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BY 4.0 license Open Access Published by De Gruyter November 11, 2021

Multiple positive solutions for a class of Kirchhoff type equations with indefinite nonlinearities

  • Guofeng Che and Tsung-fang Wu EMAIL logo

Abstract

We study the following Kirchhoff type equation:

a+bRN|u|2dxΔu+u=k(x)|u|p2u+m(x)|u|q2uinRN,

where N=3, a,b>0 , 1<q<2<p<min{4,2} , 2≤=2N/(N − 2), k ∈ C (ℝN) is bounded and m ∈ Lp/(pq)(ℝN). By imposing some suitable conditions on functions k(x) and m(x), we firstly introduce some novel techniques to recover the compactness of the Sobolev embedding H1(RN)Lr(RN)(2r<2) ; then the Ekeland variational principle and an innovative constraint method of the Nehari manifold are adopted to get three positive solutions for the above problem.

MSC 2010: 35J20; 35B09

1 Introduction

We investigate the existence of multiple positive solutions to the Kirchhoff type equation with indefinite nonlinearities:

(p) a+bRN|u|2dxΔu+u=k(x)|u|p2u+m(x)|u|q2uinRN,

where N ≥ 3, a,b>0 , 1<q<2<p<min{4,2} , 2≤=2N/(N − 2), and functions k(x) and m(x) satisfy the following conditions.

(H1) k ∈ C (ℝN) is a bounded function in ℝN;

(H2) k is sign–changing in ℝN and Ω1={xRN:k(x)>0} is a bounded domain;

(H3) mLq(RN) and m+=max{m(x), 0} ≡ ≠ 0, where q=ppq .

Problem (P) is a variant type of the Kirchhoff problem as follows:

(1.1) a+bΩ|u|2dxΔu=f(x,u),inΩ.u=0,onΩ,

which is related to the stationary analogue of the equation:

(1.2) utta+bΩ|u|2dxΔu=f(x,u),inΩ,u=0,onΩ,

where Ω is a bounded domain in ℝN. Such problems are nonlocal owing to the appearance of Ω|u|2dxΔu , which makes Eq. (1.1) and Eq. (1.2) are no longer pointwise identities. In [19], Kirchhoff firstly introduced Eq. (1.2) as an extension of the classical D'Alembert’s wave equation for free vibrations of elastic strings. The changes in length of the string produced by transverse vibrations are considered by the Kirchhoff’s model. Moreover, such nonlocal problem are also appeared in other fields, for instance, biological systems. For more details about the backgrounds of problems (1.1) and (1.2), one can be referred to [2,4,12,20].

In [22], Lions firstly proposed an abstract framework of Eq. (1.1), from then on, lots of researchers began to study Eq. (1.1) in general dimension, see [11,16,25,26,28,39] and the references therein. More precisely, Zhang and Perera [39] obtained the existence of a positive solution, a negative solution and a sign–changing solutions for Eq. (1.1) with N ≥ 1 by using invariant sets of descent flow. Pei and Ma [26] got three positive solutions for Eq. (1.1) with N=1, 2, 3 via the minimax method and the Morse theory. By using the theory developed in [27], Ricceri [28] obtained three positive solutions for Eq. (1.1) with N ≥ 4.

Recently, many papers [6, 8,29,30,38,32,15,31] study the Kirchhoff equation on the whole space:

(1.3) a+bRN|u|2dxΔu+V(x)u=f(x,u),inRN,uH1(RN),inRN.

When the potential function V(x) satisfies the following assumptions:

(V1) V ∈ C (ℝN) with V(x) ≥ 0 in ℝN and there exists c0>0 such that the set {V<c0}:={xRN|V(x)<c0} has finite positive Lebesgue measure, where ∣·∣ is the Lebesgue measure;

(V2) Ω=int{xRN V(x)=0} is nonempty and has smooth boundary with Ω=V1(0) . Sun and Wu [29] obtained the existence, nonexistence and concentration of nontrivial solutions for Eq. (1.4) with N=3 and V being replaced by λV,λ>0 . Later, when N ≥ 4, V is replaced by λV,λ>0 , and f satisfies the superlinear condition, Sun et al. [30] proved that Eq. (1.4) possessed two positive solutions. When N=3, V satisfies (V1)−(V2) and f satisfies the classical Ambrosetti–Rabinowitz type condition, Xie and Ma [38] proved that Eq. (1.4) has at least a positive solution. Moreover, the concentration behavior is also studied by the authors. In [32], Sun and Zhang obtained the uniqueness of the positive ground state solution for Eq. (1.4) with N=3, f(x,u)=d|u|p1u,3<p<5 , and V(x) ≡ c, where c and d are positive constants. Besides, by using the uniqueness result and the concentration–compactness lemma [23], the authors also got the existence and concentration theorems for the following Kirchhoff problem:

(1.4) ε2a+εbR3|u|2dxΔu+V(x)u=K(x)|u|p1u,inR3.

In [15], by using the Schwartz symmetric arrangement, Guo proved that Eq. (1.4) possesses a positive ground state solution when V is continuous and f does not satisfies the classical Ambrosetti–Rabinowitz type condition.

For the elliptic problems with concave–convex nonlinearities, there are also several results. For example, one can be referred to [1,37,35,24,36] and the references therein. Indeed, in [1], Ambrosetti et al. firstly introduced the elliptic problem involving the concave–convex nonlinearities:

(1.5) Δu=λuq1+up1inΩ,u>0inΩ,uH01(Ω),

where 1<q<2<p2 (2≤=∞ if N=1, 2; 2≤=2N/(N − 2) if N ≥ 3), λ>0 and Ω⊂ℝN(N ≥ 1) is bounded. With the aid of variational methods, the authors established the existence, nonexistence and multiplicity of solutions for Eq. (1.6). Later, Wu [37] obtained the existence of multiple positive solutions for the following problem:

(1.6) Δu+u=f(x)uq1+g(x)up1inRN,u>0inRN,uH1(RN),

where 1<q<2<p2 (2≤=∞ if N=1, 2; 2≤=2N/(N − 2) if N ≥ 3), f is sign–changing and g is positive in ℝN.

To the best of our knowledge, for the Kirchhoff problem with concave–convex terms, there are few results [5, 8,9,10, 21]. In [9], Chen et al. considered a class of Kirchhoff problem on the bounded domain Ω⊂ℝN(N ≥ 1):

(1.7) aΩ|u|2dx+bΔu=λf(x)|u|q2u+g(x)|u|p2uinΩ,u=0onΩ,

where 1<q<2<p2 (2≤=∞ if N=1, 2; 2≤=2N/(N − 2) if N ≥ 3). By giving different scopes on a and λ, the authors proved that Eq. (1.8) possesses multiple positive solutions. By using the Nehari manifold technique, Liao et al. [21] obtained two nonnegative solutions for Eq. (1.8). When f(x)=g(x) ≡ 1 in Eq. (1.8), by constraining the energy functional of the problem on a subset of the nodal Nehari set, Chen and Ou [10] proved that there exists a constant λ>0 such that for any λ<λ , Eq. (1.8) has a nodal solution uλ with positive energy. In [8], we obtained three positive solutions for the Kirchhoff type problem with steep potential well.

Motivated by the above facts, more precisely by [9], it is quite natural for us to ask that whether Eq.(P) can have multiple positive solutions when 1<q<2<pmin{4,2} and the domain is unbounded? As far as we know, such problem has never been discussed in the available literature. In our paper, we will give a definite answer to the above question. Under some suitable conditions on functions k and m, we will obtain three positive solutions for Eq.(P). It is worthy pointing out, in [8], we considered Eq.(P) with steep potential well and positive nonlinearities, which we overcome the compactness of the Sobolev embedding by giving some inequalities. While, in the present paper, since the potential function is a constant and the nonlinearities are sigh–changing, the standard method of recovering the compactness does not work, motivated by [17], a novel method will be used to get the Palais–Smale(PS for short) condition.

Before introducing our main conclusions, we first recall a well known result (c.f. [34]). Suppose that wΩ1 is the positive ground state solution to the nonlinear elliptic equation:

(1) Δu+u=k(x)|u|p2uuH01Ω1 in Ω1

where Ω1 is defined by the condition (H2).

Obviously,

(1.8) wΩ1H12:=Ω1|wΩ1|2+wΩ12dx=Ω1k(x)wΩ1pdx,

and

infuMΩ1JΩ1(u)=JΩ1(wΩ1)=p22pΩ1k(x)wΩ1pdx,

where JΩ1 is the energy functional of problem (EΩ1) , defined by

(1.9) JΩ1(u)=12Ω1|u|2+u2dx1pΩ1k(x)|u|pdx,

and the corresponding Nehari manifold is given by:

MΩ1={uH01(Ω1){0}|JΩ1(u),u=0}.

In order to simplify the calculation, in the present paper, we hypothesize that a=1 in problem (P). Denote kmax:=supxΩ1k(x) and A(p)=24p1/(p2) . Let Sr, Sr,Ω1 and S be the best Sobolev constants for the embedding of H1(ℝN) in Lr(ℝN), H01(Ω1) in Lr1) and D1,2(ℝN) in L2(RN) with 2 ≤ r ≤ 2≤, respectively. For any 2 ≤ r ≤ +∞, we shall also denote by ∣·∣r the Lr–norm. If we take a subsequence of a sequence {un}, we may denote it again by {un},

We now summarize our main results below.

Theorem 1.1

Assume that functions k and m satisfy hypotheses (H1)−(H3). Then there exists a constant Π1>0 such that for each 0<b+|m|q<Π1 , Eq.(P) admits two positive solutions ub1 , ub , satisfying

ub1H1<(2q)Spp(pq)kmax1/(p2)<ubH1<A(p)wΩ1H1,

and

Ib(ub1)<0<Ib(ub),

where Ib is the corresponding energy functional of Eq.(P) defined in Section 2.

Theorem 1.2

Assume that functions k and m satisfy hypotheses (H1)−(H3). In addition, we assume that (H4) The weight function m changes sign inN and Ω2={xRN:m(x)>0} is a bounded domain. Then there exists a constant Π2>0 such that for each 0<b+|m|q<Π2 , Eq.(P) admits three positive solutions ub1 , ub2 and ub satisfying

ub1H1<(2q)Spp(pq)kmax1/(p2)<ubH1<ub2H1,

and

maxIb(ub1),Ib(ub2)<0<Ib(ub).

Remark 1.1

Theorem 1.2 seems to be the first result about the Kirchhoff type equation with constant potential and indefinite nonlinearities, which has three positive solutions in the whole space ℝN, N ≥ 3. We also remark that, in the previous papers, to obtain three positive solutions for the Kirchhoff problem, the domain is usually required to be bounded. For the unbounded domain ℝN, N ≥ 3, as far as we are concerned, there are few results except [8], which we obtained three positive solutions for the Kirchhoff type equation with steep potential well.

In order to obtain our main results, we will make use of variational methods. Since we consider Eq.(P) in the whole space ℝN, the embedding from H1(ℝN) into Ls(RN)(2<s<2) is not compact. Moreover, the appearance of the nonlocal term RN|u|2dxΔu makes it hard to prove that the (PS) condition holds. Precisely, for any (PS) sequence {un}, if unu in H1(ℝN), we do not know whether there holds

RN|un|2dxRNunvdxRN|u|2dxRNuvdx,for~allvH1(RN).

Besides, the weight function k is sign–changing, which also makes it much more complicated to recover the compactness. Note that in order to overcome the lack of compactness of the Sobolev embedding, there are some existed strategies, such as constructing a convergent Cerami sequence [18]. While in this paper, inspired by [17, 20], we will make use of a novel method to verify that the (PS) condition holds. It is worthy pointing out, due to 2<p<min{4,2} , the energy functional of Eq.(P) is usually not coercive and bounded below, while, in this paper, under the hypotheses (H1)–(H4), we can prove that Ib is coercive and bounded below in ℝN, thus, two positive solutions are obtained. To get the third positive solution for Eq.(P), motivated by [31], the filtration of the Nehari manifold will be utilized:

M b ( c ) = { u M b : I b ( u ) < c } for some  c > 0 depending on p , q , k , m ,

where Mb is the corresponding Nehari manifold that can be divided into Mb(1)(c) and Mb(2)(c) , where

M b ( 1 ) ( c ) = { u M b ( c ) : u H 1 < c 1 } and M b ( 2 ) ( c ) = { u M b ( c ) : u H 1 > c 2 } ,

for some ci>0,i=1,2 . By a simple computation, we know that Ib is bounded from below on Mb(1)(c) under the assumptions (H1)–(H3), and then the third positive solution for Eq.(P) is obtained.

The remainder of this paper is organized as follows. In Section 2, we recall some preliminaries. In Section 3, we first propose some novel techniques to recover the compactness of the Sobolev embedding, then obtain two positive solutions with negative energy for Eq.(P). After introducing the filtration of Nehari manifold in Section 4, we obtain the third positive solution with positive energy for Eq.(P) and prove our results in Section 5.

2 Preliminaries

For any u ∈ H1(ℝN), let

(2.1) Ib(u)=12uH12+b4RN|u|2dx21pRNk(x)|u|pdx1qRNm(x)|u|qdx.

Then Ib is a well defined C1 functional with the following derivative:

(2.2) Ib(u),v=u,vH1+bRN|u|2dxRNuvdxRNk(x)|u|p2uvdxRNm(x)|u|q2uvdx.

Define the Nehari manifold

Mb:=uH1(RN){0}:Ib(u),u=0.

Thus uMb if and only if

uH12+bRN|u|2dx2=RNk(x)|u|pdx+RNm(x)|u|qdx.

Note that Mb is closely related to the behavior of the fibering map [13,3]: hb,u:tIb(tu),t>0 . From (2.1), it is easy to know that

hb,u(t)=Ib(tu)=t22uH12+bt44RN|u|2dx2tppRNk(x)|u|pdxtqqRNm(x)|u|qdx.

Thus

hb,u(t)=tuH12+bt3RN|u|2dx2tp1RNk(x)|u|pdxtq1RNm(x)|u|qdx,

and

hb,u′′(t)=uH12+3bt2RN|u|2dx2(p1)tp2RNk(x)|u|pdx(q1)tq2RNm(x)|u|qdx.

A straightforward calculation shows that

thb,u(t)=tuH12+bRN|tu|2dx2RNk(x)|tu|pdxRNm(x)|tu|qdx.

Obviously, tuMb holds if and only if hb, u(t)=0. Hence, points in Mb correspond to the stationary points of the maps hb, u, then it is natural to divide Mb into three parts corresponding to the local minima, local maxima and points of inflection. Following [33], we define

Mb+:=uMb:hb,u′′(1)>0,
Mb0:=uMb:hb,u′′(1)=0,

and

Mb:=uMb:hb,u′′(1)<0.

Similar to the arguments of Brown and Zhang [3, Theorem 2.3], we may get the following conclusion.

Lemma 2.1

Suppose that u0 is a local minimizer for Ib on Mb and u0Mb0 . Then Ib(u0)=0 .

For each uMb , there holds

(2.3) h b , u ( 1 ) = 2 u H 1 2 + 4 p R N k ( x ) | u | p d x + 4 q R N m ( x ) | u | q d x

(2.4) = 2 q u H 1 2 + 4 q b R N | u | 2 d x 2 p q R N k ( x ) | u | p d x = p 2 u H 1 2 + ( 4 p ) b R N | u | 2 d x 2 + p q R N m ( x ) | u | q d x .

For each uMb , in view of (2.4), there holds

(2.5) 2quH12<2quH12+4qbRN|u|2dx2<pqkmaxSppuH1p,

since 1<q<2<p<min{4,2} . From (2.5), for every uMb , we derive

(2.6) uH1>ρ0:=2qSpppqkmax1/p2.

From (2.3) and (2.6), for any uMb , there holds

Ib(u)>p24puH124qpq4pqRNm(x)|u|qdxp24p2qSpppqkmax2q/p24qpq|m|q4pqSpquH1q>0,

provided that

(2.7) |m|q<Γ:=qp24qSp2pq(pq)/(p2)2qkmax(2q)/(p2).

Therefore, we may have the following result.

Lemma 2.2

Assume that hypotheses (H1)−(H3) hold. Then for every 0<|m|q<Γ , Ib is coercive and bounded from below on Mb , where Γ is defined by (2.7). Moreover, for every uMb , there holds Ib(u)>C0 for some C0:=C0(p,q,k,m)>0 .

3 Existence of two negative–energy positive solutions

Lemma 3.1

Under the conditions (H1)−(H3), there exists a constant Γ0>0 such that for every 0<|m|qΓ0 , we obtain

inf{Ib(u) |uH1(RN)withuH1=ρ0}0.

Proof. For any u ∈ H1(ℝN) with uH1=ρ0 , in view of (2.1) and the Hölder inequality, there holds

(3.1) Ib(u)12uH12kmaxpSppuH1p|m|qqSpquH1q=ρ0q12ρ02qkmaxpSppρ0pq|m|qqSpq.

Define a function l:[0,+)R as follows:

l(t)=12t2qkmaxpSpptpq.

A straightforward calculations implies that

maxt0l(t)=l(t0)=12p2pqp(2q)Spp2(pq)kmax(2q)/(p2),

where

t0=p(2q)Spp2(pq)kmax1/(p2)>0.

Note that

ρ0=(2q)Spp(pq)kmax1/(p2)<t0,

and

l(ρ0)=(p2)(p+2q)2p(pq)(2q)Spp(pq)kmax(2q)/(p2)>0.

Thus, for every u ∈ H1(ℝN) satisfying uH1=ρ0 , there hold

Ib(u)ρ0ql(ρ0)|m|qqSpq0,

and

(3.2) |m|qΓ0:=q(p2)(p+2q)Spq2p(pq)(2q)Spp(pq)kmax(2q)/(p2).

Consequently, the proof is complete. □

Lemma 3.2

Assume that hypotheses (H1)−(H4) hold. Then Ib is bounded from below and coercive on H1(ℝN). Furthermore, for every b>0 , there exists Rb>ρ0 such that inf{Ib(u) ∣ u ∈ H1(ℝN) with uH1Rb}0 .

Proof. Denote |u|r,Ωir=Ωi|u|rdx for i=1, 2 and 1 ≤ r ≤ +∞, where Ω1 and Ω2 are given by conditions (H2) and (H4). From the Sobolev and Hölder inequalities, for any 2 ≤ r ≤ 2≤ and i=1, 2, we obtain

(3.3) Ωi|u|rdx|u|2,Ωir|Ωi|2r/2Sr|Ωi|2r/2uD1,2r.

In view of (3.3) and the Hölder inequality again, we derive

(3.4) Ib(u)=b4uD1,24+12uH121pRNk(x)|u|pdx1qRNm(x)|u|qdxb4uD1,24+12uH121pΩ1k(x)|u|pdx1qΩ2m(x)|u|qdxb4uD1,24+12uH12kmaxpΩ1|u|pdx|m|qqΩ2|u|pdxq/pb4uD1,24+12uH12kmax|Ω1|2p/2pSpuD1,2p|m|q|Ω2|q2p/p2qSquD1,2q.

Thus, for every b>0 , there exists Rˆb>0 such that

Ib(u)12uH12 for every uD1,2Rˆb.

Let

(3.5) Rb=Rˆb2+4kmax|Ω1|2p/2pSpRˆbp+|m|q|Ω2|q2p/p2qSqRˆbq.

Next, we prove that

(3.6) Ib(u)14uH12 for every uH1Rb.

For any u ∈ H1(ℝN) satisfying uH1Rb . If uD1,2Rˆb , (3.6) holds. If uD1,2<Rˆb , then it suffices to prove that Ib(u)14uH12 when

(3.7) R N u 2 d x 4 k max | Ω 1 | 2 p / 2 p S p R ^ b p + | m | q | Ω 2 | q 2 p / p 2 q S q R ^ b q .

Note that, by (3.3), for any r ∈ [2, 2≤] and i=1, 2, there holds

(3.8) Ωi|u|rdxSr|Ωi|2r/2uD1,2rSr|Ωi|2r/2Rˆbr.

Then from (3.7) and (3.8), we obtain

(3.9) Ib(u)b4uD1,24+12uD1,22+12RNu2dx1pΩ1k(x)|u|pdx1qΩ2m(x)|u|qdx14uH12+14RNu2dxkmaxpΩ1|u|pdx|m|qqΩ2|u|pdxq/p14uH12+14RNu2dxkmax|Ω1|2p/2pSpRˆbp|m|q|Ω2|q2p/p2qSqRˆbq14uH12.

From (3.9), it is obvious that Ib is bounded from below and coercive on H1(ℝN) and inf{Ib(u) ∣ u ∈ H1(ℝN) with uH1Rb}0 . The proof is complete. □

Lemma 3.3

Under the conditions (H1)−(H3), if {un} is a bounded (PS) sequence, then {un} has a convergent sequence.

Proof. Let {un} be a (PS) sequence satisfying unH1M , where M is a positive constant. Then there exists u ∈ H1(ℝN) such that

(3.10) unu in H1RN
(3.11) unu in LlocrRN for 2r<2,
(3.12) unu a.e. in RN.

Denote ωn=k(x)∣unp−2un, and ω=k(x)∣up−2u. Then ωn → ω a.e. in ℝN. Moreover, since {un}n∈ℕ is bounded in Lp(ℝN) for 2<p<min{4,2} and k(x) is bounded in ℝN, thus {ωn}n∈ℕ is bounded in Lp/(p−1)(ℝN) and ωnω in Lp/(p−1)(ℝN) for 2<p<min{4,2} . Then for any v ∈ H1(ℝN), it follows from (H3) and (3.10) that

(3.13) RNk(x)|un|p2unvdxRNk(x)|u|p2uvdx,

and

(3.14) RNunv+unvdxRNuv+uvdx,

as n → ∞. Similarly, we obtain

(3.15) RNm(x)|un|q2unvdxRNm(x)|u|q2uvdxasn.

Note that {un} is bounded in H1(ℝN), thus we obtain

RN|un|2dxAasn,

for some A>0 . Define

(3.16) Jb(u)=12uH12+bA2RN|u|2dx1pRNk(x)|u|pdx1qRNm(x)|u|qdx.

Then it follows from (3.15)(3.16) that

(3.17) o(1)=Ib(un),v1+bARNuvdx+RNuvdxRNk(x)|u|p2uvdxRNm(x)|u|q2uvdx=Jb(un),v=0for~any~vH1(RN).

Note that from (2.1) and (3.16), we can also easily obtain

(3.18) Ib(un),un+o(1)=Jb(un),un=o(1).

Denote vn=unu, then by (3.10) vn⇀ 0 in H1(ℝN), it follows from Brézis–Lieb Lemma that

(3.19) unH12=uH12+vnH12+o(1),
(3.20) RNk(x)|un|pdx=RNk(x)|u|pdx+RNk(x)|vn|pdx+o(1),

and

(3.21) RNm(x)|un|qdx=RNm(x)|u|qdx+RNm(x)|vn|qdx+o(1).

In view of (3.19)(3.21), there holds

Jb(un),un=Jb(u),u+vnH12+bARN|vn|2dxRNk(x)|vn|pdxRNm(x)|vn|qdx+o(1),

which, together with (3.17)(3.18), implies that

(3.22) unuH12+bARNunu2dxRNk(x)unupdxRNm(x)unuqdx0 as n.

From 2<p<min{4,2} , (3.11) and (H2), we have

(3.23) k(x)0for anyxRNΩ1andΩ1k(x)|unu|pdx0asn,

On the other hand, since mLq(RN) , thus for every ε>0 , there exists R>0 such that

(3.24) |x|>R|m|qdx<εq.

Then from (3.24), we obtain

(3.25) |x|>Rm(x)|unu|qdx|x|>R|m|qdx1q|x|>R|unu|pdxqpCε,

since un and u are bounded in Lp(ℝN) for 2p<min{4,2} .

Note that from (3.11), we may also have

(3.26) |x|Rm(x)|unu|qdx|x|R|m|qdx1q|x|R|unu|pdxqp0

as n → ∞. In view of (3.25) and (3.26), there holds

(3.27) RNm(x)|unu|qdx=|x|Rm(x)|unu|qdx+|x|>Rm(x)|unu|qdx0asn.

It is now deduced from (3.21)(3.23) and (3.27) that

(3.28) 0lim supnunuH12lim supnunuH12+bARN|unu|2dxlim infnRNk(x)|unu|pdx+RNm(x)|unu|qdxlimnΩ1k(x)|unu|pdx+RNm(x)|unu|qdx=0.

Thus, un → u in H1(ℝN) by (3.028). The proof is complete. □

Theorem 3.4

Assume that hypotheses (H1)−(H3) hold. Then the functional Ib has a local minimizer ub1H1(RN) for each 0<|m|qΓ0 , where Γ0 is given by (3.2). Moreover,

(i) ub1 is a positive solution of Eq.(P);

(ii) Ib(ub1)<0 and ub1H1<ρ0 .

Proof. In view of the hypothesis (H3), there exists ω ∈ H1(ℝN)\{0} satisfying

RNm(x)|ω|qdx>0.

When t>0 is sufficiently small, there holds

(3.29) Ib(tω)=t22ωH12+bt44RN|ω|2dx2tppRNk(x)|ω|pdxtqqRNm(x)|ω|qdx<0.

Thus, from (3.29) and Lemma 3.1, we obtain

(3.30) θm:=inf{Ib(u) | uBρ0(0)}<0.

From the Ekeland variational principle [14], there exists a sequence {un}Bρ0(0) satisfying

Ibun=θm+o(1) and Ibun=o(1).

It is easy to know that Ib satisfies the (PS)θm –condition in Bρ0(0) by Lemma 3.3. As a result, there exists ub1Bρ0(0) satisfying unub1 as n → +∞. This implies that ub1 is a local minimizer on Bρ0(0) that satisfies Ib(ub1)=θm<0 and ub1H1<ρ0 by Lemma 3.1. Since Ib(ub(1))=Ib(|ub1|)=θm , we may hypothesize that ub1 is a positive solution of Eq.(P). The proof is complete. □

Theorem 3.5

Assume that hypotheses (H1)−(H4) hold. Then there exists a constant b0>0 such that for any 0<b<b0 , Eq.(P) admits a positive solution ub2 satisfying Ib(ub2)<0 and ρ0<ub2H1<Rb .

Proof. Since k ∈ C (ℝN) and kmax:=supxΩ1k(x) , thus there exists a domain Ω0⊂Ω1 such that for every x ∈ Ω0, there holds k(x)kmax2 . Let I0(u)≔ Ib(u) with b=0. Then for every ξH01(Ω1) , we know that limt+I0(tξ)tpkmax2pΩ0|ξ|pdx<0 , which shows that I0(tξ) → −∞ as t → +∞. Hence, there exists ζ ∈ H1(ℝN) with ζH1>ρ0 such that I0(ζ)<0 . Since Ib(ζ) → I0(ζ) as b → 0+, then there exists b0>0 such that for every 0<b<b0 , there holds Ib(ζ)<0 . It is easy to know that ζH1<Rb by Lemma 3.2. The rest proof is analogous to that of Theorem 3.4, we omit it here. □

4 The filtration of Nehari manifold

Set

B(p,q,k,m)=p(4q)2q/(p2)|m|qp2qSp,Ω1qΩ1k(x)wΩ1pdx(2q)/2.

For any uMb with

(4.1) uH1>ρ0=2qSpppqkmax1/p2,

and

Ib(u)<p22pA2(p)(1+B(p,q,k,m))Ω1k(x)wΩ1pdx,

the Sobolev and Hölder inequalities imply that

p22pA2(p)(1+B(p,q,k,m))Ω1k(x)wΩ1pdx>Ib(u)=12uH12+b4RN|u|2dx21pRNk(x)|u|pdx1qRNm(x)|u|qdxp22puH124pb4puH14pqpqSpq|m|quH1q>p22puH124pb4p+pqpqSpqρ0q4|m|quH14,

since uH1q<ρ0q4uH14 . This implies that for

0<(4p)4pb+pqpqSpqρ0q4|m|q<Γ1,

there are two constants C1, C2 satisfying 0<C1<A(p)wΩ1H1<C2 such that

uH1<C1oruH1>C2,

where

(4.2) Γ1:=minp28pA2(p)(1+B(p,q,k,m))Ω1k(x)wΩ1pdx,p24pA2(p)Ω1k(x)wΩ1pdx.

Thus, it holds

(4.3) Mbp22pA2(p)(1+B(p,q,k,m))Ω1k(x)wΩ1pdx=uMb|uH1>ρ0andIb(u)<p22pA2(p)(1+B(p,q,k,m))Ω1k(x)wΩ1pdx=Mb(1)Mb(2),

where

Mb(1):=uMbp22pA2(p)(1+B(p,q,k,m))Ω1k(x)wΩ1pdx|uH1<C1,

and

Mb(2):=uMbp22pA2(p)(1+B(p,q,k,m))Ω1k(x)wΩ1pdx|uH1>C2.

Moreover, we get

(4.4) u H 1 < C 1 < A ( p ) w Ω 1 H 1  for all  u M b ( 1 ) ,

and

u H 1 > C 2 > A ( p ) w Ω 1 H 1  for all  u M b ( 2 ) .

It follows from (2.3),(4.1) and (4.4) that

(4.5) hb,u(1)(p2)uH12+(4p)buH14+pqSpq|m|quH1q<(p2)uH12+(4p)b+pqSpqρ0q4|m|quH14<0,

provided that

(4p)b4+pqqSpqρ0q4|m|q<p24A2(p)wΩ1H12.

Hence, we get the following result.

Lemma 4.1

Under the conditions (H1)−(H3), there is a constant Π0>0 such that for every 0<b+|m|q<Π0 , Mb(1)Mb is a C1 sub–manifold. Furthermore, the local minimizer of the functional Ib in Mb(1) is a critical point of Ib in H1(ℝN).

Lemma 4.2

Under the conditions (H1)−(H3), there exists a constant Π1 ∈ (0, Π0) such that for any 0<b+|m|q<Π1 , there is a constant tb satisfying tb<tˉb such that tbwΩ1(x)Mb(1) , where

tˉb=24p1/(p2)ifΩ1m(x)wΩ1qdx0,24p1+Ωm(x)wΩ1qdxΩ1k(x)wΩ1pdx1/(p2)ifΩ1m(x)wΩ1qdx<0.

Proof. Let

f(t)=t2wΩ1H12tp4Ω1k(x)wΩ1pdxtq4Ω1m(x)wΩ1qdxfort>0.

Then

hb,wΩ1(t)=twΩ1H12+bt3Ω1|wΩ1|2dx2tp1Ω1k(x)wΩ1pdxtq1Ω1m(x)wΩ1qdx=t3f(t)+bΩ1|wΩ1|2dx2.

It is obvious that twΩ1Mb if and only if hb,wΩ1(t)=0 , i.e., f(t)+bΩ1|wΩ1|2dx2=0 . In view of (1.8), we obtain

f(t)=t2Ω1|wΩ1|2+wΩ12dxtp4Ω1k(x)wΩ1pdxtq4Ω1m(x)wΩ1qdx=tq4t2qtpqCk,mΩ1k(x)wΩ1pdxfort>0,

where

Ck,m:=Ω1m(x)wΩ1qdxΩ1k(x)wΩ1pdx|m|qSp,Ω1qΩ1k(x)wΩ1pdx(2q)/2.

Define f1(t)=t2−qtpqCk, m, then f(t)>0 if and only if f1(t)>0 , and f1(t)=0 exactly at

t1=2qpq1/(p2),

and

f(t1)=t1q4t12qt1pqCk,mΩ1k(x)wΩ1pdx=t1q4p2pq2qpq(2q)/(p2)Ck,mΩ1k(x)wΩ1pdx>0,

provided that

|m|q<Γ2:=(p2)Sp,Ω1qpq2qpq(2q)/(p2)Ω1k(x)wΩ1pdx(2q)/2.

Next, we consider the problem in the following two cases.

Case (I):Ω1m(x)wΩ1qdx0 . Let

f˜(t)=t2wΩ1H12tp4Ω1k(x)wΩ1pdx,t>0.

Obviously, f(t)f˜(t) for t > 0. In view of (1.8), there holds

f˜(t)=t2Ω1|wΩ1|2+wΩ12dxtp4Ω1k(x)wΩ1pdx=t2tp4Ω1k(x)wΩ1pdx.

Evidently,

f ~ 1 = 0 , lim t 0 + f ~ ( t ) =  and  lim t f ~ ( t ) = 0.

Moreover

inft>0f˜(t)=f˜(t˜0)=p24p4p22/(p2)Ω1k(x)wΩ1pdx<0,

where

t˜0:=24p1/(p2)>1.

Observe that

p24p4p22/(p2)Ω1k(x)wΩ1pdxΩ1|wΩ1|2dx2p24p4p22/(p2)Ω1k(x)wΩ1pdxwΩ1H14=p24p4p22/(p2)1Ω1k(x)wΩ1pdx.

Thus, for each

0<b<b(1):=p24p4p22/(p2)1Ω1k(x)wΩ1pdx,

we obtain

inft>0f˜(t)=p24p4p22/(p2)Ω1k(x)wΩ1pdx<bΩ1|wΩ1|2dx2.

This shows that there exist two constants tb+ and tb that satisfy

tb<24p1/(p2)<tb+

such that

f(tb±)+bΩ1|wΩ1|2dx2=0,f(tb)<0 and f(tb+)>0.

That is, tb±wΩ1Mb . Besides, it is evident that

hb,tbwΩ1(1)=(tb)5f(tb)<0,

and

hb,tb+wΩ1(1)=(tb+)5f(tb+)>0.

These imply that tb±wΩ1Mb± .

Case (II):Ω1m(x)wΩ1qdx<0 . Note that

f(t)=t2Ω1|wΩ1|2+wΩ12dxtp4Ω1k(x)wΩ1pdx+tq4Ω1m(x)wΩ1qdx=t2tp4Ω1k(x)wΩ1pdx+tq4Ω1m(x)wΩ1qdx.

Define

fˆ(t)=t2Ω1k(x)wΩ1pdx+Ω1m(x)wΩ1qdxtp4Ω1k(x)wΩ1pdx.

It is obvious that

(4.6) f(t)<fˆ(t) for t>1,

and fˆ(1)=f(1)=|Ω1m(x)wΩ1qdx|>0 and limtfˆ(t)=0 . Furthermore, we obtain

(4.7) inf t > 0 f ^ ( t ) = f ^ ( t ^ 0 ) = p 2 ( 4 p ) t ^ 0 2 Ω 1 k ( x ) w Ω 1 p d x + Ω 1 m ( x ) w Ω 1 q d x < 0 ,

where

(4.8) tˆ0:=24p1+Ω1m(x)wΩ1qdxΩ1k(x)wΩ1pdx1/(p2)>1.

Observe that

p 2 ( 4 p ) t ^ 0 2 Ω 1 k ( x ) w Ω 1 p d x + Ω 1 m ( x ) w Ω 1 q d x 1 Ω 1 | w Ω 1 | 2 d x 2 p 2 ( 4 p ) t ^ 0 2 Ω 1 k ( x ) w Ω 1 p d x + Ω 1 m ( x ) w Ω 1 q d x 1 w Ω 1 H 1 4 = p 2 ( 4 p ) t ^ 0 2 Ω 1 k ( x ) w Ω 1 p d x + Ω 1 m ( x ) w Ω 1 q d x 1 Ω 1 k ( x ) w Ω 1 p d x 2 = p 2 4 p 4 p 2 2 / ( p 2 ) 1 + Ω 1 m ( x ) w Ω 1 q d x Ω 1 k ( x ) w Ω 1 p d x ( p 4 ) / ( p 2 ) 1 Ω 1 k ( x ) w Ω 1 p d x := b ( 2 ) .

It is obvious that b(2)<b(1) since 2<p<min{4,2} . Thus, for each 0<b<b(2) , it follows from (4.7) that

(4.9) fˆ(tˆ0)=p2(4p)tˆ02Ω1k(x)wΩ1pdx+Ω1m(x)wΩ1qdx<bΩ1|wΩ1|2dx2.

From (4.6), (4.8) and (4.9), we derive

(4.10) f(tˆ0)<fˆ(tˆ0)<bΩ1|wΩ1|2dx2.

In view of (4.6) and (4.10), there are two constants tb+ and tb that satisfy

(4.11) 1<tb<tˆ0<tb+

such that

f(tb±)+bΩ1|wΩ1|2dx2=0,f(tb)<0 and f(tb+)>0.

That is, tb±wΩ1Mb . A straightforward calculation shows that

hb,tbwΩ1(1)=(tb)5f(tb)<0,

and

hb,tb+wΩ1(1)=(tb+)5f(tb+)>0.

These imply that tb±wΩ1Mb± .

Next we prove that tbwΩ1Mb(1) .

If Ω1m(x)wΩ1qdx<0 , it follows from (4.11) that

(4.12) I b ( t b w Ω 1 ) = 1 4 ( t b ) 2 1 4 p p ( t b ) p 2 Ω 1 k ( x ) w Ω 1 p d x + 4 q 4 q ( t b ) q Ω 1 m ( x ) w Ω 1 q d x 1 4 ( t b ) 2 1 4 p p ( t b ) p 2 Ω 1 k ( x ) w Ω 1 p d x
+4q4q24p1+Ω1m(x)wΩ1qdxΩ1k(x)wΩ1pdxq/(p2)Ω1m(x)wΩ1qdx:=I1+I2,

where

I1=14(tb)214pp(tb)p2Ω1k(x)wΩ1pdx,

and

I2=4q4q24p1+Ω1m(x)wΩ1qdxΩ1k(x)wΩ1pdxq/(p2)Ω1m(x)wΩ1qdx.

Define a function by

g(t)=t2414pptp2, for 2<p<min{4,2},

where

1 < t < t ^ 0 := 2 4 p 1 + Ω 1 m ( x ) w Ω 1 q d x Ω 1 k ( x ) w Ω 1 p d x 1 / ( p 2 ) .

By a direct calculation, we obtain

max0<ttˆ0g(t)=gt=p24p24p2/(p2),

where

(4.13) t=24p1/(p2)(1,tˆ0).

Hence,

max0<ttˆ0g(t)Ω1k(x)wΩ1pdx(p2)A2(p)2pwΩ1H12,

that is,

(4.14) I1(p2)A2(p)2pwΩ1H12.

On the other hand, by 0<b+|m|q<Π1 and (4.11), it holds

(4.15) I 2 = 4 q 4 q 2 4 p 1 + Ω 1 m ( x ) w Ω 1 q d x Ω 1 k ( x ) w Ω 1 p d x q / ( p 2 ) Ω 1 m ( x ) w Ω 1 q d x 4 q 4 q 2 4 p 1 + | m | q | w Ω 1 | p q Ω 1 k ( x ) w Ω 1 p d x q / ( p 2 ) | m | q | w Ω 1 | p q 4 q 2 q 2 4 p q / ( p 2 ) | m | q S p , Ω 1 q Ω 1 k ( x ) w Ω 1 p d x q / 2 if  q p 2 4 q 4 q 4 4 p q / ( p 2 ) | m | q S p , Ω 1 q Ω 1 k ( x ) w Ω 1 p d x q / 2 if  q > p 2
4q2q44pq/(p2)|m|qSp,Ω1qΩ1k(x)wΩ1pdxq/2p22pA2(p)B(p,q,k,m)Ω1k(x)wΩ1pdx.

It follows from (4.12), (4.14) and (4.15) that tbwΩ1Mb1 .

Moreover, if Ω1m(x)wΩ1qdx0 , then by (H2), we have

Ib(tbwΩ1)=14(tb)2wΩ1H124p4p(tb)pΩ1k(x)wΩ1pdx4q4q(tb)qΩ1m(x)wΩ1qdx14(tb)2wΩ1H124p4p(tb)pΩ1k(x)wΩ1pdx=14(tb)214pp(tb)p2Ω1k(x)wΩ1pdx.

Let

η(t)=t2414pptp2,2<p<min{4,2},

where

1<tt˜0:=24p1/(p2).

A simple calculation shows that

max0<tt˜0η(t)=ηt˜0=p24p24p2/(p2).

Then,

max0<tt˜0η(t)Ω1k(x)wΩ1pdx<p22pA2(p)Ω1k(x)wΩ1pdx,

that is,

Ib(tbwΩ1)<p22pA2(p)Ω1k(x)wΩ1pdxp22pA2(p)(1+B(p,q,k,m))Ω1k(x)wΩ1pdx.

The proof is complete. □

5 Proofs of main results

Lemma 5.1

Under the conditions (H1)−(H3), there exist a constant 𝜎 > 0 and a differentiable function t*:B (0, 𝜎)⊂ H1(ℝN) → ℝ+ such that for any uMb(1) , there holds

t(0)=1 and t(v)(uv)Mb(1)

for all v ∈ B (0, σ), and

(t)(0),φ=ψ(u,φ)Γ(u),

where

(5.1) ψ(u,φ)=2RN(uφ+uφ)dx+4bRN|u|2dxRNuφdxpRNk(x)|u|p2uφdxqRNm(x)|u|q2uφdx,

and

(5.2) Γ(u)=uH12(p1)RNk(x)|u|pdx(q1)RNm(x)|u|qdx

for all φ ∈ H1(ℝN).

Proof. The proof of Lemma 5.1 is analogous to [7, Lemma 3.1] and [24, Lemma 3.1], we omit it here. □

Define

α b = inf u M b ( 1 ) I b ( u ) .

In view of Lemma 2.2 and Mb(1)Mb , we derive

(5.3) 0<C0αb<p22pA2(p)(1+B(p,q,k,m))Ω1k(x)wΩ1pdx.

Proposition 5.2

Assume that hypotheses (H1)−(H3) hold. Then there exists a sequence {un}Mb(1) such that

(5.4) Ib(un)=αb+o(1) and I b(un)=o(1).

Proof. The proof of Proposition 5.2 is similar to [7, Proposition 3.2] and [24, Proposition 2], we omit it here. □

By similar arguments to those of Lemma 3.3, we may get the following conclusion.

Proposition 5.3

Assume that hypotheses (H1) −(H3) hold. Then for any 0<d<p22pA2(p)(1+B(p,q,k,m))Ω1k(x)wΩ1pdx , Ib satisfies the (PS)d–condition in Mb(1) .

Theorem 5.4

Assume that hypotheses (H1)−(H3) hold. Then for any 0<b+|m|q<Π1 , where Π1>0 is given by Lemma 4.2, Eq.(P) admits a positive solution ubMb(1) satisfying Ib(ub)>0 and

(2q)Spp(pq)kmax1/(p2)<ubH1<A(p)wΩ1H1.

Proof. By Proposition 5.2, there is a sequence {un}Mb(1) satisfying

Ib(un)=αb+o(1) and ~Ib(un)=o(1).

Clearly, 0<αb<p22pA2(p)(1+B(p,q,k,m))Ω1k(x)wΩ1pdx , and for small b, Ib satisfies the (PS)αb –condition in Mb(1) by Proposition 5.3. Thus there exists ubH1(RN){0} such that unub in H1(ℝN) . Observe that {un}Mb(1) and

unH1<C1<A(p)wΩ1H1 for all n=1,2,...,

then from Fatou’s lemma, we have

(5.5) u b H 1 lim inf n u n H 1 C 1 < A ( p ) w Ω 1 H 1 .

(4.5) and (5.5) imply that

hb,ub(1)<0,

provided that

4p4b+pqqSpqρ04q|m|q<Γ1,

where Γ1 is defined by (4.2). Thus, ubMb . On the other hand, observe that

αb=Ib(ub)Ib(tbwΩ1)p22pA2(p)1+B(p,q,k,m)wΩ1H12,

then ubMb(1) . Furthermore, Ib(ub)=Ib(|ub|)=αb . Thus, from Lemma 2.1, we may assume that ub is a positive solution of Eq.(P). The proof is complete. □

Proof of Theorem 1.1: From Theorems 3.4 and 5.4, there exists a constant Π1>0 such that for any 0<b+|m|q<Π1 , Eq.(P) admits two positive solutions ub1 and ub that satisfy

ub1H1<(2q)Spp(pq)kmax1/(p2)<ubH1<A(p)wΩ1H1,

and

Ib(ub1)<0<Ib(ub).

The proof is complete.

Proof of Theorem 1.2: From Theorems 3.4, 3.5 and 5.4, there exists a constant Π2=min{b0,Π1}>0 such that for any 0<b+|m|q<Π2 , Eq.(P) admits three positive solutions ub1 , ub2 and ub that satisfy

ub1H1<ρ0:=(2q)Spp(pq)kmax1/(p2)<minubH1,ub2H1,

and

maxIb(ub1),Ib(ub2)<0<Ib(ub).

Then by Lemma 2.2 and (4.3), ubMb(2) . Hence,

ub1H1<ρ0<ubH1<ub2H1.

The proof is complete.

Acknowledgments

G. Che was supported by the National Natural Science Foundation of China (Grant No. 12001114) and the Guangdong Basic and Applied Basic Research Foundation (Grant No. 2019A1515110275). T. F. Wu was supported by the Ministry of Science and Technology, Taiwan (110-2115-M-390-006-MY2).

  1. Conflict of interest

    Conflict of interest statement: Authors state no conflict of interest.

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Received: 2021-07-30
Accepted: 2021-10-10
Published Online: 2021-11-11

© 2021 Guofeng Che and Tsung-fang Wu, published by De Gruyter

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

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