ON THE SOLUTION OF SOME HIGHER-ORDER INTEGRO-DIFFERENTIAL EQUATIONS OF SPECIAL FORM Текст научной статьи по специальности «Математика»

Scientific article

DOI: 10.18287/2541-7525-2020-26-1-14-22 Submited: 18.12.2019

Revised: 20.01.2020

Accepted: 28.02.2020

E. Providas

University of Thessaly, Larissa, Greece E-mail: [email protected]. ORCID: https://orcid.org/0000-0002-0675-4351

I.N. Parasidis University of Thessaly, Larissa, Greece E-mail: [email protected]. ORCID: https://orcid.org/0000-0002-7900-9256

ON THE SOLUTION OF SOME HIGHER-ORDER INTEGRO-DIFFERENTIAL EQUATIONS OF SPECIAL FORM

ABSTRACT

The article is devoted to the solution of boundary value problems for higher-order linear integro-differential equations of Fredholm type with differential and integral operators encompassing powers of an ideal bijective linear differential operator whose inverse is known explicitly. The conditions for existence and uniqueness of solutions are derived and the solutions are delivered in closed form. The approach is based on the view that an integro-differential operator is a perturbed differential operator. The results obtained are of both theoretical and practical importance. The method is elucidated by solving two illustrative problems.

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

Citation. Providas E., Parasidis I.N. On the solution of some higher-order integro-differential equations of special form. Vestnik Samarskogo universiteta. Estestvennonauchnaia seriia = Vestnik of Samara University. Natural Science Series, 2020, vol. 26, no. 1, pp. 14-22. DOI: http://doi.org/10.18287/2541-7525-2020-26-1-14-22. (In Russ.)

Information about the conflict of interests: authors and reviewers declare no conflict of interests.

Information about the authors: © Providas Efthimios — Candidate of Technical Sciences, associate professor, University of Thessaly, Larissa, 41110, Greece.

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

Introduction

Integro-differential equations are used to model physical phenomena and processes in engineering, physics, biology and economics [1-4]. Higher-order Integro-differential equations are also play an important role in mathematical modeling [5-7]. In most cases, they cannot be solved analytically and therefore approximate solutions are persuaded [8; 9]. However, exact solutions are always attractive and in some instances essential [10-12]. Based on the theory of extensions of operators [13-15], the authors have obtained closed form solutions of several types of boundary value problems for integro-differential equations with classical and nonlocal boundary conditions [16-21]. The present paper is a sequel to work of [19] and deals with the exact solution of one more class of boundary value problems of special form.

Let X be a Banach space of complex valued functions of x defined on Q, A : X ^ X a bijective linear differential operator incorporating some initial or boundary conditions. First, we consider the integro-differential equation

m ,,

Bu(x ) = Au(x) / Kj(x, t)A u(t)dt = f (x), x G Q,

j=lJn

D(B) = D(A) C X, (1)

where B : X ^ X is a linear operator, Kj(x,t) G X(Q x Q) are known kernel functions, f (x) G X is an input function and u(x) G D(B) is the sought function describing the response of the system modeled by (1).

Next, we contemplate the more involved integro-differential equation

B4u(x) D(BA)

A4u(x) - Kij (x,t)Aiu(x)dt\ = f (x),

¿=i V J

x), x G Q,

D(A4),

(2)

where B4 : X ^ X is a linear operator, Ai, i = 1,2, 3,4, are powers, self-compositions, of the operator A and the kernels Kij(x,t) e X(Q x Q) are known functions. We assume that the inverse operator I = A-1 is known explicitly, and that the kernels Kj(x,t), Kij(x,t) are separable functions. We establish existence and uniqueness criteria and provide the solutions of the problems (1) and (2) in closed form. Lastly, we investigate under which conditions B4 = B4 and develop a decomposition technique for obtaining the exact solution to the problem B4u(x) = f (x).

The outline of the paper is as follows. In Section 1., the two problems are put in a convenient matrix form and exact solution formulae are obtained by a direct method. In Section 2., the conditions for factorizing problem (2) and a decomposition solution method are presented. Finally, two examples are solved in Section 3.

1. The Direct Method

First, we deal with problem (1). We assume that the kernels Kj(x,t) are separable functions, i.e. Kj(x,t) = = gj(x)hj(t), j = 1,... ,m, where gj(x), hj(x) e X, and introduce the vector of functions

g = ( gi ... gm ) e Xm, gj = gj(x) e X, j = l,...,m, (3)

and the vector of linear and bounded functionals

/

Ф

ex:

V

Ф] = Ф] (•) = I hj (t) • dt G x *, j = l,...,m. JQ

(4)

Also, by Ф(f) and Ф(е) we denote the vector and the m x m matrix

/ ад) \ Г $i(gi) ••• Qi(gm)

; I, *(&)= ; ;

V $m(f) ) L ®:(gi) ••• ®m(gm)

*(f ) = respectively, and note that

(5)

(6)

$(gN) = $(g)N,

where N is an m x k, k e N, constant matrix. Finally, 1m symbolizes the m x m identity matrix and 0 the zero column vector.

Thus, the problem (1) can be put in the convenient form

Bu = An - gФ(A4u) = f, D(B) = D(A).

(7)

The conditions for existence and uniqueness, the solution to problem and the correctness of the operator B are provided by the next theorem.

It is recalled here that an operator P : X ^ X is said to be correct if P is bijective and its inverse P-1 is bounded on X.

Theorem 1. Let X be a complex Banach space, A : X ^ X a bijective linear operator and I = A-1 its inverse, g e Xm and $ e Xm as defined in (3) and (4), respectively, and B : X ^ X the linear operator

Bu = Au - g$(Au), D(B) = D(A). (8)

Then the following statements are true:

(i) The operator B is bijective on X if and only if

det W = det[1m - $(g)]=0, (9)

and the unique solution to boundary value problem (7), for any f e X, is given by the formula

u = B-1f = If + IgW-1$(f). (10)

(ii) If in addition the inverse operator A-1 is bounded on X, then the operator B is correct.

Proof. (i) Let det W — 0 and u G ker B. Then,

Bu = Au - g$(Au) = 0, and after applying the vector $ on both sides of (11) and utilizing (6),

[1m - $(g)] $(Au) = W$(Au) = 0,

(11)

(12)

which implies that $(Au) = 0. Substitution into (11) yields Bu = Au = 0 and hence u = 0. This means that ker B = {0} and therefore the operator B is injective. Conversely, we prove that if B is an injective operator then det W = 0, or equivalently, if det W = 0, then B is not injective. Let det W = 0. Then there exists a nonzero vector c = col(ci,..., cm) such that Wc = 0. Consider the element u0 = A-1gc and note that u0 = 0; otherwise u0 = 0 implies gc = 0 and then Wc = [1m — $(g)]c = c — $(gc) = c = 0 which contradicts the hypothesis that c is a nonzero vector. From equation (8), we get

Buo = gc — g$(g)c = g[1m — $(g)]c = gWc = g0 = 0,

which means that ker B = 0 and so B is not injective.

Further, by multiplying from the left both sides of (7) by A-1, we get

u — A-1g$(Au) = A-1 f,

while acting by the vector $ on (7), we have

[1m — $(g)] $(Au) = W$(Au) = $(f).

(13)

(14)

(15)

By hypothesis det W = 0 and therefore equation (15) can be solved with respect to Ф(Аи). Substituting Ф(Аи) = W-^(/) into (14), we obtain the solution formula (10).

Finally, since the input function / in (7) and (10) is any arbitrary / e X, we have R(B) = X which means that B is bijective.

(ii) Suppose that (9) is true and that the operator A-1 is bounded. Then by (i) the operator B is bijective and the unique solution to problem (7) is given by (10). In (10) the operator A-1 and the functionals Ф1;..., Фт are bounded. This means that the operator B-1 is bounded and hence the operator B is correct. □ To find the solution of the boundary value problem (2), we now introduce the vectors of functions

( q1 ... qm ) , Г = ( Г1 ... rm ) , s = ( S1 ... sm ) ,

q

Z — ( Z\ .. . zm

and write the problem (2) in the form

qj,

Sj, Zj G X, j — 1,.. ., m,

B4 u

A4u - qФ(Âu) - ^(Â2u) - бФ^^ - zФ(Â4u) — f,

D(B4) — D(A4),

(16)

(17)

and prove the following theorem.

Theorem 2. Let X be a complex Banach space, A : X ^ X a bijective operator and I — A-1 its inverse, q, r, s, z G Xm and Ф G Xm as in (16) and (4), respectively, and the operator B4 : X ^ X defined by

B4u

A4u - qФ(Au) - ^(A2u) - бФ^^ - zФ(A4u),

D(B4) — D(A4). Then the following statements are true:

(i) If

where

V

Ф(13q) - 1m

Ф(12q)

Ф(1 q) Ф^)

det V — 0,

Ф(13г) Ф(13б) Ф(13z)

Ф(12r) - 1m ф(12б) ф(12z)

Ф(1 r) Ф(*з) - 1m ф(1 z)

Ф(г) Ф(б) Ф(z) - 1Г

(18)

(19)

(20)

then the operator B4 is bijective on X and the unique solution to problem (17) is given by

u — B4-1 f

I4 f - ( 14q I4r I4s I4z ) V-

f ф(I3f) \ Ф(I2f) ф(If )

V ф(f) J

(21)

1

(ii) Conversely, if the operator B4 is injective and also the vectors q, r, s, z are linearly independent, then det V = 0.

(iii) If in addition the inverse I = Ai-1 is bounded on X, then the operator B4 is correct. Proof. (i) Let det V = 0 and u e ker B4, i.e.

A4u - qФ(Au) - гФ(А2ч) - бФ(Аu) - z^(AAu) = 0

(22)

where u G D(A4). We introduce the vectors

q = ( q

and write (22) conveniently in the matrix form

), ФМ

( Ф(А^ \

фА2^

у ФА4^ )

(23)

A4u - q$(u) = 0.

By applying the inverse operator I = A-1 four times on both sides of (24), we get

A4-nu - Inq$(u) = 0, n = 1, 2, 3, u - I4q$(u) = 0.

Acting by the vector $ on both sides of (24) and the three equations (25), we obtain

ФА4^^ - Ф(Inq)Ф(u) = 0,

0,1, 2, 3,

where 10 = 1. Equation (27) can be rearranged in the following configuration

Ф(13q) - U Ф(12q)

ф(1 q)

Ф^)

Ф(13г) Ф(12г) - 1 Ф(1 г) Ф(г)

Ф(13s) Ф(12s)

Ф(1 s) - 1П Ф(8)

Ф(13z) Ф(12z) Ф(1 z) Ф^) - 1

Ф(u) = 0,

or

(24)

(25)

(26)

(27)

(28)

(29)

V$(u) = 0,

where V is the 4m x 4m matrix in (20). Since det V = 0, it is implied from (29) that $(u) = 0 and then from (26) that u = 0. This means that ker B = {0} and therefore the operator B is injective. To obtain the solution of (17), we write it in the form

A4u - q$(u) = f, f e X, By applying the inverse operator I = A-1 four times on both sides of (30), we get

A4-nu - Inqф(u) = Inf,

u - I4qФ(u) = I4f,

1, 2, 3,

and then acting by the vector Ф on both sides of (30) and (31), we have

ФА4-'^) - Ф(Inq)Ф(u) = Ф(Inf ), n = 0,1, 2, 3,

or in matrix form

уфн = -

( Ф(I3f) \

ФЦ 2f )

ф(и )

(30)

(31)

(32)

(33)

(34)

V *(f) /

By inverting (34) and substituting into (32), we obtain (21). In (17) and (30), f is an arbitrary element of X and therefore R(B4) = X and so the operator B4 is bijective and the problem (17) everywhere solvable.

(ii) Let the vectors q, r, s, z be linearly independent. We will 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) such that Vc = 0 where Oj = col (сц,..., cim), i = 1,2, 3,4. Consider the element u0 = I4(qc1 + rc2 + sc3 + zc4) G D(A4).

r s z

n

n

Substituting this element into (18), we obain

Buo = qci + rc2 + SC3 + ZC4

-q ^(/3q)ci + Ф(13г)с2 + Ф(1^^з + Ф^^] -r [Ф(/2q)ci + Ф(12r)c2 + Ф(/2s)cз + Ф(1^^4] -s [Ф(1 q)ci + Ф(1 r)c2 + Ф(Л^з + Ф^^] -z ^(q)ci + Ф(r)c2 + Ф(s)cз + Ф(z)C4] = - ( q r s z ) Vc = 0. (35)

This means that u0 G ker B4. Note that u0 = 0, because by hypothesis q, r, s, z are linearly independent and c = 0. Hence, B4 is not injective.

(iii) Let additionally I = A~i is bounded, i.e. A is correct. Then the operators 1г, i = 2,3,4, are bounded and since the functionals Ф-,j = 1,...,m, are bounded on X, it is concluded that the operator B-i is bounded too by means of (21). Hence, if the operator A is correct and the condition (19) is satisfied then the operator B4 is correct. □

2. Decomposition Method

Let the operator B : X ^ X be defined by

Bu = Au — z$(Au), D(B) = D(A), (36)

where z G D(A3)m, and let the vectors

q = Ax — z$(Ar), r = As — z$(As), s = Az — z$(Az). (37)

In this case, problem (2) can be formulated as

B4 u = B4u = f, D(B4) = D(A4). (38)

It is recognized that conditions (37) are seldom met in real life problems, but when it happens the solution process is simplified further. Moreover, the results derived here are of theoretical importance and valuable for extra developments.

Theorem 3. Let X be a complex Banach space, A : X ^ X a bijective operator and I = A-1 its inverse and $ G Xm as in (4). Let q, r, s, z G Xm satisfy (37) and the operator B : X ^ X be defined by (36). Then the following statements are true:

(i) The operator

B4u = !4u — q$(Au) — r$(A2u) — s$(A3u) — z$(A4u), D(B4) = D(A4), (39)

is biquadratic, meaning B4u = B4u.

(ii) If

det W = det[1m — $(z)] = 0, (40)

then the operator B4 is bijective and the unique solution to boundary value problem (38) is obtained by

u = B4 1 f = u4,

uk+1 = Iuk + IzW-1$(uk), k = 0,1, 2, 3, u0 = f. (41)

(iii) Conversely, if the operator B4 is injective and the components of the vector z are linearly independent, then det W = 0.

(iv) Finally, if (40) holds true and in addition the inverse operator I = A-1 is bounded on X, then B4 is correct.

Proof. (i) From (37) and (36), we have

s = Bz, r = Bs = B2z,

q = Br = B3z, (42)

where the operators B2,B3 : X ^ X with D(B2) = D(A2) and D(B3) = D(A3). Substituting (42) into (39), we get

B4u = A4u - B3z^(Au) - B2z$(A2u) - Bz^(Aiu) - z<^(A4u)

■ л3„

= BA?u- B

A3u - B'2z$(Au) - Bz$(A2u) - z$(A3u)

B'2z^(Au) + Bz®(A?u) + z<^(A3u)

B

B

B2

B2 B3 B4u.

•

BA2u - B2z§(Au) - Bz^(A2u)

i2„

A u - Bz^(Au) - zФ(A BAu - Bz$(Au) Au - z§(Au)

(43)

(ii) Let det W = 0. Then by Theorem 1 the operator B in (36) is bijective and consequently the operator B4 = B4 is again bijective as composition of bijective linear operators. Thus, the boundary value problem (38) may be solved as follows

B4u B3u B2u

Bu

uo, B-B-B-B-

uo uo = lui = u2 = u3 =

= f,

ui , u2,

u3, u4,

u1 = Iu0 + IzW-1Ф(u0), u2 = Iu1 + IzW-1§(u1), u3 = Iu2 + IzW-1§(u2), u4 = Iu3 + IzW-1§(u3),

(44)

which yields (41).

(iii) Suppose that the components of the vector z are linearly independent and we will prove that if B4 is an injective operator then det W = 0, or equivalently, if det W = 0 then B4 is not injective. Let det W = 0. Then there exists a nonzero vector c = col(c1,... ,cm) such that Wc = 0. Consider the element u0 = Izc and notice that u0 =0 because the components of the vector z are linearly independent and c is nonzero. From (39) and (42), we obtain

B4uo

A4u0 - B3z$(Au0) - B2zФ(A2u0) - Bz$(A3u0) - z$(A4u0)

i3„

A'3zc - B'3z^(z)c - B2z§(Az)c - Bz^(A?z)c - zФ(A'3z)c

BA2z - B3z<^(z) - B2z§(Az) - Bz§(A2z A2z - B2z<^(z) - Bz%(Az) - z^(A2z BAz - B2z$(z) - Bz$(Az) Az - Bz^(z) - zФ(Az)

B2 [Bz - Bz^(z)] c B3z [1m - Фф] c B3zWc = 0.

'2

B

B

B2

(45)

From (45) it is implied that u0 e ker B4 and hence B4 is not injective.

(iv) Assume (40) holds true. Then the operator B4 is bijective and its inverse is given by (41). If in addition A-1 is bounded and since the functionals $1,..., $m in (41) are bounded, it is implied that the operator B-1 is bounded and hence B4 is correct. □

u

c

3. Examples

In this section, we select and solve two representative example problems to explain the implementation as well as to show the efficiency of the techniques presented in the previous sections.

Example 1. Let Q = {x e R3 : \x\ < l} and dQ = {x e R3 : \x\ = l}. Consider the boundary value problem

△u(x) - E2=1 gj(x) In vj (v)^u(v)dy = f (x), x e Q,

u\8Q = 0, (46)

where A is the Laplace operator in R3. We take X = L2(Q) and

Au(x) = Au(x), D(A) = {u(x) G W2(Q) : u|dn = 0},

$j(-) = / Vi(x) ■ dx, i = 1, 2, ( Q

g = ( 51 (x) 92(x) ), (47)

and thus the problem (46) is put in the form

Bu(x) = Au(x) — g$(Au(x)) = f (x), D(B) = D(A). (48)

It is known that the problem

Au(x) = Au(x) = f (x), u(x)|dn = 0, u G W2(Q), f (x) G L2(Q). (49)

is uniquely and everywhere solvable on L2(Q) and

u(x) = A-1f(x)= / G(x,y)f(y)dy, Wf G L2(Q), (50)

■Jn

where G(x,y) = G1(x,y) + g(x,y) is the Green's function, G1(x,y) = 4n\l,-y\ and g(x,y) is a harmonic function

in Q and belongs to CTO(Q). Also, the operator A-1 is bounded and hence A is correct. We compute,

1 — $1(51 (x)) —$1(g2(x)) —$2(51 (x)) 1 — $2(g2(x))

If det W = 0, then the problem (46) admits a unique solution which is obtained by substituting into (10).

Finally, the operator B by Theorem 1 is correct.

Example 2. Consider the fourth order integro-differential equation

3 !■ 1

det W = det [12 - Ф^)] = det

(51)

3 /

(t) + — (5101t3 - 32978t2 + 102042t - 148458)/ xu'(x)dx 64 Jo

1 Г1

--(593t3 - 3834t2 + 11874t - 17266) / xu''(x)dx

16 o

1 Г1

+ ^(23t3 - 150t2 + 462t - 670) J xu'''(x)dx

-(t3 - 6t2 + 18t - 26) / xu(4)(x)dx o

279

= - g4g (5101t3 - 32978t2 + 102042t - 148458), (52)

subject to boundary conditions

u(i)(0) = 2u(i)(1), i = 0,1, 2, 3. (53)

We take X = C[0,1],

Au = u'(t), D(A) = {u(t) G C 1[0,1] : u(0) = 2u(1)}, (54)

consequently,

a42u = u''(t), D(a42) = {u G C2[0,1] : u(i)(0) = 2u(i)(1), i = 0д} , Â3u = u'''(t), D(A3) = {u G C3[0,1] : u(i)(0) = 2u(i)(1), i = 0,2} , a44u = u(4)(t), D(l4) = {u G C4[0,1] : u(i)(0) = 2u(i)(1), i = 0,3} , (55)

and

Ф^^) = ^ xu(i) (x)dx)j , i = 1, 2, 3,4,

q = 64 (5101t3 - 32978t2 + 102042t - 148458^ ,

r = ^ (593t3 - 3834t2 + 11874t - 17266)^ ,

s = 1 (23t3 - 150t2 + 462t - 670^ ,

z = (t3 - 6t2 + 18t - 26) ,

279

f = - — (5101t3 - 329782 + 102042t - 148458). (56)

640

It is easy to verify that the operator A is correct and its inverse is given by

A-1f (t) = Jo f (x)dx - 2/0 f (x)dx, f G C[0,1].

(57)

and that the functional $ is bounded, i.e. $ e C[0, lj* = X*. Thus, the boundary value problem (52), (53) can be expressed now in the operator form (17).

To find its solution we can apply Theorem 2. However, by inspecting the operator A, the vector of functionals $ and the vectors q, r, s, z, we can verify that the conditions (37) are satisfied, and therefore the decomposition Theorem 3 is more appropriate, which is much easier to implement. It is straightforward to show that

det W = det [1 - Ф^)]

1 - i x(x3 - 6x2 + 18x - 26)dx

o

= -=0. 10 ^

(58)

Thus, the problem (52), (53) is correct and it can be formulated as B4u = f, where B is the first order operator in (36). By substituting into (41), we obtain the unique solution of the problem (52), (53), which is

t4

75

u(t) =--2t3 + 9t2 - 26t +--.

( ) 4 + +2

(59)

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Научная статья DOI: 10.18287/2541-7525-2020-26-1-14-22

УДК 629.7.05 Дата: поступления статьи: 18.12.2019

после рецензирования: 20.01.2020 принятия статьи: 28.02.2020

E. Провидас

Университет Фессалии, г. Ларисса, Греция E-mail: [email protected]. ORCID: https://orcid.org/0000-0002-0675-4351

И.Н. Парасидис Университет Фессалии, г. Ларисса, Греция E-mail: [email protected]. ORCID: https://orcid.org/0000-0002-7900-9256

О РЕШЕНИИ НЕКОТОРЫХ ИНТЕГРО-ДИФФЕРЕНЦИАЛЬНЫХ УРАВНЕНИЙ ВЫСШЕГО ПОРЯДКА СПЕЦИАЛЬНОГО ВИДА

АННОТАЦИЯ

Статья посвящена решению краевых задач для линейных интегро-дифференциальных уравнений высшего порядка типа Фредгольма с дифференциальными и интегральными операторами, охватывающими степени идеального биективного линейного дифференциального оператора, обратный которому явно известен. Выводятся условия существования и единственности решений, и решения изложены в закрытом виде. Подход основан на представлении о том, что интегро-дифференциальный оператор является возмущенным дифференциальным оператором. Полученные результаты имеют как теоретическое, так и практическое значение. Метод поясняется решением двух иллюстративных задач.

Ключевые слова: интегро-дифференциальные уравнения, задачи с начальными условиями, граничные задачи, дифференциальные операторы, энергетические операторы, смешанные продукты, точные решения.

Цитирование. Providas E., Parasidis I.N. On the solution of some higher-order integro-differential equations of special form // Вестник Самарского университета. Естественнонаучная серия. 2020. Т. 26, № 1. С. 14-22. DOI: http://doi.org/10.18287/ 2541-7525-2020-26-1-14-22.

Информация о конфликте интересов: авторы и рецензенты заявляют об отсутствии конфликта интересов.

Информация об авторах: © Провидас Евтимиос — кандидат технических наук, доцент, Университет Фессалии, Греция, г. Ларисса, Гайополис, 41110.

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