# A Nonlinear Schrödinger Equation Resonating at an Essential Spectrum

Authors: Shaowei Chen; Haijun Zhou
Journal: Advances in Mathematical Physics
 (2016)

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## Abstract

We consider the nonlinear Schrödinger equation-Δu+f(u)=V(x)uinRN. The potential function V satisfies that the essential spectrum of the Schrödinger operator -Δ-V is [0,+∞) and this Schrödinger operator has infinitely many negative eigenvalues accumulating at zero. The nonlinearity f satisfies the resonance type condition limt→∞f(t)/t=0. Under some additional conditions on V and f, we prove that this equation has infinitely many solutions.


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## Body

## 1. Introduction and Statement of Results

In this paper, the following nonlinear Schrödinger problem inRN (N≥3) is considered: (1)-Δu=Vxu+guinRN.The nonlinearity g satisfies the resonance type condition: (2)limt→∞⁡gtt∈σessT.Here, σess(T) denotes the essential spectrum (see [1, Chapter 7] or [2, Chapter 7.4]) of the Schrödinger operator T defined as follows: (3)T:L2RN⟶L2RN,u⟼-Δu-Vu,with  domainDT≔u∈L2RN∣Tu∈L2RN.Equation (1) arises in quantum mechanics and is related to the study of the nonlinear Schrödinger equation for a particle in an electromagnetic field and it has attracted considerable attention from researchers in recent decades. One can see [3–6] and the references therein. However, there are very few results on (1) with the resonance type condition (2).The following nonlinear elliptic problem with resonance type conditions in bounded domain has been studied by many authors and numerous existence and multiplicity results have been obtained in the past forty years:(4)-Δu=hx,uinΩ,u=0on∂Ω.Here, Ω is a bounded domain in RN, and g satisfies the resonance type condition: (5)limt→∞⁡hx,tt=νk.νk is the kth eigenvalue of -Δ with 0-Dirichlet boundary condition on ∂Ω. We refer to [7–10] and references therein for more detailed discussions of some historical results. This problem has deeply inspired developments in critical point theory in the last forty years, such as the Landesman-Lazer type conditions (e.g., [8, 10]), Morse theory (e.g., [9]), and the variational reduction method (e.g., [11]). The difficulty of this problem lies in the proof of the boundedness of the Palais-Smale sequence or the (C)c sequence (see Definition 5 in Section 2) of its corresponding functional.The resonant Schrödinger problem (1) is much more difficult than the bounded domain case and it has fewer studies. Unlike the case of bounded domains for which the linear operators -Δ in bounded domain are compact, there are continuous spectra of the linear operator T. Moreover, the proofs of boundedness and compactness of the (C)c sequence of the corresponding functional of (4) are greatly different from and more difficult than the case of bounded domain (see [4, 5, 12]).More precisely, this paper considers the nonlinear Schrödinger equation:(6)-Δu+fu=Vxu,u∈H1RN.Here, N≥3 and H1(RN) is the Sobolev space: (7)H1RN≔u:RN⟶R∣∫RN∇u2dx+∫RNu2dx<+∞.ForV:RN→R, the following is assumed:(V1)
ConsiderV∈C1(RN∖{0}), V≥0 in RN∖{0}, limx→∞V(x)=0, V∈Llocs(RN) with s=max⁡{2,N/2} if N≥3 and N≠4, V∈Llocs(RN) for some s>2 if N=4, and there exist R>0, 0<β<2, and C>0 such that, for x≥R,(8)Vx≥Cx-β.(V2)
For anyx∈RN∖{0}, x·∇V(x)≤0 and there exists 0<γ<2 such that (9)γVx+x·∇Vx≥0,∀x∈RN∖0.Forf, the following is assumed:(f1)
f:R→R is a continuous function:(10)a0≔limt→0⁡ftt>0,(11)limt→∞⁡ftt=0,(12)tft≥0,∀t∈R.(f2)
There existsD>0 such that, for any t∈R,(13)Ft≤Dtft.Here,(14)Ft=∫0tfsds,t∈R.Fort∈R, let(15)gt=a0t-ft,(16)F~t=Ft-12tft.(f3)
For anyt∈R, F~(t)≥0 and there exists δ∈(0,a0) such that if gt≥a0-δt, then(17)F~t≥δ.(f4)
For anyt∈R, f(-t)=-f(t).The main result of this paper is as follows.Theorem 1.
Suppose thatV1-(V2) and (f1)–(f4) are satisfied; then, problem (6) has infinitely many solutions.Remark 2.
(i) Under the assumptions(V1) and (V2), the essential spectrum of T defined by (3) is [0,+∞) (see Lemma 3 in Section 2). Therefore, condition (11) indicates that f satisfies a resonance type condition at the essential spectrum of T.
(ii) A typical example forV which satisfies V1-(V2) is (18)Vr,θ=Pθrβ.Here, r=x, θ=x/x∈SN-1≔{y∈RN∣y=1}, 0<β<min⁡{2,N/2}, P∈C1(SN-1), and P>0 in SN-1.
(iii) There are many functions satisfying(f1)–(f4). Let G(t)=∫0tg(s)ds. Using (19)ddtGtt2=2F~tt3,we can construct a function f satisfying (f1)–(f4) from a given F~. For example, let (20)F~t=σt31+t5/2,t∈R,where (21)σ=a04τ,τ=∫0+∞ds1+s5/2.Then, by (19), for t≥0, (22)Gt=2σt2∫0tds1+s5/2and, for t<0, G(t)=G(-t). It follows that (23)gt=4σt∫0tds1+s5/2+2σt21+t5/2ift≥0,gt=-g-tift<0,and f(t)=a0t-g(t) for any t∈R. Because the even function g(t)/t is strictly increasing in (0,+∞), g(t)/t→a0 as t→+∞, it can be deduced that f satisfies tf(t)>0, ∀t∈R∖{0}. Moreover, it is easy to verify that f(t)=a0t+o(t) as t→0, f(t)=(2/3)σt-1/2+o(t-1/2) as t→∞, F(t)=(1/2)a0t2+o(t2) as t→0, and F(t)=(4/3)σt1/2+o(t1/2) as t→∞. Because the even function g(t)/t is strictly increasing in (0,+∞), g(t)/t→a0 as t→+∞, and F~(t)→+∞ as t→+∞, it can be deduced that, for sufficiently small δ>0, if gt/t≥a0-δ, then F~(t)≥δ. Therefore, f satisfies (f1)–(f4).Let(24)X=H1RN.The inner product and the norm in X are (25)u,v=∫RN∇u∇v+a0uvdx,u=u,u1/2,respectively. Here, a0 is the constant in (10).Let(26)Φu=12∫RN∇u2dx-12∫RNVxu2dx+∫RNFudx,u∈X.Then, for any u,v∈X, the Gateaux derivative of Φ is as follows: (27)Φ′u,v=∫RN∇u∇vdx-∫RNVxuvdx+∫RNfuvdx.It follows that u∈X is a solution of (6) if and only if u is a critical point of Φ. In Section 2, it is shown that Φ satisfies the Cerami condition. In Section 3, an abstract critical point theorem from [13] is used to show that Φ has infinitely many critical points.Notation. Let E be Banach space. We denote the dual space of E by E′ and denote strong and weak convergence in E by → and ⇀, respectively. For φ∈C1(E,R), we denote the Fréchet derivative of φ at u by φ′(u). The Gateaux derivative of φ is denoted by 〈φ′(u),v〉,∀u,v∈E. Lp(RN) denotes the standard Lp space (1≤p≤∞). We use O(h) and o(h) to mean Oh≤Ch and o(h)/h→0, respectively.


## 2. The Cerami Condition forΦ

Lemma 3.
IfV satisfies the assumptions (V1), then the essential spectrum of the Schrödinger operator T defined by (3) is [0,+∞) and T has infinitely many negative eigenvalues accumulating at zero.Proof.
BecauseV satisfies (V1), by [14, Theorem XIII.15(b)] or [2, Theorem 10.29(b)], the essential spectrum of the Schrödinger operator T defined by (3) is [0,+∞). The second result of this lemma comes from [2, Theorem 10.31] or [14, Theorem XIII.6(a)].Lemma 4.
IfV satisfies the assumptions (V1) and (V2), then T has no eigenvalue in [0,+∞).Proof.
BecauseV satisfies the assumptions (V1) and (V2), by [15, Corollary 2], to prove this lemma, it suffices to prove that V satisfies the conditions I and II in [15, Corollary 2]. For this, it suffices to prove that, for 0<α<2, the functions (28)Mx=∫x-y≤1Vy·x-y4-N-αdy,Lx=∫x-y≤1y∇Vy·x-y4-N-αdyare bounded functions in RN. Because limx→∞V(x)=0 and 0<α<2, there exists R′>0 such that, for x>R′, (29)Mx≤∫x-y≤1x-y4-N-α=∫y≤1y4-N-αdy<+∞.Because V∈LlocN/2(RN), for x≤R′, the following is true: (30)Mx≤∫x-y≤1VyN/2dy2/N∫x-y≤1x-yN4-N-α/N-2N-2/N=∫y≤R′+1VyN/2dy2/N∫y≤1yN4-N-α/N-2N-2/N.Because 0<α<2, the following holds: (31)∫y≤1yN4-N-α/N-2<+∞.Combining (29) and (30) yields that M is a bounded function in RN. From (V2), we have L(x)≤γM(x), ∀x∈RN. It follows that L is also a bounded function in RN.Definition 5 (see [16]).
LetE be a Banach space. Let J∈C1(E,R) and c∈R. One can call the fact that J satisfies the Cerami condition at c, denoted by (C)c condition, if for any {un}⊂E satisfying (32)limn→∞⁡Jun=c,limn→∞⁡1+unJ′un=0,there exists u∈E such that, up to a subsequence, un-u→0 as n→∞. If J satisfies the (C)c condition for every c∈R, then J is said to satisfy the Cerami condition. A sequence {un}⊂E satisfying (32) is called a (C)c sequence of J.It was shown in [13] that the Cerami condition actually suffices to get a deformation theorem (see [8]) and then, by standard minimax arguments, it allows rather general minimax results.Lemma 6.
Suppose thatV satisfies that V∈LlocN/2(RN) and limx→∞V(x)=0. If un⇀u in X, then (33)limn→∞⁡∫RNVxun2dx=∫RNVxu2dx.Proof.
Becauseun⇀u in X, {un} is bounded. It follows that (34)M≔supn∫RNun-u2dx<+∞.From limx→∞V(x)=0, we deduce that, for any ϵ>0, there exists Rϵ>0 such that (35)supx≥Rϵ⁡Vx<ϵM.It follows that (36)∫x≥RϵVx·un-u2dx≤supx≥Rϵ⁡Vx·∫RNun-u2dx<ϵ.For y∈RN and r>0, let (37)BRNy,r=x∈RN∣x-y<r.Because V∈LN/2(BRN(0,Rϵ)) and un⇀u in X, by Lemma 2.13 of [6], we have (38)limn→∞⁡∫x<RϵVxun2dx=∫x<RϵVxu2dx.Combining (36) and (38) yields (33).Lemma 7.
Let{un}⊂X be a (C)c sequence of Φ. Then, up to a subsequence, {un} is bounded.Proof.
By Definition5, we have (39)limn→∞⁡Φun=c,limn→∞⁡1+unΦ′un=0.Arguing indirectly, assume that un→∞ and set wn=un/un.
Case 1. {wn} is nonvanishing; that is, there exist yn∈RN and α>0 such that (40)∫BRNyn,1wn2dx≥α,n=1,2,….Set w~n=wn(·+yn). Then, by (40), w~n⇀w≠0 in X.
By (39) and (27), for any ϕ∈C0∞(RN), we have, as n→∞, (41)∫RN∇un∇ϕdx-∫RNVxunϕdx+∫RNfunϕdx=o1.It follows that, as n→∞, (42)∫RN∇w~n∇ϕdx-∫RNVx+ynw~nϕdx+∫RNfu~nu~nw~nϕdx=o1,where u~n=un(·+yn). If lim⁡yn=+∞, sending n→∞ in (42), by (11), limx→∞V(x)=0 and the fact that, for x∈x∈RN∣w≠0, u~n(x)→+∞, we get that (43)∫RN∇w∇ϕdx=0,∀ϕ∈C0∞RN. This is impossible, because 0≠w∈X. If {yn} has a subsequence bounded in RN, up to a subsequence, it can be assumed that y0=limn→∞yn, and sending n→∞ in (42), we get that (44)∫RN∇w∇ϕdx-∫RNVx+y0wϕdx=0,∀ϕ∈C0∞RN.It follows that (45)∫RN∇w·-y0∇ϕdx-∫RNVxw·-y0ϕdx=0,∀ϕ∈C0∞RN.This implies that w(·-y0) is a nonzero eigenfunction of T corresponding to eigenvalue 0. It contradicts Lemma 4.
Case 2. {wn} is vanishing; that is, (46)limn→∞⁡supy∈RN⁡∫BRNy,1wn2dx=0.The Lions lemma (see, e.g., [6, Lemma 1.21]) shows that, for any p∈(2,2N/(N-2)), (47)limn→∞⁡wnLp=0.Here, ·Lp denotes Lp(RN) norm. Equations (39) and (27) show that, as n→∞, (48)o1=Φ′un,un=∫RN∇un2dx-∫RNVxun2dx+∫RNfunundx=∫RN∇un2dx+a0∫RNun2dx-∫RNVxun2dx-∫RNgunundx=un2-∫RNVxun2dx-∫RNgunundx=un21-∫RNVxwn2dx-∫RNgununwn2dx,where g is defined in (15). It follows that (49)∫RNgununwn2dx+∫RNVxwn2dx⟶1,n⟶∞.By Lemma 6 and wn⇀0 in X, we have (50)limn→∞⁡∫RNVxwn2dx=0.Together with (49), this implies (51)∫RNgununwn2dx⟶1,n⟶∞.
By (39), (52)o1+c=Φun-12Φ′un,un=∫RNF~undx.Let (53)En=x∈RN∣gunxunx≥a0-δ.By F~≥0, (17), and (52), the following is true: (54)δmesEn≤∫EnF~undx≤∫RNF~undx≤c+o1.Here, mes⁡(A) denotes the Lebesgue measure of A. It follows that (55)supn⁡mesEn<+∞.
Let(56)Fn=x∈RN∣gunxunx<a0-δ.Because δ∈(0,a0) (see (f3)), we have (57)∫Fngununwn2dx≤a0-δ∫RNwn2dx≤a0-δwnL22≤a0-δa0wn2=a0-δa0<1.Under the assumption (f1), there exists C>0 such that (58)ft≤Ct,∀t∈R.It follows that gun≤Cun for some constant C>0. Then, by (55) and (47), for 2<p<2N/(N-2), (59)∫Engununwn2dx≤C∫Enwn2dx≤CmesEnp-2/pwnLp2⟶0,n⟶∞.Combining (57) and (59) yields (60)limsupn→∞⁡∫RNgununwn2dx≤limsupn→∞⁡∫Fngununwn2dx+limsupn→∞⁡∫Engununwn2dx<1.This contradicts (51).
From the two cases given above, it can be deduced that{un} is bounded in X.Lemma 8.
The functionalΦ satisfies the (C)c condition for every c∈R.Proof.
Let{un} be a (C)c sequence of Φ. By Lemma 7, un is bounded. It follows that there exists u∈X such that, up to a subsequence, un⇀u in X. Because the embedding X↪Lloc2(RN) is compact, we get that, up to a subsequence, (61)un⟶ua.e.  inRN.As the proof of Lemma 6.15 of [6], we have Φ′(u)=0. It follows that (62)∫RN∇u2dx+∫RNfuudx-∫RNVxu2dx=0.
We are going to prove that, up to a subsequence,un-u→0 as n→∞.
By (39), as n→∞, (63)Φ′un,un=∫RN∇un2dx+∫RNfunundx-∫RNVxun2dx=o1.By Lemma 6, (64)limn→∞⁡∫RNVxun2dx=∫RNVxu2dx.Combining (62)–(64) yields (65)limn→∞⁡∫RN∇un2dx+∫RNfunundx=∫RN∇u2dx+∫RNfuudx.Because un⇀u in X, by [17, Theorem 1.6], we have (66)liminfn→∞⁡∫RN∇un2dx≥∫RN∇u2dx.Because tf(t)≥0 for any t∈R (see (12)), by (61) and the Fatou lemma, we have (67)liminfn→∞⁡∫RNfunundx≥∫RNfuudx.Using (65)–(67), it can be deduced that, up to a subsequence, (68)limn→∞⁡∫RN∇un2dx=∫RN∇u2dx,(69)limn→∞⁡∫RNfunundx=∫RNfuudx.
By (13), {Dunf(un)-F(un)} is a sequence of nonnegative functions. Because (70)Dunfun-Fun⟶Dufu-Fu,a.e.  inRN,by the Fatou lemma and (69), the following is true: (71)∫RNDuxfux-Fuxdx≤liminfn→∞⁡∫RNDunxfunx-Funxdx=D∫RNuxfuxdx-limsupn→∞⁡∫RNFunxdx.It follows that (72)∫RNFuxdx≥limsupn→∞⁡∫RNFunxdx.By (12), we deduce that F≥0 in R. Then, by (73)Fun⟶Fu,a.e.  inRN,and the Fatou lemma, the following holds: (74)∫RNFuxdx≤liminfn→∞⁡∫RNFunxdx.Combining (72) and (74) leads to (75)limn→∞⁡∫RNFunx=∫RNFuxdx.
By (10) and (11), it can be deduced that there exists C>0 such that (76)ft≤Ct,∀∈R.Then, by the mean value theorem, the Hölder inequality, and (76), we get that, for any a,b∈R and for any sufficiently small ϵ>0, there exists 0<θ<1 such that (77)Fa+b-Fa=bfa+θb≤Cb·a+θb≤Ca·b+Cb2≤Cϵ22a2+C1+12ϵ2b2.Because (61), (75), and (77) hold and {un} is bounded, Theorem 2 of [18] shows that (78)limn→∞⁡∫RNFun-udx=0.It follows that, for every sufficiently small ϵ>0, (79)limn→∞⁡∫En,ϵFun-udx=0.Here, (80)En,ϵ=x∣unx-ux≤ϵ-1.
Equations (10) and (12) show that (81)Ft>0,t≠0,(82)liminft→0⁡Ftt2>0.Combining (81) and (82) yields that, for any sufficiently small ϵ>0, there exists Cϵ>0 such that (83)t2≤CϵFt,∀t∈-ϵ-1,ϵ-1.By (79) and (83), we get that, for every sufficiently small ϵ>0, (84)limn→∞⁡∫En,ϵun-u2dx=0.From (85)ϵ-2mesRN∖En,ϵ≤∫RN∖En,ϵun-u2dx≤M≔supn∫RNun-u2dx<∞,we get that (86)mesRN∖En,ϵ≤Mϵ2,n=1,2,….By (86), the following holds: (87)∫RN∖En,ϵun-u2dx≤mesRN∖En,ϵ2/N∫RNun-u2∗dx2/2∗≤M′mesRN∖En,ϵ2/N≤M′M2/Nϵ4/N,where 2∗=2N/(N-2) and (88)M′=supn⁡∫RNun-u2∗dx2/2∗<∞.Combining (84) and (87) leads to (89)limn→∞⁡∫RNun-u2dx=0.Together with (68), this implies un→u in X. This completes the proof.


## 3. Proof of Theorem1

For the proof of Theorem1, the following abstract theorem, which is a corollary of [13, Theorem 2.4], is used.Theorem 9.
LetE be a real Hilbert space with norm ·. Suppose that J is a C1-functional defined in E and satisfies the following conditions:(A1)
J satisfies the (C)c condition for every c>0, and J(0)≥0.(A2)
J is an even functional; that is, J(-u)=J(u) for every u∈E.(A3)
There existc0>J(0) and a closed subspace Z of E with finite codimension such that there exists ρ>0 such that (90)infSρ∩Z⁡J≥c0.Here,Sρ={u∈E∣u=ρ}.(A4)
For anyn∈N, there exists a subspace V⊂X with dimension larger than or equal to n such that (91)supV⁡J<+∞.Then, J has infinitely many pairs of critical points.Proof.
Arguing indirectly, assume thatJ has only a finite number of critical points, say n0. The codimension of Z is here denoted by m0. By (A4), there exists a subspace V⊂Y with dimension larger than or equal to n0+m0+1 such that (92)supV⁡J<+∞.This indicates that there exists c∞>c0 such that (93)supV⁡J<c∞.Then, by [13, Theorem 2.4], J has at least (n0+m0+1)-m0=n0+1 critical points. This contradicts the assumption that J has only n0 critical points.Let(94)λ1<λ2≤λ3≤⋯≤λn≤⋯be the eigenvalues of T below the essential spectrum of T. Each λn has been repeated in the sequence according to its finite multiplicity. Then, by Lemma 3, the following is true:(95)limn→∞λn=0,λn<0,∀n.Let (96)n∗=min⁡n∈N∣a0+λn>0.Here, a0 is the constant in (10). From this definition, (97)a0+λn∗>0≥a0+λn∗-1,ifn∗>1.Let {Eλ∣λ∈R} denote the spectral family of the operator T defined by (3).Let(98)Z=X,ifn∗=1id-EνL2RN∩X,ifn∗>1.Here, id:L2(RN)→L2(RN) is the identity map and (99)ν=12λn∗+λn∗-1.It is easy to see that the codimension of Z equals n∗-1.Lemma 10.
There existsς>0 such that (100)∫RN∇u2dx+a0∫RNu2dx-∫RNVxu2dx≥ςu2,∀u∈Z.Proof.
From (96) and (98), the following holds: (101)∫RN∇u2dx+a0∫RNu2dx-∫RNVxu2dx≥κ∫RNu2dx,∀u∈Z.Here, (102)κ=a0+λn∗>0.It follows that (103)ς≔∫RN∇u2dx+a0∫RNu2dx-∫RNVxu2dx∣u∈Z,u=1≥0.
Ifς=0, then there exists {un}⊂Z satisfying un=1, n=1,2,…, and (104)limn→∞⁡∫RN∇un2dx+a0∫RNun2dx-∫RNVxun2dx=0.Up to a subsequence, it can be assumed that un⇀u0 in X and un→u0 a.e. in RN. Since Z is a closed subspace of X, by un⇀u0 in X, we have u0∈Z. Because un⇀u0 in X, by [17, Theorem 1.6], we have (105)liminfn→∞⁡∫RN∇un2dx≥∫RN∇u02dx.Because un→u0 a.e. in RN, by the Fatou lemma, the following holds: (106)liminfn→∞⁡∫RNun2dx≥∫RNu02dx.Moreover, by Lemma 6, (107)limn→∞⁡∫RNVxun2dx=∫RNVxu02dx.Combining (104)–(107) yields (108)∫RN∇u02dx+a0∫RNu02dx-∫RNVxu02dx≤0.Because u0∈Z, by (101) and (108), we have u0=0. Then, by (104) and (107), (109)limn→∞⁡∫RN∇un2dx+a0∫RNun2dx=0.This contradicts un=1, n=1,2,…. Therefore, ς>0.Proof of Theorem1.
We are going to prove that the functionalΦ satisfies the conditions (A1)–(A4) in Theorem 9.
By Lemma8, the functional Φ satisfies the condition (A1). By (f4), Φ satisfies the condition (A2).
By (10) and (11), it can be deduced that, for any ϵ>0, there exists Cϵ>0 such that(110)Ft-a0t22≤ϵt2+Cϵtμ,∀t∈R.Here, (111)μ=2N-2N-2∈2,2∗.Let Z be the space defined by (98). It has a finite codimension. By Lemma 10 and (110), we have, for any u∈Z, (112)Φu=12∫RN∇u2dx+a02∫RNu2dx-12∫RNVxu2dx+∫RNFu-a02u2dx≥ς2u2-ϵuL22-CϵuLμμ≥ς2-ϵu2-CϵuLμμ≥ς2-ϵu2-C′Cϵuμ,where the positive constant C′ comes from the Sobolev inequality uLμ≤(C′)1/μu, ∀u∈X. Choosing ϵ=ς/4 and ρ=(ς/8C′Cϵ)1/(μ-2), by (112), the following holds: (113)infSρ∩Z⁡Φ≥c0≔ς8ρ2>Φ0.It follows that Φ satisfies the assumption (A3).
Letϑ∈C0∞(RN) such that ϑL2=1 and supp⁡ϑ⊂{x∈RN∣1<x<2}. Then, for the function ϑλ(x)=λ-N/2ϑ(λ-1x), ϑλL2=1 and supp⁡ϑλ⊂{x∈RN∣λ<x<2λ}. Then, by (8), it follows that, for λ>R,(114)-∫RNVxϑλ2dx+∫RN∇ϑλ2dx≤-Cλ-βϑL22+Cλ-2∇ϑL22,where C is a positive constant independent of λ. Because β<2, it follows that there exists λ0>R such that (115)∫RN∇ϑλ2dx-∫RNVxϑλ2dx<0forλ≥λ0.Let ψn=ϑ2nλ0, n∈N. These functions have mutually disjoint compact supports. Let (116)Vn=spanψ1,ψ2,…,ψn.Then, Vn is an n-dimensional subspace of X. By (10) and (11), it can be deduced that, for any ϵ>0, there exist Lϵ>0 and Kϵ>0 such that (117)Ft≤ϵt2,ift>Lϵ,Ft≤Kϵt3/2,ift≤Lϵ.Because {ψi∣i∈N} have mutually disjoint compact supports, by (115), we have, for u=∑i=1ntiψi∈Vn, (118)Φu=12∑i=1nαiti2+∫RNFudx,where (119)αi=∫RN∇ψi2dx-∫RNVxψi2dx<0,i=1,2,…,n.Let (120)Ωn=⋃i=1nsupp⁡ψi.Then, Ωn is a bounded set in RN, since every ψi has a compact support. By (117), (121)∫RNFudx=∫x∣ux>LϵFudx+∫x∣ux≤LϵFudx≤ϵ∫RNu2dx+Kϵ∫Ωnu3/2dx≤ϵ∑i=1nϖiti2+KϵmesΩn1/4∫Ωnu2dx3/4=ϵ∑i=1nϖiti2+KϵmesΩn1/4∑i=1nϖiti23/4,where (122)ϖi=∫RNψi2dx>0,i=1,…,n.Choosing (123)ϵ=min⁡-αi∣1≤i≤n4max⁡ϖi∣1≤i≤n,by (118) and (121), we get that (124)Φu≤14∑i=1nαiti2+KϵmesΩn1/4∑i=1nϖiti23/4.Since αi<0, for every i, (124) implies (125)supVn⁡Φ<∞.Therefore, Φ satisfies the condition (A4) of Theorem 9.
BecauseΦ satisfies the conditions (A1)–(A4) of Theorem 9, by Theorem 9, it has infinitely many critical points. It follows that (6) has infinitely many solutions.



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*Source: 3042493-2016-03-02.xml*