Monotone and convex positive solutions for fourth-order multi-point boundary value problems

Yang Liu; Zhang Weiguo; Shen Chunfang

doi:10.1186/1687-2770-2011-21

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Monotone and convex positive solutions for fourth-order multi-point boundary value problems

Yang Liu^{1, 2}Email author,
Zhang Weiguo¹ and
Shen Chunfang²

Boundary Value Problems20112011:21

https://doi.org/10.1186/1687-2770-2011-21

Received: 27 December 2010

Accepted: 5 September 2011

Published: 5 September 2011

Abstract

The existence results of multiple monotone and convex positive solutions for some fourth-order multi-point boundary value problems are established. The nonlinearities in the problems studied depend on all order derivatives. The analysis relies on a fixed point theorem in a cone. The explicit expressions and properties of associated Green's functions are also given.

MSC: 34B10; 34B15.

Keywords

multi-point boundary value problem positive solution cone fixed point

1 Introduction

Boundary value problems for second and higher order nonlinear differential equations play a very important role in both theory and applications. For example, the deformations of an elastic beam in the equilibrium state can be described as a boundary value problem of some fourth-order differential equations. Owing to its importance in application, the existence of positive solutions for nonlinear second and higher order boundary value problems has been studied by many authors. We refer to recent contributions of Ma [1–3], He and Ge [4], Guo and Ge [5], Avery et al. [6, 7], Henderson [8], Eloe and Henderson [9], Yang et al. [10], Webb and Infante [11, 12], and Agarwal and O'Regan [13]. For survey of known results and additional references, we refer the reader to the monographs by Agarwal [14] and Agarwal et al. [15].

When it comes to positive solutions for nonlinear fourth-order ordinary differential equations, two point boundary value problems are studied extensively, see [16–24]. Few papers deal with the multi-point cases. Furthermore, for nonlinear fourth-order equations, only the situation that the nonlinear term does not depend on the first, second and third order derivatives are considered, see [16–23]. Few paper deals with the situation that lower order derivatives are involved in the nonlinear term explicitly. In fact, the derivatives are of great importance in the problem in some cases. For example, in the linear elastic beam equation (Euler-Bernoulli equation)

{(EI u^{″} (t))}^{″} = f (t), t \in (0, L),

where u(t) is the deformation function, L is the length of the beam, f(t) is the load density, E is the Young's modulus of elasticity and I is the moment of inertia of the cross-section of the beam. In this problem, the physical meaning of the derivatives of the function u(t) is as follows: u⁽⁴⁾(t) is the load density stiffness, u'''(t) is the shear force stiffness, u''(t) is the bending moment stiffness and the u'(t) is the slope. If the payload depends on the shear force stiffness, bending moment stiffness or the slope, the derivatives of the unknown function are involved in the nonlinear term explicitly.

In this paper, we are interested in the positive solution for fourth-order nonlinear differential equation

x^{(4)} (t) = f (t, x (t), x^{'} (t), x^{″} (t), x^{‴} (t)), t \in [0, 1],

(1.1)

subject to multi-point boundary condition

x^{‴} (1) = 0, x^{″} (1) = 0, x^{'} (0) = 0, x (1) = \sum_{i = 1}^{m - 2} β_{i} x (ξ_{i})

(1.2)

or

x^{‴} (1) = 0, x^{″} (1) = 0, x^{'} (1) = 0, x (0) = \sum_{i = 1}^{m - 2} β_{i} x (ξ_{i})

(1.3)

where 0 < ξ₁ < ξ₂ < ⋯ < ξ_m-2< 1, β_i > 0, 1 = 1, 2, ..., m - 2, $\sum_{i = 1}^{m - 2} β_{i} > 1$ , and f ∈ C([0, 1] × R⁴, [0, +∞)).

One can see that all lower order derivatives are involved in the nonlinear term explicitly and the BCs are the m-point cases. In this sense, the problems studied in this paper are more general than before. In the paper, multiple monotone and convex positive solutions for problems (1.1), (1.2) and (1.1), (1.3) are established. The results presented extend the study for fourth-order boundary value problems of nonlinear ordinary differential equations.

This paper is organized as follows. In Section 2, we present some preliminaries and lemmas. Section 3 is devoted to the existence of at least three convex and increasing positive solutions for problem (1.1), (1.2). In Section 4, we prove that there exist at least three convex and decreasing positive solutions for problem (1.1), (1.3).

2 Preliminaries and lemmas

In this section, some preliminaries and lemmas used later are presented.

Definition 2.1 The map α is said to be a nonnegative continuous convex functional on cone P of a real Banach space E provided that α : P → [0, +∞) is continuous and

α (t x + (1 - t) y) \leq t α (x) + (1 - t) α (y) for all x, y \in P and t \in [0, 1] .

Definition 2.2 The map β is said to be a nonnegative continuous concave functional on cone P of a real Banach space E provided that β : P → [0, +∞) is continuous and

β (t x + (1 - t) y) \geq t β (x) + (1 - t) β (y), for all x, y \in P and t \in [0, 1] .

Let γ, θ be nonnegative continuous convex functionals on P, α be a nonnegative continuous concave functional on P and ψ be a nonnegative continuous functional on P. Then for positive numbers a, b, c and d, we define the following convex sets:

\begin{gathered} P (γ, d) = {x \in P | γ (x) < d}, \\ P (γ, α, b, d) = {x \in P | b \leq α (x), γ (x) \leq d}, \\ P (γ, θ, α, b, c, d) = {x \in P | b \leq α (x), θ (x) \leq c, γ (x) \leq d} \end{gathered}

and a closed set

R (γ, ψ, a, d) = {x \in P | a \leq ψ (x), γ (x) \leq d} .

Lemma 2.1 [25] Let P be a cone in Banach space E. Let γ, θ be nonnegative continuous convex functionals on P, α be a nonnegative continuous concave functional and ψ be a nonnegative continuous functional on P satisfying

ψ (λ x) \leq λ ψ (x), f o r 0 \leq λ \leq 1,

such that for some positive numbers l and d,

α (x) \leq ψ (x), ∥ x ∥ \leq l γ (x)

for all $x \in \bar{P (γ, d)}$ . Suppose $T : \bar{P (γ, d)} \to \bar{P (γ, d)}$ is completely continuous and there exist positive numbers a, b, c with a < b such that

(S₁) {x ∈ P (γ, θ, α, b, c, d)|α(x) > b} ≠ ∅ and α(Tx) > b for x ∈ P (γ, θ, α, b, c, d);

(S₂) α(Tx) > b for x ∈ P (γ, α, b, d) with θ(Tx) > c;

(S₃) 0 ∉ R(γ, ψ, a, d) and ψ(Tx) < a for x ∈ R(γ, ψ, a, d) with ψ(x) = a.

Then T has at least three fixed points x₁, x₂,

x_{3} \in \bar{P (γ, d)}

such that:

γ (x_{i}) \leq d, i = 1, 2, 3; b < α (x_{1}); a < ψ (x_{2}), α (x_{2}) < b; ψ (x_{3}) < a .

3 Positive solutions for problem (1.1), (1.2)

We begin with the fourth-order m-point boundary value problem

x^{(4)} (t) = y (t), t \in [0, 1],

(3.1)

x^{‴} (1) = 0, x^{″} (1) = 0, x^{'} (0) = 0, x (1) = \sum_{i = 1}^{m - 2} β_{i} x (ξ_{i}),

(3.2)

where 0 < ξ₁ < ξ₂ < ⋯ < ξ_m-2< 1, β_i > 0, i = 1, 2, ..., m - 2.

The following assumption will stand throughout this section:

(H_{1}) f \in C ([0, 1] \times R^{4}, [0, + \infty)), \sum_{i = 1}^{m - 2} β_{i} > 1, \sum_{i = 1}^{m - 2} β_{i} ξ_{i} < 1 .

Lemma 3.1 Denote ξ₀ = 0, ξ_{m -1}= 1, β₀ = β_{m -1}= 0, and y(t) ∈ C[0, 1]. Problem (3.1), (3.2) has the unique solution

x (t) = \int_{0}^{1} G (t, s) y (s) d s,

where

G (t, s) = \{\begin{matrix} - \frac{1}{6} t^{3} + \frac{1}{2} s t^{2} + \frac{1}{6} s^{3} + \frac{\sum_{k = 0}^{i - 1} β_{k} (\frac{1}{6} ξ_{k}^{3} - \frac{1}{2} ξ_{k}^{2} s - \frac{1}{6} s^{3}) + \frac{1}{2} (1 - \sum_{k = i}^{m - 2} β_{k} ξ_{k}) s^{2}}{\sum_{k = 0}^{m - 1} β_{k} - 1}, & t \leq s, \\ \frac{1}{2} s^{2} t + \frac{\sum_{k = 0}^{i - 1} β_{k} (\frac{1}{6} ξ_{k}^{3} - \frac{1}{2} ξ_{k}^{2} s - \frac{1}{6} s^{3}) + \frac{1}{2} (1 - \sum_{k = i}^{m - 2} β_{k} ξ_{k}) s^{2}}{\sum_{k = 0}^{m - 1} β_{k} - 1}, & t \geq s, \end{matrix}

for ξ_i-1≤ s ≤ ξ_i, i = 1, 2, ..., m -1.

Proof Let G(t, s) be the Green's function of problem x⁽⁴⁾(t) = 0 with boundary condition (3.2). We can suppose

G (t, s) = \{\begin{matrix} a_{3} t^{3} + a_{2} t^{2} + a_{1} t + a_{0}, t \leq s, ξ_{i - 1} \leq s \leq ξ_{i}, i = 1, 2, \dots, m - 1, \\ b_{3} t^{3} + b_{2} t^{2} + b_{1} t + b_{0}, t \geq s, ξ_{i - 1} \leq s \leq ξ_{i}, i = 1, 2, \dots, m - 1 . \end{matrix}

Considering the definition and properties of Green's function together with the boundary condition (3.2), we have

\{\begin{matrix} a_{3} s^{3} + a_{2} s^{2} + a_{1} s + a_{0} = b_{3} s^{3} + b_{2} s^{2} + b_{1} s + b_{0}, \\ 3 a_{3} s^{2} + 2 a_{2} s + a_{1} = 3 b_{3} s^{2} + 2 b_{2} s + b_{1}, \\ 6 a_{3} s + 2 a_{2} = 6 b_{3} s + 2 b_{2}, \\ 6 a_{3} - 6 b_{3} = - 1, \\ 6 b_{3} = 0, 6 b_{3} + 2 b_{2} = 0, \\ a_{1} = 0, \\ b_{3} + b_{2} + b_{1} + b_{0} = \sum_{k = 0}^{i = 1} β_{k} (a_{3} ξ_{k}^{3} + a_{2} ξ_{k}^{2} + a_{1} ξ_{k} + a_{0}) + \sum_{k = i}^{m - 2} β_{k} (b_{3} ξ_{k}^{3} + b_{2} ξ_{k}^{2} + b_{1} ξ_{k} + b_{0}) . \end{matrix}

A straightforward calculation shows that

\begin{gathered} a_{3} = - \frac{1}{6}, a_{2} = \frac{s}{2}, a_{1} = 0, a_{0} = \frac{\sum_{k = 0}^{i - 1} β_{k} (\frac{1}{6} ξ_{k}^{3} - \frac{1}{2} ξ_{k}^{2} s - \frac{1}{6} s^{3}) + \frac{1}{2} (1 - \sum_{k = i}^{m - 2} β_{k} ξ_{k}) s^{2}}{\sum_{k = 0}^{m - 1} β_{k} - 1} + \frac{1}{6} s^{3}, \\ b_{3} = b_{2} = 0, b_{1} = \frac{s^{2}}{2}, b_{0} = \frac{\sum_{k = 0}^{i - 1} β_{k} (\frac{1}{6} ξ_{k}^{3} - \frac{1}{2} ξ_{k}^{2} s - \frac{1}{6} s^{3}) + \frac{1}{2} (1 - \sum_{k = i}^{m - 2} β_{k} ξ_{k}) s^{2}}{\sum_{k = 0}^{m - 1} β_{k} - 1} \end{gathered}

These give the explicit expression of the Green's function and the proof of Lemma 3.1 is completed.

Lemma 3.2 One can see that G(t, s) ≥ 0, t, s ∈ [0, 1].

Proof For ξ_i-1≤ s ≤ ξ_i, i = 1, 2, ..., m - 1,

\frac{\partial G (t, s)}{\partial t} = \{\begin{matrix} \frac{1}{2} t (2 s - t), & t \leq s, ξ_{i - 1} \leq s \leq ξ_{i}, \\ \frac{1}{2} s^{2}, & t \geq s, ξ_{i - 1} \leq s \leq ξ_{i} . \end{matrix}

Then

\frac{\partial G (t, s)}{\partial t} \geq 0

, 0 ≤ t, s ≤ 1. Thus G(t, s) is increasing on t. By a simple computation, we see

G (0, s) = \frac{1}{6} s^{3} + \frac{\sum_{k = 0}^{i - 1} β_{k} (\frac{1}{6} ξ_{k}^{3} - \frac{1}{2} ξ_{k}^{2} s - \frac{1}{6} s^{3}) + \frac{1}{2} (1 - \sum_{k = i}^{m - 2} β_{k} ξ_{k}) s^{2}}{\sum_{k = 0}^{m - 1} β_{k} - 1} \geq 0 .

These ensures that G(t, s) ≥ 0, t, s ∈ [0, 1].

Lemma 3.3 Suppose x(t) ∈ C³[0, 1] and

x^{‴} (1) = 0, x^{″} (1) = 0, x^{'} (0) = 0, x (1) = \sum_{i = 1}^{m - 2} β_{i} x (ξ_{i}) .

Furthermore x⁽⁴⁾(t) ≥ 0 and there exist t₀ such that x⁽⁴⁾(t₀) > 0. Then x(t) has the following properties:

(1)

$min_{0 \leq t \leq 1} | x (t) | \geq δ max_{0 \leq t \leq 1} | x (t) |,$
(2)

$max_{0 \leq t \leq 1} | x (t) | \leq γ max_{0 \leq t \leq 1} | x^{'} (t) |,$
(3)

$max_{0 \leq t \leq 1} | x^{'} (t) | \leq max_{0 \leq t \leq 1} | x^{″} (t) | max_{0 \leq t \leq 1} | x^{″} (t) | \leq max_{0 \leq t \leq 1} | x^{‴} (t) |,$

where $δ = (1 - \sum_{i = 1}^{m - 2} β_{i} ξ_{i}) ∕ \sum_{i = 1}^{m - 2} β_{i} (1 - ξ_{i}), γ = \sum_{i = 1}^{m - 2} β_{i} (1 - ξ_{i}) ∕ (\sum_{i = 1}^{m - 2} β_{i} - 1)$ are positive constants.

Proof Since x⁽⁴⁾(t) ≥ 0, t ∈ [0, 1], then x'''(t) is increasing on [0, 1]. Considering x'''(1) = 0, we have x'''(t) ≤ 0, t ∈ [0, 1]. Thus x''(t) is decreasing on [0, 1]. Considering this together with the boundary condition x''(1) = 0, we conclude that x''(t) ≥ 0. Then x(t) is convex on [0, 1]. Taking into account that x'(0) = 0, we get that

max_{0 \leq t \leq 1} x (t) = x (1), min_{0 \leq t \leq 1} x (t) = x (0) .

(1)

From the concavity of x(t), we have
$ξ_{i} (x (1) - x (0)) \geq x (ξ_{i}) - x (0) .$

Multiplying both sides with β_i and considering the boundary condition, we have

(1 - \sum_{i = 1}^{m - 2} β_{i} ξ_{i}) x (1) \leq \sum_{i = 1}^{m - 2} β_{i} (1 - ξ_{i}) x (0) .

(3.3)

Thus

min_{0 \leq t \leq 1} | x (t) | \geq δ max_{0 \leq t \leq 1} | x (t) | .

(2)

Considering the mean-value theorem together with the concavity of x(t), we have
$x (1) - x (ξ_{i}) \leq (1 - ξ_{i}) x^{'} (1) .$

(3.4)

Multiplying both sides with β_i and considering the boundary condition, we have

(\sum_{i = 1}^{m - 2} β_{i} - 1) x (1) \leq \sum_{i = 1}^{m - 2} β_{i} (1 - ξ_{i}) x^{'} (1),

(3.5)

which yields that

x (1) \leq \sum_{i = 1}^{m - 2} β_{i} (1 - ξ_{i}) ∕ (\sum_{i = 1}^{m - 2} β_{i} - 1) | x^{'} (1) | = γ max_{0 \leq t \leq 1} | x^{'} (t) |

.

(3)

For $x^{'} (t) = x^{'} (0) + \int_{0}^{t} x^{″} (s) d s$ and x'(0) = 0, we get
$| x^{'} (t) | = | \int_{0}^{t} x^{″} (s) d s | \leq \int_{0}^{1} | x^{″} (s) | d s .$

For

x^{″} (t) = x^{″} (1) - \int_{t}^{1} x^{‴} (s) d s

and x"(1) = 0, we get

| x^{″} (t) | = | \int_{t}^{1} x^{‴} (s) d s | \leq \int_{0}^{1} | x^{‴} (s) | d s .

Consequently

max_{0 \leq t \leq 1} | x^{'} (t) | \leq max_{0 \leq t \leq 1} | x^{″} (t) |, max_{0 \leq t \leq 1} | x^{″} (t) | \leq max_{0 \leq t \leq 1} | x^{‴} (t) | .

These give the proof of Lemma 3.3.

Remark Lemma 3.3 ensures that

max {max_{0 \leq t \leq 1} | x (t) |, max_{0 \leq t \leq 1} | x^{'} (t) |, max_{0 \leq t \leq 1} | x^{″} (t) |, max_{0 \leq t \leq 1} | x^{‴} (t) |} \leq γ max_{0 \leq t \leq 1} | x^{‴} (t) | .

Let Banach space E = C³[0, 1] be endowed with the norm

∥ x ∥ = max {max_{0 \leq t \leq 1} | x (t) |, max_{0 \leq t \leq 1} | x^{'} (t) |, max_{0 \leq t \leq 1} | x^{″} (t) |, max_{0 \leq t \leq 1} | x^{‴} (t) |}, x \in E .

Define the cone P ⊂ E by

P = {x \in E | x (t) \geq 0, x^{‴} (1) = 0, x^{″} (1) = 0, x^{'} (0) = 0, x (1) = \sum_{i = 1}^{m - 2} β_{i} x (ξ_{i}), x (t) is convex on [0,1]} .

Let the nonnegative continuous concave functional α, the nonnegative continuous convex functionals γ, θ and the nonnegative continuous functional ψ be defined on the cone by

γ (x) = max_{0 \leq t \leq 1} | x^{‴} (t) |, θ (x) = ψ (x) = max_{0 \leq t \leq 1} | x (t) |, α (x) = min_{0 \leq t \leq 1} | x (t) | .

By Lemma 3.3, the functionals defined above satisfy

δ θ (x) \leq α (x) \leq θ (x) = ψ (x), ∥ x ∥ \leq γ γ (x) .

(3.6)

Denote

m = \int_{0}^{1} G (0, s) d s, N = \int_{0}^{1} G (1, s) d s, λ = min {m, δ γ} .

Assume that there exist constants 0 < a, b, d with a < b < λd such that

\begin{array}{l} (A_{1}) f (t, u, v, w, p) \leq d, & (t, u, v, w, p) \in [0,1] \times [0, γ d] \times [0, d] \times [0, d] \times [- d, 0], \\ (A_{2}) f (t, u, v, w, p) > b / m, & (t, u, v, w, p) \in [0, 1] \times [b, b / δ] \times [0, d] \times [0, d] \times [- d, 0], \\ (A_{3}) f (t, u, v, w, p) < a / N, & (t, u, v, w, p) \in [0, 1] \times [0, a] \times [0, d] \times [0, d] \times [- d, 0] . \end{array}

Theorem 3.1 Under assumptions (A₁)-(A₃), problem (1.1), (1.2) has at least three positive solutions x₁, x₂, x₃ satisfying

\begin{gathered} max_{0 \leq t \leq 1} | {x^{‴}}_{i} (t) | \leq d, i = 1, 2, 3; b < min_{0 \leq t \leq 1} | x_{1} (t) |; \\ a < max_{0 \leq t \leq 1} | x_{2} (t) |, min_{0 \leq t \leq 1} | x_{2} (t) | < b; \\ max_{0 \leq t \leq 1} | x_{3} (t) | \leq a . \end{gathered}

Proof Problem (1.1, 1.2) has a solution x = x(t) if and only if x solves the operator equation

x (t) = \int_{0}^{1} G (t, s) f (s, x (s), x^{'} (s), x^{″} (s), x^{‴} (s)) d s = (T x) (t) .

Then

{(T x)}^{‴} (t) = - \int_{t}^{1} f (s, x, x^{'}, x^{″}, x^{‴}) d s .

For

x \in \bar{P (γ, d)}

, considering Lemma 3.3 and assumption (A₁), we have f(t, x(t), x'(t), x''(t), x'''(t)) ≤ d. Thus

γ (T x) = | {(T x)}^{‴} (0) | = | - \int_{0}^{1} f (s, x, x^{'}, x^{″}, x^{‴}) d s | = \int_{0}^{1} | f (s, x, x^{'}, x^{″}, x^{‴}) | d s \leq d .

Hence

T : \bar{P (γ, d)} \to \bar{P (γ, d)}

. An application of the Arzela-Ascoli theorem yields that T is a completely continuous operator. The fact that the constant function x(t) = b/δ ∈ P(γ, θ, α, b, c, d) and α(b/δ) > b implies that

{x \in P (γ, θ, α, b, c, d) | α (x) > b} \neq \emptyset .

For x ∈ P(γ, θ, α, b, c, d), we have b ≤ x(t) ≤ b/δ and |x'''(t)| < d. From assumption (A₂), we see

f (t, x, x^{'}, {x^{'}}^{'}, {x^{'}}^{'}^{'}) > b / m .

Hence, by definition of α and the cone P, we can get

α (T x) = (T x) (0) = \int_{0}^{1} G (0, s) f (s, x, x^{'}, x^{″}, x^{‴}) d s \geq \frac{b}{m} \int_{0}^{1} G (0, s) d s > \frac{b}{m} m = b,

which means α(Tx) > b, ∀x ∈ P(γ, θ, α, b, b/δ, d). This ensures that condition (S1) of Lemma 2.1 is fulfilled.

Second, with (3.4) and b < λd, we have

α (T x) \geq δ θ (T x) > δ \times \frac{b}{δ} = b

for all x ∈ P(γ, α, b, d) with $θ (T x) > \frac{b}{δ}$ .

Thus, condition (S₂) of Lemma 2.1 holds. Finally we show that (S₃) also holds. We see ψ(0) = 0 < a and 0 ∉ R(γ, ψ, a, d). Suppose that x ∈ R(γ, ψ, a, d) with ψ(x) = a, then by the assumption of (A₃),

ψ (T x) = max_{0 \leq t \leq 1} | (T x) (t) | = \int_{0}^{1} G (1, s) f (s, x, x^{'}, x^{″}, x^{‴}) d s < \frac{a}{N} \int_{0}^{1} G (1, s) d s = a,

which ensures that condition (S₃) of Lemma 2.1 is fulfilled. Thus, an application of Lemma 2.1 implies that the fourth-order m-point boundary value problem (1.1, 1.2) has at least three positive convex increasing solutions x₁, x₂, x₃ with the properties that

\begin{gathered} max_{0 \leq t \leq 1} | x_{i}^{‴} (t) | \leq d, i = 1, 2, 3; b < min_{0 \leq t \leq 1} | x_{1} (t) |; \\ a < max_{0 \leq t \leq 1} | x_{2} (t) |, min_{0 \leq t \leq 1} | x_{2} (t) | < b; \\ max_{0 \leq t \leq 1} | x_{3} (t) | \leq a . \end{gathered}

4 Positive solutions for problem (1.1), (1.3)

The following assumption will stand throughout this section:

(H_{2}), f \in C ([0, 1] \times R^{4}, [0, + \infty)), \sum_{i = 1}^{m - 2} β_{i} > 1, \sum_{i = 1}^{m - 2} β_{i} ξ_{i} + 1 - \sum_{i = 1}^{m - 2} β_{i} > 0 .

Lemma 4.1 Denote ξ₀ = 0, ξ_m-1= 1, β₀ = β_m-1= 0, the Green's function of problem

x^{(4)} (t) = 0,

(4.1)

x^{‴} (1) = 0, x^{″} (1) = 0, x^{'} (1) = 0, x (0) = \sum_{i = 1}^{m - 2} β_{i} x (ξ_{i}),

(4.2)

is

H (t, s) = \{\begin{matrix} - \frac{1}{6} t^{3} + \frac{1}{2} s t^{2} - \frac{1}{2} s^{2} t + \frac{\sum_{k = 0}^{i - 1} β_{k} (\frac{1}{6} ξ_{k}^{3} - \frac{1}{2} ξ_{k}^{2} s + \frac{1}{2} ξ_{k} s^{2}) + \sum_{k = i}^{m - 2} \frac{1}{6} β_{k} s^{3}}{\sum_{k = 0}^{m - 1} β_{k} - 1}, & t \leq s, ξ_{i - 1} \leq s \leq ξ_{i}, \\ - \frac{1}{6} s^{3} + \frac{\sum_{k = 0}^{i - 1} β_{k} (\frac{1}{6} ξ_{k}^{3} - \frac{1}{2} ξ_{k}^{2} s + \frac{1}{2} ξ_{k} s^{2}) + \sum_{k = i}^{m - 2} \frac{1}{6} β_{k} s^{3}}{\sum_{k = 0}^{m - 1} β_{k} - 1}, & t \geq s, ξ_{i - 1} \leq s \leq ξ_{i}, \end{matrix}

for i = 1, 2, ..., m - 1.

Proof Suppose that

H (t, s) = \{\begin{matrix} a_{3} t^{3} + a_{2} t^{2} + a_{1} t + a_{0} & t \leq s, ξ_{i - 1} \leq s \leq ξ_{i}, i = 1, 2, \dots, m - 1, \\ b_{3} t^{3} + b_{2} t^{2} + b_{1} t + b_{0} & t \geq s, ξ_{i - 1} \leq s \leq ξ_{i}, i = 1, 2, \dots, m - 1 . \end{matrix}

Considering the definition and properties of Green's function together with the boundary condition (4.2), we have

\{\begin{matrix} a_{3} s^{3} + a_{2} s^{2} + a_{1} s + a_{0} = b_{3} s^{3} + b_{2} s^{2} + b_{1} s + b_{0}, \\ 3 a_{3} s^{2} + 2 a_{2} s + a_{1} = 3 b_{3} s^{2} + 2 b_{2} s + b_{1}, \\ 6 a_{3} s + 2 a_{2} = 6 b_{3} s + 2 b_{2}, \\ 6 a_{3} - 6 b_{3} = - 1, \\ b_{3} = 0, \\ 6 b_{3} + 2 b_{2} = 0, \\ 3 b_{3} + 2 b_{2} + b_{1} = 0, \\ a_{0} = \sum_{k = 0}^{i = 1} β_{k} (a_{3} ξ_{k}^{3} + a_{2} ξ_{k}^{2} + a_{1} ξ_{k} + a_{0}) + \sum_{k = i}^{m - 2} β_{k} (b_{3} ξ_{k}^{3} + b_{2} ξ_{k}^{2} + b_{1} ξ_{k} + b_{0}) . \end{matrix}

Consequently

\begin{gathered} a_{3} = - \frac{1}{6}, a_{2} = \frac{1}{2} s, a_{1} = - \frac{1}{2} s^{2}, b_{3} = b_{2} = b_{1} = 0, \\ a_{0} = \frac{\sum_{k = 0}^{i - 1} β_{k} (\frac{1}{6} ξ_{k}^{3} - \frac{1}{2} ξ_{k}^{2} s + \frac{1}{2} ξ_{k} s^{2}) + \sum_{k = i}^{m - 2} \frac{1}{6} β_{k} s^{3}}{\sum_{k = 0}^{m - 1} β_{k} - 1}, \\ b_{0} = \frac{\sum_{k = 0}^{i - 1} β_{k} (\frac{1}{6} ξ_{k}^{3} - \frac{1}{2} ξ_{k}^{2} s + \frac{1}{2} ξ_{k} s^{2}) + \sum_{k = i}^{m - 2} \frac{1}{6} β_{k} s^{3}}{\sum_{k = 0}^{m - 1} β_{k} - 1} - \frac{1}{6} s^{3} . \end{gathered}

The proof of Lemma 4.1 is completed.

Lemma 4.2 One can see that H(t, s) ≥ 0, t, s ∈ [0, 1].

Proof For ξ_i-1≤ s ≤ ξ_i, i = 1, 2, ..., m - 1,

\frac{\partial H (t, s)}{\partial t} = \{\begin{matrix} - \frac{1}{2} {(t - s)}^{2}, & t \leq s, ξ_{i - 1} \leq s \leq ξ_{i}, \\ 0, & t \geq s, ξ_{i - 1} \leq s \leq ξ_{i} . \end{matrix}

Then

\frac{\partial H (t, s)}{\partial t} \leq 0

, 0 ≤ t, s ≤ 1, which implies that H(t, s) is decreasing on t. The fact that

H (1, s) = \frac{\sum_{k = 0}^{i - 1} β_{k} (\frac{1}{6} ξ_{k}^{3} - \frac{1}{2} ξ_{k}^{2} s + \frac{1}{2} ξ_{k} s^{2}) + \sum_{k = i}^{m - 2} \frac{1}{6} β_{k} s^{3}}{\sum_{k = 0}^{m - 1} β_{k} - 1} - \frac{1}{6} s^{3} \geq 0

ensures that H(t, s) ≥ 0, t, s ∈ [0, 1].

Lemma 4.3 If x(t) ∈ C³[0, 1],

x^{‴} (1) = 0, x^{″} (1) = 0, x^{'} (1) = 0, x (0) = \sum_{i = 1}^{m - 2} β_{i} x (ξ_{i}),

and x⁽⁴⁾(t) ≥ 0, there exists t₀ such that x⁽⁴⁾(t₀) > 0, then

(1)

$min_{0 \leq t \leq 1} | x (t) | \geq δ_{1} max_{0 \leq t \leq 1} | x (t) |,$
(2)

$max_{0 \leq t \leq 1} | x (t) | \leq γ_{1} max_{0 \leq t \leq 1} | x^{'} (t) |,$
(3)

$max_{0 \leq t \leq 1} | x^{'} (t) | \leq max_{0 \leq t \leq 1} | x^{″} (t) |, max_{0 \leq t \leq 1} | x^{″} (t) | \leq max_{0 \leq t \leq 1} | x^{‴} (t) |$

where $δ_{1} = (1 - \sum_{i = 1}^{m - 2} β_{i} (1 - ξ_{i})) ∕ \sum_{i = 1}^{m - 2} β_{i} ξ_{i}, γ_{1} = \sum_{i = 1}^{m - 2} β_{i} ξ_{i} ∕ (\sum_{i = 1}^{m - 2} β_{i} - 1)$ are positive constants.

Proof It follows from the same methods as Lemma 3.3 that x(t) is convex on [0, 1]. Taking into account that x'(1) = 0, one can see that x(t) is decreasing on [0, 1] and

max_{0 \leq t \leq 1} x (t) = x (0), min_{0 \leq t \leq 1} x (t) = x (1) .

(1)

From the concavity of x(t), we have
$ξ_{i} (x (1) - x (0)) \geq x (ξ_{i}) - x (0) .$

Multiplying both sides with β_i and considering the boundary condition, we have

\sum_{i = 1}^{m - 2} β_{i} ξ_{i} x (1) \geq (1 - \sum_{i = 1}^{m - 2} β_{i} (1 - ξ_{i})) x (0) .

(4.3)

Thus

min_{0 \leq t \leq 1} | x (t) | \geq δ_{1} max_{0 \leq t \leq 1} | x (t) | .

(2)

Considering the mean-value theorem, we get
$(0) - x (ξ_{i}) \leq ξ_{i} | x^{'} (0) | .$

From the concavity of x similarly with above we know

(\sum_{i = 1}^{m - 2} β_{i} - 1) x (0) < \sum_{i = 1}^{m - 2} β_{i} ξ_{i} | x^{'} (0) | .

(4.4)

Considering (4.3) together with (4.4) we have

x (0) \leq γ_{1} | x^{'} (0) | = γ_{1} max_{0 \leq t \leq 1} | x^{'} (t) |

.

(3)

For $x^{'} (t) = x^{'} (1) - \int_{t}^{1} x^{″} (s) d s, x^{″} (t) = x^{″} (1) - \int_{t}^{1} x^{‴} (s) d s$ and x'(1) = 0, x''(1) = 0, we get
$| x^{'} (t) | = | \int_{t}^{1} x^{″} (s) d s | \leq \int_{0}^{1} | x^{″} (s) | d s, | x^{″} (t) | = | \int_{t}^{1} x^{‴} (s) d s | \leq \int_{0}^{1} | x^{‴} (s) | d s .$

Thus

max_{0 \leq t \leq 1} | x^{'} (t) | \leq max_{0 \leq t \leq 1} | x^{″} (t) |, max_{0 \leq t \leq 1} | x^{″} (t) | \leq max_{0 \leq t \leq 1} | x^{‴} (t) | .

Remark We see that

max {max_{0 \leq t \leq 1} | x (t) |, max_{0 \leq t \leq 1} | x^{'} (t) |, max_{0 \leq t \leq 1} | x^{″} (t) |, max_{0 \leq t \leq 1} | x^{‴} (t) |} \leq γ_{1} max_{0 \leq t \leq 1} | x^{‴} (t) | .

Let Banach space E = C³[0, 1] be endowed with the norm

∥ x ∥ = max {max_{0 \leq t \leq 1} | x (t) |, max_{0 \leq t \leq 1} | x^{'} (t) |, max_{0 \leq t \leq 1} | x^{″} (t) |, max_{0 \leq t \leq 1} | x^{‴} (t) |}, x \in E .

Define the cone P ⊂ E by

P_{1} = \{x \in E | x (t) \geq 0, x^{‴} (1) = 0, x^{″} (1) = 0, x^{'} (1) = 0, x (0) = \sum_{i = 1}^{m - 2} β_{i} x (ξ_{i}), x (t) is convex on [0, 1]\} .

Denote

m_{1} = \int_{0}^{1} H (1, s) d s, N_{1} = \int_{0}^{1} H (0, s) d s, λ_{1} = min {m_{1}, δ_{1} γ_{1}} .

Assume that there exist constants 0 < a, b, d with a < b < λ₁d such that

\begin{array}{l} (A_{4}) f (t, u, v, w, p) \leq d, & (t, u, v, w, p) \in [0,1] \times [0, γ_{1} d] \times [- d, 0] \times [0, d] \times [- d, 0], \\ (A_{5}) f (t, u, v, w, p) > b / m_{1}, & (t, u, v, w, p) \in [0, 1] \times [b, b / δ_{1}] \times [- d, 0] \times [0, d] \times [- d, 0], \\ (A_{6}) f (t, u, v, w, p) < a / N_{1}, & (t, u, v, w, p) \in [0, 1] \times [0, a] \times [- d, 0] \times [0, d] \times [- d, 0] . \end{array}

Theorem 4.1 Under assumptions (A₄)-(A₆), problem (1.1), (1.3) has at least three positive solutions x₁, x₂, x₃ with the properties that

max_{0 \leq t \leq 1} | x_{i}^{‴} (t) | \leq d, i = 1, 2, 3; b < min_{0 \leq t \leq 1} | x_{1} (t) |; a < max_{0 \leq t \leq 1} | x_{2} (t) |, min_{0 \leq t \leq 1} | x_{2} (t) | < b; max_{0 \leq t \leq 1} | x_{3} (t) | \leq a .

Proof Problem (1.1), (1.3) has a solution x = x(t) if and only if x solves the operator equation

x (t) = \int_{0}^{1} H (t, s) f (s, x (s), x^{'} (s), x^{″} (s), x^{‴} (s)) d s = (T_{1} x) (t) .

Then

{(T_{1} x)}^{‴} (t) = - \int_{t}^{1} f (s, x, x^{'}, x^{″}, x^{‴}) d s .

For

x \in \bar{P_{1} (γ, d)}

, considering Lemma 4.3 and assumption (A₄), we have

f (t, x (t), x^{'} (t), x^{″} (t), x^{‴} (t)) \leq d .

Thus

γ (T_{1} x) = | {(T_{1} x)}^{‴} (0) | = | - \int_{0}^{1} f (s, x, x^{'}, x^{″}, x^{‴}) d s | = \int_{0}^{1} | f (s, x, x^{'}, x^{″}, x^{‴}) | d s \leq d .

Hence $T_{1} : \bar{P_{1} (γ, d)} \to \bar{P_{1} (γ, d)}$ and T₁ is a completely continuous operator obviously. The fact

that the constant function x(t) = b/δ₁ ∈ P₁(γ, θ, α, b, c, d) and α(b/δ₁) > b implies that

{x \in P_{1} (γ, θ, α, b, c, d | α (x) > b)} \neq \emptyset .

This ensures that condition (S1) of Lemma 2.1 holds.

For x ∈ P₁(γ, θ, α, b, c, d), we have b ≤ x(t) ≤ b/δ₁ and |x'''(t)| < d. From assumption (A₄),

f (t, x, x^{'}, {x^{'}}^{'}, {x^{'}}^{'}^{'}) > b / m_{1} .

Hence, by definition of α and the cone P₁, we can get

α (T_{1} x) = (T_{1} x) (1) = \int_{0}^{1} H (1, s) f (s, x, x^{'}, x^{″}, x^{‴}) d s \geq \frac{b}{m_{1}} \int_{0}^{1} H (1, s) d s > \frac{b}{m_{1}} m_{1} = b,

which means α(T₁x) > b, ∀x ∈ P₁(γ, θ, α, b, b/δ, d).

Second, with (4.4) and b < λ₁d, we have

α (T_{1} x) \geq δ_{1} θ (T_{1} x) > δ_{1} \times \frac{b}{δ_{1}} = b

for all x ∈ P₁(γ, α, b, d) with $θ (T_{1} x) > \frac{b}{δ_{1}}$ .

Thus, condition (S₂) of Lemma 2.1 holds. Finally we show that (S₃) also holds. We see ψ(0) = 0 < a and 0 ∉ R(γ, ψ, a, d). Suppose that x ∈ R(γ, ψ, a, d) with ψ(x) = a, then by the assumption of (A₆),

ψ (T_{1} x) = max_{0 \leq t \leq 1} | (T_{1} x) (t) | = \int_{0}^{1} H (0, s) f (s, x, x^{'}, x^{″}, x^{‴}) d s < \frac{a}{N_{1}} \int_{0}^{1} H (0, s) d s = a,

which ensures that condition (S₃) of Lemma 2.1 is satisfied. Thus, an application of Lemma 2.1 implies that the fourth-order m-point boundary value problem (1.1), (1.3) has at least three positive convex decreasing solutions x₁, x₂, x₃ satisfying the conditions that

max_{0 \leq t \leq 1} | x_{i}^{‴} (t) | \leq d, i = 1, 2, 3; b < min_{0 \leq t \leq 1} | x_{1} (t) |; a < max_{0 \leq t \leq 1} | x_{2} (t) |, min_{0 \leq t \leq 1} | x_{2} (t) | < b; max_{0 \leq t \leq 1} | x_{3} (t) | \leq a .

Declarations

Acknowledgements

We are indebted to the anonymous referee for a detailed reading and useful comments and suggestions, which allowed us to improve this work. This work was supported by the Anhui Provincial Natural Science Foundation (10040606Q50), National Natural Science Foundation of China (No.11071164), Shanghai Natural Science Foundation (No.10ZR1420800).

Authors’ Affiliations

(1)

College of Science, University of Shanghai for Science and Technology

(2)

Department of Mathematics, Hefei Normal University

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