School of Financial Mathematics Using critical point theory, we obtain the existence and multiplicity of nonzero and Statistics, Guangdong University of Finance, Guangzhou, solutions to anti-periodic boundary value problems with p-Laplacian in the case P.R. China where the nonlinearities are p-sublinear at zero. Some examples are given to illustrate Full list of author information is the results. available at the end of the article MSC: 37J45; 39A12; 39A70 Keywords: Anti-periodic boundary value problem; Discrete p-Laplacian; Critical points theorem 1 Introduction Diﬀerence equations occur in many ﬁelds [1, 20], such as economics, discrete optimiza- tion, computer science. In the past decade, discrete p-Laplacian problems and diﬀerence equations have become a hot topic; see [11–19, 21, 22]and [25, 26]. Among the methods used are the method of upper and lower solutions, ﬁxed point theory, Leray–Schauder de- gree, mountain pass lemma and the linking theorem. Recently, a lot of new results [5–11, 16, 23, 24] have been established by using variational methods. In these last years, the existence and multiplicity of solutions for nonlinear discrete prob- lems subject to various boundary value conditions have been widely studied by using dif- ferent methods (see, e.g. [2–4]and [12–19, 21]). Bai et al. [2, 3] studied the second-order diﬀerence equations with Neumann boundary value conditions. D’Aguì et al. [16]inves- tigated the existence of positive solutions for a discrete two point nonlinear boundary value problem with p-Laplacian in the case where the nonlinear term is p-sublinear at zero. However, little work has been done that has referred to anti-periodic boundary value problems with the discrete p-Laplacian operators in the case where the nonlinearities are p-sublinear at zero. The idea of this paper comes from the method in [6, 9, 16]. One obtained two distinct critical points for functionals unbounded from below without p-superlinear nonlinearities at zero. The loss of p-superlinear condition at zero puts some critical points theorems cannot be immediately used. Therefore, In this paper, we mainly deal with the existence © The Author(s) 2018. This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, pro- vided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made. Kuang and Yang Boundary Value Problems (2018) 2018:86 Page 2 of 11 and multiplicity of solutions for anti-periodic boundary value problems –[a(k –1)φ (u(k – 1))] = λf (k, u(k)), (1.1) u(0) = –u(T), u(1) = –u(T +1), p–2 for k ∈ [1, T], where p >1 is a ﬁxed real number and φ (t)= |t| t for all t ∈ R. a(k)>0 and a(0) = a(T), f :[1, T] × R → R, is continuous and is p-sublinear at zero in the sec- ond variable for all k ∈ [1, T]. Moreover, is the forward diﬀerence operator deﬁned by u(k)= u(k +1)– u(k), u(k)= (u(k)). The rest of this paper is organized as follows. In Sect. 2, we establish the variational structure associated with (1.1), and provide some preliminary results. In Sect. 3,westate our main results and give examples. In Sect. 4, we provide the proofs of the main results. 2 Variational structure and some preliminaries In this section, we establish a variational structure which reduces the existence of solutions for (1.1) to the existence of critical points of the corresponding functional. Throughout this paper, we always assume that the following conditions are satisﬁed: (a) a(k)>0 for all k ∈ [1, T] and a(0) = a(T).Let a ¯ and a be the maximum and minimum of {a(k)},respectively. (f) f (k, u) is continuous in u and F(k, u)= f (k, s) ds for u ∈ R and k ∈ [1, T]. We deﬁne the set E as E = u = u(k) |u(T +1) = –u(1), u(k) ∈ R for k ∈ [1, T +1] . Then E is a vector space with au + bv = {au(k)+ bv(k)} for u, v ∈ E and a, b ∈ R.Obviously, E is isomorphic to R and hence E can be equipped with the norm · as T p u = u(k) for u ∈ E. k=1 We also deﬁne norms · and · in E by u = max u(k) :1 ≤ k ≤ T and 1/p T–1 p p u = a(T) u(1) + u(T) + a(k) u(k) k=1 respectively. Consider the functionals (u), (u)and I (u)on E deﬁned by T–1 a(T) a(k) p p (u)= u(1) + u(T) + u(k) , (2.1) p p k=1 (u)= F k, u(k) (2.2) k=1 Kuang and Yang Boundary Value Problems (2018) 2018:86 Page 3 of 11 and I (u)= (u)– λ(u) T–1 T a(T) a(k) p p = u(1) + u(T) + u(k) – λ F k, u(k) . (2.3) p p k=1 k=1 Then the partial derivatives of (u)are given by ∂(u) =–a(1)φ (u(1)) + a(T)φ (u(1) + u(T)), ⎪ p p ∂u(1) ∂(u) = a(1)φ (u(1)) – a(2)φ (u(2)), p p ∂u(2) (2.4) ..., ∂(u) ⎪ = a(T –2)φ (u(T –2)) – a(T –1)φ (u(T – 1)), p p ⎪ ∂u(T–1) ∂(u) = a(T –1)φ (u(T –1)) + a(T)φ (u(1) + u(T)). p p ∂u(T) This, combined with a(0) = a(T), u(0) = –u(T)and u(1) = –u(T +1), gives us ∂(u) =–[a(0)φ (u(0))], ⎪ p ∂u(1) ⎪ ∂(u) =–[a(1)φ (u(1))], ⎪ p ∂u(2) (2.5) ..., ∂(u) ⎪ =–[a(T –2)φ (u(T – 2))], ∂u(T–1) ∂(u) =–[a(T –1)φ (u(T – 1))]. ∂u(T) Then has continuous Gâteaux derivatives in ﬁnite dimensional space and ∈ C (E, R), the Fréchet derivative is given by (u), v =– a(k –1)φ u(k –1) v(k) (2.6) k=1 for u, v ∈ E. By direct computation, we have (u), v = a(k)φ u(k) v(k). (2.7) k=1 Similarly, we have ∈ C (E, R). The Fréchet derivative is given by (u), v = f k, u(k) v(k) (2.8) k=1 for u, v ∈ E. Therefore, I ∈ C (E, R), the Fréchet derivative is given by I (u), v = (u), v – λ (u), v . λ Kuang and Yang Boundary Value Problems (2018) 2018:86 Page 4 of 11 The partial derivatives of I are given by ∂I =–[a(0)φ (u(0))] – λf (1, u(1)), ⎪ p ∂u(1) ⎪ ∂I =–[a(1)φ (u(1))] – λf (2, u(2)), ⎪ p ∂u(2) (2.9) ..., ∂I ⎪ λ ⎪ =–[a(T –2)φ (u(T – 2))] – λf (T –1, u(T – 1)), ∂u(T–1) ∂I =–[a(T –1)φ (u(T – 1))] – λf (T, u(T)). ∂u(T) Equations (2.3)and (2.9) imply that a nonzero critical point of the functional I on E is a nontrivial solution of (1.1). Deﬁnition 2.1 Let I ∈ C (H, R). A sequence {x }⊂ H is called a Palais–Smale sequence (P.S. sequence) for I if {I(x )} is bounded and I (x ) → 0as j → +∞.Wesay I satisﬁes the j j Palais–Smale condition (P.S. condition) if any P.S. sequence for I possesses a convergent subsequence. Our main tool is taken from [9], whichwerecallherefor thereader’sconvenience. Theorem 2.1 ([9]) Let X be a real Banach space and let , : X → Rbe two functionals 1 ∗ of class C such that inf = (0) = (0) = 0. Assume that there are r ∈ Rand u ∈ X, with 0< (u )< r, such that 1 (u ) sup (u)< , (2.10) r (u ) –1 u∈ (–∞,r] and for each –1 (u ) λ ∈ = , r sup (u) , (u ) –1 u∈ (–∞,r] the functional I = – λ satisﬁes the P.S. condition and it is unbounded from below. Then for each I it admits at least two nonzero critical points u , u such that I (u )< λ λ1 λ2 λ λ1 0< I (u ). λ λ2 3 Main results and examples Theorem 3.1 Assume that the conditions (a) and (f) hold. Thereexist twopositivecon- stants b and ρ such that F(k, u) ≥ b|u| for k ∈ [1, T] and |u|≥ ρ. (3.1) There also exist two positive constants c and d with ∗ ∗ 1/p 1/q a 1 d < c ∗ ∗ a ¯ T such that T T p–1 pT p pb max F(k, ξ) < min F(k, d ), , (3.2) p p (p+1) a (2c ) |ξ |≤c a ¯(2d ) a ¯2 ∗ ∗ ∗ ∗ k=1 k=1 Kuang and Yang Boundary Value Problems (2018) 2018:86 Page 5 of 11 where 1/p +1/q =1. Then, for each λ ∈ with –1 –1 T T (p+1) p p a ¯2 a ¯(2d ) a (2c ) ∗ ∗ ∗ = max , F(k, d ) , max F(k, ξ) , p–1 pb p pT |ξ |≤c k=1 k=1 (1.1) admits at least two nonzero solutions u , u such that I (u )<0< I (u ). λ1 λ2 λ λ1 λ λ2 Remark 3.1 If all the conditions of Theorem 3.1 are satisﬁed and f (k, u) is odd in u for each k ∈ [1, T], then (1.1) admits at least four nonzero solutions ±u , ±u such that λ1 λ2 I (–u )= I (u )<0< I (u )= I (–u ). λ λ1 λ λ1 λ λ2 λ λ2 Corollary 3.1 Assume that the conditions (a) and (f) hold. If f (k, u) is odd in u for each k ∈ [1, T], and F(k, s) F(k, s) lim =+∞, lim =+∞ (3.3) p + p |s|→+∞ s→0 |s| s for all k ∈ [1, T], then, for each λ ∈ with –1 a (2c ) ∗ ∗ = 0, max F(k, ξ) , p–1 |ξ |≤c pT ∗ k=1 (1.1) admits at least four nonzero solutions ±u and ±u . λ1 λ2 Example 3.1 Let p =4, T =2, a(k)= k and f (k, u)= u + u for all k ∈ [1, T]. Then, for each λ ∈ (0, 3/8), it is easy to check that all the conditions of Corollary 3.1 are satisﬁed, (1.1) admits at least four nonzero solutions. Theorem 3.2 Assume that the conditions (a) and (f) hold, T =2, a = a ¯ = a, f (k, x) ≥ 0 for all x <0, k ∈ [1, T]. Put F(k, s) + + + L (k)= lim , L = min L (k). (3.4) ∞ ∞ ∞ s→+∞ s k∈[1,T] If L >0 and there exist two positive constants c and d with ∗ ∗ 1/q d < c ∗ ∗ such that T T p p pL max F(k, ξ) < min F(k, d ), , (3.5) p ∗ p (p+1) 2ac |ξ |≤c a(2d ) a2 ∗ ∗ k=1 k=1 Kuang and Yang Boundary Value Problems (2018) 2018:86 Page 6 of 11 where 1/p +1/q =1. Then, for each λ ∈ with –1 –1 T p T (p+1) p a2 a(2d ) 2ac ∗ ∗ = max , F(k, d ) , max F(k, ξ) , |ξ |≤c pL p p ∗ k=1 k=1 (1.1) admits at least two nonzero solutions u , u such that I (u )<0< I (u ). λ1 λ2 λ λ1 λ λ2 Corollary 3.2 Assume that the conditions (a) and (f) hold, T =2, a = a = a, f (k, x) ≥ 0 for all x <0, k ∈ [1, T]. If F(k, s) F(k, s) lim =+∞ and lim =+∞, (3.6) p + p s→+∞ s→0 s s for all k ∈ [1, T], then, for each λ ∈ with –1 2ac = 0, max F(k, ξ) , |ξ |≤c p ∗ k=1 (1.1) admits at least two nonzero solutions. Example 3.2 Let p =2, T =2, a(k)=4 and f (k, u)= e for k ∈ [1, T]. Then, for each λ ∈ (0, ), it is easy to check that all the conditions of Corol- e–1 lary 3.2 are satisﬁed, (1.1) admits at least two nonzero solutions. 4 Proofsofmainresults In ordertoprove Theorem 3.1, we need the following lemmas. Lemma 4.1 If u ∈ Eand p >1, then T T (p+1) a 2 a ¯2 ∗ p p u(k) ≤ (u) ≤ u(k) p T p k=1 k=1 and 1/p 1/p a u ≤u≤ 2(2a ¯) u . p p Proof On the one hand, T–1 a(T) a(k) p p (u)= u(1) + u(T) + u(k) p p k=1 T–1 a ¯ p p p p p p ≤ 2 u(1) + u(T) + 2 u(k) + u(k +1) k=1 T T (p+1) a ¯ a ¯2 p p ≤ 2 2 u(k) = u(k) . (4.1) p p k=1 k=1 Kuang and Yang Boundary Value Problems (2018) 2018:86 Page 7 of 11 On the other hand, u(1) = –u(T +1), for each k ∈ [1, T], 2u(k)= u(2) – u(1) + ··· + u(k)– u(k –1)+ u(k)– u(k +1)+ ··· + u(T)– u(T +1) ≤ u(2) – u(1) + ··· + u(k)– u(k –1) + u(k)– u(k +1) + ··· + u(T)+ u(1) = u(2) – u(1) + ··· + u(k)– u(k –1) + u(k +1)– u(k) + ··· + u(T)+ u(1) 1/p T–1 p p 1/q ≤ u(T)+ u(1) + u(k) T , (4.2) k=1 where 1/p +1/q =1, that is, 1/p T–1 p p 1/q u ≤ u(T)+ u(1) + u(k) T . (4.3) k=1 Since p p u(k) ≤ T u , k=1 this, combined with (4.3), gives us T T–1 p p p u(k) ≤ u(T)+ u(1) + u(k) k=1 k=1 and T–1 p T a a 2 p p p ∗ ∗ (u) ≥ u(T)+ u(1) + u(k) ≥ u(k) . p p T k=1 k=1 The proof is complete. Lemma 4.2 If the condition (3.1) holds, then the functional I satisﬁes the P.S. condition (p+1) a ¯2 and it is unbounded from below for all λ ∈ ( ,+∞). pb Proof Let {I (u )} be a bounded sequence and {u } be a sequence in E, i.e., there exists a λ j j positive constant M such that I (u ) ≤ M for j ∈ Z . λ j Let M = max F(k, u)– b|u| : |u|≤ ρ . 1≤k≤T It is easy to check that F(k, u) ≥ b|u| – M for k ∈ [1, T]and u ∈ R. (4.4) ρ Kuang and Yang Boundary Value Problems (2018) 2018:86 Page 8 of 11 By (4.4)and (4.1), we have (p+1) a ¯2 I (u )= (u )– λ(u ) ≤ – λb u + T λM (4.5) λ j j j j ρ (p+1) a ¯2 for j ∈ Z .Now,weclaim {u } is bounded. In fact, u → +∞, u → +∞ and – j j j p λb <0, one has I (u ) → –∞ and this is absurd. Hence, I satisﬁes the P.S. condition. Next, λ j λ we prove that I is unbounded from below. By (4.5), we have I (u ) → –∞ as u → λ λ n n +∞. Proof of Theorem 3.1 Put and as in (2.1)and (2.2), it is easily checked that and satisfy all regularity assumptions required in Theorem 2.1. So, our end is to verify condi- –1 tion (2.10)inTheorem 2.1.Let u ∈ (–∞, r]; this means that T–1 T–1 a a(T) a(k) ∗ p p p p u(T)+ u(1) + u(k) ≤ u(1) + u(T) + u(k) ≤ r, p p p k=1 k=1 this, combined with (4.2), gives us 1/p 1 rp 1/q u(k) ≤ T 2 a for k ∈ [1, T]. Let 1/p 1 rp 1/q c = T , 2 a then a (2c ) ∗ ∗ r = p–1 pT and 1 1 sup (u) ≤ max F(k, ξ) r r |ξ |≤c –1 ∗ u∈ (–∞,r] k=1 p–1 pT = max F(k, ξ). (4.6) a (2c ) |ξ |≤c ∗ ∗ k=1 Now, we deﬁne u ∈ E by u = {u (k)} and ∗ ∗ ∗ 1/p 1/q a 1 u (k)= d < c ∗ ∗ ∗ a ¯ T for k ∈ [1, T]. It is easy to check that (u )< r and (u ) p ≥ F(k, d ). (u ) a ¯(2d ) ∗ ∗ k=1 Kuang and Yang Boundary Value Problems (2018) 2018:86 Page 9 of 11 This, combined with (4.6)and (3.2), produces at once (2.10). Therefore, Theorem 2.1 en- sures that (1.1) has at least two nonzero critical points u and u .The proofiscom- λ1 λ2 plete. Proof of Theorem 3.2 Let {u } be a sequence in E such that {I (u )} is bounded and I (u ) → j λ j j + – 0as j → +∞.Put u (k)= max{u (k), 0} and u (k)= max{–u (k), 0} for all k ∈ [1, T], then j j j j + + – – + – u = {u (k)} and u = {u (k)} for k ∈ [1, T]. Therefore, u = u – u for all j ∈ Z . Consid- j j j j j j (p+1) (p+1) (p+1) + a2 a2 + a2 ering that L >0 and λ ∈ ( ,+∞), we ﬁx λ > and ﬁx l such that L > l > . + + ∞ ∞ pλ pL pL ∞ ∞ Now, we claim {u } is bounded. By direct computation, we have T–1 p p p – – – – u = a u (1) + u (T) + a u (k) j j j j k=1 T–1 p p – – – = a –u (T +1)+ u (T) + a u (k) j j j k=1 ≤ – a φ u (k) u (k) p j k=1 =– (u ), u , (4.7) – – + – where u (T +1) = –u (1), for all j ∈ Z . Moreover, by deﬁnition of u and since f (k, x) ≥ 0 j j j for all x <0, we have – – – u ≤ – (u ), u + λ f k, u (k) u (k) j j j j j k=1 – – =– (u ), u + λ (u ), u j j j j =– I (u ), u (4.8) λ j for all j ∈ Z . This, combined with the formulas –I (u ), u λ j lim I (u )=0, lim =0, j→+∞ j→+∞ u gives us lim u =0. j→+∞ – + Hence, our claim is proved. Therefore, there exists Q >0 such that u ≤ Q for all j ∈ Z . Usingasimilarargumentto(4.2)producesatonce 1/p Q 1 u (k) ≤ = L (4.9) a 2 for all k ∈ [1, T]and j ∈ Z . Kuang and Yang Boundary Value Problems (2018) 2018:86 Page 10 of 11 Now, arguing by contradiction, assume that {u } is unbounded, that is, {u } is un- bounded. From F(k, s) lim ≥ L > l, s→+∞ there exists δ >0 such that F(k, s)> ls for all s > δ and k ∈ [1, T]. Let M = max F(k, u)– l|u| :–L ≤ u ≤ δ . 1≤k≤T Then it is easy to check that F k, u (k) ≥ l u (k) – M for k ∈ [1, T]and j ∈ Z . (4.10) j j δ This, combined with (4.1), gives us (p+1) 1 a2 p p p I (u )= u – λ(u ) ≤ u – λlu + T λM , λ j j j j j δ p p p p that is, (p+1) a2 I (u ) ≤ – λl u + T λM λ j j δ (p+1) + a ¯2 for j ∈ Z .Since u → +∞ and – λl <0, one has I (u ) → –∞ andthisisabsurd. j λ j Hence I satisﬁes the P.S. condition. Finally, we prove that I is unbounded from below. Arguing as before, we have I (u ) → λ λ n –∞ as u → +∞. The rest of the proof is similar to Theorem 3.1 and is omitted. The proof is complete. Acknowledgements The authors would like to thank the editor and the referees for their valuable comments and suggestions, which improved the quality of our manuscript. Funding This work is supported by the Natural Science Foundation of Guangdong Province, China (No. 2014A030310334 and No. 2016A030313005) and the Innovation Project of Department of Education of Guangdong Province, China (No. 2015KTSCX147). Abbreviations Not applicable. Availability of data and materials Data sharing not applicable to this article as no datasets were generated or analysed during the current study. Competing interests The authors declare that they have no competing interests. Authors’ contributions The two authors contributed equally to this work. They both read and approved the manuscript. Author details 1 2 School of Mathematics and Computational Sciences, Wuyi University, Jiangmen, P.R. China. School of Financial Mathematics and Statistics, Guangdong University of Finance, Guangzhou, P.R. China. Kuang and Yang Boundary Value Problems (2018) 2018:86 Page 11 of 11 Publisher’s Note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional aﬃliations. Received: 5 January 2018 Accepted: 22 May 2018 References 1. Agarwal, R.P.: Diﬀerence Equations and Inequalities. Dekker, New York (2000) 2. Bai, D., Henderson, J., Zeng, Y.: Positive solutions of discrete Neumann boundary value problems with sign-changing nonlinearities. Bound. Value Probl. 2015, Article ID 231 (2015) 3. Bai, D., Lian, H., Wang, H.: Exact multiplicity of solutions for discrete second order Neumann boundary value problems. Bound. Value Probl. 2015, Article ID 229 (2015) 4. Bisci, G.M., Repovs, D.: Existence of solutions for p-Laplacian discrete equations. Appl. 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Published: May 30, 2018
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