Open Access

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

Boundary Value Problems20132013:184

DOI: 10.1186/1687-2770-2013-184

Received: 20 May 2013

Accepted: 30 July 2013

Published: 14 August 2013

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 [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 , 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
  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 ) d s .

Lemma 3 [10]

Let X be a Banach space, K a cone and Ω a bounded open subset of X. 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]

Let X be a Banach space, let K be a cone of X. 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 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 ) 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 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 ( 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 ,

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 + ( 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 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 ) 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 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 ˜ + ( 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 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
( 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 ( 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., Q u = Q 1 , 0 u = H F u .

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 ) ( Q u ) ( 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 ) d v d τ .

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 ) d τ d v , 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 ( ρ u ) 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 T F u ( t ) T F u ( t )
G ( T F u ) ( t ) G ( T F u ) ( t ) for  t [ 0 , 1 ] .
(34)
Finally, since f ( s , u ) is nondecreasing in u, then by using form (34), f ( s , G ( T F u ) ( t ) ) f ( s , G ( T F 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.,
( T G ) T F ( u ) ( T G ) T F u .
By induction, it is easy to see that
( T G ) n T F ( u ) ( T G ) n T F 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 ( ρ u ) 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 T F ( ρ u ) ρ α T F ( 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.,
( T G ) ( T F ρ u ) ( t ) ρ α 2 ( T G ) T F ( u ) ( t ) .
By induction, it is easy to see that
( T G ) n ( T F ρ u ) ( t ) ρ α n + 1 ( T G ) T F ( ρ u ) ( 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 ) Q g ( 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 = λ Q u , 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 < λ 1 .

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 < λ 1 . □

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 ) = 1 0.5 t 2 , B ( t ) = 4.5 0.5 sin π t , C ( t ) = 5 + cos 0.5 π t , D ( t ) = 0.5 t ( 1 t ) and f ( t , u ) = 1 + sin π t + u 2 . It is easy to see that π 6 + a π 4 b π 2 + c = 1 , 015.3 > 0 , a = sup t [ 0 , 1 ] A ( t ) , b = inf t [ 0 , 1 ] B ( t ) and c = sup t [ 0 , 1 ] C ( t ) . Note also that K = max 0 t 1 [ A ( t ) + B ( t ) C ( t ) ( a + b c ) ] = 2 , D 2 = max t [ 0 , 1 ] 0 1 G 2 ( t , v ) d v = 0.15768 , C = max t [ 0 , 1 ] D ( t ) = 0.125 , d 1 = max t [ 0 , 1 ] 0 1 G ( t , s ) d s = 0.10336 , μ = 1 D 2 K D 2 C d 1 = 336.1 and D 2 K = 0.3153 < 1 . Thus, if 0 < μ < 336.1 , then the conditions of Theorem 1 (note L = D 2 ( K + μ C d 1 ) < 1 ) are fulfilled (in particular, (H3)-(H5) are satisfied). As a result, Theorem 1 can be applied to (43).

Declarations

Authors’ Affiliations

(1)
Department of Mathematics, Texas A&M University-Kingsville
(2)
Department of Analysis, University of Miskolc
(3)
School of Mathematics, Statistics and Applied Mathematics, National University of Ireland

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