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.

1. Introduction

We are concerned with the existence of positive T-periodic solution for the second-order nonautonomous singular coupled systems:

()

where f_i ∈ C((0, T)×[0, +∞), [0, +∞)), c_i : (0, T)→[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

()

were proved by using some fixed point theorems in cones for completely continuous operators, where a₁, a₂, c₁, c₂ ∈ L¹(0, T), f₁, f₂ ∈ Car([0, T]×(0, +∞), (0, +∞)). When the Green′s function G_i(t, s) (i = 1, 2), associated with the periodic boundary problem

()

is nonnegative for all (t, s)∈[0, T]×[0, T] and f_i (i = 1,2) satisfies weak singularities

()

where 0 < α_i < 1,

, b_i ≥ 0, and

, 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 Owing to the disappearing of the terms a_i(t) (i = 1,2) in (2), the methods in [9, 10] are no longer valid.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

()

which can be expressed by

()

By a direct computation, we can get

()

Assume f_i (i = 1,2) satisfies the following conditions.

(H₁)
For t ∈ (0, T), f_i(t, 1) > 0 and there exist constants μ₁ ≥ μ₂ > 1, λ₁ ≥ λ₂ > 1 such that, for any constants 0 ≤ ρ_i ≤ 1, i = 1,2,
()

()
(H₂)
and
()

where

, i = 1, 2.

Remark 1. For any ρ₁ ≥ 1, we get from (8) that

()

For any ρ₂ ≥ 1, we get from (9) that

()

Typical functions that satisfy (8) or (9) are those taking the form

()

where a_ik ∈ C(0, T), a_ik(t) > 0, l_ik > 1, i = 1,2; k = 1,2, …, n.

Definition 2. Supposing that (u, v) ∈ C¹[0, T]∩C²(0, T) × C¹[0, T]∩C²(0, T) satisfies (1) and u(t) > 0, v(t) > 0 for any t ∈ [0, T], then one says that (u, v) is a C¹[0, T] × C¹[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, Ω₁, Ω₂ two nonempty bounded open sets in K, . is a completely continuous operator. If

(i)
T(x) ≠ λx, x ∈ ∂Ω₁, λ > 1,
(ii)
T(x) ≠ λx, x ∈ ∂Ω₂, 0 < λ < 1, ,

then T has a fixed point in

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),

()

uniformly with respect to t ∈ [α, β].

Proof. We only deal with f₁. Without loss of generality, let 0 ≤ x ≤ y. If y = 0, we get f₁(t, x) ≤ f₁(t, y). If y ≠ 0, let c₀ = x/y, then 0 ≤ c₀ ≤ 1. From (8), we get

()

which means f₁(t, u) is increasing on u.

Assume u > 1. It follows from (11) that . Thus

()

From (H₁), for any [α, β]⊂(0, T), we obtain

()

Therefore

()

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

()

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,

()

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

()

Then the solution of periodic boundary value problem

()

can be expressed by

, and the solution of periodic boundary value problem

()

can be expressed by

. Obviously,

, i = 1, 2. Then

()

From (11), (12), and Lemma 5, we have

()

Then, for any fixed (u, v) ∈ P × P, it follows from (H₂) that

()

Thus, we can define the operator T : P × P → P × P, T(u, v) = (T₁u, T₂v) by

()

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

()

has one positive T-periodic solution.

Lemma 7. Assuming that (H₁), (H₂) hold, then T(Q × Q) ⊂ Q × Q and T : Q × Q → Q × Q is completely continuous.

Proof. For (u, v) ∈ Q × Q, we have

()

Then, we can get

()

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,

()

From (27), we have

()

Let

()

Thus

()

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

()

From (25) and (H₂), we have

()

Using the same method, we can obtain . Therefore, T(B) is equicontinuous. According to Ascoli-Arzela theorem, T(B) is a relatively compact set.

Next, we prove that T : Q × Q → Q × Q is continuous. Suppose (u_n, v_n), (u₀, v₀) ∈ Q × Q, (u_n, v_n)→(u₀, v₀), n → +∞, that is, u_n → u₀, v_n → v₀, n → +∞. We know that there exists a constant L > 0 such that

()

We shall prove T(u_n, v_n) → T(u₀, v₀), n → +∞, that is,

()

We first deal with T₁u_n → T₁u₀, n → +∞. Otherwise, there exist ε₀ > 0, {t_n}∈[0, T] such that |T₁u_n(t_n) − T₁u₀(t_n)| ≥ ε₀. Without loss of generality, we can assume t_n → t₀ ∈ [0, T]. We know

()

Next, we show T₁u_n(t₀) − T₁u₀(t₀) → 0, n → ∞. In fact,

()

Let

()

Since

()

and f₁ is continuous, we know r_n(s) → 0, n → ∞. From (24), we know

()

It can be inferred from (8) that

()

Set . Thus, we get

()

From (H₂), we know

. Using Lebesgue-dominated convergence theorem, we get

()

From (40) and (47), we obtain

()

which is a contradiction. Thus, we know T₁u_n → T₁u₀, n → +∞. Using the same method, we can obtain T₂v_n → T₂v₀, n → +∞. Then

()

which means T : Q × Q → Q × Q is continuous. Therefore T : Q × Q → Q × Q is a completely continuous operator.

3. Proof of Theorem 3

We proceed to prove Theorem 3 in two steps.

(1) Let Ω₁ = {(u, v) ∈ Q × Q : ∥u∥ < 2M²r₁/m, ∥v∥ < 2M²r₂/m}. We can get

()

For (u, v) ∈ ∂Ω₁, we have the following two cases.

Case I. One has {(u, v) ∈ Q × Q : ∥u∥ = 2M²r₁/m, ∥v∥ ≤ 2M²r₂/m}. Under this condition, we can get

()

Otherwise, there exist (u₀, v₀) ∈ ∂Ω₁, λ₀ > 1 such that T₁u₀ = λ₀u₀. As

()

then

. By a direct computation, we know u₀ = (1/λ₀)T₁u₀ satisfies

()

Then, we get

()

Since , integrating both sides of (54) on [0, T], we get

()

Then

()

That is,

()

which contradicts with (H₂).

Case II. One has {(u, v) ∈ Q × Q : ∥u∥ ≤ 2M²r₁/m, ∥v∥ = 2M²r₂/m}. Under this condition, we can get

()

Otherwise, there exist (u₀, v₀) ∈ ∂Ω₁, λ₀ > 1 such that T₂v₀ = λ₀v₀. As

()

then

. By a direct computation, we know v₀ = (1/λ₀)T₂v₀ satisfies

()

For , using the same method as condition I, we obtain

()

which is also a contradiction.

(2) Choose an interval [α, β]⊂(0, T) satisfying β − α = T/2. Set

. From Lemma 5, there exists R₁ > 2M²r/m, r = max {r₁, r₂} such that

()

Let R ≥ (2M/m)R₁ > R₁ > 2M²r/m. Define Ω₂ = {(u, v) ∈ Q × Q : ∥u∥ < R, ∥v∥ < R}. We can get

()

For (u, v) ∈ ∂Ω₂, we have the following two cases.

Case I. One has {(u, v) ∈ Q × Q : ∥u∥ = R, ∥v∥ ≤ R}. Under this condition, we know

()

Thus

. Furthermore, we obtain from (62) that

()

On the other hand, for 0 < λ < 1, we know λv(t) < v(t) ≤ R. From the choice of

, we get T₂v ≠ λv.

Case II. One has {(u, v) ∈ Q × Q : ∥u∥ ≤ R, ∥v∥ = R}. Under this condition, we get

()

Thus

. From (62), we get

()

For 0 < λ < 1, λu(t) < u(t) ≤ R, from the choice of

, we know T₁u ≠ λu.

Furthermore, we can obtain

()

It implies

From Lemma 4, we know T has a fixed point in . For , we have the following three cases.

Case 1. One has

()

Case 2. One has

()

Case 3. One has

()

Next, we show Cases 1 and 2 are impossible. In Case 1, we have

()

It follows that

. By a direct computation, we know

satisfies

()

Then, we get

()

Since

, integrating both sides of (74) on [0, T], we get

()

Then

()

That is,

()

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

satisfying

()

Since

()

from Lemma 6, we know

satisfy

()

Let . For

()

we obtain

()

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:

()

Choosing

, i = 1,2, and μ₁ = 2 > μ₂ = 5/4 > 1, λ₁ = 2 > λ₂ = 5/4 > 1, then (H₁) is satisfied. Notice (H₂) also holds, since

()

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:

()

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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Positive Periodic Solution of Second-Order Coupled Systems with Singularities

Abstract

1. Introduction

2. Preliminaries

3. Proof of Theorem 3

4. Applications of Theorem 3

Acknowledgments

References

References

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Positive Periodic Solution of Second-Order Coupled Systems with Singularities

Abstract

1. Introduction

2. Preliminaries

3. Proof of Theorem 3

4. Applications of Theorem 3

Acknowledgments

References

References

Related

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