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

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.

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 [27].

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 [812]. 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 ) ) ,1in.

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

Throughout this paper we assume that following conditions hold:

(C1) a,b,c,d[0,) with ac+ad+bc>0; α j , β j [0,), ξ j ( 0 , 1 ) T for 1jm2,

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

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

(C4) each of f i 0 := lim x 0 + f i ( x ) x and f i := lim x f i ( x ) x , 1in, 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 hC[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+at,φ(t)=d+c(1t).
(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:=ad+ac+bc.
(2.6)

Lemma 2.1 Let (C1) hold. Assume that

(C5) Δ0.

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+at)+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+at)+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 2.2 Let (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 h0, the solution u 1 of the problem (2.1) satisfies

u 1 (t)0for t [ 0 , 1 ] T .

Proof It is an immediate subsequence of the facts that G0 on [ 0 , 1 ] T × [ 0 , 1 ] T and A(h)0, B(h)0. □

Lemma 2.3 Let (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 2.4 Let (C1) and h0 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

0G(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]

Let B be a Banach space, and let PB be a cone in B. Assume Ω 1 and Ω 2 are open subsets of B with 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

  1. (i)

    Tuu, uP Ω 1 , and Tuu, uP Ω 2 , or

  2. (ii)

    Tuu, uP Ω 1 , and Tuu, uP Ω 2 .

Then T has 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 PB 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:PB, 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:PP. 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 P. 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 3.1 Suppose conditions (C1)-(C7) are satisfied. Then, for each λ 1 , λ 2 ,, λ n satisfying

M 1 < λ i < M 2 ,1in,
(3.3)

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

Proof Let λ k , 1kn, 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:PP defined by (3.1). Now, from the definitions of f i 0 , 1in, there exists an H 1 >0 such that, for each 1in,

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 0t1,

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 0t1,

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 , 1in, there exists H ¯ 2 >0 such that, for each 1in,

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

M 3 < λ i < M 4 ,1in,
(3.6)

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

Proof Let λ k , 1kn, 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 , 1in, there exists H ¯ 3 >0 such that, for each 1in,

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, 1in, 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 ,3in

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 , 1in, is unbounded at ∞.

For each 1in, set

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

Then, for each 1in, 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 , 1in, there exists H ¯ 4 such that, for each 1in,

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 1in,

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 , 1in. 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+2t, φ(t)=32t, 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.

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Karaca, I.Y., Tokmak, F. Eigenvalues for iterative systems of nonlinear m-point boundary value problems on time scales. Bound Value Probl 2014, 63 (2014). https://doi.org/10.1186/1687-2770-2014-63

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