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

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.

1 Introduction

In this paper we consider the problem

x=Ax+λBx,

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, qC(0,1) with q>0 on (0,1) and 0 1 t(1t)q(t)dt<. 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, ΩE bounded and open. The following theorems are needed in our paper.

Theorem 2.1 ([8])

Suppose θΩ, A:P Ω ¯ P is continuous and compact and

Axμx,xPΩ,μ1.

Then

i(A,PΩ,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:PΩP such that

  1. (1)

    inf x P Ω Kx>0;

  2. (2)

    xAxλKx, xPΩ, λ0,

then

i(A,PΩ,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 } ) Bx=+

and B({xP|rxR}) 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)ds,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 xP{θ}, let

(Bx)(t):= 0 1 G(t,s)a(s) x γ (s)ds,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 θΩ, A:P Ω ¯ P is continuous and compact and B:P{θ}P is singular continuous and compact. Assume that

Axμx,xPΩ,μ1.
(2.1)

Then there exists a λ >0 such that, for any λ(0, λ ), there exist x λ PΩ{θ} with

x λ =A x λ +λB x λ .

Proof Choose x 0 P{θ}, and define

B n x=B ( x + 1 n x 0 ) ,xP,nN.

Set

γ:= inf ( μ , x ) [ 1 , + ) × P Ω μxAx.

Now we claim that

γ>0.
(2.2)

If γ=0, there exists {( μ n , x n )}[1,+)×PΩ 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 xA x n +,

and this is a contradiction.

Next, we show that there exists a ( μ 0 , x 0 )[1,+)×PΩ 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 PΩ 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 }PΩ such that

lim n + B n x n = lim n + B ( x n + 1 n x 0 ) =+.
(2.5)

For all xPΩ, 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:rxR} 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 θPΩ. This contradicts θΩ. Hence, (2.4) holds.

Set

β:= sup n N β n <+

and

λ := γ β >0.

For 0<λ< λ , xPΩ, μ1, we have

Ax+λ B n xμxAxμxλ B n xγβλ>0.

Theorem 2.1 guarantees that

i(A+λ B n ,PΩ,P)=1.
(2.6)

Note (2.6) guarantees that, for any λ(0, λ ), there exists a { x n }PΩ such that

x n =A x n +λ B n x n ,nN.
(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 <+,nN.

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 +λB y 1
(2.11)

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

Corollary 2.1 Suppose that θΩ, A:P Ω ¯ P is continuous and compact and B:P{θ}P is singular continuous and compact. Assume that

Ax<x,xPΩ
(2.12)

or

Axx,xPΩ.
(2.13)

Then there exists a λ >0 such that, for any λ(0, λ ), there exist x λ PΩ with

x λ =A x λ +λB 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) Axμx, xP Ω 1 , μ1;

(C2) inf x P Ω 2 Kx>0;

(C3) xAxμKx, xP Ω 2 , μ0.

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 +λB x λ , 1 , x λ , 2 =A x λ , 2 +λB x λ , 2 .

Proof Choose x 0 P{θ}, and define

B n x=B ( x + 1 n x 0 ) ,xP,nN.

Set

γ 1 := inf ( μ , x ) [ 1 , + ) × P Ω 1 μxAx

and

γ 2 := inf ( μ , x ) [ 0 , + ) × P Ω 2 xAxμKx.

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 Kx=0.
(2.17)

Now since

x n A x n μ n Kx μ n inf x Ω 2 Kx 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 , xP Ω 1 , μ1, we have

Ax+λ B n xμxAxμxλ B n xγβλ>0,

which guarantees that

i(A+λ B n ,P Ω 1 ,P)=1,nN,
(2.18)

and for xP Ω 2 , μ0, we have

x ( A x + λ B n x ) μ K x xAxμKxλ B n xγβλ>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 ),nN.

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 +λB y 1 , y 2 =A y 2 +λB 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)

Ax<x,xP Ω 1

or

Axx,xP Ω 1 ;

(C5)

Ax>x,xP Ω 2

or u 0 P{θ} such that

xAxμ u 0 ,xP Ω 2 ,μ0

or

Axx,xP 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 +λB x λ , 1

and

x λ , 2 =A x λ , 2 +λB 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,,m2, c i 0, i=1,2,,n2, 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,gC([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 uP, define an operator

(Au)(t)= 0 T G 1 (t,s)f ( s , 0 T G 2 ( s , τ ) g ( τ , u ( τ ) ) d τ ) ds,t[0,T]

and for uP{θ}, define an operator

(Bu)(t)= 0 T G 1 (t,s) u γ (s)ds,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

Ax<x,xP 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=Au+λBu.

Let

v(t)= 0 T G 2 (t,s)g ( s , u ( s ) ) ds,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 +λB u 1

and

u 2 =A u 2 +λB u 2 .

Let

v 1 (t)= 0 T G 2 (t,s)g ( s , u 1 ( s ) ) ds,t[0,T]

and

v 2 (t)= 0 T G 2 (t,s)g ( s , u 2 ( s ) ) ds,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].

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Acknowledgements

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

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Correspondence to Ravi P Agarwal.

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All authors contributed equally to the paper. All authors read and approved the final manuscript.

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Yan, B., O’Regan, D. & Agarwal, R.P. Fixed-point theorems for nonlinear operators with singular perturbations and applications. J Inequal Appl 2014, 37 (2014). https://doi.org/10.1186/1029-242X-2014-37

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Keywords

  • boundary value problems
  • singularity
  • fixed-point index