Factorization of linear operators and some eigenvalue problems of special operators Текст научной статьи по специальности «Математика»

UDC 51.0

FACTORIZATION OF LINEAR OPERATORS AND SOME EIGENVALUE PROBLEMS OF SPECIAL OPERATORS

© I. N. Parasidis1*, E. Providas1, P. C. Tsekrekos

Technological Educational Institute of Larissa 41110 Larissa, Greece.

Phone: +30 2410 684383.

E-mail: paras@teilar.gr E-mail: providas@teilar.gr 2National Technical University 42 28 Oktovriou, 10682 Athens, Greece.

An exact solution to a boundary value problem Bj x = f is presented where the linear operator Bj has a decomposition of the form Bj = BGBG with BG and BG being two linear

operators of a special form. If the operator Bj is correct then the solution can be found. Also, the

eigenvalues and eigenvectors of the operator Bj as well as the quadratic operator B2 = B2 and

biquadratic operator B4 = B4 , where B is a known operator of special form, are derived analytically.

Keywords: correct operator, quadratic and biquadratic operators, factorization of operators, eigenvalue problems, integro-differential equations, boundary value problems.

1. Introduction

Integral and differential operators model many situations from science and engineering. The integral and differential operators describing these problems are complicated and the exact solution of the corresponding boundary value problems is a difficult task. In some cases the initial problem can be transformed to a simpler one involving simpler operators and thus the solution can be found easier. (See for example |l-3]). Furthermore, the eigenvalues and eigenvectors of an operator are important in many applications (See for example [4-5]). Within this frame, the present paper discusses the exact solution of boundary value problems with linear operators of a special form and the analytical computation of the corresponding eigenvalues and eigenvectors.

In particular, this work is a sequel of a series of papers by the authors [6-!0]. In [6] some quadratic correct extensions of minimal operators in Banach space and their exact solutions have been presented. In [7-9] the correctness and self-adjointness of the

quadratic operator B2 = B2 and biquadratic operator

B4 = B4 as well as the solution of the integral and differential boundary value problems have been examined. Reference [i0] deals with quadratic correct self-adjoint extensions of minimal symmetric operators in Hilbert space and the solution of the corresponding boundary value problem. In the present paper an exact solution to a boundary value problem Bj x = f is presented where the linear operator Bj has a decomposition of the form Bj = BGBG with BG and

BG being two linear operators of special form. If the

operator Bj is correct then the solution can be found. Also, the eigenvalues and eigenvectors of the operator Bj as well as the quadratic operator B2 = B2 and the

biquadratic operator B4 = B4 , where B is a known operator of special form, are derived analytically.

The paper is organized as follows. In Section 2 we develop the theory for the solution of the problem Bjx = f when Bj = BGBG . In Section 3 the

eigenvalues and eigenvectors of the operators B2, B4 and Bj are derived in the closed form. Finally, in Section 4 we give two examples of integral and differential equations demonstrating the power and usefulness of the methods presented.

Throughout this paper we use the following terminology and notation. We denote by (x, f) H the inner product of elements x, f of a complex Hilbert space H . We write D( A) and R( A) for the domain and range of the operator A respectively. An operator A is called correct if R(A) = H and the inverse

^ —I

A exists and is continuous. If an operator B : H ^ H is correct, then we say that the problem Bx = f is correct. An operator Bj is called quadratic (resp. biquadratic) if there exists an operator B such that Bj = B2 (resp. Bj = B4). If for

operator Bj : H ^ H there exist two operators BG

* автор, ответственный за переписку

and BGsuch that Bj = BGBGthen we say that BgBg0 is a decomposition (factorization) of Bj. If Ft, gi e H,i = i,K,m then F = (Fj,...,Fm),

G = ( gl,к, gm ) and AF = (AF1,••., AFm ) are

vectors of Hm. Denote by A~i =(¿4_1)!, i = j,2.

We also write Ft and (Ax, Ft) m for the column

vectors col (Fj,..., Fm) and

col ((Ax, Fj)H,..., (Ax, Fm )H) respectively. We denote by M (resp.Mt) the conjugate (resp. transpose) matrix of M and by {Gt, F) m the

Hm

m X m matrix whose i, j -th entry is the inner product (gi, Fj) H. Notice that (G‘, F) m defines the

j H

matrix inner product and has the properties:

(CGt, F)h = C(G‘, F)h , <Gt, FC)h = <Gt, F)hC and (G‘,F)H = (F‘,G)H t, where C is a mXm constant matrix. Finally, we denote by I m and [0]m the identity mXm and the zero mXm matrix

respectively.

2. Decomposition of linear operators We first rewrite the following theorem from [6] for Hilbert space.

Theorem 2.1 Let the operator B : H ^ H be defined by

Bx = Ax — G{Ax,F‘) m =

H (2.1)

= f, D(B) = D(A), f e H.

where A is a correct operator on H,

F = (Fj,..., Fm) , and G = ( gj,..., gm) e Hm. We

suppose

?k

components

of

F =(Fj,K,Fk) (k < m) are linearly

independent elements while the components of

F„

—k

(Fk+І,к,Fm) are linear combinations of

F^..., F,.

Write

F = (Fk, Fm—k )

Fmt —k = M — F kt where M

' m—k ,k

m—k ,k

is a

(m — k) X k matrix, and G = (Gk, Gm—k ) with Gk =(g!,...,gk ) (k < m)

and

G

—k

(i)

Bx = Ax — GM <x, Fk )Hk = f, D(B) = D( ,4),

(2.2)

where G^^ = G + Gm—kM m—k ,k .

(ii) B is correct if and only if

detL = det[Im — <Ft, G)Hm ] ^ 0

or equivalently

(2.3)

detLk = det[7k - <Fk‘,>Hk ] ^ 0. (2.4)

(iii) If B is correct, then the unique solution of (2.1 ) or (2.2) is given by

x = B—f = A-1 f + ( A “G) X

—І

—І

X [Im — F , G) Hm ]—І { f, F‘ )

or by

x = B ~1f = A ~1f + ( A-GM ) X

X[I, — (Fk ,GM)Hk ]—І'(Fk , f )

(2.5)

(2.6)

k

Remark 2.2. The previous theorem shows that the correctness of the operator B and the solution of Bx = f do not depend on the linear independence of the elements of the vectors F and G . The correctness condition of Bx = f is det L ^ 0 or det Lk ^ 0.

Corollary 2.3 Let A be a correct operator on H and F = ( Fj, k , Fm) and

G = (gj,...,gm)e Hm. Then:

(i) the operator B defined by (2.1) is correct if and only if (2.3) holds true.

(ii) If B is correct, then the unique solution of (2.1) is given by (2.5)

Remark 2.4 It is noted that ker B = {0} if and only if (2.3) holds.

Our main interest is to solve more complicated problems involving an operator of the form

ВІ x = Ax — S <AO x, Ф ф >^

— G<A, Fl > Hm =

= f, D(B^ = D(A)

(2.7)

where A, A0 are correct operators in H and the

vectors S, G, ^ and F e Hm . The case where Bj is a quadratic operator, i.e.

A = A2, $ = F,

S = A0G — G(Ft, A0QHm,

G e D(A0))m has been studied extensively by the authors in [6], [7] and [8]. Here we investigate the problem where Bj is not quadratic but it can be written as a product of two other correct operators BG , BG, i.e. Bj = BGBG .

m

H

m

In this case the solvability condition and the solution are essentially simpler than in the general case.

Theorem 2.5 Let the operators

BG , BG, Bj : H ^ H defined by

BG0 x = A.x - G0 (A.^ Ф ' > Hm = f,

D(BGo) = D( ¿4.),

BgX = Ax - G< ) = f,

H

d( Bg ) = d( A),

Bj x = ( AA0) x - S ( A0 x, Ф t >

G<( AA o) x, Ft > m = f,

(2.s)

(2.9)

(2.70)

D( Bj) = D( AAq)

and there exists a set of linearly independent elements

xj,...,xm e D(AA0) such that

<A)xi, $ j ) Hm = $ij ,i, j m, where 4 and

A are linear operators on H, the vectors G,S,F e Hm and G0 e D(A)m and the

components of the vector $ = ($j,..., $m) are linearly independent elements of H . Then:

(i) the operator Bj can be decomposed in Bj = BG BG if and only if

S g Я(BG )m, and

(2.11)

S = BgG0= A G0 — G< Ff, AG0 > Hm.

(ii) The operator Bj defined by (2.10), where S = BgG0 , 4 and A are correct operators and A is densely defined on H , is correct if and only if the operators BG and BG are correct which means

det[Im - (Фt, G0 ) Hm ] * 0 and

det[Im - < F( , G) Hm ] * 0

(2.12)

(iii) If Bj defined as in (ii) is correct, then the unique solution of (2.10) is given by x = B—' f = ( A—j A-1) f +

+ [(A—A-j)G + AqGqLQj <$(, A-lG)hm ] X (2J3) X L-1 < f,F1)um + A^G,LQj <A—j f,$')um.

Proof. (i) Notice that D(BGBG ) = D(yi/iQ).

We put y = B^ x. Then for each x e D() since (2.9) and (2.8) we have

BGBG0 x = BGy = АУ - G(Ay> F‘ >Hm =

= A( A0 x - G0 < A0 x, Ф ' >„m ) -

G{ F‘, A( a x - G0 < A x, Ф t > Hm )> =

= ( AA0 )x - [ AG0 - G<F‘, A G0 > Hm ] x (2.14)

x < 4 x, Ф t > - G<F‘, ( AA0 ) x> Hm =

H

= (4A0)x - BGG0 <A0x, Ф‘ >Hm -

- G<( A 4) x, F‘ >Tr m

where the relation BgG0 = AG0 — G{F‘, AG0)Hm

follows by substituting x = G0 in (2.9). By comparing (2.j4) with (2.j0), it is easy to verify that BGBG^x = Bjx for each xe D(yA/i,) if and only if

(BgG0 — S)(4x, $‘) m =0. From the last G Hm equation by substituting x = xt in (4 x, $ $) m ,

i Hm since (Aq xi, $ j ) Hm = Si} ,i, j =1,..., m, we get

S e R(BG)m and S = BgG0.

(ii) Let the operator Bj be defined by (2^0)

where S = BgG0 and 4, A are correct operators

and A is densely defined on H. Then Equation (2.j0) can also be written in matrix notation as

Bj x = (44,) x — (BgGq , G) X

x

<( A4) x, A Фt >

<( AA 0) x, Ft > „

= f

(2.15)

or

Bj = Ax - G<Ax, F ) H 2m = f,

(2.16)

and

D(Bj) = D(A), where A = 440, G = ( BgG0, G)

^ &—j

Ff = col(A $f, F1). Notice that the operator

A = 44 is correct because A and 4 are correct

operators. By Corollary 2.3 the operator Bj is correct iff

detLJ = det[12m - <Ff, G)H2m ] Ф 0

By substituting

in

-J

G = (S, G), F = ( A Ф, F ),

this

we take

formula

S = AG0 -G{Ft,AG0)

0/H”

m

H

m

H

det L1 = det

<Ф1, Go — A G<F1, A Gq > h” >„m — Im <Ф1, A G)

<F1, A Gq — G< F1, A Gq > „m >

<F1, G>

det

(Ф1, Oq ) „m — (Ф1 , A -G) Hm (F1 , AOq ) „m — Im (Ф ' , A“G)

(F1, A Oq ) „m — (F1, O) „m (F1, A Oq )

0/ H”

(F1, O) и

I

Multiplying the elements of the second column by

{F‘, AG0)Hm and adding to the corresponding elements of the first column of the determinant Lj we

take

det LІ = det

{Ф1, Go >h” — I” {Ф1, A G>„m

[0]” {F1, G> „m — I”

det[Im - <S, Go>Hm ] • det[Im - <Ff, G>Hm ].

By Corollary 2.3 (i) the operator BG (resp. BG )

is correct iff det L = det[Im — (F‘, G)Hm ] ^ 0

(resp. det L, = det[Im — <$f, G0)Hm ] ^ 0. So det L = det L0 • det L and by Corollary 2.3 Bj is correct iff det L0 ^ 0 and det L ^ 0 or iff BG, B^ are correct operators.

(iii) Let xe D(^^¿4,) and BGBG^x = f . Then

by Corollary 2.3 (ii) since BG, B^ are correct operators, we obtain

B—x = B—f = A ~lf + (A-1G)L-1 < f, F1 >

and

Hm

further

x = BG1 ( A ~1f + ( A-G) L-1 < f, Ff > Hm ). Denoting

0 H

by g = A~lf + (A”G)L_1 <f, Ff>Hm from the last equation we get

x = bg; g = Ao-1 g + ( Ao-Go) l-1 < g, s( > Hm =

= ,4o-1(A-f + (Â-1G)L-1 < f, Ff >Hm ) +

H

j <$t, ¿4—f +

+ (AqG0) LQj , j j =

+ ( A G)L~— < f, F1 )^ ^

= (A— A_1) f + (A— A_1 )GL_ 1 < f, F1 > Hm +

+ ( A,Go) LQj[< A~lf, $1> Hm +

+ <$ 1 , ji~1G> HmL-j < f, F1 > Hm ]

which implies (2.j3). So the theorem has been proved.

Corollary 2.6 Let the operator Bj be defined by (2.10), where 4 and A are correct operators and

A is densely defined on H, then:

(i) there exists the unique operator BG , defined by (2.9).

(ii) If ker BG ^ {0} and S e R(Bg)m then for each solution G, of the equation B—G, = S there exists an operator BG defined by (2.8) such that

Bj = BGBG0 , Le. we have many decompositions of Bj . 0

(iii) If kerBG ={0} and S e R(Bg)m then

there exist a unique vector G0 = BG S G D(A)m and a unique operator BG defined by (2.8) such that

B; = BgBg , i.e. the decomposition of B; is unique.

Proof.

If

ker BG Ф {0} then,

since

S G R(BG )m, the equation BgG0 = S has many solutions G0 G D(A)m and so there exist many operators BG defined by (2.8) such that B; = BGBG . If ker BG ={0} then from BgG0 = S the vector G0 is defined uniquely by

G0 = BG jS and the factorization of Bj = BG BG is unique.

From the previous theorem and corollary the next main theorem follows, which is useful in applications and gives a criterion for the correctness and the solution of (2J0) in an elegant way.

Theorem 2.7 Let the operators

A,, A,Bj: H ^ H, the operator Bj defined by

B1 x = ( AA) x — S < A x, Ф1 >

G<( AAq) x, F‘ > „m = f,

(2. 17)

D( B^ = D( AAq),

m

H

m

m

m

H

m

H

m

m

H

where 4 and A are correct operators and A is densely defined on H, the vectors S, G, F e Hm and the components of the vector $ = ($j,..., $m ) are linearly independent elements of H. Then:

(i) the operator Bj has the unique decomposition Bj = BG BG , where BG and BG are defined by

(2.8) and (2.9) respectively, if and only if

det L = det[Im — <F‘, G) Hm ] * 0 (2.18)

and

Go = A~'S + ( A-G)l- 1WJ) Hm, (219)

Gq e D(A)m.

(ii) if Bj has a unique factorization then Bj is correct if and only if

det Lo=det[Im — <$1, Gq > Hm ] =

= det[ Im — <$1, A~XS > Hm— (2.20)

— <$1, A~lG> Hm L 1 <Fl, S> Hm ] * 0,

(iii) if Bj has a unique factorization and Bj is correct then the problem (2.17) has a unique solution given by (2.13).

Proof. (i) Let for Bj there exists the unique factorization Bj = BGBG . Then, by Theorem 2.5 (i),

BgG0 = S holds. From the uniqueness of BG

defined by (2.8), the correctness of 4 and linear independence of $j,...,$m follows the uniqueness of the solution G0 of equation BgG0 = S. Then kerBG ={0} and G0 = BG1S. By Remark 2.4, det L * 0 and by Corollary 2.3 (ii) B~lf = A~lf + (A-G)L-1 < f, F1)Hm ,Vf e H. Using this formula, we get

G0 = B—S = A^S + (A~lG)L~j < F1, S ) Hm.

Conversely, let (2J8) and (2J9) hold true. We remind that the operator BG is defined by the formula

(2.8). From (2.9), taking into account (2J9), we get BgG0 = Bg ( A-'s + (AlO) Lj < F1, S ) Hm)

= ^( A-'S + (AlG)Lj < F1, S) Hm) —

— G(Ff, A(ii-'S + (A~lG)Lj (F1, S>Hm )>Hm = = S + GL"1 (F1, S > Hm —

— G< F1, S > Hm —G{ F1, G) HmL— {F1, S > Hm = S + + G[Im — <F1, G>Hm ]L1 <F1 , S>Hm —

— G{ F1, S > Hm = S.

So S = BGG0 and by Theorem 2.5 (i) Bj = BGBG . By Corollary 2.3 (i) and (2.j8) the

operator BG is correct. Then G0 = BG jS and by Corollary 2.6 (iii) the factorization Bj = BGBGis unique.

(ii) By Theorem 2.5 (ii) Bj is correct if and only if (2.j2) holds true. Substituting G0 from (2.j9) into

the first inequality of (2.j2) we obtain (2.20).

(iii) Is proved in Theorem 2.5 (iii). So the theorem is proved.

3. Eigenvalues and eigenvectors of operators

In this section we deal with the problem of finding the eigenvalues and eigenvectors of the

Bx = Ax — G{Ax, F‘) = f,

operators H the

D(B) = D(^4), f e H,

quadratic operator B2 , the biquadratic operator B4 and the product Bj = BG BG when the eigenvalues

and eigenvectors of the operator A are known and certain conditions are fulfilled.

Recall from [j0] the following theorem and lemma.

Theorem 3.1 We consider the operators A, B, BSG : H ^ H, where

Bx = Ax — F ) Hm = f,

Hm , (3.1)

D(B) = D(^4), f e H

BSGx = x — S (^4x, F ) Hm —

— G(,42 x, F1) Hm = f, , (3.2)

D(Bjg ) = D( ^42)

A is a correct and densely defined operator, vectors S = (sj,K, sm ) e Hm

G = (gj,K, gm ) e D(A)m and the components of the vector F = ( Fj,k, Fm) are linearly independent on H . Then:

(i) BSG is quadratic, i.e. BSG = B2 if and only if

G e D(ii)m, S = iiG — G(F‘,iiG> m. (3.3)

(ii) For each G e D(A)m and

S = AG — G(F‘, AG) m, BSG is correct if B is correct which in turn is correct if

detL = det[Im — <F‘, G)Hm ] * 0. (3.4)

(iii) If BSG is quadratic and correct, then the unique solution of (3.2) is given by

x = BS—G f = A -2 f+[ A -2g+

+ (A _1G) L"1 < F1, ii _1G> Hm ]L_1 X (3.5)

X < f, F1 > Hm + (A -G) L-1 < ii "jf, F1 > Hm.

Lemma 3.2 Let the operators ii, B and the vectors G, F are defined as in Theorem 3.1. Then the statements hold true:

(i) D( B 2) = D( A2) if and only if G e D(ii)m.

(ii) If G e D(A)m then the operator B2 : H ^ H is defined by

B2 x = A2 x — BG( A x, F1 > —

— G( A2 x, F1 > Hm = f, (3.6)

D( B 2) = D( A2),

where

BG = A G — G(Ff, A G) Hm. (3.7)

Note the correctness and density of A and the linear independence of components of vector F apply only to the statement (i), while for the statement (ii) this condition can be loosen as in the following corollary.

Corollary 3.3 Let the operator B be defined by (3.1) where now simply F = (Fj,..., Fm ) e Hm, A is a linear operator (in general can be not correct and not densely defined ), G e D(A)m and

D( B 2) = D( A2 ) . Then the operator B2 is defined by

B2 x = A2 x — BG<Ax, F1 >flm —

— G< ii2 x, F1 > Hm = f,

where

BG = AG — G<F1, AG) Hm,

x /H (3.8)

D(B 2) = D( A2).

Also, from [9] recall the following lemma which also holds true for the case where

F = (Fj,., Fm) e Hm instead of

F = (Fj,k,Fm)e D(i44)m.

Lemma 3.4 Let the operators B, B4 : H ^ H be defined by

Bx = Ax — G(A x F‘) = f,

H (3.9)

D(B) = D( ii),

B4 x = ii4 x — V < ii x, F‘) —

H

— F < ii2 x, F1) Hm — S < ii3 x, F1) —

H H (3.10)

— G< ii4 x, F1) Hm = f,

D(B4) = D( ii4), where A is a linear operator on H, G is a vector of D(i4 )m, the vectors V, F, S, G satisfy the equations

V = iiF — G(F‘,iiF>Hm ,

F = iiS — G(Ff, AS>Hm, (3.11)

S = ii G — G( F1, ii G> Hm

and the components of the vector F = (Fj, k , Fm )

belong to D(ii4) . Then B4 = B4, i.e. B4 is a

biquadratic operator.

Now we prove the following theorem concerning the eigenvalues and eigenvectors of the operators B, B2 and B4 .

Theorem 3.5 We consider the operators ii,B,B2,B4: H ^ H, where ii is linear and B, B2 and B4 are defined by

Bx = i/x — G< ^ F‘) = f,

H

D(B) = D(ii), F,Ge Hm,

B2 x = A x — S (^l1, F ) Hm — G{ A2 x, F1) Hm = f, D(B2) = D( ii2), F e Hm, Ge D(Am),

S = AG — G{AG, F1)Hm,

B4 x = A4 x — V < A x, F > Hm — F < A2 x, F1 > Hm — S < A3 x, F1 > Hm — G< ,i4 x, F1 > Hm = f,

D(B4) = D(A4), F e Hm

Ge D(A )m, 5 = AG -G<Ff, AG)

Y = AS - G(Ff, AS ) Hm,

V = A F — G{F1, A F) Hm.

If * is an eigenvalue and x0 is the corresponding eigenvector of an operator A, i.e.

A x0 = Ax0 and

G = (gl,..., gm ) = (k X0,..., kmXoX V- km e C ,

then x0 is also the eigenvector of operators B, B2 , B4 and the numbers

1(1 - <*0. F > „m ).

i=1

m

1(1 - £*, < *0. F,> )2

,=1

m

14(1 - Tf, < V F, > Hm )4

,=1

are the eigenvalues of the operators B , B2 and B4 respectively.

Proof. Notice that, by Corollary 3.3, B2 is quadratic, B2 = B2 and, by Lemma 3.4, B4 is biquadratic and B4 = B 4 . It is easy to verify that

m

Bx0 = AQ — Yki <^ F > Hm )^

and then

,=1

B 2 *0= 1(1 - fk, < *0. F > Hm )2 *0.

,=1

m

B 4 xo= l4d — ^ < xo. F, > Hm )4 x„.

i=j

So the theorem has been proved.

For the case Bj = BGBGthe following theorem

is proved, which provides the eigenvalues and eigenvectors of the operator Bj .

Theorem 3.6 Let the operators BG , BG, Bj ,4,

A and vectors F and $ be defined as in Theorem 2.5 above . We also suppose that x0 is an eigenvector

of both operators 4 and A, the numbers ao,a are

the eigenvalues of the operators 4 and A respectively. Finally, let

G = (kj Xo,..., km x0 ) = kX0,

G0 = (lj x0 ,..., lm x0 ) = lx0 and

S = (slXo,..., smxm ) = SX0, ki, li , si e C ,

i = j,...,m. Then x0 is also the eigenvector of the operator Bj and:

G) ao[a(j — k( X0, F > Hm ) — X0, $1 > Hm ]

is its corresponding eigenvalue

(ii) Bj = BG BG if and only if

r = a(j — x0,F) m )R. The number

m

aa(j — Ek<Xo,Fi>Hm)X

i=j (3.12)

m

X a—El, < * j > „m)

j=j

is the eigenvalue of Bj = BG BG .

(iii) Bj has the unique decomposition

Bj = BG BG if and only if

k(X0, F > „m * j and

j r

Go=-d — k( Xo, F1 > „m)-j S. a H

Proof. It is easy to verify that X0 is the

eigenvector of the operators BG , BG , Bj and

BGBG^. In particular, by substitution in to (2.8), (2.9) and (2JQ) we get

Bg0 xo = ao(j — R<Xo,$1 >„m )xo,

BG X0 = a(j — k( F‘ ) „m ) Xo ,

Bj xq = a0[a(j — k( XQ , F‘ ) „„ ) — r< XQ , $1 ) „„ ]xQ ,

BGBG0 X0 = aoa(j — Tm ki (X0 , F' )„m ) X

X (j — {XQ, $ j ) „m ) X0

Since $j,..., $ m are linearly independent elements then from Theorem 2.5 it follows that

Bj XQ = BgBgq x0 iff s = a(j — k{ XQ, F ) „m )l .

The vector l (i.e. G0 ) is defined uniquely as

iff

- j -

l=-a—k<xo,f1) )—j-

a H

k{X0,F) m * j. Then the operator BG is unique

H 0

and so the factorization of Bj is also unique. From the

m

H

m

decomposition B1 = BGBG follows immediately (3.12).

4. Examples

In the following H*(0,1) (resp. H 2(0,1)) denotes the Sobolev space of all complex functions of L2 (0,1) which have generalized derivatives up to the first (resp. second) order that are Lebesque integrable.

In [10] has been proved that the operator

A: L2 (0,1) ^ L2(0,1) defined by

A u = iu' = f, D( A) =

= |u(0 e H1 (0,1): u(0) + u(1) = 0} is correct and self-adjoint and that the unique solution u of the problem (4.1) is given by the formula

a~if=2 j>(x)dx -ij/(x)dx (42)

for all f e H.

^ 2

Then easily follows that the operator A defined

(4.1)

by

Au = —u " = f,

D(A2) = {u є H2(0,i) : u(0) +

(4.3)

+ u(j) = 0,u,(0) + u ,(j) = 0} is correct and self-adjoint too and for every f e L2(0,j) the unique solution u of the problem (4.3) is given by the formula

u = A—2 f = —| (t — x) f (x)dx +

j f! 0 (44 + - jo(2t — 2 x + j) f (x)dx.

Remark 4.1 It is easy to verify that Ak = (2k — j)n, k e Z are the eigenvalues of the

operator A and vk = e

—i (2 k —i)nx

are the corresponding

eigenvectors.

Example 4.2 The operator

Bj : L2(0,j) ^ L2(0,j) corresponding to the

problem:

8i

Bju(t) = — u" (t) — 2i(n — ^y)e~iM X

n

X [u'(x)(x2 — x)dx +

•Ю

+ 2e~1M [u"(x)(x2 — x)dx = f (t)

JO

D(Bi) = {u є H2(0,i) : u(0) + + u (І) = 0, u '(0) + ur (І) = 0}

(4.5)

(4.6)

is quadratic and correct. The eigenvalue of Bj is 0 / 8i , 2

A = (n--) and its corresponding eigenvector i

is

v = e m . The unique solution of the problem (4.5)-(4.6) for every f e L2(0,j) is given by the formula

u(t) = — J (t — x) f (x)dx

+

+—Jo(2t — 2 x + І) f ( x)dx + 2n4 e

• (~б—8-)2 fQ(x2 — x)f (x)dx

-7t6 л. &A. JQ

п + б4 jQ

_2 —ілї _3

+

(4.7)

+ -

in"eп + 8i fi

б п6 + б4

Jo(i + 4 x3 — б x2) f ( x)dx.

Proof. First we show that the operator Bj is a

BSG operator. If we compare equation (4.5) with

^ 2 //

equation (3.2), it is naturally to take A u = — u , with D(^i2) = D(Bj), m = j, F = t2 — t. Then A is defined by (4.j) and

(^i u, F) H = ^iu (x)( x — x)dx,

^ 2 f j 2 (^i u,F)H = —I u (x)(x — x)dx. The last

J0

equation implies

I u (x)(x — x)dx = —(A u, F)H. Substituting

0

these formulas into (4.5), we obtain:

Bju = —m" — 2(n — -82)e ~ia < ,4 u, F > „ —

п

— 2e -,пг < ,A2m, F ) H = f (t )

(4.8)

Again comparing (4.8) with (3.2), we get 8i

S = 2(n — -8T)e ~im, G = 2e -ini. (4.9)

n

It is easy to verify that G e D(^i). By simple calculation we find A G = 2ne im ,

{F‘,G)H = 8і/п , {F1, AG)H = 8і/п . Then

8i

G — G< F1, yi G> „m = 2ne"int — 2e-m^- = S

n

and by Theorem 3.j the operator Bj is quadratic. From the above and (3.4) it follows that

det L = det[Im — <F1, G)„ ] = (n3 — 8i)/n3 * 0.

By Theorem 3.j, the operator Bj is correct. Now we find the solution of (4.5). It is evident that

L—

п (п + 8i) пб + б4

If we

substitute

f = G = 2e m in (4.2) and (4.4), we receive

A lG = 2e ш/ж and A ZG = 2e !П/п/. Then

-2/

-in /_ 2

(Ft, A _iG) H =8i/n4 and

<A~lf,F)h = -^ [Vi + 4x3 -6x2) f (x)dx.

I / •'U

and

from

j2Jo

We also have (f, F )„ = J (x2 — x) f (x)dx.

From the above and (3.5), (4.4) follows the solution (4.7) of problem (4.5)-(4.6). Notice that, by Remark

4.1, A = n is the eigenvalue of A and vj = e~1M is

the corresponding eigenvector of A. Then, since G = 2e~iM = 2vj

(vj, F) „ = [ e~inx( x2 — x)dx = 4i/n3,

JO

Theorem 3.5 it follows that the number

A(1 — 2<vj, F) „ )2 = n2(1 — 8i/n3)2 =

= (n — 8i/n2)2

is the eigenvalue of the operator Bj and vj = e~m is the corresponding eigenvector.

Example 4.3 The operator

Bj : L2(0,j) ^ L2(0,j) which corresponds to the problem

Bju = u" — (ncos nt + 2e1 /n) X

f1 , t f1 2 „ (4.10)

X J^xu (x)dx — e J x u (x)dx = f (t),

D(Bj) = {u e W22(0,j): u(1) = 0,

u (0) = 0} is correct and its unique solution is given by

u(t) = ([(x — t )n( x — t) +1 — j] f (x)dx +

Jo

j r t ^ cos nt +1

+-------[e — t — e + j--------------] X

3 — e 2(n — j)

X [ x2f (x)dx — cosnt +j f (j — x)f (x)dx.

Jo n — j Jo

Proof. If we compare equation (4.10) with equation (3.2) we see that the operator Bj is not BSG operator and so is not quadratic, but it can be the decomposition of two correct operators BG , BG,

defined by (2.8) and (2.9) respectively. We refer to

Theorem 2.7. If we compare equation (4.j0) with

^ ^ //

equation (2.17) it is natural to take AA00u = u with

D( AA0) = D( Bj), A0m = u' (t), m = j, F = t2, $ = t, G = e1,.

S = n cos nt + 2c1 /n

(4.77)

(4.72)

Then, since (4.jj)

D(A0) = {u e W2 (0,j): u(j) = 0} and we can

take A to be defined by Au = u'(t),

D(A) = {m e W2 (0,1): u(0) = 0}. By simple calculations we find

det L = det[/m - <Ft, G> h ] =

= i - f x2exdx = 3 - e Ф 0,

Jo

\nx +

(F‘, S > H = [ x 2(ncos,

•Ю

+ 2ex/n)dx = (2e - 6)/n It is easy to verify that

A -lf (t ) = J/ ( x)dx, ,4o-^.f (t ) = Jf ( x)dx A ~lS = [ (ncos nx + 2ex/n)dx =

•Ю

= 2(et - i)/n+sin та

A iG = Jexdx = et - i.

Jo

Then L

-i

3-e

and by (2.19) we get:

Go = 2(et - i)/n + sin П + (et -

i 2e - 6 i)-----------------= sinnt.

3 — e n

Substituting this into (2.20) we obtain

det L0 = j — [ x sin nxdx = n—1 * 0. Hence, by 0 Jo n

Theorem 2.7 the operator Bj is correct.

Now we find the solution of (4.10)-(4.11). By

simple calculations we find L0 =

ж - i

A0 G0 = -(cos n + i)/n,

(Ao-i A-i )f (t) = J0[(z -1)n(z -1) +1 - i]f (z)dz

where

is

П( x) =

i x >0 0 x < 0

the

Heaviside’s

function,

(A-A )G(t) = et -1 - e + i,

and

From the above

<$1, A G) „ =1/2

(A -1, $1 > „ = £(1 — X) f (x)dx.

and (2.13) we get (4.12).

A comment from the first and second authors: This paper has been initiated together with our friend

i

P. C. Tsekrekos who passed away from a heart attack and to whom we express our deepest sorrow for his sudden death.

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Submitted to the Journal 02 December 2011.