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BY 4.0 license Open Access Published by De Gruyter March 5, 2021

Nontrivial solutions to the p-harmonic equation with nonlinearity asymptotic to |t|p–2t at infinity

  • Qihan He EMAIL logo , Juntao Lv and Zongyan Lv

Abstract

We consider the following p-harmonic problem

Δ(|Δu|p2Δu)+m|u|p2u=f(x,u),xRN,uW2,p(RN),

where m > 0 is a constant, N > 2p ≥ 4 and limtf(x,t)|t|p2t=l uniformly in x, which implies that f(x, t) does not satisfy the Ambrosetti-Rabinowitz type condition. By showing the Pohozaev identity for weak solutions to the limited problem of the above p-harmonic equation and using a variant version of Mountain Pass Theorem, we prove the existence and nonexistence of nontrivial solutions to the above equation. Moreover, if f(x, u) ≡ f(u), the existence of a ground state solution and the nonexistence of nontrivial solutions to the above problem is also proved by using artificial constraint method and the Pohozaev identity.

MSC 2010: 35J35; 35J60; 47J30

1 Introduction

In this paper, we deal with the existence and nonexistence of nontrivial solutions to the following p-harmonic problem

Δ(|Δu|p2Δu)+m|u|p2u=f(x,u),xRN,uW2,p(RN), (1.1)

where m > 0 denotes a constant and N > 2p ≥ 4.

Throughout this paper, we assume that f(x, t) satisfies the following conditions:

  1. f : ℝN × ℝ → ℝ satisfies the Caratheodory conditions; f(x, t)t ≥ 0, ∀ (x, t) ∈ ℝN × ℝ and f(x, 0) ≡ 0, ∀ x ∈ ℝN.

  2. limt0f(x,t)|t|p2t=0 uniformly in x ∈ ℝN.

  3. limtf(x,t)|t|p2t=l uniformly in x ∈ ℝN for some l ∈ (0, +∞).

  4. For a.e. x ∈ ℝN, f(x,t)|t|p2t is nondecreasing with respect to t > 0, and nonincreasing with respect to t < 0.

  5. (t) ∈ C1(ℝ) such that |f(x, t)| ≥ |(t)|, ∀ (x, t) ∈ ℝN × ℝ, meas {x ∈ ℝN : |f(x, t)| > |(t)|} > 0, ∀t ∈ ℝ ∖ {0} and lim|x|+ f(x, t) = (t) uniformly in t ∈ ℝ.

  6. (t) ∈ C1(ℝ) satisfies that (p – 1)(t) > ′(t)t for all t < 0 and (p – 1)(t) < ′(t)t for all t > 0.

Definition 1.1

We call uW2,p(ℝN) a (weak) solution of (1.1) if for all ϕW2,p(ℝN), we have

RN(|Δu|p2ΔuΔϕ+m|u|p2uϕ)dx=RNf(x,u)ϕdx.

It is easy to see that any solution of (1.1) corresponds to a critical point of the following energy functional defined on W2,p(ℝN):

I(u)=1pRN(|Δu|p+m|u|p)dxRNF(x,u)dx, (1.2)

where F(x, u) = 0u f(x, s)ds.

The famous Mountain Pass Theorem proposed in [1] and the constraint minimization are the useful tools to get critical points of I(u). Based on the Mountain Pass Theorem, many results about the existence of solutions to second order nonlinear elliptic problems included p-laplacian or bi-harmonic operators have been obtained (see [2, 3, 4, 5] and the references therein).

Evidently, compared with the case of bounded domain, when treating a problem in ℝN, the compactness of Sobolev imbedding is absent. Researchers have attempted various methods to study this kind of problem. For example, to the following semilinear elliptic problem

Δu+mu=f(x,u),xRN,uH1(RN), (1.3)

where f(x, u) is spherical symmetrical or autonomous , the existence of nontrivial solutions to (1.3) was considered in the radially symmetrical Sobolev space (see [6, 7, 8]). However, this method can not be used to the case of general f(x, t). To dealt with this kind of problem, the concentration-compactness principle was proved by P. L. Lions. Many researchers have studied the variational elliptic problems in ℝN by this principle (see [9, 10] and the references therein).

Yang and Zhu [11] considered the following quasilinear elliptic equation

div(|u|p2u)+a(x)|u|p2u=f(x,u),xRN,uW1,p(RN), (1.4)

where N > p ≥ 2 and a(x) satisfies the following assumption:

(a) a(x) ∈ C(ℝN), a(x) ≥ a0 > 0 and lim|x|+ a(x) = ā > 0.

Deng, Gao and Jin [12] studied the p-harmonic problem (1.1). The authors in [12] required that f(x, t) is subcritical and satisfies some assumptions similar to (C1) – (C2), (C4) – (C6) and the (AR) condition:

θ>0s.t0F(x,t)=0tf(x,s)ds1p+θf(x,t)t,(x,t)RN×R. (1.5)

This condition implies that for some C > 0,

F(x,t)C|t|p+θfor|t|>0large.

However, (C3) implies that f(x, t) is asymptotic to |t|p–2t at infinity, and (1.5) does not satisfy. During the last twenty years, Researchers have shown a lot of results about (1.3) and (1.4) without the (AR) condition(see [13, 14, 15, 16, 17]). Using the concentration-compactness principle together with a variant version of Mountain Pass Theorem, Li and Zhou in [18] proved that (1.3) has a positive solution under similar conditions on f(x, t) as (C1) – (C6). After that, He and Li in [19] proved the existence of a nontrivial solution to the p & q-Laplacian equation under similar assumptions on f(x, t) as (C1) – (C6). With the help of Ekeland variational principle and Mountain Pass Theorem, the existence of multiple solutions for boundary value problem of nonhomogeneous p-harmonic equation has been proved (see [20, 21] and the references therein).

This paper is motivated by [12], [18] and [19] . We want to consider the existence and nonexistence of nontrivial solutions to the equation problem (1.1). To our best knowledge, there are few results on problem (1.1) when f(x, t) is asymptotic to |t|p–2t at infinity.

We first define the limited problem of (1.1) as follows:

Δ(|Δu|p2Δu)+m|u|p2u=f¯(u),xRN,uW2,p(RN). (1.6)

For any uW2,p(ℝN), we define

I(u)=1pRN(|Δu|p+m|u|p)dxRNF¯(u)dx, (1.7)

where F¯(u)0uf¯(s)ds. Clearly, IC1(W2,p(ℝN), ℝ). Denote

Λ={uW2,p(RN):I(u),u=0,u ̸0}, (1.8)

where 〈 ⋅, ⋅〉 denotes the dual paring between W2,p(ℝN) and (W2,p(ℝN))–1. Λ ≠ ∅ will be shown in Lemma 2.4 below. So

J=inf{I(u):uΛ} (1.9)

is well-defined.

Our main results can be stated as follows:

Theorem 1.2

Assume that conditions (C1) – (C6) hold. If l ∈ (m, +∞), then J > 0 and it is achieved by some ũW2,p(ℝN) ∖ {0}, which is a ground state solution for problem (1.6). Moreover, if lm, then problem (1.6) has no nontrivial solutions.

Theorem 1.3

Assume that (C1) – (C6) hold. Then there exists at least a nontrivial weak solution to (1.1) if l ∈ (m, +∞) and there is no nontrivial weak solutions to (1.1) if lm.

To show the Theorems, we need to prove the following Proposition:

Proposition 1.4

(Pohozaev identity for weak solution) Suppose that is a continuous function satisfying the following growth condition:

|f¯(t)|c|t|p1+C|t|d01foralltR,

where c, C > 0 and d0=NppN2p. If u is a weak solution of the p-harmonic problem (1.6), that is

RN(|Δu|p2ΔuΔϕ+m|u|p2uϕ)dx=RNf¯(u)ϕdx,ϕW2,p(RN), (1.10)

then u satisfies the Pohozaev type identity:

N2pNpRN|Δu|pdx+1pRNm|u|pdx=RNF¯(u)dx. (1.11)

Remark 1.5

The role of exponent d0 = NppN2p is to guarantee that (2.7) is true. It is easy to verify that p* := NpNp < d0 < NpN2p =: p** and (2.7) holds naturally for d0 = p** if the weak solution u LLoc (ℝN). So the Pohozaev type identity (1.11) is also true for d0 = p** if u LLoc (ℝN).

Following from the Proposition 1.4, we can get the following Corollary directly.

Corollary 1.6

Assume that uW2,p(ℝN) is a weak solution of

Δ(|Δu|p2Δu)=λ|u|q2u,xRN

where λ ∈ ℝ and q ∈ [p, d0]. Then u ≡ 0.

In the proofs of our main results, we are faced with several difficulties:

Firstly, the method in [19] can not be applied directly to the p-harmonic problem. For example, for the quasilinear elliptic problem (1.4), uW1,p(ℝN) implies that |u|, u+, uW1,p(ℝN), where u+ = max{u, 0}, u = max{–u, 0}. From this fact, we can prove that a solution can be taken to be positive. While for the p-harmonic problem (1.1), this way fails completely since uW2,p(ℝN) does not imply that |u|, u+, uW2,p(ℝN). To this end, we need to take prolongation on f(x, t) for t < 0.

Secondly, as we can see in [18], the Pohozaev identity played an important role in the proofs of the main results, which is slightly different from what [19] did. However, we can not get the Pohozaev identity of p-harmonic problem (1.6) as usual, since there is less information on the regularity of solutions to (1.6). Due to the lack of regularity of the solutions to problem (1.6), the usual method to derive the corresponding Pohozaev identity can not work. Inspired by [22], we take ϕ:=ψj=1NxjDjhu as a test function to derive the corresponding Pohozaev identity, which relaxs the restriction on the regularity of the solution and where Djhu denotes the difference quotient and ψ is a given cut-off function. But our problem (1.6) is a quasilinear elliptic equation of fourth order or more, we need to estimate the higher order difference. Thus more delicate analysis is needed.

Thirdly, W2,p(ℝN) is not a Hilbert space in general. It is not clear that, up to a subsequence,

|Δun|p2Δun|Δu|p2ΔuinLpp1(RN),

even if the (PS) sequence un} of the functional I is bounded in W2,p(ℝN) and

unuinW2,p(RN).

Hence, more delicate analysis is needed to prove that

ΔunΔua.e. inRN.

Finally, since f(x, t) and (t) are asymptotic to |t|p–2t at infinity, the (AR) condition (1.5) does not satisfy so that showing the boundedness of any (PS)c sequence for I(u) or I(u) in W2,p(ℝN) has become one of the main difficulties for studying the existence of nontrivial weak solutions to (1.1) or (1.6) in W2,p(ℝN). To show Theorem 1.2, inspired by [18], we apply Ekeland’s variational principle to get a minimizing sequence {un} for J with I(un) → 0 in (W2,p(ℝN))–1, which guarantees that we can show J > 0. Then we can show that J is achieved by some u0. As to Theorem 1.3, motivated by [19], we would prove it by a mountain pass theorem without Cerami condition together with the concentration-compacteness principle. With the help of the ground state solution ũ to (1.6) obtained in Theorem 1.2, we construct the mountain pass level c as

c=infγΓmax0t1I(γ(t)),

where Γ = {γC([0, 1], W2,p(ℝN)) : γ(0) = 0, γ(1) = t0ũ} for some t0 > 0 large enough and which is slightly different from what [18] did. Liu and Zhou [18] use u~(xt0) instead of t0ũ in the definition of Γ and prove that I( u~(xt0) ) < 0 for t0 large enough. But for the solutions of (1.6), we can not obtain the regularity result to ensure that γ0(t)=u~(xtt0),t(0,1],0,t=0, belongs to Γ, which is a key point to show c < J. So the method in [18] does not work here. Due to limtf¯(t)|t|p2t=l, (C3) and (C5), we can show that I() < I() < 0 for t > 0 large enough, which implies that the mountain pass level c defined above is well-defined and c < J. By the mountain pass theorem without the Cerami condition, we can see that there exists a Cerami sequence {un} ⊂ W2,p(ℝN) of I(u) at the level c, that is

I(un)cand(1+un)I(un)(W2,p(RN))10,asn+. (1.12)

Then we can apply the fact that c < J and the concentration-compactness principle to prove that {un} is bounded in W2,p(ℝN) and the weak limit of such subsequence of {un} is a nontrivial weak solution of (1.1).

This paper is organized as follows. In Section 2, we first derive the Pohozaev identity for the weak solutions of problem (1.6), and some preliminary lemmas are presented. The proof of Theorem 1.2 is put into the Section 3. In Section 4, by using a variant version of Mountain Pass Theorem, we devote to prove Theorem 1.3.

2 Some notations and preliminary Lemmas

In this section, we devote to some notations and preliminary Lemmas, which are crucial in our proofs of main results.

In the sequel, C represents positive constant. We denote the norm of u ∈ {Lp(ℝN) by

up=(RN|u|pdx)1p,1p<+,

and the norm of uW2,p(ℝN) by

u=[RN(|Δu|p+m|u|p)dx]1p,2p<+.

Let N > 2p and denote p** = NpN2p . It follows from (C1) – (C5) that for any ε > 0, τ ∈ (p, p**), there exists Cε > 0 such that for all x ∈ ℝN, t ∈ ℝ,

|f(x,t)|ε|t|p1+Cε|t|τ1,|f¯(t)|ε|t|p1+Cε|t|τ1, (2.1)
F(x,t)ε|t|p+Cε|t|τ,F¯(t)ε|t|p+Cε|t|τ, (2.2)
|f(x,t)|l|t|p1,|f¯(t)|l|t|p1, (2.3)

where F¯(t)=0tf¯(s)ds. Combining (C2), (C3) and (C5), we see that

limt0f¯(t)|t|p2t=0,limtf¯(t)|t|p2t=l. (2.4)

According to (C6), ones can see that

f¯(t)|t|p2tis strictly increasing in t>0,and strictly decreasing in t<0,F¯(t)<1pf¯(t)t,t0. (2.5)

Based on the above observations, we are going to prove Proposition 1.4.

The proof of Proposition 1.4

Set g(u) = (u) – m|u|p–2u, and G(u) = 0u g(t)dt. Denote Dihu(x)=u(x+hei)u(x)h, where h ∈ ℝ∖{0} and ei denotes the unit vector along coordinate xi. We take a cut-off function ψ C0 (ℝN) satisfying that ψ(x) = 1 for |x| ≤ R, ψ(x) = 0 for |x| ≥ 2R, |∇ψ| ≤ 2R and |Δψ| ≤ 2R2 . Following the idea in [22], we set ϕ=ψj=1NxjDjhu, Taking such ϕ as a test function in ((1.10), we have

RN|Δu|p2ΔuΔ(ψj=1NxjDjhu)dx=RNg(u)ψj=1NxjDjhudx. (2.6)

A straightforward computation gives us

RN|Δu|p2ΔuΔ(ψj=1NxjDjhu)dx=j=1NRN|Δu|p2ΔuΔ(ψxjDjhu)dx=j=1NRN|Δu|p2ΔuDjhuΔ(ψxj)dx+j=1NRN|Δu|p2ΔuΔ(Djhu)ψxjdx+2i,j=1NRN|Δu|p2ΔuDi(ψxj)Di(Djhu)dxI1+I2+I3,

where Diu denotes ∇uei. Now we estimate the three terms I1, I2 and I3.

I1=j=1NRN|Δu|p2ΔuDjhuΔ(ψxj)dx=j=1NRN|Δu|p2ΔuDjhu(Δψxj+2Djψ)dx.

Since DjhupCDup, we have Djh uDju weakly in Lp(ℝN), which means that, as h → 0,

I1RN|Δu|p2Δu(u,x)Δψdx+2RN|Δu|p2Δu(u,ψ)dx.
I2=j=1NRN|Δu|p2ΔuΔ(Djhu)ψxjdx=1pj=1NRNDjh(|Δu|p)ψxjdx1pj=1NRN(Djh(|Δu|p)p|Δu|p2ΔuΔ(Djhu))ψxjdx=I21I22

and

I21=1pj=1NRNDjh(|Δu|p)ψxjdx=1pj=1NRNDjh(|Δu|pψxj)dx1pj=1NRN|Δu(x+hej)|pDjh(ψxj)dx=1pj=1NRN|Δu(x+hej)|pDjh(ψxj)dx.

It follows from Djh(ψxj)pCDj(ψxj)p that Djh (ψxj) ⇀ Dj(ψxj) weakly in Lp(ℝN), which implies that when h → 0,

I211pj=1NRN|Δu|pDj(ψxj)dx=NpRN|Δu|pψdx1pRN|Δu|p(ψ,x)dx.

Since u satisfies ((1.10), we have

I22=1pj=1NRN(Djh(|Δu|p)p|Δu|p2ΔuΔ(Djhu))ψxjdx=1pj=1NRN|Δu(x+hej)|p|Δu(x)|phψxjdx1pj=1NRNp|Δu|p2ΔuΔu(x+hej)Δu(x)hψxjdx=1pj=1NRNp|Δu~|p2Δu~(Δu(x+hej)Δu(x))hψxjdx1pj=1NRNp|Δu|p2ΔuΔu(x+hej)Δu(x)hψxjdx
=j=1NRN|Δu~|p2Δu~[Δ(u(x+hej)ψxjh)u(x+hej)Δ(ψxj)h]dxj=1NRN|Δu~|p2Δu~[Δ(u(x)ψxjh)u(x)Δ(ψxj)h]dx2i,j=1NRN|Δu~|p2Δu~Di(Djhu)Di(ψxj)dxj=1NRN|Δu|p2Δu[Δ(u(x+hej)ψxjh)u(x+hej)Δ(ψxj)h]dx+j=1NRN|Δu|p2Δu[Δ(u(x)ψxjh)u(x)Δ(ψxj)h]dx+2i,j=1NRN|Δu|p2ΔuDi(Djhu)Di(ψxj)dx=j=1NRN(g(u~)g(u))Djhuψxjdx+j=1NRN(|Δu|p2Δu|Δu~|p2Δu~)DjhuΔ(ψxj)dx+2i,j=1NRN(|Δu|p2Δu|Δu~|p2Δu~)Di(Djhu)Di(ψxj)dx,

where = λu + (1 – λ)u(⋅ + hej), λ ∈ [0, 1]. By the Hölder inequality, we have

|I22|j=1N(RN|Djhu|pdx)1p(RN|g(u~)g(u)|q|ψxj|qdx)1q+j=1N(RN|Djhu|pdx)1p(RN|Δ(ψxj)|q||Δu|p2Δu|Δu~|p2Δu~|qdx)1q+2i,j=1N(RN|Di(Djhu)|pdx)1p(RN|Di(ψxj)|q||Δu|p2Δu|Δu~|p2Δu~|qdx)1q

Note that DjhupCup<+,Di(Djhu)pCΔup<+, then we have

|I22|C(RN|g(u~)g(u)|q|ψxj|qdx)1q+C(RN|Δ(ψxj)|q||Δu|p2Δu|Δu~|p2Δu~|qdx)1q+C(RN|Di(ψxj)|q||Δu|p2Δu|Δu~|p2Δu~|qdx)1q0,ash0,

where 1p+1q=1. Additionally,

I3=2i,j=1NRN|Δu|p2ΔuDi(ψxj)Di(Djhu)dx=2i,j=1NRN|Δu|p2ΔuDi(Djhu)(Diψxj+ψδij)dx,

where δij is the Kronecker symbol, that is δij = 1 when i = j and 0 otherwise. By Di(Djhu)pCDi(Dju)p, we obtain Di( Djh u) ⇀ Di(Dju) weakly in Lp(ℝN), which means that

I32j=1NRN|Δu|p2Δu(ψ,(Dju))xjdx+2RN|Δu|pψdx.

Next we turn to estimate the right hand side of (2.6).

Since uW2,p(ℝN), DiuLd(ℝN) and uLq(ℝN), where d ∈ [p, NpNp ], q ∈ [p, NpN2p ]. It follows from the definition of Dihu(x), we have ||Dihu(x)||Ld(RN)C||Diu||Ld(RN). Therefore, Dihu(x)Diu weakly in Ld(ℝN), which implies that, as h → 0,

RN|g(u)ψxi(Dihu(x)Diu)|dxRN2R(c|u|p1+C|u|d01)|Dihu(x)Diu|dxCRN|u|p1|Dihu(x)Diu|dx+CRN|u|d01|Dihu(x)Diu|dx0, (2.7)

where we have used the facts that |u|p1Lpp1(RN)=(Lp(RN))1,d0=NppN2p and |u|d01LNpNpN+p(RN)=(LNpNp(RN))1. So we have

RNg(u)ψj=1NxjDjhudxRNg(u)ψ(u,x)dx=RNψ(G(u),x)dx=RNdiv(G(u)ψx)dxRNG(u)(ψ,x)dxNRNG(u)ψdx=RNG(u)(ψ,x)dxNRNG(u)ψdx.

Altogether, as h → 0, by (2.6), we have

RN|Δu|p2Δu(u,x)Δψdx+2RN|Δu|p2Δu(u,ψ)dxNpRN|Δu|pψdx1pRN|Δu|p(ψ,x)dx+2j=1NRN|Δu|p2Δu(ψ,(Dju))xjdx+2RN|Δu|pψdx=RNG(u)(ψ,x)dxNRNG(u)ψdx.

Thus

N2ppRN|Δu|pψdx+1pRN|Δu|p(ψ,x)dx=RNG(u)(ψ,x)dx+NRNG(u)ψdx+RN|Δu|p2Δu(u,x)Δψdx+2RN|Δu|p2Δu(u,ψ)dx+2j=1NRN|Δu|p2Δu(ψ,(Dju))xjdx.

Notice that |∇ψ| ≤ 2R , |Δψ| ≤ 2R2 and supp ∇ψ, supp △ψ ⊂ ⊂ {x ∈ ℝN : R ≤ |x| ≤ 2R}, then

RN|Δu|p(ψ,x)dx4R|x|2R|Δu|pdx0asR+.

Similarly we have

RNG(u)(ψ,x)dx0asR+,RN|Δu|p2Δu(u,x)Δψdx0asR+,RN|Δu|p2Δu(u,ψ)dx0asR+,

and

j=1NRN|Δu|p2Δu(ψ,(Dju))xjdx4(R|x|2R|Δu|pdx)p1p(R|x|2R|(u)|pdx)1p0asR+.

On the other hand, as R → +∞,

RN|Δu|pψdxRN|Δu|pdx,RNG(u)ψdxRNG(u)dx.

Therefore we obtain the Pohozaev identity

N2ppRN|Δu|pdx=NRNG(u)dx,

which gives (1.11) since G(u) = (u) – 1pm|u|p.

We list some useful Lemmas.

Lemma 2.1

(Lemma 1.1 of [9]) Let {ρn} ⊂ L1(ℝN) be a bounded sequence and ρn ≥ 0, then there exists a subsequence, still denoted by {ρn}, such that one of the following two possibilities occurs:

  1. (Vanishing): limn+supyRNBR(y)ρn(x)dx=0 for all 0 < R < +∞.

  2. (Nonvanishing): There exist α > 0, 0 < R < +∞ and {yn} ⊂ ℝN such that

    limn+BR(yn)ρn(x)dxα>0.

Lemma 2.2

(Lemma 2.1 of [12]) Let p ≥ 2, pτ < p** = NpN2p . Assuming that {un} is bounded in W2,p(ℝN) and as n → +∞

supyRNBR(y)|un|τdx0forsomeR>0,

then un → 0 in Lα(ℝN) as n → +∞, for any α ∈ (p, p**).

Lemma 2.3

For the functional I defined by (1.7), if unW2,p(ℝN) satisfiesI(un), un〉 = 0 for all n ≥ 1, then I(tun) ≤ I(un) for all t > 0.

Proof

A similar proof can be found in [19]. We omit it here.□

In the following, we give several results for the set Λ and the minimization problem (1.9).

Lemma 2.4

Assume that (C1) – (C3) hold and l > m. Then Λ ≠ ∅.

Proof

Let

[d(N)]p=RNep|x|pdx.

For α > 0, we set

ωα(x)=[d(N)]1αNp2eα|x|p

and

D(N)=pp[d(N)]pRN[p|x|2p2e|x|p(p2+N)|x|p2e|x|p]pdx.

It is easy to see that ωαW2,p(ℝN). Straightforward calculations show that

ωαp=1,Δωαpp=α2D(N).

By using (2.1), for some τ ∈ (p, p**), we have

I(tωα),tωα=tpωαpRNf¯(tωα)tωαdxtpωαpRNεtp|ωα|pdxRNCεtτ|ωα|τdx=tp(α2D(N)+mε)CεtτRN|ωα|τdx>0, (2.8)

if we take t > 0 small enough. On the other hand, if we choose α(0,[lmD(N)]12), where l is given by (C3), then

Δωαpp<lm. (2.9)

whence, by Fatou’s lemma together with (2.4), one has

limt+I(tωα),tωαtp=ωαplimt+RNf¯(tωα)tωαtpdxωαpRNlimt+f¯(tωα)ωαp(tωα)p1dx=Δωαpp+ml<0.

So 〈I(α), α〉 → –∞, as t → +∞, which, together with (2.8), implies that there exists a t* > 0 such that 〈I(t*ωα), t*ωα〉 = 0. that is t*ωαΛ.□

Now we continue to study Λ. By (t) ∈ C1(ℝ), it is easy to see that the functional g(u) ≜ 〈I(u), u〉 = ∫N(|Δu|p + m|u|p)dx – ∫N (u)udxC1( W2,p(ℝN), ℝ). Then by (C6), for any uΛ

g(u),u=pRN(|Δu|p+m|u|p)dxRN[f¯(u)u+f¯(u)u2]dx=RN[(p1)f¯(u)uf¯(u)u2]dx=RN[(p1)f¯(u)f¯(u)u]udx<0.

So Λ is a closed complete submanifold of W2,p(ℝN) with the natural Finsler structure (see [23]). Therefore, using the standard Ekeland’s variational principle on Finsler manifold (see Lemma 2.4 in [18] and Corollary 1.3 in Chapter 2 of [23]) and Lemma 2.4, we can deduce the following Lemma.

Lemma 2.5

If (C1) – (C3) and (C5) – (C6) hold, then there exists a sequence {un} ⊂ Λ such that as n → +∞

I(un)J=infuΛI(u)andI(un)0in(W2,p(RN))1.

Lemma 2.6

Assume that (C1) – (C3) and (C5) – (C6) hold. Then we have that J > 0.

Proof

If J = 0, then it follows from Lemma 2.5 that there exists a sequence {un} ⊂ Λ such that as n → +∞

I(un)=1punpRNF¯(un)dxJ=0, (2.10)
I(un),un=unpRNf¯(un)undx=0, (2.11)

and

I(un)0in(W2,p(RN))1. (2.12)

First, we show the boundedness of {un} in W2,p(ℝN). Seeking a contradiction, we assume that

un+,asn+, (2.13)

and for any fixed α > 0, let

tn=αun,ωn(x)=tnun(x)=αun(x)un(x). (2.14)

Clearly {ωn} is bounded in W2,p(ℝN). For ρn = |ωn|p, then {ρn} ⊂ L1(ℝN) is a bounded sequence and ρn ≥ 0, which implies that either (i) or (ii) of Lemma 2.1 holds. We will get contradictions by showing none of these alternatives can occur.

  1. Vanishing: In this case, it follows from Lemma 2.2 and (2.2) that

    RNF¯(ωn)dx0,asn+.

    Then by (2.14), we have

    I(ωn)=1pωnpRNF¯(ωn)dx=1pωnp+on(1)=1pαp+on(1), (2.15)

    where on(1) → 0 as n → +∞. But by Lemma 2.3 and (2.10), one has

    I(ωn)=I(tnun)I(un)J=0,asn+, (2.16)

    which contradicts to (2.15).

  2. Nonvanishing: In this case, there exist η > 0, R > 0 and {yn} ⊂ ℝN such that

    limn+BR(yn)|ωn|pdxη>0. (2.17)

Set ω̃n(x) = ωn(x + yn), then ∥ω̃n∥ = ∥ωn∥ = α and by Sobolev imbedding, we may assume that for some ω̃W2,p(ℝN), such that as n → +∞

ω~nω~inW2,p(RN),ω~nω~inLlocp(RN),ω~nω~a.e. inRN.

It follows from (2.17) that

ω~0. (2.18)

By (2.5), for n large enough, we have

tn=αun(0,1)andf¯(tnun)|tnun|p2tnunf¯(un)|un|p2un.

Hence from (2.11)

ωnpRNf¯(ωn)ωndx=tnpunpRNf¯(tnun)tnundx=tnp[unpRNf¯(tnun)|tnun|p2tnun|un|pdx]tnp[unpRNf¯(un)|un|p2un|un|pdx]=0.

Then Fatou’s Lemma and (2.5) imply that

I(ωn)=1pωnpRNF¯(ωn)dx1pRNf¯(ωn)ωndxRNF¯(ωn)dxRN[1pf¯(ω~)ω~F¯(ω~)]dx+on(1).

So by (2.5) and (2.16), we have

0RN[1pf¯(ω~)ω~F¯(ω~)]dx0,

which means ω̃ ≡ 0, contradicting to (2.18).

Thus, {un} is bounded in W2,p(ℝN).

Next, we prove that J > 0. To this end, we take ρn = |un|p. Following from Sobolev imbedding, we have that {ρn} is bounded in L1(ℝN). Therefore, Lemma 2.1 implies that for some subsequence of {ρn} either Vanishing or Nonvanishing occurs. We will show that both Vanishing and Nonvanishing are impossible if J = 0.

  1. Vanishing is impossible:

    In fact, if Vanishing occurs, by Lemma 2.2 and (2.2), we have

    I(un)=1punp+on(1). (2.19)

    Taking ε = m2 in (2.1), it follows from (2.11) that

    unp=RNf¯(un)undx12unp+Cunτ,forτ(p,p),

    which means that there exists a δ > 0 such that

    unδ. (2.20)

    So, if I(un) → J = 0 as n → +∞, (2.19) and (2.20) can deduce a contradiction.

  2. Nonvanishing is also impossible:

    In fact, if Nonvanishing occurs, there exist η > 0, R > 0, {yn} ⊂ ℝN such that

    limn+BR(yn)|un|pdxη>0. (2.21)

Let ũn(x) = un(x + yn). Since {un} is bounded in W2,p(ℝN), {ũn} is also bounded in W2,p(ℝN), then by Sobolev imbedding, we may assume that there exists 0 ≢ ũW2,p(ℝN), such that as n → +∞

u~nu~inW2,p(RN),u~nu~inLlocp(RN),u~nu~a.e. inRN. (2.22)

It is easy to see that

I(un)=I(u~n)andI(un),un=I(u~n),u~n.

So by (2.5), (2.10)-(2.11) and Fatou’s Lemma, we have

0<RN[1pf¯(u~)u~F¯(u~)]dxlim infn+RN[1pf¯(u~n)u~nF¯(u~n)]dx=lim infn+[1pu~npRNF¯(u~n)dx]=J=0,

which means that Nonvanishing is also impossible.

Above arguments show that both Vanishing and Nonvanishing are impossible if J = 0. This contradiction gives that J > 0.□

3 Proof of Theorem 1.2

This section will devote to the proof of Theorem 1.2.

Proof of Theorem 1.2

By Lemma 2.6, we have J > 0. By Lemma 2.5, there exists {un} ⊂ W2,p(ℝN) such that as n → +∞

I(un)J>0, (3.1)
I(un)0in(W2,p(RN))1, (3.2)

i.e., {un} is a (PS)J sequence. Now we divide the proof into two steps.

  1. {un} is bounded in W2,p(ℝN).

    If ∥un∥ → +∞, as n → +∞, we let

    tn=αun,ωn(x)=tnun(x)=αun(x)un, (3.3)

    where α is chosen such that αp ≥ 2pJ > 0. Set ρn = |ωn|p, if there is a subsequence, still denoted by {ρn}, such that Lemma 2.1 holds, then by the similar arguments of (2.15)-(2.16), we know that Vanishing doesn’t occur.

    If Nonvanishing occurs, i.e. there exist η > 0, R > 0 and {yn} ⊂ ℝN such that

    limn+BR(yn)|ωn|pdxη>0,ω~n=ωn(x+yn),ω~n=ωn=α.

    Hence, there exists some 0 ≢ ω̃W2,p(ℝN) satisfying as n → +∞

    ω~nω~inW2,p(RN),ω~nω~inLlocτ(RN),τ(p,p),ω~nω~a.e. inRN. (3.4)

    Set ũn(x) = un(x + yn). Then ω̃n = tnũn, and it is not difficult to verity that ∀ϕ C0 (ℝN)

    I(u~n),ϕ=on(1), (3.5)

    that is

    RN(|Δu~n|p2Δu~nΔϕ+m|u~n|p2u~nϕ)dxRNf¯(u~n)ϕdx=on(1).

    Then, ∀ϕ C0 (ℝN)

    RN(|Δω~n|p2Δω~nΔϕ+m|ω~n|p2ω~nϕ)dxRNf¯(u~n)|u~n|p2u~n|ω~n|p2ω~nϕdx=on(1). (3.6)

    Now we claim that as n → +∞

    f¯(u~n)|u~n|p2u~n|ω~n|p2ω~nl|ω~|p2ω~a.e.xRN. (3.7)

    Indeed, let

    Ω1={xRN:ω~(x)0}andΩ2={xRN:ω~(x)=0}.

    If xΩ1, then we have |ũn(x)| = α–1un(x)∥ ⋅ |ω̃n(x)| → +∞, as n → +∞. Hence by (2.4), we have as n → +∞

    f¯(u~n)|u~n|p2u~n|ω~n|p2ω~nl|ω~|p2ω~,a.e.xΩ1.

    Since |f¯(t)||t|p1l and ω̃n(x) → ω̃(x) = 0 a.e. xΩ2, it follows that as n → +∞

    f¯(u~n)|u~n|p2u~n|ω~n|p2ω~n0=l|ω~|p2ω~,a.e.xΩ2.

    Therefore, (3.7) is proved. Let ϕ C0 (ℝN) be arbitrary and fixed, and let Ω ⊂ ℝN be a compact set such that supp ϕΩ. The compactness of the Sobolev embedding W2,p(Ω) ↪ Lp–1(Ω) implies ω̃nω̃ strongly in Lp–1(Ω). Therefore, it follows from [24], by using the Lebesgue Dominated Theorem, then as n → +∞

    RNf¯(u~n)|u~n|p2u~n|ω~n|p2ω~nϕdxlRN|ω~|p2ω~ϕdx. (3.8)

    Similarly, for any ϕ C0 (ℝN), as n → +∞

    RNm|ω~n|p2ω~nϕdxRNm|ω~|p2ω~ϕdx. (3.9)

    On the other hand, by (3.4)-(3.5), we have that

    I(ω~n)I(ω~),ηρ(ω~nω~)=on(1) (3.10)

    for any cut-off function ηρ C0 (ℝN) with 0 ≤ ηρ ≤ 1, ηρ = 1 on Bρ(0) = {x ∈ ℝN : |x| ≤ ρ}. Next we are going to prove that as n → +∞

    Δω~nΔω~a.e. inRN. (3.11)

    Following the idea in [17, 25], we let

    Pn(x)=(|Δω~n|p2Δω~n|Δω~|p2Δω~)(Δω~nΔω~).

    By the well-known inequality

    (|ξ|γ2ξ|η|γ2η)(ξη)>0

    for any γ > 1 and ξ, η ∈ ℝN with ξη, we have Pn(x) ≥ 0. By (3.4) and (3.10), it is easy to see that

    limn+Bρ(0)Pn(x)dx=0for anyρ>0.

    Then as n → +∞

    Δω~nΔω~inLp(Bρ(0)).

    Since ρ > 0 is arbitrary, we know that (3.11) holds. Hence as n → +∞

    |Δω~n|p2Δω~n|Δω~|p2Δω~inLpp1(RN).

    So for any ϕ C0 (ℝN), as n → +∞

    RN|Δω~n|p2Δω~nΔϕdxRN|Δω~|p2Δω~Δϕdx. (3.12)

    Combining (3.6), (3.8), (3.9) and (3.12), we have for any ϕ C0 (ℝN)

    RN|Δω~|p2Δω~Δϕdx=(lm)RN|ω~|p2ω~ϕdx,

    which contradicts to Corollary 1.6, and Step 1 is completed.

  2. J is achieved by some ũW2,p(ℝN) ∖ {0}.

    Let ρn = |un|p. Up to a subsequence, we may assume that {ρn} satisfies Lemma 2.1.

    If Vanishing occurs, then it follows from Lemma 2.2 and (2.1)-(2.2) that

    RNF¯(un)dx0andRNf¯(un)undx0asn+.

    Since

    0=I(un),un=unpRNf¯(un)undx,

    then as n → +∞

    I(un)=1punpRNF¯(un)dx=RN[1pf¯(un)unF¯(un)]dx0,

    which contradicts to (3.1).

    So only Nonvanishing occurs. Similar to the arguments in Step 1, there exists 0 ≢ ũW2,p(ℝN) such that

    RN(|Δu~|p2Δu~Δϕ+m|u~|p2u~ϕ)dx=RNf¯(u~)ϕdx,ϕC0(RN). (3.13)

    Hence ũΛ and I(ũ) ≥ J > 0. On the other hand, since

    0=I(un),un=I(u~n),u~n=u~npRNf¯(u~n)u~ndx,

    then by (2.5) and Fatous’s Lemma, one has

    J=I(un)+on(1)=I(u~n)+on(1)=1pu~npRNF¯(u~n)dx+on(1)=RN[1pf¯(u~n)u~nF¯(u~n)]dx+on(1)RN[1pf¯(u~)u~F¯(u~)]dx+on(1)=I(u~)+on(1),

    which implies that JI(ũ).

    So I(ũ) = J with ũ ≢ 0.

    For the case lm, we assume that Problem (1.6) has a nontrivial weak solution uW2,p(ℝN). Then, following from Proposition 1.4, (2.3) and (2.5), we have

    0N2pNpRN|Δu|pdx=RN(F¯(u)1pm|u|p)dx<1pRN(f¯(u)um|u|p)dx1pRN(lm)|u|pdx0,

    which is impossible. So when lm, there is no nontrivial weak solutions to (1.6).

    We complete the proof.□

4 Proof of Theorem 1.3

We will put the proof of Theorem 1.3 into this section. First, we can verify that the functional I defined in (1.2) exhibits the Mountain Pass geometry.

Lemma 4.1

Assume that (C1) – (C3) and (C5) hold. Then the functional I satisfies

  1. there exist α0 and ρ0 > 0 such that I(u) ≥ α0 for allu∥ = ρ0;

  2. there exists wW2,p(ℝN) such thatw∥ > ρ0 and I(w) ≤ 0.

Proof

By (2.2), for any ε > 0 there exists a Cε > 0 such that for some τ ∈ (p, p**)

I(u)=1pRN(|Δu|p+m|u|p)dxRNF(x,u)dx1pRN(|Δu|p+m|u|p)dxRN(ε|u|p+Cε|u|τ)dxCupCuτ.

If we choose ∥u∥ = ρ0 small, then I(u) ≥ α0 > 0. We can get the result (i).

On the other hand, Let ũ be the ground state solution obtained in Theorem 1.2. We have that if xΩ0 := {x ∈ ℝN : ũ(x) = 0}, then

F¯(tu~)tp=0=1pf¯(u~)u~,

and if xΩ := {x ∈ ℝN : ũ(x) ≠ 0}, then

lim inft+F¯(tu~)tp=limt+f¯(tu~)u~ptp1=1p|u~|plimt+f¯(tu~)|tu~|p2tu~1p|u~|pf¯(u~)|u~|p2u~=1pf¯(u~)u~.

Thus,

lim supt+RN(1pf¯(u~)u~F¯(tu~)tp)dxRN(1pf¯(u~)u~liminft+F¯(tu~)tp)dx0,

which implies that, for t large enough,

I(tu~)=1ptpRN(|Δu~|p+m|u~|p)dxRNF¯(tu~)dx=1ptpRNf¯(u~)u~dxRNF¯(tu~)dx=tpRN(1pf¯(u~)u~F¯(tu~)tp)dx0. (4.1)

So there exists w = t0ũW2,p(ℝN)(t0 is large enough) such that

I(w)0and||w||>ρ0.

By (C5), we have

I(w)I(w)0.

We complete the proof.□

As a consequence of Lemma 4.1 and the Mountain Pass Theorem without Cerami condition, founded in [26], for the constant

c=infγΓmax0t1I(γ(t)), (4.2)

where

Γ={γC([0,1],W2,p(RN)):γ(0)=0,γ(1)=w},

there exists a Cerami sequence {un} ⊂ W2,p(ℝN) at the level c, that is

I(un)cand(1+un)I(un)(W2,p(RN))10,asn+. (4.3)

Lemma 4.2

The sequence {un} obtained in (4.3) is bounded.

Proof

Just replacing J by c, and following exactly the same procedures as in Step 1 in the proof of Theorem 1.2, we know that if ∥un∥ → +∞ as n → +∞, by Lemma 2.1, Vanishing can not happen. If Nonvanishing occurs, then there exist η > 0, R > 0 and {yn} ⊂ ℝN such that

limn+BR(yn)|ωn|pdxη>0.

Let ω̃n(x) = ωn(x + yn), then ∥ω̃n∥ = ∥ωn∥ and there exists 0 ≢ ω̃W2,p(ℝN) such that as n → +∞

ω~nω~inW2,p(RN),ω~nω~inLlocτ(RN),τ(p,p),ω~nω~a.e. inRN. (4.4)

Let ũn(x) = un(x + yn), and for any ϕ C0 (ℝN), we have

I(un),ϕ(xyn)=on(1).

Hence

RN(|Δu~n|p2Δu~nΔϕ+m|u~n|p2u~nϕ)dxRNf(x+yn,u~n)ϕdx=on(1),

that is, for any ϕ C0 (ℝN), we have

RN(|Δω~n|p2Δω~nΔϕ+m|ω~n|p2ω~nϕ)dxRNf(x+yn,u~n)|u~n|p2u~n|ω~n|p2ω~nϕdx=on(1). (4.5)

Similar to the Nonvanishing case in Step 1 of Theorem 1.2, we can also prove that for any ϕ C0 (ℝN)

RN|Δω~|p2Δω~Δϕdx=(lm)RN|ω~|p2ω~ϕdx,

which contradicts to Corollary 1.6, and the Lemma is proved.□

Lemma 4.3

Assume that the assumptions of Theorem 1.3 hold. Then c < J.

Proof

Assume that wW2,p(ℝN) is the function given in Lemma 4.1, we set γ̃(t) = tw. Then γ̃Γ. Hence, by the definition of c and Lemma 2.3,

cmaxt[0,1]I(γ~(t))maxt>0I(tu~(x))<maxt>0I(tu~(x))=I(u~(x))=J.

We complete the proof.□

Finally, we give the proof of Theorem 1.3.

The Proof of Theorem 1.3

Since {un} is bounded in W2,p(ℝN), then there exist some u0W2,p(ℝN) and a subsequence of {un}, still denoted by {un}, such that as n → +∞

unu0inW2,p(RN),unu0inLlocτ(RN),τ(p,p),unu0a.e. inRN. (4.6)

By (4.6) and similar to the proof of Step 1 in Theorem 1.2, we can prove that

ΔunΔu0a.e. inRN,asn+.

By (2.1)-(2.2), (4.3) and Lebesgue Dominated Theorem, we have that for any ϕ C0 (ℝN)

I(un)I(u0),ϕ=RN(|Δun|p2Δun|Δu0|p2Δu0)Δϕdx+RNm(|un|p2un|u0|p2u0)ϕdxRN[f(x,un)f(x,u0)]ϕdx0,asn+.

Thus I′(u0) = 0. In order to complete the proof of Theorem 1.3, we must show u0 is nontrivial. To this end, we suppose by contradiction that u0 ≡ 0.

  1. un ↛ 0 in W2,p(ℝN) as n → +∞.

    Assume that un → 0 in W2,p(ℝN) as n → +∞. Then limn+ I(un) = 0, which contradicts to limn+ I(un) = c > 0.

  2. If un ⇀ 0 in W2,p(ℝN) as n → +∞, then

    limn+[I(un)I(un)]=0,limn+[I(un),unI(un),un]=0. (4.7)

    In fact, for any given R > 0, we have that

    |I(un)I(un)|RN|F(x,un)F¯(un)|dx|x|R|F(x,un)F¯(un)|dx+|x|R|F(x,un)F¯(un)|dxIn1+In2.

    Since un ⇀ 0 in W2,p(ℝN) as n → +∞, by Sobolev imbedding and up to a subsequence, we have that un → 0 in Llocτ (ℝN) as n → +∞, where τ ∈ [p, p**). Hence (2.2) implies that limn+In1=0.

    On the other hand, by (C2) – (C3) and (C5) for any δ > 0, we have

    In2=[{|x|R:|un|<δ}+{|x|R:δ|un|1δ}+{|x|R:|un|>1δ}]|F(x,un)F¯(un)|dxε1(δ)RN|un|pdx+ε2(R)δpRN|un|pdx+ε3(δ)RN|un|pdx,

    where

    ε1(δ)=sup{|x|R:|t|<δ}|F(x,t)F¯(t)||t|p0asδ0+,ε3(δ)=sup{|x|R:|t|>1δ}|F(x,t)F¯(t)||t|p0asδ0+andε2(R)=sup{|x|R:δ|t|1δ}|F(x,t)F¯(t)||t|p0asR+,for fixedδ.

    Therefore, limn+In2=0 and hence limn+[I(un)I(un)]=0. Similarly, we can deduce that limn+ [〈I′(un), un〉 – 〈I(un), un〉] = 0. So (4.7) is proved.

  3. There is a A > 0 such that

    RN|Δun|pdxA>0,n1. (4.8)

    In fact, if (4.8) is false, then ∫N|Δun|pdx → 0 as n → +∞. Using Claim I, we may assume that for some η > 0

    unW2,p(RN){0}for alln1andlimn+un=η,

    which, together with ∫N|Δun|pdx → 0 as n → +∞, implies that

    (RNm|un|pdx)1pη>0asn+.

    By (C1) – (C3), we have that for any ε > 0, there exists Cε > 0 such that

    f(x,un)unε|un|p+Cε|un|p.

    Then

    on(1)=I(un),unRN(|Δun|p+m|un|p)dxRN(ε|un|p+Cε|un|p)dxRN(|Δun|p+m|un|p)dxεRN|un|pdxC(RN|Δun|pdx)ppCηp+on(1),

    which is a contradiction. So (4.8) is true.

    Based on the above three Claims, we would show that there is a contradiction, which implies that u0 ≠ 0. By (4.3), (4.6) and (4.7), we have

    c=I(un)+on(1)=I(un)+on(1),I(un),un=on(1). (4.9)

    Let ūn(x) = un(tnx), where tn > 0 will be determined later. Then

    I(u¯n)=1ptnN(tn2p1)RN|Δun|pdx+tnNI(un) (4.10)

    and

    I(u¯n),u¯n=tnN[(tn2p1)RN|Δun|pdx+I(un),un]. (4.11)

    Taking tn=|(1I(un),unRN|Δun|pdx)|12p, then by (4.8), (4.9) and (4.11) we know that

    tn1asn+,andu¯nΛfor nlarge enough. (4.12)

    Combining (4.9)-(4.10) and (4.12), one has

    c=I(un)+on(1)=I(u¯n)+on(1)J+on(1), (4.13)

    which contradicts to Lemma 4.3.

    Therefore u0W2,p(ℝN) ∖ {0}.

    Going on as the proof of Theorem 1.2, we can get the nonexistence of nontrivial weak solutions for the case lm. So the proof of Theorem 1.3 is complete.□

Acknowledgments

The authors would like to thank Professor Yinbin Deng for stimulating discussions and helpful suggestions. This paper was supported by the fund from NSF of China (No. 11701107, 11701108, 11926332, 11926320) and the NSF of Guangxi Province (2017GXNSFBA198190).

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Received: 2020-06-23
Accepted: 2021-01-07
Published Online: 2021-03-05

© 2021 Qihan He et al., published by De Gruyter

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

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