THE DEGREE THEORY OF A NEW CLASS OF OPERATORS AND ITS APPLICATION

THE DEGREE THEORY OF A NEW CLASS OF

OPERATORS AND ITS APPLICATION

Yan Baoqiang

(Received July 29, 1996 Revised December 24, 1997)

Abstract. This paper denes a concept of a semi-k-set-contraction operator, and establishes a degree theory for it. As its application, we discuss the existence for the solution of two-point boundary value problems for nonlinear second order integro-dierential equations in Banach spaces.

AMS 1991Mathematics Subject Classi
cation. 47H11, 34G20.

Key words and phrases. k-set-contraction operator, topological degree, two-point boundary value problems, equicontinuous set.

x

1. INTRODUCTION

It is well known that the degree theory for the strict-set-contraction operator and the condensing operator has many applications to the existence of the solutions of some equations(see 1], 2], 3], 4]). However, some important op-erators are not strict-set-contraction opop-erators or condensing opop-erators. Now we give an example.

Let

E

be a Banach space,

C

(0, 1],

E

) = f

x

,

x

is a mapping from 0, 1]

into

E

and

x

(

t

) is continuous at every

t

20, 1]g. Obviously

C

(0, 1]

E

) is a

Banach space with norm k

x

k= maxfk

x

(

t

)k,

t

20

1]g. For

x

2

C

(0, 1],

E

),

let (

Ax

)(

t

) =Z 1 0

G

1(

t
s

)

x

(

s

) +

g

(

x

(

s

))]

ds

( )

where

g

2

C

(

E

,

E

),

g

(

D

) is relatively compact for any bounded

D



E

and

G

1(

t

,

s

) = min f

t

,

s

g.

It is dicult to prove that

A

is a strict-set-contraction operator or a con-densing operator from

C

(0

1]

E

) into

C

(0

1]

E

). So it is necessary to es-tablish degree theory for the operators such as

A

dened by (*).

Now we dene a new class of operators.

Let

I

be a bounded, closed interval of real numbers. Assume that

C

m₍

_I

_,

E

) =f

x

,

x

is a mapping from

I

into

E

and

x

(

t

) is

m

-times continuously norm

dierentiable(

m

 1).g. Obviously

C

m(

I

,

E

) is a Banach space with norm k

x

km = maxfk

x

k 0, k

x

0 k 0, ,k

x

(m) k 0 g, herek

x

k 0 =max fk

x

(

t

)k,

t

2

I

g.

Assume that

A

is an operator from a bounded set

S



C

m(

I

,

E

) into

C

m(

I

,

E

), and



(

S

) denotes the Kuratowski measure of noncompactness in

C

m₍

_I

_,

E

).

Now we give a new denition.

Denition 1.

A

:

S

!

C

m(

I

,

E

)(

S

:bounded) is called a semi-

k

-set-contraction

operator if

A

is a bounded, continuous operator, (

AS

)(m) is equicontinuous on

I

, and



(

A

(

D

))

k

(

D

)

for any bounded

D



S

with equicontinuous

D

(m), where 0 

k <

1 is a constant, (

AS

)(m) = f

y

,

y

(

t

) = (

Ax

) (m)(

t

) for

t

2

I

,

x

2

S:

g. And

A

:

C

m(

I

,

E

) !

C

m(

I

,

E

) is called a semi-

k

-set-contraction operator if the restriction

A

:

S

!

C

m(

I

,

E

) is a semi-

k

-set-contraction operator for any bounded

S



C

m(

I

,

E

).

It is easy to see that this denition is dierent from that of the

k

-set-contraction operator and that of the condensing operator(see1], 5]). For example

A

dened by (*),

A

:

C

(

I

,

E

) !

C

(

I

,

E

) and for any bounded set

S



C

(

I

,

E

),

AS

is bounded and equicontinuous. Moreover, by the following

lemma 1, for any equicontinuous subset

D



S

, we have



(

AD

(

t

)) =



(f Z 1 0

G

1(

t
s

)

x

(

s

) +

g

(

x

(

s

))]

ds
x

2

D

g) = Z 1 0

G

1(

t
s

)



(

D

(

s

)) +



(

g

(

D

(

s

)))]

ds

= Z 1 0

G

1(

t
s

)



(

D

(

s

))

ds

 Z 1 0

G

1(

t
s

)

ds

(

D

)

<

3₄



(

D

)

:

By lemma 2, we have



(

AD

) 3 4



(

D

)

:

So

A

_{is a semi-34-set-contraction operator. Insection 2, we establishthe degree} theory for the semi-

k

-set-contraction operators and prove some xed point

theorems. As their application, in section 3 we discuss the existence of the solution of two-point boundary value problems for nonlinear integrodierential equations in Banach spaces.

The following lemmas are necessary.

Lemma 1

(see3]). If

S



C

(

I

,

E

) is bounded and equicontinuous, then



(f Z I

x

(

t

)

dt
x

2

S

g) Z I



(

S

(

t

))

dt:

(1)

Lemma 2

(see2]). If

S



C

m(

I

,

E

) is bounded and

S

(m) is equicontinuous

on

I

, then



(

S

) = maxfsupf



(

S

(

t

))

t

2

I

g

supf



(

S

0(

t

))

t

2

I

g



supf



(

S

(m)

(

t

))

t

2

I

gg

:

x

2. ESTABLISHMENT OF THE DEGREE THEORY

Before establishing the degree theory for the class of the semi-

k

-set-contraction operator

A

, we give some lemmas. Let 

C

m(

I

,

E

) be open and bounded,

and

A

: !

C

m(

I

,

E

) a semi-

k

-set-contraction,

f

=

id

;

A

, where

id

denotes

the

indentity

operator. Then

f

is called a semi-

k

-set-contraction eld.

Lemma 3

. Assume

A

: !

C

m(

I

,

E

) is a semi-

k

-set-contraction operator,

then

1)

f

is proper, i.e.,

f

;1(

D

) is compact for any compact set

D



C

m(

I

,

E

)

2)

f

is a closed mapping, i.e.,

f

(

S

) is closed for any closed set

S



:

Proof. 1) Let

D

1=

f

;1(

D

)(

D

1 ), then

D

1 

A

(

D

1) +

D

. Since

D

(m) and

A

(

D

1) (m) are equicontinuous on

I

,

D

(m) 1 is equicontinuous on

I

. Consequently,



(

D

1) 



(

A

(

D

1)) +



(

D

) =



(

AD

1) 

k

(

D

1)

:

It is easy to see that



(

D

1) = 0. So

D

1 is relatively compact. Consequently,

D

1 is compact. 2) Let

y

n2

f

(

S

),

y

n!

y

0 2

C

m(

I

,

E

). We will prove

y

0 2

f

(

S

). Suppose that

y

n=

f

(

x

n),

x

n 2

S

. Let

S

0 = f

y

0,

y

1,

y

2, 

:

g. Obviously

S

0 

C

m(

I

,

E

) is compact. By the proof of 1),

f

;1(

S

0) 

C

m(

I

,

E

) is compact. So there exists a subsequencef

x

n i g,

x

n i !

x

0 2

C

m(

I

,

E

). Since

S

is closed,

x

0 2

S

. By the continuity of

f

,

y

ni =

f

(

x

ni) !

f

(

x

0). Consequently,

y

0 =

f

(

x

0). So

Lemma 4

. If

D



C

(

I

,

E

) is bounded and equicontinuous on

I

, then

co

(

D

)

is bounded and equicontinuous on

I

.

The proof of lemma 4 is routine and may be omitted.

Lemma 5

. Let f

S

ig

E

be bounded, closed and

S

1

S

2

S

3 

S

n 

S

n 6= 

n

= 1, 2, 3, . If



(

S

n) ! 0, then

S

= 1 \ i=1

S

i is a nonempty compact set.

This Lemma is the exercise 4, page 53, in 1]. In what follows, we give the denition of the degree for a semi-

k

-set-contraction eld.

Denition 2.

Let  

C

m(

I

,

E

) open and bounded,

A

:  !

C

m(

I

,

E

) be

a semi-

k

-contraction operator , 0

k <

1,

f

=

id

;

A:

(1) Assume that



62

f

(

@

). Let

D

1 =

co

(

A

()) and

D

n=

co

(

A

(

D

n;1 \)),

n

= 2, 3, 

:

1) If there exists an

n

0 such that

D

n 0 =

, then we dene that deg(

f

, ,



) = 0

:

2) Now we suppose that

D

n6=,

n

= 1, 2,. So

D

n\ is bounded and

closed(

n

= 1, 2, 

:

). Let

D

= 1 \

i=1

D

n. Then

D

is bounded, convex, closed and nonempty as we show below. Obviously

D

1

D

2. If

D

n;1

D

n, then

D

n =

co

(

A

(

D

n;1 \))

co

(

A

(

D

n\)) =

D

n +1. So

D

n;1

D

n,

n

= 2, 3, . By lemma 4, (

D

n) (m) is equicontinuous on

I

and



(

D

n) =



(

A

(

D

n;1 \))

k

(

D

n ;1 \)

k

(

D

n ;1)

:

So



(

D

n) 

k

n ;1



(

D

1). By

k <

1 and lemma 5, we know

D

is a nonempty

compact set. Because of

D

n;1

\

D

n\,

D

n T 6=and



(

D

n\)!0, we know

D

\ = ( 1 \ n=1

D

n)\ is nonempty and compact. On the other hand,

from

A

(

D

n\)

co

(

A

(

D

n ;1 \)) =

D

n we have

A

(

D

\) 1 \ n=1

A

(

D

n\) 1 \ n=1

D

n=

D:

(2)

Since

D

is compact,

A

:

D

\ !

D

is completely continuous. So by the

extention theorem of completely continuous operator(see1], page 44), there exists a completely continuous operator

A

1 : 

!

D

such that

A

1

x

=

Ax

for

every

x

2

D

\. Let

f

1 =

id

;

A

1. It is easy to see that



62

f

1(

@

). So the

Leray-Schauder degree degLS(

f

1, ,



) can be dened. Let

deg(

f





) = degLS(

f

1





)

(3)

where degLS(

f

1, ,



) denotes the degree of completely continuous operator

eld

f

1 =

id

;

A

the choice of

f

1. In fact, let

A

2 : 

!

D

be another extension of

A

, and

f

2 =

id

;

A

2. Let

H

(

t

,

x

) =

x

;

tA

1

x

;(1;

t

)

A

2

x

,

x

2 , 0 

t

 1.

We will prove

H

(

t

,

x

) 6=



for

t

2 0, 1] and

x

2

@

. On the contrary,

if there exist

t

0, 0 

t

0  1, and

x

0 2

@

 such that

H

(

t

0,

x

0) =



, i.e.,

x

0 =

t

0

A

1

x

0+ (1 ;

t

0)

A

2

x

0. Since

A

1

x

0 2

D

,

A

2

x

0 2

D

and

D

is convex, we know

x

0 2

D

. So

x

0 =

t

0

A

1

x

0+ (1 ;

t

0)

A

2

x

0 =

Ax

0. This contradicts to



62

f

(

@

). Hence degLS(

f

1





) = deg LS(

f

2





)

:

(4)

(2) Suppose

p

62

f

(

@

). It is easy to see



62(

f

;

p

)(

@

) and set

deg(

f



p

) = deg(

f

;

p





)

:

(5)

Now we have successfully dened the degree deg(

f

, ,

p

) for a semi-

k

-set-contraction operator

A

.

Remark 1

: If

A

has a xed point

x

0

2, we have

x

0

2

D

n\6=,

n

= 1, 2, 

:

So the xed point set

F

is also non-void with

F



D

\.

Remark 2

: We can notice the method of establishing f

D

ngn in denition 2

is same as that off

Q

ngn appearing on page 107 in 5].

Lemma 6.

Assume that

A

is a semi-

k

-set-contration operator as in denition 2,

f

=

id

;

A

,



62

f

(

@

), and 2) of Denition 2 is satised. If

B

: !

S

is

continuous with

Bx

=

Ax

for all

x

2

S

\, where

S

D

(

D

is the same as in

the denition 1) is compact and convex with

A

(

S

\)

S

. Let

g

=

id

;

B

,

then

deg(

f





) = degLS(

g





)

:

(6) Proof. Assume that

A

1 and

f

1 are such as those of 2) in denition 2. Let

H

(

t
x

) =

x

;

tA

1

x

;(1;

t

)

Bx

for

x

2 and 0

t

1

:

Then we have

H

(

t
x

)6=



for

x

2

@

 and 0

t

 1

:

In fact, suppose that

H

(

t

0

x

0) =



for

x

0 2

@



0 

t

0  1

:

Since

S

D

is convex,

x

0 =

t

0

A

1

x

0 + (1 ;

t

0)

Bx

0 2

S:

So

Bx

0 =

Ax

0,

x

0 =

t

0

A

1

x

0 + (1 ;

t

0)

Ax

0

:

From

Ax

0 2

D

1 and

A

1

x

0 2

D



D

1, we have

x

0 =

t

0

A

1

x

0 + (1 ;

t

0)

Ax

0 2

D

1. So

Ax

0 2

D

2,

A

1

x

0 2

D



D

2. Consequently

x

0 =

t

0

A

1

x

0+(1 ;

t

0)

Ax

0 2

D

2. Proceeding as before, we have

x

0

2

D

n(

n

= 1, 2, 3, ). Therefore

x

0 2

D

. So we have

A

1

x

0 =

Ax

0,

x

0 =

t

0

Ax

0 + (1 ;

t

0)

Ax

0=

Ax

0. This contradicts



62

f

(

@

). So degLS(

g





) = degLS(

f

1





)

:

Theorem 3.

The degree of a semi-

k

-set-contraction eld dened in Denition 2 has the following properties:

1) deg(

id

, ,

p

) = 1 for

p

2

2) deg(

f

, ,

p

) = deg(

f

, 1,

p

) + deg(

f

, 2,

p

) whenever 1, 2

 are open with 1 \ 2= ,

p

62

f

(n( 1  2))

3) deg(

id

;

H

(

t

, ), ,

p

) =

const

for all

t

2 0, 1] whenever

H

(

t

, ) is a

semi-

k

-set-contraction operator for all

t

20, 1] and as

t

!

t

0 for any

t

0,

H

(

t

,

x

) converges to

H

(

t

0,

x

) in

C

m₍

_I

_,

_E

_{) uniformly in}

_x

2, where

p

62

h

t(

@

),

h

t=

id

;

H

(

t

,)

4) if deg(

f

, 

p

)6= 0, then the equation

f

(

x

) =

p

has a solution in .

Moreover, set

g

=

id

;

G

, where

G

: !

C

m(

I

,

E

) is a semi-

k

-set-contraction

operator. Then

5) deg(

f

, ,

p

) = deg(

g

, ,

p

) whenever

G

j@ =

A

j@ 

6) deg(

f

, ,

p

) = deg(

f

, 1,

p

) for every open subset 1 of  such that

p

62

f

(; 1)

7) deg(

f

, ,) is constant on every connected subset of

C

m(

I
E

);

f

(

@

). Proof. We might well suppose that

p

=



. Since 1) is same as the normality

of strict-set-contraction eld in 5], we can omit the proof. First we prove 2). We discuss three possibilities.

1' Suppose 2) of (1) in denition 2 for 1 and 2 is true. Obviously 2) of

(1) in denition 2 for  is true. Now we get

(

D

(1) \ 1 6 =

D

(2) \ 2 6 =

D

\6=

D

(1) 

D
D

(2) 

D

A

(

D

(1) \ 1) 

D

(1)

A

(

D

(2) \ 2) 

D

(2)

A

(

D

\)

D

where

D

(1) and

D

(2) are obtained as

D

in 2) of (1) in denition 2 for

A

j

1 and

A

j

2 respectivelyt.. And

D

is the same as in 2) of (1) in denition 2 . Let

A

1 : 

!

D

is the completely continuous operator as in 2) of (1) in denition

2, and

f

1=

id

;

A

1. According to (3), we get

deg(

f





) = degLS(

f

1





)

:

By virtue of lemma 6, we have

( deg(

f

1



) = deg LS(

f

1

1



)

deg(

f

2



) = deg LS(

f

1

2



)

:

()

By virtue of the degree theory of Leray-Schauder, we get degLS(

f

1





) = deg LS(

f

1

1



) + deg LS(

f

1

2



)

:

According to above conclusion, we get

2' Suppose that one of 1 and 2 satises 2) of (1) in denition 2(for

example, 1), one of 1 and 2 saties 1) of (1) in denition 2(for example,

2). Obviously  satises (1)2) in denition 2. Therefore

deg(

f

2



) = 0

:

By virtue of lemma 6, we have

deg(

f

1



) = deg LS(

f

1

1



)

where

f

1 is as in 1'. Now we will prove

degLS(

f

1

2



) = 0

:

In fact, if degLS(

f

1

2



)

6

= 0, then there exsits an

x

0

2 such that

f

1(

x

0) =

0, i.e.

x

0 =

A

1

x

0

2

D

. So

A

1

x

0 =

Ax

0,

x

0 =

Ax

0. By the Remark 1, 2

saties 2) of (1) in denition 2. This is a contradiction. Now by (**), we have deg(

f





) = deg(

f

1



) + deg(

f

2



)

:

3' Suppose 1 and 2 satisfy 1) of (1) in denition 2. Now we have

deg(

f

1



) = 0

deg(

f

2



) = 0

:

By the Remark 1,



62

f

( 1  2). Hence,



62

f

(). Then we have deg(

f





) = 0

:

So

deg(

f





) = deg(

f

1

1



) + deg(

f

1

2



)

:

And Since the proof of (2) includs that of (4), we can omit the proof of (4). Next we prove 3). First we need to prove

H

(0

1]) is bounded. In fact,

assume that there exists a sequencef

t

ng0

1] and af

x

ng such that k

H

(

t

n

x

n)km !1

n

!1

:

(8)

We might as well suppose that

t

n!

t

0. We have k

H

(

t

n

x

n)km k

H

(

t

n

x

n);

H

(

t

0

x

n) km+k

H

(

t

0

x

n) km

:

(9) Since

H

(

t

0,

) is a semi-

k

-set-contraction operator, k

H

(

t

0,

x

n) km is bounded. And becausek

H

(

t

n,

x

);

H

(

t

0,

x

) k!0(

n

!+1) uniformly in

x

2 ,k

H

(

t

n,

x

n);

H

(

t

0,

x

n)

km is bounded. So k

H

(

t

n,

x

n)k is bounded. This contradicts

(9). Consequently,

H

(0

1]) is bounded. Let

D

 1 =

co

(

H

(0

1] )), and

D

 n=

co

(

H

(0

1](\

D

 n;1)))

n

= 2, 3, . Obviously

D

 1

D

 2. If

D

 n;1

D

 n, then

D

 n=

co

(

H

(0

1](

D

 n;1 \))

co

(

H

(0

1](

D

 n\))) =

D

 n+1. So

D

 n;1

D

 n,

n

= 2, 3,. We need to prove

D

 n(m) is equicontinuous on

I

.

First we will prove that

D

 1

(m)

is equicontinuous. From lemma 4, we have only to prove that

H

(0

1])

(m)is equicontinuous. Assume that

H

(0

1] )

(m)

is not equicontinuous. Then there exists an

" >

0, a subsequence f

x

ng 

H

(0

1]) with

x

n=

H

(

t

n,

y

n), and j

t

1n ;

t

2n j

<

1 n such that k

x

(m) n (

t

1n) ;

x

(m) n (

t

2n) k

":

(10)

We might as well suppose

t

n!

t

0, then we have k

H

(

t

n

y

n) (m) (

t

1n)) ;

H

(

t

n

y

n) (m) (

t

2n)) k  k

H

(

t

n

y

n) (m)(

t

1n)) ;

H

(

t

0

y

n) (m)(

t

1n)) k +k

H

(

t

0

y

n) (m) (

t

2n)) ;

H

(

t

n

y

n) (m) (

t

2n)) k +k

H

(

t

0

y

n) (m)(

t

1n)) ;

H

(

t

0

y

n) (m)(

t

2n)) k =

I

1n+

I

2n+

I

3n

:

And because k

H

(

t

n,

x

);

H

(

t

0,

x

) k ! 0(

n

! +1) uniformly in

x

2 , we have

I

1n+

I

2n ! 0

n

! +1. Since

H

(

t

0

) is a semi-

k

-set-contraction operator, we have

I

3n ! 0

n

!+1. Then

I

1n+

I

2n+

I

3n ! 0

n

! +1.

This contradicts (10). By lemma 4,

D

 1 (m) is equicontinuous on

I

. By the monotonity of f

D

 n(m) g,

D

 n(m) is equicontinuous.

For given

t

20

1], let

D

1(

t

) =

co

(

H

(

t

))

D

n(

t

) =

co

(

H

(

t
D

n;1(

t

) \))

n

= 2

3



:

(11) Obviously

D

n(

t

)

D

n ;1(

t

)

n

= 2

3



:

If there exists an

n

0 with

D

 n0 \ = for every, then

D

n

0(

t

)

\ =

t

20

1]. Then we have

deg(

h

t





)0

t

20

1]

:

Now suppose

D



n\6=(

n

= 1

2

).

Take any

" >

0 and

t

0

2 0

1]. Then for each

n

 2 there exist a nite

coveringf

S

igri =1 such that

H

(

t

0

D

 n;1 \) r i=1

S

i with

d

(

S

i)

k

(

D

 n;1)+

"

,

i

= 1

2



r

since



(

H

(

t

0

D

 n;1 \)) 

k

(

D

 n;1 \) 

k

(

D

 n;1). On

the other hand, from the assumption, there is a

>

0 such that k

H

(

t
x

);

H

(

t

0

x

)

k

< "

for all

x

2  when j

t

;

t

0 j

<

. Let

S

"i = f

x
d

(

x
S

i)

<

"

g

I

(

t

0

) = (

t

0 ;

t

0+

) \0

1]. So

H

(

I

(

t

0

) (

D

 n;1 \)) r i=1

S

"i

d

(

S

_"i)

d

(

S

i) + 2

"



k

(

D

 n;1) + 3

":

We have



(

H

(

I

(

t

0

) (

D

 n;1 \)) 

k

(

D

 n;1) + 3

"

. By the compactness

of the interval 0

1], there exist

t

i 2 0

1],

i

>

0,

i

= 1

2



s

such that

0

1] = s i=1

I

(

t

i

i), and



(

H

(

I

(

t

i

i)(

D

 n;1 \)))

k

(

D

 n;1) + 3

"
i

= 1

2



s:

So



(

D

 n) =



((

H

(0

1](

D

 n;1 \))) =



( s i=1

H

(

I

(

t

i

i)(

D

 n;1 \))) = maxf



(

H

(

I

(

t

i

i)(

D

 n;1 \)))

i

= 1

2



s:

g 

k

(

D

 n;1) + 3

":

By the arbitrariness of

"

, we have



(

D



n) 

k

(

D

 n;1)

n

= 2

3



:

Con-sequently,



(

D

 n) 

k

n ;1



(

D

 1). This implies



(

D

 n) ! 0. By lemma 5,

D

= 1 \ n=1

D



nis nonempty, convex and compact(recall that we are now assum-ing

D



n\ 6=  for

n

= 1

2

). By the same proof,

D



\ also is shown

to be nonempty and compact. Since

H

(0

1](

D

 n\)) 

co

(

H

(0

1] (

D

 n\))) =

D

 n+1 

D

 n. So

H

(0

1](

D

 \)) 1 \ n=1

H

(0

1](

D

 n\))  1 \ n=1

D

 n=

D



:

By the extention theorem of completely continuous function, there exists a

G

: 0

1]!

D

 such that

G

(

t
x

) =

H

(

t
x

) when (

t
x

)

20

1](

D



\).

Let

g

t=

x

;

G

(

t
x

). We will prove deg(

h

t





) = deg

LS(

g

t





)

:

It is easy

to see that



62

g

t(

@

). In fact, if there exist

t

0 with 0 

t

0 1, and

x

0 2

@

 such that

g

t0(

x

0) = 0. Then

x

0 =

G

(

t

0

x

0) 2

D

. So

G

(

t

0

x

0) =

H

(

t

0

x

0),

x

0 =

H

(

t

0

x

0). This contradicts



62

h

t(

@

). So



62

g

t(

@

).

(a) If the condition 1) of denition 2 is satised for

h

t, we have deg(

h

t





) = 0. In this case, since

H

(

t
x

) has not xed points in ,

G

(

t
x

) also has not xed points in . By the theory of Leray-Schauder degree, we have

degLS(

g

t





) = 0

:

(b) If

h

t satises the condition 2) in denition 2, by lemma 6, we have deg(

h

t





) = degLS(

g

t





)

:

Therefore we have degLS(

g

t





) = const

0

t

1

:

Hence deg(

h

t





) =const

0

t

1

:

If

p

6=



, let

h

t=

id

;

H

(

t

);

p

. Then by the result proved above, we have

deg(

h

t





) =const

:

Finally since the proofs of 5), 6), 7) are similar to the proofs of the relative properties of degree throry of strict-set-contraction eld in 1], we omit the proofs. Thus proof is complete. 2

Theorem 4.

Let  be a bounded, convex open set in

C

m₍

_I
E

₎

_A

_{: } !

C

m₍

_I
E

_{) be a semi-}

_k

_{-set-contraction operator, 0}

k <

1,

A

(

@

) without

xed point in

@

, then deg(

id

;

A





) = 1. Proof. Choose an

x

0 2 arbitrarily. Let

h

t=

t

(

x

;

Ax

) + (1;

t

)(

x

;

x

0) =

x

;

H

(

t
x

), here

H

(

t
x

) =

tAx

+(1;

t

)

x

0. Obviously k

H

(

t

n

x

);

H

(

t

0

x

) k!

0(

n

! +1) uniformly in

x

2 . And

H

(

t

) is a semi-

k

-set-contraction

operator for all

t

2 0

1]. In virtue of the fact: let

A

be a convex set in a

topological vector space

E

with a interior point

x

0, then for any

x

1

2

A

, the

open segment with end points

x

0 and

x

1 is contained in 

A

(cf. N.Bourbaki, "Espace Vectoriels Topologiques", Prop.16 in Chap.2,x2,

n

6), it is easy to see

that



62

h

t(

@

), 0

t

1. By Theorem 3, deg(

id

;

A





) = deg(

id





) =

1. The proof is complete. 2

x

3. EXISTENCE OF THE SOLUTION FOR TWO-POINT

BOUNDARY VALUE PROBLEMS IN BANACH SPACES

Now we consider the following boundary value problem

8 > < > : ;

x

00(

t

) =

f

(

t
x

(

t

)

x

0(

t

)

(

Tx

)(

t

)

(

Sx

)(

t

))

0 

t

1

ax

(0);

bx

0(0) =

x

0

cx

(1) +

dx

0(1) =

x

1

(12) where (

Tx

)(

t

) =Z t 0

k

(

t
s

)

x

(

s

)

ds

(

Sx

)(

t

) =Z 1 0

h

(

t
s

)

x

(

s

)

ds:

(13) Here

k

2

C

(

D

,

R

+),

D

= f(

t
s

) 2

R

2: 0 

s



t

 1g and

h

2

C

(

D

0,

R

+),

D

0 = f(

t

,

s

)2

R

2: 0



t

,

s

1g.

E

is Banach space. And assume

a

0,

b

0,

c

0,

d

0 and

J

=

ac

+

ad

+

bc >

0 throughout this section.

In order to investigate BVP (12), we rst consider the integral operator (

Ax

)(

t

) =Z 1 0

G

(

t
s

)

f

(

s
x

(

s

)

x

0(

s

)

(

Tx

)(

s

)

(

Sx

)(

s

))

ds

+

y

(

t

)

(14) where

f

2

C

(

I



E



E



E



E

,

P

),

y

2

C

2(

I
E

) and

y

(

t

) 



for

t

2

I

we dene the relation

x



y

by

y

;

x

2

P

, then '' is an order relation in

E

.

Moreover,





x



y

impliesk

x

k

N

k

y

k). We denote the relation

y

;

x

2 

P

by

x



y

). Let

G

(

t
s

) = (

J

;1(

at

+

b

)(

c

(1 ;

s

) +

d

)

t



s



J

;1(

as

+

b

)(

c

(1 ;

t

) +

d

)

t > s

(15) here

a

0,

b

0,

c

0,

d

0 and

J

=

ac

+

ad

+

bc >

0. Moreover,

T

and

S

are dened by (13). In the following, let

B

R=f

x

2

E

:k

x

k

R

g (

R >

0)

and

k

0= max f Z t 0

k

(

t
s

)

ds
t

2

I

g

h

0 = max f Z 1 0

h

(

t
s

)

ds
t

2

I

g

:

(16) Furthermore, let

P

(

I

) = f

x

2

C

1(

I

,

E

) :

x

(

t

) 



for

t

2

I

g

:

Then

P

(

I

) is

a cone in

C

1(

I

,

E

). Usually,

P

(

I

) is not normal in

C

1(

I

,

E

) even if

P

is a

normal cone in

E

. Let

q

1= sup t201] Z 1 0

G

(

t
s

)

ds
q

2 = sup t201] Z 1 0 j

G

0 t(

t
s

)

ds

and

q

= maxf

q

1

q

2 g (17)

Then we have the following lemma 7.

Lemma 7.

Let

f

be uniformly continuous on

I



B

R

B

R

B

R

B

Rfor any

R >

0. Suppose that there exist constants

L

i 0(

i

= 1, 2, 3, 4) such that



(

f

(

t
X
Y
Z
W

))

L

1



(

X

) +

L

2



(

Y

) +

L

3



(

Z

) +

L

4



(

W

) (18)

for any bounded

X
Y

,

Z

,

W



E

,

t

2

I

and

k

=

q

(

L

1+

L

2+

k

0

L

3+

h

0

L

4)

<

1

:

(19)

Then the operator

A

dened by (14) is a semi-

k

-set-contraction operator from

C

1(

I

,

E

) into

P

(

I

).

Proof. By direct dierention of (14), we have for

x

2

C

1(

I

,

E

), (

Ax

(

t

))0 = Z 1 0

G

0 t(

t
s

)

f

(

s
x

(

s

)

x

0(

s

)

(

Tx

)(

s

)

(

Sx

)(

s

))

ds

+

y

0(

t

)

(20) where

G

0 t(

t
s

) = (

J

;1

a

(

c

(1 ;

s

) +

d

)

t < s



J

;1( ;

c

)(

as

+

b

)

t > s

(21)

and ((

Ax

)(

t

))00 =;

f

(

t
x

(

t

)

x

0 (

t

)

(

Tx

)(

t

)

(

Sx

)(

t

)) +

y

00 (

t

)

:

(22) It is easy to see that the uniform continuity of

f

on

I



B

R

B

R

B

R

B

R

implies the boundedness of

f

on

I



B

R

B

R

B

R

B

R. So

A

is bounded and

continuous from

C

1(

I
E

) into

P

(

I

). Now, let

Q



C

1(

I

,

E

) be bounded. By virtue of (22), fk(

Ax

(

t

)) 00 k:

x

2

Q

,

t

2

I

g is a bounded set of

E

. So (

A

(

Q

)) 0

is equicontinuous, and hence lemma 2 implies that



(

A

(

Q

)) = maxfsupf



(

AQ

(

t

))

t

2

I

g

supf



((

AQ

)

0(

t

))

t

2

I

gg

:

(23)

On the other hand, it is easy to see that for any bounded

Q



C

1(

I

,

E

) with equicontinuous

Q

0,

f

f

(

s

,

x

(

s

),

x

0(

s

), (

Tx

)(

s

), (

Sx

)(

s

)),

x

2

Q

g is

equicontinuous because of the uniform continuity of

f

. By lemma 1, lamma 2 and (18) we have



(

AQ

(

t

)) =



 f Z 1 0

G

(

t
s

)

f

(

s
x

(

s

)

x

0 (

s

)

(

Tx

)(

s

)

(

Sx

)(

s

))

ds

+

y

(

t

)

x

2

Q

g   Z 1 0

G

(

t
s

)



; f

f

(

s
x

(

s

)

x

0(

s

)

(

Tx

)(

s

)

(

Sx

)(

s

))

x

2

Q

g

ds

 Z 1 0

G

(

t
s

)

L

1



(

Q

(

s

)) +

L

2



(

Q

0(

s

)) +

L

3



((

TQ

)(

s

)) +

L

4



((

SQ

)(

s

)) 

ds

 Z 1 0

G

(

t
s

)

L

1



(

Q

(

s

)) +

L

2



(

Q

0(

s

)) +

L

3 Z s 0

k

(

s
r

)



(

Q

(

r

))

dr

+

L

4 Z 1 0

h

(

s
r

)



(

Q

(

r

))

dr

]

ds

 Z 1 0

G

(

t
s

)

ds



L

1+

L

2+

L

3

k

0+

L

4

h

0]



(

Q

) 

q

1

L

1+

L

2+

L

3

k

0+

L

4

h

0]



(

Q

) 

q



L

1+

L

2+

L

3

k

0+

L

4

h

0]



(

Q

)

:

(24) Similarly, we have



((

AQ

)0(

t

)) =



 f Z 1 0

G

0 t(

t
s

)

f

(

s
x

(

s

)

x

0(

s

)

(

Tx

)(

s

)

(

Sx

)(

s

))

ds

+

y

0(

t

)

x

2

Q

g   Z 1 0 j

G

0 t(

t
s

)j



; f

f

(

s
x

(

s

)

x

0(

s

)

(

Tx

)(

s

)

(

Sx

)(

s

))

x

2

Q

g

ds

 Z 1 0 j

G

0 t(

t
s

)j

L

1



(

Q

(

s

)) +

L

2



(

Q

0(

s

)) +

L

3

k

0



(

Q

) +

L

4

h

0



(

Q

) 

ds

 Z 1 j

G

0 t(

t
s

)j

ds



L

1+

L

2+

L

3

k

0+

L

4

h

0]



(

Q

)



q

2

L

1+

L

2+

L

3

k

0+

L

4

h

0]



(

Q

) 

q



L

1+

L

2+

L

3

k

0+

L

4

h

0]



(

Q

)

:

(25) From (23), we have



(

A

(

Q

)) = max supf



(

A

(

Q

(

t

))

t

2

I

g

supf



((

AQ

) 0 (

t

))

t

2

I

g 

k

(

Q

) (26)

So

A

is a semi-

k

-set-contraction operator. The proof is complete.2

Let us list some conditions for convenience: (

H

1)

x

0




x

1





,

f

2

C

(

I



E



E



E



E
P

) is uniformly continuous

on

I



B

R

B

R

B

R

B

R for any

R >

0 and there exists

L

i0(

i

= 1

2

3

4)

such that (18) and (19) hold (

H

2) limR!+1

M(R)

R

<

1

qm, where

M

(

R

) = supfk

f

(

t
x
y
z
w

)k: (

t
x
y
z
w

)2

I



B

R

B

R

B

R

B

Rg

m

= maxf1

k

0

h

0

gand

q

is dened by (17)

Theorem 5.

Let (

H

1)

(

H

2)

(

H

3) be satised. Then BVP (12) has at least

one nonnegative solution in

C

2(

I
E

). Proof. It is well known that the

C

2(

I
E

) solution of (12) is equivalent to

C

1(

I
E

) solution of the following integral equation

x

(

t

) =Z 1 0

G

(

t
s

)

f

(

s
x

(

s

)

x

0(

s

)

(

Tx

)(

s

)

(

Sx

)(

s

))

ds

+

y

(

t

)

where

G

(

t
s

) is the Green function given by (15) and

y

(

t

) denotes the unique solution of BVP (

x

00=



0 

t

1

ax

(0);

bx

0(0) =

x

0

cx

(1) +

dx

0(1) =

x

1

which is given by

y

(

t

) =

J

;1 f(

c

(1;

t

) +

d

)

x

0+ (

at

+

b

)

x

1 g

:

Evidently,

y

2

C

2(

I
E

)

\

P

(

I

). Let

A

be dened by (14). Then condition

(

H

1) and lemma 7 imply that

A

is a semi-

k

-set-contraction operator from

C

1(

I
E

) to

P

(

I

). By (

H

2), there exist

>

0 and

R >

2 k

u

0

ksuch that for any

R

0 

R

M

(

R

0)

R

0

<

1

q

(

m

+

)

(27) and

_m

m

+

+k

y

k 1

R <

1 (28)

Let

U

=f

x

2

C

1(

I
E

), k

x

k

1

< R

g. So

U

is bounded convex open set. For

x

2

U

, we have k

x

k 1 

R

and k

Ax

k 0 = maxfk Z 1 0

G

(

t
s

)

f

(

s
x

(

s

)

x

0(

s

)

(

Tx

)(

s

)

(

Sx

)(

s

))

ds

+

y

(

t

) k

t

2

I

g  maxf Z 1 0

G

(

t
s

)k

f

(

s
x

(

s

)

x

0(

s

)

(

Tx

)(

s

)

(

Sx

)(

s

)) k

ds

+k

y

(

t

)k

t

2

I

g 

M

(

mR

)maxf Z 1 0

G

(

t
s

)

ds
t

2

I

g+k

y

k 1 

mR

1

q

(

m

+

)

q

1+ k

y

k 1

< R

(

_m

m

₊

+k

y

k 1

R

)

< R

(29) and k(

Ax

) 0 k 0 = maxfk Z 1 0

G

0 t(

t
s

)

f

(

s
x

(

s

)

x

0(

s

)

(

Tx

)(

s

)

(

Sx

)(

s

))

ds

+

y

0(

t

) k

t

2

I

g  maxf Z 1 0 j

G

0 t(

t
s

)jk

f

(

s
x

(

s

)

x

0(

s

)

(

Tx

)(

s

)

(

Sx

)(

s

)) k

ds

+k

y

0(

t

) k

t

2

I

g 

M

(

mR

)maxf Z 1 0 j

G

0 t(

t
s

)j

ds
t

2

I

g+k

y

k 1 

mR

1

q

(

m

+

)

q

2+ k

y

k 1

< R

(

_m

m

₊

+k

y

k 1

R

)

< R

(30) hencek

Ax

k 1

< R

.

In virtue of (29), (30),

AU



U

. Then by theorem 4 we get

deg(

id

;

A
U


) = 1

i.e., there is a xed point

x

2

U

. The proof is complete.2

Example 1

.

We consider following system of scalar valued dierential equations

8 > < > : ;

x

00 n= 3(j

x

nj+ 1) 1 2 + 1 n+1(

x

02 n+1) 1 3 + 1 2n j Rt 0 1 1+t+s

x

2n(

s

)

ds

j 1 3 + 1 3n( R 1 0 cos(

t

;

s

)

x

3n(

s

)

ds

) 2 3 + 17

x

n(0) =

x

n(1) = 0

n

= 1

2



:

(31)

Conclusion

: equation (31) has at least one positive solution. Proof. Let

E

=f

x

= (

x

1

x

2

x

n

), sup n2N j

x

nj

<

+1

:

g with norm k

x

k= sup n2N j

x

nj, and

P

= f

x

= (

x

1

x

2

) 2

E

,

x

n  0,

n

= 1

2

g. Then

P

is a normal solid cone of

E

and (31) can be regarded as a BVP of the form (12), where

a

=

c

= 1,

b

=

d

= 0,

x

0 =

x

1 =



,

k

(

t
s

) = 1 1 +

t

+

s

,

h

(

t
s

) = cos(

t

;

s

),

x

= (

x

1

x

2

),

y

= (

y

1

y

2

),

z

= (

z

1

z

2

),

w

= (

w

1

w

2

), and

f

=

g

+

h

= (

g

1

g

2

)+(

h

1

h

2

) in which

g

n(

t
x
y
z
w

) = 3(j

x

nj+ 1) 1 2 + 17

(32) and

h

n(

y
z
w

) = 1

_n

_{+ 1(}

y

2 n+1) 1 3 + 12

nz

1 3 2n+ 13

nw

2 3 3n

:

(33) Then k

f

k3(k

x

k+ 1) 1 2 + 12(k

y

k) 2 3 + 12k

z

k 1 3 + 13k

w

k 2 3 + 17

:

(34) which implies

M

(

R

)3(

R

+ 1) 1 2 + 12

R

2 3 + 12

R

1 3 + 13

R

2 3 + 17 and consequently lim R!+1

M

(

R

)

R

= 0

:

This shows that condition (

H

2) is satised.

Obviously,

f

2

C

(

I



E



E



E



E
P

) and

f

is uniformly continuous

on

I



B

R

B

R

B

R

B

R for any

R >

0. Now for any bounded

D



E

, it

is easy to see that



(

g

(

D

)) 3

2



(

D

). And for any bounded

Y



E

,

Z



P

,

W



P

, we have



(

h

(

Y
Z
W

)) = 0. In fact, let f

y

(m) g 

Y

, f

z

(m) g 

Z

, f

w

(m) g

W

, and

v

(m) n =

h

n(

y

(m),

z

(m),

w

(m)). By (33), we get j

v

(m) n j 1

n

+ 1k

y

(m) k 2 3 + 12

n

k

z

(m) k 1 3 + 13

n

k

w

(m) k 1 3

:

Now by the diagonal method, we can select a subsequence f

v

(m i ) g  f

v

(m) g such that

v

(m i ) !

v

0 2

P:

So



(

h

(

Y
Z
W

)) = 0. On the other hand, it is easy to see that in this case

q

_{= 12}

m

= 1

:

So the condition (

H

1) is satised. Consequently, our conclusion follows from

theroem 5. 2

The operator

A

dened by (31) is not a strict-set-contraction operator or a condensing operator. So the degree theory of the condensing operator or the strict-set-contraction operator is not suitable.

Acknowledgement

The author thanks the referee for his suggestions and helps.

References

1] K.Deimling, Nonlinear Functional Analysis, Springer-verlag, 1985.

2] Guo Dajun, Nonnegative Solutions of Two-point Boundary Value Problems for Nonlinear Second Order Integro-dierential equations in Banach Spaces, J.Appl.Math.Stochastic Anal.,4(1991), 47-69.

3] Guo Dajun, Extremal Solutions of Nonlinear Fredholm Integral equation in Or-dered Banach Spaces, Northeastern Math.J.,(4), 1991, 416-425.

4] L.Vaughn, Existence and Comparison Results on the Nonlinear Volterra Integral Equation in a Banach Space, Applicable Anal., 7(1978), 334-348.

5] Wieslaw Krawcewicz & Jianhong Wu, sl Theory of Degree with Applications to Bifurcations and Dierential Equations, A Wiley-Interscience Publication, 1997.

Yan Baoqiang

Department of Mathematics, Shandong Normal University Ji-Nan, Shandong 250014, People's Republic of China