THE FIXED POINTINDEXFORACCRETIVEMAPPING* WITH K---SETCONTRACTION PERTURBATIONIN CONES YU-QINGCHEN Department ofMathemat=cs StchuanUmverstty 610064Chengdu P.R

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VOL. !9 NO. 2 (1996) 287-290

THE FIXED POINT

INDEX

FOR

ACCRETIVE

MAPPING*

WITH K---SET

CONTRACTION PERTURBATION

IN CONES

YU-QINGCHEN Department ofMathemat=cs

StchuanUmverstty 610064Chengdu

P.R. China

(Received July 21, 1993 and in revised form August 15, 1994)

ABSTRACT:LetPbe cone BanachspaceE, A,K are twomappingsinP,A accretive, Ks k-setcontrac-

tion, then fixedpointindex definedformapping-A+K, somefixed pointtheoremsarealso deduced.

KEYWORDS AND PHRAESE accretivemapping, k--setcontraction,cone,fixedpoint index.

1992 AMSSUBJECTCLASSIFICATIONCODES:47H10. 47H05. 54H25

1. INTRODUCTION

Thefixedpointindexisaimportanttoolinsolvingpositivesolutionsofnonlinear equationsmordered Banach space. Sowhat nonlinear mappingcould bedefinedaindex theory becomesaveryinterestingproblem, many authors have studiedthisproblem,see13.23.83.(103,(123.133. Inthis paper.E Banachspace PCEis closedcone, i. Pisclosedconvex,and

PCP,V 0, P(q(--P)={0};

(ZEsa nonemptyopenbounded subset. LetA: ^D(A) P--2 beamultivaluedaccretive mapping, i, e x-y

<

.z’--y+,(a--az)II ,.z’,yGDtA) ,aA:c,

K=FJP--P^is ^strict^k--setcontraction,i,e0k<l; If

(I+A)(D(A) )=P,amt.z-E--A.z’+K.z’,Y

thenafixed point indexisdefined for--A+K,whenKis compact,suchtypemappingwerestudied by 4 143,153.

2. MAIN RESULTS

LetEbeaBanachspace,PCEis aclosedcone,"" isthe order induced byPinE,i,e

xy

ifandonlyify

--xP.

PROPOSITION :A:D(A)=P-"Pisacontinuous accretivemapping,for each xP, thereexistsfl(x)O, suchthatAxfl(x). then(M+A)P--P,t20;

PROOF.:Foreach

z

P, considerthe followingdifferential equation

t[0, -k-c)

(2" 1) x(O)=uP

Foreach xGP, sinceAx[3(x)

,

sothereexists W(x)CzP, such that fl(x) ^x=Ax+W(x) Sowehavex-+-(----Ax+z)=(1--e2--fl(x))x+eW(.z)

Forsufhcientsmall^0,suchthat1--2--fl(x)O, then (1-)--fl(x))x +eW(x)+zP Hence

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288 Y.Q. CHEN

bye63.weknow ^(El) has onlvonesolunon. Let^a’(t.u) betheunique solution of (El)withx(0)--u.

Now. define mappingB:P---P following

Blu-a’(T,u),urP,T>O constant;

For v6P. Let (t)= a’(t,u)-x(t.v) ,then

(t)D (t)-(x’(t,u)--w’(t.v).x(t.u)--x(t.v))

whereD

f(t)--lm;(t)--(t--h);

see(63P.36)

(,(t)D (t) Lr(tou)-- Ax(t,u)-+-Lr(t ,v)+Ax(t.v) x(t,u)--x(t,v) Aisaccertive.so

Ax(t.u

+

^Ax(t ^,x(t^,v ^x(t^,v) (Ax(t,u) Ax(t ,v) ,x(t ,u) x(t ,v))_O

Therefore

f(t)D (t)--a(t) (t)e

Sowehave B, u--B-,v

<e

Hence, Br^has^aumque fixedpointu0

-

^P,t,e^B-^u0 ^u,. Thin mpliesx’^(t,u0) 0,t>0, SoO= Au,, Au,,+z,z6(A+AI) (P).

Thiscompletethe proof.

In^thefollowing,weassumeA:D(A)CP--2 is multivaluedaccritivemapping,(A+I) (D(A))=P, it’swell known(I+A) is nonexpanmve(see(43).

Letflbe openboundedsubset ofE,K:OI"IP--Pis strictk--setcontraction, i, ek Suppose^D(A)O=i ^,and

x

^Ax-+-Kx,^t

x

ff’ll’qD(A) then

x:(I+A) Kx,

(I+A) Kisalsoastrictk--setcontraction.sothe fixedpointindexi((I+A)-’K, I’JOP) iswelldefined, see 13, (83. Now, wedefine

(--A+K,OOD(A))=i((I+A)

THEOREM 1: ^(a)IfO=B(O,r),Kx=xoB(O,r)f]P,t xB(O,r)f]P,then t(--A+K,B(O,r)IqD(A))=

(b)Suppose

=

^O,^U^{fJ, O, f]}

,

then

A+^K,l’J^D(A) A

+

^{K .0, f]}^{D(A) +(} A-4-K

,

f]^D(A)

(c)LetH(t,x) (0,13 (Of]P)--P. ifH(t, x isuniformlycontinuous m xfor eacht, and for each 0,13.H

(t, )Of]P-Pisa strtctk--set contraction,kdoesn’tdependont, suppose

then i(--A+H(t,x),Of]D(A)) doesn’t dependont.

(d) If i(--A+K,Of]D(A))O, then

x

^--AxWKx^hasâ^solutionⁱⁿÔD(A). î,e--AA-Khasafixedpoint.

PROOF: by thedefinition,(b), (c), (d) isobvious. (see(13or Now,weprove(a). First, wehave

O^D(A)^,atwlO^AO (2.2)

Infact, (A+I)D(A)=P, sothereexistsxD(A),aAx,suchthatx-+-a--O Sincex>O,a>O, Sowemusthavex--O,a=OAO. Hence

(A+I) 0=0 (2.3)

by the definition,weneedtoprove

i((I+A)--K,OP)=I,0=B(O,r) (2.4)

Since(I+A)Kx=(I+A)Xo,xCO["lP,and

(I+A)-’Xo--<l+A)-’O

<

^x0 ^<r

So(I+A) ’XoP=B(O,r)P, by 13(seealso83).

t((I+A)-K,B(O.r)(qP)=

Soi--(--A+K,B(O,r)f’ID(A))=I.

Inthefollowing, K,A,,are^{same as}above.

LEMMAl:IfKx=x. xQP;and^0,Q,then

PROOF:Let^H(t,x)^=tKx,t^0,13^{,xl’JP. If}

x

--Ax-+-tKxforsome

x

^qfq ^D(A)^andt^0,13,

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thent:eO(otherwise.weget 0 ’oacontrad,ctmn)

So K.

----

contradictiontoK.rx.

Hencc, H(t.x)satisfyall the conditions of (c) theorem1.

AK.D(A) )=t(--A+O,D(A)

by 2.3), wehave (I+A) 006I’

So(I,4) O,P)I, andweget

A0,I)(A)) HolIc(.

t(--A+K.D(A))--1

COROLLARY 1: If0,andKa’

,

_.rl

,

then --A+KhasafixedintinD(A) PROOF: It’sobvousKax,a’P. ^By^lemma ^1,

t(--AK.D(A))-- Theorem1,(d) imphes-A+K has fixedpoint D(A).

LEMMA2: Let^u,,0,tl,,P,supposex--tu,,--A(.r--tu,,)+Kx,f

x

P, andx--tuoD(A) ,fort

0 Then

t(--A+K,D(A))=O

PRF: Supposet(I+A) K,D(A)O

For eachr0,Let H(t,x)=(I+A) K+tru,,,xP,t

It’sobvious H(t,x)isuniformlycontinuousin for eacht,and H(t, )isstrictk--setcontractionfor eacht.

By(l.(seealso 83).Weget

t((I+A) K+ruo,P)=t(I+A) K,DP)O

thereexistsx,P,suchthat

x--(I+A) Kx=ru,, (2 6)

Lettingrov,theleft rodeof(2.6) is bounded, butthemghtrode of(2,6) is unbounded,acontradiction.

Wemusthave(--A+K,D(A))=O

THEOREM2 I{A: D(A) P-2 is an accretive mapping,(I+A D(A) P,

, ,

^are^twoopenund subsetsofE.0 ,K:Pis strictk--setcontractionmapping, OuoP

(i)Foreachx,,xKx;for each

x

^P,x--tuo ^D(A),tO,x--tu,, --A(x--tuo)+Kx;

(i)Foreach

x

;xKx,for each

x

zP,x--tuoD(A),tO,x--tuo --A(x--tuo)+Kx;

Supposeeither (i)or (ii)issatisfied,then --A+Khasafixed pointin(,--)D(A) PRF: Supposecondition (i) issatmfiedby, Lemma 1, wehave

i(--A+K,,D(A))= (2.7)

byLemma2,wehave

i(--A+K,D(A))= (2.8)

by(b) o{Theorem1,and (6), (7).Weget

i(--A+K,(-- D(A)

by(d)ofTheorem1, weknow--A+K^hasafixedpoint in(,--)D(A).

(A)

(2.5)

PROOF:we maysuppose

x

^--Ax+Kx. ^xOE^D(A)

LetH(t,x)=tKx,xEzfJNP,tE(O,13;

It’sobvious H(t,x)isuniformlycontinuous inx ,and H(t, ),sstrictk--setcontractionfor eacht.

Weshow that

x

--Ax+H(t,x),xEaO["lD(A),t0,1]

Ifx--Ax+H(t,x) forsomexaOD(A) ,tO,1],thenx---(I+A) IH(t,x)

(2.9)

(2.10) If (xi)is sausfied,theproofissimilar. We complete theproof.

THEOREM3: Foreach Of’ID(A), Kx

<

,and 0/J,then --A+Khasafixed pointin

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290 Y.Q. CHEN

Since (I-+-A) is nonexpansveand (1+,,1) 0:-0.So

Thereforet=1. contradictto(8),by(c)ofTheoreml,

(--A+K,DD(A) =t(--A+0,DD(A)) and (2.5) lmphest(--A+K,DD(A))=I.

by(d)of Theorem 1,--A+Khas fxedpoint D(A).

THEOREM 4:If0, D(A).

PROOF:Wemayassume

-

^Az+Kx,^xCD(A)

LetH(t,x)=tKx,t6 0,1]

If

x

^-Ax+tKx^for^somet6[0.1],xGD(A),thentKxGx+Ax So thereexists

Bytheassumptmn (2. ll),tl,wemusthaveKx=0, x+a=0 By (2.3),x=06,a contradmrionto

Sowehave

x

--Ax+H(t,x), xCD(A),t60,1].

ThefollowingprfismmdartothatofTheorem 3. This end the prof.

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