Keywords:oupled systemofseond-order boundary value problems,nonloal boundaryondition,nonlinearboundaryondition,superlineargrowth,positive solution Classiation: Primary34B10,34B15,34B18;Seondary47H07,47H10 1

Nonloal systems of BVPs with

asymptotially superlinear boundary onditions

Christopher S. Goodrih

Abstrat.Inthispaperweonsideraoupledsystemofseond-orderboundary

valueproblemswithnonloal,nonlinear boundaryonditions,and weexamine

onditionsunderwhihsuhproblemswillhaveatleastonepositivesolution. By

imposingonlyanasymptotigrowthonditiononthenonlinearboundaryfun-

tions,weareabletoahievegeneralizationsoverexistingworksand,inpartiular,

weallowforthenonloaltermstobeabletoberealizedasLebesgue-Stieltjesin-

tegralspossessingsignedBorelmeasures.Weonludewithanumerialexample

toillustratetheuseofoneofourtwomainresults.

Keywords:oupled systemofseond-order boundary value problems,nonloal

boundaryondition,nonlinearboundaryondition,superlineargrowth,positive

solution

Classiation: Primary34B10,34B15,34B18;Seondary47H07,47H10

1. Introdution

Inthispaperweonsider asystemofnonloal boundaryvalueproblemswith

nonlinearboundaryonditions. In partiular, weonsider thenonlinear system

ofboundaryvalueproblems

(1.1)

x 00

(t)= a

1 (t)g

1

(x(t);y(t)); t2(0;1);

y 00

(t)= a

2 (t)g

2

(x(t);y(t)); t2(0;1);

x(0)=0=y(0);

x(1)=H

1

1 (x)+"

1

0 x

1

0

;

2 (y)+"

2

0 y

1

0

;

y(1)=H

2

1 (x)+"

1

0 x

2

0

;

2 (y)+"

2

0 y

2

0

;

where "

1

0 , "

2

0

> 0 are onstants, whih shall be speied later, 1

0

; 2

0

2 (0;1)

arexed,

1 ,

2

:C([0;1℄)!R arelinearfuntionals,whihapturethenonloal

natureoftheboundaryonditions,andH

1 ,H

2 :R

2

!Rareontinuousfuntions,

whih apturethenonlinearnatureoftheboundaryonditions. Wealsoassume

that the nonlinearities g

1 , g

2

: [0;+1)[0;+1) ! [0;+1) are ontinuous

Lebesgue-Stieltjesintegrals|thatis,

(1.2)

1 (x):=

Z

[0;1℄

x(t)d

1

(t) and

2 (y):=

Z

[0;1℄

y(t)d

2 (t);

with

1 ,

2

2BV([0;1℄). Sineitmaybeassumedwithoutlossthat,infat,

1 ,

2

2NBV([0;1℄),wegetthatassoiatedtoeahof

1 ,

2

thereexists aunique

Borelmeasure,say

1 and

2

, respetively. Inourontext,importantly,these

measuresmaybesigned.

Herewestudytheexisteneofatleastonepositivesolutiontoproblem(1.1).

Toaomplishthistask,weusetheperturbationtermsin(1.1)|namely,"

1

0 x(

1

0 ),

"

2

0 y(

1

0 ), "

1

0 x(

2

0

), and "

2

0 y(

2

0

) | as well as a new ondition on the nonlinear

funtions H

1 and H

2

. These novelties reveal, in awaythat shallbe delineated

momentarily,thatmanyoftherestritionspreviousauthorshaveimposedonthe

varioustermsappearinginotherproblemssimilarto(1.1)are,infat,unneessary

in oursetting. Ourprinipal onditiononthese funtions isto requirethat, for

eahi=1;2,

(1.3) lim

z1+z2!+1 H

i ( z

1

;z

2 )

z p

1

i

1 +z

q 1

i

2

=+1

holdsforsomep 1

i ,q

1

i

2(0;1℄withatleastoneofp 1

i andq

1

i

,foreahi=1;2,

ableto betakenequal tounity. Inpartiular (f., Remark 3.2), ondition(1.3)

impliesthat eah ofH

1 and H

2

mayenjoyasymptotiallysuperlinear growthin

atleastoneofthetwooordinatediretions(f.,Remark3.3). Wewillevengive

anexisteneresultassoiatedtothesomewhatmorerelaxedondition

(1.4) limsup

z1+z2!0 +

H

i (z

1

;z

2 )

z

1 +z

2

<

i

;

for eah i = 1;2, with

i

a positive onstant to beseleted later; importantly,

theresultassoiatedtoondition(1.4)willevenbeappliableintheunperturbed

ase| i.e., "

1

0

="

2

0

=0. It should be pointedout that, in fat, Yang [16℄, [17℄

introduedan asymptoti onditionsimilar to (1.4), though in the ontext of a

slightlydierentproblem. Regardless,Yangimposesanumberofotherhypotheses

|suhasompliatedonditionsontheequivalentofournonlinearities g

1 and

g

2

aswell asthe assumption that the equivalent of

1 and

2

be positive|

withwhih weompletelydispensehere.

Inanyase,to plaeproblem(1.1)inanappropriateontext, weremarkthat

it is, in fat, most losely related to reent papersboth of Kang and Wei [10℄

and of Infante and Pietramala [7℄. Regarding [10℄, Kangand Weionsidered a

problemverysimilarto (1.1). However,theywereforedtoassumethateahof

themeasures

1 and

2

waspositive. Moreover,regardingtheir equivalentof

thenonlinearitiesg

1 andg

2

appearingin(1.1),theyassumedthatthesefuntions

satisedverystritgrowthonditions. Ontheotherhand,regarding[7℄,asimilar

H

1 andH

2

werefuntionsofasinglevariableonly|forinstane,H

1

(x). Inany

ase,theauthorsthereassumedthateahofthemeasures

1 and

2

wasposi-

tive. Furthermore,theyassumedthatthenonlinearboundarynonlinearities(i.e.,

theequivalentof H

1 and H

2

) satiseduniformly lineargrowth| that is, there

were 0< suh that zH(z)z, forall z0. Thislatter ondition

is somewhatrestritive,and weremoveit ompletely in this work. Finally, our

tehniquesevenallowforthenonlinearitiesg

1 andg

2

tohaveompletelydierent

limitingbehavior|f., point(5)below.

Inadditionto[7℄,[10℄, therehavebeenmanyotherreentworks onnonloal,

nonlinear boundary value problems | see, for example, [6℄, [8℄, [9℄, [16℄, [17℄.

While ourwork here is slightlyless diretly related to these, it is, nonetheless,

nontrivially onneted to these other papers, and we provide here tehniques

and insights notfoundin anyof thoseother works. It isertainly importantto

mention that the basi one theoreti tehnique used in this paper is indebted

to theimportantpaperof Infante and Webb [12℄. Finally,wementionthat our

resultshereomplementertainoftheresultswhihwehavereentlygivenin[4℄.

Insummary,weprovideherethefollowinggeneralizationsoverpreedingworks.

(1) Weallowforeahof

1 and

2

tobesignedmeasuresratherthanmerely

positive. This isanimprovementoverthepreedingworks,asintimated

above.

(2) Wedonot assumeauniformlineargrowthonditiononeitherH

1 orH

2 .

Weinsteadassumeeithertheasymptotionditiongivenin(1.3)together

with an assumption that these funtions possess superlinear growth as

z

1 +z

2

!0orondition(1.4). Inpartiular, thisshowsthatsuperlinear

growthat (+1;+1) isallowable. Moregenerally,oneneednot assume

auniformlineargrowthonditionasseemstoappearin nearlyallworks

onthissortsofproblems|f.,[6℄,[7℄,[8℄,[9℄|sineinoursettingthere

maybeno >0suh thatH

i (z

1

;z

2 )(z

1 +z

2

),forallz

1 ,z

2 0.

(3) Speially regardingYang's works [16℄, [17℄, wepoint out that our re-

sults hereeven provide someinterestinggeneralizations of the methods

ontained therein. In partiular, while the results of [16℄, [17℄ onern

dierent problems than (1.1), those works do appear to be among the

only ones to onsider anasymptoti ondition with respet to the non-

linearboundaryfuntions,atleasttothebestoftheauthor'sknowledge.

A loseexamination of the proofs in those works, however, reveals that

they use in a very expliit way the positivity of therespetive Stieltjes

measures. Lakingthispositivity,aswedohere,wemustsearhforalter-

nativeapproahes. Consequently,wefeelthat ourresultshererepresent

aninterestingadvanementoverthosepresentedin[16℄, [17℄.

(4) Webelievethat ourtehniquesevenallowH to beonlyeventuallyposi-

tive,thoughwedonotprovesuhatheoremhere|see[3℄foranexemplar

ofthisextensioninaontextsomewhatdierentfromthisone.

(5) We show that the assumption of asymptoti superlinearity of the fun-

tionsH andH allowsforneitherg norg to haveanypartiulartype

ofgrowth(e.g.,sub-orsuperlinearity)ask(x;y)k!+1. Inpartiular,

thismeansthatg

1 andg

2

anhaveompletelydierentlimitingbehavior.

Forexample,g

1

ouldbesublinearask(x;y)k!+1,whilstg

2

issuper-

linearask(x;y)k!+1. WhileYang alsoallowedformixedasymptoti

behaviorofthenonlinearitiesin[16℄,aursoryexaminationofthatpaper

indiatesthatanumberofompliatedonditionsarerequiredtodedue

that result. By ontrast, our onditionsare quite simple and relatively

easytohekomputationally.

2. Preliminaries

WeonsiderinthisworkthespaeX:=BB,whereBrepresentstheBanah

spae C([0;1℄) when equipped with the usual supremum norm, kk := kk

1 .

Note | see Dunninger and Wang [2℄ | that X beomes a Banah spae when

equippedwiththenormk(x;y)k:=kxk+kyk. Itisthenknownthataxedpoint

inXof

(2.1)

S(x;y)(t)

:=(T

1

(x;y);T

2 (x;y))

=

tH

1

1 (x)+"

1

0 x

1

0

;

2 (y)+"

2

0 y

1

0

+ Z

1

0

G(t;s)a

1 (s)g

1

(x(s);y(s))ds;

tH

2

1 (x)+"

1

0 x

2

0

;

2 (y)+"

2

0 y

2

0

+ Z

1

0

G(t;s)a

2 (s)g

2

(x(s);y(s))ds

isasolutionofproblem(1.1),whereS :X!XandT

i

:X!B,foreahi=1;2.

Here G: [0;1℄[0;1℄!R appearingin (2.1)is theGreen's funtion assoiated

tothetwo-pointonjugateproblem|thatis,

(2.2) G(t;s):=

(

t(1 s); 0ts1;

s(1 t); 0st1;

asiswellknown|see,forexample,[11℄. Inthesequel,weshallassumethatthe

set[a;b℄isagivenxedsubintervalof(0;1). Withthisdelarationitisthenwell

knownthat thereisaonstant:=min

t2[a;b℄

ft;1 tgsuhthat

(2.3) min

t2[a:b℄

G(t;s) max

t2[0;1℄

G(t;s)=G(s;s);

foreahs2[0;1℄. Notethat2(0;1). Finally,letusalsoreallasapreliminary

lemmaKrasnosel'ski'sxedpointtheorem|see[1℄.

Lemma2.1. LetB beaBanah spaeand letK Bbeaone. Assume that

1 and

2

areboundedopensetsontainedin Bsuh that02

1 and

1

2 .

Assume,further,thatT :K\(

2 n

1

)!Kisaompletelyontinuousoperator.

Ifeither

(1) kTykkykfory2K\ andkTykkykfory2K\ ; or

(2) kTykkykfory2K\

1

andkTykkykfory2K\

2

;

thenT hasat leastonexedpointinK\(

2 n

1 ).

3. Mainresult and numerialexample

We begin by listing the various strutural onditionsweimpose on the on-

stituentpartsof problem(1.1). Theseonditionsarethefollowing.

H1: Foreahi, letH

i :R

2

![0;+1)beareal-valued,ontinuousfuntion.

Moreover, H

i

:[0;+1)[0;+1)! [0;+1) | i.e., H

i

is nonnegative

whenrestritedto [0;+1)[0;+1).

H2: Foreah i,the funtional

i

(y) appearingin (1.1) islinear and,in par-

tiular,hastherealization

(3.1)

i (y):=

Z

[0;1℄

y(t)d

i (t);

where

i

:[0;1℄!R satises

i

2BV([0;1℄).

H3: Foreahi, there is aonstant"

i

1 2[0;

1

2

)suhthat the funtional

i in

(1.1)satisestheinequality

(3.2) j

i

(y)j"

i

1 kyk

forally2C([0;1℄).

H4: Foreahi,therearep 1

i

2(0;1℄andq 1

i

2(0;1℄,whereforeahiatleast

oneofp 1

i andq

1

i

isequaltounity,suh that

(3.3) lim

z1+z2!+1 H

i ( z

1

;z

2 )

z p

1

i

1 +z

q 1

i

2

=+1

holds. Furthermore,foreahiitholdsthat

(3.4) lim

z1+z2!0 +

H

i ( z

1

;z

2 )

z

1 +z

2

=0:

H5: Wendthat

(3.5) lim

x+y!0 +

g

1 (x;y)

x+y

=0 and lim

x+y!0 +

g

2 (x;y)

x+y

=0:

H6: Theonstants"

1

0 , "

2

0 ,"

1

1 ,and"

2

1 satisfy

(3.6) 0"

1

0 +"

2

0 +"

2

1 +"

2

1

<

1

2 :

H7: Foreahi, eahof

(3.7)

Z

td

i (t)0

and

(3.8)

Z

[0;1℄

G(t;s)d

i (t)0

holds,wherethelatterholdsforeahs2[0;1℄.

Letusmakesomebriefremarksregardingertainofthepreedingonditions.

Remark 3.1. Regarding onditions (H2){(H3), we point out that a wide vari-

ety of funtions satisfy these onditions. Indeed, onsider the following pair of

funtionals.

(3.9)

i

1 (y):=

Z

F y(t)dt;

i

2 (y):=

n

X

k =1 a

k y(

k ):

Sine eah of (3:9)

1 {(3:9)

2

is linear, eah satises (H2). On the other hand, so

longasm(F)"

i

0

, say, where m is theLebesgue measure,then (3:9)

1

satises

(H3). Provided that P

n

k =1 ja

k j "

i

0

, then (3:9)

2

satises (H3). Example 3.9

ontainsanotherexample.

Remark 3.2. Regardingondition(H4) and speially (3.3)therein, this isthe

asymptotisuperlinear onditionwhih, in part,distinguishes ourmethods here

fromothers. Ontheotherhand,(3.4)appearinginondition(H4)impliesthatH

isalsosuperlinearas(x;y)!(0 +

;0 +

) . Somefuntions,H:[0;+1)[0;+1)!

[0;+1),satisfyingondition(H4), then, arethefollowing. (In eah ase,p 1

i

=

q 1

i

=1,foreahi.)

(3.10)

H( z

1

;z

2 ):=z

r1

1 +z

r2

2

; r

1

;r

2

>1;

H( z

1

;z

2 ):=( z

1 +z

2 )

r

os

1

z

1 +z

2 +1

; r>1;

H( z

1

;z

2 ):=

(

(z

1 +z

2 )

2

; 0z

1 +z

2 1;

e z1+z2 1

; z

1 +z

2

>1:

Itiseasytohekthateahof(3:10)

1 {(3:10)

3

satiseseahpartofondition(H4).

Furthermore, we should mention that eah of the funtions above annot be

inorporatedintothetheoryofeither [7℄or[10℄duetothesuperlineargrowthat

(+1;+1). Infat,suhnonlinearboundaryfuntionsouldnotbeinorporated

intoanyoftheresultsgivenin[6℄,[8℄,[9℄,[10℄forthatmatter. So,ondition(H4)

allows fora vastly dierent variety of nonlinear boundaryfuntions than other

reent works on these sorts of problems. Moreover, as shall be expliated in

the proof of Theorem 3.5, whih is our rst existene result, this asymptoti

superlineargrowthonditionalsoallowsforthemixedgrowthofthenonlinearities

g andg ,asmentionedinSetion1.

Remark 3.3. Also regarding ondition (H4), we point out that this ondition

allowsforH

i

tohavedierenttypesofgrowthinthedierentoordinatediretions

(i.e.,wheneitherz

1

=0orz

2

=0). Forexample,onsidertheontinuousfuntion

H :[0;+1)[0;+1)![0;+1)dened by

(3.11) H(z

1

;z

2 ):=

(

(z

1 +z

2 )

2

z 2

1 +

p

z

2

; 0z

1 +z

2 1;

z 2

1 +

p

z

2

; z

1 +z

2 1:

Inthez

1

-oordinate diretion,we ndthat H growssuperlinearly asz

1 +z

2

!

+1. On the other hand,in the z

2

-oordinate diretion, we nd that H grows

sublinearlyasz

1 +z

2

!+1. Finally,itholdsthat

(3.12) lim

z1+z2!0 +

H(z

1

;z

2 )

z

1 +z

2

=0 and lim

z

1 +z

2

!+1 H(z

1

;z

2 )

z

1 +z

0:3

2

=+1:

Remark 3.4. As remarked in Setion 1, we believethat the onditionsimposed

on H

i

by ondition(H4) may be hanged in amanner similar to the argument

presentedin[3℄. Butweleavesuhinvestigationsforfuturework.

Now,let

0

betheonstantdened by

(3.13)

0

:=minfa;1 bg;

where

0

2(0;1). Thentheone,K ,weshalluseinthesequelisthendenedby

(3.14)

K:=

(x;y)2X :x;y0; min

t2[a;b℄

[x(t)+y(t)℄

0

k(x;y)k;

1 (x);

2 (y)0

;

whih is a simple modiation of aone rst introdued by Infante and Webb

[12℄. Letus point outatthis junture that K doesnotontainonlythe neutral

elementof X. Indeed,ifweput, say,

1

(t):=(t;0),

2

(t):=(0;t), and

3 (t):=

(

1 +

2

)(t)=(t;t),thenitiseasytoseethat

1

;

2

;

3

2KsothatKontains

innitelymanynontrivialelementsofX.

Inanyase, withthesepreliminary observations, wenowstateand proveour

mainresult. Wenote, however,that inthestatementofthistheoremweassume

that p 1

1

= p 1

2

= 1. In other words, it is the numbers q 1

1 , q

1

2

that an be

potentiallylessthanunity. Wedothisonlyfordenitenessandeaseofexposition

inthesequel.

Theorem 3.5. Assume that 1

0

; 2

0

2 [a;b℄, where [a;b℄is a xed set satisfying

[a;b℄b(0;1)asin Setion 2. Then thereexists anumberÆ2(0;1)suh thatif

bothq 1

1

;q 1

2

2(1 Æ;1℄and(H1){(H7)hold,thenproblem(1:1)hasatleastone

positivesolution.

Proof: Tobegin,asin (2.1)above,weonsidertheoperatorS:X!Xdened

by

(3.15) S( x;y)(t):=(T (x;y);T (x;y))

where,foreahi=1;2,wehavethatT

i

:X!Bisdenedby

(3.16)

T

i

(x;y):=tH

i

1 (x)+"

1

0 x

i

0

;

2 (y)+"

2

0 y

i

0

+ Z

1

0

G(t;s)a

i (s)g

i

(x(s);y(s)) ds:

WeshallrstarguethatS:K!K . Tothisend,itisobviousthatfor(x;y)2K ,

itfollowsthatT

i

(x;y)(t)0,foreah t2[0;1℄andi=1;2. Wealsonote from

thedenitionof

0

in(3.13)that

(3.17)

min

t2[a;b℄

T

i

(x;y)

0 H

i

1 (x)+"

1

0 x

i

0

;

2 (y)+"

2

0 y

i

0

+ max

t2[0;1℄

Z

1

0

G(t;s)a

i (s)g

i

(x(s);y(s))ds

0 kT

i

(x;y)k:

Weonludethat

(3.18) min

t2[a;b℄

[( T

1

(x;y))(t)+(T

2

(x;y))(t)℄

0

kS( x;y)k:

Finally, weobservethat

(3.19)

1 (T

1

(x;y))=H

1

1 (x)+"

1

0 x

1

0

;

2 (y)+"

2

0 y

1

0

Z

[0;1℄

td

1 (t)

+ Z

[0;1℄

Z

1

0

G(t;s)a

1 (s)g

1

(x(s);y(s))dsd

1 (t)

=H

1

1 (x)+"

1

0 x

1

0

;

2 (y)+"

2

0 y

1

0

Z

[0;1℄

td

1 (t)

+ Z

1

0

"

Z

[0;1℄

G(t;s)d

1 (t)

#

a

1 (s)g

1

(x(s);y(s)) ds

0;

where the nal inequality follows from assumption (H7). In a similar way, it

followsthat

2 (T

2

(x;y))0. Thus,S:K!K ,aslaimed. Letusalsopointout

atthisjunturethat,byastandardargumentinvolvingtheArzela-Asolitheorem

(reallherethat H

i

isassumedto beontinuous, foreah i=1;2),wendthat

the operator S is ompletely ontinuous; we omit the details of this argument,

however.

Now,byondition(H5) wendthat thereisanumberr

1

>0suh that

(3.20) g (x;y) (x+y)

wheneverk(x;y)kr

1

andwhere

1

>0satises

(3.21)

1 max

Z

1

0

G(s;s)a

1 (s)ds;

Z

1

0

G(s;s)a

2 (s)ds

1

4 :

In addition, ondition (H4) | i.e., equation (3.4) | implies the existeneof a

numberr

1

>0suhthat,foreahi=1;2,

(3.22)

H

i

1 (x)+"

1

0 x

i

0

;

2 (y)+"

2

0 y

i

0

<

2

1 (x)+"

1

0 x

i

0

+

2 (y)+"

2

0 y

i

0

whenever

(3.23)

1 (x)+"

1

0 x

i

0

+

2 (y)+"

2

0 y

i

0

<r

1

;

andwhere

2

>0isdenedby

(3.24)

2 :=

1

8maxf"

1

0

;"

2

0

;"

2

1

;"

2

1 g

:

Notiethat

(3.25)

1 (x)+"

1

0 x

i

0

+

2 (y)+"

2

0 y

i

0

"

1

1 kxk+"

2

1 kyk+"

1

0 kxk+"

2

0 kyk

max

"

1

1

;"

2

1

+max

"

1

0

;"

2

0

k(x;y)k

2max

"

1

0

;"

2

0

;"

2

1

;"

2

1

k(x;y)k:

So,inpartiular,if(x;y)2K satises

(3.26) k(x;y)k<

r

1

2maxf"

1

0

;"

2

0

;"

2

1

;"

2

1 g

;

thenitfollowsthat(3.22) holds.

So,set

(3.27) r

1

:=min

r

1

;

r

1

2maxf"

1

0

;"

2

0

;"

2

1

;"

2

1 g

:

Put

(3.28)

r

:=f(x;y)2X:k(x;y)k<r

1 g:

Thenforeah(x;y)2K\

r

1

, wehavethat

(3.29)

kT

1 (x;y)k

H

1

1 (x)+"

1

0 x

1

0

;

2 (y)+"

2

0 y

1

0

+ Z

1

0

G(s;s)a

1 (s)g

1

(x(s);y(s))ds

2

1 ( x)+"

1

0 x

1

0

+

2 (y)+"

2

0 y

1

0

+

1 Z

1

0

G(s;s)a

1

(s)(x(s)+y(s))ds

2

1 ( x)+"

1

0 x

1

0

+

2 (y)+"

2

0 y

1

0

+ 1

4 k(x;y)k

1

4

k(x;y)k+ 1

4

k(x;y)k

= 1

2

k(x;y)k:

Thus,weonludethat

(3.30) kT

1

(x;y)k 1

2

k(x;y)k;

for eah (x;y) 2 K \

r

1

. A similar argument holds for the operator T

2 .

Consequently,wededuethat

(3.31) kS(x;y)kk(x;y)k;

foreah(x;y)2K\

r

1 .

On theother hand,let us assumewithoutloss ofgeneralitythat p 1

i

=1for

eahi sothat q 1

i

2(0;1℄,foreah i. Thenondition(H4)|i.e.,equation(3.3)

|impliestheexisteneofanumberr

2 :=r

2 (

3

)>0suhthat

(3.32)

H

1

1 (x)+"

1

0 x

1

0

;

2 (y)+"

2

0 y

1

0

3

1 (x)+"

1

0 x

1

0 +

2 (y)+"

2

0 y

1

0

q 1

1

whenever

(3.33)

1 (x)+"

1

0 x

1

0

+

2 (y)+"

2

0 y

1

0

r

2

for some number r

2

. Note that by piking r

2

suÆiently large, the sametype

ofestimatelikewise holdsforH

2

; weassumeheneforththat thisis so. Here,in

(3.32),wehoose

3

tobethenumber

(3.34)

3 :=

1

t

0

0 minf"

1

0

;"

2

0

;"

2

0 g

;

wheret

0

2(a;b) isxed but arbitrary;sine(a;b)b(0;1), itholdsthat t

0 6=0,

andso,

3

>0. Importantly,

3

depends neither onq 1

nor onq 1

. Now,notie

thatfor(x;y)2K sine

1 (x),

2

(y)0and

0

2E,wemayestimate

(3.35)

1 (x)+"

1

0 x

1

0

+

2 (y)+"

2

0 y

1

0

min

"

1

0

;"

2

0

x 1

0

+y 1

0

min

"

1

0

;"

2

0 min

t2[a;b℄

[x(t)+y(t)℄

0 min

"

1

0

;"

2

0

k(x;y)k:

Consequently,if(x;y)satises

(3.36) k(x;y)k

r

2

0 minf"

1

0

;"

2

0 g

;

then(3.32)holds.

Wenextinterrupttoproveaneasylemma. Supposethat x;y0withx;y

M forsome M 1 andnite. Letq satisfy 0<q 1. Choose the onstant

suhthat

(3.37) :=min

1;M q 1

;

notethat 1<q 10. Obviously,2(0;1℄sineM 1andq 10. Then

itfollowsthat

(3.38) x+y

q

(x+y);

forall(x;y)2[0;M℄[0;M℄. Indeed,wemerelynotiethat,for(x;y)2[0;M℄

[0;M℄

(3.39) xx

and

(3.40) yy

q

;

sine y 7! y q 1

is dereasing for y > 0, whereupon adding (3.39){(3.40) we

estimate

(3.41) x+yx+y

q

;

whih evidentlyprovesinequality(3.38).

Nowontinuingwiththeproof,letus put

(3.42) r

2

:=max

1;2r

1

;

r

2

0 minf"

1

0

;"

2

0 g

;

whih isindependent of eahofq 1

1 andq

1

2

. Dene

r

2 by

(3.43)

r

:=f(x;y)2X:k(x;y)k<r

2 g:

Usingestimate(3.38),then,andthefatthat

(3.44)

Z

1

0 G( t

0

;s)a

1 (s)g

1

(x(s);y(s))ds0;

wededuethatforeah (x;y)2K\

r

2

(3.45) (T

1

(x;y))(t

0 )=t

0 H

1

1 (x)+"

1

0 x

1

0

;

2 (y)+"

2

0 y

1

0

+ Z

1

0 G( t

0

;s)a

1 (s)g

1

(x(s);y(s))ds

t

0 H

1

1 (x)+"

1

0 x

1

0

;

2 (y)+"

2

0 y

1

0

t

0

3

1 (x)+"

1

0 x

1

0

+

2 (y)+"

2

0 y

1

0

q 1

1

t

0

3 h

"

1

0 x

1

0

+ "

2

0

q 1

1

y 1

0

q 1

1 i

t

0

3 h

"

1

0 x

1

0

+"

2

0

y 1

0

q 1

1 i

t

0

3 min

"

1

0

;"

2

0 h

x 1

0

+

y 1

0

q 1

1 i

t

0

3 min

"

1

0

;"

2

0

1

x 1

0

+y 1

0

t

0

3 min

"

1

0

;"

2

0

0

1 k(x;y)k

1

k(x;y)k;

wherewehaveusedthelemmaofthepreviousparagraphtogetthethird-to-last

inequality,andso,here

1

:=min f1;(r

2 )

q 1

1 1

g. Wehavealsousedboththefat

that"

2

0 2[0;

1

2

)andthat q 1

1

2(0;1℄sothat("

2

0 )

q 1

1

"

2

0

. Insummary,itfollows

that

(3.46) kT

1

(x;y)k

1

k(x;y)k:

Likewise,foreah(x;y)2K\

r

2

wededuethatfor

2

:=minf1;(r

2 )

q 1

2 1

g

(3.47) kT

2

(x;y)k

2

k(x;y)k:

Wenowonlude theargumentby onsidering ases. If q 1

1

=q 1

2

=1, then

from (3.37), itis obviousthat

1

=

2

=1. Inthis asewededue from(3.46){

(3.47)that

(3.48) kS(x;y)k2k(x;y)k>k(x;y)k;

foreah(x;y)2K\

r

2

. On theother hand,inase0<maxfq 1

1

;q 1

2 g<1,

then

(3.49)

1 :=( r

) q

1

1 1

and

2 :=( r

) q

1

2 1

:

Inorderthat

1 +

2

1besatised,ataminimumwemusthavethat

(3.50) min

n

2 1

1 q 1

1

;2 1

1 q 1

2 o

r

2 :

Evidently, sine r

2

is nite and (1 q 1

i )

1

! +1 asq 1

i

! 1 , there exists

a Æ > 0 suÆiently small suh that for eah q 1

1 , q

1

2

2 (1 Æ;1℄ we have that

(3.50)holds. Inthisase,weagaindeduethat(3.48)holdswith,say,thefator

2replaedby 1. Importantly, wepoint outthat r

2

doesnot depend onq 1

i for

eitheri. Consequently,wemay,ininequality(3.50)above,freelyinreaseq 1

i ,for

eahi, withouthangingthepreviouslyseletedandxedvalueofr

2 .

Finally, puttingthepreeding paragraphstogether, wemaketwoonlusions.

Firstly,ifq 1

1

=q 1

2

=1,thenbyLemma2.1andinequality(3.48) wededuethe

existeneof afuntion (x

0

;y

0

)2K suh that S(x

0

;y

0 ) =(x

0

;y

0

), where x

0 (t),

y

0

(t) forms apositivesolution of problem(1.1). Seondly, if q 1

1 , q

1

2

1, then

thereexistsaÆ>0suÆientlysmallsuhthatifq 1

1 ,q

1

2

2(1 Æ;1℄,thenproblem

(1.1) still has at least one positive solution. And asthese asesare exhaustive

thisompletestheproof.

We now provea seond result that demonstratesan alternativeapproah to

problem(1.1). Inpartiular,webeginbyintroduingthefollowingondition.

H8: Foreahi=1;2,thereisaonstant

i

>0suhthat

(3.51) limsup

z1+z2!0 +

H

i ( z

1

;z

2 )

z

1 +z

2

<

i

holds,where

i 2[0;

1

2maxf"

1

1

;"

2

1 g

).

Ontheonehand,ondition (H8)is ertainlymoregeneralthan ondition(H4).

Forinstane, the ontinuousfuntion H :[0;+1)[0;+1)![0;+1) dened

by

(3.52) H(z

1

;z

2 ):=

(

(z

1 +z

2 )os

1

z1+z2

; z

1 +z

2 6=0

0; z

1

=z

2

=0

satises

(3.53) limsup

z1+z2!0 +

H( z

1

;z

2 )

z

1 +z

2

=1

butlim

z

1 +z

2

!0 +

H(z

1

;z

2 )

z1+z2

doesnotexist. Ontheotherhand,inordertoprovethe

next result, we shall haveto impose growth onditionson the nonlinearities g

1

andg

2

atinnity. Thus,weintrodueondition(H9)below.

H9: Wendthat

(3.54) lim

x+y!+1 g

1 (x;y)

=+1 and lim

x+y!+1 g

2 (x;y)

=+1:

With ondition(H8)and(H9) inhand westateandprovethefollowingtheo-

rem. Werstgivetwopreliminaryremarks.

Remark 3.6. We note that ondition (H8) is more losely related to ertain of

the onditions given by Yang [16℄, [17℄, to whih wasalluded in Setion 1. In

partiular,however,wenotethatunliketheresultsYanggives,whihadmittedly

were for aslightly dierent problem than (1.1), we do not requireompliated

onditionsonthenonlinearitiesg

1 andg

2

. Indeed,onditions(H5) and(H9)are

quitestraightforwardandstandard. Moreover,themeasuresherearesigned. So,

weonsidertheseobservationstobebothinterestingandnoteworthy.

Remark 3.7. We also note, as will beome lear in the statement and proof of

Theorem3.8in thesequel,that withthispartiularassumption |namely(H8)

|wemaydispensewiththeperturbationtermsappearingin(1.1). Inpartiular

andimportantly,then, wemayset"

1

0

="

2

0

=0.

Theorem3.8. Supposethat onditions(H1){(H3) and(H5){(H9)hold. Inad-

dition, suppose that "

1

0

= "

2

0

= 0. Then theunperturbed problem (1:1) has at

leastonepositivesolution.

Proof: Duetotheassumptionsgiveninthestatementofthistheorem,itisstill

theasethat T :K!Kandthat T isaompletelyontinuousoperator. So,we

proeeddiretlytotheonetheoretipartoftheargument.

Tothisend,let

i

<

1

2maxf"

1

1

;"

2

1 g

begiven,foreahi=1;2. Evidently,wemay

seletk2N suÆientlylargesuhthat

(3.55) 0

i

<

2 k

1

2 k +1

maxf"

1

1

;"

2

1 g

<

1

2maxf"

1

1

;"

2

1 g

holdsforeahi. Moreover,foreahi,seletthenumber

i

>0suhthat

(3.56)

i Z

1

0

G(s;s)a

i

(s)ds 1

2 k +1

holds. Condition (H5) implies the existene of a number r

1

> 0 suh that

g

i

(x;y)

i

(x+y)forall0x+y<r

1

andforeahi. Ontheotherhand,from

ondition(H8),wemayseletanumber0<"<minf

1

;

2

gsuÆientsmallsuh

that

(3.57) H

i (z

1

;z

2 )<(

i

")(z

1 +z

2 )

holds whenever0z

1 +z

2

<r

1

forsomenumberr

1

>0,for eah i=1;2. In

addition,sine(3.55)holds,foreahi, itevidentlyholdsthat

(3.58) 0<

i

"<

2 k

1

2 k +1

maxf"

1

;"

2

g :

Now,ondition(H3)impliesthat

(3.59)

1 (x)"

1

1

kxk andthat

2 (y)"

2

1 kyk:

Consequently,foreah(x;y)2Ksatisfying

(3.60) 0k(x;y)k<minfr

1

;r

1 g;

itfollowsthat

(3.61)

1 (x)"

1

1

kxk"

1

1

k(x;y)k<

1

2 r

1

andthat

2 (y)"

2

1

kyk"

2

1

k(x;y)k<

1

2 r

1 :

Now,seletr

1

>0suhthat

(3.62) r

1

<minfr

1

;r

1 g

and put

r

1

:= f(x;y) 2 K : k(x;y)k < r

1

g. Then upon ombining(3.59){

(3.62),wemayestimate

(3.63) H

i (

1 (x);

2

(y))<(

i

")(

1 (x)+

2 (y));

foreah(x;y)2K\

r

1

andi=1;2. So,ombiningalloftheseestimates,we

deduethat

(3.64) kT

1

(x;y)kH

1 (

1 (x);

2 (y))+

Z

1

0

G(s;s)a

1 (s)g

1

(x(s);y(s))ds

(

1

")(

1 (x)+

2 (y))+

1

2 k +1

k(x;y)k

2

k

1

2 k +1

maxf"

1

1

;"

2

1 g

"

1

1 kxk+"

2

1 kyk

+ 1

2 k +1

k(x;y)k

2

k

1

2 k +1

maxf"

1

1

;"

2

1 g

max

"

1

1

;"

2

1

(kxk+kyk)+ 1

2 k +1

k(x;y)k

= 2

k

1

2 k +1

maxf"

1

1

;"

2

1 g

max

"

1

1

;"

2

1

k(x;y)k+ 1

2 k +1

k(x;y)k

= 1

2

k(x;y)k:

Similarly,wededuethat

(3.65) kT

2

(x;y)k 1

2

k(x;y)k

whene

foreah(x;y)2K\

r

1 .

Ontheotherhand,seletthenumber

3

>0to satisfy

(3.67)

3 max

(

Z

[a;b℄

2

0

G(s;s)a

1 (s)ds;

Z

[a;b℄

2

0

G(s;s)a

2 (s)ds

)

1

2 :

Thenbyondition(H9), wehavethat

(3.68) g

i

(x;y)

3

(x+y);

forallx+yr

2

and foreahi=1;2. Put

(3.69) r

2

:=max

r

2

0

;2r

1

:

ThensineH

1 (z

1

;z

2

)0,forall(z

1

;z

2

)2[0;+1)[0;+1),wededuethat

(3.70)

min

t2[a;b℄

(T

1

(x;y))(t)

3 Z

[a;b℄

0

G(s;s)a

1

(s)[x(s)+y(s)℄ds

k(x;y)k

3 Z

[a;b℄

2

0

G(s;s)a

1 (s)ds

1

2

k(x;y)k;

whene

(3.71) kT

1

(x;y)k 1

2

k(x;y)k;

foreah(x;y)2K\

r

2

. Similarly,

(3.72) kT

2

(x;y)k 1

2

k(x;y)k;

sothat kS(x;y)kk(x;y)k, for(x;y)2K\

r

2

. Consequently,wemayinvoke

Lemma 2.1 to dedue the existene of at least one positive solution to prob-

lem(1.1).

Weonludewith anexpliitnumerial exampletogetherwith somenal re-

Example3.9. Considertheboundaryvalueproblem

(3.73)

x 00

(t)=(2t+1)g

1

(x(t);y(t));

y 00

(t)=e 3t+1

g

2

(x(t);y(t));

x(0)=H

1

1 (x)+

1

40 x

1

2

;

2 (y)+

1

300 y

2

5

;

y(0)=H

2

1 (x)+

1

40 x

1

2

;

2 (y)+

1

300 y

2

5

;

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

wherewemakethefollowingdelarations:

(3.74)

H

1 (z

1

;z

2 ):=(z

1 +z

2 )

3

;

H

2 (z

1

;z

2 ):=z

1:1

1 e

z1

+z 2

2 e

z2

;

1 (x):=

1

8 x

1

3

1

40 x

1

2

1

12 x

3

5

+ 1

2 Z

[ 13

20

; 3

4

℄

x(s)ds;

2 (y):=

1

300 y

2

5

+ 1

15 y

9

20

1

100 y

11

20

+ 1

10 Z

[ 3

5

; 7

10

℄

y(s)ds;

g

1

(x;y):=

(

(x+y) 2

; x+y1;

p

x+y; x+y1;

g

2

(x;y):=(x+y) 3

:

Interestingly,notethatg

1

issublinearasx+y!+1,whereasg

2

issuperlinear.

Furthermore, letus observe at this junture that on aount of the denitions

of

1 and

2

given in (3.74), we may reast the boundary onditionsat t = 0

in(3.73)in thesomewhatsimplerform

(3.75)

x(0)=H

1 (

1 (x);

2

(y))=[

1 (x)+

2 (y)℄

3

;

y(0)=H

2 (

1 (x);

2

(y))=(

1 (x))

1:1

e 1

(x)

+(

2 (y))

2

e 2

(y)

;

where we have put

1

(x) :=

1 (x) +

1

40 x(

1

2 ) and

2

(y) :=

2 (y)+

1

300 y(

2

5 ).

Inidentally, though wedo notshow thisexpliitly, letus also remarkthat it is

easytoshowthattheStieltjesmeasures

1 and

2

aresigned forthisproblem.

It is now easy to hek that eah of onditions (H1){(H7) is satised. In

partiular,note that wemayselet"

1

1 :=

17

60 , "

2

1 :=

9

100 ,"

1

0 :=

1

40 ,and "

2

0 :=

1

300 .

Moreover,wenotethat R

[0;1℄

td

1 (t)=

17

1200

0andthat R

[0;1℄

td

2 (t)=

89

3000

0. In any ase, we onlude that we may invoke Theorem 3.5 to dedue that

problem(3.73)hasatleastonepositivesolution. Likewise,problem(3.75)hasat

Remark 3.10. Wenotethatproblem(3.73)ouldnotbeaddressedbyanyexisting

results. This is true for a variety of reasons, among whih are the following:

problem (3.73) involvesa systemof equations; itimposes nogrowth onditions

on g

1 and g

2

for (x;y) large in norm; it allowsfor eah of H

1 and H

2

to have

superlinear growth as x +y ! +1; and it allows for eah of

1 and

2 to

be haveassoiatedsigned Borel measures. In short,weare notaware that any

resultsintheexistingliteratureanbeappliedtoproblem(3.73). Andthisisthe

advantageoftheasymptotionditions(H4)and(H8),whihwehaveintrodued

inthiswork.

Remark 3.11. ObservethatExample3.9demonstratesthatitisnot neessaryfor

thefuntionH

2 (z

1

;z

2

)tobeabletoberealizedin theform

(3.76) H

2 (z

1

;z

2 )=

e

H( z

1 +z

2 );

for some funtion e

H. Indeed, while suh a deomposition is an easy way in

whihtosatisfyondition(H4),thefuntionH

2 (z

1

;z

2 )=z

1:1

1 e

z

1

+z 2

2 e

z

2

in(3.74)

annotberealizedin this simplerform. Of ourse,thefuntions H

1 ,g

1 , andg

2

neednotbeabletoberealizedasafuntion ofz

1 +z

2

either. Thepointisthat

ondition(H4)anstillbesatisedinspiteofthis. Infat,forexample,toensure

that(3.3)inondition(H4)issatised,itisenough,forinstane, that

(3.77) H

i (z

1

;z

2 )( z

1 +z

2 )

holdsforz

1 +z

2

suÆientlylargeandforsome>1. Evidently,(3.77)doesnot

requirethatH

i

satisfy(3.76)for some e

H. Furthermore,note thatan additional

exampleofthis sortwasprovidedinboth(3:10)

1

and(3.11).

Remark 3.12. Wehaveeleted notto give anexampleof Theorem 3.8 sineits

appliationwouldproeedinaverysimilarmannertoExample3.9. Nonetheless,

we emphasize that in the ase of Theorem 3.8, we may take the perturbation

termsin(1.1)equaltozeroand,hene,inthisasewearereoveringsolutionsto

theunperturbed (i.e.,"

1

0

="

2

0

=0)problem(1.1).

Aknowledgments. Theauthorwouldliketothanktheanonymousrefereefor

his or her arefulreading of this paper. The referee's ommentsand questions

improvedthequalityofthispaper.

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DepartmentofMathematis,UniversityofNebraska-Linoln,Linoln,

NE68588,USA

E-mail: s-goodri4math.unl.edu

(Reeived Otober3,2011 , revised November24,2011)