- Email: [email protected]
Periodic solutions for generalized high-order neutral differential equation in the critical case
Nonlinear Analysis 71 (2009) 6182–6193
Contents lists available at ScienceDirect
Nonlinear Analysis journal homepage: www.elsevier.com/locate/na
Periodic solutions for generalized high-order neutral differential equation in the critical caseI Jingli Ren a,b,∗ , Zhibo Cheng a a
Department of Mathematics, Zhengzhou University, Zhengzhou 450001, PR China
b
Department of Mathematics, Dresden University of Technology, Dresden 01069, Germany
article
info
Article history: Received 18 January 2009 Accepted 5 June 2009 MSC: 34K13 34K40
abstract By applying Mawhin’s continuation theory and some new inequalities, we obtain sufficient conditions for the existence of periodic solutions for a generalized high-order neutral differential equation in the critical case. Moreover, an example is given to illustrate the results. © 2009 Elsevier Ltd. All rights reserved.
Keywords: Periodic solution High order Neutral differential equation Critical case
1. Introduction Consider a generalized high-order neutral differential equation in the following form
(ϕp (x(t ) − cx(t − τ ))(l) )(n−l) = F (t , x(t ), x0 (t ), . . . , x(l−1) (t )),
(1.1)
where ϕp : R → R is given by ϕp (s) = |s| s with p ≥ 2 is a constant, F is a continuous function defined on R and is periodic to t with F (t , ·, . . . , ·) = F (t + 2π , ·, . . . , ·), F (t , a, 0, . . . , 0) 6≡ 0 for all a ∈ R, τ is a constant. Complicated behavior of models for technical applications is often described by nonlinear high-order differential equations [1], for example, the Lorenz model of a simplified hydrodynamic flow, the dynamo model of erratic inversion of the earth’s magnetic field, etc. Oftentimes high-order equations are a result of combinations of lower-order equations. Due to its obvious complexity, studies on high-order differential equation are rather infrequent, especially on high-order delay differential equation. Most of the results on high-order delay differential equation are concentrated in the few years. In [2], Cheng and Ren present the existence of periodic solutions for a fourth-order Rayleigh type p-Laplacian delay differential equation as follows p−2
(ϕp (x(t )00 ))00 + f (t , x0 (t − σ (t ))) + g (t , x(t − τ (t ))) = e(t ).
l
(1.2)
In [3], Pan studies the nth-order differential equation x(n) (t ) =
n −1 X
bi x(i) (t ) + f (t , x(t ), x(t − τ1 (t )), . . . , x(t − τm (t ))) + p(t ),
i=1
I Supported by AvH Foundation of Germany and NNSF of China (60504037).
∗
Corresponding author at: Department of Mathematics, Zhengzhou University, Zhengzhou 450001, PR China. E-mail address: [email protected] (J. Ren).
0362-546X/$ – see front matter © 2009 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2009.06.011
(1.3)
J. Ren, Z. Cheng / Nonlinear Analysis 71 (2009) 6182–6193
6183
and obtain the existence of periodic solutions for Eq. (1.3). Afterwards, Ren and Cheng [4] obtain sufficient conditions for the existence of periodic solutions for a general high-order delay differential equation x(n) (t ) = F (t , x(t ), x(t − τ (t )), x0 (t ), . . . , x(n−1) (t )).
(1.4)
In [5], Li and Lu consider the following high-order p-Laplacian differential equation
(ϕp (y(m) (t )))(m) = f (y(t ))y0 (t ) + h(y(t )) + β(t )g (y(t − τ (t ))) + e(t ),
(1.5)
and by using the theory of Fourier series, Bernoulli number theory and continuation theorem of coincidence degree theory, they get the existence of periodic solutions for Eq. (1.5). In [6], Wang and Lu investigate the existence for the high-order neutral functional differential equation with distributed delay
(x(t ) − cx(t − σ ))(n) + f (x(t ))x0 (t ) + g
Z
0
x(t + s)dα(s)
= p(t ).
(1.6)
−r
Inspired by these results, we consider a generalized high-order neutral differential equation (1.1). In [7], Ren, Cheung and Cheng have researched on periodic solution for (1.1) in the case |c | 6= 1. In the paper, we shall move on the research of (1.1) in the more complicated case, i.e, in the critical case |c | = 1. By citing some results on Lu [8,9] and Zhang [10], we first carry out further results on the neutral operator in the critical case and then by applying Mawhin’s continuation theorem we obtain sufficient conditions for the existence of periodic solutions for Eq. (1.1). Our results are new and our methods are different from the above works. Meanwhile, an example is given to illustrate our results. Throughout this paper, we will denote by Z the set of integers, Z1 the set of odd integers, Z2 the set of even integers, N the set of positive integers, N1 the set of odd positive integers and N2 the set of even positive integers. Let C21π = {x : x ∈ C 1 (R, R), x(t + 2π ) ≡ x(t )} with the norm |ϕ|C 1 = {maxt ∈[0,2π] |ϕ(t )|, maxt ∈[ 0,2π ] |ϕ 0 (t )|}, C2π = {x : x ∈ C (R, R), x(t + 2π
R 2π
2π) ≡ x(t )} with the norm |ϕ|0 = maxt ∈[ 0,2π ] |ϕ(t )|, C20π = {x : x ∈ C2π , 0 x(s)ds = 0}, C2−π = {x : x ∈ C (R, R), R 2π +,0 x(t + π ) ≡ −x(t )}, C2+π = {x : x ∈ C (R, R), x(t + π ) ≡ x(t )} and C2π = {x : x ∈ C2+π , 0 x(s)ds = 0} equipped with the norm | · |0 , L2 = {x : R → R is 2π periodic and its restriction to [0, 2π ] belongs to L2 ([0, π])}, under the norm
R 2π
|ϕ(t )|2 dt
12
, L2− = {x : x ∈ L2 , x(t + π ) ≡ −x(t )} and L2+ = {x : x ∈ L2 , x(t + π ) ≡ x(t )} with the norm R 2π 0 2 2− and L2+ are all Banach space. We also denote h¯ = 21π 0 h(s)ds, ∀ h ∈ L2 . | · |2 . Clearly, C21π , C2π , C2+π , C20π , C2+, π ,L ,L |ϕ|2 =
0
2. Preparation Let X and Y be real Banach spaces and 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 X1 , Y1 , of X , Y respectively, such that X = Ker L ⊕ X1 , Y = Im L ⊕ Y1 , and let P : X → Ker L and Q : Y → Y1 denote the natural projections. Clearly, Ker L ∩ (D(L) ∩ X1 ) = {0}, thus the restriction LP := L|D(L)∩X1 is invertible. Let K denote the inverse of LP . Let Ω be an open bounded subset of X with D(L) ∩ Ω 6= ∅. 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 (Gaines and Mawhin [11]). Suppose that X and Y are two Banach spaces, and L : D(L) ⊂ X → Y is a Fredholm operator with index zero. Furthermore, Ω ⊂ X is an open bounded set and N : Ω → Y is L-compact on Ω . Assume that the following conditions hold: (1) Lx 6= λNx, ∀ x ∈ ∂ Ω ∩ D(L), λ ∈ (0, 1); (2) Nx ∈ 6 Im L, ∀ x ∈ ∂ Ω ∩ Ker L; (3) deg{JQN , Ω ∩ Ker L, 0} 6= 0, where J : Im Q → Ker L is an isomorphism, then the equation Lx = Nx has a solution in Ω ∩ D(L). Lemma 2.2 ([10]). If ω ∈ C 1 (R, R) and ω(0) = ω(T ) = 0, then T
Z
|ω(t )| dt ≤ p
0
T
πp
p Z
T
|ω0 (t )|p dt 0
where p is a fixed real number with p > 1, and πp = 2
R (p−1)/p
First, we define operators A in the following form: A : C2π → C2π ,
(Ax)(t ) = x(t ) − cx(t − τ ).
0
ds p
(1− ps−1 )1/p
=
2π (p−1)1/p . p sin(π /p)
6184
J. Ren, Z. Cheng / Nonlinear Analysis 71 (2009) 6182–6193
Lemma 2.3 ([8]). The following propositions are true: (1) If c = −1, |τ | =
m n
π with m, n coprime positive integers and m even, then 1
σ1 := inf |1 − ce−ikτ | = inf [2(1 + cos kτ )] 2 > 0. k∈N
(2.1)
k∈N
(2) If c = −1, |τ | =
m n
π with m, n coprime positive integers and m odd and n even, then 1
σ2 := inf |1 − ce−ikτ | = inf [2(1 + cos kτ )] 2 > 0. k∈N1
(2.2)
k∈N1
(3) If c = −1, |τ | =
m n
π with m, n coprime positive odd integers, then 1
σ3 := inf |1 − ce−ikτ | = inf [2(1 + cos kτ )] 2 > 0. k∈N2
(4) If c = 1, |τ | =
(2.3)
k∈N2
m n
π with m, n coprime positive integers and m odd, then 1
σ4 := inf |1 − ce−ikτ | = inf [2(1 + cos kτ )] 2 > 0. k∈N1
(2.4)
k∈N1
(5) If c = 1, |τ | = π , then 1
σ5 := inf |1 − ce−ikτ | = inf [2(1 + cos kτ )] 2 = 2 > 0. k∈N1
(2.5)
k∈N1
In what follows, we will discuss the invertibility of the difference operator A : [Ax](t ) = x(t ) − cx(t − τ ). Firstly, we ¯ k∈Z1 {sin kt , cos kt }, C2+π ⊂ span ¯ k∈Z2 {1, sin kt , cos kt } and C2π ⊂ span ¯ k∈Z {1, sin kt , cos kt }. So we have the see C2−π ⊂ span following results. Since Ax(n) = (Ax)(n) , ∀ x ∈ C2π := {x : x ∈ C n (R, R), x(t + 2π ) ≡ x(t )}. Then Eq. (1.1) is transformed into
(ϕp ((Ax(l) )(t )))(n−l) = F (t , x(t ), x0 (t ), . . . , x(l−1) (t )).
(2.6)
Lemma 2.4 ([9]). The following propositions are true: (1) Suppose c = −1, |τ | = (m/n)π , where m, n are coprime positive integers with m even, then A : L22π → L22π ,
[Ax](t ) = x(t ) − c (t − τ ),
∀ t ∈ [0, 2π ],
: → satisfying kA k ≤ σ1 . 1 (2) Suppose c = −1, |τ | = (m/n)π , where m, n are coprime odd positive integers, then L22π
−1
has a unique inverse A
2+ A : L22+ π → L2 π ,
L22π
−1
[Ax](t ) = x(t ) − c (t − τ ),
∀ t ∈ [0, 2π ],
2+ 1 −1 has a unique inverse A−1 : L22+ π → L2π satisfying kA k ≤ σ3 . (3) Suppose c = −1, |τ | = (m/n)π , where m, n are coprime positive integers with m odd and n even, then 2− A : L22− π → L2 π ,
[Ax](t ) = x(t ) − c (t − τ ), L22− π
∀ t ∈ [0, 2π ],
L22− π
satisfying kA−1 k ≤ σ1 . 2 (4) Suppose c = 1, |τ | = (m/n)π , where m, n are coprime positive integers with m odd, then has a unique inverse A−1 : 2− A : L22− π → L2 π ,
−1
has a unique inverse A
→
[Ax](t ) = x(t ) − c (t − τ ), :
L22− π
→
(5) Suppose c = 1, |τ | = π , then 2− A : L22− π → L2 π ,
L22− π
−1
satisfying kA
[Ax](t ) = x(t ) − c (t − τ ),
∀ t ∈ [0, 2π ], k≤
1
σ4
.
∀ t ∈ [0, 2π ],
2− 1 −1 has a unique inverse A−1 : L22− π → L2π satisfying kA k ≤ σ5 .
Now we consider (2.6). Define the conjugate index q ∈ (1, 2] by y1 (t ) = x(t ),
y2 (t ) = x0 (t ), (l)
yl+1 (t ) = ϕp (Ax (t )),
1 p
+
1 q
y3 (t ) = x00 (t ), . . . , yl (t ) = x(l−1) (t ),
= 1. Introducing new variables
yl+2 (t ) = (ϕp (Ax(l) (t )))0 , . . . , yn (t ) = (ϕp (Ax(l) (t )))(n−l−1) ,
J. Ren, Z. Cheng / Nonlinear Analysis 71 (2009) 6182–6193
6185
and using the fact that ϕq ◦ ϕp ≡ id, Eq. (1.1) can be rewritten as
0 y1 (t ) = y2 (t ) y02 (t ) = y3 (t ) ··· y0l−1 (t ) = yl (t ) y0l (t ) = A−1 ϕq (yl+1 (t )) y0l+1 (t ) = yl+2 (t ) ··· 0 yn0 −1 (t ) = yn (t ) yn (t ) = F (t , y1 (t ), y2 (t ), . . . , yl (t )).
(2.7)
It is clear that if y(t ) = (y1 (t ), y2 (t ), . . . , yn (t ))> is a 2π -periodic solution to Eq. (2.7), then y1 (t ) must be a 2π -periodic solution to Eq. (1.1). Thus, the problem of finding a 2π -periodic solution for Eq. (1.1) reduces to finding one for Eq. (2.7). Define the linear spaces X = Y = {y = (y1 (·), y2 (·), . . . , yn (·))> ∈ C 0 (R, Rn ) : y(t + 2π ) ≡ y(t )} with norm kyk = max{ky1 k, ky2 k, . . . , kyn k}. Obviously, X and Y are Banach spaces. Define L : D(L) = {y ∈ C 1 (R, Rn ) : y(t + 2π ) = y(t )} ⊂ X → Y by
0 y1 0
y2 . . . 0 Ly = y = 0 yl . . . y0n
and N :X →Y by y2 (t ) y3 (t )
Ny =
.. .
A−1 ϕq (yl+1 (t )) yl+2 (t )
.. . F (t , y1 (t ), y2 (t ), . . . , yl (t ))
.
(2.8)
Then Eq. (2.7) can be rewritten as the abstract equation Ly = Ny. From the definition of L, one can easily see that Ker L = R 2π {y ∈ C 1 (R, Rn ) : y is constant} ' Rn , Im L = {y : y ∈ X , 0 y(s) = 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 2π
2π
Z
y(s)ds;
0
Qy =
Z
1 2π
2π
y(s)ds.
0
R 2π
It is easy to see that Ker L = Im Q = Rn . Moreover, for all y ∈ Y , we have 0 y∗ (s)ds = 0 if y∗ = y − Q (y), which means y∗ ∈ Im L. This is to say Y = Im Q ⊕ Im L and then 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) , we have
[Ky](t ) =
2π
Z
G1 (t , s)y1 (s)ds,
0
2π
Z
G2 (t , s)y2 (s)ds, . . . ,
0
2π
Z
Gn (t , s)yn (s)ds
>
0
where
s , π Gi (t , s) = 2 s − 2π , 2π
0 ≤ s < t ≤ 2π . 0 ≤ t ≤ s ≤ 2π .
i = 1, 2, . . . , n.
(2.9)
From (2.8) and (2.9), it is clearly that QN and K (I − Q )N are continuous, and QN (Ω ) is bounded and then K (I − Q )N (Ω ) is ¯ . For the functions in the domain of L we have compact for any open bounded Ω ⊂ X which means that N is L-compact on Ω
6186
J. Ren, Z. Cheng / Nonlinear Analysis 71 (2009) 6182–6193
Lemma 2.5. Assume that y(t ) ∈ C 1 (R, Rn ) and y(t + 2π ) = y(t ). (1) If c = −1, |τ | =
π with m, n coprime positive integers and m even, then p(l−i) q(n−l) Z 2π Z 2π 1 2π 2π |y0i (t )|p dt ≤ p |y0n (t )|q dt , πq σ1 πp 0 0 m n
+ 1q = 1, p ≥ 2, i = 1, 2, . . . , l − 1. (2) If c = −1, |τ | = m π with m, n coprime positive integers with m odd and n even, then n p(l−i) q(n−l) Z 2π Z 2π 1 2π 2π |y0i (t )|p dt ≤ p |y0n (t )|q dt , πq σ2 πp 0 0 where
1 p
+ 1q = 1, p ≥ 2, i = 1, 2, . . . , l − 1. (3) If c = −1, |τ | = m π with m, n coprime positive integers and m and n odd, then n p(l−i) q(n−l) Z 2π Z 2π 2π 1 2π |y0i (t )|p dt ≤ p |y0n (t )|q dt , πq σ3 πp 0 0 where
(2.11)
1 p
+ 1q = 1, p ≥ 2, i = 1, 2, . . . , l − 1. (4) If c = 1, |τ | = m π with m, n coprime positive integers and m odd, then n p(l−i) q(n−l) Z 2π Z 2π 2π 1 2π |y0n (t )|q dt , |y0i (t )|p dt ≤ p πq σ4 πp 0 0 where
(2.10)
(2.12)
1 p
(2.13)
where 1p + 1q = 1, p ≥ 2, i = 1, 2, . . . , l − 1. (5) If c = 1, |τ | = π , then 2π
Z
|yi (t )| dt ≤ 0
0
where
1 p
+
1 q
p
1
σ
p 5
2π
πp
p(l−i)
2π
q(n−l) Z
πq
2π
|y0n (t )|q dt ,
(2.14)
0
= 1, p ≥ 2, i = 1, 2, . . . , l − 1.
Proof. From y1 (0) = y1 (2π ), there exists a point t1 ∈ [0, 2π ] such that y01 (t1 ) = 0. Let ω1 (t ) = y01 (t + t1 ), and then ω1 (0) = ω1 (2π ) = 0. From y2 (0) = y2 (2π ), there exists a point t2 ∈ [0, 2π ] such that y02 (t2 ) = 0. Let ω2 (t ) = y02 (t + t2 ), and then ω2 (0) = ω2 (2π ) = 0. Continuing this way we get from yl−1 (0) = yl−1 (2π), there exists a point tl−1 ∈ [0, 2π ] such that y0l−1 (tl−1 ) = 0. Let ωl−1 (t ) = y0l−1 (t + tl−1 ), and then ωl−1 (0) = ωl−1 (2π ) = 0. From yl (t ) = yl (t + 2π ), we have
R 2π
R 2π (Ay0l )(t )dt = 0 (Ayl )0 (t )dt = (Ayl )(t ) |20π = 0, so there exists a point tl ∈ [0, 2π ] such that Ay0l (tl ) = 0, then we have 0 ϕ ((Ayl )(tl )) = 0. Let ωl (t ) = ϕp ((Ay0l )(t + tl )) = yl+1 (t + tl ), and then ωl (0) = ωl (2π ) = 0. Continuing this way we get from yn−1 (0) = yn−1 (2π ), there exists a point tn−1 ∈ [0, 2π ] such that y0n−1 (tn−1 ) = 0. Let ωn−1 (t ) = y0n−1 (t + tn−1 ), and then ωn−1 (0) = ωn−1 (2π ) = 0. By Lemma 2.2, we have Z 2π Z 2π |y01 (t )|p dt = |ω1 (t )|p dt 0 0 p Z 2 π 2π ≤ |ω10 (t )|p dt πp 0 p Z 2 π 2π = |y02 (t )|p dt πp 0 p Z 2 π 2π = |ω2 (t )|p dt πp 0 2p Z 2π 2π ≤ |ω20 (t )|p dt πp 0 ··· p(l−1) Z 2π 2π ≤ |ωl0−1 (t )|p dt πp 0 p(l−1) Z 2π 2π |y0l (t )|p dt . (2.15) = πp 0 0 p
J. Ren, Z. Cheng / Nonlinear Analysis 71 (2009) 6182–6193
6187
By Lemmas 2.4 and 2.2, we have 2π
Z
|y0l (t )|p dt =
2π
Z
0
|A−1 ϕq (yl+1 (t ))|p dt
0
p 1
σ
σ1p σ σ
=
2π
σ
σ ···
1
1
2π
|ωl0 (t )|q dt
0 2π
q Z
|yl+2 (t )|q dt
0
q(n−l) Z
πq
σ1p
=
2π
2π
q Z
πq
p 1
≤
2π
πq
p 1
1
|ωl (t )|q dt
0
1
|yl+1 (t )|q dt
0
p 1
≤
2π
Z
1
|yl+1 (t )|pq−p dt
0
p 1
=
2π
Z
1
=
|ϕq (yl+1 (t ))|p dt
0
Z
1
=
2π
Z
1
≤
2π
|ωn0 −1 (t )|q dt
0
q(n−l) Z
πq
σ1p
2π
2π
|y0n (t )|q dt .
(2.16)
0
From (2.15) and (2.16), we can get 2π
Z
|y01 (t )|p dt ≤
0
1
2π
p(l−1)
2π
πp
σ1p
q(n−l) Z
πq
2π
|y0n (t )|q dt .
0
Similarly, we get 2π
Z
|y0i (t )|p dt ≤
0
1
2π
p(l−i)
πp
σ1p
2π
q(n−l) Z
πq
2π
|y0n (t )|q dt ,
0
where i = 1, 2, . . . , l − 1. Cases (2)–(5) can be proved in the same way as in the proof of case (1). This completes the proof of Lemma 2.5. Remark 2.1. In particular, if take p = 2, then q = 2 and
πp = πq = π2 = 2
(2−1)/2
Z
ds
2π (2 − 1)1/2
= π. 2 sin(π /2) (1 − 2−1 )1/2 R 2π 2(n−i) R 2π 0 |yn (t )|2 dt . In the case, (2.10) is transformed into 0 |yi (t )|2 dt ≤ 12 2ππ 0 σ s2
0
=
1
3. Main results For the sake of convenience, we list the following assumptions which will be used repeatedly in what follows:
(H1 ) there exists a positive constant D > 0 such that z1 F (t , z1 , z2 , . . . , zl ) > 0,
∀ (t , z1 , z2 , . . . , zl ) ∈ [0, 2π ] × Rl with |z1 | > D;
(H2 ) there exists a positive constant M > 0 such that |F (t , z1 , z2 , . . . , zl )| ≤ M ,
∀ (t , z1 , z2 , . . . , zl ) ∈ [0, 2π ] × Rl ;
(H3 ) there exist non-negative constants α1 , α2 , . . . , αl , m, such that |F (t , z1 , z2 , . . . , zl )| ≤ α1 |z1 | + α2 |z2 | + · · · + αl |zl | + m, ∀ (t , z1 , z2 , . . . , zl ) ∈ [0, 2π ] × Rl .
6188
J. Ren, Z. Cheng / Nonlinear Analysis 71 (2009) 6182–6193
Theorem 3.1. If (H1 ) and (H2 ) hold, then Eq. (1.1) has at least one non-constant 2π -periodic solution. Proof. Consider the equation Ly = λNy,
λ ∈ (0, 1).
Let Ω1 = {y : Ly = λNy, λ ∈ (0, 1)}. If y(t ) = (y1 (t ), y2 (t ), . . . , yn (t ))> ∈ Ω1 , then
y0 (t ) = λy (t ) 2 1 y02 (t ) = λy3 (t ) .. . 0 y ( t ) = λyl (t ) l−10 yl (t ) = λϕq (yl+1 (t )) 0 yl+1 (t ) = λyl+2 (t ) .. . 0 y ( t ) = λyn (t ) n − 1 0 yn (t ) = λF (t , y1 (t ), y2 (t ), . . . , yl (t )).
(3.1)
We first claim that there exists a constant ξ ∈ R such that
|y1 (ξ )| ≤ D.
(3.2)
Integrating the last equation of (3.1) over [0, 2π ], we have 2π
Z
F (t , y1 (t ), y2 (t ), . . . , yl (t ))dt = 0.
0
Then from the continuity of F , we know that there exists a constant ξ ∈ [0, 2π ] such that F (ξ , y1 (ξ ), . . . , yl (ξ )) = 0. From assumption (H1 ) we get (3.2). As a consequence, we have
Z t Z |y1 (t )| = y1 (ξ ) + y01 (s)ds ≤ D + ξ
2π
|y01 (s)|ds.
(3.3)
0
On the other hand, multiplying both sides of the last equation of (3.1) by y0n (t ) and integrating over [0, 2π ], by using assumption (H2 ), we have 2π
Z
|yn (t )| dt = λ 0
2
Z
0
2π
0 2π
Z ≤
F (t , y1 (t ), y2 (t ), . . . , yl (t ))y0n (t )dt
|F (t , y1 (t ), y2 (t ), . . . , yl (t ))||y0n (t )|dt
0 2π
Z
|y0n (t )|dt
≤M 0
1/2
≤ M (2π )
2π
Z
|yn (t )| dt 0
2
1/2
.
0
It is easy to see that there exists a constant Mn0 > 0 (independent of λ) such that 2π
Z
|y0n (t )|2 dt ≤ Mn0 .
0
From yn−1 (0) = yn−1 (2π ), there exists a point t1 ∈ [0, 2π ] such that yn (t1 ) = 0, and by applying Hölder’s inequality, we have
|yn (t )| ≤
2π
Z
1/2
|yn (t )|dt ≤ (2π ) 0
2π
Z
0
|yn (t )| dt 0
2
1/2
≤ (2π )1/2 Mn01/2 := Mn .
0
From yn−2 (0) = yn−2 (2π ), there exists a point t2 ∈ [0, 2π] such that yn−1 (t2 ) = 0, we have
|yn−1 (t )| ≤
2π
Z 0
|y0n−1 (t )|dt =
Z 0
2π
|λyn (t )|dt ≤
2π
Z 0
|yn (t )|dt ≤ 2π Mn := Mn−1 .
J. Ren, Z. Cheng / Nonlinear Analysis 71 (2009) 6182–6193
6189
Similarly,
|yn−2 (t )| ≤ 2π Mn−1 := Mn−2 . And continuing this way for yn−3 , . . . , yl+1 , we get
|yl+1 (t )| ≤ 2π Ml+2 := Ml+1 . Case (1) If c = −1, |τ | =
π with m, n coprime positive integers and m even, then Z 2π Z 2π 2π 1 1 |λA−1 ϕq (yl+1 (t ))|dt ≤ |y0l (t )|dt ≤ |yl+1 (t )|q−1 dt ≤ 2π Mlq+−11 := Ml1 . |yl (t )| ≤ σ1 0 σ1 0 0 m n
Z
Case (2) If c = −1, |τ | =
π with m, n coprime positive integers and m odd and n even, then Z 2π Z 2π 2π 1 1 |λA−1 ϕq (yl+1 (t ))|dt ≤ |y0l (t )|dt ≤ |yl (t )| ≤ |yl+1 (t )|q−1 dt ≤ 2π Mlq+−11 := Ml2 . σ2 0 σ2 0 0 m n
Z
Case (3) If c = −1, |τ | =
π with m, n coprime positive odd integers, then Z 2π Z 2π 2π 1 1 |yl (t )| ≤ |y0l (t )|dt ≤ |yl+1 (t )|q−1 dt ≤ 2π Mlq+−11 := Ml3 . |λA−1 ϕq (yl+1 (t ))|dt ≤ σ σ 3 3 0 0 0 m n
Z
π with m, n coprime positive integers and m odd, then Z 2π Z 2π 2π 1 1 |yl (t )| ≤ |y0l (t )|dt ≤ |λA−1 ϕq (yl+1 (t ))|dt ≤ |yl+1 (t )|q−1 dt ≤ 2π Mlq+−11 := Ml4 . σ σ 4 4 0 0 0
Case (4) If c = 1, |τ | =
m n
Z
Case (5) If c = 1, |τ | = π , then
|yl (t )| ≤
Z
2π
|y0l (t )|dt ≤
Z
0
2π
|λA−1 ϕq (yl+1 (t ))|dt ≤
0
1
σ5
2π
Z 0
|yl+1 (t )|q−1 dt ≤
1
σ5
q−1
2π Ml+1 := Ml5 .
Define Ml = max{Ml1 , Ml2 , Ml3 , Ml4 , Ml5 }.
|yl−1 (t )| ≤ 2π Ml := Ml−1 . .. . |y2 (t )| ≤ 2π M3 := M2 . Meanwhile, from Eq. (3.3), we can get
|y1 (t )| ≤ D +
2π
Z
|y01 (t )|dt ≤ D + 2π M2 := M1 .
0
Let M0 = max{M1 , M2 , . . . , Mn }, obviously ky1 k ≤ M0 , ky2 k ≤ M0 , . . ., kyn k ≤ M0 . Let Ω2 = {y ∈ Ker L : Ny ∈ Im L}. If y ∈ Ω2 , then y ∈ Ker L which means y = constant and QNy = 0. We see that
y2 = 0, y3 = 0, .. . yn = 0, F (t , y1 , 0, . . . , 0) = 0. So
|y1 | ≤ D ≤ M0 ,
y2 = y3 = · · · = yn = 0 ≤ M0 .
Now take Ω = {y = (y1 , y2 , . . . , yn )> ∈ X : ky1 k < M0 + 1, ky2 k < M0 + 1, . . . , kyn k < M0 + 1}. By the analysis of the above, it is easy to see that Ω 1 ⊂ Ω , Ω 2 ⊂ Ω and conditions (1) and (2) of Lemma 2.1 are satisfied. Next we show that condition (3) of Lemma 2.1 is also satisfied. Define the isomorphism J : Im Q → Ker L as follows: J (y1 , y2 , . . . , yn )> = (yn , y1 , . . . , yn−1 )> .
6190
J. Ren, Z. Cheng / Nonlinear Analysis 71 (2009) 6182–6193
Let H (µ, y) = µy + (1 − µ)JQNy, (µ, y) ∈ [0, 1] × Ω , then ∀(µ, y) ∈ (0, 1) × (∂ Ω ∩ Ker L),
1−µ
µy1 +
2π
H (µ, y) =
2π
Z 0
F (t , y1 , 0, . . . , 0)dt
.
y2
.. . A ϕq (yl+1 ) .. . −1
yn From (H1 ), it is obvious that yH (µ, y) > 0, ∀(µ, y) ∈ (0, 1) × (∂ Ω ∩ Ker L). Therefore, deg{JQN , Ω ∩ Ker L, 0} = deg{H (0, y), Ω ∩ Ker L, 0}
= deg{H (1, y), Ω ∩ Ker L, 0} = deg{I , Ω ∩ Ker L, 0} 6= 0, which means that condition (3) of Lemma 2.1 is also satisfied. By applying Lemma 2.1, we conclude that equation Ly = Ny ¯ , i.e., Eq. (1.1) has a T -periodic solution y∗1 (t ) with ky∗1 k < M0 + 1. has a solution y∗ (t ) = (y∗1 (t ), y∗2 (t ), . . . , y∗n (t ))> on Ω Finally, observe that y∗1 (t ) is not constant. Otherwise, suppose y∗1 (t ) ≡ a (constant), then from Eq. (1.1) we have F (t , a, 0, . . . , 0) ≡ 0, which is contradict to assumption F (t , a, 0, . . . , 0) 6≡ 0, so the proof is complete. Theorem 3.2. If (H1 ) and (H3 ) hold, then Eq. (1.1) has at least one non-constant 2π -periodic solution, if one of the following conditions hold: (1) p > 2; (2) p = 2 and if c = −1, |τ | = 2π l−2
α3
π
2π
+ · · · + αl
π
π
(3) p = 2 and if c = −1, |τ | =
2π l−2
α3
π
π
π
· · · + αl
π
2π n−l
π
· · · + αl
π
2π n−l
π
m n
2π l−1
π
+
< 1; 2π l−1
+
2π l−2
+
2π l−2
+
π
< 1;
π with m, n coprime positive odd integers, and
1
σ3
(α1 2π + α2 )
2 π l −1
π
+ α3
π
< 1;
(5) p = 2 and if c = 1, |τ | = 2π
(α1 2π + α2 )
, n coprime positive integers with m odd and n even, and σ12 (α1 2π +α2 )
π
(4) p = 2 and if c = −1, |τ | = 2π
1
σ1
m with m n 2π n−l
2π
+ · · · + αl
π with m, n coprime positive integers and m even, and
m n n −l 2π
m n
π with m, n coprime positive integers and m odd, and
1
σ4
(α1 2π + α2 )
2 π l −1
π
+ α3
π
< 1;
(6) p = 2 and if c = 1, |τ | = π , and σ1 (α1 2π + α2 ) 5
2 π l −1
π
+ α3
2π l−2
π
+ · · · + αl
2π
π
2π n−l
π
< 1.
Proof. Let Ω1 be defined as in Theorem 3.1. If y(t ) = (y1 (t ), y2 (t ), . . . , yn (t ))> ∈ Ω1 , then from the proof of Theorem 3.1 we see that y0n (t ) = λF (t , y1 (t ), y2 (t ), . . . , yl (t )),
(3.4)
and 2π
Z |y1 |0 ≤ D +
|y01 (s)|ds.
0
We claim that |yn |0 is bounded. Multiplying both sides of Eq. (3.4) by ϕq (y0n (t )) and integrating it over [0, 2π ], by using assumption (H3 ), we have 2π
Z
|yn (t )| dt = λ 0
q
Z
0
2π
0 2π
Z ≤
F (t , y1 (t ), y2 (t ), . . . , yl (t ))ϕq (y0n (t ))dt
|F (t , y1 (t ), y2 (t ), . . . , yl (t ))kϕq (y0n (t ))|dt
0
≤ α1
2π
Z
|y1 (t )kϕq (y0n (t ))|dt + α2
0
+ · · · + αl
2π
Z
|y2 (t )kϕq (y0n (t ))|dt
0 2π
Z 0
|yl (t )kϕq (y0n (t ))|dt + m
2π
Z 0
|ϕq (y0n (t ))|dt
J. Ren, Z. Cheng / Nonlinear Analysis 71 (2009) 6182–6193
≤ α1 |y1 |0
2π
Z
|ϕq (y0n (t ))|dt + α2
2π
Z
6191
|y2 (t )kϕq (y0n (t ))|dt
0
0
+ · · · + αl Z ≤ α1 D +
Z 2π |ϕq (y0n (t ))|dt |yl (t )kϕq (y0n (t ))|dt + m 0 0 Z 2π Z 2π 2π |y2 (t )kϕq (y0n (t ))|dt |ϕq (y0n (t ))|dt + α2 |y01 (t )|dt 2π
Z
+ · · · + αl
0
0
0 2π
Z
|yl (t )kϕq (y0n (t ))|dt + m
2π
Z
|ϕq (y0n (t ))|dt .
0
0
Applying Hölder’s inequality, we have 2π
Z
|yn (t )| dt ≤ α1 D + (2π ) 0
q
2π
Z
1 p
0
|y1 (t )| dt 0
q
1q
1
(2π ) q
0
+ α2
2π
Z
2π
Z
|ϕq (y0n (t ))|p dt
1p
0
|y2 (t )| dt q
1q Z
0
2π
|ϕq (yn (t ))| dt 0
p
1p
0
+ · · · + αl
2π
Z
|yl (t )| dt q
1q Z
0
≤ α1 2π · (2π )
2π
|ϕq (yn (t ))| dt 0
p
2π
Z
|y1 (t )| dt 0
p
2π
Z
q(p1−1) Z
|ϕq (yn (t ))| dt 0
p
1p
2π
|yn (t )| dt 0
q
1p
0
|y2 (t )| dt p
q(p1−1) Z
2π
|yn (t )| dt 0
q
1p
0
0
+ · · · + αl (2π )
2π
Z 0
0
+ α2 (2π )
+ m(2π )
1 q
0
p−2 q(p−1)
p−2 q(p−1)
1p
p−2 q(p−1)
2π
Z
|yl (t )| dt p
q(p1−1)
0 2π
Z ×
|yn (t )| dt 0
q
1p
+ (α1 D + m) (2π )
1 q
0
|yn (t )| dt 0
q
1p
0
= α1 2π · (2π )
p−2 p
2π
Z
|y1 (t )| dt 0
p
1p Z
0
+ α2 (2π )
p−2 p
2π
Z
2π
×
|yn (t )| dt 0
q
1p
|y2 (t )| dt p
1p Z
2π
|yn (t )| dt 0
q
1p
+ · · · + αl (2π )
0
|yn (t )| dt 0
2π 0
0
Z
2π
Z
q
1p
p−2 p
Z
2π
|yl (t )| dt p
1p
0
+ (α1 D + m) (2π )
1 q
2π
Z
0
|yn (t )| dt 0
q
1p
.
(3.5)
0
Then by applying Lemma 2.5 and Eq. (3.5), we can get Case (1) If c = −1, |τ | = m π with m, n coprime positive integers and m even, n 2π
Z 0
|yn (t )| dt ≤ α1 2π · (2π ) 0
q
p−2 p
1
2π
l−1
2π
q(np−l) Z
2π
2p
|yn (t )| dt πp πq 0 l−1 q(np−l) Z 2π 2p p−2 1 2π 2π + α2 (2π ) p + ··· + |y0n (t )|q dt σ1 πp πq 0 q(np−l) Z 2π 2p Z 2π 1p p−2 1 1 2π 2π + αl (2π ) p |y0n (t )|q dt + (α1 D + m)(2π ) q |y0n (t )|q dt σ1 πp πq 0 0 " l−1 l −2 # q(np−l) p − 2 1 2π 2π 2π 2π ≤ (α1 2π + α2 ) + α3 + · · · + αl (2π ) p σ1 πp πp πp πq 2π
Z × 0
q
σ1
|yn (t )| dt 0
0
q
2p
+ (α1 D + m)(2π )
1 q
2π
Z 0
|yn (t )| dt 0
q
1p
.
6192
J. Ren, Z. Cheng / Nonlinear Analysis 71 (2009) 6182–6193
l−1 If p = 2 and σ1 (α1 2π + α2 ) 2ππ + α3 1 Mn0 > 0 (independent of λ) such that Z 2π |y0n (t )|q dt ≤ Mn0 .
2 π l −2
π
2π
+ · · · + αl
π
2π n−l
π
< 1, it is easy to see that there exists a constant
0
Similarly, Case (2) If p = 2 and c = −1, |τ | =
π with m, n coprime positive integers with m odd and n even, and σ12 (α1 2π 2π n−l 2π l−2 + · · · + αl 2ππ < 1, it is easy to see that there exists a constant Mn0 > 0 (independent π π m n
l−1 + α3 + α2 ) 2ππ of λ) such that Z 2π |y0n (t )|q dt ≤ Mn0 . 0
Case (3) If p = 2 and c = −1, |τ | =
· · · + αl Z
2π
π
2π
2π n−l
π
m n
π with m, n coprime positive odd integers, and σ13 (α1 2π +α2 )
2 π l −1
π
+α3
2 π l −2
π
+
< 1, it is easy to see that there exists a constant Mn > 0 (independent of λ) such that 0
|y0n (t )|q dt ≤ Mn0 .
0
Case (4) If p = 2 and c = 1, |τ | =
α3
2π l−2
π
2π
Z
+ · · · + αl
2π
π
2 π n −l
π
m n
π with m, n coprime positive integers and m odd, and
1
σ4
(α1 2π + α2 )
2 π l −1
π
+
< 1, it is easy to see that there exists a constant Mn > 0 (independent of λ) such that 0
|y0n (t )|q dt ≤ Mn0 .
0
Case (5) If p = 2 and c = 1, |τ | = π , and σ1 (α1 2π + α2 ) 2ππ + α3 5 see that there exists a constant Mn0 > 0 (independent of λ) such that
l−1
2π
Z
2π l−2
π
+ · · · + αl
2π
π
2π n−l
π
< 1, it is easy to
|y0n (t )|q dt ≤ Mn0 .
0
Case (6) If p > 2, it is easy to see that there exists a constant Mn0 > 0 (independent of λ) such that 2π
Z
|y0n (t )|q dt ≤ Mn0 .
0
From yn−1 (0) = yn−1 (2π ), there exists a point t1 ∈ [0, 2π] such that yn (t1 ) = 0, By applying Hölder’s inequality, we have
|yn (t )| ≤
2π
Z
|yn (t )|dt ≤ (2π ) 0
1 p
0
Z
2π
|yn (t )| dt 0
q
1q
1
01
≤ (2π ) p Mn q := Mn .
0
This proves the claim and the rest of the proof of the theorem is identical to that of Theorem 3.1.
Remark 3.1. If p = 2, then Eq. (1.1) is transformed into
(x(t ) − cx(t − τ ))(n) = F (t , x(t ), x0 (t ), . . . , x(n−1) (t )), then the results of Theorems 3.1 and 3.2 still hold. Remark 3.2. If the equation is taken the form as
(ϕp (x(t ) − cx(t − τ ))(l) )(n−l) = F (t , x(t ), x0 (t ), . . . , x(l−1) (t )) + e(t ), R 2π where e(t ) ∈ C (R, R), e(t + 2π ) = e(t ) and 0 e(t )dt = 0, the results of Theorems 3.1 and 3.2 still hold. Finally, we present an example to illustrate our result. Example 3.1. Consider the n-order delay differential equation
(ϕp (x(t ) − x(t − π ))(4) )0 =
1 24π
x( t ) +
1 8
sin x0 (t ) +
1 8
cos x00 (t ) sin t +
1 8
sin x000 (t ).
(3.6)
J. Ren, Z. Cheng / Nonlinear Analysis 71 (2009) 6182–6193
Here p is a constant with p ≥ 2. It is clear that T = 2π , c = 1, τ = π , F (t , z1 , z2 , z3 , z4 ) =
6193
+ 18 sin z2 + 18 cos z3 sin t + sin z4 , F (t , a, 0, 0, 0) = a + sin t 6≡ 0. Choose D = 24π such that (H1 ) hold, and it is obvious that (H2 ) is not satisfied here. Now we consider the assumption (H3 ). Since 1 8
1 24π
|F (t , z1 , z2 , z3 , z4 )| ≤
1 8
1 24π
1 z 24π 1
|z1 | + 1.
That is to say (H3 ) hold with α1 =
1 24π
, α2 = 0, α3 = 0, α4 = 0, m = 1.
Case (1) If p > 2, by Theorem 3.2, we know that Eq. (3.6) has at least one non-constant 2π -periodic solution. Case (2) If p = 2, and from Lemma 2.3 1
σ5 := inf |1 − ce−ikτ | = inf [2(1 + cos kτ )] 2 = 2 > 0. k∈N1 k∈N1 " 3 2 # 2π 2π 2π 1 2π 1 1 3 + α3 + α4 = × × 2π + 0 × 2 + 0 + 0 × 2 (α1 2π + α2 ) σ5 π π π π 2 24π =
1 2
×
2 3
× 2 < 1.
So by Theorem 3.2, we know that Eq. (3.6) has at least one non-constant 2π -periodic solution. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11]
E.A. Jackson, Perspectives of Nonlinear Dynamics, Cambridge University Press, 1991. Z.B. Cheng, J.L. Ren, Periodic solutions for a fourth-order Rayleigh type p-Laplacian delay equation, Nonlinear Anal. TMA 70 (2009) 516–523. L.J. Pan, Periodic solutions for higher order differential equations with deviating argument, J. Math. Anal. Appl. 343 (2008) 904–918. J.L. Ren, Z.B. Cheng, On high-order delay differential equation, Comput. Math. Appl. 57 (2009) 324–331. X.J. Li, S.P. Lu, Periodic solution for a kind of high-order p-Laplacian differential equation with sign-changing coefficient ahead of non linear term, Nonlinear Anal. 70 (2009) 1011–1022. K. Wang, S.P. Lu, On the existence of periodic solutions for a kind of high-order neutral functional differential equation, J. Math. Anal. Appl. 326 (2007) 1161–1173. J.L. Ren, W.S. Cheung, Z.B. Cheng, On generalized high-order neutral differential equation, preprint. S.P. Lu, W.G. Ge, Z.X. Zheng, Periodic solutions to a kind of neutral functional differential equation in the critical case, J. Math. Anal. Appl. 293 (2004) 462–475. S.P. Lu, Z.J. Gui, W.G. Ge, Periodic solutions to a second order nonlinear neutral function functional equation in the critical case, Nonlinear Anal. TMA 64 (2006) 1587–1607. M.R. Zhang, Nonuniform nonresonance at the first eigenvalue of the p-Laplacian, Nonlinear Anal. TMA 29 (1997) 41–51. R.E. Gaines, J.L. Mawhin, Coincidence Degree and Nonlinear Differential Equation, Springer, Berlin, 1977.
Contents lists available at ScienceDirect
Nonlinear Analysis journal homepage: www.elsevier.com/locate/na
Periodic solutions for generalized high-order neutral differential equation in the critical caseI Jingli Ren a,b,∗ , Zhibo Cheng a a
Department of Mathematics, Zhengzhou University, Zhengzhou 450001, PR China
b
Department of Mathematics, Dresden University of Technology, Dresden 01069, Germany
article
info
Article history: Received 18 January 2009 Accepted 5 June 2009 MSC: 34K13 34K40
abstract By applying Mawhin’s continuation theory and some new inequalities, we obtain sufficient conditions for the existence of periodic solutions for a generalized high-order neutral differential equation in the critical case. Moreover, an example is given to illustrate the results. © 2009 Elsevier Ltd. All rights reserved.
Keywords: Periodic solution High order Neutral differential equation Critical case
1. Introduction Consider a generalized high-order neutral differential equation in the following form
(ϕp (x(t ) − cx(t − τ ))(l) )(n−l) = F (t , x(t ), x0 (t ), . . . , x(l−1) (t )),
(1.1)
where ϕp : R → R is given by ϕp (s) = |s| s with p ≥ 2 is a constant, F is a continuous function defined on R and is periodic to t with F (t , ·, . . . , ·) = F (t + 2π , ·, . . . , ·), F (t , a, 0, . . . , 0) 6≡ 0 for all a ∈ R, τ is a constant. Complicated behavior of models for technical applications is often described by nonlinear high-order differential equations [1], for example, the Lorenz model of a simplified hydrodynamic flow, the dynamo model of erratic inversion of the earth’s magnetic field, etc. Oftentimes high-order equations are a result of combinations of lower-order equations. Due to its obvious complexity, studies on high-order differential equation are rather infrequent, especially on high-order delay differential equation. Most of the results on high-order delay differential equation are concentrated in the few years. In [2], Cheng and Ren present the existence of periodic solutions for a fourth-order Rayleigh type p-Laplacian delay differential equation as follows p−2
(ϕp (x(t )00 ))00 + f (t , x0 (t − σ (t ))) + g (t , x(t − τ (t ))) = e(t ).
l
(1.2)
In [3], Pan studies the nth-order differential equation x(n) (t ) =
n −1 X
bi x(i) (t ) + f (t , x(t ), x(t − τ1 (t )), . . . , x(t − τm (t ))) + p(t ),
i=1
I Supported by AvH Foundation of Germany and NNSF of China (60504037).
∗
Corresponding author at: Department of Mathematics, Zhengzhou University, Zhengzhou 450001, PR China. E-mail address: [email protected] (J. Ren).
0362-546X/$ – see front matter © 2009 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2009.06.011
(1.3)
J. Ren, Z. Cheng / Nonlinear Analysis 71 (2009) 6182–6193
6183
and obtain the existence of periodic solutions for Eq. (1.3). Afterwards, Ren and Cheng [4] obtain sufficient conditions for the existence of periodic solutions for a general high-order delay differential equation x(n) (t ) = F (t , x(t ), x(t − τ (t )), x0 (t ), . . . , x(n−1) (t )).
(1.4)
In [5], Li and Lu consider the following high-order p-Laplacian differential equation
(ϕp (y(m) (t )))(m) = f (y(t ))y0 (t ) + h(y(t )) + β(t )g (y(t − τ (t ))) + e(t ),
(1.5)
and by using the theory of Fourier series, Bernoulli number theory and continuation theorem of coincidence degree theory, they get the existence of periodic solutions for Eq. (1.5). In [6], Wang and Lu investigate the existence for the high-order neutral functional differential equation with distributed delay
(x(t ) − cx(t − σ ))(n) + f (x(t ))x0 (t ) + g
Z
0
x(t + s)dα(s)
= p(t ).
(1.6)
−r
Inspired by these results, we consider a generalized high-order neutral differential equation (1.1). In [7], Ren, Cheung and Cheng have researched on periodic solution for (1.1) in the case |c | 6= 1. In the paper, we shall move on the research of (1.1) in the more complicated case, i.e, in the critical case |c | = 1. By citing some results on Lu [8,9] and Zhang [10], we first carry out further results on the neutral operator in the critical case and then by applying Mawhin’s continuation theorem we obtain sufficient conditions for the existence of periodic solutions for Eq. (1.1). Our results are new and our methods are different from the above works. Meanwhile, an example is given to illustrate our results. Throughout this paper, we will denote by Z the set of integers, Z1 the set of odd integers, Z2 the set of even integers, N the set of positive integers, N1 the set of odd positive integers and N2 the set of even positive integers. Let C21π = {x : x ∈ C 1 (R, R), x(t + 2π ) ≡ x(t )} with the norm |ϕ|C 1 = {maxt ∈[0,2π] |ϕ(t )|, maxt ∈[ 0,2π ] |ϕ 0 (t )|}, C2π = {x : x ∈ C (R, R), x(t + 2π
R 2π
2π) ≡ x(t )} with the norm |ϕ|0 = maxt ∈[ 0,2π ] |ϕ(t )|, C20π = {x : x ∈ C2π , 0 x(s)ds = 0}, C2−π = {x : x ∈ C (R, R), R 2π +,0 x(t + π ) ≡ −x(t )}, C2+π = {x : x ∈ C (R, R), x(t + π ) ≡ x(t )} and C2π = {x : x ∈ C2+π , 0 x(s)ds = 0} equipped with the norm | · |0 , L2 = {x : R → R is 2π periodic and its restriction to [0, 2π ] belongs to L2 ([0, π])}, under the norm
R 2π
|ϕ(t )|2 dt
12
, L2− = {x : x ∈ L2 , x(t + π ) ≡ −x(t )} and L2+ = {x : x ∈ L2 , x(t + π ) ≡ x(t )} with the norm R 2π 0 2 2− and L2+ are all Banach space. We also denote h¯ = 21π 0 h(s)ds, ∀ h ∈ L2 . | · |2 . Clearly, C21π , C2π , C2+π , C20π , C2+, π ,L ,L |ϕ|2 =
0
2. Preparation Let X and Y be real Banach spaces and 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 X1 , Y1 , of X , Y respectively, such that X = Ker L ⊕ X1 , Y = Im L ⊕ Y1 , and let P : X → Ker L and Q : Y → Y1 denote the natural projections. Clearly, Ker L ∩ (D(L) ∩ X1 ) = {0}, thus the restriction LP := L|D(L)∩X1 is invertible. Let K denote the inverse of LP . Let Ω be an open bounded subset of X with D(L) ∩ Ω 6= ∅. 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 (Gaines and Mawhin [11]). Suppose that X and Y are two Banach spaces, and L : D(L) ⊂ X → Y is a Fredholm operator with index zero. Furthermore, Ω ⊂ X is an open bounded set and N : Ω → Y is L-compact on Ω . Assume that the following conditions hold: (1) Lx 6= λNx, ∀ x ∈ ∂ Ω ∩ D(L), λ ∈ (0, 1); (2) Nx ∈ 6 Im L, ∀ x ∈ ∂ Ω ∩ Ker L; (3) deg{JQN , Ω ∩ Ker L, 0} 6= 0, where J : Im Q → Ker L is an isomorphism, then the equation Lx = Nx has a solution in Ω ∩ D(L). Lemma 2.2 ([10]). If ω ∈ C 1 (R, R) and ω(0) = ω(T ) = 0, then T
Z
|ω(t )| dt ≤ p
0
T
πp
p Z
T
|ω0 (t )|p dt 0
where p is a fixed real number with p > 1, and πp = 2
R (p−1)/p
First, we define operators A in the following form: A : C2π → C2π ,
(Ax)(t ) = x(t ) − cx(t − τ ).
0
ds p
(1− ps−1 )1/p
=
2π (p−1)1/p . p sin(π /p)
6184
J. Ren, Z. Cheng / Nonlinear Analysis 71 (2009) 6182–6193
Lemma 2.3 ([8]). The following propositions are true: (1) If c = −1, |τ | =
m n
π with m, n coprime positive integers and m even, then 1
σ1 := inf |1 − ce−ikτ | = inf [2(1 + cos kτ )] 2 > 0. k∈N
(2.1)
k∈N
(2) If c = −1, |τ | =
m n
π with m, n coprime positive integers and m odd and n even, then 1
σ2 := inf |1 − ce−ikτ | = inf [2(1 + cos kτ )] 2 > 0. k∈N1
(2.2)
k∈N1
(3) If c = −1, |τ | =
m n
π with m, n coprime positive odd integers, then 1
σ3 := inf |1 − ce−ikτ | = inf [2(1 + cos kτ )] 2 > 0. k∈N2
(4) If c = 1, |τ | =
(2.3)
k∈N2
m n
π with m, n coprime positive integers and m odd, then 1
σ4 := inf |1 − ce−ikτ | = inf [2(1 + cos kτ )] 2 > 0. k∈N1
(2.4)
k∈N1
(5) If c = 1, |τ | = π , then 1
σ5 := inf |1 − ce−ikτ | = inf [2(1 + cos kτ )] 2 = 2 > 0. k∈N1
(2.5)
k∈N1
In what follows, we will discuss the invertibility of the difference operator A : [Ax](t ) = x(t ) − cx(t − τ ). Firstly, we ¯ k∈Z1 {sin kt , cos kt }, C2+π ⊂ span ¯ k∈Z2 {1, sin kt , cos kt } and C2π ⊂ span ¯ k∈Z {1, sin kt , cos kt }. So we have the see C2−π ⊂ span following results. Since Ax(n) = (Ax)(n) , ∀ x ∈ C2π := {x : x ∈ C n (R, R), x(t + 2π ) ≡ x(t )}. Then Eq. (1.1) is transformed into
(ϕp ((Ax(l) )(t )))(n−l) = F (t , x(t ), x0 (t ), . . . , x(l−1) (t )).
(2.6)
Lemma 2.4 ([9]). The following propositions are true: (1) Suppose c = −1, |τ | = (m/n)π , where m, n are coprime positive integers with m even, then A : L22π → L22π ,
[Ax](t ) = x(t ) − c (t − τ ),
∀ t ∈ [0, 2π ],
: → satisfying kA k ≤ σ1 . 1 (2) Suppose c = −1, |τ | = (m/n)π , where m, n are coprime odd positive integers, then L22π
−1
has a unique inverse A
2+ A : L22+ π → L2 π ,
L22π
−1
[Ax](t ) = x(t ) − c (t − τ ),
∀ t ∈ [0, 2π ],
2+ 1 −1 has a unique inverse A−1 : L22+ π → L2π satisfying kA k ≤ σ3 . (3) Suppose c = −1, |τ | = (m/n)π , where m, n are coprime positive integers with m odd and n even, then 2− A : L22− π → L2 π ,
[Ax](t ) = x(t ) − c (t − τ ), L22− π
∀ t ∈ [0, 2π ],
L22− π
satisfying kA−1 k ≤ σ1 . 2 (4) Suppose c = 1, |τ | = (m/n)π , where m, n are coprime positive integers with m odd, then has a unique inverse A−1 : 2− A : L22− π → L2 π ,
−1
has a unique inverse A
→
[Ax](t ) = x(t ) − c (t − τ ), :
L22− π
→
(5) Suppose c = 1, |τ | = π , then 2− A : L22− π → L2 π ,
L22− π
−1
satisfying kA
[Ax](t ) = x(t ) − c (t − τ ),
∀ t ∈ [0, 2π ], k≤
1
σ4
.
∀ t ∈ [0, 2π ],
2− 1 −1 has a unique inverse A−1 : L22− π → L2π satisfying kA k ≤ σ5 .
Now we consider (2.6). Define the conjugate index q ∈ (1, 2] by y1 (t ) = x(t ),
y2 (t ) = x0 (t ), (l)
yl+1 (t ) = ϕp (Ax (t )),
1 p
+
1 q
y3 (t ) = x00 (t ), . . . , yl (t ) = x(l−1) (t ),
= 1. Introducing new variables
yl+2 (t ) = (ϕp (Ax(l) (t )))0 , . . . , yn (t ) = (ϕp (Ax(l) (t )))(n−l−1) ,
J. Ren, Z. Cheng / Nonlinear Analysis 71 (2009) 6182–6193
6185
and using the fact that ϕq ◦ ϕp ≡ id, Eq. (1.1) can be rewritten as
0 y1 (t ) = y2 (t ) y02 (t ) = y3 (t ) ··· y0l−1 (t ) = yl (t ) y0l (t ) = A−1 ϕq (yl+1 (t )) y0l+1 (t ) = yl+2 (t ) ··· 0 yn0 −1 (t ) = yn (t ) yn (t ) = F (t , y1 (t ), y2 (t ), . . . , yl (t )).
(2.7)
It is clear that if y(t ) = (y1 (t ), y2 (t ), . . . , yn (t ))> is a 2π -periodic solution to Eq. (2.7), then y1 (t ) must be a 2π -periodic solution to Eq. (1.1). Thus, the problem of finding a 2π -periodic solution for Eq. (1.1) reduces to finding one for Eq. (2.7). Define the linear spaces X = Y = {y = (y1 (·), y2 (·), . . . , yn (·))> ∈ C 0 (R, Rn ) : y(t + 2π ) ≡ y(t )} with norm kyk = max{ky1 k, ky2 k, . . . , kyn k}. Obviously, X and Y are Banach spaces. Define L : D(L) = {y ∈ C 1 (R, Rn ) : y(t + 2π ) = y(t )} ⊂ X → Y by
0 y1 0
y2 . . . 0 Ly = y = 0 yl . . . y0n
and N :X →Y by y2 (t ) y3 (t )
Ny =
.. .
A−1 ϕq (yl+1 (t )) yl+2 (t )
.. . F (t , y1 (t ), y2 (t ), . . . , yl (t ))
.
(2.8)
Then Eq. (2.7) can be rewritten as the abstract equation Ly = Ny. From the definition of L, one can easily see that Ker L = R 2π {y ∈ C 1 (R, Rn ) : y is constant} ' Rn , Im L = {y : y ∈ X , 0 y(s) = 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 2π
2π
Z
y(s)ds;
0
Qy =
Z
1 2π
2π
y(s)ds.
0
R 2π
It is easy to see that Ker L = Im Q = Rn . Moreover, for all y ∈ Y , we have 0 y∗ (s)ds = 0 if y∗ = y − Q (y), which means y∗ ∈ Im L. This is to say Y = Im Q ⊕ Im L and then 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) , we have
[Ky](t ) =
2π
Z
G1 (t , s)y1 (s)ds,
0
2π
Z
G2 (t , s)y2 (s)ds, . . . ,
0
2π
Z
Gn (t , s)yn (s)ds
>
0
where
s , π Gi (t , s) = 2 s − 2π , 2π
0 ≤ s < t ≤ 2π . 0 ≤ t ≤ s ≤ 2π .
i = 1, 2, . . . , n.
(2.9)
From (2.8) and (2.9), it is clearly that QN and K (I − Q )N are continuous, and QN (Ω ) is bounded and then K (I − Q )N (Ω ) is ¯ . For the functions in the domain of L we have compact for any open bounded Ω ⊂ X which means that N is L-compact on Ω
6186
J. Ren, Z. Cheng / Nonlinear Analysis 71 (2009) 6182–6193
Lemma 2.5. Assume that y(t ) ∈ C 1 (R, Rn ) and y(t + 2π ) = y(t ). (1) If c = −1, |τ | =
π with m, n coprime positive integers and m even, then p(l−i) q(n−l) Z 2π Z 2π 1 2π 2π |y0i (t )|p dt ≤ p |y0n (t )|q dt , πq σ1 πp 0 0 m n
+ 1q = 1, p ≥ 2, i = 1, 2, . . . , l − 1. (2) If c = −1, |τ | = m π with m, n coprime positive integers with m odd and n even, then n p(l−i) q(n−l) Z 2π Z 2π 1 2π 2π |y0i (t )|p dt ≤ p |y0n (t )|q dt , πq σ2 πp 0 0 where
1 p
+ 1q = 1, p ≥ 2, i = 1, 2, . . . , l − 1. (3) If c = −1, |τ | = m π with m, n coprime positive integers and m and n odd, then n p(l−i) q(n−l) Z 2π Z 2π 2π 1 2π |y0i (t )|p dt ≤ p |y0n (t )|q dt , πq σ3 πp 0 0 where
(2.11)
1 p
+ 1q = 1, p ≥ 2, i = 1, 2, . . . , l − 1. (4) If c = 1, |τ | = m π with m, n coprime positive integers and m odd, then n p(l−i) q(n−l) Z 2π Z 2π 2π 1 2π |y0n (t )|q dt , |y0i (t )|p dt ≤ p πq σ4 πp 0 0 where
(2.10)
(2.12)
1 p
(2.13)
where 1p + 1q = 1, p ≥ 2, i = 1, 2, . . . , l − 1. (5) If c = 1, |τ | = π , then 2π
Z
|yi (t )| dt ≤ 0
0
where
1 p
+
1 q
p
1
σ
p 5
2π
πp
p(l−i)
2π
q(n−l) Z
πq
2π
|y0n (t )|q dt ,
(2.14)
0
= 1, p ≥ 2, i = 1, 2, . . . , l − 1.
Proof. From y1 (0) = y1 (2π ), there exists a point t1 ∈ [0, 2π ] such that y01 (t1 ) = 0. Let ω1 (t ) = y01 (t + t1 ), and then ω1 (0) = ω1 (2π ) = 0. From y2 (0) = y2 (2π ), there exists a point t2 ∈ [0, 2π ] such that y02 (t2 ) = 0. Let ω2 (t ) = y02 (t + t2 ), and then ω2 (0) = ω2 (2π ) = 0. Continuing this way we get from yl−1 (0) = yl−1 (2π), there exists a point tl−1 ∈ [0, 2π ] such that y0l−1 (tl−1 ) = 0. Let ωl−1 (t ) = y0l−1 (t + tl−1 ), and then ωl−1 (0) = ωl−1 (2π ) = 0. From yl (t ) = yl (t + 2π ), we have
R 2π
R 2π (Ay0l )(t )dt = 0 (Ayl )0 (t )dt = (Ayl )(t ) |20π = 0, so there exists a point tl ∈ [0, 2π ] such that Ay0l (tl ) = 0, then we have 0 ϕ ((Ayl )(tl )) = 0. Let ωl (t ) = ϕp ((Ay0l )(t + tl )) = yl+1 (t + tl ), and then ωl (0) = ωl (2π ) = 0. Continuing this way we get from yn−1 (0) = yn−1 (2π ), there exists a point tn−1 ∈ [0, 2π ] such that y0n−1 (tn−1 ) = 0. Let ωn−1 (t ) = y0n−1 (t + tn−1 ), and then ωn−1 (0) = ωn−1 (2π ) = 0. By Lemma 2.2, we have Z 2π Z 2π |y01 (t )|p dt = |ω1 (t )|p dt 0 0 p Z 2 π 2π ≤ |ω10 (t )|p dt πp 0 p Z 2 π 2π = |y02 (t )|p dt πp 0 p Z 2 π 2π = |ω2 (t )|p dt πp 0 2p Z 2π 2π ≤ |ω20 (t )|p dt πp 0 ··· p(l−1) Z 2π 2π ≤ |ωl0−1 (t )|p dt πp 0 p(l−1) Z 2π 2π |y0l (t )|p dt . (2.15) = πp 0 0 p
J. Ren, Z. Cheng / Nonlinear Analysis 71 (2009) 6182–6193
6187
By Lemmas 2.4 and 2.2, we have 2π
Z
|y0l (t )|p dt =
2π
Z
0
|A−1 ϕq (yl+1 (t ))|p dt
0
p 1
σ
σ1p σ σ
=
2π
σ
σ ···
1
1
2π
|ωl0 (t )|q dt
0 2π
q Z
|yl+2 (t )|q dt
0
q(n−l) Z
πq
σ1p
=
2π
2π
q Z
πq
p 1
≤
2π
πq
p 1
1
|ωl (t )|q dt
0
1
|yl+1 (t )|q dt
0
p 1
≤
2π
Z
1
|yl+1 (t )|pq−p dt
0
p 1
=
2π
Z
1
=
|ϕq (yl+1 (t ))|p dt
0
Z
1
=
2π
Z
1
≤
2π
|ωn0 −1 (t )|q dt
0
q(n−l) Z
πq
σ1p
2π
2π
|y0n (t )|q dt .
(2.16)
0
From (2.15) and (2.16), we can get 2π
Z
|y01 (t )|p dt ≤
0
1
2π
p(l−1)
2π
πp
σ1p
q(n−l) Z
πq
2π
|y0n (t )|q dt .
0
Similarly, we get 2π
Z
|y0i (t )|p dt ≤
0
1
2π
p(l−i)
πp
σ1p
2π
q(n−l) Z
πq
2π
|y0n (t )|q dt ,
0
where i = 1, 2, . . . , l − 1. Cases (2)–(5) can be proved in the same way as in the proof of case (1). This completes the proof of Lemma 2.5. Remark 2.1. In particular, if take p = 2, then q = 2 and
πp = πq = π2 = 2
(2−1)/2
Z
ds
2π (2 − 1)1/2
= π. 2 sin(π /2) (1 − 2−1 )1/2 R 2π 2(n−i) R 2π 0 |yn (t )|2 dt . In the case, (2.10) is transformed into 0 |yi (t )|2 dt ≤ 12 2ππ 0 σ s2
0
=
1
3. Main results For the sake of convenience, we list the following assumptions which will be used repeatedly in what follows:
(H1 ) there exists a positive constant D > 0 such that z1 F (t , z1 , z2 , . . . , zl ) > 0,
∀ (t , z1 , z2 , . . . , zl ) ∈ [0, 2π ] × Rl with |z1 | > D;
(H2 ) there exists a positive constant M > 0 such that |F (t , z1 , z2 , . . . , zl )| ≤ M ,
∀ (t , z1 , z2 , . . . , zl ) ∈ [0, 2π ] × Rl ;
(H3 ) there exist non-negative constants α1 , α2 , . . . , αl , m, such that |F (t , z1 , z2 , . . . , zl )| ≤ α1 |z1 | + α2 |z2 | + · · · + αl |zl | + m, ∀ (t , z1 , z2 , . . . , zl ) ∈ [0, 2π ] × Rl .
6188
J. Ren, Z. Cheng / Nonlinear Analysis 71 (2009) 6182–6193
Theorem 3.1. If (H1 ) and (H2 ) hold, then Eq. (1.1) has at least one non-constant 2π -periodic solution. Proof. Consider the equation Ly = λNy,
λ ∈ (0, 1).
Let Ω1 = {y : Ly = λNy, λ ∈ (0, 1)}. If y(t ) = (y1 (t ), y2 (t ), . . . , yn (t ))> ∈ Ω1 , then
y0 (t ) = λy (t ) 2 1 y02 (t ) = λy3 (t ) .. . 0 y ( t ) = λyl (t ) l−10 yl (t ) = λϕq (yl+1 (t )) 0 yl+1 (t ) = λyl+2 (t ) .. . 0 y ( t ) = λyn (t ) n − 1 0 yn (t ) = λF (t , y1 (t ), y2 (t ), . . . , yl (t )).
(3.1)
We first claim that there exists a constant ξ ∈ R such that
|y1 (ξ )| ≤ D.
(3.2)
Integrating the last equation of (3.1) over [0, 2π ], we have 2π
Z
F (t , y1 (t ), y2 (t ), . . . , yl (t ))dt = 0.
0
Then from the continuity of F , we know that there exists a constant ξ ∈ [0, 2π ] such that F (ξ , y1 (ξ ), . . . , yl (ξ )) = 0. From assumption (H1 ) we get (3.2). As a consequence, we have
Z t Z |y1 (t )| = y1 (ξ ) + y01 (s)ds ≤ D + ξ
2π
|y01 (s)|ds.
(3.3)
0
On the other hand, multiplying both sides of the last equation of (3.1) by y0n (t ) and integrating over [0, 2π ], by using assumption (H2 ), we have 2π
Z
|yn (t )| dt = λ 0
2
Z
0
2π
0 2π
Z ≤
F (t , y1 (t ), y2 (t ), . . . , yl (t ))y0n (t )dt
|F (t , y1 (t ), y2 (t ), . . . , yl (t ))||y0n (t )|dt
0 2π
Z
|y0n (t )|dt
≤M 0
1/2
≤ M (2π )
2π
Z
|yn (t )| dt 0
2
1/2
.
0
It is easy to see that there exists a constant Mn0 > 0 (independent of λ) such that 2π
Z
|y0n (t )|2 dt ≤ Mn0 .
0
From yn−1 (0) = yn−1 (2π ), there exists a point t1 ∈ [0, 2π ] such that yn (t1 ) = 0, and by applying Hölder’s inequality, we have
|yn (t )| ≤
2π
Z
1/2
|yn (t )|dt ≤ (2π ) 0
2π
Z
0
|yn (t )| dt 0
2
1/2
≤ (2π )1/2 Mn01/2 := Mn .
0
From yn−2 (0) = yn−2 (2π ), there exists a point t2 ∈ [0, 2π] such that yn−1 (t2 ) = 0, we have
|yn−1 (t )| ≤
2π
Z 0
|y0n−1 (t )|dt =
Z 0
2π
|λyn (t )|dt ≤
2π
Z 0
|yn (t )|dt ≤ 2π Mn := Mn−1 .
J. Ren, Z. Cheng / Nonlinear Analysis 71 (2009) 6182–6193
6189
Similarly,
|yn−2 (t )| ≤ 2π Mn−1 := Mn−2 . And continuing this way for yn−3 , . . . , yl+1 , we get
|yl+1 (t )| ≤ 2π Ml+2 := Ml+1 . Case (1) If c = −1, |τ | =
π with m, n coprime positive integers and m even, then Z 2π Z 2π 2π 1 1 |λA−1 ϕq (yl+1 (t ))|dt ≤ |y0l (t )|dt ≤ |yl+1 (t )|q−1 dt ≤ 2π Mlq+−11 := Ml1 . |yl (t )| ≤ σ1 0 σ1 0 0 m n
Z
Case (2) If c = −1, |τ | =
π with m, n coprime positive integers and m odd and n even, then Z 2π Z 2π 2π 1 1 |λA−1 ϕq (yl+1 (t ))|dt ≤ |y0l (t )|dt ≤ |yl (t )| ≤ |yl+1 (t )|q−1 dt ≤ 2π Mlq+−11 := Ml2 . σ2 0 σ2 0 0 m n
Z
Case (3) If c = −1, |τ | =
π with m, n coprime positive odd integers, then Z 2π Z 2π 2π 1 1 |yl (t )| ≤ |y0l (t )|dt ≤ |yl+1 (t )|q−1 dt ≤ 2π Mlq+−11 := Ml3 . |λA−1 ϕq (yl+1 (t ))|dt ≤ σ σ 3 3 0 0 0 m n
Z
π with m, n coprime positive integers and m odd, then Z 2π Z 2π 2π 1 1 |yl (t )| ≤ |y0l (t )|dt ≤ |λA−1 ϕq (yl+1 (t ))|dt ≤ |yl+1 (t )|q−1 dt ≤ 2π Mlq+−11 := Ml4 . σ σ 4 4 0 0 0
Case (4) If c = 1, |τ | =
m n
Z
Case (5) If c = 1, |τ | = π , then
|yl (t )| ≤
Z
2π
|y0l (t )|dt ≤
Z
0
2π
|λA−1 ϕq (yl+1 (t ))|dt ≤
0
1
σ5
2π
Z 0
|yl+1 (t )|q−1 dt ≤
1
σ5
q−1
2π Ml+1 := Ml5 .
Define Ml = max{Ml1 , Ml2 , Ml3 , Ml4 , Ml5 }.
|yl−1 (t )| ≤ 2π Ml := Ml−1 . .. . |y2 (t )| ≤ 2π M3 := M2 . Meanwhile, from Eq. (3.3), we can get
|y1 (t )| ≤ D +
2π
Z
|y01 (t )|dt ≤ D + 2π M2 := M1 .
0
Let M0 = max{M1 , M2 , . . . , Mn }, obviously ky1 k ≤ M0 , ky2 k ≤ M0 , . . ., kyn k ≤ M0 . Let Ω2 = {y ∈ Ker L : Ny ∈ Im L}. If y ∈ Ω2 , then y ∈ Ker L which means y = constant and QNy = 0. We see that
y2 = 0, y3 = 0, .. . yn = 0, F (t , y1 , 0, . . . , 0) = 0. So
|y1 | ≤ D ≤ M0 ,
y2 = y3 = · · · = yn = 0 ≤ M0 .
Now take Ω = {y = (y1 , y2 , . . . , yn )> ∈ X : ky1 k < M0 + 1, ky2 k < M0 + 1, . . . , kyn k < M0 + 1}. By the analysis of the above, it is easy to see that Ω 1 ⊂ Ω , Ω 2 ⊂ Ω and conditions (1) and (2) of Lemma 2.1 are satisfied. Next we show that condition (3) of Lemma 2.1 is also satisfied. Define the isomorphism J : Im Q → Ker L as follows: J (y1 , y2 , . . . , yn )> = (yn , y1 , . . . , yn−1 )> .
6190
J. Ren, Z. Cheng / Nonlinear Analysis 71 (2009) 6182–6193
Let H (µ, y) = µy + (1 − µ)JQNy, (µ, y) ∈ [0, 1] × Ω , then ∀(µ, y) ∈ (0, 1) × (∂ Ω ∩ Ker L),
1−µ
µy1 +
2π
H (µ, y) =
2π
Z 0
F (t , y1 , 0, . . . , 0)dt
.
y2
.. . A ϕq (yl+1 ) .. . −1
yn From (H1 ), it is obvious that yH (µ, y) > 0, ∀(µ, y) ∈ (0, 1) × (∂ Ω ∩ Ker L). Therefore, deg{JQN , Ω ∩ Ker L, 0} = deg{H (0, y), Ω ∩ Ker L, 0}
= deg{H (1, y), Ω ∩ Ker L, 0} = deg{I , Ω ∩ Ker L, 0} 6= 0, which means that condition (3) of Lemma 2.1 is also satisfied. By applying Lemma 2.1, we conclude that equation Ly = Ny ¯ , i.e., Eq. (1.1) has a T -periodic solution y∗1 (t ) with ky∗1 k < M0 + 1. has a solution y∗ (t ) = (y∗1 (t ), y∗2 (t ), . . . , y∗n (t ))> on Ω Finally, observe that y∗1 (t ) is not constant. Otherwise, suppose y∗1 (t ) ≡ a (constant), then from Eq. (1.1) we have F (t , a, 0, . . . , 0) ≡ 0, which is contradict to assumption F (t , a, 0, . . . , 0) 6≡ 0, so the proof is complete. Theorem 3.2. If (H1 ) and (H3 ) hold, then Eq. (1.1) has at least one non-constant 2π -periodic solution, if one of the following conditions hold: (1) p > 2; (2) p = 2 and if c = −1, |τ | = 2π l−2
α3
π
2π
+ · · · + αl
π
π
(3) p = 2 and if c = −1, |τ | =
2π l−2
α3
π
π
π
· · · + αl
π
2π n−l
π
· · · + αl
π
2π n−l
π
m n
2π l−1
π
+
< 1; 2π l−1
+
2π l−2
+
2π l−2
+
π
< 1;
π with m, n coprime positive odd integers, and
1
σ3
(α1 2π + α2 )
2 π l −1
π
+ α3
π
< 1;
(5) p = 2 and if c = 1, |τ | = 2π
(α1 2π + α2 )
, n coprime positive integers with m odd and n even, and σ12 (α1 2π +α2 )
π
(4) p = 2 and if c = −1, |τ | = 2π
1
σ1
m with m n 2π n−l
2π
+ · · · + αl
π with m, n coprime positive integers and m even, and
m n n −l 2π
m n
π with m, n coprime positive integers and m odd, and
1
σ4
(α1 2π + α2 )
2 π l −1
π
+ α3
π
< 1;
(6) p = 2 and if c = 1, |τ | = π , and σ1 (α1 2π + α2 ) 5
2 π l −1
π
+ α3
2π l−2
π
+ · · · + αl
2π
π
2π n−l
π
< 1.
Proof. Let Ω1 be defined as in Theorem 3.1. If y(t ) = (y1 (t ), y2 (t ), . . . , yn (t ))> ∈ Ω1 , then from the proof of Theorem 3.1 we see that y0n (t ) = λF (t , y1 (t ), y2 (t ), . . . , yl (t )),
(3.4)
and 2π
Z |y1 |0 ≤ D +
|y01 (s)|ds.
0
We claim that |yn |0 is bounded. Multiplying both sides of Eq. (3.4) by ϕq (y0n (t )) and integrating it over [0, 2π ], by using assumption (H3 ), we have 2π
Z
|yn (t )| dt = λ 0
q
Z
0
2π
0 2π
Z ≤
F (t , y1 (t ), y2 (t ), . . . , yl (t ))ϕq (y0n (t ))dt
|F (t , y1 (t ), y2 (t ), . . . , yl (t ))kϕq (y0n (t ))|dt
0
≤ α1
2π
Z
|y1 (t )kϕq (y0n (t ))|dt + α2
0
+ · · · + αl
2π
Z
|y2 (t )kϕq (y0n (t ))|dt
0 2π
Z 0
|yl (t )kϕq (y0n (t ))|dt + m
2π
Z 0
|ϕq (y0n (t ))|dt
J. Ren, Z. Cheng / Nonlinear Analysis 71 (2009) 6182–6193
≤ α1 |y1 |0
2π
Z
|ϕq (y0n (t ))|dt + α2
2π
Z
6191
|y2 (t )kϕq (y0n (t ))|dt
0
0
+ · · · + αl Z ≤ α1 D +
Z 2π |ϕq (y0n (t ))|dt |yl (t )kϕq (y0n (t ))|dt + m 0 0 Z 2π Z 2π 2π |y2 (t )kϕq (y0n (t ))|dt |ϕq (y0n (t ))|dt + α2 |y01 (t )|dt 2π
Z
+ · · · + αl
0
0
0 2π
Z
|yl (t )kϕq (y0n (t ))|dt + m
2π
Z
|ϕq (y0n (t ))|dt .
0
0
Applying Hölder’s inequality, we have 2π
Z
|yn (t )| dt ≤ α1 D + (2π ) 0
q
2π
Z
1 p
0
|y1 (t )| dt 0
q
1q
1
(2π ) q
0
+ α2
2π
Z
2π
Z
|ϕq (y0n (t ))|p dt
1p
0
|y2 (t )| dt q
1q Z
0
2π
|ϕq (yn (t ))| dt 0
p
1p
0
+ · · · + αl
2π
Z
|yl (t )| dt q
1q Z
0
≤ α1 2π · (2π )
2π
|ϕq (yn (t ))| dt 0
p
2π
Z
|y1 (t )| dt 0
p
2π
Z
q(p1−1) Z
|ϕq (yn (t ))| dt 0
p
1p
2π
|yn (t )| dt 0
q
1p
0
|y2 (t )| dt p
q(p1−1) Z
2π
|yn (t )| dt 0
q
1p
0
0
+ · · · + αl (2π )
2π
Z 0
0
+ α2 (2π )
+ m(2π )
1 q
0
p−2 q(p−1)
p−2 q(p−1)
1p
p−2 q(p−1)
2π
Z
|yl (t )| dt p
q(p1−1)
0 2π
Z ×
|yn (t )| dt 0
q
1p
+ (α1 D + m) (2π )
1 q
0
|yn (t )| dt 0
q
1p
0
= α1 2π · (2π )
p−2 p
2π
Z
|y1 (t )| dt 0
p
1p Z
0
+ α2 (2π )
p−2 p
2π
Z
2π
×
|yn (t )| dt 0
q
1p
|y2 (t )| dt p
1p Z
2π
|yn (t )| dt 0
q
1p
+ · · · + αl (2π )
0
|yn (t )| dt 0
2π 0
0
Z
2π
Z
q
1p
p−2 p
Z
2π
|yl (t )| dt p
1p
0
+ (α1 D + m) (2π )
1 q
2π
Z
0
|yn (t )| dt 0
q
1p
.
(3.5)
0
Then by applying Lemma 2.5 and Eq. (3.5), we can get Case (1) If c = −1, |τ | = m π with m, n coprime positive integers and m even, n 2π
Z 0
|yn (t )| dt ≤ α1 2π · (2π ) 0
q
p−2 p
1
2π
l−1
2π
q(np−l) Z
2π
2p
|yn (t )| dt πp πq 0 l−1 q(np−l) Z 2π 2p p−2 1 2π 2π + α2 (2π ) p + ··· + |y0n (t )|q dt σ1 πp πq 0 q(np−l) Z 2π 2p Z 2π 1p p−2 1 1 2π 2π + αl (2π ) p |y0n (t )|q dt + (α1 D + m)(2π ) q |y0n (t )|q dt σ1 πp πq 0 0 " l−1 l −2 # q(np−l) p − 2 1 2π 2π 2π 2π ≤ (α1 2π + α2 ) + α3 + · · · + αl (2π ) p σ1 πp πp πp πq 2π
Z × 0
q
σ1
|yn (t )| dt 0
0
q
2p
+ (α1 D + m)(2π )
1 q
2π
Z 0
|yn (t )| dt 0
q
1p
.
6192
J. Ren, Z. Cheng / Nonlinear Analysis 71 (2009) 6182–6193
l−1 If p = 2 and σ1 (α1 2π + α2 ) 2ππ + α3 1 Mn0 > 0 (independent of λ) such that Z 2π |y0n (t )|q dt ≤ Mn0 .
2 π l −2
π
2π
+ · · · + αl
π
2π n−l
π
< 1, it is easy to see that there exists a constant
0
Similarly, Case (2) If p = 2 and c = −1, |τ | =
π with m, n coprime positive integers with m odd and n even, and σ12 (α1 2π 2π n−l 2π l−2 + · · · + αl 2ππ < 1, it is easy to see that there exists a constant Mn0 > 0 (independent π π m n
l−1 + α3 + α2 ) 2ππ of λ) such that Z 2π |y0n (t )|q dt ≤ Mn0 . 0
Case (3) If p = 2 and c = −1, |τ | =
· · · + αl Z
2π
π
2π
2π n−l
π
m n
π with m, n coprime positive odd integers, and σ13 (α1 2π +α2 )
2 π l −1
π
+α3
2 π l −2
π
+
< 1, it is easy to see that there exists a constant Mn > 0 (independent of λ) such that 0
|y0n (t )|q dt ≤ Mn0 .
0
Case (4) If p = 2 and c = 1, |τ | =
α3
2π l−2
π
2π
Z
+ · · · + αl
2π
π
2 π n −l
π
m n
π with m, n coprime positive integers and m odd, and
1
σ4
(α1 2π + α2 )
2 π l −1
π
+
< 1, it is easy to see that there exists a constant Mn > 0 (independent of λ) such that 0
|y0n (t )|q dt ≤ Mn0 .
0
Case (5) If p = 2 and c = 1, |τ | = π , and σ1 (α1 2π + α2 ) 2ππ + α3 5 see that there exists a constant Mn0 > 0 (independent of λ) such that
l−1
2π
Z
2π l−2
π
+ · · · + αl
2π
π
2π n−l
π
< 1, it is easy to
|y0n (t )|q dt ≤ Mn0 .
0
Case (6) If p > 2, it is easy to see that there exists a constant Mn0 > 0 (independent of λ) such that 2π
Z
|y0n (t )|q dt ≤ Mn0 .
0
From yn−1 (0) = yn−1 (2π ), there exists a point t1 ∈ [0, 2π] such that yn (t1 ) = 0, By applying Hölder’s inequality, we have
|yn (t )| ≤
2π
Z
|yn (t )|dt ≤ (2π ) 0
1 p
0
Z
2π
|yn (t )| dt 0
q
1q
1
01
≤ (2π ) p Mn q := Mn .
0
This proves the claim and the rest of the proof of the theorem is identical to that of Theorem 3.1.
Remark 3.1. If p = 2, then Eq. (1.1) is transformed into
(x(t ) − cx(t − τ ))(n) = F (t , x(t ), x0 (t ), . . . , x(n−1) (t )), then the results of Theorems 3.1 and 3.2 still hold. Remark 3.2. If the equation is taken the form as
(ϕp (x(t ) − cx(t − τ ))(l) )(n−l) = F (t , x(t ), x0 (t ), . . . , x(l−1) (t )) + e(t ), R 2π where e(t ) ∈ C (R, R), e(t + 2π ) = e(t ) and 0 e(t )dt = 0, the results of Theorems 3.1 and 3.2 still hold. Finally, we present an example to illustrate our result. Example 3.1. Consider the n-order delay differential equation
(ϕp (x(t ) − x(t − π ))(4) )0 =
1 24π
x( t ) +
1 8
sin x0 (t ) +
1 8
cos x00 (t ) sin t +
1 8
sin x000 (t ).
(3.6)
J. Ren, Z. Cheng / Nonlinear Analysis 71 (2009) 6182–6193
Here p is a constant with p ≥ 2. It is clear that T = 2π , c = 1, τ = π , F (t , z1 , z2 , z3 , z4 ) =
6193
+ 18 sin z2 + 18 cos z3 sin t + sin z4 , F (t , a, 0, 0, 0) = a + sin t 6≡ 0. Choose D = 24π such that (H1 ) hold, and it is obvious that (H2 ) is not satisfied here. Now we consider the assumption (H3 ). Since 1 8
1 24π
|F (t , z1 , z2 , z3 , z4 )| ≤
1 8
1 24π
1 z 24π 1
|z1 | + 1.
That is to say (H3 ) hold with α1 =
1 24π
, α2 = 0, α3 = 0, α4 = 0, m = 1.
Case (1) If p > 2, by Theorem 3.2, we know that Eq. (3.6) has at least one non-constant 2π -periodic solution. Case (2) If p = 2, and from Lemma 2.3 1
σ5 := inf |1 − ce−ikτ | = inf [2(1 + cos kτ )] 2 = 2 > 0. k∈N1 k∈N1 " 3 2 # 2π 2π 2π 1 2π 1 1 3 + α3 + α4 = × × 2π + 0 × 2 + 0 + 0 × 2 (α1 2π + α2 ) σ5 π π π π 2 24π =
1 2
×
2 3
× 2 < 1.
So by Theorem 3.2, we know that Eq. (3.6) has at least one non-constant 2π -periodic solution. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11]
E.A. Jackson, Perspectives of Nonlinear Dynamics, Cambridge University Press, 1991. Z.B. Cheng, J.L. Ren, Periodic solutions for a fourth-order Rayleigh type p-Laplacian delay equation, Nonlinear Anal. TMA 70 (2009) 516–523. L.J. Pan, Periodic solutions for higher order differential equations with deviating argument, J. Math. Anal. Appl. 343 (2008) 904–918. J.L. Ren, Z.B. Cheng, On high-order delay differential equation, Comput. Math. Appl. 57 (2009) 324–331. X.J. Li, S.P. Lu, Periodic solution for a kind of high-order p-Laplacian differential equation with sign-changing coefficient ahead of non linear term, Nonlinear Anal. 70 (2009) 1011–1022. K. Wang, S.P. Lu, On the existence of periodic solutions for a kind of high-order neutral functional differential equation, J. Math. Anal. Appl. 326 (2007) 1161–1173. J.L. Ren, W.S. Cheung, Z.B. Cheng, On generalized high-order neutral differential equation, preprint. S.P. Lu, W.G. Ge, Z.X. Zheng, Periodic solutions to a kind of neutral functional differential equation in the critical case, J. Math. Anal. Appl. 293 (2004) 462–475. S.P. Lu, Z.J. Gui, W.G. Ge, Periodic solutions to a second order nonlinear neutral function functional equation in the critical case, Nonlinear Anal. TMA 64 (2006) 1587–1607. M.R. Zhang, Nonuniform nonresonance at the first eigenvalue of the p-Laplacian, Nonlinear Anal. TMA 29 (1997) 41–51. R.E. Gaines, J.L. Mawhin, Coincidence Degree and Nonlinear Differential Equation, Springer, Berlin, 1977.