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Multiple positive solutions for first-order impulsive singular integro-differential equations on the half line in a Banach space

Abstract

In this paper, the author discusses the multiple positive solutions for an infinite three-point boundary value problem of first-order impulsive superlinear singular integro-differential equations on the half line in a Banach space by means of the fixed-point theorem of cone expansion and compression with norm type.

MSC:45J05, 34G20, 47H10.

1 Introduction

In recent years, multiple solutions of boundary value problems for impulsive differential equations in scalar spaces had been extensively studied (see, for example, [13]). In recent papers [4] and [5], Professor D. Guo discussed two infinite boundary value problems for n th-order impulsive nonlinear singular integro-differential equations of mixed type on the half line in a Banach space. By constructing a bounded closed convex set, apart from the singularities, and using the Schauder fixed-point theorem, he obtained the existence of positive solutions for the infinite boundary value problems. But such equations are sublinear, and there are no results on existence of two positive solutions. Now, in this paper, we shall discuss the existence of two positive solutions for first-order superlinear singular equations by means of a different method, i.e., by using the fixed-point theorem of cone expansion and compression with norm type (see [6, 7]), and the key point is to introduce a new cone Q.

Let E be a real Banach space and P be a cone in E, which defines a partial ordering in E by xy if and only if yxP. P is said to be normal if there exists a positive constant N such that θxy implies xNy, where θ denotes the zero element of E, and the smallest N is called the normal constant of P. If xy and xy, we write x<y. Let P + =P{θ}, i.e., P + ={xP:x>θ}. For details on cone theory, see [7].

Consider the infinite three-point boundary value problem for a first-order impulsive nonlinear singular integro-differential equation of mixed type on the half line in E:

{ u ( t ) = f ( t , u ( t ) , ( T u ) ( t ) , ( S u ) ( t ) ) , t J + , Δ u | t = t k = I k ( u ( t k ) ) ( k = 1 , 2 , 3 , ) , u ( ) = γ u ( η ) + β u ( 0 ) ,
(1)

where J=[0,), J + =(0,), 0< t 1 << t k < , t k , J + = J + { t 1 ,, t k ,}, fC[ J + × P + ×P×P,P], I k C[ P + ,P] (k=1,2,3,), 0γ<1, β+γ>1, t m 1 <η< t m (for some m), u()= lim t u(t) and

(Tu)(t)= 0 t K(t,s)u(s)ds,(Su)(t)= 0 H(t,s)u(s)ds,
(2)

KC[D, R + ], D={(t,s)J×J:ts}, HC[J×J, R + ], R + denotes the set of all nonnegative numbers. Δu | t = t k denotes the jump of u(t) at t= t k , i.e.,

Δu | t = t k =u ( t k + ) u ( t k ) ,

where u( t k + ) and u( t k ) represent the right and left limits of u(t) at t= t k , respectively. In the following, we always assume that

lim t 0 + f ( t , u , v , w ) =,u P + ,v,wP
(3)

and

lim u θ + f ( t , u , v , w ) =,t J + ,v,wP,
(4)

(where u θ + means u>θ, u0), i.e., f(t,u,v,w) is singular at t=0 and u=θ. We also assume that

lim u θ + I k ( u ) =(k=1,2,3,),
(5)

i.e., I k (u) (k=1,2,3,) are singular at u=θ.

Let PC[J,E] = {u:u is a map from J into E such that u(t) is continuous at t t k , left continuous at t= t k , and u( t k + ) exists, k=1,2,3,} and BPC[J,E]={uPC[J,E]: sup t J u(t)<}. It is clear that BPC[J,E] is a Banach space with norm

u B = sup t J u ( t ) .

Let BPC[J,P]={uBPC[J,E]:u(t)θ,tJ} and Q={uBPC[J,P]:u(t) β 1 (1γ)u(s),t,sJ}. Obviously, BPC[J,P] and Q are two cones in space BPC[J,E] and QBPC[J,P]. uBPC[J,P] C 1 [ J + ,E] is called a positive solution of the infinite three-point boundary value problem (1) if u(t)>θ for tJ and u(t) satisfies (1). Let Q + ={uQ: u B >0} and Q p q ={uQ:p u B q} for q>p>0.

2 Several lemmas

Let us list some conditions.

( H 1 ) sup t J 0 t K(t,s)ds<, sup t J 0 H(t,s)ds< and

lim t t 0 | H ( t , s ) H ( t , s ) | ds=0,tJ.

In this case, let

k = sup t J 0 t K(t,s)ds, h = sup t J 0 H(t,s)ds.

( H 2 ) There exist aC[ J + , R + ] and gC[ R + + × R + × R + , R + ] such that

f ( t , u , v , w ) a(t)g ( u , v , w ) ,t J + ,u P + ,v,wP,

and

a = 0 a(t)dt<,

where R + + ={x R + :x>0}.

( H 3 ) There exist γ k 0 (k=1,2,3,) and FC[ R + + , R + ] such that

I k ( u ) γ k F ( u ) ,u P + (k=1,2,3,),

and

γ = k = 1 γ k <.

( H 4 ) For any t J + and r>p>0, f(t, P p r , P r , P r )={f(t,u,v,w):u P p r ,v,w P r } and I k ( P p r )={ I k (u):u P p r } (k=1,2,3,) are relatively compact in E, where P r ={uP:ur} and P p r ={uP:pur}.

Remark Obviously, condition ( H 4 ) is satisfied automatically when E is finite dimensional.

Remark It is clear: If condition ( H 1 ) is satisfied, then the operators T and S defined by (2) are bounded linear operators from BPC[J,E] into BPC[J,E] and T k , S h ; moreover, we have T(BPC[J,P])BPC[J,P] and S(BPC[J,P])BPC[J,P].

We shall reduce the infinite three-point boundary value problem (1) to an impulsive integral equation. To this end, we consider the operator A defined by

( A u ) ( t ) = 1 β + γ 1 { η f ( s , u ( s ) , ( T u ) ( s ) , ( S u ) ( s ) ) d s + ( 1 γ ) 0 η f ( s , u ( s ) , ( T u ) ( s ) , ( S u ) ( s ) ) d s + k = m I k ( u ( t k ) ) + ( 1 γ ) k = 1 m 1 I k ( u ( t k ) ) } + 0 t f ( s , u ( s ) , ( T u ) ( s ) , ( S u ) ( s ) ) d s + 0 < t k < t I k ( u ( t k ) ) , t J .
(6)

In what follows, we write J 1 =[0, t 1 ], J k =( t k 1 , t k ] (k=2,3,4,).

Lemma 1 Let cone P be normal and conditions ( H 1 )-( H 4 ) be satisfied. Then operator A defined by (6) is a continuous operator from Q + into Q; moreover, for any q>p>0, A( Q p q ) is relatively compact.

Proof Let u Q + and u B =r. Then r>0 and

u(t) β 1 (1γ)u(s)θ,t,sJ,

so,

u ( t ) N 1 β 1 (1γ) u B ,tJ,
(7)

where N denotes the normal constant of cone P, and consequently,

N 1 β 1 (1γ)r u ( t ) r,tJ.
(8)

By condition ( H 2 ) and (8), we have

f ( t , u ( t ) , ( T u ) ( t ) , ( S u ) ( t ) ) Ma(t),tJ,
(9)

where

M=max { g ( x , y , z ) : N 1 β 1 ( 1 γ ) r x r , 0 y k r , 0 z h r } ,

which implies the convergence of the infinite integral

0 f ( t , u ( t ) , ( T u ) ( t ) , ( S u ) ( t ) ) dt
(10)

and

0 f ( t , u ( t ) , ( T u ) ( t ) , ( S u ) ( t ) ) d t 0 f ( t , u ( t ) , ( T u ) ( t ) , ( S u ) ( t ) ) dtM a .
(11)

On the other hand, by condition ( H 3 ) and (8), we have

I k ( u ( t k ) ) D γ k (k=1,2,3,),
(12)

where

D=max { F ( x ) : N 1 β 1 ( 1 γ ) r x r } ,

which implies the convergence of the infinite series

k = 1 I k ( u ( t k ) )
(13)

and

k = 1 I k ( u ( t k ) ) k = 1 I k ( u ( t k ) ) D γ .
(14)

It follows from (6), (11), and (14) that

( A u ) ( t ) 1 β + γ 1 { η f ( s , u ( s ) , ( T u ) ( s ) , ( S u ) ( s ) ) d s + ( 1 γ ) 0 η f ( s , u ( s ) , ( T u ) ( s ) , ( S u ) ( s ) ) d s + k = m I k ( u ( t k ) ) + ( 1 γ ) k = 1 m 1 I k ( u ( t k ) ) } + 0 t f ( s , u ( s ) , ( T u ) ( s ) , ( S u ) ( s ) ) d s + 0 < t k < t I k ( u ( t k ) ) 1 β + γ 1 { 0 f ( s , u ( s ) , ( T u ) ( s ) , ( S u ) ( s ) ) d s + k = 1 I k ( u ( t k ) ) } + 0 f ( s , u ( s ) , ( T u ) ( s ) , ( S u ) ( s ) ) d s + k = 1 I k ( u ( t k ) ) = β + γ β + γ 1 { 0 f ( s , u ( s ) , ( T u ) ( s ) , ( S u ) ( s ) ) d s + k = 1 I k ( u ( t k ) ) } β + γ β + γ 1 ( M a + D γ ) , t J ,

which implies that AuBPC[J,P] and

A u B β + γ β + γ 1 ( M a + D γ ) .
(15)

Moreover, by (6), we have

( A u ) ( t ) 1 β + γ 1 { η f ( s , u ( s ) , ( T u ) ( s ) , ( S u ) ( s ) ) d s + ( 1 γ ) 0 η f ( s , u ( s ) , ( T u ) ( s ) , ( S u ) ( s ) ) d s + k = m I k ( u ( t k ) ) + ( 1 γ ) k = 1 m 1 I k ( u ( t k ) ) } , t J
(16)

and

( A u ) ( t ) 1 β + γ 1 { η f ( s , u ( s ) , ( T u ) ( s ) , ( S u ) ( s ) ) d s + ( 1 γ ) 0 η f ( s , u ( s ) , ( T u ) ( s ) , ( S u ) ( s ) ) d s + k = m I k ( u ( t k ) ) + ( 1 γ ) k = 1 m 1 I k ( u ( t k ) ) } + 0 f ( s , u ( s ) , ( T u ) ( s ) , ( S u ) ( s ) ) d s + k = 1 I k ( u ( t k ) ) , t J .
(17)

It is clear,

0 f ( s , u ( s ) , ( T u ) ( s ) , ( S u ) ( s ) ) d s + k = 1 I k ( u ( t k ) ) 1 1 γ { η f ( s , u ( s ) , ( T u ) ( s ) , ( S u ) ( s ) ) d s + ( 1 γ ) 0 η f ( s , u ( s ) , ( T u ) ( s ) , ( S u ) ( s ) ) d s + k = m I k ( u ( t k ) ) + ( 1 γ ) k = 1 m 1 I k ( u ( t k ) ) } ,
(18)

so, (17) and (18) imply

( A u ) ( t ) { 1 β + γ 1 + 1 1 γ } { η f ( s , u ( s ) , ( T u ) ( s ) , ( S u ) ( s ) ) d s + ( 1 γ ) 0 η f ( s , u ( s ) , ( T u ) ( s ) , ( S u ) ( s ) ) d s + k = m I k ( u ( t k ) ) + ( 1 γ ) k = 1 m 1 I k ( u ( t k ) ) } , t J .
(19)

It follows from (16) and (19) that

( A u ) ( t ) 1 β + γ 1 ( 1 β + γ 1 + 1 1 γ ) 1 ( A u ) ( s ) = β 1 ( 1 γ ) ( A u ) ( s ) , t , s J .
(20)

Hence, AuQ, i.e., A maps Q + into Q.

Now, we are going to show that A is continuous. Let u n , u ¯ Q + , u n u ¯ B 0 (n). Write u ¯ B =2 r ¯ ( r ¯ >0) and we may assume that

r ¯ u n B 3 r ¯ (n=1,2,3,).

So, by (7),

N 1 β 1 (1γ) r ¯ u n ( t ) 3 r ¯ ,tJ(n=1,2,3,)
(21)

and

N 1 β 1 (1γ) r ¯ <2 N 1 β 1 (1γ) r ¯ u ¯ ( t ) 2 r ¯ <3 r ¯ ,tJ.
(22)

Similar to (15), it is easy to get

A u n A u ¯ B β + γ β + γ 1 { 0 f ( s , u n ( s ) , ( T u n ) ( s ) , ( S u n ) ( s ) ) f ( s , u ¯ ( s ) , ( T u ¯ ) ( s ) , ( S u ¯ ) ( s ) ) d s + k = 1 I k ( u n ( t k ) ) I k ( u ¯ ( t k ) ) } ( n = 1 , 2 , 3 , ) .
(23)

It is clear that

f ( t , u n ( t ) , ( T u n ) ( t ) , ( S u n ) ( t ) ) f ( t , u ¯ ( t ) , ( T u ¯ ) ( t ) , ( S u ¯ ) ( t ) ) as n,tJ,
(24)

and, similar to (9) and observing (21) and (22), we have

f ( t , u n ( t ) , ( T u n ) ( t ) , ( S u n ) ( t ) ) f ( t , u ¯ ( t ) , ( T u ¯ ) ( t ) , ( S u ¯ ) ( t ) ) 2 M ¯ a ( t ) = d ( t ) , t J ( n = 1 , 2 , 3 , ) ; d L [ J , R + ] ,
(25)

where

M ¯ =max { g ( x , y , z ) : N 1 β 1 ( 1 γ ) r ¯ x 3 r ¯ , 0 y 3 k r ¯ , 0 z 3 h r ¯ } .

It follows from (24), (25), and the dominated convergence theorem that

lim n 0 f ( t , u n ( t ) , ( T u n ) ( t ) , ( S u n ) ( t ) ) f ( t , u ¯ ( t ) , ( T u ¯ ) ( t ) , ( S u ¯ ) ( t ) ) dt=0.
(26)

On the other hand, for any ϵ>0, we can choose a positive integer j such that

D ¯ k = j + 1 γ k <ϵ,
(27)

where

D ¯ =max { F ( x ) : N 1 β 1 ( 1 γ ) r ¯ x 3 r ¯ } .

And then, choose an positive integer n 0 such that

k = 1 j I k ( u n ( t k ) ) I k ( u ¯ ( t k ) ) <ϵ,n> n 0 .
(28)

From (27), (28), and observing condition ( H 3 ) and (21), (22), we get

k = 1 I k ( u n ( t k ) ) I k ( u ¯ ( t k ) ) <ϵ+2 D ¯ k = j + 1 γ k <3ϵ,n> n 0 ,

hence,

lim n k = 1 I k ( u n ( t k ) ) I k ( u ¯ ( t k ) ) =0.
(29)

It follows from (23), (26), and (29) that A u n A u ¯ B 0 as n, and the continuity of A is proved.

Finally, we prove that A( Q p q ) is relatively compact, where q>p>0 are arbitrarily given. Let v n A( Q p q ) (n=1,2,3,). Then, by (7),

N 1 β 1 (1γ)p v n ( t ) q,tJ(n=1,2,3,).
(30)

Similar to (9), (12), (15), and observing (30), we have

f ( t , v n ( t ) , ( T v n ) ( t ) , ( S v n ) ( t ) ) M 1 a(t),t J + (n=1,2,3,),
(31)
I k ( v n ( t k ) ) D 1 γ k (k,n=1,2,3,)
(32)

and

A v n B β + γ β + γ 1 ( M 1 a + D 1 γ ) (n=1,2,3,),
(33)

where

M 1 =max { g ( x , y , z ) : N 1 β 1 ( 1 γ ) p x q , 0 y k q , 0 z h q }

and

D 1 =max { F ( x ) : N 1 β 1 ( 1 γ ) p x q } .

Consider J i =( t i 1 , t i ] for any fixed i. By (6) and (31), we have

( A v n ) ( t ) ( A v n ) ( t ) t t f ( s , v n ( s ) , ( T v n ) ( s ) , ( S v n ) ( s ) ) d s M 1 t t a ( s ) d s , t , t J i , t > t ( n = 1 , 2 , 3 , ) ,
(34)

which implies that the functions { w n (t)} (n=1,2,3,) defined by

w n (t)={ ( A v n ) ( t ) , t J i = ( t i 1 , t i ] , ( A v n ) ( t i 1 + ) , t = t i 1 (n=1,2,3,)
(35)

((A u n )( t i 1 + ) denotes the right limit of (A u n )(t) at t= t i 1 ) are equicontinuous on J ¯ i =[ t i 1 , t i ]. On the other hand, for any ϵ>0, choose a sufficiently large τ>η and a sufficiently large positive integer j>m such that

M 1 τ a(s)ds<ϵ, D 1 k = j + 1 γ k <ϵ.
(36)

We have, by (35), (6), (31), (32), and (36),

w n ( t ) = 1 β + γ 1 { η τ f ( s , v n ( s ) , ( T v n ) ( s ) , ( S v n ) ( s ) ) d s + τ f ( s , v n ( s ) , ( T v n ) ( s ) , ( S v n ) ( s ) ) d s + ( 1 γ ) 0 η f ( s , v n ( s ) , ( T v n ) ( s ) , ( S v n ) ( s ) ) d s + k = m j I k ( v n ( t k ) ) + k = j + 1 I k ( v n ( t k ) ) + ( 1 γ ) k = 1 m 1 I k ( v n ( t k ) ) } + 0 t f ( s , v n ( s ) , ( T v n ) ( s ) , ( S v n ) ( s ) ) d s + k = 1 i 1 I k ( v n ( t k ) ) , t J ¯ i ( n = 1 , 2 , 3 , )
(37)

and

τ f ( s , v n ( s ) , ( T v n ) ( s ) , ( S v n ) ( s ) ) d s <ϵ(n=1,2,3,),
(38)
k = j + 1 I k ( v n ( t k ) ) <ϵ(n=1,2,3,).
(39)

It follows from (37), (38), (39), and ([8], Theorem 1.2.3) that

α ( W ( t ) ) 1 β + γ 1 { 2 η τ α ( f ( s , V ( s ) , ( T V ) ( s ) , ( S V ) ( s ) ) ) d s + 2 ϵ + 2 ( 1 γ ) 0 η α ( f ( s , V ( s ) , ( T V ) ( s ) , ( S V ) ( s ) ) ) d s + k = m j α ( I k ( V ( t k ) ) ) + 2 ϵ + ( 1 γ ) k = 1 m 1 α ( I k ( V ( t k ) ) ) } + 2 0 t α ( f ( s , V ( s ) , ( T V ) ( s ) , ( S V ) ( s ) ) ) d s + k = 1 i 1 α ( I k ( V ( t k ) ) ) , t J ¯ i ,
(40)

where W(t)={ w n (t):n=1,2,3,}, V(s)={ v n (s):n=1,2,3,}, (TV)(s)={(T v n )(s):n=1,2,3,}, (SV)(s)={(S v n )(s):n=1,2,3,} and α(U) denotes the Kuratowski measure of noncompactness of bounded set UE (see [[8], Section 1.2]). Since V(s) P p q and (TV)(s),(SV)(s) P q for sJ, where p = N 1 β 1 (1γ)p and q =max{q, k q, h q}, we see that, by condition ( H 4 ),

α ( f ( s , V ( s ) , ( T V ) ( s ) , ( S V ) ( s ) ) ) =0,sJ
(41)

and

α ( I k ( V ( t k ) ) ) =0(k=1,2,3,).
(42)

It follows from (40) to (42) that

α ( W ( t ) ) 4 ϵ β + γ 1 ,t J ¯ i ,

which implies by virtue of the arbitrariness of ϵ that α(W(t))=0 for t J ¯ i .

By the Ascoli-Arzela theorem (see [[8], Theorem 1.2.5]), we conclude that W={ w n :n=1,2,3,} is relatively compact in C[ J ¯ i ,E], hence, { w n (t)} has a subsequence which is convergent uniformly on J ¯ i , so, {(A v n )(t)} has a subsequence which is convergent uniformly on J i . Since i may be any positive integer, so, by diagonal method, we can choose a subsequence {(A v n i )(t)} of {(A v n )(t)} such that {(A v n i )(t)} is convergent uniformly on each J k (k=1,2,3,). Let

lim i (A v n i )(t)=w(t),tJ.

It is clear that wPC[J,P]. By (33), we have

A v n i B β + γ β + γ 1 ( M 1 a + D 1 γ ) (i=1,2,3,),

which implies that wBPC[J,P] and

w B β + γ β + γ 1 ( M 1 a + D 1 γ ) .

Let ϵ>0 be arbitrarily given and choose a sufficiently large positive number τ such that

M 1 τ a(s)ds+ D 1 t k τ γ k <ϵ.
(43)

For any τ<t<, we have, by (6),

( A v n i ) ( t ) ( A v n i ) ( τ ) = τ t f ( s , v n i ( s ) , ( T v n i ) ( s ) , ( S v n i ) ( s ) ) d s + τ t k < t I k ( v n i ( t ) ) ( i = 1 , 2 , 3 , ) ,

which implies by virtue of (31), (32), and (43) that

( A v n i ) ( t ) ( A v n i ) ( τ ) M 1 τ t a(s)ds+ D 1 τ t k < t γ k <ϵ(i=1,2,3,).
(44)

Letting i in (44), we get

w ( t ) w ( τ ) ϵ,t>τ.
(45)

On the other hand, since {(A v n i )(t)} converges uniformly to w(t) on [0,τ] as i, there exists a positive integer i 0 such that

( A v n i ) ( t ) w ( t ) <ϵ,t[0,τ],i> i 0 .
(46)

It follows from (44) to (46) that

( A v n i ) ( t ) w ( t ) ( A v n i ) ( t ) ( A v n i ) ( τ ) + ( A v n i ) ( τ ) w ( τ ) + w ( τ ) w ( t ) < 3 ϵ , t > τ , i > i 0 .
(47)

By (46) and (47), we have

A v n i w B 3ϵ,i> i 0 ,

hence, A v n i w B 0 as i, and the relative compactness of A( Q p q ) is proved. □

Lemma 2 Let cone P be normal and conditions ( H 1 )-( H 4 ) be satisfied. Then u Q + C 1 [ J + ,E] is a positive solution of the infinite three-point boundary value problem (1) if and only if u Q + is a solution of the following impulsive integral equation:

u ( t ) = 1 β + γ 1 { η f ( s , u ( s ) , ( T u ) ( s ) , ( S u ) ( s ) ) d s + ( 1 γ ) 0 η f ( s , u ( s ) , ( T u ) ( s ) , ( S u ) ( s ) ) d s + k = m I k ( u ( t k ) ) + ( 1 γ ) k = 1 m 1 I k ( u ( t k ) ) } + 0 t f ( s , u ( s ) , ( T u ) ( s ) , ( S u ) ( s ) ) d s + 0 < t k < t I k ( u ( t k ) ) , t J ,
(48)

i.e., u is a fixed point of operator A defined by (6) in Q + .

Proof For uPC[J,E] C 1 [ J + ,E], it is easy to get the following formula:

u(t)=u(0)+ 0 t u (s)ds+ 0 < t k < t [ u ( t k + ) u ( t k ) ] ,tJ.
(49)

Let u Q + C 1 [ J + ,E] be a positive solution of the infinite three-point boundary value problem (1). By (1) and (49), we have

u(t)=u(0)+ 0 t f ( s , u ( s ) , ( T u ) ( s ) , ( S u ) ( s ) ) ds+ 0 < t k < t I k ( u ( t k ) ) ,tJ.
(50)

We have shown in the proof of Lemma 1 that the infinite integral (10) and the infinite series (13) are convergent, so, by taking limits as t in both sides of (50), we get

u()=u(0)+ 0 f ( s , u ( s ) , ( T u ) ( s ) , ( S u ) ( s ) ) ds+ k = 1 I k ( u ( t k ) ) .
(51)

On the other hand, by (1) and (50), we have

u()=γu(η)+βu(0)
(52)

and

u(η)=u(0)+ 0 η f ( s , u ( s ) , ( T u ) ( s ) , ( S u ) ( s ) ) ds+ k = 1 m 1 I k ( u ( t k ) ) .
(53)

It follows from (51) to (53) that

u ( 0 ) = 1 β + γ 1 { η f ( s , u ( s ) , ( T u ) ( s ) , ( S u ) ( s ) ) d s + ( 1 γ ) 0 η f ( s , u ( s ) , ( T u ) ( s ) , ( S u ) ( s ) ) d s + k = m I k ( u ( t k ) ) + ( 1 γ ) k = 1 m 1 I k ( u ( t k ) ) } ,

and, substituting it into (50), we see that u(t) satisfies equation (48), i.e., u=Au.

Conversely, assume that u Q + is a solution of Equation (48). We have, by (48),

u ( 0 ) = 1 β + γ 1 { η f ( s , u ( s ) , ( T u ) ( s ) , ( S u ) ( s ) ) d s + ( 1 γ ) 0 η f ( s , u ( s ) , ( T u ) ( s ) , ( S u ) ( s ) ) d s + k = m I k ( u ( t k ) ) + ( 1 γ ) k = 1 m 1 I k ( u ( t k ) ) }
(54)

and

u ( η ) = 1 β + γ 1 { η f ( s , u ( s ) , ( T u ) ( s ) , ( S u ) ( s ) ) d s + ( 1 γ ) 0 η f ( s , u ( s ) , ( T u ) ( s ) , ( S u ) ( s ) ) d s + k = m I k ( u ( t k ) ) + ( 1 γ ) k = 1 m 1 I k ( u ( t k ) ) } + 0 η f ( s , u ( s ) , ( T u ) ( s ) , ( S u ) ( s ) ) d s + k = 1 m 1 I k ( u ( t k ) ) .
(55)

Moreover, by taking limits as t in (48), we see that u() exists and

u ( ) = 1 β + γ 1 { η f ( s , u ( s ) , ( T u ) ( s ) , ( S u ) ( s ) ) d s + ( 1 γ ) 0 η f ( s , u ( s ) , ( T u ) ( s ) , ( S u ) ( s ) ) d s + k = m I k ( u ( t k ) ) + ( 1 γ ) k = 1 m 1 I k ( u ( t k ) ) } + 0 f ( s , u ( s ) , ( T u ) ( s ) , ( S u ) ( s ) ) d s + k = 1 I k ( u ( t k ) ) .
(56)

It follows from (54) to (56) that

γu(η)+βu(0)=u().

On the other hand, direct differentiation of (48) gives

u (t)=f ( t , u ( t ) , ( T u ) ( t ) , ( S u ) ( t ) ) ,t J + ,

and, it is clear, by (48),

Δu | t = t k = I k ( u ( t k ) ) (k=1,2,3,).

Hence, u C 1 [ J + ,E] and u(t) satisfies (1). Since u Q + , so (7) holds and u B >0, hence u(t)>θ for tJ. Consequently, u(t) is a positive solution of the infinite three-point boundary value problem (1). □

Lemma 3 (The fixed-point theorem of cone expansion and compression with norm type; see [[6], Theorem 3] or [[7], Theorem 2.3.4])

Let P be a cone in real Banach space E and Ω 1 , Ω 2 be two bounded open sets in E such that θ Ω 1 , Ω ¯ 1 Ω 2 , and operator A:P( Ω ¯ 2 Ω 1 )P be completely continuous, where θ denotes the zero element of E and Ω ¯ i denotes the closure of Ω i (i=1,2). Suppose that one of the following two conditions is satisfied:

(a)Axx,xP Ω 1 ;Axx,xP Ω 2 ,

where Ω i denotes the boundary of Ω i (i=1,2).

(b)Axx,xP Ω 1 ;Axx,xP Ω 2 .

Then A has at least one fixed point in P( Ω ¯ 2 Ω 1 ).

Remark 1 Lemma 3 is different from the Krasnoselskii fixed-point theorem of cone expansion and compression (see [[9], Theorem 44.1]). In Krasnoselskii’s theorem, the condition corresponding to (a) is

( a ) Axx,xP Ω 1 ,Axx,xP Ω 2 .

It is clear, conditions (a) and (a′) are independent each other. On the other hand, in Krasnoselskii’s theorem, Ω 1 and Ω 2 are balls with center θ.

3 Main theorems

Let us list more conditions.

( H 5 ) There exist u 0 P + , bC[ J + , R + + ] and τC[ P + , R + ] such that

f(t,u,v,w)b(t)τ(u) u 0 ,t J + ,u P + ,v,wP,

and

τ ( u ) u as u P + ,u,

and

b = 0 b(t)dt<.

Remark 2 Condition ( H 5 ) means that f(t,u,v,w) is superlinear with respect to u.

( H 6 ) There exist u 1 P + , cC[ J + , R + + ] and σC[ P + , R + ] such that

f(t,u,v,w)c(t)σ(u) u 1 ,t J + ,u P + ,v,wP,

and

σ(u)as u P + ,u0,

and

c = 0 c(t)dt<.

Theorem 1 Let cone P be normal and conditions ( H 1 )-( H 6 ) be satisfied. Assume that there exists a ξ>0 such that

N ( β + γ ) β + γ 1 ( M ξ a + D ξ γ ) <ξ,
(57)

where N denotes the normal constant of P, and

M ξ =max { g ( x , y , z ) : N 1 β 1 ( 1 γ ) ξ x ξ , 0 y k ξ , 0 z h ξ } ,
(58)
D ξ =max { F ( x ) : N 1 β 1 ( 1 γ ) ξ x ξ }
(59)

(for g(x,y,z), F(x), a and γ ; see conditions ( H 2 ) and ( H 3 )). Then the infinite three-point boundary value problem (1) has at least two positive solutions u , u Q + C 1 [ J + ,E] such that 0< u B <ξ< u B .

Proof By Lemma 1 and Lemma 2, operator A defined by (6) is continuous from Q + into Q and we need to prove that A has two fixed points u and u in Q + such that 0< u B <ξ< u B .

By condition ( H 5 ), there exists a r 1 >0 such that

τ(u) β ( β + γ 1 ) N 2 ( 1 γ ) 2 b u 0 u,u P + ,u r 1 ,
(60)

so,

f(t,u,v,w) β ( β + γ 1 ) N 2 u ( 1 γ ) 2 b u 0 b(t) u 0 ,t J + ,u P + ,v,wP,u r 1 .
(61)

Choose

r 2 >max { N β ( 1 γ ) 1 r 1 , ξ } .
(62)

For uQ, u B = r 2 , we have by (7) and (62),

u ( t ) N 1 β 1 (1γ) u B = N 1 β 1 (1γ) r 2 > r 1 ,tJ,
(63)

so, (6), (63), (61), and (7) imply

( A u ) ( t ) 1 γ β + γ 1 ( 0 f ( s , u ( s ) , ( T u ) ( s ) , ( S u ) ( s ) ) d s ) β N 2 ( 1 γ ) b u 0 ( 0 u ( s ) b ( s ) d s ) u 0 N u B b u 0 ( 0 b ( s ) d s ) u 0 = N u B u 0 u 0 , t J ,
(64)

and consequently,

A u B u B ,uQ, u B = r 2 .
(65)

By condition ( H 6 ), there exists r 3 >0 such that

σ(u) ( β + γ 1 ) N ξ ( 1 γ ) c u 1 ,u P + ,0<u< r 3 ,
(66)

so,

f(t,u,v,w) ( β + γ 1 ) N ξ ( 1 γ ) c u 1 c(t) u 1 ,t J + ,u P + ,v,wP,0<u< r 3 .
(67)

Choose

0< r 4 <min{ r 3 ,ξ}.
(68)

For uQ, u B = r 4 , we have by (68) and (7),

r 3 > u ( t ) N 1 β 1 (1γ) u B = N 1 β 1 (1γ) r 4 >0,
(69)

so, we get by (6), (69), and (67),

( A u ) ( t ) 1 γ β + γ 1 ( 0 f ( s , u ( s ) , ( T u ) ( s ) , ( S u ) ( s ) ) d s ) N ξ c u 1 ( 0 c ( s ) d s ) u 1 = N ξ u 1 u 1 , t J ,

which implies

( A u ) ( t ) ξ> r 4 ,tJ,

and consequently,

A u B > u B ,uQ, u B = r 4 .
(70)

On the other hand, for uQ, u B =ξ, by condition ( H 2 ), condition ( H 3 ), (58), and (59), we have

f ( t , u ( t ) , ( T u ) ( t ) , ( S u ) ( t ) ) M ξ a(t),t J +
(71)

and

I k ( u ( t k ) ) D ξ γ k (k=1,2,3,).
(72)

It is clear, by (17),

(Au)(t) β + γ β + γ 1 ( 0 f ( s , u ( s ) , ( T u ) ( s ) , ( S u ) ( s ) ) d s + k = 1 I k ( u ( t k ) ) ) ,tJ.
(73)

It follows from (71) to (73) that

A u B N ( β + γ ) β + γ 1 ( M ξ a + D ξ γ ) .
(74)

Thus, (74) and (57) imply

A u B < u B ,uQ, u B =ξ.
(75)

From (62) and (68), we know 0< r 4 <ξ< r 2 , and by Lemma 1, A: Q r 4 r 2 Q is completely continuous, where Q r 4 r 2 ={uQ: r 4 u B r 2 }, hence, (65), (70), (75), and Lemma 3 imply that A has two fixed points u , u Q + such that r 4 < u B <ξ< u B r 2 . The proof is complete. □

Theorem 2 Let cone P be normal and conditions ( H 1 )-( H 4 ) and ( H 6 ) be satisfied. Assume that

g ( x , y , z ) x + y + z 0 as x
(76)

uniformly for y,z R + , and

F ( x ) x 0 as x
(77)

(for g(x,y,z) and F(x), see conditions ( H 2 ) and ( H 3 )). Then the infinite three-point boundary value problem (1) has at least one positive solution u Q + C 1 [ J + ,E].

Proof As in the proof of Theorem 1, we can choose r 4 >0 such that (70) holds (in this case, we put ξ=1 in (66) and (68)). On the other hand, by (76) and (77), there exists r 5 >0 such that

g(x,y,z) ϵ 0 (x+y+z),x> r 5 ,y0,z0
(78)

and

F(x) ϵ 0 x,x> r 5 ,
(79)

where

ϵ 0 = β + γ 1 N ( β + γ ) [ ( 1 + k + h ) a + γ ] .
(80)

Choose

r 6 >max { N β ( 1 γ ) 1 r 5 , r 4 } .
(81)

For uQ, u B = r 6 , we have by (7) and (81),

u ( t ) N 1 β 1 (1γ) r 6 > r 5 ,tJ

so, (78) and (79) imply

g ( u ( t ) , ( T u ) ( t ) , ( S u ) ( t ) ) ϵ 0 ( u ( t ) + ( T u ) ( t ) + ( S u ) ( t ) ) ϵ 0 ( 1 + k + h ) r 6 , t J
(82)

and

F ( u ( t k ) ) ϵ 0 u ( t k ) ϵ 0 r 6 (k=1,2,3,).
(83)

It follows from (73), conditions ( H 2 ), condition ( H 3 ), (82), (83), and (80) that

( A u ) ( t ) N ( β + γ ) β + γ 1 { ϵ 0 ( 1 + k + h ) r 6 0 a ( s ) d s + ϵ 0 r 6 k = 1 γ k } = N ( β + γ ) ϵ 0 r 6 β + γ 1 { ( 1 + k + h ) a + γ } = r 6 , t J ,

and consequently,

A u B u B ,uQ, u B = r 6 .
(84)

Since r 6 > r 4 by virtue of (81), we conclude from (70), (84), and Lemma 3 that A has a fixed point u Q + such that r 4 < u B r 6 . The theorem is proved. □

Example 1

Consider the infinite system of scalar first-order impulsive singular integro-differential equations of mixed type on the half line:

{ u n ( t ) = e 2 t 20 n 2 t { 1 8 ( u n + 1 ( t ) + m = 1 u m ( t ) ) 2 + 1 9 ( m = 1 u m ( t ) ) 1 } u n ( t ) = + e 3 t 18 n 3 t { ( 0 t e ( t + 1 ) s u n ( s ) d s ) 2 + 1 2 ( 0 u n + 2 ( s ) d s ( 1 + t + s ) 2 ) 3 } , 0 < t < , t k ( k = 1 , 2 , 3 , ; n = 1 , 2 , 3 , ) , Δ u n | t = k = 5 k 16 n 2 { u 2 n ( k ) + 1 3 ( m = 1 u m ( k ) ) 1 2 } ( k = 1 , 2 , 3 , ; n = 1 , 2 , 3 , ) , u n ( ) = 1 2 u n ( 9 2 ) + 6 u n ( 0 ) ( n = 1 , 2 , 3 , ) .
(85)

Conclusion Infinite system (85) has at least two positive solutions { u n (t)} (n=1,2,3,) and { u n (t)} (n=1,2,3,) such that

0 < inf 0 t < n = 1 u n ( t ) sup 0 t < n = 1 u n ( t ) < 1 < sup 0 t < n = 1 u n ( t ) , inf 0 t < n = 1 u n ( t ) > 0 .

Proof Let E= l 1 ={u=( u 1 ,, u n ,): n = 1 | u n |<} with norm u= n = 1 | u n | and P={( u 1 ,, u n ,): u n 0,n=1,2,3,}. Then P is a normal cone in E with normal constant N=1, and infinite system (85) can be regarded as an infinite three-point boundary value problem of form (1). In this situation, u=( u 1 ,, u n ,), v=( v 1 ,, v n ,), w=( w 1 ,, w n ,), t k =k (k=1,2,3,), K(t,s)= e ( t + 1 ) s , H(t,s)= ( 1 + t + s ) 2 , η= 9 2 , γ= 1 2 , β=6, f=( f 1 ,, f n ,) and I k =( I k 1 ,, I k n ,), in which

f n ( t , u , v , w ) = e 2 t 20 n 2 t { 1 8 ( u n + 1 + m = 1 u m ) 2 + 1 9 ( m = 1 u m ) 1 } + e 3 t 18 n 3 t ( v n 2 + 1 2 w n + 2 3 ) , t J + = ( 0 , ) , u P + = { u P : u > 0 } , v , w P ( n = 1 , 2 , 3 , )
(86)

and

I k n (u)= 5 k 16 n 2 { u 2 n + 1 3 ( m = 1 u m ) 1 2 } ,u P + (k=1,2,3,;n=1,2,3,).
(87)

It is easy to see that fC[ J + × P + ×P×P,P], I k C[ P + ,P] (k=1,2,3,) and condition ( H 1 ) is satisfied and k 1, h 1. We have, by (86),

0 f n ( t , u , v , w ) e 2 t 20 n 2 t { 1 8 ( 2 u ) 2 + 1 9 u 1 } + e 3 t 18 n 3 t ( v 2 + 1 2 w 3 ) e 2 t n 2 t ( 1 40 u 2 + 1 180 u 1 + 1 18 v 2 + 1 36 w 3 ) , t J + , u P + , v , w P ( n = 1 , 2 , 3 , ) ,
(88)

so, observing the inequality n = 1 1 n 2 <2, we get

f ( t , u , v , w ) = n = 1 f n ( t , u , v , w ) e 2 t t ( 1 20 u 2 + 1 90 u 1 + 1 9 v 2 + 1 18 w 3 ) , t J + , u P + , v , w P ,

which implies that condition ( H 2 ) is satisfied for

a(t)= e 2 t t

and

g(x,y,z)= 1 20 x 2 + 1 90 x + 1 9 y 2 + 1 18 z 3

with

a = 0 e 2 t t dt< 0 1 d t t + 1 e 2 t dt=2+ 1 2 e 2 < 29 14 .

By (87), we have

0 I k n (u) 5 k 16 n 2 ( u + 1 3 u 1 2 ) ,u P + (k=1,2,3,;n=1,2,3,),
(89)

so,

I k ( u ) 1 8 5 k ( u + 1 3 u 1 2 ) ,u P + (k=1,2,3,),

which implies that condition ( H 3 ) is satisfied for γ k = 1 8 5 k ( γ = 1 32 ) and

F(x)=x+ 1 3 x .

On the other hand, (86) implies

f n (t,u,v,w) e 2 t 160 n 2 t u 2 ,t J + ,u P + ,v,wP(n=1,2,3,)

and

f n (t,u,v,w) e 2 t 180 n 2 t u 1 ,t J + ,u P + ,v,wP(n=1,2,3,),
(90)

so, we see that condition ( H 5 ) is satisfied for b(t)= e 2 t 160 t ( b < 29 2 , 240 ), τ(u)= u 2 and u 0 =(1,, 1 n 2 ,) and condition ( H 6 ) is satisfied for c(t)= e 2 t 180 t ( c < 29 2 , 520 ), σ(u)= u 1 and u 1 =(1,, 1 n 2 ,). In addition, from (90), we have

f ( t , u , v , w ) ( n = 1 1 n 2 ) e 2 t 160 t u 1 > e 2 t 160 t u 1 ,t J + ,u P + ,v,wP,

which implies that (3) and (4) hold, i.e., f(t,u,v,w) is singular at t=0 and u=θ. Moreover, from (87), we get

I k n (u) 5 k 48 n 2 u 1 2 ,u P + (k=1,2,3,;n=1,2,3,),

and so,

I k ( u ) ( n = 1 1 n 2 ) 5 k 48 u 1 2 > 5 k 48 u 1 2 ,u P + (k=1,2,3,),

which implies that (5) holds, i.e., I k (u) (k=1,2,3,) are singular at u=θ. Now, we check that condition ( H 4 ) is satisfied. Let t J + and r>p>0 be fixed, and { z ( m ) } be any sequence in f(t, P p r , P r , P r ), where z ( m ) =( z 1 ( m ) ,, z n ( m ) ,). Then, we have, by (86) and (88),

0 z n ( m ) e 2 t n 2 t ( 29 360 r 2 + 1 180 p + 1 36 r 3 ) (n,m=1,2,3,).
(91)

So, { z n ( m ) } is bounded, and, by diagonal method, we can choose a subsequence { m i }{m} such that

z n ( m i ) z ¯ n as i(n=1,2,3,),
(92)

which implies by virtue of (91) that

0 z ¯ n e 2 t n 2 t ( 29 360 r 2 + 1 180 p + 1 36 r 3 ) (n=1,2,3,).
(93)

Consequently, z ¯ =( z ¯ 1 ,, z ¯ n ,) l 1 =E. Let ϵ>0 be given. Choose a positive integer n 0 such that

e 2 t t ( n = n 0 + 1 1 n 2 ) ( 29 360 r 2 + 1 180 p + 1 36 r 3 ) < ϵ 3 .
(94)

By (92), we see that there exists a positive integer i 0 such that

| z n ( m i ) z ¯ n | < ϵ 3 n 0 ,i> i 0 (n=1,2,, n 0 ).
(95)

It follows from (91) to (95) that

z ( m i ) z ¯ = n = 1 | z n ( m i ) z ¯ n | n = 1 n 0 | z n ( m i ) z ¯ n | + n = n 0 + 1 | z n ( m i ) | + n = n 0 + 1 | z ¯ n | < ϵ 3 + ϵ 3 + ϵ 3 = ϵ , i > i 0 ,

hence, z ( m i ) z ¯ in E as i. Thus, we have proved that f(t, P p r , P r , P r ) is relatively compact in E. Similarly, by using (89), we can prove that I k ( P p r ) is relatively compact in E. Hence, condition ( H 4 ) is satisfied. Finally, we check that inequality (57) is satisfied for ξ=1. In this case,

M 1 max { g ( x , y , z ) : 1 12 x 1 , 0 y 1 , 0 z 1 } 1 20 + 12 90 + 1 9 + 1 18 = 7 20

and

D 1 =max { F ( x ) : 1 12 x 1 } 1+ 2 3 3 <2.2,

so,

N ( β + γ ) β + γ 1 ( M 1 a + D 1 γ ) < 13 11 ( 7 20 × 29 14 + 2.2 × 1 32 ) = 1 , 651 1 , 760 <1,

i.e., inequality (57) is satisfied for ξ=1. Hence, our conclusion follows from Theorem 1. □

Example 2

Consider the infinite system of scalar first order impulsive singular integro-differential equations of mixed type on the half line:

{ u n ( t ) = 1 n 3 t 1 3 ( 1 + t ) 3 { u n ( t ) + 2 u n + 1 ( t ) + ( m = 1 u m ( t ) ) 2 } u n ( t ) = + 1 n 4 t 1 3 ( 1 + t ) 4 { ( 0 t u 2 n ( s ) d s 1 + t s + s 2 ) 1 2 + ( 0 e s sin 2 ( t s ) u 3 n ( s ) d s ) 1 3 } , 0 t < , t 2 k ( k = 1 , 2 , 3 , ; n = 1 , 2 , 3 , ) ; Δ u n | t = 2 k = e k n 2 ( u 2 n + 1 ( 2 k ) ) 1 3 + 2 k n 3 ( m = 1 u m ( 2 k ) ) 3 ( k = 1 , 2 , 3 , ; n = 1 , 2 , 3 , ) , 4 u n ( ) = 3 u n ( 7 ) + 2 u n ( 0 ) ( n = 1 , 2 , 3 , ) .
(96)

Conclusion Infinite system (96) has at least one positive solution { u n (t)} (n=1,2,3,) such that

inf 0 t < n = 1 u n (t)>0.

Proof Let E= l 1 ={u=( u 1 ,, u n ,): n = 1 | u n |<} with norm u= n = 1 | u n | and P={u=( u 1 ,, u n ,) l 1 : u n 0,n=1,2,3,}. Then P is a normal cone in E with normal constant N=1, and infinite system (96) can be regarded as an infinite three-point boundary value problem of form (1) in E. In this situation, u=( u 1 ,, u n ,), v=( v 1 ,, v n ,), w=( w 1 ,, w n ,), t k =2k (k=1,2,3,), K(t,s)= ( 1 + t s + s 2 ) 1 , H(t,s)= e s sin 2 (ts), η=7, γ= 3 4 , β= 1 2 , f=( f 1 ,, f n ,) and I k =( I k 1 ,, I k n ,), in which

f n ( t , u , v , w ) = 1 n 3 t 1 3 ( 1 + t ) 3 { u n + 2 u n + 1 + ( m = 1 u m ) 2 } + 1 n 4 t 1 3 ( 1 + t ) 4 ( v 2 n 1 2 + w 3 n 1 3 ) , t J + , u P + , v , w P ( n = 1 , 2 , 3 , )
(97)

and

I k n (u)= e k n 2 u 2 n + 1 1 3 + 2 k n 3 ( m = 1 u m ) 3 (k=1,2,3,;n=1,2,3,).
(98)

It is clear that fC[ J + × P + ×P×P,P], I k C[ P + ,P] (k=1,2,3,) and condition ( H 1 ) is satisfied and k π 2 , h 1. We have, by (97) and (98),

0 f n ( t , u , v , w ) 1 n 3 t 1 3 ( 1 + t ) 3 ( 3 u + u 2 + v 1 2 + w 1 3 ) , t J + , u P + , v , w P ( n = 1 , 2 , 3 , )

and

0 I k n (u) 2 k n 2 ( u 1 3 + u 3 ) ,u P + (k=1,2,3,;n=1,2,3,),

so, observing

n = 1 1 n 3 < n = 1 1 n 2 <2,

we get

f ( t , u , v , w ) 1 t 1 3 ( 1 + t ) 3 ( 2 3 u + 2 u 2 + 2 v 1 2 + 2 w 1 3 ) , t J + , u P + , v , w P

and

I k ( u ) 2 k + 1 ( u 1 3 + u 3 ) ,u P + (k=1,2,3,),

which imply that conditions ( H 2 ) is satisfied for

a(t)= 1 t 1 3 ( 1 + t ) 3

and

g(x,y,z)=2 3 x +2 x 2 +2 y 1 2 +2 z 1 3

with

a = 0 d t t 1 3 ( 1 + t ) 3 < 0 1 d t t 1 3 + 1 d t ( 1 + t ) 3 = 13 8

and ( H 3 ) is satisfied for γ k = 2 k + 1 ( γ =2) and

F(x)= x 1 3 + x 3 .

By (97), we have

f n (t,u,v,w) 1 n 3 t 1 3 ( 1 + t ) 3 u 2 ,t J + ,u P + ,v,wP(n=1,2,3,)
(99)

so, condition ( H 6 ) is satisfied for

c(t)= 1 t 1 3 ( 1 + t ) 3 ( c = a < 13 8 ) ,

σ(u)= u 2 and u 1 =(1,, 1 n 3 ,). Moreover, (99) implies

f ( t , u , v , w ) ( n = 1 1 n 3 ) 1 t 1 3 ( 1 + t ) 3 u 2 > 1 t 1 3 ( 1 + t ) 3 u 2 , t J + , u P + , v , w P ,

so, (3) and (4) are satisfied, i.e., f(t,u,v,w) is singular at t=0 and u=θ. Similarly, (98) implies

I k ( u ) ( n = 1 1 n 3 ) 2 k u 3 > 2 k u 3 ,u P + (k=1,2,3,),

so, (5) is satisfied, i.e., I k (u) (k=1,2,3,) are singular at u=θ. Similar to the discussion in Example 1, we can prove that f(t, P p r , P r , P r ) and I k ( P p r ) (for fixed t J + and r>p>0; k=1,2,3,) are relatively compact in E= l 1 , so, condition ( H 4 ) is satisfied. On the other hand, we have

0 < g ( x , y , z ) x + y + z = 2 3 ( x x + y + z ) 1 2 ( x + y + z ) 1 2 + x 2 ( x + y + z ) 1 + 2 ( y x + y + z ) 1 2 ( x + y + z ) 1 2 + 2 ( z x + y + z ) 1 3 ( x + y + z ) 2 3 0 2 3 x 1 2 + x 3 + 2 x 1 2 + 2 x 2 3 , x > 0 , y 0 , z 0 ,

so, (76) is satisfied. Moreover, it is clear that (77) is satisfied. Hence, our conclusion follows from Theorem 2. □

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The author would like to thank Professor D. Guo for his valuable suggestions.

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Chen, Y., Qin, B. Multiple positive solutions for first-order impulsive singular integro-differential equations on the half line in a Banach space. Bound Value Probl 2013, 69 (2013). https://doi.org/10.1186/1687-2770-2013-69

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Keywords

  • impulsive singular integro-differential equation in a Banach space
  • infinite three-point boundary value problem
  • fixed-point theorem of cone expansion and compression with norm type