# Existence and Multiplicity of Solutions for Sublinear Schrödinger Equations with Coercive Potentials

Authors: Dong-Lun Wu
Journal: Mathematical Problems in Engineering
 (2019)

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## Abstract

In this study, we consider the following sublinear Schrödinger equations−Δu+Vxu=fx,u,forx∈ℝN,ux⟶0,asu⟶∞, where fx,u satisfies some sublinear growth conditions with respect to u and is not required to be integrable with respect to x. Moreover, V is assumed to be coercive to guarantee the compactness of the embedding from working space to LpℝN for all p∈1,2∗. We show that the abovementioned problem admits at least one solution by using the linking theorem, and there are infinitely many solutions when fx,u is odd in u by using the variant fountain theorem.


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## Body

## 1. Introduction and Main Results

In this study, we consider the following nonlinear Schrödinger equations:(1)−Δu+Vxu=fx,u,forx∈ℝN,ux⟶0,asu⟶∞,where fx,u∈CℝN×ℝ,ℝ. The existence and multiplicity of solutions for (1) have attracted much attention of mathematicians. As we know, the existence of solutions for problem (1) is important in finding the solutions for(2)iℏ∂φ∂t=−ℏ22mΔφ+Wxφ−bx,φφ.With the development of the variational methods in past decades, many mathematicians used this tool to show the existence and multiplicity of solutions for differential equations with superlinear, sublinear, asymptotic linear, and mixed nonlinearities (see [1–25]). In this study, we mainly consider the sublinear case. Many mathematicians have considered this topic with fx,t being integrable in x. For example, in 2012, Zhang and Wang [23] assumed V to be coercive and fx,t satisfies the following: F0fx,t=μhxtμ−2t, where μ∈1,2 is a constant and h:ℝN⟶ℝ is a positive continuous function such that h∈L2/2−μℝN.In addition, they obtained infinitely many nontrivial solutions for (1). By using the dual fountain theorem, Bahrouni [1] showed (1) admitted infinitely many solutions when fx,t=axup−1u+bxuq−1u, where 0<q<p<1 and a, b satisfy the following conditions:F1a∈L∞ℝN∩​L2/1−pℝN and there exists β>0 such that ax<−β for any x∈ℝN.F2b∈Lp+1/p−qℝN and b>0.Later, Chen and Tang [6] relaxed assumption F0, and they obtained the following theorem.Theorem 1.
(see [6])Suppose the following conditions holdF3V∈CℝN,ℝ and infℝNVx>0.F4 For any x,t∈ℝN×ℝ, there exist 1<r1<r2<2, a1∈L2/2−r1ℝN,0,+∞, and a2∈L2/2−r2ℝN,0,+∞ such that(3)fx,t≤r1a1xtr1−1+r2a2xtr2−1,forx,t∈ℝN×ℝ.F5 There exist an open set J∈ℝN and δ,η>0, r3∈1,2 such that Fx,t≥ηtr3,∀x,t∈J×−δ,δ, where Fx,t≔∫0tfx,sds, x∈ℝN,t∈ℝ.Then, problem (1) possesses at least one nontrivial solution. Moreover, if fx,−u=−fx,u for all x,u∈ℝN×ℝ, then problem (1) possesses infinitely many nontrivial solutions.In Theorem 1,V is not required to be coercive, and the authors took advantage of the decay of aix to show the convergence of the asymptotic critical points sequence. Recently, Zhang et al. [24] introduced some different sublinear conditions of f to obtain infinitely many nontrivial solutions for (1). They assumed the weight functions to be integrable in L2∗/2∗−ζℝN∩​LpℝN with ζ∈1,2 and p>N/2.After then, Hou et al. [7] considered the fractional Schrödinger-type equations with the similar conditions to F4 and F5. In the abovementioned studies, the results depend on the integrability of the weight functions. There are still some mathematicians considered the case without any integrable weight functions. In 2009, Benrhouma et al. [3] considered the following problem:(4)−Δu+Vxu+axupsgnu=gx,forx∈ℝN,u∈H1ℝN∩​Lp+1ℝN,where Vx=ax=1 and g∈L2ℝN/0. They used the minimizing method on a Nehari manifold and concentration-compactness principle due to Lions [10, 11] to show the existence of unique positive solution for (2). Then, Benrhouma [2] improved this result by assuming the following:F6V∈L∞ℝN and limx⟶+∞=v∞≥0.F7a∈L∞ℝN, limx⟶+∞=a∞>0 ax>α>0 a.e. x∈ℝN.In 2013, Bao and Han [5] considered (2), where g≡0 and V is a sign-changing function while 0 lies in the spectrum of −Δ+Vx and obtained infinitely many solutions for this resonant problem. In a recent study, Isernia [8] generalizes the results in [2] to a class of sublinear fractional Schrödinger equations. Motivated by the abovementioned studies, we introduce some new sublinear growth conditions without weight functions to obtain the existence and multiplicity of solutions for (1). First of all, we recall the following coercive condition which is proposed by Wang and Han [14] to show the existence of infinitely many small solutions for a class of Kirchhoff equations with local sublinear nonlinearities by using the symmetric mountain pass theorem.V There are constants ρ>0 and α>N such that for any b>0(5)limy⟶+∞measx∈Bρy:Vxxα≤b=0.Remark 1.
ConditionV is introduced by Wang and Han in [14] to obtain a compactness embedding theorem. In addition, the authors assumed that V possesses the global positive lower bound. Although we do not assume that V is positive in the whole space in this study, the compactness embedding theorem still holds in the same way.Now, we state our main results.Theorem 2.
Suppose thatV∈CℝN,ℝ satisfies V and f satisfies the following:f1 There exist b1>0 and 1<β<2 such that(6)Fx,t≥b1tβ,for allx,t∈ℝN×ℝ.f2Fx,0=0 for all x∈ℝN and there exists M0>0 such that(7)fx,t≤M01+t,for allx∈ℝNandt∈ℝ.f3Limt⟶∞fx,t/t=0 uniformly in x.f4 Letting F˜x,t=2Fx,t−fx,tt, there exist b2, r1>0, and μ>1 such that(8)F˜x,t≥b2tμ,fort≥r1andx∈ℝN.f5 There exists b3>0 such that(9)F˜x,t≥−b3t,fort≤r1andx∈ℝN.Then, there exists at least one nontrivial solution for problem (1).Theorem 3.
Suppose that the conditions of Theorem 2 hold andf6Limt⟶∞∫ℝNFx,tdx=+∞.f7Fx,−t=Fx,t for any x∈ℝN and t∈ℝ.
Then, there exist infinitely many solutions for problem (1).Remark 2.
In Theorem 2,Fx,t is sublinear at infinity and can be a sign-changing function. Let(10)Fx,t=t5/4−3t3/2+t7/4.
Then,Fx,t satisfies the conditions of Theorem 2, but the conditions in [1–8, 20, 23–25].


## 2. Preliminaries

First, we show the decomposition of the space. Similar to [15], we denote by A the self-adjoint extension of the operator −Δ+Vx with the domain DA⊂L2ℝN,ℝ. Let ℰϑ:−∞<ϑ<∞ be the spectral family of A and A1/2 be the square root of A. Define u=id−ℰ0−ℰ−0. u commutes with A, and A=uA is the polar decomposition of A. Let E=DA1/2 be the domain of A1/2. Define the norm on E as(11)u,wo=A1/2u,A1/2wL2+u,wL2,uo=u,uo1/2,where and in the sequel, ⋅,⋅L2 denotes the usual L2ℝN inner product. Then, E is a Hilbert space. The following compact embedding theorem is obtained with the condition V.Lemma 1.
Suppose thatVholds. Then, E is compact embedded into LpℝN for any p∈1,2∗.Proof.
The proof of this lemma is similar to the proof of Lemma 1 in [14].Remark 3.
From Lemma 1, for anyp∈1,2∗, there exists a constant ηp>0 such that(12)up≤ηpu,for  allu∈E.
It follows from Lemma 1 that the spectrumσA consists of eigenvalues numbered in λ1≤λ2≤⋯⟶∞, and a corresponding system of eigenfunctions enAen=λnen forms an orthogonal basis in L2. Let n−=#iλi<0, n0=#iλi=0, and n¯=n−+n0. Set E−=spane1,…,en−, E0=spanen−+1,…,en¯, and E+=spanen¯+1,…¯. Then, E=E−⊕E0⊕E+ with(13)u,w=A1/2u,A1/2wL2+u0,w0L2,u=u,u1/2,where u=u−+u0+u+ and w=w−+w0+w+∈E=E−⊕E0⊕E+.
The following lemma can easily be obtained byf2. We omit its proof here.Lemma 2.
Suppose thatf2 holds, then for any ε>0, there exists dε>0 such that(14)Fx,t≤2M0t+t2,for allx∈ℝNandt∈ℝ.
LetI:E⟶ℝ be the corresponding functional defined by(15)Iu=12u+2−u+2−∫ℝNFx,udx.
It is known that the critical points ofI are the solutions for problem (1). The following lemma shows that I is differentiable in E.Lemma 3.
Suppose the conditions of Theorem 2 hold, thenI∈C1E,ℝand(16)I′u,v=∫ℝNu˙,v˙+Vxu,vdx−∫ℝNfx,uvdx,∀u,v∈E,which yields that(17)I′u,u=u+2−u+2−∫ℝNfx,uudx,∀u∈E.Proof.
By (12), it is easy to see that I is well defined. Subsequently, we show (16) holds. For any u, v∈E, ε>0, and 0<θ<1, it follows from Lemma 2 and (12) that(18)∫ℝNmaxs∈0,1fx,u+sθvvdx≤ε+dε∫ℝNmaxs∈0,11+u+sθvvdx≤ε+dε∫ℝN1+u+vvdx≤ε+dεv1+u2v2+v22≤ε+dεη1v+η22uv+η22v2<∞.
It follows from the definition ofI, (18), the mean value theorem, and the Lebesgue’s dominated convergence theorem that there exists 0<θ0<1 such that(19)I′u,v=lims⟶0+Iu+sv−Ius=∫ℝN∇u∇vdx+∫ℝNVxuvdx−lims⟶0+1s∫ℝNFx,u+sv−Fx,udx=∫ℝN∇u∇vdx+∫ℝNVxuvdx−lims⟶0+∫ℝNfx,u+sθ0vvdx=∫ℝN∇u∇vdx+∫ℝNVxuvdx−∫ℝNfx,uvdx.
Then, we obtain (16). Subsequently, we show I′ is continuous. Let(20)ψ′u,v=∫ℝNfx,uvdx,for anyv∈E.
Assume thatun⇀u in E as n⟶+∞. By Lemma 1, we obtain that(21)un⟶uinLpℝN,for allp∈1,2∗,which implies that for any ε>0 and v∈E with v=1, there exists T>0 such that(22)∫x≥Tun2dx≤ε,∫x≥Tu2dx≤ε,∫x≥Tvdx≤ε,∫x≥Tun2dx≤2∫x≥Tun−u2dx+∫x≥Tu2dx≤3ε,for n large enough. Since fx,t is continuous in t, we get(23)∫x<Tfx,un−fx,u2dx⟶0asn⟶∞.
Then, forn large enough, we have(24)supv=1ψ′unv−ψ′uv=supv=1∫ℝNfx,un−fx,uvdx≤supv=1∫ℝNfx,un−fx,uvdx=supv=1∫x<Tfx,un−fx,uvdx+∫x≥Tfx,un−fx,uvdx≤supv=1η2∫x<Tfx,un−fx,u2dx1/2+M0∫x≥T2+un+uvdx≤2M0+1η2ε+2M0ε,which yields that ψ′unv−ψ′uv⟶0 as un⟶u. Hence, ψ′ is weak continuous. Therefore, ψ is compact and I′ is continuous. The proof of this lemma is finished.


## 3. Proof of Theorem 1

In order to apply the link theorem by Rabinowitz (see [6], Theorem 5.29) to obtain the critical points of I, we assume J=−I in the following proof. Obviously, I and J share the same critical points.Lemma 4.
Suppose the conditions of Theorem 1 hold, thenJ satisfies the PS condition.Proof.
Letun⊂E be a PS sequence, then there exists a constant M1>0 such that(25)Jun≤M1,J′un≤M1.
First, we show thatun is bounded in E. Arguing in an indirect way, we assume that un⟶+∞ as n⟶∞. It follows from (25), f4 and f5 that(26)M12+un≥2Jun−J′un,un≥∫ℝNF˜x,undx≥−b3∫un≤r1undx+b2∫un≥r1unμdx.
Let(27)vn=un,forun≤r1,0,forun≥r1,wn=un−vn,then, we have(28)b2wnμμ≤M12+un+b3η1un.
SinceE−⊕E0 is a finite-dimensional subspace of E, there exists M2>0 such that(29)un−+un022=∫un≤r1un−+un02dx+∫un≥r1un−+un02dx≤∫un≤r1un−+un0undx+∫un≥r1un−+un0undx≤r1∫un≤r1un−+un0dx+∫un≥r1un−+un0wndx≤r1un−+un01+un−+un0μ/μ−1wnμ≤M2un−+un021+wnμ.
Moreover, there existM3, M4>0 such that(30)un−+un0≤M3un−+un02≤M2M31+wnμ≤M41+un1/μ,which implies that(31)un−+un0un⟶0asn⟶∞.
Choosing0<s<min1,μ/2, we can deduce from (26) and (12) that(32)∫un≥r1un2dx=∫un≥r1un2sun21−sdx≤∫un≥r1unμdx2s/μ∫un≥r1un2μ1−s/μ−2sdxμ−2s/μ≤M51+∫un≤r1undx+un2s/μun2μ1−s/μ−2s21−s≤M61+un2s/μun21−s≤M7un21−s+un2s+μ−sμ/μ,for some M5, M6, M7>0, where max21−s,2s+μ−sμ/μ<2. By (25), we obtain(33)M1un≥J′un1+un+≥un+2−∫ℝNfx,unun+dx≥un+2−∫ℝNfx,unun+dx.
Then, it follows fromf2, (32), and (33) that(34)un+2≤∫ℝNfx,unun+dx+M1un≤M0∫ℝN1+unun+dx+M1un≤M8un+∫un≤r1unun+dx+∫un≥r1unun+dx+M1un≤M8un+r1∫un≤r1un+dx+∫un≥r1un2dx1/2∫un≥r1un+2dx1/2≤M9un+un+un1−s+uns+μ−sμ/μ≤M9un+un2−s+uns+2μ−sμ/μ,for some M8, M9>0, which implies that un+/un⟶0 as n⟶∞. Together with (31), we obtain 1=un/un⟶0 as n⟶∞, which is a contradiction. Hence, un is bounded in E. Then, there exists u0∈E such that un⇀u0, which implies that un⟶u0 uniformly in LpℝN for any p∈1,2∗. Similar to (4), we deduce that(35)un++u0+2=−J′un−J′u0,un+−u0++∫ℝNfx,un−fx,u0un+−u0+dx≤J′unun++u0++J′u0,un+−u0++∫ℝNfx,un−fx,u0un+−u0+dx≤o1+M0∫ℝN2+un+u0un+−u0+dx≤o1+M0un22+u022∫ℝNun+−u0+2dx+2M0∫ℝNun+−u0+dx⟶0asn⟶∞,which shows that un⟶u0 as n⟶∞ since E−⊕E0 is of finite dimensions. Hence, J satisfies the PS condition.Lemma 5.
Suppose the conditions of Theorem 1 hold, then for anym∈ℕ, Em=E−⊕E0⊕spanen¯+1,…,en¯+m, there are ρ=ρm and δ=δm>0 such that J≥0 on Bρ∩​Em and J∂Bρ∩​Em≥δ, where Bρ=u∈E:u≤ρ.Proof.
Fromf1, for any u=u−+u0+se with u−+u0∈E−⊕E0 and s>0, there exists M10>0 such that(36)s2=se,seL2=se,uL2≤seβ/β−1uβ≤M10suβ.
SincedimEm<+∞, for u∈Em, one has(37)Ju=∫ℝNFx,udx+12u−2−12se2≥b1∫ℝNuβdx−12s2≥M11sβ−12s2,for some M11>0. Then, we can easily find the desired ρ and δ.Lemma 6.
Suppose the conditions of Theorem 1 hold, thenJu⟶−∞ as u∈E+.Proof.
For anyu∈E+, from f3, there exists r∞>0 such that(38)Fx,u≤14η22u2,for allx∈ℝNandt≥r∞,which, together with Lemma 2, implies that(39)Ju≤∫ℝNFx,udx−12u2≤14η22∫u≥r∞u2dx+2M0∫u≤r∞u+u2dx−12u2≤14u2+2M01+r∞∫u≤r∞udx−12u2≤2M0η11+r∞u−14u⟶−∞asϱ⟶+∞.
We finish the proof of this lemma.Proof of Theorem 2.
IfE−⊕E0=0, we can deduce from Lemma 4–6 that J possesses a minimax-type critical point, which is a solution for (1). If E−⊕E0≠0, let X=E−⊕E0⊕ℝen¯+1. Similar to the proof of Lemma 5, there exists s0>0 small and τ>0 such that(40)Ju≥τ,∀u∈S≡E−⊕E0+s0e.
In addition, by Lemma 6, we can chooser>s0 large enough such that(41)fu≤0,∀u∈E+withu≥r.
LetQ=Br∩​E+. Since S and ∂Q link, by the linking theorem (see [6], Theorem 5.29), J has a critical point u¯ with Ju¯≥δ.Proof of Theorem 3.
In order to obtain infinitely many solutions for problem (1), we need the variant fountain theorem by Zou in [25]. Let E be a Banach space with the norm ⋅ and E=⊕j∈ℕXj¯ with dimXj for any j∈ℕ. Set Yk=⊕j=0kXj and Zk=⊕j=k∞Xj¯. Consider the following C1−functional Iλ:E⟶ℝ defined by(42)Iλu=Au−λBu,λ∈1,2.Lemma 7.
(see [25])Suppose that the functional Iλu defined above satisfies the following:A1Iλ maps bounded sets to bounded sets uniformly for λ∈1,2. Furthermore,Iλ−u=Iλufor allλ,u∈1,2×E;A2Bu≥0; Bu⟶∞ as u⟶∞ on any finite dimensional subspace of E;A3 There exists ρk>rk>0such that(43)akλ≔infu∈Zk,u=ρkIλu≥0>bkλ≔maxu∈Yk,u=rkIλuforλ∈1,2;dkλ≔infu∈Zk,u≤ρkIλu⟶0ask⟶∞uniformlyforλ∈1,2.
Then, there existλn⟶1, uλn∈Yn such that Iλn′∣Ynuλn=0, Iuλn⟶ck∈dk2,bk1 as n⟶∞. In particular, if uλn has a convergent subsequence for every k, then I1 has infinitely many nontrivial critical points uk⊂E\0 satisfying I1uk⟶0− as k⟶∞.
Subsequently, we show I possesses infinitely many critical points whenFx,t is even in t, which are obtained by the variant fountain theorem. Define Xj=spanej for any j∈ℕ. Let E be decomposed as follows E=⊕j∈ℕXj¯. Set Yk=⊕j=1kXj and Zk=⊕j=k∞Xj¯. In the following proof, we show that I possesses geometrical structure of the variant fountain theorem. In order to make use of the variant fountain theorem, we set(44)Au=12u+2,Bu=12u−2+∫ℝNFx,udx,Iλu=Au−λBu.
Then, we have the following lemmas.Lemma 8.
There exists a sequenceρk⟶0+ as k⟶∞ such that(45)akλ≔infu∈Zk,u=ρkIλu≥0.Proof.
Set(46)βkp≔supu∈Zk,u=1up,for any p∈1,2∗. Since Zk+1⊂Zk⊂E+ for k>n¯+1, we can see that 0≤βk+1p≤βkp. So βkp⟶βp as k⟶∞ for some βp≥0 for any p∈1,2∗. For every k≥0, there exists uk∈Zk such that uk=1 and ukp>βkp/2. Furthermore, for any v∈E, we can see that v=∑i=1+∞liei, which implies that(47)uk,v=uk,∑i=k+∞liei≤uk⋅∑i=k+∞liei=∑i=k+∞li2⟶0ask⟶+∞.
Then, we haveuk⇀0 as k⟶∞ in E. It follows from Lemma 1 that uk⟶0 in LpℝN. Hence, we conclude that βp=0 for any p∈1,2∗. For any u∈Zk, by the definitions of Iλ and Lemma 2, we obtain(48)Iλu=12u2−∫ℝNFx,udx≥12u2−2M0∫ℝNu+u2dx≥12u2−2M0βk1u+βk22u2.
Letρk=16M0βk1. Obviously, we have ρk⟶0+ as k⟶∞. When k large enough, we can conclude that(49)akλ≔infu∈Zk,u=ρkIλu≥12ρk2−2M0βk1ρk+βk22ρk2≥14ρk2−2M0βk1ρk≥18ρk2≥0.
Moreover, it follows from (48) that(50)Iλu≥−2M0βk1ρk+βk22ρk2.
Therefore, whenk large enough, we can deduce that(51)0≥infu∈Zk,u≤ρkIλu≥−2M0βk1ρk+βk22ρk2.
Sinceβkp, ρk⟶0 as k⟶∞, we have(52)dkλ≔infu∈Zk,u≤ρkIλu⟶0ask⟶∞uniformly,forλ∈1,2.
Then, we obtain our conclusion.Lemma 9.
There exists a sequence0<rk<ρk for all k∈ℕ such that(53)bkλ≔maxu∈Yk,u=rkIλu<0,forλ∈1,2.Proof.
For anyk∈ℕ, dimYk=k. It follows from f1 that(54)Iλu≤12u+2−∫ℝNFx,udx≤12u2−b1uββ≤12u2−M12βuβ,for some M12>0 and any λ∈1,2. Let(55)0<rk<minρk,2M12β1/2−β.
Then, we obtain that(56)bkλ=maxu∈Yk,u=rkIλu<0,forλ∈1,2.
The proof is completed.
It is easy to see conditionA1 is satisfied in Lemma 4 since Fx,t is even in t. Similar to Lemma 4 in [13], it can easily be deduced from f6 that Bu⟶∞ as u⟶∞ on any finite dimensional subspace of E. Then, the condition A2 in Lemma 7 is satisfied. From Lemma 8 and Lemma 9, we can see that all the conditions of Lemma 1 are satisfied. Then, there exist λn⟶1, uλn∈Yn such that(57)Iλn′Ynuλn=0,Iλnuλn⟶ck∈dk2,bk1asn⟶∞.
Similar to Lemma 4, there existsuk such that uλn⟶uk in E as n⟶∞ for any k∈ℕ. Then, we can deduce that I=I1 possesses infinitely many nontrivial critical points. Therefore, (1) has infinitely many nontrivial solutions.



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