## 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

(1)

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

Many authors consider the problem

(2)

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 [39] 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 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, . 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 be completely continuous and α be a nonnegative continuous concave functional on P such that α (x) = ||x|| for all . 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 , and 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 .

## 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.   □

Let

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

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

(H4)

(H5) There exists a constant 2 ≤ p1 such that

(H6) There exists a constant p2 with such that

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

Proof. From (H4), there exists a 0 < η < ∞ such that

(3)

Choosing R1 ∈ (0, η), set Ω1 = {uE : ||u|| < R1}. We now prove that

(4)

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 such that

(5)

From (H6), there exists such that

(6)

Choosing , set Ω2 = {uE : ||u|| < R2}. We now prove that

(7)

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 , which is a positive solution of Equation 1.   □

Example. Let q = 2, h(t) = 1, Φ(s) = 2 + s, φ(t) = 2t, and , 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 such that

(H9)

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

Proof. From (H7), there exists η1 > 0 such that

(8)

From (H8), there exists η2 > 0 such that

(9)

Choosing , set Ω1 = {uE : ||u|| < R1}. We now prove that

(10)

If uP ∩ ∂Ω1, we have

From Equations 8, 9, we can prove

Then, Equation 10 holds.

On the other hand, from (H7), there exists such that

(11)

Choosing , set Ω2 = {uE : ||u|| < R2}. We now prove that

(12)

If uP ∩ ∂Ω2, Since minθt≤1-θu(t) ≥ θ ||u|| and ||u|| = R2, we have

(13)

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 , which is a positive solution of Equation 1.   □

Example. Let q = 2, h(t) = t, Φ(s) = 2 + s, φ(t) = 2t, 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:

(H10)

(H11)

(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 such that .

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

Set

Taking

If , then

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

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

Take . Then, for each , we have

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

In fact,

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, and, , 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 and (H1), then (H12) holds.