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Scientific article
DOI: 10.18287/2541-7525-2021-27-1-29-43 Submited: 15.01.2021
Revised: 17.02.2021 Accepted: 28.02.2021
E. Providas
University of Thessaly, Larissa, Greece E-mail: providas@uth.gr. ORCID: https://orcid.org/0000-0002-0675-4351
L.S. Pulkina
Samara National Research University, Samara, Russian Federation E-mail: louise@samdiff.ru. ORCID: https://orcid.org/0000-0001-7947-612
I.N. Parasidis University of Thessaly, Larissa, Greece E-mail: paras@teilar.gr. ORCID: https://orcid.org/0000-0002-7900-9256
FACTORIZATION OF ORDINARY AND HYPERBOLIC INTEGRO-DIFFERENTIAL EQUATIONS WITH INTEGRAL BOUNDARY
CONDITIONS IN A BANACH SPACE
ABSTRACT
The solvability condition and the unique exact solution by the universal factorization (decomposition) method for a class of the abstract operator equations of the type
Biu = Au - ^(Aow) - GF(Au) = f, u e D(Bi),
where A, A0 are linear abstract operators, G, S are linear vectors and F are linear functional vectors is investigagted. This class is useful for solving Boundary Value Problems (BVPs) with Integro-Differential Equations (IDEs), where A,A0 are differential operators and F(Au), $(A0u) are Fredholm integrals. It was shown that the operators of the type Bi can be factorized in the some cases in the product of two more simple operators BG, BGo of special form, which are derived analytically. Further the solvability condition and the unique exact solution for Bi u = f easily follow from the solvability condition and the unique exact solutions for the equations BGv = f and BGou = v.
Key words: correct operator; factorization (decomposition) method; Fredholm integro-differential equations; initial problem; nonlocal boundary value problem with integral boundary conditions.
Citation. Providas E., Pulkina L.S., Parasidis I.N. Factorization of ordinary and hyperbolic integro-differential equations with integral boundary conditions in a Banach space. Vestnik Samarskogo universiteta. Estestvennonauchnaia seriia = Vestnik of Samara University. Natural Science Series, 2021, vol. 27, no. 1, pp. 29-43. DOI: http://doi.org/10.18287/2541-7525-2021-27-1-29-43.
Information about the conflict of interests: authors and reviewers declare no conflict of interests.
© Providas E., 2021
Efthimios Providas — Candidate of Technical Sciences, associate professor, University of Thessaly, Larissa, 41110, Greece.
©c Pulkina L.S., 2021
Ludmila Stepanovna Pulkina — Doctor of Physical and Mathematical Sciences, professor, Department of Differential Equations and Control Theory, Samara National Research University, 34, Moskovskoye shosse, 443086, Russian Federation. ©c Parasidis I.N., 2021
Ioannis Nestorios Parasidis — Candidate of Technical Sciences, associate professor, University of Thessaly, Larissa, 41110, Greece.
1. Initial Position
Integro-differential equations play an important role in characterizing many physical, biological, social and engineering problems and are often solved by factorization (decomposition) methods. The factorization methods have applications in biology, ecology, population dynamics, mathematics of financial derivatives, quantum physics, hydrodynamics, gas dynamics, in transport theory, electromagnetic theory, mechanics and chemistry [1-10]. Factorization Methods successfully are used in pure mathematics for solving linear and nonlinear ordinary and partial differential and Volterra-Fredholm integro-differential equations, integro-differential equations of fractional order, fuzzy Volterra-Fredholm integral equations and delay differential equations [11-20]. There are well-known decomposition (factorization) methods: Domain decomposition method, the natural transform decomposition method, the Adomian decomposition method, Modified Adomian decomposition method and the Combined Laplace transform-Adomian decomposition method, which use the so-called Adomyan polynomials or iterations to obtain an n-term approximation of solution, whereas the proposed in this paper factorization method gives the unique exact solution in the closed form. Furthermore it is universal because can be applied in the investigation of Fredholm integro-differential equations, both ordinary and partial.
There are many papers are devoted to investigation of the uniqueness of the solution to nonlocal boundary value problems with integral boundary conditions for hyperbolic differential equation [21-25]. Finding of the exact solution in the general case is the difficult task. We by the universal factorization method find the solvability condition and a unique solution to a nonlocal BVP with integral boundary conditions for Fredholm ordinary integro-differential and integro-hyperbolic differential equations of the type Biu = f. It is the aim of this paper to reappraise the factorization method for integro-differential equations of type Biu = f. This paper is a generalization of the article [26], where by factorization method were stydied the solvability condition and a unique solution to the correct self-adjoint abstract equation of the type Biu = f in terms of a Hermitian matrix in a Hilbert space.
The quadratic factorization methods was applied to some BVPs with integro-differential equations in the case of a Banach space in [27-29].
It is well known that the class of the operators which can be factorized as a superposition of two more simple operators is not wide. But if the operator can be factorized, then the solvability condition and the solution of the given problem are essentially simpler than in the general case without factorization. The paper is organized as follows. In Section 2 we develop the theory for the solution of the problem Bix = f when Bi = BB0. Further by factorization method we solve a nonlocal boundary value problem with integral boundary conditions for Fredholm integro-hyperbolic differential equation. Finally, we give two examples of integro-differential equations demonstrating the power and usefulness of the methods presented. Throughout this paper we use the following terminology and notation. By X,Y we denote the complex Banach spaces and by X* the adjoint space of X, i.e. the set of all complex-valued linear and bounded functionals on X. We denote by f (u) the value of f on u e X. We write D(A) and R(A) for the domain and the range of the operator A : X ^ Y, respectively. An operator A2 is said to be an extension of an operator Ai, or Ai is said to be a restriction of A2, in symbol Ai C A2, if D(A2) 2 D(Ai) and Aix = A2x, for all x e D(Ai). An operator A : X ^ Y is said to be injective or uniquelly solvable if for all ui,u2 e D(A) such that Aui = Au2, follows that ui = u2. Remind that a linear operator A is injective if and only if ker A = {0}. An operator A : X ^ Y is called surjective or everywhere solvable if R(A) = Y. The operator A : X ^ Y is called bijective if A is both injective and surjective. Lastly, A is said to be correct if A is bijective and its inverse A-i is bounded on Y. If gi e X and ^ e X*,i = l,...,m, then we denote by g = (gi,...,gm), ^ =col(^i,..., ^m) and ^(u) =col(^i(u),..., ^m(u)) and write g e Xm, ^ e Xm. We will denote by ^(g) the m x m matrix whose i,j-th entry 1ii(gj) is the value of functional ^ on element gj. Note that ^(gC) = ty(g)C, where C is a m x k constant matrix. We will also denote by 0m the zero and by Im the identity mxm matrices. By 0 we will denote the zero column vector.
2. Factorization of integro-differential equations in a Banach space
We remind first the following Theorem 1 from [29]. Theorem 2.1. Let A be a bijective operator on a Banach space X, the components of the vectors G = = (gi, ...,gm), F = col(Fi, ...Fm) arbitrary elements of X and X*, respectively and the operator BG : X ^ X be defined by
BGu = Au - GF(Au) = f, D(Bg) = D(A), f e X. (2.1)
Then the following statements are true:
(i) The operator BG is bijective on X if and only if
det L = det[Im - F(G)] = 0, (2.2)
and the unique solution to boundary value problem (2.1), for any f e X, is given by the formula
u = B-1 f = A-1f + A-1G[Im - F(G)]-1F(f). (2.3)
(ii) If in addition the operator A is correct, then BG is correct.
Now, by using the above theorem we prove the following theorem which is useful for solving integro-differential equations by factorization method.
Theorem 2.2. Let X be a Banach space, the vectors G0 = (g^,..., gin1), G = (g1,..., gm), S = (s1,...,sm) e Xm, the components of the vectors F = col(F1,..., Fm) and $ = col(^1,..., ^m) belong to X* and the operators BGo ,BG,B1 : X ^ X defined by
BGou = A0u - G0$Au) = f, D(Bgo) = DA), (2.4)
BGu = Au - GF(Au) = f, D(Bg)= D(A), (2.5)
B1u = AA0u - S$(A0u) - GF(AA0u) = f, D(B1) = D(AA0) (2.6)
where A0 and A are linear correct operators on X and G0 e D(A)m. Then the following statements are satisfied:
(i) If
S e R(BG)m and S = BgG0 = AG0 - GF(AG0), (2.7)
then the operator B1 can be factorised in B1 = BGBGo.
(ii) If in addition the components of the vector $ = ^^..., $m) are linearly independent elements of X* and the operator B1 can be factorised in B1 = BGBGo, then (2.7) is fulfilled.
(iii) If the operator B1 can be factorised in B1 = BGBGo, then B1 is correct if and only if the operators BGo and BG are correct which means that
det L0 = det[Im - $(G0)] = 0 and det L = det[Im - F(G)] = 0. (2.8)
(iv) If the operator B1 has the factorization in B1 = BGBGo and is correct, then the unique solution of (2.6) is
u = B-1f = A-1A-1f + [A-1A-1G + A-1G0L-1$(A-1G)] x
xL-1F (f) + A-1G0L-1$(A-1f). (2.9)
Proof. (i) Taking into account that G0 e D(A)m and (2.4)-(2.6) we get
D(BgBgo) = {u e D(Bgo): Bg0u e D(Bg)} =
= {u e D(A0) : A0u - G0$(A0u) e D(A)} = = {u e D(A0) : A0u e D(A)} = D(AA0) = DB).
So D(B1) = D(BgBg0). Let y = Bg0u. Then for each u e D(AA0) since (2.5) and (2.4) we have
BgBgo u = BGy = Ay - GF (Ay) =
= A[A0u - G0$(A0u)] - GF (A[A0u - G0$(A0u)]) =
= AA0u - AG0$(A0u) - GF(AA0u) + GF(AG0)$(A0u) = (2.10)
= AA0u - [AG0 - GF(AG0)]$(A0u) - GF(AA0u) = = AA0u - BGG0$(A0u) - GF(AA0u),
where the relation BGG0 = AG0 - GF(AG0) follows from (2.5) if instead of u we take G0. By comparing (2.10) with (2.6), it is easy to verify that B1u = BGBGou for each u e D(AA0) if a vector S satisfies (2.7).
(ii) Let the operator B1 can be factorized in B1 = BGBGo. Then by comparing (2.10) with (2.6) we obtain
(BgG0 - S)$(A0u) = 0. (2.11)
Because of the correctness of operators A,A0 and the linear independence of $1,..., $m, there exists a system u1, ...,um e D(AA0) such that $(A0u0) = Im where u0 = (u1, ...,um). By substituting u = u0 into (2.11) we get S = BgG0. Hence S e R(BG)m and S = BgG0 = AG0 - GF(AG0).
(iii) Let the operator B1 be defined by (2.6) where S = BgG0. Then Equation (2.6) can be equivalently represented in matrix form:
B1u = AA0u - (BgG0, G)( $AAAAU)U)) = f (2.12)
or
B1u = Au — GF(Au) = f, D(B1) = D(A), (2.13)
where A = AAq, G = (BqG0 ,G), F =со1(Ф ,F ), F (v) = ( = ( Ф(А V
(BgGo ,G), F =col($ ,F ), F (v) = ^ F(V) J = ^ F(v) ) ' Notice that the
operator A = AA0 is correct, because of A and A0 are the correct operators and that the functional vector F is bounded, since the vector $ is bounded as a superposition of a bounded functional $ and a bounded operator A_1. Then we apply Theorem 2.1. By this theorem the operator B1 is correct if and only if
det L1 = det[l2m — F (G )] = det
Im 0m \ / <^(BgGq) &(G) 0m Im J V F (BgGq) F (G)
= d.( Im - $(AGo - GF (AGo)) -$(G) \ = aei V -[F (AGo - GF (AGo))] Im - F (G) J
= d t ( Im - $G - A-1GF(AGo)) -$(A-1G) ) = t \ -[F (AGo - GF (AGo))] Im - F (G) )
= det ( Im - $G) + $(A-1G)F(AGo) -$(-A-iG) ) = 0 t \ -F(AGo) + F(G)F(AGo) Im - F(G) J = °
Multiplying from the left the elements of the second column by F(AG0) and adding to the corresponding elements of the first column of the determinant Li, by Remark 1, [31] we get
det Li = det 0m - $(Go) (G) ) = det[Im - $Go)] det[Im - F(G)]
= det L0 det L = 0.
So we proved that the operator Bi is correct if and only if (2.8) is fulfilled.
(iv) Let u e D(AA0) and BGBGou = f. By Theorem 2.1 (ii) since BG,BGo are correct operators, we obtain
Bgo u = B-if = A-if + A-i GL-iF (f),
u = B-1 (A-1f + A-1GL-1F(f)) .
In the last equation we denote by g = A-if + A-iGL-iF(f). Bu using again Theorem 2.1 (ii), with A0,G0, $,L0, in place of A,G,F,L respectively, we get
u = B-lg = A- ig + A-GoL-^g) = A-1 (A-i f + A-iGL-iF (f)) +
+A-1GoL-1$ (A-1 f + A-1GL-1F(f)) = A0-1A-1 f + A--1 A-1 GL-1F(f)+
+A-1GoL-1 [$(A-1f) + $(A-1 G)L-1F (f)]
which implies (2.9). The theorem is proved. □
The next theorem is useful for applications. Theorem 2.3. Let the space X and the vectors F, $ be defined as in Theorem 2.2, the vectors G = = (g1,...,gm), S = (s1,..., sm) e Xm and the operator B1 : X ^ X by
B1u = Au - S$(A0u) - GF(Au) = f, x e D(B1) (2.14)
where A0 : X ^ X is a correct m-order differential operator and A is a n-order differential operator, m < n. Then the next statements are fulfilled:
(i) if there exist a n - m order differential operator A : X ^ X, such that
A = AAo, D(B1) = D(AAo), (2.15)
and a vector G0 e D(A), satisfying
AG0 - GF(AG0) = S, (2.16)
then the operator B1 can be factorized into B1 = BGBGo, where BGo and BG are given by (2.4) and (2.5) respectively, BG is determined by A and G,F from (2.14), (2.15) and lastly, the operator BGo by A0, $ and G0 from (2.14) and (2.16),
(ii) if there exists a bijective n - m order differential operator A : X ^ X, satisfying (2.15) and
det L = det[Im - F(G)] = 0, (2.17)
then the operator B1 is factorized in B1 = BGBGo, where the operators BGo, BG, A0,A, the vectors G,F, $ are determined as in (i) and
G0 = A-1S + A-1 GL-1F (S). (2.18)
(iii) if in addition to (ii) A is correct, then B1 is correct if and only if
det L0 = det[Im - $(G0)] = det[Im - $(A-1S) - $(A-1G)L-1F(S)] = 0, (2.19)
and the problem (2.14)-(2.16) has the unique solution given by (2.9).
Proof (i) If there there exist a n - m order differential operator A and a vector G0 satisfying (2.15) and (2.16), then from (2.14) we get
B1u = AA0u - S$(A0u) - GF(AA0u) = f, u e D(AA0). (2.20)
From (2.20) we take a triple of elements, the operator A and vectors G, F, and construct the operator BG according to the formula (2.5). To determine the operator BGo by formula (2.4), we take from (2.20) the operator A0 and the vector $, whereas as G0 we take any solution G0 of Equation (2.16). We proved in the previous theorem (i) that D(BGBGo) = D(AA0) = D(B1). Substituting (2.16) into (2.20), for every u e D(B1) we get
B1 u = AA0u - [AG0 - GF(AG0)] $(A0u) - GF(AA0u) = BGA0u - BGG0$(A0u) = = Bg [A0u - G0$(A0u)] = BgBg0u.
Thus B1 = BgBg0 .
(ii) As in the proof of (i) we construct the operators BG,BGo. By Theorem 2.1, since (2.17), the operator BG is correct and Equation (2.16) can be presented by BgG0 = S. Then G0 = B-1S. The last equation by Corollary 2.1, implies the unique vector G0 by (2.18). Further as in the proof of (i) we get the factorization B1 = BGBGo, where BGo is unique.
(iii) If (2.17), (2.18) hold true, then by statements (i), (ii), B1 can be factorized in B1 = BGBGo. By Theorem 2.2 (iii), B1 is correct if and only if (2.8) holds or, taking into account (2.17) and (2.18), if and only if det L0 = det[Im - $(G0)] = 0, or if and only if (2.19) is fulfilled. The last inequality immediately follows by substitution (2.18) into det L0 = det[Im - $(G0)]. Since B1 is correct and factorized in B1 = BGBGo, by Theorem 2.2 (iv), we obtain the unique solution (2.9) to the problem (2.14)-(2.16). So the theorem is proved. □ Example 2.4. Let u(x) e C2[0,1]. Then the problem
u''(t) - t f01 tu'(t)dt - t2 f01 t3u''(t)dt = 2t + 1, (2.21)
u(0)+ u(1) = 0, u'(0) - 2u'(1) = 0,
is correct on C[0,1] and its unique solution is given by the formula
, N 261377 - 665232t + 103608t2 + 30080t3 + 8790t4
u(t) =-------. (2.22)
v 7 207216 v 7
Proof. First we need to find the operators B1,A,A0 and check the condition D(B1) = D(AA0). If we compare equation (2.21) with equation (2.14), (2.15), it is natural to take
B1u(t) = u''(t) - t f01 tu'(t)dt - t2 f01 t3u''(t)dt = 2t + 1, (2.23)
D(B1) = {u(t) e C2[0, 1] : u(0) + u(1) = 0, u'(0) - 2u'(1) = 0},
Au = AA0u = u''(t),
D(B1) = {u(t) e C2[0, 1] : u(0)+ u(1) = 0, u'(0) - 2u'(1) = 0}, (2.24)
A0u(t) = u'(t), D(A0) = {u(t) e C 1[0,1] : u(0) = -u(1)},
$(A0u) = [ tu'(t)dt, F(AA0u) = i t3u''(t)dt, (2.25)
00
S = t, G = t2. Denote A0u(t) = u'(t) = y(t) = y. Then from (2.24) we have y e D(A), AA0u = (u'(t))' = = y'(t) = Ay(t), y(0) - 2y(1) = 0. So we proved that
Ay = y'(t), D(A) = {y(t) e C1 [0,1] : y(0) - 2y(1) = 0}.
Then by definition
D(AA0) = {u(t) e D(A0) : A0u(t) e D(A)} =
= {u(t) e C 1[0,1] : u(0) = -u(1), u'(t) e C 1[0,1], u'(0) - 2u'(1) = 0} = = {u(t) e C2[0, 1] : u(0) + u(1) = 0, u'(0) - 2u'(1) = 0} = D(B1).
So D(B1) = D(AA0). It is easy to verify that the operators A,A0 are correct on C[0,1] and for every f (t) e C[0,1] the corresponding inverse operators are defined by
A-1f (t) = /o f (s)ds - 2 fo1 f (s)ds, (2.26)
A-f (t) = /0 f (s)ds - f /1 f (s)ds. (2.27)
From (2.25) we have
$(f) = f sf(s)ds, F(f) = f s3f(s)ds. (2.28)
Jo Jo
Then F (G) = f01 s3s2ds = 1, F (S) = f01 s3sds = 1, det L = det[Im - F(G)] = 1 - 1/6 = 5/6, L-1 = 6/5,
ft r 1 t2 ft r 1 t3 2
A-1S = sds - 2 sds = — - 1, A-1G = s2ds - 2 s2ds = — --, Jo jo 2 jo jo 33
g=a-1s+a-1gl-1f(s)=if -1 ^t3 - 3) H =+25(2 - «>.
Taking into account (2.28) we obtain
1 f1 ,. q „ o , 439
1 i1
$(Gq) = 1 s(4«3 + 25s2 — 58)ds = —50 J q
T0) 50 J0 ' " " 1000'
Since det L0 = det[Im - $(Go)] = 1400 =0 then, by Theorem 2.3 (iii), Problem (2.23) or (2.21) is correct. By (2.27) we calculate
. ^ 73 29t t3 t4 . ^ ^ 7 2t t4
A-1G0 =----1---1--, A-1A-1G =----1--
0 0 150 25 6 50' 0 24 3 12
and for f (t) = 2t + 1 by (2.26)-(2.28) we obtain
10 t2 t3 13 17
A-1f = -4 +1 +12, A-1A-1f =19 - 4t + \ + F(f ) = 20, $(A-1f ) =
Substituting these values into (2.9) we obtain the unique solution of (2.23), which is given by (2.22). □ Example 2.5. Let u(x) G ^3[0,1]. Then the problem
u'''(x) - 8x2 f01 tu'(t)dt - (3x + 1) f01 t2u'''(t)dt = 2x2 - 6x + 4, (2.29)
u(0) = 2 f01 u(t)dt, u'(0) = -u'(1), u''(0) = -u''(1),
is uniquely solvable on C[0,1] and its unique solution is given by
31
u(x) = x3 - 2x2 + 2 ■ (2.30)
Proof. First we must determine the operators B1:A and A0. By comparing Problem (2.29) with (2.14) it is natural to take X = C[0,1],
, 1 , 1 -2 ' tu'(t)dt — (3x +1) t2u'"<
B1u = u'''(x) - 8x2 ( tu'(t)dt - (3x +1) i t2u'''(t)dt = 2x2 - 6x + 4, (2.31)
Jo Jo
D(B1) = {u(x) e C3[0,1] : u(0) = 2 i u(t)dt, u'(0) = -u'(1), u''(0) = -u''(1)},
Jo
$(A0u) = /1 /1 tu'(t)dtdy, F(AA0u) = /1 /1 t2u'''(t)dtdy, (2.32)
Ax = AAou(x) = u''' (x), Aou = u' (x).
Denote v(x) = u'(x). Then AA0u(x) = u'''(x) = (u'(x))" = Av(x) = v''(x). From boundary conditions (2.31) follws that v(0) = u'(0) = -u'(1) = -v(1), v'(0) = u''(0) = -u''(1) = -v'(1). So the operators A,Ao are defined by
Av(x) = v'' (x), D(A) = {v(x) e C2\0, 1] : v(0) = -v(1), v'(0) = -v'(1)},
A0u(x) = u'(x), D(A0) = {u(x) e C 1[0,1] : u(0) = 2 /1 u(x)dx}. Now we make sure that D(B1) = D(AA0). Using the definition of the product operators we get D(AA0) = {u(x) e D(A0) : A0u e D(A)} = {u(x) e C1 [0,1] : u(0) =
= 2 ¡0 u(x)dx, u'(x) G C2[0, 1], u'(0) = —u'(1), u''(0) = —u"(1)} = D(B1).
Since D(B1) = D(AA0), we can apply Theorem 2.3. It is easy to verify that the operators A and A0 are correct and their inverse operators for all f (t) e C[0,1] are given by
A^f (x) = 2 J'0 (t - 1)f (t)dt + f0x f (t)dt, (2.33)
A-1f (x) = 1 fo1 (t - x - 2) f (t)dt + fX(x - t)f (t)dt. (2.34)
By comparing again (2.31) with (2.14) it is natural to take S = S(x) = 8x2, G = G(x) = 3x + 1. From (2.32) we get
$(f) = f tf(t)dt, F(f) = f t2f(t)dt. (2.35)
Jo Jo
Let f (x) = A-1 f (x) and AA0u(x) = f (x). Then, since A0,A are invertible, by means of (2.33) and (2.34) we have f f
u(x) = A-1 A- 1f (x) = A-f(x) = 2 f0 (t - 1)f (t)dt + ] f (t)dt =
= 2 f[(t - 1) [2 fo1 (s -1 - 1) f (s)ds + ft(t - s)f (s)ds] dt+ + foX [2 fo (s - t - 2) f (s)ds + ft(t - s)f (s)ds] dt. Further using Fubini theorem we obtain
A-1A-1f (x) = -1 f01[3x2 + 3x(1 - 2s) - 4s3 + 12s2 - 6s - 1]f (s)ds +
+1 fX(x - s)2f (s)ds. (2.36)
Using (2.36) for f = f (x) =2x2 - 6x + 4 and G = 3x +1 we get
A-1A-1f = -12 f'1 [3x2 + 3x(1 - 2s) - 4s3 + 12s2 - 6s - 1](2s2 - 6s + 4)ds + (2.37)
+1 fX(x - s)2(2s2 - 6s + 4)ds = 1 (2x5 - 15x4 + 40x3 - 25x2 - 10x + 12), A-1A-1 = -12 f1[3x2 + 3x(1 - 2s) - 4s3 + 12s2 - 6s - 1](3s + 1)ds + (2.38)
+1 fX(x - s)2(3s + 1)ds = 110(15x4 + 20x3 - 75x2 + 15x + 19). Using (2.34) for S = S(x) = 8x2, G = G(x) = 3x + 1, f (x) = 2x2 - 6x + 4 we find A-1S = 1 fo (t - x - 1) (8t2)dt + f0X(x - t)(8t2)dt = 2x4-34x+1, A-1 G = 2 f1 (t - x - 1) (3t + 1)dt + f0X(x - t)(3t + 1)dt = 4x2(x+18>-10X+1, A-1f = 1 f'1 (t - x - 2) (2t2 - 6t + 4)dt + f0X(x - t)(2t2 - 6t + 4)dt
= x2(x2-6x+12)-5x-1
= 6 .
Then by using (2.35) we arrive at
$(A-1G) = 1 f1 t[4t2(t +1) - 10t + 1]dt = -,
F(G) = f! t2(3t +1)dt = 12, $(A-1f) = 1 f1 t[t2(t2 - 6t +12) - 5t - 1]dt = -1,
F(f) = ¡1 t2(2t2 - 6t + 4)dt = , F(S) = ¡1 t2(8t2)dt = §. Further by (2.17), (2.18) and (2.19) we find
det L = det[Im - F(G)j = 1 - 13/Y2 = -1/12,
Go = G0(x, y) = A-lS + A-1GL-1F(S) = 2x4-4x+1 + 4x"(x+1^-10x+1 (-12)§
= 10x4-144x3-144x2+34 0 x-31 1§ '
$(G0) = 1§ ¡1 t(10t4 - 144t3 - 144t2 + 340t - 31)dt = 3§7, det Lo =det[Im - $(Go)] = 1 - = - . Since det L, det L0 = 0, by Theorem 2.3, Problem (2.31) or (2.29) is correct. Applying (2.33) we thus have
A-1Go = 2 ¡1(t - 1)Go(t)dt + ¡0x Go(t)dt =
= 1§ ¡01(t - 1)(10t4 - 144t3 - 144t2 + 340t - 31)dt+
+1 Io(10t4 - 144t3 - 144t2 + 340t - 31) dt =
= _ 223 + x(2x4-36x3-48x2 + 170x-31) = 75 + 15 .
Substituting the above values into (2.9) we get the solution (2.30). □
3. Factorization of hyperbolic integro-differential equations with integral boundary conditions
Everywhere below ll = {(x,y) e R2 : 0 ^ x,y ^ 1}. Lemma 3.1. Let a(x),c(x) e C[0,1], K(y) e C[0,1]. Then the operator A : C(i) ^ C(i) corresponding to the problem:
Au(t) = u'y (x, y) + c(x)u(x, y) = f (x, y), (3.1)
D(A) = |u(x, y) e C(i) : u'y (x, y) e C(i), u(x, 0) = a(x) f^ K(y)u(x, y)dy j is correct if and only if
a(x) K (y)e -yc(x)dy = 1, (3.2)
Jo
and the unique solution of the above problem is given by the formula
u(x, y) = A-1f (x, y) = a(x)e-yc(x) (1 - a(x) f^ K(y)e-yc(x) dy^j 1 x
x f01 K(y)e-yc(x) fy f (x,t)etc(x)dtdy + e-yc(x) fy f (x,t)etc(x)dt. (3.3)
Proof. Assume that u(x,y) e ker A and (3.2) hold. Then from (3.1) we get
u'y(x,y) + c(x)u(x, y) = 0, u(x, 0) = a(x) K(y)u(x,y)dy. (3.4)
Jo
From the above equation by integration on y we obtain
u(x, y) = u(x, 0)e-yc(x), u(x, y) = a(x)e-yc(x) f^ K(y)u(x, y)dy, (3.5)
/0 K(y)u(x, y)dy = a(x)fo K(y)e-yc(x)dy ¡0 K(У)u(x, У)dУ,
1 - a(x) f! K(y)e-y<x)dy\ fj K(y)u(x, y)dy = 0.
From the last equation, since (3.2), follows that K(y) u(x,y)dy = 0. Substitution of this value into (3.5) implies u(x,y) =0. This means that the operator A is injective.
Conversaly. Let u(x,y) e ker A and a(x) f^ K(y)e-yc(x)dy = 1. Then (3.4) holds. It is easy to verify that u(x,y) = e-yc(x> satisfies problem (3.4). Thus we prove that u(x,
y) = e-yc(x) e ker A and so A is not
injective.
We will find the solution to (3.1). Let a(x Uo K(y)e-yc(x)dy = 1. Then A is injective and problem (3.1) has a unique solution. From (3.1) by integration on y we obtain
u(x, y) = e-yc(x)a(x) fj K(y)u(x, y)dy + e-yc(x) fy f (x, t)etc(x)dt, (3.6)
/0 K(y)u(x, y)dy = a(x) fg K(y)e-yc(x)dy /J K(y)u(x, y)dy + + f1 K(y)e-yc(x) fy f (x, t)etc(x)dtdy, 1 - a(x) fj K(y)e-yc(x^dy] fj K(y)u(x, y)dy = = £ K(y)e-yc(x) f y0 f (x,t)etc(x)dtdy.
Then since (3.2) we obtain
f K(y)u(x, y)dy =(1 - a(x) I1 K(y)e-yc(x)d^j - x
x f! K(y)e-yc(x) f0y f (x, t)etc(x)dtdy. (3.7)
Substituting (3.7) into (3.6), we obtain the unique solution (3.3) to (3.1) for every f e C(Q). Since f in (3.3) is an arbitrary element of C(Q), then R(A) = C(Q). It is easy to verify that A-1 is bounded. Hence A is correct. □
Lemma 3.2. Let b(y), d(y) e C[0,1], K0(x) e C[0,1]. Then the operator A0 : C(Q) ^ C(Q) corresponding to the problem:
Aq u(t) = u'x(x, y) + d(y)u(x, y) = f (x, y), (3.8)
f!
is correct if and only if
D(Aq) = {u(x, y) G C(Q) : u'x G C(Q), u(0, y) = b(y) fQ1 KQ(x)u(x, y)dx j
b(y) K0(x)e xd(y)dx = 1 (3.9)
0
and the unique solution of the above problem is given by the formula
u(x, y) = A-1f (x, y) = b(y)e-xd(y) (1 - b(y) fj K0(x)e-xd(y)dx) * x (3.10)
x f01 K0(x)e-xd(y) f0x f (s, y)esd(y)dsdx + e-xd(y) fQx f (s,y)esd(y)ds. Proof. Assume that u(x,y) G ker A0 and (3.9) hold. Then from (3.8) we get
u'x(x,y) + d(y)u(x,y) = 0, u(0,y) = b(y) K0(x)u(x,y)dx■ (3.11)
0
From the last equation by integration on x we obtain
u(x,y)= u(0, y)e-xd(y), u(x,y) = e-xd(y)b(y) f01 K0(x)u(x,y)dx, (3.12)
¡0 K0(x)u(x,y)dx = b(y) ¡0 K0(x)e-xd(y)dx f0 K(x)u(x,y)dx,
1 - b(y) fg1 K0(x)e-xd(y)dx1 f! K0(x)u(x,y)dx = 0.
If b(y) /0 K0(x)e-xd(y)dx = 1, we get /0 K0(x)u(x,y)dx = 0. Substitution of this value into (3.12) implies u(x,y) =0. This means that A0 is injective.
Conversaly. Let u(x,y) G ker A0 and b(y) /0 K0(x)e-xd(y)dx = 1■ Then (3.11) holds. It is easy to verify that u(x,y) = e-xd(y) = 0 satisfies (3.11). Thus we prove that kerA0 = {0} and so A0 is not injective. We will find the solution to (3.8). Let b(y) f01 K0(x)e-xd(y)dx = ±1. Then A0 is injective and Problem (3.8) has a unique solution. From (3.8) by integration on x for every f G C(Ù) we obtain
u(x, y) = e-xd(y)b(y) f01 K0(x)u(x, y)dx + e-xd(y) f0x f (s, y)esd(y)ds, (3.13)
/0 K0(x)u(x,y)dx = b(y) ¡0 K0(x)e-xd(y)dx f0 K0(x)u(x,y)dx + + f01 K0(x)e-xd(y) f0x f (s, y)esd(y)dsdx, 1 - b(y) /0 K0(x)e-xd(y)dx /0 K0(x)u(x, y)dx = = f1 K0(x)e-xd(y) fx f (s,y)esd(y)dsdx.
Then since (3.9) we obtain
J K0(x)u(x,y)dx = (1 - b(y) /01 K0(x)e-xd(y)dx) 1 x
x f01 Ko (x)e- xd(y)f0X f (s,y)esd(y)dsdx. (3.14)
Substituting (3.14) into (3.13), we obtain the unique solution (3.10) to (3.8) for every f e C(Q). Since f in (3.10) is an arbitrary element of C(Q), then R(A0) = C(Q). It is easy to verify that A--1 is bounded. Hence A0 is correct. □
Theorem 3.3. Let a(x),c(x),K0(x) e C[0,1],b(y),K(y) e C [0,1],d(y) e C1 [0,1], h(x,y),u(x,y) e C1 (Q), u'Xy(x,y) e C(QQ). Then the problem
u'Xy(x, y) + c(x)ux(x, y) + d(y)u'y(x, y) + h(x, y)u(x, y) = f (x, y), (3.15)
u(0,y) = b(y) ¡0 Ko(x)u(x,y)dx,
u'X(x, 0) + d(0)u(x, 0) = a(x) Jo K(y)[u'X(x, y) + d(y)u(x, y)]dy
is correct if
h(x, y) = d'(y) + c(x)d(y), (3.16)
a(x) £ K(y)e-yc(x)dy =1, b(y) f1 K0(x)e-xd(y)dx = 1 (3.17)
and its unique solution is given by the formula
u(x, y) = b(y)e-xd(y) (1 - b(y) J^ K0(x)e-xd(y)dx^j 1 x (3.18)
x J'1 K0(x)e-xd(y) f0X v(s, y)esd(y)dsdx + e-xd(y) fj v(s, y)esd(y)ds,
where
v(x, y) = a(x)e-yc(x) (1 - a(x) fj K(y)e-yc(x)dyj 1 x (3.19)
x ^ K(y)e-yc(x) f y f (x, t)etc(x)dtdy + e-yc(x) f y f (x,t)etc(x)dt.
Proof. Let the operator A be defined by (3.1) and the operator A0 by (3.8), where we suppose that d(y) e C 1[0,1]. Denote by A1 the operator corresponding to Problem (3.15), namely:
A1u(x, y) = u'Xy(x, y) + c(x)uX(x, y) + d(y)u'y(x, y) + h(x, y)u(x, y), (3.20)
D(A1) = {u(x, y) e C(i) : uX(x, y), u'y(x, y),u'Xy(x, y) e C(i), (3.21)
u(0,y) = b(y) fg1 K0(x)u(x, y)dx,
uX(x, 0) + d(0)u(x, 0) = a(x) f^ K(y)[uX(x,y) + d(y)u(x,y)]dy}.
We will prove that A1 = AA0, i.e. D(A1) = D(AA0), A1u = AA0u for all u e D(A1) if h(x,y) = d'(y) + + c(x)d(y). Using the definition of a superposition of two operators, we find
D(AA0) = {u e D(A0) : A0u e D(A)} = (3.22)
= {u(x,y) e C(i) : u'x e C(ii), u(0,y) = b(y) Jo K0(x)u(x,y)dx, A0u e D(A)} = = {u(x, y) e C(i) : uX(x, y) e C(i), (uX(x, y) + d(y)u(x, y))'y e C(i), u(0,y) = b(y) Jo K0(x)u(x, y)dx, u'X(x, 0) + d(0)u(x, 0) = a(x) f^ K(y)[u'X(x,y) + d(y)u(x, y)]dy}, AA0u(x,y) = (uX(x,y) + d(y)u(x,y))'y + c(x)[uX(x,y) + d(y)u(x,y)]. (3.23)
Since d(y) e C 1[0,1], from (u'X(x,y) + d(y)u(x,y))'y e C(i) follows that u'Xy e C(i) and
(u'x(x, y) + d(У)u(x, y))'y = uXy(x, y) + d'(У)u(x, y) + d(y)u'y(x, y) e C(ii).
Then from (3.22) follows that D(AA0) = D(A1). Furthermore if the condition (3.16) is additionally satisfied then (3.23) implies A1u = AA0u for all u e D(A1). Thus we proved that if (3.16) holds, then A1 = AA0. Now we find the solvability condition and solution of A1u = f,u e D(A1) for the case when (3.16) holds. Denote by v(x,y) = A0u(x,y) = u'X(x,y) + d(y)u(x,y). Then A1u = AA0u = Av = f. The last equation is correct by Lemma 3.1 if and only if (3.2) is satisfied. Then v = A0u = A-1 f where A-1 f is calculated by (3.3) which is (3.19). The equation A0u = v is correct by Lemma 3.2 if and only if (3.9) is satisfied. Then u = A-1v where A-1v is calculated by (3.10) which is (3.18). Thus we proved that if (3.16), (3.17) hold true then the operator A1 or Problem (3.15) is correct and its unique solution is (3.18) where v(x,y) is given by (3.19). The theorem is proved. □
From Theorem 3.3 for c(x) = d(y) = h(x, y) = 0 follows the next Corollary 3.4. Let a(x),K0(x) e C[0,1], b(y),K(y) e C[0,1], u(x,y) e C 1(i), u 'y(x,y) e C(i). Then the problem
u Xy (x,y )= f (x,y), (3.24)
u(0,y) = b(y) fg1 K0(x)u(x, y)dx,
10
u' Jx, °) = a(x) Jo K(v)u'x(x, v)dv
is correct on C(Q) if
i(x) i K(y)dy = 1, b(y) i K0(x)dx = 1. (3.25)
Jo Jo
and the unique solution of Problem (3.24) is given by the formula
u(x,y) = (3.26)
b(y)
_ml_ f1 Kn(x)
1-b(y) /J K0(x)dx Jo Ko(x)
foX 1-a(S)a/jK(y)dy fo1 K(y)J0 f (s, t)dtdyds +
+ foXfoy f (s,t)dtds
dx + So 1-a(S)af0S)K(y)dy So1 K(y)Sy f (s> t)dtdyds +
+ SoX Soy f (s,t)dtds.
The following problem is solved by Theorem 2.3.
Example 3.5. Let u(x,y),u'X(x,y),u'y(x,y),u'Xy £ C(Q). Then the problem
u'Xy -(x + y)fofo xu'x(x, y)dxdy — 3x3 Jo fo y2u'Xy(x,y)dxdy (3-27)
„3
15x3 — 2x — 2y,
1
u'X(x, 0) = 0, u(0, y) = (y + 1) /0 u(x, y)dx,
is uniquelly solvable if y = 0 and the unique solution of (3.27) is given by the formula
u(x, y) = 5x4y — y — 1. (3.28)
Proof. Denote by B the operator corresponding to Problem (3.27). First we must determine the operators A and A0 and make sure that D(B1) = D(AA0). Comparing (3.27) with (2.14) it is natural to take X = C(Q),
ou) = fg1 fg xu'x(x, y)dxdy, F(AA0u) = f^ f^ y2u'Xy(x, y)dxdy, (3.29)
AA0u(x, y) = u'Xy (x, y), Aou = uX(x, y).
Denote v(x,y) = u'X(x,y). Then AA0u(x,y) = u'Xy(x,y) = (u'X(x,y))'y = Av(x,y) = v'y(x,y). From boundary conditions (3.27) follws that v(x, 0) = 0. So the operators A, A0 are defined by
Av(x, y) = vy (x, y), D(A) = {v{x, y) £ C(Q) : v'y £ C(Q), v(x, 0) = 0},
Aou(x, y) = uX(x, y), D(Ao) = {u(x, y) £ C(Q) : uX £ C(Q),
u(0, y) = (y + 1) fa u(x, y)dxj.
Then
D(AAo) = {u(x,y) £ D(Ao) : Aou £ D(A)} = {u(x,y) £ C(Q) :
uX, u'Xy £ C(Q), u(0, y) = (y + 1) fQ1 u(x, y)dx, uX(x, 0) = 0} = D(Bi).
Since D(B1) = D(AAo), we can apply Theorem 2.3. Note that the operator A coincides with the operator A from Lemma 3.1 if a(x) = c(x) = 0 and the operator Ao coincides with the operator Ao from Lemma 3.2 if b(y) = y + 1, d(y) = 0, Ko(x) = 1. Then by Lemma 3.1, the operator A is correct and
r y
A-1f (x,y)= f (x,t)dt, (3.30)
Jo
by Lemma 3.2 the operator Ao is correct if and only if y = 0 and its inverse is defined by
+ 1 f1 rX
A-1f (x,y) = — y—~ (1 — s)f (s,y)ds + f (s,y)ds. (3.31)
y
Notice that the operator AA coincides with the operator corresponding to Problem (3.24) and, by Corollary 3.4, is correct if y = 0 and its inverse is defined by
1 X y X y
-1 A-1ff \ y +1
T (x,y) =--, ,
y
Comparing again (3.27) with (2.14) it is natural to take S = x + y, G = 3x3, f =15x3 — 2x — 2y. From (3.29) follows that
r-1 r 1 r 1 r 1
.2 ,
y + 1 f1 rx fy cx fy
A-1A-1f (x,y) = f (s,t)dtdsdx +/ f (s,t)dtds. (3.32)
y
$(f) = i f xf(x,y)dxdy, F(f)= f f y2f(x,y)dxdy. (3.33)
Jo Jo Jo Jo
Using (3.32) for f = 15x3 — 2x — 2y and G = 3x3 we find
A-1A-1f (x, y) = - tffZ J0y (15s3 - 2s - 2t)dtdsdx+
С x rVde Я 45x4.. 12x2.. iOx..2
+ fQx fy (15s3 - 2s - 2t)dtds = ^у-^у-1^+(y+1)(6y-§),
+ 1 f1 fx ry y JO Jo Jo
Jo JO V-1-"0 12
A-1A-1G = - ^ f1 f0x f0y 3s3dtdsdx+
+ f0x f0y 3s3dtds = - (y + 1) + 4x4y + 5x y+6x y+6;2xy -3(y+1} . By means (3.30) for G = 3x3, S = x + y we get
A-1G = fyy G(x, t)dt = fyy 3x3dt = 3x3y,
A-1S = f0y S(x,t)dt = f0y (x + t)dt = xy + y2/2,
A-1f = f0y f (x,t)dt = f0y(15x3 - 2x - 2y)dt = 15x3y - 2xy - y2.
Using (3.33) we get
О "Vй-; Jo
y f (x,t)dt = f0y (
F(S) = fo Jo y2S(x, y)dxdy = f^ f^ y2(x + y)dxdy = ,
F(G) = Jo fo y2G(x, y)dxdy = f^ f^ y23x3dxdy = 1,
F(f) = Jo Jo y2f (x, y)dxdy = fgfg y2(15x3 - 2x - 2y)dxdy = ,
Ф(A-1G) = f 1 fg x3x3ydxdy = Ц,
Ф(A-1f) = fg fg x(15x3y - 2xy - y2)dxdy = 1. Further by (2.17)-(2.19) we find
L = Im - F(G) = 1 - 1/4 = 3/4,
G0 = G0(x, y) = A-1S + A-1GL-1F(S) = xy + + §x3y, L0 = Im - &(Go) = 1 - fo fo xGo(x, y)dxdy 1 r1 „ („„. y2 §ji„\ - 7
1 - fo fo x (xy + \ + §x3y^J dxdy = 12.
Since det L = 3/4 = 0 and det L0 = 7/12 = 0, by Theorem 2.3, Problem 3.27 is correct. Applying (3.31) obtain
A-1G0 = -^ f!(1 - s)G0(s,y)ds + fQX G0(s,y)ds
_ 5x4y+6x2y+6xy2-3(y+1)2
= 12 .
Substituting the above values into (2.9) we get (3.28). □
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Научная статья DOI: 10.18287/2541-7525-2021-27-1-29-43
УДК 629 Дата: поступления статьи: 15.01.2021
после рецензирования: 17.02.2021 принятия статьи: 28.02.2021
E. Провидас
Университет Фессалии, г. Ларисса, Греция E-mail: providas@uth.gr. ORCID: https://orcid.org/0000-0002-0675-4351
Л.С. Пулькина
Самарский национальный исследовательский университет имени академика С.П. Королева, г. Самара, Российская Федерация E-mail: louise@samdiff.ru. ORCID: https://orcid.org/0000-0001-7947-612
И.Н. Парасидис Университет Фессалии, г. Ларисса, Греция E-mail: paras@teilar.gr. ORCID: https://orcid.org/0000-0002-7900-9256
ФАКТОРИЗАЦИЯ ОБЫКНОВЕННЫХ И ГИПЕРБОЛИЧЕСКИХ ИНТЕГРО-ДИФФЕРЕНЦИАЛЬНЫХ УРАВНЕНИЙ С ИНТЕГРАЛЬНЫМИ УСЛОВИЯМИ В БАНАХОВОМ ПРОСТРАНСТВЕ
АННОТАЦИЯ
В статье исследованы условия существования единственного точного решения для одного класса абстрактных операторных уравнений вида Biu = Au — <S^(Aou) — GF(Au) = f, u £ D(B\), где A, Ao — линейные абстрактные операторы; G, S — линейные векторы; Ф, F — линейные функциональные векторы. Этот класс уравнений полезен для решения краевых задач для интегро-дифференциальных уравнений в случае, когда A,Ao — дифференциальные операторы, а F(Au), Ф^^) — интегральные операторы Фредгольма. Показано, что операторы типа B1 могут быть в некоторых случаях представлены как произведения двух более простых операторов BG, BGo специального вида, что позволяет получить условие существования единственного точного решения уравнения B1 u = f из условий однозначной разрешимости уравнений BGv = f и BGou = v.
Ключевые слова: корректная (по Адамару) разрешимость; метод факторизации (декомпозиции); интегро-дифференциальные уравнения Фредгольма; начальная задача; нелокальная краевая задача с интегральными условиями.
Цитирование. Providas E., Pulkina L.S., Parasidis I.N. Factorization of ordinary and
hyperbolic integro-differential equations with integral boundary conditions in a Banach space //
Вестник Самарского университета. Естественнонаучная серия. 2021. Т. 27, № 1. С. 29-43. DOI: http://doi.org/10.18287/2541-7525-2021-27-1-29-43.
Информация о конфликте интересов: авторы и рецензенты заявляют об отсутствии конфликта интересов.
© Провидас Е., 2021
Евтимиос Провидас — кандидат технических наук, доцент, Университет Фессалии, Греция, г. Ларисса, Гайополис, 41110.
© Пулькина Л.С., 2021
Людмила Степановна Пулькина — доктор физико-математических наук, профессор кафедры дифференциальных уравнений и теории управления, Самарский национальный исследовательский университет имени академика С.П. Королева, 443086, Российская Федерация, г. Самара, Московское шоссе, 34.
© Парасидис И.Н., 2021
Иван Нестерович Парасидис — кандидат технических наук, доцент, Университет Фессалии, Греция, г. Ларисса, Гайополис, 41110.
