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Advances in Nonlinear Analysis

Editor-in-Chief: Radulescu, Vicentiu / Squassina, Marco


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Critical growth elliptic problems involving Hardy-Littlewood-Sobolev critical exponent in non-contractible domains

Divya Goel / Konijeti Sreenadh
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  • Department of Mathematics, Indian Institute of Technology Delhi, Hauz Khaz, New Delhi-110016, India
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Published Online: 2019-08-06 | DOI: https://doi.org/10.1515/anona-2020-0026

Abstract

The paper is concerned with the existence and multiplicity of positive solutions of the nonhomogeneous Choquard equation over an annular type bounded domain. Precisely, we consider the following equation

Δu=Ω|u(y)|2μ|xy|μdy|u|2μ2u+finΩ,u=0 on Ω,

where Ω is a smooth bounded annular domain in ℝN(N ≥ 3), 2μ=2NμN2, fL(Ω) and f ≥ 0. We prove the existence of four positive solutions of the above problem using the Lusternik-Schnirelmann theory and varitaional methods, when the inner hole of the annulus is sufficiently small.

Keywords: Hardy-Littlewood-Sobolev inequality; critical problems; non-contractible domains

MSC 2010: 35A15; 35J60; 35J20

1 Introduction

In the pioneering work, Tarantello [31] studied the nonhomogeneous elliptic equation

Δu=|u|22u+f in Ω,u=0 on Ω,(1.1)

where 2=2NN2 is the critical Sobolev exponent and Ω is a bounded domain in ℝN with smooth boundary. If fH−1 then it is shown that there exists at least two solutions of (1.1) by using variational methods. Cao and Zhou [9] proved the existence of two positive solutions of the following nonhomogeneous elliptic equation

Δu=f(x,u(x))+h in RN(1.2)

where f(x, u) is a Carathéodory function with subcritical grotwh at ∞. Further, many researchers investigated (1.1) and (1.2) for the existence and multiplicity of solutions. For details, we refer [10, 11, 20, 21, 33] and references therein. Recently, Gao and Yang [30] proved the existence of two positive solutions of the nonhomogeneous Choquard equation involving Hardy-Littlewood-Sobolev critical exponent using the splitting Nehari manifold method of Tarantello [31].

The existence, uniqueness, and multiplicity of positive solutions of the nonlocal elliptic equation, precisely the Choquard equation both for mathematical analysis and in perspective of physical models has recently gained significant attention amongst researchers. As an instance, in 1954 Pekar [28] proposed the equation

Δu+u=1|x||u|2u in R3(1.3)

to study the quantum theory of polaron. Later in 1976, Ph. Choquard [22] examined the steady state of one component plasma approximation in Hartee-Fock theory using (1.3). In [22], Leib proved the existence and uniqueness of the ground state of (1.3). The work of Moroz and Schaftingen enriches the literature of Choquard equations. In [25] authors studied the following Choquard equation

Δu+Vu=IαF(u)F(u),in RN,(1.4)

where α ∈ (0, N), N ≥ 3, Iα is the Riesz Potential and F(u) ∈ C1(ℝ, ℝ) with sub critical growth. In this work authors established the existence of ground state soloutions of (1.4) and assuming some suitable growth conditions on F and V, they studied the properties like constant sign solutions and radial symmetry of the solution. Moreover, authors proved the Pohožaev identity and nonlocal Brezis-Kato type estimate. Interested readers are referred to [16, 24, 26, 27] and references therein for the study of Choquard equation on the unbounded domain.

Concerning the boundary value problems of Choquard equation, Gao and Yang [15] studied the Brezis-Nirenberg type existence results for the following critical equation

Δu=λh(u)+Ω|u(y)|2μ|xy|μdy|u|2μ2u in Ω,u=0 on Ω,

where λ > 0, 0 < μ < N, h(u) = u, Ω is a smooth bounded domain in ℝN. Later in [14] authors proved the existence and multiplicity of positive solutions for convex and convex-concave type nonlinearities (h(u) = uq, 0 < q < 1) using variational methods.

The geometry of the domain Ω plays an essential and significant role on the existence and multiplicity of the elliptic boundary value problems. Indeed, in [12], Coron proved the existence of a high energy positive solution of the problem

Δu=|u|22uinΩ,u=0 on Ω,(1.5)

where Ω is a bounded domain in ℝN(N ≥ 3), precisely an annulus with a small hole. Later in [3], Bahri and Coron, proved that a positive solution always exists as long as the domain has non-trivial homology with ℤ2-coefficients. In [6], Benci and Cerami studied the following equation

εΔu+u=f(u)inΩ,u=0 on Ω,(1.6)

where ε ∈ ℝ+, Ω is a bounded domain in ℝN(N ≥ 3) and f : ℝ+ → ℝ is a C1,1 function. Here authors proved that there exists ϵ* > 0 such that for all ε ∈ (0, ϵ*), (1.6) has cat(Ω)+1 solutions under some growth conditions on the function f. Since then, the study of existence and multiplicity of solutions of elliptic equations over non-contractible domain has been substantially studied, for instance, [4, 5, 13, 20, 29, 32] and references therein.

The existence of high energy solution of (1.5) is a much more delicate issue. In this spirit, recently Goel, Rădulescu and Sreenadh [19] studied the Coron problem for Choquard equations. Here authors proved the existence of a positive high energy solution for the problem (Pf) when f(x) ≡ 0 and Ω is a smooth bounded domain in ℝN(N ≥ 3) satisfying the following condition

  1. There exists constants 0 < R1 < R2 < ∞ such that

    {xRN:R1<|x|<R2}Ω,{xRN:|x|<R1}Ω¯.

In the light of above works, in this article, we study following problem

(Pf)Δu=Ω|u+(y)|2μ|xy|μdy|u+|2μ2u++f,in Ω,u=0 on Ω,

where 2μ=2NμN2, is the critical exponent in the sense of Hardy-Littlewood-Sobolev inequality (2.1) and f with := {f : fL(Ω), f ≥ 0, f ≢ 0}. The domain Ω ⊂ ℝN(N ≥ 3) satisfies the condition (A). Here we prove the existence of four solutions of the problem (Pf). To achieve this, we first seek the help of Nehari manifold associated with (Pf) to prove the existence of the first solution (say u1). To proceed further, we prove many new estimates on the convolution terms involving the minimizers of best constant SH,L (see Lemma 4.1, 4.3 and 4.4). With the help of these estimates we prove that the minima of the functional over 𝓝f is below the first critical level where the first critical level is

Jf(u1)+Nμ+22(2Nμ)SH,L2NμNμ+2.

Here 𝓙f is the energy functional associated to (Pf) (defined in (2.3)). Moreover, 𝓙f satisfies the Palais-Smale condition below the first critical level. Subsequently, we show the existence of the second and the third solution of (Pf), in Nf (a closed subset of the Nehari manifold) by using a well-known result of Ambrosetti [2](see Lemma 5.2) and assumption (A). To study the existence of the fourth solution, a high energy solution, we prove that the functional 𝓙f satisfies the Palais-Smale condition between the first and the second critical levels, where the second critical level is

infuNfJf(u)+Nμ+22(2Nμ)SH,L2NμNμ+2.

To prove the existence of fourth solution, we use the minmax Lemma (See Lemma 6.6). To the best of our knowledge, there is no work on the existence and multiplicity of solutions to Choquard equations (Pf) in non-contractible domains. With this introduction, we state our main result.

Theorem 1.1

Assume μ < min{4, N}, fL(Ω) and f ≥ 0 and Ω be a bounded domain satisfying the conditon (A). Then there exists e* > 0 such that (Pf) has at least three positive solutions whenever 0 < ∥fH−1 < e*. Moreover, if R1 is small enough then there exists e** > 0 such that (Pf) has at least four positive solutions whenever 0 < ∥fH−1 < e**.

The paper is organized as follows: In Section 2, we give the variational framework and preliminary results. In section 3, using the Nehari manifold technique, we prove the existence of the first solution. In section 4, we prove some crucial estimates of the minimizer of SH,L(defined in (2.2)) and analyze the Palais-Smale sequences. In section 5, we prove the existence of the second and third solution. In section 6, we prove the existence of the fourth solution.

2 Variational framework and preliminary results

We start with the familiar Hardy-Littlewood-Sobolev Inequality which leads to the study of nonlocal Choquard equation using variational methods.

Proposition 2.1

[23](Hardy-Littlewood-Sobolev Inequality) Let t, r > 1 and 0 < μ < N with 1/t + μ/N + 1/r = 2, fLt(ℝN) and hLr(ℝN). There exists a sharp constant C(t, r, μ, N) independent of f and h such that

RNRNf(x)h(y)|xy|μdxdyC(t,r,μ,N)fLt(RN)hLr(RN).(2.1)

If t = r = 2N/(2Nμ), then

C(t,r,μ,N)=C(N,μ)=πμ2Γ(N2μ2)Γ(Nμ2)Γ(N2)Γ(μ2)1+μN.

Equality holds in (2.1) if and only if f ≡ (constant)h and

h(x)=A(y2+|xa|2)(2Nμ)/2,

for some A ∈ ℂ, 0 ≠ y ∈ ℝ and a ∈ ℝN.□

The best constant for the embedding D1,2(ℝN) into L2*(ℝN) (where 2=2NN2)is defined as

S=infuD1,2(RN){0}RN|u|2dx:RN|u|2dx=1.

Consequently, we define

SH,L=infuD1,2(RN){0}RN|u|2dx:RNRN|u(x)|2μ|u(y)|2μ|xy|μdxdy=1.(2.2)

Lemma 2.2

[15] The constant SH,L defined in (2.2) is achieved if and only if

u=Cbb2+|xa|2N22

where C > 0 is a fixed constant, a ∈ ℝN and b ∈ (0, ∞) are parameters. Moreover,

S=SH,L(C(N,μ))N22Nμ.

Lemma 2.3

[15] For N ≥ 3 and 0 < μ < N. Then

.NL:=RNRN|.|2μ|.|2μ|xy|μdxdy12.2μ

defines a norm on L2*(ℝN).

The energy functional 𝓙f : H01 (Ω) → ℝ associated with the problem (Pf) is

Jf(u)=12Ω|u|2dx12.2μΩΩ|u+(x)|2μ|u+(y)|2μ|xy|μdxdyΩfudx,(2.3)

where u+ = max(u, 0). By using Hardy-Littlewood-Sobolev inequality (2.1), we have

ΩΩ|u+(x)|2μ|u+(y)|2μ|xy|μdxdy12μC(N,μ)2NμN2|u|22.

It is not difficult to show that the functional 𝓙fC1(H01 (Ω), ℝ) and moreover, if μ < min {4, N} then 𝓙fC2(H01(Ω), ℝ).

Definition 2.4

A function uH01(Ω) is called a weak solution of the problem (Pf) if for all vH01(Ω) the following holds

ΩuvdxΩΩ|u+(x)|2μ|u+(y)|2μ1v(y)|xy|μdxdyΩfvdx=0.

Definition 2.5

For c ∈ ℝ, {un} is a (PS)c sequence in H01(Ω) for 𝓙f if 𝓙f = c + o(1) and Jf (un) = o(1) strongly in H−1 as n → ∞. We say 𝓙f satisfies the (PS)c condition in H01(Ω) if every (PS)c sequence in H01(Ω) has a convergent subsequence.

Since 𝓙f is not bounded below on H01(Ω), it is worth to consider the Nehari manifold

Nf:={uH01(Ω){0}|u+0 and Jf(u),u=0},

where 〈, 〉 denotes the usual duality. We define

Υf=infuNfJf(u).

Note that when f(x) ≡ 0, Υ0(Ω) is independednt of Ω and Υ0(Ω):=Υ0=Nμ+22(2Nμ)SH,L2NμNμ+2.

Notations: Throughout the paper we will use the notation 𝓙0 = 𝓙, 𝓝0 = 𝓝, ∥.∥ = .H01(Ω)

a(u)=Ω|u|2dxandb(u)=ΩΩ(u+(x))2μ(u+(y))2μ|xy|μdxdy.

An easy consequence of (2.1) gives 𝓙f is coercive and bounded below on 𝓝f.

Proposition 2.6

For any u, vH01(Ω), we have

ΩΩ|u(x)|2μ|v(y)|2μ|xy|μdxdyΩΩ|u(x)|2μ|u(y)|2μ|xy|μdxdy12ΩΩ|v(x)|2μ|v(y)|2μ|xy|μdxdy12.

Proof

For details of the proof, see [17, Lemma 2.3].□

Lemma 2.7

For each uH01(Ω), there exists a unique t > 0 such that t u ∈ 𝓝. Moreover, there holds Υ0Nμ+22(2Nμ)a(u)2μb(u)12μ1.

Proof

Let mu(t)=t22a(u)t2.2μ2.2μb(u) then on solving mu(t) = 0, we get unique t(u)=a(u)b(u)12(2μ1) such that t(u)u ∈ 𝓝. From the definition of Υ0, we have

Υ0J(t(u)u)=1212.2μa(u)b(u)12μ1a(u)=Nμ+22(2Nμ)a(u)2μb(u)12μ1.

Remark 2.8

We remark that by [15, Lemma 1.3], SH,L is never achieved on bounded domain. Therefore if u is a solution of the following equation

Δu=Ω|u(y)|2μ|xy|μdy|u|2μ2uinΩ,u=0onΩ,

then J(u)>Υ0=Nμ+22(2Nμ)SH,L2NμNμ+2.

Lemma 2.9

A sequence {un} is a (PS)Υ0- sequence for 𝓙 in H01(Ω) if and only if 𝓙(un) = Υ0 + on(1) and a(un) = b(un) + on(1).

Proof

Clearly, any (PS)Υ0- sequence satisfies a(un) = b(un) + on(1) and 𝓙(un) = Υ0 + on(1). Conversely, let 𝓙(un) = Υ0 + on(1) and a(un) = b(un) + on(1) then Υ0 = 𝓙(un) = Nμ+22(2Nμ)b(un) + on(1) and hence we have

b(un)=DΥ0+on(1) where D=2(2Nμ)Nμ+2.(2.4)

Define Tn(ψ)=ΩΩ(un+(x))2μ(un+(y))2μ1ψ(y)|xy|μdxdy for ψH01(Ω) and n = 1, 2, ….

Claim: TnH1=(DΥ0)12+on(1).

Let ψH01(Ω) such that ∥ψ∥ = 1 then by Lemma 2.7, we know that there exists a t > 0 such that a() = b(). Therefore, t=ψNL2μ2μ1andΥ01DψNL2.2μ2μ1. This implies,

ψNL1DΥ02μ12.2μ.(2.5)

Taking into account (2.4), (2.5), Proposition 2.6 and employing Hölder’s inequality, for each n, we have

|Tn(ψ)|ΩΩ(un+(x))2μ(un+(y))2μ|xy|μdxdy2.2μ12.2μΩΩ|ψ(x)|2μ|ψ(y)|2μ|xy|μdxdy12.2μ=b(un)2.2μ12.2μψNL1DΥ02μ12.2μ(DΥ0+on(1))2.2μ12.2μ=(DΥ0)12+on(1) as n.

So, we get TnH1(DΥ0)12+on(1). Moreover, Tnunun=(b(un))12=(DΥ0)12+on(1). This implies ∥TnH−1 = (DΥ0)12+on(1). Hence the proof of claim follows. Now, by Riesz representation theorem, for each n, there exists vnH01(Ω) such that

Tn(ψ)=vn,ψ=Ωvnψdx and vn=TnH1=(DΥ0)12+on(1).

Thus, 〈vn, un〉 = Tn(un) = b(un) = 0 + on(1). Hence,

unvn2=un22un,vn+vn2=DΥ02DΥ0+DΥ0+on(1)=on(1) as n.

For any ψH01(Ω) with ∥ψ∥ = 1, we have

J(un),ψ=ΩunψdxTn(ψ)=un,ψvn,ψ=unvn,ψ.

Therefore, ∥𝓙(un)∥H−1 ≤ ∥unvn∥ = on(1). It implies 𝓙(un) → 0 in H−1.□

Clearly, 𝓝f contains every non zero solution of (Pf) and we know that the Nehari manifold is closely related to the behavior of the fibering maps ϕu : ℝ+ → ℝ defined as ϕu(t) = 𝓙f(tu). It is easy to see that tu ∈ 𝓝f if and only if ϕu (t) = 0 and elements of 𝓝f correspond to stationary points of the fibering maps. It is natural to divide 𝓝f into the following sets

Nf+=:{uNf|ϕu(1)>0},Nf=:{uNf|ϕu(1)<0},andNf0=:{uNf|ϕu(1)=0}.

We also denote the infimum over Nf+andNf as

Υf+=infuNf+Jf(u)Υf=infuNfJf(u).

3 Existence of First Solution

In this section we prove the existence of first solution by showing the existence of minimizer for 𝓙f over the Nehari manifold 𝓝f. First we state some Lemmas whose proof can be found in [30]. We further prove some properties of the manifold Nf+.

Lemma 3.1

If fF̂ andfH−1 < e00 := CN,μSH,L2μ2.2μ2 where CN,μ=12.2μ12.2μ12.2μ2(2.2μ2) then α0:=infuE{CN,μu2.2μ12μ1Ωfudx} is acheived, where

E:={uH01(Ω):ΩΩ|u(x)|2μ|u(y)|2μ|xy|μdxdy=1}.

Proof

Proof follows from [30, Lemma 4.1]. Since we consider λ = 0 in equation (4.1) of [30], our result holds for all N ≥ 3.□

Lemma 3.2

For every u ∈ 𝓝f, u ≢ 0 we have a(u)(2.2μ1)b(u)0. In particular, Nf0 = {0}.

Lemma 3.3

For each uH01(Ω) with u+ ≢ 0 the following holds:

  1. There exists a unique t = t(u) > 0 such that t u Nf(Ω). In particular,

    t>a(u)(2.2μ1)b(u)12.2μ2:=tmax

    and Jf(tu)=maxttmaxJf(tu).

  2. If Ωfu>0, then there exists unique t+ ∈ (0, tmax) such that t+ uNf+ (Ω) and

    Jf(t+u)=min0<ttJf(tu).

  3. t(u) is a continuous function.

  4. Nf={uH01(Ω){0}|u+0and1ut(uu)=1}.

Lemma 3.4

For each uNf+ (Ω), we haveΩ fu dx > 0 and 𝓙f(u) < 0. In particular, Υf(Ω)Υf+(Ω)<0.

Lemma 3.5

Let u ∈ 𝓝f(Ω) be such that Jf(u)=minwNf(Ω)Jf(w) = Υf(Ω) thenΩ fu dx > 0 and u is a solution of (Pf).

Lemma 3.6

𝓙f has Palais-Smale sequences at each of the levels Υf(Ω),Υf+(Ω)andΥf(Ω).

Lemma 3.7

Let {un} ∈ 𝓝f be a (PS)Υf(Ω) sequence for 𝓙f, then there exists a subsequence of {un}, still denoted by {un}, and a non-zero u1H01(Ω) such that unu1 strongly in H01(Ω). Moreover, u1 ∈ 𝓝f and is a solution to (Pf).

Proof

𝓙f is bounded below and coercive implies {un} is bounded in H01(Ω). So, there exists a subsequence still denoted by {un} such that unu1 weakly in H01(Ω). By [19, Lemma 4.2], we have Jf(u1) = 0. In particular, u1 ∈ 𝓝f and Jf(u1)=1212.2μa(u1)112.2μΩfu1dx. Now, using the fact that a is weakly lower semi continuous we have

Υf(Ω)Jf(u1)lim infn1212.2μa(un)limn112.2μΩfundx=Υf(Ω).

Consequently, we have Υf(Ω) = 𝓙f(u1). Let wn = unu1 then by [19, Lemma 4.1], [15, Lemma 2.2] and the fact that Jf (u1) = 0, we obtain 𝓙f(wn) = 𝓙f(un) − 𝓙f(u1) = on(1) and Jf(wn),ϕ=Jf(un),ϕJf(u1),ϕ+on(1) = on(1). Therefore, 〈Jf(wn), wn〉 = on(1). It implies Jf(wn)=1212.2μa(wn)Ωfwndx=on(1) and since ∫Ω fwn dx = on(1), we get a(wn) = on(1). Hence unu strongly in H01(Ω).□

Lemma 3.8

If u be a solution of (Pf) then uC2(Ω). Moreover, u is a positive solution.

Proof

Let u be a solution of (Pf) and G(x, u) = Ω|u+(y)|2μ|xy|μdy|u+|2μ2u+f. By using same assertions and arguments as in [25, Proposition 3.1 and Theorem 2], we have Ω|u+(y)|2μ|xy|μdyL (Ω) and since f, we have |G(x, u)| ≤ C(1 + |u|2*−1). Then by the standard elliptic regularity uC2(Ω). Since f ≥ 0, we get u ≥ 0 and by using strong maximum principle, u is a positive solution of (Pf).□

Lemma 3.9

Let μ < min{ 4, N} and k0=12.2μ112(2μ1)SH,L2μ2(2μ1) and f, ∥fH−1e00 (where e00 is defined in Lemma 3.1) then

  1. Nf+(Ω) ⊂ Bk0(0) := {uH01(Ω) | ∥u∥ < k0}.

  2. 𝓙f is strictly convex in Bk0(0).

Proof

  1. Let uNf+(Ω) then ϕu(1)=0andϕu(1)>0. That is, a(u) = b(u) + ∫Ω fu dx and a(u) > (2.2μ1)b(u). Therefore, a(u) = b(u) + ∫Ω fu dx < 1(2.2μ1)a(u) + ∫Ω fu dx. It implies 11(2.2μ1)a(u)fH1u. So,

    u(2.2μ1)2(2μ1)fH1(2.2μ1)2(2μ1)CN,μSH,L2μ2.2μ2=12.2μ112(2μ1)SH,L2μ2(2μ1)=k0.

  2. By using Hölders inequality and equation (2.2), we have

    ΩΩ(u+(x))2μ1(u+(y))2μ1z(x)z(y)|xy|μdxdyb(u)2μ12μzNL2SH,L(2μ1)a(u)(2μ1)SH,L1a(z)=SH,L2μa(u)(2μ1)a(z).(3.1)

Again using Hölders inequality, Proposition 2.6 and (2.2), we have

ΩΩ(u+(x))2μ(u+(y))2μ2z2(y)|xy|μdxdyb(u)2μ12μzNL2SH,L2μa(u)(2μ1)a(z).(3.2)

From equations (3.1), (3.2) and definition of Jf(u)(z, z), we get

Jf(u)(z,z)=a(z)2μΩΩ(u+(x))2μ1(u+(y))2μ1z(x)z(y)|xy|μdxdy(2μ1)ΩΩ(u+(x))2μ(u+(y))2μ2z2(y)|xy|μdxdya(z)12μSH,L2μa(u)(2μ1)(2μ1)SH,L2μa(u)(2μ1)=a(z)1(2.2μ1)SH,L2μa(u)(2μ1)>a(z)1(2.2μ1)(2.2μ1)=0

for uBk0(0) ∖ {0}. Then Jf (u) is positive definite for uBk0(0) and 𝓙f(u) is strictly convex on Bk0(0).□

Lemma 3.10

It holds that u1Nf+ and 𝓙f(u1) = Υf+(Ω) = Υf(Ω). Moreover, u1 is the unique critical point of 𝓙f in Bk0(0) and u1 is a local minimum of 𝓙f in H01(Ω).

Proof

Using the proof of [30, Theorem 1.3], we have Ωfu1dx>0. Now if u1Nf then there exists a unique t(u1) = 1 > tmax > t+(u1) > 0 such that t+(u1)u1Nf+ then by Lemma 3.3 (b) we have

Υf(Ω)Υf+(Ω)Jf(t+(u1)u1)<Jf(t(u1)u1)=Jf(u1)=Υf(Ω).

which is a contradiction. It implies u1Nf+andΥf+(Ω)Jf(u1)=Υf(Ω)Υf+(Ω) that is, 𝓙f(u1) = Υf(Ω) = Υf+(Ω). Using Lemma 3.5 and Lemma 3.9, we get u1 is the unique critical point of 𝓙f in Bk0(0) and the proof of local minimum follows from [30, Lemma 3.2].□

Lemma 3.11

Let μ < min{4, N} and uH01(Ω) be a critical point of 𝓙f then either uNf or u = u1.

Proof

If uH01(Ω) be a critical point of 𝓙f then u ∈ 𝓝f = Nf+Nf. Now using the fact that Nf+Nf = ∅ and Nf+Bk0(0) we have either uNf or u = u1.□

4 Asymptotic estimates and Palais-Smale Analysis

In this section we shall prove that the functional 𝓙f satisfies Palais-Smale condition strictly below the first critical level and (strictly) between the first and second critical levels. To start with, we shall prove several new estimates on the nonlinearity.

It is known from Lemma 2.2 that the best constant SH,L is achieved by the function

u(x)=S(Nμ)(2N)4(Nμ+2)(C(N,μ))2N2(Nμ+2)(N(N2))N24(1+|x|2)N22,

which is a solution of the problem Δu=(|x|μ|u|2μ)|u|2μ1 in ℝN with

RN|u|2dx=RNRN|u(x)|2μ|u(y)|2μ|xy|μdxdy=SH,L2NμNμ+2.

We may assume R1 = ρ, R2 = 1/ρ for ρ ∈ (0, 12). Now, define υρCc(ℝN) such that 0 ≤ υρ(x) ≤ 1 for all x ∈ ℝN, radially symmetric and

υρ(x)=00<|x|<3ρ2,12ρ|x|12ρ,0|x|34ρ,

and

uσϵ(x)=S(Nμ)(2N)4(Nμ+2)C(N,μ)2N2(Nμ+2)(N(N2)ϵ2)N24(ϵ2+|x(1ϵ)σ|2)N22,

where σ ∈ 𝕊N−1 := {x ∈ ℝN : |x| = 1}, 0 < ϵ ≤ 1. Set

gρϵ,σ(x):=υρ(x)uσϵ(x)H01(Ω).(4.1)

Lemma 4.1

  1. a(gρϵ,σ)=b(gρϵ,σ)=SH,L2NμNμ+2+oϵ(1) uniformly in σ as ϵ → 0.

  2. J(gρϵ,σ)=Nμ+22(2Nμ)SH,L2NμNμ+2+oϵ(1) uniformly in σ as ϵ → 0.

  3. gρϵ,σ ⇀ 0 weakly in H01(Ω) uniformly in σ as ϵ → 0.

Proof

  1. Observe the fact that there exist constants d1, d2 > 0 such that

    d1<|x(1ϵ)σ|<d2 for all xB2ρ whenever ϵ<12ρ.(4.2)

    gρϵ,σL2(RN)uϵσL2(RN)(RNB12ρ)B2ρ|uϵσ|2dx+ρ2B2ρ|uϵσ|2dx+ρ2B34ρB12ρ|uϵσ|2dxCϵN2(RNB12ρ)B2ρ|x(1ϵ)σ|2|x(1ϵ)σ|2Ndx+CϵN2B2ρB34ρB12ρdx|x(1ϵ)σ|2(N2)=O(ϵN2).

    Thus, gρϵ,σL2(RN)=uϵσL2(RN)+oϵ(1)=SH,L2NμNμ+2+oϵ(1).

    Next we will prove that b(gρϵ,σ)=SH,L2NμNμ+2+oϵ(1) uniformly in σ as ϵ → 0. For this consider

    RNRN|gρϵ,σ(x)|2μ|gρϵ,σ(y)|2μ|xy|μdxdyRNRN|uϵσ(x)|2μ|uϵσ(y)|2μ|xy|μdxdy=RNRN(|υρ(x)|2μ|υρ(y)|2μ1)|uϵσ(x)|2μ|uϵσ(y)|2μ|xy|μdxdyCB2ρB2ρ+B12ρB2ρB2ρ+B12ρB2ρRNB12ρ+RNB12ρB2ρ+RNB12ρRNB12ρ|uϵσ(x)|2μ|uϵσ(y)|2μ|xy|μdxdy,=Ci=1i=5Ji,(4.3)

    Let ξϵ(x)=ϵN(ϵ2+|x(1ϵ)σ|2)N then taking into account the definition of uϵσ, (4.2) and Hardy-Littlewood-Sobolev inequality, we have the following estimates:

    J1C(N,μ)B2ρSN(Nμ)2(Nμ+2)C(N,μ)N(Nμ+2)(N(N2))N2ξϵ(x)dx2NμNCϵ2NμB2ρdx|x(1ϵ)σ|2N2NμNCϵ2NμB2ρdx2NμN=O(ϵ2Nμ),J2CB12ρB2ρξϵ(x)dx2Nμ2NB2ρξϵ(x)dx2Nμ2NCϵ2Nμ2B2ρdx|x(1ϵ)σ|2N2Nμ2N=O(ϵ2Nμ2),J3CB12ρB2ρξϵ(x)2Nμ2NRNB12ρξϵ(x)dx2Nμ2NCϵ2Nμ2RNB12ρdx|x(1ϵ)σ|2N2Nμ2N=O(ϵ2Nμ2),

    J4CRNB12ρξϵ(x)dx2Nμ2NB2ρξϵ(x)dx2Nμ2NCϵ2NμRNB12ρdx|x(1ϵ)σ|2NB2ρdx|x(1ϵ)σ|2N2Nμ2N=O(ϵ2Nμ),J5CRNB12ρξϵ(x)dx2NμNCϵ2NμRNB12ρdx|x(1ϵ)σ|2N2NμN=O(ϵ2Nμ).

    Therefore, b(gρϵ,σ)RNRN|uϵσ(x)|2μ|uϵσ(y)|2μ|xy|μdxdy0 as ϵ → 0 that is, b(gρϵ,σ)SH,L2NμNμ+2 as ϵ → 0 and completes the proof of (i).

  2. Result follows from the definition of 𝓙 and by (i).

  3. Assume by contradiction, gρϵ,σg1 ≢ 0 weakly in H01(Ω) then gρϵ,σg1 strongly in L2(Ω). Then by using the inequality r2(N−2) + s2(N−2) ≤ (r2 + s2)N−2 for all r, s ≥ 0, we have

    0Ω|gρϵ,σ|2dxC3ρ2|x|34ρϵN2(ϵ2+|x(1ϵ)σ|2)N2dx=C3ρ2|y+(1ϵ)σ|34ρϵN2ϵ2(N2)+|y|2(N2)dyC034ρ+(1ϵ)ϵN2rN1ϵ2(N2)+r2(N2)dy0.

It yields a contradiction. Hence results follows.□

Lemma 4.2

Let σ ∈ 𝕊N−1 and ϵ ∈ (0, 1), then the following holds:

  1. limρ0supσSN1,ϵ(0,1](gρϵ,σuϵσ)L2(RN)2=0.

  2. limρ0supσSN1,ϵ(0,1]gρϵ,σNL2.2μ=uϵσNL2.2μ.

Proof

  1. Consider

    RN|gρϵ,σuϵσ|2dx2RN|uϵσ(x)υρ(x)|2dx+2RN|uϵσ(x)υρ(x)uϵσ(x)|2dxCρ2B2ρ|uϵσ(x)|2dx+B2ρ|uϵσ(x)|2dx+Cρ2B34ρB12ρ|uϵσ(x)|2dx+RNB12ρ|uϵσ(x)|2dx.(4.4)

    From the definition of uϵσ, we have the following estimates

    ρ2B2ρ|uϵσ(x)|2dxCρ2B2ρdxCρN2,B2ρ|uϵσ(x)|2dxCB2ρ|xtσ|dxCB2ρdxCρN,ρ2B34ρB12ρ|uϵσ(x)|2dxCρ2B34ρB12ρ1|x|2N4dxCρN2,RNB12ρ|uϵσ(x)|2dxCRNB12ρ1|x|2N2dxCρN2.

    Therefore, from above estimates and (4.4), we obtain desired result.

  2. Consider

    gρϵ,σNL2.2μuϵσNL2.2μ=RNRN(υρ2μ(x)υρ2μ(y)1)|uϵσ(x)|2μ|uϵσ(y)|2μ|xy|μdxdyCi=15Ji,

    where Ji are defined in (4.3). Using the Hardy-Littlewood-Sobolev inequality and the definition of ξϵ, we have the following estimates:

    J1C(N,μ)B2ρξϵ(x)dx2NμNCB2ρdx2NμNCρ2Nμ,J2C(N,μ)B12ρB2ρξϵ(x)dx2Nμ2NB2ρξϵ(x)dx2Nμ2NCB2ρdx2NμNCρ2Nμ2,

    J3C(N,μ)B12ρB2ρξϵ(x)dx2Nμ2NRNB12ρξϵ(x)dx2Nμ2NCRNB12ρdx|x(1ϵ)σ|2N2Nμ2N=|y+(1ϵ)σ|12ρdy|y|2N2Nμ2N|y|12ρ1dy|y|2N2Nμ2NC(2ρ)N1(2ρ)N2Nμ2N,

Now using the same estimates as above we can easily obtain

J4Cρ2Nμ2 and J5C(2ρ)N1(2ρ)N2NμN.

Hence supσSN1,ϵ(0,1]gρϵ,σNL2.2μuϵσNL2.2μ0 as ρ → 0 and completes the proof.□

Lemma 4.3

The following asymptic estimates hold:

  1. a(gρϵ,σ)SH,L2NμNμ+2+O(ϵN2).

  2. b(gρϵ,σ)SH,L2NμNμ+2+O(ϵN).

  3. b(gρϵ,σ)SH,L2NμNμ+2O(ϵ2Nμ2).

Proof

Part (i) follows from Lemma 4.1 (i). For part (ii) we will first estimate the integral Ω|gρϵ,σ|2dx. Since

Ω|gρϵ,σ|2dxCB34ρB3ρ2|uϵσ|2dxB34ρB12ρ|uϵσ|2dx+B12ρB3ρ2|uϵσ|2dx

and

B34ρB12ρ|uϵσ|2dxCϵNB34ρB12ρdx|x(1ϵ)σ|2N=O(ϵN),B12ρB3ρ2|uϵσ|2dxRN|uϵσ|2dx=SNNμ+2C(N,μ)NNμ+2.

It implies Ω|gρϵ,σ|2dxSNNμ+2C(N,μ)NNμ+2+O(ϵN) and now using this and Hardy-Littlewood-Sobolev inequality we have

b(gρϵ,σ)=ΩΩ|gρϵ,σ(x)|2μ|gρϵ,σ(y)|2μ|xy|μdxdyC(N,μ)Ω|gρϵ,σ|2dx2NμNC(N,μ)SNNμ+2C(N,μ)NNμ+2+O(ϵN)2NμNSH,L2NμNμ+2+O(ϵN).

This proves part (ii). Now to prove part (iii), consider

b(gρϵ,σ)=ΩΩ|gρϵ,σ(x)|2μ|gρϵ,σ(y)|2μ|xy|μdxdyB12ρB2ρB12ρB2ρ|gρϵ,σ(x)|2μ|gρϵ,σ(y)|2μ|xy|μdxdy=RNRN|uϵσ(x)|2μ|uϵσ(y)|2μ|xy|μdxdyi=1i=5Ji,

where Ji are defined in equation (4.3). Using the proof of Lemma 4.1(i) and the fact that uϵσNL2.2μ=SH,L2NμNμ+2 + oϵ(1), we obtain the required result.□

Now we will give a Lemma which is taken from [18]. For the sake of completeness, we provide a complete proof.

Lemma 4.4

If μ < min{4, N} then

b(u1+tgρϵ,σ)b(u1)+b(tgρϵ,σ)+C^t2.2μ1ΩΩ(gρϵ,σ(x))2μ(gρϵ,σ(y))2μ1u1(y)|xy|μdxdy+2.2μtΩΩ(u1(x))2μ(u1(y))2μ1gρϵ,σ(y)|xy|μdxdyO(ϵ(2Nμ4)Θ)forallΘ<1,

where u1 is the local minimum obtained in Lemma 3.10.

Proof

We will divide the proof in two cases:

  • Case 1

    2μ > 3.

    It is easy to see that there exists  > 0 such that

    (a+b)pap+bp+pap1b+A^abp1 for all a,b0 and p>3,

    which implies that

    b(u1+tgρϵ,σ)b(u1)+b(tgρϵ,σ)+C^t2.2μ1ΩΩ(gρϵ,σ(x))2μ(gρϵ,σ(y))2μ1u1(y)|xy|μdxdy+2.2μtΩΩ(u1(x))2μ(u1(y))2μ1gρϵ,σ(y)|xy|μdxdy, where C^=min{A^,2.2μ}.

  • Case 2

    2 < 2μ ≤ 3.

    We recall the inequality from [7, Lemma 4]: there exist C(depending on 2μ) such that, for all a, b ≥ 0,

    (a+b)2μa2μ+b2μ+2μa2μ1b+2μab2μ1Cab2μ1 if ab,a2μ+b2μ+2μa2μ1b+2μab2μ1Ca2μ1b if ab,(4.5)

    Consider Ω × Ω = O1O2O3O4, where

    O1={(x,y)Ω×Ωu1(x)tgρϵ,σ(x) and u1(y)tgρϵ,σ(y)},O2={(x,y)Ω×Ωu1(x)tgρϵ,σ(x) and u1(y)<tgρϵ,σ(y)},O3={(x,y)Ω×Ωu1(x)<tgρϵ,σ(x) and u1(y)tgρϵ,σ(y)},O4={(x,y)Ω×Ωu1(x)<tgρϵ,σ(x) and u1(y)<tgρϵ,σ(y)}.

    Also, define the b(u)|Oi=Oi(u(x))2μ(u(y))2μ|xy|μdxdy, for all uH01(Ω) and i = 1, 2, 3, 4.

  • Subcase 1

    when (x, y) ∈ O1.

    Employing (4.5), we have the following inequality:

    b(u1+tgρϵ,σ)|O1(b(u1)+b(tgρϵ,σ))|O1+2.2μt2.2μ1O1(gρϵ,σ(x))2μ(gρϵ,σ(y))2μ1u1(y)|xy|μdxdy+2.2μtO1(u1(x))2μ(u1(y))2μ1gρϵ,σ(y)|xy|μdxdyAϵ1,

    where Aϵ1 is sum of eight non-negative integrals and each integral has an upper bound of the form CO1u1(x)(tgρϵ,σ(x))2μ1(u1(y))2μ|xy|μdxdyorCO1u1(y)(tgρϵ,σ(y))2μ1(u1(x))2μ|xy|μdxdy. Write (tgρϵ,σ(x))2μ1 = (tgρϵ,σ(x))r.(tgρϵ,σ(x))s with 2μ1=r+s,0<s<2μ2. Then utilizing the definition of O1, u1L(Ω) and Hardy-Littlewood-Sobolev inequality, we have

    O1u1(x)(tgρϵ,σ(x))2μ1(u1(y))2μ|xy|μdxdyCO1(u1(x))1+r(tgρϵ,σ(x))s(u1(y))2μ|xy|μdxdyCΩΩ(tgρϵ,σ(x))s(u1(y))2μ|xy|μdxdyCΩΩϵs(N2)2|xy|μ|x(1ϵ)σ|s(N2)dxdyCϵs(N2)2Ωdx|x(1ϵ)σ|s(2N)(N2)2Nμ2Nμ2NCϵs(N2)2Ωdx|x(1ϵ)σ|s(2N)(N2)2Nμ2Nμ2N.

    By the choice of s we have Ωdx|x(1ϵ)σ|s(2N)(N2)2Nμ<. As a result, we get

    O1u1(x)(tgρϵ,σ(x))2μ1(u1(y))2μ|xy|μdxdyO(ϵ(2Nμ4)Θ) for all Θ<1.

    In a similar manner, we have

    CO1u1(y)(tgρϵ,σ(y))2μ1(u1(x))2μ|xy|μdxdyO(ϵ(2Nμ4)Θ) for all Θ<1.

  • Subcase 2

    when (x, y) ∈ O2.

    Once again using (4.5), we have the following inequality:

    b(u1+tgρϵ,σ)|O2[b(u1)+b(tgρϵ,σ)]|O2+2.2μt2.2μ1O2(gρϵ,σ(x))2μ(gρϵ,σ(y))2μ1u1(y)|xy|μdxdy+2.2μtO2(u1(x))2μ(u1(y))2μ1gρϵ,σ(y)|xy|μdxdyAϵ2,

    where Aϵ2 is sum of eight non-negative integrals and each integral has an upper bound of the form CO2u1(x)(tgρϵ,σ(x))2μ1(gρϵ,σ(y))2μ|xy|μdxdyorCO2(u1(y))2μ1(tgρϵ,σ(y))(u1(x))2μ|xy|μdxdy. By the similar estimates as in Subcase 1, definition of O2, the fact that tgρϵ,σH01(Ω) and regularity of u1, we have

    O2u1(x)(tgρϵ,σ(x))2μ1(gρϵ,σ(y))2μ|xy|μdxdyO(ϵ(2Nμ4)Θ) for all Θ<1.

    Write (u1(y))2μ1=(u1(y))r.(u1(y))s with 2μ1=r+s,0<1+s<2μ2. Then utilizing the definition of O2, u1L(Ω) and Hardy-Littlewood-Sobolev inequality, we have

    O2(u1(y))2μ1(tgρϵ,σ(y))(u1(x))2μ|xy|μdxdyO2(u1(y))r(tgρϵ,σ(y))1+s(u1(x))2μ|xy|μdxdyCΩΩ(tgρϵ,σ(y))1+s(u1(x))2μ|xy|μdxdyCΩΩϵ(1+s)(N2)2|xy|μ|y(1ϵ)σ|(1+s)(N2)dxdyCϵ(1+s)(N2)2Ωdy|y(1ϵ)σ|(1+s)(2N)(N2)2Nμ2Nμ2NCϵ(1+s)(N2)2Ωdy|y(1ϵ)σ|(1+s)(2N)(N2)2Nμ2Nμ2N.

    By the choice of s we have Ωdx|x(1ϵ)σ|(1+s)(2N)(N2)2Nμ<. Hence we obtain

    O2(u1(y))2μ1(tgρϵ,σ(y))(u1(x))2μ|xy|μdxdyO(ϵ(2Nμ4)Θ) for all Θ<1.

  • Subcase 3

    when (x, y) ∈ O3.

    Using (4.5), we have

    b(u1+tgρϵ,σ)|O3(b(u1)+b(tgρϵ,σ))|O3+2.2μt2.2μ1O3(gρϵ,σ(x))2μ(gρϵ,σ(y))2μ1u1(y)|xy|μdxdy+2.2μtO3(u1(x))2μ(u1(y))2μ1gρϵ,σ(y)|xy|μdxdyAϵ3,

    where Aϵ3 is sum of eight non-negative integrals and each integral has an upper bound of the form CO3(u1(x))2μ1(tgρϵ,σ(x))(u1(y))2μ|xy|μdxdyorCO3u1(y)(tgρϵ,σ(y))2μ1(gρϵ,σ(x))2μ|xy|μdxdy. By the similar estimates as in Subcase 1, Subcase 2, definition of O3 and regularity of u1, we get Aϵ3O(ϵ(2Nμ4)Θ) for all Θ < 1.

  • Subcase 4

    when (x, y) ∈ O4.

    Using (4.5), we have

    b(u1+tgρϵ,σ)|O4(b(u1)+b(tgρϵ,σ))|O4+2.2μt2.2μ1O4(gρϵ,σ(x))2μ(gρϵ,σ(y))2μ1u1(y)|xy|μdxdy+2.2μtO4(u1(x))2μ(u1(y))2μ1gρϵ,σ(y)|xy|μdxdyAϵ4,

    where Aϵ4 is sum of eight non-negative integrals and each integral has an upper bound of the form CO4(u1(x))2μ1(tgρϵ,σ(x))(tgρϵ,σ(y))2μ|xy|μdxdyorCO4u1(y)(tgρϵ,σ(y))2μ1(gρϵ,σ(x))2μ|xy|μdxdy. By the similar estimates as in Subcase 2, we have

    Aϵ4O(ϵ(2Nμ4)Θ) for all Θ<1.

    From all subcases we obtain AϵiO(ϵ(2Nμ4)Θ) for all Θ < 1 and i = 1, 2, 3, 4. Combining all sub cases we conclude Case 2. From Case 1 and Case 2 we have the required result.□

Proposition 4.5

Let μ < min{4, N} then there exists ϵ0 > 0 such that for every 0 < ϵ < ϵ0 we have

supt0Jf(u1+tgρϵ,σ)<Jf(u1)+Nμ+22(2Nμ)SH,L2NμNμ+2uniformlyinσSN1,

where u1 is the local minimum in Lemma 3.10.

Proof

By Lemma 3.8, uL(Ω) and u > 0 in Ω. This implies

b(u1+tgρϵ,σ)=ΩΩ(u1+tgρϵ,σ(x))2μ(u1+tgρϵ,σ(y))2μ|xy|μdxdy.

Claim 1: There exists a R0 > 0 such that

I=ΩΩ(gρϵ,σ(x))2μ(gρϵ,σ(y))2μ1u1(y)|xy|μdxdyC^R0ϵN22.

Clearly,

IB12ρB2ρB12ρB2ρ(gρϵ,σ(x))2μ(gρϵ,σ(y))2μ1u1(y)|xy|μdxdyCB12ρB2ρB12ρB2ρ(uϵσ(x))2μ(uϵσ(y))2μ1|xy|μdxdyCB12ρB2ρB12ρB2ρϵ3N2+1μdxdy|xy|μ(ϵ2+|x(1ϵ)σ|2)2Nμ2(ϵ2+|y(1ϵ)σ|2)Nμ+22.

For any ϵ < 1 – 2ρ there exists c > 0 such that 1 – ϵ > c > 2ρ so we get

ICϵ3N2+1μBcBcdzdw|zw|μ(ϵ2+|z|2)2Nμ2(ϵ2+|w|2)Nμ+22CϵN22BcBcdzdw|zw|μ(1+|z|2)2Nμ2(1+|w|2)Nμ+22=O(ϵN22).

This proves the claim 1. Now using Lemma 4.4, we have

Jf(u1+tgρϵ,σ)12a(u1)+12a(tgρϵ,σ)+tu1,gρϵ,σH01(Ω)12.2μb(u1)12.2μb(tgρϵ,σ)C^t2.2μ1ΩΩ(gρϵ,σ(x))2μ(gρϵ,σ(y))2μ1u1(y)|xy|μdxdyΩfu1dxtΩfgρϵ,σdxtΩΩ(u1(x))2μ(u1(y))2μ1gρϵ,σ(y)|xy|μdxdy+O(ϵ(2Nμ4)Θ).

for all Θ < 1. Taking Θ=22μ, we have

Jf(u1+tgρϵ,σ)12a(u1)+12a(tgρϵ,σ)+tu1,gρϵ,σH01(Ω)12.2μb(u1)12.2μb(tgρϵ,σ)C^t2.2μ1ΩΩ(gρϵ,σ(x))2μ(gρϵ,σ(y))2μ1u1(y)|xy|μdxdyΩfu1dxtΩfgρϵ,σdxtΩΩ(u1(x))2μ(u1(y))2μ1gρϵ,σ(y)|xy|μdxdy+o(ϵN22).

This on utilizing Lemma 4.3 and claim 1 gives

Jf(u1+tgρϵ,σ)12a(u1)+12a(tgρϵ,σ)+tu1,gρϵ,σH01(Ω)12.2μb(u1)12.2μb(tgρϵ,σ)C^t2.2μ1ΩΩ(gρϵ,σ(x))2μ(gρϵ,σ(y))2μ1|xy|μdxdyΩfu1dxtΩfgρϵ,σdxtΩΩ(u1(x))2μ(u1(y))2μ1gρϵ,σ(y)|xy|μdxdy+o(ϵN22)=Jf(u1)+J(tgρϵ,σ)C^t2.2μ1ΩΩ(gρϵ,σ(x))2μ(gρϵ,σ(y))2μ1|xy|μdxdy+o(ϵN22)Jf(u1)+t22SH,L2NμNμ+2+O(ϵN2)t2.2μ2.2μSH,L2NμNμ+2O(ϵ2Nμ2)t2.2μ1C^R0ϵN22+o(ϵN22).

Now define K(t):=t22SH,L2NμNμ+2+O(ϵN2)t2.2μ2.2μSH,L2NμNμ+2O(ϵ2Nμ2)t2.2μ1C^R0ϵN22 then K(t) → ∞ as t → ∞ and limt0+ K(t) > 0 so there exists a tϵ > 0 such that supt>0 K(t) is attained and tϵ<SH,L2NμNμ+2+O(ϵN2)SH,L2NμNμ+2O(ϵ2Nμ2)12.2μ2 := SH,L(ϵ). Moreover there exists a t1 > 0 such that for sufficiently small ϵ > 0 we have tϵ > t1. Clearly the function

tt22SH,L2NμNμ+2+O(ϵN2)t2.2μ2.2μSH,L2NμNμ+2O(ϵ2Nμ2)

is an increasing function in [0, SH,L(ϵ)]. Therefore,

supt0Jf(u1+tgρϵ,σ)Jf(u1)+Nμ+22(2Nμ)SH,L2NμNμ+2+O(ϵmin{2Nμ2,N2})t12.2μ1C^R0ϵN22+o(ϵN22).

Hence there exits a ϵ0 > 0 such that for every 0 < ϵ < ϵ0 we have

supt0Jf(u1+tgρϵ,σ)<Jf(u1)+Nμ+22(2Nμ)SH,L2NμNμ+2 uniformly in σSN1.

Lemma 4.6

The following holds:

  1. H01(Ω)Nf=U1U2, where

    U1:=uH01(Ω){0}|u+0,u<tuu{0},U2:=uH01(Ω){0}|u+0,u>tuu.

  2. Nf+U1.

  3. For each 0 < ϵϵ0, there exists t0 > 1 and such that u1 + t0gρϵ,σU2.

  4. For each 0 < ϵ < ϵ0, there exists s0 ⊂ (0, 1) and such that u1+s0t0gρϵ,σNf.

  5. Υf<Υf+Nμ+22(2Nμ)SH,L2NμNμ+2.

Proof

  1. It holds by Lemma 3.3 (d).

  2. Let uNf+ then t+(u) = 1 and 1<t+(u)<tmax<t(u)=1utuu that is, Nf+U1.

  3. First, we will show that there exists a constant c > 0 such that 0<tu1+tgρϵ,σu1+tgρϵ,σ<c for all t > 0. On the contrary, let there exist a sequence {tn} such that tn → ∞ and tu1+tngρϵ,σu1+tngρϵ,σ as n → ∞. Define un:=u1+tngρϵ,σu1+tngρϵ,σ so there exists t(un) such that t(un)unNf. By dominated convergence theorem,

    b(un)=b(u1+tngρϵ,σ)u1+tngρϵ,σ2.2μ=b(u1tn+gρϵ,σ)u1tn+gρϵ,σ2.2μb(gρϵ,σ)gρϵ,σ2.2μ as n.

    Hence, 𝓙f(t(un)un) → –∞ as n → ∞, contradicts the fact that 𝓙f is bounded below on 𝓝f. Therefore, there exists c > 0 such that 0<tu1+tgρϵ,σu1+tgρϵ,σ<c for all t > 0. Let t0=|c2u12|12gρϵ,σ+1 then

    u1+t0gρϵ,σ2=u12+t02gρϵ,σ2+2t0u1,gρϵ,σu12+|c2u12|c2tu1+tgρϵ,σu1+tgρϵ,σ2.

    It implies that u1 + t0gρϵ,σU2.

  4. For each 0 < ϵ < ϵ0, define a path ξϵ(s) = u1 + st0gρϵ,σ for s ∈ [0, 1]. Then

    ξϵ(0)=u1andξϵ(1)=u1+t0gρϵ,σU2.

    Since 1utuu is a continuous function and ξϵ([0, 1]) is connected. So, there exists s0 ∈ [0, 1] such that ξϵ(s0)=u1+s0t0gρϵ,σNf.

  5. Using part (iv) and Proposition 4.5.□

At this point we will state Global compactness Lemma for the functional 𝓙f which is a version of Theorem 4.4 of [19].

Lemma 4.7

Let {un} ⊂ H01(Ω) be such that 𝓙f(un) → c, Jf(un) → 0. Then passing if necessary to a subsequence, there exists a solution v0H01(Ω) of

Δu=Ω|u+(y)|2μ|xy|μdy|u+|2μ1+finΩ

and (possibly) k ∈ ℕ ∪ {0}, non-trivial solutions {v1, v2, …, vk} of

Δu=(|x|μ|u+|2μ)|u+|2μ1inRN

with viD1,2(ℝN) and k sequences {yni}nΩ and {λni}n ⊂ ℝ+ i = 1, 2, … k, satisfying

1λnidist(yni,Ω),andunv0i=1k(λni)2N2vi((.yni)/λni)D1,2(RN)0,n,unD1,2(RN)2i=0kviD1,2(RN)2,asn,Jf(v0)+i=1kJ(vi)=c,

where J(u):=12RN|u|2dx12.2μRNRN|u+(x)|2μ|u+(y)|2μ|xy|μdxdy,uD1,2(RN).

Lemma 4.8

  1. Let {un} be a (PS)c sequence for 𝓙f with c<Υf(Ω)+Nμ+22(2Nμ)SH,L2NμNμ+2 then there exists a subsequence still denoted by {un} and a nonzero u0H01(Ω) such that unu0 strongly in H01(Ω) and 𝓙f(u0) = c.

  2. Let {un} ⊂ Nf be a (PS)c sequence for 𝓙f with

    Υf(Ω)+Nμ+22(2Nμ)SH,L2NμNμ+2<c<Υf(Ω)+Nμ+22(2Nμ)SH,L2NμNμ+2

    then there exists subsequence still denoted by {un} and a nonzero u0Nf such that unu0 strongly in H01(Ω) and 𝓙f(u0) = c.

Proof

Proof of (i) follows from [30, Lemma 3.4]. To prove (ii), Let {un} be a (PS)c sequence then by standard arguments {un} is bounded in H01(Ω) and there exists a subsequence of {un} still denoted by {un} and u0H01(Ω) such that unu0 in H01(Ω) and Jf(u0) = 0. Then by Lemma 3.11, we have either u0Nf or u0 = u1. Now using Lemma 4.7 we obtain

Υf(Ω)+Nμ+22(2Nμ)SH,L2NμNμ+2c=Jf(u0)+i=1kJ(vi)Υf(Ω)+kNμ+22(2Nμ)SH,L2NμNμ+2.

which on using Lemma 4.6(e), gives k ≤ 1. By [19, corollary 3.3], we get v1 is a constant multiple of Talenti function that is, J(v1)=Nμ+22(2Nμ)SH,L2NμNμ+2. If k = 0 then we are done and if k = 1 and u0 = u1, then

c=Jf(u0)+Nμ+22(2Nμ)SH,L2NμNμ+2=Υf(Ω)+Nμ+22(2Nμ)SH,L2NμNμ+2,

a contradiction. If k = 1 and u0Nf, we get

c=Jf(u0)+Nμ+22(2Nμ)SH,L2NμNμ+2Υf(Ω)+Nμ+22(2Nμ)SH,L2NμNμ+2,

which is again a contradiction. Hence k = 0 and result follows.□

5 Existence of Second and third Solution

In this section we will show the existence of second and third solution of problem (Pf). To prove this, we shall show that for a sufficiently small δ > 0,

cat({uNf:JfΥf(Ω)+Nμ+22(2Nμ)SH,L2NμNμ+2δ})2,

where cat(X) is the category of the set X is defined in the Definition 5.1. And then employing Lemma 5.2, we conclude the existence of second and third solutions. We shall first gather some preliminaries.

For c ∈ ℝ, we define

bc(u)=cb(u),Jc(u)=12a(u)12.2μbc(u),Mc={uH01(Ω){0}|Jc(u),u=0}.

We denote

[Jfc]={uNf|Jf(u)c}.

Definition 5.1

  1. For a topological space X, we say that a non-empty, closed subset YX is contractible to a point in X if and only if there exists a continuous mapping ϱ : [0, 1] × YX such that for some x0X, ϱ(0, x) = x for all xY and ϱ(1, x) = x0 for all xY.

  2. We define

    cat(X)=min{kN|thereexistsclosedsubsetsY1,Y2,YkXsuchthatYjiscontractibletoapointinXforalljandj=1kYj=X}

Lemma 5.2

[2] Suppose that X is a Hilbert manifold and GC1(X, ℝ). Assume that there are c1 ∈ ℝ and k ∈ ℕ, such that

  1. G satisfies the Palais-Smale condition for energy level cc1;

  2. cat({xX | G(x) ≤ c1}) ≥ k.

Then G has at least k critical points in {xX | G(x) ≤ c1}.□

Lemma 5.3

[1, Theorem 2.5] Let X be a topological space. Suppose that there are two continuous maps Φ : 𝕊N–1X and Ψ : X → 𝕊N–1 such that ΨoΦ is homotopic to the identity map of 𝕊N–1. Then cat(X) ≥ 2.□

Now we will proof a Lemma which will relate the functional 𝓙f and 𝓘c. Note that for each uH01(Ω) there exists a unique t > 0 and a unique t* > 0 such that t uNf and t* u ∈ 𝓝.

Lemma 5.4

  1. For each uΣ := {uH01(Ω)| ∥u∥ = 1}, there exists a unique tc(u) > 0 such that tc(u)u ∈ 𝓜c and

    maxt0Jc(tu)=Jc(tc(u)u)=Nμ+22(2Nμ)bc(u)N2Nμ+2.

  2. For each uH01(Ω) with u+ ≢ 0 and 0 < ω < 1, we have

    (1ω)J11ω(u)12ωfH12Jf(u)(1+ω)J11+ω(u)+12ωfH12

  3. For each uΣ and 0 < ω < 1, we have

    (1ω)2NμNμ+2J(tu)12ωfH12Jf(tu)(1+ω)2NμNμ+2J(tu)+12ωfH12.

  4. There exists e11 > 0 such that if 0 < ∥fH–1 < e11 then Υf>0.

Proof

  1. For each uΣ, define k(t)=12t2t2.2μ2.2μbc(u), then if

    tc(u)=1bc(u)12(2μ1),

    we obtain k(tc(u)) = 0 and k(tc(u)) < 0. Therefore, there exists a unique tc(u) > 0 such that

    maxt0Jc(tu)=Jc(tc(u)u)=Nμ+22(2Nμ)bc(u)N2Nμ+2.

  2. For 0 < ω < 1, we have

    |Ωfudx|fH1uω2u2+12ωfH12,

    and for uH01(Ω) with u+ ≢ 0 by the above inequality, we get

    1ω2u212.2μb(u)12ωfH12Jf(u)1+ω2u212.2μb(u)+12ωfH12.

    This implies that

    (1ω)J11ω(u)12ωfH12Jf(u)(1+ω)J11+ω(u)+12ωfH12.

  3. Using part (ii), we obtain the following estimate for each uΣ and 0 < ω < 1

    (1ω)J11ω(t11ω(u)u)12ωfH12Jf(t(u)u)(1+ω)J11+ω(t11+ω(u)u)+12ωfH12.(5.1)

    Using (5.1) in part (i) we get

    J11ω(t11ω(u)u)=Nμ+22(2Nμ)b11ω(u)N2Nμ+2=(1ω)N2Nμ+2Nμ+22(2Nμ)b(u)N2Nμ+2=(1ω)N2Nμ+2J(tu).

    Therefore, we get

    (1ω)2NμNμ+2J(tu)12ωfH12Jf(tu)(1+ω)2NμNμ+2J(tu)12ωfH12.

  4. Combining part (iii) with the fact that Υ0=Nμ+22(2Nμ)SH,L2NμNμ+2>0 contributes that

    Υf(Ω)>(1ω)2NμNμ+2Υ012ωfH12=(1ω)2NμNμ+2Nμ+22(2Nμ)SH,L2NμNμ+212ωfH12.

    Thus, there exists e11 > 0 such that Υf(Ω) > 0 whenever ∥fH–1 < e11

Lemma 5.5

If Ω satisfies condition (A) then there exists a δ0 > 0 such that if u ∈ 𝓝 with J(u)Nμ+22(2Nμ)SH,L2NμNμ+2+ δ0, then RNx|x||u|2dx0

Proof

Let {un} ∈ 𝓝 such that J(un)=Nμ+22(2Nμ)SH,L2NμNμ+2+o(1) and RNx|x||un|2dx=0. Since {un} ∈ 𝓝 therefore by Lemma 2.9, {un} is a Palais-Smale sequence of 𝓙 at level Nμ+22(2Nμ)SH,L2NμNμ+2. Now using [19, Theorem 4.4] and Remark 2.8, we have

un(λn1)2N2v1((.yn1)/λn1)D1,2(RN)0,

where v1 is a minimizer of SH,L, λn1 ∈ ℝ+, yn1Ω. Moreover, if n → ∞ then λn1 → 0, yn1|yn1|y0 is the unit vector in ℝN. Thus we obtain

0=RNx|x||un|2dx=RNx|x||un|2|(λn1)2N2v1((.yn1)/λn1)|2dx+RNx|x||(λn1)2N2v1((.yn1)/λn1)|2dx=on(1)+RNyn1+λn1z|yn1+λn1z||v1(z)|2dz=on(1)+y0SH,L2NμNμ+2,

as n → ∞, which is not possible.□

For 0 < ϵϵ0 (defined in Proposition 4.5), define Hϵ : 𝕊N–1H01(Ω) as

Hϵ(σ)=u1+s0t0gρϵ,σ,

where the function u1 + s0t0gρϵ,σ defined in Lemma 4.6.

Lemma 5.6

There exists a δϵ ∈ ℝ+ such that

Hϵ(SN1)JfΥf(Ω)+Nμ+22(2Nμ)SH,L2NμNμ+2δϵ.

Proof

Trivially, Hϵ(σ)=u1+s0t0gρϵ,σNf. So we only have to prove that 𝓙f(u1 + s0t0gρϵ,σ) ≤ Υf(Ω) + Nμ+22(2Nμ)SH,L2NμNμ+2δϵ for some δϵ > 0. Since by Proposition 4.5,

supt0Jf(u1+tgρϵ,σ)<Jf(u1)+Nμ+22(2Nμ)SH,L2NμNμ+2=Υf(Ω)+Nμ+22(2Nμ)SH,L2NμNμ+2.

Hence there exists a δϵ > 0 such that

Jf(u1+s0t0gρϵ,σ)supt0Jf(u1+tgρϵ,σ)Υf(Ω)+Nμ+22(2Nμ)SH,L2NμNμ+2δϵ.

Lemma 5.7

There exists a e22 > 0 such thatfH–1 < e22 then for any

uJfΥf(Ω)+Nμ+22(2Nμ)SH,L2NμNμ+2wehaveRNx|x||u|2dx0.

Proof

Let uJfΥf(Ω)+Nμ+22(2Nμ)SH,L2NμNμ+2 then Jf(u)Υf(Ω)+Nμ+22(2Nμ)SH,L2NμNμ+2 and uNf, that is, 1utuu=1. Since Υf(Ω) < 0 we have Jf(u)Nμ+22(2Nμ)SH,L2NμNμ+2. So for uuΣ there exits a t* > 0 such that tuuN which on using Lemma 5.4 (iii) implies

(1ω)2NμNμ+2Jtuu12ωfH12Jftuu=Jf(u).

Now using Lemma 3.4, we have

Jtuu(1ω)2NμNμ+2Jf(u)+12ωfH12(1ω)2NμNμ+2Nμ+22(2Nμ)SH,L2NμNμ+2+12ωfH12=(1ω)2NμNμ+21Nμ+22(2Nμ)SH,L2NμNμ+2+Nμ+22(2Nμ)SH,L2NμNμ+2+12ω(1ω)2NμNμ+2fH12.

Choose ω0 > 0 such that for 0 < ω < ω0, we have (1ω)2NμNμ+21Nμ+22(2Nμ)SH,L2NμNμ+2<δ02 where δ0 is defined in Lemma 5.5. Now for 0 < ω < ω0 choose e22 such that if ∥fH–1 < e22 then 12ω(1ω)2NμNμ+2fH12<δ02. Therefore, we obtain

JtuuNμ+22(2Nμ)SH,L2NμNμ+2+δ0

Using Lemma 5.5 we conclude the result.□

Define G:[JfΥf(Ω)+Nμ+22(2Nμ)SH,L2NμNμ+2]SN1 by

G(u)=RNx|x||u|2dx|RNx|x||u|2dx|.

Note that from Lemma 5.5, G is well defined.

Lemma 5.8

For 0 < ϵ < ϵ0 andfH–1 < e22, the map

GoHϵ:SN1SN1

is homotopic to the identity.

Proof

Define K:={uH01(Ω){0}|RNx|x||u|2dx0} and G : 𝓚 → 𝕊N–1 by

G¯(u)=RNx|x||u|2dx/|RNx|x||u|2dx|

as an extension of G. This on using Lemma 4.1 and Lemma 5.5, gives RNx|x||gρϵ,σ|2dx0 for sufficiently small ϵ. Thus, G¯(gρϵ,σ) is well defined. Now let y : [s1, s2] → 𝕊N–1 be a regular geodesic between G¯(gρϵ,σ) and G(Hϵ(σ)) such that y(s1) = G¯(gρϵ,σ) and y(s2) = G(Hϵ(σ)). Moreover, by a analogous argument as in Lemma 4.1, for δ0 > 0 there exists a ϵ0 > 0 such that

J(gρ2(1λ)ϵ)<Nμ+22(2Nμ)SH,L2NμNμ+2+δ0 for all 0<ϵ<ϵ0 and σSN1,λ[12,1),

where δ0 is defined in Lemma 5.5. Now define ςϵ(λ, σ) : [0, 1] × 𝕊N–1 → 𝕊N–1 by

ςϵ(λ,σ)=y(2λ(s1s2)+s2) if λ[0,12),G¯(gρ2(1λ)ϵ) if λ[12,1),σ if λ=1.

Clearly, ςϵ is well defined. We claim that limλ1 ςϵ(λ, σ) = σ and limλ12ςϵ(λ, σ) = G¯(gρϵ,σ).

  • (i)

    limλ1 ςϵ(λ, σ) = σ: Indeed

    RNx|x||gρ2(1λ)ϵ|2dx=SH,L2NμNμ+2σ+o(1) as λ1

    then limλ1 ςϵ(λ, σ) = σ.

  • (b)

    limλ12 ςϵ(λ, σ) = G¯(gρϵ,σ): Indeed

    limλ12ςϵ(λ,σ)=limλ12y(2λ(s1s2)+s2)=y(s1)=G¯(gρϵ,σ).

Hence, ςϵC([0, 1] × 𝕊N–1, 𝕊N–1) and ςϵ(0, σ) = G(Hϵ(σ)) and ςϵ(1, σ) = σ for σ ∈ 𝕊N–1 provided 0 < ϵ < ϵ0 and ∥fH–1 < e22. Thus the result follows.□

Proposition 5.9

Let e* := min{e00, e11, e22} where e00; e11 and e22 defined in Lemma 3.1, Lemma 5.4 and Lemma 5.7 respectively. LetfH–1 < e* then 𝓙f has two critical points in

[JfΥf(Ω)+Nμ+22(2Nμ)SH,L2NμNμ+2].

Equivalently, (Pf) have another two different solutions which are different from u1.

Proof

Using Lemma 5.8 and Lemma 5.3, we have

cat([JfΥf(Ω)+Nμ+22(2Nμ)SH,L2NμNμ+2δϵ])2.

Now the proof follows from Lemma 4.8(i) and Lemma 5.2.□

6 Existence of Fourth solution

In this section we will prove the existence of high energy solution by using Brouwer’s degree theory and minmax theorem given by Brezis and Nirenberg [8].

Let V:={uH01(Ω):ΩΩ|u+(x)|2μ|u+(y)|2μ|xy|μdxdy=1},hρϵ,σ(x)=gρϵ,σ(x)gρϵ,σNL where gρϵ,σ is defined in (4.1).

Lemma 6.1

hρϵ,σD1,2(RN)2SH,L as ϵ → 0 uniformly in σ ∈ 𝕊N–1.

Proof

Proof follows from Lemma 4.1(i).□

Lemma 6.2

There exists a ρ0 > 0 such that for 0 < ρ < ρ0, supσSN1,ϵ(0,1]hρϵ,σ2<2Nμ+22NμSH,L.

Proof

Since we know that uϵσL2(RN)2=uϵσNL2.2μ=SH,L2NμNμ+2 and this on using Lemma 4.2 we get supσSN1,ϵ(0,1]hρϵ,σ2SH,L as ρ → 0. So there exists a ρ0 such that 0 < ρ < ρ0, we obtain supσSN1,ϵ(0,1]hρϵ,σ2<2Nμ+22NμSH,L.

Now for any uH01(Ω), by extending it to be zero outside Ω, we define Barycenter mapping β : 𝓥 → ℝN as

β(u)=RNRNx|u+(x)|2μ|u+(y)|2μ|xy|μdxdy,

and also let 𝓠 := {u ∈ 𝓥 : β(u) = 0}.

Lemma 6.3

There holds limϵ0β(hρϵ,σ)=σ.

Proof

If there exists η > 0 and a sequence ϵn → 0+ such that |β(hρϵn)σ|η. Then

β(hρϵn)=RNRNx|hρϵn(x)|2μ|hρϵn(y)|2μ|xy|μdxdyhρϵnNL2.2μ=σ+ϵnRNRN(zσ)|υρ(ϵnz+(1ϵn)σ)|2μ|υρ(ϵnw+(1ϵn)σ)|2μ|zw|μ[1+|z|2]2Nμ2[1+|w|2]2Nμ2dzdwRNRN|υρ(ϵnz+(1ϵn)σ)|2μ|υρ(ϵnw+(1ϵn)σ)|2μ|zw|μ[1+|z|2]2Nμ2[1+|w|2]2Nμ2dzdwσ+ϵnsupzsupp(υρ)|zσ|σ+Cϵn, for some C>0.

It implies that 0 < η|β(hρϵn)σ|n → 0+ as ϵn → 0+, a contradiction.□

Lemma 6.4

Let m0 = infuQu2 then SH,L < m0.

Proof

Obviously SH,Lm0, so let if possible, SH,L = infuQu2 then there exists a sequence {vn} ∈ H01(Ω) such that ∥vnNL = 1, β(vn) = 0, ∥vn2SH,L as n → ∞. Setting wn=SH,LN22(Nμ+2)vn we get wnNL2.2μ=SH,L2NμNμ+2 and wn2SH,L2NμNμ+2. Therefore, J(wn)Nμ+22(2Nμ)SH,L2NμNμ+2 and 𝓙(wn)(wn) = o(1). Using Lemma 2.9, we obtain {wn} is a Palais-Smale sequence of 𝓙 at level Nμ+22(2Nμ)SH,L2NμNμ+2. Subsequently, by [19, Theorem 4.4] and Remark 2.8, there exist sequences ynΩ, λn ∈ ℝ+ such that yny0Ω and λn → 0, for the functions

vn=SH,LN22(Nμ+2)wn, where wn=Cλnλn2+|xyn|2N22 for some C>0.

Thus if

C1=CRNRNz|zw|μ[1+|z|2]2Nμ2[1+|w|2]2Nμ2dzdw and C2=CRNRN1|zw|μ[1+|z|2]2Nμ2[1+|w|2]2Nμ2dzdw,

then

0=β(vn)=CRNRNx|vn(x)|2μ|vn(y)|2μ|xy|μdxdy=λnC1+ynC2C2y0.

This is a contradiction. Hence SH,L < m0.□

Lemma 6.5

There exists ϵ0 > 0 such that for 0 < ϵ < ϵ0 and |σ| = 1 we have

SH,L<hρϵ,σD1,2(RN)2<m0+SH,L2.

Proof

Apparently SH,Lhρϵ,σD1,2(RN)2 and we know that SH,L is not attained on a bounded domain. Thus, SH,L < hρϵ,σD1,2(RN)2. Since SH,L < m0, there exists δ0 such that SH,L2+δ0<m02 and from Lemma 6.1 we know that hρϵ,σD1,2(RN)2SH,L as ϵ → 0. Therefore for δ0 > 0 there exists a ϵ0 > 0 such that hρϵ,σD1,2(RN)2 < SH,L + δ0 whenever 0 < ϵ < ϵ0. Hence we have the desired result.□

Now we will state the minimax lemma given by Brezis and Nirenberg [8].

Lemma 6.6

Let Y be a Banach space and ϕC1(Y, ℝ). Let A be a compact metric space, A0A be a closed set and yC(A0, Y). Define

Γ={gC(A,Y):g(s)=y(s)ifsA0},c¯=infgΓsupsAϕ(g(s)),c^=supy(A0)ϕ.

If c > ĉ then there exists a sequence {un} ∈ Y satisfying ϕ(un) → c and ϕ(un) → 0.. Further, if ϕ satisfies (PS)c condition then there exists u0Y such that ϕ(u0) = c and ϕ(u0) = 0.

Let r0 = 1 – ϵ0 and Br0 = {(1 – ϵ)σ ∈ ℝN : |(1 – ϵ)σ| ≤ r0, σ ∈ 𝕊N–1, 0 < ϵ ≤ 1}, where ϵ0 is defined in Lemma 6.5. Then we set F={qC(B¯r0,V);q|B¯r0=hρϵ,σ} and

c¯=infqFsup(1ϵ)σB¯r0q((1ϵ)σ)2,c^=supB¯r0hρϵ,σ2

Lemma 6.7

For each qF, we have q(Br0) ∩ 𝓠 ≠ ∅.

Proof

It is enough to show there exist > 0 and σ̃ ∈ 𝕊N–1 such that β(q((1 – )σ̃)) = 0. Define ψ : Br0 → ℝN by ψ((1 – ϵ)σ) = β(q((1 – ϵ)σ)). We claim that

d(ψ,B¯r0,0)=d(I,B¯r0,0)0, where d is Brouwer's topological degree.

If (1 – ϵ)σBr0 then q((1 – ϵ)σ) = hρϵ,σ which implies

ψ((1ϵ)σ)=β(q((1ϵ)σ))=β(hρϵ,σ)=σ+o(1) as ϵ0.

Now define the homotopy 𝓗 : [0, 1] × Br0 → ℝN by

H(t,(1ϵ)σ)=(1t)ψ((1ϵ)σ)+tI((1ϵ)σ)

then for (1 – ϵ)σBr0 and t ∈ [0, 1] we have

H(t,(1ϵ)σ)=(1t)σ+o(1)+t(1ϵ0)σ=o(1)+(1ϵ0t)σ0, as ϵ0.

So by Brouwer’s degree theory, claim holds. It implies that there exist > 0 and σ̃ ∈ 𝕊N–1 such that ψ((1 – )σ̃) = 0 that is, β(q((1 – )σ̃)) = 0.□

Using above Lemma we have m0sup(1ϵ)σB¯r0q((1ϵ)σ)2 for all qF. Hence

m0infqFsup(1ϵ)σB¯r0q((1ϵ)σ)2=c¯.

Also, by the definition of c, and Lemma 6.2, we have c¯<2Nμ+22NμSH,L for 0 < ρ < ρ0. Combining all these and using Lemma 6.4 we have

SH,L<m0c¯<2Nμ+22NμSH,L for ρ sufficiently small .(6.1)

In addition, from Lemma 6.5, we get

c^=supB¯r0hρϵ,σ2<m0+SH,L2<m0c¯.

Now we define

J˘f(u)=maxt>0Jf(tu):VRNand J˘(u)=maxt>0J(tu):VRN,yf=infqFsup(1ϵ)σB¯r0J˘f(q((1ϵ)σ))and y0=infqFsup(1ϵ)σB¯r0J˘(q((1ϵ)σ)).

We remark that the conclusion of Lemma 5.4 (iii) holds true for 𝓙̆f. Moreover, 𝓙̆f(u) = maxt>0 𝓙f(tu) = 𝓙f(t(u)u), where t(u) is defined in Lemma 3.3.

Lemma 6.8

The following holds:

  1. 𝓙̆fC1(𝓥, ℝ) and J˘f(u),h=t(u)Jf(t(u)u),h for all hTu(𝓥) where Tu(V):={hH01(Ω)|ΩΩ|u+(x)|2μ|u+(y)|2μ1h(y)|xy|μ=0}.

  2. If u ∈ 𝓥 is a critical point of 𝓙̆f then t(u)uNf is a critical point of 𝓙f.

  3. If {un}n∈ℕ is a (PS)c sequence of 𝓙̆f then {t(un)un}n∈ℕNf is a (PS)c sequence for 𝓙f.

Proof

  1. For every uH01(Ω), t(u)uNf that is, 〈 Jf(t(u)u), u〉 = 0 and d2dt2|t=t(u)Jf(tu)<0. Therefore, by implicit function theorem, we get t(u) ∈ C1(𝓥, (0, ∞)). As a result, 𝓙̆f(u) = 𝓙f(t(u)u) ∈ C1(𝓥, ℝ) and for all hTu(𝓥), we have

    J˘f(u),h=t(u)Jf(t(u)u),h+Jf(t(u)u),u(t(u)),h=t(u)Jf(t(u)u),h.

  2. Combining the fact that u ∈ 𝓥 is a critical point of 𝓙̆f and 〈 Jf(t(u)u), u〉 = 0, we get the desired result.

  3. Let {un}n∈ℕ is a (PS)c sequence of 𝓙̆f, that is, un ∈ 𝓥, 𝓙̆f(un) → c and

    J˘f(u)Tun(V)=sup{|J˘f(un),h|:hTun(V),h=1}0 as n.

    By Lemma 3.3 we have t(un)>u22.2μ112.2μ2>SH,L2.2μ112.2μ2>C for some C > 0. Since H01(Ω) = RunTun(𝓥) so 〈Jf(un), v〉 = 〈Jf(un), hv 〉, where hv is the projection of v in Tun(𝓥). Hence,

    Jf(t(un)un)=supvH01(Ω),v=1|Jf(t(un)un),v|=supvH01(Ω),v=1|Jf(t(un)un),hv|=supvH01(Ω),v=11t(un)|J˘f(un),hv|1CJ˘f(u)Tun(V)0.

    Clearly, 𝓙f(t(un)un) → c. Therefore, {t(un)un}n∈ℕNf is a (PS)c sequence for 𝓙f.□

Lemma 6.9

If 0 < ρ < ρ0, then Nμ+22(2Nμ)SH,L2NμNμ+2<y0<Nμ+22NμSH,L2NμNμ+2.

Proof

For u ∈ 𝓥, solving J(tu)=ta(u)t2.2μ1=0 we get t = 0 and t=(a(u))12.2μ2. Therefore,

J˘(u)=maxt>0J(tu)=Nμ+22(2Nμ)u2(2Nμ)Nμ+2.

From the definition of c, we obtain

y0=Nμ+22(2Nμ)infqFsup(1ϵ)σB¯r0q((1ϵ)σ)2(2Nμ)Nμ+2=Nμ+22(2Nμ)c¯2NμNμ+2

which on using (6.1) yields the desired result.□

Lemma 6.10

J˘f(hρϵ,σ)=Nμ+22(2Nμ)SH,L2NμNμ+2+o(1) as ϵ → 0.

Proof

By Lemma 4.1, hρϵ,σ ⇀ 0 in H01(Ω) as ϵ → 0. On solving

Jf(thρϵ,σ)=ta(hρϵ,σ)t2.2μ1Ωfhρϵ,σdx=0,

we conclude tf=gρϵ,σNL+o(1). Hence again from the Lemma 4.1 we obtain

J˘f(hρϵ,σ)=maxt>0Jf(thρϵ,σ)=Jf(tfhρϵ,σ)=Jf(gρϵ,σ)=Nμ+22(2Nμ)SH,L2NμNμ+2+o(1) as ϵ0.

Lemma 6.11

There exists e0 > 0 such that if 0 < ∥fH−1 < e0,

Υf(Ω)+Nμ+22(2Nμ)SH,L2NμNμ+2<yf<Υf(Ω)+Nμ+22(2Nμ)SH,L2NμNμ+2.

Proof

Analogous to the proof of Lemma 5.4(iii) we can have

(1ω)2NμNμ+2J(tu)12ωfH12Jf(tu)(1+ω)2NμNμ+2J(tu)+12ωfH12.

Using the above inequality with the definition of 𝓙̆ and 𝓙̆f, we get

(1ω)2NμNμ+2J(tu)12ωfH12Jf(tu)(1+ω)2NμNμ+2J(tu)+12ωfH12.

For δ > 0 there exists e1(δ) such that if ∥fH−1 < e1(δ) then

y0δ<yf<y0+δ.(6.2)

Now from Lemma 5.4(iii) for each 0 < ω < 1, we have

(1ω)2NμNμ+2Nμ+22(2Nμ)SH,L2NμNμ+212ωfH12Υf(Ω)(1+ω)2NμNμ+2Nμ+22(2Nμ)SH,L2NμNμ+2+12ωfH12.

So for δ > 0 there exists e2(δ) > 0 such that whenever ∥fH−1 < e2(δ) then

Nμ+22(2Nμ)SH,L2NμNμ+2δΥf(Ω)Nμ+22(2Nμ)SH,L2NμNμ+2+δ.

It implies

Nμ+22NμSH,L2NμNμ+2δΥf(Ω)+Nμ+22(2Nμ)SH,L2NμNμ+2Nμ+22NμSH,L2NμNμ+2+δ.(6.3)

Moreover, from Lemma 6.9

Nμ+22(2Nμ)SH,L2NμNμ+2<y0<Nμ+22NμSH,L2NμNμ+2.

Hence for fix small 0 < ϵ < min{Nμ+22NμSH,L2NμNμ+2y02,y0Nμ+22(2Nμ)SH,L2NμNμ+2} such that if ∥fH−1 < e0 = min{e2(ϵ), e2(ϵ)} then using (6.2) and (6.3), we obtain

Υf(Ω)+Nμ+22(2Nμ)SH,L2NμNμ+2<Nμ+22(2Nμ)SH,L2NμNμ+2<y0ϵyf and yf<y0+2ϵϵ<Nμ+22NμSH,L2NμNμ+2ϵΥf(Ω)+Nμ+22(2Nμ)SH,L2NμNμ+2.

That is, Υf(Ω)+Nμ+22(2Nμ)SH,L2NμNμ+2<yf<Υf(Ω)+Nμ+22(2Nμ)SH,L2NμNμ+2.

Proposition 6.12

If 0 < ρ < ρ0, 0 < ∥fH−1 < e0 (defined in Lemma 6.11) then there exists a critical point u4Nf of 𝓙f with 𝓙f(u4) = yf.

Proof

Let cΥf(Ω)+Nμ+22(2Nμ)SH,L2NμNμ+2,Υf(Ω)+Nμ+22(2Nμ)SH,L2NμNμ+2 and {un}n∈ℕ is a (PS)c sequence of 𝓙̆f. Then by Lemma 6.8, {t(un)un}n∈ℕNf is a (PS)c sequence for 𝓙f which on using Lemma 4.8 gives that {un}n∈ℕ is compact. Moreover, from Lemma 6.10, yf>J˘f(hρϵ,σ)=Nμ+22(2Nμ)SH,L2NμNμ+2+o(1) as ϵ sufficiently small. Using Lemma 6.6 we have yf is a critical value of 𝓙̆f. Therefore, there exists v4 ∈ 𝓥 such that 𝓙̆f(v4) = yf and J˘f(v4) = 0. Thus by Lemma 6.8, u4 := t(v4) v4Nf is a critical point of 𝓙f and 𝓙f(u4) = yf.□

Proof of Theorem 1.1

First note that by Lemma 3.8, we have all solutions of (Pf) are positive in Ω and from Lemma 3.7, we have u1Nf+H01(Ω) such that 𝓙f(u1) = Υf whenever 0 < ∥fH−1 < e00. By Proposition 5.9 we have two more critical point u2, u3Nf of 𝓙f such that in 𝓙f(u2), 𝓙f(u3) < Υf(Ω) + Nμ+22(2Nμ)SH,L2NμNμ+2. Therefore we get three positive solutions of (Pf) whenever 0 < ∥fH−1 < e* where e* is defined in Proposition 5.9. Let e** = min{e*, e0} then by Proposition 6.12, we get u4Nf 𝓙f(u4) = yf.

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About the article

Received: 2018-12-18

Accepted: 2019-03-08

Published Online: 2019-08-06

Published in Print: 2019-03-01


Citation Information: Advances in Nonlinear Analysis, Volume 9, Issue 1, Pages 803–835, ISSN (Online) 2191-950X, ISSN (Print) 2191-9496, DOI: https://doi.org/10.1515/anona-2020-0026.

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© 2020 D. Goel and K. Sreenadh, published by De Gruyter. This work is licensed under the Creative Commons Attribution 4.0 Public License. BY 4.0

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