Báo cáo hóa học: " Research Article Periodic Solutions for Subquadratic Discrete Hamiltonian Systems" pot

Hindawi Publishing Corporation Advances in Difference Equations Volume 2007, Article ID 13916, 16 pages doi:10.1155/2007/13916 Research Article Periodic Solutions for Subquadratic Discrete Hamiltonian Systems Xiaoqing Deng Received 4 February 2007; Accepted 26 April 2007 Recommended by Ondrej Dosly Some existence conditions of periodic solutions are obtained for a class of nonautono- mous subquadratic first-order discrete Hamiltonian systems by the minimax methods in the critical point theory. Copyright © 2007 Xiaoqing Deng. This is an open access article distributed under the Creative Commons Attribution License, which p ermits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. 1. Introduction and statement of main results We denote N, Z, R, C by the set of all natural numbers, integers, real, and complex num- bers, respectively. For a,b ∈Z,defineZ(a) ={a,a+1, }, Z(a,b) ={a,a+1, ,b} when a ≤ b. Consider the nonautonomous first-order discrete Hamiltonian systems JΔx(n)+ ∇H  n,Lx(n)  = 0, n ∈ Z, (1.1) where J =  0 −I N I N 0  , x(n) =  x 1 (n) x 2 (n)  , x i (n) ∈ R N , i = 1,2,N is a given positive integer and I N denotes the N ×N identity matrix, Δx(n) = x(n +1)−x(n), Lx(n) =  x 1 (n+1) x 2 (n)  ,forall n ∈ Z,andH ∈ C 1 (Z ×R 2N ,R).ForagivenintegerT>0, we suppose that H(n + T, z) = H(n,z)foralln ∈ Z and z ∈ R 2N ,and∇H(n,z) denotes the gr adient of H(n,z)inz ∈ R 2N . Our purpose is to establish the existence of T-periodic solutions of (1.1)whereH is subquadratic. 2AdvancesinDifference Equations Let H(n,Lx(n)) = H(n,x 1 (n +1),x 2 (n)) = (1/2)|x 1 (n +1)| 2 + V(n,x 2 (n)) with x 1 (n + 1) = Δx 2 (n), where V ∈ C 1 (Z ×R N ,R)isT-periodic in n,and∇V(n,z) denotes the gra- dient of V (n,z)inz ∈ R N .Then,from(1.1)weobtain Δ 2 x 2 (n −1) + ∇V  n,x 2 (n)  = 0, n ∈ Z, x 2 (n) ∈ R N . (1.2) As the author knows, in the past two decades, there has been a large number of papers devoted to the existence of periodic and subharmonic solutions for subquadratic first- order (see [1–3]) or second-order (see [4–8]) continuous Hamiltonian systems by using the critical point theory. On the other hand, in the last five years, by using the critical point theory, the study of existence conditions of periodic and subharmonic solutions for discrete Hamiltonian systems developed rapidly, such as the superquadratic condition for (1.1) (see [9, 10]) or (1.2) (see [11, 12]), the subquadratic condition for (1.1) in [13] or (1.2) in [14, 15], neither superquadratic nor subquadratic condition for (1.2) in [16]. As for the existence of positive solutions of (1.2) with boundary value condition, we can refer to [17, 18]. Recently, in [19] Xue and Tang established the existence of periodic solution for the second-order subquadratic discrete Hamiltonian system (1.2) and generalized the results in [14]. Here, we extend their results to the first-order subquadratic discrete Hamiltonian system (1.1). Our results are more general than those in the literature [13]. Now, we state our main results below. Theorem 1.1. Suppose that H(n,z) satisfies the following. (H 1 ) There exists an integer T>0 such that H(n + T,z) = H(n,z) for all (n,z) ∈ Z ×R 2N , (H 2 ) there are constants M 0 > 0, M 1 > 0, and 0 ≤α<1 such that   ∇ H(n,z)   ≤ M 0 |z| α + M 1 , ∀(n,z) ∈ Z ×R 2N , (1.3) (H 3 ) |z| −2α  T n =1 H(n,z) → +∞ as |z|→∞. Then problem (1.1) possesses at least one T-periodic solution. Remark 1.2. Theorem 1.1 extends [13, Theorem 1.1] which is the special case of this theorem by letting α = 0. Corollary 1.3. If H(n,z) satisfies (H 1 )-(H 2 )and (H  3 ) |z| −2α  T n =1 H(n,z) →−∞as |z|→∞, then the conclusion of Theorem 1.1 holds. Remark 1.4. Corollary 1.3 extends [13, Corollary 1.1] which is the special case of this corollary by letting α = 0. Theorem 1.5. Suppose that H(n,z) satisfies (H 1 )and (H 4 )lim |z|→∞ (H(n,z)/|z| 2 ) = 0 for all n ∈Z(1,T), (H 5 )lim |z|→∞ [(∇H(n,z),z) −2H(n,z)] =−∞for all n ∈Z(1,T). Then problem (1.1) has at least one T-periodic solution. Corollary 1.6. If H(n,z) satisfies (H 1 ), (H 4 ), and (H  5 )lim |z|→∞ [(∇H(n,z),z) −2H(n,z)] = +∞ for all n ∈Z(1,T), then the conclusion of Theorem 1.5 holds. Xiaoqing Deng 3 Corollary 1.7. If H(n,z) satisfies (H 1 ), (H 5 ), or (H  5 ), and (H  4 )lim |z|→∞ (|∇H(n,z)|/|z|) = 0 for all n ∈ Z(1, T), then the conclusion of Theorem 1.5 holds. Corollary 1.8. If H(n,z) satisfies (H 1 )and (H 6 ) there exist constants 0 <β<2 and R 1 > 0 such that for all (n,z) ∈ Z ×R 2N ,  ∇ H(n,z),z  ≤ βH(n,z), ∀|z|≥R 1 , (1.4) (H 7 ) H(n,z) → +∞ as |z|→∞for all n ∈ Z(1,T), then the conclusion of Theorem 1.5 holds. Remark 1.9. Comparing [13, Theorem 1.3] wi th Corollary 1.8, we extend the interval in which β is and delete the constraint of ( ∇H(n,z),z) > 0. Furthermore, condition (H 7 )is more general than condition (H 6 )of[13, Theorem 1.3]. 2. Variational structure and some lemmas In order to apply critical point theory, we need to state the corresponding Hilbert space and to construct a variational functional. Then we reduce the problem of finding the T-periodic solutions of (1.1) to the one of seeking the critical points of the functional. First we give some notations. Let N be a given positive integer, and S =  x =  x(n)  : x(n) =  x 1 (n) x 2 (n)  ∈ R 2N , x i (n) ∈ R N , i =1,2, n ∈ Z  . (2.1) For any x, y ∈ S, a,b ∈ R, ax + by is defined by ax + by   ax(n)+by(n)  . (2.2) Then S is a vector space. ForanygivenpositiveintegerT>0, E T is defined as a subspace of S by E T =  x =  x(n)  ∈ S : x(n +T) = x(n), n ∈Z  (2.3) with the inner product ·,· and norm ·as follows: x, y= T  n=1  x(n), y(n)  , x=  T  n=1   x(n)   2  1/2 , ∀x, y ∈ E T , (2.4) where ( ·,·)and|·|denote the usual inner product and norm in R 2N , respectively. It is easy to see that (E T ,·,·) is a finite dimensional Hilbert space with dimen- sion 2NT, which can be identified with R 2NT . For convenience, we identify x ∈ E T with x = (x τ (1),x τ (2), ,x τ (T)) τ ,wherex(n) =  x 1 (n) x 2 (n)  ∈ R 2N , n ∈ Z(1,T), and (·) τ is the transpose of a vector or a m atrix. Define another norm in E T as x r =  T  n=1   x(n)   r  1/r , ∀x ∈ E T (2.5) 4AdvancesinDifference Equations for r>1. Obviously, x 2 =x and (E T ,·)isequivalentto(E T ,· r ). Hence there exist C 1 > 0andC 2 ≥ C 1 > 0suchthat C 1 x r ≤x≤C 2 x r , ∀x ∈ E T . (2.6) Let C 1 = T −1 , C 2 = T, one can see that the above inequality holds. In fact, define x ∞ = sup n∈Z(1,T) |x(n)|, since T is a positive integer and r>1, one can get that x ∞ ≤x r ≤ T 1/r x ∞ ≤ Tx ∞ . (2.7) Then we can obtain x ∞ ≤x≤ √ Tx ∞ ≤ Tx ∞ ≤ Tx r , T −1 x r ≤x ∞ ≤x. (2.8) For T>0, we define the functional F(x)onE T as F(x) = 1 2 T  n=1  JΔLx(n −1),x(n)  + T  n=1 H  n,Lx(n)  , ∀x ∈ E T . (2.9) Then we have F ∈ C 1 (E T ,R)and  F  (x), y  = T  n=1  JΔLx(n −1), y(n)  + T  n=1  ∇ H  n,Lx(n)  ,Ly(n)  = T  n=1  JΔx(n),Ly(n)  + T  n=1  ∇ H  n,Lx(n)  ,Ly(n)  (2.10) for all x, y ∈ E T . Obviously, for any x ∈ E T , F  (x) = 0ifandonlyif JΔx(n)+ ∇H  n,Lx(n)  = 0 (2.11) for all n ∈ Z(1,T). Therefore, the problem of finding the T-periodic solution for (1.1)is reduced to the one of seeking the critical point of functional F. Next, we construct a variational structure by using the operator theory which is differ- entfromtheonein[9, 10, 13]. Consider the eigenvalue problem JΔLx(n −1) = λx(n), x(n + T) =x(n). (2.12) Setting A(λ) =  I N λI N −λI N  1 −λ 2  I N  , (2.13) then the problem (2.12)isequivalentto x(n +1) = A(λ)x(n), x(n + T) =x(n). (2.14) Xiaoqing Deng 5 As we all know, the solution of problem (2.14) is denoted by x(n) = μ n C with C = x(0) ∈ R 2N ,whereμ is the eigenvalue of A(λ)andμ T = 1. Then it follows from μ T k = 1and |A(λ k ) −μ k I 2N |=0thatμ k = exp(kωi)withω = 2π/T and λ k = 2sin(kπ/T)withλ T−k = λ k for all k ∈ Z(−[T/2],[T/2]), where [·] is Gauss function. Now we give some lemmas which will be important in the proofs of the results of the paper. Lemma 2.1. Set H k ={x ∈ E T : JΔLx(n −1) =λ k x(n) for all k ∈Z(−[T/2],[T/2])}. Then H k ⊥ H j , ∀k, j ∈Z  −  T 2  ,  T 2  , k = j, (2.15) E T = [T/2]  k=−[T/2] H k . (2.16) Proof. For all x ∈ H k , y ∈H j ,wehave λ k x, y= T  n=1  λ k x(n), y(n)  = T  n=1  JΔLx(n −1), y(n)  = T  n=1  x(n),JΔLy(n −1)  = λ j x, y. (2.17) Since λ k = λ j ,wehavex, y=0, that is, H k ⊥ H j ,then(2.15)holds. Next we consider the elements of H k for all k ∈ Z(−[T/2],[T/2]). Case 1. For all x ∈ H 0 ,itfollowsfromμ 0 = 1that H 0 =  x ∈ E T : x(n) ≡ x(0) = C ∈R 2N  , (2.18) and dimH 0 = 2N. Case 2. T is even.Fork = [T/2] = T/2, it follows from λ T/2 = 2, μ T/2 =−1, and (A(2) + I N )C = 0thatC = (ρ τ ,−ρ τ ) τ with ρ ∈ R N . Therefore, H [T/2] =  x ∈ E T : x(n) = (−1) n  ρ τ ,−ρ τ  τ , ρ ∈ R N  , (2.19) and dimH [T/2] = N. Similarly, for k =−[T/2] =−T/2, we have H −[T/2] =  x ∈ E T : x(n) = (−1) n  ρ τ ,ρ τ  τ , ρ ∈ R N  , (2.20) and dimH −[T/2] = N. T is odd. Similarly, for k = [T/2] = (T −1)/2, we have H [T/2] =  x ∈ E T : x(n) = exp  n(T −1)πi T  ρ τ ,−exp  − πi 2T  ρ τ  τ , ρ ∈ C N  , (2.21) 6AdvancesinDifference Equations and dimH [T/2] = 2N.Fork =−[T/2] =−(T −1)/2, we have H −[T/2] =  x ∈ E T : x(n) = exp  − n(T −1)πi T  ρ τ ,exp  πi 2T  ρ τ  τ , ρ ∈ C N  , (2.22) and dimH −[T/2] = 2N. Case 3. For k ∈ Z(1,[T/2] −1) ∪Z(−[T/2] + 1,−1), it follows from λ k = 2sin(kπ/T), μ k = exp(2kπi/T), and (A(λ k ) −μ k I 2N )C = 0that H k =  x ∈ E T : x(n) = exp  2knπi T  ρ τ ,−exp  −  π 2 − kπ T  i  ρ τ  τ , ρ ∈ C N  , (2.23) and dimH k = 2N. Thus, from Cases 1, 2,and3,wehave dim [T/2]  k=−[T/2] H k = 2N +2  T 2  − 1  × 2N + N + N = 2NT (2.24) when T is even, and dim [T/2]  k=−[T/2] H k = 2N +2  T 2  × 2N = 2NT (2.25) when T is odd. Since dimE T = 2NT and  [T/2] k =−[T/2] H k ⊆ E T , E T =  [T/2] k =−[T/2] H k . Lemma 2.1 is com- pleted.  Let E 0 T = H 0 , E + T =  [T/2] k =1 H k ,andE − T =  −1 k =−[T/2] H k , then it is easy to obtain the following lemma. Lemma 2.2. T  n=1  JΔLx(n −1),x(n)  = 0, ∀x ∈ E 0 T , λ 1 x 2 ≤ T  n=1  JΔLx(n −1),x(n)  ≤ λ [T/2] x 2 , ∀x ∈ E + T , −λ [T/2] x 2 ≤ T  n=1  JΔLx(n −1),x(n)  ≤− λ 1 x 2 , ∀x ∈ E − T , (2.26) where 0 <λ 1 <λ 2 < ···<λ [T/2] . Xiaoqing Deng 7 3. Proofs of the main theorems Proof of Theorem 1.1. Let F(x)bedefinedas(2.9), clearly, F ∈ C 1 (E T ,R). We will first show that F satisfies the Palais-Smale condition, that is, any sequence {x (k) }⊂E T for which |F(x (k) )|≤M 2 with constant M 2 > 0andF  (x (k) ) → 0(k →∞) possesses a convergent subsequence in E T .RecallthatE T is identified with R 2NT .Conse- quently, in order to prove that F satisfies Palais-Smale condition, we only need to prove that {x (k) } is bounded. Suppose that {x (k) } is unbounded, then we can assume, going to a subsequence if necessary, that x (k) →∞as k →∞. Let x (k) = u (k) + v (k) + w (k) = y (k) + w (k) ,whereu (k) ∈ E + T , v (k) ∈ E − T , w (k) ∈ E 0 T with w (k) (n) ≡ C (k) for all n ∈Z. In view of (H 2 ), we have      T  n=1  H  n,Lx (k) (n)  − H  n,Lw (k) (n)       ≤ T  n=1  1 0   ∇ H  n,Lw (k) (n)+sLy (k) (n)      Ly (k) (n)   ds ≤ T  n=1  1 0  M 0   Lw (k) (n)+sLy (k) (n)   α + M 1    Ly (k) (n)   ds ≤ 2M 0 T  n=1    Lw (k) (n)   α +   Ly (k) (n)   α    Ly (k) (n)   + T  n=1 M 1   Ly (k) (n)   ≤ 2M 2 0 λ 1 T  n=1   Lw (k) (n)   2α + M 0 λ 1 2M 0 T  n=1   Ly (k) (n)   2 +2M 0 T  n=1   Ly (k) (n)   α+1 + T  n=1 M 1   Ly (k) (n)   ≤ 2M 2 0 T λ 1   C (k)   2α + λ 1 2   y (k)   2 + 2M 0 C α+1 1   y (k)   α+1 + M 1 √ T   y (k)   . (3.1) By using the same method, we can obtain      T  n=1  ∇ H  n,Lx (k) (n)  ,Lu (k) (n)       ≤ 2M 2 0 λ 1 C 2α 1   x (k)   2α + λ 1 2   u (k)   2 + M 1 √ T   u (k)   , (3.2)      T  n=1  ∇ H  n,Lx (k) (n)  ,Lv (k) (n)       ≤ 2M 2 0 λ 1 C 2α 1   x (k)   2α + λ 1 2   v (k)   2 + M 1 √ T   v (k)   . (3.3) It follows from inequality (3.2)and  F  (x), y  = T  n=1  JΔLx(n −1), y(n)  + T  n=1  ∇ H  n,Lx(n)  ,Ly(n)  , ∀x, y ∈ E T (3.4) 8AdvancesinDifference Equations that λ 1   u (k)   2 ≤ T  n=1  JΔLx (k) (n −1),u (k) (n)  =  F   x (k)  ,u (k)  − T  n=1  ∇ H  n,Lx (k) (n)  ,Lu (k) (n)  ≤   u (k)   + 2M 2 0 λ 1 C 2α 1   x (k)   2α + λ 1 2   u (k)   2 + M 1 √ T   u (k)   (3.5) for sufficiently large k. That is, λ 1 2   u (k)   2 −  1+M 1 √ T    u (k)   ≤ 2M 2 0 λ 1 C 2α 1   x (k)   2α (3.6) for k large enough. Since x (k) →∞as k →∞, we can assume that x (k) ≥1forsuffi- ciently large k. Therefore, for sufficiently large k, from the above inequalit y (3.6), there exists a constant M 3 > 0suchthat   u (k)   ≤ M 3   x (k)   α . (3.7) In fact, if (3.7) is false, then there exists some subsequence of {x (k) }, still denoted by {x (k) },suchthat   x (k)   α   u (k)   −→ 0, k −→ ∞. (3.8) Thanks to the inequality (3.6), one has λ 1 2 ≤ 2M 2 0 λ 1 C 2α 1    x (k)   α   u (k)    2 + 1+M 1 √ T   u (k)   (3.9) for k large enough. Obviously, the above two inequalities imply that x (k)  is bounded for sufficiently large k, which is contradictory with the assumption that x (k) →∞as k →∞. Therefore, (3.7)istrue,andthenwehave   u (k)     x (k)   −→ 0, k −→ ∞. (3.10) Similarly, from inequality (3.3) and equality (2.10), there exists a constant M  3 > 0such that   v (k)   ≤ M  3   x (k)   α (3.11) for sufficiently large k, and then   v (k)     x (k)   −→ 0, k −→ ∞. (3.12) Xiaoqing Deng 9 It follows from (3.10)and(3.12)that   w (k)     x (k)   −→ 1, k −→ ∞, (3.13) and then (3.7 )and(3.11) mean that there exists M 4 > 0suchthat   y (k)   =   u (k)   +   v (k)   ≤ 2M 4 T α/2   C (k)   α (3.14) for sufficiently large k. Therefore, from (3.1), we have      T  n=1  H  n,Lx (k) (n)  − H  n,Lw (k) (n)       ≤  2M 2 0 T λ 1 +2λ 1 M 2 4 T α    C (k)   2α + 2 α+2 M 0 M α+1 4 T α(α+1)/2 C α+1 1   C (k)   α(α+1) +2M 1 M 4 T (α+1)/2   C (k)   α . (3.15) Then there exists M 5 > 0suchthat   C (k)   −2α      T  n=1  H  n,Lx (k) (n)  − H  n,Lw (k) (n)       ≤ M 5 (3.16) as |C (k) |→∞. By using Lemma 2.2 and the boundedness of F(x (k) ), we have M 2 ≥ F  x (k)  = 1 2 T  n=1  JΔLx (k) (n −1),x (k) (n)  + H  n,Lx (k) (n)  = 1 2 T  n=1  JΔLu (k) (n −1),u (k) (n)  + 1 2 T  n=1  JΔLv (k) (n −1),v (k) (n)  + T  n=1  H  n,Lx (k) (n)  − H  n,Lw (k) (n)  + T  n=1 H  n,Lw (k) (n)  ≥ λ 1 2   u (k)   2 − λ [T/2] 2   v (k)   2 + T  n=1  H  n,Lx (k) (n)  − H  n,Lw (k) (n)  + T  n=1 H  n,Lw (k) (n)  . (3.17) It follows from (3.14)and(3.16), by multiplying |C (k) | −2α with both sides of above in- equality, that there exists M 6 > 0suchthat   LC (k)   −2α T  n=1 H  n,LC (k)  =   C (k)   −2α T  n=1 H  n,Lw (k) (n)  ≤ M 6 (3.18) as |C (k) |→∞. This is a contradiction with (H 3 ), consequently, x (k)  is bounded. Thus we conclude that the Palais-Smale condition is satisfied. 10 Advances in Difference Equations In order to use the saddle point theorem (see [20, Theorem 4.6]), we only need to verify the following: (F 1 ) F(x) →−∞as x→∞in X 1 = E − T , (F 2 ) F(x) → +∞ as x→∞in X 2 = E 0 T ⊕E + T . In fact, for v ∈ E − T , there exists M 7 > 0suchthat F(v) = 1 2 T  n=1  JΔLv(n −1), v(n)  + T  n=1  H  n,Lv(n)  − H(n,0)  + T  n=1 H(n,0) ≤− λ 1 2 v 2 + T  n=1  1 0   ∇ H  n,sLv(n)    ·   Lv(n)   ds+ T  n=1 H(n,0) ≤− λ 1 2 v 2 + M 0 C α+1 1 v α+1 + M 1 √ Tv+ M 7 −→ − ∞ (3.19) as v→∞.Thus(F 1 )isverified. Next, for all u +w ∈ E + T ⊕E 0 T ,wehave F(u + w) = 1 2 T  n=1  JΔLu(n −1),u(n)  + T  n=1  H  n,Lu(n)+Lw(n)  − H  n,Lw(n)  + T  n=1 H  n,Lw(n)  ≥ λ 1 2 u 2 − T  n=1  1 0   ∇ H  n,Lw(n)+sLu(n)    ·   Lu(n)   ds+ T  n=1 H  n,Lw(n)  ≥ λ 1 4 u 2 − M 0 C α+1 1 u α+1 −M 1 √ Tu− 4M 2 0 T λ 1 |C| 2α + T  n=1 H(n,LC). (3.20) Since 1 ≤ α +1< 2, λ 1 4 u 2 − 2M 0 C α+1 1 u α+1 −M 1 √ Tu−→+∞, u−→∞. (3.21) By (H 3 )wehave |LC| −2α  T  n=1 H(n,LC) − 4M 2 0 T λ 1 |C| 2α  =| LC| −2α T  n=1 H(n,LC) − 4M 2 0 T λ 1 −→ +∞, |C|−→∞. (3.22) [...]... 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(3.30) Then T F(v) = T 1 JΔLv(n − 1),v(n) + H n,Lv(n) 2 n =1 n =1 λ1 ≤− v 2 2 λ + 1 v 4 2 (3.31) + TC6 − −∞ → − as v → ∞ for v ∈ X1 = ET Therefore, (F1 ) is verified Conditions (H4 ) and (H5 ) imply that H(n,z) → +∞ as |z| → ∞ for all n ∈ Z(1,T) In fact, let s > 1, from (H5 ), for all > 0 there exists constant C7 > 0 such that 1 (∇H(n,sz),sz − 2H(n,sz) ≤ − , ∀|z| ≥ C7 , (3.32) then we have (∇H(n,sz),sz . Corporation Advances in Difference Equations Volume 2007, Article ID 13916, 16 pages doi:10.1155/2007/13916 Research Article Periodic Solutions for Subquadratic Discrete Hamiltonian Systems Xiaoqing Deng Received. conditions of periodic and subharmonic solutions for discrete Hamiltonian systems developed rapidly, such as the superquadratic condition for (1.1) (see [9, 10]) or (1.2) (see [11, 12]), the subquadratic. of periodic solution for the second-order subquadratic discrete Hamiltonian system (1.2) and generalized the results in [14]. Here, we extend their results to the first-order subquadratic discrete