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Positive solutions of a singular semipositone boundary value problems for fourth-order coupled difference equations

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Abstract

In this article, we study the boundary value problems for the fourth-order nonlinear coupled difference equations

{ 4 u ( i 2 ) = f 1 ( i , v ( i ) ) + e 1 ( i ) , i [ 2 , T + 2 ] , 4 v ( i 2 ) = f 2 ( i , u ( i ) ) + e 2 ( i ) , i [ 2 , T + 2 ] , u ( 0 ) = u ( 1 ) = v ( 1 ) = v ( 0 ) = 0 , u ( T + 3 ) = u ( T + 4 ) = v ( T + 4 ) = v ( T + 3 ) = 0 .

Throughout our nonlinearity may be singular. We establish the existence of positive solutions for the fourth-order coupled systems. The proof relies on Schauder’s fixed point theorem.

MSC:39A11.

Introduction

In this article, we consider the following boundary value problems of difference equations

{ 4 u ( i 2 ) = f 1 ( i , v ( i ) ) + e 1 ( i ) , i [ 2 , T + 2 ] , 4 v ( i 2 ) = f 2 ( i , u ( i ) ) + e 2 ( i ) , i [ 2 , T + 2 ] , u ( 0 ) = u ( 1 ) = v ( 1 ) = v ( 0 ) = 0 , u ( T + 3 ) = u ( T + 4 ) = v ( T + 4 ) = v ( T + 3 ) = 0 ,
(1)

here [2,T+2]={2,3,,T+2} and u,v, e j :[0,T+4]R, f j :[0,T+4]×R R + , j=1,2. We will let [a,b] denote the discrete integer set [a,b]={a,a+1,,b}, C([a,b]) denote set of continuous function on [a,b] (discrete topology) with norm = max k [ a , b ] ||.

Recently, many literature on the boundary value of difference equations have appeared. We refer the reader to [117] and the references therein, which include work on Agarwal, Eloe, Erber, O’Regan, Henderson, Merdivenci, Yu, Ma et al., concerning the existence of positive solutions and the corresponding eigenvalue problems. Recently, the existence of positive solutions of fourth-order discrete boundary value problems have been studied by several authors, for examples, see [1618] and the references therein.

The fourth-order boundary value problems of ordinary value problems have important application in various branches of pure and applied science. As in entrepreneurial network evolution studies, the research paradigm of ‘U related to V’ was always applied, while U or V was the measurement of the attitudes of the entrepreneurs who answered the questionnaire. According to cognitive psychology studies [19, 20], questionnaire shows the change of attitude, which is the fourth-order dependent of original signal. So whether this paradigm will lead to meaningful causal outcome, which is basically depended on difference equations.

The fourth-order boundary value problems of ordinary value problems have important application in various branches of pure and applied science. They arise in the mathematical modeling of viscoelastic and inelastic flows, deformation of beams and plate deflection theory [2123]. For examples, the deformations of an elastic beam can be described by the boundary value problems of the fourth-order ordinary differential equations. There have been extensive studies on fourth-order boundary value problems with diverse boundary conditions via many methods, for example [2427] and the references therein.

The remaining part of the article is organized as follows. In Section “Preliminaries”, some preliminary results will be given. In the remaining sections, by employing a basic application of Schauder’s fixed point theorem, we state and prove the existence results for (1). Our view point sheds some new light on problems with weak force potentials and prove that in some situations weak singularities may stimulate the existence of positive solutions.

Preliminaries

In this section, we state the preliminary information that we need to prove the main results. From [4, 5], we have the following lemma.

Lemma 1 x(i) is a solution of equation

{ 4 x ( i 2 ) = h ( i ) , i [ 2 , T + 2 ] , x ( 0 ) = x ( 1 ) = x ( T + 3 ) = x ( T + 4 ) = 0
(2)

if only and if

x(i)= j = 2 T + 2 G(i,j)h(j),i[0,T+4],
(3)

where

G(i,j)={ ( T + 4 i ) 2 ( j 1 ) 2 2 ( i ( T + 3 ) 2 ( T + 4 + 2 i ) ( j 1 ) 3 ( T + 4 ) 3 ) , 2 j i + 1 , i 2 ( T + 4 j ) 2 6 ( ( T + 3 i ) ( T + 4 + 2 j ) ( T + 4 ) 3 T + 4 j ( T + 3 ) 2 ) , i + 1 < s T + 2 .
(4)

Lemma 2 The Green’s functionG(i,j)defined by (4) have properties

where C 0 = 1 3 ( T + 4 ) 7 .

For our constructions, we shall consider the Banach space E=C([0,T+4]) equipped with the standard norm x= max 0 i T + 4 |x(i)|, xE. We define a cone PE by

P= { x E | x ( i ) C 0 i 2 ( T + 4 i ) 2 x , i [ 0 , T + 4 ] } .

Define an operator A(u,v)=( A 1 (u), A 2 (v)):P×PE×E by

A 1 u ( i ) = j = 2 T + 2 G ( i , j ) ( f 1 ( i , v ( i ) ) + e 1 ( i ) ) , i [ 0 , T + 4 ] , v P , A 2 v ( i ) = j = 2 T + 2 G ( i , j ) ( f 2 ( i , u ( i ) ) + e 2 ( i ) ) , i [ 0 , T + 4 ] , u P .
(5)

Notice from (5) and Lemma 2 that, for uP, we have

A 1 u ( i ) = j = 2 T + 2 G ( i , j ) ( f 1 ( i , v ( i ) ) + e 1 ( i ) ) j = 2 T + 2 ( j 1 ) 2 ( T + 4 j ) 2 ( f 1 ( i , v ( i ) ) + e 1 ( i ) ) , A 2 v ( i ) = j = 2 T + 2 G ( i , j ) ( f 2 ( i , u ( i ) ) + e 2 ( i ) ) j = 2 T + 2 ( j 1 ) 2 ( T + 4 j ) 2 ( f 2 ( i , u ( i ) ) + e 2 ( i ) ) ,

then A 1 v j = 2 T + 2 ( j 1 ) 2 ( T + 4 j ) 2 ( f 1 (i,v(i))+ e 1 (i)) and A 2 u j = 2 T + 2 ( j 1 ) 2 ( T + 4 j ) 2 ( f 2 (i,u(i))+ e 2 (i)).

On the other hand, we have

Thus, A(P×P)P×P. In addition, standard arguments show that A is completely continuous.

Now, we define the function γ k :EE by

γ k (i)= j = 2 T + 2 G ( i , j ) i 2 ( T + 4 i ) 2 e k (j),i[0,T+4],k=1,2,

which is the unique solution of

{ 4 x ( i 2 ) = e k ( i ) , i [ 2 , T + 2 ] , x ( 0 ) = x ( 1 ) = x ( T + 3 ) = x ( T + 4 ) = 0 .

Throughout this article, we use the following notations

γ k = min k , i γ k (i), γ k = max k , i γ k (i).

The case γ 1 0, γ 2 0

Theorem 1 Assume that there exists b k 0, b ˆ k 0and0< α k <1such that

0 b ˆ k ( i ) x α k f k (i,x) b k ( i ) x α k ,for allx>0,i[2,T+2],k=1,2.
(H)

If γ 1 0, γ 2 0, then there exists a positive solution of (1).

Proof A solution of (1) is just a fixed point of the completely continuous map A(x,y)=( A 1 x, A 2 y):P×PP×P, from (5) we have

By a direct application of Schauder’s fixed point theorem, the proof is finished if we prove that A maps the closed convex set defined as

K = { ( u , v ) P × P : r 1 i 2 ( T + 4 i ) 2 u ( i ) R 1 i 2 ( T + 4 i ) 2 , r 2 i 2 ( T + 4 i ) 2 v ( i ) R 2 i 2 ( T + 4 i ) 2 , i [ 0 , T + 4 ] } ,

into itself, where R 1 > r 1 >0, R 2 > r 2 >0 are positive constants to be fixed properly. For convenience, we introduce the following notations

β k (i)= j = 2 T + 2 G ( i , j ) i 2 ( T + 4 i ) 2 b k (j), β ˆ k (i)= j = 2 T + 2 G ( i , j ) i 2 ( T + 4 i ) 2 b ˆ k (j),k=1,2.

Given (u,v)K, by the nonnegative sign of G and f k , k=1,2, we have

( A 1 u ) ( i ) = i 2 ( T + 4 i ) 2 j = 2 T + 2 G ( i , j ) i 2 ( T + 4 i ) 2 f 1 ( j , v ( j ) ) + γ 1 ( i ) i 2 ( T + 4 i ) 2 j = 2 T + 2 G ( i , j ) i 2 ( T + 4 i ) 2 b ˆ 1 ( j ) v α 1 ( j ) + i 2 ( T + 4 i ) 2 γ 1 i 2 ( T + 4 i ) 2 j = 2 T + 2 G ( i , j ) i 2 ( T + 4 i ) 2 b ˆ 1 ( j ) R 2 α 1 i 2 ( T + 4 i ) 2 β ˆ 1 1 R 2 α 1 .

Note for every (u,v)K

( A 1 u ) ( i ) = i 2 ( T + 4 i ) 2 j = 2 T + 2 G ( i , j ) i 2 ( T + 4 i ) 2 f 1 ( j , v ( j ) ) + γ 1 ( i ) i 2 ( T + 4 i ) 2 j = 2 T + 2 G ( i , j ) i 2 ( T + 4 i ) 2 b 1 ( j ) v α 1 ( j ) + i 2 ( T + 4 i ) 2 γ 1 i 2 ( T + 4 i ) 2 j = 2 T + 2 G ( i , j ) i 2 ( T + 4 i ) 2 b 1 ( j ) r 2 α 1 + i 2 ( T + 4 i ) 2 γ 1 i 2 ( T + 4 i ) 2 ( β 1 1 r 2 α 1 + γ 1 ) .

Similarly, by the same strategy, we have

Thus ( A 1 u, A 2 v)K if r 1 , r 2 , R 1 and R 2 are chosen so that

Note that β ˆ i , β i >0 and taking R= R 1 = R 2 , r= r 1 = r 2 , r= 1 R , it is sufficient to find R>1 such that

and these inequalities hold for R big enough because α i <1. The proof is complete. □

The case γ 1 0, γ 2 0

The aim of this section is to show that the presence of a weak singular nonlinearity makes it possible to find positive solutions if γ 1 0, γ 2 0.

Theorem 2 Assume that there exist b k , b ˆ k 0and0< α k <1, such that(H)is satisfied. If γ 1 0, γ 2 0, and

γ 1 [ α 1 α 2 β ˆ 1 ( β 2 ) α 1 ] 1 1 α 1 α 2 ( 1 1 α 1 α 2 ) , γ 2 [ α 1 α 2 β ˆ 2 ( β 1 ) α 2 ] 1 1 α 1 α 2 ( 1 1 α 1 α 2 ) ,
(6)

then there exists a positive solution of (1).

Proof We follow the same strategy and notation as in the proof of ahead theorem. In this case, to prove that A:KK, it is sufficient to find 0< r 1 < R 1 , 0< r 2 < R 2 such that

(7)
(8)

If we fix R 1 = β 1 r 2 α 1 , R 2 = β 2 r 1 α 2 , then the first inequality of (8) holds if r 2 satisfies

β ˆ 2 ( β 1 ) α 2 r 2 α 1 α 2 + γ 2 r 2 ,

or equivalently

γ 2 g( r 2 ):= r 2 β ˆ 2 ( β 1 ) α 2 r 2 α 1 α 2 .

The function g( r 2 ) possesses a minimum at

r 20 := [ α 1 α 2 β ˆ 2 ( β 1 ) α 2 ] 1 1 α 1 α 2 .

Taking r 2 = r 20 , then (6) holds if

γ 2 g( r 20 )= [ α 1 α 2 β ˆ 2 ( β 1 ) α 2 ] 1 1 α 1 α 2 ( 1 1 α 1 α 2 ) .

Similarly,

γ 1 h( r 1 ):= r 1 β ˆ 1 ( β 2 ) α 1 r 1 α 1 α 2 ,

h( r 1 ) possesses a minimum at

Taking r 1 = r 10 , r 2 = r 20 , then the first inequalities in (7) and (8) hold if γ 1 g( r 1 ) and γ 2 g( r 2 ), which are just condition (6). The second inequalities hold directly from the choice of R 1 and R 2 , so it remains to prove that R 1 = β 1 r 20 α 1 > r 10 , R 2 = β 2 r 10 α 2 > r 20 . This is easily verified through elementary computations:

R 1 = β 1 r 20 α 1 = β 1 { [ α 1 α 2 β ˆ 2 ( β 1 ) α 2 ] 1 1 α 1 α 2 } α 1 = β 1 [ α 1 α 2 β ˆ 2 ( β 1 ) α 2 ] α 1 1 α 1 α 2 = ( β 1 ) 1 + α 1 α 2 1 α 1 α 2 ( α 1 α 2 β ˆ 2 ) α 1 1 α 1 α 2 = ( β 1 ) 1 1 α 1 α 2 [ ( α 1 α 2 β ˆ 2 ) α 1 ] 1 1 α 1 α 2 = [ β 1 ( α 1 α 2 β ˆ 2 ) α 1 ] 1 1 α 1 α 2 = [ 1 ( α 1 α 2 ) α 1 β 1 ( β ˆ 2 ) α 1 ] 1 1 α 1 α 2 > [ α 1 α 2 β ˆ 1 ( β 2 ) α 1 ] 1 1 α 1 α 2 = r 10 ,

since β ˆ k β k , k=1,2. Similarly, we have R 2 > r 20 . The proof is complete. □

The case γ 1 0, γ 2 0 ( γ 1 0, γ 2 0)

Theorem 3 Assume (H) is satisfied. If γ 1 0, γ 2 0and

γ 2 r 21 β ˆ 2 r 21 α 1 α 2 ( β 1 + γ 1 r 21 α 1 ) α 2 ,
(9)

where 0< r 21 <+ is a unique positive solution of the equation

r 2 1 α 1 α 2 ( β 1 + γ 1 r 2 α 1 ) 1 + α 2 = α 1 α 2 β 1 β ˆ 2 ,

then there exists a positive solution of (1).

Proof In this case,to prove that A:KK, it is sufficient to find r 1 < R 1 , r 2 < R 2 such that

(10)
(11)

If we fix R 2 = β 2 r 1 α 2 , then the first inequality of (10) holds if r 1 satisfies

β ˆ 1 ( β 2 ) α 1 r 1 α 1 α 2 r 1 ,

or equivalently

0< r 1 [ β ˆ 1 ( β 2 ) α 1 ] 1 1 α 1 α 2 .
(12)

If we chose r 1 >0 small enough, then (12) holds, and R 2 is big enough.

If we fix R 1 = β 1 r 2 α 1 + γ 1 then the first inequality of (11) holds if r 2 satisfies

γ 2 r 2 β ˆ 2 R 1 α 2 = r 2 β ˆ 2 1 ( β 1 r 2 α 1 + γ 1 ) α 2 = r 2 β ˆ 2 1 ( β 1 + γ 1 r 2 α 1 r 2 α 1 ) α 2 = r 2 β ˆ 2 r 2 α 1 α 2 ( β 1 + γ 1 r 2 α 1 ) α 2 ,

or equivalently

γ 2 f( r 2 ):= r 2 β ˆ 2 r 2 α 1 α 2 ( β 1 + γ 1 r 2 α 1 ) α 2 .

According to

f ( r 2 ) = 1 β ˆ 2 1 ( β 1 + γ 1 r 2 α 1 ) 2 α 2 [ α 1 α 2 r 2 α 1 α 2 1 ( β 1 + γ 1 r 2 α 1 ) α 2 r 2 α 1 α 2 α 2 ( β 1 + γ 1 r 2 α 1 ) α 2 1 α 1 γ 1 r 2 α 1 1 ] = 1 β ˆ 2 α 1 α 2 r 2 α 1 α 2 1 ( β 1 + γ 1 r 2 α 1 ) α 2 [ 1 r 2 α 1 γ 1 β 1 + γ 1 r 2 α 1 ] = 1 α 1 α 2 β 1 β ˆ 2 r 2 α 1 α 2 1 ( β 1 + γ 1 r 2 α 1 ) 1 α 2 ,

we have f (0)=, f (+)=1, then there exists r 21 such that f ( r 21 )=0, and

f ( r 2 ) = [ α 1 α 2 β 1 β ˆ 2 ( α 1 α 2 1 ) r 2 α 1 α 2 2 ( β 1 + γ 1 r 2 α 1 ) 1 α 2 + α 1 α 2 β 1 β ˆ 2 r 2 α 1 α 2 1 ( 1 α 2 ) ( β 1 + γ 1 r 2 α 1 ) 2 α 2 γ 1 α 1 r 2 α 1 1 ] > 0 .

Then the function f( r 2 ) possesses a minimum at r 21 , i.e., f( r 21 )= min r 2 ϵ ( 0 , + ) f( r 2 ).

Note f ( r 21 )=0, then we have

1 α 1 α 2 β 1 β ˆ 2 r 21 α 1 α 2 1 ( β 1 + γ 1 r 21 α 1 ) 1 α 2 =0,

or equivalently,

r 21 1 α 1 α 2 ( β 1 + γ 1 r 21 α 1 ) 1 + α 2 = α 1 α 2 β 1 β ˆ 2 .

Taking r 2 = r 21 , then the first inequality in (11) holds if γ 2 f( r 21 ), which is just condition (9). The second inequality holds directly by the choice of R 2 , and it would remain to prove that r 21 < R 2 and r 10 < R 1 . These inequalities hold for R 2 big enough and r 1 small enough. The proof is complete. □

Similarly, we have the following theorem.

Theorem 4 Assume (H) are satisfied. If γ 1 0, γ 2 0and

γ 1 r 11 β ˆ 1 r 11 α 1 α 2 ( β 2 + γ 2 r 11 α 2 ) α 1 ,

where 0< r 11 <+ is a unique positive solution of the equation

r 1 1 α 1 α 2 ( β 2 + γ 2 r 1 α 2 ) 1 + α 1 = α 1 α 2 β 2 β ˆ 1 ,

then there exists a positive solution of (1).

The case γ 1 <0< γ 1 , γ 2 <0< γ 2

Theorem 5 Assume (H) is satisfied. If γ 1 <0< γ 1 , γ 2 <0< γ 2 and

(13)
(14)

where 0< r 10 <+ is a unique positive solution of the equation

r 1 1 α 1 α 2 ( β 2 + γ 2 r 1 α 2 ) 1 + α 1 = α 1 α 2 β 2 β ˆ 1 ,

and 0< r 20 <+ is a unique positive solution of the equation

r 2 1 α 1 α 2 ( β 1 + γ 1 r 2 α 1 ) 1 + α 2 = α 1 α 2 β 1 β ˆ 2 ,

then there exists a positive solution of (1).

Proof We follow the same strategy and notation as in the proof of ahead theorem. In this case, to prove that A:KK, it is sufficient to find r 1 < R 1 , r 2 < R 2 such that

(15)
(16)

If we fix R 1 = β 1 r 2 α 1 + γ 1 , R 2 = β 2 r 1 α 2 + γ 2 , then the first inequality of (15) holds if r 2 satisfies

γ 2 g( r 2 ):= r 2 β ˆ 2 r 2 α 1 α 2 ( β 1 + γ 1 r 2 α 1 ) α 2 .

According to

g ( r 2 ) = 1 β ˆ 2 1 ( β 1 + γ 1 r 2 α 1 ) 2 α 2 [ α 1 α 2 r 2 α 1 α 2 1 ( β 1 + γ 1 r 2 α 1 ) α 2 r 2 α 1 α 2 α 2 ( β 1 + γ 1 r 2 α 1 ) α 2 1 α 1 γ 1 r 2 α 1 1 ] = 1 β ˆ 2 α 1 α 2 r 2 α 1 α 2 1 ( β 1 + γ 1 r 2 α 1 ) α 2 [ 1 r 2 α 1 γ 1 β 1 + γ 1 r 2 α 1 ] = 1 α 1 α 2 β 1 β ˆ 2 r 2 α 1 α 2 1 ( β 1 + γ 1 r 2 α 1 ) 1 α 2 ,

we have g (0)=, g (+)=1, then there exists r 20 such that g ( r 20 )=0, and

g ( r 2 ) = [ α 1 α 2 β 1 β ˆ 2 ( α 1 α 2 1 ) r 2 α 1 α 2 2 ( β 1 + γ 1 r 2 α 1 ) 1 α 2 + α 1 α 2 β 1 β ˆ 2 r 2 α 1 α 2 1 ( 1 α 2 ) ( β 1 + γ 1 r 2 α 1 ) 2 α 2 γ 1 α 1 r 2 α 1 1 ] > 0 .

Then the function g( r 2 ) possesses a minimum at r 20 , i.e., g( r 20 )= min r 2 ϵ ( 0 , + ) g( r 2 ).

Note g ( r 20 )=0, then we have

1 α 1 α 2 β 1 β ˆ 2 r 21 α 1 α 2 1 ( β 1 + γ 1 r 21 α 1 ) 1 α 2 =0,

or equivalently,

r 20 1 α 1 α 2 ( β 1 + γ 1 r 20 α 1 ) 1 + α 2 = α 1 α 2 β 1 β ˆ 2 .

Similarly,

γ 1 g( r 1 ):= r 1 β ˆ 1 r 1 α 1 α 2 ( β 2 + γ 2 r 1 α 2 ) α 1 .

g( r 10 )= min r 1 ϵ ( 0 , + ) g( r 1 ), and

r 10 1 α 1 α 2 ( β 2 + γ 2 r 10 α 2 ) 1 + α 1 = α 1 α 2 β 2 β ˆ 1 .

Taking r 1 = r 10 and r 2 = r 20 , then the first inequality in (15) and (16) hold if γ 1 g( r 10 ), γ 2 g( r 20 ), which are just condition (13) and (14). The second inequalities hold directly by the choice of R 1 and R 2 , and it would remain to prove that r 10 < R 1 and r 20 < R 2 . This is easily verified through elementary computations.

R 1 = β 1 r 20 α 1 + γ 1 = β 1 + γ 1 r 20 α 1 r 20 α 1 = ( α 1 α 2 β 1 β ˆ 2 ) 1 1 + α 2 r 20 α 1 α 2 1 1 + α 2 r 20 α 1 = ( α 1 α 2 β 1 β ˆ 2 ) 1 1 + α 2 r 20 1 + α 1 1 + α 2 .

The proof is the same as that in R 1 , R 2 = ( α 1 α 2 β 2 β ˆ 1 ) 1 1 + α 1 r 10 1 + α 2 1 + α 1 .

Next, we will prove r 10 < R 1 , r 20 < R 2 , or equivalently,

Namely,

r 10 1 + α 2 r 20 1 + α 1 < α 1 α 2 β 1 β ˆ 2 , r 20 1 + α 1 r 10 1 + α 2 < α 1 α 2 β 2 β ˆ 1 .

On the other hand,

r 20 1 α 1 α 2 ( β 1 ) 1 + α 2 α 1 α 2 β 1 β ˆ 2 .

Then

r 20 ( α 1 α 2 ( β 1 ) α 2 β ˆ 2 ) 1 1 α 1 α 2 .

Similarly

r 10 ( α 1 α 2 ( β 2 ) α 1 β ˆ 1 ) 1 1 α 1 α 2 .

By (13) and (14),

r 10 1 + α 2 r 20 1 + α 1 ( α 1 α 2 ( β 2 ) α 1 β ˆ 1 ) 1 + α 2 1 α 1 α 2 ( α 1 α 2 ( β 1 ) α 2 β ˆ 2 ) 1 + α 1 1 α 1 α 2 .

Now if we can prove

( α 1 α 2 ( β 2 ) α 1 β ˆ 1 ) 1 + α 2 1 α 1 α 2 ( α 1 α 2 ( β 1 ) α 2 β ˆ 2 ) 1 + α 1 1 α 1 α 2 < α 1 α 2 β 1 β ˆ 2

then

r 10 1 + α 2 r 20 1 + α 1 < α 1 α 2 β 1 β ˆ 2 .

In fact,

( α 1 α 2 ) 2 + α 2 + α 1 1 1 α 1 α 2 ( β ˆ 1 β 1 ) 1 + α 2 1 α 1 α 2 ( β ˆ 2 β 2 ) α 1 ( 1 + α 2 ) 1 α 1 α 2 <1,

since β ˆ i β i , i=1,2. Similarly,we have r 20 1 + α 1 r 10 1 + α 2 < α 1 α 2 β 2 β ˆ 1 , we omit the details. Now we can obtain r 10 < R 1 , r 20 < R 2 . The proof is complete. □

The case γ 1 0, γ 2 <0< γ 2 ( γ 2 0, γ 1 <0< γ 1 )

Theorem 6 Assume (H) are satisfied. If γ 1 0, γ 2 <0< γ 2 and

(17)
(18)

where 0< r 11 <+ is a unique positive solution of the equation

r 1 1 α 1 α 2 ( β 2 + γ 2 r 1 α 2 ) 1 + α 1 = α 1 α 2 β 2 β ˆ 1 ,
(19)

then there exists a positive solution of (1).

Proof In this case,to prove that A:KK, it is sufficient to find r 1 < R 1 , r 2 < R 2 such that

(20)
(21)

If we fix R 1 = β 1 r 2 α 1 , R 2 = β 2 r 1 α 2 + γ 2 , then the first inequality of (21) holds if r 2 satisfies

γ 2 r 2 β ˆ 2 R 1 α 2 = r 2 β ˆ 2 ( β 1 ) α 2 r 2 α 1 α 2 ,
(22)

or equivalently

γ 2 f( r 2 ):= r 2 β ˆ 2 ( β 1 ) α 2 r 2 α 1 α 2 .
(23)

Then the function f( r 2 ) possesses a minimum at

r 21 = [ α 1 α 2 β ˆ 2 ( β 1 ) α 2 ] 1 1 α 1 α 2 ,
(24)

i.e., f( r 21 )= min r 2 ϵ ( 0 , + ) f( r 2 ).

On the analogy of (22), we obtain

γ 1 r 1 β ˆ 1 r 1 α 1 α 2 ( β 2 + γ 2 r 1 α 2 ) α 1 ,
(25)

or equivalently,

γ 1 h( r 1 ):= r 1 β ˆ 1 r 1 α 1 α 2 ( β 2 + γ 2 r 1 α 2 ) α 1 .
(26)

According to

h ( r 1 ):=1 α 1 α 2 β 2 β ˆ 1 r 1 α 1 α 2 1 ( β 2 + γ 2 r 1 α 2 ) 1 α 1 ,
(27)

we have h (0)=, h (+)=1, then there exists r 11 such that h ( r 11 )=0, and

h ( r 1 ) = [ α 1 α 2 β 2 β ˆ 1 ( α 1 α 2 1 ) r 1 α 1 α 2 2 ( β 2 + γ 2 r 1 α 2 ) 1 α 1 + α 1 α 2 β 2 β ˆ 1 r 1 α 1 α 2 1 ( 1 α 1 ) ( β 2 + γ 2 r 1 α 2 ) 2 α 1 γ 2 α 2 r 1 α 2 1 ] > 0 .
(28)

Then the function h( r 1 ) possesses a minimum at r 11 , i.e., h( r 11 )= min r 1 ϵ ( 0 , + ) f( r 1 ).

Note h ( r 11 )=0, then we have

1 α 1 α 2 β 2 β ˆ 1 r 11 α 1 α 2 1 ( β 2 + γ 2 r 11 α 2 ) 1 α 1 =0.

Namely,

r 11 1 α 1 α 2 ( β 2 + γ 2 r 11 α 2 ) 1 + α 1 = α 1 α 2 β 2 β ˆ 1 .

Taking r 2 = r 21 and r 1 = r 11 then the first inequality in (20) and (21) hold if γ 2 h( r 21 ) and γ 1 h( r 11 ) which are just condition (17) and (18). The second inequalities hold directly by the choice of R 2 and R 1 , so it would remain to prove that R 1 = β 1 r 21 α 1 > r 11 , R 2 = β 2 r 11 α 2 + γ 2 > r 21 . Now we turn to prove that R 1 > r 11 , R 2 > r 21 .

First,

R 1 = β 1 r 21 α 1 = β 1 { [ α 1 α 2 β ˆ 2 ( β 1 ) α 2 ] 1 1 α 1 α 2 } α 1 = β 1 [ α 1 α 2 β ˆ 2 ( β 1 ) α 2 ] α 1 1 α 1 α 2 = ( β 1 ) 1 + α 1 α 2 1 α 1 α 2 ( α 1 α 2 β ˆ 2 ) α 1 1 α 1 α 2 = ( β 1 ) 1 1 α 1 α 2 [ ( α 1 α 2 β ˆ 2 ) α 1 ] 1 1 α 1 α 2 = [ β 1 ( α 1 α 2 β ˆ 2 ) α 1 ] 1 1 α 1 α 2 = [ 1 ( α 1 α 2 ) α 1 β 1 ( β ˆ 2 ) α 1 ] 1 1 α 1 α 2 > [ α 1 α 2 β ˆ 1 ( β 2 ) α 1 ] 1 1 α 1 α 2 = r 11 ,

since β ˆ i β i , i=1,2.

On the other hand,

R 2 = β 2 r 11 α 2 + γ 2 = β 2 + γ 2 r 11 α 2 r 11 α 2 .
(29)

By (18), we have

β 2 + γ 2 r 11 α 2 = ( α 1 α 2 β 2 β ˆ 1 ) 1 1 + α 1 r 11 α 1 α 2 1 1 + α 1 .
(30)

Combing (29) and (30),

R 2 = ( α 1 α 2 β 2 β ˆ 1 ) 1 1 + α 1 r 11 1 + α 2 1 + α 1 .
(31)

In what follows, we will verify that R 2 > r 21 . In fact,

( α 1 α 2 ) 2 + α 2 + α 1 1 α 1 α 2 1 ( β ˆ 2 β 2 ) 1 + α 1 1 α 1 α 2 ( β ˆ 1 β 1 ) α 2 ( 1 + α 1 ) 1 α 1 α 2 <1,

since β ˆ i β i , i=1,2. Thus

( α 1 α 2 β 1 ( α 2 ) β ˆ 2 ) 1 + α 1 1 α 1 α 2 ( α 1 α 2 β 2 ( α 1 ) β ˆ 1 ) 1 + α 2 1 α 1 α 2 < α 1 α 2 β 2 β ˆ 1 .
(32)

On the other hand,

Thus one can see easily note that

(33)
(34)

From (33) and (34),

r 11 1 + α 2 r 21 1 + α 1 ( α 1 α 2 β 2 ( α 1 ) β ˆ 1 ) 1 + α 2 1 α 1 α 2 ( α 1 α 2 β 1 ( α 2 ) β ˆ 2 ) 1 + α 1 1 α 1 α 2 .
(35)

Combing (32) and (35),

r 11 1 + α 2 r 21 1 + α 1 < α 1 α 2 β 2 β ˆ 1 .

Therefore,

r 21 r 11 1 + α 2 1 + α 1 < ( α 1 α 2 β 2 β ˆ 1 ) 1 1 + α 1 .

Recall (31), we obtain r 21 < R 2 immediately. The proof is complete. □

Similarly, we have the following theorem.

Theorem 7 Assume (H) are satisfied. If γ 2 0, γ 1 <0< γ 1 and

where 0< r 21 <+ is a unique positive solution of the equation

r 2 1 α 1 α 2 ( β 1 + γ 1 r 2 α 1 ) 1 + α 2 = α 1 α 2 β 1 β ˆ 2 ,

then there exists a positive solution of (1).

The case γ 1 0, γ 2 <0< γ 2 ( γ 2 0, γ 1 <0< γ 1 )

Theorem 8 Assume (H) are satisfied. If γ 1 0, γ 2 <0< γ 2 and

γ 2 r 22 β ˆ 2 r 22 α 1 α 2 ( β 1 + γ 1 r 22 α 1 ) α 2 ,

where 0< r 22 <+ is a unique positive solution of the equation

r 2 1 α 1 α 2 ( β 1 + γ 1 r 2 α 1 ) 1 + α 2 = α 1 α 2 β 1 β ˆ 2 ,

then there exists a positive T-periodic solution of (1).

Proof In this case, to prove that A:KK, it is sufficient to find r 1 < R 1 , r 2 < R 2 such that

(36)
(37)

If we fix R 1 = β 1 r 2 α 1 + γ 1 , R 2 = β 2 r 1 α 2 + γ 2 , then the first inequality of (37) satisfies

β ˆ 2 ( β 1 r 2 α 1 + γ 1 ) α 2 + γ 2 r 2 ,

or equivalently

γ 2 l( r 2 ):= r 2 β ˆ 2 ( β 1 + γ 1 r 2 α 1 ) α 2 r 2 α 1 α 2 .

Then the function l( r 2 ) possesses a minimum at r 22 , i.e., l( r 22 )= min r 2 ϵ ( 0 , + ) l( r 2 ).

Note l ( r 22 )=0, then we have

1 α 1 α 2 β 1 β ˆ 2 r 22 α 1 α 2 1 ( β 1 + γ 1 r 22 α 1 ) 1 α 2 =0.

Therefore,

r 22 1 α 1 α 2 ( β 1 + γ 1 r 22 α 1 ) 1 + α 2 = α 1 α 2 β 1 β ˆ 2 .

Note that β ˆ i, β i >0, i=1,2. And taking r 2 = r 22 , R 1 = β 1 r 22 α 1 + γ 1 , r 1 = 1 R 2 , it is sufficient to find r 1 < R 1 , r 2 < R 2 such that

R 2 α 1 1 β 1 , R 2 α 2 β 2 + γ 2 R 2

and these inequalities hold for R 2 big enough because α i <1. The proof is completed. □

Similarly, we have the following theorem.

Theorem 9 Assume (H) are satisfied. If γ 2 0, γ 1 <0< γ 1 and

γ 1 r 12 β ˆ 1 r 12 α 1 α 2 ( β 2 + γ 2 r 12 α 2 ) α 1 ,

where 0< r 12 <+ is a unique positive solution of the equation

r 1 1 α 1 α 2 ( β 2 + γ 2 r 1 α 2 ) 1 + α 1 = α 1 α 2 β 2 β ˆ 1 ,

then there exists a positive solution of (1).

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Acknowledgement

This study was supported by the National Science Founds of China (NO. 71172019).

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Correspondence to Yupeng Fei.

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The authors declare that they have no competing interests.

Authors’ contributions

WL, YF, BS, and YP have critically revised the manuscript and have made substantial contributions to conception. WL designed the research and drafted the manuscript. YF, BS and YP revised it critically for important intellectual content. All authors read and approved the final manuscript.

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Keywords

  • positive solution
  • positone and semipositone boundary value problem
  • singular difference equation
  • fixed point theorem