# Infinitely Many Solutions of Schrödinger-Poisson Equations with Critical and Sublinear Terms

Authors: Xianzhong Yao; Xia Li; Fuchen Zhang; Chunlai Mu
Journal: Advances in Mathematical Physics
 (2019)

---

## Abstract

In this paper, we study the following Schrödinger-Poisson equations−Δu+u+ϕu=u5+λaxup−1u,x∈ℝ3,−Δϕ=u2,x∈ℝ3, where the parameter λ>0 and p∈0,1. When the parameter λ is small and the weight function ax fulfills some appropriate conditions, we admit the Schrödinger-Poisson equations possess infinitely many negative energy solutions by using a truncation technology and applying the usual Krasnoselskii genus theory. In addition, a byproduct is that the set of solutions is compact.


---

## Body

## 1. Introduction and Main Results

In the present paper, we are interested in the existence of infinitely many negative energy solutions of the following Schrödinger-Poisson equations:(1)−Δu+u+ϕu=u5+λaxup−1u,x∈ℝ3,−Δϕ=u2,x∈ℝ3,where the parameter λ>0, p∈0,1, and ax is a positive continuous weight function satisfying a∈L2/1−pℝ3.In recent decades, the Schrödinger-Poisson system has been studied widely by many authors, because it has strong physical background and interesting meaning. It arises in nonlinear quantum mechanics models and semiconductor theory. From a physical viewpoint, the system describes the interaction between identical charged particles, when the magnetic effects could be ignored in the interaction with each other and its solution is a standing wave for such a stationary system. The nonlinearity models the mutual interaction between many charged particles. The system consists of a Schrödinger equation coupled with a Poisson equation, which implies that the potential is determined by the charge of the wave function. The nonlocal termϕu means that the particles interact with its own electric field. For more information about the mathematical and physical background of the system, we refer the readers to see papers [1–4] and the references therein.The studies of the Schrödinger-Poisson system have been focused on the existence of positive solutions, ground state solutions, multiplicity of solutions, radial solutions and the semiclassical limit solutions, concentration behavior of solutions, and sign-changing solutions. See references [5–17] and the references therein.When the nonlinear term is presented as a subcritical growth, there are many results in the literature. Ruiz [18] studied the following system:
(2)−Δu+u+λϕu=up,x∈ℝ3,−Δϕ=u2,x∈ℝ3,where the parameter λ>0 and p∈1,5. When λ is small, the author showed that there exists at least one positive radial solutions for p∈2,5, and at least two positive radial solutions for p∈1,2. In particular, if λ≥1/4, the author proved that p=2 is a threshold of existence and nonexistence of positive radial solutions. When λ=1 in system (2), Azzollini and Pomponio [19] established the existence of ground state solution for p∈2,5. For related system and more results, please refer readers to see [20–28].In the paper, we are concerned with a critical growth of nonlinearity term and perturbation of low order terms. In this case, there are some results in the references. As regards the following relevant system,(3)−Δu+Vxu+λϕu=u5+μup,x∈ℝ3,−Δϕ=u2,x∈ℝ3,where the parameter λ,μ>0 and p∈1,5, under some suitable conditions, existence of a nontrivial solution was proved in [19] for 3<p<5 and in [29] for 1<p<5. Here, we would like to mention some other papers [30–34] for related results. We note that the existence of solutions is very seriously depending on the range of the p. As far as we know, there is no result of Schrödinger-Poisson system involving the combination with a critical nonlinearity and sublinear terms.The compactness of the imbeddingH1ℝ3 into Lpℝ32≤p≤6 does not hold, and the nonlocal term ϕuu and the critical nonlinear term u5 appear in the system, which cause many difficulties for us using the variational methods in a standard way to solve the Schrödinger-Poisson system.Motivated by works mentioned above, particularly, by the results in [16, 24, 29, 35], we overcome these difficulties mentioned above and obtain the existence of infinitely many negative energy solutions to system (1) for p∈0,1 and small λ>0.We denote byS the best constant for the Sobolev space D1,2ℝ3 imbedding into the Lebesgue space L6ℝ3, namely, D1,2ℝ3↪L6ℝ3,
(4)S≔infu∈D1,2ℝ3\0∫ℝ3∇u2dx∫ℝ3u6dx1/3.Now, we give our main result as follows.Theorem 1.
Assumep∈0,1. Then, there exists a positive constant Λ such that system (1) possesses infinitely many negative energy solutions for any λ∈0,Λ. Moreover, the set of solutions obtained above is compact.Remark 2.
Whenλ≤0, the system (1) has no solution, which follows from Pohožaev’s identity (see [36]). To some extent, we extend the results in [16, 24, 29, 35].Remark 3.
The key ingredient in the proof of Theorem1 is the genus theory, which plays an important role in obtaining infinitely many solutions of Schrödinger-Poisson equations (1). We followed the methods of Yao and Mu in [37], where the authors studied nonlocal problem of Kirchhoff-type in high dimension (N≥4).The remainder of this paper is organized as follows. In Section2, we present the abstract framework of the problem as well as some preliminary results. Theorem 1 shall be proved in Section 3.


## 2. Preliminaries and Functional Setting

In this section, we will define some notations and establish the variational setting for Schrödinger-Poisson equations (1) and list some fundamental results.
(i)
LetH1ℝ3 be the usual Sobolev space endowed with the standard inner product and induced norm
(5)u,v=∫ℝ3∇u⋅∇v+uvdx,u=∫ℝ3∇u2+u2dx1/2
(ii)
Lpℝ3 is the usual Lebesgue space equipped with the norm
(6)up=∫ℝ3updx1/p,for1≤p<∞,u∞=esssupuxx∈ℝ3
(iii)
D1,2ℝ3 is the completion of C0∞ℝ3 with respect to the norm
(7)uD1,2ℝ3≔∫ℝ3∇u2dx1/2
(iv)
The lettersC and Cii=1,2,⋯ denote various positive constants which may vary from line to line and whose exact values are irrelevant
(v)
The notations→ and ⇀ mean strong convergence and weak convergence in corresponding to functional setting, respectively
(vi)
We useo1 to denote any infinitely small quantity that tends to zero as n→∞For anyu∈H1ℝ3, the Lax-Milgram theorem implies that there exists a unique ϕu∈D1,2ℝ3 such that, for any v∈D1,2ℝ3,
(8)∫ℝ3∇ϕu⋅∇vdx=∫ℝ3u2vdx,that is, ϕu is the weak solution of −Δϕ=u2. Furthermore,
(9)ϕux=14π∫ℝ3u2yx−ydy.We then can rewrite Schrödinger-Poisson equations (1) as the following:
(10)−Δu+u+ϕuu=u5+λaxup,x∈ℝ3,and energy functional associated with equation (10) is
(11)Iu=12∫ℝ3∇u2+u2dx+14∫ℝ3ϕuu2dx−16∫ℝ3u6dx−λp+1∫ℝ3axup+1dx.It is readily seen that the energy functionalI belongs to C1H1ℝ3,ℝ and that
(12)I′u,v=∫ℝ3∇u⋅∇v+uvdx+∫ℝ3ϕuuvdx−∫ℝ3u4uvdx−λ∫ℝ3axup−1uvdx,for any v∈H1ℝ3. Hence, if u∈H1ℝ3 is a critical point of functional I, then u is a solution of equation (10) and u,ϕu∈H1ℝ3×D1,2ℝ3 is a solution of system (1). We denote Φu≔ϕu for simple expressions.In what follows, we start to state our preliminary results.Lemma 4.
Φ satisfies the following results:
(1)
Φ is continuous in H1ℝ3 and Φ≥0
(2)
Ifun⇀u in H1ℝ3, then Φun⇀Φu in D1,2ℝ3
(3)
Φtu=t2Φu for any t∈ℝ
(4)
ϕuD1,2ℝ3≤Cu2. Furthermore, ∫ℝ3ϕuu2dx≤C~u4The proof is omitted here and refers to [18, 29].Lemma 5.
Assumeun is a PSc for functional I in H1ℝ3. Then, un is bounded in H1ℝ3.Proof.
Arguing by contradiction, assumeun→∞ as n→∞. According to the Sobolev and Hölder inequality and p∈0,1, we find that
(13)c+01=Iun−14I′un,un=14un2−λ1p+1−14∫ℝ3axunp+1dx+112∫ℝ3un6dx≥14un2−λ1p+1−14a2/1−pun2p+1≥14un2−λCunp+1→n→∞∞.
This is a contradiction, andun is bounded in H1ℝ3.Lemma 6.
Supposeun is a PSc for functional I in H1ℝ3. Then, un has a convergent subsequence in H1ℝ3 provided that c<1/3S3/2.Proof.
By Lemma5, we know that un is bounded in H1ℝ3, and up to a subsequence, there exists a u∈H1ℝ3 such that
(14)un⇀uinH1ℝ3,un→uinLlocpℝ3forp∈1,6,un→ua.einℝ3.
In light of Lions’ second concentration compactness lemma [38], there exist an at most countable index set J, a sequence of points xjj∈J⊂ℝ3, and values μjj∈J, νjj∈J⊂ℝ+ such that
(15)∇un2⇀dμ≥∇u2+∑j∈Jμjδxj,un6⇀dv=u6+∑j∈Jvjδxj,Svj1/3≤μj,in the measure sense, where δxj is the Dirac mass at xj. We next shall prove that the index set J is empty. By the reduction to absurdity, let us suppose that there exists a j0 such that νj0≠0. Consider that some cut-off function ϕ∈C0∞ℝ3:ℝ3→0,1 such that
(16)ϕ≡1onBxj0,ε,ϕ=0onBxj0,2εc,∣∇ϕ∣≤2ε.
Obviously, the sequenceϕun is bounded in H1ℝ3, and because un is a PSc sequence of functional I, then limn→∞I′un,ϕun=0, i.e., for large n, there is
(17)∫ℝ3un∇un∇ϕdx+∫ℝ3∇un2ϕdx+∫ℝ3ϕun∇un2ϕdx=λ∫ℝ3axunp+1ϕdx+∫ℝ3un6ϕdx+o1.
Let us start to estimate each terms in the equation above. Using the Hölder inequality and (15), we compute simply
(18)0≤∫ℝ3un∇un∇ϕdx≤∫B2εxj0un2dx1/2∫B2εxj0un2∇ϕ2dx1/2≤C1∫B2εxj0un2∇ϕ2dx1/2≤C1∫B2εxj0un6dx1/6∫B2εxj0∇ϕ3dx1/3≤C2∫B2εxj0un6dx1/6→ε→00,limε→0limn→∞∫ℝ3∇un2ϕdx≥μj0,(19)limε→0limn→∞∫ℝ3un6ϕdx=limε→0∫ℝ3u6ϕdx+νj0=νj0,(20)limε→0limn→∞λ∫ℝ3axunp+1ϕdx=limε→0limn→∞λ∫B2εxj0axunp+1ϕdx=λlimε→0∫B2εxj0axup+1ϕdx=0.
Together with formulas above, we note(21)νj0≥μj0.
By combining this inequality with the third formula in (15), we show that
(22)νj0≥S3/2.
Take a cut-off functionψ∈C0∞ℝ3:ℝ3→0,1 such that
(23)ψ≡1onB0,R,ψ=0onB0,2Rc,∣∇ψ∣≤3R.
For simplicity of computation, denote(24)α≔14∫ℝ3u2dx−λ1p+1−14∫ℝ3axup+1dx.
By Hölder’s inequality, we easily know that(25)α≥14u22−λ1p+1−14a2/1−pu2p+1,since ax∈L2/1−pℝ3 and p∈0,1, there exists some positive constant Λ such that α>0 for λ∈0,Λ. Using (15), (21), and (22), we obtain
(26)c=limn→∞Iun−14I′un,un=limn→∞14∫ℝ3∇un2dx+14∫ℝ3un2dx−λ1p+1−14∫ℝ3axunp+1dx+112∫ℝ3un6dx>limR→∞limn→∞14∫ℝ3∇un2ψdx+14∫ℝ3un2ψdx−λ3−p4p+1∫ℝ3axunp+1ψdx+112∫ℝ3un6ψdx>limR→∞14∫ℝ3∇u2ψdx+14μj0+112∫ℝ3u6ψdx+112vj0+α>14μj0+112vj0>14Svj0+112vj0>14S3/2+112S3/2>13S3/2.
This is a contradiction with the hypothesis, so the index setJ is empty.
TakeT>0 and define
(27)μ∞≔limT→∞limsupn→∞∫x>T∇un2dx,ν∞≔limT→∞limsupn→∞∫x>Tun6dx.
In line with lemma 1.40 in [39], μ∞ and ν∞ fulfill the following formulas
(28)limsupn→∞∫ℝ3∇un2dx=∫ℝ3dμ+μ∞,limsupn→∞∫ℝ3un6dx=∫ℝ3dν+ν∞,Sν∞1/3≤μ∞.
Take a cut-off functionφR∈C0∞ℝ3:ℝ3→0,1 such that
(29)φR≡0onB0,R,φR=1onB0,2Rc,∣∇φR∣≤3R.
As above, the sequenceφRun is bounded in H1ℝ3 and limn→∞I′un,φRun=0, namely,
(30)∫ℝ3un∇un∇φRdx+∫ℝ3∇un2φRdx+∫ℝ3ϕunun2φRdx=λ∫ℝ3axunp+1φRdx+∫ℝ3un6φRdx+o1.
We need again to estimate every terms in the equation above. By the Hölder inequality and the definition ofφR, it follows that
(31)0≤limR→∞limsupn→∞∫ℝ3un∇un∇φRdx≤limR→∞limsupn→∞∫B0,2R/B0,R∇un2dx1/2∫B0,2R/B0,R∇φR2un2dx1/2≤ClimR→∞∫B0,2R/B0,R∇φR2u2dx1/2≤ClimR→∞∫B0,2R/B0,R∇φR3dx1/3∫B0,2R/B0,Ru6dx1/6≤C1limR→∞∫B0,2R/B0,Ru6dx1/6=0,(32)limR→∞limsupn→∞∫ℝ3∇un2φRdx≥limR→∞limsupn→∞∫x>R∇un2dx=μ∞,(33)limR→∞limsupn→∞∫ℝ3un6φRdx=limR→∞limsupn→∞∫x>Run6φRdx≤limR→∞limsupn→∞∫x>Run6dx=v∞.
According to the estimates above, we easily show thatμ∞≤ν∞. Combination with (28), we know that ν∞=0 or ν∞≥S3/2. If ν∞≥S3/2 holds, then we have
(34)C=limn→∞Iun−14I′un,un>limn→∞14∫ℝ3∇un2dx+14∫ℝ3un2dx−λ4∫ℝ3axunp+1dx+112∫ℝ3un6dx>14∫ℝ3∇u2dx+14μ∞+112∫ℝ3u6dx+112v∞+α>14μ∞112v∞>14Sv∞1/3112v∞>14S3/2112S3/2>13S3/2.
This is a contradiction with the assumption, thusν∞=0 holds. Together (15) and (28) with the empty index set J, we obtain that
(35)limsupn→∞∫ℝ3un6dx=∫ℝ3u6dx.
By Fatou’s lemma, we prove that(36)∫ℝ3u6dx≤liminfn→∞∫ℝ3un6dx≤limsupn→∞∫ℝ3un6dx=∫ℝ3u6dx.
Therefore,un→u in L6ℝ3. Set wn≔un−u, and wn⇀0 in H1ℝ3. Suppose limn→∞wn=w. Since I′ is sequentially weakly continuous in H1ℝ3, we have that limn→∞I′wn→0 and
(37)0=limn→∞I′wn,wn=limn→∞∫ℝ3∇wn2+wn2dx+∫ℝ3ϕwnwn2dx−λ∫ℝ3axwnp+1dx−∫ℝ3wn2dx≥limn→∞∫ℝ3∇wn2+wn2dx−λ∫ℝ3axwnp+1dx≥limn→∞wn2−Cλwnp+1≥w2−Cλwp+1.
Thus for smallλ>0, this formula forces w=0 and wn→0 as n→∞, and un→u in H1ℝ3. The proof is finished.


## 3. The Proof of Theorem1

In this section, we shall utilize the Krasnoselskii genus theory to establish a minimax class of critical points for proving that Schrödinger-Poisson equations (1) possess infinitely many negative energy solutions. And we next need to introduce the classical concept and some properties of the genus.LetE be a real Banach space and Γ denote the family of sets A⊂E\0 such that A is closed in E and symmetric with respect to 0, namely, x∈A implies −x∈A. For any A∈Γ, we define the genus of A as follows:
(38)γA≔mink∈ℕ:∃φ∈CA,ℝk\0suchthatφxisodd.If there is no finitek, then define γA=∞, and γ∅=0 by definition of the genus. For A∈Γ and δ>0, we denote by NδA a uniform δ-neighborhood of A, that is,
(39)NδA=x∈E:distx,A≤δ.In what follows, we shall list some properties of the genus that prepare for showing our results. More detail content about the genus may be seen in the references ([40], Propositions 7.5-7.8).Proposition 7.
AssumeA,B∈Γ. Then, following several results hold.
(1)
Normalization: ifx≠0, then γx∪−x=1
(2)
Mapping property: if there is an odd mapf∈CA,B, then γA≤γB
(3)
Monotonicity property: ifA⊂B, then γA≤γB
(4)
Subadditivity:γA∪B≤γA+γB
(5)
Continuity property: ifA is compact, then γA<∞ and there exists a δ>0 such that NδA∈Γ and γNδA=γA
(6)
If there is an odd homeomorphism betweenA and B, then γA=γB
(7)
IfSN−1 is the sphere in ℝN, then γSN−1=N
(8)
IfγB<∞, then γA−B¯≥γA−γB
(9)
IfY is a subspace of X with codimension k and γA≥k, then A∩Y≠∅.Proposition 8.
AssumeA∈Γ. If γA≥2, then set A includes infinitely many points.We define presently an auxiliary function, which essentially follows the idea and method in [41]. Denote
(40)ft≔12t2−λβtp+1−16S3t6,where β is a positive constant and determined below. By the Hölder and Sobolev inequality, it follows that
(41)Iu=12∫ℝ3∇u2+u2dx+14∫ℝ3ϕuu2dx−λp+1∫ℝ3axup+1dx−16∫ℝ3u6dx≥12u2−λp+1a2/1−pu2p+1−16S3u6≥12u2−λβup+1−16S3u6=fu.Forp∈0,1, we observe that f gets its positive unique maximum (see Figure 1).Figure 1To find the critical points of the energy functionalI, we truncate the functional I as the following:
(42)I~u≔12∫ℝ3∇u2+u2dx+14∫ℝ3ϕuu2dx−λp+1∫ℝ3axup+1dx−ωu6∫ℝ3u6dx,where ωu=τ∥u∥ and τ∈C∞:ℝ+→0,1 is a decreasing function and satisfies
(43)τt=1,t≤t0,τt=0,t≥T,where t0 and T are two roots of the function f (see Figure 1) and t0<T. As above, we have that I~u≥f~u with
(44)f~t≔12t2−λβtp+1−τt6S3t6,and see Figure 2. Obviously, I~∈C1 and if I~u≤0, then we observe that ∥u∥≤t0 and Iu=I~u.Figure 2Therefore, we only need to find some negative critical values for the truncating functionalI~. To do it, we next shall give some lemmas for constructing the minimax sequence of negative critical values of the truncating functional I~.Before to state the following results, we give some definitions. Denote(45)Γk≔A∈Γ:γA≥k,Kc≔u∈H1ℝ3:I~u=c,I~′u=0,(46)I~d≔u∈H1ℝ3:I~u≤d,and define
(47)ck≔infA∈Γksupu∈AI~u.Lemma 9.
For any positive integern∈ℕ, there is a ε=εn>0 satisfying γI~−ε≥n.Proof.
To show this lemma, we adopt to the argument used in [41], which handles with the local problem. Fix any n and suppose En is an n-dimensional subspace of H1ℝ3. Taking v∈En with v=1 and t∈0,t0, by Figures 1 and 2, we observe
(48)I~tv=Itv<0,then, we can choose ε>0 and t<t0 such that I~tu≤−ε for u∈En and u=1. Denote
(49)S≔u∈En:u=t,for t<t0, it is easy to know γS=n and note
(50)S⊂u∈H1ℝ3:I~u≤−ε.
BecauseEn and ℝn are isomorphic and S and Sn−1 are homeomorphic, it follows from Proposition 7 that
(51)γu∈H1ℝ3:I~u≤−ε≥γS=n.Lemma 10.
Assumeλ∈0,Λ. If c=ck=ck+1=⋯=ck+r for some integer r∈ℕ, then γKc≥r+1.Proof.
We first claim that eachck is negative. Indeed, by Lemma 4, for every integer k∈ℕ, there exists ε>0 such that γI~−ε≥k. Because functional I~ is continuous in H1ℝ3 and even I~−ε∈Γk. In virtue of the definition of ck and low boundedness of I~, we know
(52)−∞<ck=infA∈Γksupu∈AI~u≤supu∈I~−εI~u≤−ε<0.
Then, we easily show thatKc is compact by Lemma 6. In the following argument, we will prove the expected result by contradiction. Assume γKc≤r, according to the fifth in Proposition 7, then there exists a closed and symmetric set U with Kc⊂U such that γU=γKc≤r. Due to c<0, we can choose U such that U⊂I~0. Owing to the deformation lemma [42], there is an odd homeomorphism
(53)η:H1ℝ3→H1ℝ3,such that ηI~c+δ−U⊂I~c−δ for some δ with 0<δ<−c. So I~c+δ⊂I~0 and then by the definition of c=ck=cc+k, there exists A∈Γk+r such that supu∈AI~<c+δ, that is, A⊂I~c+δ and
(54)ηA−U⊂ηI~c+δ−U⊂I~c−δ.
It means(55)I~ηA−U≤c−δ.
On the other hand, because ofγAU¯≥γA−γU≥k and γηAU¯≥γAU¯≥k, then ηAU¯∈Γk, which implies
(56)supu∈ηAU¯I~u≥ck=c.
This is a contradiction with (54). Therefore, the proof is finished.Proof of Theorem 1.
By the analysis above, ifck<0, then we know that I~u=Iu. Thus, it is sufficient to show that the truncating functional I~ possesses infinitely many negative critical values. Noting that ck is nondecreasing with respect to k, there are only the following two cases:Case I.
There are1≤k1<⋯<ki<⋯ satisfying −∞<ck1<ck2<⋯<cki<⋯<0. In this case, obviously, the functional I~ has infinitely many negative critical values ck. It is done.Case II.
There is a positivel such that c=ck<0 for all k≥l. In the case, by Lemma 10, we know γKc=∞, which implies that the set Kc has infinitely many points, which means that the functional I~ possesses infinitely many negative critical values. Finally, we shall see that the set of solutions is compact.
Letuk be a sequence of solutions to the equations (1). In Case I, we know that 0≥ck=I~uk and I~′uk=0; by Lemma 5, we then easily get that there is a u∈H1ℝ3 satisfying uk→u as k→∞. Then, it follows conclusion. In Case II, it is trivial to see that the set Kc is compact. Thereupon, the proof is completed.



---

*Source: 8453176-2019-12-29.xml*