Research

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

Ravi P Agarwal1*, B Kovacs2 and D O’Regan3

Author Affiliations

1 Department of Mathematics, Texas A&M University-Kingsville, 700 University Blvd., Kingsville, 78363-8202, USA

2 Department of Analysis, University of Miskolc, Egyetemvaros, 3515, Hungary

3 School of Mathematics, Statistics and Applied Mathematics, National University of Ireland, Galway, Ireland

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Boundary Value Problems 2013, 2013:184  doi:10.1186/1687-2770-2013-184

 Received: 20 May 2013 Accepted: 30 July 2013 Published: 14 August 2013

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

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.

##### Keywords:
positive solutions; variable parameters; fixed point theorem; operator spectral theorem

### 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 [1-4] 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 , c R , 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 , D C [ 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 + b c ) ] 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 + = { u Y : 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 χ 0 and ν 0 , we denote the norm χ , ν by

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

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 u X , u 0 u 0 u ( 4 ) 0 .

For h Y , 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 + c u = λ u , u ( 0 ) = u ( 1 ) = u ( 0 ) = u ( 1 ) = u ( 4 ) ( 0 ) = u ( 4 ) ( 1 ) = 0 , and ϕ 1 ( t ) = sin π t 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 + c u = 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 + β λ α , where β < 2 π 2 , α 0 . 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 + c u = ( 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

(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 } ;

(ii) when λ i = 0 ,

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

(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

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

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

(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 ) d s .

Lemma 3[10]

LetXbe a Banach space, Ka cone and Ω a bounded open subset ofX. Let θ Ω and T : K Ω ¯ K be condensing. Suppose that T x υ x for all x K Ω and υ 1 . Then i ( T , K Ω , K ) = 1 .

Lemma 4[10]

LetXbe a Banach space, letKbe a cone ofX. Assume that T : K ¯ r K (here K r = { x K x < r } , r > 0 ) is a compact map such that T x x for all x K r . If x T x 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 ) d τ d s d v , t [ 0 , 1 ] . (9)

Thus, for every given h Y , 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

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

Throughout this article, we shall denote T h = 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 = T h , then u X 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 , ρ u ) 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 + ϰ u = 0 , u ( 0 ) = u ( 1 ) = 0 . (11)

Then, the boundary value problem

φ + ϰ φ = μ u , φ ( 0 ) = φ ( 1 ) = 0 ,

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

φ ( t ) = μ 0 1 G ( t , s ) u ( s ) d s , 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 u 0 , t [ 0 , 1 ] , and u 0 .

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 u X , let

G u = ( 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 .

The operator G : X Y is linear. By Lemmas 2 and 3 in [8], u X , 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 + b c ) ] , d 1 = max t [ 0 , 1 ] 0 1 G ( t , s ) d s , χ = λ 1 + λ 3 0 , ν = λ 1 λ 3 0 . Hence G u 0 ( K + μ C d 1 ) u χ , ν , and so G ( K + μ C d 1 ) . Also u C 4 [ 0 , 1 ] C 6 ( 0 , 1 ) is a solution of (13) if u X satisfies u = T ( G u + h 1 ) , where h 1 ( t ) = μ u ( t ) 0 1 G ( t , s ) u ( s ) d s + h ( t ) , i.e.,

u X , ( I T G ) u = T h 1 . (16)

The operator I T G maps X into X. From T D 2 together with G ( K + μ C d 1 ) and the condition D 2 ( K + μ C 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 + μ C d 1 ) < 1 is fulfilled. Let L = D 2 ( K + μ C 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 + ( T G ) T + + ( T G ) n T + . (17)

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

Now h Y + , let u = T h , then u X 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 + , ( G T h ) ( t ) 0 , t [ 0 , 1 ] , (18)

and so, ( T G ) ( T h ) ( t ) = T ( G T h ) ( t ) 0 , t [ 0 , 1 ] .

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

1 1 L ( T h ) ( t ) ( H h ) ( t ) ( T h ) ( 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 ( T h ) ( t ) ( H h ) ( t ) ( T h ) ( t ) , t [ 0 , 1 ] , and, moreover, T h 0 ( 1 L ) H h 0 .

For any u Y + , define F u = μ u ( t ) 0 1 G ( t , s ) u ( s ) d s + λ f ( 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 = H F u . (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 = H F u . 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 ( 1 L ) 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 : P P is completely continuous.

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

( T F u ) ( t ) C 1 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 , t [ 0 , 1 ] .

Thus,

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 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 8Let f ( t , u ) be nondecreasing inufor 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 ) d s d τ d v .

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 ) d s , 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 ) d τ d v , 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 ) d v , 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 ) ) d τ d v 0 ,

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 9Suppose 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 ( 1 L ) 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 u for  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 ( T F u N 0 ) ( 1 2 ) > u N 0 0 ,

Lemma 10Suppose that L < 1 , (H3) and (H4) hold and that the operator Q λ , μ has a positive fixed point inPat 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 + ( T G ) 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 11Suppose 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 ) d p , 0 < t < 1 . (29)

Multiplying (29) by sin π t 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 ) 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 .

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 12Suppose 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 ˜ + ( T G ) 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 1Suppose 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

( T G ) T λ ̲ , μ ̲ ( u λ , μ + δ ) ( T G ) 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 = { u C [ 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 ̲ u υ u for υ 1 and u P D u λ , μ . Indeed set u P 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 u for  u > l 3  and  t [ 1 4 , 3 4 ] .

Set R 1 = l 3 k and P R 1 = { u P : 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 (