Open Access

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

Boundary Value Problems20142014:38

https://doi.org/10.1186/1687-2770-2014-38

Received: 16 April 2013

Accepted: 20 December 2013

Published: 7 February 2014

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.

Keywords

nonlocal boundary conditionspositive solutionfixed point index

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 ) d α ( s )

involving a Stieltjes integral with a signed measure.

In [2], Goodrich considered the following problem:
y = λ g ( 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 ) d α ( 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 , θ Ω and A : Ω ¯ P P continuous and compact. Suppose λ A x x , x Ω P , λ ( 0 , 1 ] . Then
i ( A , Ω P , P ) = 1 .

Lemma 2.2 (see [25])

Let Ω be a bounded open set in real Banach space E, P a cone of E , θ Ω and A : Ω ¯ P P continuous and compact. Suppose A x x , x Ω P . Then
i ( A , Ω P , P ) = 0 .

Lemma 2.3 (see [25, 26])

Let E be a Banach space, R > 0 , B R = { x E : x R } , and F : B R E a continuous compact operator. If x λ F ( x ) for any x E 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 y P , 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 B V ( [ 0 , 1 ] ) with
    0 1 ( 1 t ) d α 1 ( t ) 0 , 0 1 ( 1 t ) 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 : R R 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)
    q C ( 0 , 1 ) , q > 0 on  ( 0 , 1 ) and 0 1 ( 1 t ) q ( t ) d t < .

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 y P (defined in (2.1)). Then
y ( t ) t ( 1 t ) y for  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 y P , 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 : P P is continuous and compact for all 1 ϵ > 0 .

Proof It is easy to prove that T ϵ is well defined and ( T ϵ y ) ( t ) 0 for all t P . For y P , 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 : P P . A standard argument shows that T ϵ : P P 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 ( 1 t ) , 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 ( 1 t ) x t ( 1 t ) , t [ 0 , 1 ] .
(3.6)
(2) 0 < x 1 . 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 ( 1 t ) 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 ( 1 t ) , t [ 0 , 1 ] ,
which means that
r x 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 ( 1 t ) > 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 , 1 a ] . Then, there exists an R > 1 such that for all R R
i ( T ϵ , Ω R P , P ) = 0 , 0 < ϵ 1 .
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 y y for  y P Ω R , 0 < ϵ 1 .
(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 ( 1 t ) y 0 t ( 1 t ) R for t [ 0 , 1 ] . For t [ a , 1 a ] , one has
y 0 ( t ) a 2 R a 2 R = R 1 , t [ a , 1 a ] ,
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 , 1 a ] .
(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 < ϵ 1 .

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 < ϵ 1 .
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 y y for  y P Ω R , 0 < ϵ 1 .
(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 < ϵ 1 .

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 ( 1 s ) q ( s ) d s , 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 : P P is continuous and compact.

Now we show that
y λ 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 ( 1 s ) q ( s ) d s ,
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 ( 1 t ) , 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 ) d y ( 1 + h ( r ) g ( r ) ) t n i 1 ( 1 s ) q ( s ) d s ,
which implies that
y n i ( t n i ) 0 , as  n i + .

This contradicts y n i ( t ) δ r ( 1 t ) 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 ( 1 t ) , 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 ) d s ,
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 ) d s | 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 ) d s , 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 ( 1 t ) , t [ 0 , 1 ] .
(3.27)
Since y n ( n N ) 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 ) } ) d s ,
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 , 1 a ] . 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 ( 1 t ) , 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 y P (defined in (2.1)), we have
ϕ 2 ( y ) = 0 1 y ( t ) 1 8 e s d s y 0 1 s ( 1 s ) 1 8 e s d s ,
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 ) = ( 1 s ) 0 s t d α 2 ( t ) + s s 1 ( 1 t ) 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 ) d y = 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 cos 2 π s d s , 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 B V ( [ 0 , 1 ] ) ;

  • (H2) a ( t ) C ( [ 0 , 1 ] , ( 0 , + ) ) , ( 1 t ) 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 b C ( ( 0 , 1 ) , ( 0 , + ) ) such that
    f ( t , y ) a ( t ) , 0 < y b ( 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 ( 1 s ) q ( s ) d s
    and
    max y [ 0 , r c 0 ] H ( y ) < r , R r > 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 ) ) d s , t [ 1 n , 1 1 n ] .
Proof Let Ω = { y C n : y < R } . For y Ω , 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 y C [ 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 ) = λ q ( 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 < t 1 1 n : y ( t ) = λ y ( 1 1 n ) } and t 1 = sup { t < t 1 1 n : y ( t ) = λ y ( 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 ) d s ( 1 + G ¯ ( R ) F ( R ) ) t 2 i t q ( s ) F ( y ( s ) ) d s F ( y ( t ) ) ( 1 + G ¯ ( R ) F ( R ) ) t 2 i t q ( s ) d s ,
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 ) d s , 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 ) d s ( 1 + G ¯ ( R ) F ( R ) ) 0 t q ( s ) d s ,
(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 ) d s d t , 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 ) d s d t .
(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 ) d s d t ( 1 + G ¯ ( R ) F ( R ) ) z t 1 0 t q ( s ) d s d t .
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 ) d s d t .
(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 ) d s d t .
(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 , 1 h ] } . 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 , 1 h ] such that x n ( t n ) = min { x n ( t ) , t [ h , 1 h ] } . 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 , 1 h ] } , 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 n N (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 n N ; 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 , 1 h ] 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 , 1 h ) 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 ) d s d t
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 ) d s d t .
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 ) d s d t , 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 ) d s d t , 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 ) d s d t , 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 ) d s d t , 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 cos 4 π s d s , d α 2 ( s ) = 1 9 ( e s 2 ) d s .

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 < y b ( 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.

Declarations

Acknowledgements

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

Authors’ Affiliations

(1)
Department of Mathematics, Shandong Normal University

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© Yan; licensee Springer. 2014

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