Báo cáo sinh học: " Research Article Existence and Lyapunov Stability of Periodic Solutions for Generalized Higher-Order Neutral Differential Equations" potx

Hindawi Publishing Corporation Boundary Value Problems Volume 2011, Article ID 635767, 21 pages doi:10.1155/2011/635767 Research Article Existence and Lyapunov Stability of Periodic Solutions for Generalized Higher-Order Neutral Differential Equations Jingli Ren, 1 Wing-Sum Cheung, 2 and Zhibo Cheng 1 1 Department of Mathematics, Zhengzhou University, Zhengzhou 450001, China 2 Department of Mathematics, The University of Hong Kong, Pokfulam Road, Hong Kong Correspondence should be addressed to Wing-Sum Cheung, wscheung@hku.hk Received 17 May 2010; Accepted 23 June 2010 Academic Editor: Feliz Manuel Minh´os Copyright q 2011 Jingli Ren et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Existence and Lyapunov stability of periodic solutions for a generalized higher-order neutral differential equation are established. 1. Introduction In recent years, there is a good amount of work on periodic solutions for neutral differential equations see 1–11 and the references cited therein. For example, the following neutral differential equations d du  u  t  − ku  t − τ   g 1  u  t   g 2  u  t − τ 1   p  t  ,  x  t   cx  t − r    f  x   t    g  x  t − τ  t   p  t  ,  xt − cx  t − σ  n  f  x  t  x   t   g   0 −r x  t  s  dα  s    p  t  1.1 have been studied in 1, 3, 8, respectively, and existence criteria of periodic solutions were established for these equations. Afterwards, along with intensive research on the p-Laplacian, 2 Boundary Value Problems some authors 4, 11 start to consider the following p-Laplacian neutral functional differential equations:  φ p  x  t  − cx  t − σ      g  t, x  t − τ  t   p  t  ,  φ p   x  t  − cx  t − σ     f  x   t    g  x  t − τ  t   e  t  , 1.2 and by using topological degree theory and some analysis skills, existence results of periodic solutions for 1.2 have been presented. In general, most of the existing results are concentrated on lower-order neutral functional differential equations, while studies on higher-order neutral functional differential equations are rather infrequent, especially on higher-order p-Laplacian neutral functional differential equations. In this paper, we consider the following generalized higher-order neutral functional differential equation:  ϕ p  xt − cx  t − σ  l  n−l  F  t, x  t  ,x   t  , ,x l−1  t   , 1.3 where ϕ p : R → R is given by ϕ p s|s| p−2 s with p ≥ 2 being a constant, F is a continuous function defined on R l and is periodic with respect to t with period T,thatis,Ft, ·, ,· Ft  T, ·, ,·,Ft, a, 0, ,0 / ≡ 0 for all a ∈ R,andc, σ are constants. Since the neutral operator is divided into two cases |c| /  1and|c|  1, it is natural to study the neutral differential equation separately according to these two cases. The case |c|  1 has been studied in 5. Now we consider 1.3 for the case |c| /  1. So throughout this paper, we always assume that |c| /  1, and the paper is organized as follows. We first transform 1.3 into a system of first-order differential equations, and then by applying Mawhin’s continuation theory and some new inequalities, we obtain sufficient conditions for the existence of periodic solutions for 1.3. The Lyapunov stability of periodic solutions for the equation will then be established. Finally, an example is given to illustrate our results. 2. Preparation First, we recall two lemmas. Let X and Y be real Banach spaces and let L : DL ⊂ X → Y be a Fredholm operator with index zero; here DL denotes the domain of L. This means that Im L is closed in Y and dim Ker L  dimY/ Im L < ∞. Consider supplementary subspaces X 1 , Y 1 of X, Y, respectively, such that X  Ker L ⊕ X 1 , Y  Im L ⊕ Y 1 .LetP : X → KerL and Q : Y → Y 1 denote the natural projections. Clearly, Ker L ∩ DL ∩ X 1 {0} and so the restriction L P : L| DL∩X 1 is invertible. Let K denote the inverse of L P . Let Ω be an open bounded subset of X with DL ∩ Ω /  ∅. A map N : Ω → Y is said to be L-compact in Ω if QNΩ is bounded and the operator KI − QN : Ω → X is compact. Lemma 2.1 see 12. Suppose that X and Y are two Banach spaces, and suppose that L : DL ⊂ X → Y is a Fredholm operator with index zero. Let Ω ⊂ X be an open bounded set and let N : Ω → Y be L-compact on Ω. Assume that the following conditions hold: 1 Lx /  λNx, for all x ∈ ∂Ω ∩ DL,λ∈ 0, 1, 2 Nx / ∈ Im L, for all x ∈ ∂Ω ∩ Ker L, Boundary Value Problems 3 3 deg{JQN,Ω ∩ Ker L, 0} /  0,whereJ :ImQ → Ker L is an isomorphism. Then, the equation Lx  Nx has a solution in Ω ∩ DL. Lemma 2.2 see 13. If ω ∈ C 1 R, R and ω0ωT0,then  T 0 | ω  t  | p dt ≤  T π p  p  T 0   ω   t    p dt, 2.1 where p is a fixed real number with p>1 and π p  2  p−1/p 0 ds  1 − s p /  p − 1  1/p  2π  p − 1  1/p p sin  π/p  . 2.2 For the sake of convenience, throughout this paper we denote by T a positive real number, and for any continuous function u, we write | u | 0 : max t∈  0,T  | u  t  | . 2.3 Let A : C T → C T be the operator on C T : {x ∈ CR, R : xt  Txt for all t ∈ R} given by  Ax  t  : x  t  − cx  t − σ  , ∀x ∈ C T ,t∈ R. 2.4 Lemma 2.3. The operator A has a continuous inverse A −1 on C T satisfying the following:  1   A −1 f   t   ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ f  t   ∞  j1 c j f  t − jσ  , for | c | < 1, ∀f ∈ C T , − f  t  σ  c − ∞  j1 1 c j1 f  t   j  1  σ  , for | c | > 1, ∀f ∈ C T , 2.5  2      A −1 f   t     ≤   f   0 | 1 − | c || , ∀f ∈ C T , 2.6  3   T 0     A −1 f   t     dt ≤ 1 | 1 − | c ||  T 0   f  t    dt, ∀f ∈ C T . 2.7 Remark 2.4. This lemma is basically proved in 3, 10. For the convenience of the readers, we present a detailed proof here as follows. Proof. We split it into the following two cases. Case 1 |c| < 1. Define an operator B : C T → C T by  Bx  t  : cx  t − σ  , ∀x ∈ C T ,t∈ R. 2.8 4 Boundary Value Problems Clearly, B j xtc j xt − jσ and A  I − B.NotealsothatB < |c| < 1. Therefore, A has a continuous inverse A −1 : C T → C T with A −1 I − B −1   ∞ j0 B j ; here B 0 xt : xt. Hence,  A −1 f   t   ∞  j0  B j f   t   f  t   ∞  j1 c j f  t − jσ  , 2.9 and so     A −1 f  t             ∞  j0  B j f   t               ∞  j0 c j f  t − jσ        ≤   f   0 1 − | c | ,  T 0     A −1 f   t     dt ≤ ∞  j0  T 0     B j f   t     dt  ∞  j0  T 0    c j f  t − jσ     dt ≤ 1 1 − | c |  T 0   f  t    dt. 2.10 Case 2 |c| > 1. Define operators E : C T −→ C T ,  Ex  t  : x  t  − 1 c x  t  σ  , B 1 : C T −→ C T ,  B 1 x  t  : 1 c x  t  σ  . 2.11 From the definition of the linear operator B 1 , we have  B j 1 f   t   1 c j f  t  jσ  , ∞  j0  B j 1 f   t   f  t   ∞  j1 1 c j f  t  jσ  . 2.12 Since B 1  < 1, the operator E has a bounded inverse E −1 : C T → C T with E −1   I − B 1  −1  I  ∞  j1 B j 1 , 2.13 and so, for any f ∈ C T ,  E −1 f   t   f  t   ∞  j1  B j 1 f   t  . 2.14 On the other hand, from Axtxt − cxt − σ, we have  Ax  t   x  t  − cx  t − σ   −c  x  t − σ  − 1 c x  t   . 2.15 Boundary Value Problems 5 That is,  Ax  t   −c  Ex  t − σ  . 2.16 Now, for any f ∈ C T ,ifxt satisfies  Ax  t   f  t  , 2.17 then we have −c  Ex  t − σ   f  t  , 2.18 or  Ex  t   − f  t  σ  c  f 1  t  . 2.19 So, we have x  t    E −1 f 1   t   f 1  t   ∞  j1 B j 1 f 1  t   − f  t  σ  c − ∞  j1 B j 1 f  t  σ  c . 2.20 So, A −1 exists and satisfies  A −1 f   t   − f  t  σ  c − ∞  j1 B j 1 f  t  σ  c  − f  t  σ  c − ∞  j1 1 c j1 f  t   j  1  σ  ,     A −1 f   t            − f  t  σ  c − ∞  j1 1 c j1 f  t   j  1  σ        ≤   f   0 | c | − 1 . 2.21 This proves 1 and 2 of Lemma 2.3. Finally, 3 is easily verified. By Hale’s terminology 14,asolutionxt of 1.3 is that xt ∈ C 1 R, R such that Ax ∈ C 1 R, R and 1.3 is satisfied on R. In general, xt does not belong to C 1 R, R. But we can see easily from Ax  tAx  t that a solution xt of 1.3 must belong to C 1 R, R. Equation 1.3 is transformed into  ϕ p  Ax l   t   n−l  F  t, x  t  ,x   t  , ,x l−1  t   . 2.22 Lemma 2.5 see 4. If p>1,then  T 0     A −1 f   t     p dt ≤ 1 | 1 − | c || p  T 0   f  t    p dt, ∀f ∈ C T . 2.23 6 Boundary Value Problems Now we consider 2.22. Define the conjugate index q ∈ 1, 2 by 1/p  1/q  1. Introducing new variables y 1  t   x  t  ,y 2  t   x   t  ,y 3  t   x   t  , ,y l  t   x l−1  t  , y l1  t   ϕ p  Ax l  t   ,y l2  t    ϕ p  Ax l  t    , ,y n  t    ϕ p  Ax l  t   n−l−1 . 2.24 Using the fact that ϕ q ◦ ϕ p ≡ id and by Lemma 2.3, 1.3 can be rewritten as y  1  t   y 2  t  , y  2  t   y 3  t  , . . . y  l−1  t   y l  t  , y  l  t   A −1 ϕ q  y l1  t   , y  l1  t   y l2  t  , . . . y  n−1  t   y n  t  , y  n  t   F  t, y 1  t  ,y 2  t  , ,y l  t   . 2.25 It is clear that, if yty 1 t,y 2 t, ,y n t  is a T-periodic solution to 2.25, then y 1 t must be a T-periodic solution to 1.3. Thus, the problem of finding a T-periodic solution for 1.3 reduces to finding one for 2.25. Define the linear spaces X  Y   y   y 1  ·  ,y 2  ·  , ,y n  ·    ∈ C 0  R, R n  : y  t  T  ≡ y  t   2.26 with norm y  max{y 1 , y 2 , ,y n }. Obviously, X and Y are Banach spaces. Define L : D  L    y ∈ C 1  R, R n  : y  t  T   y  t   ⊂ X −→ Y 2.27 Boundary Value Problems 7 by Ly  y   ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ y  1 y  2 . . . y  l . . . y  n ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ . 2.28 Moreover, define N : X −→ Y 2.29 by Ny  ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ y 2  t  y 3  t  . . . A −1 ϕ q  y l1  t   y l2  t  . . . F  t, y 1  t  ,y 2  t  , ,y l  t   ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ . 2.30 Then, 2.25 can be rewritten as the abstract equation Ly  Ny. From the definition of L, one can easily see that Ker L  {y ∈ C 1 R, R n  : y is constant}R n and Im L  {y : y ∈ X,  T 0 ysds  0}.So,L is a Fredholm operator with index zero. Let P : X → Ker L and Q : Y → Im Q be defined by Py  1 T  T 0 y  s  ds, Qy  1 T  T 0 y  s  ds. 2.31 It is easy to see that Ker L  Im Q  R n . Moreover, for all y ∈ Y, if we write y ∗  y − Qy, we have  T 0 y ∗ sds  0andsoy ∗ ∈ Im L.ThisistosayY  Im Q ⊕ Im L and dimY/ Im L dim Im Q  dim Ker L. So, L is a Fredholm operator with index zero. Let K denote the inverse of L| Ker p∩DL , then we have  Ky   t     T 0 G 1  t, s  y 1  s  ds,  T 0 G 2  t, s  y 2  s  ds, ,  T 0 G n  t, s  y n  s  ds   , 2.32 8 Boundary Value Problems where G i  t, s   ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ s T , 0 ≤ s<t≤ T, s − T T , 0 ≤ t ≤ s ≤ T, i  1, 2, ,n. 2.33 From 2.30 and 2.33, it is clear that QN and KI − QN are continuous, and QN Ω is bounded, and so KI − QN Ω is compact for any open bounded Ω ⊂ X. Hence, N is L- compact on Ω. For the function yty 1 t,y 2 t, ,y n t  defined as 2.24, we have the following. Lemma 2.6. If yt ∈ C 1 R, R n  and yt  Tyt,then  T 0   y  i  t    p dt ≤ 1 | 1− | c p  T π p  pl−i  T π q  qn−l  T 0   y  n t   q dt, 2.34 where 1/p  1/q  1,p≥ 2,i 1, 2, ,l− 1. Proof. From y 1 0y 1 T, there is a point t 1 ∈ 0,T such that y  1 t 1 0. Let ω 1 ty  1 tt 1 . Then, ω 1 0ω 1 T0. From y 2 0y 2 T, there is a point t 2 ∈ 0,T such that y  2 t 2 0. Let ω 2 ty  2 t  t 2 . Then, ω 2 0ω 2 T0. Continuing this way, we get from y l−1 0 y l−1 T apointt l−1 ∈ 0,T such that y  l−1 t l−1 0. Let ω l−1 ty  l−1 t  t l−1 . Then, ω l−1 0 ω l−1 T0. From y l ty l t  T, we have  T 0 Ay  l tdt   T 0 Ay l   tdt Ay l t| T 0  0, so there is a point t l ∈ 0,T such that Ay  l t l 0; hence, we have ϕ p Ay  l t l   0. Let ω l tϕ p Ay  l t  t l   y l1 t  t l . Then, ω l 0ω l T0. Continuing this way, we get from y n−1 0y n−1 T that there is a point t n−1 ∈ 0,T such that y  n−1 t n−1 0. Let ω n−1 ty  n−1 t  t n−1 . Then, ω n−1 0ω n−1 T0. By Lemma 2.2, we have  T 0   y  1 t   p dt   T 0 | ω 1  t  | p dt ≤  T π p  p  T 0   ω  1 t   p dt   T π p  p  T 0   y  2 t   p dt   T π p  p  T 0 | ω 2 t | p dt ≤  T π p  2p  T 0   ω  2 t   p dt . . . Boundary Value Problems 9 ≤  T π p  pl−1  T 0   ω  l−1  t    p dt   T π p  pl−1  T 0   y  l  t    p dt. 2.35 By Lemma 2.5 and Lemma 2.2, we have  T 0   y  l  t    p dt   T 0    A −1 ϕ q  y l1  t      p dt ≤ 1 | 1 − | c || p  T 0   ϕ q  y l1  t     p dt  1 | 1 − | c || p  T 0   y l1  t    pq−p dt  1 | 1 − | c || p  T 0   y l1 t   q dt  1 | 1 − | c || p  T 0 | ω l t | q dt ≤ 1 | 1 − | c || p  T π q  q  T 0   ω  l  t    q dt  1 | 1 − | c || p  T π q  q  T 0   y l2  t    q dt . . . ≤ 1 | 1 − | c || p  T π q  qn−l  T 0   ω  n−1  t    q dt  1 | 1 − | c || p  T π q  qn−l  T 0   y  n  t    q dt. 2.36 Combining 2.35 and 2.36,weget  T 0   y  1  t    p dt ≤ 1 | 1 − | c || p  T π p  pl−1  T π q  qn−l  T 0   y  n  t    q dt. 2.37 10 Boundary Value Problems Similarly, we get  T 0   y  i  t    p dt ≤ 1 | 1 − | c || p  T π p  pl−i  T π q  qn−l  T 0   y  n  t    q dt. 2.38 This completes the proof of Lemma 2.6. Remark 2.7. In particular, if we take p  2, then q  2and π p  π q  π 2  2  2−1/2 0 ds  1 − s 2 /  2 − 1  1/2  2π  2 − 1  1/2 2sin  π/2   π. 2.39 In this case, 2.34  is transformed into  T 0   y i  t    2 dt ≤ 1 | 1− | c 2  T π  2n−i  T 0   y  n  t    2 dt. 2.40 3. Main Results For the sake of convenience, we list the following assumptions which will be used repeatedly in the sequel. H 1  There exists a constant D>0 such that z 1 F  t, z 1 ,z 2 , ,z l  > 0, ∀  t, z 1 ,z 2 , ,z l  ∈  0,T  × R l , with | z 1 | >D. 3.1 H 2  There exists a constant M>0 such that | F  t, z 1 ,z 2 , ,z l  | ≤ M, ∀  t, z 1 ,z 2 , ,z l  ∈  0,T  × R l . 3.2 H 3  There exist nonnegative constants α 1 ,α 2 , ,α l ,msuch that | F  t, z 1 ,z 2 , ,z l  | ≤ α 1 | z 1 |  α 2 | z 2 |  ··· α l | z l |  m, ∀  t, z 1 ,z 2 , ,z l  ∈  0,T  × R l . 3.3 H 4  There exist nonnegative constants γ 1 ,γ 2 , ,γ n such that | F  t, u 1 ,u 2 , ,u n  − F  t, v 1 ,v 2 , ,v n  | ≤ γ 1 | u 1 − v 1 |  γ 2 | u 2 − v 2 |  ··· γ n | u n − v n | 3.4 for all t, u 1 ,u 2 , ,u n , t, v 1 ,v 2 , ,v n  ∈ 0,T × R n . Theorem 3.1. If H 1  and H 2  hold, then 1.3 has at least one nonconstant T-periodic solution. [...]... Natural Science Foundation of China 10971202 , and the Research Grant Council of Hong Kong SAR, China project no HKU7016/07P References 1 S Lu and W Ge, “On the existence of periodic solutions for neutral functional differential equation,” Nonlinear Analysis: Theory, Methods & Applications, vol 54, no 7, pp 1285–1306, 2003 2 S Lu, Existence of periodic solutions for a p-Laplacian neutral functional differential... no 3, pp 977–984, 2008 8 K Wang and S Lu, “On the existence of periodic solutions for a kind of high-order neutral functional differential equation,” Journal of Mathematical Analysis and Applications, vol 326, no 2, pp 1161–1173, 2007 9 J Wu and Z Wang, “Two periodic solutions of second-order neutral functional differential equations,” Journal of Mathematical Analysis and Applications, vol 329, no 1,... 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W Ge, and Z Zheng, Periodic solutions to neutral differential equation with deviating arguments,” Applied Mathematics and Computation, vol 152, no 1, pp 17–27, 2004 4 S Peng, Periodic solutions for p-Laplacian neutral Rayleigh equation with a deviating argument,” Nonlinear Analysis: Theory, Methods & Applications, vol 69, no 5-6, pp 1675–1685, 2008 5 J Ren and Z Cheng, Periodic solutions for generalized. .. high-order neutral differential equation in the critical case,” Nonlinear Analysis: Theory, Methods & Applications, vol 71, no 12, pp 6182–6193, 2009 6 J Shen and R Liang, Periodic solutions for a kind of second order neutral functional differential equations,” Applied Mathematics and Computation, vol 190, no 2, pp 1394–1401, 2007 7 Q Wang and B Dai, “Three periodic solutions of nonlinear neutral functional... , , x n−1 t , 3.32 and the results of Theorems 3.1 and 3.2 still hold Next, we study the Lyapunov stability of the periodic solutions of 3.32 Theorem 3.5 Assume that H4 holds Then every T -periodic solution of 3.32 is Lyapunov stable Proof Let z1 t x t , z2 t x t , , zn t x n−1 t 3.33 18 Boundary Value Problems Then, system 3.32 is transformed into z1 t z2 t , z2 t z3 t , 3.34 zn t A−1 F... claim, and the rest of the proof of the theorem is identical to that of Theorem 3.1 Remark 3.3 If 1.3 takes the form l ϕp x t − cx t − σ where e t ∈ C R, R , e t 3.2 still hold Remark 3.4 If p T n−l e t and F t, x t , x t , , x l−1 t T 0 e t dt e t , 3.31 0, then the results of Theorems 3.1 and 2, then 1.3 is transformed into x t − cx t − σ n F t, x t , x t , , x n−1 t , 3.32 and the results of Theorems... constant For, if y1 ≡ a constant , then from 1.3 we have F t, a, 0, , 0 ≡ 0, which contradicts the assumption that F t, a, 0, , 0 / 0 The proof ≡ is complete Theorem 3.2 If H1 and H3 hold, then 1.3 has at least one nonconstant T -periodic solution if one of the following conditions holds: 1 p > 2, 2 p 2 and 1/|1 − |c|| α1 T α2 T/π l−1 Proof Let Ω1 be defined as in Theorem 3.1 If y t the proof of Theorem... Hence, H1 holds On the other hand, 1 cos z3 sin 2t 8 1 sin z4 8 3.44 1, 1/3π, α2 0, α3 0, α4 0, and m 1 Case 1 If p > 2, then by 1 of Theorem 3.2, 3.40 has at least one nonconstant π -periodic solution Case 2 If p 2, then 1 |1 − |c|| α1 T 1 × 2 α2 T π 1 ×π 3π 3 α3 0 ×1 T π 2 0×1 α4 T π T π n−4 0×1 ×1 1 1 × < 1 2 3 So by 2 of Theorem 3.2, 3.40 has at least one nonconstant π -periodic solution 3.45 Boundary... | e−βt u2 t uk t ··· un t k 1 e−βt γ 1 u1 t |1 − |c|| −β ··· γ1 u1 t e−βt |1 − |c|| γn un t n −β k 2 1 γk uk t e−βt |1 − |c|| < 0 Hence, V is a Lyapunov function for nonautonomous 3.32 the T -periodic solution z∗ of 3.32 is Lyapunov stable see 15, page 50 , and so Finally, we present an example to illustrate our result Example 3.6 Consider the n-order delay differential equation ϕp x t − 3x t − σ 4 n−4 . Problems Volume 2011, Article ID 635767, 21 pages doi:10.1155/2011/635767 Research Article Existence and Lyapunov Stability of Periodic Solutions for Generalized Higher-Order Neutral Differential Equations Jingli. cited. Existence and Lyapunov stability of periodic solutions for a generalized higher-order neutral differential equation are established. 1. Introduction In recent years, there is a good amount of. new inequalities, we obtain sufficient conditions for the existence of periodic solutions for 1.3. The Lyapunov stability of periodic solutions for the equation will then be established. Finally,