# Multiple Standing Waves for Nonlinear Schrödinger-Poisson Systems

Authors: Jian Zhou; Yunshun Wu
Journal: Journal of Function Spaces
 (2021)

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## Abstract

In this paper, we consider the following nonlinear Schrödinger-Poisson systems. Under suitable conditions onV, K, g, and h, when 1<s<6, we obtain two nontrivial solutions for the problem and when gx,· is odd and 6<s<∞, we obtain infinitely many solutions for the problem.


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## Body

## 1. Introduction

In this paper, we consider the following nonlinear Schrödinger-Poisson equations onℝ3(1)−Δu+Vxu+Kxϕu=gx,u−hxus−2u,inℝ3,−Δϕ=Kxu2,inℝ3.Such problem arises when one is looking for standing wave solutionsψt,x=e−iωt/ℏux for the nonlinear Schrödinger equation(2)iℏ∂ψ∂t=−ℏ2mΔψ+Uxψ+Kxϕψ−fx,ψψ,t,x∈ℝ+×ℝ3,coupled with the Poisson equation(3)−Δϕ=Kxu2,x∈ℝ3.It is well known that the Schrödinger-Poisson systems have a strong physical meaning because they appear in quantum mechanics models and in semiconductor theory (see [1–3]). Problem (1) is a special case of the following Schrödinger-Poisson system:(4)−Δu+Vxu+Kxϕu=fx,u,inℝ3,−Δϕ=Kxu2,inℝ3.Cerami and Vaira in [4] studied the existence of positive solutions for the problem where they considered Vx=1, fx,u=kxup−2u, and 4<p<6. After that, many researchers have focused on the problem under various conditions (see [5, 6]).In recent years, the Schrödinger-Poisson system has been studied widely under variant assumptions onV, K, and f (see [7–10]). Because the problem is set on the whole space ℝ3, it is well known that the main difficulty of this problem is the lack of compactness for Sobolev embedding, and then, it is usually difficult to prove that a minimizing sequence or a PS sequence is strongly convergent if we seek solutions by variational methods. In order to overcome this difficulty, most of them dealt with the situation where V is a positive constant or being radially symmetric (see [11–13]). When V is not a constant and not radially symmetric, there have been many works by developing various variational techniques (see [4, 14–17]).In paper [18], Sun et al. considered the system(5)−Δu+axu+ϕu=kxuq−2u−hxup−2u,inℝ3,−Δϕ=u2,limx⟶∞ϕx=0,inℝ3,where 1<q<2<p<+∞, ax, kx, and hx are measurable functions satisfying suitable assumptions. They obtained infinitely many solutions in H1ℝ3×D1,2ℝ3 with negative energy. Since 2<p<∞ is allowed to be supercritical, the usual Sobolev space H1ℝ3 cannot be used for the study; to overcome this difficulty we introduce a new space which is motivated by [19].We should also mention another recent paper [20]; Wang et al. considered a similar problem(6)−Δu+Vxu+ϕxu=kxuq−2u−hxup−2u+gx,inℝ3,−Δϕ=u2,inℝ3,and a nontrivial solution is obtained (see [20], Theorem 2). Note that there is an inhomogeneous term on the right-hand side of the first equation. Here 1<q<2<p<4 and the potential V satisfies a coercive condition so that the working space can be compactly embedded into Lebesgue spaces.Now, we turn to our problem (1). Since the space dimension is N=3, the critical Sobolev exponent 2∗=6. For p∈1,6 we denote(7)p0=2N2N−pN−2=66−p,p′=pp−1.Note thatp′ is the conjugate exponent in Hölder inequality. To state our results on the problem (1), we make the following assumptions:(V) V∈Cℝ3,ℝ satisfies 0<infℝ3Vx≤supℝ3Vx<+∞.(K) K∈L∞ℝ3∩L2ℝ3, Kx≥0 for a.e. x∈ℝ3.(g1) g∈Cℝ3×ℝ,ℝ; for some r∈1,2 and b∈L∞ℝ3∩Lr0ℝ3, we have(8)gx,u≤bxur−1,x,u∈ℝ3×ℝ.(g2) There exist δ>0, θ∈1,2, and a∈L∞ℝ3∩Lθ0ℝ3; ax≥0 such that(9)Gx,u=∫0ugx,τdτ≥axuθ,x,u∈ℝ3×−δ,δ.(h1) s∈θ,6 and h∈L∞ℝ3∩Ls0ℝ3; hx≥0 for x∈ℝ3.(h2) s>θ, h∈L∞ℝ3, and hx≥0 for x∈ℝ3.Remark 1.
It follows from (g1) and (g2) that r≤θ for ax>0 and δ small enough.Under these assumptions, it is clear that the zero functionux=0 is the trivial solution of problem (1). Our main results on the existence of multiple nontrivial solutions are the following theorems.Theorem 2.
Suppose (V), (K), (g1), (g2), and (h1) are satisfied; then, problem (1) has two nontrivial solutions.Theorem 3.
Suppose (V), (K), (g1), (g2), and (h2) are satisfied. If gx,· is odd, then problem (1) admits infinitely many solutions.The paper is organized as follows. In Section2, we give some useful notions and set up the variational framework of the problem. In Section 3, we prove Theorem 2, and the proof of Theorem 3 is given in Section 4. For simplicity, throughout this paper, we denote the norm on Lθ=Lθℝ3 with 1<θ<∞ by uθθ=∫ℝ3uθdx.


## 2. Preliminary

Thanks to condition (V), the norm(10)u=∫∇u2+Vxu21/2is an equivalent norm on H1=H1ℝ3. Let q∈2,6; then, we have a continuous embedding H1↪Lq. Hence, there is a constant Sq such that(11)uq≤Squ,u∈H1.For later use, we also denote byS the best Sobolev constant for the continuous embedding D1,2↪L6.Foru∈H1, it is well known that the Poisson equation(12)−Δϕ=Kxu2has a unique solution ϕ=ϕu in D1,2ℝ3. Note that according to [21], Theorem 2.2.1,(13)ϕux=14π∫Kyu2yx−ydy.Consequently,ϕu≥0 in ℝ3. We also know that there exists a1>0 such that(14)0≤14∫Kxϕuu2≤a1u4,forallu∈H1.See [14], Lemma 1.1.Define a functionalΦ:H1⟶ℝ,(15)Φu=12∫∇u2+Vxu2+14∫Kxϕuu2−∫Gx,u+1s∫hxus.Under our assumptions, it is easy to see thatΦ∈C1H1. According to Benci and his collaborators [1, 22], it is well known that if u is a critical point of Φ, then u,ϕu is a (weak) solution of (1).To find critical points ofΦ, some compactness conditions, such as the well-known Palais-Smale condition (PS for short), are crucial. To establish the PS condition for Φ, we need the following results.Proposition 4.
Letf∈CℝN×ℝ,ℝ such that(16)fx,u≤αxur−1+βxus−1,x,u∈ℝN×ℝ,for 1≤r≤2≤s≤2∗, α∈L∞ℝN∩Lr0ℝN, and β∈L∞ℝN∩Ls0ℝN. Then, the functional Ψ:D1,2ℝN⟶ℝ,(17)Ψu=∫Fx,u,whereFx,u=∫0ufx,τdτ,is of class C1 and ∇Ψ:D1,2ℝN⟶D1,2ℝN∗ is compact.Remark 5.
Proposition4 is a special case of do Ó ([23], Lemma 1). Note that since the embedding H1ℝN↪D1,2ℝN is continuous, the statement of Proposition 4 remains valid if we replace D1,2ℝN with H1ℝN.Motivated by Ruiz ([13], Lemma 2.1), we also have the following result.Lemma 6.
SupposeK∈L2ℝN and un⇀u in H1. Then, ϕun⟶ϕu in D1,2ℝN.Proof.
As in the proof of [13], Lemma 2.1, we define linear functionals Tn,T:D1,2⟶ℝ,(18)Tnv=∫Kxun2v,Tv=∫Kxu2v.
It can be shown thatTn and T are continuous. Note that(19)∫∇ϕun·∇v=Tnv,∫∇ϕu·∇v=Tv,by the isometry between D1,2 and D1,2∗ via the Riesz representation theorem; it suffices j. Let ε>0; we choose R>0 such that(20)∫x≥RK2x<ε2.
Now, letv∈D1,2 with vD1,2≤1; using the Hölder inequality, we have(21)Tnv−Tv≤∫x≥R+∫x<RKxun2−u2v≤∫x≥RK2x1/2un2−u23v6+supℝ3K·∫x<Run2−u26/55/6∫x<Rv61/6≤ε·S3supnunD1,22+uD1,22+supℝ3K·S∫x<Run2−u26/55/6.
Sinceun is bounded in D1,2 and(22)∫x<Run2−u26/5⟶0by the compactness of the Sobolev embedding, letting n⟶∞ in (21), we deduce(23)lim¯nTnv−Tv≤Cεuniformly for vD1,2≤1. It follows that Tn⟶T in D1,2∗. The proof is completed.☐Remark 7.
Even though we have this lemma in hand, we do not know how to deduce(24)∫Kxϕunun−ϕuuun−u⟶0from un⇀u. However, we can still deduce the PS condition for our functional (see the proof of Lemma 9).


## 3. Proofs of Theorem2

To prove Theorem2, we will apply the truncated method, see e.g. [24]. For a function u:ℝ3⟶ℝ, we set u±=max±u,0. Then, we define the truncated functional Φ+:H1⟶ℝ,(25)Φ+u=12∫∇u2+Vxu2+14∫Kxϕuu2−∫Gx,u++1s∫hxu+s.It is well known thatΦ+ is of class C1; the critical points of Φ+ are solutions of the truncated problem(26)−Δu+Vxu+Kxϕu=gx,u+−hxu+s−1,inℝ3,−Δϕ=Kxu2,inℝ3.Moreover, supposeu∈H1 is a critical point of Φ+, then u≥0. Hence, u is a solution of (1).Lemma 8.
Under the assumptions of Theorem2, the functional Φ+ is coercive. As a consequence, Φ+ is bounded from below.Proof.
By (g1), we have(27)Gx,u≤1rbxur,x,u∈ℝ3×ℝ.
Note thatKxϕuu2≥0 and hx≥0 on ℝ3; using the Hölder inequality and the fact(28)r·r0′=6,1r0′=r6,we have(29)Φ+u=12u2+14∫Kxϕuu2−∫Gx,u++1s∫hxu+s≥12u2−1r∫bxur≥12u2−1r∫br01/r0∫ur·r0′1/r0′≥12u2−1rbr0u6r≥12u2−1rbr0·S6rur.
Sincer<2, we get Φ+u⟶+∞ as u⟶∞. Thus, we have proved Lemma 8.☐Lemma 9.
Under the assumptions of Theorem2, the functional Φ+ satisfies the PS condition.Proof.
Letun⊂H1 be a PS sequence of Φ+. By Lemma 8, un is bounded. Hence, up to a subsequence, we have un⇀u in H1. Firstly, we have(30)Φ+′un−Φ+′u,un−u⟶0.
To apply Proposition4, let Ψ1,Ψ2:H1⟶ℝ,(31)Ψ1u=∫Gx,u+,Ψ2u=1s∫hxu+s.
By our assumptions (g1 ) and (h1 ), we can apply Proposition 4 (see also Remark 5) and deduce(32)∇Ψ1un⟶∇Ψ1uinH1,∇Ψ2un⟶∇Ψ2uinH1.
Therefore, sinceun is bounded in H1, we have(33)∫gx,un+−gx,u+un−u=∇Ψ1un−∇Ψ1u,un−u⟶0,∫hxun+s−1−u+s−1un−u=∇Ψ2un−∇Ψ2u,un−u⟶0.
Using Lemma6 and the continuous embedding D1,2↪L6, we see that ϕun⟶ϕu in L6. Hence, by the boundedness of un in L12/5, we deduce(34)∫Kxϕun−ϕuunun−u≤K∞ϕun−ϕu6un12/5un−u12/5⟶0.
Combining (30)–(34), we obtain(35)un−u2=Φ+′un−Φ+′u,un−u−∫Kxϕunun−ϕuuun−u+∫gx,un+−gx,u+un−u−∫hxun+s−1−u+s−1un−u=o1−∫Kxϕunun−ϕuuun−u=o1−∫Kxϕun−ϕuunun−u−∫Kxϕuun−u2=o1−∫Kxϕuun−u2.
Since the integral in the final line is nonnegative, we deduce thatun⟶u in H1; thus, we have proved Lemma 9.☐Now we are ready to present the proof of Theorem2.Proof of Theorem 2.
Firstly, we claim that the zero function0 is not a minimizer of Φ+. For this purpose, we choose a nonnegative function v∈C0∞\0. Suppose t∈0,v∞−1δ; then, for all x∈ℝ3, we have 0≤tvx<δ. Thus, using (g2) and noting that ϕtv=t2ϕv, we deduce(36)Φ+tv=t22∫∇v2+Vxv2+14∫Kxϕtvtv2−∫Gx,tv+tss∫hxvs≤t22∫∇v2+Vxv2+t44∫Kxϕvv2−∫axtvθ+tss∫hxvs=t22∫∇v2+Vxv2+t44∫Kxϕvv2−tθ∫axvθ+tss∫hxvs.
Becauseθ∈1,2 and s∈θ,6, we see that(37)Φ+tv<0=Φ+0for t>0 small enough. So 0 is not a minimizer of Φ+.
By Lemma8 and Lemma 9, we know that Φ+ is bounded from below and satisfies the PS condition. It is well known that there exists a minimizer u+ of Φ+ (see e.g., [25], Corollary 2.5). By the claim above, we see that u+≠0 and it is a critical point of Φ+. As mentioned at the beginning of this section, u+≥0 and it is a nontrivial solution of (1).
In a similar manner, by consideringΦ−:H1⟶ℝ,(38)Φ−u=12∫∇u2+Vxu2+14∫Kxϕuu2−∫Gx,u−+1s∫hxu−s,we can obtain another nontrivial solution u−, which is nonpositive on ℝ3. This completes the proof of Theorem 2.☐


## 4. Proofs of Theorem3

In this section, we prove Theorem3. Note that, in our Theorem 3, since s∈1,+∞ is allowed to be supercritical, the usual space H1ℝ3 cannot be used as our framework for the study of problem (1). For this reason, motivated by [18, 19], we introduce a new space as our working space. Let D1,2 be the completion of C0∞ℝ3 under the norm(39)uD2=∫∇u2.For a nonnegative measurable functionlx and 1<q<+∞, we define the weighted Lebesgue space(40)Llqℝ3≔uismeasurable:∫lxuq<∞,and it is associated with the seminorm(41)uq,l=∫lxuq1/q.Motivated by [18, 19], let E be the completion of C0∞ℝ3 with respect to the norm(42)uE=u+us,h.Then,E is a Banach space.Lemma 11.
If (V) and (g1 ) are satisfied, then we have the compact embedding D1,2ℝ3↪Lbrℝ3. Furthermore, we also have the compact embedding E↪Lbrℝ3.Proof.
By our assumption onb, using the results in [26] (see ([26], page 255)), the embedding D1,2↪Lbr is well defined and compact. The compactness of E↪Lbr follows from the continuity of E↪D1,2.☐Lemma 12 ([19], Lemma 2.2).
Givenα,β>0, there is C>0 such that, for any u∈E(43)α∫∇u2+Vxu2+β∫hxus≤CuE2+uEs,and for uE≥1, we have(44)α∫∇u2+Vxu2+β∫hxus≥CuE2.Now, let us define the variational functional corresponding to problem (1). We set Φ:E⟶ℝ as(45)Φu=12∫∇u2+Vxu2+14∫Kxϕuu2−∫Gx,u+1s∫hxus.By Lemma11, all the integrals in (45) are well defined and converge; we know that the weak solutions of problem (1) correspond to the critical points of C1 functional Φ:E⟶ℝ with derivative given by(46)Φ′u,v=∫∇u∇v+Vxuv+∫Kxϕuuv−∫gx,uv+∫hxus−2uv.Lemma 13.
Under the assumptions of Theorem3, the functional Φ is coercive.Proof.
BecauseS is the best Sobolev constant(47)S=infu∈D1,2\0uD2u62,using (g1 ) and Hölder inequality, we get(48)∫Gx,u≤1r∫bxur≤1rS−r2br0uDr≤CuEr.
ForuE large enough, (48) together with Lemma 12 give that(49)Φu=12∫∇u2+Vxu2+14∫Kxϕuu2−∫Gx,u+1s∫hxus≥CuE2−CuEr⟶∞,asuE⟶+∞,because r<2. This implies that Φ is coercive on E.☐In general, to prove thePS condition, the reflexivity of the space is needed. However, we do not know whether E is reflexive, but we can still prove the following.Lemma 14.
Under the assumptions of Theorem3, the functional Φ satisfies the PS condition.Proof.
From Lemma13, we can deduce that every PS sequence un of Φ is bounded in E, and un is also bounded in H1. Therefore, we can assume that for some u∈E, up to a subsequence(50)un⇀uinH1,un⟶uinLloctℝ3,t∈2,6,un⟶ua.e.inℝ3
First, we show thatΦ′u=0. For any φ∈C0∞ℝ3, since(51)Φ′un,φ=o1φE,we have(52)∫∇un∇φ+Vxunφ+∫Kxϕununφ−∫gx,unφ+∫hxuns−2unφ=o1φE.
Now, we claim that(53)∫Kxϕununφ⟶∫Kxϕuuφ,(54)∫hxuns−2unφ⟶∫hxus−2uφ,(55)∫gx,unφ⟶∫gx,uφ.
Indeed, by (14), we see that ϕun is bounded in D1,2, hence up to a subsequence ϕun⇀ϕu in D1,2, we have(56)∫Kxϕunuφ⟶∫Kxϕuuφ,asn⟶∞.
Moreover, by the Hölder inequality, we get(57)∫Kxϕunun−uφ≤K∞ϕun6un−uL12/5Ωφ12/5⟶0,where Ω=suppφ. Therefore, (56) and (57) give that(58)∫Kxϕununφ−∫Kxϕuuφ=∫Kxϕun−ϕuuφ+∫Kxϕunun−uφ⟶0.
Thus, (53) holds.
Next, we verify (54) and (55). It is easy to see that the sequence hs−1/suns−2un is bounded in Ls/s−1ℝ3. Since un⟶u a.e. in ℝ3; applying the Brezis-Lieb lemma, up to a subsequence, we have(59)hs−1/suns−2un⇀hs−1/sus−2uinLs/s−1ℝ3.
Moreover,h1/sφ∈Lsℝ3; thus, we have (54). Similarly, using (g1 ) and the Lebesgue theorem, we have (55). Letting n⟶∞ in (52), we have(60)∫∇u∇φ+Vxuφ+∫Kxϕuuφ−∫gx,uφ+∫hxus−2uφ=0,that is, Φ′u=0.
Next, we proveun⟶uinH1. By Lemma 11 and the fact that Φ′un,un=o1unE and Φ′u,u=0, we have(61)limn⟶∞un2+∫Kxϕunun2+∫hxuns=limn⟶∞∫gx,unun=∫gx,uu=u2+∫Kxϕuu2+∫hxus.
On the other hand, by−Δϕu=Kxu2, the Hölder inequality, and Sobolev inequality, we get ϕunD≤Cun12/52, so we can see that ϕun is bounded in D1,2ℝ3; hence, we can assume(62)ϕun⇀ϕuinD1,2ℝ3.
Hence, by the weak lower semicontinuity of the norm·D, we get(63)∫Kxϕuu2=ϕuD2≤liminfn⟶∞ϕunD2=liminfn⟶∞∫Kxϕunun2.
By Fatou’s lemma, we have(64)∫hxus≤liminfn⟶∞∫hxuns.
It follows from (63) and (64) that(65)limn⟶∞un2+∫Kxϕunun2+∫hxuns≥liminfn⟶∞un2+liminfn⟶∞∫Kxϕunun2+liminfn⟶∞∫hxuns≥liminfn⟶∞un2+∫Kxϕuu2+∫hxus.
From (61) and (65), we see that(66)u2≥liminfn⟶∞un2.
From this and the weak lower semicontinuity of the norm·, we see that up to a subsequence,(67)un2⟶u2,then un⟶u in H1.
Finally, we show thatun⟶uinE. It suffices to show that(68)∫hxuns⟶∫hxus.
In fact, fromun⟶u in H1, we have un⟶u in L12/5ℝ3 and ϕun⟶ϕu in L6ℝ3. Hence, by Hölder inequality and Sobolev inequality, we get(69)∫Kxϕunun2−∫Kxϕuu2≤∫Kxϕunun2−u2+∫Kxϕun−ϕuu2≤K∞ϕun6un−u125un+u125+∫Kxϕun−ϕuu2≤Cun−u12/5+K∞∫ϕun−ϕuu2⟶0.
Using (61), (67), and (69) we get (68). Thus,(70)un−uE=un−u+un−us,h⟶0.
The proof is completed.☐Having verified thePS condition, we investigate the geometry of Φ. First, we note that obviously, Φ is even (by our assumption on g).Let(71)Ω≔x∈ℝ3:bx=0,and we define(72)Y≔u∈E:ux=0a.e.x∈Ω.Then,Y is an infinitely dimensional Banach space under the norm ∥·∥E. Therefore, from [19], Lemma 3.3, we know that the seminorm ur,b=∫bxur1/r is a norm on Y.LetΣ be the class of the closed and symmetric (with respect to the origin) subsets of E\0. For A∈Σ, we define the genus γA by(73)γA≔minm∈ℕ∣∃φ∈CA,ℝm\0,suchthatφx=−φ−x.If such a minimum does not exist, we defineγA=+∞. The main properties of the genus can be found in [27, 28]; we omit them here.Lemma 15.
Givenm∈ℕ, there is ε=εm such that(74)γu∈E∣Φu≤−ε≥m.Proof.
Givenm∈ℕ, let Xm be a m-dimensional subspace of E; as in the proof of Theorem 2, we can choose ε=εm>0 and η>0 such that Φu≤−ε, if u=η. (for η small enough).
DenoteS≔u∈Xm∣u=η; S is a sphere in Xm. Then, we have(75)S⊂u∈E∣Φu≤−ε,so, by the monotonicity property of genus, we have(76)γu∈E∣Φu≤−ε≥γS=m.
The proof is completed.☐Let(77)Σm≔A∈Σ∣γA≥m.From Lemma15, we can define a sequence of real numbers(78)cm≔infA∈Σmsupu∈AΦu.Then,(79)c1≤c2≤⋯≤cm≤cm+1≤⋯.By Lemma13, Φ is coercive and bounded from below. That is, cm>−∞, for ∀m∈ℕ. For c∈ℝ, denote Kc≔u∈E∣Φu=c,Φ′u=0. Then, a standard argument (see [29]) gives the following.Lemma 16.
All thecm are critical values of Φ. Furthermore, if c=cm=cm+1=⋯=cm+τ, then γKc≥τ+1.Proof of Theorem 3.
BecauseΦ is even and note that by Lemma 15, Φ−ε∈Σm, thus cm≤−εm<0, for ∀m∈ℕ. This final Lemma 16 gives the existence of infinitely many critical points of Φ. So problem (1) has infinitely many solutions. This completed the proof of Theorem 3.☐



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