We deal with the existence of Nehari-type ground state positive solutions for the nonlinear Schrödinger equation -Δu+Vxu=fx,u,x∈RN,u∈H1RN. Under a weaker Nehari condition, we establish some existence criteria to guarantee that the above problem has Nehari-type ground state solutions by using a more direct method in two cases: the periodic case and the asymptotically periodic case.

1. Introduction

Consider the following semilinear Schrödinger equation:
(1)-Δu+V(x)u=f(x,u),x∈RN,u∈H1(RN),
where V:RN→R and f:RN×R→R.

The Schrödinger equation has found a great deal of interest last years because not only it is important in applications but also it provides a good model for developing mathematical methods. Many authors have studied the existence of entire solutions of Schrödinger equations under various stipulations (cf., e.g., [1–28] and the references quoted in them).

When infRNV(x)>0 and V(x) is periodic, Li et al. [12] made use of a combination of the techniques in [13, 14] with applications of Lions’ concentration compactness principle [26, 29, 30] to establish the following theorem.

Theorem 1 (see [<xref ref-type="bibr" rid="B13">12</xref>]).

Assume that V and f satisfy the following assumptions:

V∈C(RN,R) and infRNV(x)>0;

V(x) is 1-periodic in each of x1,x2,…,xN;

f∈C1, ft′ is a Caratheodory function, and there exists a constant C>0 such that
(2)|ft′(x,t)|≤C(1+|t|2*-2),lim|t|→∞|ft′(x,t)||t|2*-2=0;

f(x,t) is 1-periodic in each of x1,x2,…,xN;

f(x,t)=o(|t|), as |t|→0, uniformly in x∈RN;

lim|t|→∞(|F(x,t)|/|t|2)=∞, uniformly in x∈RN;

f(x,t)/|t| is strictly increasing in t on R∖{0} for every x∈RN.

Then problem (1) has a solution u0∈H1(RN) such that Φ(u0)=infNΦ>0, where
(3)Φ(u)=12∫RN(|∇u|2+V(x)u2)
d
x-∫RNF(x,u)
d
x,∀u∈H1(RN),(4)〈Φ′(u),v〉=∫RN(∇u·∇v+V(x)uv)
d
x-∫RNf(x,u)v
d
x,∀u,v∈H1(RN)
with F(x,t)≔∫0tf(x,s)
d
s, and
(5)N={u∈H1(RN):〈Φ′(u),u〉=0,u≠0}.

The set N is the Nehari manifold, which contains infinitely many elements of H1(RN). In fact, for any u∈H1(RN)∖{0}, there exists t=t(u)>0 such that tu∈N; see Lemma 9. Since u0 is a solution at which Φ has least “energy” in set N, we will call it a Nehari-type ground state solution.

We must point out that “the least energy solution” (which is sometimes also called the ground state solution in some references) is in fact a nontrivial solution u0 which satisfies Φ(u0)=infMΦ, where
(6)M={u∈H1(RN)∖{0}:Φ′(u)=0}
is a very small subset of N; it may contain only one element. In general, it is much more difficult to find a solution u0 for problem (1) with a constraint condition Φ(u0)=infNΦ than with one Φ(u0)=infMΦ.

To establish the existence of Nehari-type ground state solutions, the so-called Nehari-type condition (S4) seems to be always necessary in the proof of the existence of ground states solutions for problem (1).

In recent paper [15, 20], Theorem 1 has been extended to the case where 0 is in the gap of the spectrum σ(-Δ+V), but an additional assumption on the nonlinearity f is assumed in [15].

Motivated by [12, 17, 20], in the present paper, we will develop a more direct method to generalize Theorem 1 by relaxing assumptions (V0), (S3), and (S4) in two cases: the periodic case and the asymptotically periodic case.

In the periodic case, we establish the following two theorems.

Theorem 2.

Assume that V and f satisfy (V0), (V1), (S1), (S2), and the following assumptions:

f∈C(RN×R,R), and there exists a constant C>0 such that
(7)|f(x,t)|≤C(1+|t|2*-1),lim|t|→∞|f(x,t)||t|2*-1=0,uniformlyinx∈RN;

lim|t|→∞(|F(x,t)|/|t|2)=∞, a.e.x∈RN;

f(x,t)/|t| is nondecreasing in t on R∖{0} for every x∈RN.

Then problem (1) has a solution u0∈H1(RN) such that Φ(u0)=infNΦ>0.Theorem 3.

Assume that V and f satisfy (V0), (V1), (S0′), (S1), (S2), (S3′), and (S4′). Then problem (1) has a positive solution u¯∈H1(RN) such that Φ+(u¯)=infN+Φ+>0, where
(8)Φ+(u)=12∫RN(|∇u|2+V(x)u2)
d
x-∫RNF(x,u+)
d
x,∀u∈H1(RN)
with u+(x)=max{u(x),0} and u-(x)=max{-u(x),0}, and
(9)N+={u∈H1(RN):〈Φ+′(u),u〉=0,u≠0}.

In the asymptotically periodic case, V(x) is allowed to be negative in some bounded domain in RN. Precisely, we use the following condition instead of (V0).

V∈C(RN)∩L∞(RN), V(x)≤V¯∈(0,∞) for all x∈RN and there exists a constant ζ0>0 such that
(10)∫RN(|∇u|2+V(x)u2)dx≥ζ0∫RN[V¯-V(x)]|u|2dx,∀u∈H1(RN).

Was first introduced by Deng et al. [6]; it is satisfied if the following assumption holds (see Lemma 7):

V∈C(RN)∩L∞(RN), and there exist two constants a1,a2>0 and a bounded measurable set Ω⊂RN such that (a1+a2)|Ω|2/N≤S, and
(11)V(x)≥{-a1,x∈Ω,a2,x∉Ω,

where S=infu∈D1,2(RN),∥u∥2*=1∥∇u∥22.

In this case, we establish the following two theorems.

Theorem 4.

Assume that V and f satisfy (V0′), (S0′), (S2), (S3), (S4′), and the following assumptions:

V(x)=V0(x)+V1(x), V0∈C(RN,(0,+∞)), V0(x) is 1-periodic in each of x1,x2,…,xN, V1(x)<0 for |x|<1+N, and V1(x)≤0 for |x|≥1+N, lim|x|→∞|V1(x)|=0;

f(x,t)=f0(x,t)+f1(x,t); f0∈C(RN×R,R) satisfies (S1), (S2), and (S4′); f1∈C(RN×R,R) satisfies that
(12)0≤tf1(x,t)≤a(x)(t2+|t|p0),∀(x,t)∈RN×R,

where p0∈(2,2*) and a∈C(RN,R+) with lim|x|→∞a(x)=0.

Then problem (1) has a solution u0∈H1(RN) such that Φ(u0)=infNΦ>0.Theorem 5.

Assume that V and f satisfy (V0′), (V2), (S0′), (S1′), (S2), (S3), and (S4′). Then problem (1) has a positive solution u¯∈H1(RN) such that Φ+(u¯)=infN+Φ+>0.

In our theorems, we give a new condition (S4′) which weakens Nehari-type condition (S3) considerably.

2. PreliminariesLemma 6 (see [<xref ref-type="bibr" rid="B7">6</xref>, Lemma 2.3]).

Assume that V satisfies (V0′). Then there exist two positive constants C1,C2>0 such that
(13)C1∥u∥H1(RN)2≤∫RN[|∇u|2+V(x)u2]
d
x≤C2∥u∥H1(RN)2,∀u∈H1(RN),
where ∥u∥H1(RN) is the usual norm of H1(RN)(14)∥u∥H1(RN)=[∫RN(|∇u|2+u2)
d
x]1/2,u∈H1(RN).

By Lemma 6, we define an inner product
(15)(u,v)=∫RN[(∇u·∇v)+V(x)uv]dx,u,v∈H1(RN),
associated with the norm
(16)∥u∥={∫RN[|∇u|2+V(x)u2]dx}1/2,u∈H1(RN).
Then H1(RN) is a Hilbert space with this inner product. Moreover, under assumptions (V0′) and (S0′), the functional Φ defined by (3) is of class C1(H1(RN),R).

Lemma 7.

If (V0′′) holds, then (V0′) does.

Proof.

By virtue of (V0′′), the Hölder inequality, and the Sobolev inequality, we have
(17)∫RN[|∇u|2+V(x)u2]dx≥∫RN|∇u|2dx-(a1+a2)∫Ωu2dx+a2∫RNu2dx≥∫RN|∇u|2dx-(a1+a2)|Ω|2/N∥u∥2*2+a2∫RNu2dx≥∥∇u∥22-(a1+a2)|Ω|2/NS-1∥∇u∥22+a2∫RNu2dx≥a2∫RNu2dx≥a2V¯+a1∫RN[V¯-V(x)]u2dx,∀u∈H1(RN).
This shows that (V0′) holds with ζ0=a2/(V¯+a1).

Lemma 8.

Let X be a Banach space. Let M0 be a closed subspace of the metric space M and Γ0⊂C(M0,X). Define
(18)Γ={γ∈C(M,X):γ|M0∈Γ0}.
If Ψ∈C1(X,R) satisfies
(19)∞>c≔infγ∈Γsupt∈MΨ(γ(t))>a≔supγ0∈Γ0supt∈M0Ψ(γ0(t)),
then there exists a sequence {un}⊂X satisfying
(20)Ψ(un)⟶c,∥Ψ′(un)∥(1+∥un∥)⟶0.

Proof.

For any γ∈Γ, define set Kγ={γ(t):t∈M} in X and the collection K={Kγ:γ∈Γ}. Let A={γ0(t):γ0∈Γ0,t∈M0},
(21)Λ={φ∈C(X,X):φ-1∈C(X,X),bothφandφ-1areboundedonboundedsets(X,X):φ-1},Λ(A)={φ∈Λ:φ(u)=u,u∈A}.
For any γ∈Γ and φ∈Λ(A), let γ0=γ|M0 and γ~(t)=φ(γ(t)),t∈M. Then γ0∈Γ0 and γ~∈C(M,X). Hence,
(22)γ~(t)=φ(γ0(t))=γ0(t),∀t∈M0;
that is, γ~|M0=γ0∈Γ0. Therefore,
(23)φ(K)∈K,∀φ∈Λ(A),K∈K.
These show that the collection K is a minimax system for A. Since (19) implies
(24)∞>c≔infK∈KsupKΨ>a≔supAΨ,
it follows from Theorem 2.4 in [18] that the result is true.

Lemma 9.

Under assumptions (V0′), (S0′), (S2), and (S3′), for any u∈H1(RN)∖{0}, there exists t(u)>0 such that t(u)u∈N.

Proof.

Let u∈H1(RN)∖{0} be fixed and define the function g(t)≔Φ(tu) on [0,∞). Clearly we have
(25)g′(t)=0⟺tu∈N⟺∥u∥2=1t∫RNf(x,tu)udx.
It is easy to verify, using (S2) and (S3′), that g(0)=0, g(t)>0 for t>0 small and g(t)<0 for t large. Therefore maxt∈[0,∞)g(t) is achieved at a t0=t(u) so that g′(t0)=0 and t(u)u∈N.

Lemma 10.

Under assumptions (V0′), (S0′), and (S4′),
(26)Φ(u)≥Φ(tu)+1-t22〈Φ′(u),u〉,∀u∈H1(RN),t≥0.

Proof.

For τ≠0, (S4′) yields
(27)f(x,s)≤f(x,τ)|τ||s|,s≤τ;f(x,s)≥f(x,τ)|τ||s|,s≥τ.
It follows that
(28)1-t22τf(x,τ)≥∫tττf(x,s)
d
s,t≥0.
Note that
(29)〈Φ′(u),u〉=∥u∥2-∫RNf(x,u)udx.
Thus, by (3), (28), and (29), one has
(30)Φ(u)-Φ(tu)=1-t22∥u∥2+∫RN[F(x,tu)-F(x,u)]dx=1-t22〈Φ′(u),u〉+∫RN[1-t22f(x,u)u+F(x,tu)-F(x,u)]dx=1-t22〈Φ′(u),u〉+∫RN[1-t22f(x,u)u-∫tuuf(x,s)
d
s]dx≥1-t22〈Φ′(u),u〉,t≥0.
This shows that (26) holds.

Corollary 11.

Under assumptions (V0′), (S0′), and (S4′), for u∈N,
(31)Φ(u)≥Φ(tu),∀t≥0.

We define
(32)c1≔infNΦ,c2≔infu∈E,u≠0maxt≥0Φ(tu),c≔infγ∈Γsupt∈[0,1]Φ(γ(t)),
where
(33)Γ={γ∈C([0,1],E):γ(0)=0,Φ(γ(1))<0}.

Lemma 12.

Under assumptions (V0′), (S0′), (S2), (S3′), and (S4′), one has that c1=c2=c>0 and there exists a sequence {un}⊂H1(RN) satisfying
(34)Φ(un)⟶c,∥Φ′(un)∥(1+∥un∥)⟶0.

Proof.

(1) Both Lemma 9 and Corollary 11 imply that c1=c2. Next, we prove that c=c1=c2. By the definition of c2, we choose a sequence {vn}⊂E∖{0} such that
(35)c2≤maxt≥0Φ(tvn)<c2+1n,n∈N.
Since Φ(tu)<0 for u∈E∖{0} and t large, there exist tn=t(vn)>0 and sn>tn such that
(36)Φ(tnvn)=maxt≥0Φ(tvn),Φ(snvn)<0,n∈N.
Let γn(t)=tsnvn for t∈[0,1]. Then γn∈Γ, and it follows from (35) and (36) that
(37)supt∈[0,1]Φ(γn(t))=maxt≥0Φ(tvn)<c2+1n,n∈N,
which implies that c≤c2. On the other hand, the manifold N separates H1(RN) into two components E+={u∈H1(RN):〈Φ′(u),u〉>0}∪{0} and E-={u∈H1(RN):〈Φ′(u),u〉<0}. By (S4′), one has
(38)f(x,t)t≥2F(x,t),∀(x,t)∈RN×R.
It follows that Φ(u)≥0 for u∈E+. By (S0′) and (S2), E+ contains a small ball around the origin. Thus every γ∈Γ has to cross N, because γ(0)∈E+ and γ(1)∈E-, and so c1≤c.

(2) In order to prove the second part of Lemma 12, we apply Lemma 8 with M=[0,1], M0={0,1}, and
(39)Γ0={γ0:{0,1}⟶H1(RN):γ0(0)=0,Φ(γ0(1))<0}.
By (S0′) and (S2), there exists r>0 such that
(40)min∥u∥≤rΦ(u)=0,inf∥u∥=rΦ(u)>0.
Hence we obtain
(41)c≥inf∥u∥=rΦ(u)>0=supγ0∈Γ0supt∈M0Φ(γ0(t)).
These show that all assumptions of Lemma 8 are satisfied. Therefore there exists a sequence (un)⊂H1(RN) satisfying (34).

Lemma 13.

Under assumptions (V0′), (S0′), (S2), (S3), and (S4′), any sequence {un}⊂H1(RN) satisfying (34) is bounded in H1(RN).

Proof.

To prove the boundedness of {un}, arguing by contradiction, suppose that ∥un∥→∞. Let vn=un/∥un∥. Then ∥vn∥=1. By Sobolev embedding theorem, there exists a constant C3>0 such that
(42)∥vn∥2+∥vn∥2*≤C3.
Passing to a subsequence, we may assume that vn⇀v in H1(RN), vn→v in Llocs(RN), 2≤s<2*, and vn→v a.e. on RN.

If δ≔limsupn→∞supy∈RN∫B1(y)|vn|2dx=0, then by Lions’ concentration compactness principle [23, Lemma 1.21], vn→0 in Ls(RN) for 2<s<2*. Fix p∈(2,2*) and R>2c. By (S0′) and (S2), for ɛ=c/4[(RC3)2+(RC3)2*]>0 there exists Cɛ>0 such that
(43)|F(x,u)|≤ɛ(|u|2+|u|2*)+Cɛ|u|p.
It follows that
(44)limsupn→∞∫RNF(x,Rvn)dx≤ɛ[(RC3)2+(RC3)2*]+RpCɛlimn→∞∥vn∥pp=c4.
Hence, by using (34), (44), and Lemma 10, one has
(45)c+o(1)=Φ(un)≥Φ(Rvn)+(12-R22∥un∥2)〈Φ′(un),un〉=R22-∫RNF(x,Rvn)dx+(12-R22∥un∥2)〈Φ′(un),un〉≥R22-c4+o(1)>7c4+o(1),
which is a contradiction. Thus, δ>0.

Going if necessary to a subsequence, we may assume the existence of kn∈ZN such that ∫B1+N(kn)|vn|2dx>δ/2. Let wn(x)=vn(x+kn). Then it follows that
(46)∫B1+N(0)|wn|2dx>δ2.
Now we define u~n(x)=un(x+kn), then u~n/∥un∥=wn, and ∥wn∥H1(RN)=∥un∥H1(RN)/∥un∥≤C4 for some C4>0. Passing to a subsequence, we have wn⇀w in H1(RN), wn→w in Llocs(RN), 2≤s<2*, and wn→w a.e. on RN. Thus, (46) implies that w≠0. Hence, it follows from (34), (S3), and Fatou’s lemma that
(47)0=limn→∞c+o(1)∥un∥2=limn→∞Φ(un)∥un∥2=limn→∞[12-∫RNF(x+kn,u~n)u~n2wn2dx]≤12-liminfn→∞∫RNF(x+kn,u~n)u~n2wn2dx≤12-∫RNliminfn→∞F(x+kn,u~n)u~n2wn2dx=-∞,
which is a contradiction. Thus {un} is bounded in H1(RN).

Remark 14.

In the proof of Lemma 13, (S3) is used only in (47). Hence, it can be weakened to (S3′) if f(x,t) is 1-periodic in each of x1,x2,…,xN.

3. The Proofs of TheoremsProof of Theorem <xref ref-type="statement" rid="thm1.2">2</xref>.

Lemma 12 implies the existence of a sequence {un}⊂H1(RN) satisfying (34), by a standard argument; we can prove Theorem 2.

Proof of Theorem <xref ref-type="statement" rid="thm1.3">3</xref>.

In view of the proofs of Lemmas 12 and 13, we can show that the conclusions of Lemmas 12 and 13 still hold if Φ and N are replaced by Φ+ and N+, respectively. Hence, there exists a bounded sequence {un}⊂H1(RN) satisfying
(48)Φ+(un)⟶c+,∥Φ+′(un)∥(1+∥un∥)⟶0,
where c+=infu∈N+Φ+(u). The rest of the proof is standard, so we omit it.

To prove Theorems 4 and 5, we define functional Φ0 and Φ+,0 as follows:
(49)Φ0(u)=12∫RN(|∇u|2+V0(x)u2)dx-∫RNF0(x,u)dx,u∈H1(RN),(50)Φ+,0(u)=12∫RN(|∇u|2+V0(x)u2)dx-∫RNF0(x,u+)dx,u∈H1(RN),
where F0(x,t)≔∫0tf0(x,s)ds. Then (V2), (S0′), and (S1′) imply that Φ0∈C1(H1(RN),R) and
(51)〈Φ0′(u),v〉=∫RN(∇u∇v+V0(x)uv)dx-∫RNf0(x,u)vdx,u,v∈H1(RN).

Proof of Theorem <xref ref-type="statement" rid="thm1.4">4</xref>.

Lemma 12 implies the existence of a sequence {un}⊂H1(RN) satisfying (34). By Lemma 13, {un} is bounded in H1(RN). Passing to a subsequence, we have un⇀u0 in H1(RN). Next, we prove u0≠0.

Arguing by contradiction, suppose that u0=0; that is, un⇀0 in H1(RN), and so un→0 in Llocs(RN), 2≤s<2*, and un→0 a.e. on RN. For any ɛ>0, it follows from (V2) that there exists Rɛ>0 such that |V1(x)|≤ɛ for |x|≥Rɛ. Hence,
(52)∫RN|V1(x)|un2dx=∫BRɛ(0)|V1(x)|un2dx+∫RN∖BRɛ(0)|V1(x)|un2dx≤supRN|V1(x)|∫BRɛ(0)un2dx+ɛ∫RN∖BRɛ(0)un2dx≤o(1)+ɛ∥un∥22≤o(1)+C3ɛ.
Since ɛ>0 is arbitrary, we have
(53)limn→∞∫RNV1(x)un2dx=0.
Similarly, by (S1′), one has
(54)limn→∞∫RNF1(x,un)dx=0,limn→∞∫RNf1(x,un)undx=0.
Note that
(55)Φ0(u)=Φ(u)-12∫RNV1(x)u2dx+∫RNF1(x,u)dx,∀u∈H1(RN),〈Φ0′(u),v〉=〈Φ′(u),v〉-∫RNV1(x)uvdx+∫RNf1(x,u)vdx,∀u,v∈H1(RN).
From (34) and (53)–(55), one has
(56)Φ0(un)⟶c,∥Φ0′(un)∥(1+∥un∥)⟶0.

By a standard argument, we may prove that there exists kn∈ZN, going if necessary to a subsequence, such that
(57)∫B1+N(kn)|un|2dx>δ2>0.
Let vn(x)=un(x+kn). Then ∥vn∥H1(RN)=∥un∥H1(RN), and
(58)∫B1+N(0)|vn|2dx>δ2.
Since V0(x) and f0(x,u) are periodic, we have
(59)Φ0(vn)⟶c,∥Φ0′(vn)∥(1+∥vn∥)⟶0.
Since {vn} is bounded in H1(RN), passing to a subsequence, we have vn⇀v¯ in H1(RN), vn→v¯ in Llocs(RN), 2≤s<2*, and vn→v¯ a.e. on RN. Obviously, (58) implies that v¯(x)≢0 for x∈B1+N(0). By a standard argument, we can prove that Φ0′(v¯)=0 and Φ0(v¯)≤c by using (59).

Since v¯≠0, it follows from Lemma 9 that there exists t0=t(v¯) such that t0v¯∈N, and so Φ(t0v¯)≥c. On the other hand, from (49), (51), (V2), (S1′), and (S4′), we have
(60)c≥Φ0(v¯)=Φ0(t0v¯)+∫RN[1-t022f0(x,v¯)v¯+F0(x,t0v¯)-F0(x,v¯)]dx≥Φ0(t0v¯)=Φ(t0v¯)-t022∫RNV1(x)v¯2dx+∫RNF1(x,t0v¯)dx>Φ(t0v¯)≥c.
This contradiction implies that u0≠0. By a standard argument, we can prove that Φ′(u0)=0 and Φ(u0)=c=infNΦ. This shows that u0∈H1(RN) is a solution for problem (1) with Φ(u0)=infNΦ>0.

Proof of Theorem <xref ref-type="statement" rid="thm1.5">5</xref>.

Similar to the proof of Theorem 3, there exists a bounded sequence {un}⊂H1(RN) satisfying (48). Passing to an appropriate subsequence, we have that un⇀u¯ in H1(RN). Next, we prove u¯≠0.

Arguing by contradiction, suppose that u¯=0; that is, un⇀0 in H1(RN). Then, un→0 in Llocs(RN), 2≤s<2*, and un→0 a.e. on RN. Analogous to the proof of Theorem 4, we can demonstrate that there exists a v¯∈H1(RN)∖{0} with v¯(x)≢0 for x∈B1+N(0) such that Φ+,0′(v¯)=0 and Φ+,0(v¯)≤c. By a standard argument, we can show that v¯≥0.

Since v¯≥(≢)0, it follows from Lemma 9 that there exists t0=t(v¯) such that t0v¯∈N+, and so Φ+(t0v¯)≥c. On the other hand, from (49), (51), (V2), (S1′), and (S4′), we have
(61)c≥Φ+,0(v¯)=Φ+,0(t0v¯)+∫RN[1-t022f0(x,v¯)v¯+F0(x,t0v¯)-F0(x,v¯)1-t022]dx≥Φ+,0(t0v¯)=Φ+(t0v¯)-t022∫RNV1(x)v¯2dx+∫RNF1(x,t0v¯)dx>Φ+(t0v¯)≥c.
This contradiction shows that u¯≠0. In the same way as the last part of the proof of Theorem 1, we can deduce that Φ+′(u¯)=0 and Φ+(u¯)=c=infN+Φ+. By a standard argument, we can demonstrate that u¯≥0. Therefore, u¯∈H1(RN) is a positive solution for problem (1) with Φ+(u¯)=infN+Φ+>0.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgment

This work is partially supported by the NNSF (no. 11171351) of China.

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