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Memoirs on Dierential Equations and Mathematical Physics

Volume 8. 1996, 1{102

N. V. Azbelev and L. F. Rakhmatullina

THEORY OF LINEAR ABSTRACT

FUNCTIONAL DIFFERENTIAL EQUATIONS

AND APPLICATIONS

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functional dierential equation Lx = f, where L : D ! B is a linear operator,B is a Banach space, and D is isomorphic to the direct product B Rn. The Green operator is constructed, continuous dependence on parameters is studied. The obtained results are applied to ordinary, impulse and singular equations.

Mathematics Subject Classication. 34K10.

Key words and phrases. Linear operator, Green operator, Noether ope- rator, boundary value problem.

reziume. naSromSi ganxilulia sasazGvro amocanaLx=f abstraq-

tuli Punqcionalur-diPerencialuri gantolebisaTvis, sadacL:D! B CrPivi operatoria, B -banaxis sivrcea, xoloD izomorPuliaBRn

pirdapiri namravlisa. agebulia grinis operatori, SesCavlilia amonax- snis parametrze uCKvetad damokidebuleba. miGebuli Sedegebi gamoKe- nebulia hveulebrivi, impulsuri da singularuli gantolebebisaTvis.

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PREFACE

The equation Lx = f with a linear operator L acting from the space

Dn of absolutely continuous functions x : [a;b] ! Rn into the space Ln of summable functions f : [a;b]!Rn has been thoroughly studied in the works of the Perm Seminar. The results of these investigations are system- atized in the book [1]. Such a generalization of the ordinary dierential equation

(Lx)(t)def= _x(t) +P(t)x(t) =f(t); t2[a;b]; (0.1) covers many classes of equations containing the derivative of the unknown function, for instance, integro-dierential equations and equations with de- viated argument as well as their hybrids and so on. The theory of the equationLx=f with the linear operatorL:Dn !Ln is based upon the isomorphism between the spaceDn and the direct productLnRn. The isomorphism may be dened by

x(t) = t

Z

a z(s)ds+; x2Dn; fz;g2LnRn:

As it turned out, the replacement of the space Ln by an arbitrary Banach space B does not violate validity of the fundamental theorems. Thus, a further generalization arises in the form of the theory of abstract functional dierential equation

Lx=f (0.2)

with a linear operator L : D ! B, where B is a Banach space and D is isomorphic to the direct productBRn (D'BRn).

The spaceWn of the functions x : [a;b] !R1 with absolutely contin- uous derivative x(n 1) is isomorphic to the direct product L1Rn. The isomorphism may be dened on the base of the representation

x(t) = t

Z

a

(t s)n 1

(n 1)! x(n)(s)ds+nX1

k=0

(t a)k k! x(k)(a) of the elementx2Wn. Thus, the equation of then-th order

(Lx)(t)def= x(n)(t) +nX1

k=0pk(t)x(k)[hk(t)] =f(t); t2[a;b]; x(k)() = 0; if 62[a;b]; k= 0;1;:::;n 1; as well as its generalization of the form

(Lx)(t)def= x(n)(t) +q(t)x(n)[g(t)] +

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+nX1

k=0 b

Z

a x(k)(s)dsrk(t;s) =f(t); t2[a;b]; x(n)() = 0; if 62[a;b];

are the equations (0.2) in the spaceWn.

The space DSn(m) of the functions x : [a;b] ! Rn permitting nite discontinuity at the xed pointst1;:::;tm2(a;b) and absolutely continuous on the intervals [a;t1), [t1;t2);:::, [tm;b] is isomorphic to LnRn(m+1). The isomorphism is dened by

x(t) =Zt

a z(s)ds+0+Xm

i=1[ti;b](t)i; z2Ln; f0;1;:::;mg2Rn(m+1); wheree denotes the characteristic function of the sete.

Any spaceDisomorphic to BRn forms its own proper class of equa- tions. Some examples of nontraditional spaces isomorphic to the product

BRn are provided in [2]. It is shown there in particular that the space of functions x : [a;b]! R1 which have \quasi-derivatives" up to the n-th order inclusively is isomorphic toL1Rn. Thus, the linear equation with quasi-derivatives is one of the form (0.2).

The theory of abstract functional dierential equations considers wide classes of equations from a single point of view. During the last ten-year period, this theory found various applications in studying old and new prob- lems due to possibility of choosing proper spaceD'BRn for each class of problems. A successful choice of the space permits in virtue of the general theory direct using of standard schemes and theorems of analysis in such cases, where we have been forced before to invent special devices and put severe restrictions connected with application of these devices.

It is relevant to emphasize the principal dierence between the gener- alization of the ordinary dierential equation in the form of the abstract functional dierential equation and the \ordinary dierential equation in Banach spaces". The equation (0.1) is dened by the operatorL:Dn!Ln belonging to the class of the so called \local operators" [3, 4]. An operatorL in a functional space is called local, if the value of the imagef(t) = (Lx)(t) in a neighborhood of each pointt depends only on the value of the preim- age x() in the neighborhood of the same point t. The generalization in the theory of ordinary dierential equations in Banach spaces consists in replacement of the nite dimensional space Rn of the values x(t) of the unknown function xby an arbitrary Banach space. In this connection, the property of Lto be a local operator keeps. In the theory of abstract func- tional dierential equation the generalization consists in replacement of the

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spaceLn by an arbitrary Banach spaceBand in replacement of the local operatorL:Dn!Ln by an arbitrary linear operatorL:D!B.

The main notation.

Rn space ofn-dimensional real vectors with the normjj.

kk

X norm of an element of the normed spaceX.

kAkX!Y norm of an operatorA:X!Y. Usually the symbol

\X !Y" is ommited.

A operator adjoint to the operatorA. R(A) range of values of the operatorA. D(A) domain of denition of the operatorA. dimM dimension of the linear set M.

kerA null-set (the kernel) of the operatorA.

indA index of the operator A : indA = dimkerA dimkerA.

[A1;A2] linear operator acting from the space X into the product Y1Y2 by [A1;A2]x=fA1x;A2xg, x2X, A1x2Y1, A2x2Y2.

fA1;A2g linear operator acting from the product of the spaces X1X2intoY byfA1;A2gfx1;x2g=A1x1+A2x2, x12X1,x22X2.

I identity operator.

E identity matrix orEn under the necessity to empha- size the dimension of the identitynn-matrix.

< ';x >,'x value of the functional'on the elementx. Sr composition operator dened by

(Srx)(t) =

8

<

:

x[r(t)]; if r(t)2[a;b]; 0; if r(t)62[a;b]: e() characteristic function of the set e:

e=

8

<

:

1; if t2e;

0; if t62e:

ij Kronecker symbol: ij =

8

<

:

1; if i=j;

0; if i6=j:

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CHAPTER I

LINEAR ABCTRACT

FUNCTIONAL DIFFERENTIAL EQUATIONS

x1. Preliminary Knowledge from the Theory of Linear Equations in Banach Spaces

The main assertions of the theory of linear abstract functional-dierential equations are based on the theorems about linear equations in Banach spaces. We give here without proofs certain results of the book [5] which we will need below. We formulate some of these assertions not in the most general form, but in the form satisfying our aims. The enumeration of the theorems in brackets means that the assertion either coincides with the corresponding result of the book [5] or is only an extraction from this result.

We will use the following notation.

X,Y,Zare Banach spaces;A,Bare linear operators;D(A) is a domain of denition ofA;R(A) is a range of values ofA;Ais an operator adjoined toA. The set of solutions of the equation Ax= 0 is said to be a null space or a kernel ofAand is denoted by kerA. The dimension of a linear setM is denoted by dimM.

LetAbe acting fromXinto Y. The equation

Ax=y (1.1)

(the operator A) is said to be normal solvable, if the set R(A) is closed;

(1.1) the operator A is said to be a Noether equation, if it is a normal solvable one and besides dimkerA <1and dimkerA<1. The number indA= dimkerA dimkerA is said to be an index of the operatorA(the equation (1.1)). If A is a Noether operator and indA = 0, the equation (1.1) (the operatorA) is said to be a Fredholm one. The equationA'=g is said to be an equation, adjoined to (1.1).

Theorem 1.1 (Theorem 3.2). An operator A is normal solvable if and only if the equation(1:1) is solvable for such and only such right hand sidey which is orthogonal to all the solutions of the homogenous adjoined equation A'= 0.

Theorem 1.2 (Theorem 16.4). The property of being Noether operator is stable in respect to completely continuous perturbations. By such perturba- tions, the index of the operator does not change.

Theorem 1.3 (Theorem 12.2). LetA be acting fromXintoY andD(B) be dense inY. IfAandBare Noether operators,BAis also a Noether one andind(BA) = indA+ indB.

Theorem 1.4 (Theorem 15.1). LetBAbe a Noether operator andD(B) R(A). ThenB is a Noether operator.

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Theorem 1.5 (Theorem 2.4 and Lemma 8.1). LetAbe dened on Xand acting into Y. A is normal solvable and dimkerA =n if and only if the spaceYis representable in the form of a direct sumY=R(A)Mn, where Mn is a nite-dimensional subspace of the dimension n.

Theorem 1.6 (Theorem 12.2). Let D(A) X, Mn be a n-dimensional subspace of X and D(A)\Mn = f0g. If A is a Noether operator, then its linear extension Ae on D(A)Mn is also a Noether operator. Besides indAe= indA+n.

Theorem 1.7. Let a Noether operator Abe dened on Xand acting into

Y, D(B) = Y, BA : X ! Z is a Noether operator. Then B is also a Noether operator.

Proof. Thanks to Theorem 1.4, we are in need only of the proof of the case R(A)6=Y. From this Theorem 1.4 we obtain also that restriction B of B onR(A) is a Noether operator.

Let dimkerA=n. Then we have from Theorem 1.5 that

Y=R(A)Mn=D( B)Mn; where dimMn=n.

From Theorem 1.6 we see that B is a Noether operator as a linear ex- tension of B onY.

Let a linear operatorAacting from a direct productX1X2intoY be dened by a pair of operatorsA1 :X1 !Y and A2 : X2 ! Y in such a way, that

Afx1;x2g=A1x1+A2x2; x12X1; x22X2;

whereA1x1=Afx1;0g,A2x2=Af0;x2g. We will denote such an operator byA=fA1;A2g.

Let a linear operatorAacting fromXinto a direct productY1Y2 be dened by a pair of operators A1 : X! Y1 and A2 : X !Y2 in such a way, that Ax=fA1x;A2xg, x 2 X. We will denote such an operator by A= [A1;A2].

The theory of linear abstract functional dierential equation is using some operators dened on a productBRn or acting in such a product.

We will formulate here certain assertions about such operators, reserving as far as possible the notations from [1].

A linear operator f; g acting from a direct product BRn of the Banach spaces B and Rn into a Banach space D is dened by a pair of linear operators :B!DandY :Rn!Din such a way that

f;Ygfz;g= z+Y ; z2B; 2Rn:

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A linear operator [;r] acting from a spaceDinto a direct productBRn is dened by a pair of linear operators:D!B andr:D!Rn in such a way that

[;r]x=fx;rxg; x2D:

If the norm in the space BRn is dened in an appropriate way, for instance by

fz;gBRn=kzkB+jj; the spaceBRn will be a Banach one.

If a bounded operatorf;Yg:BRn!Dis an inverse to a bounded operator [;r] :D!BRn, then

x= x+Y rx; x2D; (1.2) (z+Y ) =z,r(z+Y ) =beta,fz;g2BRn.

Hence

+Y r=I; =I; Y = 0; r = 0; rY =I:

We will identify the nite-dimensional operatorY :Rn !D with such a vector (y1;:::;yn),yi2D, that

Y =Xn

i=1yii; = col1;:::;n :

We denote the components of a vector-functionalrbyr1;:::;rn.

If l = fl1;:::;lmg : D ! Rm is a linear vector-functional and X = (x1;:::;xn) is a vector with components xi 2 D, then lX denotes the mn-matrix whose columns are the values of the vector-functionallon the components ofX :lX = (lixj),i= 1;:::;m;j= 1;:::;n.

The operators ,Y,,rfor the spaces suggested above in the Introduc- tion have the following forms.

For the spaceDn, (z)(t) =Zt

a z(s)ds; Y =E; x= _x; rx=x(a); whereE is the identicalnn-matrix.

For the spaceWn, (z)(t) = t

Z

a

(t s)n 1

(n 1)! z(s)ds; Y =

1;t a;:::;(t a)n 1 (n 1)!

; x=x(n); rx=x(a);x_(a);:::;x(n 1)(a) :

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For the spaceDSn(m), (z)(t) =Zt

a z(s)ds; Y =E;E[t

1;b](t);:::;E[tm;b](t); x= _x; rx=x(a);x(t1);:::;x(tm) ;

x(ti) =x(ti) x(ti 0):

Theorem 1.8. A linear bounded operator f;Yg:BRn !D has the bounded inverse if and only if the following conditions are satised.

a) The operator :B!D is a Noether one andind = n. b) dim ker = 0.

c) If1;:::;nis a basis forkerand= [1;:::;n], then detY 6= 0.

Proof. Suciency. From a) and b) it follows that dim ker=n. By virtue of Theorem 1.5,D=R()Mn, where dimMn=n. It follows from c) that any nontrivial linear combination of the elementsy1;:::;yndoes not belong toR(), thereforeMn =R(Y). ThusD=R()R(Y) and consequently the operatorf;Yghas its inverse by virtue of Banach theorem.

Necessity. From invertibility off;Yg, we haveD=R()R(Y). Con- sequently the operator is normally solvable by virtue of Theorem 1.5 and dimker=n. Besides dimker = 0. Therefore ind = n. Assumption detY = 0 leads to the conclusion that a nontrivial combination of the elementsy1;:::;yn belongs toR().

Theorem 1.9. A linear bounded operator [;r] : D ! B Rn has a bounded inverse if and only if the following conditions are satised.

a) The operator :D!B is a Noether one andind=n. b) dim ker=n.

c) Ifx1;:::;xnis a basis ofkerandX = (x1;:::;xn), then detrX 6= 0.

Proof. Suciency. From a) and b) it follows that dimker = 0. Thus R() =B. Each solution of the equationx=z has the form

x=Xn

i=1cixi+v;

where ci = const, i= 1;:::;n, and v is any solution of this equation. By virtue of c), the system

x=z; rx=

has a unique solution for each pair z 2 B, 2 Rn. Consequently the operator [;r] has the bounded inverse.

Necessity. Let [;r] 1 = f;Yg. From the equality =I, by virtue of Theorem 1.7 it follows that is a Noether operator and by virtue of

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Theorem 1.3, ind =n. As far asR() = B, we have dim ker = 0 and consequently dimker=n. If detrX = 0, then the homogeneous system

x= 0; rx= 0

has a nontrivial solution. This gives a contradiction to the invertibility of the operator [;r].

x 2. Linear Equation and Linear Boundary Value Problem The Cauchy problem

(Lx)(t)def= _x(t) P(t)x(t) =f(t); x(a) =; t2[a;b];

is uniquely solvable for2Rn and any summablef if the elements of the nn-matrixP are summable. Thus, the representation of the solution

x(t) =X(t) t

Z

a X 1(s)f(s)ds+X(t)

of the problem (the Cauchy formula), where X is a fundamental matrix such thatX(a) is the identity matrix, is also a representation of the general solution of the equation Lx = f. The Cauchy formula is the base for investigations on various problems in the theory of ordinary dierential equations. The Cauchy problem for functional dierential equation is not solvable generally speaking, but some boundary value problems may be solvable. Therefore the boundary value problem plays the same role in the theory of functional dierential equations as the Cauchy problem does in the theory of ordinary dierential equations.

We will call the equation

Lx=f (2.1)

a linear abstract functional-dierential equation if L : D ! B is a linear operator,DandBare Banach spaces and the spaceDis isomorphic to the direct product BRn (D'BRn).

LetJ =f;Yg:BRn!Dbe a linear isomorphism andJ 1= [;r].

Everywhere below the norms in the spacesBRn andDare dened by

fz;gBRn=kzkB+jj; kxkD =kxkB+jrxj:

By such a denition of the norms, the isomorphismJ is an isometrical one.

Therefore

f;YgBRn!D= 1; [;r]D!BRn= 1: Since

kzkD=f;Ygfz;0gD

f;Ygfz;0gBRn=kzkB;

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kkB!D= 1. Analogously it is established that kYkRn!D = 1. Next we have

kxkB kxkD and, ifrx= 0,

kxkB=kxkD: ThereforekkD!B= 1. AnalogouslykrkD!Rn= 1.

We will assume that the operatorL:D!Bis bounded. ApplyingLto the both parts of (1.2), we get the decomposition

Lx=Qx+Arx: (2.2)

Here Q = L : B ! B is a principal part, and A = LY : Rn ! B is a nite-dimensional part ofL.

As examples of (2.1) in the case where D is a space Dn of absolutely continuous functions x : [a;b] ! Rn and B is a space Ln of summable functions z: [a;b]!Rn, we can take an ordinary dierential equation

x_(t) P(t)x(t) =f(t); t2[a;b]; (2.3) where the columns of the matrixP belong toLn, or an integro-dierential equation

x_(t) b

Z

a H1(t;s) _x(s)ds Zb

a H(t;s)x(s)ds=f(t); t2[a;b]: (2.4) We will assume the elementshij(t;s) of the matrixH(t;s) to be measurable on [a;b][a;b], the functions Rabhij(t;s)ds to be summable on [a;b], and the integral operator

(H1z)(t) = b

Z

a H1(t;s)z(s)ds

onLn into Ln to be completely continuous. The corresponding operators

Lfor these equations have the representation in the form (2.2) (Lx)(t) = _x(t) P(t) t

Z

a x_(s)ds P(t)x(a) for (2.3), and

(Lx)(t) = _x(t) b

Z

a

H1(t;s) + b

Z

s H(t;)dx_(s)ds Zb

a H(t;s)ds x(a) for (2.4).

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Theorem 2.1. An operator L : D ! B is a Noether one if and only if the principal part Q : B ! B of L is a Noether operator. In this case, indL= indQ+n.

Proof. IfL is a Noether operator,Q=L is also Noether as the product of Noether operators and indL= indQ+n(Theorems 1.8 and 1.3).

IfQis a Noether operator,Qis also Noether. ConsequentlyL=Q+Ar is also Noether (Theorem 1.9 and 1.2).

Due to Theorem 2.1, the equality indL=nis equivalent to the fact that Q is a Fredholm operator. The operatorQ:B !B is a Fredholm one if and only if it is representable in the form Q=P 1+V (Q=P1 1+V1), where P 1 is an inverse to a bounded operator P and V is a completely continuous operator (P11is an inverse to the boundedP1andV1is a nite- dimensional operator) [6]. An operatorQ= (I+V) :B!Bis a Fredholm one, if a certain degreeVmofV is completely continuous [6]. If the operator V itself is completely continuous, the operatorQ =I+V is said to be a canonical Fredholm operator.

In the above given examplesQ=I K, whereKis an integral operator.

For (2.3),

(Kz)(t) = t

Z

a P(t)z(s)ds and it is a completely continuous operator. For (2.4),

(Kz)(t) =Zb

a

H1(t;s) +Zb

s H(t;)dz(s)ds:

Here K2is a completely continuous operator. The property of being com- pletely continuous of these operators may be established by V. Maksimov's Theorem 6.1 [7, 1] which is given below.

Theorem 2.2. Let L:D!B be a Noether operator, indL = n. Then dimkerLnanddimkerL=nif and only if the equation(2:1) is solvable for eachf 2B.

Proof. Recall that dim kerL dimkerL=n. Besides the equationLx=f is solvable for eachf 2Bif and only if dimkerL = 0 (Theorem 1.1).

We call the vectorX = (x1;:::;x) whose components constitute a basis for the kernel ofL the fundamental vectorof the equationLx = 0 and the components x1;:::;x we call the fundamental system of solutions of this equation.

Let l = [l1;:::;lm] : D ! Rm be a linear bounded vector-functional, = colf1;:::;mg2Rm. The system

Lx=f; lx= (2.5)

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is called a linear boundary value problem.

IfR(L) =Band dimkerL=n, the question of solvability of (2.5) is the one of solvability of a linear system of algebraic equations with the matrix lX = (lixj),i = 1;:::;m; j = 1;:::;n. Indeed, since the general solution of the equationLx=f has the form

x=Xn

j=1cjxj+v;

wherev is any solution of this equation, c1;:::;cn are arbitrary constants, the problem (2.5) is solvable if and only if the algebraic system

n

X

j=1lixjcj=i liv; i= 1;::: ;m;

in respect of c1;:::;cn is solvable. In such a way the problem (2.5) has a unique solution for eachf 2B,2Rmif and only ifm=nand detlX 6= 0.

The determinant detlX is said to be the determinant of the problem (2:5).

By applying the operatorl to the two parts of the equality (1.2), we get the decomposition

lx= x+ rx; (2.6)

where :B!Rm is a linear bounded vector-functional. We will denote the matrix dened by the linear operator :Rn!Rm also by .

Using the representations (2.2) and (2.6), we can rewrite the problem (2.5) in the form of the equation

Q A

xrx

=

f

: (2.5)

The operator

Q

A

: B(Rm)!B(Rn) is adjoint to the operator

Q A

: BRn!BRm:

Taking into account the isomorphism between the spaces B(Rn) and

D

, we therefore call the equation

Q

A

!

=

g

to be adjoint to the problem (2.5).

Lemma 2.1. The operator [;l] : D ! BRm is a Noether one with ind[;l] =n m.

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Proof. We have [;l] = [;0] + [0;l], where the symbol \0" denotes a null- operator on the corresponding space. The operator [0;l] :D!BRm is compact because the nite-dimensional operator l :D !Rm is compact.

Compact perturbations does not change the index of the operator (Theorem 1.2). Therefore it is sucient to prove Lemma only for the operator [;0].

The direct product Bf0g is the range of values of [;0]. The homo- geneous adjoint equation to the problem [;0]x=ff;0gis reducible to one equation!= 0 in the spaceB(Rm). The solutions of this equation are the pairsf0;g. Therefore dimker[;0]=m.

Thus [;0] :D!B Rmis a Noether operator with ind[;0] =n m. Rewrite the problem (2.5) in the form of one equation

[L;l]x=ff;g: (2.5)

Theorem 2.3. The problem (2:5) is a Noether one if and only if the prin- cipal part Q : B ! B of L is a Noether operator and also ind[L;l] = indQ+n m.

Proof. The operator [L;l] has the representation [L;l] =

Q 0 0 I

[;l] + [Ar;0];

where I :Rm!Rmis an identical operator, and the symbol \0" denotes the null operator on the corresponding space. Indeed,

Q 0 0 I

[;l]x+ [Ar;0]x=

Q 0 0 I

colfx;lxg+ colfArx;0g=

= colfQx+Arx;lxg:

The operatorQ:B!Bis Noether if and only if the operator

Q 0 0 I

: BRm!BRm is Noether with

ind

Q 0 0 I

= indQ:

Therefore the operator

Q 0 0 I

[;l] : D!BRm

is a Noether one if and only ifQis a Noether operator and also ind

Q 0 0 I

[;l] = ind

Q 0 0 I

+ ind[;l] = indQ+n m

(Theorems 1.3 and 1.7). The product Ar:D!Bis compact. Hence the operator [Ar;0] :D!BRmis also compact. Now we get the conclusion

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of the Theorem from the fact that compact perturbation does not violate the property of being Noether operator and does not change the index.

It should be noted the following Corollaries from Theorem 2.3 under the assumption ofLbeing a Noether operator.

Corollary 2.1. The problem (2:5) is a Fredholm one if and only if indQ= m n.

Corollary 2.2. The problem (2:5) is solvable if and only if the right hand sideff;gis orthogonal to all the solutions f!;gof the homogeneous ad- joint equation

Q!+ = 0; A!+ = 0: The condition of being orthogonal has the form

h!;fi+h;i= 0:

Everywhere below we assume that the operatorLis Noether with indL= nwhich means thatQis a Fredholm operator. Under such an assumption, by virtue of Corollary 2.1 the problem (2.5) is a Fredholm one if and only ifm=n.

The functionalsl1;:::;lm are assumed to be linear independent.

We will call the special case of (2.5) when l = r a principal boundary value problem. The equation [;r]x = ff;gis just the problem which is the base of the isomorphismJ 1= [;r] betweenDandBRn.

Theorem 2.4. The principal boundary value problem

Lx=f; rx= (2.7)

is uniquely solvable if and only if the principal part Q: B ! B of L has the bounded inverse Q 1 : B ! B. The solution x of (2.7) admits the representation

x= Q 1f+ (Y Q 1A): (2.8)

Proof. Using the decomposition (2.2), we can rewrite (2.7) in the form Qx+Arx=f; rx=:

IfQis invertible, then

x=Q 1f Q 1A:

An application to this equality of the operator yields (2.8) because = I Y r.

IfQis not invertible andyis a nontrivial solution of the equationQy= 0, the homogeneous problem

Lx= 0; rx= 0

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has a nontrivial solutionx, for instance x= y.

From (2.8) one can see that the vectorX=Y Q 1A is fundamental and alsorX =E (HereAdenotes the vector defying the nite-dimensional operatorA:Rn!B).

Theorem 2.5. The following assertions are equivalent.

a) R(L) =B. b) dim kerL=n.

c) There exists a vector-functional l : D ! Rn such that the problem (2:5) is uniquely solvable for eachf 2B, 2Rn.

Proof. The equivalence of the assertions a) and b) was established while proving Theorem 2.2.

Let dimkerL=nandl= [l1;:::;ln], the systeml1;:::;ln be biorthogo- nal to the basisx1;:::;xn of the kernel ofL:lixj =ij,i;j= 1;:::;nand ij be the Kronecker symbol. Then the problem (2.5) with such l has the unique solution

x=X( lv) +v;

where X = (x1;:::;xn) andv is any solution of Lx =f. This is seen by taking into account thatlX =E. Conversely, if (2.5) is uniquely solvable for eachf and, then the solutions of the problems

Lx= 0; lx=i; i2Rn; i= 1;:::;n;

can be taken as the basis x1;:::;xn provided the matrix (1;:::;n) is invertible. Thus the equivalence of the assertions b) and c) is proved.

x 3. Green Operator We will consider here the boundary value problem

Lx=f; lx= (3.1)

under the assumption that the dimensionmofl(the number of the bound- ary conditions) is equal to n. By virtue of Corollary 2.2, such a condition is necessary for unique solvability of the problem (3.1). Recall that we as- sume L to be a Noether operator with indL =n (indQ= 0). If m =n, then the problem (3.1) is Fredholm ([L;l] : D ! BRn is a Fredholm operator). Consequently, for this problem the assertions \the problem has a unique solution for some right hand sideff;g(the problem is uniquely solvable)", \the problem is solvable for each ff;g (the problem is solv- able everywhere)", \the problem is everywhere and uniquely solvable" are equivalent.

Let (3.1) be uniquely solvable and denote [L;l] 1 = fG;Xg. Then the solutionxof the problem ( 3.1) has the representation

x=Gf+X:

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The operatorG:B!Dis called the Green operator of the problem (3:1), the vectorX = (x1;:::;xn) is a fundamental vector for the equationLx= 0 and alsolX=E.

It should be noted that is the Green operator of the problemx=f, rx=.

Theorem 3.1. A linear bounded operatorG:B!Dis Green operator of the boundary value problem(3:1) if and only if the following conditions are fullled.

a) Gis a Noether operator withindG= n. b) kerG=f0g.

Proof. fG;Xg : BRn ! D is a one-to-one mapping if G is the Green operator of (3.1). So a) and b) are fullled by virtue of Theorem 1.8.

Conversely, letGbe such that a) and b) are fullled. Then dimkerG=n. If l1;:::;ln constitute a basis of kerG and l = [l1;:::;ln], then R(G) = kerl. Gis the Green operator of (3.1), where

Lx=G 1(x Ulx) +V lx;

G 1 is the inverse to G : B ! kerl, U = (u1;:::;un) with ui 2 D is a vector such thatlU =E, andV = (v1;:::;vn) withvi2Bis an arbitrary vector.

Theorem 3.2. Let the problem (3:1) be uniquely solvable and G be the Green operator of this problem. Let further U = (u1;:::;un), ui 2 D, lU =E. Then the vector

X=U GLU is a fundamental one to the equationLx= 0.

Proof. dimkerL=nby virtue of Theorem 2.5 and the unique solvability of (3.1). The components of X are linear independent becauselX =E. The equalityLX = 0 can be checked immediately.

Theorem 3.3. LetG andG1 be the Green operators of the problems

Lx=f; lx= and

Lx=f; l1x=:

Let further X be the fundamental vector ofLx= 0. Then G=G1 X(lX) 1lG1:

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Proof. The general solution ofLx=f has the representation x=Xc+G1f;

where c 2Rn is an arbitrary vector. Denec in such a way that lx= 0.

We have

0 =lx=lXc+lG1f:

Hence

c= (lX) 1lG1f

and the solutionxof the half homogeneous problemLx=f,lx= 0 has the form x= (G1 X(lX) 1lG1)f =Gf:

In the investigation of particular boundary value problems and some properties of Green operator, it is useful to employ the \elementary Green operator" Wl which is possible to construct for any boundary condition lx=. Beforehand we will prove the following proposition.

Lemma 3.1. For any linear bounded vector-functional l = [l1;:::;ln] :

D ! Rn with linear independent components, there exists a vector U = (u1;:::;un) with ui2D such thatdetrU 6= 0 and detlU6= 0.

Proof. LetU1 and U2 be n-dimensional vectors such that detrU1 6= 0 and lU2=E. Let further

U =U1+U2

whereis a numerical parameter. The function () = detrU is continuous and (0) 6= 0. Hence ()6= 0 on an interval ( 0;0). The polynomial P() = detlU = det(lU1+E) has no more thann roots. Consequently, there exists such a 12( 0;0) thatP(1)6= 0. ForU =U1+1U2 we have: detrU 6= 0 and detlU 6= 0.

LetU = (u1;:::;un),ui2D, detrU 6= 0,lU =E. Dene the operator Wl:B!Dby:

Wl= U; (3.2)

whereU :Rn!Dis a nite-dimensional operator corresponding toU and :B!Rnis the principal part of the vector-functionall (see the equality (2.6)). Let furtherL0:D!Bbe dened by

L0x=x U(rU) 1rx: (3.3)

Theorem 3.4. Wl is the Green operator of the boundary value problem

L0x=f; lx=: (3.4)

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Proof. The principal boundary value problem for the equation L0x=f is uniquely solvable. Consequently, the dimension of the fundamental vector forL0x= 0 is equal ton. By immediate substitution, we getL0U= 0. The problem (3.4) is solvable becauselU =E. We have

L0Wlf =(f Uf) U(rU) 1r(f Uf) =

=f Uf+U(rU) 1rUf =f;

lWlf = (f Uf) + r(f Uf) =

= f lUf = 0:

The collection of all Green operators corresponding to a given vector- functionall:D!Rn is of the form

G=Wl ; (3.5)

where is a linear homeomorphism ofB intoB. Indeed, if :B!Bis a homeomorphism, then by virtue of Theorem 3.1Wl is a Green operator of (3.1). Conversely, any Green operatorG:B!kerlmay be represented by (3.5), where =Wl 1G,Wl 1: kerl!Bis the inverse toWl:B!kerl.

Theorem 3.5. The collection of all Green operatorsG:B!Dis dened

by G= ( Uv) ;

where U = (u1;:::;un), ui 2 D, detrU 6= 0, v : B ! Rn is a linear bounded vector-functional, and is a linear homeomorphism of the spaceB intoB.

Proof. W = Uv is the Green operator of the problem (3.4), where lx=vx+ [E vU](rU) 1rx. Indeed,

L0Wf =( Uv)f U(rU) 1r( Uv)f =

=f Uvf+U(rU) 1rUvf=f;

lWf =v( Uv)f + [E vU](rU) 1r( Uv)f =

=vf vUvf [E vU]vf = 0:

Now the assertion of the Theorem follows from the representation (3.5).

In the investigation of boundary value problems, an important part be- longs to the so called \W-method" [8] which is based on an expedient choice of an auxiliary \model" equationL1x=f. The foundation to this method is laid by the following assertion.

Theorem 3.6. Let a model boundary value problem

L1x=f; lx= 0

be uniquely solvable and W :B!Dbe the Green operator of this problem.

The problem(3:1) is uniquely solvable, if and only if the operatorLW :B!

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B has the continuous inverse [LW] 1. In this case, the Green operator G of the problem(3:1) has the representation

G=W[LW] 1:

Proof. There is a one-to-one correspondence between the set of solutionsz2

Bof the equationLWz=f and the set of solutionsx2Dof the problem (3.1) with homogeneous boundary conditions lx = 0 the correspondence is dened by x = Wz and z = L1x. Consequently the problem (3.1) is uniquely solvable and also the solutionxof the problem (3.1) for= 0 has the representationx=W[LW] 1f. ThusG=W[LW] 1.

In the applications of Theorem 3.6 one may putW =Wl, where Wl is dened by (3.2). Let the operator U : Rn ! D be dened as above by the vector U = (u1;:::;un), ui 2 D, detrU 6= 0, lU = E. Let further : B ! Rn be the principal part of l : D ! Rn. Dene the operator F :B!BbyF=LU.

Corollary 3.1. The boundary value problem (3:1) is uniquely solvable if and only if the operator (Q F) : B ! B has the bounded inverse. The Green operator of this problem has the representation

G=Wl(Q F) 1: (3.6)

The proof follows from the fact that Wl is the Green operator of the model problemL0x=z,lx= 0, whereL0 is dened by (3.3) and

LWl=L LU =Q Ar LU =Q LU =Q F:

The following assertions characterize some properties of the Green op- erator of the problem (3.1) connected with the properties of the principal partQofL.

Theorem 3.7. Assume that a boundary value problem (3:1) is uniquely solvable. LetP :B!Bbe a linear bounded operator with bounded inverse P 1. The Green operator of this problem has the representation

G=Wl(P+H); (3.7)

whereH :B!B is compact, if and only if the principal partQof L may be represented in the form Q=P 1+V, whereV :B!B is compact.

Proof. LetG=Wl(Q F) 1(see (3.6)),Q=P 1+V. DeneV1=V F. Then

(Q F) 1= (P 1+V1) 1= (I+PV1) 1P = (I+H1)P =P+H;

whereH :B!BandH1:B!Bare compact operators.

Conversely, if (Q F) 1=P+H then

Q=F+ (P+H) 1=F+ (I+P 1H) 1P 1=

=F+ (I+V1)P 1=P 1+V;

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whereV :B!BandV1:B!Bare compact operators.

Theorem 3.8. A linear bounded operatorG:B!Dis the Green operator of the problem(3:1), whereQ=P 1+V, if and only ifkerG=f0gand

G= P+T (3.8)

with a compact operator T :B!D.

Proof. IfGis the Green operator andQ=P 1+V then (3.8) immediately follows from (3.7) and (3.2).

Conversely, if Ghas the form (3.8), then Gis a Noether operator with indG = n. By virtue of Theorem 3.1, G is the Green operator of the problem (3.1). From LG =I it follows that QP +LT = I. Hence Q = P 1+V, whereV = LTP 1.

We now state two Corollaries of Theorem 3.8.

Corollary 3.2. The representation G = P +H, where H : B ! B is a compact operator and P : B ! B is a linear bounded operator with a bounded inverse P 1 is possible if and only if G is the Green operator of the problem (3:1), where Q=P 1+V, V is a compact operator.

Proof. IfG=P+H, then

G= P+ H+Y rG and due to Theorem 3.8,Q=P 1+V.

Conversely, if Q = P 1 +V, then G = P +T and, consequently, G=P+T.

Corollary 3.3. The operator G is canonical Fredholm if and only if the principal partQof L is canonical Fredholm.

x 4. Boundary Value Problems which are Not Everywhere and Uniquely Solvable

We assume as above that indL=n(indQ= 0) and, in addition, that the equationLx= 0 has ann-dimensional fundamental vectorX. By Theorem 2.5, the equationLx=f is solvable for eachf 2B.

We will consider the boundary value problem

Lx=f; lx= (4.1)

without the assumption that the numbermof boundary conditions is equal ton.

Denote = ranklX. In the case >0, we may assume without loss of generality that the determinant of the rankcomposed of the elements in the left top of the matrixlX does not vanish. Let us choose a fundamental vector as follows. In the case >0, the elementsx1;:::;xare selected such thatlixj=ij,i;j= 1;:::;(ijis the symbol of Kronecker). If 0 < n,

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