# Remarks on a Class of Nonlinear Schrödinger Equations with Potential Vanishing at Infinity

Authors: Hongbo Zhu
Journal: Discrete Dynamics in Nature and Society
 (2013)

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## Abstract

We study the following nonlinear Schrödinger equation−Δu+V(x)u=K(x)f(u),x∈ℝN,u∈H1(ℝN), where the potential V(x) vanishes at infinity. Working in weighted Sobolev space, we obtain the ground states of problem (𝒫) under a Nahari type condition. Furthermore, if V(x),K(x) are radically symmetric with respect to x∈ℝN, it is shown that problem (𝒫) has a positive solution with some more general growth conditions of the nonlinearity. Particularly, if f(u)=up, then the growth restriction σ≤p≤N+2/N-2 in Ambrosetti et al. (2005) can be relaxed to σ~≤p≤N+2/N-2, where σ~<σ if 0<β<α<2.


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## Body

## 1. Introduction

The motivation of the paper is concerned with the existence of standing waves of the following nonlinear Schrödinger equation:(1)i∂ψ∂t=-Δxψ+V~ψ-f(x,ψ),(t,x)∈ℝ+×ℝN,
where i is the imaginary unit, V~ is a real function on ℝN, ψ(x,t):ℝN×[0,+∞)→ℂ, and f is supposed to satisfy that f(x,αeiθ)=f(x,α)eiθ for all α,θ∈ℝ. Problem (1) arises in many applications. For example, in some problems arising in nonlinear optics, in plasma physics, and in condensed matter physics, the presence of many particles leads one to consider nonlinear terms which simulates the interaction effect among them.For problem (1), we are interested in looking for a stationary solution; that is, ψ(x,t)=u(x)eiωt with u(x)>0 in ℝN and ω>0 (the frequency); then it is not difficult to see that u must satisfy
(2)-Δu+V(x)u=f(x,u),x∈ℝN,…………..u∈H1(ℝN).
Here and below, N≥3. Variational approach to (2) was initiated by Rabinowtiz [1], and since then several authors have studied (2) under different assumptions on V(x) and the nonlinearity. If V(x) is positive and bounded away from zero, then, by the well-known concentration compactness principle [2, 3], it is shown that there is bound states for problem (2); we mention here the work by Jeanjean and Tanaka [4, 5], Liu and Wang [6], Li et al. [7], Zhu [8], and the references therein.If the potentialV(x) decays to zero at infinity, the methods used in the proceeding papers cannot be employed because the variational theory in H1(ℝN) cannot be used here. The earlier work on (2) we know of where V(x) decays at infinity, is that by Ambrosetti et al. [9]; the authors proved that problem (2) has bounded states for f(x,u)=K(x)up with
(3)a1+|x|α≤V(x)≤A,0<K(x)≤k1+|x|β,σ<p<N+2N-2,
where
(4)σ={N+2N-2-4βα(N-2),0<β<α,1,otherwise.
Following [9], by requiring some further assumptions on V(x),K(x), in [10], the authors showed that there exist bound states of equation -ϵ2Δu+V(x)u=K(x)up, x∈ℝN, u>0, for all p satisfying 1<p<(N+2)/(N-2), provided that ϵ is sufficiently small. Motivated by the works [9, 10], in paper [11], the authors extended the results to potentials V(x) that can both vanish and decay to zero at infinity. And since then, there are many papers on problem (2) with potential V(x) vanishing at infinity; see, for example, [12–16].In this paper, more precisely we will focus on the following model equation:(5)-Δu+V(x)u=K(x)f(u),x∈ℝN,u∈H1(ℝN).
To our best knowledge, it seems that there are few results on problem (5), where f(u) does not satisfy (AR) condition; that is, for some θ>0,
(AR)0≤F(u)≜∫0uf(t)dt≤12+θf(u)u∀u∈ℝ,
and simultaneously V(x) decays to zero at infinity.The main aim of the paper is to extend the result of [9] to problem (5) with much more general classes of f(u). Moreover, if V(x),K(x) are radically symmetric with respect to x∈ℝN, problem (5) will be proved to also have a positive solution with some more general growth conditions of the nonlinearity. Particularly, the result can exactly extend the growth restriction of the special case of f(u)=up to a new one, and the range of which is bigger than the usual σ≤p≤(N+2)/(N-2) in [9]; for more details please see Theorem 6.Throughout the paper, we make the following assumptions onV(x), K(x), and f(t).(V)
V:ℝN→ℝ is continuous and there exist a,A,α>0 such that
(6)a1+|x|α≤V(x)≤A.(K)
K:ℝN→ℝ is smooth and there exist β,k>0 such that
(7)0<K(x)≤k1+|x|β.(F1)
ft(t) is a Caratheodory function and there exists C>0 such that
(8)|ft(t)|≤C(1+|t|2*-2),lim|t|→∞f(t)|t|2*-1=0.(F2)
lim|t|→0(f(t)/|t|)=0 and lim|t|→∞(F(t)/t2)=∞.(F3)
f(t)/|t| is nondecreasing in t∈ℝ∖{0}.Remark 1.
Ifβ>α>0, then liminf|x|→∞(V(x)/K(x))=+∞, which implies that K(x)≤CV(x).Throughout this paper, we define the following weighted Sobolev space:(9)H={∫ℝN(|∇u|2+V(x)|u|2)dx<∞u∈𝒟1,2(ℝN):∫ℝN(|∇u|2+V(x)|u|2)dx<∞}.
Clearly, H1(ℝN)⊂H. H is a Hilbert space with norm and scalar product, respectively,
(10)∥u∥=[∫ℝN(|∇u|2+V(x)|u|2)dx]1/2,(u,v)=∫ℝN(∇u∇v+V(x)uv)dx.
Denote by LKq the weighted space of measurable u:ℝN→ℝ such that
(11)|u|q,K=(∫ℝNK(x)|u(x)|qdx)1/q<∞.H and LKq are particular cases of weighted spaces in discussions in [17], where the following result is proved.Proposition 2.
LetN≥3, and suppose that (V), (K) hold with α∈(0,2) and β>0. Then
(12)H↪LKp+1,
where σ≤p≤(N+2)/(N-2), and there is C>0 such that
(13)|u|p+1,K≤C∥u∥,∀u∈H.
Furthermore, the embedding of H into LKp+1 is compact if p∈(σ,(N+2)/(N-2)).Remark 3.
For the case ofβ≥α>0, σ=1, and consequently H↪LKp+1, for all1≤p≤(N+2)/(N-2), and also the embedding is compact if 1<p<(N+2)/(N-2).Furthermore, we define the energy functional associated with problem (5) by
(14)I(u)=12∫ℝN(|∇u|2+V(x)|u|2)dxI(u)=-∫ℝNK(x)F(u)dx,F(u)=∫0uf(t)dt.
By Proposition 2, I is well defined on H and I∈C1(H,ℝ) with
(15)〈I′(u),v〉=∫ℝN(∇u∇v+V(x)uv)dx-∫ℝNK(x)f(u)vdx∀v∈H.Definition 4.
A functionu∈H is said to be a solution of problem (5) provided that u≢0 satisfies
(16)∫ℝN(∇u∇φ+V(x)uφ)dx-∫ℝNK(x)f(u)φdx=0∀φ∈H.Notation. Hereafter we use the following notation.(i)
2*=2N/N-2 if N≥3 and for some 2<q<2*=∞ if N=2.(ii)
Lq(Ω), 1≤q≤+∞, Ω⊆ℝN, denotes a Lebesgue space; the norm in Lq(Ω) is denoted by |u|q,Ω.(iii)
For anyR>0 and for any y∈ℝN, BR(y) denotes the ball of radius R centered at y.(iv)
C,Ci are various positive constants.(v)
o(1) denotes the quantity which tends to zero as n→∞.Our main results are the following.Theorem 5.
Let(F1)–(F3) and (V),(K) hold with β>α, α∈(0,2). Then problem (5) has a ground state u∈H1(ℝN).IfV(x),K(x) are radically symmetric with respect to x∈ℝN and f(t) satisfies the following conditions:(F4)
f∈C(ℝ,ℝ+), f(t)≡0 for all t≤0;(F5)
f(t)/t is nondecreasing with respect to t>0;(F6)
limt→0(f(t)/tp)=l1∈[0,∞); limt→∞(f(t)/tq)=l2∈(0,∞), where q*-1<p≤q<N+2/N-2, and
(17)q*={4(N-1-β)+2α2(N-1)-α,0<β<α<2,2,otherwise,then we have the following.Theorem 6.
Assume thatV(x), K(x) are radically symmetric with respect to x∈ℝN, and let (V), (K), and (F4)–(F6) hold. Then problem (5) has a positive solution u∈Hr1(ℝN).Remark 7.
Iff(u)=uk, then the growth restriction σ≤k≤N+2/N-2 in [9] can be relaxed to q*-1≤k≤N+2/N-2, where q*-1<σ if 0<β<α<2.


## 2. Variational Setting and Some Preliminaries

In this section, we describe the variational framework for the study of the critical points of the functional defined in (14).Set(18)c:=inf{I(u):u∈H},
where 𝒩={u∈H∖{0}:γ(u)≜〈I′(u),u〉=0}. First, it is necessary to show that c is a positive number. Now we give the following two lemmas.Lemma 8.
Suppose that(F1)–(F3) hold. Then for each u∈H∖{0}, there exists a unique t=t(u)>0 such that t(u)u∈𝒩 and maxt≥0I(tu)=I(t(u)u).Proof.
The proof of this lemma is similar to the case of assuming(AR) condition, which can be found in [18], so we omit it here.Lemma 9.
c
>
0.Proof.
Let(19)c1:=infu∈ℋ∖{0}maxt≥0I(tu),c2:=infγ∈Γmaxt∈[0,1]I(γ(t)),
where Γ:={γ∈C(([0,1]),H):γ(0)=0,I(γ(1))<0}. By Lemma 8, it is easy to see c=c1.
For anyu∈H∖{0}, by (F2), there exists t>0 large enough such that I(tu)<0. Let γ~(s)=stu; then γ~(s)∈Γ, and
(20)c2≤I(γ~(s))=I(stu)≤maxt≥0I(tu),
so c1≥c2. From (F1), (F2), there is ρ>0 small enough such that inf∥u∥=ρI(u)>α>0. Since γ~(0)=0, for large enough t>0, ∥γ~(1)∥>ρ, there exists s0∈[0,1] such that ∥γ~(s0)∥=ρ, so
(21)c2≥I(γ~(s0))≥inf∥u∥=ρI(u)>α>0.
Thus, c=c1≥c2>0.Lemma 10.
Let{un} be a minimizing sequence of c defined in (18). Then(i)
there isρ>0 such that liminfn→∞∥un∥≥ρ>0,(ii)
{un} is bounded in H,(iii)
for a subsequence,{un} converges weakly to u≢0.Proof.
(i) By(F1), (F2), for any ϵ>0, there exists Cϵ>0 such that
(22)|f(t)|≤ϵ(|t|+|t|2*-1)+Cϵ|t|q-1,q∈(2,2*).
From γ(un)=0, H↪LKq(ℝN), it is easy to get that liminfn→∞∥un∥≥ρ>0.
(ii) If{un} is not bounded, we define vn=un/∥un∥, so ∥vn∥=1. Passing to a subsequence we have
(23)vn⇀v∈H,vn⟶v∈LKq(ℝN),2<q<2*,.vn⟶va.e.onℝN.
Ifv≢0, we have
(24)12-∫ℝNK(x)F(un)|un|2|vn|2dx=c+o(1)∥un∥2>0.
By Fatou’s lemma and (F2), we have a contradiction as follows:
(25)12≥liminfn→∞∫ℝNK(x)F(un)|un|2|vn|2dx≥∫ℝNliminfn→∞K(x)F(un)|un|2|vn|2dx=∞.
If v≡0, by Proposition 2, vn⟶0 in LKq(ℝN). By (22) and for any R>0 which is determined later, we get
(26)∫ℝNK(x)F(Rvn)dx⟶0asn⟶∞.
By Lemma 8, I(tun)≤I(un) for any t>0; then we have
(27)c+o(1)⟵I(un)≥I(Rvn)=12R2-∫ℝNK(x)F(Rvn)⟶12R2-o(1),
which is a contradiction if we take R>(2c)1/2. Thus {un} is bounded in H.
(iii) We can assume{un} weakly converges to u. If u≡0, then by (i) and noting the fact that H↪LKq, q∈(2,2*) is compact, we have a contradiction as follows:
(28)0<ρ2≤∥un∥2=∫ℝNK(x)f(un)undx⟶0.


## 3. Proof of the Main Results

The aim of this section is to prove Theorems5 and 6. For Theorem 5, we will take two steps; the first is to show the existence of nonzero critical point u∈H of I, and the second is to prove the critical point u is a bound state; that is u∈L2(ℝN). To prove Theorem 6, we use a weighted Sobolev embedding theorem, which is based upon the results discussed in [19].Proof of Theorem5.
Step  1. Let {un}⊂𝒩 be the sequence minimizing for c given in (18). By Lemma 10, {un} is bounded in H, and there is some 0≢u∈H such that
(29)un⇀nuweakly  inH,
and by Proposition 2 and Remark 3, we get un⟶u in LKq(ℝN), q∈(2,2*), and consequently
(30)∫ℝN‍K(x)F(un)dx⟶∫ℝNK(x)F(u)dx……..asn⟶∞.
Using the lower semicontinuity, we have I(u)≤c. If u∈𝒩, we have I(u)=c. If u∉𝒩, by Lemma 8, there exists t=t(u)>0 such that t(u)u∈𝒩; then
(31)c≤I(t(u)u)≤liminfn→∞I(t(u)un)≤liminfn→∞I(un)=c.
Since 𝒩 is smooth, the minimizer is a critical point of I.
Next we will show that the solution found above inH belongs indeed to L2(ℝN). For this purpose, we require some preliminary decay estimates on u, which are essentially motivated by [9]. However, we should mention that these proofs are partly different from those in [9]. First we will give the following proposition, which is a special case of Proposition 11 in [9].Proposition 11.
Let0<α<2 and 2<q<2*. Then for all ϵ>0 there exists R¯>0 such that for all R>R¯(32)∫|x|>RK(x)|u(x)|qdx≤ϵ(∫|x|>R[|∇u(x)|2+V(x)u2(x)]dx)q/2,………………………………∀u∈H.Lemma 12.
Suppose that(V), (K), (F1), and (F2) hold. Let β>α, α∈(0,2) and let u be the critical point of I on H. Then there exist t0,δ1∈(0,1) and n(α)>0 such that, for all n≥n(α),
(33)∫Ωn+1(|∇u|2+V(x)|u|2)dx≤[(1-t0)+δ1t0]×∫Ωn(|∇u|2+V(x)|u|2)dx,
where
(34)Ωn=ℝN∖BRn,Rn=bn,b=(11-t0)1/4.Proof.
Sinceβ>α, there is δ∈(0,1) such that β>α+δ. Let γ=α+δ; then γ∈(0,3), so there exists δ1∈(0,1) such that (γ+1)/4<1-δ1.
Noting that(35)limt→0+ln[(1-t)+δ1t](1/4)ln(1/(1-t))=-4(1-δ1)<-(γ+1),
we may choose t0∈(0,1) such that
(36)ln[(1-t0)+δ1t0](1/4)ln(1/(1-t0))<-(γ+1).
Let Rn be given by (34), and let χn(x) be a cut-off function such that
(37)χn(x)={0,if0≤|x|≤Rn,1,if|x|≥Rn+1.
Moreover, there exists A0>0, independent of n, such that |∇χn(x)|≤A0/(Rn+1-Rn) for all x∈ℝN. Noting that α∈(0,2) and b>1, we see that
(38)limn→∞1+bα(n+1)b2(n+1)(A0bb-1)2=0.
So there exists n(α)>0 such that, for all n≥n(α),
(39)1+bα(n+1)b2(n+1)(A0bb-1)2≤(2δ1t0)2a,
where a is given by (V). From (39) and (V), for Rn≤|x|≤Rn+1, we have
(40)|∇χn(x)|2≤A02|Rn+1-Rn|-2=b-2(n+1)(A0bb-1)2≤a1+(Rn+1)α[2δ(1-η0)]2≤V(x)(2δ1t0)2.
Then by the definition of χn(x), for all n≥n(α) and x∈ℝN,
(41)|∇χn(x)|2≤V(x)(2δ1t0)2.
Since〈I′(u),uχn〉=0, by (F1), (F2), and Young inequality, for any ϵ>0, ϵ1>0, we have
(42)∫ℝNχn(|∇u|2+V(x)|u|2)dx=∫ℝNK(x)f(u)uχndx-∫ℝN∇u∇χnudx≤∫ΩnK(x)[ϵ(|u|2+|u|2*)+Cϵ|u|q]dx+∫Ωn‍ϵ1|∇u|2dx+∫Ωn14ϵ1|∇χn|2|u|2dx,
where q∈(2,2*). Taking ϵ1=δ1t0, then by (41), (42), for all n≥n(α),
(43)∫Ωn+1‍(|∇u|2+V(x)|u|2)dx≤∫Ωn‍χn[|∇u|2+V(x)|u|2]dx≤∫Ωn‍K(x)[ϵ(|u|2+|u|2*)+Cϵ|u|q]dx+δ1t0∫Ωn‍[|∇u|2+V(x)|u|2]dx.
Suppose that ∥u∥2≤Γ, Γ>0; then, from Proposition 11 and Remark 3, for the above ϵ>0 given and R sufficiently large, we deduce that
(44)∫Ωn+1‍(|∇u|2+V(x)|u|2)dx≤[(ϵ+ϵΓ(2*-2)/2+ϵCϵΓ(q-2)/2)+δ1t0]≤×∫Ωn‍[|∇u|2+V(x)|u|2]dx.
Choosing sufficiently small ϵ>0 (and hence for R large) such that ϵ+ϵΓ(2*-2)/2+ϵCϵΓ(q-2)/2<1-t0, we obtain the assertion.Lemma 13.
Letn(α) be given in Lemma 12. Then there exists R(α)>n(α) such that, for any ρ>R(α),
(45)∫|x|>ρ‍(|∇u|2+V(x)|u|2)dx≤∥u∥2exp{C(δ1,η0,θ)lnρ},
where C(δ1,t0,θ)<-(γ+1).Proof.
From (36), there is θ∈(0,1) such that
(46)C(δ1,t0,θ):=θln[(1-t0)+δ1t0](1/4)ln(1/(1-t0))<-(γ+1).
Since
(47)lim[((lnR-lnn(α))/lnb)-2](θlnR)/lnb=1θ>1.
There exists a constant R(α)>n(α) such that
(48)lnR(α)-lnn(α)>2lnb,(49)lnρ-lnn(α)lnb-2≥θlnρlnb>0,∀ρ>R(α).
Given ρ>R(α), there exist two integers n~, n¯ such that
(50)Rn¯≤n(α)≤Rn¯+1,Rn~-1≤ρ≤Rn~.
Thus
(51)n~-n¯>lnρ-lnn(α)lnb>2.
By Lemma 12, (49), and (51), we deduce that
(52)∫|x|>ρ‍(|∇u|2+V(x)|u|2)dx≤∫|x|>Rn~-1‍(|∇u|2+V(x)|u|2)dx≤[(1-t0)+δ1t0]n~-n¯-2≤×∫|x|>Rn¯+1‍[|∇u|2+V(x)|u|2]dx≤∥u∥2exp{ln[(1-t0)+δ1t0](lnρ-lnn(α)lnb-2)}≤∥u∥2exp{C(δ1,t0,θ)lnρ}.
That is, (45) holds.Proof of Theorem5.
Step 2. Let u be the weak solution found in H, and let r≥2, and |y|≥2r; then
(53)sup{1+|x|γa|y|γ:x∈Br(y)}≤1+(r+|y|)γa|y|γ≤sup{1+((3/2)|y|)γa|y|γ:|y|≥4},
where γ=α+δ1. Denote
(54)C1(γ)=sup{1+((3/2)|y|)γa|y|γ:|y|≥4}=14γa+3γ2γa.
Then
(55)sup{1+|x|γa:x∈Br(y)}≤C1(γ)|y|γ.
SinceBr(y)⊂{x∈ℝN:|x|≥|y|/2}, γ>α, by Lemma 13 and (55), for all |y|≥2R(α), we have
(56)∫Br(y)‍u2dx≤∫Br(y)‍1+|x|γa1+|x|α1+|x|γV(x)u2dx≤C1(γ)|y|γ∫Br(y)‍V(x)u2dx≤C1(γ)|y|γ∫|x|≥|y|/2‍V(x)u2dx≤C1(γ)2-C(δ1,t0,θ)|y|γ∥u∥2|y|C(δ1,t0,θ)=C2∥u∥2|y|γ+C(δ1,t0,θ),
where C2=C1(γ)2-C(δ1,t0,θ).
With similar arguments of [15], let m∈ℕ+ with |yi|≥2(i=1⋯m) be such that
(57)B5∖B2⊂⋃i=1m‍B1(yi).
LetK0 denote a positive integral such that 2K0>R(α); by (56) we deduce that
(58)∫|x|≥2‍u2dx≤∑k=0∞‍∫2k(B5∖B2)u2dx≤∑i=1m‍∑k=0∞‍∫B2k(2kyi)‍u2dx≤∑i=1m‍∑k=0K0-1‍∫B2k(2kyi)‍u2dx+C2∥u∥2∑i=1m‍∑k=K0∞‍|2kyi|γ+C(δ1,t0,θ).
Since γ+C(δ1,t0,θ)<-1, then the above inequality implies that ∫|x|≥2‍u2dx<+∞ and u∈L2(ℝN). Thus, u∈H1(ℝN) is a bound state of problem (5).Before the proof of Theorem6, let us introduce some functional spaces. Denote by ℋ the weighted Sobolev space
(59)ℋ:={Dr1,2(ℝN):∫ℝN(|∇u|2+V(|x|)u2)dx<∞},
where 𝒟r1,2={u∈L2*(ℝN):u(x)=u(|x|),∇u∈L2(ℝN)}.Denote byℒKq the weighted space of measurable u:ℝN⟶ℝ such that
(60)|u|q,K=(∫ℝN‍K(|x|)|u(x)|qdx)1/q<∞.Now we give the following embedding result, whose proof can be derived from Theorem 1 of [19].Proposition 14.
LetN≥3, and suppose that V(x), K(x) are radial, and the assumptions (V), (K) hold with α∈(0,2) and β>0. Then
(61)ℋ↪ℒKq,
where q*≤q≤2N/(N-2). Furthermore, the embedding is compact if q∈(q*,2N/(N-2)).Proof of Theorem6.
Based upon Proposition14, the proof of Theorem 6 can be followed from some standard techniques; we leave the details to the readers.



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