DECOMPOSITIO N OF ABSTRACT LINEAR OPERATORS ON BANACH SPACES Текст научной статьи по специальности «Математика»

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udc 338.984 Articles

doi:10.31799/1684-8853-2021-2-2-12

Decomposition of abstract linear operators on Banach spaces

K. D. Tsilikaa, PhD, Assistant Professor, orcid.org/0000-0002-9213-3120, [email protected] aUniversity of Thessaly, Hellenic Open University, 78, 28hs Octovriou St., 38333 Volos, Greece

Introduction: The majority of the known decomposition methods for solving boundary value problems (Adomian decomposition method, natural transform decomposition method, modified Adomian decomposition method, combined Laplace transform — Adomian decomposition method, and Domain decomposition method) use so-called Adomian polynomials or iterations to get approximate solutions. To our knowledge, a direct method for obtaining an exact analytical solution is not yet proposed. Purpose: Developing, in an arbitrary Banach space, a new universal decomposition method for the class of ordinary or partial integro-dif-ferential equations with non-local and initial boundary conditions in terms of the abstract operator equation B1x = f. Results: A class of integro-differential equations in a Banach space with non-local and initial boundary conditions in terms of an abstract operator equation B1x = Ax - S0F(Ax) - G00(Ax) = f, x = D(Bj)has been studied, where A, A are linear abstract operators, Sg, G0 are vectors and 0, F the functional vectors. Usually, A, A are linear ordinary or partial differential operators, and F(Ax), &(Ax) are Fredholm integrals. The existence and uniqueness are proved under the assumption that the operator B1 has a decomposition of the form B1 = B0B with B and B0 being different abstract linear operators of special forms. The proposed decomposition method is universal and essentially different from other decomposition methods in the relevant literature. This method can be applied to either ordinary integro-differential or partial integro-differential equations, providing a unique exact solution in closed analytical form in a Banach space. The stages of the method are illustrated by numerical examples corresponding to specific problems. Computer algebra system Mathematica is used to demonstrate the solution outcomes and to assess the effectiveness of the analysis. Practical relevance: The main advantage of the proposed solution method is that it can be integrated in the interface of any CAS software in an easy, programing-free way.

Keywords — correct operator, decomposition (factorization) of operators (equations), integro-differential equations, boundary value problems, exact solution.

For citation: Tsilika K. D. Decomposition of abstract linear operators on Banach spaces. Informatsionno-upravliaiushchie sistemy [Information and Control Systems], 2021, no. 2, pp. 2-12. doi:10.31799/1684-8853-2021-2-2-12

Preliminaries and auxiliary results

Integro-differential equations are used in many problems from science and engineering. The inte-gro-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 boundary value problem (BVP) can be transformed into a simpler one involving simpler operators and thus the solution can be found easier.

The decomposition (factorization) methods were used in many applications in gas dynamics, transport theory, electromagnetism, quantum physics, mechanics, hydrodynamics and cosmology [1-14]. In pure mathematics, decomposition (factorization) method continues to be a very successful tool for solving variational inequalities, linear and nonlinear ordinary and partial differential and Volterra — Fredholm integro-differential equations as well as systems of partial differential equations. This method is very important for solving fuzzy Volterra — Fredholm integral equations, integro-differential equations of fractional order and delay differential equations [15-32]. However, almost all the approaches of the literature listed above do not give exact solutions in their closed analytical forms and the corresponding problems are

not formulated in terms of abstract operator equations. Thus, the decomposition methods proposed and employed in these problems are not universal.

Exact solutions in their analytical form for abstract operator equations in Hilbert and Banach spaces were obtained by quadratic and biquadratic decompositions of the integro-differential equations in [33-37]. The universal decomposition method for the abstract linear operator equation

B1x = A2x - SF(Ax) - GF(A2x) = f, x e D(B1)

was given in [38] on a Hilbert space. We note that Banach spaces play a central role in functional analysis and it is important to study the exact solutions of correct BVPs in the context of Banach spaces. This work is a natural continuation of [38] to a Ba-nach space and introduces the universal decomposition method for the similar linear abstract operator equation

B1x = Ax - S0F(Ax) - G0O(Ax) = f, x e D(B1), (1)

where <A, A are linear abstract operators; S0, G0 are vectors and O, F — functional vectors. The decomposition method proposed here is different than the well-known decomposition methods (namely the Adomian decomposition method, the natural trans-

form decomposition method, modified Adomian decomposition method, the combined Laplace transform — Adomian decomposition method and domain decomposition method). In the relevant literature, the so-called Adomian polynomials or iterations were used to obtain numerical solutions (see [1-32]). The class of integro-differential equations with nonlocal boundary conditions described by an abstract operator equation is studied in [39], where all calculations are reproducible in any program of symbolic calculations and the computer codes in Mathematica are given.

In the sections that follow we use the following notations, definitions and statements.

We denote by X a complex Banach space and by X the adjoint space of X, i. e. the set of all complex-valued linear and bounded functionals f on X. We denote by f(x) the value of f on x.

We write D(A) and R(A) for the domain and the range of the operator A, respectively. An operator A: X ^ X is called correct if R(A) = X and the inverse A-1 exists and is continuous on X. If for an operator B1 there are two operators B0, B such that B1 can be written as a product B1 = B0B, then we say that B0B is a decomposition (factorization) of B1 and write B1 = B0B. An operator B1: X ^ X is called quadratic (biquadratic) if there exists an operator B: X ^ X such that B1 = B2, (B1 = B4) and the corresponding decomposition B1 = B2, (B1 = B4) is called quadratic (biquadratic). Recall that the problem Ax = f is called correct, if the operator A is correct. If x, gi e X and O; e X*, i = 1, ..., m then we denote by g = (g1, ..., gm), O = col(O1, ..., Om) and O(x) = col(O1(x), ..., Om(x)) and we write g e Xm, O e Xm. We will denote by O(g) the m x m matrix whose i, j-th entry Oi(gy) is the value of functional Oj on element g. Note that O(gC) = O(g)C, where C is a m x k constant matrix. We will also denote by 0m and Im the zero and identity m x m matrices.

Next, we state some useful outcomes. Specifically, Theorem 1 from [40] and Corollary 3.11 from [33].

Theorem 1. Let X, Y and Z be Banach spaces and A0: X ^ Y be a correct operator with D(Aq) c Z c X.

Further let the vector G0 g^,..., gl-^eYm

and the column vector O = col(^, ..., (m), where ..., (m e Z* and their restrictions on D(A0) are linearly independent. Then:

(i) The operator B0: X ^ X defined by

BqX = AqX - GqO(x) = f, D(Bq) = D(Aq), f e X, (2) is correct if and only if

detL 0 = det

I -O

■"■m ^

(Vg o)

* 0.

(3)

(ii) If B0 is correct, then for any f e Y, the unique solution of (2) is given by

x = B-f = A-f + A-1GoL-1o(A-1f. (4)

Corollary 1. Let A be a correct operator on a Banach space X and the components of the vectors G = (g1, ..., gm), F = col(F1, ..., Fm) are arbitrary elements of X and X*, respectively. Then the operator B: X ^ X defined by

Bx = Ax - GF(Ax) = f, D(B) = D(A), f e X (5) is correct if and only if

detL = det[Im - F(G)] * 0. (6)

If B is correct, then the unique solution of (5) for every f e X is given by

X = B-1f = A-1f + A-1GL-1F(f). (7)

Decomposition of abstract linear operators on a Banach space

In this section we investigate problem (1) where B1 is not quadratic but it can be written as a product of two other correct operators B0, B i. e. B1 = B0B. In this case the solvability condition and the solution formulation are essentially simpler than in the general case.

We will prove the following theorem using the technique that was first applied for the case of Hilbert space in Theorem 2.5 [38], where a given operator B0 of the type B0x = A0x - G0O(A0x) = f, x e D(B0) and an operator A is densely defined. We use a different operator B0 without the assumption of density of D(A) on X.

Theorem 2. Let X and Z be Banach spaces, ZcX,

the vectors G = (gp ..., gm), G- g^,...,

S- =^s(-^,...,s^jeXm, the components of the

vectors F = col(F1, ..., Fm) and O = col(O1, ..., Om) belong to X* and Z*, respectively, and the operators B0, B, B1: X ^ X defined by

Box = Aqx - GqO(x) = f, D(Bq) = D(Aq) c Z; (8)

Bx = Ax - GF(Ax) = f, D(B) = D(A); (9) B1x = A0Ax - S0F(Ax) - G0O(Ax) = f,

D(B1) = D(AqA), (10)

where A0 and A are linear correct operators on X; G e D(A0)m and the restrictions of O1, ..., Om on D(A0) are linearly independent. Then the following statements are satisfied:

(i) If

S0 e R(B0)m and S0 = B0G = =A0G - G0®(G),

(11)

then the operator B1 can be decomposed in B1 = B0B.

(ii) If in addition the components of the vector F = col(Fp ..., Fm) are linearly independent elements of X* and since the operator B1 can be decomposed in B1 = B0B, then (11) is fulfilled.

(iii) If the operator B1 can be decomposed in B1 = B0B then B1 is correct if and only if the operators B0, and B are correct which means that

det L o = det

I -O

Am *

(a~-1g 0)

* 0 and

detL = det[Im - F(G)] * 0.

(12)

(iv) If the operator B1 has the decomposition in B1 = B0B and is correct, then the unique solution of (10) is

x = B[lf = A~lA0lf + A"1GL"1F (A0lf j + + + A"1GL"1F(A10-1G0j^Cf). (13)

Proof: (i) Taking into account that G e D(A0)m and (8)-(10) we get 0

D(B0B) = {x e D(B): Bx e D(B0)} = = {x e D(A): Ax - GF(Ax) e DA)} = = {x e D(A): Ax e D(A0)} = D(A0A) = D(BX).

So D(B1) = D(B0B). Let y = Bx. Then for each x e D(A0A) and taking into account (8) and (9) we have

BqBX = B0y = A^y - G0®(y) = = A0[Ax - GF(Ax)] - G0O(Ax - GF(Ax)) = = AoAx - AqGF(Ax) - G0O(Ax) + G0O(G)F(Ax) = = AqAx - G0O(Ax) - [A0G - G0O(G)]F(Ax) = = A0Ax - B0GF(Ax) - G0O(Ax), (14)

where the relation B0G = AqG - G0O(G) results naturally from (8) by substituting x = G.

By comparing (14) with (10), it is easy to verify that B1x = BqBx for each x e D(A0A) if a vector S0 satisfies (11).

(ii) Let the operator B1 can be decomposed in B1 = B0B. Then by comparing (14) with (10) we obtain

(B0G - S0)F(Ax) = 0.

(15)

Because of the correctness of operators A, A0 and the linear independence of F1, ..., Fm, there exists a system x1, ..., xm e D(A0A) such that F(Aqx0) = Im where x0 = (x1, ..., xm). By substituting x = x0 into (15) we get S0 = B0G. Hence S0 e R(B0)m and S0 = B0G = A^G - G0®(G).

(iii) Let the operator B1 be defined by (10) where S0 = B0G. Then equation (10) can be equivalently represented as a matrix equation:

(

B^x = Ao Ax - ((G, Go )

F (A-1 Ao Ax) Ao Ax)

\

= f, (16)

or

B1 = Ax - GF (Ax) = f, D(B1) = D(A), (17)

where

A = AAq; G = (B0G, G0);

( F (Ax)

IF = col (F, 0 ), FF (Ax )--

0 (Ax )

then

F (v ) =

F (v )

0 (v ),

(

F (A-1v) 0 (A-1v)

A

Notice that the operator A = AA0 is correct, because of A and A are correct operators, and the functional vector F is bounded, since the vectors F, 0 are bounded as a superposition of a bounded functional F, O respectively and a bounded operator A-1. Then we apply Corollary 1. By this corollary the operator B1 is correct if and only if

detLn = detl I2m - F

= det

( I 0 ^

V0m ImJ

[l2m - F (G)] = ( F(B-G) F(G-)A

= det

(im -f(g - Ao-1go0

= det

O (BoG ) O (Go ) (g)) -f(Ao-1go ) 0(g-Ao-1go0(g)) im-®(a--1 go) l-f(g)+f(Ao-1go )0(g) -f(Ao-1go ) -0(g) + 0(Ao-1go )0(g) im -0(a--1g-)

According to properties of determinants of matrices (Remark 1, [34]), taking L1 in the last formulation from above and adding O(G) times the second

column of L1 to its first column, the determinant is unchanged. We then get

(Im - F (G) -F ((-1G-) ^ 0m Im -O(A-1Go) = det[Im - F(G)]det

detL1 = det

Im -O(a0-1G-)

= detL0 detL ^ 0.

So we proved that the operator B1 is correct if and only if (12) is fulfilled.

(iv) Let x e D(A0A) and B0Bx = f. Then by Theorem 1 (ii) since B0, B are correct operators, we obtain

Bx = B-1f = A-1f + A-1G-L-)1o(A01f), B"1 ( A-1f + A-1G-L--1o(A-1f )).

x =

In the last equation we denote by g = A-1f + + A-1G0L"-)1o(A01f. Following strictly Corollary 1 (ii), we get

x = B_1g = A _1g + A _1GL-1F (g) = = A-1 (A-1f + A01G-L-)1o(A01f )) + + A "1GL-1F (A-1f + A01G-L-)1o(A01f)) = = A-1 A-1f + A-1 A01G-L-)1o(A01f ) +

+ A _1GL-1

F (A-1f) + F (a-1G- )L"01o(A01f)

which implies (13). Thus, the theorem has been proved.

The next theorem is useful for applications. Theorem 3. Let X and Z be Banach spaces, ZcX the vectors G0 =^g1-),..., g1-) S0 =^s1-),...,s1-)^|eXm, the components of the

vectors F = col(F1, ..., Fm) and O = col(O1, ..., Om) belong to X* and Z*, respectively, the operators A, A, B1: X ^ X and the operator B1 defined by

B1x = Ax - S0F(Ax) - G0O(Ax) = f, x e D(B1), (18)

where A is a correct m-order differential operator and <A is a re-order differential operator, m < n. Then the next statements are fulfilled:

(i) If there exist a bijective n - m order differential operator A0: X ^ X and the vector G such that

A = A0A, D(B1) = D(A0A), D(A0) c Z; (19)

detL 0 = det

I 0®

(a-1G -)

* 0;

(20)

g = ao1so + vgolo

1o(a-1s-),

(21)

and the restrictions of O1, ..., Om are linearly independent on D(A0), then the operator B1 is decomposed in B1 = B0B, where B0, B are given by (8), (9), respectively, the operator B0 is constructed by the triple of elements A0, O, G0 from (18)-(20), and the operator B by the operator A and vector F from (18) and the vector G from (21).

(ii) If in addition to (i) A0 is correct, then B1 is correct if and only if

= det

detL = det[Im - F (G)] = - F (a-1S-)-F (a-1G-)

-1o(ao1s-)

* 0,

(22)

and the problem (18), (19) has the unique solution given by (13).

Proof: (i) If a bijective n - m order differential operator A0 and a vector G exist satisfying (19)-(21), then from (18) we get

B1x = A0Ax - S0F(Ax) - G0O(Ax) = f,

x e D(A0A). (23)

From (23) we take the operator A and vector F, whereas from (21) we take a vector G and construct the operator B according to the formula (9). To determine the operator B0 by the formula (8), we take from (23) the operator A0 and the vectors O, G0. We proved in the previous theorem (i) that D(B0B) = D(A0A) = D(B1). Substituting (21) into (8) we obtain

bog = B0 = A- G-O

A01S- + ao1g-l-o1o(ao1s-) A01S- + ao1g-l-o1o(ao1s-) o (A-1S- + A-1G-L--1®(a01S- ) S- + g-L-)1o(ao1So )-G-o(a-1S-) - g-o(a-1g- )l-o1o(ao1s- ) =

= S- + G-

I oO

(a-1g-)

x l-)1o(ao1So )-g-o(a-1s- ) = S-.

S0 = B0G and from (23) for S0 = B0G and every x e D(B1) we get

B1x = A0Ax - B0GF(Ax) - G0O(Ax) = = B0Ax - B0GF(Ax) = B0[Ax - GF(Ax)] = B0Bx.

Thus we obtained the decomposition B1 = B0B.

(iii) If the statement (i) holds, then B1 can be decomposed in B1 = B0B. By Theorem 3 (iii), B1 is correct if and only if (12) holds or, taking into account (20) and (21), if and only if detL = det[Im - F(G)] * 0, or if and only if (22) is fulfilled. The last inequality immediately follows by substitution (21) into detL = det[Im - F(G)]. Since B1 is correct and decomposed in B1 = B0B, by Theorem 2 (iv), we obtain the unique solution (13). So, the theorem is proved.

Remark. Usually as a Banach space X we have C[a, b] or Lp(a, b) and as a Banach space Z we have Ck[a, b] or Wpk = (a, b), k = 1, ..., n.

Numerical examples

Let us examine several examples where our findings are applied and validated (the Mathematica notebook solving each example is available upon request).

Example 1. The operator B1: C[0, 1] ^ C[0, 1] corresponding to the problem

x

"(f)-f2 J—f3x'(f)df - fj"0fx'(f)df = 2f + 1,

x(0) + x(1) = 0, x'(0) - 2x'(1) = 0 (24)

is correct. The unique solution of problem (24) is given by the formula

31990f4 - 158464f3 -451860f2 +

+2502304f-961985

x (f ) =--. (25)

v ' 903720

Proof: If we compare equation (24) with equations (18), (19), it is natural to denote O = O1 = O,

F = F1 = F, G- = g1-)= G-, S- = s1-)= S-, Im = 1, and to take X = C[0, 1],

B1x (f ) = x" (f)-f2 j1f3x'(f )d f -

- f j1fx'(f )d f = 2f +1;

(26)

D(B1) = {x(t) e C2[0, 1] : x(0) + x(1) = 0,

x'(0) - 2x'(1) = 0}; (27)

.Ax = A0Ax = x"(t); (28)

Ax(t) = x'(t), D(A) = {x(t) e C1[0, 1] : x(0) = -x(1)}; (29)

O(Ax) = J— fx' (f )d f, F (Ax ) = J—f3x'(f )d f, (30)

G0 = t, S0 = t2. Let us denote Ax(t) = x'(t) = y(t) = y. Then from (28) and (27) we have y e D(A0), A0Ax =

= (x'(t))' = y'(t) = Aoy(t), y(0) - 2y(1) = 0. So we proved that

Aoy = y'(t), D(Aq) = {y(t) e C1[0, 1]: y(0) - 2y(1) = 0}.

Now we check the condition D(B1) = D(A0A). By definition

D(A0A) = {x(t) e D(A): Ax(t) e D(A0)} = = {x(t) e C1[0, 1]: x(0) = -x'(1), x'(t) e C1[0, 1], x'(0) - 2x'(1) = 0} = = {x(t) e C2[0, 1]: x(0) + x(1) = 0, x'(0) - 2x'(1) = 0} = D(B1).

So D(B1) = D(A0A). It is easy to verify that the operators A0, A are correct on C[0, 1] and for every f(t) e C[0, 1] the following equations hold true

A--1f (f )=J—f (s )ds - 2J—f (s )ds;

A--1f (f ) = J0 f (s )d s -1 J—f (s )d s. From (30) we have

O(f ) = J—sf (s)d s, F (f ) = J—s3f (s)d s.

(31)

(32)

(33)

It is evident that O, F e C*[0, 1]. Consequently, we can take Z = C[0, 1] = X. Using (33) and (21) we find

F (So ) = J—s3s2 d s = 6, F (G- ) = J—s3s d s = 1,

f 1 f 2 A-1G- =Jo s d s - 2Jo s d s = --1,

/ 2 \

°(VG0 j-i—1«

£2 o,1

2

v

d s = -—, 8

A-1S0 = J fs2 d s - 2fV ds = - - 2 0 0 Jo JO 3 3'

(A--1S- ) = J-1

O

(s3 o 21 2 3

d s =--,

15

L— = Im-O

(a—1g-)

=11 r-1= 8

_-, Li —-,

8 1 11

G = ao1so + a-1g- lo

1®(a-1s- )

=o 2 = 33

o 11

2

i-14-1=M

^ 15J 165V

—I--I =-(55f3 - 16f2 - 78

8

Taking into account (33) we obtain

F (G ) = — J-s3 (55s3 -16s2 -78 )ds =

6o1

693o.

7531

Since det L = det[1- F (G )] =-* 0 then

v ' 6930

L1 = 7531' and by Theorem 3 (ii), problem (26), (27) or (24) is correct. By (32) we calculate

„_i 330t4 - 128t3 - 1872t + 835

A G =-'

3960

A-1 A-1G0 = — -1 + — 0 0 6 12

and for f(t) = 2t + 1 by (31)-(33) we obtain

2

A-1f = -4 + t + t2, A"1 A-1f =19 - 4t + + T—, 19

12 2 3 17

12

Substituting these values into (13) we obtain the unique solution of (26), (27) or (24), which is given by (25).

Example 2. The operator B^ C[0, ft] ^ C[0, ft] corresponding to the problem

x'"(t)- sin tjjftt2x"(t )d t -- cost2(t + 1)x"(t)dt = sin2t, (34)

x(0) + x(ft) = 0, x'(0) + 3x'(ft) = 0, x"(0) + x"(ft) = 0,

is correct. The unique solution of the problem (34) is given by the formula

x(t) = 418 3(-2 + ft2 -6ftt + 4t2 + 2cos2t)-ft(2 -3)(8cost + ft(ft-2t-4sint))

„3 Q ft - 2

3(2 + ft)2

4(tt2 -4)cost-(2ft- 1) x (71 - 2t - 4 sin t )

2(A - 2)

(35)

Proof: If we compare (34) with equations (18), (19), it is natural to denote O = O1 = O, F = F1 = F,

G0 = ^10)= G0' S0 = s10)= S0' Im = 1, and to take X = C[0, ft],

Blx (t) = x"'(t)- sin tj'It2x"(t )d t -

ft/2

- costJo' (t +1)x"(t)dt = sin2t; D(B,) = {x(t) e C3[0, ft]: x(0) + x(ft) = 0,

(36)

x'(0) + 3x'(ft) = 0, x"(0) + x"(ft) = 0}; (37) Ax = A0Ax = x"'(t); (38)

Ax(t) = x"(t); D(A) = {x(t) e C2[0, ft]: x(0) = -x(ft),

x'(0) + 3x'(ft) = 0}; (39)

0( Ax) = jft 2 (t + 1)x"(t )d t F (Ax ) = Jftt2x"(t )d t,

(40)

S0 = sin t, G0 = cos t, f = sin 2t. Denote Ax(t) = x"(t) = = y(t) = y. Then from (37) and (38) we have y e D(A0), AqAx = (x"(t))' = y'(t) = A^t), y(0) + y(ft) = 0. So we proved that

A0y = y'(t), D(A0) =

= {y(t) e C1[0, ft]: y(0) + y(ft) = 0}. (41)

Now we check the condition D(B1) = D(A0A). By definition

D(A0A) = {x(t) e D(A): Ax(t) e D(Aq)} = = {x(t) e C2[0, ft]: x(0) + x(ft) = 0, x'(0) + 3x'(ft) = 0, x"(t) e C1[0, ft], x"(0) + x"(ft) = 0} = = {x(t) e C3[0, ft]: x(0) + x(ft) = 0, x'(0) + 3x'(ft) = 0, x"(0) + x"(ft) = 0} = D(B1).

So D(B1) = D(AA0). It is easy to verify that the operators A, A are correct on C[0, ft] and for every f(t) e C[0, ft] from (39) and (41) follows that

A-1f (t ) = J- (t - s)f (s )d s +

Joft(2s - 3t -ft/ 2)) (s)d s; (42)

1 eft

+ —

4->o

Ao-1f (t) = J-f (s)ds -1 Joft f (s)ds.

2J-

(43)

From (40) we have

®(f) = ift2(s + 1)f (s)ds' F (f) = jVf (s)ds. (44)

It is evident that F, O e C*[0, ft]. Consequently we can take Z = C[0, ft] = X. From (43), (44), (20), (21) we get

-1 ft ..1 fft A- G- =J cos s ds — J coss d s = sin t

2Jo

+

®(aô1Go ) = j0n/ 2 (s +1)

sin s d s = 2,

Ao1So = -cos t,

det Lo = det

®(a-1So ) = J— 2 (s + l)(-cos s)d s = - —,

®(a-1Go ) = l - 2 = -l * 0, L01 =-G = A-1S0 + A-1GoL'01®(a-1So ) = 2 sin t - cos t, F (G ) = J s21— sin s - cos si d s = —,

then

det L = det [l - F (G)] =

2-—3 -l = 2

-, L —-

2-—3

Since detL ^ 0, by Theorem 3 (ii), problem (36)-(39) or (34) is correct. Further by using (42) and tak-

ing into account that Ao1Go = sin t, we find

—sin t —(2t-—)

A lG = cost--

A 1 A0lG0 = J0 (t - s )sin s d s

l r — — 1

— I I 2s - 3t--i sin s d s = -sin t -

AJ0 9

4 0

7—1 .

2t- —

For f(t) = sin 2t by (42), (43) we calculate l - cos2t

ao-1/ = -

2 F

®(A-lf ) = (— + 2)2 /16,

Kf )=—6--4

A_lA-1f = 16(4t2 -6—t + —2 -2 + 2cos2t

Substituting these values into (13) we obtain the unique solution of (34), which is given by (35).

Example 3. Let Q = {(t, s) e R: 0 < t, s <1. The operator B1: C(Q) ^ C(Q) corresponding to the problem

x'is (t, s) - t3sJ01 J0 s2x't (t, s)dt d s -- ts2 J0 Jltxt(t, s)d t d s = 5t2 + s,

i1 fl

t't, x'ls e C(Q), x(0, s) = s2 J0 J0x(t, s)dtds,

x't(t, 0 ) = tJ1 J1sxt(t, s)dtds, (45)

is correct. The unique solution of problem (45) is given by the formula

85148684t2 + 31416680s3t2 +

+ x

(t,s ) = "

287762400st

3

172657440 s2 (l23613741 + 86328720t + 15746840t4)

172657440

(46)

Proof: If we compare (45) with (18), (19), it is natural to denote

O = ®1 = ®, F = F1 = F, Go = 40)= GO , SO = s^0) = = S0, Im = 1, and to take X = C(Q),

B1x (t ) = x's (t, s ) - t3sJ0 Jo s2x't (t, s )dt d s -

- ts2 J0 Jltxt(t, s)d t d s = 5t2 + s; (47) D(B1 ) = {x(t, s) e C(Q), xt, x's e C (Q), x (0, s) = s2 J0 J^x (t, s )d t d s;

xt(t, 0) = tJ0 Jlsxt(t, s )d t d s}; (48)

A0Ax = x'ts (, s); (49)

Ax (t, s) = xt(t, s); (50) D (A) = {x(t, s) e C(Q) : xt (t, s) e C(Q),

x(0, s) = s2 Jo Jo x(t, s)dtds,

F (Ax ) = J0 Jjs2xt(t, s) d t d s,

O(Ax ) = J1 Jq tx't (t, s) d t d s, (51)

S0 = t3s, G0 = ts2, f = 5t2 + s. We denote Ax (t, s) = xt(t, s) = y (t, s) = y• Then from (48), (49) we have

y e D(Ao), AqAx =( (t, s))s = y'3 (t, s) = A0y(t, s),

y(t, 0) = tJO JQsy(t, s)dtds.

So we proved that

Aoy = У's(t, s), D (Ao ) = {y (t, s) e C (Q) : y's e C (Q), y (t, O) =

=t JO J1 sy(t,s)d t d s}.

Now we check the condition D(B1) = D(A^). By definition

D (A0A) = {x (t's) e D (A): Ax (t's) e D (A0 )} = = {x(t's) e C(Q): x't e C(Q) x(0's) =

= s2 J- x(t's)dtds' x's (t's) e C(Q)

xt(t' 0) = tJ- J-sxt(t's)dtds} = = {x (t's ) e C (Q): e C (Q)'

x(0' s) = s2J J x(t' s)dtds}'

t (t' 0) = tjJ Jjsxt (t's)dtds} = D(B1).

So D(B1) = D(A0A). It is easy to verify that the operators A, A are correct on C(Q) and for every f(t, s) e C(Q) the following hold true

A-f (t' s) = J0sf (t' s1 )d s1 +

+f J1 JO sJ0sf (t' s1)d s1d t d s;

A-f (t' s) = J0f (t1' s )d t1 +

(52)

+

32 Jo1 Jo1 J0f (t1's )d t1d t d s.

2 jOJOJO

From (51) for every f(t, s) e C(Q) we get

(53)

F (f ) = Jo1 JO s2f (t' s)d t d s'

O(f ) = J0 JOtf (t' s)dt d s.

(54)

It is evident that F, O e C*(Q). Consequently we can take Z = C(Q) = X.

Further by using (52), (54), (20), (21) for S0 = t3s, G0 = ts2 we get

A-1So =Jost3s1 d s1

t s2t3

4t fl fl rs 3 t

+— I I si t si ds-| dtds = — +

3 JoJo Jo 1 1 24 2 A-1 Go =J0ts2 d s1 +

4t r1r1 rs 2 2t s3t

+ — J J s! ts-f d si d t d s = — +-'

3 JoJo Jo 1 1 45 3

45

O

(a-1go )=j0 JO s2 (a-1go ) = j0 Jo1t

(ao1go)

2t ^

45 3

v /

3. ^

d t d s =

2t s3t 45 ~3

d t d s =

19 540'

23 540'

det Lo = det

1 -O(

517 -O _540 "' Lo ="

540

517

G = A-1 So + A-1 Go L01o(A-1SO )

•(9

t (907 + 340s3 + 10340s2t2) 20680 '

F (G) = det L = det |"l - F (G)l= 39967

v ' 41360 L

41360

1 =41360

L —-.

39967

Since detL ^ 0 then, by Theorem 3 (ii), problem (47), (48) or (45) is correct. By (53) we calculate

907t2 + 340s3t2 + s2 (1013 + 5170t4

A" OG =-

41360

„-1 120s3t2 + 23s2 + 16t2 A Ao G =-

720

and for f(t, s) = 5t2 + s by (52)-(54) we obtain

„--u s2 49t 2

A01f = — +-+ 5st2'

0 2 54

^ ^^ 2 (t 287^ 5st3 49t2

A 1 A-1f = s2 I- +-1+-+-'

0 l,2 432J 3 108

f (f 841 ■ »((o-1^655

648

Substituting these terms into (13) we obtain the unique solution of (45), which is given by (46).

Conclusion

The main research result of this paper is the existence and uniqueness of the operator equation B1u = f in the space setting of Banach spaces, given that B1 = BqB. The necessary and sufficient conditions for the correctness of the operator B1 are intermediate, secondary results. The solution procedure follows the universal decomposition method and provides a unique exact solution in closed form. This method can be also applied in more complex problems, as of the type B1u = f, where B1 = BqB2 or B1 = B2B and for B0, B given by (8), (9), respectively.

The entire approach is given in an algorithmic procedure that is reproducible in any program of symbolic calculations.

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УДК 338.984

doi:10.31799/1684-8853-2021-2-2-12

Разложение абстрактных линейных операторов на банаховых пространствах

К. Д. Тсиликаа, PhD, доцент, orcid.org/0000-0002-9213-3120, [email protected] ^Университет Фессалии, 38221, Волос, Греция

Введение: большинство известных методов декомпозиции для решения краевых задач (метод декомпозиции Адомяна, естественное преобразование метода декомпозиции, модифицированный метод декомпозиции Адомяна, комбинированный метод преобразования Лапласа — декомпозиции Адомяна и метод декомпозиции области) используют так называемые полиномы Адомяна или итерации для получения приближенных решений. Насколько нам известно, прямой метод получения точного аналитического решения пока не предложен. Цель: разработать в произвольном банаховом пространстве новый универсальный метод разложения для класса обыкновенных интегро-дифференциальных уравнений или интегро-дифференциальных уравнений в частных производных с нелокальными и начальными граничными условиями в терминах абстрактного операторного уравнения Blx = f. Результаты: исследован класс интегро-дифференциальных уравнений в банаховом пространстве с нелокальными и начальными граничными условиями в терминах абстрактного операторного уравнения Blx = Лх - S0F(Ax) - G00(Ax) = f, x e D(Bj),

где Л, А — линейные абстрактные операторы; S0, G0 — векторы, а Ф, F — функциональные векторы. Обычно Л, А — это линейные обыкновенные дифференциальные операторы или дифференциальные операторы в частных производных, а F(Ax), Ф(Ах) — интегралы Фредгольма. Основным результатом нашего исследования является теорема существования и единственности уравнения B1x = f при условии, что оператор B1 имеет разложение вида B1 = B0B, где B и B0 — различные абстрактные линейные операторы специального вида. Предлагаемый метод разложения универсален и существенно отличается от других методов разложения в соответствующей литературе. Этот метод может быть применен как к обыкновенным интегро-дифференциальным уравнениям, так и к интегро-дифференциальным уравнениям в частных производных, и дает единственное точное решение в замкнутой аналитической форме в банаховом пространстве. Этапы метода решения иллюстрируются численными примерами, соответствующими конкретным задачам. Система компьютерной алгебры Mathematica используется для демонстрации результатов решения и оценки эффективности анализа. Практическая значимость: основным преимуществом настоящего метода решения является легкость его интеграции в интерфейс любого программного обеспечения CAS.

Ключевые слова — корректный оператор, разложение (факторизация, декомпозиция) операторов (уравнений), интегро-диф-ференциальные уравнения, краевые задачи, точное решение.

Для цитирования: Tsilika K. D. Decomposition of abstract linear Operators on Banach spaces. Информационно-управляющие системы, 2021, № 2, с. 2-12. doi:10.31799/1684-8853-2021-2-2-12

For citation: Tsilika K. D. Decomposition of abstract linear operators on Banach spaces. Informatsionno-upravliaiushchie sistemy [Information and Control Systems], 2021, no. 2, pp. 2-12. doi:10.31799/1684-8853-2021-2-2-12

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