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Positive solutions for the singular nonlocal boundary value problems involving nonlinear integral conditions

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Abstract

In this paper, using the theory of fixed point index on a cone and the Leray-Schauder fixed point theorem, we present the multiplicity of positive solutions for the singular nonlocal boundary-value problems involving nonlinear integral conditions and the existence of at least one positive solution for the singular nonlocal boundary-value problems with sign-changed nonlinearities.

MSC:34B10, 34B15, 34B18.

1 Introduction

Nonlocal boundary-value problems with linear and nonlinear integral conditions have seen a great deal of study lately (see [116], and references therein) because of their interesting theory and their applications to various problems, such as heat flow in a bar of finite length [4, 11]. In this paper, we consider the existence of positive solutions of the nonlinear boundary-value problem (BVP) of the form

y =q(t)f ( t , y ( t ) ) ,t(0,1)
(1.1)

with integral boundary conditions

y(0)=H ( ϕ ( y ) ) ,y(1)=0,
(1.2)

where ϕ(y) is a linear functional on C[0,1] given by

ϕ(y)= 0 1 y(s)dα(s)

involving a Stieltjes integral with a signed measure.

In [2], Goodrich considered the following problem:

y =λg ( t , y ( t ) ) ,t(0,1)
(1.3)

with integral boundary conditions

y(0)=H ( ϕ ( y ) ) ,y(1)=0
(1.4)

and deduced the existence of at least one positive solution to the BVP (1.3)-(1.4) in which H(ϕ(y)) has either asymptotically sublinear or asymptotically superlinear growth, and in [3] Goodrich demonstrated that if the nonlinear functional H(ϕ(y)) satisfies a certain asymptotic behavior, then the BVP (1.3)-(1.4) possesses at least one positive solution. For the case that H is linear and ϕ(y)= 0 1 y(s)dα(s) involves a signed measure, Webb and Infante discussed the multiplicity of positive solutions for nonlocal boundary-value problems [1214]. For the case that H is linear and the Borel measure associated with the Lebesgue-Stieltjes integral is positive, we can find some results on the existence of positive solutions [7, 8, 16, 17]. The results in the above literature are obtained under the condition that f(t,x) is continuous on (0,1)×[0,+), i.e., f has no singularity at x=0. And it is well known that study of singular two-point boundary-value problems for the second-order differential equation (1.1) (singular in the dependent variable) is very important and there are many results on the existence of positive solutions [15, 1824]. But there are fewer results on the existence of positive solutions for the singular BVP (1.1)-(1.2) [5, 6]. One goal in this paper is to consider the existence of positive solutions under the condition that f(t,x) is singular at x=0. Our paper has the following features.

Firstly, in order to overcome the difficulties of the singularity of f we establish a new cone and get the new condition (3.13) which is different from that in [5, 6]. Moreover, we get a multiplicity of positive solutions for BVP (1.1)-(1.2) different from that in [2, 3, 1214] under the condition that H(y) or f(t,y) is superlinear at y=+.

Secondly, when f is singular and sign-changed, we get the existence of at least one positive solution to the BVP (1.1)-(1.2) which is different from that in [2, 3, 5, 6, 1214] where f is nonnegative and continuous at x=0. Moreover, the results are different from that in [7, 8, 16, 17] where integral boundary conditions are linear and the Borel measure is positive.

Our paper is organized as follows. In Section 2, we present some lemmas and preliminaries. Section 3 discusses the existence of multiple positive solutions for the BVP (1.1)-(1.2) when f is positive. In Section 4, we discuss the existence of at least one positive solution of BVP (1.1)-(1.2) when f is singular and sign-changed.

2 Preliminaries

In this paper, the following lemmas are needed.

Lemma 2.1 (see [25])

Let Ω be a bounded open set in real Banach space E, P a cone of E,θΩ and A: Ω ¯ PP continuous and compact. Suppose λAxx, xΩP, λ(0,1]. Then

i(A,ΩP,P)=1.

Lemma 2.2 (see [25])

Let Ω be a bounded open set in real Banach space E, P a cone of E,θΩ and A: Ω ¯ PP continuous and compact. Suppose Axx, xΩP. Then

i(A,ΩP,P)=0.

Lemma 2.3 (see [25, 26])

Let E be a Banach space, R>0, B R ={xE:xR}, and F: B R E a continuous compact operator. If xλF(x) for any xE with x=R and 0<λ<1, then F has a fixed point in B R .

Let us begin by stating the hypotheses which we shall impose on the BVP (1.1)-(1.2).

  • (C1) Assume that there are three linear functionals ϕ, ϕ 1 , ϕ 2 :C([0,1])R such that

    ϕ(y)= ϕ 1 (y)+ ϕ 2 (y).

    Moreover, assume that there exists a constant ε 0 >0 such that

    ϕ 2 (y) ε 0 y

    holds for each yP, where P is the cone introduced in (2.1) below [2].

  • (C2) The functionals ϕ 1 (y) and ϕ 2 (y) are linear and, in particular, have the form

    ϕ 1 (y):= 0 1 y(t)d α 1 (t), ϕ 2 (y):= 0 1 y(t)d α 2 (t),

    where α 1 , α 2 :[0,1]R satisfy α 1 , α 2 BV([0,1]) with

    0 1 (1t)d α 1 (t)0, 0 1 (1t)d α 2 (t)0

    and

    0 1 k(t,s)d α 1 (t)0, 0 1 k(t,s)d α 2 (t)0

    hold, where the latter holds for each s[0,1] and k(t,s) is defined in (3.2) below [2].

  • (C3) Let H:RR be a real-valued, continuous function. Moreover, H:(0,+)(0,+).

  • (C4)

    { f : [ 0 , 1 ] × ( 0 , ) ( 0 , )  is continuous and there exists a function  ψ 1 continuous on  [ 0 , 1 ]  and positive on  ( 0 , 1 )  such that f ( t , y ) ψ 1 ( t )  on  ( 0 , 1 ) × ( 0 , 1 ] .
  • (C5)

    qC(0,1),q>0on (0,1)and 0 1 (1t)q(t)dt<.

Let C[0,1]={y:[0,1]R:y(t) is continuous on [0,1]} with norm y= max t [ 0 , 1 ] |y(t)|. It is easy to see that C[0,1] is a Banach space.

Assume that (C2) hold. Define

P = { y C [ 0 , 1 ] : y  is concave on  [ 0 , 1 ]  with  y ( t ) 0  for all  t [ 0 , 1 ] , ϕ 1 ( y ) 0 , ϕ 2 ( y ) 0 } .
(2.1)

It is easy to prove P is a cone of C[0,1] and we have the following lemma.

Lemma 2.4 (see [20])

Let yP (defined in (2.1)). Then

y(t)t(1t)yfor t[0,1].

3 Multiplicity of positive solutions for the singular boundary-value problems with positive nonlinearities

In this section, we consider the existence of multiple positive solutions for the BVP (1.1)-(1.2). To show that the BVP (1.1)-(1.2) has a solution, for yP, we define

( T ϵ y ) ( t ) = ( 1 t ) H ( ϕ ( y ) ) + 0 1 k ( t , s ) q ( s ) f ( s , max { ϵ , y ( s ) } ) d s , t [ 0 , 1 ] , 1 ϵ > 0 ,
(3.1)

where

k(t,s)= { ( 1 t ) s , 0 s t 1 , ( 1 s ) t , 0 t s 1 .
(3.2)

Lemma 3.1 Suppose (C1)-(C5) hold. Then T ϵ :PP is continuous and compact for all 1ϵ>0.

Proof It is easy to prove that T ϵ is well defined and ( T ϵ y)(t)0 for all tP. For yP, we have

{ ( T ϵ y ) ( t ) 0 on  ( 0 , 1 ) , ( T ϵ y ) ( 0 ) = H ( ϕ ( y ) ) , ( T ϵ y ) ( 1 ) = 0 ,

and so

( T ϵ y)(t) is concave on [0,1].
(3.3)

Moreover, from (C2), we may estimate

ϕ 1 ( T ϵ y ) = 0 1 ( 1 t ) d α 1 ( t ) H ( ϕ ( y ) ) + 0 1 0 1 k ( t , s ) q ( s ) f ( s , max { ϵ , y ( s ) } ) d s d α 1 ( t ) = 0 1 ( 1 t ) d α 1 ( t ) H ( ϕ ( y ) ) + 0 1 q ( s ) f ( s , max { ϵ , y ( s ) } ) 0 1 k ( t , s ) d α 1 ( t ) d s 0
(3.4)

and

ϕ 2 ( T ϵ y ) = 0 1 ( 1 t ) d α 2 ( t ) H ( ϕ ( y ) ) + 0 1 0 1 k ( t , s ) q ( s ) f ( s , max { ϵ , y ( s ) } ) d s d α 2 ( t ) = 0 1 ( 1 t ) d α 2 ( t ) H ( ϕ ( y ) ) + 0 1 q ( s ) f ( s , max { ϵ , y ( s ) } ) 0 1 k ( t , s ) d α 2 ( t ) d s 0 .
(3.5)

Combining (3.3), (3.4), and (3.5), T ϵ :PP. A standard argument shows that T ϵ :PP is continuous and compact [9, 18, 26]. □

Define

Φ r : = { x P C 2 ( ( 0 , 1 ) , R ) : x r  and  x  satisfies x ( t ) + q ( t ) f ( t , max { ϵ , x ( t ) } ) = 0 , 0 < t < 1 , x ( 0 ) = H ( ϕ ( x ) ) , x ( 1 ) = 0 , 1 ϵ > 0 } .

Lemma 3.2 If Φ r and (C2) hold, there exists a δ r >0 such that

x(0) δ r t(1t),t[0,1],x Φ r .

Proof Suppose x Φ r . There are two cases to consider.

(1)x>1. Lemma 2.4 implies that

x(t)t(1t)xt(1t),t[0,1].
(3.6)

(2) 0<x1. Condition (C4) guarantees that

x ( t ) = ( 1 t ) H ( ϕ ( x ) ) + 0 1 k ( t , s ) q ( s ) f ( s , max { ϵ , x ( s ) } ) d s 0 1 k ( t , s ) q ( s ) ψ 1 ( s ) d s : = γ 0 ( t ) , t [ 0 , 1 ] .

Since γ 0 (t)0, γ 0 (0)=0, and γ 0 (1)=0, we know that γ 0 is concave on [0,1] and γ 0 (t)0 for all t[0,1]. And from (C2), a similar argument as (3.4) and (3.5) shows that ϕ 1 ( γ 0 )0 and ϕ 2 ( γ 0 )0. Then γ 0 P and Lemma 2.4 implies that

γ 0 (t)t(1t) γ 0 ,t[0,1].
(3.7)

Let δ 1 =min{1, γ 0 }. From (3.6) and (3.7), one has

x(t) δ 1 t(1t),t[0,1],

which means that

rx δ 1 .

Thus

ϕ(x)= 0 1 x(s)d α 1 (s)+ 0 1 x(s)d α 2 (s) c 0 x c 0 r,

where

c 0 = d e f . 0 1 | d α 1 ( s ) | + 0 1 | d α 2 ( s ) |

and (C1) guarantees that

ϕ(x) ϕ 2 (x) ε 0 x.

And so

x(0)=H ( ϕ ( x ) ) min s [ ε 0 δ 1 , c 0 r ] H(s):= δ r >0.

The concavity x(t) yields

x(t) δ r (1t)>0,t[0,1],x Φ r .

The proof is complete. □

For R>0, let

Ω R = { x C [ 0 , 1 ] : x < R } .

We have the following lemmas.

Lemma 3.3 Suppose that (C1)-(C5) hold and there exists an a(0, 1 2 ) such that

lim y + f ( t , y ) y =+
(3.8)

uniformly on [a,1a]. Then, there exists an R >1 such that for all R R

i( T ϵ , Ω R P,P)=0,0<ϵ1.

Proof From (3.8), there exists an R 1 >1 such that

f(t,y) N y,y R 1 ,
(3.9)

where

N > 2 a 2 a 1 a k ( a , s ) q ( s ) d s .

Let R = R 1 a 2 and

Ω R := { x C [ 0 , 1 ] : x < R } ,R R .

Now we show

T ϵ yyfor yP Ω R ,0<ϵ1.
(3.10)

Suppose that there exists a y 0 P Ω R with T ϵ y 0 y 0 . Then, y 0 =R. Since y 0 (t) is concave on [0,1] (since y 0 P) we find from Lemma 2.4 that y 0 (t)t(1t) y 0 t(1t)R for t[0,1]. For t[a,1a], one has

y 0 (t) a 2 R a 2 R = R 1 ,t[a,1a],

which together with (3.9) yields

f ( t , max { ϵ , y 0 ( t ) } ) =f ( t , y 0 ( t ) ) N y 0 (t) N a 2 R,t[a,1a].
(3.11)

Then we have, using (3.11),

y 0 ( a ) T ϵ y 0 ( a ) = ( 1 a ) H ( ϕ ( y 0 ) ) + 0 1 k ( a , s ) q ( s ) f ( s , max { ϵ , y 0 ( s ) } ) d s a 1 a k ( a , s ) q ( s ) f ( s , max { ϵ , y 0 ( s ) } ) d s = a 1 a k ( a , s ) q ( s ) f ( s , y 0 ( s ) ) d s N R a 2 a 1 a k ( a , s ) q ( s ) d s > R = y 0 ,

which is a contradiction. Hence equation (3.10) is true. Lemma 2.2 guarantees that

i( T ϵ , Ω R P,P)=0,0<ϵ1.

The proof is complete. □

Lemma 3.4 Suppose that (C1)-(C5) hold and

lim s + H ( s ) s =+.
(3.12)

Then, there exists an R >1 such that for all R R

i( T ϵ , Ω R P,P)=0,0<ϵ1.

Proof From equation (3.12), there exists an R 1 >1 such that

H(s) N s,s R 1 ,
(3.13)

where

N > 2 ε 0 ( ε 0  defined in (C ) 1 ) .

Let R = R 1 ε 0 and

Ω R = { x C [ 0 , 1 ] : x < R } ,R R .

Now we show

T ϵ yyfor yP Ω R ,0<ϵ1.
(3.14)

Suppose that there exists a y 0 P Ω R with T ϵ y 0 y 0 . Then, y 0 =R. Now (C1) guarantees that

ϕ( y 0 )= ϕ 1 ( y 0 )+ ϕ 2 ( y 0 ) ε 0 y 0 = ε 0 R R 1 ,

which together with equation (3.13) implies that

y 0 (0) T ϵ y 0 (0)=H ( ϕ ( y 0 ) ) N ϕ( y 0 )> 2 ε 0 ε 0 y 0 > y 0 .

This is a contradiction. Hence (3.14) is true. Lemma 2.2 guarantees that

i( T ϵ , Ω R P,P)=0,0<ϵ1.

The proof is complete. □

Theorem 3.1 Suppose (C1)-(C5) hold and the following conditions are satisfied:

{ 0 f ( t , y ) g ( y ) + h ( y )  on  [ 0 , 1 ] × ( 0 , )  with g > 0  continuous and nonincreasing on  ( 0 , ) , h 0  continuous on  [ 0 , ) ,  and h g  nondecreasing on  ( 0 , )
(3.15)

and

sup r ( 0 , + ) min { 1 1 + h ( r ) g ( r ) 0 r d y g ( y ) , r max y [ 0 , c 0 r ] H ( y ) } >max{1, b 0 }
(3.16)

hold; here

b 0 = 0 1 (1s)q(s)ds, c 0 = 0 1 | d α 1 ( s ) | + 0 1 | d α 2 ( s ) | .

Then the BVP (1.1)-(1.2) has at least one positive solution.

Proof From equation (3.16), choose ϵ>0 and r>0 with ϵ<min{1,r} such that

min { 1 1 + h ( r ) g ( r ) 0 r d y g ( y ) , r max y [ 0 , c 0 r ] H ( y ) } >max{1, b 0 }.
(3.17)

Let

Ω 1 = { y C [ 0 , 1 ] : y < r } ,

and n 0 > 1 ϵ . For n{ n 0 , n 0 +1,}, we define T 1 n as in equation (3.1). Lemma 3.1 guarantees that T 1 n :PP is continuous and compact.

Now we show that

yλ T 1 n y,y Ω 1 P,λ(0,1].
(3.18)

Suppose that there is a y 0 Ω 1 P and λ 0 [0,1] with y 0 = λ 0 T 1 n y 0 , i.e., y 0 satisfies

{ y 0 ( t ) + λ 0 q ( t ) f ( t , max { 1 n , y 0 ( t ) } ) = 0 , 0 < t < 1 , y 0 ( 0 ) = λ 0 H ( ϕ ( y ) ) , y 0 ( 1 ) = 0 .
(3.19)

Then y 0 (t)0 on (0,1). From equation (3.17), we have y 0 (0)= λ 0 H(ϕ( y 0 )) max y [ 0 , c 0 r ] H(y)<r, which together with y 0 (1)=0<r implies that there exists a t 0 (0,1) with y 0 ( t 0 )= y 0 =r, y 0 ( t 0 )=0 and y 0 (t)0 for all t( t 0 ,1). For t(0,1), from equations (3.15) and (3.19), we have

y 0 ( t ) g ( max { 1 n , y 0 ( t ) } ) { 1 + h ( max { 1 n , y 0 ( t ) } ) g ( max { 1 n , y 0 ( t ) } ) } q ( t ) g ( max { 1 n , y 0 ( t ) } ) { 1 + h ( r ) g ( r ) } q ( t ) .
(3.20)

We integrate equation (3.20) from t 0 ( t 0 <t) to t to obtain

y 0 ( t ) g ( max { 1 n , y 0 ( t ) } ) { 1 + h ( r ) g ( r ) } t 0 t q ( s ) d s g ( y 0 ( t ) ) { 1 + h ( r ) g ( r ) } t 0 t q ( s ) d s
(3.21)

and then integrate equation (3.21) from t 0 to 1 to obtain

y 0 ( 1 ) y 0 ( t 0 ) d y g ( y ) { 1 + h ( r ) g ( r ) } t 0 1 t 0 s q ( τ ) d τ d s = { 1 + h ( r ) g ( r ) } t 0 1 ( 1 s ) q ( s ) d s { 1 + h ( r ) g ( r ) } 0 1 ( 1 s ) q ( s ) d s ,

i.e.,

0 r d y g ( y ) { 1 + h ( r ) g ( r ) } 0 1 (1s)q(s)ds,

which contradicts equation (3.17). Therefore, equation (3.18) is true. Lemma 2.1 implies that

i( T 1 n , Ω 1 P,P)=1,

which yields the result that there exists a y n Ω 1 P such that

T 1 n y n = y n ,

i.e., Φ r in Lemma 3.2. Now Lemma 3.2 guarantees that there exists a δ r >0 such that

y n (0) δ r , y n (t) δ r (1t),t[0,1],x{ n 0 , n 0 +1,}.
(3.22)

Now we consider the set { y n } n = n 0 . Obviously, y n r means that

the functions belonging to  { y n ( t ) }  are uniformly bounded on [0,1].
(3.23)

Now we show that

the functions belonging to  { y n ( t ) }  are equicontinuous on [0,1].
(3.24)

There are two cases to consider.

(1) There exists a subsequence { y n i } of { y n } with y n i (0)=H(ϕ( y n i ))< y n i . Without loss of generality, we assume that y n (0)=H(ϕ( y n ))< y n , n{ n 0 , n 0 +1,}, which together with y n (1)=0 implies that there exists a t n satisfying that y n ( t n )=0 with y n (t)0 for t(0, t n ) and y n (t)0 for t( t n ,1). Let t =sup{ t n ,n n 0 }. Now we show that t <1. To the contrary, suppose that t =1. Then there exists a subsequence { n i } of {n} such that t n i 1 as n i +. From equation (3.21), using y n in place of y 0 , we have

0 y n i ( t n i ) 1 g ( y ) dy ( 1 + h ( r ) g ( r ) ) t n i 1 (1s)q(s)ds,

which implies that

y n i ( t n i )0,as  n i +.

This contradicts y n i (t) δ r (1t) for all t[0,1].

Let t 0 ( t ,1). From equation (3.22), we have

y n (t) k 0 := min t [ 0 , t 0 ] δ r (1t),t[0, t 0 ].

Similarly as the proof in equation (3.21), one has

y n (t)g( k 0 ) ( 1 + h ( r ) g ( r ) ) 0 1 q(s)ds,

which means that

the functions belonging to  { y n ( t ) }  are equicontinuous on [0, t 0 ].
(3.25)

For t 1 , t 2 [ t 0 ,1), from equation (3.21), using y n in place of y 0 , we have

| y n ( t 1 ) y n ( t 2 ) 1 g ( y ) d y | ( 1 + h ( r ) g ( r ) ) 0 1 q(s)ds| t 1 t 2 |,

which yields

the functions belonging to  { y n ( t ) }  are equicontinuous on [ t 0 ,1].
(3.26)

Combining equations (3.25) and (3.26), we find that equation (3.24) holds.

(2) There exists a k 1 >0 such that y n (0)= y n and y n (t) is nonincreasing on [0,1] for all n> k 1 . From y n (0)=H(ϕ( y n ))= y n and y n (1)=0, there exists t n (0,1) such that y n ( t n )=H(ϕ( y n )). Now y n (t)0 implies that y n (0) y n ( t n )=H(ϕ( y n )). Hence, from equation (3.20), using y n in place of y 0 , we have

y n (t)+ y n (0)g ( y n ( t ) ) ( 1 + h ( r ) g ( r ) ) 0 t q(s)ds,t(0,1)

and so

y n ( t ) g ( y n ( t ) ) ( 1 + h ( r ) g ( r ) ) 0 t q ( s ) d s y 0 ( 0 ) g ( y n ( t ) ) ( 1 + h ( r ) g ( r ) ) 0 t q ( s ) d s + H ( ϕ ( y n ) ) g ( y n ( t ) ) ( 1 + h ( r ) g ( r ) ) 0 t q ( s ) d s + 1 g ( r ) max s [ 0 , c 0 r ] H ( r ) , t ( 0 , 1 ) .

Then

| y n ( t 1 ) y n ( t 2 ) 1 g ( y ) d y | = | t 1 t 2 y n ( s ) g ( y n ( s ) ) d s | ( 1 + h ( r ) g ( r ) ) | t 1 t 2 0 s q ( τ ) d τ d s | + 1 g ( r ) max s [ 0 , c 0 r ] H ( r ) | t 1 t 2 | , t 1 , t 2 [ 0 , 1 ] ,

which implies that (3.24) hold.

Now Arzela-Ascoli theorem guarantees that { y n (t)} has a convergent subsequence. Without loss of generality, we assume that there is a y C[0,1] such that

lim n + y n = y ,

which together with equation (3.22) and y n (1)=0 implies that

y (1)=0, y (t) δ r (1t),t[0,1].
(3.27)

Since y n (nN) satisfies y n = T 1 n y n , we have

y n (t)=q(t)f ( t , max { 1 n , y n ( t ) } ) =0,0<t<1.

We integrate the above equation from 1 2 to t to yield

y n (t)= y n ( 1 2 ) 1 2 t q(s)f ( s , max { 1 n , y n ( s ) } ) ds,

and so

y n ( t ) = y n ( 1 2 ) + y n ( 1 2 ) ( t 1 2 ) 1 2 t 1 2 s q ( τ ) f ( τ , max { 1 n , y n ( τ ) } ) d τ d s = y n ( 1 2 ) + y n ( 1 2 ) ( t 1 2 ) + 1 2 t ( s t ) q ( s ) f ( s , max { 1 n , y n ( s ) } ) d s

for t(0,1) and

y n (0)=H ( ϕ ( y n ) ) =H ( 0 1 y n ( s ) d α 1 ( s ) + 0 1 y n ( s ) d α 2 ( s ) ) ,

and the Lebesgue Dominated Convergent theorem together with equation (3.27) implies that

y ( t ) = lim n + y n ( t ) = lim n + [ y n ( 1 2 ) + y n ( 1 2 ) ( t 1 2 ) + 1 2 t ( s t ) q ( s ) f ( s , max { 1 n , y n ( s ) } ) d s ] = y ( 1 2 ) + y ( 1 2 ) ( t 1 2 ) + 1 2 t ( s t ) q ( s ) f ( s , y ( s ) ) d s
(3.28)

for t(0,1) and

y ( 0 ) = lim n + y n ( 0 ) = lim n + H ( ϕ ( y n ) ) = lim n + H ( 0 1 y n ( s ) d α 1 ( s ) + 0 1 y n ( s ) d α 2 ( s ) ) = H ( ϕ 1 ( y ) + ϕ 2 ( y ) ) = H ( ϕ ( y ) ) .
(3.29)

We differentiate equation (3.28) to get

y (t)+q(t)f ( t , y ( t ) ) =0,t(0,1),

which together with equations (3.27) and (3.29) means that the BVP (1.1)-(1.2) has at least one positive solution. The proof is complete. □

Theorem 3.2 Suppose the conditions of Theorem  3.1 hold and there exists an a(0, 1 2 ) such that

lim y + f ( t , y ) y =+

uniformly on [a,1a]. Then the BVP (1.1)-(1.2) has at least two positive solutions.

Proof Choose r>0 as in (3.17), n 0 >0 with 1 n 0 <min{1,r}, and R>max{r, R } in Lemma 3.3. Set N n 0 ={ n 0 , n 0 +1,}, and

Ω 1 = { y C [ 0 , 1 ] : y < r } , Ω 2 = { y C [ 0 , 1 ] : y < R } .

By the proof of Theorem 3.1 and Lemma 3.3, we have

i( T 1 n , Ω 1 P,P)=1

and

i( T 1 n , Ω 2 P,P)=0,

which implies that

i ( T 1 n , ( Ω 2 Ω ¯ 1 ) P , P ) =1.

Then, there exist x 1 , n Ω 1 P and x 2 , n ( Ω 2 Ω ¯ 1 )P such that

T 1 n x 1 , n = x 1 , n , T 1 n x 2 , n = x 2 , n .

By the proof of Theorem 3.1, there exist a subsequence { x 1 , n i } of { x 1 , n } and x 1 P such that

lim n i + x 1 , n i (t)= x 1 (t),t[0,1].

And moreover, x 1 (t) is a positive solution to the BVP (1.1)-(1.2) with r> x 1 (t) δ r (1t), t[0,1].

A similar argument shows that there exist a subsequence { x 2 , n j } of { x 2 , n } and x 2 P( Ω 2 Ω ¯ 1 ) such that

lim n i + x 2 , n j (t)= x 2 (t),t[0,1].

And moreover, x 2 (t) is a positive solution to the BVP (1.1)-(1.2) and equation (3.18) guarantees that x 2 >r. Hence, x 1 (t) and x 2 (t) are two positive solutions for the BVP (1.1)-(1.2). The proof is complete. □

Theorem 3.3 Suppose the conditions of Theorem  3.1 hold and

lim s + H ( s ) s =+.

Then the BVP (1.1)-(1.2) has at least two positive solutions.

Proof Choose r>0 as in (3.17), n 0 >0 with 1 n 0 <min{1,r}, and R>max{r, R } in Lemma 3.4. Set N n 0 ={ n 0 , n 0 +1,}, and

Ω 1 = { y C [ 0 , 1 ] : y < r } , Ω 2 = { y C [ 0 , 1 ] : y < R } .

By the proof of Theorem 3.1 and Lemma 3.4, we have

i( T 1 n , Ω 1 P,P)=1

and

i( T 1 n , Ω 2 P,P)=0,

which implies that

i ( T 1 n , ( Ω 2 Ω ¯ 1 ) P , P ) =1.

Then, there exist x 1 , n Ω 1 P and x 2 , n ( Ω 2 Ω ¯ 1 )P such that

T 1 n x 1 , n = x 1 , n , T 1 n x 2 , n = x 2 , n .

A similar argument to that in Theorem 3.2 shows that the BVP (1.1)-(1.2) has at least two positive solutions. The proof is complete. □

Example 3.1 Consider

y (t)+μ 1 1 t ( 1 200 + 1 300 sin t 2 + 1 100 y δ 1 ( t ) + 1 100 y δ 2 ( t ) ) =0,0<t<1,
(3.30)

with

y(0)=H ( ϕ ( y ) ) ,y(1)=0,
(3.31)

where

H(t)= 1 2 t+ 1 3 t 1 3 ,ϕ(y)= ϕ 1 (y)+ ϕ 2 (y)= 0 1 y(s)d α 1 (s)+ 0 1 y(s)d α 2 (s),

with

d α 1 ( s ) = 1 8 cos 2 π s d s , d α 2 ( s ) = 1 8 d e s , δ 1 > 0 , δ 2 > 1 , 100 ( δ 1 + 1 ) 3 > 1 .
(3.32)

Then equations (3.30)-(3.31) have at least two positive solutions.

To prove that the BVP (3.30)-(3.31) has at least two positive solutions, we use Theorem 3.2. Let q(t)=μ 1 1 t , f(t,y)= 1 200 + 1 300 sin t 2 + 1 100 y δ 1 + 1 100 y δ 2 , g(y)= 1 100 y δ 1 , h(y)= 1 100 + 1 100 y δ 2 , c 0 = 0 1 |d α 1 (s)|+ 0 1 |d α 2 (s)|= 1 4 π + e 1 8 , b 0 = 2 3 μ. For yP (defined in (2.1)), we have

ϕ 2 (y)= 0 1 y(t) 1 8 e s dsy 0 1 s(1s) 1 8 e s ds,

which means that (C1) holds. Since

0 1 ( 1 t ) d α 1 ( t ) = 0 , 0 1 ( 1 t ) d α 2 ( t ) > 0 , 0 1 k ( t , s ) d α 1 ( t ) = ( 1 s ) 0 s t d α 1 ( t ) + s s 1 ( 1 t ) d α 1 ( t ) = 1 cos 2 π s 32 π 2 0 ,

and

0 1 k(t,s)d α 2 (t)=(1s) 0 s td α 2 (t)+s s 1 (1t)d α 2 (t)0,

(C2) is true. Since c 0 <1, we have max y [ 0 , c 0 r ] H(y)= 1 2 c 0 r+ 1 3 ( c 0 r ) 1 3 1 2 r+ 1 3 r 1 3 . Then

1 max y [ 0 , c 0 1 ] H ( y ) = 1 1 2 c 0 1 + 1 3 ( c 0 1 ) 1 3 >1.

Equation (3.32) guarantees that

1 1 + h ( 1 ) g ( 1 ) 0 1 1 g ( y ) dy= 100 3 ( 1 + δ 1 ) >1.

Letting μ 0 <3, we have

sup r ( 0 , + ) min { 1 1 + h ( r ) g ( r ) 0 r d y g ( y ) , r max y [ 0 , c 0 r ] H ( y ) } >max{1, b 0 },

for all μ μ 0 , which means that equations (3.15)-(3.16) hold. Since

f(t,x) 1 200 + 1 300 sin t 2 ,(t,x)[0,1]×(0,1],

we get (C4). Moreover, since

lim y + f ( t , y ) y =+

uniformly on [0,1], all conditions of Theorem 3.2 hold, which implies that equations (3.30)-(3.31) have at least two positive solutions.

Example 3.2 Consider

y (t)+μ y δ 1 (t)=0,0<t<1,
(3.33)

with

y(0)=H ( ϕ ( y ) ) ,y(1)=0,
(3.34)

where

H(t)= 1 2 t 3 + 1 3 t 1 3 ,ϕ(y)= ϕ 1 (y)+ ϕ 2 (y)= 0 1 y(s)d α 1 (s)+ 0 1 y(s)d α 2 (s),

with

d α 1 (s)= 1 8 cos2πsds,d α 2 (s)= 1 8 d e s , δ 1 >0.

Then equations (3.33)-(3.34) have at least two positive solutions.

To prove that the BVP (3.33)-(3.34) has at least two positive solutions, we use Theorem 3.3. Let q(t)=μ, f(t,y)= y δ 1 , g(y)= y δ 1 , h(y)=0, c 0 = 1 4 π + e 1 8 , b 0 = 1 2 μ. Since c 0 <1, we have max y [ 0 , c 0 r ] H(y)= 1 2 ( c 0 r ) 3 + 1 3 ( c 0 r ) 1 3 1 2 r 3 + 1 3 r 1 3 . Then

1 max y [ 0 , c 0 1 ] H ( y ) = 1 1 2 ( c 0 1 ) 3 + 1 3 ( c 0 1 ) 1 3 >1.

Also we have

lim r + 0 r d y g ( y ) ( 1 + h ( r ) g ( r ) ) 1 =+.

Then, letting μ 0 2, we get

sup r ( 0 , + ) min { 1 1 + h ( r ) g ( r ) 0 r d y g ( y ) , r max y [ 0 , c 0 r ] H ( y ) } >max{1, b 0 },

for all μ μ 0 , which means that equations (3.15)-(3.16) hold. Since

f(t,x)1,(t,x)[0,1]×(0,1],

we get (C4). Obviously, (C1)-(C3), and (C5) hold. Moreover, since

lim y + H ( s ) s =+

uniformly on [0,1], all conditions of Theorem 3.3 hold, which implies that equations (3.30)-(3.31) have at least two positive solutions.

4 Positive solutions for singular boundary-value problems with sign-changing nonlinearities

  • (H1) Assume that there are three linear functionals ϕ, ϕ 1 , ϕ 2 :C([0,1])R

    ϕ(y)= ϕ 1 (y)+ ϕ 2 (y), ϕ 1 (y):= 0 1 y(t)d α 1 (t), ϕ 2 (y):= 0 1 y(t)d α 2 (t),

    where α 1 , α 2 :[0,1]R satisfy α 1 , α 2 BV([0,1]);

  • (H2) a(t)C([0,1],(0,+)), (1t)q(t) L 1 ((0,1]);

  • (H3) Let H:R[0,+) be a real-valued, continuous function. Moreover, H:(0,+)(0,+);

  • (H4) f(t,y)C([0,1]×(0,+),(,+)), there exists a decreasing function F(y)C((0,+),(0,+)), and a nonnegative function G(y)C([0,+),[0,+)) such that f(t,y)F(y)+G(y) and there exists a bC((0,1),(0,+)) such that

    f(t,y)a(t),0<yb(t),t(0,1);
  • (H5) there exist R>1 such that

    0 R d y F ( y ) ( 1 + G ¯ ( R ) F ( R ) ) 1 > 0 1 (1s)q(s)ds

    and

    max y [ 0 , r c 0 ] H(y)<r,Rr>0,where  c 0 = 0 1 | d α 1 ( s ) | + 0 1 | d α 2 ( s ) | ,

    where G ¯ (R)= max s [ 0 , R ] G(s).

For n>3, let b n =min{ 1 n , min t [ 1 n , 1 1 n ] b(t)}. Obviously, b n >0. For y C n =C[ 1 n ,1 1 n ], we define T n as

( T n y ) ( t ) = ( 1 1 n t ) H ( ϕ n ( y ) ) + b n + 1 n 1 1 n k n ( t , s ) q ( s ) f ( s , max { b n , y ( s ) } ) d s , t [ 1 n , 1 1 n ] ,

where

k n (t,s)= { ( s 1 n ) ( 1 1 n t ) , 1 n s t 1 1 n , ( t 1 n ) ( 1 1 n s ) , 1 n t s 1 1 n

and

ϕ n (y)= 1 n 1 1 n y(s)d α 1 (s)+ 1 n 1 1 n y(s)d α 2 (s).

From a standard argument (see [18, 25, 26]), we have the following result.

Lemma 4.1 Suppose (H1)-(H4) hold. Then the operator T n is continuous and compact from C n to C n .

From (H3) and (H5), there exists ϵ 0 >0 such that

ϵ 0 R d y F ( y ) ( 1 + G ¯ ( R ) F ( R ) ) 1 > 0 1 ( 1 s ) q ( s ) d s , max y [ 0 , c 0 R ] H ( y ) + ϵ 0 < R .
(4.1)

Choose n 0 >3 with 1 n 0 < ϵ 0 and let N n 0 ={ n 0 , n 0 +1,}. Now we have the following lemmas.

Lemma 4.2 Suppose (H1)-(H5) hold. Then, for n N 0 , there exists a x n C n with b n x n (t)R such that

x n (t)= ( 1 1 n t ) H ( ϕ n ( x n ) ) + b n + 1 n 1 1 n k n (t,s)q(s)f ( s , x n ( s ) ) ds,t [ 1 n , 1 1 n ] .

Proof Let Ω={y C n :y<R}. For yΩ, we now prove that

y ( t ) λ ( T n y ) ( t ) = λ ( ( 1 1 n t ) H ( ϕ n ( y ) ) + b n ) + λ 1 n 1 1 n k n ( t , s ) q ( s ) f ( s , max { b n , y ( s ) } ) d s , t [ 1 n , 1 1 n ]
(4.2)

for any λ(0,1].

Suppose equation (4.2) is not true. Then there exists yC[ 1 n ,1 1 n ] with y=R and 0<λ<1 such that

y ( t ) = λ ( T y ) ( t ) = λ ( ( 1 1 n t ) H ( ϕ n ( y ) ) + b n ) + λ 1 n 1 1 n k n ( t , s ) q ( s ) f ( s , max { b n , y ( s ) } ) d s , t [ 1 n , 1 1 n ] .
(4.3)

We first claim that y(t)λ b n for any t[ 1 n ,1 1 n ].

Suppose there exists a η(0,1) with y(η)<λ b n . Let γ 0 =inf{ t 1 :y(s)<λ b n ,s[ t 1 ,η]} and γ 1 =sup{ t 1 :y(s)<λ b n ,s[η, t 1 ]}. Since y( 1 n )λ b n and y(1 1 n )=λ b n , we have γ 0 1 n , γ 1 1 1 n , y( γ 0 )=y( γ 1 )=λ b n , and y(t)<λ b n for all t( γ 0 , γ 1 ), which implies that

y (t)=λq(t)f(t, b n )<0,t( γ 0 , γ 1 )

and so y(t) is concave down on [ γ 0 , γ 1 ]. This is a contradiction.

Now (H5) guarantees that

y ( 1 n ) =λ ( ( 1 2 n ) H ( ϕ n ( y ) ) + b n ) max r [ 0 , c 0 R ] h(r)+ ϵ 0 <R,

which together with y(1 1 n )=λ b n <R means that there is a t( 1 n ,1 1 n ) with y (t)=0 and y(t)=R. Let t =sup{t:y(t)=R, y (t)=0} and t =inf{t:y(t)=R, y (t)=0}. Obviously, 1 n < t t <1 1 n , y( t )=R, y ( t )=0, y( t )=R, y ( t )=0, y(t)<R for all t( t ,1 1 n ] and y(t)<R for all t( 1 n , t ]. Let t 1 =inf{ t <t1 1 n :y(t)=λy(1 1 n )} and t 1 =sup{t< t 1 1 n :y(t)=λy( 1 n )}. It is easy to see that t < t 1 1 1 n , y(t)>y( t 1 ) for all t( t , t 1 ), t 1 < t and y(t)>y( t 1 ) for all t( t 1 , t ).

Now we consider the properties of y on ( t , t 1 ). We get a countable set { t i } of ( t , t 1 ] such that

  1. 1.

    t > t 2 m > t 2 m 1 >> t 5 t 4 > t 3 t 2 > t 1 = t 1 , t 2 m t ,

  2. 2.

    y( t 2 i )=y( t 2 i + 1 ), y ( t 2 i )=0, i=1,2,3, ,

  3. 3.

    y(t) is strictly decreasing in [ t 2 i , t 2 i 1 ], i=1,2,3, (if y(t) is strictly decreasing in [ t , t 1 ], put m=1; i.e, [ t 2 , t 1 ]=[ t , t 1 ]).

Differentiating equation (4.3) and using the assumptions (H2) and (H4), we obtain

y ( t ) = λ q ( t ) f ( t , max { b n , y ( t ) } ) λ q ( t ) ( F ( max { b n , y ( t ) } ) + G ( max { b n , y ( t ) } ) ) = λ q ( t ) F ( max { b n , y ( t ) } ) ( 1 + G ( max { b n , y ( t ) } ) F ( max { b n , y ( t ) } ) ) < q ( t ) F ( max { b n , y ( t ) } ) ( 1 + G ¯ ( R ) F ( R ) ) q ( t ) F ( y ( t ) ) ( 1 + G ¯ ( R ) F ( R ) ) , t [ t 2 i , t 2 i 1 ) , i = 1 , 2 , 3 , .
(4.4)

Integrating (4.4) from t 2 i to t, we have, by the decreasing property of F(y),

t 2 i t y (s)ds ( 1 + G ¯ ( R ) F ( R ) ) t 2 i t q(s)F ( y ( s ) ) dsF ( y ( t ) ) ( 1 + G ¯ ( R ) F ( R ) ) t 2 i t q(s)ds,

for t[ t 2 i , t 2 i 1 ), i=1,2,3, ; that is to say,

y (t)F ( y ( t ) ) ( 1 + G ¯ ( R ) F ( R ) ) t 2 i t q(s)ds,t[ t 2 i , t 2 i 1 ),i=1,2,3,.
(4.5)

It follows from equation (4.5) that

y ( t ) F ( y ( t ) ) ( 1 + G ¯ ( R ) F ( R ) ) t 2 i t q(s)ds ( 1 + G ¯ ( R ) F ( R ) ) 0 t q(s)ds,
(4.6)

for t[ t 2 i , t 2 i 1 ), i=1,2,3, .

On the other hand, for any z( 1 n ,1 1 n ) with y(z)>λ b n , we can choose i 0 and z ( t , t 1 ) such that z [ t 2 i 0 , t 2 i 0 1 ), y( z )=y(z) and z z . Integrating equation (4.6) from t 2 i to t 2 i 1 , i=1,2,3,, i 0 1 and from t 2 i 0 to z , we have

y ( t 2 i 1 ) y ( t 2 i ) d y F ( y ) ( 1 + G ¯ ( R ) F ( R ) ) t 2 i t 2 i 1 0 t q(s)dsdt,i=1,2,3,, i 0 1,
(4.7)

and

y ( t 2 i 0 ) y ( z ) d y F ( y ) ( 1 + G ¯ ( R ) F ( R ) ) z t 2 i 0 0 t q(s)dsdt.
(4.8)

Summing equation (4.7) from 1 to i 0 1, we have by equation (4.8) and y( t 2 i )=y( t 2 i + 1 )

y ( t 1 ) y ( z ) d y F ( y ) ( 1 + G ¯ ( R ) F ( R ) ) z t 1 0 t q(s)dsdt ( 1 + G ¯ ( R ) F ( R ) ) z t 1 0 t q(s)dsdt.

Since y(z)=y( z ),

y ( t 1 ) y ( z ) d y F ( y ) ( 1 + G ¯ ( R ) F ( R ) ) z t 1 0 t q(s)dsdt.
(4.9)

For the properties of y on ( t 1 , t ), a similar argument shows that for any z> t 1

y ( t 1 ) y ( z ) d y F ( y ) ( 1 + G ¯ ( R ) F ( R ) ) t 1 z 0 t q(s)dsdt.
(4.10)

Letting z t in (4.9), we have

ϵ 0 R d y F ( y ) y ( t 1 ) R d y F ( y ) ( 1 + G ¯ ( R ) F ( R ) ) t t 1 0 t q ( s ) d s d t ( 1 + G ¯ ( R ) F ( R ) ) 0 1 0 t q ( s ) d s d t = ( 1 + G ¯ ( R ) F ( R ) ) 0 1 ( 1 s ) q ( s ) d s ,

which contradicts equation (4.1). Hence equation (4.2) holds.

It follows from Lemma 3.2 that T n has a fixed point x n in C n . Using x n and 1 in place of y and λ in (4.3), we obtain easily b n x n (t)R, t[ 1 n ,1 1 n ]. And x n satisfies

x n ( t ) = ( 1 1 n t ) H ( ϕ n ( x n ) ) + b n + 0 1 k n ( t , s ) q ( s ) f ( s , x n ( s ) ) d s , t [ 1 n , 1 1 n ] .
(4.11)

The proof is complete. □

Lemma 4.3 Suppose that all conditions of Lemma  4.2 hold and x n satisfies (4.11). For a fixed h(0,min{ 1 2 ,η}), let m n , h =min{ x n (t),t[h,1h]}. Then m h =inf{ m n , h }>0.

Proof Since x n (t) b n >0, we get m h 0. For any fixed natural number n (n> n 0 defined in Lemma 4.2), let t n [h,1h] such that x n ( t n )=min{ x n (t),t[h,1h]}. If m h =0, there exists a countable set { n i } such that

lim n i + x n i ( t n i )=0.

So there exists N 0 such that x n i ( t n i )min{b(t),t[ h 2 ,1h]}, n i > N 0 . Let N ¯ 0 ={ n 0 > N 0 :n N 0  with  lim n i + x n i ( t n i )=0}. Then we have two cases.

Case 1. There exist n k N ¯ 0 and t n k [ h 2 ,h] such that x n k ( t n k ) x n k ( t n k ). By the same argument in Lemma 4.2, we can get t n k , t n k [ h 2 ,1], t n k < t n k such that

x n k ( t ) min { b ( t ) , t [ h 2 , 1 ] } , t [ t n k , t n k ] , x n k ( t ) x n k ( t n k ) , x n k ( t ) x n k ( t n k ) , t ( t n k , t n k ) ,
(4.12)

and

x n k (t)=q(t)f ( t , x n k ( t ) ) <0,t ( t n k , t n k ) .
(4.13)

The inequality (4.13) shows that x n k (t) is concave down in [ t n k , t n k ], which contradicts equation (4.12).

Case 2. x n i (t)< x n i ( t n i ), t[ h 2 ,h] for any n i N ¯ 0 . And so we have

lim n i + x n i (t)=0,t [ h 2 , h ] .
(4.14)

On the other hand, for any t[ h 2 ,h],

x n i ( t ) = 2 h h 2 t ( s h 2 ) ( h t ) q ( s ) f ( s , x n i ( s ) ) d s + 2 h t h ( t h 2 ) ( h s ) q ( s ) f ( s , x n i ( s ) ) d s + x n i ( h 2 ) + x n i ( h ) 2 h [ h 2 t ( s h 2 ) ( h t ) a ( s ) d s + t h ( t h 2 ) ( h s ) a ( s ) d s ] > 0 ,

which contradicts equation (4.14). Hence, m h >0. The proof is complete. □

Theorem 4.1 If (H1)-(H5) hold, then BVP (1.1)-(1.2) has at least one positive solution.

Proof For any natural number nN (defined in Lemma 4.2), it follows from Lemma 4.2 that there exist x n C n , b n x n (t)R for all t[ 1 n ,1 1 n ] satisfying (4.11). Now we divide the proof into three steps.

Step 1. There exists a convergent subsequence of { x n } in (0,1). For a natural number k n 0 in Lemma 4.2, it follows from Lemma 4.3 that 0< m 1 k x n (t)R, t[ 1 k ,1 1 k ] for any natural numbers nN; i.e., { x n } is uniformly bounded in [ 1 k ,1 1 k ]. Since x n also satisfies

x n ( t ) = 1 1 2 k 1 k t ( s 1 k ) ( 1 1 k t ) q ( s ) f ( s , x n ( s ) ) d s + 1 1 2 k t 1 1 k ( t 1 k ) ( 1 1 k s ) q ( s ) f ( s , x n ( s ) ) d s + x n ( 1 k ) + x n ( 1 1 k ) ,

we have

x n ( t ) = 1 1 2 k 1 k t ( s 1 k ) q ( s ) f ( s , x n ( s ) ) d s + 1 1 2 k t 1 1 k ( 1 1 k s ) q ( s ) f ( s , x n ( s ) ) d s .

Obviously

| x n ( t ) | 2 ( 1 2 k ) max { q ( t ) | f ( t , x n ( t ) ) | : ( t , x ) [ 1 k , 1 1 k ] × [ m 1 k , R ] } ,
(4.15)

for t[ 1 k ,1 1 k ]. It follows from inequality (4.15) that { x n } is equicontinuous in [ 1 k ,1 1 k ]. The Ascoli-Arzela theorem guarantees that there exists a subsequence of { x n (t)} which converges uniformly on [ 1 k ,1 1 k ]. Then, for k= n 0 , we choose a convergent subsequence of { x n } on [ 1 n 0 ,1 1 n 0 ],

x n 1 ( n 0 ) (t), x n 2 ( n 0 ) (t), x n 3 ( n 0 ) (t),, x n k ( n 0 ) (t),;

for k= n 0 +1, we choose a convergent subsequence of { x n k ( n 0 ) } on [ 1 n 0 + 1 ,1 1 n 0 + 1 ],

x n 1 ( n 0 + 1 ) (t), x n 2 ( n 0 + 1 ) (t), x n 3 ( n 0 + 1 ) (t),, x n k ( n 0 + 1 ) (t),;

for k= n 0 +2, we choose a convergent subsequence of { x n k ( n 0 + 1 ) } on [ 1 n 0 + 2 ,1 1 n 0 + 2 ],

x n 1 ( n 0 + 2 ) ( t ) , x n 2 ( n 0 + 2 ) ( t ) , x n 3 ( n 0 + 2 ) ( t ) , , x n k ( n 0 + 2 ) ( t ) , ; , , , ;

for k= n 0 +j, we choose a convergent subsequence of { x n k ( n 0 + j 1 ) } on [ 1 n 0 + j ,1 1 n 0 + j ],

x n 1 ( n 0 + j ) ( t ) , x n 2 ( n 0 + j ) ( t ) , x n 3 ( n 0 + j ) ( t ) , , x n k ( n 0 + j ) ( t ) , ; , , , .

We may choose the diagonal sequence { x n k + 1 ( n 0 + k ) (t)} which converges everywhere in (0,1) and it is easy to verify that { x n k + 1 ( n 0 + k ) (t)} converges uniformly on any interval [c,d](0,1). Without loss of generality, let { x n k + 1 ( n 0 + k ) (t)} be { x n (t)} in the rest. Putting x(t)= lim n + x n (t), t(0,1), we have x(t) continuous in (0,1) and x(t) m h >0, t[h,1h] for any h(0, 1 2 ) by Lemma 4.3.

Step 2. x(t) satisfies equation (1.1). Fixed t(0,1), we may choose h(0, 1 2 ) such that t(h,1h) and

x n ( t ) = 1 1 2 h h t ( s h ) ( 1 h t ) q ( s ) f ( s , x n ( s ) ) d s + 1 1 2 h t 1 h ( t h ) ( 1 h s ) q ( s ) f ( s , x n ( s ) ) d s + x n ( h ) + x n ( 1 h ) .

Letting n+ in above equation, we have

x ( t ) = 1 1 2 h h t ( s h ) ( 1 h t ) q ( s ) f ( s , x ( s ) ) d s + 1 1 2 h t 1 h ( t h ) ( 1 h s ) q ( s ) f ( s , x ( s ) ) d s + x ( h ) + x ( 1 h ) .
(4.16)

Differentiating equation (4.16), we get the desired result.

Step 3. x(t) satisfies equation (1.2). Let

t n =sup { t : x n ( t ) = x n , x n ( t ) = 0 , t [ 1 n , 1 1 n ] }

and

t n =inf { t : x n ( t ) = x n , x n ( t ) = 0 , t [ 1 n , 1 1 n ] } ,

where x n = max 1 n t 1 1 n x n (t)R. Then

t n , t n [ 1 n , 1 1 n ] , x n ( t n )= x n ( t n ) = x n , x n ( t n )= x n ( t n ) =0.

Using x n (t), 1, t n in place of y(t), λ and t in Lemma 4.2, from equation (4.9); we have

b n x n d x F ( x ) ( 1 + G ¯ ( R ) F ( R ) ) t n 1 1 n 0 t q(s)dsdt

and using x n (t), 1, t n in place of y(t), λ and t in Lemma 4.2, from equation (4.10), we obtain easily

x n ( 1 n ) + b n x n d x F ( x ) ( 1 + G ¯ ( R ) F ( R ) ) 1 n t n 0 t q(s)dsdt.

It follows from the above inequalities that a=inf{ t n }>0 and b=sup{ t n }<1.

  1. (1)

    Fixing z(b,1), we get b n < x n (z)< x n R. From equation (4.9) of the proof in Lemma 4.2, one easily has

    b n x n ( z ) d x F ( x ) ( 1 + G ¯ ( R ) F ( R ) ) z 1 1 n 0 t q(s)dsdt,z(b,1).

Letting n+ in the above inequality and noticing b n 0, we have

0 x ( z ) d x F ( x ) ( 1 + G ¯ ( R ) F ( R ) ) z 1 0 t q(s)dsdt,z(b,1).
(4.17)

It follows from equation (4.17) that x(1)= lim z 1 x(z)=0.

  1. (2)

    Fixing z(0,a), we get x n ( 1 n )+ b n < x n (z)< x n R. From equation (4.10) in the proof of Lemma 4.2, we easily get

    x n ( 1 n ) + b n x n ( z ) d x F ( x ) ( 1 + G ¯ ( R ) F ( R ) ) 1 n z 0 t q(s)dsdt,z(0,a).
    (4.18)

Since lim n + x n (t)=x(t) and x n R, the Lebesgue Dominated Convergent theorem guarantees that

lim n + 1 n 1 1 n x n (t)d α 1 (t)= 0 1 x(t)d α 1 (t), lim n + 1 n 1 1 n x n (t)d α 2 (t)= 0 1 x(t)d α 2 (t).

Since H is continuous, we have

lim n + x n ( 1 n ) = lim n + ( 1 2 n ) H ( ϕ n ( x n ) ) =H ( ϕ ( x ) ) .
(4.19)

Letting n+ in equation (4.18) and noticing b n 0 and equation (4.19), we have

H ( ϕ ( x ) ) x ( z ) d x F ( x ) ( 1 + G ¯ ( R ) F ( R ) ) 0 z 0 t q(s)dsdt,z(0,a).
(4.20)

It follows from equation (4.20) that x(0)= lim z 0 + x(z)=H(ϕ(x)). This complete the proof. □

Example 4.1 Consider

y (t)+ 1 8 ( 1 217 y 2 ( t ) + 1 100 ( 1 y 2 ( t ) y 3 ( t ) t 10 3 t 4 ) ) =0,0<t<1,
(4.21)

with boundary conditions

y(0)= 1 100 | 0 1 y(s)d α 1 (s)+ 0 1 y(s)d α 2 (s) | 3 ,y(1)=0,
(4.22)

where

d α 1 (s)= 1 10 cos4πsds,d α 2 (s)= 1 9 ( e s 2 ) ds.

Then the BVP (4.21)-(4.22) has at least one positive solution.

Let q(t)= 1 8 , f(t,y)= 1 217 y 2 + 1 100 ( 1 y 2 y 3 t 10 3 t 4 ), G(y)= 1 217 y 2 , F(y)= 1 100 y 2 , b(t)= 1 2 t 2 , a(t)= 7 8 t 4 . Let R=2 and H(y)= 1 100 | y | 3 . We have

0 2 1 F ( y ) d y ( 1 + G ( 2 ) F ( 2 ) ) 1 > 200 9 > 1 16 = 0 1 ( 1 s ) q ( s ) d s , max y [ 0 , c 0 r ] H ( r ) = 1 100 ( c 0 r ) 3 < r , r ( 0 , 2 ] ,

where c 0 = 0 1 |d α 1 (s)|+ 0 1 |d α 2 (s)|<1 and

f(t,y)a(t),0<yb(t),t(0,1).

Then (H1)-(H5) hold. Now Theorem 4.1 guarantees that the BVP (4.21)-(4.22) has at least one positive solution.

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Acknowledgements

The author thanks the referees for their suggestions and this research is supported by Young Award of Shandong Province (ZR2013AQ008).

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Correspondence to Baoqiang Yan.

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The author finished the paper himself.

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Keywords

  • nonlocal boundary conditions
  • positive solution
  • fixed point index