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Positive solutions for a sixth-order boundary value problem with four parameters

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Abstract

This paper investigates the existence and multiplicity of positive solutions of a sixth-order differential system with four variable parameters using a monotone iterative technique and an operator spectral theorem.

MSC:34B15, 34B18.

1 Introduction

It is well known that boundary value problems for ordinary differential equations can be used to describe a large number of physical, biological and chemical phenomena. In recent years, boundary value problems for sixth-order ordinary differential equations, which arise naturally, for example, in sandwich beam deflection under transverse shear have been studied extensively, see [14] and the references therein. The deformation of the equilibrium state of an elastic circular ring segment with its two ends simply supported can be described by a boundary value problem involving a sixth-order ordinary differential equation

u ( 6 ) + 2 u ( 4 ) + u = f ( t , u ) , 0 < t < 1 , u ( 0 ) = u ( 1 ) = u ( 0 ) = u ( 1 ) = u ( 4 ) ( 0 ) = u ( 4 ) ( 1 ) = 0 .
(1)

Liu and Li [5] studied the existence and nonexistence of positive solutions of the nonlinear fourth-order beam equation

u ( 4 ) ( t ) + β u ( t ) α u ( t ) = λ f ( t , u ( t ) ) , 0 < t < 1 , u ( 0 ) = u ( 1 ) = u ( 0 ) = u ( 1 ) = 0 .
(2)

They showed that there exists a λ >0 such that the above boundary value problem has at least two, one, and no positive solutions for 0<λ< λ , λ= λ and λ> λ , respectively.

In this paper, we discuss the existence of positive solutions for the sixth-order boundary value problem

u ( 6 ) + A ( t ) u ( 4 ) + B ( t ) u + C ( t ) u = ( D ( t ) + u ) φ + λ f ( t , u ) , 0 < t < 1 , φ + ϰ φ = μ u , 0 < t < 1 , u ( 0 ) = u ( 1 ) = u ( 0 ) = u ( 1 ) = u ( 4 ) ( 0 ) = u ( 4 ) ( 1 ) = 0 , φ ( 0 ) = φ ( 1 ) = 0 .
(3)

For this, we shall assume the following conditions throughout

(H1) f(t,u):[0,1]×[0,)[0,) is continuous;

(H2) a,b,cR, a= λ 1 + λ 2 + λ 3 > π 2 , b= λ 1 λ 2 λ 2 λ 3 λ 1 λ 3 >0, c= λ 1 λ 2 λ 3 <0 where λ 1 0 λ 2 π 2 , 0 λ 3 < λ 2 and π 6 +a π 4 b π 2 +c>0, and A,B,C,DC[0,1] with a= sup t [ 0 , 1 ] A(t), b= inf t [ 0 , 1 ] B(t) and c= sup t [ 0 , 1 ] C(t).

Let K= max 0 t 1 [A(t)+B(t)C(t)(a+bc)] and Γ= π 6 +a π 4 b π 2 +c.

Assumption (H2) involves a three-parameter nonresonance condition.

More recently Li [6] studied the existence and multiplicity of positive solutions for a sixth-order boundary value problem with three variable coefficients. The main difference between our work and [6] is that we consider boundary value problem not only with three variable coefficients, but also with two positive parameters λ and μ, and the existence of the positive solution depends on these parameters. In this paper, we shall apply the monotone iterative technique [7] to boundary value problem (3) and then obtain several new existence and multiplicity results. In the special case, in [8] by using the fixed point theorem and the operator spectral theorem, we establish a theorem on the existence of positive solutions for the sixth-order boundary value problem (3) with λ=1.

2 Preliminaries

Let Y=C[0,1] and Y + ={uY:u(t)0,t[0,1]}. It is well known that Y is a Banach space equipped with the norm u 0 = sup t [ 0 , 1 ] |u(t)|. Set X={u C 4 [0,1]:u(0)=u(1)= u (0)= u (1)=0}. For given χ0 and ν0, we denote the norm χ , ν by

χ , ν = sup t [ 0 , 1 ] { | u ( 4 ) ( t ) | + χ | u ( t ) | + ν | u ( t ) | } ,uX.

We also need the space X, equipped with the norm

u 2 =max { u 0 , u 0 , u ( 4 ) 0 } .

In [8], it is shown that X is complete with the norm χ , ν and u 2 , and moreover uX, u 0 u 0 u ( 4 ) 0 .

For hY, consider the linear boundary value problem

u ( 6 ) + a u ( 4 ) + b u + c u = h ( t ) , 0 < t < 1 , u ( 0 ) = u ( 1 ) = u ( 0 ) = u ( 1 ) = u ( 4 ) ( 0 ) = u ( 4 ) ( 1 ) = 0 ,
(4)

where a, b, c satisfy the assumption

π 6 +a π 4 b π 2 +c>0,
(5)

and let Γ= π 6 +a π 4 b π 2 +c. Inequality (5) follows immediately from the fact that Γ= π 6 +a π 4 b π 2 +c is the first eigenvalue of the problem u ( 6 ) +a u ( 4 ) +b u +cu=λu, u(0)=u(1)= u (0)= u (1)= u ( 4 ) (0)= u ( 4 ) (1)=0, and ϕ 1 (t)=sinπt is the first eigenfunction, i.e., Γ>0. Since the line l 1 ={(a,b,c): π 6 +a π 4 b π 2 +c=0} is the first eigenvalue line of the three-parameter boundary value problem u ( 6 ) +a u ( 4 ) +b u +cu=0, u(0)=u(1)= u (0)= u (1)= u ( 4 ) (0)= u ( 4 ) (1)=0, if (a,b,c) lies in l 1 , then by the Fredholm alternative, the existence of a solution of the boundary value problem (4) cannot be guaranteed.

Let P(λ)= λ 2 +βλα, where β<2 π 2 , α0. It is easy to see that the equation P(λ)=0 has two real roots λ 1 , λ 2 = β ± β 2 + 4 α 2 with λ 1 0 λ 2 > π 2 . Let λ 3 be a number such that 0 λ 3 < λ 2 . In this case, (4) satisfies the decomposition form

u ( 6 ) +a u ( 4 ) +b u +cu= ( d 2 d t 2 + λ 1 ) ( d 2 d t 2 + λ 2 ) ( d 2 d t 2 + λ 3 ) u,0<t<1.
(6)

Suppose that G i (t,s) (i=1,2,3) is the Green’s function associated with

u + λ i u=0,u(0)=u(1)=0.
(7)

We need the following lemmas.

Lemma 1 [5, 9]

Let ω i = | λ i | , then G i (t,s) (i=1,2,3) can be expressed as

  1. (i)

    when λ i >0,

    G i (t,s)= { sinh ω i t sinh ω i ( 1 s ) ω i sinh ω i , 0 t s 1 , sinh ω i s sinh ω i ( 1 t ) ω i sinh ω i , 0 s t 1 } ;
  2. (ii)

    when λ i =0,

    G i (t,s)= { t ( 1 s ) , 0 t s 1 , s ( 1 t ) , 0 s t 1 } ;
  3. (iii)

    when π 2 < λ i <0,

    G i (t,s)= { sin ω i t sin ω i ( 1 s ) ω i sin ω i , 0 t s 1 , sin ω i s sin ω i ( 1 t ) ω i sin ω i , 0 s t 1 } .

Lemma 2 [5]

G i (t,s) (i=1,2,3) has the following properties

  1. (i)

    G i (t,s)>0, t,s(0,1);

  2. (ii)

    G i (t,s) C i G i (s,s), t,s[0,1];

  3. (iii)

    G i (t,s) δ i G i (t,t) G i (s,s), t,s[0,1],

where C i =1, δ i = ω i sinh ω i , if λ i >0; C i =1, δ i =1, if λ i =0; C i = 1 sin ω i , δ i = ω i sin ω i , if π 2 < λ i <0.

In what follows, we let D i = max t [ 0 , 1 ] 0 1 G i (t,s)ds.

Lemma 3 [10]

Let X be a Banach space, K a cone and Ω a bounded open subset of X. Let θΩ and T:K Ω ¯ K be condensing. Suppose that Txυx for all xKΩ and υ1. Then i(T,KΩ,K)=1.

Lemma 4 [10]

Let X be a Banach space, let K be a cone of X. Assume that T: K ¯ r K (here K r ={xKx<r}, r>0) is a compact map such that Txx for all x K r . If xTx for x K r , then i(T, K r ,K)=0.

Now, since

u ( 6 ) + a u ( 4 ) + b u + c u = ( d 2 d t 2 + λ 1 ) ( d 2 d t 2 + λ 2 ) ( d 2 d t 2 + λ 3 ) u = ( d 2 d t 2 + λ 2 ) ( d 2 d t 2 + λ 1 ) ( d 2 d t 2 + λ 3 ) u = h ( t ) ,
(8)

the solution of boundary value problem (4) can be expressed as

u(t)= 0 1 0 1 0 1 G 1 (t,v) G 2 (v,s) G 3 (s,τ)h(τ)dτdsdv,t[0,1].
(9)

Thus, for every given hY, the boundary value problem (4) has a unique solution u C 6 [0,1], which is given by (9).

We now define a mapping T:C[0,1]C[0,1] by

(Th)(t)= 0 1 0 1 0 1 G 1 (t,v) G 2 (v,s) G 3 (s,τ)h(τ)dτdsdv,t[0,1].
(10)

Throughout this article, we shall denote Th=u the unique solution of the linear boundary value problem (4).

Lemma 5 [8]

T:Y(X, χ , ν ) is linear completely continuous, where χ= λ 1 + λ 3 , ν= λ 1 λ 3 and T D 2 . Moreover, h Y + , if u=Th, then uX Y + , and u 0, u ( 4 ) 0.

We list the following conditions for convenience

(H3) f(t,u) is nondecreasing in u for t[0,1];

(H4) f(t,0)> c ˆ >0 for all t[0,1];

(H5) f = lim u f ( t , u ) u = uniformly for t[0,1];

(H6) f(t,ρu) ρ α f(t,u) for ρ(0,1) and t[0,1], where α(0,1) is independent of ρ and u.

Suppose that G(t,s) is the Green’s function of the linear boundary value problem

u +ϰu=0,u(0)=u(1)=0.
(11)

Then, the boundary value problem

φ +ϰφ=μu,φ(0)=φ(1)=0,

can be solved by using Green’s function, namely,

φ(t)=μ 0 1 G(t,s)u(s)ds,0<t<1,
(12)

where ϰ> π 2 . Thus, inserting (12) into the first equation in (3), yields

u ( 6 ) + A ( t ) u ( 4 ) + B ( t ) u + C ( t ) u = μ ( D ( t ) + u ( t ) ) 0 1 G ( t , s ) u ( s ) d s + λ f ( t , u ) , u ( 0 ) = u ( 1 ) = u ( 0 ) = u ( 1 ) = u ( 4 ) ( 0 ) = u ( 4 ) ( 1 ) = 0 .
(13)

Let us consider the boundary value problem

u ( 6 ) + A ( t ) u ( 4 ) + B ( t ) u + C ( t ) u = h ( t ) , 0 < t < 1 , u ( 0 ) = u ( 1 ) = u ( 0 ) = u ( 1 ) = u ( 4 ) ( 0 ) = u ( 4 ) ( 1 ) = 0 .
(14)

Now, we consider the existence of a positive solution of (14). The function u C 6 (0,1) C 4 [0,1] is a positive solution of (14), if u0, t[0,1], and u0.

Let us rewrite equation (13) in the following form

u ( 6 ) + a u ( 4 ) + b u + c u = ( A ( t ) a ) u ( 4 ) ( B ( t ) b ) u ( C ( t ) c ) u + μ ( D ( t ) + u ( t ) ) 0 1 G ( t , s ) u ( s ) d s + h ( t ) .
(15)

For any uX, let

Gu= ( A ( t ) a ) u ( 4 ) ( B ( t ) b ) u ( C ( t ) c ) u+μD(t) 0 1 G(t,s)u(s)ds.

The operator G:XY is linear. By Lemmas 2 and 3 in [8], uX, t[0,1], we have

| ( G u ) ( t ) | [ A ( t ) + B ( t ) C ( t ) ( a + b c ) ] u 2 + μ C d 1 u 0 ( K + μ C d 1 ) u 2 ( K + μ C d 1 ) u χ , ν ,

where C= max t [ 0 , 1 ] D(t), K= max t [ 0 , 1 ] [A(t)+B(t)C(t)(a+bc)], d 1 = max t [ 0 , 1 ] 0 1 G(t,s)ds, χ= λ 1 + λ 3 0, ν= λ 1 λ 3 0. Hence G u 0 (K+μC d 1 ) u χ , ν , and so G(K+μC d 1 ). Also u C 4 [0,1] C 6 (0,1) is a solution of (13) if uX satisfies u=T(Gu+ h 1 ), where h 1 (t)=μu(t) 0 1 G(t,s)u(s)ds+h(t), i.e.,

uX,(ITG)u=T h 1 .
(16)

The operator ITG maps X into X. From T D 2 together with G(K+μC d 1 ) and the condition D 2 (K+μC d 1 )<1, and applying the operator spectral theorem, we find that ( I T G ) 1 exists and is bounded. Let μ(0, 1 D 2 K D 2 C d 1 ), where 1 D 2 K>0, then the condition D 2 (K+μC d 1 )<1 is fulfilled. Let L= D 2 (K+μC d 1 ), and let μ = 1 D 2 K D 2 C d 1 .

Let H= ( I T G ) 1 T. Then (16) is equivalent to u=H h 1 . By the Neumann expansion formula, H can be expressed by

H= ( I + T G + + ( T G ) n + ) T=T+(TG)T++ ( T G ) n T+.
(17)

The complete continuity of T with the continuity of ( I T G ) 1 guarantees that the operator H:YX is completely continuous.

Now h Y + , let u=Th, then uX Y + , and u 0, u ( 4 ) 0. Thus, we have

( G u ) ( t ) = ( A ( t ) a ) u ( 4 ) ( B ( t ) b ) u ( C ( t ) c ) u + μ D ( t ) 0 1 G ( t , s ) u ( s ) d s 0 , t [ 0 , 1 ] .

Hence

h Y + ,(GTh)(t)0,t[0,1],
(18)

and so, (TG)(Th)(t)=T(GTh)(t)0, t[0,1].

It is easy to see [11] that the following inequalities hold: h Y + ,

1 1 L (Th)(t)(Hh)(t)(Th)(t),t[0,1],
(19)

and, moreover,

( H h ) 0 1 1 L ( T h ) 0 .
(20)

Lemma 6 [8]

H:Y(X, ϰ , ν ) is completely continuous, where χ= λ 1 + λ 3 , ν= λ 1 λ 3 and h Y + , 1 1 L (Th)(t)(Hh)(t)(Th)(t), t[0,1], and, moreover, T h 0 (1L) H h 0 .

For any u Y + , define Fu=μu(t) 0 1 G(t,s)u(s)ds+λf(t,u). From (H1), we have that F: Y + Y + is continuous. It is easy to see that u C 4 [0,1] C 6 (0,1), being a positive solution of (13), is equivalent to u Y + , being a nonzero solution of

u=HFu.
(21)

Let us introduce the following notations

T λ , μ u ( t ) : = T F u ( t ) = 0 1 0 1 0 1 G 1 ( t , v ) G 2 ( v , s ) G 3 ( s , τ ) × ( μ u ( τ ) 0 1 G ( τ , s ) u ( s ) d s + λ f ( τ , u ( τ ) ) ) d τ d s d v , Q λ , μ u : = H F u = T F u + ( T G ) T F u + ( T G ) 2 T F u + + ( T G ) n T F u + = T λ , μ u + ( T G ) T λ , μ u + ( T G ) 2 T λ , μ u + + ( T G ) n T λ , μ u + ,

i.e., Q λ , μ u=HFu. Obviously, Q λ , μ : Y + Y + is completely continuous. We next show that the operator Q λ , μ has a nonzero fixed point in Y + .

Let P={u Y + :u(t) δ 1 (1L) g 1 (t) u ( t ) 0 ,t[ 1 4 , 3 4 ]}, where g 1 (t)= 1 C 1 G 1 (t,t). It is easy to see that P is a cone in Y, and now, we show Q λ , μ (P)P.

Lemma 7 Q λ , μ (P)P and Q λ , μ :PP is completely continuous.

Proof It is clear that Q λ , μ :PP is completely continuous. Now uP, let h 1 =Fu, then h 1 Y + . Using Lemma 6, i.e., ( Q λ , μ u)(t)=(HFu)(t)(TFu)(t), t[0,1] and by Lemma 2, for all uP, we have

(TFu)(t) C 1 0 1 0 1 0 1 G 1 (v,v) G 2 (v,s) G 3 (s,τ)(Fu)(τ)dτdsdv,t[0,1].

Thus,

0 1 0 1 0 1 G 1 (v,v) G 2 (v,s) G 3 (s,τ)(Fu)(τ)dτdsdv 1 C 1 T F u 0 .
(22)

On the other hand, by Lemma 6 and (22), we have

( T F u ) ( t ) δ 1 G 1 ( t , t ) 0 1 0 1 0 1 G 1 ( v , v ) G 2 ( v , s ) G 3 ( s , τ ) ( F u ) ( τ ) d τ d s d v δ 1 G 1 ( t , t ) 1 C 1 T F u 0 δ 1 G 1 ( t , t ) 1 C 1 ( 1 L ) Q u 0 , t [ 0 , 1 ] .

Thus, Q λ , μ (P)P. □

3 Main results

Lemma 8 Let f(t,u) be nondecreasing in u for t[0,1] and f(t,0)> c ˆ >0 for all t[0,1], where c ˆ is a constant and L<1. Then there exists λ >0 and μ >0 such that the operator Q λ , μ has a fixed point u at ( λ , μ ) with u P{θ}.

Proof Set u ˆ 1 (t)=( Q λ , μ u )(t), where

u (t)= 2 1 L 0 1 0 1 0 1 G 1 (t,v) G 2 (v,τ) G 3 (τ,s)dsdτdv.

It is easy to see that u (t)P. Let λ = M f u 1 and μ =min( N f u 1 ; μ ), where M f u = max t [ 0 , 1 ] f(t, u (t)) and N f u = max t [ 0 , 1 ] u (t) 0 1 G(t,s) u (s)ds, respectively. Then M f u >0 and N f u >0, and from Lemma 6, we obtain

u ˆ 0 ( t ) = u ( t ) = 2 1 L 0 1 0 1 0 1 G 1 ( t , v ) G 2 ( v , τ ) G 3 ( τ , s ) d s d τ d v 1 1 L 0 1 0 1 0 1 G 1 ( t , v ) G 2 ( v , τ ) G 3 ( τ , s ) × ( λ f ( s , u ( s ) ) + μ u ( t ) 0 1 G ( t , s ) u ( s ) d s ) d s d τ d v = 1 1 L ( T λ , μ u ) ( t ) ( Q λ , μ u ) ( t ) = u ˆ 1 ( t ) .

It is easy to see that

u ˆ n ( t ) = ( Q λ , μ u ˆ n 1 ) ( t ) = ( T λ , μ u ˆ n 1 + ( T G ) T λ , μ u ˆ n 1 + ( T G ) 2 T λ , μ u ˆ n 1 + + ( T G ) n T λ , μ u ˆ n 1 + ) ( T λ , μ u ˆ n 2 + ( T G ) T λ , μ u ˆ n 2 + ( T G ) 2 T λ , μ u ˆ n 2 + + ( T G ) n T λ , μ u ˆ n 2 + ) = u ˆ n 1 ( t ) .

Indeed, for h 1 , h 2 Y + , let h 1 (t) h 2 (t), then from (10), we have u 1 (t)=T h 1 T h 2 = u 2 (t). Using equation (4) and (6), we obtain

u + λ 2 u= 0 1 0 1 G 1 (t,v) G 3 (v,τ)h(τ)dτdv,t[0,1]
(23)

and

u ( 4 ) ( λ 2 + λ 3 ) u + λ 2 λ 3 u= 0 1 G 1 (t,v)h(v)dv,t[0,1].
(24)

Then by (23), we have for t[0,1]

u 1 (t) u 2 (t)= λ 2 ( u 1 ( t ) u 2 ( t ) ) 0 1 0 1 G 1 (t,v) G 3 (v,τ) ( h 1 ( t ) h 2 ( t ) ) dτdv0,

because λ 2 <0, and finally, from (24), we have

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

because λ 2 + λ 3 0 and λ 2 λ 3 0. From the equation

( G u ) ( t ) = ( A ( t ) a ) u ( 4 ) ( B ( t ) b ) u ( C ( t ) c ) u + μ D ( t ) 0 1 G ( t , s ) u ( s ) d s 0 , t [ 0 , 1 ]

we have

( G u 1 ) ( t ) ( G u 2 ) ( t ) = ( A ( t ) a ) ( u 1 ( 4 ) ( t ) u 2 ( 4 ) ( t ) ) ( B ( t ) b ) ( u 1 ( t ) u 2 ( t ) ) ( C ( t ) c ) ( u 1 ( t ) u 2 ( t ) ) + μ D ( t ) 0 1 G ( t , s ) ( u 1 ( t ) u 2 ( t ) ) d s 0 , t [ 0 , 1 ]
(25)

i.e., (G u 1 )(t)(G u 2 )(t) for all t[0,1]. Finally, if h 1 =F u 1 and h 2 =F u 2 , then

( H h 1 ) ( t ) = T ( h 1 ) + ( T G ) ( T h 1 ) + ( T G ) 2 ( T h 1 ) + + ( T G ) n ( T h 1 ) + ( H h 2 ) ( t ) = T ( h 2 ) + ( T G ) ( T h 2 ) + ( T G ) 2 ( T h 2 ) + + ( T G ) n ( T h 2 ) +

i.e.,

Q λ , μ u 1 = ( H F u 1 ) ( t ) = T ( F u 1 ) + ( T G ) ( T F u 1 ) + ( T G ) 2 ( T F u 1 ) + + ( T G ) n ( T F u 1 ) + ( H F u 2 ) ( t ) = T ( F u 2 ) + ( T G ) ( T F u 2 ) + ( T G ) 2 ( T F u 2 ) + + ( T G ) n ( T F u 2 ) + = Q λ , μ u 2 ,
(26)

and from (26), it follows that for u 1 , u 2 Y + , if u 1 (t) u 2 (t) then, we have

Q λ , μ u 1 Q λ , μ u 2 .
(27)

Set u ˆ 0 (t)= u (t) and u ˆ n (t)=( Q λ , μ u ˆ n 1 )(t), n=1,2, , t[0,1]. Then

u ˆ 0 (t)= u (t) u ˆ 1 (t) u ˆ n (t) L 1 G 1 (t,t),

where

L 1 = λ δ 1 δ 2 δ 3 c ˆ C 23 C 12 C 3 .

 □

Indeed, by Lemma 6, we have

u ˆ n ( t ) = Q λ , μ u ˆ n 1 = ( H F ) ( u ˆ n 1 ) ( T F ) ( u ˆ n 1 ) λ 0 1 0 1 0 1 G 1 ( t , v ) G 2 ( v , τ ) G 3 ( τ , s ) f ( s , u ˆ n 1 ( s ) ) d s d τ d v λ c ˆ 0 1 0 1 0 1 G 1 ( t , v ) G 2 ( v , τ ) G 3 ( τ , s ) d s d τ d v λ c ˆ δ 1 δ 2 δ 3 C 12 C 23 C 3 G 1 ( t , t ) .

Now, f(t,u) nondecreasing in u for t[0,1], Lemma 2, and the Lebesgue convergence theorem guarantee that { u n } n = 0 = { Q λ , μ u ˆ 0 } n = 0 decreases to a fixed point u P{θ} of the operator Q λ , μ .

Lemma 9 Suppose that (H3)-(H5) hold, and L<1. Set

S λ , μ = { u P : Q λ , μ u = u , ( λ , μ ) A } ,

where A[a,)×[b,) for some constants a>0, b>0. Then there exists a constant C A such that u 0 < C A for all u S λ , μ .

Proof Suppose, to the contrary, that there exists a sequence { u n } n = 1 such that lim n u n 0 =+, where u n P is a fixed point of the operator Q λ , μ at ( λ n , μ n )A (n=1,2,). Then

u n (t)k u n 0 for t [ 1 4 , 3 4 ] ,

where k= δ 1 C 1 (1L) min t [ 1 4 , 3 4 ] G 1 (t,t).

Choose J 1 >0, so that

J 1 a δ 1 δ 2 δ 3 C 12 C 23 m 1 m 2 k>2,

and l 1 >0 such that

f(t,u) J 1 ufor u> l 1  and t [ 1 4 , 3 4 ] ,

and N 0 , so that u N 0 > l 1 k . Now,

( Q λ N 0 , μ N 0 u N 0 ) ( 1 2 ) ( T F u N 0 ) ( 1 2 ) λ N 0 0 1 0 1 0 1 G 1 ( 1 2 , v ) G 2 ( v , τ ) G 3 ( τ , s ) f ( s , u N 0 ( s ) ) d s d τ d v λ N 0 δ 1 δ 2 δ 3 C 12 C 23 G 1 ( 1 2 , 1 2 ) 0 1 G 3 ( s , s ) f ( s , u N 0 ( s ) ) d s λ N 0 δ 1 δ 2 δ 3 C 12 C 23 G 1 ( 1 2 , 1 2 ) 1 4 3 4 G 3 ( s , s ) f ( s , u N 0 ( s ) ) d s 1 2 a δ 1 δ 2 δ 3 C 12 C 23 m 1 m 2 J 1 u N 0 ( t ) 1 2 a δ 1 δ 2 δ 3 C 12 C 23 m 1 m 2 J 1 k u N 0 0 > u N 0 0 ,

and so,

u N 0 0 = Q λ N 0 , μ N 0 u N 0 0 ( T F ) u N 0 0 (TF u N 0 ) ( 1 2 ) > u N 0 0 ,

which is a contradiction. □

Lemma 10 Suppose that L<1, (H3) and (H4) hold and that the operator Q λ , μ has a positive fixed point in P at λ ˆ >0 and μ ˆ >0. Then for every ( λ , μ )(0, λ ˆ )×(0, μ ˆ ) there exists a function u P{θ} such that Q λ , μ u = u .

Proof Let u ˆ (t) be a fixed point of the operator Q λ , μ at ( λ ˆ , μ ˆ ). Then

u ˆ (t)= Q λ ˆ , μ ˆ u ˆ (t) Q λ , μ u ˆ (t),

where 0< λ < λ ˆ , 0< μ < μ ˆ . Hence

0 1 0 1 0 1 G 1 ( t , v ) G 2 ( v , τ ) G 3 ( τ , s ) ( λ ˆ f ( s , u ˆ ( s ) ) + μ ˆ u ˆ ( s ) 0 1 G ( s , p ) u ˆ ( p ) d p ) d s d τ d v 0 1 0 1 0 1 G 1 ( t , v ) G 2 ( v , τ ) G 3 ( τ , s ) × ( λ f ( s , u ˆ ( s ) ) + μ u ˆ ( s ) 0 1 G ( s , p ) u ˆ ( p ) d p ) d s d τ d v .

Set

( T λ , μ u ) ( t ) = 0 1 0 1 0 1 G 1 ( t , v ) G 2 ( v , τ ) G 3 ( τ , s ) × ( λ f ( s , u ( s ) ) + μ u ( s ) 0 1 G ( s , p ) u ( p ) d p ) d s d τ d v

and

( Q λ , μ u)(t)= T λ , μ u+(TG) T λ , μ u+ ( T G ) 2 T λ , μ u++ ( T G ) n T λ , μ u+

u 0 (t)= u ˆ (t) and u n (t)= Q λ , μ u n 1 . Then

u 0 ( t ) = u ˆ ( t ) = T λ ˆ , μ ˆ u ˆ + ( T G ) T λ ˆ , μ ˆ u ˆ + ( T G ) 2 T λ ˆ , μ ˆ u ˆ + + ( T G ) n T λ ˆ , μ ˆ u ˆ + T λ , μ u ˆ + ( T G ) T λ , μ u ˆ + ( T G ) 2 T λ , μ u ˆ + + ( T G ) n T λ , μ u ˆ + = u 1 ( t )

and

u n ( t ) = Q λ , μ u n 1 = T λ , μ u n 1 + ( T G ) T λ , μ u n 1 + ( T G ) 2 T λ , μ u n 1 + + ( T G ) n T λ , μ u n 1 + T λ , μ u n 2 + ( T G ) T λ , μ u n 2 + ( T G ) 2 T λ , μ u n 2 + + ( T G ) n T λ , μ u n 2 + = u n 1 ( t )

because f(t,u) is nondecreasing in u for t[0,1] and T λ , μ u is also nondecreasing in u. Thus

u 0 (t) u 1 (t) u n (t) u n + 1 (t) L 2 G 1 (t,t),
(28)

where

L 2 = λ c ˆ δ 1 δ 2 δ 3 C 12 C 23 C 3 .

 □

Indeed, by Lemma 6, we have

u n ( t ) = Q λ , μ u n 1 = ( H F ) ( u n 1 ) T λ , μ ( u n 1 ) λ 0 1 0 1 0 1 G 1 ( t , v ) G 2 ( v , τ ) G 3 ( τ , s ) f ( s , u n 1 ( s ) ) d s d τ d v λ c ˆ 0 1 0 1 0 1 G 1 ( t , v ) G 2 ( v , τ ) G 3 ( τ , s ) d s d τ d v λ c ˆ δ 1 δ 2 δ 3 C 12 C 23 C 3 G 1 ( t , t ) .

Lemma 2 implies that { Q λ n u } n = 1 decreases to a fixed point u P{θ}.

Lemma 11 Suppose that L<1, (H3)-(H5) hold. Let

Λ= { λ > 0 , μ > 0 : Q λ , μ  have at least one fixed point at  ( λ , μ )  in  P } .

Then Λ is bounded.

Proof Suppose, to the contrary, that there exists a fixed point sequence { u n } n = 0 P of Q λ , μ at ( λ n , μ n ) such that lim n λ n = and 0< μ n < μ . Then there are two cases to be considered: (i) there exists a subsequence { u n i } n = 0 such that lim i u n i 0 =, which is impossible by Lemma 9, so we only consider the next case: (ii) there exists a constant H>0 such that u n 0 H, n=0,1,2,3, . In view of (H3) and (H4), we can choose l 0 >0 such that f(t,0)> l 0 H, and further, f(t, u n )> l 0 H for t[0,1]. We know that

u n = Q λ n , μ n u n T λ n , μ n u n .

Let v n (t)= T λ n , μ n u n , i.e., u n (t) v n (t). Then it follows that

v n ( 6 ) +a v n ( 4 ) +b v n +c v n = λ n f(t, u n )+ μ n u n (t) 0 1 G(t,p) u n (p)dp,0<t<1.
(29)

Multiplying (29) by sinπt and integrating over [0,1], and then using integration by parts on the left side of (29), we have

Γ 0 1 v n (t)sinπtdt= λ n 0 1 f(t, u n )sinπtdt+ μ n 0 1 u n (t)sinπt 0 1 G(t,p) u n (p)dpdt.

Next, assume that (ii) holds. Then

Γ 0 1 u n ( t ) sin π t d t Γ 0 1 v n ( t ) sin π t d t = λ n 0 1 f ( t , u n ) sin π t d t + μ n 0 1 u n ( t ) sin π t 0 1 G ( t , p ) u n ( p ) d p d t

and

Γ H 0 1 sin π t d t Γ u n 0 0 1 sin π t d t Γ 0 1 u n ( t ) sin π t d t Γ 0 1 v n ( t ) sin π t d t = λ n 0 1 f ( t , u n ) sin π t d t + μ n 0 1 u n ( t ) sin π t 0 1 G ( t , p ) u n ( p ) d p d t λ n l 0 H 0 1 sin π t d t

lead to Γ λ n l 0 , which is a contradiction. The proof is complete. □

Lemma 12 Suppose that L<1, (H3)-(H4) hold. Let

Λ μ = { λ > 0 : ( λ , μ ) Λ  and  μ  is fixed } ,

and let λ ˜ μ =sup Λ μ . Then Λ μ =(0, λ ˜ μ ], where Λ is defined in Lemma 11.

Proof By Lemma 10, it follows that (0, λ ˜ )×(0,μ)Λ. We only need to prove ( λ ˜ μ ,μ)Λ. We may choose a distinct nondecreasing sequence { λ n } n = 1 Λ such that lim n λ n = λ ˜ μ . Set u n P as a fixed point of Q λ , μ at ( λ n ,μ), n=1,2, , i.e., u n = Q λ n , μ u n . By Lemma 9, { u n } n = 1 , is uniformly bounded, so it has a subsequence, denoted by { u n k } k = 1 , converging to u ˜ P. Note that

u n = T λ n , μ u n + ( T G ) T λ n , μ u n + ( T G ) 2 T λ n , μ u n + + ( T G ) n T λ n , μ u n + = Q λ n , μ u n .
(30)

Taking the limit as n on both sides of (30), and using the Lebesgue convergence theorem, we have

u ˜ = T λ ˜ , μ u ˜ +(TG) T λ ˜ , μ u ˜ n + ( T G ) 2 T λ ˜ , μ u ˜ ++ ( T G ) n T λ ˜ , μ u ˜ +

which shows that Q λ , μ has a positive fixed point u ˜ at ( λ ˜ ,μ). □

Theorem 1 Suppose that (H3)-(H5) hold, and L<1. For fixed μ (0, μ ), then there exists at λ >0 such that (3) has at least two, one and has no positive solutions for 0<λ< λ , λ= λ for λ> λ , respectively.

Proof Suppose that (H3) and (H4) hold. Then there exists λ >0 and μ >0 such that Q λ , μ has a fixed point u λ , μ P{θ} at λ= λ and μ= μ . In view of Lemma 12, Q λ , μ also has a fixed point u λ ̲ , μ ̲ < u λ , μ , u λ ̲ , μ ̲ P{θ}, and 0< λ ̲ < λ , 0< μ ̲ < μ , μ (0, μ ). For 0< λ ̲ < λ , there exists δ 0 >0 such that

f(t, u λ , μ +δ)f(t, u λ , μ )f(t,0) ( λ λ ̲ 1 )

for t[0,1], 0<δ δ 0 . In this case, it is easy to see that

T λ ̲ , μ ̲ ( u λ , μ , + δ ) = λ ̲ 0 1 0 1 0 1 G 1 ( t , v ) G 2 ( v , τ ) G 3 ( τ , s ) f ( s , u λ , μ ( s ) + δ ) d s d τ d v + μ ̲ 0 1 0 1 0 1 G 1 ( t , v ) G 2 ( v , τ ) G 3 ( τ , s ) ( u λ , μ ( s ) + δ ) × 0 1 G ( s , p ) ( u λ , μ ( p ) + δ ) d p d s d τ d v λ 0 1 0 1 0 1 G 1 ( t , v ) G 2 ( v , τ ) G 3 ( τ , s ) f ( s , u λ ( s ) ) d s d τ d v + μ 0 1 0 1 0 1 G 1 ( t , v ) G 2 ( v , τ ) G 3 ( τ , s ) u λ , μ ( s ) × 0 1 G ( s , p ) u λ , μ ( p ) d p d s d τ d v = T λ , μ u λ , μ .

Indeed, we have

λ ̲ 0 1 0 1 0 1 G 1 ( t , v ) G 2 ( v , τ ) G 3 ( τ , s ) f ( s , u λ ( s ) + δ ) d s d τ d v λ 0 1 0 1 0 1 G 1 ( t , v ) G 2 ( v , τ ) G 3 ( τ , s ) f ( s , u λ ( s ) ) d s d τ d v = λ ̲ 0 1 0 1 0 1 G 1 ( t , v ) G 2 ( v , τ ) G 3 ( τ , s ) { f ( s , u λ ( s ) + δ ) f ( s , u λ ( s ) ) } d s d τ d v ( λ λ ̲ ) 0 1 0 1 0 1 G 1 ( t , v ) G 2 ( v , τ ) G 3 ( τ , s ) f ( s , u λ ( s ) ) d s d τ d v ( λ λ ̲ ) 0 1 0 1 0 1 G 1 ( t , v ) G 2 ( v , τ ) G 3 ( τ , s ) f ( s , 0 ) d s d τ d v ( λ λ ̲ ) 0 1 0 1 0 1 G 1 ( t , v ) G 2 ( v , τ ) G 3 ( τ , s ) f ( s , u λ ( s ) ) d s d τ d v = ( λ λ ̲ ) 0 1 0 1 0 1 G 1 ( t , v ) G 2 ( v , τ ) G 3 ( τ , s ) { f ( s , 0 ) f ( s , u λ ( s ) ) } d s d τ d v 0 .

Similarly, it is easy to see that

μ ̲ 0 1 0 1 0 1 G 1 ( t , v ) G 2 ( v , τ ) G 3 ( τ , s ) ( u λ , μ ( s ) + δ ) 0 1 G ( s , p ) ( u λ , μ ( p ) + δ ) d p d s d τ d v μ 0 1 0 1 0 1 G 1 ( t , v ) G 2 ( v , τ ) G 3 ( τ , s ) u λ , μ ( s ) 0 1 G ( s , p ) u λ , μ ( p ) d p d s d τ d v 0 .

Moreover, from (25), it follows that for T λ ̲ , μ ̲ ( u λ , μ +δ) T λ , μ u λ , μ we have

G ( T λ ̲ , μ ̲ ( u λ , μ + δ ) ) G( T λ , μ u λ , μ ).

Finally, we have

(TG) T λ ̲ , μ ̲ ( u λ , μ +δ)(TG) T λ , μ u λ , μ .

By induction, it is easy to see that

( T G ) n T λ ̲ , μ ̲ ( u λ , μ +δ) ( T G ) n T λ , μ u λ , μ ,n=1,2,.
(31)

Hence, using (31), we have

Q λ ̲ , μ ̲ ( u λ , μ + δ ) = T λ ̲ , μ ̲ ( u λ , μ + δ ) + ( T G ) T λ ̲ , μ ̲ ( u λ , μ + δ ) + ( T G ) 2 T λ ̲ , μ ̲ ( u λ , μ + δ ) + + ( T G ) n T λ ̲ , μ ̲ ( u λ , μ + δ ) + T λ , μ u λ , μ + ( T G ) T λ , μ u λ , μ + ( T G ) 2 T λ , μ u λ , μ + + ( T G ) n T λ , μ u λ , μ + = Q λ , μ ( u λ , μ )

i.e.,

Q λ ̲ , μ ̲ ( u λ , μ +δ) Q λ , μ ( u λ , μ )0,

so that

Q λ ̲ , μ ̲ ( u λ , μ +δ) Q λ , μ ( u λ , μ )= u λ , μ < u λ , μ +δ.

Set D u λ , μ ={uC[0,1]:δ<u(t)< u λ , μ +δ}. Then Q λ ̲ , μ ̲ :P D u λ , μ P is completely continuous. Furthermore, Q λ ̲ , μ ̲ uυu for υ1 and uP D u λ , μ . Indeed set uP D u λ , μ . Then there exists t 0 [0,1] such that u( t 0 )= u 0 = u λ , μ + δ 0 and

( Q λ ̲ , μ ̲ u ) ( t 0 ) = ( T λ ̲ , μ ̲ ( u ) + ( T G ) T λ ̲ , μ ̲ ( u ) + ( T G ) 2 T λ ̲ , μ ̲ ( u ) + + ( T G ) n T λ ̲ , μ ̲ ( u ) + ) ( t 0 ) ( T λ ̲ , μ ̲ ( u λ , μ + δ ) + ( T G ) T λ ̲ , μ ̲ ( u λ , μ + δ ) + ( T G ) 2 T λ ̲ , μ ̲ ( u λ , μ + δ ) + + ( T G ) n T λ ̲ , μ ̲ ( u λ , μ + δ ) + ) ( t 0 ) = Q λ ̲ , μ ̲ ( u λ , μ + δ ) ( t 0 ) < u λ , μ ( t 0 ) + δ = u ( t 0 ) υ u ( t 0 ) , υ 1 .

By Lemma 3, i( Q λ ̲ , μ ̲ ,P D u λ , μ ,P)=1.

Let k be such that

u(t)k u 0 for t [ 1 4 , 3 4 ] .

We know that lim u f ( t , u ) u = uniformly for t[0,1], so we may choose J 3 >0, so that

λ ̲ J 3 δ 1 δ 2 δ 3 C 12 C 23 m 1 C 3 k>2,

l 3 > u λ , μ + δ 0 >0, so that

f(t,u) J 3 ufor u> l 3  and t [ 1 4 , 3 4 ] .

Set R 1 = l 3 k and P R 1 ={uP: u 0 < R 1 }. Then Q λ ̲ , μ ̲ : P ¯ R 1 P is completely continuous. It is easy to obtain

( Q λ ̲ , μ ̲ u ) ( t ) ( T λ ̲ , μ ̲ u ) ( t ) λ ̲ 0 1 0 1 0 1 G 1 ( t , v ) G 2 ( v , τ ) G 3 ( τ , s ) f ( s , u ( s ) ) d s d τ d v λ ̲ δ 1 δ 2 δ 3 C 12 C 23 G 1 ( t , t ) 0 1 G 3 ( s , s ) f ( s , u ( s ) ) d s λ ̲ δ 1 δ 2 δ 3 C 12 C 23 G 1 ( t , t ) 1 4 3 4 G 3 ( s , s ) f ( s , u ( s ) ) d s 1 2 λ ̲ δ 1 δ 2 δ 3 C 12 C 23 m 1 C 3 J 3 u ( t ) 1 2 λ ̲ δ 1 δ 2 δ 3 C 12 C 23 m 1 C 3 J 3 k u 0 > u 0

for t[0,1] and u P R 1 . Now u(t)k u 0 =k R 1 = l 3 , and so

Q λ ̲ , μ ̲ u 0 > u 0 .

In view of Lemma 4, i( Q λ ̲ , μ ̲ , P R 1 ,P)=0. By the additivity of the fixed point index,

i( Q λ ̲ , μ ̲ , P R 1 P D ¯ u λ , μ ,P)=i( Q λ ̲ , μ ̲ , P R 1 ,P)i( Q λ ̲ , μ ̲ ,P D u λ , μ ,P)=1.

Thus Q λ ̲ , μ ̲ has a fixed point in {P D u λ , μ }{θ} and has another fixed point in P R 1 P D u λ , μ by choosing λ = λ ˜ . □

Let us introduce the notation μ=0 in the equation of (13), then we have

u ( 6 ) + A ( t ) u ( 4 ) + B ( t ) u + C ( t ) u = λ f ( t , u ) , u ( 0 ) = u ( 1 ) = u ( 0 ) = u ( 1 ) = u ( 4 ) ( 0 ) = u ( 4 ) ( 1 ) = 0 .
(32)

In this case, we can prove the following theorem, which is similar to Theorem 1.

Theorem 2 Suppose that (H3)-(H5) hold, and L<1. Then there exists at λ >0 such that (32) has at least two, one and has no positive solutions for 0<λ< λ , λ= λ for λ> λ , respectively.

We follow exactly the same procedure, described in detail in the proof of Theorem 1 for μ=0.

Let us introduce the following notations for μ=0 and λ=1

T F u ( t ) = 0 1 0 1 0 1 G 1 ( t , v ) G 2 ( v , s ) G 3 ( s , τ ) f ( τ , u ( τ ) ) d τ d s d v , Q u : = H F u = T F u + ( T G ) T F u + ( T G ) 2 T F u + + ( T G ) n T F u + ,
(33)

i.e., Qu= Q 1 , 0 u=HFu.

Lemma 13 Suppose that (H3), (H4) and (H6) hold, and L<1. Then for any u C + [0,1]{θ}, there exist real numbers S u s u >0 such that

s u g(t)(Qu)(t) S u g(t),for t[0,1],

where g(t)= 0 1 0 1 G 1 (t,τ) G 2 (τ,v) G 3 (v,v)dvdτ.

Proof For any u C + [0,1]{θ} from Lemma 6, we have

( Q u ) ( t ) = ( H F u ) ( t ) 1 1 L 0 1 0 1 0 1 G 1 ( t , v ) G 2 ( v , τ ) G 3 ( τ , s ) f ( s , u ( s ) ) d s d τ d v C 3 1 L max s [ 0 , 1 ] f ( s , u ( s ) ) 0 1 0 1 G 1 ( t , τ ) G 2 ( τ , v ) G 3 ( v , v ) d v d τ = C 3 1 L max s [ 0 , 1 ] f ( s , u ( s ) ) g ( t ) = S u g ( t ) for  t [ 0 , 1 ] .

Note that for any u C + [0,1]{θ}, there exists an interval [ a 1 , b 1 ](0,1) and a number p>0 such that u(t)p for t[ a 1 , b 1 ]. In addition, by (H6), there exists s 0 >0 and u 0 (0,) such that f(t, u 0 ) s 0 for t[ a 1 , b 1 ]. If p u 0 , then f(t,u)f(t,p)f(t, u 0 ) s 0 ; if p< u 0 , then f(t,u)f(t,p)f(t, p u 0 p) ( p u 0 ) α s 0 . Hence

( Q u ) ( t ) ( T F u ) ( t ) = 0 1 0 1 0 1 G 1 ( t , v ) G 2 ( v , τ ) G 3 ( τ , s ) f ( s , u ( s ) ) d s d τ d v δ 3 0 1 0 1 0 1 G 1 ( t , v ) G 2 ( v , τ ) G 3 ( τ , τ ) G 3 ( s , s ) f ( s , u ( s ) ) d s d τ d v δ 3 g ( t ) a 1 b 1 G 3 ( s , s ) f ( s , u ( s ) ) d s d τ d v ( b 1 a 1 ) δ 3 g ( t ) m G ( p u 0 ) α = s u g ( t ) ,

where m G = min s [ a 1 , b 1 ] G 3 (s,s), g(t)= 0 1 0 1 G 1 (t,v) G 2 (v,τ) G 3 (τ,τ)dτdv, s u =( b 1 a 1 ) δ 3 m G ( p u 0 ) α . □

Theorem 3 Suppose that (H3), (H4) and (H6) hold, L<1 and λ=1. Then

  1. (i)

    (32) has a unique positive solution u C + [0,1]{θ} satisfying

    m u g(t) u (t) M u g(t)for t[0,1],

where 0< m u < M u are constants.

  1. (ii)

    For any u 0 (t) C + [0,1]{θ}, the sequence

    u n ( t ) = ( Q u n 1 ) ( t ) = ( H F u n 1 ) ( t ) = T F u n 1 + ( T G ) T F u n 1 + ( T G ) 2 T F u n 1 + + ( T G ) n T F u n 1 +

(n=1,2,) converges uniformly to the unique solution u , and the rate of convergence is determined by

u n ( t ) u ( t ) =O ( 1 d α n ) ,

where 0<d<1 is a positive number.

Proof In view of (H3), (H4) and (H6), Q: C + [0,1] C + [0,1] is a nondecreasing operator and satisfies Q(ρu) ρ α Q(u) for t[0,1] and u C + [0,1]. Indeed, let u (t) u (t), u , u C + [0,1], since f(s,u) is nondecreasing in u, then by using f(s, u (s))f(s, u (s)), for t[0,1], it follows that

T F u ( t ) = 0 1 0 1 0 1 G 1 ( t , v ) G 2 ( v , τ ) G 3 ( τ , s ) f ( s , u ( s ) ) d s d τ d v 0 1 0 1 0 1 G 1 ( t , v ) G 2 ( v , τ ) G 3 ( τ , s ) f ( s , u ( s ) ) d s d τ d v = T F u ( t ) .

Moreover, from (25), it follows that for TF u (t)TF u (t)

G(TF u )(t)G(TF u )(t)for t[0,1].
(34)

Finally, since f(s,u) is nondecreasing in u, then by using form (34), f(s,G(TF u )(t))f(s,G(TF u )(t)), for t[0,1], we have

( T G ) T F ( u ) = 0 1 0 1 0 1 G 1 ( t , v ) G 2 ( v , τ ) G 3 ( τ , s ) f ( s , G ( T F u ) ( s ) ) d s d τ d v 0 1 0 1 0 1 G 1 ( t , v ) G 2 ( v , τ ) G 3 ( τ , s ) f ( s , G ( T F u ) ( s ) ) d s d τ d v = ( T G ) T F u ,

i.e.,

(TG)TF( u )(TG)TF u .

By induction, it is easy to see that

( T G ) n TF( u ) ( T G ) n TF u ,n=1,2,.
(35)

Hence, using (35), we have

Q ( u ) = T F ( u ) + ( T G ) T F ( u ) + ( T G ) 2 T F ( u ) + + ( T G ) n T F ( u ) + T F ( u ) + ( T G ) T F ( u ) + ( T G ) 2 T F ( u ) + + ( T G ) n T F ( u ) + = Q ( u ) .
(36)

Now, we show that Q: C + [0,1] C + [0,1] satisfies Q(ρu) ρ α Q(u) for t[0,1] and u C + [0,1]. Note that

T F ( ρ u ) = 0 1 0 1 0 1 G 1 ( t , v ) G 2 ( v , τ ) G 3 ( τ , s ) f ( s , ρ u ( s ) ) d s d τ d v ρ α 0 1 0 1 0 1 G 1 ( t , v ) G 2 ( v , τ ) G 3 ( τ , s ) f ( s , u ( s ) ) d s d τ d v = ρ α T F ( u ) .

Moreover, from (25), it follows that for TF(ρu) ρ α TF(u),

G ( T F ρ u ) ( t ) G ( ρ α T F ( u ) ) ( t ) = ρ α G ( T F ( u ) ) ( t ) for  t [ 0 , 1 ] .

Finally, we have

( T G ) T F ( ρ u ) ( t ) = 0 1 0 1 0 1 G 1 ( t , v ) G 2 ( v , τ ) G 3 ( τ , s ) f ( s , G ( T F ρ u ) ( s ) ) d s d τ d v 0 1 0 1 0 1 G 1 ( t , v ) G 2 ( v , τ ) G 3 ( τ , s ) f ( s , ρ α G ( T F ( u ) ) ( s ) ) d s d τ d v ρ α 2 0 1 0 1 0 1 G 1 ( t , v ) G 2 ( v , τ ) G 3 ( τ , s ) f ( s , G ( T F ( u ) ) ( s ) ) d s d τ d v = ρ α 2 ( T G ) T F ( u ) ( t ) ,

i.e.,

(TG)(TFρu)(t) ρ α 2 (TG)TF(u)(t).

By induction, it is easy to see that

( T G ) n (TFρu)(t) ρ α n + 1 (TG)TF(ρu)(t),n=1,2,.
(37)

Hence, using (35) and ρ(0,1), α(0,1), we have

Q ( ρ u ) = T F ( ρ u ) + ( T G ) T F ( ρ u ) + ( T G ) 2 T F ( ρ u ) + + ( T G ) n T F ( ρ u ) + ρ α T F ( u ) + ρ α 2 ( T G ) T F ( u ) + ρ α 3 ( T G ) 2 T F ( u ) + + ρ α n + 1 ( T G ) n T F ( u ) + ρ α T F ( u ) + ρ α ( T G ) T F ( u ) + ρ α ( T G ) 2 T F ( u ) + + ρ α ( T G ) n T F ( u ) + = ρ α ( T F ( u ) + ( T G ) T F ( u ) + ( T G ) 2 T F ( u ) + + ( T G ) n T F ( u ) + ) = ρ α Q ( u ) .
(38)

By Lemma 13, there exists 0< s g S g such that

s u g(t)Qg(t) S u g(t).

Let

s=sup { s g : s u g ( t ) Q g ( t ) } ,S=inf { S g : Q g ( t ) S u g ( t ) } .

Pick m s and M s such that

0< m s <min { 1 , s 1 1 α }
(39)

and

max { 1 , S 1 1 α } = M s <.
(40)

Set u 0 (t)= m s g(t), v 0 (t)= M s g(t), u n =Q u n 1 , and v n =Q v n 1 , n=1,2, . From (36) and (38), we have

m s g(t)= u 0 (t) u 1 (t) u n (t) v n (t) v 1 (t) v 0 (t)= M s g(t).
(41)

Indeed, from (39) m s <1, and m s α 1 s>1, we have

u 1 ( t ) = Q ( u 0 ) = Q ( m s g ( t ) ) m s α Q ( g ( t ) ) m s α s g ( t ) = m s α 1 s m s g ( t ) = m s α 1 s u 0 ( t ) u 0 ( t ) ,

and by induction

u n + 1 (t)=Q( u n )Q( u n 1 )= u n (t).

From (40), M s >1, and M s α 1 S<1, we have

v 1 ( t ) = Q ( v 0 ) M s α Q ( g ( t ) ) = M s α Q ( 1 M s v 0 ) = M s α Q ( g ) M s α S g S M s α 1 M s g = S M s α 1 v 0 ( t ) v 0 ( t ) ,

and by induction

v n + 1 (t)=Q( v n )Q( v n 1 )= v n (t).

Let d= m s M s . Then

u n d α n v n .
(42)

In fact u 0 =d v 0 is clear. Assume that (42) holds with n=k (k is a positive integer), i.e., u k d α k v k . Then

u k + 1 =Q( u k )Q ( d α k v k ) ( d α k ) α Q( v k )= d α k + 1 Q( v k )= d α k + 1 v k + 1 .

By induction, it is easy to see that (42) holds. Furthermore, in view of (38), (41) and (42), we have

0 u n + z u n v n u n ( 1 d α n ) v 0 = ( 1 d α n ) M s g(t)

and

u n + z u n v n u n ( 1 d α n ) M s g,

where z is a nonnegative integer. Thus, there exists u C + [0,1] such that

lim n u n (t)= lim n v n (t)= u (t)for t[0,1]

and u (t) is a fixed point of Q and satisfies

m g g(t) u (t) M g g(t).

This means that u C + [0,1], where C + [0,1]={u C + [0,1],u(t)>0 for t(0,1)}.

Next we show that u is the unique fixed point of Q in C + [0,1]. Suppose, to the contrary, that there exists another u ¯ C + [0,1] such that Q u ¯ = u ¯ . We can suppose that

u (t) u ¯ (t), u (t) u ¯ (t)for t[0,1].

Let τ ˆ =sup{0<τ<1:τ u u ¯ τ 1 u }. Then 0< τ ˆ 1 and τ ˆ u u ¯ τ ˆ 1 u . We assert τ ˆ =1. Otherwise, 0< τ ˆ <1, and then

u ¯ = Q u ¯ Q ( τ ˆ u ) τ ˆ α Q ( u ) = τ ˆ α u , u = Q u Q ( τ ˆ u ¯ ) τ ˆ α Q ( u ¯ ) = τ ˆ α u ¯ .

This means that τ ˆ α u u ¯ ( τ ˆ α ) 1 u , which is a contradiction of the definition of τ ˆ , because τ ˆ < τ ˆ α .

Let us introduce the following notations for μ=0

T λ u ( t ) : = T F u ( t ) = λ 0 1 0 1 0 1 G 1 ( t , v ) G 2 ( v , s ) G 3 ( s , τ ) f ( τ , u ( τ ) ) d τ d s d v , Q λ u : = H F u = T F u + ( T G ) T F u + ( T G ) 2 T F u + + ( T G ) n T F u + = T λ u + ( T G ) T λ u + ( T G ) 2 T λ u + + ( T G ) n T λ u + ,

i.e., Q λ u=λQu, where Q is given by (33). □

Theorem 4 Suppose that (H3), (H4), (H6) and L<1 hold. Then (32) has a unique positive solution u λ (t) for any 0<λ1.

Proof Theorem 3 implies that for λ=1, the operator Q λ has a unique fixed point u 1 C + [0,1], that is Q 1 u 1 = u 1 . Then from Lemma 10, for every λ (0,1), there exists a function u P{θ} such that Q λ u = u .

Thus, u λ is a unique positive solution of (32) for every 0<λ1. □

4 Application

As an application of Theorem 1, consider the sixth-order boundary value problem

u ( 6 ) + ( 1 0.5 t 2 ) u ( 4 ) + ( 4.5 0.5 sin π t ) u + C ( 5 + cos 0.5 π t ) u = ( 0.5 t ( 1 t ) + u ) φ + λ ( 1 + sin π t + u 2 ) , 0 < t < 1 , φ + 2 φ = μ u , 0 < t < 1 , u ( 0 ) = u ( 1 ) = u ( 0 ) = u ( 1 ) = u ( 4 ) ( 0 ) = u ( 4 ) ( 1 ) = 0 , φ ( 0 ) = φ ( 1 ) = 0 ,
(43)

for a fixed λ 1 =2, λ 2 =2, λ 3 =1 and ϰ=2. In this case, a= λ 1 + λ 2 + λ 3 =1, b= λ 1 λ 2 λ 2 λ 3 λ 1 λ 3 =4, and c= λ 1 λ 2 λ 3 =4. We have A(t)=10.5 t 2 , B(t)=4.50.5sinπt, C(t)=5+cos0.5πt, D(t)=0.5t(1t) and f(t