Yanjun LIU Chungen LIU

Abstract In this paper, the authors consider the following singular Kirchhoff-Schrdinger problem

where 0 < η < N, M is a Kirchhoff-type function and V(x) is a continuous function with positive lower bound, f(x,t) has a critical exponential growth behavior at infinity.Combining variational techniques with some estimates, they get the existence of ground state solution for(Pη). Moreover, they also get the same result without the A-R condition.

Keywords Ground state solutions, Singular elliptic equations, Critical exponential growth, Kirchhoff-Schrdinger equations, Singular Trudinger-Moser inequality

1 Introduction

The research of nonlinear Kirchhoffequations has attracted a lot of attention and a classical Kirchhoffequation is given by

where ? ?RNis a bounded domain, a,b > 0 and f : ?×R →R is a continuous function. It is related to the stationary analogue of the following equation

which was proposed by Kirchhoffin [21] as an extension of the classical D’Alembert’s wave equation for free vibrations of elastic strings. The problem(1.1)is nonlocal since the appearance of the integration termdx. In[28], Lions proposed an abstract framework for this kind problems and after that, the Kirchhoffproblem began to receive a lot of attention, see [3–4,6, 8, 11, 32] and the references therein. For Kirchhofftype equation with a singular nonlinear term, in [33], by proving the mountain pass structure of the related functional and using the concentration compactness principle,the authors obtained the existence of a nontrivial solution of the following polyharmonic Kirchhofftype problem

where ?pwith p ≥2 is the p-Laplacian operator, ? ?RNis a bounded domain with smooth boundary, m is an integer and 2 ≤2m ≤N, 0 < η < N, M(·) is a Kirchhoff-type function,f(x,t) has critical exponential growth behavior at infinity. Moreover, they also discuss the above problem with convex–concave type sign changing nonlinearity. All theses results are based on Trudinger-Moser inequality (see [34, 36]) and critical point theory. In the case of M =1 and 2m=N, it becomes to the following Polyharmonic problem

When m = 1, the above problem was investigated in [1, 13, 15]. In [22], Lam and Lu studied the above polyharmonic equation in both cases of f satisfying the well-known Ambrosetti-Rabinowitz condition and without the Ambrosetti-Rabinowitz condition. One can also refer to[24–25] for some relevant results.

where N ≥2, 0 ≤η < N, V : RN→R is a continuous function, f(x,s) is continuous in RN×R and behaves likeas |s| →∞. When η = 0, (1.5) is a usual elliptic equation with no singular term and it was first studied by Cao [7] for the case N =2. It was studied by Panda [35], do[16] and Alves [5] for the general dimensional cases. When 0<η

For the following perturbation problem

when η = 0, using Ekeland variational principle and mountain-pass theorem, doet al [17]got the multiplicity of solutions. When η > 0, the existence of nontrivial weak solution of the problem(1.6)was proved by Adimurthi and Yang[2]. In[37],Yang derived some similar results for the equations with bi-Laplacian operator in dimensional four, and proved the existence and multiplicity of weak solutions for N-Laplacian elliptic equations in his another paper (see[38]).Lam and Lu [23] considered the existence and multiplicity of nontrivial weak solution for nonuniformly N-Laplacian elliptic equations. Motivated by the work of Lions [29], doet al. [18]improved the Trudinger-Moser inequality in RNand obtained a ground state solution for the quasilinear elliptic equation

with the function f(x,s) satisfying the so called exponential critical growth condition at +∞,i.e., there exists α0>0 such that

where ?Nu = div(|?u|N?2?u), k > 0, 1 < p < N, V : RN→(0,∞) is a continuous function with positive lower bound and coercive, λ > 0 is a real parameter, Q(x) is a positive function inand f(t) satisfies exponential growth condition. Recently, Furtado and Zanata[20] studied the following Schrdinger-Kirchhofftype equation

where M(t)is a Kirchhoff-type function and V(x) may vanish on a set of positive measure and may take negative values in somewhere. The function A(x) is locally bounded and the function f(t) has critical exponential growth. Applying variational method, they got the existence of ground state solution. Moreover, in the local case M ≡1, they also got some relevant results.In [12], the existence and multiplicity of solutions were investigated for the elliptic systems involving Kirchhoffequations.

In this paper, we consider the following singular Kirchhoff-Schringer

where N ≥2, 0<η

(M1) for any d>0 there exists κ:=κ(d)>0 such that M(t)≥κ for all t ≥d;

(M2) for any t1, t2≥0, it holds

(M3) there exists θ >1 such thatis decreasing in (0,∞).

Remark 1.1A typical example of M(t) is given by M(t) = a+btθ?1for all t ≥0 and some θ >1, where a,b ≥0 and a+b>0.

Remark 1.2By (M3), we can obtain that θM(t)?M(t)t is nondecreasing for t > 0. In particular,

Since it is concerned with nonnegative weak solutions, we require that f(x,t) = 0 for all(x,t)∈RN×(?∞,0]. Furthermore, we assume the function f satisfying:

(f0) f is a continuous function and f(x,t)>0 for all t>0 and x ∈RN.

(f1) There exist positive constants t0and M0such that

(f2) There exist constants α0,c1,c2>0 such that for all (x,t)∈RN×R+,

where

(f3) There exists μ>θN such that

where x ∈RNand t ∈R+.

(f3) is the well known Ambrosetti-Rabinowitz condition (A-R condition, for short).

We also give the following conditions on the potential V(x):

(V1) V is a continuous function satisfying V(x)≥V0>0.

Define a function space

equipped with the norm

The condition (V1) implies that E is a reflexive Banach space. For any p ≥N, we define

and

The continuous embedding of EW1,N(RN)Lp(RN) (p ≥N) and Hlder inequality imply

Theorem 1.1Suppose V satisfies (V1) and f satisfies (f0)–(f6). Then problem (1.10)possesses a positive ground state solution.

Remark 1.3The main difficulty is how to obtain a strong convergence subsequence from a (PS) sequence and prove that the limit is a ground state solution of problem (1.10), which can be overcome by the concentration compactness principle and the singular Trudinger-Moser inequality.

Now instead the conditions (f1) and (f3), we assume that

Remark 1.4In Theorem 1.2, we study the ground state solution of Kirchhoff-Schrdinger equation without the A-R condition. For Schrdinger equation with exponential growth and singular term, the A-R condition was weakened in [31]. Instead of using the mountain-pass theorem of (PS) sequence, we use the mountain-pass theorem of Cerami sequence and obtain the boundedness of Cerami subsequence for the energy functional.

The rest of the paper is organized as follows. In Section 2, some preliminary results are introduced. In Section 3, we study the functionals related to (1.10). In Section 4, we give a proof of Theorem 1.1. In Section 5, we give a proof of Theorem 1.2, which is a key step to prove the boundedness of (C)csequence.

2 Preliminaries

In this section, we give some preliminaries for our use later.

Lemma 2.1(see [39, p.5]) Suppose q ≥N and 0 < η < N. Then E can be compactly embedded into Lq(RN,|x|?ηdx).

Notice that 0<η

Lemma 2.2(see[2,p.2397]) For all α>0, 0<η

Next, we claim that from the condition (f6) one can get the following condition.

Lemma 2.4If (f6) holds, then for all x ∈RN, we have that tf(x,t)?θNF(x,t) is strictly increasing in t>0.

ProofLet 0

On the other hand,

From (2.3)–(2.4), we derive that

This completes the proof.

3 Functionals and Compactness Analysis

We say that u ∈E is a positive weak solution of problem (1.10) if u>0 in RN, and for all φ ∈E,

Define the functional I :E →R by

for all u,φ ∈E. Hence a critical point of I defined in (3.1) is a weak solution of (1.10).

Next, we will check the geometry of the functional I.

Lemma 3.1Assume that (V1), (f2) and (f4) hold. Then there exist positive constants δ and r such that

ProofFrom (f4), there exist σ,ε>0, such that if |u|≤ε,

for all x ∈RN. On the other hand, using (f2) for each q >θN, we have

for |u|≥ε and x ∈RN. Combining the above estimates, we obtain

We now choose sufficiently small r >0 such that

So we derive that

This completes the proof.

Lemma 3.2If the condition (f3) is satisfied, then there exists e ∈E with ‖e‖E> r such that

where r is given in Lemma 3.1.

ProofFrom (1.11), we have M(t)≤M(1)tθ, t ≥1. Let u ∈E{0}, u ≥0 with compact support ? = supp(u) and ‖u‖E= 1. From (f3) or (1.12), for μ > θN, there exist C1,C2> 0 such that for all (x,t)∈?×R+,

Then for all t ≥1, there holds

Hence, I(tu)→?∞as t →∞. Setting e=tu with t sufficiently large,the proof of Lemma 3.2 is completed.

From Lemmas 3.1–3.2, we get a (PS)csequence {un}?E, i.e.,

where

and

Γ=:{γ ∈C([0,1]:E):γ(0)=0,γ(1)=e}.

Lemma 3.3Suppose that the condition(f5)is satisfied,then for the min–max level c defined in (3.4), there holds c

ProofFirstly,we claim the best constant Spdefined in(1.13)can be achieved at an element u0∈E. In fact, since

we can choose unsuch that

so unis bounded in E. From Lemma 2.1, there exists u0∈E such that up to a subsequence un?u0in E, un→u0in Lq(RN,|x|?ηdx)and un(x)→u0(x)almost everywhere in RN. This implies

The inequality c ≥δ can be easily proved as in the proof of Lemma 3.1. From the definition of c, take γ : [0,1] →E,γ(t) = te, where e = t0u0with I(t0u0) < 0 as explained above. We have γ ∈Γ and by using the condition (f5), we have

The proof of Lemma 3.3 is completed.

Lemma 3.4Suppose that the conditions (V1), (f1)–(f4) and (f6) are satisfied. Let {un}?E be an arbitrary (PS)csequence of I. Then there exist a subsequence of {un} (still denoted by {un}) and u ∈E such that

ProofLet {un}?E be an arbitrary (PS)csequence of I, i.e.,

We shall prove that the sequence {un} is bounded in E. Arguing by contradiction, suppose that {un} is unbounded in E. Then up to a subsequence, we have ‖un‖E→∞and d :=>0. Since μ>θN, combining (1.11) and (M1), we have

c+o(1)+o(1)‖un‖E

Divide the above inequality byand let n →∞, we have

which is impossible. Therefore {un} is bounded in E.

It then follows from (3.3) that

By [13, Lemma 2.1, p.143], we get

By (f1) and (f2), there exists C >0 such that

From Lemma 2.1 and the generalized Lebesgue’s dominated convergence theorem in [27, p.20],arguing as [39, Lemma 4.6, p.13], we derive that

This completes the proof of Lemma 3.4.

4 Proof of Theorem 1.1

Proof of Theorem 1.1By the process as in the proof of Lemma 3.4, we have that the (PS)csequence {un} is bounded in E. We claim that I(u) ≥0. Indeed, suppose by contradiction that I(u) < 0. Then u0, set r(t) := I(tu), t ≥0, we have r(0) = 0 and r(1)<0. As the proof of Lemma 3.1, for t>0 small enough, it holds r(t)>0. So there exists t0∈(0,1) such that

By Remark 1.1 and Lemma 2.4, we have

Furthermore, by the weak lower semi-continuity of the norm and Fatou’s lemma, we have

which is impossible. Thus the claim is true.

Thus

Choosing q >1 sufficiently close to 1 and β0>0 such that for large n,

By using concentration compactness principle(see[19,p.230],[39,p.3]),together with singular Trudinger-Moser inequality, it holds

From (f2) and Hlder inequality, there holds

Since I′(un)(un?u)→0, there holds

On the other hand, by un?u in E, we have

(4.4) minus (4.5) and applying with the next inequality

and

we can derive

Thus by using singular Trudinger-Moser inequality, it holds

here we have used Lemma 2.1 in the last estimate. From I′(un)un→0, we have

From the condition (M1), we can get ‖un‖ →0. Then I(un) →0, which contradicts the fact that I(un)→c>0, so u is nonzero.

Finally, we are ready to show the existence of positive ground state solution for (1.10).Setting

Let c be the mountain-pass level, obviously m ≤c.

On the other hand, for any u ∈Λ, there holds u>0. In fact, denote u?:=min{u,0},from I′(u)u?=0, we get ‖u?‖=0. Since f is nonnegative, by applying the Harnack inequality, we have u > 0 in RN. Define h : (0,+∞) →R by h(t) = I(tu). We have that h is differentiable and

From I′(u)u=0, we get

so

By (M3), (f6) and u > 0, we conclude that h′(t) > 0 for 0 < t < 1 and h′(t) < 0 for t > 1.From h′(1)=0, we have

From the above argument, we see that h′(t) < 0 is strongly decreasing in t ∈(1,+∞), so h(t) →?∞as t →+∞. Now, define γ : [0,1] →E, γ(t) = tt0u, where t0is a real number which satisfies I(t0u)<0, we have γ ∈Γ, and therefore

Since u ∈Λ is arbitrary, we have c ≤m, thus c=m. This ends the proof of Theorem 1.1.

5 The Ground State Solution Without the A-R Condition

In this section, insteading the conditions (f1) and (f3), we assume following condition on the function f.

We will use a Cerami’s mountain pass theorem which was introduced in[9–10]. For readers’convenience, we give a brief introduction here.

Definition ALet (E,‖·‖E) be a real Banach space with its dual space (E?,‖·‖E?). Suppose I ∈C1(E,R). For c ∈R, we call {un}?E a (C)csequence of the functional I, if

Proposition ALet(E,‖·‖E)be a real Banach space. I ∈C1(E,R), I(0)=0 and satisfies:

(i) There exist positive constants δ and r such that

and

(ii) there exists e ∈E with ‖e‖E>r such that

Define c by

where

Then I possesses a (C)csequence.

In order to prove Theorem 1.2, along the line of the proof of Theorem 1.1, we get into two steps. Firstly, we check the mountain pass geometry of the functional I under the weak condition. Secondly, it is the key step to establish that any (C)csequence is bounded.

Lemma 5.1Assume that (V1), (f2), (, (f4) hold. Then

(i) there exist positive constants δ and r such that

(ii) there exists e ∈E with ‖e‖E>r such that

ProofThe proof of (i) is similar as that in Lemma 3.1. From (M3), we have M(t) ≤M(1)tθ, t ≥1. Let u ∈E {0}, u ≥0 with compact support ? = supp(u), by (), for any L>0, there exists d>0 such that for all (x,s)∈?×R+,

Then

From Lemmas 3.1, 5.1 and Proposition A, we get a (C)csequence {un}?E, i.e.,

Lemma 5.2Suppose that the conditions (f2), (f5) and (f6) are satisfied. Let {un}?E be an arbitrary (C)csequence of I. Then {un} is bounded up to a subsequence.

ProofLet {un}?E be a (C)csequence of I, i.e.,

and

where τn→0 as n →∞. We shall prove that the sequence {un} is bounded in E. Indeed,suppose by contradiction that

and set

Combining with Lemma 2.1 and the similar argument as in [30, Lemma 5, p.12], we can get?0 in E.

Let tn∈[0,1] be such that

In the following argument we will takeand so we have ε →0.

By condition (f2), there exists C >0 such that

By using Young inequality, for=1, p,q >1, there holds

So we have

Since I(0)=0 and I(un)→c,we can assume tn∈(0,1),and so I′(tnun)tnun=0,it follows from Lemma 2.4 and Remark 1.1,

which is a contradiction to (5.5). This proves that {un} is bounded in E.

Proof of Theorem 1.2From Lemma 5.2, we have that the (C)csequence {un} of I is bounded in E. Applying the same procedure as in proof of Theorem 1.1, we derive that I′(u) = 0 and I(u) = c. Moreover, we also get that u is positive and u is a ground state solution of (1.10).