Research

# Eigenvalues for iterative systems of nonlinear m-point boundary value problems on time scales

Ilkay Y Karaca1* and Fatma Tokmak12

Author Affiliations

1 Department of Mathematics, Ege University, Bornova, Izmir, 35100, Turkey

2 Department of Mathematics, Gazi University, Teknikokullar, Ankara, 06500, Turkey

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Boundary Value Problems 2014, 2014:63  doi:10.1186/1687-2770-2014-63

 Received: 3 October 2013 Accepted: 3 March 2014 Published: 21 March 2014

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 credited.

### Abstract

In this paper, we determine the eigenvalue intervals of the parameters λ 1 , λ 2 , , λ n for which there exist positive solutions of the iterative systems of m-point boundary value problems on time scales. The method involves an application of Guo-Krasnosel’skii fixed point theorem. We give an example to demonstrate our main results.

MSC: 34B18, 34N05.

##### Keywords:
Green’s function; iterative system; eigenvalue interval; time scales; boundary value problem; fixed point theorem; m-point; positive solution

### 1 Introduction

The study of dynamic equations on time scales goes back to Stefan Hilger [1]. Theoretically, this new theory has not only unify continuous and discrete equations, but it has also exhibited much more complicated dynamics on time scales. Moreover, the study of dynamic equations on time scales has led to several important applications, for example, insect population models, biology, neural networks, heat transfer, and epidemic models; see [2-7].

There has been much interest shown in obtaining optimal eigenvalue intervals for the existence of positive solutions of the boundary value problems on time scales, often using Guo-Krasnosel’skii fixed point theorem. To mention a few papers along these lines, see [8-12]. On the other hand, there is not much work concerning the eigenvalues for iterative system of nonlinear boundary value problems on time scales; see [13,14].

In [15], Ma and Thompson are concerned with determining values λ, by using the Guo-Krasnosel’skii fixed point theorem for which there exist positive solutions of the m-point boundary value problem

{ ( p ( t ) u ) q ( t ) u + λ f ( t , u ) = 0 , 0 < t < 1 , a u ( 0 ) b p ( 0 ) u ( 0 ) = i = 1 m 2 α i u ( ξ i ) , c u ( 1 ) + d p ( 1 ) u ( 1 ) = i = 1 m 2 β i u ( ξ i ) .

In [13], Benchohra et al. studied the eigenvalues for iterative system of nonlinear boundary value problems on time scales,

u i ( t ) + λ i a i ( t ) f i ( u i + 1 ( σ ( t ) ) ) = 0 , 1 i n , t [ 0 , 1 ] T , u n + 1 ( t ) = u 1 ( t ) , t [ 0 , 1 ] T ,

satisfying the boundary conditions,

u i ( 0 ) = 0 = u i ( σ 2 ( 1 ) ) , 1 i n .

The method involves application of Guo-Krasnosel’skii fixed point theorem for operators on a cone in a Banach space.

In [14], Prasad et al. studied the eigenvalues for iterative system of nonlinear boundary value problems on time scales,

y i ( t ) + λ i p i ( t ) f i ( y i + 1 ( t ) ) = 0 , 1 i n , t [ t 1 , t m ] T , y n + 1 ( t ) = y 1 ( t ) , t [ t 1 , t m ] T ,

satisfying the m-point boundary conditions,

y i ( t 1 ) = 0 , α y i ( σ ( t m ) ) + β y i ( σ ( t m ) ) = k = 2 m 1 y i ( t k ) , 1 i n .

They used the Guo-Krasnosel’skii fixed point theorem.

Motivated by the above results, in this study, we are concerned with determining the eigenvalue intervals of λ i , 1 i n , for which there exist positive solutions for the iterative system of nonlinear m-point boundary value problems on time scales,

{ u i ( t ) + λ i q i ( t ) f i ( u i + 1 ( t ) ) = 0 , t [ 0 , 1 ] T , 1 i n , u n + 1 ( t ) = u 1 ( t ) , t [ 0 , 1 ] T , (1.1)

satisfying the m-point boundary conditions,

{ a u i ( 0 ) b u i ( 0 ) = j = 1 m 2 α j u i ( ξ j ) , c u i ( 1 ) + d u i ( 1 ) = j = 1 m 2 β j u i ( ξ j ) , 1 i n , (1.2)

where is a time scale, 0 , 1 T , [ 0 , 1 ] T = [ 0 , 1 ] T .

Throughout this paper we assume that following conditions hold:

(C1) a , b , c , d [ 0 , ) with a c + a d + b c > 0 ; α j , β j [ 0 , ) , ξ j ( 0 , 1 ) T for 1 j m 2 ,

(C2) f i : R + R + is continuous, for 1 i n ,

(C3) q i C ( [ 0 , 1 ] T , R + ) and q i does not vanish identically on any closed subinterval of [ 0 , 1 ] T , for 1 i n ,

(C4) each of f i 0 : = lim x 0 + f i ( x ) x and f i : = lim x f i ( x ) x , 1 i n , exists as positive real number.

In fact, our results are also new when T = R (the differential case) and T = Z (the discrete case). Therefore, the results can be considered as a contribution to this field.

This paper is organized as follows. In Section 2, we construct the Green’s function for the homogeneous problem corresponding to (1.1)-(1.2) and estimate bounds for the Green’s function. In Section 3, we determine the eigenvalue intervals for which there exist positive solutions of the boundary value problem (1.1)-(1.2) by using the Guo-Krasnosel’skii fixed point theorem for operators on a cone in a Banach space. Finally, in Section 4, we give an example to demonstrate our main results.

### 2 Preliminaries

We need the auxiliary lemmas that will be used to prove our main results.

We define B = C [ 0 , 1 ] , which is a Banach space with the norm

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

Let h C [ 0 , 1 ] , then we consider the following boundary value problem:

{ u 1 ( t ) = h ( t ) , t [ 0 , 1 ] T , a u 1 ( 0 ) b u 1 ( 0 ) = j = 1 m 2 α j u 1 ( ξ j ) , c u 1 ( 1 ) + d u 1 ( 1 ) = j = 1 m 2 β j u 1 ( ξ j ) . (2.1)

Denote by θ and φ, the solutions of the corresponding homogeneous equation

u 1 ( t ) = 0 , t [ 0 , 1 ] T , (2.2)

under the initial conditions

{ θ ( 0 ) = b , θ Δ ( 0 ) = a , φ ( 1 ) = d , φ Δ ( 1 ) = c . (2.3)

Using the initial conditions (2.3), we can deduce from equation (2.2) for θ and φ the following equations:

θ ( t ) = b + a t , φ ( t ) = d + c ( 1 t ) . (2.4)

Set

Δ : = | j = 1 m 2 α j ( b + a ξ j ) ρ j = 1 m 2 α j ( d + c ( 1 ξ j ) ) ρ j = 1 m 2 β j ( b + a ξ j ) j = 1 m 2 β j ( d + c ( 1 ξ j ) ) | (2.5)

and

ρ : = a d + a c + b c . (2.6)

Lemma 2.1Let (C1) hold. Assume that

(C5) Δ 0 .

If u 1 C [ 0 , 1 ] is a solution of the equation

u 1 ( t ) = 0 1 G ( t , s ) h ( s ) s + A ( h ) ( b + a t ) + B ( h ) ( d + c ( 1 t ) ) , (2.7)

where

G ( t , s ) = 1 ρ { ( b + a σ ( s ) ) ( d + c ( 1 t ) ) , σ ( s ) t , ( b + a t ) ( d + c ( 1 σ ( s ) ) ) , t s , (2.8)

A ( h ) : = 1 Δ | j = 1 m 2 α j 0 1 G ( ξ j , s ) h ( s ) s ρ j = 1 m 2 α j ( d + c ( 1 ξ j ) ) j = 1 m 2 β j 0 1 G ( ξ j , s ) h ( s ) s j = 1 m 2 β j ( d + c ( 1 ξ j ) ) | (2.9)

and

B ( h ) : = 1 Δ | j = 1 m 2 α j ( b + a ξ j ) j = 1 m 2 α j 0 1 G ( ξ j , s ) h ( s ) s ρ j = 1 m 2 β j ( b + a ξ j ) j = 1 m 2 β j 0 1 G ( ξ j , s ) h ( s ) s | , (2.10)

then u 1 is a solution of the boundary value problem (2.1).

Proof Let u 1 satisfy the integral equation (2.7), then we have

u 1 ( t ) = 0 1 G ( t , s ) h ( s ) s + A ( h ) ( b + a t ) + B ( h ) ( d + c ( 1 t ) ) ,

i.e.,

u 1 ( t ) = 0 t 1 ρ ( b + a ( σ ( s ) ) ) ( d + c ( 1 t ) ) h ( s ) s u 1 ( t ) = + t 1 1 ρ ( b + a t ) ( d + c ( 1 σ ( s ) ) ) h ( s ) s u 1 ( t ) = + A ( h ) ( b + a t ) + B ( h ) ( d + c ( 1 t ) ) , u 1 ( t ) = 0 t c ρ ( b + a ( σ ( s ) ) ) h ( s ) s u 1 ( t ) = + t 1 a ρ ( d + c ( 1 σ ( s ) ) ) h ( s ) s u 1 ( t ) = + A ( h ) a B ( h ) c .

Hence

u 1 ( t ) = 1 ρ ( c ( b + a ( σ ( t ) ) ) a ( d + c ( 1 σ ( t ) ) ) ) h ( t ) u 1 ( t ) = 1 ρ ( ( a d + a c + b c ) ) h ( t ) = h ( t ) , u 1 ( t ) = h ( t ) .

Since

u 1 ( 0 ) = 0 1 b ρ ( d + c ( 1 σ ( s ) ) ) h ( s ) s + A ( h ) b + B ( h ) ( d + c ) , u 1 ( 0 ) = 0 1 a ρ ( d + c ( 1 σ ( s ) ) ) h ( s ) s + A ( h ) a B ( h ) c ,

we have

a u 1 ( 0 ) b u 1 ( 0 ) = B ( h ) ρ = j = 1 m 2 α j [ 0 1 G ( ξ j , s ) h ( s ) s + A ( h ) ( b + a ξ j ) + B ( h ) ( d + c ( 1 ξ j ) ) ] . (2.11)

Since

u 1 ( 1 ) = 0 1 d ρ ( b + a ( σ ( s ) ) ) h ( s ) s + A ( h ) ( b + a ) + B ( h ) d , u 1 ( 1 ) = 0 1 c ρ ( b + a ( σ ( s ) ) ) h ( s ) s + A ( h ) a B ( h ) c ,

we have

c u 1 ( 1 ) + d u 1 ( 1 ) = A ( h ) ρ = j = 1 m 2 β j [ 0 1 G ( ξ j , s ) h ( s ) s + A ( h ) ( b + a ξ j ) + B ( h ) ( d + c ( 1 ξ j ) ) ] . (2.12)

From (2.11) and (2.12), we get

{ [ j = 1 m 2 α j ( b + a ξ j ) ] A ( h ) + [ ρ i = 1 m 2 α j ( d + c ( 1 ξ j ) ) ] B ( h ) = i = 1 m 2 α j 0 1 G ( ξ j , s ) h ( s ) s , [ ρ j = 1 m 2 β j ( b + a ξ j ) ] A ( h ) + [ i = 1 m 2 β j ( d + c ( 1 ξ j ) ) ] B ( h ) = j = 1 m 2 β j 0 1 G ( ξ j , s ) h ( s ) s ,

which implies that A ( h ) and B ( h ) satisfy (2.9) and (2.10), respectively. □

Lemma 2.2Let (C1) hold. Assume

(C6) Δ < 0 , ρ j = 1 m 2 β j ( b + a ξ j ) > 0 , a j = 1 m 2 α j > 0 .

Then for u 1 C [ 0 , 1 ] with h 0 , the solution u 1 of the problem (2.1) satisfies

u 1 ( t ) 0 for  t [ 0 , 1 ] T .

Proof It is an immediate subsequence of the facts that G 0 on [ 0 , 1 ] T × [ 0 , 1 ] T and A ( h ) 0 , B ( h ) 0 . □

Lemma 2.3Let (C1) and (C6) hold. Assume

(C7) c j = 1 m 2 β j < 0 .

Then the solution u 1 C [ 0 , 1 ] of the problem (2.1) satisfies u 1 ( t ) 0 for t [ 0 , 1 ] T .

Proof Assume that the inequality u 1 ( t ) < 0 holds. Since u 1 ( t ) is nonincreasing on [ 0 , 1 ] T , one can verify that

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

From the boundary conditions of the problem (2.1), we have

c d u 1 ( 1 ) + 1 d i = 1 m 2 β i u 1 ( ξ i ) u 1 ( t ) < 0 .

The last inequality yields

c u 1 ( 1 ) + i = 1 m 2 β i u 1 ( ξ i ) < 0 .

Therefore, we obtain

i = 1 m 2 β i u 1 ( 1 ) < i = 1 m 2 β i u 1 ( ξ i ) < c u 1 ( 1 ) ,

i.e.,

( c i = 1 m 2 β i ) u 1 ( 1 ) > 0 .

According to Lemma 2.2, we have u 1 ( 1 ) 0 . So, c i = 1 m 2 β i > 0 . However, this contradicts to condition (C7). Consequently, u 1 ( t ) 0 for t [ 0 , 1 ] T . □

Lemma 2.4Let (C1) and h 0 hold. Let μ ( 0 , 1 / 2 ) T be a constant. Then the unique solution u 1 of the problem (2.1) satisfies

min t [ μ , 1 μ ] T u 1 ( t ) γ u 1 ,

where u 1 = sup t [ 0 , 1 ] T u 1 ( t ) and

γ : = min { b + a μ b + a , d + c μ d + c } . (2.13)

Proof We have from (2.8) that

0 G ( t , s ) G ( σ ( s ) , s ) , t [ 0 , 1 ] T , (2.14)

which implies

u 1 ( t ) 0 1 G ( σ ( s ) , s ) h ( s ) s + A ( h ) ( b + a ) + B ( h ) ( d + c ) (2.15)

for all t [ 0 , 1 ] T . Applying (2.8), we have for t [ μ , 1 μ ] T ,

G ( t , s ) G ( σ ( s ) , s ) = { d + c ( 1 t ) d + c ( 1 σ ( s ) ) , 0 σ ( s ) t 1 , b + a t b + a σ ( s ) , 0 t s 1 { d + c μ d + c , 0 σ ( s ) t 1 μ , b + a μ b + a , μ t s 1 γ , (2.16)

where

γ : = min { b + a μ b + a , d + c μ d + c } .

Thus for t [ μ , 1 μ ] T ,

u 1 ( t ) = 0 1 G ( t , s ) h ( s ) s + A ( h ) ( b + a t ) + B ( h ) ( d + c ( 1 t ) ) γ ( 0 1 G ( σ ( s ) , s ) h ( s ) s + A ( h ) ( b + a ) + B ( h ) ( d + c ) ) γ u 1 .

So, the proof is completed. □

We note that an n-tuple ( u 1 ( t ) , u 2 ( t ) , , u n ( t ) ) is a solution of the boundary value problem (1.1)-(1.2) if and only if

u 1 ( t ) = λ 1 0 1 G ( t , s 1 ) q 1 ( s 1 ) f 1 ( λ 2 0 1 G ( s 1 , s 2 ) q 2 ( s 2 ) u 1 ( t ) = f n 1 ( λ n 0 1 G ( s n 1 , s n ) q n ( s n ) f n ( u 1 ( s n ) ) s n ) s 2 ) s 1 u 1 ( t ) = + A ( λ 1 q 1 ( ) f 1 ( u 2 ( ) ) ) ( b + a t ) + B ( λ 1 q 1 ( ) f 1 ( u 2 ( ) ) ) ( d + c ( 1 t ) ) , t [ 0 , 1 ] T , u i ( t ) = λ i 0 1 G ( t , s ) q i ( s ) f i ( u i + 1 ( s ) ) s + A ( λ i q i ( ) f i ( u i + 1 ( ) ) ) ( b + a t ) u i ( t ) = + B ( λ i q i ( ) f i ( u i + 1 ( ) ) ) ( d + c ( 1 t ) ) , 2 i n , t [ 0 , 1 ] T

and

u n + 1 ( t ) = u 1 ( t ) , t [ 0 , 1 ] T ,

where

A ( λ i q i ( ) f i ( u i + 1 ( ) ) ) : = 1 Δ | j = 1 m 2 α j λ i 0 1 G ( ξ j , s ) q i ( s ) f i ( u i + 1 ( s ) ) s ρ j = 1 m 2 α j ( d + c ( 1 ξ j ) ) j = 1 m 2 β j λ i 0 1 G ( ξ j , s ) q i ( s ) f i ( u i + 1 ( s ) ) s j = 1 m 2 β j ( d + c ( 1 ξ j ) ) | , B ( λ i q i ( ) f i ( u i + 1 ( ) ) ) : = 1 Δ | j = 1 m 2 α j ( b + a ξ j ) j = 1 m 2 α j λ i 0 1 G ( ξ j , s ) q i ( s ) f i ( u i + 1 ( s ) ) s ρ j = 1 m 2 β j ( b + a ξ j ) j = 1 m 2 β j λ i 0 1 G ( ξ j , s ) q i ( s ) f i ( u i + 1 ( s ) ) s | .

To determine the eigenvalue intervals of the boundary value problem (1.1)-(1.2), we will use the following Guo-Krasnosel’skii fixed point theorem [16].

Theorem 2.1[16]

Letbe a Banach space, and let P B be a cone in. Assume Ω 1 and Ω 2 are open subsets ofwith 0 Ω 1 and Ω ¯ 1 Ω 2 , and let

T : P ( Ω ¯ 2 Ω 1 ) P

be a completely continuous operator such that either

(i) T u u , u P Ω 1 , and T u u , u P Ω 2 , or

(ii) T u u , u P Ω 1 , and T u u , u P Ω 2 .

ThenThas a fixed point in P ( Ω ¯ 2 Ω 1 ) .

### 3 Positive solutions in a cone

In this section, we establish criteria to determine the eigenvalue intervals for which the boundary value problem (1.1)-(1.2) has at least one positive solution in a cone. We construct a cone P B by

P = { u B : u ( t ) 0  on  [ 0 , 1 ] T  and  min t [ μ , 1 μ ] T u ( t ) γ u } ,

where γ is given in (2.13).

Now, we define an integral operator T : P B , for u 1 P , by

T u 1 ( t ) = λ 1 0 1 G ( t , s 1 ) q 1 ( s 1 ) f 1 ( λ 2 0 1 G ( s 1 , s 2 ) q 2 ( s 2 ) f n 1 ( λ n 0 1 G ( s n 1 , s n ) q n ( s n ) f n ( u 1 ( s n ) ) s n ) s 2 ) s 1 + A ( λ 1 q 1 ( ) f 1 ( u 2 ( ) ) ) ( b + a t ) + B ( λ 1 q 1 ( ) f 1 ( u 2 ( ) ) ) ( d + c ( 1 t ) ) . (3.1)

Notice from (C1)-(C6) and Lemma 2.2 that, for u 1 P , T u 1 ( t ) 0 on t [ 0 , 1 ] T . Also, we have from (2.8), that

T u 1 ( t ) λ 1 0 1 G ( σ ( s 1 ) , s 1 ) q 1 ( s 1 ) f 1 ( λ 2 0 1 G ( s 1 , s 2 ) q 2 ( s 2 ) f n 1 ( λ n 0 1 G ( s n 1 , s n ) q n ( s n ) f n ( u 1 ( s n ) ) s n ) s 2 ) s 1 + A ( λ 1 q 1 ( ) f 1 ( u 2 ( ) ) ) ( b + a ) + B ( λ 1 q 1 ( ) f 1 ( u 2 ( ) ) ) ( d + c ) ,

so that

T u 1 λ 1 0 1 G ( σ ( s 1 ) , s 1 ) q 1 ( s 1 ) f 1 ( λ 2 0 1 G ( s 1 , s 2 ) q 2 ( s 2 ) f n 1 ( λ n 0 1 G ( s n 1 , s n ) q n ( s n ) f n ( u 1 ( s n ) ) s n ) s 2 ) s 1 + A ( λ 1 q 1 ( ) f 1 ( u 2 ( ) ) ) ( b + a ) + B ( λ 1 q 1 ( ) f 1 ( u 2 ( ) ) ) ( d + c ) . (3.2)

Next, if u 1 P , we have from Lemma 2.4 and (3.2) that

min t [ μ , 1 μ ] T T u 1 ( t ) = min t [ μ , 1 μ ] T { λ 1 0 1 G ( t , s 1 ) q 1 ( s 1 ) f 1 ( λ 2 0 1 G ( s 1 , s 2 ) q 2 ( s 2 ) f n 1 ( λ n 0 1 G ( s n 1 , s n ) q n ( s n ) f n ( u 1 ( s n ) ) s n ) s 2 ) s 1 + A ( λ 1 q 1 ( ) f 1 ( u 2 ( ) ) ) ( b + a t ) + B ( λ 1 q 1 ( ) f 1 ( u 2 ( ) ) ) ( d + c ( 1 t ) ) } γ ( λ 1 0 1 G ( σ ( s 1 ) , s 1 ) q 1 ( s 1 ) f 1 ( λ 2 0 1 G ( s 1 , s 2 ) q 2 ( s 2 ) f n 1 ( λ n 0 1 G ( s n 1 , s n ) q n ( s n ) f n ( u 1 ( s n ) ) s n ) s 2 ) s 1 + A ( λ 1 q 1 ( ) f 1 ( u 2 ( ) ) ) ( b + a ) + B ( λ 1 q 1 ( ) f 1 ( u 2 ( ) ) ) ( d + c ) ) γ T u 1 .

Hence, T u 1 P and T : P P . In addition, the operator T is completely continuous by an application of the Arzela-Ascoli theorem.

Now, we investigate suitable fixed points of T belonging to the cone . For convenience we introduce the following notations.

Let

M 1 = max 1 i n { [ γ 2 μ 1 μ G ( σ ( s ) , s ) q i ( s ) s f i ] 1 }

and

M 2 = min 1 i n { [ ( 0 1 G ( σ ( s ) , s ) q i ( s ) s + A ( q i ( ) ) ( b + a ) + B ( q i ( ) ) ( d + c ) ) f i 0 ] 1 } .

Theorem 3.1Suppose conditions (C1)-(C7) are satisfied. Then, for each λ 1 , λ 2 , , λ n satisfying

M 1 < λ i < M 2 , 1 i n , (3.3)

there exists ann-tuple ( u 1 , u 2 , , u n ) satisfying (1.1)-(1.2) such that u i ( t ) > 0 , 1 i n , on [ 0 , 1 ] T .

Proof Let λ k , 1 k n , be as in (3.3). Now, let ϵ > 0 be chosen such that

max 1 i n { [ γ 2 μ 1 μ G ( σ ( s ) , s ) q i ( s ) s ( f i ϵ ) ] 1 } min 1 k n λ k

and

max 1 k n λ k min 1 i n { [ ( 0 1 G ( σ ( s ) , s ) q i ( s ) s + A ( q i ( ) ) ( b + a ) + B ( q i ( ) ) ( d + c ) ) ( f i 0 + ϵ ) ] 1 } .

We investigate fixed points of the completely continuous operator T : P P defined by (3.1). Now, from the definitions of f i 0 , 1 i n , there exists an H 1 > 0 such that, for each 1 i n ,

f i ( x ) ( f i 0 + ϵ ) x , 0 < x H 1 .

Let u 1 P with u 1 = H 1 . We have from (2.14) and the choice of ϵ, for 0 s n 1 1 ,

λ n 0 1 G ( s n 1 , s n ) q n ( s n ) f n ( u 1 ( s n ) ) s n λ n 0 1 G ( σ ( s n ) , s n ) q n ( s n ) f n ( u 1 ( s n ) ) s n λ n 0 1 G ( σ ( s n ) , s n ) q n ( s n ) ( f n 0 + ϵ ) u 1 ( s n ) s n λ n 0 1 G ( σ ( s n ) , s n ) q n ( s n ) s n ( f n 0 + ϵ ) u 1 u 1 H 1 .

It follows in a similar manner from (2.14), for 0 s n 2 1 , that

λ n 1 0 1 G ( s n 2 , s n 1 ) q n 1 ( s n 1 ) f n 1 ( λ n 0 1 G ( s n 1 , s n ) q n ( s n ) f n ( u 1 ( s n ) ) s n ) s n 1 λ n 1 0 1 G ( σ ( s n 1 ) , s n 1 ) q n 1 ( s n 1 ) s n 1 ( f n 1 , 0 + ϵ ) u 1 u 1 = H 1 .

Continuing with this bootstrapping argument, we have, for 0 t 1 ,

λ 1 0 1 G ( t , s 1 ) q 1 ( s 1 ) f 1 ( λ 2 0 1 G ( s 1 , s 2 ) q 2 ( s 2 ) f n ( u 1 ( s n ) ) s n s 2 ) s 1 λ 1 0 1 G ( σ ( s 1 ) , s 1 ) q 1 ( s 1 ) s 1 ( f 10 + ϵ ) H 1 , A ( λ 1 q 1 ( ) f 1 ( u 2 ( ) ) ) λ 1 Δ | j = 1 m 2 α j 0 1 G ( ξ j , s ) q 1 ( s ) s ρ j = 1 m 2 α j ( d + c ( 1 ξ j ) ) j = 1 m 2 β j 0 1 G ( ξ j , s ) q 1 ( s ) s j = 1 m 2 β j ( d + c ( 1 ξ j ) ) | f 1 ( u 2 ) λ 1 A ( q 1 ( ) ) f 1 ( u 2 ) , B ( λ 1 q 1 ( ) f 1 ( u 2 ( ) ) ) λ 1 Δ | j = 1 m 2 α j ( b + a ( ξ j ) ) j = 1 m 2 α j 0 1 G ( ξ j , s ) q 1 ( s ) s ρ j = 1 m 2 β j ( b + a ( ξ j ) ) j = 1 m 2 β j 0 1 G ( ξ j , s ) q 1 ( s ) s | f 1 ( u 2 ) λ 1 B ( q 1 ( ) ) f 1 ( u 2 ) ,

so that, for 0 t 1 ,

T u 1 ( t ) λ 1 ( 0 1 G ( σ ( s 1 ) , s 1 ) q 1 ( s 1 ) s 1 ( f 10 + ϵ ) H 1 + A ( q 1 ( ) ) f 1 ( u 2 ) ( b + a ) + B ( q 1 ( ) ) f 1 ( u 2 ) ( d + c ) ) λ 1 ( 0 1 G ( σ ( s 1 ) , s 1 ) q 1 ( s 1 ) s 1 + A ( q 1 ( ) ) ( b + a ) + B ( q 1 ( ) ) ( d + c ) ) ( f 10 + ϵ ) H 1 H 1 = u 1 .

Hence, T u 1 H 1 = u 1 . If we set

Ω 1 = { u B | u < H 1 } ,

then

T u 1 u 1 for  u 1 P Ω 1 . (3.4)

Next, from the definitions of f i , 1 i n , there exists H ¯ 2 > 0 such that, for each 1 i n ,

f i ( x ) ( f i ϵ ) x , x H ¯ 2 .

Let

H 2 = max { 2 H 1 , H ¯ 2 γ } .

Let u 1 P and u 1 = H 2 . Then, we have from Lemma 2.4

min t [ μ , 1 μ ] T u 1 ( t ) γ u 1 H ¯ 2 .

Consequently, from Lemma 2.4 and the choice of ϵ, for 0 s n 1 1 , we have

λ n 0 1 G ( s n 1 , s n ) q n ( s n ) f n ( u 1 ( s n ) ) s n γ λ n μ 1 μ G ( σ ( s n ) , s n ) q n ( s n ) f n ( u 1 ( s n ) ) s n γ λ n μ 1 μ G ( σ ( s n ) , s n ) q n ( s n ) ( f n ϵ ) u 1 ( s n ) s n γ 2 λ n μ 1 μ G ( σ ( s n ) , s n ) q n ( s n ) s n ( f n ϵ ) u 1 u 1 = H 2 .

It follows in a similar manner from Lemma 2.4 and the choice of ϵ, for 0 s n 2 1 ,

λ n 1 0 1 G ( s n 2 , s n 1 ) q n 1 ( s n 1 ) f n 1 ( λ n 0 1 G ( s n 1 , s n ) q n ( s n ) f n ( u 1 ( s n ) ) s n ) s n 1 γ λ n 1 μ 1 μ G ( σ ( s n 1 ) , s n 1 ) q n 1 ( s n 1 ) s n 1 ( f n 1 , ϵ ) H 2 γ 2 λ n 1 μ 1 μ G ( σ ( s n 1 ) , s n 1 ) q n 1 ( s n 1 ) s n 1 ( f n 1 , ϵ ) H 2 H 2 .

Again, using a bootstrapping argument, we have

λ 1 0 1 G ( t , s 1 ) q 1 ( s 1 ) f 1 ( λ 2 0 1 G ( s 1 , s 2 ) q 2 ( s 2 ) f n ( u 1 ( s n ) ) s n s 2 ) s 1 H 2 ,

so that

T u 1 ( t ) H 2 = u 1 .

Hence, T u 1 u 1 . So if we set

Ω 2 = { u B | u < H 2 } ,

then

T u 1 u 1 for  u 1 P Ω 2 . (3.5)

Applying Theorem 2.1 to (3.4) and (3.5), we see that T has a fixed point u 1 P ( Ω ¯ 2 Ω 1 ) . Therefore, setting u n + 1 = u 1 , we obtain a positive solution ( u 1 , u 2 , , u n ) of (1.1)-(1.2) given iteratively by

u k ( t ) = λ k 0 1 G ( t , s ) q k ( s ) f k ( u k + 1 ( s ) ) s + A ( λ k q k ( ) f k ( u k + 1 ( ) ) ) ( b + a t ) + B ( λ k q k ( ) f k ( u k + 1 ( ) ) ) ( d + c ( 1 t ) ) , k = n , n 1 , , 1 .

The proof is completed. □

For our next result, we define the positive numbers M 3 and M 4 by

M 3 = max 1 i n { [ γ 2 μ 1 μ G ( σ ( s ) , s ) q i ( s ) s f i 0 ] 1 }

and

M 4 = min 1 i n { [ ( 0 1 G ( σ ( s ) , s ) q i ( s ) s + A ( q i ( ) ) ( b + a ) + B ( q i ( ) ) ( d + c ) ) f i ] 1 } .

Theorem 3.2Suppose conditions (C1)-(C7) are satisfied. Then, for each λ 1 , λ 2 , , λ n satisfying

M 3 < λ i < M 4 , 1 i n , (3.6)

there exists ann-tuple ( u 1 , u 2 , , u n ) satisfying (1.1)-(1.2) such that u i ( t ) > 0 , 1 i n , on [ 0 , 1 ] T .

Proof Let λ k , 1 k n , be as in (3.6). Now, let ϵ > 0 be chosen such that

max 1 i n { [ γ 2 μ 1 μ G ( σ ( s ) , s ) q i ( s ) s ( f i 0 ϵ ) ] 1 } min 1 k n λ k

and

max 1 k n λ k min 1 i n { [ ( 0 1 G ( σ ( s ) , s ) q i ( s ) s + A ( q i ( ) ) ( b + a ) B ( q i ( ) ) ( d + c ) ) ( f i + ϵ ) ] 1 } .

Let T be the cone preserving, completely continuous operator that was defined by (3.1). From the definition of f i 0 , 1 i n , there exists H ¯ 3 > 0 such that, for each 1 i n ,

f i ( x ) ( f i 0 ϵ ) x , 0 < x H ¯ 3 .

Also, from the definition of f i 0 , it follows that f i 0 ( 0 ) = 0 , 1 i n , and so there exist 0 < K n < K n 1 < < K 2 < H ¯ 3 such that

λ i f i ( t ) K i 1 0 1 G ( σ ( s ) , s ) q i ( s ) s , t [ 0 , K i ] T , 3 i n

and

λ 2 f 2 ( t ) H ¯ 3 0 1 G ( σ ( s ) , s ) q 2 ( s ) s , t [ 0 , K 2 ] T .

Choose u 1 P with u 1 = K n . Then we have

λ n 0 1 G ( s n 1 , s n ) q n ( s n ) f n ( u 1 ( s n ) ) s n λ n 0 1 G ( σ ( s n ) , s n ) q n ( s n ) f n ( u 1 ( s n ) ) s n 0 1 G ( σ ( s n ) , s n ) q n ( s n ) K n 1 s n 0 1 G ( σ ( s n ) , s n ) q n ( s n ) s n = K n 1 .

Continuing with this bootstrapping argument, we get

λ 2 0 1 G ( s 1 , s 2 ) q 2 ( s 2 ) f 2 ( λ 3 0 1 G ( s 2 , s 3 ) q 3 ( s 3 ) f n ( u 1 ( s n ) ) s n s 3 ) s 2 H ¯ 3 .

Then

T u 1 ( t ) λ 1 0 1 G ( t , s 1 ) q 1 ( s 1 ) f 1 ( λ 2 0 1 G ( s 1 , s 2 ) q 2 ( s 2 ) f n ( u 1 ( s n ) ) s n s 2 ) s 1 γ 2 λ 1 μ 1 μ G ( σ ( s 1 ) , s 1 ) q 1 ( s 1 ) ( f 10 ϵ ) u 1 s 1 u 1 .

So, T u 1 u 1 . If we put

Ω 3 = { u B | u < K n } ,

then

T u 1 u 1 for  u 1 P Ω 3 . (3.7)

Since each f i is assumed to be a positive real number, it follows that f i , 1 i n , is unbounded at ∞.

For each 1 i n , set

f i ( x ) = sup 0 s x f i ( s ) .

Then, for each 1 i n , f i is a nondecreasing real-valued function, f i f i , and

lim x f i ( x ) x = f i .

Next, by definition of f i , 1 i n , there exists H ¯ 4 such that, for each 1 i n ,

f i ( x ) ( f i + ϵ ) x , x H ¯ 4 .

It follows that there exists H 4 > max { 2 H ¯ 3 , H ¯ 4 } such that, for each 1 i n ,

f i ( x ) f i ( H 4 ) , 0 < x H 4 .

Choose u 1 P with u 1 = H 4 . Then, using the bootstrapping argument, we have

λ 1 0 1 G ( t , s 1 ) q 1 ( s 1 ) f 1 ( λ 2 ) s 1 λ 1 0 1 G ( t , s 1 ) q 1 ( s 1 ) f 1 ( λ 2 ) s 1 λ 1 0 1 G ( σ ( s 1 ) , s 1 ) q 1 ( s 1 ) f 1 ( H 4 ) s 1 λ 1 0 1 G ( σ ( s 1 ) , s 1 ) q 1 ( s 1 ) s 1 ( f 1 + ϵ ) H 4 .

So we have

T u 1 ( t ) λ 1 ( 0 1 G ( σ ( s 1 ) , s 1 ) q 1 ( s 1 ) s 1 ( f 1 + ϵ ) H 4 + A ( q 1 ( ) ) f 1 ( u 2 ) ( b + a ) + B ( q 1 ( ) ) f 1 ( u 2 ) ( d + c ) ) λ 1 ( 0 1 G ( σ ( s 1 ) , s 1 ) q 1 ( s 1 ) s 1 ( f 1 + ϵ ) H 4 + A ( q 1 ( ) ) f 1 ( u 2 ) ( b + a ) + B ( q 1 ( ) ) f 1 ( u 2 ) ( d + c ) ) λ 1 ( 0 1 G ( σ ( s 1 ) , s 1 ) q 1 ( s 1 ) s 1 + A ( q 1 ( ) ) ( b + a ) + B ( q 1 ( ) ) ( d + c ) ) × ( f 1 + ϵ ) H 4 H 4 = u 1 .

Hence, T u 1 u 1 . So, if we set

Ω 4 = { u B | u < H 4 } ,

then

T u 1 u 1 for  u 1 P Ω 4 . (3.8)

Applying Theorem 2.1 to (3.7) and (3.8), we see that T has a fixed point u 1 P ( Ω ¯ 4 Ω 3 ) , which in turn with u n + 1 = u 1 , we obtain an n-tuple ( u 1 , u 2 , , u n ) satisfying (1.1)-(1.2) for the chosen values of λ i , 1 i n . The proof is completed. □

### 4 An example

Example 4.1 In BVP (1.1)-(1.2), suppose that T = [ 0 , 1 ] , n = m = 3 , q 1 ( t ) = q 2 ( t ) = q 3 ( t ) = 1 , a = c = 2 , b = d = 1 , ξ 1 = 1 2 , μ = 1 4 , α 1 = 1 2 and β 1 = 3 i.e.,

{ u i ( t ) + λ i f i ( u i + 1 ( t ) ) = 0 , t [ 0 , 1 ] , 1 i 3 , u 4 ( t ) = u 1 ( t ) , t [ 0 , 1 ] T , (4.1)

satisfying the following boundary conditions:

{ 2 u i ( 0 ) u i ( 0 ) = 1 2 u i ( 1 2 ) , 2 u i ( 1 ) + u i ( 1 ) = 3 u i ( 1 2 ) , 1 i 3 , (4.2)

where

f 1 ( u 2 ) = u 2 ( 1 , 000 999 e u 2 ) ( 520 512 e 2 u 2 ) , f 2 ( u 3 ) = u 3 ( 700 698 e 2 u 3 ) ( 1 , 500 1 , 498 e 3 u 2 ) , f 3 ( u 1 ) = u 1 ( 800 796 e u 1 ) ( 400 396 e 5 u 1 ) .

It is easy to see that (C1)-(C7) are satisfied. By simple calculation, we get ρ = 8 , θ ( t ) = 1 + 2 t , φ ( t ) = 3 2 t , Δ = 8 , γ = 1 2 , A ( 1 ) = 9 , B ( 1 ) = 3 2 and

G ( t , s ) = 1 8 { ( 1 + 2 s ) ( 3 2 t ) , s t , ( 1 + 2 t ) ( 3 2 s ) , t s .

We obtain

f 10 = 8 , f 20 = 4 , f 30 = 16 , f 1 = 520 , 000 , f 2 = 1 , 050 , 000 , f 3 = 320 , 000 , M 1 = max { 0.0000039279869 , 0.00000194528875 , 0.000006382978723 }

and

M 2 = min { 0.00355450236 , 0.007109004739 , 0.001777251184 } .

Applying Theorem 3.1, we get the optimal eigenvalue interval 0.000006382978723 < λ i < 0.001777251184 , i = 1 , 2 , 3 , for which the boundary value problem (4.1)-(4.2) has a positive solution.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

All authors contributed equally to the manuscript and typed, read and approved the final manuscript.

### Acknowledgements

The authors would like to thank the referees for their valuable suggestions and comments.

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