GENERALIZATIONS TO SOME INTEGRO-DIFFERENTIAL EQUATIONS EMBODYING POWERS OF A DIFFERENTIAL OPERATOR Текст научной статьи по специальности «Математика»

УДК 517.968.7 Дата поступления статьи: 4/Х/2019

Б01: 10.18287/2541-7525-2019-25-4-14-21 Дата принятия статьи: 18/Х/2019

М.М. Байбурин

ОБОБЩЕНИЯ ДЛЯ НЕКОТОРЫХ ИНТЕГРО-ДИФФЕРЕНЦИАЛЬНЫХ УРАВНЕНИЙ, СОДЕРЖАЩИХ СТЕПЕНИ ДИФФЕРЕНЦИАЛЬНОГО

ОПЕРАТОРА

© Байбурин Мерхасыл Мукеевич — кандидат физико-математических наук, доцент кафедры фундаментальной математики, Евразийский национальный университет им. Л.Н. Гумилева, 010008, Республика Казахстан, г. Нур-Султан, ул. Сатпаева, 2.

E-mail: [email protected]. ORCID: https://orcid.org/0000-0003-0083-6142

АННОТАЦИЯ

В данной статье исследуются абстрактные уравнения, содержащие операторы второй, третьей и четвертой степени.

Необходимые условия разрешимости для абстрактных уравнений, содержащих операторы второй и четвертой степени, доказаны без применения линейной независимости векторов, входящих в данные уравнения. Некоторые авторы существенно использовали линейную независимость векторов для доказательства необходимого условия разрешимости.

В данной статье также дается критерий корректности для абстрактного уравнения, содержащего операторы третьей степени с произвольными векторами, и его решение в терминах этих операторов в банаховом пространстве.

Теория, представленная здесь, может быть полезна для исследования интегро-дифференциальных уравнений Фредгольма, содержащих степени некоторого обыкновенного дифференциального оператора или дифференциального оператора в частных производных.

Ключевые слова: интегро-дифференциальные уравнения Фредгольма, начальные задачи, краевые задачи, дифференциальные операторы, степенные операторы, точные решения.

Цитирование. Baiburin M.M. Generalizations to some Integro-differential equations embodying powers of a differential operator // Вестник Самарского университета. Естественнонаучная серия. 2019. Т. 25. № 4. С. 14-21. DOI: http://doi.org/10.18287/2541-7525-2019-25-4-14-21.

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

UDC 517.968.7 Submitted: 4/X/2019

DOI: 10.18287/2541-7525-2019-25-4-14-21 Accepted: 18/X/2019

M.M. Baiburin

GENERALIZATIONS TO SOME INTEGRO-DIFFERENTIAL EQUATIONS EMBODYING POWERS OF A DIFFERENTIAL OPERATOR

© Baiburin Merkhassyl Mokeevich — Candidate of Physical and Mathematical Sciences, associate professor of the Department of Fundamental Mathematics, L.N. Gumilyov Eurasian National University, 2, Satpayev street, Nur-Sultan, 010008, Republic of Kazakhstan.

E-mail: [email protected]. ORCID: https://orcid.org/0000-0003-0083-6142

ABSTRACT

The abstract equations containing the operators of the second, third and fourth degree are investigated in this work.

The necessary conditions for the solvability of the abstract equations, containing the operators of the second and fourth degree, are proved without using linear independence of the vectors included in these equations. Previous authors have essentially used the linear independence of the vectors to prove the necessary solvability condition.

The present paper also gives the correctness criterion for the abstract equation, containing the operators of the third degree with arbitrary vectors, and its exact solution in terms of these vectors in a Banach space.

The theory presented here, can be useful for investigation of Fredholm integro-differential equations embodying powers of an ordinary differential operator or a partial differential operator.

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

Citation. Baiburin M.M. Obobshcheniya dlya nekotorykh integro-differentsial'nykh uravnenii, soderzhashchikh stepeni differentsial'nogo operatora [Generalizations to some 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. 4, pp. 14-21. DOI: http://doi.org/10.18287/2541-7525-2018-25-4-14-21 [in Russian].

Introduction

Boundary value problems (BVPs) for integro-differential equations (IDEs) with initial and nonlocal boundary conditions arise in various fields of mechanics, physics, biology, biotechnology, chemical engineering, medical science, finance and others. (see [1; 3-5; 12; 14].) Finding an exact solution of BVPs for Fredholm IDEs is a difficult problem and is given in [2; 6-11; 13; 15-17]. IDEs embodying powers of a differential operator of the type

B2u = A2u - p$(u) - q$(Au) - r$(A^u) = f,

D(B2) = D(P), (1)

B3u = A3u - p$(u) - q$(Au) - r$(A^u) - s$(Ai3u) = f,

D(Bs) = D(A3), (2)

B4u = A4u - p$(u) - q$(Au) - r$(APu) - s$(Ai3u) - z$(A4u) = f,

D(B4) = D(A4). (3)

are solved usually by more simple methods. Such equations are used for solving Fredholm ordinary and partial

IDEs with initial and boundary conditions, when A is a differential operator, functionals $(u), $(A®u), i =

= 1,2, 3,4 are Fredholm integral operators with separable kernels. For example, the problem [13]

z''(x) - X Jo1 x[z''(y)+ z(y)]dy = x, z(0) = 1, z'(0) = 0,

by substitution u(x) = z(x) - 1 is reduced to

u''(x) - Xx /0V(y) + u(y)]dy =(1 + 2X)x, u(0) = 0, u'(0) = 0,

which is of the type (1), where

D(A) = {u(x) e C 1[0,1] : u(0) = 0}, D(A2) = {u(x) e C2[0,1] : u(0) = u'(0) =0}, p = r = Xx, q = = 0, $(u) = Jo1 u(y)dy, $(A2u) = Jo1 u '' (y)dy.

The operator B3 with the third degree operator A which we study in (2), has a more complex shape then in [9]. The main result of this paper is Theorem 1.3. Finally, we give one example of integro-differential equation demonstrating the power and usefulness of the method presented. By C we denote the set of all complex numbers and by X, Y the complex Banach spaces. The domain and range of a linear operator P : X ^ Y will be designated by D(A) and R(A), respectively. We recall that a linear operator P : X ^ Y is said to be injective (or uniquely solvable) if for all u1,u2 € 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. If an operator P is injective (correct), then the corresponding equation Pu = f is called uniquely solvable (correct).

If € X *,i = 1,... ,m, then we denote by ^ = co/(^1,..., and ^(x) =col(^1(x),..., ^m(x)).

If g1,... ,gm € X, then we write g = (g1,... ,gm) € Xm. We will denote by ^(g) the n x n matrix whose i,j-th entry ^i(gj) is the value of functional ^ on element gj. It is easy to verify that for a constant m x k matrix C holds ^(gC) = ^(g)C. We denote below by 0m the zero matrix and by Im the identity m x m matrix. By 0 we will denote the zero column vector.

1. Main Results

First we generalize Theorem 1 [7], where prove the necessary solvability condition of the operator B2 without to clame the linear independence of the vectors p, q, r.

Theorem 1.1 Let X be a complex Banach space, A a bijective linear operator and I = A-1 its inverse,

$ = co/($i, ..., $m) G Xm, and p = (pi, ...,pm), q = (qi,..., qm), r = (ri, B2 : X ^ X be defined by

,rm) G Xm. Let the operator

B2u D(B2)

A2u - p$(u) - q$(Au) - r$(APu) = f, D(A?).

The following statements are true:

(i) The operator B2 is injective (uniquely solvable) if and only if

det W = det

$(r) - U $(/ r) $(/2r)

$(q)

$(/q) - 1

$(/2q)

$(p)

$(/p) $(/2p) - 1r

= 0.

(4)

(5)

(ii) If the operator B2 is injective, then it is bijective and the unique solution to (4) for any f € X is given by

( *(f) \

u = B-1 f = I2f -{ I2r I2q I2p ) W-1 ( $(If) ) . (6)

V *(I2f) )

(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) The sufficient solvability condition is proved as in [7]. We prove now the neccesary solvability condition, i.e. 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 = co/(c1, c2, c3), where ci = co/(ci1,..., cim), i = 1, 2, 3, such that Wc = 0. Consider the element u0 = I2(rc1 + qc2 + pc3) € D(A2). Note that u0 = 0, alternatively rc1 + qc2 + pc3 = 0 and from Wc = 0 and the linearity of a functional vector $ follows that

$(r) - 1m $(q) $(p) ] r c1

Wc = $(Ir) $(Iq) - 1m $(Ip) c2

$(I2r) $(I2q) $(I2p) - c3

$(r)c1 - c1 + $(q)c2 + $(p)c3 $(Ir)c1 + $(Iq)c2 - c2 + $(Ip)c3 _ $(I2r)c1 + $(I2q)c2 + $(I2p)c3 - c3 $(rc1 + qc2 + pc3) - c1 $ (I(rc1 + qc2 + pc3)) - c2 _ $ (I2(rc1 + qc2 + pc3)) - c3 _

Then ci = 0, i = 1,2,3 and so c = 0. But by hypothesis c = 0. So u0 = 0. That u0 € ker B2 is proved as in [7]. Thus we proved that, if B2 is an injective operator then det W = 0. Statments (ii), (iii) are proved as in Theorem 1 [7].

ci " 0 "

=- 02 = 0

. C3 _ 0

We generalize Theorem 2 [7], where prove the necessary solvability condition of the operator B4 without to clame the linear independence of the vectors p, q, r, s, z.

— 1

Theorem 1.2 Let the space X, the operator A and its inverse operator I = A , and the vectors $, p, q, r, as above. Let s = (si;..., sm), z = (zi,.., zm) G Xm and the operator B4 : X ^ X be defined by

B4u = A4u - p$(u) - q$(Au) - r$(A2u) - s$(A3u) - z$(A4u) = f

D(B4) = D(A4),

where f € X. Then the following statements are true: (i) The operator B4 is injective if and only if

(7)

det V = 0,

where

V

$(z) - 1

$(I z) $(I2z) $(I3z) $(I4z)

$(s) $(Is) - 1m

$(I2s) $(I3s)

$(I4s) $(q)

$(Iq) $(I2 q)

$(I3q) - 1 $(I4 q)

(ii) If the operator B4 is injective, then it is bijective and the unique solution to (7) for any f € X is given by

/ $(f) \ $(If)

u = I4f - ( I4z 14s 14r 14q /4p ) V-1

$(r) $(I r) $(I2r) - 1m $(I3r)

$(I4r )

$(p)

$(I p) $(I2p) $(I3p) $(I4p) -1,

$(I2f) $(I3f)

V $(I4f) y

(8)

(iii) If the inverse operator I = A-1 is bounded on X, that is, A is correct, then the operator B4 is correct. Proof. (i) The sufficient solvability condition is proved as in [7]. We prove now the neccesary solvability condition, i.e. 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 of constants c = col(c1, c2, c3, c4, c5), where c = col(ci1,..., cim), i = 1,..., 5 such that Vc = 0. Consider the element u0 = I4(zc1 + sc2 + rc3 + + qc4 + pc5) € D(A4). Note that u0 = 0, alternatively zc1 + sc2 + rc3 + qc4 + pc5 = 0 and from Vc = 0 and the linearity of a functional vector $ follows that

" $(r)

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

$(13r)

Vc

$(z) - 1 $(I z) $(I2z) $(I3z) $(I4z)

$(s) $(Is) - 1m

$(I2s) $(I3s)

$(I4s) $(q)

$(I q) $(I2q)

$(I3q) - 1, $(I4q)

ci

c2

c3

c4

c5

$(I4r)

$(p)

$(I p) $(I2p) $(I3p) $(I4p) - 1

$(z)ci - ci + $(s)c2 + $(r)c3 + $(q)c4 + $(p)c5 $(Iz)ci + $(Is)c2 - c2 + $(Ir)c3 + $(Iq)c4 + $(Ip)c5 $(I2z)ci + $(I2s)c2 + $(I2r)c3 - c3 + $(I2q)c4 + $(I2p)c5 $(I3z)ci + $(I3s)c2 + $(I3r)c3 + $(I3q)c4 - c4 + $(I3p)c5 $(I4z)ci + $(I4s)c2 + $(I4r)c3 + $(I4q)c4 + $(I4p)c5 - c5

$(zci + sc2 + rc3 + qc4 + pc5) - ci $ (I(zci + sc2 + rc3 + qc4 + pc5 ) $ (12 (zci + sc2 + rc3 + qc4 + pc5 $ (13(zci + sc2 + rc3 + qc4 + pc5 $ (14(zci + sc2 + rc3 + qc4 + pc5

ci ci " 0 "

- c2 c2 0

- c3 = - c3 = 0

- c4 c4 0

- c5 c5 0

Then ci = 0, i = 1,..., 5 and so c = 0. But by hypothesis c = 0. So u0 = 0. That u0 € ker B4 is proved as in [7]. Thus if B4 is an injective operator then det V = 0. Statments (ii), (iii) are proved as in Theorem 2

[7].

Below we generalize Theorem 3.4 [9], where in the case of a Hilbert space H was proved the correctness and selfadjointness of the operator B3 corresponding to the boundary value problem:

B3x = A3x - Y {Ax, F t)H m - S{Azx, F t)H m - G(A3 x, F t)Hm = f.

72,

■Â3,

D(Bs) = D( Â3 ),

where A is a correct selfadjoint operator, Y = a42G - S {F t,AG)Hm - G {F t,A2G)

) Hm

S = AG - G(Fl,AG)Hm and C is a Hermitian m x m matrix.

Theorem 1.3 Let the space X, the operator A and its inverse operator I = A-1, $, p, q, r, s as above. Let the operator B3 : X ^ X be defined by

B3u = A3u - p$(u) - q$(Au) - r$(A2u) - s$(A3u) = f,

D(B3)

D(A3),

and the vectors

(10)

where f £ X. Then the following statements are true: (i) The operator B3 is injective if and only if

where

L

det L = 0,

Ф(8) - 1m Ф(Г) Ф(ч) Ф(Р)

Ф(/8) Ф(1 Г) - 1m Ф(1 q) Ф(/р)

ф(/2б) ф(/2г) ф(1 2q) - im Ф(/2р)

ф(/3б) ф(/3г) Ф(13q) Ф(/3р) - 1„

(11)

(ii) If the operator B3 is injective, then B3 is bijective and the unique solution to (10) for any f G X is given by

( Ф(Л \

B-lf = I3f -{ I3s I 3г 13q I3p ) L-

Ф(If ) Ф(12f)

V Ф(!^) }

(12)

(iii) If the inverse operator I = A 1 is bounded on X, that is, A is correct, then the operator B3 is correct. Proof. (i) Let det L = 0 and u € ker B3. Then

A3u - рФ(и) - qФ(An) - гФ(A2u) - вФ(A3u) = 0, u G D(A3).

(13)

By applying the inverse operator I = A 1 on the both sides of (13) and on the equations following from (13), we get

Â2u - 1рФ^) - IqФ(Au) - IгФ(Â2u) - !вФ(A3u) = 0, Au - 12pФ(u) - I^Ф(Au) - 12гФ(l2u) - 12яФ(I3u) = 0, u - I^(u) - I^Ф(Au) - I3гФ(l2u) - 13яФ(l3u) = 0.

(14)

Now applying the functional $ on the both sides of (13) and the above system, we obtain the system

[$(s) - 1m]$(23u) + $(r)$(l2u) + $(q)$(lu) + $(p)$(u) = 0, $(Is)$(A3u) + [$(Ir) - 1m]$(A2u) + $(Iq)$(Au) + $(Ip)$(u) = 0, $(I2s)$(A3u) + $(I2r)$(A2u) + [$(12q) - 1m]$(Au) + $(I2p)$(u) = 0, $(I3s)$(23u) + $(I3r)$(l2u) + $(I3q)$(lu) + [$(I3p) - 1m]$(u) = 0,

or

( Ф(Â3u) \

L

where the matrix L is given in (11)

Ф( Â3u)

Ф( A2u) Ф( Au) \ ФН j Then, since det L = 0, we get

Ф( Au) = Ф(u)

0 0

0 0

Ф( Â2u)

0.

(15)

(16)

1

u

C1

C2

C3

. C4 -

C1 " 0 "

C2 0

C3 0

. C4 _ 0

Substitution (16) into (14) implies that u = 0. Thus ker B3 = {0} and hence B3 is an injective operator. Conversely. We prove that if B3 is an injective operator then det L = 0, or equivalently, if det L = 0, then B3 is not injective. Let det L = 0. Then there exists a nonzero vector of constants c = col(c1, c2, c3, c4), where c = col(ci1,..., cim), i = 1,..., 4 such that Lc = 0. Consider the element u0 = I3(sc1 + rc2 + qc3 + + pc4) € D(A4). Note that u0 = 0, because alternatively sc1 + rc2 + qc3 + pc4 = 0 and then from Lc = 0 follows that Lc =

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

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

$(I2s) $(I2r) $(I2q) - 1m $(I2p) $(l3s) $(l3r) $(I3q) $(I3p) - 1

$(s)c1 - c1 + $(r)c2 + $(q)c3 + $(p)c4 $(Is)c1 + $(Ir)c2 - c2 + $(Iq)c3 + $(Ip)c4 $(I2s)c1 + $(I2r)c2 + $(I2q)c3 - c3 + $(I2p)c4 $(13s)c 1 + $(l3r)c2 + $(l3q)c3 + $(I3p)c4 - c4

$(sc1 + rc2 + qc3 + pc4) - c1 $ (I(sc1 + rc2 + qc3 + pc4)) - c2 $ (12(sc1 + rc2 + qc3 + pc4)) - c3 $ (I3 (sc1 + rc2 + qc3 + pc^ - c4

Then c = 0, i =1,..., 4 and thus we obtain c = 0. Remind that by hypothesis c = 0. So u0 = 0. We will prove that u0 € ker B3. Indeed

B3u0 = A3u0 - p$(u0) - q$(Au0) - r$(A2u0) - s$(A3u0) = sc1 + rc2 + qc3 + pc4 - p$ (I3 (sc1 + rc2 + qc3 + pc4)) -q$ (I2(sc1 + rc2 + qc3 + pc4)) - r$ (I(sc1 + rc2 + qc3 + pc4)) s$ (sc1 + rc2 + qc3 + pc4)

= -(s, r, q, p)Lcol(c1, c2, c3, c4) = -(s, r, q, p)Lc = -(s, r, q, p)0 = 0.

Then ker B3 = {0} and B3 is not injective. Thus we proved that B3 is an injective operator if and only if det L = 0.

(ii) From (i) follows that det L = 0. Then problem 10 has a unique solution. Acting as in the proof of (i) for any f € X we get

A2u - Ip$(u) - Iq$(Au) - Ir$(A2u) - Is$(A3u) = If, Au - 12p$(u) - I2q$(Au) - I2r$(A2u) - I2s$(A3u) = I2f, u - 13p$(u) - 13q$(Au) - I3r$(A2u) - 13s$(A3u) = I3f.

Now applying the functional $ on the both sides of (13) and (18), we obtain

[$(s) - 1m]$(A3u) + $(r)$(A2u) + $(q)$(Au) + $(p)$(u) = -$(f), -$(Is)$(A3u) + [$(Ir) - 1m]$(A2u) + $(Iq)$(Au) + $(Ip)$(u) = -$(If), -$(I2s)$(A3u) + $(I2r)$(A2u) + [$(12q) - 1m]$(Au) + $(I2p)$(u) = -$(I2f), -$(I3s)$(A3u) + $(I3r)$(A2u) + $(13q)$(Au) + [$(I3p) - 1m]$(u) = -$(I3f),

(17)

(18)

or

The last equation gives

V

/ $(A3w) \ $(A2w)

$(Aw) \ $(«)

/

( $(A3w) \ $(A2w) $(Aw)

V $(u) )

-L-

( *(f) \ *(If)

$(/2 f)

V ^(i3 f)

/ 3(f) \

w)

$(/2f)

V *(I3f) J

(19)

Substituting (19) into (18), we get the unique solution (12) to problem (10). Since this solution holds for any f € X, then R(B3) = X, which means that B3 is bijective.

(iii) If the inverse operator I = A-1 is bounded, then the operator B-1, defined by (12) is bounded, since the operator A-1 and the components of the vector $ are bounded. Hence, B3 is correct.

i

Example 1.1 The operator BI : C[0,1] ^ C[0,1] which corresponds to the problem

u'''(t) — 30t xu(x)dx — 10(t2 + 1) / xu' (x)dx — t3 xu''(x)dx Jo Jo Jo

= t3 + It2 + 9t — 5, u(0) = u' (0) = u' '(0) = 0,

is correct and the unique solution to ( 20), for every f G C[0,1], is given by

u = t4 - 2t3.

(21)

Proof We refer to Theorem 1.3. If we compare equation ( 20) with equation ( 10) it is natural to take A3u = u'''(t), with D(A3) = {u(t) :G C3[0,1] : u(0) = u'(0) = u''(0) = 0}, and so A is defined by Au = u'(t), D(A) = {u(t) G Cx[0,1] : u(0) = 0}. Then

A2u = u''(t), D(A2) = {u(t) G C2[0,1] : u(0) = u'(0) = 0}. It is easy to verify that for any f G C[0,1]

Also we can take f-1

A-1f (t) = If = /0 f (x)dx, A-2 f = I2f = I(If), A-3f = I3f = I(I2f). p = p = 30t, q = 10(t2 + 1), r = r = t3, s = s = 0,

$(u) = /0 xu(x)dx, $(Au) = /0 xu'(x)dx, $(A2u) = /Q xu' ' (x)dx, f = t3 + 7t2 + 31t - 5. Then by Derive programm, we compute Is = I2s = I3s = $(s) = $(Is) = $(I2s) = $(I3s) = 0, Ir = J4 x3dx = , Iq = ^f- + 10t, Ip

12r = I (Ir) = J0 x3dx = 20, 12 q = I (Iq) = ^ + 5t2, I2p = I (Ip) = 5t3,

15t2

5t4

6

I3r = I (I2r) = 120, 13q = I (12q) = J + 513, I3p -$(r) = JO,1 x(x3)dx = 1/5, $(q) = 10 ¡0 x(x2 + 1)dx $(p) = 30 J01 x2dx = 10,

I(I2p) = 15/2,

5 .4

$(Ir) = 1 /o x(x4)dx = 24, $(Ip) = 15 ¡0 x3dx = 15/4,

$(Iq)

JO,1 x( ^ + 10x)da

4,

r) = 140,

^(13 r) = 960,

$(I2q) = 25/18, $(I2p) = 1,

$(l3q) = 5/14, $(I3p) = 5/24. Since

det L

$(s) - !m $(r) $(q) $(p)

$(I s) $(Ir) - 1m $(Iq) $(Ip)

$(I2 s) $(12r) $(I2q) - 1m $(I2p)

$(13s) $(I3r) $(I3q) $(I3p) -

-1 1/5 15/2 10

= 0 0 1/24-1 1/140 4 15/4 25/18 - 1 1 = 0,

0 1/960 5/14 5/24 - 1

problem (20), by Theorem 1.3, is correct. Further we compute

If = JOt(x3 + 7x2 + 31x - 5)dx = 414 + 713 + f-12 - 5t,

I (If) = 2015 +17214 + 3113 - 512,

I( 12 f) = _L_t6 + JLt5 + 3114 _ 513

1 (I J ) = 1201 + 601 + 241 at

I f

— ^ J J — I201 __ __ „

$(f) = 587/60, $(If) = 163/60, $(I2f) into (12), we obtain (21).

323/630, $(I3f) = 191/2880. Substituting the abow values

4

m

References

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