sLq+1(RN) , we deduce

(2.21) limk+Ωf(x,uk+)(uku)dx=0.

A joint use of (2.20),(2.21) and the fact that uku in X0s , lead to ∣uks → ∣us as k → +∞ and, by classical arguments, to ∥ukus → 0 as k → +∞, as desired.  □

Proof of Theorem 4. It is straightforward by the mountain pass theorem and joining Lemmas 15, 16 and 17. By Theorem 3, the solution is nonnegative.

Proof of Theorem 7.For any uX0s , define the fibrering map ϕu:[0,+)R by ϕu(t):=F(tu) . Of course, uN if and only if

(2.22) us2Ω|u|pdx=0,

and tuN iff ϕu(t) = 0. By direct computations,

ϕu(t)=tus2tp1Ω|u|pdx,ϕu′′(t)=us2(p1)tp2Ω|u|pdx,

and so ϕu has a unique critical point (a local maximum), namely

(2.23) t=t(u)=us2Ω|u|pdx1p2.

Moreover ϕu is increasing in (0, t(u)), decreasing in (t(u), +∞) and t(u)uN .

Let us split N as follows

N+:={uN:ϕu′′(t)>0},N:={uN:ϕu′′(t)<0},N0:={uN:ϕu′′(t)=0}

to obtain N=N . F is of course bounded below on N , as

F(u)=121pupp0,uN.

Setting

cˉ:=infuNF(u),

we easily derive that cˉ>0 . Indeed, if v = u/∥us and t(v) is as in (2.23), we get

F(u)=F(t(v)v)=121pt2(v)vs2121pvs2pp2vp2pp2121pcp2pp2>0,

for any uN , being cp > 0 the best constant of the embedding X0sLp(Rn) .

The next step is to show that cˉ is attained at some function within N . Let {uk}N be a minimizing sequence, i.e. such that F(uk)cˉ as k → +∞. Clearly {uk} is bounded, so, up to a subsequence, uku0X0s and uku0Lp(RN) . So,

0<cˉ=limk+F(uk)=limk+121pΩ|uk|pdx=121pΩ|u0|pdx,

which yields u0≠ 0. If uk → ≠ u0 in X0s , we would obtain

ϕu0(1)=u0s2Ω|u0|pdx<lim infk+uks2Ω|uk|pdx=0.

We know that there exists a unique t0 ∈ (0, 1) such that t0u0N and that tϕuk(t) admits its maximum at t = 1, so

F(t0u0)<lim infk+F(t0uk)limk+F(uk)=cˉ,

a contradiction. So uku0 and ϕu0(1)=0 , namely u0N . We finally get

F(u0)=limk+F(uk)=cˉ

and u0 is the minimizer sought for.

To conclude the proof, we need to give a minimax characterization of cˉ . If ψ1:X0s{0}(0,+) , ψ2:{uX0s:uϵ=1}N are the maps defined by ψ1(u)≔ t(u), ψ2(u)≔ t(u)u, it is easy to show that ψ1 is continuous, while ψ2 is a homeomorphism.

Then, arguing as in [28 Theorem 4.1], define

c:=infuX0s{0}maxt0ϕu(t),c:=infyΓmaxt[0,1]F(y(t)),

where

Γ:={yC0([0,1],X0s):y(0)=0,F(y(1))<0}.

The above considerations yield cˉ=c . Moreover, if uX0s{0} ,

F(tu)=t22us2tppΩ|u|pdx

as t → +∞ and hence cc*. If X1 and X2 are the two components of X0s induced by N , with 0 ∈ X2, there will exist r > 0 such that B (0, r) ⊂ X2. Since ϕu(t)=F(tu),u0 for all t ∈ [0, t(u)], one has F(u)0 for all uX2. As a result, any γ ∈ Γ will have a non-empty intersection with N and then cˉc .

This implies that cˉ=c=c and the proof is complete.

Remark

It is worth pointing out that each ground state solution u0 of (1.19) does not change sign. Indeed, denoting as usual u0+=max{u0,0} and u0=min{u0,0} the positive and the negative part of u0, respectively, multiplying (1.19) by u0± and integrating, we get u0±N . As a result

cˉ=F(u0)=121pu0s2=121pu0+u0,u0+u0s=121pu0+s2+u0+s2=F(u0+)+F(u0)2cˉ,

a contradiction.

Before proving Theorem 8, it is necessary to introduce some notations and some preliminary results, namely an antisymmetric maximum principle holding for ̏narrow̎ regions. This is needed to apply the moving plane method directly to problem (1.20) (cf. [7]).For λR , denote by

Tλ:={xRn:x1=λ}

the moving planes, by

Σλ:={xRn:x1<λ}

the region to the left of the plane and by

xλ:=(2λx1,x2,,xn)

the reflection of xRN with respect to Tλ, and set

wλ(x):=u(xλ)u(x).

Lemma 19

Let V be a bounded subset of Σλ contained in {xRn:λl<x1<λ} , l > 0 sufficiently small. Suppose that uLsC2(V) is lower semicontinuous on Vˉ . If c:RnR is bounded from below in V and

(2.24) (Δ)su(x)Δu(x)+c(x)u(x)0inV,u(x)0inΣλV,u(xλ)=u(x)inΣλ,

then u(x) = 0 in V for l sufficiently small. Moreover, if u(x) = 0 for some xV, then u(x) ≡ 0 for a.e. xRn .

Proof

Arguing by contradiction, assume that u solves (2.24) and u < 0 somewhere in V. Since u is lower semicontinuous, there exists x0 ∈ int(V) such that

u(x0)=minuVˉu(x)<0.

Since u is anti-symmetric in Σλ, following [7] one has

(Δ)su(x0)=cN,sPVRnu(x0)u(y)|x0y|N+2sdy=cN,sPVΣλu(x0)u(y)|x0y|N+2sdy+{xRN:xλΣλ}u(x0)u(y)|x0y|N+2sdy=cN,sPVΣλu(x0)u(y)|x0y|N+2sdy+Σλu(x0)+u(y)|x0yλ|N+2sdycN,sΣλ2u(x0)|x0yλ|N+2sdy,

and, after a change of variables, it is not difficult to see that the last term diverges to −∞ as l → 0+. Being Δu(x0) ≥ 0 and c bounded below in V, we get

(Δ)su(x0)ϵΔu(x0)+c(x0)u(x0)<0asl0+,

which contradicts (2.24).  □

Proof of Theorem 8. Since u is a positive solution of (1.20), we get

(2.25) (Δ)swλ(x)Δwλ(x)+cλ(x)wλ(c)=0,

for every x ∈ Σλ, where

cλ(x):=(uλ(x))p1u(x)p1uλ(x)u(x),cλ(x):=f(uλ(x))f(u(x))uλ(x)u(x),

and ∣cλ(x)∣ ≤ L, for some L > 0. Applying Theorem 2.24, for λ → −1+ one has wλ(x) = 0 in Σλ. Now, setting

λ0:=sup{λ0:wμ(x)0,xΣμ,μλ},

we claim that λ0 = 0. Assuming the contrary, then it should be

wλ0(x)>0xΣλ0,

otherwise there would exist x0Σλ0 :

wλ0(x0)=minuΣλ0wλ0(x)=0.

Let us estimate

(Δ)swλ0(x0)=cN,sPVRNwλ0(y)|x0y|N+2sdy=cN,sPVΣλ0wλ0(y)|x0y|N+2sdy+RNΣλ0wλ0(y)|x0y|N+2sdy=cN,sPVΣλ01|x0yλ0|N+2s1|x0y|N+2swλ0(y)dy<0

which together with the fact that Δwλ0(x0)0 obviously contradicts (2.25). So, there exists δ>0 such that, for any λ ∈ (λ0, λ0+δ) we have wλ(x) = 0, x ∈ Σλ, which contradicts the definition of λ0.

Therefore

w0(x)0,xΣ0,

or equivalently

u(x1,x2,,xn)u(x1,x2,,xn),0<x1<1.

Since the x1- direction is arbitrary, the previous relation implies that u is radially symmetric with respect to the origin. The monotonicity follows from the fact that wλ(x) = 0 for all λ ∈ (−1, 0].

Finally, we deal with the regularity of solutions to the problem

(2.26) (Δ)suΔu=f(x)inΩ,u=0inRNΩ,

where ΩRN is a bounded domain, N > 2. To be precise, we will study the Lp regularity, including the boundedness of solution via Stampacchia’s method, exponential summability and a Calderón-Zygmund type result (for the purely nonlocal case we refer to [16]).

Let us recall following lemma by Stampacchia

Lemma 20

Let ψ:R+R+ be a nonincreasing function such that

ψ(h)Mψ(k)δ(hk)γ,h>k>0,

where M > 0, δ>1 and y>0. Then ψ(d) = 0, where dy=Mψ(0)δ12δγδ1 .

Proof of Theorem 9 (1) We follow Stampacchia’s method. For k ≥ 0, we define the function

Gk(σ)=σk,ifσk,0,if|σ|<k,σ+k,ifσk.

Clearly, Gk(0) = 0 and Gk(σ) is Lipschitz continuous. For uX0s , we get Gk(u)X0s . Using Gk(u) as a test function, we get

(2.27) ΩuGk(u)dx+RN×RN(u(x)u(y))(Gk(u(x))Gk(u(y)))|xy|N+2sdxdy=Ωf(x)Gk(u)dx.

Since (u(x) − u(y))(Gk(u(x))−Gk(u(y))) ≥ (Gk(u(x))−Gk(u(y)))2, from (2.27) we get

Ω|Gk(u)|2dxAk|f(x)Gk(u(x))|dx,

where Ak≔ {xΩ: ∣u(x)∣ ≥ k}. By applying Sobolev and Hardy inequalities, we have

S2Gk(u)22fqGk(u)2|Ak|1121q,

where S is the best constant of the embedding X0sL2(RN) .

For any h > k, we have AhAk and Gk(σ)χAhhk , thus

S2(hk)|Ah|12fq|Ak|1121q,

which implies

|Ah|S22fq2|Ak|21121q(hk)2.

The assumption q>N2 implies 2(1121q)>1 and so, by Lemma 20, there exists k0 > 0 such that ψ(k0)=|Ak0|=0 . Consequently, we get ∣u(x)∣ ≤ k0 a.e. in Ω.

(2) For any T > 0, we consider the function

Φ(σ)=eασ1if0σ<T,αeαT(σT)+eαT1ifσT,

where α>0 will be fixed later. Note that Φ(0) = 0, Φ ∈ C1, is convex and its first derivative is Lipschitz continuous.Thus, if uX0s , then Φ(u)X0s as well.

From

Φ(u(x))Φ(u(y))Φ(u(x))[u(x)u(y)],

one has

Ω[ΔΦ(u)+(Δ)sΦ(u)]Φ(u)dxΩ[Δu+(Δ)su]Φ(u)Φ(u)dx=Ωf(x)Φ(u)Φ(u)dx,

which implies

S2Φ(u)22α{xΩ:u(x)<T}[Φ2(u)f+Φ(u)f]dx+αeαT{xΩ:u(x)T}Φ(u)fdxαΦ(u)22fN2(1+|Ω|N22N)+Cf+αeαTΦ1(T){xΩ:u(x)T}Φ2(u)fdxαΦ(u)22fN2(3+|Ω|N22N)+Cf,

provided that α T > 1. So, fixing α small enough and taking T large, we obtain

Φ(u)22C,

and the conclusion follows.

(3) For any T > 0, consider the function

Φ(σ)=σβif0σ<T,βTβ1(σT)+TβifσT,

where β=m2>1 . It is straightforward to observe that Φ enjoys the same properties as in item (2). Then, following the same steps of item (2), we have

Ω[ΔΦ(u)+(Δ)sΦ(u)]Φ(u)dxΩf(x)Φ(u)Φ(u)dx,

and so

S2Φ(u)22fqΦ(u)Φ(u)qCfqΩ|Φ(u)|2dx1q,

where 1/q + 1/q′ = 1 and C > 0 is independent of T. Thus,

(2.28) S2Ω|Φ(u)|2dx221qCfq.

and letting T → +∞ and recalling that 2*β = q′(2β−1) we arrive at

umCfq,

as desired.

Finally, we prove the boundedness of the solution u obtained in Corollary 6.

Lemma 21

The solution u of (1.19) stated in Corollary 6 satisfiesu<C for some constant C > 0.

Proof. Let β, L > 0 and let Y:R0+R be the function defined by

(2.29) Y(t)=t2β+1if0t<L1β,(2β+1)L2t2βL2+1βiftL1β.

Note that Y is convex, differentiable, Y(u)X0s and

(Δ)sY(u)Y(u)(Δ)su.

Besides, we get

(2.30) YY=(2β+1)u4β+1if0t<L1β,(2β+1)2uL42β(2β+1)L4+1βiftL1β.

Clearly, Y(u)Y(u)H01(Ω) . Multiplying Y′(u)Y(u) both sides of (1.19) and integrating, we get

(2.31) 0ΩY(u)(Δ)sY(u)dxΩY(u)Y(u)(Δ)sudx=ΩY(u)Y(u)Δudx+ΩY(u)Y(u)|u|p2udx,

which yields

(2.32) Ω(Y(u)Y(u))udxΩY(u)Y(u)|u|p2udx.

Direction calculations show that

(2.33) Ω|(umin{u2β,L2})|2dxΩY(u)Y(u)|u|p2udx.

Then, arguing as in [27 Lemma B.3], we get uLq(Ω) for any q < +∞ and, in particular, for q>N2 . The conclusion then is a consequence of Theorem 9-(1).  □

3 Proof of Theorems 10 and 11

Here we study the asymptotic behavior of ground states of problem (1.3). Let uϵ be a minimizer of Sϵ, as well as a ground state of (1.3), as shown in Section 1.

It is immediate to estimate the ϵ-norm and the Lp-norm of such minimizers.

Lemma 22

uϵϵ=Sϵp2(p2) and uϵp=Sϵ1p2 .

Proof. By direct calculations, we get

uϵϵ=Sϵ12pwϵϵ=Sϵ12p+12=Sϵp2(p2)

and

uϵp=Sϵ12pwϵp=Sϵ1p2.

Next, we prove that the function {Sϵ} is bounded. To be more precise, we show that {Sϵ} converges to different constants according to the range of p. We consider three different cases.

3.1 The case 2<p<2s

Let

S=infuH0s{0}[u]s2up2=inf{[u]s2:up=1}.

By Sobolev embedding theorems, for 2<p<2s there exists a non-negative minimizer w0 of S. Analogously to the proof carried out in Section 1, the function u0=S12pw0 is a non-negative ground state of

(3.1) (Δ)su=|u|p2uinΩ,u=0inRNΩ.

We point out that (3.1)) also admits a mountain pass type solution, see e.g. [24].

In the same flavor of Lemma 22, we also have

Lemma 23

[u0]s=Sp2(p2) , u0p=S1p2 .

The asymptotic behavior of Sϵ as ϵ → 0 is established in the next

Lemma 24

limϵ → 0Sϵ = S.

Proof. Since X0sH0s , we get

SinfuX0s{0}[u]s2up2Sϵ,

and thus

Slim infϵ0Sϵ.

Assume w0 is a minimizer of S. By density, there exists a sequence {wk}C0(RN) such that wkw0 in H0s(RN) and Lp(RN) . Thus, we have

(3.2) Sϵwkϵ2wnp2

and

(3.3) lim supϵ0Sϵlimklim supϵ0wkϵ2wkp2=limk[wk]s2wkp2=S.

This completes the proof.

From Lemmas 22, 23 and 24, we get

uϵϵ[u0]s,uϵpu0p,asϵ0.

Note that since [uϵ]s is bounded, up to a subsequence we may assume that uϵ⇀ũ in H0s(Ω) and thus uϵ → ũ in Lp(Ω). On the other hand, from ∣∣uϵ∣∣p → ∣∣u0∣∣p, we obtain uϵu0 in Lp(Ω). This gives ũ = u0 a.e. in ℝN. Therefore, we obtain

Lemma 25

limϵ → 0 ϵ ∣∣∇ uϵ∣∣2 = 0, limϵ → 0[uϵ]s = [u0]s, limϵ → 0∣∣uϵ∣∣p = ∣∣u0∣∣p.

3.2 The case 2sp<2

For a fixed ϵ>0, let us consider the problem

(3.4) (Δ)suΔu=|u|p2uinΩϵ,u=0inRNΩϵ,

where Ωϵ:=ϵ12s2Ω and 2sp<2 . Note that ΩϵRN as ϵ → 0.

Define

S˜ϵ=inf{us2:uX0s(Ωϵ),uLp(Ωϵ)=1}=infuX0s(Ωϵ){0}us2up2.

Let w˜ϵ be a minimizer of S˜ϵ satisfying w˜ϵs2=S˜ϵ and w˜ϵLp(Ωϵ)=1 . Then, u˜ϵ=S˜ϵ12pw˜ϵ is a ground state of (3.4). Moreover, one has

(3.5) Sϵ=ϵp(2sN)+2N2p(s1)S˜ϵ.

Now let us introduce the scaled function vϵ(x)=ϵs(s1)(p2)u˜ϵ(ϵ12s2x) , xΩ. We show that it is a solution of problem (1.3). In fact, one has

(3.6) Ω|vϵ|pdx=ϵsp(s1)(p2)Ω|u˜ϵ(ϵ12s2x)|pdx=ϵsp(s1)(p2)N2s2Ωϵ|u˜ϵ(y)|pdy=ϵ2spN(p2)2(s1)(p2)Ωϵ|u˜ϵ(y)|pdy=S˜ϵp2p if p=2s,ϵ2N+p(2sN)2(s1)(p2)S˜ϵp2p if p>2s,=Sϵp2p for p2s

and, clearly,

(3.7) vϵϵ2=ϵ2s(s1)(p2)+2sN2s2Ωϵ|u˜ϵ|2dx+[u˜ϵ]s2=ϵ2N+p(2sN)2(s1)(p2)u˜ϵs2=S˜ϵp2p if p=2s,ϵ2N+p(2sN)2(s1)(p2)S˜ϵp2p if p>2s,=Sϵp2p for p2s.

By (3.5), (3.6) and (3.7), we have

(3.8) Pϵ(vϵ)=Sϵforp2s,

and thus vϵ is a minimizer of Sϵ and also a ground state of (1.3), with J(vϵ)=p22pSϵpp2 .

3.2.1 The case p=2s

Lemma 26

For p=2s , S˜ϵS˜ , where S˜ is the Sobolev constant

S˜:=infuDs,2(RN){0}[u]s,RN2uL2s(RN)2,

or equivalently

S˜:=inf[u]s,RN2:uL2s(RN)=1.

Proof

It is well-known that S˜ is achieved by the function

U(x)=CN,s(1+|x|2)2sN2,

where CN,s=2N2s2[Γ((N+2s)/2)Γ((N2s)/2)1]N2s4s . Direct calculations show that U2s=1 . Define

Uϵ(x)=ϵ2sN8(s1)U(ϵ144sx),ϕϵ(x)=ϕ(ϵ122sx),

where ϕC0(RN) satisfies supp(ϕ)Ω , ϕ = 1 in 12ΩΩ with diam(12Ω)=12diam(Ω) , 0 ≤ ϕ≤1, |ϕ|Cdiam(Ω) . Setting vϵ(x) = Uϵ(x)ϕϵ(x) and vˆϵ(x)=vϵ(x)vϵ2s for all xΩϵ, then vˆϵ2s=1 and, by definition,

S˜ϵvˆϵ22.

Now, we deduce that

(3.9) limϵ0Ωϵvϵ2sdx=limϵ0ϵN4(1s)ΩϵU2s(ϵ144sx)ϕ2s(ϵ122sx)dx=limϵ0ϵ14s4ΩU2s(x)ϕ2s(ϵ144sx)dx=limϵ012ϵ14s4ΩU2s(x)dx+limϵ0ϵ14s4Ω12ϵ14s4ΩU2s(x)ϕ2s(ϵ144sx)dx=limϵ012ϵ14s4ΩU2s(x)dx+CN,s2slimϵ0ϵN44sΩ12Ω(ϵ122s+|x|2)Nϕ2s(x)dx=RNU2s(x)dx=1.

Next, one has

(3.10) Ωϵ|vˆϵ|2dx=1vϵ2s2Ωϵ|vϵ|2dx2vϵ2s2ΩϵUϵ2ϕϵ2(x)dx+Ωϵ|ϕϵ|2Uϵ2(x)dx.

By direct calculations, we also have

(3.11) limϵ0Uϵ2ϕϵ2(x)dx=limϵ0ϵ12ϵ14s4ΩU2ϕ2(ϵ144sx)dxlimϵ0ϵ12RNU2dx=0,

and

(3.12) limϵ0Ωϵϕϵ2Uϵ2(x)dx=limϵ0ϵ42sN4(1s)Ωϕ2U2(ϵ14s4x)dx=CN,s2limϵ0ϵ46s+N4(1s)Ωϕ2(ϵ122s+|x|2)2sNdx=0,

where we used the facts that ∣∇ U2 < ∞ and N > 2s.

By combining (3.9), (3.13), (3.11)) and (3.12), we deduce

(3.13) limϵ0Ωϵ|vˆϵ|2dx=0.

Finally, let us estimate the term [vϵ]s2 . To this end, we rewrite it asfollows

(3.14) [vϵ]s2=Iϵ1+Iϵ2+Iϵ3,

where

Iϵ1:=RN×RN|Uϵ(x)Uϵ(y)|2|xy|N+2sϕϵ2(x)dydx,
Iϵ2:=RN×RN|ϕϵ(x)ϕϵ(y)|2|xy|N+2sUϵ2(y)dydx,
Iϵ3:=RN×RN(ϕϵ(x)Uϵ(y)(ϕϵ(x)ϕϵ(y))Uϵ(y)ϕϵ(x)|xy|N+2sdydx.

Then, as in the proof of [2, Theorem A.1], we get

(3.15) limϵ0Iϵ1=S˜,limϵ0Iϵ2=limϵ0Iϵ3=0.

Thus, from (3.9), (3.14) and (3.15), we obtain

(3.16) limϵ0[v^ϵ]s2=limϵ0[vϵ]s2=S~

and therefore

(3.17) lim supϵ0S˜ϵlimϵ0[vˆϵ]s2=S˜.

We have

(3.18) S˜lim infϵ0S˜ϵ.

By definition of S˜ϵ , for any δ>0, there exists wϵ,δX0s(Ωϵ) with ∣wϵ, δLp(Ωϵ) = 1 such that

(3.19) wϵ,δs2<S˜ϵ+δ.

Let η be the standard mollifier function, ησ(x)=σNη(xσ) . Define wϵ,δσ=wϵ,δησ and vϵ,σσ=wϵ,δσwϵ,δσL2s(RN) . Then vϵ,σσwϵ,δ in Ds,2(RN) as σ → 0. Thus, we have

(3.20) S˜[vϵ,σσ]s,RN2[wϵ,δ]s,RN2,asσ0.

This gives

S˜<S˜ϵ+δ

and the arbitrariness of δ yields (3.18) This ends the proof.  □

Combining (3.6) and (3.7), one has

Lemma 27

limϵ0vϵϵ2=limϵ0vϵ2s2s=S~2s22s .

3.2.2 The case 2s<p<2

Now, we consider the minimization problem

(3.21) S=infI(u):uDs,2(RN)D1,2(RN):up=1

with 2s<p<2 , where

(3.22) I(u)=RN|u|2dx+RN×RN|u(x)u(y)|2|xy|N+2sdxdy.

Lemma 28

S is attained.

Proof

Let unDs,2(RN)D1,2(RN) be a minimizing sequence with ∣unp = 1 such that

limnI(un)=S.

Thus, {un} is bounded in Ds,2(RN)D1,2(RN) and there exists uDs,2(RN)D1,2(RN) such that unu in Ds,2(RN)D1,2(RN) and Lp(RN) with 2s<p<2 .

By means of symmetric rearrangement techniques, we may assume that un is radially symmetric (cf. [20]). Thus, we get unu in Lp(RN) , which yields that ∣up = 1.  □

It is easy to check that u is radially symmetric. Moreover, applying the Lagrange multiplier rule, u satisfies

(3.23) (Δ)suΔu=μ|u|p2uinRN

for some μ>0.

Lemma 29

For 2s<p<2 , then limϵ0S˜ϵ=S .

Proof

Let w be a minimizer of S and set wϵ(x) = w(x)ϕϵ(x), wˆϵ(x)=wϵwϵp for all xΩϵ. Then wˆϵp=1 and by definition,

S˜ϵwˆϵs2.

Now,

(3.24) limϵ0Ωϵ|wϵ|pdx=limϵ0RN|wϕϵ|pdxlimϵ0RNΩϵ|wϕϵ|pdx=wpp=1.

and

(3.25) limϵ0Ωϵ|wϵ|2dx=limϵ0Ωϵw2ϕϵ2dx+2Ωϵϕϵwdx+Ωϵ|ϕϵ|2w2dx=RN|w|2dx.

Similarly to the proof of (3.14), (3.15) and (3.16), we deduce that

(3.26) limϵ0[wϵ]s2=[w]s2.

Thus, we have

(3.27) lim supϵ0S˜ϵlim supϵ0wˆϵ22=limϵ0wϵ22+[wϵ]s,RN2wϵp2=S,

while the inequality S¯lim infϵ0S~ϵ can be obtained by the same arguments as Lemma 26.  □

From (3.5), (3.6) and (3.7), one has

Lemma 30

For 2s<p<2 , limϵ → 0Sϵ = limϵ → 0vϵϵ = limϵ → 0vϵp = 0.

3.3 The case ϵ → +∞

Setting vϵ(x)=ϵ1p2uϵ(x) , then vϵ satisfies

ϵ1(Δ)svΔv=|v|p2vinΩ,v>0inΩ,v=0inRNΩ,

and is a minimizer of

Sϵ:=inf{Qϵ(v):vp=1},

or, equivalently, of

Sϵ=infvX0s{0}Qϵ(v)vp2,

where

Qϵ(v)=v22+ϵ1[v]s2.

Then, Sϵ=ϵ1Sϵ . We also define

(3.28) Sˆ=infuH01(Ω){0}u22up2for2<p<2.

It is well known that Sˆ is achieved by a positive classical minimizer.

Lemma 31

limϵ+Sϵ=S^ .

Proof

It closely follows the one of Lemma 24. Since X0sH01 , we get

SˆinfvX0s{0}v22vp2Sϵ,

which yields that

Sˆlim infϵ+Sϵ.

Assume that w0 is a minimizer of Sˆ . By density, there exists a sequence {wn}C0(Ω) such that wnw0 in H01(Ω) and Lp(Ω). Thus, we have

(3.29) Sϵwn22+ϵ1[wn]s2wnp2.

Then,

(3.30) lim supϵ+Sϵlimnlim supϵ+wn22+ϵ1[wn]s2wnp2=limnwn22wnp2=Sˆ,

and this completes the proof.  □

Lemma 32

limϵ → +∞Sϵ = 0 and limϵ → +∞uϵϵ = +∞.

Proof

This is a direct consequence of Lemma 31, as

uϵϵ2=ϵuϵ22+[uϵ]s2=ϵpp2Qϵ(vϵ)=ϵpp2Sϵ.

 □



Acknowledgements

The authors join to thank the anonymous reviewers for their valuable work which has improved the presentation of the paper.

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Received: 2020-10-29
Accepted: 2020-11-24
Published Online: 2020-12-31

© 2021 D. Cassani et al., published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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