Boundary Value Problems

, 2011:5 | Cite as

Existence and multiplicity of positive solutions for a nonlocal differential equation

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Research

Abstract

In this paper, the existence and multiplicity results of positive solutions for a nonlocal differential equation are mainly considered.

Keywords

Nonlocal boundary value problems Cone Fixed point theorem 

Introduction

In this paper, we are concerned with the existence and multiplicity of positive solutions for the following nonlinear differential equation with nonlocal boundary value condition

where α, β, γ, δ are nonnegative constants, ρ = αγ + αδ + βγ > 0, q ≥ 1; Open image in new window , Open image in new window denote the Riemann-Stieltjes integrals.

Many authors consider the problem
because of the importance in numerous physical models: system of particles in thermodynamical equilibrium interacting via gravitational potential, 2-D fully turbulent behavior of a real flow, one-dimensional fluid flows with rate of strain proportional to a power of stress multiplied by a function of temperature, etc. In [1, 2], the authors use the Kras-noselskii fixed point theorem to obtain one positive solution for the following nonlocal equation with zero Dirichlet boundary condition

when the nonlinearity f is a sublinear or superlinear function in a sense to be established when necessary. Nonlocal BVPs of ordinary differential equations or system arise in a variety of areas of applied mathematics and physics. In recent years, more and more papers were devoted to deal with the existence of positive solutions of nonlocal BVPs (see [3, 4, 5, 6, 7, 8, 9] and references therein). Inspired by the above references, our aim in the present paper is to investigate the existence and multiplicity of positive solutions to Equation 1 using the Krasnosel'skii fixed point theorem and Leggett-Williams fixed point theorem.

This paper is organized as follows: In Section 2, some preliminaries are given; In Section 3, we give the existence results.

Preliminaries

Lemma 2.1[3]. Let y(t) ∈ C([0, 1]), then the problem
has a unique solution
where the Green function G(t, s) is
It is easy to see that

and there exists a Open image in new window such that G(t, s) ≥ θ G(s, s), θt ≤ 1 - θ, 0 ≤ s ≤ 1.

For convenience, we assume the following conditions hold throughout this paper:

(H1) f, g, Φ: R+R+ are continuous and nondecreasing functions, and Φ (0) > 0;

(H2) φ(t) is an increasing nonconstant function defined on [0, 1] with φ(0) = 0;

(H3) h(t) does not vanish identically on any subinterval of (0, 1) and satisfies
Obviously, uC2(0, 1) is a solution of Equation 1 if and only if uC(0, 1) satisfies the following nonlinear integral equation

At the end of this section, we state the fixed point theorems, which will be used in Section 3.

Let E be a real Banach space with norm || · || and PE be a cone in E, P r = {xP : ||x|| < r}(r > 0). Then, Open image in new window . A map α is said to be a nonnegative continuous concave functional on P if α: P → [0, +∞) is continuous and
for all x, yP and t ∈ [0, 1]. For numbers a, b such that 0 < a < b and α is a nonnegative continuous concave functional on P, we define the convex set
Lemma 2.2[10]. Let Open image in new window be completely continuous and α be a nonnegative continuous concave functional on P such that α (x) = ||x|| for all Open image in new window . Suppose there exists 0 < d < a < b = c such that
  1. (i)

    {xP (α, a, b): α (x) > a} ≠ ∅ and α (Ax) > a for xP (α, a, b);

     
  2. (ii)

    ||Ax|| < d for ||x|| ≤ d;

     

(iii) α(Ax) > a for xP (α, a, c) with ||Ax|| > b.

Then, A has at least three fixed points x1, x2, x3 satisfying
Lemma 2.3[10]. Let E be a Banach space, and let PE be a closed, convex cone in E, assume Ω1, Ω2 are bounded open subsets of E with Open image in new window , and Open image in new window be a completely continuous operator such that either
  1. (i)

    ||Au|| ≤ ||u||, uP ∩ ∂Ω1 and ||Au|| ≥ ||u||, uP ∩ ∂Ω2; or

     
  2. (ii)

    ||Au|| ≥ ||u||, uP ∩ ∂Ω1 and ||Au|| ≤ ||u||, uP ∩ ∂Ω2.

     

Then, A has a fixed point in Open image in new window .

Main result

Let E = C[0, 1] endowed norm ||u|| = max0≤t≤1|u|, and define the cone PE by

Then, it is easy to prove that E is a Banach space and P is a cone in E.

Define the operator T: EE by

Lemma 3.1. T: EE is completely continuous, and Te now prove thatPP.

Proof. For any uP, then from properties of G(t, s), T (u)(t) ≥ 0, t ∈ [0, 1], and it follows from the definition of T that
Thus, it follows from above that

From the above, we conclude that TPP. Also, one can verify that T is completely continuous by the Arzela-Ascoli theorem.   □

Then, it is clear to see that 0 < lL < L.

Theorem 3.2. Assume (H1) to (H3) hold. In addition,

(H5) There exists a constant 2 ≤ p1 such that
(H6) There exists a constant p2 with Open image in new window such that

Then, problem (Equation 1) has one positive solution.

Proof. From (H4), there exists a 0 < η < ∞ such that
Choosing R1 ∈ (0, η), set Ω1 = {uE : ||u|| < R1}. We now prove that
Let uP ∩ ∂Ω1. Since minθt≤1-θu(t) ≥ θ ||u|| and ||u|| = R1, from Equation 3, (H1) and (H3), it follows that

Then, Equation 4 holds.

On the other hand, from (H5), there exists Open image in new window such that
From (H6), there exists Open image in new window such that
Choosing Open image in new window , set Ω2 = {uE : ||u|| < R2}. We now prove that
If uP ∩ ∂Ω2, we have
From Equations 5, 6, we can prove

Then, Equation 7 holds.

Therefore, by Equations 4 and 7 and the second part of Lemma 2.3, T has a fixed point in Open image in new window , which is a positive solution of Equation 1.   □

Example. Let q = 2, h(t) = 1, Φ(s) = 2 + s, φ(t) = 2t, Open image in new window and Open image in new window , namely,
It is easy to see that (H1) to (H3) hold. We also can have
Take p1 = 2, then it is clear to see that (H4) and (H5) hold. Since

then (H6) hold.

Theorem 3.3. Assume (H1) to (H3) hold. In addition,

(H7) There exists a constant 2 ≤ p1 such that
(H8) There exists a constant p2 with Open image in new window such that

Then, problem (Equation 1) has one positive solution.

Proof. From (H7), there exists η1 > 0 such that
From (H8), there exists η2 > 0 such that
Choosing Open image in new window , set Ω1 = {uE : ||u|| < R1}. We now prove that
If uP ∩ ∂Ω1, we have
From Equations 8, 9, we can prove

Then, Equation 10 holds.

On the other hand, from (H7), there exists Open image in new window such that
Choosing Open image in new window , set Ω2 = {uE : ||u|| < R2}. We now prove that
If uP ∩ ∂Ω2, Since minθt≤1-θu(t) ≥ θ ||u|| and ||u|| = R2, we have
By Equation 11, (H1) and (H3), it follows that

Then, Equation 12 holds.

Therefore, by Equations 10 and 12 and the first part of Lemma 2.3, T has a fixed point in Open image in new window , which is a positive solution of Equation 1.   □

Example. Let q = 2, h(t) = t, Φ(s) = 2 + s, φ(t) = 2t, Open image in new window and g(s) = s2.

Theorem 3.4. Assume that (H1) to (H3) hold. In addition, φ(1) ≥ 1, and the functions f, g satisfy the following growth conditions:

(H12) There exists a constant a > 0 such that

Then, BVP (Equation 1) has at least three positive solutions.

Proof. For the sake of applying the Leggett-Williams fixed point theorem, define a functional σ(u) on cone P by

Evidently, σ: PR+ is a nonnegative continuous and concave. Moreover, σ(u) ≤ ||u|| for each uP.

Now, we verify that the assumption of Lemma 2.2 is satisfied.

Firstly, it can verify that there exists a positive number c with Open image in new window such that Open image in new window .

By (H10), it is easy to see that there exists τ > 0 such that

by (H1) to (H3) and (H10).

Next, from (H11), there exists d' ∈ (0, a) such that

Finally, we will show that {uP (σ, a, b): σ(u) > a} ≠ ∅ and σ(Tu) > a for all uP(σ, a, b).

For uP (σ, a, b), we have
for all t ∈ [θ, 1 -θ]. Then, we have
by (H1) to (H3), (H12). In addition, for each uP (θ, a, c) with ||Tu|| > b, we have
Above all, we know that the conditions of Lemma 2.2 are satisfied. By Lemma 2.2, the operator T has at least three fixed points u i (i = 1, 2, 3) such that

The proof is complete.   □

Example. Let q = 2, h(t) = t, Φ(s) = 2 + s, φ(t) = 2t, Open image in new window and, Open image in new window , namely,
From a simple computation, we have

Then, it is easy to see that (H1) to (H3) and (H10) to (H11) hold. Especially, take a = 1, by Open image in new window and (H1), then (H12) holds.

Notes

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Copyright information

© Wang et al; licensee Springer. 2011

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Authors and Affiliations

  1. 1.College of Aeronautics and AstronauticsNanjing University of Aeronautics and AstronauticsNanjingPeople's Republic of China
  2. 2.College of ScienceHohai UniversityNanjingPeople's Republic of China
  3. 3.Department of MathematicsNanjing University of Aeronautics and AstronauticsNanjingPeople's Republic of China

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