Open Access

Fixed-point theorems for nonlinear operators with singular perturbations and applications

Journal of Inequalities and Applications20142014:37

https://doi.org/10.1186/1029-242X-2014-37

Received: 11 November 2013

Accepted: 3 January 2014

Published: 24 January 2014

Abstract

In this paper, using fixed-point index theory and approximation techniques, we consider the existence and multiplicity of fixed points of some nonlinear operators with singular perturbation. As an application we consider the existence and multiplicity of positive solutions of singular systems of multi-point boundary value problems, which improve the results in the literature.

Keywords

boundary value problemssingularityfixed-point index

1 Introduction

In this paper we consider the problem
x = A x + λ B x ,

where A is continuous and compact and B is a singular continuous and compact operator (defined in Section 2).

In the study of nonlinear phenomena many models give rise to singular boundary value problems (singular in the dependent variable) (see [13]). In [4], Taliaferro showed that the singular boundary value problem
{ y + q ( t ) y α = 0 , 0 < t < 1 , y ( 0 ) = 0 = y ( 1 ) ,

has a C [ 0 , 1 ] C 1 ( 0 , 1 ) solution; here α > 0 , q C ( 0 , 1 ) with q > 0 on ( 0 , 1 ) and 0 1 t ( 1 t ) q ( t ) d t < . For more recent work we refer the reader to [514] and the references therein.

In this paper we consider abstract singular operators (defined in Section 2) and we consider the existence and multiplicity of fixed points of some nonlinear operators with singular perturbations. As an application we discuss the existence and multiplicity of positive solutions of singular systems of multi-point boundary value problems.

2 Fixed-point theorems

Let E be a Banach space, P a cone of E, Ω E bounded and open. The following theorems are needed in our paper.

Theorem 2.1 ([8])

Suppose θ Ω , A : P Ω ¯ P is continuous and compact and
A x μ x , x P Ω , μ 1 .
Then
i ( A , P Ω , P ) = 1 .

Theorem 2.2 ([8])

Assume that A : P Ω ¯ P is continuous and compact. If there exists a compact and continuous operator K : P Ω P such that
  1. (1)

    inf x P Ω K x > 0 ;

     
  2. (2)

    x A x λ K x , x P Ω , λ 0 ,

     
then
i ( A , P Ω , P ) = 0 .

Now we give a new definition.

Definition 2.1 If B : P { θ } P is continuous with
lim x θ , x ( P { θ } ) B x = +

and B ( { x P | r x R } ) is relatively compact, for any 0 < r < R < + , then B : P { θ } P is called a singular continuous and compact operator.

Remark Consider
x ( t ) + a ( t ) x γ ( t ) = 0 , t ( 0 , 1 ) , x ( 0 ) = 0 , x ( 1 ) = 0 ,
where 1 > γ > 0 and a ( t ) C ( ( 0 , 1 ) , ( 0 , + ) ) L 1 ( 0 , 1 ) or equivalently
x ( t ) = 0 1 G ( t , s ) a ( s ) x γ ( s ) d s , t [ 0 , 1 ] ,
where
G ( t , s ) = { s ( 1 t ) , 0 s t 1 , t ( 1 s ) , 0 t s 1 .
Set
P : = { x C [ 0 , 1 ] : x ( t ) t ( 1 t ) x } ,
where x = max t [ 0 , 1 ] | x ( t ) | . For x P { θ } , let
( B x ) ( t ) : = 0 1 G ( t , s ) a ( s ) x γ ( s ) d s , t [ 0 , 1 ] .

It is easy to see that B : P { θ } P is a singular continuous and compact operator (see [7, 14]).

Theorem 2.3 Suppose that θ Ω , A : P Ω ¯ P is continuous and compact and B : P { θ } P is singular continuous and compact. Assume that
A x μ x , x P Ω , μ 1 .
(2.1)
Then there exists a λ > 0 such that, for any λ ( 0 , λ ) , there exist x λ P Ω { θ } with
x λ = A x λ + λ B x λ .
Proof Choose x 0 P { θ } , and define
B n x = B ( x + 1 n x 0 ) , x P , n N .
Set
γ : = inf ( μ , x ) [ 1 , + ) × P Ω μ x A x .
Now we claim that
γ > 0 .
(2.2)
If γ = 0 , there exists { ( μ n , x n ) } [ 1 , + ) × P Ω such that
lim n + μ n x n A x n = 0 .
(2.3)

First, we show { μ n } is bounded.

To see this suppose { μ n } is unbounded. Without loss of generality, we assume that lim n + μ n = + . Then
0 μ n x n A x n | μ n | x n A x n μ n inf x P Ω x A x n + ,

and this is a contradiction.

Next, we show that there exists a ( μ 0 , x 0 ) [ 1 , + ) × P Ω such that
μ 0 x 0 A x 0 = 0 .
The boundedness of { μ n } means that { μ n } has a convergent subsequence. Without loss of generality, we assume that μ n μ 0 1 . Since { x n } is bounded and A is continuous and compact, { A x n } has a convergent subsequence { A x n i } with lim n i + A x n i y 0 . From (2.3), we have
lim n i + μ n i x n i A x n i = 0 ,
which implies that
μ n i x n i y 0 , n i + .
Then
x n i 1 μ 0 y 0 , as  n i + .
Let x 0 = 1 μ 0 y 0 . Clearly, x 0 P Ω and
μ 0 x 0 A x 0 = lim n i + μ n i x n i A x n i = 0 ,

which contradicts (2.1).

Let
β n = sup x P Ω B n x .
Now we claim that
sup n N β n < + .
(2.4)
To see this suppose that
sup n N β n = + .
Without loss of generality, assume that
lim n + β n = + ,
which implies that there exists a sequence { x n } P Ω such that
lim n + B n x n = lim n + B ( x n + 1 n x 0 ) = + .
(2.5)
For all x P Ω , we have
x + 1 n x 0 x + x 0 sup x P Ω x + x 0 : = R < + .
Since, for any r > 0 , B : P { x : r x R } is relatively compact, (2.5) guarantees that there exists a subsequence { x n i } { x n } such that
lim n i + x n i + 1 n i x 0 = 0 .
Thus
lim n i + x n i = 0 ,

which implies that θ P Ω . This contradicts θ Ω . Hence, (2.4) holds.

Set
β : = sup n N β n < +
and
λ : = γ β > 0 .
For 0 < λ < λ , x P Ω , μ 1 , we have
A x + λ B n x μ x A x μ x λ B n x γ β λ > 0 .
Theorem 2.1 guarantees that
i ( A + λ B n , P Ω , P ) = 1 .
(2.6)
Note (2.6) guarantees that, for any λ ( 0 , λ ) , there exists a { x n } P Ω such that
x n = A x n + λ B n x n , n N .
(2.7)
Now we show that
inf n N x n > 0 ,
(2.8)
which implies that
inf n N x n + 1 n x 0 > 0 .
To see this suppose that
inf n N x n = 0 .
Then there exists a { x n i } such that
lim n i + x n i = 0 ,
(2.9)
and so
lim n i + x n i + 1 n i x 0 = 0 .
Thus
lim n i + B n i x n i = lim n i + B ( x n i + 1 n i x 0 ) = + .
(2.10)
The compactness of A guarantees that { A x n i } has a convergent subsequence. Without loss of generality, we assume that lim n i + A x n i = y 0 . From (2.10), we have
lim n i + x n i = lim n i + A x n i + λ B n i x n i lim n i + λ B n i x n i lim n i + A x n i = + ,

which contradicts (2.9).

Now (2.8) guarantees that
0 < inf n N x n + 1 n x 0 x n + 1 n x 0 sup x P Ω x + x 0 < + , n N .
Then { A x n + λ B n x n } has a convergent subsequence. Without loss of generality, we assume that
A x n + λ B n x n y 1 , as  n + .
Then
x n y 1 , as  n + .
Now (2.8) guarantees that y 1 θ . Letting n + in (2.7), and we have
y 1 = A y 1 + λ B y 1
(2.11)

and y 1 P Ω { θ } . The proof is complete. □

Corollary 2.1 Suppose that θ Ω , A : P Ω ¯ P is continuous and compact and B : P { θ } P is singular continuous and compact. Assume that
A x < x , x P Ω
(2.12)
or
A x x , x P Ω .
(2.13)
Then there exists a λ > 0 such that, for any λ ( 0 , λ ) , there exist x λ P Ω with
x λ = A x λ + λ B x λ .

It is easy to see that (2.12) or (2.13) guarantees that (2.1) holds (see [8]).

Theorem 2.4 Suppose that Ω 1 , Ω 2 are bounded open sets and θ Ω 1 Ω 2 , A , K : P Ω ¯ 2 P are continuous and compact and B : P { θ } P is singular continuous and compact. Assume that

(C1) A x μ x , x P Ω 1 , μ 1 ;

(C2) inf x P Ω 2 K x > 0 ;

(C3) x A x μ K x , x P Ω 2 , μ 0 .

Then there exists a λ > 0 such that, for any λ ( 0 , λ ) , there exist x λ , 1 ( P Ω 1 { θ } ) and x λ , 2 P ( Ω 2 Ω ¯ 1 ) with
x λ , 1 = A x λ , 1 + λ B x λ , 1 , x λ , 2 = A x λ , 2 + λ B x λ , 2 .
Proof Choose x 0 P { θ } , and define
B n x = B ( x + 1 n x 0 ) , x P , n N .
Set
γ 1 : = inf ( μ , x ) [ 1 , + ) × P Ω 1 μ x A x
and
γ 2 : = inf ( μ , x ) [ 0 , + ) × P Ω 2 x A x μ K x .
We claim that
γ 1 > 0 , γ 2 > 0 .
(2.14)
An argument similar to that in (2.2) shows that
γ 1 > 0 .
(2.15)
Now we show that
γ 2 > 0 .
(2.16)
To see this suppose that γ 2 = 0 . Then there exists { ( μ n , x n ) } [ 0 , + ) × ( P Ω 2 ) such that
lim n + x n A x n μ n K x = 0 .
(2.17)
Now since
x n A x n μ n K x μ n inf x Ω 2 K x x n A x n ,
we have { μ n } is bounded, which means that { μ n } has a convergent subsequence. Without loss of generality, we assume that
lim n + μ n = μ 0 .
Since { x n } is bounded and A and K are compact, { A x n } and { K x n } have convergent subsequences { A x n i } and { K x n i } with lim n i + A x n i = y 0 and lim n i + K x n i = z 0 . Now
lim n + x n A x n μ n K x n = 0 ,
which implies that
lim n i + x n i y 0 μ 0 z 0 = 0 .
Let x 0 = y 0 + μ 0 z 0 . Clearly, x 0 P Ω 2 . Now
x 0 A x 0 μ 0 K x 0 = lim n i + x n i A x n i μ n i K x n i = 0 ,

which contradicts condition (C3).

Consequently, (2.16) is true, which together with (2.15) yields (2.14).

Let γ = min { γ 1 , γ 2 } . Obviously, γ > 0 .

Let
β n , 1 = sup x P Ω 1 B n x , β n , 2 = sup x P Ω 2 B n x .
An argument similar to that in (2.4) shows that
sup n N β n , 1 < + , sup n N β n , 2 < + .
Let
β : = max { sup n N β n , 1 , sup n N β n , 1 } < +
and
λ : = γ β > 0 .
For 0 < λ < λ , x P Ω 1 , μ 1 , we have
A x + λ B n x μ x A x μ x λ B n x γ β λ > 0 ,
which guarantees that
i ( A + λ B n , P Ω 1 , P ) = 1 , n N ,
(2.18)
and for x P Ω 2 , μ 0 , we have
x ( A x + λ B n x ) μ K x x A x μ K x λ B n x γ β λ > 0 ,
which guarantees that
i ( A + λ B n , P Ω 2 , P ) = 0 .
Thus
i ( A + λ B n , P ( Ω 2 Ω ¯ 1 ) , P ) = 1 .
(2.19)
Now (2.18) and (2.19) guarantee that there exist { x n , 1 } P Ω 1 and { x n , 2 } P ( Ω 2 Ω ¯ 1 ) such that
x n , 1 = A x n , 1 + λ B n x n , 1 , x n , 2 = A x n , 2 + λ B n x n , 2 , λ ( 0 , λ ) , n N .
An argument similar to that in (2.11) shows that there exist y 1 P Ω 1 { θ } and y 2 P ( Ω 2 Ω ¯ 1 ) with
y 1 = A y 1 + λ B y 1 , y 2 = A y 2 + λ B y 2 , λ ( 0 , λ ) .

The proof is complete. □

Corollary 2.2 Suppose that Ω 1 , Ω 2 are bounded open sets and θ Ω 1 Ω 2 , A : P Ω ¯ 2 P is continuous and compact and B : P { θ } P is singular continuous and compact. Assume that

(C4)
A x < x , x P Ω 1
or
A x x , x P Ω 1 ;
(C5)
A x > x , x P Ω 2
or u 0 P { θ } such that
x A x μ u 0 , x P Ω 2 , μ 0
or
A x x , x P Ω 2 .
Then there exists a λ > 0 such that, for any λ ( 0 , λ ) , there exist x λ , 1 P Ω 1 { θ } and x λ , 2 P ( Ω 2 Ω ¯ 1 ) with
x λ , 1 = A x λ , 1 + λ B x λ , 1
and
x λ , 2 = A x λ , 2 + λ B x λ , 2 .

It is easy to see that (C4) and (C5) guarantee that (C1)-(C3) hold (see [8]).

3 Applications for singular systems of multi-point boundary value problems

In [9], Henderson and Luca considered the system of nonlinear second-order ordinary differential equations
{ u ( t ) + f ( t , v ( t ) ) = 0 , t ( 0 , T ) , v ( t ) + g ( t , u ( t ) ) = 0 , t ( 0 , T )
(3.1)
with multi-point boundary conditions
{ u ( 0 ) = 0 , u ( T ) = i = 1 m 2 b i u ( ξ i ) , m 3 , v ( 0 ) = 0 , v ( T ) = i = 1 n 2 c i v ( η i ) , n 3 .
(3.2)

The following conditions come from [9]:

(H1) 0 < ξ 1 < < ξ m 2 < T , 0 < η 1 < < η n 2 < T , b i > 0 , i = 1 , 2 , , m 2 , c i 0 , i = 1 , 2 , , n 2 , d = T i = 1 m 2 b i ξ i > 0 , e = T i = 1 n 2 c i η i > 0 , i = 1 m 2 b i ξ i > 0 , i = 1 n 2 c i η i > 0 ,

(H2) we have the functions f , g C ( [ 0 , T ] × [ 0 , + ) , [ 0 , + ) ) and f ( t , 0 ) = 0 , g ( t , 0 ) = 0 for all t [ 0 , T ] ,

(H3) there exists a positive constant p ( 0 , 1 ] such that
  1. (1)

    f i = lim u + inf inf t [ 0 , T ] f ( t , u ) u p ( 0 , + ] ;

     
  2. (2)

    g i = lim u + inf inf t [ 0 , T ] g ( t , u ) u 1 p = + ,

     
(H4) there exists a r ( 0 , + ) such that
  1. (1)

    f s = lim u + sup sup t [ 0 , T ] f ( t , u ) u r ( 0 , + ] ;

     
  2. (2)

    g s = lim u + sup sup t [ 0 , T ] g ( t , u ) u 1 r = 0 ,

     
(H5)
  1. (1)

    f 0 i = lim u 0 + inf inf t [ 0 , T ] f ( t , u ) u ( 0 , + ] ;

     
  2. (2)

    g 0 i = lim u 0 + inf inf t [ 0 , T ] g ( t , u ) u = + ,

     
(H6) for each t [ 0 , T ] , f ( t , u ) and g ( t , u ) are nondecreasing with respect to u, and there exists a constant N > 0 such that
f ( t , m 0 0 T g ( s , N ) d s ) < N m 0 , t [ 0 , T ] ,

where m 0 = T 2 4 max { a 1 T , a ˜ 1 } and a 1 , a ˜ 1 are defined in [9].

Theorem 3.1 ([9])

Assume that (H1)-(H2) and (H4)-(H5) hold. Then Problem (3.1), (3.2) has at least one positive solution ( u ( t ) , v ( t ) ) , t [ 0 , T ] .

Theorem 3.2 ([9])

Assume that (H1)-(H3) and (H5)-(H6) hold. Then Problem (3.1), (3.2) has at least two positive solutions ( u 1 ( t ) , v 1 ( t ) ) , ( u 2 ( t ) , v 2 ( t ) ) , t [ 0 , T ] .

Here we consider
{ u ( t ) + f ( t , v ( t ) ) + λ u γ = 0 , t ( 0 , T ) , v ( t ) + g ( t , u ( t ) ) = 0 , t ( 0 , T )
(3.3)

with multi-point boundary conditions (3.2), where 1 > γ > 0 .

Let C [ 0 , T ] : = { x : [ 0 , T ] R : x ( t )  is continuous on  [ 0 , T ] } with norm
x = max t [ 0 , T ] | x ( t ) | .
Obviously, C [ 0 , T ] is a Banach space. Let
P : = { x C [ 0 , T ] : x ( t ) 0  is concave and  inf t [ θ 0 , T ] x ( t ) γ x } ,

where θ 0 and γ are defined in Section 2 in [9].

For u P , define an operator
( A u ) ( t ) = 0 T G 1 ( t , s ) f ( s , 0 T G 2 ( s , τ ) g ( τ , u ( τ ) ) d τ ) d s , t [ 0 , T ]
and for u P { θ } , define an operator
( B u ) ( t ) = 0 T G 1 ( t , s ) u γ ( s ) d s , t [ 0 , T ] ,

where G 1 ( t , s ) and G 2 ( t , s ) are defined in [9].

It is easy to see that B : P { θ } P is a singular continuous and compact operator (see [9, 11]).

Theorem 3.3 Assume that (H1)-(H2) and (H4) hold. Then there exists a λ > 0 such that Problem (3.3), (3.2) has at least one positive solution ( u ( t ) , v ( t ) ) , t [ 0 , T ] for all λ ( 0 , λ ) .

Proof Let B R be defined as that in Theorem 3.2 of [9]. From the proof in [9], it is easy to see that
A x < x , x P B R .
Now Corollary 2.1 guarantees that there exists a λ > 0 such that, for any λ ( 0 , λ ) , there exist u ( P B R { θ } ) such that
u = A u + λ B u .
Let
v ( t ) = 0 T G 2 ( t , s ) g ( s , u ( s ) ) d s , t [ 0 , T ] .

Then ( u ( t ) , v ( t ) ) is a positive solution for (3.3), (3.2). The proof is complete. □

Theorem 3.4 Assume that (H1)-(H3) and (H6) hold. Then Problem (3.3), (3.2) has at least two positive solutions ( u 1 ( t ) , v 1 ( t ) ) , ( u 2 ( t ) , v 2 ( t ) ) , t [ 0 , T ] .

Proof Let B N and B L be defined as in Theorem 3.3 of [9]. From the proof in [9], it is easy to see that
A x < x , x P B N , x A x + λ u 0 , x P B L , λ 0 , L > N .
Now Corollary 2.2 guarantees that there exists a λ > 0 , for any λ ( 0 , λ ) , such that there exist u 1 ( P B N { θ } ) and u 2 P ( B L B ¯ N ) with
u 1 = A u 1 + λ B u 1
and
u 2 = A u 2 + λ B u 2 .
Let
v 1 ( t ) = 0 T G 2 ( t , s ) g ( s , u 1 ( s ) ) d s , t [ 0 , T ]
and
v 2 ( t ) = 0 T G 2 ( t , s ) g ( s , u 2 ( s ) ) d s , t [ 0 , T ] .

Then ( u 1 ( t ) , v 1 ( t ) ) and ( u 2 ( t ) , v 2 ( t ) ) are two positive solutions for (3.3), (3.2). The proof is complete. □

Remark Note that f and g have no singularity at u = 0 and v = 0 in [9, 10], so Theorems 3.3 and 3.4 improve the results in [9, 10].

Declarations

Acknowledgements

This research is supported by Young Award of Shandong Province (ZR2013AQ008).

Authors’ Affiliations

(1)
Department of Mathematics, Shandong Normal University
(2)
School of Mathematics, Statistics and Applied Mathematics, National University of Ireland
(3)
Department of Mathematics, Faculty of Science, King Abdulaziz University
(4)
Department of Mathematics, Texas A&M University-Kingsville

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© Yan et al.; licensee Springer. 2014

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