TheSpaeofSolutionsDp Considertheboundary-valueproblem 8 &lt

M.Alves

ABOUTAPROBLEMARISINGINCHEMICALREACTORTHEORY

(ReportedonDeember23,1998)

1. Notation

ThroughoutthispaperCC[0;1℄denotestheBanahspaeofontinuousfuntions

x:[0;1℄ !R 1

,

kxk

C def

= max

0t1 jx(t)j;

L

p L

p

[0;1℄(1p<1)denotestheBanahspaeofsummableinp-thdegreefuntions

x:[0;1℄ !R 1

,

kxk

L

p def

=

1

Z

0 jx(t)j

p

dt

1=p

;

L1L1[0;1℄denotes theBanah spaeof essentiallyboundedmeasurablefuntions

x:[0;1℄ !R 1

,

kxk

L1 def

= vraisup

0t1 jx(t)j;

W 2

p

W

2

p

[0;1℄denotestheBanahspaeofontinuousfuntionsx:[0;1℄ !R 1

with

theabsolutelyontinuousderivativex_ suhthatx2L

p ,

kxk

W 2

p def

= kxk

L

p

+jx(0)j+jx(0)j:_

2. TheSpaeofSolutionsDp

Considertheboundary-valueproblem

8

<

:

(=0x)(t) def

= x(t) + k

t _

x(t)=f(t); t=2[0;1℄;

_

x(0)=0; x(1)=;

(1)

wherek>

1

p 0

,f2Lp,1<p1,2R 1

, 1

p +

1

p 0

=1,p 0

=1ifp=1.

ConsideringthisproblemonthetraditionalspaeW 2

p

,weseethat=

0

isnotdened

asanoperatoratingfromW 2

p intoL

p

. Followingthe shemegiveninthemonograph

[1℄,wewillinvestigatethisproblemonthespaeDpW 2

p

offuntionsx:[0;1℄ !R 1

;

suhthatx(0)_ =0anddenedby

x(t)= t

Z

0

(t s)z(s)ds+

1991MathematisSubjetClassiation. 34B15,34K10.

Key words and phrases. Boundaryvalue problem,funtional dierential equation,

foreahpairfz;g2LpR 1

.ThespaeDpisisomorphitothediretprodutLpR 1

.

TheisomorphismsJ :LpR 1

!Dpand J 1

:Dp !LpR 1

wedenebythe

equalitiesJ=f;Yg,J 1

=[Æ;r℄,where

8

>

>

>

<

>

>

>

:

(z)(t)= t

Z

0

(t s)z(s)ds; (Y)(t)=;

Æx=x; rx=x(0):

ThespaeDpbeomesaBanahoneunderthenorm

kxk

Dp def

=kxk

Lp +jx(0)j:

Theprinipalpartoftheoperator=0:Dp !Lpis

(Qz)(t) def

= (=

0

z)(t)=z(t)+(Pz)(t);

where(Pz)(t) def

= k

t t

Z

0

z(s)dsistheCesarooperator[2℄onthespaeL

p

. Thefuntions

u(t)1and v(t) =t 1 k

satisfytheequation =

0

x=0. Neverthelessthe fundamental

systemof =

0

x =0 onsists onlyof u(t) 1, suh as the otherelement v(t) =t 1 k

doesnotbelongtothespaeD

p

.Byvirtueoftheresultsof[5,p.102℄itfollowsthat,if

k>

1

p 0

,theoperatorQ:Lp !Lphastheboundedinverse

(Q 1

z)(t)=z(t) kt (1+k )

t

Z

0 s

k

z(s)ds:

Thesolutionoftheproblem(1)onthespaeD

p

isgivenbytheexpression

x=Wf+;

wheretheGreenoperatorW:L

p

!D

p

isdenedby

(Wf)(t) def

= 1

Z

0

W(t;s)f(s)ds;

W(t;s) def

= 8

>

>

<

>

>

: s

k

(t 1 k

1)

1 k

if0st1;

s k

(s 1 k

1)

1 k

if0t<s1;

fork>

1

p 0

; k6=1;or

W(t;s) def

= (

slnt if0st1;

fork=1. Really,usingtheequality

_ x(t)=

t

Z

0

 x(s)ds

forx2D

p

werewritetheproblem(1)onthespaeD

p

intheform



x(t)=f(t) kt (1+k )

t

Z

0 s

k

f(s)ds; t2[0;1℄; x(1)=:

Immediateomputationsshowthat

x(t) = t

Z

0 (t s)

h

f(s) ks (1+k )

s

Z

0

k

f()d i

ds+x(0)=

= t

Z

0 h

t s ks k

t

Z

s

(t ) (1+k )

d i

f(s)ds+x(0):

Theonditionx(1)=0 gives

x(0)= 1

Z

0 h

1 s ks k

1

Z

s

(1 )

(1+k )

d i

f(s)ds:

Consequently

x(t)= 1

Z

0

W(t;s)f(s)ds+; t2[0;1℄:

Bellowwewilluseresultsof[4℄aboutestimationofthespetralradius(H)ofthe

isotonioperatorH:C !C. Weformulatethisresultintheformsatisfyingouraims:

Lemma1. SupposethattheisotonioperatorHenjoystheproperty(H)(1)=0for

eah2C.Thefollowingstatementsareequivalent:

1)Thereexistsy2C suhthat

y(t)>0; y(t) (Hy)(t)>0; t2[0;1);

2) (H)<1.

Lemma2. The integral operator W :L

p

!C is ompletely ontinuous, for all

1<p1.

Proof. We onsider only the ase k >

1

p 0

, k 6= 1, the ase k = 1 an be proved

analogously. Toprovethe ompatnessoftheoperatorWitsuÆestoshow[5,p.102℄

that,foranyt

0

2[0;1℄theequality

lim

t!t

0 1

Z

jW(t;s) W(t0;s)j p

0

ds=0

holds.For1p 0

<1,0<t

0

<t1wehavethat

1

Z

0

jW(t;s) W(t0;s)j p

0

ds= t

0

Z

0

s

k

(t 1 k

1)

1 k s

k

(t 1 k

0 1)

1 k

p

0

ds+

+ t

Z

t

0

s

k

(s 1 k

1)

1 k s

k

(t 1 k

0 1)

1 k

p

0

ds

t

p 0

k +1

0

j1 kj p

0

(p 0

k+1) (t

1 k

t 1 k

0 )

p 0

+O(t p

0

+1

t p

0

+1

0

)!0; t!t +

0 :

Analogouslyweprovetherespetivestatementfor0=t

0

<t1and0t<t

0

1.

3. ThedelaV all

ee-PoussinLikeTheorem

Considertheboundary-valueproblem

8

<

: (=x)(t)

def

= (=

0

x)(t) (Tx)(t)=f(t); t2[0;1℄;

_

x(0)=0; x(1)=;

(2)

wherek>

1

p 0

,T:C !Lpisalinearantitonioperator,f2Lp.DenoteA def

= WT :

C !C.

Lemma3. Thefollowingstatementsareequivalent:

1)Thereexistsanelementy2Dpsuh that

y(t)>0; (t) def

= (=

0

y)(t) (Ty)(t)0; t2[0;1); and

y(1) 1

Z

0

(s)ds>0;

2)(A)<1;

3)Theboundaryvalue-problem(2) isuniquelysolvable onD

p

foreahf 2L

p ,2

R 1

,anditsGreenoperatorGisantitoni;

4)There existsa positive solution u 2 Dp on [0;1℄ of the homogeneous equation

=x=0.

Proof. Siney()satises

(=0x)(t) (Tx)(t)=(t); t2[0;1℄; x(1)=y(1)

onthespaeDp,itfollowsthat

y Ay=W+y(1)>0

on the spae C. Byvirtue of Lemma1 it follows that (A) <= 1. The impliation

1)=)2)isproved.

Supposing0weonsidertheproblem(2),whihisequivalenttotheequation

onthespaeC. Here

g() def

= 1

Z

0

W(;s)f(s)ds+:

Sine(A)<1,itfollowsthat

G=(I+A+A 2

+)W:

Consequentlytheimpliation2)=)3)isproved.

Theproblem

=0x Tx=0; x(1)=

isequivalenttotheequation

x=Ax+:

Sine(A)<1,wehavex=+A+A 2

+0if>0. Thusthe impliation

3)=)4)isproved.

Theimpliation4)=)1),followsfromLemma1beausethepositivesolutionu(t)

oftheequation=x=0satisestheinequalities

u(t)>0; u(t) (Au)(t)=>0; t2[0;1℄:

4. TheMainResult

Considerthenonlinearboundary-valueproblem

=

0

x=f(;x); x(0)_ =0; x(1)=; (3)

where :C !L

p

isalinearisotoni operator,1<p1, k>

1

p 0

,thefuntion

f(;)satisestheCaratheodoryonditions.Bydenitionputv=v;z=z,[v;z℄

def

=

fx2Lp:vxzg.

Following[5℄,wewillsaythatthefuntionf(;) satisestheonditionL i

[v;z℄, i=

1;2;ifitispossiblethedeomposition

f[t;u(t)℄=q

i

(t)u(t)+M

i

[t;u(t)℄; u2[v;z℄;

whereq

i 2L

1

,i=1;2;the operatorM

i :[v;z℄

L

p

!L

p

isdenedby(M

i u)()

def

=

M

i

[;u()℄,M

1

isisotoniandM

2

isantitoni.

Theorem1. Letv;z2D

p

beapairoffuntionssuhthatv(t)<z(t),t2[0;1℄,and

=

0

vf(;v); =

0

zf(;z); v(1)z(1): (4)

Supposethatthefuntionf(;) satisestheonditionL 2

[v;z℄ withq2 2L1, q2()

0. Thentheproblem(3)hasatleastonesolutionx2[v;z℄

Dp .

IfbesidestheL 1

[v;z℄ onditionisfullledwitha oeÆientq

1 2L

1

,andtheGreen

operatorof theauxiliaryproblem

=

1 x

def

= =

0 x q

1

x='; x(1)=0; (5)

Proof. Rewrite(3)intheform

(=

2 x)()

def

= (=

0 x)() q

2

()(x)()=M

2

[;(x)()℄; x(1)=

onthespaeDp. Thisproblemisequivalenttotheequation

x=A

2

x (6)

withtheompletelyontinuousisotonioperatorA

2 :[v;z℄

C

!C,denedby

(A

2 x)()

def

= 1

Z

0 G

2 (;s)M

2

[s;(x)(s)℄ds+u

2 ();

whereu

2

()isthesolutionofthesemi-homogeneousproblem

(=

2

x)(t)=0; t2[0;1℄; x(1)=;

G2(;)istheGreenfuntionoftheproblem

=

2

x=; x(1)=0: (7)

WeuseherethefatthattheGreenoperatorG

2

oftheproblem(7)hastherepresentation

G

2

=W [1,p.19℄,where :L

p

!L

p

isalinear homeomorphism,onsequently G

2

isaompletelyontinuous operatorbeause ofLemma2. Eahontinuous solutionof

the equation (6)belongsto the spaeDp,beause the operator A

2

isdened on the

orderinterval[v;z℄

C

ofthespaeCandmapsthisintervalintothespaeDp.Obviously

the isotoni operator :C !Lp mapsthe orderinterval[v;z℄

C

intoorderinterval

[v;z℄

Lp

. The operator M2 :[v;z℄

Lp

!Lp isantitoni, therefore itmaps the order

interval[v;z℄

Lp into[M

2 z;M

2 v℄

Lp .Lety

def

= z v.Theny(t)>0,t2[0;1℄,

=

2 yM

2

z M

2 v0;

beauseoftheantitoniityofM2and

y(1) 1

Z

0

(=2y)(s)ds>0:

Consequently,byLemma3wehavethattheGreenoperatorG

2

:Lp !DpCofthe

problem(7)isantitoni. Thus

[G

2 M

2 v;G

2 M

2 z℄

D

p [G

2 M

2 v;G

2 M

2 z℄

C :

Thereforetheequation(6)maybeonsideredintheorderinterval[v;z℄

C

ofthespae

C.Byvirtueoftheonditions(4)itfollowsthatz(t)(A2z)(t)andv(t)(A2v)(t)for

allt2[0;1℄. BeauseoftheisotoniityoftheoperatorA2:[v;z℄

C

!Cthisguarantees

A2[v;z℄

C [v;z℄

C

. For 1<p1the operatorA2 :[v;z℄

C

![v;z℄

C

isompletely

ontinuous as a produt of the operators : [v;z℄

C

! [v;z℄

Lp

, M2 : [v;z℄

Lp

!

[M2z;M2v℄

Lp

andtheompletelyontinuousG2:Lp !C.

Thus,theoperatorA

2

mapsthelosedonvexset[v;z℄

C

oftheBanahspaeC into

itself. Inaordane withthe Shauder's xed point theoremthe equation (6) has at

leastonesolutionx2[v;z℄

C .

Let us show that the set of all solutions x 2 [v;z℄

C

has a superior element x 2

[v;z℄

C

(the upper solution) and an inferior element x 2 [v;z℄

C

(the lower solution).

Let x 2 [v;z℄

C

be a solutionof the equation (6). The sequene fz i

g, z i+1

= A

2 z

i

;

z 0

= z monotonially dereasesand is bounded by x 2 [v;z℄

C

, beause the operator

A2 maps the set [v;z℄ into itself. Aompat monotone sequene fz i

gonverges [2,

p.38℄tox= lim

i!1 z

i

.Sinethislimitisasolution,theinequalityxxforanysolution

x2[v;z℄

C

isproved. Analogouslyweshowtheexisteneoftheinferiorsolutionx.

Now wehaveto show that ifthe ondition L 1

[v;z℄ isfullled, the solutionof the

problem(3)isunique,i.e. x=x. Using theL 1

[v;z℄, onditionwerewritethe problem

(3)intheform

(=1x)()=M1[;(x)()℄; x(1)=:

Thisproblemisequivalenttotheequation

x=A

1 x

onthe orderinterval[v;z℄

C

ofthespaeC withantitonioperatorA1 :[v;z℄

C

!C,

denedby

(A

1 x)()

def

= 1

Z

0 G

1 (;s)M

1

[s;(=x)(s)℄ds+u

1 ();

whereG1(;)istheGreenfuntionoftheproblem(5),u1()isthe solutionofsemi-ho-

mogeneousproblem

(=

1

x)(t)=0; t2[0;1℄; x(1)=:

Considertheequalityx x=A1x A1x. Theleft-handsideofthe equalityisnon

negativeandtheright-handsideisnonpositive,thuswegetx=x.

5. Examples

Example1.Considertheboundary-valueproblem

8

<

:

 x(t)+

1

t _

x(t)= exp

1

jx(t)j

; t2[0;1℄;

_

x(0)=0; x(1)=0;

(8)

where0e 2

. Thisproblemdesribesproessesarisinginhemialreatortheory

withilindrialsymmetry[7,p.326℄,undertheArrheniuslaw.Weonsiderthisproblem

onthespaeD1.

Asomparisonfuntionswehoose

v(t)0; z(t)=

4 (1 t

2

)+ 1

2 :

Atrivialveriationshowsthattheonditions(4)arefullled:

 v(t)+

_ v(t)

t

+exp

1

jv(t)j

=0;

 z(t)+

_ z(t)

t

+exp

1

jz(t)j

+=0; t2[0;1℄;

v(1)=0=x(1)<z(1)= 1

2 :

Thefuntionf(;x)= exp

1

jxj

satisestheonditionL 2

[v;z℄withtheoeÆ-

ientq20. Theboundary-valueproblem

=0x=; x(1)=0;

hasforeah2L1 auniquesolutionx2D1,anditsGreenfuntion W(t;s)0on

Besides,thefuntionf(;x)= exp 1

jxj

satisestheonditionL 1

[v;z℄withthe

oeÆientq

1

= 4e 2

.

Takingthefuntiony(t)=

4 (1 t

2

)wehave:

(=

1

y)(t)=(=

0

y)(t)+4e 2

y(t)= + 2

e 2

(1 t 2

)<( 1+e 2

)0

and

y(1) 1

Z

0

(=1y)(s)ds= 1

Z

0

[

2

e 2

(1 s 2

)℄ds=( 2

3

2

e 2

)>0;

sinee 2

:Consequently,byLemma3,theGreenoperatorG

1

oftheproblem

=

0 x+4e

2

x=; x(1)=0

isantitoni. Then,beauseofTheorem1theproblem(8)hasauniquesolutionx2D1

suhthat

0x(t)

4 (1 t

2

)+ 1

2

; t2[0;1℄:

Example2.Let

8

<

:

 x(t)+

2

t _

x(t)= exp

1

jx(t)j

; t2[0;1℄;

_

x(0)=0; x(1)=0;

(9)

beanonlinearboundary-valueproblem,37:28 12

17 e

17=2

. Thisproblemdesribes

proessesarisinginhemialreatorwithspherialsymmetry[7,p.326℄.

Theproblem(9)withsuhhasmorethanonesolutiononthespaeD

p

,1<p1.

Indeed,thereareatleasttwopairsoffuntions

v

1

(t)0; z

1 (t)=

2(1 t 2

)

17

; v

2

(t)=4[erf(1) erf(t 2

)℄; z

2 (t)=

6 (1 t

2

):

Theonditions(4)arefullled:

 v1(t)+

2v_

1 (t)

t

+exp

1

jv

1 (t)j

=0;

 z1(t)+

2z1(t)_

t

+exp

1

jz1(t)j

=exp

17

2(1 t 2

)

12

17

<0;

 v

2 (t)+

2v2(t)_

t

+exp

1

jv

2 (t)j

=

16exp( t 4

)

p

(4t

4

3)+

+exp

0:25

erf(1) erf(t 2

)

>0;

 z

2 (t)+

2z2(t)_

t

+exp

1

jz

2 (t)j

=exp

6

(1 t 2

)

<0;

sine37:28 12

17 e

17=2

,t2[0;1℄. Theexistene ofsolutionoftheproblem(9) on

eahinterval[v

i

;z

i

℄,i=1;2,followsfromTheorem1.Sinetheintervals[v

1

;z

1

℄,[v

2

;z

2

℄

aredisjoint,the problem(9)has atleasttwo solutionsx

1 ,x

2 2D

p

,1<p1,suh

thatv

1 x

1 z

1 ,v

2 x

2 z

2 .

Referenes

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ential equationsandappliations. Mem. DierentialEquationsMath. Phys. 8(1996),

1{102.

2.P.Zabre

ikoandothers,Integralequations. (Russian)Nauka,Mosow,1968.

3.I.Muntean,ThespetrumoftheCesarooperator.Mathematia{Revued'analyse

numeriqueetdetheoriedel'approximation45(1980),97{105.

4.N.AzbelevandL.Rakhmatullina,Ontheestimateforthespetralradiusofa

linearoperatorinthespaeofontinuousfuntions. (Russian)Izv. Vuzov. Matematika

11(1996),23{28.

5. N. Azbelev,V.Maksimov,andL.Rakhmatullina,Introdutiontothetheory

offuntionaldierentialequations. (Russian)Nauka,Mosow,1991.

6.M.

Krasnosel'ski

,Positivesolutionsofoperatorequations. (Russian)Izv. Vuzov.

Matematika11(1996),23{28.

7. D.

Frank-Kamenetski

,Diusionandheatexhange inhemialkinetis.(Rus-

sian)Nauka,Mosow,1987.

Author'saddress:

EduardoMondlaneUniversity

DepartmentofMathematis

P.O.Box257-Maputo

Mozambique