Integro-differential equations embodying powers of a differential operator Текст научной статьи по специальности «Математика»

12

Вестник Самарского университета. Естественнонаучная серия. Том 25 № 3 2019

УДК 629.7.05 Дата поступления статьи: 2/^77/20 1 9

Б01: 10.18287/2541-7525-2019-25-3-12-21 Дата принятия статьи: 15/^777/2019

И.Н. Парасидис, Е. Провидас

ИНТЕГРО-ДИФФЕРЕНЦИАЛЬНЫЕ УРАВНЕНИЯ, СОДЕРЖАЩИЕ СТЕПЕНИ НЕКОТОРОГО ДИФФЕРЕНЦИАЛЬНОГО ОПЕРАТОРА

© Парасидис Иван Нестерович — кандидат технических наук, доцент, Университет Фессалии, Греция, Ларисса, Гайополис, 41110 .

E-mail: paras@teilar.gr. ORCID: https://orcid.org/0000-0002-7900-9256

Провидас Евтимиос — кандидат технических наук, доцент, Университет Фессалии, Греция, Ларисса, Гайополис, 411 10.

E-mail: providas@uth.gr. ORCID: https://orcid.org/0000-0002-0675-4351

АННОТАЦИЯ

Установлены критерии разрешимости и корректности для двух линейных интегро-дифференциальных операторов типа Фредгольма B2, B4, включающих вплоть до второй и четвертой степени, соответственно дифференциального оператора A с неизвестным обратным I = Л-1. Представлены явные формулы решения соответствующих начальных и краевых задач при использовании обратного дифференциального оператора. Подход основан на теории продолжения линейных операторов в банаховых пространствах. В качестве примера решены три задачи для обыкновенных и частичных интегро-дифференциальных операторов.

Ключевые слова: интегро-дифференциальные уравнения, задачи Коши, краевые задачи, дифференциальные операторы, степенные операторы, составные произведения, точные решения.

Цитирование. Parasidis I.N., Providas E. Integro-differential equations embodying powers of a differential operator // Вестник Самарского университета. Естественнонаучная серия. 2019. Т. 25. № 3. С. 12-21. DOI: http://doi.org/ 10.18287/2541-7525-2019-25-3-12-21.

This work is licensed under a Creative Commons Attribution 4.0 International License.

UDC 629.7.05 Submitted: 2/VII/2019

DOI: 10.18287/2541-7525-2019-25-3-12-21 Accepted: 15/VIII/2019

I.N. Parasidis, E. Providas

INTEGRO-DIFFERENTIAL EQUATIONS EMBODYING POWERS OF A DIFFERENTIAL OPERATOR

© Parasidis Ioannis Nestorios — Candidate of Technical Sciences, associate professor, University of Thessaly, Greece, Larissa, Gaiopolis, 41110.

E-mail: paras@teilar.gr. ORCID: https://orcid.org/0000-0002-7900-9256

Providas Efthimios — Candidate of Technical Sciences, associate professor, University of Thessaly, Greece, Larissa, Gaiopolis, 411 10.

E-mail: providas@uth.gr. ORCID: https://orcid.org/0000-0002-0675-4351

ABSTRACT

We establish solvability and correctness criteria for two Fredholm type linear integro-differential operators B2, B4 encompassing up to second and fourth powers, respectively, of a differential operator A with a known inverse I = A-1. We also derive explicit solution formulae to corresponding initial and boundary value problems by using the inverse of the differential operator. The approach is based on the theory of the extensions of linear operators in Banach spaces. Three example problems for ordinary and partial integro-differential operators are solved.

Key words: integro-differential equations, initial value problems, boundary value problems, differential operators, power operators, composite products, exact solutions.

Citation. Parasidis I.N., Providas E. Integro-differentsial'nye uravneniya, osushchestvlyayushchie moshchnost' differentsial'nogo operatora [Integro-differential equations embodying powers of a differential operator]. Vestnik Samarskogo universiteta. Estestvennonauchnaya seriya [Vestnik of Samara University. Natural Science Series], 2019, no. 25, no. 3, pp. 12-21. DOI: http://doi.org/10.18287/2541-7525-2018-25-3-12-21 [in English].

Introduction

Let C denote the set of all complex numbers and X, Y be complex Banach spaces. Let P : X ^ Y be a linear operator and D(P) and R(P) its domain and range, respectively. We recall that a linear operator P : X ^ Y is said to be injective (or uniquely solvable) if for all u1,u1 G D(P) such that Pu1 = Pu2, follows that u1 = u2; alternatively, the operator P is injective if and only if ker P = {0}. A linear operator P : X ^ Y is called surjective (or everywhere solvable) if R(P) = Y. The operator P is called bijective if P is both injective and surjective. Lastly, P is said to be correct if P is bijective and its inverse P-1 is bounded on Y.

Let X = Y and let the bijective operator A : X ^ X. We consider the power operators A1 = AA and A4 = A1 A1 defined as composite products, and the perturbed linear operators B2 : X ^ X, B4 : X ^ X defined by

B2u = Â2u - p$(u) - q$(Àu) - u),

B4U

A4u - p$(u) - q$(Au) - u) - s$(a43u) - z$(A4u),

with D(B2) = D(A2) and D(B4) = D(A4), respectively. The column vector

I $1 \ ( $i(u)

, $(u) =

a3„

\

(1) (2)

(3)

\ $m(u)

is a set of complex-valued, linear and bounded functionals : X ^ C, j = 1,... ,m, i.e. G X* and

r = ( ri ■ ■■ rm ) ,

$ G X*m, where X* is the adjoint space of X. The row vectors

p = ( pi ■■ ■ Pm ), q = ( qi ■■ ■ qm ),

( -1

(-1

(4)

(5)

s

s

z

z

m

m

are sets of elements pj, qj, rj, Sj, Zj G X, j = 1, . . . , m, i.e. p, q, r, s, z G Xm. For later usage, we mention here that

$(p)

$m(pi)

$l(Pm) ^m(Pm)

is the m x m matrix whose i, j-th entry $i(pj) is the value of the functional $ on element pj. Also, we note that $(pN) = $(p)N, where N is a m x k, k = 1,2, . . . , constant matrix. Lastly, 1m symbolizes the m x m identity matrix and 0 the zero column vector.

In the case where A is a linear differential operator of order n and the functionals , j = = 1,...,m, designate Fredholm integral operators with separable kernels, then B2, B4 describe Fredholm linear integro-differential operators. Integro-differential equations play an important role in modeling physical phenomena and processes in various disciplines in engineering, physics, biology, population dynamics, epidemiology, finance and others. Initial and boundary value problems for integro-differential equations are usually solved by numerical methods due to their complexity. Closed form solutions are obtained only for a limited number of problems, see for example in [5], [6], [9], [10] and the recent works by the authors [2], [3] [4], [7], [8].

In this paper, we are concerned with the solvability and the construction of the solution in closed form of the following two ordinary or partial integro-differential equations subject to initial or boundary conditions, which have not been studied before, namely

B2u = f, D(B2) = D(A2), (7)

Bin = f, D(B4) = D(A4), (8)

for any f G X. Our approach is based on the theory of the extensions of linear operators in Banach spaces [1]. Problems (7), (8) are solved by using the inverse I = A-1.

The rest of the paper is organized as follows. In Section 1., the theory is developed and two main theorems are shown. In Section 2., the theory is applied to solve several example problems. Finally, some conclusions are stated in Section 2..

1. Main Results

We first derive solvability and correctness criteria for the operator B2 and construct the exact solution to initial and boundary value problems involving B2. We state the following theorem. Theorem 1. Let X be a complex Banach space, A : X ^ X a bijective linear operator and I = A-1 its inverse, $ G Xm, and p, q, r G Xm. Let the operator B2 : X ^ X be defined by

B2u = A2u - p$(u) - q$(Aw) - r$(Al2M) = f,

D(B2) = D(P), where f G X. The following statements are true:

(9)

(i) If

det W = det

= 0,

(10)

$(r) - 1m $(q) $(p)

$(Ir) $(Iq) - 1m $(Ip) $(I2r) $(I2q) $(I2p) -

then the operator B2 is injective and everywhere solvable (bijective). The unique solution to (9) for any f G X is given by

B-1 f =

I2 f - ( 12r 12q I2p ) W

i

*(f) *(If)

$(I2f)

(11)

(ii) If the operator B2 is injective and the vectors p, q, r are linearly independent, then det W = 0. (iii) If the inverse operator I = A-1 is bounded on X, that is, A is correct, then the operator B2 is correct. Proof. (i) Let det W = 0 and u G ker B2, i.e.

A2u - p$(u) - q$(Au) - r$(A2u) = 0,

(12)

u=

where u G D(A2). By applying the inverse operator I = A 1 twice on both sides of (12), we obtain in succession

Au - Ip$(u) - Iq$(Au) - Ir$(A2u) = 0, u - I2p$(u) - I2q$(Au) - I2r$(A2u) = 0.

(13)

(14)

Acting now by the functional vector $ on the both sides of (12)-(14), we obtain the following system of equations

/ $(A2u)

W I $(Au) I = 0, (15)

V $(u)

where the 3m x 3m matrix

W

$(r) - 1m $(q) $(p)

$(Ir) $(Iq) - 1m $(Ip) $(I2r) $(I2q) $(I2p) - 1r

Since det W = 0, it is concluded that

$(Ai2u) = $(Au) = $(u) = 0.

(16)

(17)

Substitution of (17) into (14) yields u = 0. Thus, ker B2 = {0} and therefore B2 is an injective operator.

To find the solution to problem (9), we work as above. By applying the inverse operator I = A-1 twice on both sides of (9), we get successively

Au - Ip$(u) - Iq$(Au) - Ir$(A2u) = If, u - I1 p$(u) - I1 q$(Au) - 12r$(A2u) = I2f.

Then acting by the vector of functionals $ on both sides of (9), (18), (19), we acquire the system

W

$(A2u) $(Au) $(u)

*(f) - | *(If)

$(I2f)

By inverting (20), we obtain

Putting (19) into the form

$(A2 u) $(Au) $(u)

-W"

$(f) *(If)

$(12f)

(18) (19)

(20)

(21)

(22)

( $(A1u)

u = I2f +( I1 r 11q I1p ) I $(Alu)

$(u)

and then substituting (21) into (22), we obtain formula (11) which is the unique solution of the problem (9).

Finally, because f in (11) is an arbitrary element of X, it is implied that R(B2) = X. Hence B2 is surjective.

(ii) We prove that if B2 is an injective operator then det W = 0, or equivalently, if det W = 0 then B2 is not injective. Let det W = 0. Then there exists a nonzero vector of constants c = col(c1, c2, c3), where Oj = col(ci1,..., cim), i = 1, 2, 3, such that

Wc = 0. (23)

Consider the element u0 = I1(rc1 + qc2 + pc3) G D(A2). It follows that

B2u0 = Ai2uo - p$(u0) - q$(Au0) - r$(Ai2u0) = - ( r q p ) Wc = 0,

(24)

by taking into account (23). This means that u0 G ker B2. Note that u0 =0, because by hypothesis p, q, r are linearly independent and c = 0. Therefore B1 is not injective.

(iii) In (11), the functionals of the vector $ are bounded. From the hypothesis that A is correct, it is implied that I = A-1 and I1 are bounded on X. Therefore the operator B-1 is bounded on X, and from (i) follows that B2 is correct. The theorem is proved. □

Remark 1. In the cases where one or two of the vectors p, q, r are equal to zero vector, then analogous results to Theorem 1 can be derived. In practice, we can obtain the solution formula directly from (11)

i

by removing the corresponding columns and rows. For instance, let us assume that r = 0. In this case the problem (9) degenerates to the following one,

B2u = A2u - рФ(и) - qФ(Au) = f, D(B2) = D(A2). Then the operator B2 is injective if

Ф(1 q) - 1m Ф(1р) ф(1 2q) Ф(12p) - 1

and the unique solution of (25) for any f £ X is given by

u = B-f =

2

det W = det

= 0,

I2f - ( 12 q I2p ) W-1( .

(25)

(26)

(27)

Next, we elaborate on the solvability and correctness of the operator B4 and the exact solution of initial and boundary value problems incorporating B4. We show the subsequent theorem.

Theorem 2. Let X be a complex Banach space, A : X ^ X a bijective operator and I = A-1 its inverse, $ G Xm, and p, q, r, s, z G Xm. Let the operator B4 : X ^ X be defined by

B4u

A4u - рФ(и) - qФ(Au) - гФ(Л2и) - бФ^и) - zФ(A4u) = f

D(Bi) = D(A4), where f G X. Then the following statements are true:

(i) If

(28)

det V = det

$(z) - U Ф(1 z) Ф(12z) ф(13z) Ф(14z)

Ф(8) Ф(1б) - 1 Ф(12s) ф(13s) Ф(14s)

Ф(г) Ф(1 г) Ф(12г) - 1т Ф(13 г) Ф(14 г)

$(q) Ф(р)

ф(1 q) ф(1р)

ф(12q) ф(12 р)

ф(13q) - im ф(13р)

ф(14q) ф(14 р) - ir

= 0, (29)

then the operator B4 is injective and everywhere solvable on X (bijective). The unique solution to the problem (28) for any f G X is given by

u = B4-1 f =

( *(f) \ *(If )

= I4 f - ( 14z 14s 14г I4q !4р ) V-1 Ф(I2f )

Ф(!3f)

V ф(I4f) j

(30)

(ii) If the operator B4 is injective and the vectors p, q, r, s, z are linearly independent, then det W = 0. (iii) If the inverse I = A-1 is bounded on X, i.e. A is correct, then the operator B4 is correct. Proof. (i) Let det V = 0 and u G ker B4, i.e.

A4u - рФ(и) - qФ(Au) - гФ(Л12и) - бФ(л43и) - zФ(A4u) = 0,

(31)

where u G D(A4). By applying the inverse operator I = A 1 four times on both sides of (31), we get consecutively

A3u - I (рФ(и) - qФ(Лu) - ^(A2u) - бФ^и) - zФ(Л4u)) = 0, A2u - I2 ^рФ(и ) - qФ(Au) - ^(A2u) - sФ(A3u) - zФ(A4 u) = 0, Au - I3 (рФ(и) - qФ(Лu) - гФ^и) - бФ^и) - zФ(Л4u)) = 0, и - I4 (рФ(и) - qФ(Лu) - ^(A2u) - БФ^и) - zФ(Л4и)) = 0.

(32)

(33)

(34)

Implementing the vector of functionals $ on both sides of (31)-(35), we obtain the system of equations

/ $(A4u) \ $(A3 u) V $(A2 u) $(Au) V $(u) J

where the 5m x 5m matrix V is given in (29). Then, since det V = 0, we acquire

(36)

Ф(Л4 u) = Ф^^ = Ф(!2м) = Ф(!и) = Ф(и) = 0. (37)

Substitution of (37) into (35) yields u = 0. Thus, ker B4 = {0} and hence B4 is an injective operator.

To obtain the solution of (28) we work in similar manner. By applying the inverse operator I = A-1 on both sides of (28) four successive times, we get

' Â2

A3u - I (рф(м) - qФ(Au) - гФ(12и) - 8ф(13и) - zФ(l4u)) = If, A2u - I2 (рф(м) - qФ(Au) - гФ(12и) - 8ф(13и) - zФ(l4u)) = I2f, Au - I3 (рф(м) - qФ(Au) - ^(Â2u) - 8ф(13и) - zФ(l4u)) = I3f, u - I4 (pф(u) - qФ(Au) - гФ(^42u) - sФ(AÎ3u) - zФ(AÎ4u^ = I4f.

(38)

(39)

(40)

(41)

Acting by the functional vector $ on both sides of (28) and (38)-(41), we obtain the system

( $(A4u) \ $(a43u) $(Ai2u) $(Alu) $(u)

V

f Ф(f) \ Ф(If ) Ф(I2)f Ф(13 f ) V Ф(!4f) )

(42)

We write (41) in matrix form

I4f ^ 14z 14s 14r 14q I4p )

( Ффп) \

Ф(A42u) Ф(A4u)

V ф(u) )

(43)

By inverting (42) and substituting into (43), we get the unique solution (30) of the problem (28).

Lastly, f in (30) is an arbitrary element of X and therefore R(B4) = X. Hence the operator B4 is surjective.

(ii) We prove that if B4 is an injective operator then det V = 0, or equivalently, if det V = 0 then B4 is not injective. Let det V = 0. Then there exists a nonzero vector c = col(ci, c2, c3, c4, c5), where Oj = = col(cji,..., cim), i = !,..., 5, such that

Vc = 0. (44)

Consider the element u0 = I4(zc1 + sc2 + rc3 + qc4 + pc5) G D(A4). Then, we have

B4uo

A4u0 - рФ^0) - qФ(Au0) - rФ(A\?u0) - sФ(A43u0) - zФ(A44u0) - ( z s r q p) Vc = 0,

Î3,

(45)

by making use of (44). This means that u0 G ker B4. Note that u0 = 0 since by hypothesis the vectors p, q, r, s, z are linearly independent and c = 0. Hence B4 is not injective.

(iii) Since the functionals in vector $ and the operators I, Ij, i = 2, 3,4, are bounded on X, it is concluded that the operator B-1 is bounded too and from (i) follows that B2 is correct. The theorem is proved. □ Remark 2. In the cases where one or more of the vectors p, q, r, s, z are zero, then similar results to Theorem 2 can be obtained. Actually, we can have the corresponding solution formula directly from (30) by removing the like columns and rows. For example, suppose that p = 0. Then the problem (28) reduces to

B4 u D(B4)

A4u - qФ(Au) - гФ(^42u) - sФ(A3u) - zФ(A4u) = f, D(A4).

Î3„

u

Then the operator B4 is injective if det V = 0, where

$(z) - 1m $(s) $(r) $(q)

$(Iz) $(Is) - 1m $(Ir) $(Iq)

$(I2z) $(I2s) $(I2r) - 1m $(I2q)

$(i3z) $(i3s) $(I3r) $(I3q) - 1r

The unique solution of the problem (46) for every f £ X is given by

V

I4/ — ( I4z I4s I4r 14q ) V-

( Ф(/) \ Ф(1/)

Ф(12/)

V ф(13/) У

(48)

2. Applications

Example 1. Consider the following problem

z''(x) - aJ x[z'(t)+ z(t)]dt = x, x G [-1,1],

Z(0) = 1, Z (0) = 0. Making the substitution u(x) = z(x) - 1, we get

u''(x) - \x J [u'(t) + u(t)]dt =(1 + 2A)x, x G [-1,1], u(0) = u (0) = 0.

We take X = C[-1,1], accordingly X1 = C 1[-1,1], X2 = C2[-1,1], and

Au = u', D(A) = {u : u G X1, u(0) = 0}, A2u = u'', D(A2') = {u : u GX2, u(0) = u'(0) = 0},

$(u) = ($1(u)) = ^J u(t)dt^j ,

$(Au) = (^(Au)) = ^ u (t)dt^ ,

p = q = (Ax), r = 0, f = (1 + 2A)x, m =1, and the operator B2 : X ^ X as

B2u = A?u - p$(u) - q$(Au) =

- \x J u(t)dt — \x J u'(t)dy = (1 + 2Л)

D(B2) = D(A2) = {u : u £ X2, u(0) = u'(0) = 0}.

Note that A is injective and R(A) = X, Ф1 £ X*, and p, q are linearly dependent. It is known that the inverse operator I and its composite I2 are given by

(49)

(50)

(51)

I- = A-

■dt, I2- = A-

(x — t)dt,

(52)

and they are bounded on X. As stated by Remark 1 and by using the results in (51) and (52), we compute

" $(Iq) - 1 $(Ip) ^ $(I2q) $(I2p) - 1

det W

det

= det

A _ 1 A

3 1 3

0

1

Л

1 — Л •

From Theorem 1, it is implied that the operator B2 is correct if A = 3. In this case the unique solution to the problem (50) is

u(x)=Ia-3)x3, (53)

by means of (27), while the solution to the problem (49) follows from u(x) = z(x) - 1.

u

x

x

1

2

0

0

Example 2. Let Q = {x € R3 : \x\ < 1}, dQ = {x € R3 : \x\ = 1} and H4(Q) the Sobolev space of all functions of L2(Q) which have their partial generalized derivatives up to the fourth order Lebesgue integrable. Consider the problem,

A2u(x) — g(x) I v(y)Au(y)dy = f (x), x € Q, JQ

u\dn =0, Au\dn = 0, (54)

where g(x),v(x), f (x) € L2(Q) are given functions and u(x) € H4(Q) is the unknown function. Comparing (54) with (9) in Theorem 1, we take X = L2(Q) and

Au = Au, D(A) = {u : u € H2(Q), u\dn = 0},

A2u = A2u, D(A2) = {u : u € H4(Q), u\dn = 0, Audn = 0},

$(Au) = ($i(Au)j = ^J v(x)Au(x)d^j ,

P = r = 0, q = (g(x)), (55)

m =1, and the operator B2 : X ^ X defined as

B2u = Aí2u — q$(Aiu), D(B2 ) = D(A2). (56)

Since v(x) € X, it is concluded that the functional is bounded on X, i.e. € X*. The Dirichlet problem for Poisson equation

Au(x) = f (x), u(x)\dQ = 0, u € H2(Q), f € X, (57)

is known to be everywhere solvable and admits a unique solution for almost all x € Q, viz.

u(x) = A-1f(x)= / G(x,y)f(y)dy, yf € X, (58)

J n

where G(x,y) is Green's function. Thus, the operator A is bijective and

I• = A-1- = f G(x,y) • dy, ■Jn

I2^ = A-2• = i G(x,y) í G(y,t) • dtdy. (59)

Jn Jn

From Theorem 1, Remark 1 and the use of (55) and (59) we have

detW = det[$(Iq) — 1] = / v(x) G(x,y)g(y)dydx — 1. (60)

nn

If fn v(x) fn G(x,y)g(y)dydx = 1, then the operator B2 is injective and the unique solution of the problem (54), for any f € X, is obtained by substitution into

u = B-f = I2f — 12qW-1 $(If). (61)

Example 3. Let the problem

■1

3

x(4)(x) — 48x2 f (1 — t)u'(t)dt — 15x3 [ (1 — t)u''(t)dt — Jo Jo

—8xf (1 — t)u'''(t)dt — 3x4 Í (1 — t)u(4)(t)dt =

oo

oo

= x4 + 1 x3 + 1 x2 + 2x - 1, 2 3 '

u(0) = u'(0) = u'' (0) = u'''(0) = 0. (62)

We take X = C[0,1], suitably X1 = C 1[0,1], X2 = C2[0,1], X3 = C3[0,1], X4 = C4[0,1], and

Au = u', D(A) = {u : u € X1, u(0) = 0},

Á2u = u'', D(A2) = {u : u € X2, u(0) = u'(0) = 0},

Aí3u = u''', D(A3) = {u : u € X3, u(0) = u'(0) = u''(0) = 0},

A4u = u(4), D(A4) = {u : u € X4, u(0) = u'(0) = u''(0) = u'''(0) = 0},

$(Au) = ^^ (1 — t)u'(t)dtj

§(A2u) =

(1 - t)u''(t)dt ,

'0

$(A3u) = 43u)J = ^ J (1 - t)u'"(t)dt ) ,

r-1

1

$(A4u) = №(A4u)

(1 - t)u(4)(t)dt

0

p = 0, q = (48x2) , r = (15x3) , s = (8x), z = (3x4)

f = x4 + 1 x3 + 3 x2 + 2x - 1, m =1, and the operator B4 : X ^ X defined by

(63)

B4u = A4u - q$(Au) - r$(A2u) - s$(Ai3u) - z$(A4u), D(B4) = D(A4).

Notice that A is injective and R(A) = X, G X*, and q, r, s, z are linearly independent. It is known from elementary books of differential equations that

Ik-

A

-k

1

(k - 1)!

(x -1)

k1

•dt, k = 1, 2, 3, 4,

(64)

and they are bounded on X. According to Theorem 2 and specifically to Remark 2, we form the matrix V and compute its determinant, viz.

det V = det

$(z) - 1 $(I z) $(12z)

$(s) $(Is) - 1 $(12s)

$(r) $(I r) $(I2r) - 1

$(q)

$(I q) $(12q)

$(I3z) $(I3s) $(I3r) $(I3q) - 1

= det

70

1

560

1

5040

240388859 444528000

1

15

90

= 0.

1

448

_2_

15

103 105

Hence, from Theorem 2, Remark 2 and the use of (63) and (64) follows that the operator B4 is correct and the unique solution to the problem (62) is given analytically by

u(x)

(x - 5)x4 120

(65)

1

x

0

4

Conclusions

We have presented a method for constructing the exact solution to initial and boundary value problems for a class of integro-differential operators embodying powers of a correct differential operator. We have included several problems to demonstrate the applicability and efficiency of the method. The proposed solution technique can be easily incorporated to any computer algebra system and therefore it may be a useful tool to researchers and students.

In closing, we state without elaborating that under certain conditions the two problems discussed in the present paper can become of the kind,

B2u = B2u = f, D(B2) = D(A), B4u = B4u = f, D(B4) = D(A4),

where the operator B : X ^ X is defined by

Bu = Au - p$(u) - q$(Au), D(B) = D(A), (66)

to facilitate further the solution process by employing decomposition techniques.

Литература / References

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