Volume 2013, Issue 1 687595

Research Article

Open Access

On the Existence of Positive Solutions of Resonant and Nonresonant Multipoint Boundary Value Problems for Third-Order Nonlinear Differential Equations

Liu Yang,

Corresponding Author

Liu Yang

[email protected]

Department of Mathematics, Hefei Normal University, Hefei, Anhui Province 230601, China hftc.edu.cn

College of Science, University of Shanghai for Science and Technology, Shanghai 200093, China usst.edu.cn

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Chunfang Shen,

Chunfang Shen

Department of Mathematics, Hefei Normal University, Hefei, Anhui Province 230601, China hftc.edu.cn

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Dapeng Xie,

Dapeng Xie

Department of Mathematics, Hefei Normal University, Hefei, Anhui Province 230601, China hftc.edu.cn

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Xiping Liu,

Xiping Liu

College of Science, University of Shanghai for Science and Technology, Shanghai 200093, China usst.edu.cn

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Liu Yang,

Corresponding Author

Liu Yang

[email protected]

Department of Mathematics, Hefei Normal University, Hefei, Anhui Province 230601, China hftc.edu.cn

College of Science, University of Shanghai for Science and Technology, Shanghai 200093, China usst.edu.cn

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Chunfang Shen,

Chunfang Shen

Department of Mathematics, Hefei Normal University, Hefei, Anhui Province 230601, China hftc.edu.cn

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Dapeng Xie,

Dapeng Xie

Department of Mathematics, Hefei Normal University, Hefei, Anhui Province 230601, China hftc.edu.cn

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Xiping Liu,

Xiping Liu

College of Science, University of Shanghai for Science and Technology, Shanghai 200093, China usst.edu.cn

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First published: 15 December 2013

https://doi.org/10.1155/2013/687595

Academic Editor: Zhanbing Bai

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Abstract

Positive solutions for a kind of third-order multipoint boundary value problem under the non-resonant conditions and the resonant conditions are considered. In the nonresonant case, by using Leggett-Williams fixed-point theorem, the existence of at least three positive solutions is obtained. In the resonant case, by using Leggett-Williams norm-type theorem due to O’Regan and Zima, existence result of at least one positive solution is established. The results obtained are valid and new for the problem discussed. Two examples are given to illustrate the main results.

1. Introduction

We consider the third-order m-point boundary value problem given by

()

where 0 < ξ₁ < ξ₂ < ⋯<ξ_m−2 < 1, 0 ≤ α_i, β_i ≤ 1, i = 1,2, …, m − 2,

, and f ∈ C([0,1]×[0, ∞), R). For the convenience of writing later, we denote ξ₀ = 0,ξ_m−1 = 1, and α₀ = α_m−1 = β₀ = β_m−1 = 0.

If condition

holds, the problem is nonresonant; that is, the associated linear problem

()

has only zero solution, and the differential operator with boundary conditions is invertible. Otherwise, the problem is at resonance.

Third-order differential equations arise in many different areas of applied mathematics and physics, as the varying cross-section or deflection of a curved beam having a constant, three-layer beam and so on [1]. Recently, there have been extensive studies on positive solutions for nonresonant two-point or three-point boundary value problems for nonlinear third-order ordinary differential equations. For examples, Anderson [2] established the existence of at least three positive solutions to problem

()

where f : R → [0, +∞) is continuous and 1/2 ≤ t₂ < 1.

By using the well-known Guo-Krasnosel’skii fixed-point theorem [3], Palamides and Smyrlis [4] proved that there exists at least one positive solution for the third-order three-point problem:

()

In another paper [5], Graef and Kong studied the existence of positive solutions for the third-order semipositone boundary value problem:

()

where 1/2 < p < q < 1 are constants and λ > 0 is a parameter. Also f : (0,1)×[0, ∞) → R, e : (0,1) → R, and e ∈ L(0,1). Moreover w : [q, 1]→[0, ∞) are continuous functions. By using the Guo-Krasnosel’skii fixed-point theorem, the author established the existence of positive solutions. For more existence results of positive solutions for boundary value problems of third-order ordinary differential equations, one can see [6–12] and references therein.

For boundary value problems of second-order or higher-order differential equations at resonance, many existence results of solutions have been established; see [13–25]. In [25], the authors considered the following problem:

()

By using Mawhin continuation theorem [26], the existence results of solutions were obtained under the resonant conditions β = 1,

, and

and β = 1/η,

, and

, respectively.

It is well known that the problem of existence of positive solution for nonlinear BVP is very difficult when the resonant case is considered. Only few works gave the approach in this area for first- and second-order differential equations [27–34]. To our best knowledge, few paper dealt with the existence result of positive solution for resonant third-order boundary value problems. Motivated by the approach in [28–30, 35], we study the positive solution for problem (1) under nonresonant condition

and resonant condition

, respectively. By using Leggett-Williams fixed-point theorem and its generalization [27, 29], we establish the existence results of positive solutions. The results obtained in this paper are interesting in the following aspects.

(1)
In the nonresonant case, Green’s function is established and the results obtained are more general than those of earlier work.
(2)
It is the first time that the positive solution is considered for third-order boundary value problem at resonance.

2. Background Definitions and Lemmas

Let X, Y be real Banach spaces. A nonempty closed convex set C ⊂ X is said to be a cone provided that ax ∈ C, if x ∈ C, a ≥ 0 and x, −x ∈ C implies x = 0.

Definition 1. The map ψ is a nonnegative continuous concave functional on C if ψ : C → +∞ is continuous and

()

Definition 2. Let constants 0 < a < b be given and let ψ be a nonnegative continuous concave functional on the cone C. Define the convex sets C_r and C(ψ, a, b) as follow:

()

Lemma 3 (Leggett-Williams fixed-point theorem [35]). Let be a completely continuous operator and let ψ be a nonnegative continuous concave functional on C such that ψ(x) ≤ ∥x∥ for all . Suppose that there exist 0 < a < b < d ≤ c such that

(H₁) {x ∈ C(ψ, b, d)∣ψ(x) > b} ≠ ⌀ and ψ(Tx) > b for x ∈ C(ψ, b, d),
(H₂) ∥Tx∥ < a for ∥x∥ ≤ a,
(H₃) ψ(Tx) > b for x ∈ C(ψ, b, c) with ∥Tx∥ ≥ d.

Then operator T has at least triple fixed-points x₁, x₂, and x₃ with ∥x₁∥ < a, b < ψ(x₂), ∥x₃∥ > a, and ψ(x₃) < b.

Operator L : dom L ⊂ X → Y is called a Fredholm operator with index zero, which means that Im L is closed and dim Ker L = codim Im L < ∞, and there exist continuous projections P : X → X and Q : Y → Y such that Im P = Ker L and Ker Q = Im L. Furthermore, for dim Im Q = codim Im L, there exists an isomorphism J : Im Q → Ker L. Denote by L_P the restriction of L to Ker P∩dom L to Im L and its inverse by K_P, so K_P : Im L → Ker P∩dom L and the coincidence equation Lx = Nx is equivalent to the operator equation:

()

Let γ : X → C be a retraction, which means a continuous mapping such that γx = x for all x ∈ C and

()

Lemma 4 (Leggett-Williams norm-type theorem [28]). Assume that C is a cone in X and that Ω₁ and Ω₂ are open bounded subsets of X with and . Suppose that L : dom L ⊂ X → Y is a Fredholm operator of index zero and that

(C1) QN : X → Y is bounded and continuous and K_P(I − Q)N : X → X is compact on every bounded subset of X,
(C2) Lx ≠ λNx for all x ∈ C∩∂Ω₂∩dom L and λ ∈ (0,1),
(C3) γ maps subsets of into bounded subsets of C,
(C4) d_B([I − (P + JQN)γ]|_ker L, Ker L∩Ω₂, 0) ≠ 0, where d_B stands for the Brouwer degree,
(C5) there exists u₀ ∈ C∖{0} such that ∥x∥ ≤ σ(u₀)∥Ψx∥ for x ∈ C(u₀)∩∂Ω₁, where C(u₀) = {x ∈ C : μu₀ ≤ x} for some μ > 0 and σ(u₀) is such that ∥x + u₀∥ ≥ σ(u₀)∥x∥ for every x ∈ C,
(C6) (P + JQN)γ(∂Ω₂) ⊂ C,
(C7) .

Then the equation Lx = Nx has a solution in the set

3. Main Results for Nonresonant Case

In this section we consider the positive solution for the nonresonant case with the condition and we always suppose that f ∈ C([0,1]×[0, ∞), [0, ∞)).

Firstly, we consider the third-order m-point boundary value problem given by

()

Lemma 5. Suppose y(t) ∈ C[0,1]. Then problem (11) and (12) is equivalent to

()

where for i = 1, 2, …, m − 1

()

Proof. Let G(t, s) be Green’s function of problem −x^′′′(t) = 0 with boundary condition (12). We can suppose

()

where a₀, a₁, a₂, b₀, b₁, and b₂ are undetermined coefficients.

Considering the properties of Green′s function together with boundary condition (12), we have

()

The explicit expression of Green’s function is obtained by solving the linear function systems.

Lemma 6. Green’s function G(t, s) satisfies that G(t, s) ≥ 0, t, s ∈ [0,1].

Proof. For ξ_i−1 ≤ s ≤ ξ_i, i = 1, 2, …, m − 1, and t ≤ s,

()

which means that G(t, s) is decreasing on variable t. Thus

()

On the other hand, for ξ_i−1 ≤ s ≤ ξ_i, i = 1,2, …, m − 1, and t ≥ s,

()

Thus,

()

Furthermore,

()

Thus, for t, s ∈ [0, 1],

()

This gives that G(t, s) ≥ 0, t, s ∈ [0,1].

Lemma 7. If y(t) ≥ 0, t ∈ [0,1], and x(t) is the solution of problem (11) and (12), then

()

where

is a positive constant.

Proof. From

()

we see that x^′′(t) is decreasing on [0,1]. Considering x^′′(0) = 0, we have x^′′(t) ≤ 0, t ∈ (0,1). Next we claim that x^′(0) ≤ 0. Suppose that, on the contrary, x^′(0) > 0. We have

()

Thus,

()

From the concavity of x(t), we have

()

Multiplying left and right sides by β_i and considering

, we have

()

This completes the proof of Lemma 7.

Let the Banach space E = C[0, 1] be endowed with the maximum norm. We define the cone C ⊂ E by

()

Define the continuous nonnegative concave functional ψ : C → [0, ∞) by

()

Define the constants m^*, m_* by

()

Theorem 8. Suppose that there exist constants 0 < a < b < b/δ ≤ c such that

(A1) f(t, x) < a/m^*, (t, x)∈[0,1]×[0, a],
(A2) f(t, x) > b/m_*, (t, x)∈[0,1]×[b, b/δ],
(A3) f(t, x) < c/m^*, (t, x) ∈ [0,1] × [0, c].

Then problem (1) has at least three positive solutions x₁, x₂, and x₃ satisfying

()

Proof. The operator T : C → E is defined by

()

It is clear that T : C → C and it is completely continuous.

Next, the conditions of Lemma 3 are checked. If , then ∥x∥ ≤ c and condition (A3) implies that

()

Then

()

Thus,

Similar to the proof above, we obtain that . Hence, condition (H₂) of Lemma 3 is satisfied.

The fact that the constant function x(t) = b/δ ∈ {P(ψ, b, b/δ)∣ψ(x) > b} implies that {P(ψ, b, b/δ)∣ψ(x) > b} ≠ Ø. If x ∈ P(ψ, b, b/δ), from assumption (A2),

()

Thus

()

which ensures that condition (H₁) of Lemma 3 is satisfied. Finally we show that condition (H₃) of Lemma 3 also holds. Suppose that x ∈ P(ψ, b, c) with ∥Tx∥ > b/δ. Then

()

So, condition (H₃) of Lemma 3 is satisfied. Thus, an application of Lemma 3 implies that the nonresonant third-order boundary value problem (1) has at least three positive solutions x₁, x₂, and x₃ satisfying

()

Here an example is given to illustrate the main results of this section. We consider the following nonresonant three-point boundary value problem:

()

where

()

Here α₁ = 1/2, β₁ = 1/2, ξ₁ = 1/3, and

()

By a simple computation, we can get that

()

We choose a = 1, b = 4, and c = 32. It is easy to check that

(1)
f(t, x) < 81/32, [t, x]∈[0, 1]×[0, 1],
(2)
f(t, x) > 81/4, [t, x]∈[0, 1]×[4, 10],
(3)
f(t, x) < 81, [t, x]∈[0, 1]×[0, 32].

Thus all conditions of Theorem 8 hold. This ensures that problem (40) has at least three positive solutions x₁, x₂, and x₃ satisfying

()

4. Main Results for Resonant Case

In this section the condition is considered. Obviously, problem (1) is at resonance under this condition. The norm-type Leggett-Williams fixed-point theorem will be used to establish the existence results of positive solution.

We define the spaces X = Y = C[0,1] endowed with the maximum norm. It is well known that X and Y are the Banach spaces.

Define the linear operator L : dom L ⊂ X → Y and

()

where

()

and the nonlinear operator N : X → Y with

()

It is obvious that Ker L = {x ∈ dom L : x(t) ≡ c, t ∈ [0,1]}. Denote the function G(s), s ∈ [0,1] as follow:

()

Define the function k(t, s) as follow:

()

for i = 1,2, …, m − 1 and

The function U(t, s) and positive number κ are given by

()

Theorem 9. Suppose that there exists positive constant R ∈ (0, ∞) such that f : [0,1]×[0, R]→(−∞, + ∞) is continuous and satisfies the following conditions:

(S1) f(t, x)≥−κx, for (t, x)∈[0,1]×[0, R],
(S2) f(t, x) < 0 for [t, x]∈[0, 1]×[(1 − (κσ/2))R, R],
(S3) there exist r ∈ (0, R), M ∈ (0,1), t₀ ∈ [0,1], a ∈ (0,1], and continuous functions
()

such that

()

and h(x)/x^a is nonincreasing on (0, r] with

()

Then resonant problem (1) has at least one positive solution.

Proof. Firstly we prove that

()

Indeed, for each

, we choose

()

We can check that

()

which means x(t) ∈ dom L. Thus

()

On the other hand, for each y(t) ∈ Im L, there exists x(t) ∈ dom L such that

()

Integrating both sides on [0, t], we have

()

Considering the boundary condition together with the resonant condition

, we have

()

Thus,

()

It is obvious that dim Ker L = 1 and Im L is closed.

Secondly we see Y = Y₁ ⊕ Im L, where

()

In fact, for each y(t) ∈ Y, we have

()

This induces that y − y₁ ∈ Im L. Since Y₁∩Im L = {0}, we have Y = Y₁⨁Im L. Thus L is a Fredholm operator with index zero.

Define two projections P : X → X and Q : Y → Y by

()

Clearly, Im P = Ker L and Ker Q = Im L. Note that, for y ∈ Im L, the inverse K_P of L_P is given by

()

In fact, It is easy to check that

()

Next we will check that every condition of Lemma 4 is fulfilled. Remark that f can be extended continuously on [0,1]×(−∞, +∞) and condition (C1) of Lemma 4 is fulfilled.

Define the set of nonnegative functions C and subsets of X Ω₁, Ω₂ by

()

Remark that Ω₁ and Ω₂ are open and bounded sets. Furthermore

()

Let the isomorphism J = I and (γx)(t) = |x(t)| for x ∈ X. Then γ is a retraction and maps subsets of into bounded subsets of C, which ensures that condition (C3) of Lemma 4 is fulfilled.

Then we prove that (C2) of Lemma 4 is fulfilled. For this purpose, suppose that there exist x₀ ∈ C∩∂Ω₂∩dom L and λ₀ ∈ (0,1) such that Lx₀ = λ₀Nx₀. Then

()

for all t ∈ [0,1]. Thus

()

Let x₀(t₀) = ∥x₀∥ = R. The proof is divided into three cases.

(1)
We show that t₀ ≠ 1. Suppose, on the contrary, that x₀(t) achieves maximum value R only at t₀ = 1. Then the boundary condition in combination with the resonant condition yields that max _{1≤i≤m−2}x₀(ξ_i) ≥ R, which is a contradiction.
(2)
We claim that t₀ ≠ 0. Suppose, on the contrary, that x₀(t) achieves maximum value R at t₀ = 0. From condition (S2), there exists t₁ > 0 near to zero such that
()
Then
()
which contradicts the fact that x₀(t) achieves maximum value at t₀ = 0.
(3)
Thus there exists t₀ ∈ (0,1) such that x₀(t₀) = R = max _0≤t≤1x₀(t). We may choose η < t₀ nearest to t₀ with . From the mean value theory, we claim that there exists ξ ∈ (η, t₀) such that
()

However,

()

Thus

()

Then, considering assumption (S2), we have

()

which is a contradiction. Thus (C2) of Lemma 4 is fulfilled.

Remark 10. The sign of third-order derivative of a function h(t) at point t₀ cannot be confirmed when t₀ is a maximal value of h(t). Thus the methods in [28] are not applicable directly to this problem.

For x ∈ Ker L ∩ Ω₂, define the projection H(x, λ) as follows:

()

where λ ∈ [0,1] and x ∈ Ker L∩Ω₂. Suppose H(x, λ) = 0. In view of (S1) we obtain

()

Hence H(x, λ) = 0 implies c ≥ 0. Hence, if H(R, λ) = 0, we get

()

contradicting (S2). Thus H(x, λ) ≠ 0 for x ∈ ∂Ω₂ and λ ∈ [0,1]. Therefore

()

This ensures

()

Let

and t ∈ [0,1]. From condition (S1), we see

()

Hence

. Moreover, for x ∈ ∂Ω₂, we have

()

which means (P + JQN)γ(∂Ω₂) ⊂ C. These ensure that (C6) and (C7) of Lemma 4 hold.

At last, we confirm that (C5) is satisfied. Taking u₀(t) ≡ 1 on [0,1], we see

()

and we can choose σ(u₀) = 1. For x ∈ C(u₀)∩∂Ω₁, we have

()

Therefore, in view of (S3), we obtain, for all x ∈ C(u₀)∩∂Ω₁,

()

So ∥x∥ ≤ σ(u₀)∥Ψx∥ for all x ∈ C(u₀)∩∂Ω₁, which means (C5) of Lemma 4 holds.

Thus with the application of Lemma 4, we confirm that the equation Lx = Nx has a solution , which implies that the resonant problem (1) has at least one positive solution.

Finally an example is given to illustrate the main results of the resonance case. We investigate the resonant third-order three-point boundary value problem:

()

where α₁ = 1/2, β = 1, ξ = 2/3, and

()

Here

()

By a simple computation, we have

()

Choose R = 1, r = 1/4, t₀ = 0, a = 1, and M = 1/2.

We take

()

Then,

()

It is easy to check that

(1)
f(t, x)>−(31/81)x, for all (t, x)∈[0,1]×[0, 1],
(2)
f(t, x) < 0, for all (t, x)∈[0,1]×[1055/1458, 1],
(3)
, and is nonincreasing on (0, 1/4] with
()

Then all conditions of Theorem 9 are satisfied. This ensures that the resonant problem has at least one solution, positive on [0,1].

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

The project is supported by the National Natural Science Foundation of China (11201109) and the Natural Science Foundation of Anhui Educational Department (KJ2012B144 and KJ2012Z335).

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On the Existence of Positive Solutions of Resonant and Nonresonant Multipoint Boundary Value Problems for Third-Order Nonlinear Differential Equations

Abstract

1. Introduction

2. Background Definitions and Lemmas

3. Main Results for Nonresonant Case

4. Main Results for Resonant Case

Conflict of Interests

Acknowledgments

References

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On the Existence of Positive Solutions of Resonant and Nonresonant Multipoint Boundary Value Problems for Third-Order Nonlinear Differential Equations

Abstract

1. Introduction

2. Background Definitions and Lemmas

3. Main Results for Nonresonant Case

4. Main Results for Resonant Case

Conflict of Interests

Acknowledgments

References

References

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