Open Access

Periodic solutions of second-order differential equations with multiple delays

Advances in Difference Equations20122012:43

DOI: 10.1186/1687-1847-2012-43

Received: 14 October 2011

Accepted: 11 April 2012

Published: 11 April 2012

Abstract

By using the critical point theory and S1 index theory, we obtain a new result for the existence and multiplicity of periodic solutions for a class of second-order delay differential equations x" (t) = f (x(t))-[f (x(t- 1))+f (x(t- 2))+...+f (x(t-(N -1)))].

Keywords

delay differential equations multiple periodic so1utiois critical point theory index theory

1 Introduction

Inspired by the excellent study in [1], many authors [216] studied the following differential delay equations
x ( t ) = - f x ( t - 1 ) + f x ( t - 2 ) + + f x t - ( N - 1 ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equ1_HTML.gif
(1.1)

where f C (, ),and N ≥ 2 is an integer.

Kaplan and Yorke [2] introduced a technique of couple system which allows them to reduce the search for periodic solutions of a differential delay equation to the problem of finding periodic solutions for a related system of ordinary differential equations. They study periodic solutions of (1.1) with N = 2, f C (, ) is odd, xf (x) > 0 for x ≠ 0 and f satisfies some suitable conditions near 0 and ∞. More precisely, if the solution x(t) of (1.1) with N = 2 satisfies x(t) = -x(t-2), let
x 1 ( t ) = x ( t ) , x 2 t = x ( t - 1 ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equ2_HTML.gif
(1.2)
then X(t) = (x1(t), x2(t)) T satisfies
X t = A 2 H ( X ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equ3_HTML.gif
(1.3)

where, A 2 = 0 - 1 1 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_IEq1_HTML.gif

i.e., A2 is a skew symmetric matrix, and
H ( X ) = 0 x 1 f ( s ) d s + 0 x 2 f ( s ) d s . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equ4_HTML.gif
(1.4)

H(X) is the gradient of H.

In fact, by direct computation, one has the following proposition:

Proposition 1.1. (i) Any solution x(t) of (1.1) with N = 2, and x(t) = -x(t - 2) will give a solution of (1.3) X(t) = (x1(t), x2(t)) T by (1.2). Moreover, X(t) has the following symmetric structure
x 1 ( t ) = - x 2 t - 1 , x 2 t = x 1 ( t - 1 ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equ5_HTML.gif
(1.5)

(ii) Any solution X(t) = (x1(t), x2(t)) T of (1.3) with the symmetric structure (1.5) will give a solution of (1.1) by letting x(t) = x1 (t). Moreover, x(t + 2) = -x(t).

Kaplan and Yorke proved that (1.3) has periodic solutions with the symmetric structure (1.5), which give the Kaplan-Yorke type periodic solutions of (1.1) with period 4, i.e., x(t) satisfying x(t) = -x(t - 2). They further conjectured that similar result should be true for the general case N ≥ 2, i.e., under similar conditions for f, (1.1) has a 2N-periodic Kaplan-Yorke type periodic solution x(t), i.e., x(t) satisfying x(t) = -x(t - N).

Li and He [68], in an attempt to reuse Kaplan and Yorke's original idea, applied Lyapunov Center Theorem and some known results about convex Hamiltonian systems [17, Theorem 7.2] to obtain 4-periodic solutions of (1.4). But those 4-periodic solutions obtained by [17, Theorem 7.2] give no information about the symmetric structure (1.5) or the minimal period. The solutions of (1.3), which will not generate noncontact solutions of (1.1), see [14, Remark 3.3].

Herz [12] study (1.1) with N = 2 by Lyapunov direct method. And, Jekel and Johnston [13], proved the existence of a 2N-periodic Kaplan-Yorke type periodic solution for (1.1) by Kaplan-Yorke original method and homotopic method.

Fei [14, 15] applied the pseudo-index theory [1720] to obtain periodic solution in a subspace, which surely have the required symmetric structure (1.5) and give solutions to (1.1).

In recent years, Guo and Yu [16] considered (1.1) with N = 2 by variational methods directly, and they obtain the Kaplan-Yorke type periodic solutions. That is to say that they do not necessarily transform the existence problem of (1.1) to existence problems for related systems (1.3). Afterwards, Cheng and Hu [21] studied (1.1) with N = 2 by Guo-Yu's method in [16]. Guo [22] studied the following second-order differential delay equation by Guo-Yu's method in [16]
x t = - f x t - r , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equ6_HTML.gif
(1.6)

they obtained the multiplicity results for periodic solutions, but the solutions are not Kaplan- Yorke type.

The authors [23, 24] considered the Kaplan-Yorke type periodic solutions of (1.1) with N = 2 by Maslov-type index [25] and Morse theory [26], respectively.

Recently, some researchers [2731] have begun to study the existence of solutions for second-order differential delay equation by using a variational method. However, to the best of authors' knowledge, the study of Kaplan-Yorke type periodic solutions of second- order differential delay equation using a variational method has received considerably less attention. We find the method apply in [615], such as the structure of variational does not directly apply to second-order differential delay equation.

Motivated by the study in [716, 2124, 2732], in this article we are concerned with the existence of Kaplan-Yorke type periodic solutions of the following second-order differential delay equations
x t = f x t - f x t - 1 + f x t - 2 + + f x t - N - 1 , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equ7_HTML.gif
(1.7)

where f C (,) is odd, and N ≥ 2 is an integer

If x(t) = x(t+N), let
x 1 t = x ( t ) , x 2 t = x ( t - 1 ) , . . . , x N t = x t - N - 1 , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equ8_HTML.gif
(1.8)
then z(t) = (x1(t), x2(t),..., x N (t)) T satisfies
z ( t ) = A N H ( z ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equ9_HTML.gif
(1.9)

where A N = 1 - 1 - 1 - 1 1 - 1 - 1 - 1 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_IEq2_HTML.gif

i.e., A N is a n × n symmetric matrix, and
H z = 0 x 1 f s d s + 0 x 2 f s d s + + 0 x N f s d s . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equ10_HTML.gif
(1.10)

In fact, by direct computation, one has the following proposition.

Proposition 1.2. (i) Any solution x(t) of (1.7) with x(t) = x(t - N) will give a solution of (1.9) X(t) = (x1(t), x2 (t), . . . , x N (t)) T by (1.8). Moreover, X(t) has the following symmetric structure
x 1 ( t ) = x N t - 1 , x 2 t = x 1 ( t - 1 ) , x 3 t = x 2 ( t - 1 ) , . . . , x N t = x N - 1 ( t - 1 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equ11_HTML.gif
(1.11)

(ii) Any solution X(t) = (x1(t), x2(t), . . . , x N (t)) T of (1.9) with the symmetric structure (1.11) will give a solution of (1.7) by letting x(t) = x1(t). Moreover, x(t + N) = x(t).

Throughout this article, we always assume that:

(f 1) f C (, ) is odd and 0 , β < +
lim x 0 f ( x ) x = α , lim x f ( x ) x = β ; https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equa_HTML.gif

(f 2±) |f (x) - βx| is bounded and G β (x) → ± ∞, as |x|;

(f 3±) ±Gα (x) > 0 for |x| > 0 being small,

where F x = 0 x f s d s https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_IEq3_HTML.gif, and
G β x = F x - 1 2 β x 2 , G α ( x ) = F x - 1 2 α x 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equ12_HTML.gif
(1.12)
Similarly to the argument in [14, 20, 33], for given a number α , and N ≥ 2, k ≥ 1 being two integer, we set:
i α , N = numbers of elements of  k 2 π N 2 k - 1 2 - 2 α < 0 , 2 k - 1  mod  N 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equb_HTML.gif
and
v α , N = numbers of elements of  k 2 π N 2 k - 1 2 - 2 α = 0 , 2 k - 1  mod  N 0 . . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equc_HTML.gif

For convenience, denote #(1.7) = the number of geometrically different nonconstant periodic solutions of (1.7) which satisfy x(t - N/2) = -x(t), t R.

Our main result reads as:

Theorem 1.1. Suppose f satisfies (f 1) and N ≥ 2 being an integer in (1.7). We have the following conclusions:

(i) #(1.7) ≥ i(α, N) - i(β, N) provided v(β, N) = 0 or (f 2-) holds.

(ii) #(1.7) ≥ i(α, N) - i(β, N) + v(α, N) provided (f 3+)holds and either v(β, N) = 0 or (f 2-) holds.

(iii) #(1.7) ≥ i(β, N) + v(β, N) - i(α, N) provided (f 3-) and (f 2+) holds.

2 Variational structure

For S1 = /(N ),let E = H1(S1, N ). Then E is a Hilbert space with norm ||·|| and inner product〈,〉, and E consists of those z(t) in L2(S1, N ) whose Fourier series
z t = a 0 + m = 1 a m cos 2 π N m t + b m sin 2 π N m t , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equd_HTML.gif
satisfies
z 2 = N a 0 2 + N 2 m = 1 1 + β m 2 a m 2 + b m 2 < , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Eque_HTML.gif

where a m , b m N and β m = 2 π m N https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_IEq4_HTML.gif.

We can define an operator
L 0 z , y = 0 N A N - 1 ż , d t , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equ13_HTML.gif
(2.1)
on E. By direct computation, L0 is a bounded self-adjoint linear operator on E and
L 0 z t = m = 1 β m 2 1 + β m 2 A N - 1 a m cos β m t + b m sin β m t . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equ14_HTML.gif
(2.2)
By (f1), one can show that H(z) C1 (, ) and satisfies
H z d 1 z 2 + d 2 , z , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equf_HTML.gif
where d1, d2 > 0. By using similar arguments as in [14, 21], we know that
φ z = 1 2 L 0 z , z - 0 N H z d t C 1 E , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equ15_HTML.gif
(2.3)

and critical points of φ in E are classic solutions of (1.9).

Let T N be the N × N matrix given by
T N = 0 0 0 1 1 0 0 0 0 1 0 0 0 0 1 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equg_HTML.gif
For z(t) E, define
δ z ( t ) = T N z ( t - 1 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equ16_HTML.gif
(2.4)

Then we have δ N z(t) = z(t - N), and G = {δ, δ2, . . . , δ N } is a compact group action over E. Moreover, if δz(t) = z(t) holds, z(t) has the symmetric structure (1.11).

Lemma 2.1. Denote S E = { z E : δ z ( t ) = z ( t ) , z ( t - N 2 ) = - z ( t ) } https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_IEq5_HTML.gif Then we have
S E = z t = m = 1 a m cos 2 π N 2 m - 1 t + b m sin 2 π N 2 m - 1 t : a m b m span u m w m , - w m u m , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equ17_HTML.gif
(2.5)
where θ m = 2 π N 2 m - 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_IEq6_HTML.gif and
u m = ( 1 , cos θ m , . . . , cos ( N - 1 ) θ m ) T , w m = ( 0 , sin θ m , . . . , sin ( N - 1 ) θ m ) T . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equ18_HTML.gif
(2.6)
Proof. For any, z t = a 0 + m = 1 a m cos 2 π N m t + b m sin 2 π N m t S E https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_IEq7_HTML.gif, we must have z(t-N/2) = -z(t), which implies that
a 0 = - a 0 , a m = ( - 1 ) m + 1 a m , b m = ( - 1 ) m + 1 b m ,  i . e . , a m = b m = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equh_HTML.gif

for even m.

this means that for any z SE,
z t = m = 1 a m cos 2 π N 2 m - 1 t + b m sin 2 π N 2 m - 1 t . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equi_HTML.gif
Note that δz(t) = z(t) T N z(t - 1) = z(t), i.e.,
m = 1 a m cos 2 π N 2 m - 1 t + b m sin 2 π N 2 m - 1 t = m = 1 T N a m cos 2 π N 2 m - 1 t - 1 + T N b m sin 2 π N 2 m - 1 t - 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equj_HTML.gif
This implies that for k ≥ 1
a m = T N a m cos θ m - T N b m sin θ m b m = T N a m sin θ m + T N b m cos θ m https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equ19_HTML.gif
(2.7)
If we introduce complex vector C m = a m + ib m , the above (2.7) becomes
C m = e i θ m T N C m . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equk_HTML.gif
Note that
det ( T N - λ I N ) = λ N - 1 , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equl_HTML.gif
the eigenvalues of T N are
λ = e - i 2 j π N , j = 1 , 2 , 3 , , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equm_HTML.gif
Equation (2.7) implies that C m must be the eigenvector associated with the eigenvalue
λ = e - i θ m , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equn_HTML.gif
We obtain that
C m = ( 1 , e i θ m , , e i ( N - 1 ) θ m ) T , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equ20_HTML.gif
(2.8)
{ u m = R e ( C m ) = ( 1 , cos θ m , cos 2 θ m , , cos ( N 1 ) θ m ) T , w m = I m ( C m ) = ( 0 , sin θ m , sin 2 θ m , , sin ( N 1 ) θ m ) T . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equ21_HTML.gif
(2.9)

By direct computation, iC m is also the eigenvector associated with the eigenvalue λ = e - i θ m https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_IEq8_HTML.gif.

Therefore, we have the conclusion. The proof is complete.   □

Lemma 2.2. Let φ be given in (2.3) and φ | SE be the restriction of φ on SE. Then critical points of φ | SE over SE are critical points of φ over E.

Proof. By (1.9) and direct computation, we have
A N T N = T N A N , H ( T N z ) = H ( z ) , H ( T N z ) = T N H ( z ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equo_HTML.gif
Combining these with (2.3) and the fact that any z(t) E is N-periodic, one can easily verify that
φ ( δ z ) = φ ( z ) , φ ( δ z ) = δ φ ( z ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equp_HTML.gif

i.e., φ is G-invariant, and φ', is G-equivariant. The conclusion follows directly.   □

For any α , define an operator L α by extending the bilinear form
L α z , y = L 0 z , y - 0 N α z , y d t , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equ22_HTML.gif
(2.10)
on E. Moreover, L α is G-equivariant. By direct computation, L α is a bounded self-adjoint linear operator on E and if z(t) SE
L α z ( t ) = m = 1 1 1 + β m 2 ( β m 2 A N - 1 - α I ) ( a m cos β m t + b m sin β m t ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equ23_HTML.gif
(2.11)
For m ≥ 1, denote
S E ( m ) = z ( t ) = a m cos 2 π N ( 2 m - 1 ) t + b m sin 2 π N ( 2 m - 1 ) t : a m b m satisfies ( 2 . 5 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equq_HTML.gif

Then S E = j = 1 S E ( j ) https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_IEq9_HTML.gif. Since A N = I N - ( T N + T N 2 + + T N N - 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_IEq10_HTML.gif, and A N T = A N https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_IEq11_HTML.gif, so SE(m) is the eigen-subspace of L0 corresponding to eigenvalue λ m , here λ m = 2 - N, as (2m - 1 mod N) = 0; λ m = 2, as (2m - 1 mod N) ≠ 0. Denote by M -(·), M +(·) and M0(·) the positive definite, negative definite and null subspaces of the self-adjoint linear operator defining it, respectively.

Lemma 2.3. For k ≥ 1, γ k α = ( 2 π N ( 2 k - 1 ) ) 2 λ k - 1 - α https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_IEq12_HTML.gif. Then L α , as an operator on SE, has the following properties on SE.
M - ( L α ) = k = 1 , γ k α < 0 S E ( k ) , M + ( L α ) = k = 1 , γ k α > 0 S E ( k ) , M 0 ( L α ) = k = 1 , γ k α = 0 S E ( k ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equ24_HTML.gif
(2.12)
Proof. For m ≥ 1 and z m = a m cos 2 π N ( 2 m - 1 ) t + b m sin 2 π N ( 2 m - 1 ) t https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_IEq13_HTML.gif, consider the eigenvalue problem
L α z m = λ m α z m . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equr_HTML.gif
By (2.7) and (2.11), we have
1 1 + θ m 2 ( θ m 2 A N - 1 - α I ) a m = λ m α a m , 1 1 + θ m 2 ( θ m 2 A N - 1 - α I ) b m = λ m α b m . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equs_HTML.gif

Since a m , b m are the eigenvector of A N corresponding to eigenvalue λ m , then λ m α = ( 1 + θ m 2 ) - 1 ( θ m 2 λ m - 1 - α ) https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_IEq14_HTML.gif. Therefore L α is positive definite, negative definite, or null on SE(k) if and only if γ k α = 2 π N ( 2 k - 1 ) 2 λ k - 1 - α https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_IEq15_HTML.gif is positive, negative, or zero, respectively.

This implies (2.12) directly.   □

For S1 = /(N ), there is a natural S1-action over SE, defined by
T ( θ ) z ( t ) = z ( t + θ ) , θ S 1 , z S E . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equt_HTML.gif
It is easy to see that φ is S1-invariant, φ' is S1-equivariant, and
Fix ( S 1 ) = { u S E : T ( θ ) u = u , θ S 1 } = { 0 } . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equu_HTML.gif

By directly applying [20, Theorem 2.4] to φ over SE, we have the following lemma.

Lemma 2.4. Assume there exist two closed S1-invariant linear subspaces, SE+ and SE-, of SE and r > 0 such that (a) (SE+ + SE-) is closed and of finite codimension in SE,

(b) L(SE - ) SE-with L = L α or L = L β ,

(c) there exist c0 , c0 > -∞ such that
inf z S E + φ ( z ) = c 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equv_HTML.gif
(d) there exists c such that
φ ( z ) c < φ ( 0 ) , z ( S E - S r ) = { z S E - : z = r } , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equw_HTML.gif

(e) φ satisfies (PS)c condition for c0cc , i.e., every sequence {z m } SE with φ (z m ) → c and φ'(z m ) → 0 possesses a convergent subsequence. Then φ possesses at least 1 2 [ dim ( S E - S E + ) - codi m S E ( S E - + S E + ) ] https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_IEq16_HTML.gif geometrically different critical orbits in φ-1([c0, c ]).

3 Proof of main results

Proof of Theorem 1.1 As we already proved in Lemma 2.2, critical points of φ over SE are critical points of φ over E. Hence they are nonconstant classic N -periodic solutions of (1.7) with the symmetric structure (1.11). By Proposition 1.2, they give solutions of (1.7) with the property x(t - N/2) = -x(t). Therefore, we can seek critical points of φ on SE directly.

Set:
ψ β ( z ) = 0 N H ( z ) - 1 2 β z , z d t , ψ α ( z ) = 0 N H ( z ) - 1 2 α z , z d t https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equx_HTML.gif
Then
φ ( z ) = 1 2 L α z , z - ψ α ( z ) , φ ( z ) = 1 2 L β z , z - ψ β ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equy_HTML.gif

Case (i): If i(α, N ) > i(β, N ). We shall carry out the proof in several steps.

Step 1: let
ψ α ( 0 ) = 0 , ψ α ( z ) z 0 , a s z 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equz_HTML.gif
Then Lemma 2.4(a)and (b) hold with L = L α . By (f1), using the same argument as [18, Lemma 5.5], it is easy to show that
ψ α ( 0 ) = 0 , ψ α ( z ) z 0 , a s z 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equaa_HTML.gif
We denote
λ - = min { | γ m α | ( 1 + θ m 2 ) - 1 γ m α < 0 , m } , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equab_HTML.gif
and
λ + = min { ( 1 + θ m 2 ) - 1 γ m β γ m β > 0 , m } . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equac_HTML.gif
Since ψ α (0) = 0, | | ψ α ( z ) | | | | z | | 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_IEq17_HTML.gif, as ||z|| → 0, for ε = λ - 4 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_IEq18_HTML.gif, there exists a constant r > 0 such that for any z H1 (S1, N )
ψ α ( z ) λ - 4 z , a s z r . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equad_HTML.gif
Furthermore,
ψ α ( z ) = ψ α ( z ) - ψ α ( 0 ) = 0 1 ψ α ( τ z ) τ d τ 0 1 ψ α ( τ z ) , z d τ λ - 4 z 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equae_HTML.gif
Therefore, we have for z (E-S r ) = {z E-| ||z|| = r}
φ ( z ) = 1 2 L α z , z - ψ α ( z ) - λ - 2 z 2 + λ - 4 z 2 = - λ - 4 z 2 = - λ - 4 r 2 < 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equaf_HTML.gif

Thus (d) of Lemma 2.4 holds.

If M 0 ()= {0}, by (f1), using the same argument as [18, Lemma 5.5], it is easy to show that
ψ α ( z ) z 0 , a s z , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equag_HTML.gif
it follows that given ε = λ + 2 > 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_IEq19_HTML.gif there exists r > 0, such that
ψ α ( z ) λ + 2 z , z > r . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equah_HTML.gif
Moreover, there is d1 > 0 such that
ψ α ( z ) d 1 , z r , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equai_HTML.gif
thus
ψ α ( z ) λ + 2 z + d 1 , z H 1 ( S 1 , N ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equaj_HTML.gif
Using the above formula we get
| H ( z ) - β z | λ + 2 | z | + d 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equak_HTML.gif
Since for any z SE + = M+()M0 ()
ψ β ( z ) = ψ β ( z ) - ψ β ( 0 ) = 0 1 ψ β ( τ z ) τ d τ = 0 1 ψ β ( τ z ) , z d τ 0 1 λ + τ 2 z 2 + d 1 z d τ λ + 4 z 2 + d 1 z . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equal_HTML.gif
Then for any z SE+ = M+() M 0()= M + () {0}
φ ( z ) = 1 2 L β z , z - ψ β ( z ) λ + 2 z 2 - λ + 4 z 2 - d 1 z = λ + 4 z 2 - d 1 z . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equam_HTML.gif

Thus there exists c0 > -∞ such that (c) of Lemma 2.4 holds.

If M0(L β ) ≠ {0}, by (f 2-) and (1.12), for any M > 0, there exists a constant d2 > 0 such that
G β ( z ) - d 2 , | z | > M . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equ25_HTML.gif
(3.1)
Let
Ω 1 = { t [ 0 , N ] | | z | > M } https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equ26_HTML.gif
(3.2)
and
Ω 2 = { t [ 0 , N ] | | z | M } . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equ27_HTML.gif
(3.3)
Since F (z) C2 ( N , ), by (3.3), there exists a constant d3 > 0 such that
Ω 2 [ G β ( z ( t ) ) , z ( t ) ] d t d 3 , | z | M . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equ28_HTML.gif
(3.4)
By (3.1) - (3.4), we have
ψ β ( z ) = 0 N [ H ( z ) - ( 1 2 β z , z ) ] d t d 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equan_HTML.gif
Then we have for z = z+ + z0 M+(L β ) M 0 (L β )
φ ( z ) = 1 2 L β z , z - ψ β ( z ) = 1 2 L β z + , z + - ψ β ( z ) - d 3 . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equao_HTML.gif

Thus φ is bounded from below on E+. Therefore, condition (c) of Lemma 2.4 is verified.

Step 2: Using the same argument as [22, 25] and [26, Lemma 4.2], one can prove that φ satisfies (PS)c condition for any c under the condition either (f 1) with v(β, N) = 0 holds or (f 2±) holds.

Step 3: By Lemma (2.4), φ has at least σ = 1 2 [ dim ( S E - S E + ) - codi m S E ( S E - + S E + ) ] https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_IEq20_HTML.gif geometrically different critical orbits in φ-1 ([c0, c]).

Now by Lemma (2.3), it is easy to show that
σ = 1 2 [ dim ( S E - S E + ) - codi m S E ( S E - + S E + ) ] = 1 2 [ dim M - ( L α ) - dim M - ( L β ) ] = 1 2 [ 2 i ( α , N ) - 2 i ( β , N ) ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equap_HTML.gif

This means #(1.7) ≥ i(α, N)-i(β, N), and this completes the proof of case (i).

For case (ii), (iii), using the same idea and similar arguments, one can show that the conclusions hold. We omit the details. The proof is complete.   □

Example. Consider the following equation
x ( t ) = f ( x ( t ) ) - f ( x ( t - 1 ) ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equ29_HTML.gif
(3.5)
and its coupled system
x ( t ) = f ( x ) - f ( y ) , y ( t ) = - f ( x ) + f ( y ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equ30_HTML.gif
(3.6)
with f C (, ) being odd and
f ( x ) = α x + α 0 x 1 3 , f o r | x | 100 f ( x ) = β x + β 0 x 3 , f o r | x | 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equaq_HTML.gif
Here N = 2, and by Definition
i ( α , 2 ) = { k ( π ( 2 k - 1 ) ) 2 - 2 α < 0 , k = 1 , 2 , } , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equar_HTML.gif
and
v ( α , 2 ) = { k ( π ( 2 k - 1 ) ) 2 - 2 α = 0 , k = 1 , 2 , } . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equas_HTML.gif
Let α = 16, β = 1. Then it is easy to see that
i ( 16 , 2 ) = 2 , v ( 16 , 2 ) = 0 , i ( 1 , 2 ) = 0 , v ( 1 , 2 ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2012-43/MediaObjects/13662_2011_Article_158_Equat_HTML.gif

By Theorem 1.1, Equation (3.5) has at least i(16, 2) - i(1, 2) = 2 geometrically different nonconstant 2-periodic solutions which satisfy x(t - 1) = -x(t).

Declarations

Acknowledgements

The authors thank the referees for valuable comments and suggestions which improved the presentation of this manuscript. This study was partially supported by NFSC (10871206, 10961011, 60964006, 11161013) Guangxi Natural Science Foundation (2011GXNSFA018134).

Authors’ Affiliations

(1)
School of Mathematical Science and Computing Technology, Central South University
(2)
School of Mathematics and Computing Science, Guilin University of Electronic Technology
(3)
School of Science, Shandong University of Technology

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