Research

# The existence and multiplicity of positive solutions of nonlinear sixth-order boundary value problem with three variable coefficients

Wanjun Li

Author Affiliations

Department of Mathematics, Longdong University, Qingyang 745000, Gansu, P. R. China

Boundary Value Problems 2012, 2012:22  doi:10.1186/1687-2770-2012-22

 Received: 22 November 2011 Accepted: 22 February 2012 Published: 22 February 2012

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.

### Abstract

In this article, we discuss the existence and multiplicity of positive solutions for the sixth-order boundary value problem with three variable parameters as follows:

u ( 6 ) + A ( t ) u ( 4 ) + B ( t ) u ( 2 ) + C ( t ) u + f ( x , u ) = 0 , u ( 0 ) = u ( 1 ) = u ( 0 ) = u ( 1 ) = u ( 4 ) ( 0 ) = u ( 4 ) ( 1 ) = 0 ,

where A(t), B(t), C(t) ∈ C[0,1], f(t, u) : [0,1] × [0, ∞) → [0. ∞) is continuous. The proof of our main result is based upon spectral theory of operators and fixed point theorem in cone.

##### Keywords:
sixth-order differential equation; positive solution; fixed point theorem; spectral theory of operators

### 1 Introduction

In this article, we study the existence and multiplicity of positive solution for the following nonlinear sixth-order boundary value problem (BVP for short) with three variable parameters

- u ( 6 ) - C ( t ) u ( 4 ) + B ( t ) u - A ( t ) u = f ( t , u ) , t ( 0 , 1 ) , u ( 0 ) = u ( 1 ) = u ( 0 ) = u ( 1 ) = u ( 4 ) ( 0 ) = u ( 4 ) ( 1 ) = 0 , (1.1)

where A(t), B(t), C(t) ∈ C[0,1], f(t, u) : [0,1] × [0, ∞) → [0. ∞) is continuous.

In recent years, BVPs for sixth-order ordinary differential equations have been studied extensively, see [1-7] and the references therein. For example, Tersian and Chaparova [1] have studied the existence of positive solutions for the following systems (1.2):

u ( 6 ) + A u ( 4 ) + B u + C u - f ( t , u ) = 0 . 0 < x < L , u ( 0 ) = u ( L ) = u ( 0 ) = u ( L ) = u ( 4 ) ( 0 ) = u ( 4 ) ( L ) = 0 , (1.2)

where A, B, and C are some given real constants and f(x, u) is a continuous function on R2, is motivated by the study for stationary solutions of the sixth-order parabolic differential equations

u t = 6 u x 6 + A 4 u x 4 + B 2 u x 2 + f ( x , u ) .

This equation arose in the formation of the spatial periodic patterns in bistable systems and is also a model for describing the behaviour of phase fronts in materials that are undergoing a transition between the liquid and solid state. When f(x, u) = u - u3, it was studied by Gardner and Jones [2] as well as by Caginalp and Fife [3]. In [1], existence of nontrivial solutions for (1.2) is proved using a minimization theorem and a multiplicity result using Clarks theorem when C = 1 and f(x, u) = u3. The authors have studied also the homoclinic solutions for (1.2) when C = -1 and f(x, u) = -a(x)u|u|σ, where a(x) is a positive periodic function and σ is a positive constant by the mountain-pass theorem of Brezis-Nirenberg and concentration-compactness arguments. In [4], by variational tools, including two Brezis-Nirenbergs linking theorems, Gyulov et al. have studied the existence and multiplicity of nontrivial solutions of BVP (1.2).

Recently, in [5], the existence and multiplicity of positive solutions of sixth-order BVP with three parameters

- u ( 6 ) - γ u ( 4 ) + β u - α u = f ( t , u ) , t [ 0 , 1 ] , u ( i ) ( 0 ) = u ( i ) ( 1 ) = 0 , i = 0 , 1 , 2 , 3 , 4 , 5 (1.3)

has been studied under the hypothesis of

(A1) f : [0,1] × [0, ∞) → [0. ∞) is continuous.

(A2) α, β, γ R and under the condition of satisfying

α π 6 + β π 4 + γ π 2 < 1 , 3 π 4 - 2 γ π 2 - β > 0 , γ < 3 π 2 , 18 α β γ - β 2 γ 2 + 4 α γ 3 + 27 α 2 - 4 β 3 0 ,

the existence and multiplicity for positive solution of BVP (1.3) are established by using fixed point index theory. In this article, we consider more general BVP (1.1), based upon spectral theory of operators and fixed point theorem in cone, we will establish the existence and multiplicity positive solution of BVP (1.1) and extend the result of [5] under appropriate conditions. Our ideas mainly come from [5,8-10].

We list the following conditions for convenience:

(H1) f : [0,1] × [0, +∞) → [0. +∞) is continuous.

(H2) A(t), B(t), C(t) ∈ C[0,1], α = min0≤t≤1 A(t), β = min0≤t≤1 B(t), γ = min0≤t≤1 C(t), and satisfies

α π 6 + β π 4 + γ π 2 < 1 , 3 π 4 - 2 γ π 2 - β > 0 , γ < 3 π 2 , 18 α β γ - β 2 γ 2 + 4 α γ 3 + 27 α 2 - 4 β 3 0 .

Let Y = C[0,1], Y+ = {u Y : u(t) ≥ 0, t ∈ [0,1]}. It is well known that Y is a Banach space equipped with the norm ||u||0 = sup0≤t≤1 |u(t)|, u Y. Set X = { u C4[0,1] : u(0) = u(1) = u''(0) = u''(1) = 0}, then X also is a Banach space equipped with the norm ||u||X = max {||u(t)||0, ||u"(t)||0, ||u(4)(t)||0}. If u C4[0,1] ∩ C6(0,1) fulfils BVP (1.1), then we call u is a solution of BVP (1.1). If u is a solution of BVP (1.1), and u(t) > 0, t ∈ (0, 1), then we say u is a positive solution of BVP (1.1).

### 2 Preliminaries

In this section, we will make some preliminaries which are needed to show our main results.

Lemma 2.1. Let u X, then ||u||0 ≤ ||u"||0 ≤ ||u(4)||0 ≤ ||u||X.

Proof. The proof is similar to the Lemma 1 in [8], so we omit it. □

Lemma 2.2. [5] Let λ1, λ2, and λ3 be the roots of the polynomial P (λ) = λ3 + γλ2 - βλ + α. Suppose that condition (H2) holds, then λ1, λ2, and λ3 are real and greater than -π2.

Note : Based on Lemma 2.2, it is easy to learn that when the three parameters satisfy the condition of (H2), they satisfy the condition of non-resonance.

Let Gi(t, s)(i = 1, 2, 3) be the Green's function of the linear BVP

-u"(t) + λiu(t) = 0, u(0) = u(1) = 0,

Lemma 2.3. [10]Gi(t, s)(i = 1, 2, 3) has the following properties

(c1) Gi(t, s) > 0, ∀t, s ∈ (0, 1).

(c2) Gi(t, s) <CiGi(s, s), ∀t, s ∈ [0,1], in which Ci > 0 is constant.

(c3) Gi(t, s) ≥ δiGi(t, t)Gi(s, s), ∀t, s ∈ [0,1], in which δi > 0 is constant.

We set

M i = max 0 t 1 G i ( s , s ) , m i = min 1 4 t 3 4 G i ( s , s ) , i = 1 , 2 , 3 . (2.1)

C i j = 0 1 G i ( δ , δ ) G j ( δ , δ ) d δ , c i j = 1 4 3 4 G i ( δ , δ ) G j ( δ , δ ) d δ , i , j = 1 , 2 , 3 . (2.2)

D i = max 0 t 1 0 1 G i ( t , s ) d s , d i = max 1 4 t 3 4 1 4 3 4 G i ( t , s ) d s i = 1 , 2 , 3 , (2.3)

then starting from Lemma 2.3 we know Mi, mi, Cij > 0.

For any h Y, take into consideration of linear BVP:

- u ( 6 ) - γ u ( 4 ) + β u - α u = h ( t ) , t [ 0 , 1 ] , u ( 0 ) = u ( 1 ) = u ( 0 ) = u ( 1 ) = u ( 4 ) ( 0 ) = u ( 4 ) ( 1 ) = 0 , (2.4)

where α, β, γ satisfy assumption (H2). Since

- u ( 6 ) - γ u ( 4 ) + β u - α u = - d 2 d t 2 + λ 1 - d 2 d t 2 + λ 2 - d 2 d t 2 + λ 3 u , (2.5)

then for any h Y, the LBVP(2.4) has a unique solution u, which we denoted by Ah = u. The operator A can be expressed by

u ( t ) = ( A h ) ( t ) : = 0 1 0 1 0 1 G 1 ( t , δ ) G 2 ( δ , τ ) G 3 ( τ , s ) h ( t ) d s d τ d δ . (2.6)

Lemma 2.4. The linear operator A : Y X is completely continuous and ||A|| ≤ ϖ, where ϖ = |λ23|(C1C2C3M1M2M3|λ3|+C1C2M1M2)+| λ2λ3|(C1C2C3M1M2M3+C1M1).

Proof. It is easy to show that the operator A : Y X is linear operator. ∀h Y, u = Ah X, u(0) = u(1) = u"(0) = u"(1) = u(4)(0) = u(4)(1) = 0. Let v = - d 2 d t 2 + λ 2 - d 2 d t 2 + λ 3 u , that is

v = - d 2 d t 2 + λ 2 - d 2 d t 2 + λ 3 u = u ( 4 ) - ( λ 2 + λ 3 ) u + λ 2 λ 3 u , (2.7)

by (2.5) and (2.7), we have

- v + λ 1 v = h ( t ) , t ( 0 , 1 ) , v ( 0 ) = v ( 1 ) = 0 ,

and v ( t ) = 0 1 G 1 ( t , s ) h ( s ) d s , t [ 0 , 1 ] , so

u ( 4 ) - ( λ 2 + λ 3 ) u + λ 2 λ 3 u = 0 1 G 1 ( t , s ) h ( s ) d s , t [ 0 , 1 ] . (2.8)

By (2.6), for any t ∈ [0,1], we have

| u ( t ) | 0 1 0 1 0 1 G 1 ( t , δ ) G 2 ( δ , τ ) G 3 ( τ , s ) | h ( t ) | d s d τ d δ C 1 C 2 C 3 M 1 M 2 M 3 | | h | | 0 . (2.9)

Again, let ω = -u" + λ3u, then ω(0) = ω(1) = ω"(0) = ω"(1) = 0, by (2,5), we have

ω ( 4 ) - ( λ 1 + λ 2 ) ω + λ 1 λ 2 ω = h ( t ) , t ( 0 , 1 ) , ω ( 0 ) = ω ( 1 ) = ω ( 0 ) = ω ( 1 ) = 0 . (2.10)

Then ω ( t ) = 0 1 0 1 G 1 ( t , τ ) G 2 ( τ , s ) h ( s ) d s d τ ,   t [ 0 , 1 ] , that is

- u + λ 3 u = 0 1 0 1 G 1 ( t , τ ) G 2 ( τ , s ) h ( s ) d s d τ , t [ 0 , 1 ] . (2.11)

So

| u | C 1 C 2 M 1 M 2 ( 1 + | λ 3 | C 3 M 3 ) | | h | | 0 , t [ 0 , 1 ] . (2.12)

Based on (2.8), (2.9), and (2.12), we have

| u ( 4 ) ( t ) | | λ 2 + λ 3 | | u ( t ) | + | λ 2 λ 3 | | u ( t ) | + 0 1 G 1 ( t , s ) | h ( s ) | d s | λ 2 + λ 3 | ( C 1 C 2 C 3 M 1 M 2 M 3 | λ 3 | + C 1 C 2 M 1 M 2 ) | | h | | 0 + | λ 2 λ 3 | ( C 1 C 2 C 3 M 1 M 2 M 3 | | h | | 0 + C 1 M 1 ) | | h | | 0 ϖ | | h | | 0 , t [ 0 , 1 ] , (2.13)

where

ϖ = | λ 2 + λ 3 | ( C 1 C 2 C 3 M 1 M 2 M 3 | λ 3 | + C 1 C 2 M 1 M 2 ) + | λ 2 λ 3 | ( C 1 C 2 C 3 M 1 M 2 M 3 + C 1 M 1 ) . (2.14)

So, ||u(4)(t)|| ≤ ϖ||h||0, by Lemma2.1, ||u||X ϖ||h||0, then

| | A h | | X ϖ | | h | | 0 , (2.15)

so A is continuous, and ||A|| ≤ ϖ.

Next, we will show that A is compact with respect to the norm ||·||X on X.

Suppose {hn}(n = 1, 2, . . .) an arbitrary bounded sequence in Y, then there exists K0 >0 such that ||hn||0 K0, n = 1, 2, . . . . Let un = Ahn, 1, 2, ...By (2.8), ∀t1, t2 ∈ [0, 1], t1 < t2, we have

u n ( 4 ) ( t 2 ) - u n ( 4 ) ( t 1 ) | λ 2 + λ 3 | | u n ( t 2 ) - u n ( t 1 ) | + | λ 2 λ 3 | | u n ( t 2 ) - u n ( t 1 ) | + 0 1 | G 1 ( t 2 , s ) - G 1 ( t 1 , s ) | | h n ( s ) | d s . | λ 2 + λ 3 | | λ 3 | | u n ( t 2 ) - u n ( t 1 ) | + 0 1 0 1 | G 1 ( t 2 , τ ) - G 1 ( t 1 , τ ) | | G 2 ( τ , s ) | | h n ( s ) | d s d τ + | λ 2 λ 3 | | u n ( t 2 ) - u n ( t 1 ) | + 0 1 | G 1 ( t 2 , s ) - G 1 ( t 1 , s ) | | h n ( s ) | d s . ( λ 3 2 + 2 | λ 2 λ 3 | ) 0 1 0 1 0 1 | G 1 ( t 2 , δ ) - G 1 ( t 1 , δ ) | | G 2 ( δ , τ ) | | G 3 ( τ , s ) | | h n ( s ) | d s d τ d δ . + | λ 2 + λ 3 | 0 1 0 1 | G 1 ( t 2 , τ ) - G 1 ( t 1 , τ ) | | G 2 ( τ , s ) | | h n ( s ) | d s d τ + 0 1 | G 1 ( t 2 , s ) - G 1 ( t 1 , s ) | | h n ( s ) | d s . ( λ 3 2 + 2 | λ 2 λ 3 | ) 0 1 0 1 0 1 | G 1 ( t 2 , δ ) - G 1 ( t 1 , δ ) | G 2 ( δ , τ ) G 3 ( τ , s ) d s d τ d δ . + | λ 2 + λ 3 | 0 1 0 1 | G 1 ( t 2 , τ ) - G 1 ( t 1 , τ ) G 2 ( τ , s ) d s d τ + 0 1 | G 1 ( t 2 , s ) - G 1 ( t 1 , s ) | d s K 0 .

Because Gi(t, s)(i = 1, 2, 3) is uniform continuity on [0,1] × [0,1], based on the above demonstration, it is easy to proof that u n ( 4 ) n = 1 is equicontinuous on [0,1]. From (2.15), we know ||u||0, ||u"||0, ||u(4)||0 ≤ ||u||X ϖ||hn||0 ϖK0, so u n ( t ) , { u n ( t ) } and u n ( 4 ) ( t ) are relatively compact in R. Based on Lemma 1.2.7 in [11], we know { u n } n = 1 is the relatively compact in X, so A is compact operator. □

The main tools of this article are the following well-known fixed point index theorems.

Let E be a Banach Space and K E be a closed convex cone in E. Assume that Ω is a bounded open subset of E with boundary ∂Ω, and K ∩ Ω ≠ ∅. Let A : K Ω ̄ K be a completely continuous mapping. If Au u for every u K ∩ ∂Ω, then the fixed point index i(A, K ∩ Ω, K) is well defined. We have that if i(A, K ∩ Ω, K) ≠0, then A has a fixed point in K ∩ Ω.

Let Kr = {u K |||u|| <r} and ∂Kr = {u K |||u|| <r} for every r >0.

Lemma 2.5. [12] Let A : K K be a completely continuous mapping. If μAu u for every u ∈ ∂Kr and 0 < μ ≤ 1, then i(A, Kr, K) = 1.

Lemma 2.6. [12] Let A : K K be a completely continuous mapping. Suppose that the following two conditions are satisfied:

(i) inf u K r | | A u | | > 0 ,

(ii) μAu u for every u ∈ ∂Kr and μ ≥ 1,

then i(A, Kr, K) = 0.

Lemma 2.7. [12] Let X be a Banach space, and let K X be a cone in X. For p >0, define K p = { u K | | | u | | < p } . Assume that A : Kp K is a completely continuous mapping such that Au u for every u ∈ ∂Kp = {u K|||u|| = p}.

(i) If ||u|| ≤ ||Au||, for every u ∈ ∂Kp, then i(A, Kp, K) = 0.

(ii) If ||u|| ||Au||, for every u ∈ ∂Kp, then i(A, Kp, K) = 1.

### 3 Main results

We bring in following notations in this section:

f 0 = lim u 0 + inf min 0 t 1 ( f ( t , u ) / u ) , f ̄ = lim u + sup max 0 t 1 ( f ( t , u ) / u ) , f ̄ 0 = lim u 0 + sup max 0 t 1 ( f ( t , u ) / u ) , f = lim u + inf min 0 t 1 ( f ( t , u ) / u ) . a ( t ) = A ( t ) - α , b ( t ) = B ( t ) - β , c ( t ) = C ( t ) - γ , Γ = π 6 - γ π 4 - β π 2 - α , K = max 0 t 1 [ a ( t ) + b ( t ) + c ( t ) ] ,

Suppose that:

(H3) L = ϖK <1, where ϖ is defined as in (2.14).

Theorem 3.1. Assume that (H1)-(H3) hold, and b ( t ) ( λ 2 + λ 3 ) c ( t ) , λ 3 b ( t ) - a ( t ) λ 3 2 c ( t ) , then in each of the following cases:

(i) f 0 > Γ , f ̄ < ( 1 - L ) Γ , (ii) f ̄ 0 < ( 1 - L ) Γ , f > Γ , the BVP (1.1) has at least one positive solution.

Proof. ∀h Y, consider the LBVP

- u ( 6 ) - C ( t ) u ( 4 ) + B ( t ) u - A ( t ) u = h ( t ) , u ( 0 ) = u ( 1 ) = u ( 0 ) = u ( 1 ) = u ( 4 ) ( 0 ) = u ( 4 ) ( 1 ) = 0 , 0 < t < 1 (3.1)

It is easy to prove (3.1) is equivalent to the following BVP

- u ( 6 ) - γ u ( 4 ) + β u - α = G u + h ( t ) , u ( 0 ) = u ( 1 ) = u ( 0 ) = u ( 1 ) = u ( 4 ) ( 0 ) = u ( 4 ) ( 1 ) = 0 , 0 < t < 1 (3.2)

where Gv := (C(t) - γ)v(4) - (B(t)- β)v" + (A(t) - α)v, ∀v X. Obviously, the operator G : X Y is linear, and ∀v X, t ∈ [0,1], we have |Gv(t)| ≤ K ||v||X. Hence ||Gv||0 K ||v||X, and so ||G|| ≤ K. On the other hand, u C4[0,1]⋂C6(0,1), t ∈ [0,1] is a solution of (3.2) iff u X satisfies u = A(Gu + h), i.e.,

u X , ( I - A G ) u = A h . (3.3)

Owing to G : X Y and A : Y X, the operator I - AG maps X into Y. From A ϖ (by Lemma 2.4) together with ||G|| ≤ K and condition (H3), applying operator spectral theorem, we have that the operator (I - AG)-1 exists and is bounded. Let H = (I - AG)-1A, then (3.3) is equivalent to u = Hh. By the Neumann expansion formula, H can be expressed by

H = ( I + A G + + ( A G ) n + ) A = A + ( A G ) A + + ( A G ) n A + .   (3.4)

The complete continuity of A with the continuity of (I - AG)-1 yields that the operator H : Y X is completely continuous. If we restrict H : Y+ Y, ∀h Y+ and mark u = Ah, then u X Y+. Based on equation (2.8), (2.11) and Lemma 2.4, we have

u = λ 3 u - 0 1 0 1 G 1 ( t , τ ) G 2 ( τ , s ) h ( s ) d s d τ λ 3 u , t [ 0 , 1 ] , u ( 4 ) = ( λ 2 + λ 3 ) u - λ 2 λ 3 u + 0 1 G 1 ( t , s ) h ( s ) d s ( λ 2 + λ 3 ) u - λ 2 λ 3 u , t [ 0 , 1 ] ,

by b(t) ≥ (λ2 + λ3)c(t) and λ 3 b ( t ) - a ( t ) λ 3 2 c ( t ) , we have

( G u ) ( t ) = c ( t ) u ( 4 ) b ( t ) u ' ' + a ( t ) u [ ( λ 2 + λ 3 ) c ( t ) b ( t ) ] u ' ' [ λ 2 λ 3 c ( t ) a ( t ) ] u λ 3 [ ( λ 2 + λ 3 ) c ( t ) b ( t ) ] u [ λ 2 λ 3 c ( t ) a ( t ) ] u [ λ 3 2 c ( t ) λ 3 b ( t ) + a ( t ) ] u 0 , t [ 0 , 1 ] .

Hence

h Y + , ( G A h ) ( t ) 0 , t 0 ,  1 , (3.5)

and so (AG)(Ah)(t) = A(GAh)(t) ≥ 0, ∀t ∈ [0,1]. Suppose that ∀h Y+, (AG)k (Ah)(t) ≥ 0, ∀t ∈ [0,1]. For any h Y+, let h1 = GAh, by (3.5) we have h1 Y+, and so

( A G ) k + 1 ( A h ) ( t ) = ( A G ) k ( A G A h ) ( t ) = ( A G ) k ( A h 1 ) ( t ) 0 , t [ 0 , 1 ] .

Thus by induction it follows that ∀n ≥ 1, ∀h Y+, (AG)n (Ah)(t) ≥ 0, ∀t ∈ [0,1]. By (3.4), we have

h Y + , ( H h ) ( t ) = ( A h ) ( t ) + ( A G ) ( A h ) ( t ) + + ( A G ) n ( A h ) ( t ) + ( A h ) ( t ) , t [ 0 , 1 ] . (3.6)

So H : Y+ Y+ X.

On the other hand, we have

h Y + , ( H h ) ( t ) ( A h ) ( t ) + | | ( A G ) | | ( A h ) ( t ) + + | | ( A G ) n | | ( A h ) ( t ) + ( 1 + L + + L n + ) ( A h ) ( t ) 1 1 - L ( A h ) ( t ) , t [ 0 , 1 ] . (3.7)

So the following inequalities hold

| | H h | | 0 1 1 - L | | A h | | 0 , t [ 0 , 1 ] . (3.8)

For any u Y+, define Fu = f(t, u). Based on condition (H1), it is easy to show F : Y+ Y+ is continuous. By (3.1)-(3.3), It is easy to see that u C4[0,1] ∩ C6(0, 1) is a positive solution of BVP (1.1) iff u Y+ is a nonzero solution of an operator equation as follows

u = H F u . (3.9)

Let Q = HF. Obviously, Q : Y+ Y+ is completely continuous. We next show that the operator Q has at least one nonzero fixed point in Y+.

Let

P = { u Y + | u ( t ) σ | | u | | 0 , t [ 0 , 1 ] } .

In which

σ = δ 1 δ 2 δ 3 C 12 C 23 C 1 C 2 C 3 M 1 M 2 ( 1 - L ) G 1 ( t , t ) . (3.10)

Here M1 and M2 can be defined as that in (2.1), C12 and C23 can be defined as that in (2.2), Ci,δi(i = 1, 2, 3) can be defined as that in Lemma 2.3. It is easy to prove that P is a cone in Y. We will prove QP P next.

For any u P, let h = Fu, then h Y+. By (3.6) and Lemma 2.3, we have

( Q u ) ( t ) = ( H F u ) ( t ) ( A F u ) ( t ) , t [ 0 , 1 ] } .

By Lemma 2.3, for all u P, we have

( A F u ) ( t ) = 0 1 0 1 0 1 G 1 ( t , δ ) G 2 ( δ , τ ) G 3 ( τ , s ) ( F u ) ( s ) d s d τ d δ C 1 C 2 C 3 M 1 M 2 0 1 G 3 ( s , s ) ( F u ) ( s ) d s .

And accordingly we have | | A F u | | 0 C 1 C 2 C 3 M 1 M 2 0 1 G 3 ( s , s ) ( F u ) ( s ) d s , that is

0 1 G 3 ( s , s ) ( F u ) ( s ) d s | | A F u | | 0 C 1 C 2 C 3 M 1 M 2 . (3.11)

By using (c3) in Lemma 2.3, (3.8) and (3.11), we have

( A F u ) ( t ) δ 1 δ 2 δ 3 C 12 C 23 G 1 ( t , t ) 0 1 G 3 ( s , s ) ( F u ) ( s ) d s δ 1 δ 2 δ 3 C 12 C 23 G 1 ( t , t ) C 1 C 2 C 3 M 1 M 2 | | A F u | | 0 δ 1 δ 2 δ 3 C 12 C 23 G 1 ( t , t ) C 1 C 2 C 3 M 1 M 2 ( 1 - L ) | | H F u | | 0 δ 1 δ 2 δ 3 C 12 C 23 G 1 ( t , t ) C 1 C 2 C 3 M 1 M 2 ( 1 - L ) | | Q u | | 0 .

So ( Q u ) ( t ) δ 1 δ 2 δ 3 C 12 C 23 G 1 ( t , t ) C 1 C 2 C 3 M 1 M 2 ( 1 - L ) | | Q u | | 0 = σ | | Q u | | 0 . Thus QP P.

Let

ρ = δ 1 δ 2 δ 3 C 12 C 23 m 1 ( 1 - L ) C 1 C 2 C 3 M 1 M 2 , (3.12)

in which m1 can be defined as that in (2.1). It's easy to prove

u P u ( t ) ρ | | u | | 0 , t 1 4 , 3 4 . (3.13)

Case (i), since f 0 > Γ , there exist ε >0 and r0 >0 such that f(t, x) ≥ (Γ + ε)x, 0 ≤ t ≤ 1, 0 < × r0. Let r ∈ (0, r0) and Ωr = {u P | ||u||0 r}, then for every u ∈ ∂Ωr, we have ||u||0 = r, 0 < u(t) ≤ r, t ∈ (0, 1), and so f(t, u(t)) ≥ (Γ + ε)u(t), t ∈ (0,1). By (3.13), it follows that

f ( t , u ( t ) ) > ( Γ + ε ) u ( t ) ( Γ + ε ) ρ r , t 1 4 , 3 4 . (3.14)

From (3.6) and (3.14), we have

| | Q u | | 0 Q u 1 2 = ( H F u ) 1 2 ( A F u ) 1 2 = 0 1 0 1 0 1 G 1 1 2 , δ G 2 ( δ , τ ) G 3 ( τ , s ) f ( s , u ( s ) ) d s d τ d δ 1 4 3 4 1 4 3 4 1 4 3 4 G 1 1 2 , δ G 2 ( δ , τ ) G 3 ( τ , s ) ( Γ + ε ) ρ r d s d τ d δ δ 1 δ 2 δ 3 m 1 C 12 C 23 ( Γ + ε ) ρ r 1 4 3 4 G 3 ( s , s ) d s . 1 2 δ 1 δ 2 δ 3 m 1 m 3 C 12 C 23 ( Γ + ε ) ρ r > 0 .

Therefore, inf u Ω r Q u 0 > 0 . Now we shall prove ∀u ∈ ∂Ωr, μ ≥ 1, μQu u. In fact, suppose the contrary, then there exist u 0 Ω r , and μ0 ≥ 1 such that μ0Qu0 = u0. By (3.6), we have u 0 ( t ) 1 μ 0 u 0 ( t ) = Q u 0 ( t ) ( A F u 0 ) ( t ) . Let ω0 = AFu0, then u0 ω0 and ω0(t) satisfies BVP (2.4) with h = Fu0. Hence

- ω 0 ( 6 ) - γ ω 0 ( 4 ) + β ω 0 - α ω 0 = f ( t , u 0 ) , t ( 0 , 1 ) , ω 0 ( 0 ) = ω 0 ( 1 ) = ω 0 ( 0 ) = ω 0 ( 1 ) = ω 0 ( 4 ) ( 0 ) = ω 0 ( 4 ) ( 1 ) = 0 , (3.15)

After multiplying the two sides of the first equation in (3.15) by sin &#960t and integrating on [0,1], we have

Γ 0 1 ω 0 ( t ) sin π t d t = 0 1 f ( t , u 0 ( t ) ) sin π t d t ,

then

( Γ + ε ) 0 1 u 0 ( t ) sin π t d t 0 1 f ( t , u 0 ( t ) ) sin π t d t = Γ 0 1 ω 0 ( t ) sin π t d t Γ 0 1 u 0 ( t ) sin π t d t . (3.16)

Since u 0 ( t ) ρ u 0 0 = ρ r , t 1 4 , 3 4 , so 0 1 u 0 ( t ) sin π t d t > 0 and we see that Γ + ε < Γ, which is a contradiction. Then based on Lemma 2.6, we come to

i ( Q , Ω r , P ) = 0 . (3.17)

On the other hand, since f ̄ < ( 1 - L ) Γ , there exist ε ∈ (0, (1 - L)Γ) and R0 >0 such that f(t, x) ≤ [(1-L)Γ - ε] x, 0 ≤ t ≤ 1, x >R0. Let M R 0 = sup 0 t 1 , 0 x R 0 f ( t , x ) . Then

f ( t , x ) < ( 1 - L ) Γ - ε x + M R 0 , 0 t 1 , x 0 .

We choose R > max R 0 , r , 2 M R 0 ρ ε and let Ω R = { u P | | | u | | 0 < R } . Next we prove ∀u ∈ ∂ΩR, μ ≥ 1, μu Qu. Assume on the contrary that ∃μ0 ≥ 1, u0 ∈ ∂ΩR, such that μ0u0 = Qu0. Let ω1 = AFu0, by (3.6), we have u 0 μ 0 u 0 = Q u 0 1 1 - L A F u 0 1 1 - L ω 1 and ω1(t) satisfies BVP (2.4) with h = Fu0. Similarly to (3.16), we can prove

( 1 - L ) Γ 0 1 u 0 ( t ) sin π t d t Γ 0 1 ω 1 ( t ) sin π t d t = 0 1 f ( t , u 0 ( t ) ) sin π t d t ( 1 - L ) Γ - ε 0 1 u 0 ( t ) sin π t d t + M R 0 0 1 sin π t d t , (3.18)

and so

M R 0 0 1 sin π t d t ε 0 1 u 0 ( t ) sin π t d t ε 1 4 3 4 u 0 ( t ) sin π t d t ρ ε | | u 0 | | 0 0 1 sin π t d t , (3.19)

Thus, by (3.19), we have R = | | u 0 | | 0 2 M R 0 ρ ε which is contradictory with R > 2 M R 0 ρ ε .

Then by Lemma 2.5 we know

i ( Q , Ω R , P ) = 1 . (3.20)

Now, by the additivity of fixed point index, combine (3.17) and (3.20) to conclude that

i ( Q , Ω R \ Ω ̄ r , P ) = i ( Q , Ω R , P ) - i ( Q , Ω r , P ) = 1 .

Therefore Q has a fixed point in Ω R \ Ω ̄ r , which is the positive solution of BVP (1.1).

Case (ii), since f ̄ 0 < ( 1 - L ) Γ , based on the definition of f ̄ 0 , we may choose ε >0 and ω >0, so that

f ( t , u ) ( 1 - L ) Γ - ε u , 0 t 1 , 0 u ω . (3.21)

Let r ∈ (0, ω), we now prove that μQu u for every u ∈ ∂Ωr, and 0 < μ ≤ 1. In fact, suppose the contrary, then there exist u0 ∈ ∂Ωr, and 0 < μ0 ≤ 1 such that μ0Qu0 = u0. Let ω2 = AFu0, by (3.6), we have u 0 = μ 0 Q u 0 1 1 - L A F u 0 1 1 - L ω 2 and ω2(t) satisfies BVP (2.4) with h = Fu0. Similarly to (3.18), we have

( 1 - L ) Γ 0 1 u 0 ( t ) sin π t d t Γ 0 1 ω 2 ( t ) sin π t d t = 0 1 f ( t , u 0 ( t ) ) sin π t d t ( 1 - L ) Γ - ε 0 1 u 0 ( t ) sin π t d t . (3.22)

Since 0 1 u 0 ( t ) sin π t d t > 0 , We see that (1 - L)Γ ≤ (1 - L)Γ - ε, which is a contradiction. By Lemma 2.5, we have

i ( Q , Ω r , P ) = 1 . (3.23)

On the other hand, because f > Γ , there exist ε ∈ (0, Γ) and H >0 such that

f ( t , x ) ( Γ + ε ) x , t [ 0 , 1 ] , x > H . (3.24)

Let C = max0≤t≤1,0≤xH|f(t,x) - (Γ + ε)x| + 1, then it is clear that

f ( t , x ) ( Γ + ε ) x - C , t [ 0 , 1 ] , x 0 . (3.25)

Choose R > R0 = max {H/ρ, ω}, ∀u ∈ ∂ΩR. By (3.13) and (3.25), we have

u ( s ) ρ | | u | | 0 > H , s 1 4 , 3 4 .

And so

f ( s , u ( s ) ) ( Γ + ε ) u ( s ) ( Γ + ε ) ρ | | u | | 0 , s 1 4 , 3 4 . (3.26)

From (3.6) and (3.26), we get

| | Q u | | 0 Q u 1 2 = ( H F u ) 1 2 ( A F u ) 1 2 = 0 1 0 1 0 1 G 1 1 2 , δ G 2 ( δ , τ ) G 3 ( τ , s ) f ( s , u ( s ) ) d s d τ d δ 1 4 3 4 1 4 3 4 1 4 3 4 G 1 1 2 , δ G 2 ( δ , τ ) G 3 ( τ , s ) ( Γ + ε ) ρ | | u | | 0 d s d τ d δ δ 1 δ 2 δ 3 m 1 C 12 C 23 ( Γ + ε ) ρ | | u | | 0 1 4 3 4 G 3 ( s , s ) d s . 1 2 δ 1 δ 2 δ 3 m 1 m 3 C 12 C 23 ( Γ + ε ) ρ | | u | | 0 > 0 ,

from which we see that inf u Ω r | | Q u | | 0 > 0 , namely the hypotheses (i) of Lemma 2.6 holds. Next, we show that if R is large enough, then μQu u for any u ∈ ∂ΩR and μ ≥ 1. In fact, suppose the contrary, then there exist u0 ∈ ∂ΩR and μ0 ≥ 1 such that μ0Qu0 = u0, then by (3.6), A F u 0 Q u 0 u 0 = μ 0 Q u 0 μ 0 1 - L A F u 0 . Let ω0 = AFu0, then ω 0 u 0 μ 0 1 - L ω 0 , and ω0 satisfies BVP (2.4), in which h = Fu0, consequently,

- ω 0 ( 6 ) - γ ω 0 ( 4 ) + β ω 0 - α ω 0 = f ( t , u 0 ) , t [ 0 , 1 ] , ω 0 ( 0 ) = ω 0 ( 1 ) = ω 0 ( 0 ) = ω 0 ( 1 ) = ω 0 ( 4 ) ( 0 ) = ω 0 ( 4 ) ( 1 ) = 0 , (3.27)

After multiplying the two sides of the first equation in (3.27) by sin πt and integrating on [0,1], we have

Γ 0 1 ω 0 ( t ) sin π t d t = 0 1 f ( t , u 0 ( t ) ) sin π t d t ( Γ + ε ) 0 1 u 0 ( t ) sin π t d t - 2 C π ( Γ + ε ) 0 1 ω 0 ( t ) sin π t d t - 2 C π .

Consequently, we obtain that

0 1 ω 0 ( t ) sin π t d t 2 C π ε . (3.28)

It's easy to prove that ω0(t), the solution of LBVF (3.27) satisfies

ω 0 ( t ) δ 1 δ 2 δ 3 C 12 C 23 C 1 C 2 C 3 M 1 M 2 G 1 ( t , t ) | | ω 0 | | 0 ,

and accordingly,

0 1 ω 0 ( t ) sin π t d t δ 1 δ 2 δ 3 C 12 C 23 | | ω 0 | | 0 C 1 C 2 C 3 M 1 M 2 0 1 G 1 ( t , t ) sin π t d t , (3.29)

by (3.28), we get

| | ω 0 | | 0 2 C C 1 C 2 C 3 M 1 M 2 δ 1 δ 2 δ 3 C 12 C 23 π ε 0 1 G 1 ( t , t ) sin π t d t - 1 : = R ̄ , (3.30)

Consequently, | | u 0 | | 0 μ 0 1 - L | | ω 0 | | 0 μ 0 1 - L R ̄ .

We choose R > max { μ 0 1 - L R ̄ , R 0 } , then to any u ΩR, μ ≥ 1, there is always μQu u. Hence, hypothesis (ii) of Lemma 2.6 also holds. By Lemma 2.6, we have

i ( Q , Ω R , P ) = 0 . (3.31)

Now, by the additivity of fixed point index, combine (3.23) and (3.31) to conclude that

i ( Q , Ω R \ Ω ̄ r , P ) = i ( Q , Ω R , P ) - i ( Q , Ω r , P ) = - 1 .

Therefore, Q has a fixed poind in Ω R \ Ω ̄ r , which is the positive solution of BVP (1.1). The proof is completed. □

From Theorem 3.1, we immediately obtain the following.

Corollary 3.1. Assume (H1)-(H3) hold, then in each of the following cases:

(i) f 0 = , f ̄ = 0 , (ii) f ̄ 0 = 0 , f = ,

the BVP (1.1) has at least one positive solution.

### 4 Multiple solutions

Next, we study the multiplicity of positive solutions of BVP (1.1) and assume in this section that

(H4) there is a p >0 such that 0 ≤ u p and 0 ≤ t ≤ 1 imply f(t, u) < ηp, where η = C 1 C 2 C 3 M 1 M 2 1 - L 0 1 G 1 ( s , s ) d s - 1 .

(H5) there is a p >0 such that σp u p and 0 ≤ t ≤ 1 imply f (t, u) ≥ λp, where λ - 1 = δ 1 δ 2 δ 3 m 1 C 12 C 23 1 4 3 4 G 3 ( s , s ) d s . Here, σ can be defined as (3.10).

Theorem 4.1. Assume (H1)-(H4) hold. If f 0 > Γ and f > Γ , then BVP (1.1) has at least two positive solution u1 and u2 such that 0 ≤ ||u1||0 p ≤ ||u2||0.

Proof. According to the proof of Theorem 3.1, there exists 0 < r0 < p < R1 <+∞, such that 0 < r < r0 implies i(Q, Ωr, P) = 0 and R R1 implies i(Q, ΩR, P) = 0.

Next we prove i(Q, Ωp, P) = 1 if (H4) is satisfied. In fact, for every u ∈ ∂Ωp, based on the preceding definition of Q we come to

( Q u ) ( t ) = ( H F u ) ( t ) 1 1 - L | | A F u | | 0 = 1 1 - L max 0 t 1 0 1 0 1 0 1 G 1 ( t , δ ) G 2 ( δ , τ ) G 3 ( τ , s ) ( F u ) ( s ) d s d τ d δ C 1 C 2 C 3 M 1 M 2 1 - L 0 1 G 3 ( s , s ) f ( s , u ( s ) d s .

Consequently,

| | Q u | | 0 C 1 C 2 C 3 M 1 M 2 1 - L 0 1 G 3 ( s , s ) f ( s , u ( s ) ) d s C 1 C 2 C 3 M 1 M 2 1 - L 0 1 G 3 ( s , s ) η p d s = p = | | u | | 0 .

Therefore, by (ii) of Lemma 2.7 we have

i ( Q , Ω p , P ) = 1 . (4.1)

Combined with (3.17), (3.31), and (4.1), we have

i ( Q , Ω R \ Ω ̄ p , P ) = i ( Q , Ω R , P ) - i ( Q , Ω p , P ) = - 1 . i ( Q , Ω p \ Ω ̄ r , P ) = i ( Q , Ω p , P ) - i ( Q , Ω r , P ) = 1 .

Therefore, Q has fixed points u1 and u2 in Ω p \ Ω ̄ r and Ω R \ Ω ̄ p , respectively, which

means that u1(t) and u2(t) are positive solutions of BVP (1.1) and 0 ≤ ||u1||0 p ≤ ||u2||0. The proof is completed. □

Theorem 4.2. Assume (H1)-(H3) and (H5) can be established, and f ̄ 0 < ( 1 - L ) Γ and f ̄ < ( 1 - L ) Γ , then BVP (1.1) has at least two positive solution u1 and u2 such that 0 ≤ ||u1||0 p ≤ ||u2||0.

Proof. According to the proof of Theorem 3.1, there exists 0 < ω < p < R2 <+ ∞, such that 0 < r < ω implies i(Q, Ωr, P) = 1 and R R2 implies i(Q, ΩR, P) = 1.

We now prove that i(Q, Ωp, P) = 0 if (H5) is satisfied. In fact, for every u Ωp, by (3.13) we come to ρp ρ||u||0 u(t) ≤ ||u||0 = p, t ∈ [1/4, 3/4], accordingly, by (H5), we have

f ( t , u ) λ p , t 1 4 , 3 4 , u Ω p .

from the proof of (ii) of Theorem 3.1, we have

| | Q u | | 0 Q u 1 2 = ( H F u ) 1 2 ( A F u ) 1 2 = 0 1 0 1 0 1 G 1 1 2 , δ G 2 ( δ , τ ) G 3 ( τ , s ) f ( s , u ( s ) ) d s d τ d δ 1 4 3 4 1 4 3 4 1 4 3 4 G 1 1 2 , δ G 2 ( δ , τ ) G 3 ( τ , s ) λ p d s d τ d δ δ 1 δ 2 δ 3 m 1 C 12 C 23 1 4 3 4 G 3 ( s , s ) λ p d s = p = | | u | | 0 .

Therefore, | | Q u | | 0 Q u 1 2 | | u | | 0 , according to (i) of Lemma 2.7, we come to

i ( Q , Ω p , P ) = 0 . (4.2)

Combined with (3.20), (3.23), and (4.2), there exist

i ( Q , Ω R \ Ω ̄ p , P ) = i ( Q , Ω R , P ) - i ( Q , Ω p , P ) = 1 . i ( Q , Ω p \ Ω ̄ r , P ) = i ( Q , Ω p , P ) - i ( Q , Ω r , P ) = - 1 .

Therefore, Q has fixed points u1 and u2 in Ω p \ Ω ̄ r and Ω R \ Ω ̄ p , respectively, which means that u1(t) and u2(t) are positive solutions of BVP (1.1) and 0 ≤ ||u1||0 p ≤ ||u2||0. The proof is completed. □

Theorem 4.3. Assume that (H1)-(H3) hold. If f 0 > Γ and f ̄ < ( 1 - L ) Γ , and there exists p2 > p1 >0 that satisfies

(i) f(t, u) < ηp1 if 0 ≤ t ≤ 1 and 0 ≤ u p1,

(ii) f(t, u) ≥ λp2 if 0 ≤ t ≤ 1 and σp2 u p2,

where η, σ, λ are just as the above, then BVP (1.1) has at least three positive solutions u1, u2, and u3 such that 0 ≤ ||u1||0 p1 ≤ ||u2||0 p2 ≤ ||u3||0.

Proof. According to the proof of Theorem 3.1, there exists 0 < r0 < p1 < p2 < R3 <+∞, such that 0 < r < r0 implies i(Q, Ωr, P) = 0 and R R3 implies i(Q, ΩR, P) = 1.

From the proof of Theorems 4.1 and 4.2, we have i ( Q , Ω p 1 , P ) = 1 , i ( Q , Ω p 2 , P ) = 0 . Combining the four afore-mentioned equations, we have

i ( Q , Ω R \ Ω ̄ p 2 , P ) = i ( Q , Ω R , P ) - i ( Q , Ω p 2 , P ) = 1 . i ( Q , Ω p 2 \ Ω ̄ p 1 , P ) = i ( Q , Ω p 2 , P ) - i ( Q , Ω p 1 , P ) = - 1 . i ( Q , Ω p 1 \ Ω ̄ r , P ) = i ( Q , Ω p 1 , P ) - i ( Q , Ω r , P ) = 1 .

Therefore, Q has fixed points u1, u2 and u3 in Ω R \ Ω ̄ p 2 , Ω p 2 \ Ω ̄ p 1 and Ω p 1 \ Ω ̄ r , respectively, which means that u1(t), u2(t) and u3(t) are positive solutions of BVP (1.1) and 0 ≤ ||u1||0 p1 ≤ ||u2||0 p2 ≤ ||u3||0. The proof is completed. □

### Competing interests

The author declares that they have no competing interests.

### Authors' contributions

WL conceived of the study, and participated in its design and coordination. The author read and approved the final manuscript.

### Acknowledgements

The author is very grateful to the anonymous referees for their valuable suggestions, and to be sponsored by the Tutorial Scientific Research Program Foundation of Education Department of Gansu Province P.R.China(1110-05).

### References

1. Tersian, S, Chaparova, J: Periodic and homoclinic solutions of some semilinear sixth-order ODE. J Math Anal Appl. 272, 223–239 (2002). Publisher Full Text

2. Gardner, RA, Jones, CKRT: Traveling waves of a perturbed diffusion equation arising in a phase field model. Indiana Univ Math J. 39(4), 1197–1222 (1990). Publisher Full Text

3. Caginalp, G, Fife, P: Higher order phase field models and detailed anisotropy. Phys Rev B. 34, 4940–4943 (1986). Publisher Full Text

4. Gyulov, T, Morosanu, G, Tersian, S: Existence for a semilinear sixth-order ODE. J Math Anal Appl. 321, 86–98 (2006). Publisher Full Text

5. Zhang, L, An, Y: Existence and multiplicity of positive solutions of a boundary-value problem for sixth-order ode with three parameters. Bound Value Probl. 2010, 13 (Article ID 878131) doi:10.1155/2010/878131 (2010)

6. Bates, P, Fife, P, Gardner, R, Jones, CKRJ: The existence of traveling wave solutions of a generalized phase-field model. SIAM J Math Anal. 28, 60–93 (1997). Publisher Full Text

7. Peletier, LA, Troy, WC, Van der Vorst, RCAM: Stationary solutions of a fourth-order nonlinear diffusion equation. Diff Equ. 31, 301–314 (1995)

8. Chai, G: Existence of positive solutions for fourth-order boundary value problem with variable parameters. Nonlinear Anal. 66, 870–880 (2007). Publisher Full Text

9. Zhang, M, Wei, Z: Existence of positive solutions for fourth-order m-point boundary value problem with variable parameters. Appl Math Comput. 190, 1417–1431 (2007). Publisher Full Text

10. Li, Y: Positive solutions of fourth-order boundary value problems with two parameters. J Math Anal Appl. 281, 477–484 (2003). Publisher Full Text

11. Guo, D, Lakshmikantham, V, Liu, X: Nonlinear Integral Equtions in Abstract Spaces, Kluwer Academic Publishers, Dordrecht-Boston-Loudon (1996)

12. Guo, D, Lakshmikantham, V: Nonlinear Problems in Abstract Cones, Academic Press, New York (1998)