Positive Periodic Solution of Second-Order Coupled Systems with Singularities

Abstract

This paper establishes the existence of periodic solution for a kind of second-order singular nonautonomous coupled systems. Our approach is based on fixed point theorem in cones. Examples are given to illustrate the main result.

(ProQuest: ... denotes non-US-ASCII text omitted.)

Academic Editor:Chuanzhi Bai

1, School of Mathematical Sciences, Capital Normal University, Beijing 100048, China

Received 1 March 2013; Revised 11 June 2013; Accepted 13 June 2013

This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

1. Introduction

We are concerned with the existence of positive T -periodic solution for the second-order nonautonomous singular coupled systems: [figure omitted; refer to PDF] where _{f i} ∈ C ( ( 0 , T ) × [ 0 , + ∞ ) , [ 0 , + ∞ ) ) , _{c i} : ( 0 , T ) [arrow right] [ 0 , + ∞ ) , i = 1,2 , are Lebesgue integrable, _{f i} may be singular at t = 0 , T and _{c i} can have finitely many singularities.

Singular differential equations or systems arise from many branches of applied mathematics and physics such as gas dynamics, Newtonian fluid mechanics, and nuclear physics, which have been widely studied by many authors (see [1-7] and references therein). Some classical-tools have been used to study the positive solutions for two point nonperiodic boundary value problems of coupled systems [8, 9]. However, there are few works on periodic solutions of second order nonautonomous singular coupled systems of type (1).

In the recent papers [10, 11], the periodic solutions of singular coupled systems [figure omitted; refer to PDF] were proved by using some fixed point theorems in cones for completely continuous operators, where _{a 1} , _{a 2} , _{c 1} , _{c 2} ∈ ^{L 1} ( 0 , T ) , _{f 1} , _{f 2} ∈ Car ( [ 0 , T ] × ( 0 , + ∞ ) , ( 0 , + ∞ ) ) . When the Green's function _{G i} ( t , s ) ( i = 1 , 2 ), associated with the periodic boundary problem [figure omitted; refer to PDF] is nonnegative for all ( t , s ) ∈ [ 0 , T ] × [ 0 , T ] and _{f i} ( i = 1,2 ) satisfies weak singularities [figure omitted; refer to PDF] where 0 < _{α i} < 1 , _{b ^ i} , _{b i} ∈ ^{L 1} ( 0 , T ) , _{b ^ i} ...5; 0 , _{b i} ...5; 0 , and _{b ^ i} , _{b i} are strictly positive on some positive measure subsets of ( 0 , T ) , some sufficient conditions for the existence of periodic solutions of (2) were obtained in [10, 11].

Motivated by the papers [9-11], we consider the existence of positive T -periodic solution of (1). Owing to the disappearing of the terms _{a i} ( t ) ( i = 1,2 ) in (2), the methods in [9, 10] are no longer valid. In present paper, we will deal with the periodic solutions of (1) under new conditions. Let k be a constant satisfying 0 < k < π / T . Denote by G ( t , s ) the Green function of [figure omitted; refer to PDF] which can be expressed by [figure omitted; refer to PDF] By a direct computation, we can get [figure omitted; refer to PDF] Assume _{f i} ( i = 1,2 ) satisfies the following conditions.

(H₁ ): For t ∈ ( 0 , T ) , _{f i} ( t , 1 ) > 0 and there exist constants _{μ 1} ...5; _{μ 2} > 1 , _{λ 1} ...5; _{λ 2} > 1 such that, for any constants 0 ...4; _{ρ i} ...4; 1 , i = 1,2 , [figure omitted; refer to PDF]

(H₂ ): ^{∫ 0 T} ... ^{c 1 -} ( t ) d t = _{r 1} > 0 , ^{∫ 0 T} ... ^{c 2 -} ( t ) d t = _{r 2} > 0 and [figure omitted; refer to PDF]

where ^{c i +} ( t ) = max { _{c i} ( t ) , 0 } , ^{c i -} ( t ) = max { - _{c i} ( t ) , 0 } , i = 1 , 2 .

Remark 1.

For any _{ρ 1} ...5; 1 , we get from (8) that [figure omitted; refer to PDF]

For any _{ρ 2} ...5; 1 , we get from (9) that [figure omitted; refer to PDF]

Typical functions that satisfy (8) or (9) are those taking the form [figure omitted; refer to PDF] where _{a i k} ∈ C ( 0 , T ) , _{a i k} ( t ) > 0 , _{l i k} > 1 , i = 1,2 ; k = 1,2 , ... , n .

Definition 2.

Supposing that ( u , v ) ∈ ^{C 1} [ 0 , T ] ∩ ^{C 2} ( 0 , T ) × ^{C 1} [ 0 , T ] ∩ ^{C 2} ( 0 , T ) satisfies (1) and u ( t ) > 0 , v ( t ) > 0 for any t ∈ [ 0 , T ] , then one says that ( u , v ) is a ^{C 1} [ 0 , T ] × ^{C 1} [ 0 , T ] positive solution of system (1).

By using fixed point theorem in cones, we are able to prove the following result.

Theorem 3.

Assume that (H₁ ), (H₂ ) hold. Then (1) has at least one positive T -periodic solution.

The proof of Theorem 3 will be given in Section 3 of this paper.

2. Preliminaries

Lemma 4 (see [12]).

Let X be a Banach space, K a cone in X , _{Ω 1} , _{Ω 2} two nonempty bounded open sets in K , θ ∈ _{Ω 1} ⊂ _{Ω 1} ¯ ⊂ _{Ω 2} . T : _{Ω 2} ¯ / _{Ω 1} [arrow right] K is a completely continuous operator. If

(i) T ( x ) ...0; λ x , x ∈ ∂ _{Ω 1} , λ > 1 ,

(ii) T ( x ) ...0; λ x , x ∈ ∂ _{Ω 2} , 0 < λ < 1 , _{inf x ∈ ∂ Ω 2} || T x || > 0 ,

then T has a fixed point in _{Ω 2} ¯ / _{Ω 1} .

Lemma 5.

If _{f i} ( t , u ) satisfy (H₁ ), then, for t ∈ ( 0 , T ) , _{f i} ( t , u ) are increasing on u and, for any [ α , β ] ⊂ ( 0 , T ) , [figure omitted; refer to PDF] uniformly with respect to t ∈ [ α , β ] .

Proof.

We only deal with _{f 1} . Without loss of generality, let 0 ...4; x ...4; y . If y = 0 , we get _{f 1} ( t , x ) ...4; _{f 1} ( t , y ) . If y ...0; 0 , let _{c 0} = x / y , then 0 ...4; _{c 0} ...4; 1 . From (8), we get [figure omitted; refer to PDF] which means _{f 1} ( t , u ) is increasing on u .

Assume u > 1 . It follows from (11) that f 1 ( t , u ) ...5; u μ 2 f 1 ( t , 1 ) . Thus [figure omitted; refer to PDF] From (H1 ), for any [ α , β ] ⊂ ( 0 , T ) , we obtain [figure omitted; refer to PDF] Therefore [figure omitted; refer to PDF] uniformly with respect to t ∈ [ α , β ] .

Let X = C [ 0 , T ] . We know X is a Banach space with the norm || u || = _{max t ∈ [ 0 , T ]} | u ( t ) | . Define the sets [figure omitted; refer to PDF] It is easy to check that P , Q are cones in X and Q ⊂ P . Throughout this paper, we consider the space X × X . It is easy to see X × X is a Banach space with the norm, [figure omitted; refer to PDF]

We can get the conclusion that P × P , Q × Q are cones in X × X and Q × Q ⊂ P × P .

For any u ∈ X , define the function [figure omitted; refer to PDF] Then the solution of periodic boundary value problem [figure omitted; refer to PDF] can be expressed by _{x ~ 1} ( t ) = - ^{∫ 0 T} ... G ( t , s ) ^{c 1 -} ( s ) d s , and the solution of periodic boundary value problem [figure omitted; refer to PDF] can be expressed by _{x ~ 2} ( t ) = - ^{∫ 0 T} ... G ( t , s ) ^{c 2 -} ( s ) d s . Obviously, _{x ~ i} ( t ) < 0 , i = 1 , 2 . Then [figure omitted; refer to PDF] From (11), (12), and Lemma 5, we have [figure omitted; refer to PDF] Then, for any fixed ( u , v ) ∈ P × P , it follows from ( _{H 2} ) that [figure omitted; refer to PDF]

Thus, we can define the operator T : P × P [arrow right] P × P , T ( u , v ) = ( _{T 1} u , _{T 2} v ) by [figure omitted; refer to PDF] for ( u , v ) ∈ P × P . Then, we have the following lemma.

Lemma 6.

Assuming that (H₁ ), (H₂ ) hold, then T has a fixed point if and only if [figure omitted; refer to PDF] has one positive T -periodic solution.

Lemma 7.

Assuming that (H₁ ), (H₂ ) hold, then T ( Q × Q ) ⊂ Q × Q and T : Q × Q [arrow right] Q × Q is completely continuous.

Proof.

For ( u , v ) ∈ Q × Q , we have [figure omitted; refer to PDF] Then, we can get [figure omitted; refer to PDF] which means T ( Q × Q ) ⊂ Q × Q .

Let B ⊂ Q × Q be any bounded set. Then there exists a constant N such that, for any ( u , v ) ∈ B , [figure omitted; refer to PDF] From (27), we have [figure omitted; refer to PDF] Let [figure omitted; refer to PDF] Thus [figure omitted; refer to PDF] which implies that T ( B ) is bounded.

Next we prove that T ( B ) is equicontinuous. For any ( u , v ) ∈ B , t ∈ [ 0 , T ] , we know [figure omitted; refer to PDF] From (25) and (H₂ ), we have [figure omitted; refer to PDF]

Using the same method, we can obtain | ^{( _{T 2} v ) [variant prime]} ( t ) | < + ∞ . Therefore, T ( B ) is equicontinuous. According to Ascoli-Arzela theorem, T ( B ) is a relatively compact set.

Next, we prove that T : Q × Q [arrow right] Q × Q is continuous. Suppose ( _{u n} , _{v n} ) , ( _{u 0} , _{v 0} ) ∈ Q × Q , ( _{u n} , _{v n} ) [arrow right] ( _{u 0} , _{v 0} ) , n [arrow right] + ∞ , that is, _{u n} [arrow right] _{u 0} , _{v n} [arrow right] _{v 0} , n [arrow right] + ∞ . We know that there exists a constant L > 0 such that [figure omitted; refer to PDF]

We shall prove T ( _{u n} , _{v n} ) [arrow right] T ( _{u 0} , _{v 0} ) , n [arrow right] + ∞ , that is, [figure omitted; refer to PDF]

We first deal with _{T 1 u n} [arrow right] _{T 1 u 0} , n [arrow right] + ∞ . Otherwise, there exist _{[straight epsilon] 0} > 0 , { _{t n} } ∈ [ 0 , T ] such that | _{T 1 u n} ( _{t n} ) - _{T 1 u 0} ( _{t n} ) | ...5; _{[straight epsilon] 0} . Without loss of generality, we can assume _{t n} [arrow right] _{t 0} ∈ [ 0 , T ] . We know [figure omitted; refer to PDF]

Next, we show _{T 1 u n} ( _{t 0} ) - _{T 1 u 0} ( _{t 0} ) [arrow right] 0 , n [arrow right] ∞ . In fact, [figure omitted; refer to PDF] Let [figure omitted; refer to PDF] Since [figure omitted; refer to PDF] and _{f 1} is continuous, we know _{r n} ( s ) [arrow right] 0 , n [arrow right] ∞ . From (24), we know [figure omitted; refer to PDF] It can be inferred from (8) that [figure omitted; refer to PDF]

Set F ( s ) = 2 M ^{k 2} L + 2 M ^{( L + 1 ) _{μ 1}} ( _{f 1} ( s , 1 ) + ^{c 1 +} ( s ) ) . Thus, we get [figure omitted; refer to PDF] From (H₂ ), we know ^{∫ 0 T} ... F ( s ) d s < + ∞ . Using Lebesgue-dominated convergence theorem, we get [figure omitted; refer to PDF]

From (40) and (47), we obtain [figure omitted; refer to PDF] which is a contradiction. Thus, we know _{T 1 u n} [arrow right] _{T 1 u 0} , n [arrow right] + ∞ . Using the same method, we can obtain _{T 2 v n} [arrow right] _{T 2 v 0} , n [arrow right] + ∞ . Then [figure omitted; refer to PDF] which means T : Q × Q [arrow right] Q × Q is continuous. Therefore T : Q × Q [arrow right] Q × Q is a completely continuous operator.

3. Proof of Theorem 3

We proceed to prove Theorem 3 in two steps.

(1) Let _{Ω 1} = { ( u , v ) ∈ Q × Q : || u || < 2 ^{M 2} _{r 1} / m , || v || < 2 ^{M 2} _{r 2} / m } . We can get [figure omitted; refer to PDF] For ( u , v ) ∈ ∂ _{Ω 1} , we have the following two cases.

Case I.

One has { ( u , v ) ∈ Q × Q : || u || = 2 ^{M 2} _{r 1} / m , || v || ...4; 2 ^{M 2} _{r 2} / m } . Under this condition, we can get [figure omitted; refer to PDF] Otherwise, there exist ( _{u 0} , _{v 0} ) ∈ ∂ _{Ω 1} , _{λ 0} > 1 such that _{T 1 u 0} = _{λ 0 u 0} . As [figure omitted; refer to PDF] then [ _{u 0} ( t ) + _{x ~ 1} ( t ) ^{] *} = _{u 0} ( t ) + _{x ~ 1} ( t ) . By a direct computation, we know _{u 0} = ( 1 / _{λ 0} ) _{T 1 u 0} satisfies [figure omitted; refer to PDF] Then, we get [figure omitted; refer to PDF]

Since [ _{v 0} ( t ) + _{x ~ 2} ( t ) ^{] *} ...4; _{v 0} ( t ) < 2 ^{M 2} _{r 2} / m + 1 , integrating both sides of (54) on [ 0 , T ] , we get [figure omitted; refer to PDF] Then [figure omitted; refer to PDF] That is, [figure omitted; refer to PDF] which contradicts with (H₂ ).

Case II.

One has { ( u , v ) ∈ Q × Q : || u || ...4; 2 ^{M 2} _{r 1} / m , || v || = 2 ^{M 2} _{r 2} / m } . Under this condition, we can get [figure omitted; refer to PDF] Otherwise, there exist ( _{u 0} , _{v 0} ) ∈ ∂ _{Ω 1} , _{λ 0} > 1 such that _{T 2 v 0} = _{λ 0 v 0} . As [figure omitted; refer to PDF] then [ _{v 0} ( t ) + _{x ~ 2} ( t ) ^{] *} = _{v 0} ( t ) + _{x ~ 2} ( t ) . By a direct computation, we know _{v 0} = ( 1 / _{λ 0} ) _{T 2 v 0} satisfies [figure omitted; refer to PDF]

For [ _{u 0} ( t ) + _{x ~ 1} ( t ) ^{] *} ...4; _{u 0} ( t ) < 2 ^{M 2} _{r 1} / m + 1 , using the same method as condition I, we obtain [figure omitted; refer to PDF] which is also a contradiction.

(2) Choose an interval [ α , β ] ⊂ ( 0 , T ) satisfying β - α = T / 2 . Set M ~ > 4 M / T ^{m 2} . From Lemma 5, there exists _{R 1} > 2 ^{M 2} r / m , r = max { _{r 1} , _{r 2} } such that [figure omitted; refer to PDF]

Let R ...5; ( 2 M / m ) _{R 1} > _{R 1} > 2 ^{M 2} r / m . Define _{Ω 2} = { ( u , v ) ∈ Q × Q : || u || < R , || v || < R } . We can get [figure omitted; refer to PDF]

For ( u , v ) ∈ ∂ _{Ω 2} , we have the following two cases.

Case I.

One has { ( u , v ) ∈ Q × Q : || u || = R , || v || ...4; R } . Under this condition, we know [figure omitted; refer to PDF] Thus [ u ( t ) + _{x ~ 1} ( t ) ^{] *} = u ( t ) + _{x ~ 1} ( t ) . Furthermore, we obtain from (62) that [figure omitted; refer to PDF] On the other hand, for 0 < λ < 1 , we know λ v ( t ) < v ( t ) ...4; R . From the choice of M ~ , we get _{T 2} v ...0; λ v .

Case II.

One has { ( u , v ) ∈ Q × Q : || u || ...4; R , || v || = R } . Under this condition, we get [figure omitted; refer to PDF] Thus [ v ( t ) + _{x ~ 2} ( t ) ^{] *} = v ( t ) + _{x ~ 2} ( t ) . From (62), we get [figure omitted; refer to PDF] For 0 < λ < 1 , λ u ( t ) < u ( t ) ...4; R , from the choice of M ~ , we know _{T 1} u ...0; λ u .

Furthermore, we can obtain [figure omitted; refer to PDF] It implies _{inf ( u , v ) ∈ ∂ Ω 2} || T ( u , v ) _{|| 0} > 0 .

From Lemma 4, we know T has a fixed point ( u ~ , v ~ ) in _{Ω 2} ¯ / _{Ω 1} . For ( u ~ , v ~ ) ∈ _{Ω 2} ¯ / _{Ω 1} , we have the following three cases.

Case 1.

One has [figure omitted; refer to PDF]

Case 2.

One has [figure omitted; refer to PDF]

Case 3.

One has [figure omitted; refer to PDF]

Next, we show Cases 1 and 2 are impossible. In Case 1, we have [figure omitted; refer to PDF] It follows that [ u ~ ( t ) + _{x ~ 1} ( t ) ^{] *} = u ~ ( t ) + _{x ~ 1} ( t ) . By a direct computation, we know u ~ = _{T 1} u ~ satisfies [figure omitted; refer to PDF] Then, we get [figure omitted; refer to PDF] Since [ v ~ ( t ) + _{x ~ 2} ( t ) ^{] *} ...4; v ~ ( t ) < 2 ^{M 2} _{r 2} / m + 1 , integrating both sides of (74) on [ 0 , T ] , we get [figure omitted; refer to PDF] Then [figure omitted; refer to PDF] That is, [figure omitted; refer to PDF] which contradicts with (H₂ ). Using the same method, we can prove that Case 2 is also impossible. Therefore, Case 3 is satisfied and T has a fixed point ( u ~ , v ~ ) in _{Ω 2} ¯ / _{Ω 1} satisfying [figure omitted; refer to PDF] Since [figure omitted; refer to PDF] from Lemma 6, we know u ~ ( t ) , v ~ ( t ) satisfy [figure omitted; refer to PDF]

Let ( ^{u *} ( t ) , ^{v *} ( t ) ) = ( u ~ ( t ) + _{x ~ 1} ( t ) , v ~ ( t ) + _{x ~ 2} ( t ) ) . For [figure omitted; refer to PDF] we obtain [figure omitted; refer to PDF]

This means ( ^{u *} , ^{v *} ) is one positive T -periodic solution of (1).

4. Applications of Theorem 3

Finally, we give some examples as the applications of Theorem 3: [figure omitted; refer to PDF]

Choosing _{f 1} ( t , v ) = ^{k 2} T m ^{v 3 / 2} / ^{( 2 M 2 T / m + 1 ) 2} t , _{f 2} ( t , u ) = ^{k 2} T m ^{u 3 / 2} / ^{( 2 M 2 T / m + 1 ) 2} T - t , ^{c i -} ( t ) = 2 / t , ^{c i +} ( t ) ...1; 0 , _{r i} = ^{∫ 0 T} ... ^{c i -} ( t ) d t 4 T , i = 1,2 , and _{μ 1} = 2 > _{μ 2} = 5 / 4 > 1 , _{λ 1} = 2 > _{λ 2} = 5 / 4 > 1 , then (H₁ ) is satisfied. Notice (H₂ ) also holds, since [figure omitted; refer to PDF]

Existence of the positive T -periodic solutions is guaranteed from Theorem 3. We can also consider the following examples and the same result can be obtained: [figure omitted; refer to PDF]

Acknowledgments

The author is grateful to the referees for valuable comments and useful remarks on the paper. Research supported by China Postdoctoral Science Foundation (2012M510341), Beijing Natural Science Foundation (1112006) and the Grant of Beijing Education Committee Key Project (KZ20130028031).

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