# FACTORIZATION OF ORDINARY AND HYPERBOLIC INTEGRO-DIFFERENTIAL EQUATIONS WITH INTEGRAL BOUNDARY CONDITIONS IN A BANACH SPACE

**Authors:**Providas E.^{1}, Pulkina L.S.^{2}, Parasidis I.N.^{1}-
**Affiliations:**- University of Thessaly
- Department of Equations of Mathematical Physics, Samara National Research University

**Issue:**Vol 27, No 1 (2021)**Pages:**29-43**Section:**Статьи**URL:**https://journals.ssau.ru/est/article/view/10059**DOI:**https://doi.org/10.18287/2541-7525-2021-27-1-29-43

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## Full Text

## 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 B1u = Au − SΦ(A0u) − GF(Au) = f, u ∈ D(B1),

where A,A0 are linear abstract operators, G, S are linear vectors and Φ, 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), Φ(A0u) are Fredholm integrals. It was shown that the operators of the type B1 can be factorized in the some cases in the product of two more

simple operators BG, BG0 of special form, which are derived analytically. Further the solvability condition and the unique exact solution for B1u = f easily follow from the solvability condition and the unique exact solutions for the equations BGv = f and BG0u = v.

## Full Text

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 B1u = f . It is the aim of this paper to reappraise the factorization method for integro-diﬀerential equations of type B1u = 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 B1u = 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 B1x = f when B1 = 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 ∈ 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 A1, or A1 is said to be a restriction of A2, in symbol A1 ⊂ A2, if D(A2) ⊇ D(A1) and A1x = A2x, for all x ∈ D(A1). An operator A : X → Y is said to be injective or uniquelly solvable if for all u1, u2 ∈ D(A) such that Au1 = Au2, follows that u1 = 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−1 is bounded on Y . If gi ∈ X and Ψi ∈ X∗, i = 1, . . . , m,

then we denote by g = (g1, . . . , gm), Ψ =col(Ψ1, . . . , Ψm) and Ψ(u) =col(Ψ1(u), . . . , Ψm(u)) and write

m

g ∈ Xm, Ψ ∈ X∗ . We will denote by Ψ(g) the m × m matrix whose i, j-th entry Ψi(gj ) is the value of

Ψi on element gj . Note that Ψ(gC) = Ψ(g)C, where C is a m× k constant matrix. We will also

functional

denote by

0m the zero and by Im the identity m×m matrices. By 0 we will denote the zero column vector.

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 =

= (g1, ..., gm), F = col(F1, ...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 ∈ X. (2.1)

Then the following statements are true:

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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 ∈ X, is given by the formula

G

u = B−1f = A−1f + A−1G[Im − F (G)]−1F (f ). (2.3)

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(0), ..., g(0)), G = (g , ..., g

), S = (s , ..., s ) ∈

1 m 1 m 1 m

Xm ∗

, the components of the vectors F = col(F1, ..., Fm) and Φ = col(ϕ1, ..., ϕm) belong to X BG0 , BG, B1 : X → X deﬁned by

and the operators

BG0 u = A0u − G0Φ(A0u) = f, D(BG0 ) = D(A0), (2.4)

BGu = Au − GF (Au) = f, D(BG) = D(A), (2.5)

B1u = AA0u − SΦ(A0u) − GF (AA0u) = f, D(B1) = D(AA0) (2.6)

where A0 and A are linear correct operators on X and G0 ∈ D(A)m. Then the following statements are satisﬁed:

If

S ∈ R(BG)m and S = BGG0 = AG0 − GF (AG0), (2.7)

then the operator B1 can be factorised in B1 = BGBG0 .

If in addition the components of the vector Φ = (Φ1, ..., Φm) are linearly independent elements of X∗ and

the operator B1 can be factorised in B1 = BGBG0 , then (2.7) is fulﬁlled.

If the operator B1 can be factorised in B1 = BGBG0 , then B1 is correct if and only if the operators

BG0 and BG are correct which means that

det L0 = det[Im − Φ(G0)] ̸= 0 and det L = det[Im − F (G)] ̸= 0. (2.8)

If the operator B1 has the factorization in B1 = BGBG0 and is correct, then the unique solution of (2.6) is

u = B−1f = A−1A−1f + [A−1A−1G + A−1G0L−1Φ(A−1G)] ×

1 0 0 0 0

×L−1F (f ) + A−1G0L−1Φ(A−1f ). (2.9)

0 0

Proof. (i) Taking into account that G0 ∈ D(A)m and (2.4)–(2.6) we get

D(BGBG0 ) = {u ∈ D(BG0 ) : BG0 u ∈ D(BG)} =

= {u ∈ D(A0) : A0u − G0Φ(A0u) ∈ D(A)} =

= {u ∈ D(A0) : A0u ∈ D(A)} = D(AA0) = D(B1).

So D(B1) = D(BGBG0 ). Let y = BG0 u. Then for each u ∈ D(AA0) since (2.5) and (2.4) we have

BGBG0 u = BGy = Ay − GF (Ay) =

= A[A0u − G0Φ(A0u)] − GF (A[A0u − G0Φ(A0u)]) =

= AA0u − AG0Φ(A0u) − GF (AA0u) + GF (AG0)Φ(A0u) =

= AA0u − [AG0 − GF (AG0)]Φ(A0u) − GF (AA0u) =

= AA0u − BGG0Φ(A0u) − GF (AA0u),

(2.10)

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 = BGBG0 u for each u ∈ D(AA0) if a vector S satisﬁes (2.7).

Let the operator B1 can be factorized in B1 = BGBG0 . Then by comparing (2.10) with (2.6) we obtain

(BGG0 − S)Φ(A0u) = 0. (2.11)

Because of the correctness of operators A, A0 and the linear independence of Φ1, ..., Φm, there exists a system u1, ..., um ∈ D(AA0) such that Φ(A0u0) = Im where u0 = (u1, ..., um). By substituting u = u0 into (2.11) we get S = BGG0. Hence S ∈ R(BG)m and S = BGG0 = AG0 − GF (AG0).

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)

( Φ(A−1AA0u) )

F (AA0u)

= f (2.12)

Providas E., Pulkina L.S., Parasidis I.N. Factorization of ordinary and hyperbolic integro-diﬀerential equations...

32Провидас E., Пулькина Л.С., Парасидис И.Н. Факторизация обыкновенных и гиперболических...

or

B1u = Au − GF (Au) = f, D(B1) = D(A), (2.13)

(

where A = AA0, G = (BGG0, G), F =col(Φˆ , F ), F (v) =

Φˆ (v)

F (v)

) ( Φ(A−1v) )

= 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 Φˆ

is bounded as a superposition of a bounded functional Φ and a bounded

operator A−1. Then we apply Theorem 2.1. By this theorem the operator B1 is correct if and only if

[( Im 0m )

( Φˆ (BGG0)

Φˆ (G) )]

det L1 = det[I2m − F (G)] = det

0m Im

— F (BG

=

G0) F (G)

= det

( Im − Φˆ (AG0 − GF (AG0)) −Φˆ (G) ) =

−[F (AG0 − GF (AG0))] Im − F (G)

= det

=

( Im − Φ(G0 − A−1GF (AG0)) −Φ(A−1G) )

−[F (AG0 − GF (AG0))] Im − F (G)

= det

( Im − Φ(G0) + Φ(A−1G)F (AG0) −Φ(A−1G) )

−F (AG0) + F (G)F (AG0) Im − F (G)

̸= 0.

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 L1, by Remark 1, [31] we get

( Im − Φ(G0) −Φ(A−1G) )

det L1 = det

0m Im − F (G)

= det[Im − Φ(G0)] det[Im − F (G)]

= det L0 det L ̸= 0.

So we proved that the operator B1 is correct if and only if (2.8) is fulﬁlled.

Let u ∈ D(AA0) and BGBG0 u = f . By Theorem 2.1 (ii) since BG, BG0 are correct operators, we obtain

BG −1

1

0 u = BG f = A−

f + A−1

GL−1

F (f ),

G

u = B−1 (A−1f + A−1GL−1F (f )) .

0

In the last equation we denote by g = A−1f + A−1GL−1F (f ). Bu using again Theorem 2.1 (ii), with

A0, G0, Φ, L0, in place of A, G, F, L respectively, we get

u = B−1g = A−1g + A−1G0L−1Φ(g) = A−1(A−1f + A−1GL−1F (f ))+

G0 0 0 0 0

0 G0L0 Φ (A

f + A GL

F (f )) = A0 A

f + A0 A GL

F (f )+

+A−1 −1 −1

−1 −1

−1 −1

−1 −1 −1

0 G0L0 [Φ(A

f ) + Φ(A

G)L

F (f )]

+A−1 −1 −1

−1 −1

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, Φ be deﬁned as in Theorem 2.2, the vectors G =

= (g1, ..., gm), S = (s1, ..., sm) ∈ Xm and the operator B1 : X → X by

B1u = Au − SΦ(A0u) − GF (Au) = f, x ∈ 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:

if there exist a n − m order diﬀerential operator A : X → X, such that

A = AA0, D(B1) = D(AA0), (2.15)

and a vector G0 ∈ D(A), satisfying

AG0 − GF (AG0) = S, (2.16)

then the operator B1 can be factorized into B1 = BGBG0 , where BG0 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 BG0 by A0, Φ and G0 from (2.14) and (2.16),

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 = BGBG0 , where the operators BG0 , BG, A0, A, the vectors G, F, Φ are determined as in (i) and

G0 = A−1S + A−1GL−1F (S). (2.18)

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if in addition to (ii) A is correct, then B1 is correct if and only if

det L0 = det[Im − Φ(G0)] = det[Im − Φ(A�−1S) − Φ(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 − SΦ(A0u) − GF (AA0u) = f, u ∈ 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 BG0 by formula (2.4), we take from (2.20) the operator A0 and the vector Φ, whereas as G0 we take any solution G0 of Equation (2.16). We proved in the previous theorem (i) that D(BGBG0 ) = D(AA0) = D(B1). Substituting (2.16) into (2.20), for every

u ∈ D(B1) we get

B1u = AA0u − [AG0 − GF (AG0)] Φ(A0u) − GF (AA0u) = BGA0u − BGG0Φ(A0u) =

= BG [A0u − G0Φ(A0u)] = BGBG0 u.

Thus B1 = BGBG0 .

As in the proof of (i) we construct the operators BG, BG0 . By Theorem 2.1, since (2.17), the operator

G

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 = BGBG0 , where BG0 is unique.

If (2.17), (2.18) hold true, then by statements (i), (ii), B1 can be factorized in B1 = BGBG0 . 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 − Φ(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 − Φ(G0)]. Since B1 is correct and factorized in B1 = BGBG0 ,

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) ∈ C2[0, 1]. Then the problem

u′′(t) − t ∫ 1 tu′(t)dt − t2 ∫ 1 t3u′′(t)dt = 2t + 1, (2.21)

0 0

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

−

261377 665232t + 103608t2 + 30080t3 + 8790t4

u(t) =

207216

. (2.22)

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

1

B1u(t) = u′′(t) − t ∫

tu′(t)dt − t2 ∫ 1

t3u′′(t)dt = 2t + 1, (2.23)

0 0

D(B1) = {u(t) ∈ C2[0, 1] : u(0) + u(1) = 0, u′(0) − 2u′(1) = 0},

Au = AA0u = u′′(t),

D(B1) = {u(t) ∈ C2[0, 1] : u(0) + u(1) = 0, u′(0) − 2u′(1) = 0}, (2.24)

A0u(t) = u′(t), D(A0) = {u(t) ∈ C1[0, 1] : u(0) = −u(1)},

Φ(A0u) =

∫ 1

tu′(t)dt, F (AA0u) =

0

∫ 1

t3u′′(t)dt, (2.25)

0

S = t, G = t2. Denote A0u(t) = u′(t) = y(t) = y. Then from (2.24) we have y ∈ 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) ∈ C1[0, 1] : y(0) − 2y(1) = 0}.

Then by deﬁnition

D(AA0) = {u(t) ∈ D(A0) : A0u(t) ∈ D(A)} =

= {u(t) ∈ C1[0, 1] : u(0) = −u(1), u′(t) ∈ C1[0, 1], u′(0) − 2u′(1) = 0} =

= {u(t) ∈ C2[0, 1] : u(0) + u(1) = 0, u′(0) − 2u′(1) = 0} = D(B1).

Providas E., Pulkina L.S., Parasidis I.N. Factorization of ordinary and hyperbolic integro-diﬀerential equations...

34Провидас E., Пулькина Л.С., Парасидис И.Н. Факторизация обыкновенных и гиперболических...

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) ∈ C[0, 1] the corresponding inverse operators are deﬁned by

A−1f (t) = ∫ t f (s)ds − 2 ∫ 1 f (s)ds, (2.26)

0

A−1 t

0

1 ∫ 1

From (2.25) we have

0

0 f (t) = ∫

f (s)ds − 2

0 f (s)ds. (2.27)

Φ(f ) =

∫ 1

sf (s)ds, F (f ) =

0

∫ 1

s3f (s)ds. (2.28)

0

Then F (G) = ∫ 1 s3s2ds = 1 , F (S) = ∫ 1 s3sds = 1 ,

0 6 0 5

det L = det[Im − F (G)] = 1 − 1/6 = 5/6, L−1 = 6/5,

A−1S =

∫ t ∫ 1

sds − 2

t2

sds =

− 1, A−1G =

∫ t ∫ 1

s2ds − 2

3

s2ds = t − 2 ,

0 0 2

t2

(

*t*30 0 3 3

2 ) 6 1 1

G0 = A−1S + A−1GL−1F (S) =

Taking into account (2.28) we obtain

2 − 1 +

3 − 3

=

5 5 50

(4t3 + 25t2 − 58).

1

Φ(G ) = 1 ∫

0

0 50

s(4s3

+ 25s

2 439

− 58)ds = − 1000 .

1000

Since det L0 = det[Im − Φ(G0)] = 1439 ̸= 0 then, by Theorem 2.3 (iii), Problem (2.23) or (2.21) is correct.

By (2.27) we calculate

A−1

73

29t t3 t4

1 −1

7 2t t4

0 G0 = 150 −

−

25 + 6 + 50 , A0 A

G = 24 −

+

3 12

and for f (t) = 2t + 1 by (2.26)-(2.28) we obtain

2

A−1f = −4 + t + t2, A−1A−1f = 19 − 4t + t

t3 13 17

1

+ , F (f ) = , Φ(A− f ) = −

0 12 2 3 20 12

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) ∈ C3[0, 1]. Then the problem

u′′′(x) − 8x2 ∫ 1 tu′(t)dt − (3x + 1) ∫ 1 t2u′′′(t)dt = 2x2 − 6x + 4, (2.29)

0 0

0

u(0) = 2 ∫ 1 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

u(x) = x3 − 3 x2 + 1 . (2.30)

2 2

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],

B1u = u′′′(x) − 8x2

∫ 1 ∫ 1

tu′(t)dt − (3x + 1)

0 0

∫ 1

t2u′′′(t)dt = 2x2 − 6x + 4, (2.31)

D(B1) = {u(x) ∈ C3[0, 1] : u(0) = 2

u(t)dt, u′(0) = −u′(1), u′′(0) = −u′′(1)},

0

0 0

Φ(A u) = ∫ 1 ∫ 1

1

tu′(t)dtdy, F (AA0u) = ∫

∫ 1 t2u

′′′

(t)dtdy, (2.32)

0 0 0

Ax = AA0u(x) = u′′′(x), A0u = 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, A0 are

deﬁned by

Av(x) = v′′(x), D(A) = {v(x) ∈ C2[0, 1] : v(0) = −v(1), v′(0) = −v′(1)},

1

0

A0u(x) = u′(x), D(A0) = {u(x) ∈ C1[0, 1] : u(0) = 2 ∫

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) ∈ D(A0) : A0u ∈ D(A)} = {u(x) ∈ C1[0, 1] : u(0) =

= 2 ∫ 1 u(x)dx, u′(x) ∈ C2[0, 1], u′(0) = −u′(1), u′′(0) = −u′′(1)} = D(B ).

0 1

Вестник Самарского университета. Естественнонаучная серия. 2021. Том 27, № 1. С. 29–43

Vestnik of Samara University. Natural Science Series. 2021, vol. 27, no. 1, pp. 29–43 35

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) ∈ C[0, 1] are given by

A−1 1 x

0 f (x) = 2 ∫0 (t − 1)f (t)dt + ∫0 f (t)dt, (2.33)

A−1f (x) = 1 ∫ 1 (t − x − 1 ) f (t)dt + ∫ x(x − t)f (t)dt. (2.34)

2 0 2 0

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

Φ(f ) =

∫ 1

tf (t)dt, F (f ) =

0

∫ 1

t2f (t)dt. (2.35)

0

Let

fˆ(x) = A−1f (x) and AA0u(x) = f (x). Then, since A0, A are invertible, by means of (2.33) and (2.34)

we have

u(x) = A−1A−1f (x) = A−1fˆ(x) = 2 ∫ 1(t − 1)fˆ(t)dt + ∫ x fˆ(t)dt =

0 0 0 0

= 2 ∫ 1(t − 1) [ 1 ∫ 1 (s − t − 1 ) f (s)ds + ∫ t(t − s)f (s)ds] dt+

0 2 0 2 0

]

+ ∫ x [ 1 ∫ 1 (s − t − 1 ) f (s)ds + ∫ t(t − s)f (s)ds

dt.

0 2 0 2 0

Further using Fubini theorem we obtain

A−1 −1

1 1 2 3 2

0

0 A f (x) = − 12 ∫

[3x

+ 3x(1 − 2s) − 4s

+ 1 ∫ x 2

+ 12s

− 6s − 1]f (s)ds +

2 0 (x − s) f (s)ds. (2.36)

Using (2.36) for f = f (x) = 2x2 − 6x + 4 and G = 3x + 1 we get

0 A

A−1 −1

12

f = − 1 ∫

1[3x2

+ 3x(1 − 2s) − 4s3

+ 12s2

− 6s − 1](2s2

− 6s + 4)ds + (2.37)

0

+ 1 ∫ x 2 2

1 5

4 3 2

2 0 (x − s) (2s

− 6s + 4)ds = 60 (2x

− 15x

+ 40x

− 25x

− 10x + 12),

A−1 −1

1 1 2 3 2

0 A = − 12 ∫0 [3x

+ 3x(1 − 2s) − 4s

+ 12s

− 6s − 1](3s + 1)ds + (2.38)

+ 1 ∫ x 2

1 4 3 2

2 0 (x − s) (3s + 1)ds = 120 (15x

+ 20x

− 75x

+ 15x + 19).

Using (2.34) for S = S(x) = 8x2, G = G(x) = 3x + 1, f (x) = 2x2 − 6x + 4 we ﬁnd

4

A−1S = 1 ∫ 1 (t − x − 1 ) (8t2)dt + ∫ x(x − t)(8t2)dt = 2x −4x+1 ,

2 0 2 0 3

2

A−1G = 1 ∫ 1 (t − x − 1 ) (3t + 1)dt + ∫ x(x − t)(3t + 1)dt = 4x (x+1)−10x+1 ,

2 0 2 0 8

A−1f = 1 ∫ 1 (t − x − 1 ) (2t2 − 6t + 4)dt + ∫ x(x − t)(2t2 − 6t + 4)dt

2 0 2 0

2 2

6 .

= x (x −6x+12)−5x−1

Then by using (2.35) we arrive at

Φ(A−1G) = 1 ∫ 1 t[4t2(t + 1) − 10t + 1]dt = − 31 ,

8 0 240

F (G) = ∫ 1 t2(3t + 1)dt = 13 ,

0 12

Φ(A−1f ) = 1 ∫ 1 t[t2(t2 − 6t + 12) − 5t − 1]dt = − 1 ,

6 0 30

F (f ) = ∫ 1 t2(2t2 − 6t + 4)dt = 7 , F (S) = ∫ 1 t2(8t2)dt = 8 .

0 30 0 5

Further by (2.17), (2.18) and (2.19) we ﬁnd

det L = det[Im − F (G)] = 1 − 13/12 = −1/12,

4 2

G0 = G0(x, y) = A−1S + A−1GL−1F (S) = 2x −4x+1 + 4x (x+1)−10x+1 (−12) 8

4 3 2

3 8 5

15 ,

= 10x −144x −144x +340x−31

1

Φ(G0) = 1 ∫

t(10t4 − 144t3 − 144t2 + 340t − 31)dt = 347 ,

15 0

150

det L0 = det[Im − Φ(G0)] = 1 − 347 = − 197 .

150

150

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−1

0 G0 = 2 ∫

1

x

(t − 1)G0(t)dt + ∫

G0(t)dt =

0

= 2 ∫ 1

0

4 3 2

15 0 (t − 1)(10t

− 144t

− 144t

+ 340t − 31)dt+

Providas E., Pulkina L.S., Parasidis I.N. Factorization of ordinary and hyperbolic integro-diﬀerential equations...

36Провидас E., Пулькина Л.С., Парасидис И.Н. Факторизация обыкновенных и гиперболических...

+

__1__∫ x 4 3 215 0 (10t

− 144t

4 3

− 144t

2

+ 340t − 31)dt =

75 +

= − 223

.

x(2x −36x −48x +170x−31)

15

Substituting the above values into (2.9) we get the solution (2.30). ✷

Factorization of hyperbolic integro-diﬀerential equations with integral boundary conditions

Everywhere below Ω = {(x, y) ∈ R2 : 0 � x, y � 1}.

Lemma 3.1. Let a(x), c(x) ∈ C[0, 1], K(y) ∈ C[0, 1]. Then the operator A : C(Ω¯ ) → C(Ω¯ ) corresponding to

the problem:

y

Au(t) = u′ (x, y) + c(x)u(x, y) = f (x, y), (3.1)

D(A) = {u(x, y) ∈ C(Ω¯ ) : u′ (x, y) ∈ C(Ω¯ ), u(x, 0) = a(x) ∫ 1 K(y)u(x, y)dy}

is correct if and only if

y

∫ 1

a(x)

0

0

K(y)e−yc(x)dy ̸= 1, (3.2)

and the unique solution of the above problem is given by the formula

0

−1

u(x, y) = A−1f (x, y) = a(x)e−yc(x) (1 − a(x) ∫ 1 K(y)e−yc(x)dy) ×

× ∫ 1

yc(x) ∫ y

tc(x)

−yc(x) ∫ y

tc(x)

0 K(y)e−

0 f (x, t)e

dtdy + e

0 f (x, t)e

dt. (3.3)

Proof. Assume that u(x, y) ∈ ker A and (3.2) hold. Then from (3.1) we get

∫ 1

u′

y (x, y) + c(x)u(x, y) = 0, u(x, 0) = a(x)

From the above equation by integration on y we obtain

K(y)u(x, y)dy. (3.4)

0

0

u(x, y) = u(x, 0)e−yc(x), u(x, y) = a(x)e−yc(x) ∫ 1 K(y)u(x, y)dy, (3.5)

∫ 1

0 K(y)u(x, y)dy = a(x)

∫ 1

0 K(y)e−

yc(x)

0

dy ∫ 1 K(y)u(x, y)dy,

[1 − a(x) ∫ 1 K(y)e−yc(x)dy] ∫ 1 K(y)u(x, y)dy = 0.

0 0

0

From the last equation, since (3.2), follows that ∫ 1 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.

0

Conversaly. Let u(x, y) ∈ ker A and a(x) ∫ 1 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) ∈ ker A and so A is not

injective.

0

We will ﬁnd the solution to (3.1). Let a(x) ∫ 1 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) ∫ 1 K(y)u(x, y)dy + e−yc(x) ∫ y f (x, t)etc(x)dt, (3.6)

0

∫ 1 ∫ 1

0

yc(x) 1

0 K(y)u(x, y)dy = a(x)

+ ∫ 1

0 K(y)e−

yc(x) ∫ y

0

dy ∫ K(y)u(x, y)dy +

tc(x)

0 K(y)e−

0 f (x, t)e

dtdy,

[1 − a(x) ∫ 1 K(y)e−yc(x)dy] ∫ 1 K(y)u(x, y)dy =

0

= ∫ 1

yc(x) ∫ y

0

tc(x)

Then since (3.2) we obtain

∫ 1

0 K(y)e−

(

0 f (x, t)e

dtdy.

)−1

K(y)u(x, y)dy =

0

0

1 − a(x) ∫ 1 K(y)e−yc(x)dy ×

× ∫ 1

0 K(y)e−

yc(x) ∫ y

0 f (x, t)e

tc(x)

dtdy. (3.7)

Вестник Самарского университета. Естественнонаучная серия. 2021. Том 27, № 1. С. 29–43

Vestnik of Samara University. Natural Science Series. 2021, vol. 27, no. 1, pp. 29–43 37

Substituting (3.7) into (3.6), we obtain the unique solution (3.3) to (3.1) for every f ∈ C(Ω¯ ). Since f in (3.3) is an arbitrary element of C(Ω¯ ), then R(A) = C(Ω¯ ). It is easy to verify that A−1 is bounded. Hence A is correct. ✷

Lemma 3.2. Let b(y), d(y) ∈ C[0, 1], K0(x) ∈ C[0, 1]. Then the operator A0 : C(Ω¯ ) → C(Ω¯ ) corresponding to

the problem:

x

A0u(t) = u′ (x, y) + d(y)u(x, y) = f (x, y), (3.8)

x

D(A0) = {u(x, y) ∈ C(Ω¯ ) : u′

∈ C(Ω¯ ), u(0, y) = b(y) ∫ 1

K0(x)u(x, y)dx}

is correct if and only if

∫ 1

b(y)

0

0

K0(x)e−xd(y)dx ̸= 1 (3.9)

and the unique solution of the above problem is given by the formula

−1

u(x, y) = A−1f (x, y) = b(y)e−xd(y) (1 − b(y) ∫ 1 K (x)e−xd(y)dx)

× (3.10)

0

× ∫ 1

xd(y) ∫ x

sd(y)

0 0

−xd(y) ∫ x

sd(y)

0 K0(x)e−

0 f (s, y)e

dsdx + e

0 f (s, y)e

ds.

Proof. Assume that u(x, y) ∈ ker A0 and (3.9) hold. Then from (3.8) we get

∫ 1

u′

x(x, y) + d(y)u(x, y) = 0, u(0, y) = b(y)

From the last equation by integration on x we obtain

K0(x)u(x, y)dx. (3.11)

0

u(x, y) = u(0, y)e−xd(y), u(x, y) = e−xd(y)b(y) ∫ 1 K (x)u(x, y)dx, (3.12)

∫ 1 ∫ 1

0 0

xd(y) 1

0 K0(x)u(x, y)dx = b(y)

[

0 K0(x)e−

] ∫ 1

0

dx ∫ K0(x)u(x, y)dx,

1 − b(y) ∫ 1 K (x)e−xd(y)dx

K (x)u(x, y)dx = 0.

0 0 0 0

If b(y) ∫ 1 K (x)e−xd(y)dx ̸= 1, we get ∫ 1 K (x)u(x, y)dx = 0. Substitution of this value into (3.12) implies

0 0 0 0

u(x, y) = 0. This means that A0 is injective.

1

xd(y)

Conversaly. Let u(x, y) ∈ ker A0 and b(y) ∫0 K0(x)e−

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 ker A0 ̸= {0} and so A0 is not injective.

We will ﬁnd the solution to (3.8). Let b(y) ∫ 1 K (x)e−xd(y)dx ̸= ±1. Then A

is injective and Problem (3.8)

0 0 0

has a unique solution. From (3.8) by integration on x for every f ∈ C(Ω¯ ) we obtain

u(x, y) = e−xd(y)b(y) ∫ 1 K (x)u(x, y)dx + e−xd(y) ∫ x f (s, y)esd(y)ds, (3.13)

0 0

∫ 1 ∫ 1

0

xd(y) 1

0 K0(x)u(x, y)dx = b(y)

+ ∫ 1

0 K0(x)e−

xd(y) ∫ x

0

dx ∫

sd(y)

K0(x)u(x, y)dx +

0 K0(x)e−

0 f (s, y)e

dsdx,

[1 − b(y) ∫ 1 K (x)e−xd(y)dx] ∫ 1 K (x)u(x, y)dx =

0 0

= ∫ 1

xd(y) ∫ x

0 0

sd(y)

Then since (3.9) we obtain

∫ 1

0 K0(x)e−

(

0 f (s, y)e

1

dsdx.

)−1

K0(x)u(x, y)dx =

0

0

1 − b(y) ∫

K0(x)e−xd(y)dx ×

× ∫ 1

0 K0(x)e−

xd(y) ∫ x

0 f (s, y)e

sd(y)

dsdx. (3.14)

Substituting (3.14) into (3.13), we obtain the unique solution (3.10) to (3.8) for every f ∈ C(Ω¯ ). Since f

0

in (3.10) is an arbitrary element of C(Ω¯ ), then R(A0) = C(Ω¯ ). It is easy to verify that A−1

is bounded.

Hence A0 is correct. ✷

Theorem 3.3. Let a(x), c(x), K0(x) ∈ C[0, 1], b(y), K(y) ∈ C[0, 1], d(y) ∈ C1[0, 1], h(x, y), u(x, y) ∈

xy

C1(Ω¯ ), u′′ (x, y) ∈ C(Ω¯ ). Then the problem

u′′ ′

xy (x, y) + c(x)ux(x, y) + d(y)uy (x, y) + h(x, y)u(x, y) = f (x, y), (3.15)

u(0, y) = b(y) ∫ 1 K (x)u(x, y)dx,

0 0

u′ 1

x

x(x, 0) + d(0)u(x, 0) = a(x) ∫0 K(y)[u′ (x, y) + d(y)u(x, y)]dy

38Провидас E., Пулькина Л.С., Парасидис И.Н. Факторизация обыкновенных и гиперболических...

is correct if

h(x, y) = d′(y) + c(x)d(y), (3.16)

a(x) ∫ 1 K(y)e−yc(x)dy ̸= 1, b(y) ∫ 1 K (x)e−xd(y)dx ̸= 1 (3.17)

0 0 0

and its unique solution is given by the formula

−1

u(x, y) = b(y)e−xd(y) (1 − b(y) ∫ 1 K (x)e−xd(y)dx)

× (3.18)

× ∫ 1

xd(y) ∫ x

sd(y)

0 0

−xd(y) ∫ x

sd(y)

where

0 K0(x)e−

0 v(s, y)e

dsdx + e

0 v(s, y)e

−1

ds,

0

v(x, y) = a(x)e−yc(x) (1 − a(x) ∫ 1 K(y)e−yc(x)dy)

× (3.19)

× ∫ 1

0 K(y)e−

yc(x) ∫ y

0 f (x, t)e

tc(x)

dtdy + e

f (x, t)e

−yc(x) ∫ y

0

tc(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) ∈ C1[0, 1]. Denote by A1 the operator corresponding to Problem (3.15), namely:

A1u(x, y) = u′′ (x, y) + c(x)u′ (x, y) + d(y)u′ (x, y) + h(x, y)u(x, y), (3.20)

xy x y

D(A1) = {u(x, y) ∈ C(Ω¯ ) : u′ (x, y), u′ (x, y), u′′ (x, y) ∈ C(Ω¯ ), (3.21)

x y xy

u(0, y) = b(y) ∫ 1 K (x)u(x, y)dx,

0

u′

1

x(x, 0) + d(0)u(x, 0) = a(x) ∫

0

K(y)[u′ (x, y) + d(y)u(x, y)]dy}.

0 x

We will prove that A1 = AA0, i.e. D(A1) = D(AA0), A1u = AA0u for all u ∈ 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 ∈ D(A0) : A0u ∈ D(A)} = (3.22)

= {u(x, y) ∈ C(Ω¯ ) : u′

∈ C(Ω¯ ), u(0, y) = b(y) ∫ 1 K (x)u(x, y)dx, A u ∈ D(A)} =

x 0 0 0

= {u(x, y) ∈ C(Ω¯ ) : u′ (x, y) ∈ C(Ω¯ ), (u′ (x, y) + d(y)u(x, y))′

∈ C(Ω¯ ),

x x y

u(0, y) = b(y) ∫ 1 K (x)u(x, y)dx,

u′

x(x, 0) + d(0)u(x, 0) = a(x) ∫

0 0

1

K(y)[u′ (x, y) + d(y)u(x, y)]dy},

0 x

AA0u(x, y) = (u′ (x, y) + d(y)u(x, y))′ + c(x)[u′ (x, y) + d(y)u(x, y)]. (3.23)

x y x

Since d(y) ∈ C1[0, 1], from (u′ (x, y) + d(y)u(x, y))′

∈ C(Ω¯ ) follows that u′′

∈ C(Ω¯ ) and

x y xy

(u′ (x, y) + d(y)u(x, y))′

= u′′ (x, y) + d′(y)u(x, y) + d(y)u′ (x, y) ∈ C(Ω¯ ).

x y xy y

x

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 ∈ 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 ∈ D(A1) for the case when (3.16) holds. Denote by v(x, y) = A0u(x, y) = u′ (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−1f where A−1f 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),

0 0

(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

xy

Corollary 3.4. Let a(x), K0(x) ∈ C[0, 1], b(y), K(y) ∈ C[0, 1], u(x, y) ∈ C1(Ω¯ ), u′′ (x, y) ∈ C(Ω¯ ). Then the

problem

u′′

xy (x, y) = f (x, y), (3.24)

u(0, y) = b(y) ∫ 1 K (x)u(x, y)dx,

0 0

u′ 1

is correct on C(Ω¯ ) if

x

x(x, 0) = a(x) ∫0 K(y)u′ (x, y)dy

∫ 1

a(x)

0

∫ 1

K(y)dy ̸= 1, b(y)

0

K0(x)dx ̸= 1. (3.25)

Вестник Самарского университета. Естественнонаучная серия. 2021. Том 27, № 1. С. 29–43

Vestnik of Samara University. Natural Science Series. 2021, vol. 27, no. 1, pp. 29–43 39

and the unique solution of Problem (3.24) is given by the formula

u(x, y) = (3.26)

[

= b(y) ∫ 1

∫ x a(s) ∫ 1 ∫ y

0

1−b(y) ∫ 1 K0 (x)dx

0 K0(x)

]

0

0 1−a(s) ∫ 1 K(y)dy

0 K(y)

0 f (s, t)dtdyds +

+ ∫ x ∫ y

∫ x a(s) ∫ 1 ∫ y

0 0 f (s, t)dtds

dx +

0

0 1−a(s) ∫ 1 K(y)dy

0 K(y)

0 f (s, t)dtdyds +

+ ∫ x ∫ y

0 0 f (s, t)dtds.

The following problem is solved by Theorem 2.3.

Example 3.5. Let u(x, y), u′ (x, y), u′ (x, y), u′′

∈ C(Ω). Then the problem

x

u′′

y xy

1 1 1 1

0

0

xy −(x + y) ∫ ∫

x

0

xu′ (x, y)dxdy − 3x3 ∫

xy

0

∫ y2u′′ (x, y)dxdy (3.27)

= 15x3 − 2x − 2y,

u′ 1

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 B1 the operator corresponding to Problem (3.27). First we must determine the op__e__rators

A and A0 and make sure that D(B1) = D(AA0). Comparing (3.27) with (2.14) it is natural to take X = C(Ω),

1

Φ(A0u) = ∫

1

1

1

∫ xu′ (x, y)dxdy, F (AA0u) = ∫ ∫

y2u′′ (x, y)dxdy, (3.29)

0 0 x

0 0 xy

AA0u(x, y) = u′′ (x, y), A0u = u′ (x, y).

xy x

Denote v(x, y) = u′ (x, y). Then AA0u(x, y) = u′′ (x, y) = (u′ (x, y))′

= Av(x, y) = v′ (x, y). From boundary

x xy x y y

conditions (3.27) follws that v(x, 0) = 0. So the operators A, A0 are deﬁned by

Av(x, y) = v′ (x, y), D(A) = {v(x, y) ∈ C(Ω¯ ) : v′ ∈ C(Ω¯ ), v(x, 0) = 0},

y y

A0u(x, y) = u′ (x, y), D(A0) = {u(x, y) ∈ C(Ω¯ ) : u′

∈ C(Ω¯ ),

Then

x x

0

u(0, y) = (y + 1) ∫ 1 u(x, y)dx}.

D(AA0) = {u(x, y) ∈ D(A0) : A0u ∈ D(A)} = {u(x, y) ∈ C(Ω¯ ) :

u′ ′′ 1 ′

x, uxy ∈ C(Ω¯ ), u(0, y) = (y + 1) ∫0 u(x, y)dx, ux(x, 0) = 0} = D(B1).

Since D(B1) = D(AA0), 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 A0 coincides with the operator A0 from Lemma 3.2 if b(y) = y + 1, d(y) = 0, K0(x) = 1. Then by Lemma 3.1, the operator A is correct and

∫ y

A−1f (x, y) =

f (x, t)dt, (3.30)

0

by Lemma 3.2 the operator A0 is correct if and only if y ̸= 0 and its inverse is deﬁned by

A−1

0 f (x, y) = −

y + 1 ∫ 1

y 0

(1 − s)f (s, y)ds +

∫ x

f (s, y)ds. (3.31)

0

Notice that the operator AA0 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

A−1 −1

y + 1 ∫

1 ∫ x ∫ y

∫ x ∫ y

y

0 A f (x, y) = −

0 0

f (s, t)dtdsdx +

0 0

f (s, t)dtds. (3.32)

0

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

Φ(f ) =

∫ 1 ∫ 1

xf (x, y)dxdy, F (f ) =

∫ 1 ∫ 1

y2f (x, y)dxdy. (3.33)

0 0 0 0

Using (3.32) for f = 15x3 − 2x − 2y and G = 3x3 we ﬁnd

40Провидас E., Пулькина Л.С., Парасидис И.Н. Факторизация обыкновенных и гиперболических...

A−1 −1

y+1

1 x y 3

0 A f (x, y) = −

∫ ∫

y 0 0

0

∫ (15*s*

2s − 2t)dtdsdx+

4 2 2

+ ∫ x ∫ y 3

45x y−12x y−12xy +(y+1)(6y−5)

0 0 (15s

2s − 2t)dtds =

12 ,

A−1 −1

y+1

1 x y 3

0 A G = −

∫ ∫ ∫

y 0 0 0

3s dtdsdx+

4 2 2 2

+ ∫ x ∫ y 3 3

3 4 5x y+6x y+6xy −3(y+1)

0 0 3s dtds = − 20 (y + 1) + 4 x y +

12 .

By means (3.30) for G = 3x3, S = x + y we get

A−1G = ∫ y G(x, t)dt = ∫ y 3x3dt = 3x3y,

0 0

A−1S = ∫ y S(x, t)dt = ∫ y (x + t)dt = xy + y2/2,

0 0

A−1f = ∫ y f (x, t)dt = ∫ y (15x3 − 2x − 2y)dt = 15x3y − 2xy − y2.

0 0

Using (3.33) we get

F (S) = ∫ 1 ∫ 1 y2S(x, y)dxdy = ∫ 1 ∫ 1 y2(x + y)dxdy = 5 ,

0 0 0 0 12

F (G) = ∫ 1 ∫ 1 y2G(x, y)dxdy = ∫ 1 ∫ 1 y23x3dxdy = 1 ,

0 0 0 0 4

F (f ) = ∫ 1 ∫ 1 y2f (x, y)dxdy = ∫ 1 ∫ 1 y2(15x3 − 2x − 2y)dxdy = 5 ,

0 0 0 0 12

Φ(A−1G) = ∫ 1 ∫ 1 x3x3ydxdy = 3 ,

0 0 10

Φ(A−1f ) = ∫ 1 ∫ 1 x(15x3y − 2xy − y2)dxdy = 1.

0 0

Further by (2.17)-(2.19) we ﬁnd

L = Im − F (G) = 1 − 1/4 = 3/4,

2

G0 = G0(x, y) = A−1S + A−1GL−1F (S) = xy + y

+ 5 x3y,

1

L0 = Im − Φ(G0) = 1 − ∫

2 3

1

∫ xG0(x, y)dxdy

0

2

= 1 − ∫ 1 ∫ 1 *x *(*xy *+ y

0

+ 5 x3y) dxdy = 7 .

0 0 2 3 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−1

y+1 1 x

0 G0 = −

y 0 (1 − s)G0(s, y)ds + 0 G0(s, y)ds

4 2 2 2

12 .

= 5x y+6x y+6xy −3(y+1)

Substituting the above values into (2.9) we get (3.28). ✷

## About the authors

### E. Providas

University of Thessaly
**Author for correspondence.**

Email: providas@uth.gr

ORCID iD: 0000-0002-0675-4351

Candidate of Technical Sciences, associate

professor

### L. S. Pulkina

Department of Equations of Mathematical Physics, Samara National Research University
Email: louise@samdiff.ru

ORCID iD: 0000-0001-7947-6121

Scopus Author ID: 6506395220

ResearcherId: C-1180-2017

Russian Federation, 34, Moskovskoye Shosse, Samara, 443086, Russian Federation.

### I. N. Parasidis

University of Thessaly
Email: paras@teilar.gr

ORCID iD: 0000-0002-7900-9256

Сandidate of Technical Sciences, associate professor

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