R E S E A R C H
Open Access
Ground state solution and multiple solutions
to asymptotically linear Schrödinger
equations
Xiang-Dong Fang
1,2*and Zhi-Qing Han
1**Correspondence: [email protected]; [email protected]
1School of Mathematical Sciences, Dalian University of Technology, Dalian, 116024, P.R. China 2State Key Laboratory of Structural Analysis for Industrial Equipment, Department of Engineering Mechanics, Dalian University of Technology, Dalian, 116024, P.R. China
Abstract
In this paper, we consider the Schrödinger equation –
u+V(x)u=f(x,u),x∈RN, whereVandfare periodic inx1,. . .,xN, asymptotically linear and satisfies a
monotonicity condition. We use the generalized Nehari manifold methods to obtain a ground state solution and infinitely many geometrically distinct solutions whenfis odd inu.
Keywords: Schrödinger equation; ground state solution; multiplicity of solutions; asymptotically linear
1 Introduction
We consider the problem
–u+V(x)u=f(x,u), u∈HRN, (.) wherefandVare periodic inx, . . . ,xN, asymptotically linear and satisfies a monotonicity condition. In the case that the nonlinear term is asymptotically linear at infinity, there are some results in the literature [–] and the references therein, where multiplicity results are considered in [–, , , ]. As far as we know, there are only a few papers concerned with the existence of infinitely many solutions for the asymptotical linear case whenf and V are also periodic inx, . . . ,xN;e.g.see []. Except for [], there seem to be few results on the existence of a ground state solution in the asymptotically linear case. Motivated by [], this paper is to present a different approach involving the critical point theory with the discreteness property of the Palais-Smale in search for a ground state solution and multiple solutions for the asymptotically linear Schrödinger equations. It should be pointed out that in [], they cannot make sure the existence of a ground state solution. Our results can be regarded as complements or different attempts of the results in [, ].
Setting F(x,u) :=uf(x,s)ds, we suppose thatV andf satisfy the following assump-tions:
(V) V is continuous, -periodic inxi,≤i≤N, and there exists a constanta> such
thatV(x)≥afor allx∈RN.
(f) f is continuous, -periodic inxi,≤i≤N. (f) f(x,u) =o(u)asu→, uniformly inx.
(f) There isq(x) >V(x),∀x∈RN, such thatf(x,u)/u→q(x), as|u| → ∞, whereqis
con-tinuous, -periodic inxi,≤i≤N.
(f) u→f(x,u)/|u|is strictly increasing on(–∞, )and(,∞).
Let∗denote the action ofZN onH(RN) given by
(k∗u)(x) :=u(x–k), k∈ZN. (.) It follows from (V) and (f) that ifuis a solution of (.), then so isk∗ufor allk∈ZN.
Set
O(u) :=
k∗u:k∈ZN
.
O(u) is calledthe orbitofuwith respect to the action ofZN, and it is calleda critical
orbitfor a functionalFifuis a critical point ofFandFisZN-invariant,i.e.,F(k∗u) =F(u)
for allk∈ZN and allu(then of course all points ofO(u
) are critical). Two solutionsu,
uof (.) are said to begeometrically distinctifO(u)=O(u).
Theorem . Suppose that(V), (f)-(f)are satisfied.Then(.)has a ground state solution.
In addition,if f is odd in u,then(.)admits infinitely many pairs±u of geometrically distinct solutions.
Notation C,C,C, . . . will denote different positive constants whose exact value is
inessential. The usual norm in the Lebesgue space Lp() is denoted byu
p,, and by up if=RN. E denotes the Sobolev space H(RN) andS is the unit sphere inE. It follows from (V) that
u:=
RN
|∇u|+V(x)u
/
is an equivalent norm inE. It is more convenient for our purposes than the standard one and will be used henceforth. For a functionalI, as in [], we put
Id:=u:I(u)≤d, Ic:=
u:I(u)≥c, Icd:=u:c≤I(u)≤d.
2 Preliminary results Consider the functional
I(u) :=
RN|∇u|
+
RNV(x)u
–
RNF(x,u). (.)
ThenIis well defined onEandI∈C(E,R) under the hypotheses (V), (f
)-(f). Note also
that (V), (f) implyIis invariant with respect to the action ofZN given by (.). It is easy
to see that
I(u),v=
RN∇u∇v+
RNV(x)uv–
RNf(x,u)v (.)
Let
M:=u∈E\ {}:I(u),u= . (.)
Recall thatMis called the Nehari manifold. We do not know whetherMis of classC
under our assumptions and therefore we cannot use minimax theory directly onM. To overcome this difficulty, we employ the arguments developed in [, , ].
We assume that (V) and (f)-(f) are satisfied from now on. First, (f) and (f) imply that
for eachε> there isCε> such that
f(x,u)≤ε|u|+Cε|u|p– for allu∈R, (.)
where <p< ∗, ∗:= N/(N– ) ifN≥, ∗:=∞ifN= or . Fort> , let
h(t) :=I(tu) =t
RN|∇u|
+V(x)u–
RNF(x,tu).
Let
E:=
u∈E:
RN|∇u|
+V(x)u<
RNq(x)u
.
It follows fromq(x) –V(x) > ,∀x∈RN, thatE=∅.
Lemma . F(x,u) > andf(x,u)u>F(x,u)if u= .
This follows immediately from (f) and (f).
Lemma .
() For eachu∈Ethere is a uniquetu> such thath(t) > for <t<tuandh(t) < fort>tu.Moreover,tu∈Mif and only ift=tu.
() Ifu∈/E,thentu∈/Mfor anyt> .
Proof () For eachu∈E, due to the Lebesgue dominated convergence theorem and (f),
(f), we get
lim
t→∞
I(tu) t =
RN|∇u|
+V(x)u– lim
t→∞
u=
F(x,tu) tu u
=
RN|∇
u|+V(x)u–
RN q(x)u
<
and
lim
t→
I(tu) t =
RN|∇u|
+V(x)u–lim
t→
u=
F(x,tu) tu u
=
RN|∇u|
Hencehhas a positive maximum. The conditionh(t) = is equivalent to
u=
u=
f(x,tu) tu u
.
By (f), the first conclusion holds. The second conclusion follows fromh(t) =t–I(tu),tu.
() Iftu∈Mfor somet> , thenI(tu),u= and therefore using (f) and (f)
u=
u=
f(x,tu) tu u
<
RNq(x)u
.
Henceu∈E.
Lemma .
() There existsρ> such thatc:=infMI≥infSρI> .
() u≥cfor allu∈M.
Proof () Using (.) and the Sobolev inequality we haveinfSρI> ifρis small enough.
The inequalityinfMI≥infSρIis a consequence of Lemma . since for everyu∈Mthere
iss> such thatsu∈Sρ(andI(tuu)≥I(su)). () Foru∈M, by Lemma . we have
c≤ u
–
RNF(x,u)≤ u
.
We do not know whetherIis coercive onM. However, we can prove the following.
Lemma . All Palais-Smale sequences(un)⊂Mare bounded.
Proof Arguing by contradiction, suppose there exists a sequence (un)⊂M such that
un → ∞ and I(un)≤d for some d∈ [c,∞). Let vn:=un/un. Then vnv and vn(x)→v(x) a.e. inRN after passing to a subsequence. Chooseyn∈RN so that
B(yn)
vn=max
y∈RN
B(y)
vn. (.)
SinceIandMare invariant with respect to the action ofZNgiven by (.), we may assume that (yn) is bounded inRN. If
B(yn)
vn→ asn→ ∞, (.)
then it follows thatvn→ inLr(RN) for <r< ∗by Lions’ lemma (cf.[], Lemma .), and therefore (.) implies thatRNF(x,svn)→ for everys∈R. Lemma . implies that
d≥I(un)≥I(svn) = s
–
RNF(x,svn)→ s
.
Letϕ∈C∞(RN). ThenI(u
n),ϕ → and hence
RN∇
vn∇ϕ+V(x)vnϕ–
RN f(x,un)
un
vnϕ→.
By the Lebesgue dominated convergence theorem we therefore have
RN∇
v∇ϕ+V(x)vϕ=
RN q(x)vϕ.
So v= and –v+V(x)v=q(x)v. This is impossible because –+V –qhas only an
absolutely continuous spectrum. The proof is complete.
Lemma . IfVis a compact subset ofE,then there exists R> such that I≤on(R+V)\ BR().
Proof We may assume without loss of generality thatV⊂S. Arguing by contradiction, suppose there existun∈Vandwn=tnun, whereun→u,tn→ ∞andI(wn)≥. We have
≤I(tnun) t
n =
RN|∇un|
+V(x)u
n–
un=
F(x,tnun) t
nun un
→
RN|∇u|
+V(x)u–
RNq(x)u
< .
LetU:=E∩Sand define the mappingm:U→Mby setting
m(w) :=tww,
wheretwis as in Lemma ..
Lemma . U is an open subset of S.
Proof Obvious becauseEis open inE.
Lemma . Assume un∈U,un→u∈∂U,and tnun∈M,then I(tnun)→ ∞.
Proof Sinceu∈∂U,
RN|∇u|+V(x)u=
RNq(x)u. Using this, we have
I(tu) =
t
RN|∇u|
+V(x)u –t
RN
F(x,tu)
tu
u
= t RN
q(x) –F(x,tu) tu
u
= t RN
q(x) –f(x,tu) tu
u
+
RN
f(x,tu)tu–F(x,tu).
Note that by (f), we have for large enoughs, there isδ> such that
(see [], Remark .). SoI(tu)→ ∞, ast→ ∞(we have used Fatou’s lemma). Given
C> , chooset> such thatI(tu)≥C. Sinceun→u,
lim
n→∞I(tnun)≥nlim→∞I(tun) =I(tu)≥C
and henceI(tnun)→ ∞.
The following lemmas are taken from [, ]. Below we shall use the notations
K:=w∈S:(w) = ,
Kd:=
w∈K:(w) =d.
Sincefis odd inu, we can choose a subsetFofKsuch thatF= –Fand each orbitO(w)⊂ K has a unique representative inF. We must show that the setF is infinite. Arguing indirectly, assume
Fis a finite set. (.)
Lemma . The mapping m is a homeomorphism between U andM,and the inverse of
m is given by m–(u) = u
u.
We consider the functional:U→Rgiven by
(w) :=Im(w).
Lemma .
() ∈C(U,R)and
(w),z=m(w)Im(w),z for allz∈Tw(U).
() If(wn)is a Palais-Smale sequence for,then(m(wn))is a Palais-Smale sequence forI.If(un)⊂Mis a bounded Palais-Smale sequence forI,then(m–(un))is a Palais-Smale sequence for.
() wis a critical point ofif and only ifm(w)is a nontrivial critical point ofI. Moreover,the corresponding values ofandIcoincide andinfU=infMI. () is even(becauseIis).
By (.), the following lemma also holds.
Lemma . Let d≥c.If(v
n), (vn)⊂dare two Palais-Smale sequences for,then either
vn–vn →as n→ ∞orlim supn→∞vn–vn ≥ρ(d) > ,whereρ(d)depends on d but not on the particular choice of Palais-Smale sequences.
flow defined by
⎧ ⎨ ⎩
d
dtη(t,w) = –H(η(t,w)), η(,w) =w,
(.)
where
G:=(t,w) :w∈U\K,T–(w) <t<T+(w)
and (T–(w),T+(w)) is the maximal existence time for the trajectoryt→η(t,w). Note that
ηis odd inwbecauseHis andt→(η(t,w)) is strictly decreasing by the properties of a pseudogradient.
LetP⊂U,δ> and defineUδ(P) :={w∈U:dist(w,P) <δ}.
Lemma . Let d≥c.Then for everyδ> there existsε=ε(δ) > such that (a) dd–+εε∩K=Kdand
(b) limt→T+(w)(η(t,w)) <d–εforw∈d+ε\Uδ(Kd).
Part (a) is an immediate consequence of (.) and (b) has been proved in []; see Lem-mas . and . there. The argument is exactly the same except thatSshould be replaced byU. We point out that an important role in the proof of Lemma . is played by the dis-creteness property of the Palais-Smale sequences expressed in Lemma ..
3 Proof of Theorem 1.1
Proof of Theorem. Taking a similar argument as in the proof of Theorem . in [], it is easy to get a ground state solution. Noting that by Lemma . and Ekeland’s variational principle, it can make sure the existence of a (PS)csequence belonging toU.
For the multiplicity the argument is the same as in Theorem . (cf.[]). However, there are details which need to be clarified.
Letη be the flow given by (.). IfT+(w) <∞, then lim
t→T+(w)η(t,w) exists (cf.[],
Lemma ., Case ) but unlike the situation in [], this limit may be a pointw∈∂U.
This possibility is ruled out by Lemma ..
Finally, we need to show thatUcontains sets of arbitrarily large genus. Since the spec-trum of –+V –q inL(RN) is absolutely continuous,E ∪ {} contains an infinite-dimensional subspaceE. HenceE∩S⊂Uandγ(E∩S) =∞.
Remark . There is a small gap in the proof of Theorem . in []. Lemma . as stated there does not exclude the possibility ofη(t,w) approaching the boundary ast→T+(w) (because we only know thatη(t,w) goes to infinity). But it is easy to prove thatI(η(t,w)) goes to infinity as well in []. In Lemma . of this paper we make some proper modifi-cations which also apply to [] and were proposed by Andrzej Szulkin.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
Acknowledgements
The authors thank the referee for providing some of the references and suggestions. The authors thank Professor Szulkin for his encouragement. This work was supported by NSFC 11171047. The first author was supported the Fundamental Research Funds for the Central Universities.
Received: 5 June 2014 Accepted: 11 September 2014 References
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doi:10.1186/s13661-014-0216-1