Multiple Standing Waves for Nonlinear Schrödinger- Poisson Systems

<jats:p>In this paper, we consider the following nonlinear Schrödinger-Poisson systems. Under suitable conditions on <jats:inline-formula>
                     <math xmlns="http://www.w3.org/1998/Math/MathML" id="M1">
                        <mi>V</mi>
                     </math>
                  </jats:inline-formula>, <jats:inline-formula>
                     <math xmlns="http://www.w3.org/1998/Math/MathML" id="M2">
                        <mi>K</mi>
                     </math>
                  </jats:inline-formula>, <jats:inline-formula>
                     <math xmlns="http://www.w3.org/1998/Math/MathML" id="M3">
                        <mi>g</mi>
                     </math>
                  </jats:inline-formula>, and <jats:inline-formula>
                     <math xmlns="http://www.w3.org/1998/Math/MathML" id="M4">
                        <mi>h</mi>
                     </math>
                  </jats:inline-formula>, when <jats:inline-formula>
                     <math xmlns="http://www.w3.org/1998/Math/MathML" id="M5">
                        <mn>1</mn>
                        <mo><</mo>
                        <mi>s</mi>
                        <mo><</mo>
                        <mn>6</mn>
                     </math>
                  </jats:inline-formula>, we obtain two nontrivial solutions for the problem and when <jats:inline-formula>
                     <math xmlns="http://www.w3.org/1998/Math/MathML" id="M6">
                        <mi>g</mi>
                        <mfenced open="(" close=")">
                           <mrow>
                              <mi>x</mi>
                              <mo>,</mo>
                              <mo>·</mo>
                           </mrow>
                        </mfenced>
                     </math>
                  </jats:inline-formula> is odd and <jats:inline-formula>
                     <math xmlns="http://www.w3.org/1998/Math/MathML" id="M7">
                        <mn>6</mn>
                        <mo><</mo>
                        <mi>s</mi>
                        <mo><</mo>
                        <mo>∞</mo>
                     </math>
                  </jats:inline-formula>, we obtain infinitely many solutions for the problem.</jats:p>


Introduction
In this paper, we consider the following nonlinear Schrödinger-Poisson equations on ℝ 3 −Δu + V x ð Þu + K x ð Þϕu = g x, u ð Þ− h x ð Þ u j j s−2 u, inℝ 3 , Such problem arises when one is looking for standing wave solutions ψðt, xÞ = e −ðiωt/ℏÞ uðxÞ for the nonlinear Schrödinger equation coupled with the Poisson equation 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][2][3]). Problem (1) is a special case of the following Schrödinger-Poisson system: Cerami and Vaira in [4] studied the existence of positive solutions for the problem where they considered VðxÞ = 1, f ðx, uÞ = kðxÞjuj p−2 u, 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 on V, K, and f (see [7][8][9][10]). Because the problem is set on the whole space ℝ 3 , 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 ðPSÞ 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][12][13]). When V is not a constant and not radially symmetric, there have been many works by developing various variational techniques (see [4,[14][15][16][17]).
In paper [18], Sun et al. considered the system where 1 < q < 2 < p < +∞, aðxÞ, kðxÞ, and hðxÞ are measurable functions satisfying suitable assumptions. They obtained infinitely many solutions in H 1 ðℝ 3 Þ × D 1,2 ðℝ 3 Þ with negative energy. Since 2 < p < ∞ is allowed to be supercritical, the usual Sobolev space H 1 ðℝ 3 Þ 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 and a nontrivial solution is obtained (see [20], Theorem 2). Note that there is an inhomogeneous term on the righthand 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 ∈ ð1, 6Þ we denote Note that p ′ is the conjugate exponent in Hölder inequality. To state our results on the problem (1), we make the following assumptions: (g 2 ) There exist δ > 0, θ ∈ ð1, 2Þ, and a ∈ L ∞ ðℝ 3 Þ ∩ L θ 0 ðℝ 3 Þ; aðxÞ ≥ 0 such that (h 1 ) s ∈ ðθ, 6Þ and h ∈ L ∞ ðℝ 3 Þ ∩ L s 0 ðℝ 3 Þ; hðxÞ ≥ 0 for Remark 1. It follows from (g 1 ) and (g 2 ) that r ≤ θ for aðxÞ > 0 and δ small enough.
Under these assumptions, it is clear that the zero function uðxÞ = 0 is the trivial solution of problem (1). Our main results on the existence of multiple nontrivial solutions are the following theorems.  The paper is organized as follows. In Section 2, 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

Preliminary
Thanks to condition (V), the norm is an equivalent norm on H 1 = H 1 ðℝ 3 Þ. Let q ∈ ½2, 6; then, we have a continuous embedding H 1 ↪L q . Hence, there is a constant S q such that For later use, we also denote by S the best Sobolev constant for the continuous embedding D 1,2 ↪L 6 .
For u ∈ H 1 , it is well known that the Poisson equation has a unique solution ϕ = ϕ u in D 1,2 ðℝ 3 Þ. Note that according to [21], Theorem 2.2.1, Consequently, ϕ u ≥ 0 in ℝ 3 . We also know that there exists a 1 > 0 such that See [14], Lemma 1.1.

Journal of Function Spaces
Define a functional Φ : Under our assumptions, it is easy to see that Φ ∈ C 1 ðH 1 Þ. According to Benci and his collaborators [1,22], it is well known that if u is a critical point of Φ, then ðu, ϕ u Þ is a (weak) solution of (1).
To find critical points of Φ, some compactness conditions, such as the well-known Palais-Smale condition (ðPSÞ for short), are crucial. To establish the ðPSÞ condition for Φ , we need the following results.
Proof. As in the proof of [13], Lemma 2.1, we define linear functionals T n , T : It can be shown that T n and T are continuous. Note that by the isometry between D 1,2 and ðD 1,2 Þ * via the Riesz representation theorem; it suffices j. Let ε > 0; we choose R > 0 such that Now, let v ∈ D 1,2 with kvk D 1,2 ≤ 1; using the Hölder inequality, we have Since fu n g is bounded in D 1,2 and by the compactness of the Sobolev embedding, letting n ⟶ ∞ in (21), we deduce lim uniformly for kvk D 1,2 ≤ 1. It follows that T n ⟶ T in ðD 1,2 Þ * .
The proof is completed. ☐ Remark 7. Even though we have this lemma in hand, we do not know how to deduce from u n ⇀ u. However, we can still deduce the ðPSÞ condition for our functional (see the proof of Lemma 9).

Journal of Function Spaces
It is well known that Φ + is of class C 1 ; the critical points of Φ + are solutions of the truncated problem Moreover, suppose u ∈ H 1 is a critical point of Φ + , then u ≥ 0 . Hence, u is a solution of (1).

Lemma 8.
Under the assumptions of Theorem 2, the functional Φ + is coercive. As a consequence, Φ + is bounded from below.
Proof. By (g 1 ), we have Note that KðxÞϕ u u 2 ≥ 0 and hðxÞ ≥ 0 on ℝ 3 ; using the Hölder inequality and the fact r · r 0 ′ = 6, we have Since r < 2, we get Φ + ðuÞ ⟶ +∞ as kuk ⟶ ∞. Thus, we have proved Lemma 8. Proof. Let fu n g ⊂ H 1 be a ðPSÞ sequence of Φ + . By Lemma 8, fu n g is bounded. Hence, up to a subsequence, we have u n ⇀ u in H 1 . Firstly, we have To apply Proposition 4, let Ψ 1 , Ψ 2 : H 1 ⟶ ℝ, By our assumptions (g 1 ) and (h 1 ), we can apply Proposition 4 (see also Remark 5) and deduce Therefore, since fu n g is bounded in H 1 , we have Using Lemma 6 and the continuous embedding D 1,2 ↪L 6 , we see that ϕ u n ⟶ ϕ u in L 6 . Hence, by the boundedness of fu n g in L 12/5 , we deduce Combining (30)-(34), we obtain Since the integral in the final line is nonnegative, we deduce that u n ⟶ u in H 1 ; thus, we have proved Lemma 9. ☐ Now we are ready to present the proof of Theorem 2.

Journal of Function Spaces
Proof of Theorem 2. Firstly, we claim that the zero function 0 is not a minimizer of Φ + . For this purpose, we choose a nonnegative function v ∈ C ∞ 0 \ f0g. Suppose t ∈ ð0, jvj −1 ∞ δÞ; then, for all x ∈ ℝ 3 , we have 0 ≤ tvðxÞ < δ. Thus, using (g 2 ) and noting that ϕ tv = t 2 ϕ v , we deduce Because θ ∈ ð1, 2Þ and s ∈ ðθ, 6Þ, we see that for t > 0 small enough. So 0 is not a minimizer of Φ + . By Lemma 8 and Lemma 9, we know that Φ + is bounded from below and satisfies the ðPSÞ 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 considering Φ − : we can obtain another nontrivial solution u − , which is nonpositive on ℝ 3 . This completes the proof of Theorem 2. ☐

Proofs of Theorem 3
In this section, we prove Theorem 3. Note that, in our Theorem 3, since s ∈ ð1,+∞Þ is allowed to be supercritical, the usual space H 1 ðℝ 3 Þ 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 D 1,2 be the completion of C ∞ 0 ðℝ 3 Þ under the norm For a nonnegative measurable function lðxÞ and 1 < q < +∞, we define the weighted Lebesgue space and it is associated with the seminorm Motivated by [18,19], let E be the completion of C ∞ 0 ðℝ 3 Þ with respect to the norm Then, E is a Banach space.
Lemma 11. If (V) and (g 1 ) are satisfied, then we have the compact embedding D 1,2 ðℝ 3 Þ↪L r b ðℝ 3 Þ. Furthermore, we also have the compact embedding E↪L r b ðℝ 3 Þ.
Proof. By our assumption on b, using the results in [26] (see ([26], page 255)), the embedding D 1,2 ↪L r b is well defined and compact. The compactness of E↪L r b follows from the continuity of E↪D 1,2 . ☐ and for kuk E ≥ 1, we have Now, let us define the variational functional corresponding to problem (1). We set Φ : E ⟶ ℝ as By Lemma 11, 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 C 1 functional Φ : E ⟶ ℝ with derivative given by Lemma 13. Under the assumptions of Theorem 3, the functional Φ is coercive.

Journal of Function Spaces
Proof. Because S is the best Sobolev constant using (g 1 ) and Hölder inequality, we get For kuk E large enough, (48) together with Lemma 12 give that because r < 2. This implies that Φ is coercive on E. ☐ In general, to prove the ðPSÞ 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 Theorem 3, the functional Φ satisfies the ðPSÞ condition.
Proof. From Lemma 13, we can deduce that every ðPSÞ sequence fu n g of Φ is bounded in E, and fu n g is also bounded in H 1 . Therefore, we can assume that for some u ∈ E, up to a subsequence u n ⇀ u in H 1 , First, we show that Φ′ðuÞ = 0. For any φ ∈ C ∞ 0 ðℝ 3 Þ, since we have Now, we claim that Indeed, by (14), we see that ϕ u n is bounded in D 1,2 , hence up to a subsequence ϕ u n ⇀ ϕ u in D 1,2 , we have Moreover, by the Hölder inequality, we get where Ω = supp φ. Therefore, (56) and (57) give that ð K x ð Þϕ u n u n φ− Thus, (53) holds. Next, we verify (54) and (55). It is easy to see that the sequence fh ðs−1Þ/s ju n j s−2 u n g is bounded in L s/ðs−1Þ ðℝ 3 Þ. Since u n ⟶ u a.e. in ℝ 3 ; applying the Brezis-Lieb lemma, up to a subsequence, we have Moreover, h 1/s φ ∈ L s ðℝ 3 Þ; thus, we have (54). Similarly, using (g 1 ) and the Lebesgue theorem, we have (55). Letting n ⟶ ∞ in (52), we have that is, Φ′ðuÞ = 0. Next, we prove u n ⟶ u in H 1 . By Lemma 11 and the fact that hΦ′ðu n Þ, u n i = oð1Þku n k E and hΦ′ðuÞ, ui = 0, we have On the other hand, by −Δϕ u = KðxÞu 2 , the Hölder inequality, and Sobolev inequality, we get kϕ u n k D ≤ Cju n j 2 12/5 , so we can see that ϕ u n is bounded in D 1,2 ðℝ 3 Þ; hence, we can assume Hence, by the weak lower semicontinuity of the norm k·k D , we get