Existence of homoclinic solutions for Duffing type differential equation with deviating argument

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1 «28 «3 Sept Communication on Applied Mathematics and Computation Vol.28 No.3 DOI /j.issn Existence of homoclinic solutions for Duffing type differential equation with deviating argument CHEN Wen-bin 1,2, GAO Fang 2 (1. College of Mathematics and Computer Science, Wuyi University, Wuyishan , Fujian Province, China; 2. School of Mathematics and Computer Science, Anhui Normal University, Wuhu , Anhui Province, China) Abstract By using the continuation theorem of the coincidence degree theory and some analysis methods, the existence of a set with 2kT-periodic solutions for a class of duffing type differential equation with delay is studied, and then a homoclinic solution is obtained as a limit of a certain subsequence of the above set. Meanwhile, β(t) is allowed to change sign. Key words homoclinic solution; periodic solution; coincidence degree 2010 Mathematics Subject Classification 34C37 Chinese Library Classification O175.7 Duffing ÊÈÄ ÎÇÆ ÑÔÐ 1,2, Ó Ò 2 (1. ¾ º ; 2.» ¾ ) ³ À ²µ ±Æ ¼ Duffing Å 2kT Á Å Á ¹ Å ¼ β(t) ½ Å ËÍÅ Á À 2010 ÉÏÌ 34C37 ÉÏÌ O175.7 à A ÂÌ (2014) Received ; Revised Corresponding author CHEN Wen-bin, research interest is functional differential equation. cwb210168@126.com

2 No. 3 CHEN Wen-bin, et al.: Existence of homoclinic solutions Introduction This paper is devoted to investigating the existence of homoclinic solutions for a kind of differential systems as follows: x (t) + β(t)x (t) + g(t, x(t τ(t))) = f(t), (1) where β C 1 (R, R), g (R R, R), τ, f C(R, R), g(t+t, ) g(t, ), β(t+t) β(t), β(t) is allowed to change sign and τ(t) is a continuous T-periodic function with τ(t) 0, and T > 0 is a given constant. As we all know, a solution x(t) of Eq. (1) is named homoclinic (to 0) if x(t) 0 and x (t) 0 as t +. In addition, if x(t) 0, then x(t) is called a nontrivial homoclinic solution. The existence of homoclinic solutions for some second-order differential equations has been extensively studied by means of the critical point theory or the methods of the bifurcation theory (see [1-4] and the references therein). Like in the work of Rabinowitz [5], Lzydorek and Janczewska [6], and Tang and Xiao [7], the existence of a homoclinic solution for the equation was obtained as a limit of a certain sequence of 2kT-periodic solutions for the following equation: x (t) + β(t)x (t) + g(t, x(t τ(t))) = f k (t), (2) where k N, f k : R R is a 2kT-periodic function such that f(t), t [, kt ε 0 ), f k (t) = f(kt ε 0 ) + f() f(kt ε (3) 0) (t kt + ε 0 ), t [kt ε 0, kt], ε 0 where ε 0 (0, T) is a constant independent of k. In this paper, we obtain the existence of 2kT-periodic solutions to Eq. (2) by using an extension of Mawhin s continuation theorem, not by using the methods of the critical point theory and the bifurcation theory. This is quite different from the approach used in [1-5]. Furthermore, the methods for getting priori bounds of periodic solutons for Eq. (2) are essentially different from those used in [8-10]. Among our results, an interesting thing is that the coefficient β(t) is allowed to change sign, which could be achieved infrequently in the previous papers. Thus, the results of this paper in the existence of homoclinic solutions are new. 1 Preliminaries Throughout this paper, denotes the absolute value and the Euclidean norm on R. For each k N, let C 2kT = {x x C(R, R), x(t + 2kT) x(t)}, C 1 2kT = {x x

3 310 Communication on Applied Mathematics and Computation Vol. 28 C 1 (R, R), x(t + 2kT) x(t)}, and x 0 = max x(t). If the norms of C 2kT and C2kT 1 are t [0,2kT] defined by C2kT = 0 and C 1 2kT = max{ x 0, x 0 }, respectively, then C 2kT and C 1 2kT are all Banach spaces. Furthermore, for ϕ C 2kT, ( ϕ r = ϕ(t) r dt r, r > 1. Lemma 1 Suppose that X and Y are two Banach spaces, and L : D(L) X Y is a Fredholm operator with index zero, Ω X is an open bounded set, and Ω Y is L-compact on Ω. If all the following conditions hold: (i) Lx λnx, x Ω Dom(L), λ (0, 1); (ii) Nx/ Im L, x Ω KerL; (iii) deg{jqn, Ω KerL, 0} 0, where J : Im Q KerL is an isomorphism, then Lx = Nx has at least one solution on Ω Dom(L). Lemma 2 If u R is continuously differentiable on R, a > 0, µ > 1, and p > 1 are constants, then for all t R, the following inequality holds: u(t) (2a) 1 µ ( t+a t a This lemma is a special case of Lemma 2.2 in [7]. Lemma 3 [11] ( u(s) µ µ ds + a(2a) t+a 1 p t a u (s) p ds Let s C(R, R) with s(t + ω) s(t) and s(t) [0, ω], t R. Suppose p (1, + ), α = max t [0,ω] s(t), and u C1 (R, R) with u(t + ω) u(t). Then, ω 2 Main results 0 ω u(t) u(t s(t)) p dt α p u (t) p dt. For the sake of convenience, we list the following conditions. [H 1 ] There are positive constants a and b such that xg(t, x) ax 2, x R; g(t, u) g(t, v) b u v, t, u, v R. [H 2 ] f C(R, R) is a bounded function with f(t) 0 = (0, 0,, 0) T and ( )1 B := f(t) 2 2 dt + sup f(t) < +. t R R Remark 1 From Eq. (3) and [H 2 ], we see that f k (t) sup f(t). Therefore, if [H 2 ] t R holds, k N, ( fk 2 (t)dt 2 < B. 0 p.

4 No. 3 CHEN Wen-bin, et al.: Existence of homoclinic solutions 311 In order to study the existence of 2kT-periodic solutions to system (2), we firstly study some properties of all possible 2kT-periodic solutions to the following system: x (t) + λβ(t)x (t) + λg(t, x(t τ(t))) = λf k (t), λ (0, 1]. (4) For all k N, let Σ C2kT 1 represent the set of all the 2kT-periodic solutions to system (4). Theorem 1 Suppose [H 1 ] and [H 2 ], β (t) > a, and τ 2 0 b2 ( 1 2 β min (t) + a) 1 < 1 hold, where β min (t) = min t R β (t). For all k N, if u Σ, then there are positive constants A 0, A 1, ρ 0, and ρ 1 which are independent of k and λ, such that u 2 A 0, u 2 A 1, u 0 ρ 0, u 0 ρ 1. Proof For all k N, if u Σ, then u must satisfy u (t) + λβ(t)u (t) + λg(t, u(t τ(t))) = λf k (t), λ (0, 1]. (5) Multiplying both sides of Eq. (5) by u (t) and integrating on the interval [, kt], we have u λ β(t)u (t)u(t)dt + λ g(t, u(t τ(t)))u(t)dt = λ f k (t)u(t)dt. (6) It follows from Eq. (6) that and furthermore λ β(t)u (t)u(t)dt = 1 2 λ β (t)u 2 (t)dt, (7) g(t, u(t τ(t)))u(t)dt = + (g(t, u(t τ(t))) g(t, u(t)))u(t)dt g(t, u(t))u(t)dt, (8) and using Lemma 3, (g(t, u(t τ(t))) g(t, u(t)))u(t) dt b u(t) u(t τ(t)) u(t) dt τ 0 b u 2 u 2. (9) Substituting Eqs. (7) (9), Remark 1, and the assumption [H 1 ] and [H 2 ] into Eq. (6), we have u λ 1 2 β min(t) u λa u 2 2 λ τ 0 b u 2 u 2 + λb u 2. (10) By Eq. (10), we can obtain ( 1 1( τ 0 u 2 2 β min(t) + a) b u 2 + B). (11)

5 312 Communication on Applied Mathematics and Computation Vol. 28 By applying Eqs. (10) and (11), we have that u 2 2 τ 0 b u 2 u 2 + B u 2 τ 0 b u 2 ( 1 2 β min (t) + a ) 1( τ 0 b u 2 + B) 1 1( τ 0 +B( 2 β min(t) + a) b u 2 + B) τ 2 0 b2( 1 ) 1 ( 1 ) 1 u 2 β min (t)+a u τ 0bB 2 β min (t)+a 2 +B 2( 1 ) 1. 2 β min(t) + a (12) Since τ 2 0b 2 ( 1 2 β min (t)+a) 1 < 1, it follows from Eq. (12) that there exists a constant A 1 > 0 such that u 2 A 1. (13) Combining Eqs. (11) and (13), we obtain u 2 ( 1 2 β min (t) + a ) 1( τ 0 ba 1 + B) := A 0. (14) Clearly, A 0 and A 1 are constants independent of k and λ. Then, by using Lemma 2, for all t [, kt], we get i.e., u(t) (2T) 1 2 (2T) 1 2 = (2T) 1 2 ( t+t ( t+kt t ( ( u(s) 2 t+t 2 ds + T(2T) 1 2 ( u(s) 2 t+kt 2 ds + T(2T) 1 2 u(s) 2 ds 2 + T(2T) 1 2 ( t u (s) 2 2 ds u (s) 2 2 ds u (s) 2 ds 2, u 0 (2T) 1 2 u 2 + T(2T) 1 2 u 2 (2T) 1 2 A0 + T(2T) 1 2 A1 := ρ 0. (15) For i = k, k + 1,, k 1, from the continuity of u (t), one can find that there is t i [it, (i + 1)T] such that u (t i ) = 1 (i+1)t T it u (s)ds = u(i + 1) u(it) T 2 T ρ 0,

6 No. 3 CHEN Wen-bin, et al.: Existence of homoclinic solutions 313 and it follows from Eq. (13) that for t [it, (i + 1)T], i = k, k + 1,, k 1, u (t) = t t u (s)ds + u (t i ) t i t i u (s) ds + 2 T ρ 0 (i+1)t it (i+1)t it β 0 T 1 2 u (s) ds + 2 T ρ 0 (i+1)t β(s)u (s) ds + ( it )1 u (s) 2 2 ds + Tg max + TB + 2 T ρ 0 β 0 T 1 2 A1 + Tg max + TB + 2 T ρ 0 := ρ 1, (i+1)t g(s, u(s τ(s))) ds + f(s) ds + 2 it T ρ 0 i.e., u 0 ρ 1, (16) where g max = max u 0 ρ 0 g(t, u(t τ(t))). From inequalities (13) (16), one can assert that the conclusion of Theorem 1 holds. Theorem 2 Assume that the conditions of Theorem 1 are satisfied. Then, for all k N, system (5) has at least one 2kT-periodic solution u k (t) such that where ρ 0 and ρ 1 are constants defined by Theorem 1. u k 0 ρ 0, u k 0 ρ 1, (17) Proof In order to use Lemma 2, for all k N, we consider the following system: x (t) + λβ(t)x (t) + λg(t, x(t τ(t))) = λf k (t), λ (0, 1). (18) Let Ω 1 C2kT 1 represent the set of all the 2kT-periodic of system (18), since (0, 1) (0, 1], Ω 1 Σ, where Σ is defined by Theorem 1. If u Ω 1, by using Theorem 1, we get Let Ω 2 = {x : x KerL, QNx = 0}, where u 0 ρ 0, u 0 ρ 1. L : D(L) C 2kT C 2kT, Lu = u, N : C 2kT C2kT 1, Nu = β(t)u (t) g(t, u(t τ(t))) + f k (t), Q : C 2kT C 2kT /Im L, Qy = 1 y(s)ds. 2kT If x Ω 2, then x = c 1 R (constant vector), and by the assumption [H 1 ], we see that 2kTa c 1 2 c 1 f k (t) dt 2kT c 1 B,

7 314 Communication on Applied Mathematics and Computation Vol. 28 i.e., ( B ) c 1 := B 0. a If we set Ω = {x : x C 1 2kT, x 0 < ρ 0 +B 0, x 0 < ρ 1 +1}, then Ω Ω 1 Ω 2. Thus, (i) and (ii) of Lemma 1 are satisfied. In order to verify (iii) of Lemma 1, define the isomorphism J : Im Q KerL as J(x) = x. Let H(x, µ) : (Ω R) [0, 1] R, H(x, µ) = µx + 1 µ 2kT 2kT 0 (g(t, x(t τ(t))) f k (t))dt. It is easy to see from [H 1 ] that H(x, µ) 0, (x, µ) [ (Ω R)] [0, 1]. Hence, deg{jqn, Ω KerL, 0} = deg{h(x, 0), Ω KerL, 0} = deg{h(x, 1), Ω KerL, 0} 0, thus, condition (iii) of Lemma 1 is satisfied. Threrfore, by using Lemma 1, we see that Eq. (2) has a 2kT-periodic solution u k Ω. Obviously, u k (t) is a 2kT-periodic solution to Eq. (4) for the case of λ = 1, and u k Σ. Thus, by using Theorem 1, we get u k 2 A 0, u k 2 A 1, u k 0 ρ 0, u k 0 ρ 1. (19) Theorem 3 Suppose that the conditions in Theorem 1 hold, then Eq.(1) has a nontrivial homoclinic solution. Proof From Theorem 2, we see that for all k N, there exists a 2kT-periodic solution u k (t) to Eq. (2). Therefore, for all k N, u k (t) is satisfied u k (t) = β(t)u k (t) g(t, u k(t τ(t))) + f k (t). (20) By Eq. (19), we know u k 0 ρ 1, and by Eq. (20), u k 0 β 0 ρ 1 + g max + B := ρ 2. Obviously, ρ 2 is a constant independent of k. Similar to the proof of Lemma 2.4 in [7], we see that there exists u 0 C 1 (R, R) and a subsequence {u kj } of {u k } such that for each interval [c, d] R, u kj (t) u 0 (t), and u k j (t) u 0 (t) uniformly on [c, d]. For all a, b R with a < b, there must be a positive integer j 0 such that for j > j 0, [ k j T, k j T ε 0 ] [a γ 0, b + γ 0 ]. Thus, for t [a γ 0, b + γ 0 ], from Eqs. (3) and (20), we see that u k j (t) = β(t)u k j (t) g(t, u kj (t τ(t))) + f(t). (21) In view of u kj (t) u 0 (t), u k j (t) u 0 (t) uniform on [a, b] and by Eq. (21), we see that u k j (t) = β(t)u k j (t) g(t, u kj (t τ(t))) + f(t) β(t)u 0 (t) g(t, u 0(t τ(t))) + f(t) := χ(t),

8 No. 3 CHEN Wen-bin, et al.: Existence of homoclinic solutions 315 which is uniform on [a, b]. By the fact that u k j (t) is continuously differential on (a, b), for j > j 0 and u k j (t) χ(t) uniformly [a, b], we have χ(t) = u 0(t), t R, in view of a, b R being arbitrary. That is, u 0 (t) is a solution to system (1). Now, we will prove u 0 (t) 0 and u 0(t) 0 for t +. Since + ( u 0 (t) 2 + u 0 (t) 2 )dt = lim i + for all i N, if k j > i, then by Eq. (16), it it ( u kj (t) 2 + u k j (t) 2 )dt Let i + and j +, we have and then + t r Therefore, by using Lemma 2, u(t) (2T) 1 2 ( t+t = lim it it lim i + j + kjt k jt ( u 0 (t) 2 + u 0 (t) 2 )dt it it ( u kj (t) 2 + u k j (t) 2 )dt, ( u kj (t) 2 + u k j (t) 2 )dt A A2 1. ( u 0 (t) 2 + u 0 (t) 2 )dt A A2 1, (22) ( u 0 (t) 2 + u 0(t) 2 )dt 0, u(s) 2 2 ds max{(2t) 1 2,T(2T) 1 2 } Finally, we will show that t+t + T(2T) 1 2 From Eq. (19), we know u 0 (t) ρ 0 for t R, and ( t+t r +. u (s) 2 2 ds ( u 0 (t) 2 + u 0(t) 2 )dt 0, t +. u 0 (t) 0, t +. (23) u 0 β 0ρ 0 + sup g(t, u) + B := M 1, t R. u ρ 0,t [0,T] If Eq. (23) does not hold, then there exists ε 0 (0, 1 2 ) and a sequence {t k} such that and u 0(t k ) 2ε 0, k = 1, 2,. t 1 < t 2 < t 3 < < t k + 1 < t k+1, k = 1, 2,, From this, we have that for t [t k, t k + ε0 1+M 1 ], u 0 (t) = u 0 (t k ) + t t k u 0 (s)ds u 0 (t k ) t t k u 0 (s)ds ε 0.

9 316 Communication on Applied Mathematics and Computation Vol. 28 It follows that + u 0 (t) 2 dt k=1 tk + ε 0 1+M 1 t k u 0 (t) 2 dt =, which contradicts Eq. (22), and therefore Eq. (23) holds. The proof is complete. References [1] Tang X H, Lin X Y. Homoclinic solutions for a class of second-order Hamiltonian system [J]. J Math Anal Appl, 2009, 354(2): [2] Tan X H, Xiao L. Homoclinic solutions for non-autonomous second-order Hamiltonian systems with a coercive potential [J]. Math Anal Appl, 2009, 351: [3] Zhang Z H, Yuan R. Homoclinic solutions for a class of non-autonomous subquadratic secondorder Hamiltonian systems [J]. Nonlinear Anal, 2009, 71(9): [4] Izydorek M, Janczewska J. Homoclinic solutions for non-autonomous second-order Hamiltonian systems with a coercive potential [J]. Math Anal Appl, 2007, 335: [5] Rabinowitz P H. Homoclinic orbits for a class of Hamiltonian systems [J]. Proc Roy Soc Edinburgh Sect A, 1990, 114: [6] Lzydorek M, Janczewska J. Homoclinic solutions for a class of the second order Hamiltonian systems [J]. J Differential Equations, 2005, 219(2): [7] Tang X H, Xiao L. Homoclinic solutions for ordinary p-laplacian systems with a coercive potential [J]. Nonlinear Anal, 2009, 71(3/4): [8] Cheung W S, Ren J L. Periodic solutions for p-laplacian Liénard equation with a deviating argument [J]. Nonlinear Anal, 2004, 59(1/2): [9] Lu S P, Ge W G, Zheng Z X. Periodic solutions for a kind of Rayleigh equation with a deviating argument [J]. Appl Math Lett, 2004, 17(4): [10] Lu S P, Ge W G. Some new results on the existence of periodic solutions to a kind of Rayleigh equation with a deviating argument [J]. Nonlinear Anal, 2004, 56(4): [11] Lu S P, Ge W G. Sufficient conditions for the existence of periodic solutions to some second order differential equations with a deviating argument [J]. J Math Anal Appl, 2005, 38(2):