Skip to main content

Positive solutions for nonlocal fourth-order boundary value problems with all order derivatives

Abstract

In this article, by the fixed point theorem in a cone and the nonlocal fourth-order BVP's Green function, the existence of at least one positive solution for the nonlocal fourth-order boundary value problem with all order derivatives

u ( 4 ) ( t ) + A u ( t ) = λ f ( t , u ( t ) , u ( t ) , u ( t ) , u ( t ) ) , 0 < t < 1 , u ( 0 ) = u ( 1 ) = 0 1 p ( s ) u ( s ) d s , u ( 0 ) = u ( 1 ) = 0 1 q ( s ) u ( s ) d s

is considered, where f is a nonnegative continuous function, λ > 0, 0 < A < π2, p, q L[0, 1], p(s) ≥ 0, q(s) ≥ 0. The emphasis here is that f depends on all order derivatives.

1 Introduction

The deformation of an elastic beam in equilibrium state, whose two ends are simply supported, can be described by a fourth-order ordinary equation boundary value problem. Owing to its significance in physics, the existence of positive solutions for the fourth-order boundary value problem has been studied by many authors using nonlinear alternatives of Leray-Schauder, the fixed point index theory, the Krasnosel'skii's fixed point theorem and the method of upper and lower solutions, in reference [110].

In recent years, there has been much attention on the question of positive solutions of the fourth-order differential equations with one or two parameters. By the Krasnosel'skii's fixed point theorem in cone [11], Bai [5] investigated the following fourth-order boundary value problem with one parameter

u ( 4 ) ( t ) + β u ( t ) = λ f ( t , u ( t ) , u ( t ) ) , 0 < t < 1 , u ( 0 ) = u ( 1 ) = 0 1 p ( s ) u ( s ) d s , u ( 0 ) = u ( 1 ) = 0 1 q ( s ) u ( s ) d s ,

where λ > 0, 0 < β < π2, f: C([0, 1] × [0, ∞) × (-∞, 0], [0, ∞)) is continuous, p, q L[0, 1], p(s) ≥ 0, q(s) ≥ 0, 0 1 p ( s ) ds<1, 0 1 q ( s ) sin β sds+ 0 1 q ( s ) sin β ( 1 - s ) ds<sin β .

By the fixed point index in cone, Ma [7] proved the existence of symmetric positive solutions for the nonlocal fourth-order boundary value problem

u ( 4 ) ( t ) = h ( t ) f ( t , u ( t ) ) , 0 < t < 1 , u ( 0 ) = u ( 1 ) = 0 1 p ( s ) u ( s ) d s , u ( 0 ) = u ( 1 ) = 0 1 q ( s ) u ( s ) d s .

All the above works were done under the assumption that all order derivatives u', u″, u‴ are not involved explicitly in the nonlinear term f. In this article, we are concerned with the existence of positive solutions for the nonlocal fourth-order boundary value problem

u ( 4 ) ( t ) + A u ( t ) = λ f ( t , u ( t ) , u ( t ) , u ( t ) , u ( t ) ) , 0 < t < 1 , u ( 0 ) = u ( 1 ) = 0 1 p ( s ) u ( s ) d s , u ( 0 ) = u ( 1 ) = 0 1 q ( s ) u ( s ) d s .
(1.1)

Throughout, we assume

(H1) λ > 0, 0 < A < π2;

(H2) f: [0, 1] × R4R+ is continuous, p, q L[0, 1], p(s) ≥ 0, q(s) ≥ 0, 0 1 p ( s ) ds<1, 0 1 q ( s ) sin A sds+ 0 1 q ( s ) sin A ( 1 - s ) ds<sin A .

We will impose all order derivatives in f and make use of two continuous convex functionals which will ensure the existence of at least one positive solution to (1.1). Bai [5] applied Krasnoselskii's fixed point theorem. Ma [8] used fixed point index in cone and Leray-Schauder degree. In this article, to show the existence of positive solutions to (1.1), we define two positive continuous convex functionals. Then, using the new fixed point theorem [12] in a cone and the nonlocal fourth-order BVP's Green function, we give some new criteria for the existence of positive solutions to (1.1).

2 The preliminary lemmas

Let Y = C[0, 1] be the Banach space equipped with the norm

| | u ( t ) | | 0 = max t [ 0 , 1 ] | u ( t ) | .

Set λ1, λ2 be the roots of the polynomial P(λ) = λ2 + , namely λ1 = 0, λ2 = -A. By (H1), it is obviously that -π2< λ2< 0.

Let Q1(t, s), Q2(t, s) be, respectively the Green's functions of the following problems

- u ( t ) + λ 1 u ( t ) = 0 , 0 < t < 1 , u ( 0 ) = u ( 1 ) = 0 1 p ( s ) u ( s ) d s , - u ( t ) + λ 2 u ( t ) = 0 , 0 < t < 1 , u ( 0 ) = u ( 1 ) = 0 1 q ( s ) u ( s ) d s .

Then, carefully calculation yield

Q 1 ( t , s ) = G 1 ( t , s ) + 0 1 G 1 ( s , x ) p ( x ) d x 1 - 0 1 p ( x ) d x ,
Q 2 ( t , s ) = G 2 ( t , s ) + sin A t + sin A ( 1 - t ) 0 1 G 2 ( s , x ) q ( x ) d x sin A - 0 1 q ( x ) sin A x d x - 0 1 q ( x ) sin A ( 1 - x ) d x ,
G 1 ( t , s ) = s ( 1 - t ) , 0 s t 1 , t ( 1 - s ) , 0 t s 1 ,
G 2 ( t , s ) = sin A s sin A ( 1 - t ) A sin A , 0 s t 1 , sin A t sin A ( 1 - s ) A sin A , 0 t s 1 .

Denote

ω 1 = 1 1 - 0 1 p ( x ) d x , ω 2 ( t ) = sin A t + sin A ( 1 - t ) sin A - 0 1 q ( x ) sin A x d x - 0 1 q ( x ) sin A ( 1 - x ) d x .

Lemma 2.1. [5] Suppose that (H1) and (H2) hold. Then for any y(t) C[0, 1], the problem

u ( 4 ) ( t ) + A u ( t ) = y ( t ) , 0 < t < 1 , u ( 0 ) = u ( 1 ) = 0 1 p ( s ) u ( s ) d s , u ( 0 ) = u ( 1 ) = 0 1 q ( s ) u ( s ) d s .
(2.1)

has a unique solution

u ( t ) = 0 1 0 1 Q 1 ( t , s ) Q 2 ( s , τ ) y ( τ ) d τ d s ,
(2.2)

where

Q 1 ( t , s ) = G 1 ( t , s ) + ω 1 0 1 G 1 ( s , x ) p ( x ) d x , Q 2 ( s , τ ) = G 2 ( s , τ ) + ω 2 ( s ) 0 1 G 2 ( τ , x ) q ( x ) d x .

By (2.2), we get

u ( t ) = 0 1 0 1 Q 2 ( s , τ ) y ( τ ) d τ d s - 0 1 0 1 s Q 2 ( s , τ ) y ( τ ) d τ d s ;
(2.3)
u ( t ) = - 0 1 Q 2 ( t , τ ) y ( τ ) d τ ,
(2.4)
u ( t ) = - 0 1 Q 2 ( t , τ ) t y ( τ ) d τ .
(2.5)

Lemma 2.2. [5] Assume that (H1) and (H2) hold. Then one has

(i) Q i (t, s) ≥ 0, t, s [0, 1]; Q i (t, s) > 0, t, s (0, 1);

(ii) G i (t, s) ≥ b i G i (t, t)G i (s, s), t, s [0, 1];

(iii) G i (t, s) ≤ c i G i (s, s), t, s [0, 1].

where b1 = 1, b 2 = A sin A ; c1 = 1, c 2 = 1 sin A .

Let

d i = min 1 4 t 3 4 b i G i ( t , t ) , ( i = 1 , 2 ) ; ξ = min 1 4 t 3 4 ω 2 ( t ) max 1 4 t 3 4 ω 2 ( t ) .

Lemma 2.3. [5] Suppose that (H1) and (H2) hold and w2, d i , ξ i are given as above. Then

one has

(i) max 0 t 1 ω 2 ( t ) = ω 2 1 2 ;

(ii) 0< d i <1,0<ξ<1.

Lemma 2.4. If y(t) C[0, 1] and y(t) ≥ 0, then the unique solution u(t) of problem (2.1)

satisfies

min 1 4 t 3 4 u ( t ) d 1 | | u | | 0 , min 1 4 t 3 4 ( - u ( t ) ) d 2 ξ c 2 | | u | | 0 .

Proof. By (2.2) and (iii) of Lemma 2.2, we get

u ( t ) = 0 1 0 1 Q 1 ( t , s ) Q 2 ( s , τ ) y ( τ ) d τ d s 0 1 0 1 c 1 G 1 ( s , s ) + ω 1 0 1 G 1 ( s , x ) p ( x ) d x Q 2 ( s , τ ) y ( τ ) d τ d s = 0 1 0 1 G 1 ( s , s ) + ω 1 0 1 G 1 ( s , x ) p ( x ) d x Q 2 ( s , τ ) y ( τ ) d τ d s = 0 1 0 1 Q 1 ( s , s ) Q 2 ( s , τ ) d τ d s .

So,

| | u | | 0 0 1 0 1 Q 1 ( s , s ) Q 2 ( s , τ ) d τ d s .

Using (ii) of Lemma 2.2, we have

min 1 4 t 3 4 u ( t ) = min 1 4 t 3 4 0 1 0 1 Q 1 ( t , s ) Q 2 ( s , τ ) y ( τ ) d τ d s min 1 4 t 3 4 0 1 0 1 [ b 1 G 1 ( t , t ) G 1 ( s , s ) + ω 1 0 1 G 1 ( s , x ) p ( x ) d x ] Q 2 ( s , τ ) y ( τ ) d τ d s = 0 1 0 1 d 1 G 1 ( s , s ) + ω 1 0 1 G 1 ( s , x ) p ( x ) d x Q 2 ( s , τ ) y ( τ ) d τ d s d 1 0 1 0 1 G 1 ( s , s ) + ω 1 0 1 G 1 ( s , x ) p ( x ) d x Q 2 ( s , τ ) y ( τ ) d τ d s = d 1 0 1 0 1 Q 1 ( s , s ) Q 2 ( s , τ ) y ( τ ) d τ d s d 1 | | u | | 0 .

By (2: 4) and (iii) of Lemma 2.2, we get

max 1 4 t 3 4 ( - u ( t ) ) = max 1 4 t 3 4 0 1 Q 2 ( t , τ ) y ( τ ) d τ 0 1 c 2 G 2 ( τ , τ ) + max 1 4 t 3 4 ω 2 ( t ) 0 1 G 2 ( τ , x ) q ( x ) d x y ( τ ) d τ c 2 max 1 4 t 3 4 ω 2 ( t ) 0 1 G 2 ( τ , τ ) + 0 1 G 2 ( τ , x ) q ( x ) d x y ( τ ) d τ .

So,

| | u | | 0 c 2 max 1 4 t 3 4 ω 2 ( t ) 0 1 G 2 ( τ , τ ) + 0 1 G 2 ( τ , x ) q ( x ) d x y ( τ ) d τ .

Using (ii) of Lemma 2.2, we have

min 1 4 t 3 4 ( - u ( t ) ) = min 1 4 t 3 4 0 1 Q 2 ( t , τ ) y ( τ ) d τ min 1 4 t 3 4 0 1 b 2 G 2 ( t , t ) G 2 ( τ , τ ) + ω 2 ( t ) 0 1 G 2 ( τ , x ) q ( x ) d x y ( τ ) d τ 0 1 b 2 G 2 ( t , t ) G 2 ( τ , τ ) + min 1 4 t 3 4 ω 2 ( t ) 0 1 G 2 ( τ , x ) q ( x ) d x y ( τ ) d τ = 0 1 d 2 G 2 ( τ , τ ) + min 1 4 t 3 4 ω 2 ( t ) 0 1 G 2 ( τ , x ) q ( x ) d x y ( τ ) d τ d 2 min 1 4 t 3 4 ω 2 ( t ) 0 1 G 2 ( τ , τ ) + 0 1 G 2 ( τ , x ) q ( x ) d x y ( τ ) d τ d 2 c 2 min 1 4 t 3 4 ω 2 ( t ) max 1 4 t 3 4 ω 2 ( t ) | | u | | 0 d 2 ξ c 2 | | u | | 0 .

The proof is completed.

Let X be a Banach space and K X a cone. Suppose α, β: ×R+ are two continuous convex functionals satisfying α(λu) = |λ|α(u), β(λu) = |λ|β(u), for u X, λ R, and ||u|| ≤ M max{α(u), β(u)}, for u X and α(u) ≤ α(v) for u, v K, uv, where M > 0 is a constant.

Theorem 2.1. [12] Let r2> r1> 0, L > 0 be constants and

Ω i = { u X : α ( u ) < r i , β ( u ) < L } , i = 1 , 2 ,

two bounded open sets in X. Set

D i = { u X : α ( u ) = r i } , i = 1 , 2 .

Assume T: KK is a completely continuous operator satisfying

(A1) α(Tu) < r1, u D1K; α(Tu) > r2, u D2K;

(A2) β(Tu) < L, u K;

(A3) there is a p 2K) \ {0} such that α(p) ≠ 0 and α(u + λp) ≥ α(u), for all u K and λ ≥ 0.

Then T has at least one fixed point in ( Ω 2 \ Ω ̄ 1 ) K.

3 The main results

Let X = C4[0, 1] be the Banach space equipped with the norm ||u||= max t [ 0 , 1 ] |u ( t ) |+ max t [ 0 , 1 ] | u ( t ) |+ max t [ 0 , 1 ] | u ( t ) |+ max t [ 0 , 1 ] | u ( t ) |, and K= u X : u ( t ) 0 , u ( t ) 0 , min 1 4 t 3 4 u ( t ) d 1 | | u | | 0 , min 1 4 t 3 4 ( - u ( t ) ) d 2 ξ c 2 | | u | | 0 is a cone in X.

Define two continuous convex functionals α ( u ) = max t [ 0 , 1 ] |u ( t ) |+ max t [ 0 , 1 ] | u ( t ) | and β ( u ) = max t [ 0 , 1 ] | u ( t ) |+ max t [ 0 , 1 ] | u ( t ) |, for each u X, then ||u|| ≤ 2 max{α(u), β(u)} and α(λu) = |λ|α(u), β(λu) = |λ|β(u), for u X, λ R; α(u) ≤ α(v) for u, v K, uv.

In the following, we denote

B = 0 1 0 1 Q 1 ( s , s ) Q 2 ( s , τ ) d τ d s , D = 0 1 G 2 ( τ , τ ) + ω 2 1 2 0 1 G 2 ( τ , x ) q ( x ) d x d τ , F = 1 sin A 0 1 sin A τ d τ + A 0 1 0 1 G 2 ( τ , x ) q ( x ) d x d τ sin A - 0 1 q ( x ) sin A x d x - 0 1 q ( x ) sin A ( 1 - x ) d x , η 0 = 1 B + c 2 D , η 1 = 1 1 4 3 4 Q 2 1 2 , τ d τ , η 2 = 2 3 c 2 D + 4 F , θ = min d 1 2 , d 2 ξ 2 c 2 .

We will suppose that there are L > b > θb > c > 0 such that f(t, u, v, u0, v0) satisfies the following growth conditions:

( H 3 ) f ( t , u , v , u 0 , v 0 ) < c η 0 λ , for ( t , u , v , u 0 , v 0 ) [ 0 , 1 ] × [ 0 , c ] × [ - L , L ] × [ - c , 0 ] × [ - L , L ] , ( H 4 ) f ( t , u , v , u 0 , v 0 ) b η 1 λ , for ( t , u , v , u 0 , v 0 ) 1 4 , 3 4 × [ θ b , b ] × [ - L , L ] × [ - b , 0 ] × [ - L , L ] 1 4 , 3 4 × [ 0 , b ] × [ - L , L ] × [ - b , - θ b ] × [ - L , L ] , ( H 5 ) f ( t , u , v , u 0 , v 0 ) < L η 2 λ , for ( t , u , v , u 0 , v 0 ) [ 0 , 1 ] × [ 0 , b ] × [ - L , L ] × [ - b , 0 ] × [ - L , L ] .

Let f 1 ( t , u , v , u 0 , v 0 ) = f 1 t , u * , v * , u 0 * , v 0 * , where

u * = min { max ( u , 0 ) , b } , v * = min { max ( v , - L ) , L } , u 0 * = min { max ( u 0 , - b ) , 0 } , v 0 * = min { max ( v , - L ) , L } .

We denote

( T u ) ( t ) = λ 0 1 0 1 Q 1 ( t , s ) Q 2 ( s , τ ) f 1 ( τ , u ( τ ) , u ( τ ) , u ( τ ) , u ( τ ) ) d τ d s ,
(3.1)
( T u ) ( t ) = λ t 1 0 1 Q 2 ( s , τ ) f 1 ( τ , u ( τ ) , u ( τ ) , u ( τ ) , u ( τ ) ) d τ d s - 0 1 0 1 s Q 2 ( s , τ ) f 1 ( τ , u ( τ ) , u ( τ ) , u ( τ ) , u ( τ ) ) d τ d s ,
(3.2)
( T u ) ( t ) = - λ 0 1 Q 2 ( t , τ ) f 1 ( τ , u ( τ ) , u ( τ ) , u ( τ ) , u ( τ ) ) d τ ,
(3.3)
( T u ) ( t ) = - λ 0 1 Q 2 ( t , τ ) t f 1 ( τ , u ( τ ) , u ( τ ) , u ( τ ) , u ( τ ) ) d τ .
(3.4)

Lemma 3.1. Suppose that (H1) and (H2) hold. Then T: KK is completely continuous.

Proof. For u K, by (3.1), (3.3) and Lemma 2.2, it is obviously that Tu ≥ 0, (Tu)″ ≤ 0. In view of c1 = 1, c2> 1, so

| | T u | | 0 = max t [ 0 , 1 ] λ 0 1 0 1 Q 1 ( t , s ) Q 2 ( s , τ ) f 1 ( t , u ( τ ) , u ( τ ) , u ( τ ) , u ( τ ) ) d τ d s λ 0 1 0 1 [ c 1 G 1 ( s , s ) + ω 1 0 1 G 1 ( s , x ) p ( x ) d x ] Q 2 ( s , τ ) f 1 ( τ , u ( τ ) , u ( τ ) , u ( τ ) , u ( τ ) ) d τ d s = λ 0 1 0 1 Q 1 ( s , s ) Q 2 ( s , τ ) f 1 ( τ , u ( τ ) , u ( τ ) , u ( τ ) , u ( τ ) ) d τ d s ,
| | ( T u ) | | 0 = max t [ 0 , 1 ] - λ 0 1 Q 2 ( t , τ ) f 1 ( τ , u ( τ ) , u ( τ ) , u ( τ ) , u ( τ ) ) d τ λ 0 1 [ c 2 G 2 ( τ , τ ) + max 1 4 t 3 4 ω 2 ( t ) 0 1 G 2 ( τ , x ) q ( x ) d x ] f 1 ( τ , u ( τ ) , u ( τ ) , u ( τ ) , u ( τ ) ) d τ max 1 4 t 3 4 ω 2 ( t ) 0 1 G 2 ( τ , τ ) + 0 1 G 2 ( τ , x ) q ( x ) d x × f 1 ( τ , u ( τ ) , u ( τ ) , u ( τ ) , u ( τ ) ) d τ .

By Lemma 2.3, (3.1) and (3.3), we have

min 1 4 t 3 4 ( T u ) ( t ) = min 1 4 t 3 4 λ 0 1 0 1 Q 1 ( t , s ) Q 2 ( s , τ ) f 1 ( τ , u ( τ ) , u ( τ ) , u ( τ ) , u ( τ ) ) d τ d s λ 0 1 0 1 [ b 1 G 1 ( t , t ) G 1 ( s , s ) + ω 1 0 1 G 1 ( s , x ) p ( x ) d x ] Q 2 ( s , τ ) f 1 ( τ , u ( τ ) , u ( τ ) , u ( τ ) , u ( τ ) ) d τ d s = λ 0 1 0 1 d 1 G 1 ( s , s ) + ω 1 0 1 G 1 ( s , x ) p ( x ) d x × Q 2 ( s , τ ) f 1 ( τ , u ( τ ) , u ( τ ) , u ( τ ) , u ( τ ) ) d τ d s d 1 λ 0 1 0 1 Q 1 ( s , s ) Q 2 ( s , τ ) f 1 ( τ , u ( τ ) , u ( τ ) , u ( τ ) , u ( τ ) ) d τ d s d 1 | | T u | | 0 ,
min 1 4 t 3 4 ( - ( T u ) ( t ) ) = min 1 4 t 3 4 0 1 Q 2 ( t , τ ) f 1 ( τ , u ( τ ) , u ( τ ) , u ( τ ) , u ( τ ) ) d τ min 1 4 t 3 4 0 1 [ b 2 G 2 ( t , t ) G 2 ( τ , τ ) + ω 2 ( t ) 0 1 G 2 ( τ , x ) q ( x ) d x ] f 1 ( τ , u ( τ ) , u ( τ ) , u ( τ ) , u ( τ ) ) d τ 0 1 [ b 2 G 2 ( t , t ) G 2 ( τ , τ ) + min 1 4 t 3 4 ω 2 ( t ) 0 1 G 2 ( τ , x ) q ( x ) d x ] f 1 ( τ , u ( τ ) , u ( τ ) , u ( τ ) , u ( τ ) ) d τ = 0 1 [ d 2 G 2 ( τ , τ ) + min 1 4 t 3 4 ω 2 ( t ) 0 1 G 2 ( τ , x ) q ( x ) d x ] f 1 ( τ , u ( τ ) , u ( τ ) , u ( τ ) , u ( τ ) ) d τ d 2 min 1 4 t 3 4 ω 2 ( t ) 0 1 [ G 2 ( τ , τ ) + 0 1 G 2 ( τ , x ) q ( x ) d x ] f 1 ( τ , u ( τ ) , u ( τ ) , u ( τ ) , u ( τ ) ) d τ d 2 c 2 min 1 4 t 3 4 ω 2 ( t ) max 1 4 t 3 4 ω 2 ( t ) | | ( T u ) | | 0 d 2 ξ c 2 | | ( T u ) | | 0 .

So we can get T(K) K: Let B K is bounded, it is clear that T(B) is bounded. Using f1, Q1(t, s), Q2(t, s) is continuous, we show that T(B) is equicontinuous. By the Arzela-Ascoli theorem, a standard proof yields T: KK is completely continuous.

Theorem 3.1. Suppose that (H1)-(H5) hold. Then BVP (1.1) has at least one positive solution u(t) satisfying

c < α ( u ) < b , β ( u ) < L .

Proof. Take

Ω 1 = { u X : α ( u ) < c , β ( u ) < L } , Ω 2 = { u X : α ( u ) < b , β ( u ) < L } ,

two bounded open sets in X, and

D 1 = { u X : α ( u ) = c } , D 2 = { u X : α ( u ) = b } .

By Lemma 3.1, T: KK is completely continuous. Let p= b 2 ( Ω 2 K ) \ { 0 } ,α ( p ) 0. It is easy to see that α(u + λp) ≥ α(u), for all u K and λ ≥ 0.

Let u D1, we have

| | T u | | 0 = max t [ 0 , 1 ] λ 0 1 0 1 Q 1 ( t , s ) Q 2 ( s , τ ) f 1 ( τ , u ( τ ) , u ( τ ) , u ( τ ) , u ( τ ) ) d τ d s λ 0 1 0 1 c 1 G 1 ( s , s ) + ω 1 0 1 G 1 ( s , x ) p ( x ) d x Q 2 ( s , τ ) d τ d s × c η 0 λ = c η 0 0 1 0 1 Q 1 ( s , s ) Q 2 ( s , τ ) d τ d s = B c η 0 ,
| | ( T u ) | | 0 = max t [ 0 , 1 ] - λ 0 1 Q 2 ( t , τ ) f 1 ( τ , u ( τ ) , u ( τ ) , u ( τ ) , u ( τ ) ) d τ < λ 0 1 c 2 G 2 ( τ , τ ) + ω 2 1 2 0 1 G 2 ( τ , x ) q ( x ) d x d τ × c η 0 λ c 2 c η 0 0 1 G 2 ( τ , τ ) + ω 2 1 2 0 1 G 2 ( τ , x ) q ( x ) d x d τ = c 2 D c η 0 ,

Hence, for u D1K, α(u) = c, we get

α ( T u ) = | | T u | | 0 + | | ( T u ) | | 0 < B c η 0 + c 2 D c η 0 = ( B + c 2 D ) c η 0 = c .

Whereas for u D2K, α(u) = b, there is ||u| | 0 b 2 or || u | | 0 b 2 , By Lemma 2.4, we get

min 1 4 t 3 4 u ( t ) d 1 | | u | | 0 d 1 b 2 or min 1 4 t 3 4 ( - u ( t ) ) d 2 ξ c 2 | | u | | 0 d 2 ξ b 2 c 2 .

Therefore, using (H4) and (3.3), we have

| ( T u ) 1 2 | = λ 0 1 Q 2 1 2 , τ f 1 ( τ , u ( τ ) , u ( τ ) , u ( τ ) , u ( τ ) ) d τ > λ 1 4 3 4 Q 2 1 2 , τ f 1 ( τ , u ( τ ) , u ( τ ) , u ( τ ) , u ( τ ) ) d τ λ × b η 1 λ 1 4 3 4 Q 2 1 2 , τ d τ = b .

Hence,

α ( T u ) ( T u ) 1 2 > b .

By (3.2), (3.4), and (H5), for u K, we have

( T u ) 0 = max t [ 0 , 1 ] | λ t 1 0 1 Q 2 ( s , τ ) f 1 ( τ , u ( τ ) , u ( τ ) , u ( τ ) , u ( τ ) ) d τ d s λ 0 1 0 1 s Q 2 ( s , τ ) f 1 ( τ , u ( τ ) , u ( τ ) , u ( τ ) , u ( τ ) ) d τ d s | max t [ 0 , 1 ] | λ t 1 0 1 Q 2 ( s , τ ) f 1 ( τ , u ( τ ) , u ( τ ) , u ( τ ) , u ( τ ) ) d τ d s | + max t [ 0 , 1 ] | λ 0 1 0 1 s Q 2 ( s , τ ) f 1 ( τ , u ( τ ) , u ( τ ) , u ( τ ) , u ( τ ) ) d τ d s | λ | 0 1 0 1 ( 1 + s ) Q 2 ( s , τ ) f 1 ( τ , u ( τ ) , u ( τ ) , u ( τ ) , u ( τ ) ) d τ d s | λ × η 2 L λ | 0 1 0 1 ( 1 + s ) [ c 2 G 2 ( τ , τ ) + ω 2 ( 1 2 ) 0 1 G 2 ( τ , x ) q ( x ) d x ] d τ d s | η 2 L × 3 2 c 2 0 1 [ G 2 ( τ , τ ) + ω 2 ( 1 2 ) 0 1 G 2 ( τ , x ) q ( x ) d x ] d τ = 3 2 c 2 D η 2 L ,
( T u ) 0 = max t [ 0 , 1 ] λ 0 1 Q 2 ( t , τ ) t f 1 ( τ , u ( τ ) , u ( τ ) , u ( τ ) , u ( τ ) ) d τ 2 λ 0 1 sin A τ sin A + A 0 1 G 2 ( τ , x ) q ( x ) d x sin A - 0 1 q ( x ) sin A x d x - 0 1 q ( x ) sin A ( 1 - x ) d x × | f 1 ( τ , u ( τ ) , u ( τ ) , u ( τ ) , u ( τ ) ) | d τ < λ 2 F × η 2 L λ = 2 F η 2 L .

So,

β ( T u ) = ( T u ) 0 + ( T u ) 0 < 3 2 c 2 D η 2 L + 2 F η 2 L = 3 2 c 2 D + 2 F η 2 L = L .

Theorem 2.1 implies there is ( Ω 2 \ Ω ̄ 1 ) K such that u = Tu. So, u(t) is a positive solution for BVP (1.1) satisfying

c < α ( u ) < b , β ( u ) < L .

Thus, Theorem 3.1 is completed.

4 Example

Example 4.1. Consider the following boundary value problem

u ( 4 ) ( t ) + π 2 9 u ( t ) = π 2 f ( t , u ( t ) , u ( t ) , u ( t ) , u ( t ) ) , 0 < t < 1 , u ( 0 ) = u ( 1 ) = 0 1 s u ( s ) d s , u ( 0 ) = u ( 1 ) = 0 ,
(4.1)

where

f ( t , u , v , u 0 , v 0 ) = 1 20 ( u - u 0 ) + 1 2 | cos ( v + v 0 ) | , ( t , u , v , u 0 , v 0 ) [ 0 , 1 ] × [ 0 , 2 ] × [ - 16000 , 16000 ] × [ - 2 , 0 ] × [ - 16000 , 16000 ] , 1 20 ( 2 - u 0 ) ( 3 - u ) + 27 2 ( 3 - u 0 ) ( u - 2 ) + 1 2 | cos ( v + v 0 ) | , ( t , u , v , u 0 , v 0 ) [ 0 , 1 ] × [ 2 , 3 ] × [ - 16000 , 16000 ] × [ - 2 , 0 ] × [ - 16000 , 16000 ] , 1 20 ( u + 2 ) ( u 0 + 3 ) - 27 2 ( u + 3 ) ( u 0 + 2 ) + 1 2 | cos ( v + v 0 ) | , ( t , u , v , u 0 , v 0 ) [ 0 , 1 ] × [ 0 , 2 ] × [ - 16000 , 16000 ] × [ - 3 , - 2 ] × [ - 16000 , 16000 ] , 1 5 ( 3 - u ) ( u 0 + 3 ) + 135 2 ( u - 2 ) ( u 0 + 3 ) - 27 2 ( u + 3 ) ( u 0 + 2 ) + 1 2 | cos ( v + v 0 ) | , ( t , u , v , u 0 , v 0 ) [ 0 , 1 ] × [ 2 , 3 ] × [ - 16000 , 16000 ] × [ - 3 , - 2 ] × [ - 16000 , 16000 ] , 27 2 ( u - u 0 ) + 1 2 | cos ( v + v 0 ) | , ( t , u , v , u 0 , v 0 ) [ 0 , 1 ] × [ 3 , 40 ] × [ - 16000 , 16000 ] × [ - 40 , 0 ] × [ - 16000 , 16000 ] , [ 0 , 1 ] × [ 0 , 40 ] × [ - 16000 , 16000 ] × [ - 40 , - 3 ] × [ - 16000 , 16000 ] .

In this problem, we know that A= π 2 9 ,λ= π 2 ,p ( t ) =t,q ( t ) =0, then we can get b 1 = 1 , b 2 = 3 π 6 , c 1 = 1 , c 2 = 2 3 3 , ω 1 = 2 , ω 2 = 2 3 sin π 3 ( 1 + t ) 3 , d 1 = 3 16 , d 2 = 3 - 1 4 , ξ = 2 + 3 2 . Further more, we obtain

B = 1944 3 - 972 π - 9 π 3 4 π 5 , D = 9 - 3 π 2 π 2 , F = 3 π .

then η 0 = 12 π 5 5832 3 - 2916 π - 27 π 3 + 36 3 π 3 - 12 π 4 , η 1 = π 2 3 6 + 3 3 - 9 , η 2 = 2 π 2 9 3 + 4 3 π - 3 π , θ=min d 1 2 , d 2 ξ 2 c 2 = 2 + 3 ( 3 - 3 ) 32 , θb ≈ 3.06 > 3.

If we take c = 2, b = 40, L = 16000, then we get

f ( t , u , v , u 0 , v 0 ) = 1 20 ( u - u 0 ) + 1 2 | cos ( v + v 0 ) | 0 . 7 < c η 0 λ 0 . 8 , for ( t , u , v , u 0 , v 0 ) [ 0 , 1 ] × [ 0 , 2 ] × [ - 16000 , 16000 ] × [ - 2 , 0 ] × [ - 16000 , 16000 ] , f ( t , u , v , u 0 , v 0 ) = 27 2 ( u - u 0 ) + 1 2 | cos ( v + v 0 ) | 40 > b η 1 λ 38 , for ( t , u , v , u 0 , v 0 ) 1 4 , 3 4 × [ θ b , 40 ] × [ - 16000 , 16000 ] × [ - 40 , 0 ] × [ - 16000 , 16000 ] [ 1 4 , 3 4 ] × [ 0 , 40 ] × [ - 16000 , 16000 ] × [ - 40 , - θ b ] × [ - 16000 , 16000 ] , f ( t , u , v , u 0 , v 0 ) 1080 . 5 < L η 2 λ 1146 , for ( t , u , v , u 0 , v 0 ) [ 0 , 1 ] × [ 0 , 40 ] × [ - 16000 , 16000 ] × [ - 40 , 0 ] × [ - 16000 , 16000 ] .

Then all the conditions of Theorem 3.1 are satisfied. Therefore, by Theorem 3.1 we know that boundary value problem (4.1) has at least one positive solution u(t) satisfying

2 < α ( u ) < 40 , β ( u ) < 16000 .

References

  1. Bai ZB: The method of lower and upper solution for a bending of an elastic beam equation. J Math Anal Appl 2000, 248: 195-202. 10.1006/jmaa.2000.6887

    Article  MathSciNet  Google Scholar 

  2. Bai ZB: The method of lower and upper solutions for some fourth-order boundary value problems. Nonlinear Anal 2007, 67: 1704-1709. 10.1016/j.na.2006.08.009

    Article  MathSciNet  Google Scholar 

  3. Chai GQ: Existence of positive solutions for fourth-order boundary value problem with variable parameters. Nonlinear Anal 2007, 66: 870-880. 10.1016/j.na.2005.12.028

    Article  MathSciNet  Google Scholar 

  4. Zhao JF, Ge WG: Positive solutions for a higher-order four-point boundary value problem with a p -Laplacian. Comput Math Appl 2009, 58: 1103-1112. 10.1016/j.camwa.2009.04.022

    Article  MathSciNet  Google Scholar 

  5. Bai ZB: Positive solutions of some nonlocal fourth-order boundary value problem. Appl Math Comput 2010, 215: 4191-4197. 10.1016/j.amc.2009.12.040

    Article  MathSciNet  Google Scholar 

  6. Li YX: Positive solutions of fourth-order boundary value problems with two parameters. J Math Anal Appl 2003, 281: 477-484. 10.1016/S0022-247X(03)00131-8

    Article  MathSciNet  Google Scholar 

  7. Ma HL: Symmetric positive solutions for nonlocal boundary value problems of fourth-order. Nonlinear Anal 2008, 68: 645-651. 10.1016/j.na.2006.11.026

    Article  MathSciNet  Google Scholar 

  8. Ma RY: Existence of positive solutions of a fourth-order boundary value problem. Appl Math Comput 2005, 168: 1219-1231. 10.1016/j.amc.2004.10.014

    Article  MathSciNet  Google Scholar 

  9. Ma R, Wang H: On the existence of positive solutions of fourth-order ordinary differential equations. Appl Anal 1995, 59: 225-231. 10.1080/00036819508840401

    Article  MathSciNet  Google Scholar 

  10. Yao QL: Local existence of multiple positive solutions to a singular cantilever beam equation. J Math Anal Appl 2010, 363: 138-154. 10.1016/j.jmaa.2009.07.043

    Article  MathSciNet  Google Scholar 

  11. Guo D, Lakshmikantham V: Nonlinear Problems in Abstract Cones. Academic Press, New York; 1988.

    Google Scholar 

  12. Guo YP, Ge WG: Positive solutions for three-point boundary value problems with dependence on the first order derivatives. J Math Anal Appl 2004, 290: 291-301. 10.1016/j.jmaa.2003.09.061

    Article  MathSciNet  Google Scholar 

Download references

Acknowledgements

The project is supported by the Natural Science Foundation of China (10971045) and the Natural Science Foundation of Hebei Province (A2009000664, A2011208012). The research item financed by the talent training project funds of Hebei Province. The authors would like to thank the referee for helpful comments and suggestions.

Author information

Authors and Affiliations

Authors

Corresponding author

Correspondence to Yongchun Liang.

Additional information

Competing interests

The authors declare that they have no competing interests.

Authors' contributions

The authors declare that the work was realized in collaboration with same responsibility.

All authors read and approved the final manuscript.

Rights and permissions

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

Reprints and permissions

About this article

Cite this article

Guo, Y., Yang, F. & Liang, Y. Positive solutions for nonlocal fourth-order boundary value problems with all order derivatives. Bound Value Probl 2012, 29 (2012). https://doi.org/10.1186/1687-2770-2012-29

Download citation

  • Received:

  • Accepted:

  • Published:

  • DOI: https://doi.org/10.1186/1687-2770-2012-29

Keywords