Available online at www.tjnsa.com J. Nonlinear Sci. Appl. 9 (2016), 652–660 Research Article Infinitely many large energy solutions for Schrödinger-Kirchhoff type problem in RN Bitao Chenga,b,∗, Xianhua Tangb a School of Mathematics and Information Science, Qujing Normal University, Qujing, Yunnan 655011, P. R. China. b School of Mathematics and Statistics, Central South University, Changsha, Hunan, 410083, P. R. China. Communicated by R. Saadati Abstract In this paper, we consider the following Schrödinger-Kirchhoff-type problem R − a + b RN |∇u|2 dx 4u + V (x)u = g(x, u), for x ∈ RN , u(x) → 0, as |x| → ∞, (1.1) where constants a > 0, b ≥ 0, N = 1, 2 or 3, V ∈ C(RN , R), g ∈ C(RN × R, R). Under more relaxed assumptions on g(x, u), by using some special techniques, a new existence result of infinitely many energy c solutions is obtained via Symmetric Mountain Pass Theorem. 2016 All rights reserved. Keywords: Schrödinger-Kirchhoff type problem, critical point, symmetric Mountain Pass Theorem, variational methods. 2010 MSC: 35J20, 35J60. 1. Introduction and main results In this paper, we consider the following Schrödinger-Kirchhoff type problem R − a + b RN |∇u|2 dx 4u + V (x)u = g(x, u), for x ∈ RN , u(x) → 0, as |x| → ∞, (1.1) where constants a > 0, b ≥ 0, N = 1, 2 or 3, V ∈ C(RN , R) and g ∈ C(RN × R, R) satisfy some further conditions. ∗ Corresponding author Email addresses: chengbitao2006@126.com (Bitao Cheng ), tangxh@csu.edu.cn (Xianhua Tang) Received 2015-09-10 B. Cheng, X. Tang, J. Nonlinear Sci. Appl. 9 (2016), 652–660 653 We note that when a = 1, b = 0, the problem (1.1) reduces to the following semilinear Schrödinger equation −4u + V (x)u = g(x, u), for x ∈ RN , (1.2) u(x) → 0, as |x| → ∞, which has been studied extensively by many authors, and there is a large body of literature on the existence and multiplicity of solutions for the equation (1.2), for example, we refer the reader to [1, 21, 22] and references therein. When V ≡ 0 and RN is replaced by a bounded domain Ω ⊂ RN , the problem (1.1) reduces to the following nonlocal Kirchhoff type problem R −(a + b Ω |∇u|2 dx)4u = g(x, u), in Ω; (1.3) u = 0, on ∂Ω. The problem (1.3) is related to the stationary analogue of the Kirchhoff equation Z |∇u|2 dx 4u = g(x, t), utt − a + b (1.4) RN which was proposed by Kirchhoff [13] as a model given by the equation of elastic strings Z L ∂2u ∂ 2 u P0 E ∂u ρ 2 − + | |2 dx = 0. ∂t h 2L 0 ∂x ∂x2 (1.5) The equation (1.5) is an extension of the classical D’Alembert’s wave equation by taking into account the changes in the length of the string during the transverse vibration. It was pointed out in [9] that Kirchhoff type problem (1.3) models several physical and biological systems, where u describes a process which depends on the average of itself (for example, population density). Moreover, a lot of interesting studies by variational methods can be found in [2, 6, 7, 8, 10, 15, 16, 18, 20, 27] for Kirchhoff type problem (1.3) on bounded domain with several growth conditions on g. Recently, Kirchhoff type problems setting on the unbounded domain or the whole space RN have also attracted a lot of attention. Many solvability conditions on the nonlinearity have been given to obtain the existence and multiplicity of solutions for Kirchhoff type problems in RN , we refer the readers to [3, 4, 11, 12, 14, 17, 23, 24, 25, 26] and references therein. Particularly, Wu obtained four results of the existence of a sequence of high energy solutions for the problem (1.1) by means of symmetric mountain pass theorem in [25]. Those results had been subsequently unified and improved by Y. Ye and C. Tang with the aid of fountain theorem in [26]. Motivated by the works mentioned above, in the present paper, under more relaxed assumptions on the nonlinear term g, we will present a new proof technique to construct infinitely many large energy solutions for the problem (1.1). In order to reduce the statements of our result, we make the following assumptions. (V1 ) V ∈ C(RN , R) satisfies inf V (x) ≥ V0 > 0 and for each M > 0, meas{x ∈ RN : V (x) ≤ M } < +∞, where V0 is a constant and meas denote the Lebesgue measure in RN . (g1 ) There exist C1 > 0 and p ∈ (4, 2∗ ) such that |g(x, t)| ≤ C1 (|t| + |t|p−1 ), ∀(x, t) ∈ RN × R. R N , G(x, t) = t g(x, s)ds. (g2 ) G(x,t) → +∞ as |t| → +∞ uniformly in x ∈ R 4 0 t (g3 ) There exists L > 0 such that 4G(x, t) − g(x, t)t ≤ d|t|2 , where 0 ≤ d ≤ V20 . (g4 ) g(x, −t) = −g(x, t) for all (x, t) ∈ RN × R. for a.e. x ∈ RN and ∀|t| ≥ L, B. Cheng, X. Tang, J. Nonlinear Sci. Appl. 9 (2016), 652–660 654 Next, we give some notations. Define the function space H 1 (RN ) = u ∈ L2 (RN ) : ∇u ∈ L2 (RN ) with the norm Z kukH 1 = Denote E= 1 N 2 2 |∇u| + u 1 2 dx . RN Z 2 2 |∇u| + V (x)u u ∈ H (R ) : dx < +∞ RN with the inner product and the norm Z hu, viE = RN 1 (∇u · ∇v + V (x)uv)dx, kukE = hu, uiE2 . Obviously, under the assumption (V1 ) on V (x), the following embedding E ,→ Ls (RN ), 2 ≤ s ≤ 2∗ is continuous. Hence, for any s ∈ [2, 2∗ ], there is a constant as > 0 such that kukLs ≤ as kukE . (1.6) It is well known that a weak solution for the problem (1.1) is a critical point of the following functional I defined on E by Z 2 Z Z Z b 1 a 2 2 2 |∇u| dx + |∇u| dx + V (x)u dx − G(x, u)dx (1.7) I(u) = 2 RN 4 2 RN RN RN for all u ∈ E. We say that a weak solutions sequence {un } ⊂ E for the problem (1.1) is a large energy solutions sequence if the energy I(un ) → +∞ as n → ∞. Now, we can state our result as follows. Theorem 1.1. Assume that (V1 ) and (g1 ) − (g4 ) hold. Then the problem (1.1) possesses infinitely many large energy solutions in E. Remark 1.2. (i) Since the problem (1.1) is defined in RN which is unbounded, the lack of compactness of the Sobolev embedding becomes more delicate by using variational techniques. To overcome the lack of compactness, the condition (V1 ), which was first introduced by Bartsch and Wang in [5], is always assumed to preserve the compactness of embedding of the working space. (ii) From Remark 1 in [26], the condition (g1 ) is much weaker than the combination of usual subcritical condition and asymptotically linear condition near zero. Furthermore, condition (g3 ) is much weaker than the following condition: (g30 ) There exists L > 0 such that tg(x, t) − 4G(x, t) ≥ 0, for a.e. x ∈ RN and ∀|t| ≥ L, which was used in Theorem 5 in [26]. Hence, Theorem 1.1 improves Theorem 5 in [26]. 2. Some lemmas In order to apply variational techniques, we first state the key compactness result. Lemma 2.1 ([28], Lemma 3.4). Under the assumption (V1 ), the embedding E ,→ Ls (RN ), 2 ≤ s < 2∗ is compact. The following lemma has been proved by Lemma 1 in [26]. B. Cheng, X. Tang, J. Nonlinear Sci. Appl. 9 (2016), 652–660 655 Lemma 2.2. Let assumptions (V1 ) and (g1 ) hold. Then I is well defined on E, I ∈ C 1 (E, R) and for any u, v ∈ E, Z Z Z Z 2 0 g(x, u)vdx. (2.1) V (x)uvdx − |∇u| dx ∇u∇vdx + hI (u), vi = a + b RN RN RN Moreover, Ψ0 : E → E ∗ is compact, where Ψ(u) = R RN RN G(x, u)dx. Recall that we say I satisfies the (P S) condition at the level c ∈ R ((P S)c condition for short) if any sequence {un } ⊂ E along with I(un ) → c and I 0 (un ) → 0 as n → ∞ possesses a convergent subsequence. If I satisfies (P S)c condition for each c ∈ R, then we say that I satisfies the (P S) condition. Lemma 2.3. Let assumption (V1 ) and (g1 ) hold. Then any bounded Palais-Smale sequence of I has a strongly convergent subsequence in E. Proof. Let {un } ⊂ E be any bounded Palais-Smale sequence of I, then, up to a subsequence, there exist c1 ∈ R such that I(un ) → c1 , I 0 (un ) → 0 and sup kun kE < +∞. (2.2) n Since the embedding E ,→ Ls (RN ), is compact, going if necessary to a subsequence, we un * u, un → u, un (x) → u(x), 2 ≤ s < 2∗ can assume that there is a u ∈ E such that weakly in E; strongly in Ls (RN ); a.e.in RN . (2.3) In view of (2.1), it has hI 0 (un ) − I 0 (u), un − ui Z Z Z 2 = a+b |∇un | dx ∇un · ∇(un − u)dx + V (x)|un − u|2 dx N N N R RZ Z ZR − a+b |∇u|2 dx ∇u · ∇(un − u)dx − [g(x, un ) − g(x, u)](un − u)dx N R RN RN Z Z Z 2 2 = a+b |∇un | dx |∇(un − u)| dx + V (x)|un − u|2 dx (2.4) RN RN RN Z Z Z Z −b |∇u|2 dx − |∇un |2 dx ∇u · ∇(un − u)dx − [g(x, un ) − g(x, u)](un − u)dx N N R R RN RN Z Z Z 2 2 2 ≥ min{a, 1}kun − ukE − b |∇u| dx − |∇un | dx ∇u · ∇(un − u)dx RN RN RN Z − [g(x, un ) − g(x, u)](un − u)dx. RN Then (2.4) implies that min{a, 1}kun − uk2E ≤ hI 0 (uRn ) − I 0 (un ), unR− ui R +bR RN |∇u|2 dx − RN |∇un |2 dx RN ∇u · ∇(un − u)dx + RN [g(x, un ) − g(x, u)](un − u)dx. Define the functional hu : E → R by Z ∇u · ∇vdx, hu (v) = RN ∀ v ∈ E. (2.5) B. Cheng, X. Tang, J. Nonlinear Sci. Appl. 9 (2016), 652–660 656 Obviously, hu is a linear functional on E. Furthermore, Z |∇u · ∇v|dx ≤ kukE kvkE , |hu (v)| ≤ RN which implies hu is bounded on E. Hence hu ∈ E ∗ . Since, un * u in E, it has lim hu (un ) = hu (u), that n→∞ R is, RN ∇u · ∇(un − u)dx → 0 as n → ∞. Consequently, by (2.3) and the boundedness of {un }, it has Z Z Z |∇un |2 dx ∇u · ∇(un − u)dx → 0 , n → +∞. (2.6) b |∇u|2 dx − RN RN RN By (g1 ), using the Hölder inequality, we can conclude Z Z N [g(x, un ) − g(x, u)](un − u)dx ≤ C1 [|un | + |u| + |un |p−1 + |u|p−1 ]|un − u|dx RN R ≤ C1 (kun kL2 + kukL2 )kun − ukL2 p−1 p + C1 (kun kp−1 Lp + kukLp )kun − ukL . Therefore, it follows from (2.6) that Z [g(x, un ) − g(x, u)](un − u)dx → 0, as n → ∞. (2.7) RN Moreover, combining (2.5) with (2.6), then < I 0 (un ) − I 0 (u), un − u >→ 0, as n → ∞. (2.8) Consequently, (2.5)–(2.8) imply that un → u, strongly in E as n → ∞. This completes the proof. Lemma 2.4. Let assumptions (V1 ), (g1 ) and (g3 ) hold. Then any Palais-Smale sequence of I is bounded. Proof. Let {un } ⊂ E be any Palais-Smale sequence of I, then, up to a subsequence, there exist c1 ∈ R such that I(un ) → c1 , and I 0 (un ) → 0. (2.9) The combination of (1.6), (1.7), (2.1), (2.9), (V1 ) with (g3 ) implies 1 c1 + 1 + kun kE ≥ I(un ) − hI 0 (un ), un i 4 Z Z Z a 1 2 2 e un )dx |∇un | dx + V (x)un dx + G(x, = 4 RN 4 RN RN Z Z Z Z a 1 d 2 2 2 e un )dx ≥ |∇un | dx + V (x)un dx − u dx + G(x, 4 RN 4 RN 4 RN n An Z Z Z Z 1 1 a 2 2 2 e un )dx ≥ |∇un | dx + V (x)un dx − V0 un dx + G(x, 4 RN 4 RN 8 RN An Z Z Z Z a 1 1 2 2 2 e un )dx ≥ |∇un | dx + V (x)un dx − V (x)un dx + G(x, 4 RN 4 RN 8 RN An Z Z 1 1 2 2 e un )dx, ≥ min{a, 1}kun kE + V (x)un dx + G(x, 16 16 RN An (2.10) B. Cheng, X. Tang, J. Nonlinear Sci. Appl. 9 (2016), 652–660 657 e un ) = 1 g(x, un )un − G(x, un ) and An = {x ∈ RN : |un | ≤ L}. For x ∈ RN and |un | ≤ L, by where G(x, 4 (g1 ), it has e un )| ≤ 1 |g(x, un )||un | + |G(x, un )| |G(x, 4 1 1 1 ≤ C1 (|un |2 + |un |p ) + C1 |un |2 + |un |p 4 2 p 3 p + 1 = C1 + |un |p−2 |un |2 4 p 3 p + 1 ≤ C1 + Lp−2 |un |2 . 4 p n o p−2 , V Take M > max 16C1 43 + p+1 0 , then p L e un ) ≥ − M |un |2 , G(x, 16 ∀ x ∈ RN , |un | ≤ L. e = {x ∈ RN : V (x) ≤ M }. By (V1 ) and (2.11), we can conclude Let A Z Z Z 1 1 2 e V (x)un dx + G(x, un )dx ≥ (V (x) − M )|un |2 dx 16 RN 16 |un |≤L An Z 1 ≥ (V (x) − M )L2 dx 16 A∩A e n 1 e ∩ An ) ≥ (V0 − M )L2 meas(A 16 1 e ≥ (V0 − M )L2 meas(A). 16 e < +∞ due to (V1 ), it follows from (2.10) and (2.12) that Note that meas(A) c1 + 1 + kun kE ≥ (2.11) (2.12) 1 1 e min{a, 1}kun k2E + (V0 − A)L2 meas(A), 16 16 which implies {un } ⊂ E is bounded in E. Hence the proof is completed. Remark 2.5. Comparing with [26], we present a new proof technique to verify the boundedness of PalaisSmale sequences, which is much clearer and simpler than them. 3. Proof of theorem 1.1 In this section we will use the classical Symmetric Mountain Pass Theorem of Rabinowitz instead of Fountain Theorem in [26] to obtain infinitely many large energy solutions for the problem (1.1) and prove Theorem 1.1. First of all, we state some notations. In view of E ,→ L2 (RN ) and L2 (RN ) is a separable Hilbert space, then E has a countable orthogonal basis {ej }∞ j=1 . Let k M Ek = Xj , Zk = Ek⊥ , j=1 where Xj = span{ej }. Thus, E = Ek L Zk and Ek is finite dimensional. Lemma 3.1. Let the assumption (V1 ) hold. Define η(k) := sup kukL2 , u∈Zk ,kukE =1 then there exists k0 ∈ N such that 0 < η(k0 ) ≤ min{a,1} 4C1 1 2 . k ∈ N, B. Cheng, X. Tang, J. Nonlinear Sci. Appl. 9 (2016), 652–660 658 Proof. Firstly, η(k) is convergent science η(k) ≥ 0 and η(k) is decreasing in k. Furthermore, for any k ∈ N , by the definition of η(k), there exists uk ∈ Zk such that kuk kE = 1 For any v ∈ E, v = ∞ P kuk kL2 > and η(k) . 2 (3.1) an en , it has n=1 ∞ ∞ ∞ X X X an e n → 0 | < uk , v >E | = < uk , an en >E ≤ kuk kE an en ≤ n=1 E n=k+1 n=k+1 E as k → ∞, which implies that uk * 0 weakly in E. By virtue of Lemma 2.1, we can conclude uk → 0 strongly in L2 (RN ). (3.2) The combination (3.1) with (3.2) implies that η(k) → 0, as k → ∞. Then there exists k0 ∈ N such that 1 2 . Hence the proof is completed. 0 < η(k0 ) ≤ min{a,1} 4C1 Lemma 3.2. Let assumptions (V1 ) and (g1 ) hold, then there exist some constants ρ, α such that I(u) ≥ α whenever u ∈ Zk0 with kukE = ρ. Proof. For any u ∈ Zk0 , by Lemma 3.1, we have kukL2 ≤ η(k0 )kukE 0 < η(k0 ) ≤ and min{a, 1} 4C1 1 2 . By (1.6), (1.7), (g1 ), (3.3) and Hölder inequality, it has I(u) = a 2 Z |∇u|2 dx + RN b 4 Z Z ZR |∇u|2 dx 2 N 1 V (x)u2 dx − G(x, u)dx 2 RN N Z R min{a, 1} kuk2E − G(x, u)dx ≥ 2 RN min{a, 1} ≥ kuk2E − C1 (kuk2L2 + kukpLp ) 2 min{a, 1} ≥ kuk2E − C1 η 2 (k0 )kuk2E − C1 app kukpE 2 h min{a, 1} i p−1 ≥ kukE kukE − C1 app kukE . 4 + Set min{a, 1} t − C1 app tp−1 , ∀t ≥ 0. 4 Note that 4 < p < 2∗ , we can conclude that there exists a constant ρ > 0 such that l(t) = l(ρ) = max l(t) > 0. t≥0 Therefore, I(u) ≥ ρl(ρ) =: α > 0, whenever u ∈ Zk0 with kukE = ρ. This completes the proof. (3.3) B. Cheng, X. Tang, J. Nonlinear Sci. Appl. 9 (2016), 652–660 659 e ⊂ E, there is Lemma 3.3. Let assumptions (g1 ) − (g2 ) hold, then for each finite dimensional subspace E e an r = r(E) > 0 such that I|E\B e r < 0. Proof. With the aid of assumptions (g1 ) − (g2 ), for any K > 0, there exists C(K) > 0 such that G(x, z) ≥ K|z|4 − C(K)|z|2 , ∀ (x, z) ∈ RN × R. (3.4) e ⊂ E, by the equivalence of norms in the finite dimensional space, For any finite dimensional subspace E there exists a constant βs > 0 such that kukLs ≥ βs kukE , for 2 ≤ s < 2∗ . Therefore, Choosing K > 0 such that with (3.4)–(3.5) implies e ∀ u ∈ E, b 4 (3.5) − Kβ44 < 0, then the combination of (1.6)–(1.7) Z 2 Z Z Z a b 1 2 2 G(x, u)dx I(u) = |∇u| dx + |∇u| dx + V (x)u2 dx − 2 RN 4 2 RN RN RN 1 b ≤ max{a, 1}kuk2E + kuk4E − Kkuk4L4 + C(K)kuk2L2 2 4 b 1 − Kβ44 kuk4E + C(K)a22 kuk2E ≤ max{a, 1}kuk2E + 2 4 ≤ C2 kuk2E − C3 kuk4E e where C2 = 1 max{a, 1} + C(K)a2 > 0, C3 = Kβ 4 − for all u ∈ E, 2 4 2 such that I|E\B e r < 0. This completes the proof. b 4 e >0 > 0. Hence there is an r = r(E) Next, we shall prove our Theorem 1.1. To begin with, for convenience to quote, we state the classical Symmetric Mountain Pass Theorem as in the following. Theorem 3.4 ([19], Theorem 9.12). Let E be an infinite dimensional Banach space, I ∈ C 1 (E, R) be even L and satisfy (P S) condition and I(0) = 0. If E = Y Z, Y is finite dimensional, and I satisfies (I1 ) There exist constants ρ, α > 0 such that I|∂Bρ ∩Z ≥ α, and e ⊂ E, there is r = r(E) e such that I ≤ 0 on E e \ Br . (I2 ) for each finite dimensional subspace E Then I possesses an unbounded sequence of critical values. Proof of TheoremL 1.1. The proof is to verify I satisfies all the conditions of Theorem 3.4. Set Y = Ek0 , Z = Zk0 , then E = Y Z and Y is finite dimensional. First, I satisfies (I1 ) and (I2 ) in Theorem 3.4 by Lemma 3.2 and 3.3, respectively. Second, I satisfies (P S) condition by virtue of Lemma 2.3 and 2.4. Finally, I(0) = 0, I is even on E due to (g4 ) and I ∈ C 1 (E, R) by Lemma 2.2. Hence, the conclusion follows from Theorem 3.4. The proof is completed. Remark 3.5. Comparing with Theorem 5 in [26], on one hand, the assumptions imposed on g are much weaker than them. 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