International Scholarly Research Notices

Volume 2011, Issue 1 612591

Research Article

Open Access

Positive Solutions for Second-Order Nonlinear Ordinary Differential Systems with Two Parameters

Lan Sun,

Corresponding Author

Lan Sun

[email protected]

Department of Mathematics, Nanjing University of Aeronautics and Astronautics, Nanjing 210016, China nuaa.edu.cn

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Yukun An,

Yukun An

Department of Mathematics, Nanjing University of Aeronautics and Astronautics, Nanjing 210016, China nuaa.edu.cn

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Min Jiang,

Min Jiang

College of Computer Science and Technology, Nanjing University of Aeronautics and Astronautics, Nanjing 210016, China nuaa.edu.cn

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Lan Sun,

Corresponding Author

Lan Sun

[email protected]

Department of Mathematics, Nanjing University of Aeronautics and Astronautics, Nanjing 210016, China nuaa.edu.cn

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Yukun An,

Yukun An

Department of Mathematics, Nanjing University of Aeronautics and Astronautics, Nanjing 210016, China nuaa.edu.cn

Search for more papers by this author

Min Jiang,

Min Jiang

College of Computer Science and Technology, Nanjing University of Aeronautics and Astronautics, Nanjing 210016, China nuaa.edu.cn

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First published: 15 December 2011

https://doi.org/10.5402/2011/612591

Academic Editor: G. C. Georgiou

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Abstract

By using fixed-point theorem and under suitable conditions, we investigate the existence and multiplicity positive solutions to the following systems: u^′′(t) + au(t) + bv(t) + λh₁(t)f(u(t), v(t)) = 0, t ∈ [0,1], v^′′(t) + cu(t) + dv(t) + μh₂(t)g(u(t), v(t)) = 0, t ∈ [0,1], u(0) = u(1) = 0, v(0) = v(1) = 0, where a, b, c, d are four positive constants and λ > 0, μ > 0, f(u, v), g(u, v) ∈ C(R⁺ × R⁺, R⁺) and h₁, h₂ ∈ C([0,1], R⁺). We derive two explicit intervals of λ and μ, such that the existence and multiplicity of positive solutions for the systems is guaranteed.

1. Introduction

In this paper, we are concerned with the existence and multiplicity of positive solutions to the following boundary value problem of second-order nonlinear differential systems with two parameters:

(1.1)

where a, b, c, d are four positive constants.

In addition, we assume the following conditions throughout this paper:

H₁ h₁(t), h₂(t) ∈ C([0,1], R⁺) does not vanish identically on any subinterval of [0,1];
H₂ 0 < a < π², 0 < d < π², c > 0, b > 0;
H₃ f, g ∈ C(R⁺ × R⁺, R⁺).

The boundary value problem of ordinary differential systems has attracted much attention, see [1–6] and the references therein. Recently, Fink and Gatica [4] and Ma [7] have studied the existence of positive solutions of the following systems:

(1.2)

In [4], a multiplicity result has been established when f(0,0) > 0, g(0,0) > 0. In [7], a multiplicity result has been given for the more general case f(0,0) ≥ 0, g(0,0) ≥ 0.

Also, by using Krasnosel′skill fixed-point theorem, Ru and An [8] considered the existence of positive solutions of the following systems:

(1.3)

where λ > , μ > 0, p, q ∈ N. f, g : [0, ∞) × [0, ∞) → [0, ∞).

More recently, Dalbono and Mckenna [5] proved the existence and multiplicity of solutions to a class of asymmetric weakly coupled systems as follows:

(1.4)

where ɛ is suitably small and the positive numbers b₁, b₂ satisfy

(1.5)

Applying a classical change of variables, the authors transformed the initial problem into an equivalent problem whose solutions can be characterized by their nodal properties. Meanwhile, in [5], there are two open questions.

(1)
Can one replace the near-diagonal matrix with something more general and use information on the eigenvalues of matrix?
(2)
Can one replace the near-diagonal matrix with something more general and use information on the eigenvalues of matrix?

Inspired by the above works and the two open questions, we consider the existence and multiplicity of positive solutions to (1.1). The paper is organized as follows. In Section 2, we state some preliminaries. In Sections 3 and 4, we prove the existence and multiplicity results of (1.1).

2. Preliminaries

In order to prove our results, we state the well-known fixed-point theorem [9]:

Lemma 2.1. Let E be a Banach space and let K ⊂ E be a cone in E. Assume Ω₁, Ω₂ are two open subsets of E with 0 ∈ Ω₁, , and let be a completely continuous operator such that either

(i)
∥Tu∥ ≤ ∥u∥, u ∈ K∩∂Ω₁ and ∥Tu∥ ≥ ∥u∥, u ∈ K∩∂Ω₂; or
(ii)
∥Tu∥ ≥ ∥u∥, u ∈ K∩∂Ω₁ and ∥Tu∥≤∥u∥, u ∈ K∩∂Ω₂.

Then, T has a fixed point in

To be convenient, we introduce the following notations:

(2.1)

and suppose that f₀, g₀, f_∞, g_∞ ∈ [0, +∞].

Let , G_i(t, s), i = 1,2 be Green^′s function of the corresponding to linear boundary value problem:

(2.2)

Then, the solution of (2.2) is given by

(2.3)

It is well known that G_i(t, s) can be expressed by

(2.4)

In addition, it can be easily to be checked that

(2.5)

See [6], we have the following lemma.

Lemma 2.2. G_i(t, s) has the following properties:

(i)
G_i(t, s) > 0, for all t, s ∈ (0,1),
(ii)
G_i(t, s) ≤ C_iG_i(s, s), for all t, s ∈ [0,1], C_i = 1/sin ω_i,
(iii)
G_i(t, s) ≥ δ_iG_i(t, t)G_i(s, s), for all t, s ∈ [0,1], δ_i = ω_isin ω_i, i = 1,2.

It is obvious that problem (1.1) is equivalent to the equation:

(2.6)

and consequently it is equivalent to the fixed-point problem:

(2.7)

with A : C[0,1] × C[0,1] → C[0,1] × C[0,1] given by

(2.8)

For convenience, denote

(2.9)

It is obvious that A is completely continuous. Let σ_i = δ_im_i/C_i = (sin ω_i/4)(sin 3ω_i/4)(sin ω_i), i = 1,2, σ = min {σ₁, σ₂} and define a cone in C[0,1] × C[0,1] by

(2.10)

where ∥(u, v)∥ = ∥u∥ + ∥v∥ = sup _t∈[0,1]u(t) + sup _t∈[0,1]v(t).

Lemma 2.3. A(K) ⊂ K.

Proof. For any (t, s) ∈ [1/4, 3/4] × [0,1], by Lemma 2.2, we have

(2.11)

Similarly, for any (t, s) ∈ [1/4, 3/4] × [0,1], we have

(2.12)

Then, for any (t, s) ∈ [1/4, 3/4] × [0,1], we have

(2.13)

Thus, min _{1/4≤ t ≤3/4}A_λ(u, v) + A_μ(u, v) ≥ σ∥A(u, v)∥. Therefore, A(K) ⊂ K.

3. Existence Results

We assume the following:

(H₄) 0 < b < 2ω₁cos ²(ω₁/2), 0 < c < 2ω₂cos ²(ω₂/2);
(H₅) ;
(H₆) .

Theorem 3.1. Assume (H₁)–(H₆) hold. Then one has the following:

(1)
If 0 < f₀, f_∞, g₀, g_∞ < ∞, 2P₁A₁f₀ < Q₁σB₁f_∞(1 − 2P₁b), then for each λ ∈ (1/Q₁σB₁f_∞, 1 − 2P₁b/2P₁A₁f₀) and μ ∈ (0, (1 − 2P₂c)/2P₂A₂g₀), (1.1) has at least one positive solution.
(2)
If 0 < f₀, f_∞, g₀, g_∞ < ∞, 2P₂A₂g₀ < Q₂σB₂g_∞(1 − 2P₂c), then for each λ ∈ (0, (1 − 2P₁b)/2P₁A₁f₀) and μ ∈ (1/Q₂σB₂g_∞, (1 − 2P₂c)/2P₂A₂g₀), (1.1) has at least one positive solution.
(3)
If f₀ = 0, f_∞ = ∞, then for each λ ∈ (0, ∞) and μ ∈ (0, (1 − 2P₂c)/2P₂A₂g₀), (1.1) has at least one positive solution.
(4)
If g₀ = 0, g_∞ = ∞, then for each λ ∈ (0, (1 − 2P₁b)/2P₁A₁f₀) and μ ∈ (0, ∞), (1.1) has at least one positive solution.
(5)
If f₀ = 0, f_∞ = ∞ and g₀ = 0, g_∞ = ∞, then for each λ ∈ (0, ∞) and μ ∈ (0, ∞), (1.1) has at least one positive solution.
(6)
If f_∞ = ∞, 0 < f₀ < ∞ or g_∞ = ∞, 0 < g₀ < ∞, then for each λ ∈ (0, (1 − 2P₁b)/2P₁A₁f₀) and μ ∈ (0, (1 − 2P₂c)/2P₂A₂g₀), (1.1) has at least one positive solution.
(7)
If f₀ = 0, 0 < f_∞ < ∞ and g₀ = 0, 0 < g_∞ < ∞, then for each λ ∈ (1/Q₁σB₁f_∞, ∞), μ ∈ (0, ∞) or λ ∈ (0, ∞), μ ∈ ((1/Q₂σB₂g_∞)∞), (1.1) has at least one positive solution.

Proof. We only prove case (1.1). The other cases can be proved similarly. In order to apply the Lemma 2.1, we construct the sets Ω₁, Ω₂.

Let λ ∈ (1/Q₁σB₁f_∞, (1 − 2P₁b)/2P₁A₁f₀), μ ∈ (0, (1 − 2P₂c)/2P₂A₂g₀), and we choose ɛ > 0 such that

(3.1)

By the definition of f₀ and g₀, there exists R₁ > 0, such that

(3.2)

Choosing (u, v) ∈ K with ∥(u, v)∥ = R₁, we have

(3.3)

namely, ∥A_λ(u, v)(t)∥ ≤ (1/2)∥(u, v)∥. In the same way, we also have

(3.4)

then ∥A_λ(u, v)(t)∥ ≤ ∥(u, v)(t)∥. Thus, if we set Ω₁ = {(u, v) ∈ K : ∥(u, v)∥ < R₁}, then

(3.5)

On the other hand, by the definition of f_∞, there exists R₂′ > 0, such that f(u, v) ≥ (f_∞ − ϵ)(u + v), for

. Let

and Ω₂ = {(u, v) ∈ K : ∥(u, v)∥ < R₂}. If we choose (u, v) ∈ K with ∥(u, v)∥ = R₂, such that min _{1/4≤ t ≤3/4}(u(t) + v(t)) ≥ σ∥(u, v)∥ ≥ R₂, then we have

(3.6)

Hence

(3.7)

Therefore, it follows from (3.5) to (3.7) and Lemma 2.2, A has a fixed point

, which is a positive solution of (1.1).

Similarly, we also have the following results.

Theorem 3.2. Assume (H₁)–(H₆) hold. Then one has the following:

(1)
If 0 < f₀, f_∞, g₀, g_∞ < ∞, 2P₁A₁f_∞ < Q₁σB₁f₀(1 − 2P₁b), then for each λ ∈ (1/Q₁σB₁f₀, (1 − 2P₁b)/2P₁A₁f_∞) and μ ∈ (0, (1 − 2P₂c)/2P₂A₂g_∞), (1.1) has at least one positive solution.
(2)
If 0 < f₀, f_∞, g₀, g_∞ < ∞, 2P₂A₂g_∞ < Q₂σB₂g₀(1 − 2P₂c), then for each λ ∈ (0, (1 − 2P₁b)/2P₁A₁f_∞) and μ ∈ (1/Q₂σB₂g₀, (1 − 2P₂c)/2P₂A₂g_∞), (1.1) has at least one positive solution.
(3)
If f₀ = ∞, f_∞ = 0, then for each λ ∈ (0, ∞) and μ ∈ (0, (1 − 2P₂c)/2P₂A₂g₀), (1.1) has at least one positive solution.
(4)
If g₀ = ∞, g_∞ = 0, then for each λ ∈ (0, (1 − 2P₁b)/2P₁A₁f₀) and μ ∈ (0, ∞), (1.1) has at least one positive solution.
(5)
If f₀ = ∞, f_∞ = 0 and g₀ = ∞, g_∞ = 0, then for each λ ∈ (0, ∞) and μ ∈ (0, ∞), (1.1) has at least one positive solution.
(6)
If f₀ = ∞, 0 < f_∞ < ∞ or g₀ = ∞, 0 < g_∞ < ∞ then for each λ ∈ (0, (1 − 2P₁b)/2P₁A₁f_∞) and μ ∈ (0, (1 − 2P₂c)/2P₂A₂g_∞), (1.1) has at least one positive solution.
(7)
If f_∞ = 0, 0 < f₀ < ∞ and g_∞ = 0, 0 < g₀ < ∞ then for each λ ∈ (1/Q₁σB₁f₀, ∞), μ ∈ (0, ∞) or λ ∈ (0, ∞), μ ∈ (1/Q₂σB₂g₀), (1.1) has at least one positive solution.

Proof. We only prove case (1). The other cases can be proved similarly.

Let λ ∈ (1/Q₁σB₁f₀, (1 − 2P₁b)/2P₁A₁f_∞), μ ∈ (0, (1 − 2P₂c)/2P₂A₂g_∞), and we choose ɛ > 0 such that

(3.8)

by the definition of f₀, there exists R₁ > 0, such that

(3.9)

Choosing (u, v) ∈ K with ∥(u, v)∥ = R₁, such that min _{1/4≤ t ≤3/4}(u(t) + v(t)) ≥ σ∥(u, v)∥, then we have

(3.10)

So, if we set Ω₁ = {(u, v) ∈ K : ∥(u, v)∥ < R₁}, then

(3.11)

On the other hand, by the definition of f_∞ and g_∞, there exists R₂ > 2R₁, such that

(3.12)

Choosing (u, v) ∈ K with ∥(u, v)∥ = R₂, we have

(3.13)

In the same way, we also have

(3.14)

Hence, if we set Ω₂ = {(u, v) ∈ K : ∥(u, v)∥ < R₂}, then

(3.15)

Therefore, it follows from (3.11) to (3.15) and Lemma 2.2 that A has a fixed point

, which is a positive solution of (1.1).

4. Multiplicity Results

Theorem 4.1. Assume (H₁)–(H₆) hold. In addition, assume that there exist three constants r, M, N, where N is sufficient small with 2P₁NA₁ ≤ Q₁B₁M(1 − 2P₁b), 2P₂NA₂ ≤ Q₂B₂M(1 − 2P₂c) such that

(i)
f₀ = f_∞ = 0, g₀ = g_∞ = 0,
(ii)
f(u, v) ≥ Mr or g(u, v) ≥ Mr, for σr ≤ ∥(u, v)∥ ≤ r.

Then, for any λ ∈ (1/Q₁B₁M, (1 − 2P₁b)/2P₁A₁N), μ ∈ (0, (1 − 2P₂c)/2P₂A₂N), or λ ∈ (0, (1 − 2P₁b)/2P₁A₁N), μ ∈ (1/Q₂B₂M, (1 − 2P₂c)/2P₂A₂N), the problem (1.1) has at least two positive solutions.

Proof. We only prove the case of λ ∈ (1/Q₁σB₁M, (1 − 2P₁b)/2P₁A₁N), μ ∈ (0, (1 − 2P₂c/2P₂A₂N)), The other case is similar.

Step 1. Since f₀ = g₀ = 0, there exists r₁ ∈ (0, r) such that

(4.1)

Set Ω₁ = {(u, v) ∈ K : ∥(u, v)∥ < r₁}, for all (u, v) ∈ K∩∂Ω₁, then we have

(4.2)

In the same way, we also have

(4.3)

Hence,

(4.4)

Step 2. Since f_∞ = g_∞ = 0, there exists such that

(4.5)

Similarly, set Ω₂ = {(u, v) ∈ K : ∥(u, v)∥ < r₂}, then

(4.6)

Step 3. Let Ω₃ = {(u, v) ∈ K : ∥(u, v)∥ < r}, we can see that

(4.7)

Then,

(4.8)

Consequently, by Lemma 2.1 and from (4.4)–(4.8), A has two fixed points: and , so (1.1) has at least two positive solutions satisfying 0 < (u₁, v₁) < r < (u₂, v₂).

Theorem 4.2. Assume (H₁)–(H₆) hold. In addition, assume that there exist constants r, M, N, where N is sufficient large with 2P₁MA₁ ≤ Q₁σB₁N(1 − 2P₁b), 2P₂MA₂ ≤ Q₂σB₂N(1 − 2P₂c) such that

(i)
f₀ = f_∞ = ∞ or g₀ = g_∞ = ∞,
(ii)
f(u, v) ≤ Mr or g(u, v) ≤ Mr, for σr ≤ ∥(u, v)∥ ≤ r,

then, for any λ ∈ (1/Q₁σB₁N, (1 − 2P₁b)/2P₁A₁M), μ ∈ (0, (1 − 2P₂c)/2P₂A₂M), or λ ∈ (0, (1 − 2P₁b)/2P₁A₁M), μ ∈ (1/Q₂σB₂N, (1 − 2P₂c)/2P₂A₂M), the problem (1.1) has at least two positive solutions.

Proof. We only prove the case of λ ∈ (1/Q₁σB₁N, (1 − 2P₁b)/2P₁A₁M), μ ∈ (0, (1 − 2P₂c)/2P₂A₂M), The other case is similar.

Step 1. Since f₀ = ∞, there exists r₁ ∈ (0, r) such that

(4.9)

Set Ω₁ = {(u, v) ∈ K : ∥(u, v)∥ < r₁}, then we have

(4.10)

Hence,

(4.11)

Step 2. Since f_∞ = ∞, there exists such that

(4.12)

Similarly, set Ω₂ = {(u, v) ∈ K : ∥(u, v)∥ < r₂}, then

(4.13)

Step 3. Let Ω₃ = {(u, v) ∈ K : ∥(u, v)∥ < r}, for all (u, v) ∈ K∩∂Ω₃, we can see that

(4.14)

In the same way, we also have

(4.15)

Hence,

(4.16)

Consequently, by Lemma 2.1 and from (4.11)–(4.16), A has two fixed points: and , so (1.1) has at least two positive solutions satisfying 0 < (u₁, v₁) < r < (u₂, v₂).

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Positive Solutions for Second-Order Nonlinear Ordinary Differential Systems with Two Parameters

Abstract

1. Introduction

2. Preliminaries

3. Existence Results

4. Multiplicity Results

References

References

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Positive Solutions for Second-Order Nonlinear Ordinary Differential Systems with Two Parameters

Abstract

1. Introduction

2. Preliminaries

3. Existence Results

4. Multiplicity Results

References

References

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