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Multiplicity and uniqueness results for the singular nonlocal boundary value problem involving nonlinear integral conditions

Abstract

In this paper, using fixed point index and the mixed monotone technique, we present some multiplicity and uniqueness results for the singular nonlocal boundary value problems involving nonlinear integral conditions. Our nonlinearity may be singular in its dependent variable and it is allowed to change sign.

1 Introduction

In this paper, we consider the existence of positive solutions of nonlinear nonlocal 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)=α[y]= 0 1 ( y ( s ) ) a dA(s),y(1)=β[y]= 0 1 ( y ( s ) ) b dB(s)
(1.2)

involving Stieltjes integrals, a0, b0.

In [1], using the Leray-Schauder alternative, Z. Yang considered the problem

y =f ( y ( t ) ) ,t(0,1)
(1.3)

with integral boundary conditions

y(0)=α[y]= 0 1 y(s)dA(s),y(1)=β[y]= 0 1 y(s)dB(s)
(1.4)

and discussed the existence and uniqueness of a positive solution for BVP (1.3)-(1.4) with a sign-changing nonlinearity f. C.S. Goodrich discussed (1.1) with nonlinear integral conditions

y(0)= H 1 ( 0 1 y ( s ) d A ( s ) ) + E H 2 ( s , y ( s ) ) ds,y(1)=0,

where E(0,1) is some measurable set (see [2]). Moreover, there are some interesting results when the measures are signed (see [3]–[5]). Using the mixed monotone technique, L. Kong considered

y =λf ( t , y ( t ) ) ,t(0,1)
(1.5)

with (1.4) and discussed the uniqueness of positive solutions (see [4]). J.R.L. Webb discussed the multiplicity of positive solutions for BVP (1.5)-(1.4) when f(t,y) is positive and continuous on (0,1)×[0,+); note that f has no singularities at y=0 (see [5]). Using the fixed point index, G. Infante discussed (1.1) with nonlinear integral boundary conditions (see [3]),

y(0)+ H 1 ( 0 1 y ( s ) d A ( s ) ) =0,y(1)+u(η)= H 2 ( 0 1 y ( s ) d B ( s ) ) .

Inspired by the above works and [6]–[16], we consider the BVP (1.1)-(1.2) when f is singular at y=0 and f may be sign changing. Using the fixed point index and the mixed monotone technique we establish some new existence results for the BVP (1.1)-(1.2).

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 BVP (1.1)-(1.2) when f is positive. In Section 4, we discuss the multiplicity of positive solutions for the semipositone BVP (1.1)-(1.2). In Section 5, using the mixed monotone technique, we discuss the uniqueness of a positive solution of BVP (1.1)-(1.2).

2 Preliminaries

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. Define

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

It is easy to prove P is a cone of C[0,1].

Lemma 2.1

(see [17])

Let Ω be a bounded open set in a real Banach space E, P be a cone of E, θΩandA: Ω ¯ PPbe continuous and compact. SupposeλAxx, xΩP, λ(0,1]. Then

i(A,ΩP,P)=1.
(2.2)

Lemma 2.2

(see [17])

Let Ω be a bounded open set in a real Banach space E, P be a cone of E, θΩandA: Ω ¯ PPbe continuous and compact. SupposeAxx, xΩP. Then

i(A,ΩP,P)=0.
(2.3)

Lemma 2.3

(see [18])

LetyP (defined in (2.1)). Then

y(t)t(1t)yfor t[0,1].
(2.4)

Now we present the following conditions for convenience:

(C1):A and B are of bounded variation with positive measures, 0< 0 1 dA(s), a>0, 0< 0 1 dB(s), b>0,

(C2):

{ 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 ] ,
(2.5)

(C3):

qC(0,1),q>0on (0,1)and 0 1 t(1t)q(t)dt<,
(2.6)

(C4):

f:[0,1]×(0,)(0,) is continuous,
(2.7)

(C5): there exists a continuous function g:[0,1]×(0,)×(0,)(0,) with g(t,x,y) nondecreasing in x and nonincreasing in y and for x>0 we have f(t,x)=g(t,x,x). Moreover, there is a constant θ with 0θ<1 such that

g ( t , λ x , 1 λ y ) λ θ g(t,x,y),x>0,y>0,0<λ<1.

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

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

( T ϵ y ) ( t ) = ( 1 t ) α [ y ] + t β [ 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 .

Lemma 3.1

Suppose (C1)-(C4) hold. Then T ϵ :PPis continuous and completely continuous for all1ϵ>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 ) = α [ y ] , ( T ϵ y ) ( 1 ) = β [ y ] ,
(3.2)

so

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

Consequently, T ϵ :PP. A standard argument shows that T ϵ :PP is continuous and completely continuous (see [4], [19], [20]). □

Lemma 3.2

Suppose that 0 1 dA(s)>0and 0 1 dB(s)>0. Then

0 1 s a ( 1 s ) a dA(s)>0, 0 1 s b ( 1 s ) b dB(s)>0.

The proof is trivial and we omit it.

Define

H = { x C ( [ 0 , 1 ] , R ) C 1 ( [ 0 , 1 ) , R ) C ( ( 0 , 1 ) , ( 0 , + ) ) C 2 ( ( 0 , 1 ) , R ) | x  satisfies x ( t ) + q ( t ) f ( t , max { ϵ , x ( t ) } ) = 0 , 0 < t < 1 , x ( 0 ) = α [ x ] , x ( 1 ) = β [ x ] , 1 ϵ > 0 } .

Lemma 3.3

IfHand (C2) hold, there exists a δ 0 >0such that

x(t) δ 0 ,t[0,1],xH.

Proof

Suppose xH. There are two cases to consider:

  1. (1)

    x>1. Lemma 2.3 implies that

    x(t)t(1t)xt(1t),t[0,1].
    (3.4)
  1. (2)

    0<x1. Condition (C2) guarantees that

    x ( t ) = ( 1 t ) α [ x ] + t β [ 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 and γ 0 (0)=0 and γ 0 (1)=0, Lemma 2.3 implies that

γ 0 (t)t(1t) γ 0 ,t[0,1].
(3.5)

Let δ 1 =min{1, γ 0 }. From (3.4) and (3.5), one has

x(t) δ 1 t(1t),t[0,1].

Lemma 3.2 implies that

x(0)=α[x]= 0 1 x a (s)dA(s) δ 1 a 0 1 s a ( 1 s ) a dA(s)>0

and

x(1)=β[x]= 0 1 x b (s)dB(s) δ 1 b 0 1 s b ( 1 s ) b dB(s)>0.

Set

δ 0 =min { δ 1 a 0 1 s a ( 1 s ) a d A ( s ) , δ 1 b 0 1 s b ( 1 s ) b d B ( s ) } .

Since x(t) is concave on [0,1], we have

x(t) δ 0 ,t[0,1],xH.

The proof is complete. □

Lemma 3.4

Suppose that there exists an a ¯ (0, 1 2 ) such that

lim y + f ( t , y ) y =+
(3.6)

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

i( T ϵ , Ω R P,P)=0,0<ϵ1.

Proof

From (3.6), there exists a R 1 >1 such that

f(t,y) N y,y R 1 ,
(3.7)

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.
(3.8)

Suppose that there exists a y 0 P Ω R with T ϵ y 0 y 0 . Then y 0 =R. Also since y 0 (t) is concave on [0,1] (since y 0 P) we have from Lemma 2.3 that y 0 (t)t(1t) y 0 t(1t)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 (3.7) yields the result that

f ( t , y 0 ( t ) ) N y 0 (t) N a ¯ 2 R,t[ a ¯ ,1 a ¯ ].
(3.9)

Then we have, using (3.9),

y 0 ( a ¯ ) T ϵ y 0 ( a ¯ ) = ( 1 a ¯ ) α [ y 0 ] + a ¯ β [ 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 (3.8) is true. Lemma 2.2 guarantees that

i( T ϵ , Ω R P,P)=0,0<ϵ1.

The proof is complete. □

Lemma 3.5

Suppose thatmax{a,b}>1. Then there exists an R >1such that for allR R

i( T ϵ , Ω R P,P)=0,0<ϵ1.

Proof

Since max{a,b}>1, without loss of generality, we suppose that a>1. Let R >1 with R a 1 0 1 s a ( 1 s ) a dA(s)>1. Set

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

Now we show

T ϵ xx,x Ω R P,0<ϵ1.

In fact, suppose that x 0 Ω R P and satisfies

T ϵ x 0 x 0 .

Lemma 2.3 guarantees that

x 0 (t) x 0 t(1t)Rt(1t),t[0,1].

Then

R x 0 ( 0 ) = 0 1 x 0 a ( s ) d A ( s ) 0 1 x 0 a s a ( 1 s ) a d A ( s ) = R R a 1 0 1 s a ( 1 s ) a d A ( s ) R R a 1 0 1 s a ( 1 s ) a d A ( s ) > R .

This is a contradiction. Lemma 2.2 guarantees that

i( T ϵ , Ω R P,P)=0,R R ,0<ϵ1.

The proof is complete. □

Theorem 3.1

Suppose (C1), (C2), (C3), and (C4) 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 , )
(3.10)

and

sup r ( 0 , + ) 1 1 + h ( r ) g ( r ) c 0 max { r a , r b } r d y g ( y ) > b 0
(3.11)

hold; here

c 0 = max { 0 1 d A ( s ) , 0 1 d B ( s ) } , b 0 = max { 2 0 1 2 t ( 1 t ) q ( t ) d t , 2 1 2 1 t ( 1 t ) q ( t ) d t } .
(3.12)

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

Proof

Choose ϵ>0 and r>0 with ϵ<min{1, c 0 { r a , r b }} and

1 1 + h ( r ) g ( r ) c 0 { r a , r b } r d y g ( y ) > b 0 .
(3.13)

Let

Ω 1 = { y C [ 0 , 1 ] | y < r } ,

and T ϵ is defined in (3.1). Lemma 3.1 guarantees that T ϵ :PP is continuous and completely continuous.

Now we show that

yλ T ϵ y,y Ω 1 P,λ(0,1].
(3.14)

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

{ y 0 + λ 0 q ( t ) f ( t , max { ϵ , y 0 ( t ) } ) = 0 , 0 < t < 1 , y 0 ( 0 ) = λ 0 α [ y ] , y 0 ( 1 ) = λ 0 β [ y 0 ] .
(3.15)

Then y 0 (t)0 on (0,1) and y 0 (0)= λ 0 α[ y 0 ] r a 0 1 dA(s) c 0 r a <r= y 0 , y 0 (1)= λ 0 β[ y 0 ] r b 0 1 dB(s) c 0 r b <r= y 0 , which guarantees that there exists a t 0 (0,1), r= y 0 = y 0 ( t 0 ) with y ( t 0 )=0 and y 0 (t)0 for all t(0, t 0 ) and y 0 (t)0 for all t( t 0 ,1). For t(0,1), we have

y 0 ( t ) g ( max { ϵ , y 0 ( t ) } ) { 1 + h ( max { ϵ , y 0 ( t ) } ) g ( max { ϵ , y 0 ( t ) } ) } q ( t ) g ( max { ϵ , y 0 ( t ) } ) { 1 + h ( r ) g ( r ) } q ( t ) .
(3.16)

Integrate from t (t< t 0 ) to t 0 to obtain

y 0 (t)g ( max { ϵ , y 0 ( t ) } ) { 1 + h ( r ) g ( r ) } t t 0 q(s)dsg ( y 0 ( t ) ) { 1 + h ( r ) g ( r ) } t t 0 q(s)ds

and then integrate from 0 to t 0 to obtain

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

which together with α[ y 0 ] c 0 r a c 0 max{ r a , r b } yields the result that

c 0 max { r α , r β } r d y g ( y ) α [ y 0 ] r d y g ( y ) { 1 + h ( r ) g ( r ) } 1 1 t 0 0 t 0 s(1s)q(s)ds.
(3.17)

Similarly if we integrate (3.16) from t 0 to t (t t 0 ) and then from t 0 to 1 we obtain

c 0 max { r a , r b } r d u g ( u ) { 1 + h ( r ) g ( r ) } 1 t 0 t 0 1 s(1s)q(s)ds.
(3.18)

Now (3.17) and (3.18) imply

c 0 max { r a , r b } r d u g ( u ) b 0 { 1 + h ( r ) g ( r ) } ,

which contradicts (3.13). Therefore, (3.14) is true. Lemma 2.1 implies that

i( T ϵ , Ω 1 P,P)=1,
(3.19)

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

T ϵ y 1 , ϵ = y 1 , ϵ ,

i.e., H in Lemma 3.3. Moreover, Lemma 3.3 is true, which guarantees that there exists a δ 0 >0 such that

x(t) δ 0 ,t[0,1],xH.
(3.20)

Now let ϵ 0 = δ 0 2 . From the above proof, there exists a x 1 , ϵ 0 Ω 1 P such that

T ϵ 0 x 1 , ϵ 0 = x 1 , ϵ 0 .

Since 0< ϵ 0 1, (3.20) implies that

x 1 , ϵ 0 (t) δ 0 > ϵ 0 ,t[0,1].
(3.21)

Moreover, since x 1 , ϵ 0 (t) satisfies

{ x 1 , ϵ 0 ( t ) + q ( t ) f ( t , max { ϵ 0 , x 1 , ϵ 0 ( t ) } ) = 0 , 0 < t < 1 , x 1 , ϵ 0 ( 0 ) = α [ x 1 , ϵ 0 ] , x 1 , ϵ 0 ( 1 ) = β [ x 1 , ϵ 0 ] ,

(3.21) guarantees that

{ x 1 , ϵ 0 ( t ) + q ( t ) f ( t , x 1 , ϵ 0 ( t ) ) = 0 , 0 < t < 1 , x 1 , ϵ 0 ( 0 ) = α [ x 1 , ϵ 0 ] , x 1 , ϵ 0 ( 1 ) = β [ x 1 , ϵ 0 ] .

Thus, BVP (1.1)-(1.2) has at least one positive solution. The proof is complete. □

Theorem 3.2

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

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

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

Proof

Choose r>0 as in (3.13), ϵ 0 >0 with ϵ 0 <min{ δ 0 ,1, c 0 { r a , r b }}, where δ 0 is defined in (3.20), and R>max{r, R } in Lemma 3.4. Set

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

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

i( T ϵ 0 , Ω 1 P,P)=1

and

i( T ϵ 0 , Ω 2 P,P)=0,

which implies that

i ( T ϵ 0 , ( Ω 2 Ω 1 ) P , P ) =1.

Thus, there exist x 1 , ϵ 0 Ω 1 P and x 2 , ϵ 0 ( Ω 2 Ω 1 )P such that

T ϵ 0 x 1 , ϵ 0 = x 1 , ϵ 0 , T ϵ 0 x 2 , ϵ 0 = x 2 , ϵ 0 .

From the proof of Theorem 3.1, x 1 , ϵ 0 (t) and x 2 , ϵ 0 (t) are two positive solutions for BVP (1.1)-(1.2). The proof is complete. □

Theorem 3.3

Suppose the conditions of Theorem  3.1hold andmax{a,b}>1. Then BVP (1.1)-(1.2) has at least two positive solutions.

Proof

Choose r>0 as in (3.13), ϵ 0 >0 with ϵ 0 <min{ δ 0 ,1, c 0 { r a , r b }}, where δ 0 is defined in (3.20), and R>max{r, R } in Lemma 3.5. Set

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

From the proof of Theorem 3.1 and Lemma 3.5, we have

i( T ϵ 0 , Ω 1 P,P)=1

and

i( T ϵ 0 , Ω 2 P,P)=0,

which implies that

i ( T ϵ 0 , ( Ω 2 Ω 1 ) P , P ) =1.

Thus, there exist x 1 , ϵ 0 Ω 1 P and x 2 , ϵ 0 ( Ω 2 Ω 1 )P such that

T ϵ 0 x 1 , ϵ 0 = x 1 , ϵ 0 , T ϵ 0 x 2 , ϵ 0 = x 2 , ϵ 0 .

From the proof of Theorem 3.1, x 1 , ϵ 0 (t) and x 2 , ϵ 0 (t) are two positive solutions for BVP (1.1)-(1.2). The proof is complete. □

Example 3.1

Consider

y (t)+μ ( y δ 1 ( t ) + y δ 2 ( t ) ) =0,0<t<1,
(3.22)

with

y(0)= 0 1 y 1 3 (s)dA(s),y(1)= 0 1 y 1 2 (s)dB(s),dA(s)= 1 2 ds,dB(s)= 1 8 d e s ,
(3.23)

where δ 1 >0, δ 2 >1.

Let q(t)=μ, f(t,y)= y δ 1 + y δ 2 , g(y)= y δ 1 , h(y)= y δ 2 , c 0 =max{ 0 1 dA(s), 0 1 dB(s)}= 1 2 , b 0 = 1 6 μ. It is easy to see that (C1)-(C4) and (3.10) hold. Since

1 1 + h ( 1 ) g ( 1 ) c 0 max { 1 1 3 , 1 1 2 } 1 1 g ( y ) dy= 1 ( 1 2 ) δ 1 + 1 2 ( 1 + δ 1 ) ,

letting μ 0 <3 1 ( 1 2 ) δ 1 + 1 2 ( 1 + δ 1 ) , we have

sup r ( 0 , + ) 1 1 + h ( r ) g ( r ) c 0 max { r 1 3 , r 1 2 } r 1 g ( y ) dy> b 0

for all μ μ 0 , which guarantees that (3.11) is true. Moreover, since

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

uniformly on [0,1], all the conditions of Theorem 3.2 hold, which implies that (3.22)-(3.23) has at least two positive solutions (for μ μ 0 ).

Example 3.2

Consider

y (t)+μ ( y δ 1 ( t ) + y δ 2 ( t ) ) =0,0<t<1,
(3.24)

with

y(0)= 0 1 y 3 (s)dA(s),y(1)= 0 1 y 1 2 (s)dB(s),dA(s)= 1 2 ds,dB(s)= 1 8 d e s ,
(3.25)

where δ 1 >0, δ 2 <1.

It is easy to see that all conditions of Theorem 3.3 hold, which implies that (3.24)-(3.25) has at least two positive solutions.

4 Multiplicity of positive solutions for the singular semipositone boundary value problem

In this section, we consider the case

f(t,y)=F(t,y)γ(t),t(0,1),

where the conditions (C1), (C3), (C4) for F(t,y) instead of f(t,y) hold and γC((0,1),(0,+)) with

w(t)= 0 1 k(t,s)q(s)γ(s)ds<+,t[0,1], c 1 = 0 1 q(t)γ(t)dt<+.

For yP, define

( T ϵ y ) ( t ) = ( 1 t ) α [ [ y w ] ] + t β [ [ y w ] ] + 0 1 k ( t , s ) q ( s ) F ( s , max { ϵ , [ y ( s ) w ( s ) ] } ) d s , t [ 0 , 1 ] , 0 < ϵ 1 ,
(4.1)

where k(t,s) is defined in (3.1) and

[ y ( t ) w ( t ) ] ={ y ( t ) w ( t ) , if  y ( t ) w ( t ) > 0 , 0 , if  y ( t ) w ( t ) 0 .

Now we present the following condition for convenience:

(C2)′:

{ there exists a function  ψ 4 c 1 continuous on  [ 0 , 1 ]  and positive on  ( 0 , 1 )  such that  F ( t , y ) ψ 4 c 1 ( t )  on  ( 0 , 1 ) × ( 0 , 4 c 1 ]  with max t [ 0 , 1 ] 0 1 k ( t , s ) q ( s ) ψ 4 c 1 ( s ) d s > 2 c 1 .

Define

H = { x C ( [ 0 , 1 ] , R ) C 1 ( [ 0 , 1 ) , R ) C ( ( 0 , 1 ) , ( 0 , + ) ) C 2 ( ( 0 , 1 ) , R ) | x  satisfies x ( t ) + q ( t ) F ( t , max { ϵ , [ x ( t ) w ( t ) ] } ) = 0 , 0 < t < 1 , x ( 0 ) = α [ [ x w ] ] , x ( 1 ) = β [ [ x w ] ] , 1 ϵ > 0 } .

Lemma 4.1

IfHand (C2)′ hold, then there exists a δ 0 >0such that

[ x ( t ) w ( t ) ] δ 0 ,t[0,1],xH.

Proof

Suppose that xH. There are two cases to consider:

  1. (1)

    x4 c 1 . Since

    w(t)t(1t) 0 1 q(s)γ(s)ds= c 1 t(1t),
    (4.2)

we have

w(t) 1 4 4 c 1 t(1t) 1 4 xt(1t).

From Lemma 2.3, we have

x(t)xt(1t)4 c 1 t(1t),t(0,1),

which implies that

x ( t ) w ( t ) x t ( 1 t ) 1 4 x t ( 1 t ) = 3 4 x t ( 1 t ) 3 4 4 c 1 t ( 1 t ) = 3 c 1 t ( 1 t ) , t [ 0 , 1 ] .

Hence

[ x ( t ) w ( t ) ] 3 c 1 t(1t),t[0,1]

and so

x(0)=α [ [ x w ] ] 0 1 ( 3 c 1 s ( 1 s ) ) a dA(s),x(1)=β [ [ x w ] ] 0 1 ( 3 c 1 s ( 1 s ) ) b dB(s).

The concavity of x(t) implies that

x(t)min { 0 1 ( 3 c 1 s ( 1 s ) ) a d A ( s ) , 0 1 ( 3 c 1 s ( 1 s ) ) b d B ( s ) } = d e f . δ 1 .

Then

[ x ( t ) w ( t ) ] 3 4 x(t) 3 4 δ 1 ,t[0,1].
(4.3)
  1. (2)

    0<x4 c 1 . Condition (C2)′ guarantees that

    x max t [ 0 , 1 ] 0 1 k ( t , s ) q ( s ) F ( s , max { ϵ , [ x ( s ) w ( s ) ] } ) d s max t [ 0 , 1 ] 0 1 k ( t , s ) q ( s ) ψ 4 c 1 ( s ) d s > 2 c 1 ,

which together with xP implies that

x(t)t(1t)x2 c 1 t(1t),t[0,1].
(4.4)

From (4.2) and (4.4), we have

w(t) c 1 t(1t)= 1 2 2 c 1 t(1t) 1 2 x(t),t[0,1],

and so

[ x ( t ) w ( t ) ] 1 2 x(t) c 1 t(1t),t[0,1].

Then

x ( t ) = ( 1 t ) α [ [ x w ] ] + t β [ [ x w ] ] + 0 1 k ( t , s ) q ( s ) F ( s , max { ϵ , [ x ( s ) w ( s ) ] } ) d s ( 1 t ) 0 1 ( c 1 s ( 1 s ) ) a d A ( s ) + t 0 1 ( c 1 s ( 1 s ) ) b d B ( s ) min t [ 0 , 1 ] [ ( 1 t ) 0 1 ( c 1 s ( 1 s ) ) a d A ( s ) + t 0 1 ( c 1 s ( 1 s ) ) b d B ( s ) ] , t [ 0 , 1 ] ,

which implies

[ x ( t ) w ( t ) ] 1 2 x ( t ) 1 2 min t [ 0 , 1 ] [ ( 1 t ) 0 1 ( c 1 s ( 1 s ) ) a d A ( s ) + t 0 1 ( c 1 s ( 1 s ) ) b d B ( s ) ] = d e f . δ 2 .
(4.5)

Let

δ 0 =min { δ 2 , 3 4 δ 1 } .

Now (4.3) and (4.5) guarantee that

[ x ( t ) w ( t ) ] δ 0 ,t[0,1].

The proof is complete. □

Lemma 4.2

Suppose there exists an a ¯ (0, 1 2 ) such that

lim y + F ( t , y ) y =+
(4.6)

uniformly on[ a ¯ ,1 a ¯ ]. Then there exists an R >0such that for allR R

i( T ϵ , Ω R P,P)=0,0<ϵ1.

Proof

From (4.6), there exists a R 1 >max{1,2 c 1 } such that

F(t,y) N y,y R 1 ,
(4.7)

where

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

Let R = 2 a ¯ 2 R 1 and

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

Now we show

T ϵ yyfor yP Ω R ,0<ϵ1.
(4.8)

Suppose that there exists a y 0 P Ω R with T ϵ y 0 y 0 . Then y 0 =R. Also since y 0 (t) is concave on [0,1] (since y 0 P) we have from Lemma 2.3 that y 0 (t)t(1t) y 0 t(1t)R for t[0,1]. For t[ a ¯ ,1 a ¯ ], we have (notice y 0 =R2 c 1 )

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

which together with (4.7) yields the result that

F ( t , max { ϵ , [ y 0 ( t ) w ( t ) ] } ) N [ y 0 ( t ) w ( t ) ] N 1 2 R a ¯ 2 ,t[ a ¯ ,1 a ¯ ].
(4.9)

Then we have, using (4.9),

y 0 ( a ¯ ) T ϵ y 0 ( a ¯ ) = ( 1 a ¯ ) α [ [ y 0 w ] ] + a ¯ β [ [ y 0 w ] ] + 0 1 k ( a ¯ , s ) q ( s ) F ( s , max { ϵ , [ y 0 ( s ) w ( s ) ] } ) d s a ¯ 1 a ¯ k ( a ¯ , s ) q ( s ) F ( s , max { ϵ , [ y 0 ( s ) w ( s ) ] } ) d s = a ¯ 1 a ¯ k ( a ¯ , s ) q ( s ) F ( s , [ y 0 ( s ) w ( s ) ] ) d s N 1 2 R a ¯ 2 a ¯ 1 a ¯ k ( a ¯ , s ) q ( s ) d s > R = y 0 ,

which is a contradiction. Hence (4.8) is true. Thus Lemma 2.2 guarantees that

i( T ϵ , Ω R P,P)=0,0<ϵ1.

The proof is complete. □

Lemma 4.3

Suppose thatmax{a,b}>1. Then there exists an R >0such that for allR R

i( T ϵ , Ω R P,P)=0,0<ϵ1.

Proof

Since max{a,b}>1, without loss of generality, we suppose that a>1. Let R >max{1,2 c 1 } with 1 2 a R a 1 0 1 s a ( 1 s ) a dA(s)>1. Set

Ω R = { x C [ 0 , 1 ] | x < R } .

Now we show that

T ϵ xx,x Ω R P,0<ϵ1.

In fact, suppose that x 0 Ω R P and satisfies

T ϵ xx.

Then y 0 =R. Also since y 0 (t) is concave on [0,1] (since y 0 P) we have from Lemma 2.3 that y 0 (t)t(1t) y 0 t(1t)R for t[0,1]. For t[0,1] we have

[ y 0 ( t ) w ( t ) ] 1 2 y 0 (t) 1 2 Rt(1t).

Then

R y ( 0 ) = 0 1 ( [ y 0 ( s ) w ( s ) ] ) a d A ( s ) 0 1 ( 1 2 y 0 s ( 1 s ) ) a d A ( s ) = 1 2 a R R a 1 0 1 s a ( 1 s ) a d A ( s ) 1 2 a R R a 1 0 1 s a ( 1 s ) a d A ( s ) > R .

This is a contradiction. Lemma 2.3 guarantees that

i( T ϵ , Ω R P,P)=0,0<ϵ1.

The proof is complete. □

Theorem 4.1

Suppose (C1), (C2)′, (C3), and (C4) 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 , ) ,
(4.10)

and

sup r ( 2 c 1 , + ) 1 1 + h ( r ) g ( r ) c 0 max { r a , r b } r d y g ( 1 2 y ) > b 0
(4.11)

holds; here

c 0 = max { 0 1 d A ( s ) , 0 1 d B ( s ) } , b 0 = max { 2 0 1 2 t ( 1 t ) q ( t ) d t , 2 1 2 1 t ( 1 t ) q ( t ) d t } .
(4.12)

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

Proof

From (4.11), choose r>2 c 1 , ϵ>0 with ϵ<min{1, 1 2 c 0 max{ r a , r b }} with

1 1 + h ( r ) g ( r ) c 0 max { r a , r b } r d y g ( 1 2 y ) > b 0 .
(4.13)

Let

Ω 1 = { y C [ 0 , 1 ] | y < r } .

Let T ϵ be defined as in (4.1). Lemma 3.1 guarantees that T ϵ :PP is continuous and completely continuous.

Now we show that

yλ T ϵ y,y Ω 1 P,λ[0,1].
(4.14)

Suppose that there is a y 0 Ω 1 P and λ 0 [0,1] with y 0 = λ 0 T ϵ y 0 . Since y 0 (t)t(1t) y 0 t(1t)2 c 1 and w(t)= 0 1 k(t,s)q(s)γ(s)dst(1t) 0 1 q(s)γ(s)ds= c 1 t(1t)= c 1 y 0 t(1t) y 0 1 2 y 0 (t), we have

y 0 (t)w(t) 1 2 y 0 (t),t[0,1].

Since y 0 satisfies

{ y 0 + λ 0 q ( t ) F ( t , max { ϵ , [ y 0 ( t ) w ( t ) ] } ) = 0 , 0 < t < 1 , y 0 ( 0 ) = λ 0 α [ [ y 0 w ] ] , y 0 ( 1 ) = λ 0 β [ [ y 0 w ] ] ,
(4.15)

y 0 (0)= λ 0 α[ [ y 0 w ] ] r a 0 1 dA(s) c 0 r a <r= y 0 and y 0 (1)= λ 0 β[ [ y 0 w ] ] r b 0 1 dB(s) c 0 r b <r= y 0 , there exists a t 0 (0,1) such that y 0 ( t 0 )=0 and y 0 (t)0 on (0, t 0 ), y 0 (t)0 on ( t 0 ,1). For t(0,1), it is easy to see that

y 0 ( t ) g ( max { ϵ , [ y 0 ( t ) w ( t ) ] } ) { 1 + h ( max { ϵ , [ y 0 ( t ) w ( t ) ] } ) g ( max { ϵ , [ y 0 ( t ) w ( t ) ] } ) } q ( t ) g ( max { ϵ , [ y 0 ( t ) w ( t ) ] } ) { 1 + h ( r ) g ( r ) } q ( t ) .
(4.16)

Integrate from t to t 0 to obtain

y 0 ( t ) g ( max { ϵ , [ y 0 ( t ) w ( t ) ] } ) { 1 + h ( r ) g ( r ) } t t 0 q ( s ) d s g ( max { ϵ , 1 2 y 0 ( t ) } ) { 1 + h ( r ) g ( r ) } t t 0 q ( s ) d s g ( 1 2 y 0 ( t ) ) { 1 + h ( r ) g ( r ) } t t 0 q ( s ) d s

and then integrate from 0 to t 0 to obtain

α [ [ y 0 w ] ] y 0 ( t 0 ) d y g ( 1 2 y ) { 1 + h ( r ) g ( r ) } 0 t 0 sq(s)ds { 1 + h ( r ) g ( r ) } 1 1 t 0 0 t 0 s(1s)ds,

which together with α[ [ y 0 w ] ]α[ y 0 ] c 0 max{ r a , r b } yields

c 0 max { r a , r β } r d y g ( 1 2 y ) α [ [ y 0 w ] ] r d y g ( 1 2 y ) { 1 + h ( r ) g ( r ) } 1 1 t 0 0 t 0 s(1s)ds.
(4.17)

Similarly if we integrate (4.16) from t 0 to t (t t 0 ) and then from t 0 to 1 we obtain

c 0 max { r a , r b } r d u g ( 1 2 u ) { 1 + h ( r ) g ( r ) } 1 t 0 t 0 1 s(1s)q(s)ds.
(4.18)

Now (4.17) and (4.18) imply

c 0 max { r a , r b } r d u g ( 1 2 u ) b 0 { 1 + h ( r ) g ( r ) } ,

which contradicts (4.13). Then (4.14) is true. Lemma 2.1 implies that

i( T ϵ , Ω 1 P,P)=1.
(4.19)

Thus, there exists an x ϵ P Ω 1 such that T ϵ x ϵ = x ϵ , which yields the result that H in Lemma 4.1 and there is a δ 0 >0 such that

[ x ( t ) w ( t ) ] δ 0 ,xH.
(4.20)

Let ϵ 0 = 1 2 δ 0 and T ϵ 0 x ϵ 0 = x ϵ 0 . Obviously x ϵ 0 H and [ x ϵ 0 ( t ) w ( t ) ] δ 0 . From

x ϵ 0 (t)=(1t)α [ [ x ϵ 0 w ] ] +tβ [ [ x ϵ 0 w ] ] + 0 1 k(t,s)q(s)F ( s , [ [ x ϵ 0 ( s ) w ( s ) ] ] ) ds,

we have

x ϵ 0 (t)=(1t)α[ x ϵ 0 w]+tβ[ x ϵ 0 w]+ 0 1 k(t,s)q(s)F ( s , x ϵ 0 ( s ) w ( s ) ) ds,t[0,1].

Let y(t)= x ϵ 0 (t)w(t), t[0,1]. It is easy to see that y(t) is a positive solution of BVP (1.1)-(1.2). The proof is complete. □

Theorem 4.2

Suppose the conditions of Theorem  4.1hold and there exists an a ¯ (0, 1 2 )such that

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

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

Proof

Choose r as in (4.13), ϵ 0 >0 with ϵ 0 <min{ δ 0 ,1, 1 2 c 0 max{ r a , r b }}, where δ 0 is defined in (4.20), and R>max{1,r} in Lemma 4.2. Set

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

From the proof of Theorem 4.1 and Lemma 4.2, we have

i( T ϵ 0 , Ω 1 P,P)=1

and

i( T ϵ 0 , Ω 2 P,P)=0,

which implies that

i ( T ϵ 0 , ( Ω 2 Ω 1 ) P , P ) =1.

Thus, there exist x 1 , ϵ 0 Ω 1 P and x 2 , ϵ 0 ( Ω 2 Ω 1 )P such that

T ϵ 0 x 1 , ϵ 0 = x 1 , ϵ 0 , T ϵ 0 x 2 , ϵ 0 = x 2 , ϵ 0 .

Let y 1 , ϵ 0 (t)= x 1 , ϵ 0 (t)w(t) and y 2 , ϵ 0 (t)= x 2 , ϵ 0 (t)w(t) for all t[0,1]. It is easy to see that y 1 , ϵ 0 (t) and y 2 , ϵ 0 (t) are two positive solutions for BVP (1.1)-(1.2). The proof is complete. □

Theorem 4.3

Suppose the conditions of Theorem  4.1hold andmax{a,b}>1. Then BVP (1.1)-(1.2) has at least two positive solutions.

Proof

Choose r as in (4.13), ϵ 0 >0 with ϵ 0 <min{ δ 0 ,1, 1 2 c 0 max{ r a , r b }}, where δ 0 is defined in (4.20), and R>max{1,r} in Lemma 4.3. Set

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

From the proof of Theorem 4.1 and Lemma 4.3, we have

i( T ϵ 0 , Ω 1 P,P)=1

and

i( T ϵ 0 , Ω 2 P,P)=0,

which implies that

i ( T ϵ 0 , ( Ω 2 Ω 1 ) P , P ) =1.

Thus, there exist x 1 , ϵ 0 Ω 1 P and x 2 , ϵ 0 ( Ω 2 Ω 1 )P such that

T ϵ 0 x 1 , ϵ 0 = x 1 , ϵ 0 , T ϵ 0 x 2 , ϵ 0 = x 2 , ϵ 0 .

Let y 1 , ϵ 0 (t)= x 1 , ϵ 0 (t)w(t) and y 2 , ϵ 0 (t)= x 2 , ϵ 0 (t)w(t) for all t[0,1]. It is easy to see that y 1 , ϵ 0 (t) and y 2 , ϵ 0 (t) are two positive solutions for BVP (1.1)-(1.2). The proof is complete. □

Example 4.1

Consider

y (t)+ 7 64 ( y 2 ( t ) + y δ 1 ( t ) 1 1000 1 t 1 2 ( 1 t ) 1 2 ) =0,0<t<1,
(4.21)
y ( 0 ) = 0 1 y 1 3 ( s ) d A ( s ) , y ( 1 ) = 0 1 ( y ( s ) ) 1 2 d B ( s ) , d A ( s ) = 1 2 d s , d B ( s ) = 1 2 d sin s ,
(4.22)

where δ 1 >1.

Let q(t)= 7 64 , F(t,y)= y 2 + y δ 1 , g(y)= y 2 , h(y)= y δ 1 , c 0 =max{ 0 1 dA(s), 0 1 dB(s)}= 1 2 , b 0 = 7 384 , γ(t)= 1 1000 1 t 1 2 ( 1 t ) 1 2 , c 1 = 0 1 q(s)γ(s)ds 1 500 . It is easy to see that (C1), (C3)-(C4) and (4.10) hold, and since F(t,y) 1 ( 4 c 1 ) 2 ( 125 ) 2 for (t,y)[0,1]×(0,4 c 1 ] and

max t [ 0 , 1 ] 0 1 k(t,s)q(s) ( 125 ) 2 ds= 7 64 125 2 max t [ 0 , 1 ] 0 1 k(t,s)ds>2 c 1 ,

we find that (C2)′ is true.

Since

1 1 + h ( 1 ) g ( 1 ) c 0 max { 1 1 3 , 1 1 2 } 1 1 g ( 1 2 y ) dy= 1 1 8 24 = 7 192 ,

we have

sup r ( 2 c 1 , + ) 1 1 + h ( r ) g ( r ) c 0 max { r 1 3 , 1 1 2 } r 1 g ( 1 2 y ) dy> b 0 ,

which guarantees that (4.11) is true. Moreover, since

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

uniformly on [0,1], all the conditions of Theorem 4.2 hold. Then (4.21)-(4.22) has at least two positive solutions.

Example 4.2

Consider

y (t)+ 7 64 ( y 2 ( t ) + y δ 1 ( t ) 1 1000 1 t 1 2 ( 1 t ) 1 2 ) =0,0<t<1,
(4.23)
y ( 0 ) = 0 1 y 3 ( s ) d A ( s ) , y ( 1 ) = 0 1 ( y ( s ) ) 1 2 d B ( s ) , d A ( s ) = 1 2 d s , d B ( s ) = 1 2 d sin s ,
(4.24)

where 0< δ 1 1.

Since a=3>1, using Theorem 4.3, we see that (4.23)-(4.24) has at least two positive solutions.

5 Uniqueness of positive solutions for the singular boundary value problem

In this section, we consider the uniqueness of positive solution for BVP (1.1)-(1.2).

Lemma 5.1

(see [20])

Suppose that E is a Banach space with a normal and solid conePEandA: P × P P is a mixed monotone operator. Moreover, suppose there is a constant θ with0θ<1such that

A ( t x , 1 t y ) t θ A(x,y),x,y P ,0<t<1.

Then A has a unique fixed point in P .

Theorem 5.1

Suppose that (C1), (C3), (C4), (C5) hold and1>max{a,b}>0. Then BVP (1.1)-(1.2) has a unique positive solution.

Proof

It is easy to see that P defined by (2.1) is a normal and solid cone. For x,y P , define

A(x,y)(t)=(1t)α[x]+tβ[y]+ 0 1 k(t,s)q(s)g ( s , x ( s ) , y ( s ) ) ds,t[0,1],

where g is given in (C5). Since x,y P , we have min t [ 0 , 1 ] x(t)>0 and min t [ 0 , 1 ] y(t)>0. Then (C1) guarantees that

0 1 x a (s)dA(s)>0, 0 1 y b (s)dB(s)>0,

which implies that

min t [ 0 , 1 ] [ ( 1 t ) α [ x ] + t β [ y ] ] >0.

Therefore, A(x,y) P .

Let θ 1 =max{θ,max{a,b}}. For 0<λ<1 and x P , y P , from (C5), we have

A ( λ x , 1 λ y ) ( t ) = ( 1 t ) α [ λ x ] + t β [ λ y ] + 0 1 k ( t , s ) g ( s , λ x ( s ) , 1 λ y ( s ) ) d s ( 1 t ) λ a α [ x ] + t λ b β [ y ] + λ θ 0 1 k ( t , s ) g ( s , x ( s ) , y ( s ) ) d s ( 1 t ) λ θ 1 α [ x ] + t λ θ 1 β [ y ] + λ θ 1 0 1 k ( t , s ) g ( s , x ( s ) , y ( s ) ) d s = λ θ 1 A ( x , y ) ( t ) , t [ 0 , 1 ] .

From Lemma 5.1, A has a unique fixed point x in P , which satisfies

x ( t ) = ( 1 t ) α [ x ] + t β [ x ] + 0 1 k ( t , s ) q ( s ) g ( s , x ( s ) , x ( s ) ) d s = ( 1 t ) α [ x ] + t β [ x ] + 0 1 k ( t , s ) q ( s ) f ( s , x ( s ) ) d s , t [ 0 , 1 ] .

Then x (t) is the unique positive solution of BVP (1.1)-(1.2). The proof is complete. □

Example 5.1

Consider

y (t)+ 7 64 ( y δ 1 ( t ) + y δ 2 ( t ) ) =0,0<t<1,
(5.1)
y ( 0 ) = 0 1 y δ 3 ( s ) d A ( s ) , y ( 1 ) = 0 1 ( y ( s ) ) δ 4 d A ( s ) , d A ( s ) = d sin s 2 , d B ( s ) = d e s ,
(5.2)

where 0<max{ δ 1 , δ 2 , δ 3 , δ 4 }<1.

It is easy to see that all conditions of Theorem 5.1 hold, which guarantees that (5.1)-(5.2) has a unique positive solution.

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Yan, B., O’Regan, D. & Agarwal, R.P. Multiplicity and uniqueness results for the singular nonlocal boundary value problem involving nonlinear integral conditions. Bound Value Probl 2014, 148 (2014). https://doi.org/10.1186/s13661-014-0148-9

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Keywords

  • Boundary Value Problem
  • Real Banach Space
  • Unique Fixed Point
  • Nonlocal Boundary
  • Singular Boundary