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FUNCTIONAL DIFFERENTIAL EQUATIONS

R. HAKL, A. LOMTATIDZE, AND I. P. STAVROULAKIS Received 24 July 2003

Theorems on the Fredholm alternative and well-posedness of the linear boundary value problemu(t)=(u)(t) +q(t),h(u)=c, where:C([a,b];R)L([a,b];R) andh:C([a, b];R)Rare linear bounded operators,qL([a,b];R), andcR, are established even in the case whenis not astrongly boundedoperator. The question on the dimension of the solution space of the homogeneous equationu(t)=(u)(t) is discussed as well.

1. Introduction

The following notation is used throughout:Nis the set of all natural numbers;Ris the set of all real numbers,R+=[0, +[; Ent(x) is an entire part ofxR;C([a,b];R) is the Banach space of continuous functionsu: [a,b]Rwith the normuC=max{|u(t)|: t[a,b]};C([a,b];R+)= {uC([a,b];R) :u(t)0 fort[a,b]};C([a, b];R) is the set of absolutely continuous functionsu: [a,b]R;L([a,b];R) is the Banach space of Lebesgue integrable functions p: [a,b]Rwith the normpL=b

a|p(s)|ds;L([a,b];

R+)= {pL([a,b];R) :p(t)0 fort[a,b]}; mesAis the Lebesgue measure of the set A;abis the set of measurable functionsτ: [a,b][a,b];ᏸabis the set of linear bounded operators:C([a,b];R)L([a,b];R);ᏸab is the set of linear strongly bounded opera- tors, that is, for each of the operatorsab, there existsηL([a,b];R+) such that

(v)(t)η(t)vC fort[a,b],vC[a,b];R; (1.1) ᏼabis the set of linear nonnegative operators, that is, operatorsabmapping the set C([a,b];R+) into the setL([a,b];R+). Ifab, then =sup{(v)L:vC1}.

Lett0[a,b]. We will say thatab is at0-Volterra operator if for arbitrarya1 [a,t0],b1[t0,b], anduC([a,b];R) such that

u(t)=0 fort a1,b1

, (1.2)

we have

(u)(t)=0 fort a1,b1

. (1.3)

Copyright©2004 Hindawi Publishing Corporation Abstract and Applied Analysis 2004:1 (2004) 45–67 2000 Mathematics Subject Classification: 34K06, 34K10 URL:http://dx.doi.org/10.1155/S1085337504309061

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On the segment [a,b], consider the boundary value problem

u(t)=(u)(t) +q(t), (1.4)

h(u)=c, (1.5)

whereab,h:C([a,b];R)Ris a linear bounded functional,qL([a,b];R), and cR.

By a solution of (1.4) we understand a functionuC([a,b]; R) satisfying the equality (1.4) almost everywhere on [a,b]. By a solution of the problem (1.4), (1.5), we understand a solutionuof (1.4) which also satisfies the condition (1.5). Together with (1.4), (1.5), we will consider the corresponding homogeneous problem

u(t)=(u)(t), (1.6)

h(u)=0. (1.7)

From the general theory of boundary value problems for functional differential equa- tions, it is known that ifab, then the problem (1.4), (1.5) has a Fredholm property (see, e.g., [1,2,7,8,10]). More precisely, the following assertion is valid.

Theorem1.1. Letab. Then the problem (1.4), (1.5) is uniquely solvable if and only if the corresponding homogeneous problem (1.6), (1.7) has only the trivial solution.

Theorem 1.1allows us to introduce the following definition.

Definition 1.2. Let ab and let the problem (1.6), (1.7) have only the trivial solu- tion. An operatorΩ:L([a,b];R)C([a,b];R) which assigns to everyqL([a,b];R) a solutionuof the problem (1.4), (1.7) is called Green operator of the problem (1.6), (1.7).

It follows fromTheorem 1.1that ifaband the problem (1.6), (1.7) has only the trivial solution, then the Green operator is well defined. Evidently, Green operator is lin- ear. Moreover, the following theorem is valid (see, e.g., [1,2,7,8]).

Theorem1.3. Letab and let the problem (1.6), (1.7) have only the trivial solution.

Then the Green operator of the problem (1.6), (1.7) is a linear bounded operator.

In [7,8] the question on the well-posedness of linear boundary value problem for systems of functional differential equations is studied.Theorem 1.3can also be derived as a consequence of more general results on well-posedness obtained therein.

Note that both Theorems1.1and1.3claim thatab. This condition covers a quite wide class of linear operators; for example, the equation with a deviating argument

u(t)=p(t)uτ(t)+q(t), (1.8) wherep,qL([a,b];R),τab, is a special case of (1.4) with

(v)(t)def= p(t)vτ(t) fort[a,b]. (1.9) More generally, it is known (see [6, page 317]) thatab if and only if the operator admits the representation by means of a Stieltjes integral.

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On the other hand, Schaefer proved that there exists an operatorabsuch thatab(see [9, Theorem 4]). Therefore, a question naturally arises to study boundary value problem (1.4), (1.5) without the additional requirement (1.1). In particular, the question whether Theorems1.1and1.3are valid for general operatorabis interesting.

The first important step in this direction was made by Bravyi (see [3]), whereTheorem 1.1was proved for ab (i.e., without the additional assumptionab). Bravyi’s proof essentially uses Nikol’ski’s theorem (see, e.g., [5, Theorem XIII.5.2, page 504]) and it is concentrated on the question of Fredholm property. The question whetherTheorem 1.3is valid for the case whenabremains open.

In the present paper, among others, we answer this question affirmatively. More pre- cisely, inSection 2we prove that the operator T:C([a,b];R)C([a,b];R) defined by T(v)(t)def=t

a(v)(s)dsfort[a,b] is compact provided thatab(seeProposition 2.9).

Based on this result and Riesz-Schauder theory, we give an alternative proof (different from that in [3]) ofTheorem 1.1forab(seeTheorem 2.1).

On the other hand, the compactness of the operatorT allows us to study a question on the well-posedness of boundary value problem (1.4), (1.5). Section 3is devoted to this question. As a special case of theorem on well-posedness, we obtain the validity of Theorem 1.3forab(seeCorollary 3.3).

InSection 4, the question on dimension of solution spaceU of homogeneous equa- tion (1.6) is discussed.Proposition 4.6 shows that if dimU2, then there exists q L([a,b];R) such that the nonhomogeneous equation (1.4) has no solution. This “patho- logical” behaviour of functional differential equations affirms the importance of the ques- tion whether the solution space of the homogeneous equation (1.6) is one dimensional.

In Theorems4.8and4.10, the nonimprovable effective sufficient conditions are estab- lished guaranteeing that dimU=1.

2. Fredholm property

Theorem2.1. Letab. Then the problem (1.4), (1.5) is uniquely solvable if and only if the corresponding homogeneous problem (1.6), (1.7) has only the trivial solution.

Analogously as inSection 1, we can introduce the notion of the Green operator of the problem (1.6), (1.7).

Definition 2.2. Let ab and let the problem (1.6), (1.7) have only the trivial solu- tion. An operatorΩ:L([a,b];R)C([a,b];R) which assigns to everyqL([a,b];R) a solutionuof the problem (1.4), (1.7) is called Green operator of the problem (1.6), (1.7).

Evidently, it follows fromTheorem 2.1that the Green operator is well defined.

Remark 2.3. From the proof ofTheorem 2.1and Riesz-Schauder theory, it follows that if the problem (1.6), (1.7) has a nontrivial solution, then for every cRthere exists qL([a,b];R), respectively, for everyqL([a,b];R) there existscR, such that the problem (1.4), (1.5) has no solution.

To proveTheorem 2.1we will need several auxiliary propositions. First we recall some definitions.

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Definition 2.4. LetXbe a linear topological space,Xits dual space. A sequence{xn}+n=1 Xis called weakly convergent if there existsxXsuch thatϕ(x)=limn+ϕ(xn) for every ϕX. The pointxis called a weak limit of this sequence.

A setMX is called weakly relatively compact if every sequence of points fromM contains a subsequence which is weakly convergent inX.

A sequence{xn}+n=1Xis called weakly fundamental if for everyϕX, a sequence {ϕ(xn)}+n=1is fundamental.

A spaceX is called weakly complete if every weakly fundamental sequence from X possesses a weak limit inX.

LetXandY be Banach spaces and letT:XY be a linear bounded operator. The operatorT is said to be weakly completely continuous if it maps a unit ball ofX into weakly relatively compact subset ofY.

Definition 2.5. A setML([a,b];R) has a property of absolutely continuous integral if for everyε >0, there existsδ >0 such that for an arbitrary measurable set E[a,b]

satisfying the condition mesEδ, the following inequality is true:

Ep(s)dsε for everypM. (2.1) Proofs of the following three assertions can be found in [4].

Lemma2.6 [4, Theorem IV.8.6]. The spaceL([a,b];R)is weakly complete.

Lemma2.7 [4, Theorem VI.7.6]. A linear bounded operator mapping the spaceC([a,b];R) into a weakly complete Banach space is weakly completely continuous.

Lemma2.8 [4, Theorem IV.8.11]. If a setML([a,b];R)is weakly relatively compact, then it has a property of absolutely continuous integral.

The following proposition plays a crucial role in the proof ofTheorem 2.1.

Proposition2.9. Letab. Then the operatorT:C([a,b];R)C([a,b];R)defined by

T(v)(t)def= t

a(v)(s)ds fort[a,b] (2.2) is compact.

Proof. LetMC([a,b];R) be a bounded set. According to Arzel´a-Ascoli lemma, it is sufficient to show that the setT(M)= {T(v) :vM}is bounded and equicontinuous.

Obviously,

T(v) C=max t

a(v)(s)ds:t[a,b]

(v) L · vC forvM,

(2.3) and thus, sinceabandMis bounded, the setT(M) is bounded.

Further, Lemmas2.6and2.7imply that the operatoris weakly completely continu- ous, that is, a set(M)= {(v) :vM}is weakly relatively compact. Therefore, according

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toLemma 2.8, for everyε >0, there existsδ >0 such that t

s(v)(ξ)dξε fors,t[a,b], |ts| ≤δ,vM. (2.4) On the other hand,

T(v)(t)T(v)(s)= t

s(v)(ξ)dξ fors,t[a,b], vC[a,b];R, (2.5) which, together with (2.4), results in

T(v)(t)T(v)(s)ε fors,t[a,b],|ts| ≤δ,vM. (2.6)

Consequently, the setT(M) is equicontinuous.

Proof ofTheorem 2.1. Let X=C([a,b];R)×R be a Banach space containing elements x=(u,α), whereuC([a,b];R) andαR, with a norm

xX= uC+|α|. (2.7)

Let

q=

t

aq(s)ds,c

(2.8) and define a linear operatorT:XXby setting

T(x)def=

α+u(a) + t

a(u)(s)ds,αh(u)

. (2.9)

Obviously, the problem (1.4), (1.5) is equivalent to the operator equation

x=T(x) +q (2.10)

in the spaceXin the following sense: ifx=(u,α)Xis a solution of (2.10), thenα=0, uC([a,b]; R), anduis a solution of (1.4), (1.5), and vice versa, ifuC([a,b]; R) is a solution of (1.4), (1.5), thenx=(u, 0) is a solution of (2.10).

According toProposition 2.9, we have that the operator T is compact. From Riesz- Schauder theory, it follows that (2.10) is uniquely solvable if and only if the corresponding homogeneous equation

x=T(x) (2.11)

has only the trivial solution (see, e.g., [11, Theorem 2, page 221]). On the other hand, (2.11) is equivalent to the problem (1.6), (1.7) in the above-mentioned sense.

Following [7,8] we introduce the following notation.

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Notation 2.10. Lett0[a,b]. Define operatorsk:C([a,b];R)C([a,b];R) and num- bersλkas follows:

0(v)(t)def=v(t), k(v)(t)def= t

t0

k1(v)(s)ds fort[a,b],kN, (2.12) λk=h0(1) +1(1) +···+k1(1) forkN. (2.13) Ifλk=0 for somekN, then let

k,0(v)(t)def=v(t) fort[a,b], k,m(v)(t)def=m(v)(t)hk(v)

λk m1

i=0

i(1)(t) fort[a,b],mN. (2.14) Theorem2.11. Letaband let there existk,mN,m0N∪ {0}, andα[0, 1[such thatλk=0and for every solutionuof the problem (1.6), (1.7), the inequality

k,m(u) Cα k,m0(u) C (2.15)

is fulfilled. Then the problem (1.4), (1.5) has a unique solution.

Remark 2.12. The proof ofTheorem 2.11is omitted since it is completely the same as the proof of [8, Theorem 1.3.1] (see also [7, Theorem 1.2]). The only difference is that instead ofTheorem 1.1,Theorem 2.1has to be used.

Theorem 2.11implies the following corollary.

Corollary2.13. Letabbe at0-Volterra operator. Then the problem u(t)=(u)(t) +q(t), ut0

=c, (2.16)

withqL([a,b];R)andcR, is uniquely solvable.

To prove this corollary we need the following lemma.

Lemma2.14. Letab be at0-Volterra operator and letk (kN∪ {0})be operators defined by (2.12). Then

klim+

k =0. (2.17)

Proof. Letε]0, 1[. According toProposition 2.9, the operator1, defined by (2.12) for k=1, is compact. Therefore, by virtue of Arzel`a-Ascoli lemma, there existsδ >0 such that

t

s(v)(ξ)dξ=1(v)(t)1(v)(s)εvC for|ts| ≤δ. (2.18)

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Let

n=Ent bt0

δ

, m=Ent

t0a δ

, ti=t0+ fori= −m,m+ 1,. . .,1, 1, 2,. . .,n,

tm1=a, tn+1=b,

(2.19)

and introduce the notation k(v) i=

k(v) C([t0,t

i];R) fori=1,n+ 1,

k(v) C([ti,t0];R) fori= −m1,1. (2.20)

We will show that

k(v) iαi(k)εkvC([a,b];R) fori=1,n+ 1,kN, (2.21) where

αi(k)=γiki1 fori=1,n+ 1,

γ1=1, γi+1=i+i+ 1 fori=1,n. (2.22) First note that

1(v) ivC([a,b];R) fori=1,n+ 1. (2.23) Indeed, according to (2.18), it is clear that

1(v) i=max t

t0

(v)(ξ)dξ:t t0,ti

i1

j=0

max t

tj

(v)(ξ)dξ:t

tj,tj+1

vC([a,b];R) fori=1,n+ 1.

(2.24)

Further, on account of (2.18) and the fact thatis at0-Volterra operator, we have k+1(v)(t)=

t

t0

k(v)(ξ)dξε k(v) 1 fort t0,t1

,kN. (2.25)

Hence, by virtue of (2.23), we get

k(v) 1εkvC([a,b];R) forkN, (2.26) that is, (2.21) holds fori=1.

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Now let the inequality (2.21) hold for somei∈ {1, 2,. . .,n}. With respect to (2.18) and the fact thatis at0-Volterra operator, we have

k+1(v) i+1=max t

t0

k(v)(ξ)dξ:t

t0,ti+1

i1

j=0

max t

tj

k(v)(ξ)dξ:t

tj,tj+1 + max

t

ti

k(v)(ξ)dξ:t ti,ti+1

k(v) i+ε k(v) i+1

i(k)εk+1vC([a,b];R)+ε k(v) i+1 forkN.

(2.27)

Hence we get

k+1(v) i+1i(k)εk+1vC([a,b];R)

+εi(k1)εkvC([a,b];R)+ε k1(v) i+1 forkN. (2.28) To continue this procedure, on account of (2.23), we obtain

k+1(v) i+1

i+ 1 +iαi(1) +···+αi(k)εk+1vC([a,b];R) forkN. (2.29) With respect to (2.22), we get

i+ 1 +i k j=1

αi(j)=i+ 1 +i

1i1+ 2i1+···+ki1i+ 1 +ikki1

=i+ 1 +iki

i+ 1 +i

ki=γi+1kiαi+1(k+ 1).

(2.30)

Therefore, from (2.29), it follows that

k+1(v) i+1αi+1(k+ 1)εk+1vC([a,b];R) forkN. (2.31)

Thus, by induction, we have proved that (2.21) holds.

In an analogous way, it can be shown that

k(v) iαi(k)εkvC([a,b];R) fori= −m1,1,kN, (2.32)

where

αi(k)=γik|i|−1 fori= −m1,1,

γ1=1, γi1= |i|γi+|i|+ 1 fori= −m,1. (2.33)

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Now from (2.21), (2.22), (2.32), and (2.33), it follows that there existsγN(indepen- dent ofk) such that

k(v) C([a,b];R) k(v) m1+ k(v) n+1

γkn+mεkvC([a,b];R) forkN. (2.34)

Hence, sinceε <1, it follows that (2.17) holds.

Proof of Corollary 2.13. Leth(v)def=v(t0). Obviously, for everyk,mN, we haveλk=1, hk(v)=0, k,m(v)(t)=m(v)(t) fort[a,b],vC[a,b];R. (2.35) According toLemma 2.14, we can choosemNsuch that

m <1. (2.36)

Thus the inequality (2.15) holds withm0=0 andα= m. Fort0-Volterra operators,Theorem 2.11can be inverted. More precisely, the following assertion is valid.

Theorem2.15. Letab be at0-Volterra operator. Then the problem (1.4), (1.5) has a unique solution if and only if there existk,mNsuch thatλk=0and

k,m <1. (2.37)

Proof. Let inequality (2.37) hold for somek,mN. Obviously, for everyuC([a,b];

R) (consequently, also for every solution of (1.6), (1.7)), we have

k,m(u) C k,m uC. (2.38) Therefore, the assumptions ofTheorem 2.11are fulfilled withm0=0 andα= k,m. Consequently, the problem (1.4), (1.5) has a unique solution.

Assume now that the problem (1.6), (1.5) is uniquely solvable. According toTheorem 2.1, the problem (1.6), (1.7) has only the trivial solution.

Letu0be a solution of the problem

u(t)=(u)(t), ut0

=1, (2.39)

the existence of which is guaranteed byCorollary 2.13. Obviously, hu0

=0, (2.40)

since otherwise the functionu0would be a nontrivial solution of the problem (1.6), (1.7).

Let

un(t)=

n1 i=0

i(1)(t) fort[a,b], nN. (2.41)

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From (2.39) it follows that

u0(t)=1 +1u0

(t) fort[a,b]. (2.42)

Hence we have

u0(t)=1 +11 +1u0

(t)=0(1)(t) +1(1)(t) +2u0

(t) fort[a,b]. (2.43) To continue this process, we obtain

u0(t)=

n1 i=0

i(1)(t) +nu0

(t) fort[a,b],nN. (2.44)

Hence, on account of (2.41) andLemma 2.14, we get

nlim+ u0un

C=0. (2.45)

Sinceλn=h(un) fornNandhis a continuous functional, we have, with respect to (2.40) and (2.45), that

nlim+λn=hu0

=0. (2.46)

Therefore, there existk0Nandδ >0 such that

λiδ forik0. (2.47)

Hence, by virtue of (2.45), it follows that there existsρ]0, +[ such that

λ1i uj Chρ forik0, jN. (2.48)

According toLemma 2.14, there existk > k0andmNsuch that k 1

2ρ, m <1

2. (2.49)

Furthermore, in view of (2.14), we have

k,m m + um C

λk h k , (2.50)

which, together with (2.48) and (2.49), implies that (2.37) holds.

Remark 2.16. For the case whenab,Theorem 2.15is proved in [8] (see also [7]).

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3. Well-posedness

Together with the problem (1.4), (1.5), for everykN, consider the perturbed boundary value problem

u(t)=k(u)(t) +qk(t), hk(u)=ck, (3.1) wherekab,hk:C([a,b];R)Ris a linear bounded functional,qkL([a,b];R), and ckR.

The question on well-posedness of general linear boundary value problem for func- tional differential equation under the assumptionsab andkab is studied in [7,8] (see also references in [8, page 70]). In this section we will show that the theo- rems on well-posedness established in [7,8] are valid also for the case whenaband kab.

Notation 3.1. Letab. Denote byM the set of functionsyC([a,b]; R) admitting the representation

y(t)=z(a) + t

a(z)(s)ds fort[a,b], (3.2) wherezC([a,b];R) andzC=1.

Theorem3.2. Let the problem (1.4), (1.5) have a unique solutionu, sup

t

a

k(y)(s)(y)(s)ds:t[a,b], yMk

−→0 ask−→+, (3.3) and let, for everyyC([a,b]; R),

klim+

1 + k t

a

k(y)(s)(y)(s)ds=0 uniformly on[a,b]. (3.4)

Let, moreover,

klim+

1 + k t

a

qk(s)q(s)ds=0 uniformly on[a,b], (3.5)

klim+hk(y)=h(y) foryC[a,b];R, (3.6)

klim+ck=c. (3.7)

Then there existsk0Nsuch that for everyk > k0the problem (3.1) has a unique solution ukand

klim+

uku C=0. (3.8)

FromTheorem 3.2, the following corollary immediately follows.

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Corollary3.3. Letab and the problem (1.6), (1.7) have only the trivial solution.

Then the Green operator of the problem (1.6), (1.7) is continuous.

To proveTheorem 3.2, we need two lemmas, the first of them immediately follows from Arzel`a-Ascoli lemma andProposition 2.9.

Lemma3.4. Letaband (y)(t) def=

t

a(y)(s)ds fort[a,b]. (3.9) Let, moreover,{xn}+n=1C([a,b];R)be a bounded sequence. Then the sequence{(x n)}+n=1

contains a uniformly convergent subsequence.

Lemma3.5. Let the problem (1.6), (1.7) have only the trivial solution and let the sequences of operatorskab and linear bounded functionalshk:C([a,b];R)Rsatisfy conditions (3.3) and (3.6). Then there existk0Nandr >0such that an arbitraryzC([a,b]; R) admits the estimate

zCk(z) fork > k0, (3.10) where

ρk(z)=hk(z)+ max1 + k t

a

z(s)k(z)(s)ds:t[a,b]

. (3.11) Proof. Note first that according to Banach-Steinhaus theorem and the condition (3.6), the sequence{hk}+k=1is bounded, that is, there existsr0>0 such that

hk(y)r0yC foryC[a,b];R. (3.12) Let, foryC([a,b];R),

(y)(t) = t

a(y)(s)ds, k(y)(t)= t

ak(y)(s)ds forkN. (3.13) Obviously,:C([a,b];R)C([a,b];R) andk:C([a,b];R)C([a,b];R) forkNare linear bounded operators and

k k forkN. (3.14) With respect to our notation, the condition (3.3) can be rewritten as follows:

sup k(y)(y) C:yMk−→0 ask−→+. (3.15) Assume on the contrary that the lemma is not valid. Then there exist an increasing sequence of natural numbers {km}+m=1 and a sequence of functions zmC([a,b]; R), mN, such that

zm C> mρkmzm formN. (3.16)

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Let

ym(t)= zm(t) zm

C

, vm(t)= t

a

ym(s)kmym(s)ds fort[a,b], (3.17) y0m(t)=ym(t)vm(t) fort[a,b], (3.18) wm(t)=km

y0m

(t)y0m

(t) +km

vm

(t) fort[a,b]. (3.19) Obviously,

ym

C=1 formN, (3.20)

y0m(t)=ym(a) +km

ym

(t) fort[a,b], mN, (3.21) y0m(t)=ym(a) +y0m

(t) +wm(t) fort[a,b],mN. (3.22) On the other hand, from (3.14) and (3.17), by virtue of (3.16), we get

vm C ρkm

zm

zm C1 + km < 1

m1 + km formN, (3.23) kmvm C km · vm C< 1

m formN. (3.24)

From (3.20) and (3.21), it follows thaty0mMkm, and therefore, in view of (3.15), we have

mlim+ km

y0m

y0m C=0. (3.25)

On account of (3.24) and (3.25), equality (3.19) implies that

mlim+ wm

C=0, (3.26)

and with respect to (3.18), (3.20), and (3.23), y0m

C ym

C+ vm

C2 formN. (3.27)

According toLemma 3.4, without loss of generality, we can assume that

mlim+y0m(t)=y0(t) uniformly on [a,b]. (3.28) With respect to (3.18), (3.20), (3.22), (3.23), and (3.26),

mlim+ ymy0

C=0, (3.29)

y0

C=1, y0(t)=y0(a) +y0

(t) fort[a,b]. (3.30) Consequently,y0is a nontrivial solution of (1.6).

(14)

On the other hand, from (3.12) and (3.16), we get hkm

y0hkm

y0ym+hkm

ym

r0 y0ym C+ 1

zm Chkmzm

r0 y0ym C+ 1

m formN.

(3.31)

Hence, on account of (3.6) and (3.29), we obtain hy0

=0. (3.32)

Thusy0is a nontrivial solution of the problem (1.6), (1.7), which contradicts the assump-

tion ofLemma 3.5.

Proof ofTheorem 3.2. Letrandk0be numbers, the existence of which is guaranteed by Lemma 3.5. Then, obviously, for everyk > k0, the problem

u(t)=k(u)(t), hk(u)=0, (3.33) has only the trivial solution. According toTheorem 2.1, for everyk > k0, the problem (3.1) is uniquely solvable.

We will show that ifu anduk are solutions of the problems (1.4), (1.5), and (3.1), respectively, then (3.8) holds. Let

vk(t)=uk(t)u(t) fort[a,b]. (3.34) Then, for everyk > k0,

vk(t)=k

vk

(t) +qk(t) fort[a,b], hk

vk

=ck, (3.35) where

qk(t)=k(u)(t)(u)(t) +qk(t)q(t) fort[a,b],

ck=ckhk(u). (3.36)

Now, by virtue of (3.4), (3.5), (3.6), and (3.7), we have δk=

1 + k max t

aqk(s)ds:t[a,b]

−→0 ask−→+, (3.37)

klim+ck=0. (3.38)

According toLemma 3.5, (3.35), and (3.37),

vk Crck+δk fork > k0. (3.39)

(15)

Hence, in view of (3.37) and (3.38), we obtain

klim+

vk

C=0, (3.40)

and, consequently, (3.8) holds.

4. On dimension of the solution set of homogeneous equation

Notation 4.1. LetUbe the solution set of the homogeneous equation (1.6). Obviously,U is a linear vector space.

According toTheorem 2.1, we haveU= {0}, that is, dimU1. Moreover, the follow- ing assertion is valid.

Theorem4.2. The spaceUis finite dimensional.

Proof. LetT:C([a,b];R)C([a,b];R) be an operator defined by T(v)(t)def=v(a) +

t

a(v)(s)ds fort[a,b]. (4.1) Evidently, the operatorTis linear. According toProposition 2.9, the operatorTis com- pact as well. Obviously, (1.6) is equivalent to the operator equation (2.11) in the following sense: ifuC([a, b];R) is a solution of (1.6), thenx=uis a solution of (2.11), and vice versa, ifxC([a,b];R) is a solution of (2.11), thenxC([a,b]; R) andu=xis a so- lution of (1.6). In other words, the setUis also a solution set of the operator equation (2.11).

On the other hand, sinceTis a linear compact operator, from Riesz-Schauder theory, it follows that the solution space of (2.11) is finite-dimensional. Therefore, dimU <+. Remark 4.3. Example 5.1below shows that dimUcan be any natural number, even in the case whenab.

Proposition4.4. The equalitydimU=1holds if and only if there existsξ[a,b]such that the problem

u(t)=(u)(t), u(ξ)=0 (4.2)

has only the trivial solution.

Proof. Let dimU=1 and let problem (4.2) have a nontrivial solutionuξ for everyξ [a,b]. Chooset0]a,b] such thatua(t0)=0. Then, obviously, functionsuaandut0are linearly independent solutions of (1.6), which contradicts the assumption dimU=1.

Now assume that there existsξ[a,b] such that the problem (4.2) has only the trivial solution and dimU2. Letu1,u2Ube linearly independent. Obviously,

u1(ξ)=0, u2(ξ)=0. (4.3)

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