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The Clark dual and multiple periodic solutions of delay differential equations

Huafeng Xiao*, Jianshe Yu and Zhiming Guo

Author Affiliations

College of Mathematics and Information Sciences, Guangzhou University, Guangzhou, 510006, PRC

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Boundary Value Problems 2011, 2011:44  doi:10.1186/1687-2770-2011-44

The electronic version of this article is the complete one and can be found online at: http://www.boundaryvalueproblems.com/content/2011/1/44


Received:18 May 2011
Accepted:11 November 2011
Published:11 November 2011

© 2011 Xiao et al; licensee Springer.

This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

We study the multiplicity of periodic solutions of a class of non-autonomous delay differential equations. By making full use of the Clark dual, the dual variational functional is considered. Some sufficient conditions are obtained to guarantee the existence of multiple periodic solutions.

2000 Mathematics Subject Classification: 34K13; 34K18.

Keywords:
periodic solution; Clark dual; Morse-Ekeland index; delay differential equation; asymptotic linearity

1 Introduction

The existence and multiplicity of periodic solutions of delay differential equations have been investigated since 1962. Various methods have been used to study such a problem [1-10]. Among those methods, critical point theory is a very important tool. By combining with Kaplan-Yorke method, it can be used indirectly to study the existence of periodic solutions of delay differential equation [11-16]. By building the variational frame for some special systems, it can be used directly to study the existence of delay differential systems [17,18]. However, the variational functionals in the above two cases are strongly indefinite. They are hard to be dealed with.

In this article, we study multiple periodic solutions of the following non-autonomous delay differential equations

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M1','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M1">View MathML</a>

(1)

where x(t) ∈ ℝn, f C (ℝ × ℝn, ℝn). We assume that

(f1) f(t, x) is odd with respect to x and π/2-periodic with respect to t, i.e.,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M2','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M2">View MathML</a>

(f2) There exists a continuous differentiable function F(t, x), which is strictly convex with respect to x uniformly in t, such that f(t, x) is the gradient of F(t, x) with respect to x;

(f3)

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M3','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M3">View MathML</a>

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M4','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M4">View MathML</a>

where A0, Aare positive definite constant matrices.

By making use of the Clarke dual, we study the dual variational functional associated with (1), which is an indefinite functional. Since the dimension of its negative space is finite, we define it as the Morse index of the dual variational functional. This Morse index is significant. Then Z2-index theory can be used and some sufficient conditions are obtained to guarantee the existence of multiple periodic solutions of (1).

The rest of this article is organized as follows: in Section 2, some preliminary results will be stated; in Section 3, linear system is discussed and the Morse index of the variational functional associated with linear system is defined; in Section 4, our main results will be stated and proved.

2 Preliminaries

Denote by ℕ, ℕ*, ℤ, ℝ the sets of all positive integers, nonnegative integers, integers, real numbers, respectively. We define S1 : = ℝ/(2πℤ).

For a matrix A, denote by σ(A) the set of eigenvalue of A. The identity matrix of order n is denoted by In and for simplicity by I.

Let x, y L2 (S1, ℝn). For every z C(S1, ℝn), if

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M5','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M5">View MathML</a>

then y is called a weak derivative of x, denoted by <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M6','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M6">View MathML</a>.

The space H1 = W1,2(S1, ℝn) consists of 2π-periodic vector-valued functions with dimension n, which possess square integrable derivative of order 1. We can choose the usual norm and inner product in H1 as follows:

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M7','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M7">View MathML</a>

where | · |, (·,·) denote the usual norm and inner product in ℝn, respectively. Then H1 is a Hilbert space.

Define the shift operator K : H1 H1 by Kx(·) = x(· + π/2), for all x H1. Clearly, K is a bounded linear operator from H1 to H1. Set

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M8','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M8">View MathML</a>

Then E is a closed subspace of H1. If x E, its Fourier expansion is

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M9','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M9">View MathML</a>

(2)

where ak, bk ∈ ℝn. In particular, E does not contain ℝn as its subspace. In addition, <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M10','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M10">View MathML</a> for all x E.

The dual variational functional corresponding to (1) defined on H1 is

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M11','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M11">View MathML</a>

(3)

By Hypothesis (f3), F(t, x)/|x| → +∞ as |x| → ∞ uniformly in t. Since F(t, ·) is strictly convex, Proposition 2.4 of [19] implies that F*(t, ·) ∈ C1(ℝn, ℝ). Since f satisfies (f3), it follows the discussion in Chapter 7 of [19] that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M12','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M12">View MathML</a>

(4)

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M13','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M13">View MathML</a>

(5)

where <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M14','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M14">View MathML</a> and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M15','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M15">View MathML</a>.

Lemma 2.1. If y E is a critical point of J, then the function x defined by <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M16','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M16">View MathML</a> is a 2π-periodic solution of (1).

Proof. Since f(·, x) is π/2-periodic, it follows that F(·, x) and then <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M17','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M17">View MathML</a> are π/2-periodic. So is <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M18','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M18">View MathML</a>. (4) implies that there exist positive constants a1, a2 such that |F*(t, y)| ≤ a1 + a2|y|s for some s > 2 and all t ∈ ℝ, y H1. We define φ by the formulas

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M19','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M19">View MathML</a>

It follows Proposition B. 37 of [20] that φ C1(H1, ℝ) and

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M20','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M20">View MathML</a>

We claim: φ'(y) ∈ E if y E.

To prove the above claim, let z H1 and y E,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M21','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M21">View MathML</a>

The arbitrary of z implies that φ'(y) ∈ E.

Since φ C1(H1, ℝ), it is easy to check that J C1(H1, ℝ) and

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M22','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M22">View MathML</a>

Assume that y E is a critical point of J. For any h H1, h = h1 + h2, where h1 E, h2 E. Then

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M23','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M23">View MathML</a>

Since φ'(y) ∈ E, it is easy to check that J'(y) ∈ E and < J'(y), h2 > = 0. Since y is a critical point of J on E, then < J'(y), h1 > = 0. Thus y is a critical point of J on H1. Applying the fundamental Lemma (cf. [19]), there exists c1 such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M24','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M24">View MathML</a>

Setting

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M25','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M25">View MathML</a>

we obtain x H1, <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M26','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M26">View MathML</a> and by duality

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M27','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M27">View MathML</a>

Thus,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M28','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M28">View MathML</a>

i.e.,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M29','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M29">View MathML</a>

Moreover, x(0) = x(2π) since x H1. It follows a regular discussion that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M30','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M30">View MathML</a> is a periodic solution of (1).   □

Let X be a Hilbert space, Φ ∈ C1(X, ℝ), i.e., Φ is a continuously Fréchet differentiable functional defined on X. If X1 is a closed subspace of X, denote by <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M31','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M31">View MathML</a> the orthogonal complement of X1 in X. Fix a prime integer p > 1. Define a map μ : X X such that ||μx|| = ||x|| for any x X and μp = idX, where idX is the identity map on X. Then μ is a linear isometric action of Zp on X, where Zp is the cyclic group with order p.

A subset A X is called μ-invariant if μ(A) ⊂ A. A continuous map h : A E is called μ-equivariant if h(μx) = μh(x) for any x A. A continuous functional H : X → ℝ is called μ-invariant if H(μx) = H(x) for any x X.

Φ is said to be satisfying (PS)-condition if any sequence {xj} ⊂ X for which {Φ(xj)} is bounded and Φ'(xj) → 0 as j → ∞, possesses a convergent subsequence. A sequence {xj} is called (PS)-sequence if {Φ(xj)} is bounded and Φ'(xj) → 0 as j → ∞.

Lemma 2.2 [21]. Let Φ ∈ C1(X, ℝ) be a μ-invariant functional satisfying the (PS)-condition. Let Y and Z be closed μ-invariant subspaces of X with codimY and dimZ finite and

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M32','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M32">View MathML</a>

Assume that the following conditions are satisfied:

(F1) Fixμ Y, Z Fixμ = {0};

(F2) infxY Φ(x) > -∞;

(F3) there exist constants r > 0 and c < 0 such that Φ(x) ≤ c whenever x Z and ||x|| = r;

(F4) if x Fixμ and Φ'(x) = 0, then Φ(x) ≥ 0.

Then there exists at least dimZ - codimY distinct Zp-orbits of critical points of Φ outside of Fixμ with critical value less than or equal to c.

3 Morse index

Let A be a positive definite constant matrix. We consider the periodic boundary value problem

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M33','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M33">View MathML</a>

(6)

Since F(t, x) = 1/2(Ax, x), it is easy to verify that its Legendre transform F*(t, y) is of the form F*(t, y) = 1/2(By, y), where B = A-1. The dual action of (6) is defined on H1 by

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M34','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M34">View MathML</a>

(7)

A is positively definite, so is B. Thus there exists δ1 > 0 such that (By, y) ≥ δ1|y|2 for all y ∈ ℝn. Wirtinger's inequality implies that the symmetric bilinear form given by

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M35','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M35">View MathML</a>

(8)

define an inner product on E. The corresponding norm || · ||1 is such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M36','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M36">View MathML</a>

(9)

Proposition 3.1. The norm || · ||1 is equivariant to the stand norm || · || of E.

Proof. Since B is positive, Wirtinger's inequality and (9) imply that there exists a positive constant δ2 such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M37','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M37">View MathML</a>

On the other hand, since B is a positive definite matrix, there exists a constant M > 0 such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M38','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M38">View MathML</a>

Thus those two norms are equivariant to each other, which completes our proof.   □

Let us define the linear operator L on E by setting

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M39','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M39">View MathML</a>

(10)

One can easily check that L is a compact self-adjoint operator. (7) can be rewritten as

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M40','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M40">View MathML</a>

(11)

It follows from the spectral theory that E can be decomposed as the orthogonal sum of Ker(I - L), E+ and E- with I - L positive definite (resp. negative definite) on E+ (resp. E-). Since L is compact, it has at most finite many eigenvalues (counting the multiplicity) greater than one. Thus the index dimE- < ∞.

Definition 3.1. The index i(A) is the Morse index of χA, i.e. the supermum of the dimension of the subspace of E on which χA is negative definite.

On the other hand, there exists a positive constant δ3 such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M41','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M41">View MathML</a>

Setting δ = δ1δ3 > 0, we reduce from (9) and (11) the estimates

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M42','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M42">View MathML</a>

(12)

Proposition 3.2. The dimension of Ker(I - L) is equal to the number of linearly independent solutions of (6).

Proof. If y E, by a Fourier argument, y Ker(I - L) if and only if

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M43','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M43">View MathML</a>

(13)

for some c2 ∈ ℝn and a.e. t ∈ [0, 2π]. Since B = A-1 is invertible, y is continuous differentiable. Thus (13) hold for all t ∈ [0, 2π], and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M44','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M44">View MathML</a>. Set

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M45','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M45">View MathML</a>

(14)

In view of (13), (14) and the Clark dual, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M46','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M46">View MathML</a>

Thus x is a solution of (6).    □

Conversely, assume now that x E ⊕ ℝn is a solution of (6). Then

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M47','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M47">View MathML</a>

and hence there exists an unique y E such that x = φy, where φ is defined by (14).

Thus

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M48','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M48">View MathML</a>

for a.e. t ∈ [0, 2π]. Integrating the above equality, we have x(t) = y(t + π/2) + c3 for some c3 ∈ ℝn and all t ∈ [0, 2π]. Consequently,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M49','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M49">View MathML</a>

for a.e. t ∈ [0, 2π], which is equivariant to (13) and shows that y Ker(I - L). Thus φ is an isomorphism between Ker(I - L) and the space of solutions of (6).

Consider the following eigenvalue problem

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M50','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M50">View MathML</a>

For any z E, we have (((I - L)y, z))1 = >λ((y, z))1. Computing directly, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M51','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M51">View MathML</a>

Denote by ei, i = 1, 2, ..., n the basic of ℝn. Choosing

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M52','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M52">View MathML</a>

it follows that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M53','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M53">View MathML</a>

Since B = A-1, then

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M54','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M54">View MathML</a>

Thus

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M55','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M55">View MathML</a>

Proposition 3.3. If σ(A) ∩ {4l + 1, l ∈ ℕ*} = , then I - L is invertible and codimE+ = i(A).

Proof. The above analysis implies that Ker (I - L) = {0} if σ(A) ∩ {4l + 1, l ∈ ℕ*} = ∅. Thus I - L is invertible. If we decompose E as E = E+ E-, then codimE+ = i(A).    □

4 Main results and their proofs

Now we consider the multiplicity of periodic solutions of (1). The main result reads as follows:

Theorem 4.1. Assume that (f1)-(f3) are satisfied. Moreover,

(A1) σ (A) ∩ {4l + 1, l ∈ ℕ*} = ∅;

(A2) i(A0) > i(A).

Then (1) has at least i(A0) - i(A) pairs of nontrivial 2π-periodic solutions.

By (f3), f (t, 0) = 0 uniformly in t. Since F(t, ·) is strictly convex, it follows that 0 is the unique equilibrium point of (1). Without loss of generality, we can assume that F (t, 0) = 0 uniformly in t. Since f (t, 0) = 0, then F*(t, 0) = 0 uniformly in t.

Lemma 4.1. The functional J satisfies (PS)-condition.

Proof. Denote by ((·, ·))1,∞ the inner product, defined by (8) replacing B by B. Let Lbe linear self-adjoint operator, defined by (10), under the inner product ((·, ·))1,∞. We define the operator N over E by setting

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M56','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M56">View MathML</a>

Let <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M57','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M57">View MathML</a> be a (PS)-sequence. Define fj : = yj - Lyj + Nyj, j ∈ ℕ. Since

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M58','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M58">View MathML</a>

then fj → 0 in E as j → ∞. Then there exists R > 0 such that for all j ∈ ℕ, || fj || ≤ R. (A1) implies that M = I - Lis invertible. Thus it follows from (4) that there exists a positive constant c4 such that, for all y E

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M59','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M59">View MathML</a>

where || · ||1,∞ denotes by the norm corresponding to ((·, ·))1,∞. Since || fj || ≤ R, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M60','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M60">View MathML</a>

The above inequality implies that {yj} is bounded. A standard argument as Lemma 4.5 in [19] shows that {yj} has a convergent sequence.    □

Lemma 4.2. The functional J is bounded from below on a closed invariant subspace Y of E of codimension i(A).

Proof. Consider the linear delay differential system

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M61','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M61">View MathML</a>

Its dual variational functional is

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M62','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M62">View MathML</a>

where ((·, ·))1,∞, Lare defined as Lemma 4.1.

Because of (A1) and Proposition 3.3, the operator I - Lis invertible. We decompose E as <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M63','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M63">View MathML</a>, where <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M64','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M64">View MathML</a> is the positive (resp. negative) definite eigenspace of I - L. Then <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M65','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M65">View MathML</a>. Let <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M66','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M66">View MathML</a>. Inequality (12) implies that there exists a positive constant δ4 such that, for each y Y,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M67','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M67">View MathML</a>

It follows (4) that there exists a positive constant c5 such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M68','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M68">View MathML</a>

for each y ∈ ℝn. Hence, by the mean value theorem,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M69','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M69">View MathML</a>

Consequently, we have, for y Y,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M70','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M70">View MathML</a>

and J is bounded from below on Y.

Lemma 4.3. There exists an invariant subspace Z of E with dimension i(A0) and some r > 0 such that J(y) < 0 whenever y Z and ||y||1 = r.

Proof. Consider the linear delay differential system

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M71','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M71">View MathML</a>

Its dual variational functional is

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M72','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M72">View MathML</a>

where ((·, ·))1,0, L0 are defined by (8) and (10) with B replaced by B0, respectively.

We decompose E as <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M73','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M73">View MathML</a>, where <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M74','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M74">View MathML</a>, <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M75','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M75">View MathML</a> is the positive (resp. negative) definite eigenspace of I - L0. Let <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M76','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M76">View MathML</a>. Then Z is a finite dimensional space. Inequality (12) implies that there exists δ5 > 0 such that, for each y Z,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M77','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M77">View MathML</a>

whenever y Z. By (5), there exists ρ > 0 such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M78','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M78">View MathML</a>

for y ∈ ℝn with |y| ≤ ρ. Hence, by the mean value theorem, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M79','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M79">View MathML</a>

(15)

whenever |y| ≤ ρ. Consequently, if y Z and 0 < |y| < ρ, we get

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M80','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M80">View MathML</a>

If J(y) = 0, then <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M81','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M81">View MathML</a>, which implies that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M82','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M82">View MathML</a> for a.e. t ∈ [0, 2π]. Thus y ∈ ℝn Z = {0}, which contradicts with 0 < |y| < ρ. Thus J(y) < 0.

Proof of Theorem 4.1. Now, we apply Lemma 2.2 to prove Theorem 4.1.

Define the action μ : E E by μx = -x. Then μ is a generator of group Z2 and Fixμ = {0}. Obviously, (F1) and (F4) hold. It is easy to check that J is μ-invariant. Lemma 4.1 implies that J satisfies (PS)-condition.

Let Y, Z define as in Lemma 4.2 and 4.3. Then Y, Z are μ-invariant subspaces and (A2) implies that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M83','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M83">View MathML</a>

It follows from Lemmas 4.2 and 4.3 that (F2) and (F3) hold. Applying Lemma 2.2, J has at least i(A0) - i(A) pairs of distinct critical points with critical values less than or equal to c. Since J(0) = 0 and E ∩ ℝn = {0}, if follows that all those i(A0) - i(A) pairs of distinct critical points are nonconstant.

If y is a critical point of J, then <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M84','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M84">View MathML</a> is a 2π-periodic solution of (1). Clearly, y ≠ 0 implies that x ≠ 0. Let x1, x2 are two periodic solutions satisfying x1 = -x2. Setting <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M85','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M85">View MathML</a> and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M86','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M86">View MathML</a>, then

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M87','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/44/mathml/M87">View MathML</a>

Integrating he above equality, we have y1 = - y2 + c6 for some c6 ∈ ℝn. Since y1, y2 E, it follows that c6 = 0. Then y1, y2 belongs the same Z2-orbit. Thus (1) has i(A0) - i(A) pairs of nontrivial periodic solutions.

Corollary 4.1. Under the hypotheses of Theorem 4.1, (1) has at least 2[i(A0) - i(A)] nontrivial 2π-periodic solutions.

Since every Z2-orbit has two elements and those two elements are different from each other, this corollary is obvious.

Competing interests

The authors declare that they have no competing interests.

Authors' contributions

The manuscript was drafted by HX and it is based on his PhD thesis. JY and ZG were the supervisors of the thesis and gave detailed comments on the manuscript. All authors read and approved the final manuscript.

Acknowledgements

This project was supported by the National Natural Science Foundation of China (Nos. 11031002 and 10871053). The authors are very grateful to the anonymous referee whose careful reading of the manuscript and valuable comments enhanced presentation of the manuscript.

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