Electronic Journal of Qualitative Theory of Differential Equations Spec. Ed. I, 2009 No. 12, 1–12; http://www.math.u-szeged.hu/ejqtde/

EXISTENCE OF PERIODIC SOLUTIONS FOR A CLASS OF EVEN ORDER DIFFERENTIAL EQUATIONS WITH

DEVIATING ARGUMENT

Chengjun Guo ^{1}, Donal O’Regan^{2}, Yuantong Xu^{3} and Ravi P.Agarwal^{4}

1Department of Applied Mathematics, Guangdong University of Technology 510006, P. R. China 2Department of Mathematics, National University of Ireland, Galway, Ireland e-mail: donal.oregan@nuigalway.ie 3Department of Mathematics, Sun Yat-sen University Guangzhou Guangdong 510275, P. R China 4Department of Mathematical Sciences, Florida Institute of Technology Melbourne, Florida 32901, USA e-mail: agarwal@fit.edu

Honoring the Career of John Graef on the Occasion of His Sixty-Seventh Birthday

Abstract Using Mawhin’s continuation theorem we establish the existence of periodic solutions for a class of even order differential equations with deviating argument.

Key words and phrases: Even order differential equation, deviating argument, Mawhin’s continuation theorem, Green’s function. AMS (MOS) Subject Classifications: 34K15; 34C25

1 Introduction

In this paper, we discuss the even order differential equation with deviating argument of the form

x^{(2n)}(t) + ∑_{i=0}^{2n−2} a_{i}(t)x^{(i)}(t) + g(x(t − τ(t))) = p(t),  (1)

where τ(t), a_{i}(t) (i = 0, 1, 2, ⋯, n), p(t) are real continuous functions defined on R with positive period T and a_{2k−2}(t) > 0 (k = 1, 2, ⋯, n) for t ∈ R, and g(x) is a real continuous function defined on R.

Periodic solutions for differential equations were studied in [2-12] and we note that most of the results in the literatue concern lower order problems. There are only a few papers [1,13,14] which discuss higher order problems.

For the sake of completeness, we first state Mawhin’s continuation theorem [3]. Let X and Y be two Banach space and L : DomL ⊂ X ⟶ Y is a linear mapping and N : X ⟶ Y is a continuous mapping. The mapping L will be called a Fredholm mapping of index zero if dimKerL = codimImL < +∞, and ImL is closed in Y. If L is a Fredholm mapping of index zero, there exist continuous projectors P : X ⟶ X and Q : Y ⟶ Y such that ImP = KerL and ImL = KerQ = Im(I − Q). It follows that L|_{DomL∩KerP} : (I − P)X ⟶ ImL has an inverse which will be denoted by K_{P}. If Ω is an open and bounded subset of X, the mapping N will be called L−compact on Ω if QN(\overline{Ω}) is bounded and \overline{K_{P} (I − Q)N(\overline{Ω})} is compact. Since ImQ is isomorphic to KerL, there exists an isomorphism J : ImQ ⟶ KerL. The following theorem is called Mawhin’s continuation theorem (see [3]).

Theorem 1.1 Let L be a Fredholm mapping of index zero, and let N be L−compact on \overline{Ω}. Suppose

(1) for each λ ∈ (0, 1) and x ∈ ∂Ω, Lx ≠ λNx, and

(2) for each x ∈ ∂Ω ∩ Ker(L), QNx ≠ 0 and deg(QN, Ω ∩ Ker(L), 0) ≠ 0.

Then the equation Lx = Nx has at least one solution in \overline{Ω} ∩ D(L).

2 Main Result

Now we make the following assumptions on a_{i}(t):

(i) M_{2k−2} = max_{t∈[0,T]} a_{2k−2}(t) ≥ a_{2k−2}(t) ≥ m_{2k−2} = min_{t∈[0,T]} a_{2k−2}(t) > 0, (k = 1, 2, ⋯, n) for each t ∈ [0, T];

(ii) M_{2n−2} < (\frac{π}{T})^{2} and \frac{M_{2n−2i}}{M_{2n−2i+2}} < (\frac{π}{T})^{2} (i = 2, 3, ⋯, n);

(iii) There exists a positive constant r with m_{0} > r, such that with A−\frac{2M_{0}+m_{0}+r}{2(m_{0}−r)}B > 0 and 1 − A^{∗} > 0, where A = 1 − A^{∗},

B = M_{1}(\frac{T}{2} )^{2n−2} + (M_{2} − m_{2})(\frac{T}{2} )^{2n−3} + M_{3}(\frac{T}{2} )^{2n−4} + (M_{4} − m_{4})(\frac{T}{2} )^{2n−5}

+ ⋯ + M_{2n−3}(\frac{T}{2} )^{2} + (M_{2n−2} − m_{2n−2})\frac{T}{2} ,

A^{∗} = [M_{2n−2}(\frac{T}{2} )^{2} + M_{2n−3}(\frac{T}{2} )^{3} + M_{2n−4}(\frac{T}{2} )^{4} + ⋯ + M_{2}(\frac{T}{2} )^{2n−2} + M_{1}(\frac{T}{2} )^{2n−1}]

and M_{2k−1} = max_{t∈[0,T]} |a_{2k−1}(t)| (k = 1, 2, ⋯, n − 1).

Our main result is the following theorem.

Theorem 2.1 Under the assumptions (i), (ii) and (iii), if

lim_{|x|→∞} sup|\frac{g(x)}{x}| ≤ r  (2)

and

lim_{|x|→∞} sgn(x)g(x) = +∞,  (3)

then Eq.(1) has at least one T −periodic solution.

In order to prove the main theorem we need some preliminaries. Set

X := {x|x ∈ C^{2n−1}(R, R), x(t + T) = x(t), ∀t ∈ R}

and x^{(0)}(t) = x(t), and define the norm on X by

||x|| = max_{0≤j≤2n−1} max_{t∈[0,T]} |x^{(j)}(t)|,

and set

Y := {y|y ∈ C(R, R), y(t + T) = y(t), ∀t ∈ R}.

We define the norm on Y by ||y||_{0} = max_{t∈[0,T]} |y(t)|. Thus both (X, ||·||) and (Y, ||·||_{0}) are Banach spaces.

Remark 2.1 If x ∈ X, then it follows that x^{(i)}(0) = x^{(i)}(T) (i = 0, 1, 2, ⋯, 2n − 1).

Define the operators L : X ⟶ Y and N : X ⟶ Y, respectively, by

Lx(t) = x^{(2n)}(t), t ∈ R,  (4)

and

Nx(t) = p(t) − ∑_{i=0}^{2n−2} a_{i}(t)x^{(i)}(t) − g(x(t − τ(t))), t ∈ R.  (5)

Clearly,

KerL = {x ∈ X : x(t) = c ∈ R}  (6)

and

ImL = {y ∈ Y : ∫_{0}^{T} y(t)dt = 0}  (7)

is closed in Y. Thus L is a Fredholm mapping of index zero.

Let us define P : X → X and Q : Y → Y/Im(L), respectively, by

P x(t) = \frac{1}{T} ∫_{0}^{T} x(t)dt = x(0), t ∈ R,  (8)

for x = x(t) ∈ X and

Qy(t) = \frac{1}{T} ∫_{0}^{T} y(t)dt, t ∈ R  (9)

for y = y(t) ∈ Y. It is easy to see that ImP = KerL and ImL = KerQ = Im(I − Q). It follows that L|_{DomL∩KerP} : (I − P)X ⟶ ImL has an inverse which will be denoted by K_{P}.

Furthermore for any y = y(t) ∈ ImL, if n = 1, it is well-known that

K_{P} y(t) = −\frac{t}{T} ∫_{0}^{T} du ∫_{0}^{u} y(s)ds + ∫_{0}^{t} du ∫_{0}^{u} y(s)ds.  (10)

4 C. Guo, D. O’Regan, Y. Xu & R. P. Agarwal If n > 1, let x(t) ∈ DomL ∩ KerP be such that K_{P} y(t) = x(t). Then x^{(2n)}(t) = y(t),

x^{(2n−1)}(t) = x^{(2n−1)}(0) + ∫_{0}^{t} x^{(2n)}(s)ds  (11)

and

x^{(2n−2)}(t) = x^{(2n−2)}(0) + x^{(2n−1)}(0)t + ∫_{0}^{t} du ∫_{0}^{u} x^{(2n)}(s)ds.  (12)

Since x^{(2n−2)}(T) = x^{(2n−2)}(0), we have

x^{(2n−1)}(0)T + ∫_{0}^{T} du ∫_{0}^{u} x^{(2n)}(s)ds = 0

or

x^{(2n−1)}(0) = −\frac{1}{T} ∫_{0}^{T} du ∫_{0}^{u} x^{(2n)}(s)ds.

From (12), we have

x^{(2n−2)}(t) = x^{(2n−2)}(0) − \frac{t}{T} ∫_{0}^{T} du ∫_{0}^{u} x^{(2n)}(s)ds + ∫_{0}^{t} du ∫_{0}^{u} x^{(2n)}(s)ds.  (13)

Now since ∫_{0}^{T} x^{(2n−2)}(s)ds = 0, from (13) we have

x^{(2n−2)}(0)T − \frac{T}{2} ∫_{0}^{T} du ∫_{0}^{u} x^{(2n)}(s)ds + ∫_{0}^{T} dw ∫_{0}^{w} du ∫_{0}^{u} x^{(2n)}(s)ds = 0, or

x^{(2n−2)}(0) = \frac{1}{2} ∫_{0}^{T} du ∫_{0}^{u} x^{(2n)}(s)ds − \frac{1}{T} ∫_{0}^{T} dw ∫_{0}^{w} du ∫_{0}^{u} x^{(2n)}(s)ds.  (14)

From (13) and (14), we have

\begin{aligned}x^{(2n−2)}(t) & = \frac{1}{2} ∫_{0}^{T} du ∫_{0}^{u} x^{(2n)}(s)ds − \frac{1}{T} ∫_{0}^{T} dw ∫_{0}^{w} du ∫_{0}^{u} x^{(2n)}(s)ds \\ & −\frac{t}{T} ∫_{0}^{T} du ∫_{0}^{u} x^{(2n)}(s)ds + ∫_{0}^{t} du ∫_{0}^{u} x^{(2n)}(s)ds \\ & = (\frac{1}{2} − \frac{t}{T}) ∫_{0}^{T} du ∫_{0}^{u} x^{(2n)}(s)ds + ∫_{0}^{t} du ∫_{0}^{u} x^{(2n)}(s)ds \\ & −\frac{1}{T} ∫_{0}^{T} dw ∫_{0}^{w} du ∫_{0}^{u} x^{(2n)}(s)ds. \end{aligned}  (15)

Let y_{0}(t) = y(t) and y_{1}(t) = x^{(2n−2)}(t). Since y(t) = x^{(2n)}(t), we have from (15) that

\begin{aligned}x^{(2n−2)}(t) = y_{1}(t) & = (\frac{1}{2} − \frac{t}{T}) ∫_{0}^{T} du ∫_{0}^{u} y_{0}(s)ds \\ & + ∫_{0}^{t} du ∫_{0}^{u} y_{0}(s)ds − \frac{1}{T} ∫_{0}^{T} dw ∫_{0}^{w} du ∫_{0}^{u} y_{0}(s)ds. \end{aligned}  (16)

From (16), we obtain

x^{(2n−3)}(t) = x^{(2n−3)}(0) + ∫_{0}^{t} y_{1}(s)ds

and

x^{(2n−4)}(t) = x^{(2n−4)}(0) + x^{(2n−3)}(0)t + ∫_{0}^{t} du ∫_{0}^{u} y_{1}(s)ds.  (17)

EJQTDE Spec. Ed. I, 2009 No. 12

Since x^{(2n−4)}(T) = x^{(2n−4)}(0), we have from (17) that

x^{(2n−3)}(0) = −\frac{1}{T} ∫_{0}^{T} du ∫_{0}^{u} y_{1}(s)ds.  (18)

Since ∫_{0}^{T} x^{(2n−4)}(s)ds = 0, we have from (17) that

x^{(2n−4)}(0) = \frac{1}{2} ∫_{0}^{T} du ∫_{0}^{u} y_{1}(s)ds − \frac{1}{T} ∫_{0}^{T} dw ∫_{0}^{w} du ∫_{0}^{u} y_{1}(s)ds.  (19)

Let y_{2}(t) = x^{(2n−4)}(t) and we have from (17)-(19) that

\begin{aligned}x^{(2n−4)}(t) = y_{2}(t) & = (\frac{1}{2} − \frac{t}{T}) ∫_{0}^{T} du ∫_{0}^{u} y_{1}(s)ds \\ & + ∫_{0}^{t} du ∫_{0}^{u} y_{1}(s)ds − \frac{1}{T} ∫_{0}^{T} dw ∫_{0}^{w} du ∫_{0}^{u} y_{1}(s)ds. \end{aligned}

Let y_{i}(t) = x^{(2n−2i)}(t) (i = 1, 2, ⋯, n − 1) and as above it is easy to check that

\begin{aligned}x^{(2n−2i)}(t) = y_{i}(t) & = (\frac{1}{2} − \frac{t}{T}) ∫_{0}^{T} du ∫_{0}^{u} y_{i−1}(s)ds \\ & + ∫_{0}^{t} du ∫_{0}^{u} y_{i−1}(s)ds − \frac{1}{T} ∫_{0}^{T} dw ∫_{0}^{w} du ∫_{0}^{u} y_{i−1}(s)ds, \end{aligned}

for (i = 1, 2, ⋯, n − 1), and

y_{n}(t) = y_{n}(0) − \frac{t}{T} ∫_{0}^{T} du ∫_{0}^{u} y_{n−1}(s)ds + ∫_{0}^{t} du ∫_{0}^{u} y_{n−1}(s)ds.

Note that y_{n}(t) = x(t) ∈ DomL ∩ KerP. Thus y_{n}(0) = x(0) = 0, and

K_{P} y(t) = −\frac{t}{T} ∫_{0}^{T} du ∫_{0}^{u} y_{n−1}(s)ds + ∫_{0}^{t} du ∫_{0}^{u} y_{n−1}(s)ds.  (20)

Let Ω be an open and bounded subset of X. In view of (5), (9) and (10) (or (20)), we can easily see that QN(\overline{Ω}) is bounded and \overline{K_{P} (I − Q)N(\overline{Ω})} is compact. Thus the mapping N is L−compact on \overline{Ω}. That is, we have the following result.

Lemma 2.1 Let L, N, P and Q be defined by (4), (5), (8) and (9) respectively. Then L is a Fredholm mapping of index zero and N is L−compact on \overline{Ω}, where Ω is any open and bounded subset of X.

In order to prove our main result, we need the following Lemmas [6, 7]. The first result follows from [6 and Remark 2.1] and the second from [7].

Lemma 2.2 Let x(t) ∈ C^{(n)}(R, R) ∩ C_{T}. Then

||x^{(i)}||_{0} ≤ \frac{1}{2} ∫_{0}^{T} |x^{(i+1)}(s)|ds, i = 1, 2, ⋯, n − 1,

where n ≥ 2 and C_{T} := {x|x ∈ C(R, R), x(t + T) = x(t), ∀t ∈ R}.

Lemma 2.3 Suppose that M, λ are positive numbers and satisfy 0 < M < (\frac{π}{T})^{2} and 0 < λ < 1 , then for any function φ defined in [0, T], the following problem

\{ \begin{matrix}x′′(t) + λMx(t) = λφ(t), \\ x(0) = x(T), x′(0) = x′(T), \end{matrix}

has a unique solution

x(t) = ∫_{0}^{T} G(t, s)λφ(s)ds,

where α = \sqrt{λM}, and

G(t, s) = \{ \begin{matrix}w(t − s), (k − 1)T ≤ s ≤ t ≤ kT, \\ w(T + t − s), (k − 1)T ≤ t ≤ s ≤ kT, (k ∈ N), \end{matrix}

with

w(t) = \frac{cos α(t− \frac{T}{2} )}{2α sin \frac{αT}{2}}.

Now, we consider the following auxiliary equation

x^{(2n)}(t) + λ ∑_{i=0}^{2n−2} a_{i}(t)x^{(i)}(t) + λg(x(t − τ(t))) = λp(t),  (21)

where 0 < λ < 1. We have

Lemma 2.4 Suppose the conditions of Theorem 2.1 are satisfied. If x(t) is a T −periodic solution of Eq.(21), then there are positive constants D_{i} (i = 0, 1, ⋯ 2n−1), which are independent of λ, such that

||x^{(i)}||_{0} ≤ D_{i}, t ∈ [0, T] for i = 0, 1, ⋯, 2n − 1.  (22)

Proof.Suppose that x(t) is a T −periodic solution of (21). By (2) of Theorem 2.1 we know that there exists a \overline{M_{1}} > 0, such that

|g(x(t))| ≤ r|x(t)|, |x(t)| > \overline{M_{1}}, t ∈ R.  (23)

Set

E_{1} = \{t : |x(t)| > \overline{M_{1}}, t ∈ [0, T]\},  (24)
E_{2} = [0, T]\E_{1}  (25)

and

ρ = max_{|x|≤\overline{M_{1}}} |g(x)|.  (26)

Periodic Solutions For a Class of Even Order Differential Equations 7 Let ε = \frac{m_{0}−r}{2}. By (21), (23), (24), (25), (26) and Lemma 2.2, we obtain

||x^{(2n−1)}||_{0} ≤ \frac{1}{2} ∫_{0}^{T} |x^{(2n)}(s)|ds

≤ \frac{λ}{2} ∫_{0}^{T} [| ∑_{i=0}^{2n−2} a_{i}(t)x^{(i)}(t)| + |g(x(t − τ(t)))| + |p(t)|]dt

≤ \frac{λT}{2} [M_{2n−2}||x^{(2n−2)}||_{0} + M_{2n−3}||x^{(2n−3)}||_{0} + ⋯ + M_{2}||x^{(2)}||_{0} + M_{1}||x^{(1)}||_{0} +M_{0}||x||_{0}] + \frac{λ}{2} ∫_{0}^{T} |g(x(t − τ(t)))|dt + \frac{λT}{2} ||p||_{0} ≤ \frac{T}{2} [M_{2n−2} \frac{T}{2} + M_{2n−3}(\frac{T}{2} )^{2} + ⋯ + M_{2}(\frac{T}{2} )^{2n−3} + M_{1}(\frac{T}{2} )^{2n−2}]||x^{(2n−1)}||_{0}+

\begin{matrix}+\frac{T}{2} M_{0}||x||_{0} + \frac{1}{2}[∫_{E_{1}} |g(x(t − τ(t)))|dt + ∫_{E_{2}} |g(x(t − τ(t)))|dt] + \frac{T}{2} ||p||_{0} \\ ≤ A^{∗}||x^{(2n−1)}||_{0} + \frac{T}{2} (M_{0} + r + ε)||x||_{0} + \frac{T}{2} C \\ = A^{∗}x^{(2n−1)}||_{0} + \frac{T}{4} (2M_{0} + r + m_{0})||x||_{0} + \frac{T}{2} C, \end{matrix}  (27)

where C = (ρ + ||p||_{0}) and

\begin{aligned}A^{∗} & = [M_{2n−2}(\frac{T}{2} )^{2} + M_{2n−3}(\frac{T}{2} )^{3} + M_{2n−4}(\frac{T}{2} )^{4} \\ & + ⋯ + M_{2}(\frac{T}{2} )^{2n−2} + M_{1}(\frac{T}{2} )^{2n−1}]. \end{aligned}

Now from (27), we have

||x^{(2n−1)}||_{0} ≤ (1 − A^{∗})^{−1}[\frac{T}{4} (2M_{0} + r + m_{0})||x||_{0} + \frac{T}{2} C].  (28)

On the other hand, from (21) and Lemma 2.3, we get

\begin{matrix}x^{(2n−2)}(t) \\ = ∫_{0}^{T} G_{1}(t, t_{1})λ[(M_{2n−2} − a_{2n−2}(t_{1}))x^{(2n−2)}(t_{1}) + p(t_{1}) \\ −g(x(t − τ(t_{1})))]dt_{1} − λ ∫_{0}^{T} G_{1}(t, t_{1})[ ∑_{i=0}^{2n−3} a_{i}(t_{1})x^{(i)}(t_{1})]dt_{1}, \end{matrix}  (29)

where α_{1} = \sqrt{λM_{2n−2}}, and

G_{1}(t, t_{1}) = \{ \begin{matrix}w_{1}(t − t_{1}), (k − 1)T ≤ t_{1} ≤ t ≤ kT, \\ w_{1}(T + t − t_{1}), (k − 1)T ≤ t ≤ t_{1} ≤ kT, (k ∈ N), \end{matrix}  (30)

with

w_{1}(t) = \frac{cos α_{1}(t− \frac{T}{2} )}{2α_{1} sin \frac{α_{1}T}{2}}  (31)

and

∫_{0}^{T} G_{1}(t, t_{1})dt_{1} = \frac{1}{λM_{2n−2}} .  (32)

EJQTDE Spec. Ed. I, 2009 No. 12

From (29) and Lemma 2.3, we have

\begin{matrix}x^{(2n−4)}(t) \\ = λ ∫_{0}^{T} G_{2}(t, t_{1}) ∫_{0}^{T} G_{1}(t_{1}, t_{2})[p(t_{2}) − g(x(t − τ(t_{2})))]dt_{2}dt_{1} \\ +λ ∫_{0}^{T} G_{2}(t, t_{1}) ∫_{0}^{T} G_{1}(t_{1}, t_{2})(M_{2n−2} − a_{2n−2}(t_{2}))x^{(2n−2)}(t_{2})dt_{2}dt_{1} \\ + ∫_{0}^{T} G_{2}(t, t_{1})[\frac{M_{2n−4}}{M_{2n−2}} x^{(2n−4)}(t_{1}) − λ ∫_{0}^{T} G_{1}(t_{1}, t_{2})a_{2n−4}(t_{2})x^{(2n−4)}(t_{2})dt_{2}]dt_{1} \\ −λ ∫_{0}^{T} G_{2}(t, t_{1}) ∫_{0}^{T} G_{1}(t_{1}, t_{2})[ ∑_{i=0}^{2n−5} a_{i}(t_{1})x^{(i)}(t_{2}) + a_{2n−3}(t_{2})x^{(2n−3)}(t_{2})]dt_{2}dt_{1}, \end{matrix} (33)

where α_{2} = \sqrt{\frac{M_{2n−4}}{M_{2n−2}}} , and

G_{2}(t, t_{2}) = \{ \begin{matrix}w_{2}(t − t_{2}), (k − 1)T ≤ t_{2} ≤ t ≤ kT, \\ w_{2}(T + t − t_{2}), (k − 1)T ≤ t ≤ t_{2} ≤ kT, (k ∈ N), \end{matrix}  (34)

with

w_{2}(t) = \frac{cos α_{2}(t− \frac{T}{2} )}{2α_{2} sin \frac{α_{2}T}{2}}  (35)

and

∫_{0}^{T} G_{2}(t, t_{2})dt_{2} = \frac{M_{2n−2}}{M_{2n−4}} .  (36)

By induction, we have x(t) = λ ∫_{0}^{T} G_{n}(t, t_{1}) ⋯ ∫_{0}^{T} G_{1}(t_{n−1}, t_{n})[p(t_{n}) − g(x(t_{n} − τ(t_{n})))]dt_{n} ⋯ dt_{1} +λ ∫_{0}^{T} G_{n}(t, t_{1}) ⋯ ∫_{0}^{T} G_{1}(t_{n−1}, t_{n})(M_{2n−2} − a_{2n−2}(t_{n}))x^{(2n−2)}(t_{n})dt_{n} ⋯ dt_{1} + ∫_{0}^{T} G_{n}(t, t_{1}) ⋯ ∫_{0}^{T} G_{2}(t_{n−2}, t_{n−1})[\frac{M_{2n−4}}{M_{2n−2}} x^{(2n−4)}(t_{n−1})−

\begin{matrix}λ ∫_{0}^{T} G_{1}(t_{n−1}, t_{n})a_{2n−4}x^{(2n−4)}(t_{n})dt_{n}]dt_{n−1} ⋯ dt_{1} \\ + ∫_{0}^{T} G_{n}(t, t_{1}) ⋯ ∫_{0}^{T} G_{3}(t_{n−3}, t_{n−2})[\frac{M_{2n−6}}{M_{2n−4}} x^{(2n−6)}(t_{n−2})− \\ λ ∫_{0}^{T} G_{2}(t_{n−2}, t_{n−1}) ∫_{0}^{T} G_{1}(t_{n−1}, t_{n})[a_{2n−6}x^{(2n−6)}(t_{n})dt_{n}dt_{n−1}]dt_{n−2} ⋯ dt_{1} \\ + ⋯ + ⋯ \\ + ∫_{0}^{T} G_{n}(t, t_{1})[\frac{M_{0}}{M_{2}} x(t_{1}) − λ ∫_{0}^{T} G_{n−1}(t_{1}, t_{2}) ∫_{0}^{T} G_{n−2}(t_{2}, t_{3}) \\ ⋯ ∫_{0}^{T} G_{1}(t_{n−1}, t_{n})a_{0}(t_{n})x(t_{n})dt_{n} ⋯ dt_{2}]dt_{1} \\ −λ ∫_{0}^{T} G_{n}(t, t_{1}) ⋯ ∫_{0}^{T} G_{1}(t_{n−1}, t_{n})[ ∑_{k=1}^{n−1} a_{2k−1}(t_{n})x^{(2k−1)}(t_{n})]dt_{n} ⋯ dt_{1} \end{matrix}  (37)

where α_{i} = \sqrt{\frac{M_{2n−2i}}{M_{2n−2i+2}}} (2 ≤ i ≤ n), and

G_{i}(t, t_{i}) = \{ \begin{matrix}w_{i}(t − t_{i}), (k − 1)T ≤ t_{i} ≤ t ≤ kT, \\ w_{i}(T + t − t_{i}), (k − 1)T ≤ t ≤ t_{i} ≤ kT, (k ∈ N), \end{matrix}  (38)

with

w_{i}(t) = \frac{cos α_{i}(t− \frac{T}{2} )}{2α_{i} sin \frac{α_{i}T}{2}}  (39)

and

∫_{0}^{T} G_{i}(t, t_{i})dt_{i} = \frac{M_{2n−2i+2}}{M_{2n−2i}} (2 ≤ i ≤ n).  (40)

From (32), (37), (40) and Lemma 2.2, we obtain ||x||_{0} ≤ max_{t∈[0,T]} λ ∫_{E_{1}} |G_{n}(t, t_{1})| ⋯ ∫_{0}^{T} |G_{1}(t_{n−1}, t_{n})||p(t_{n}) − g(x(t_{n} − τ(t_{n})))|dt_{n} ⋯ dt_{1} + max_{t∈[0,T]} λ ∫_{E_{2}} |G_{n}(t, t_{1})| ⋯ ∫_{0}^{T} |G_{1}(t_{n−1}, t_{n})||p(t_{n}) − g(x(t_{n} − τ(t_{n})))|dt_{n} ⋯ dt_{1}+ max_{t∈[0,T]} λ ∫_{0}^{T} |G_{n}(t, t_{1})| ⋯ ∫_{0}^{T} |G_{1}(t_{n−1}, t_{n})|(M_{n−1} − a_{n−1}(t_{n}))|x^{(2n−2)}(t_{n})|dt_{n} ⋯ dt_{1} + max_{t∈[0,T]} ∫_{0}^{T} |G_{n}(t, t_{1})| ⋯ ∫_{0}^{T} |G_{2}(t_{n−2}, t_{n−1})||\frac{M_{n−2}}{M_{n−1}} x^{(2n−4)}(t_{n−1})− λ ∫_{0}^{T} G_{1}(t_{n−1}, t_{n})a_{n−2}x^{(2n−4)}(t_{n})dt_{n}|dt_{n−1} ⋯ dt_{1} + max_{t∈[0,T]} ∫_{0}^{T} |G_{n}(t, t_{1})| ⋯ ∫_{0}^{T} |G_{3}(t_{n−3}, t_{n−2})||\frac{M_{n−3}}{M_{n−2}} x^{(2n−6)}(t_{n−2})− λ ∫_{0}^{T} G_{2}(t_{n−2}, t_{n−1}) ∫_{0}^{T} G_{1}(t_{n−1}, t_{n})[a_{n−3}x^{(2n−6)}(t_{n})dt_{n}dt_{n−1}|dt_{n−2} ⋯ dt_{1} + ⋯ + ⋯ + max_{t∈[0,T]} ∫_{0}^{T} |G_{n}(t, t_{1})||\frac{M_{0}}{M_{1}} x(t_{1}) − λ ∫_{0}^{T} G_{n−1}(t_{1}, t_{2}) ∫_{0}^{T} G_{n−2}(t_{2}, t_{3}) ⋯ ∫_{0}^{T} G_{1}(t_{n−1}, t_{n})a_{0}(t_{n})x(t_{n})dt_{n} ⋯ dt_{2}|dt_{1}

+ max_{t∈[0,T]} λ ∫_{0}^{T} |G_{n}(t, t_{1})| ⋯ ∫_{0}^{T} |G_{1}(t_{n−1}, t_{n})|[ ∑_{k=1}^{n−1} a_{2k−1}(t_{n})x^{(2k−1)}(t_{n})]|dt_{n} ⋯ dt_{1}

≤ \frac{1}{M_{0}} [||p||_{0} + ρ + (r + ε)||x||_{0}] + \frac{M_{0}−m_{0}}{M_{0}} ||x||_{0} + \frac{1}{M_{0}} [(M_{2n−2} − m_{2n−2})||x^{(2n−2)}||_{0} +(M_{2n−4} − m_{2n−4})||x^{(2n−4)}||_{0} + ⋯ + (M_{2} − m_{2})||x^{(2)}||_{0}] +\frac{1}{M_{0}} [M_{1}||x^{(1)}||_{0} + M_{3}||x^{(3)}||_{0} + ⋯ + M_{2n−3}||x^{(2n−3)}||_{0}] ≤ \frac{1}{M_{0}} [C + (M_{0} − m_{0} + r + ε)||x||_{0}] + \frac{1}{M_{0}} [M_{1}(\frac{T}{2} )^{2n−2} + (M_{2} − m_{2})(\frac{T}{2} )^{2n−3} +M_{3}(\frac{T}{2} )^{2n−4} + (M_{4} − m_{4})(\frac{T}{2} )^{2n−5} + ⋯ + M_{2n−3}(\frac{T}{2} )^{2} +(M_{2n−2} − m_{2n−2})\frac{T}{2} ]||x^{(2n−1)}||_{0}.

(41)

EJQTDE Spec. Ed. I, 2009 No. 12

Now (41) and ε = \frac{m_{0}−r}{2} give

||x||_{0} ≤ \frac{2(B||x^{(2n−1)}||_{0}+C)}{(m_{0}−r)},  (42)

where

\begin{aligned}B = M_{1}(\frac{T}{2} )^{2n−2} & +(M_{2} − m_{2})(\frac{T}{2} )^{2n−3} + M_{3}(\frac{T}{2} )^{2n−4} + (M_{4} − m_{4})(\frac{T}{2} )^{2n−5} \\ & + ⋯ + ⋯ + M_{2n−3}(\frac{T}{2} )^{2} + (M_{2n−2} − m_{2n−2})\frac{T}{2} \end{aligned}

and M_{2k−1} = max_{t∈[0,T]} |a_{2k−1}(t)| (k = 1, 2, ⋯ n − 1).

Thus combining (27) and (42), we see that

\begin{aligned}(A − \frac{2M_{0}+m_{0}+r}{2(m_{0}−r)}B)||x^{(2n−1)}||_{0} & ≤ \frac{T C}{2} (\frac{2M_{0}+m_{0}+r}{(m_{0}−r)} + 1) \\ & = \frac{(M_{0}+m_{0})T C}{(m_{0}−r)} , \end{aligned}  (43)

where A = 1 − A^{∗}.

From (42) and (43), we have

\begin{aligned}||x^{(2n−1)}||_{0} & ≤ \frac{(M_{0}+m_{0})T C}{(m_{0}−r)} (A − \frac{2M_{0}+m_{0}+r}{2(m_{0}−r)}B)^{−1} \\ & = D_{2n−1} \end{aligned}  (44)

and

||x||_{0} ≤ \frac{2(BD_{2n−1} + C)}{(m_{0} − r)} = D_{0}.  (45)

Finally from (44), (45) and Lemma 2.2, we get

||x^{(i)}||_{0} ≤ D_{i} (1 ≤ i ≤ 2n − 2).  (46)

The proof of Lemma 2.4 is complete.

Proof of Theorem 2.1. Suppose that x(t) is a T-periodic solution of Eq.(21). By Lemma 2.4, there exist positive constants D_{i} (i = 0, 1, ⋯, 2n − 1) which are independent of λ such that (22) is true. By (3), we know that there exists a M_{2} > 0, such that

sgn(x)g(x(t)) > 0, |x(t)| > M_{2}, t ∈ R.

Consider any positive constant \overline{D} > max_{0≤i≤2n−1}\{D_{i}\} + M_{2}.

Set

Ω := \{x ∈ X : ||x|| < \overline{D}\}.

We know that L is a Fredholm mapping of index zero and N is L-compact on \overline{Ω} (see [3]).

Recall

Ker(L) = {x ∈ X : x(t) = c ∈ R}

and the norm on X is

||x|| = max_{0≤j≤2n−1} max_{t∈[0,T]} |x^{(j)}(t)|.

Then we have

x = \overline{D} or x = −\overline{D} for x ∈ ∂Ω ∩ Ker(L).  (47)

From (3) and (47), we have (if \overline{D} is chosen large enough)

a_{0}(t)\overline{D} + g(\overline{D}) − ∥p∥_{0} > 0 for t ∈ [0, T]  (48)

and

x^{(i)}(t) = 0, ∀x ∈ ∂Ω ∩ Ker(L)(i = 1, 2, ⋯, 2n − 1).  (49)

Finally from (5), (9) and (47)-(49), we have

(QNx) = \frac{1}{T} ∫_{0}^{T} [− ∑_{i=0}^{2n−2} a_{i}(t)x^{(i)}(t) − g(x(t − τ(t))) + p(t)]dt
= −\frac{1}{T} ∫_{0}^{T} [a_{0}(t)x(t) + g(x(t − τ(t))) − p(t)]dt
≠ 0, ∀x ∈ ∂Ω ∩ Ker(L).

Then, for any x ∈ KerL ∩ ∂Ω and η ∈ [0, 1], we have

\begin{aligned}xH(x, η) & = −ηx^{2} − \frac{x}{T} (1 − η) ∫_{0}^{T} [ ∑_{i=0}^{2n−2} a_{i}(t)x^{(i)}(t) + g(x(t − τ(t))) − p(t)]dt \\ & ≠ 0. \end{aligned}

Thus

deg\{QN, Ω ∩ Ker(L), 0\} = deg\{−\frac{1}{T} ∫_{0}^{T} [ ∑_{i=0}^{2n−2} a_{i}(t)x^{(i)}(t)
+g(x(t − τ(t))) − p(t)]dt, Ω ∩ Ker(L), 0}
= deg{−x, Ω ∩ Ker(L), 0}
≠ 0.

From Lemma 2.4 for any x ∈ ∂Ω ∩ Dom(L) and λ ∈ (0, 1) we have Lx ≠ λNx. By Theorem 1.1, the equation Lx = Nx has at least a solution in Dom(L) ∩ \overline{Ω}, so there exists a T-periodic solution of Eq.(1). The proof is complete.

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