ON SOLVABILITY OF SOME BOUNDARY-VALUE PROBLEMS FOR THE NON-LOCAL POISSON EQUATION WITH FRACTIONAL-ORDER BOUNDARY OPERATORS Текст научной статьи по специальности «Математика»

118

Probl. Anal. Issues Anal. Vol. 13(31), No3, 2024, pp. 118-134

DOI: 10.15393/j3.art.2024.16550

UDC 517.954

B. Kh. Turmetov

ON SOLVABILITY OF SOME BOUNDARY-VALUE PROBLEMS FOR THE NON-LOCAL POISSON EQUATION WITH FRACTIONAL-ORDER BOUNDARY OPERATORS

Abstract. In this paper, a non-local analogue of the Laplace operator is introduced using involution-type mappings. For the corresponding non-local analogue of the Poisson equation in the unit ball, two types of boundary-value problems are considered. In the studied problems, the boundary conditions involve fractional-order operators with derivatives of the Hadamard type. The first problem generalizes the well-known Dirichlet, Neumann, and Robin problems for fractional-order boundary operators. The second problem is a generalization of periodic and antiperiodic boundary-value problems for circular domains. Theorems on the existence and uniqueness of solutions to the studied problems are proved. Exact conditions for solvability of the studied problems are found, and integral representations of the solutions are obtained.

Key words: non-local equation, fractional derivative, Hadamard operator, periodic problem, Dirichlet problem, Neumann problem

2020 Mathematical Subject Classification: 35J05,35J25

1. Introduction. This paper is devoted to the study of correct formulations of boundary-value problems for equations with transformed arguments. In the equations, the considered transformation of arguments is carried out using involution-type mappings. A mapping S is called an involution if S2 = E, where E is the identity mapping. For example, such a mapping is the Dunkl transformation [5]. Some applications of Dunkl-type mappings are considered in [7], [8].

Note that one of the first published papers for equations with involutive transformations is the work of T. Carleman [4], where equations with shifts of arguments of the type a = apt), a2it) = t were studied. Some

© Petrozavodsk State University, 2024

issues of the application of equations with shifts of the Carleman type are considered in [6].

Further, let us consider statement of the problems studied in this work. Let Q = {x e Rn : < 1} be a unit ball, n ^ 2, BQ be a unit sphere, u(x)

x d

be a smooth function in the domain Q, r = |x|, 9 = —, 5 = r— be the

r dr

Dirac operator, where r— =) Xj-—.

dr T-i dxj

3 = 1

Let us consider the modified Hadamard integro-differential operators ( [11], p. 116)

u(x), a = 0

Ju M (s) =

1 1

—— (ln-) (tx) dr, a > 0, ^ ^ 0,

r M J \ r j

0

DP [u] (x) = r-^Jm-a [8m [t11 ■ u]](x) , m — 1 < a ^ m, m ^ 1.

For any xe Rn consider the mappings Six = (x1,..., Xi-1,-Xi, Xi+1,..., xn), 1 ^ i ^ n. For the index i, we will use not only the usual notation, but also its representation in the binary number system i = (in... i2ii)2 " " in ■ 2n~1 + ■ ■ ■ + i2 ■ 21 + i1 ■ 20. Using this notation, we can consider mappings of the type ... Sz22 S1 x, where ik = 0 or ik = 1. The total number of such mappings is 2n.

Let a,i,i = 0,1,..., 2n — 1 be some real numbers, A be a Laplace operator. Let us introduce the operator

2n —1

LnU(x) = J] az (-A) u (Si? ... S22S1 x),

i"0

which we will call a non-local Laplace operator.

Then, for any point x = (x1,x2,... ,xn) e Q, we match the «opposite» point x* = (o1x1,a2x2,... ,anxn) e Q, where a1 = —1, and Oj, j = 2,... ,n take one of the values +1. Let us denote

dQ+ = {x e dQ : x1 ^ 0} , dQ_ = {x e dQ : x1 ^ 0} , I = {x e dQ: x1 = 0} .

Note that the point x* can be represented as x* = S^ ... S^2 S}x. Moreover, if x e BQ+, then x* e BQ_.

In the domain Q, we consider the equation

Lnu(x) = f (x),x e Q. (1)

In [20], for equation (1) the main boundary-value problems with the Dirichlet and Neumann conditions were investigated. Spectral issues for the operator Ln were studied in [21]. The present work is a continuation of these studies, and for equation (1) we will consider the following problems.

Problem 1. Let 0 ^ ^, 0 ^ a ^ 1. Find a function u(x) from the class C2(Q) x C(Q), for which D^[u](x) e C(Q), satisfying equation (1) and the condition

D^[u](x)= g(x),x e BQ. (2)

Problem 2. Let 0 ^ ¡3 < a ^ 1. Find a function u(x) from the class C2(Q) x C(Q), for which Dq[u](x) e C(Q), satisfying equation (1) and the conditions

^o H(x) - (-1)k[u](x*) = g0(x),x e BQ+, (3)

D«[u](x) + (-1)kD«[u](x*)= 9l(x),x e BQ+, (4)

where k takes one of the values k = +1.

As in the case x e BQ_ there is an inclusion x* e BQ+; then, from condition (3) it follows that

Dl[u](x*) - (-1)k[u](x) = g0(x*),x e BQ_.

In this case, if x p I ô x = (0,x2,... ,xn) p B^+, then for the point x* p B^+ corresponding to it we get: x* = (0,a2x2,... ,<7nxn) p ô x* e I. Therefore, for points x p I it is necessary to fulfill the conditions of agreement:

go(x) = Dl[u](x) - (-l)kDl[u](x*)

xel

-iy [u](x*) - (-l)kDO[u](x) " -(-l)k90(x*).

x*el

Let

m

B

Bmu(x) " Bxm1...dxm-,m = (mi,...,m").

Further, we will find solutions to Problem 2 from the class Cx+2 (Q) , 0 < A < 1. Then a necessary condition for the existence of a solution to Problem 2 from this class is the fulfillment of the following matching conditions:

dmgo(0,X2,...,xn) = (—1)k Bm go(0,a2X2,...,anxn),x e I,m = 0,1, 2, (5) dmg1(0,x2,... ,xn) = —(—1)kBmg1(0,a2x2,... ,anxn),x e I,m = 0,1, 2.

(6)

In what follows, we assume that conditions (5), (6) are satisfied. As J°°u(x) = u(x), then in the case a = 1 the operator D^ coincides with

B

the operator r-—K In this case, Problem 1 for ^ > 0 coincides with or

the Robin problem, and in the case ^ = 0 it coincides with the Neumann problem.

Note that boundary-value problems for an elliptic equation with fractional-order boundary operators were studied in [1], [10], [12], [13], [18], [24]. In these works, operators with Hadamard, Riemann-Liouville, Ca-puto derivatives and some of their modifications were considered as boundary operators.

Boundary-value problems with periodic and antiperiodic conditions for the Poisson equation in circular domains were first studied in [14], [15], and for the non-local analogue of the Poisson equation in the case n = 2 they were considered in [23]. Later, some generalizations of these problems with conditions of the Dirichlet, Neumann, and Robin type, as well as the Samarskii-Ionkin type, were studied in [16], [17], [19], [25].

Also note that the boundary conditions in the considered problems are specified as a relationship between the values of the unknown function at different points of the boundary. Problems of this type are usually called non-local problems of the Bitsadze-Samarskii type [2], [3].

2. Properties of integro-differentiation operators. In this section, we present some known properties of operators J^ and D™ in the class of smooth functions. The statements below clarify the conditions for reversibility and action of these operators in the Holder class. Note that studies in this direction were conducted in [9], where the properties of fractional differential operators associated with the derivative q were studied in the class of harmonic functions. The following statements were proved in [18]:

Lemma 1. Let a > 0, p > 0, 0 < A < 1 and u (x) e Cx+P ),p ^ 0. Then

1) if " > 0, then J; [u] (x) e Cx+P (fi) ;

2) if " = 0 and the condition u (0) = 0 is satisfied, the function Jg [u] (x) also belongs to the class Cx+p (Q) and the equality Jg[u] (0) = 0 is satisfied.

Lemma 2. Let " ^ 0, p — 1 < a ^ p, p = 1, 2,..., 0 <A < 1, and u (x) e Cx+9 (Q), q ^ p. Then the function D; [u] (x) belongs to the class Cx+q-p (Q + and the equality D; [u] (0) = 0 is satisfied.

Lemma 3. Let " ^ 0, p — 1 < a ^ p, p = 1, 2,..., 0 < A < 1 and u (x) e Cx+q (Q), q ^ p. Then for any x e Q the equality

j; [Di [u]] (x)- (u ' m, " > 0'

lu (x) — u (0), " = 0.

is valid.

Lemma 4. Let " ^ 0, p — 1 < a ^ p, p = 1, 2,..., 0 <A < 1, and u (x) e Cx+q (Q), q ^ p. Then for any x e Q if " > 0 the equality

d; "j; [u]] (x) = u (x) (7)

is valid; in the case " = 0, equality (7) is also valid under the additional condition u (0) = 0.

Lemma 5. Let " ^ 0, p — 1 < a ^ p, p = 1, 2,..., f (x) be a smooth function in the domain Q and —Au (x) = f (x), x e Q. Then the equality

— A D; [u] (x) = F (x), x e Q, (8)

is valid, where

F (x) = Dg+ 2 [f] (x). (9)

Lemma 6. If " = 0,0 < a ^ 1, then for the function F (x) from equality (9) there is a representation

F (x) = (r^ + 2) h-a (x),

where /i_a (x) = Jl~a [f] (x).

3. Existence and uniqueness of a solution to Problem 1. Let

us introduce the notation

2n —1

ek = S (—1)k b la*,k " 0,1,..., 2™ — 1, ¿"0

where k b i = knin + ■ ■ ■ + k1i1 is a "scalar" product of numbers (k)2 and (i)2, (i)2 = (in... i-i)2 is the notation of the index i in the binary number system.

3 .

Note that in case n = 2 the numbers £k = XI (—1) b%ai, ^ " 0,1, 2, 3,

i-0

are written as:

\ £o " £(oo)2 = oo + a,1 + a,2 + 03,£ 1 " S{o\)2 = 00 — 01 + a2 — 03, [ £2 " £(1ti)2 " ao + a1 — Q>2 — 03,63 " £(11)2 = Oo — 01 — a,2 + 03.

As we have already noted, in the case of the classical Dirichlet (a = 0) and Neumann (a = 1, ^ = 0) boundary conditions, Problem 1 was investigated in [20]. The following statement was proved:

Theorem 1. Let ^ = 0, the coefficients of the operator Ln be such that the conditions £k ^ 0, k = 0,1,..., 2ra — 1 are satisfied, and f (x) e Cx , g(x) e Cx+2 (BQ), 0 < A < 1. Then

1) if a = 0, then a solution to Problem 1 exists and is unique;

2) if a = 1, then for the solvability of Problem 1 it is necessary and

sufficient that the condition

f (y)dy + ( 2 \g(y)dsy " 0 (10)

v ' i

is satisfied.

If a solution to the problem ex(sts, then it is unique up to a constant term and belongs to the class Cx+2 (Q).

Example. Let x* = (—x1, —x2,..., —xn), Hm(x) be a homogeneous harmonic polynomial of degree m and u(x) = (1 — ^^H^x). It is obvious that u(x)|дi = 0. Moreover, Au(x) = 2 (2m + n) Hm(x). Hence,

aoAu(x) + a1Au(x*) = —2 (2m + n) [aoHm(x) + aoHm(x*)].

If the polynomial Hm(x) has the property Hm(x) = Hm(x*) and ao + a1 = 0, then we see that the function u(x) = (1 — |x|2) Hm(x) is a solution to the following homogeneous problem

aoAu(x) + a1Au(x*) = 0, x e Q; u(x)|дi = 0.

It follows from this example that if the coefficients of the operator Ln are such that the condition ek = 0, k = 0,1,..., 2n — 1, is satisfied, then the homogeneous Problem 1 can have infinitely many solutions. In the general case, the following statement is valid:

Theorem 2. Let 0 < a < 1, ^ ^ 0, the coefficients of the operator Ln be such that the conditions £k ^ 0, k = 0,1,..., 2n — 1, are satisfied, and f (x) e Cx+1 (H), g(x) e Cx+2(BQ), 0 < A < 1. Then the following statements hold:

1) if ^ > 0, then a solution to Problem 1 exists and is unique;

2) if ^ = 0, then for the solvability of Problem 1 it is necessary and sufficient that the condition

fi-a(y)dy + V a^l I g(y)dsy = 0 (11)

(I *) i'

„ U

is satisfied. If a solution to the problem exists, it is unique up to a constant term;

3) if a solution to the problem exists, it is represented in the form

u {x) = J™ M Or), (12)

where the function v(x) is a solution to the problem

Lnv(x) = F(x), x e q, (13)

v ML = g(x), (14)

where F(x) = D™+2[f ](x). In case ^ = 0, the function v(x) satisfies the additional condition v(0) = 0;

4) if a solution to the problem exists, then u(x) e Cx+2 (Q).

Proof. Let ^ > 0 and the function u (x) be a solution to Problem 1. Let us consider the function v (x) = D^u (x). If we apply the operator A to this function, then, by virtue of equality (8), we obtain Av(x) = D^+2[Au](x). Let S be an orthogonal matrix and Isu(x) = u(Sx). Then the operators Is and A, as well as Is and D™, commute. Therefore, for all i = 0,..., 2n — 1 we get Av(Szn" ... S^x) = D%[Au](Sznn ... S^x) and, hence,

2n-1

i-0

Lnv(x) = Yi aiADau(S™ ... Si'x) =

2n-1

S alA2u(Sn ... Si'x)

= Da

DH+2

i-0

= Dg+2[f](x), x e Q.

Moreover, from the boundary condition (2) it follows that

v(x) Ian = Dg[u](x) |Bq = 9(x).

Thus, if u (x) is a solution to Problem 1, then for the function v (x) = D^\u\(x) we obtain the Dirichlet problem (13), (14) with the function F(x) = 2\f](x). If f (x) e Cx+1 (H), then, due to Lemma 2, we get 2 \f] (x) e Cx (h). Then, by Theorem 1, for the functions F(x) = D2\f](x) e Cx (h) and g (x) e Cx+2 (BQ) a solution to problem (13), (14) exists, is unique, and v(x) e Cx+2 (H). If we apply the operator J^, to the equality v (x) = D^ \u] (x) on both sides, then, by virtue of the assertion of Lemma 3, we obtain u (x) = [w] (x), i.e., the solution to the problem is represented in the form (12). The inverse assertion is also valid, i.e., if the function v (x) is a solution to problem (13), (14), then the function u (x) = J^ \w] (x) satisfies all the conditions of Problem 1. Indeed, as v(x) e Cx+2 (H), by the assertion of Lemma 1 the function u (x) = \w] (x) also belongs to the class Cx+2 (H). Further, if we apply the operator Ln to the function u (x) = \w] (x), we get

L™u (x) = Ln [ Jg [w^ (x) = J;+ 2 [LnV] (x) =

= J;+ 2 [D;+2 [f]] (x) = f(x), x e Q.

Therefore, function (12) satisfies equation (1). In addition, from equality (7) it follows that

Dg[u] (x)|an = D; [j; M] (x)|an = u (x)|5n = 9(x),

i.e., the boundary condition is also satisfied.

Further, we will consider the case " = 0. In this case, for the function v (x) = Dgu (x) we also obtain problem (13), (14) with the function F(x) = Dg[f](x). In addition, by virtue of the assertion of Lemma 2, the function v (x) = Dgu (x) must satisfy the additional condition w(0) = 0. From Lemma 6 it also follows that the function F(x) = Dg [f ] (x) can be

represented as F (x) = (5 + 2) f1_a (x). Further, in [20] it is proved (see problem (5.4)) that for the equality w(0) = 0 to be satisfied, it is necessary and sufficient that

(2n-1 \

Yj aij I 9{y)dy =

Thus, the necessity of fulfilling condition (11) for the existence of a solution to Problem 1 is proved. The other part of the theorem is proved in the same way as in the case ^ > 0. The theorem is proved. □

Remark 1. If a = 1 and ^ = 0, then f° {x) = J2° [f] {x) " f {x) and then the solvability condition (11) coincides with condition (10).

4. Existence and uniqueness of a solution to Problem 2. First, let us study the uniqueness of the solution to Problem 2. The following assertion is valid:

Theorem 3. Let coefficients of the operator Ln be such that the conditions £p ^ 0, p = 0,1,..., 2n — 1, are satisfied and a solution to Problem 2 exists. Then

1) if k = 1 and ft = 0, then the solution is unique;

2) in other cases, the solution is unique up to a constant term.

Proof. Let the function u(x) be a solution to the homogeneous Problem 2. In [20] it is proved (see Lemma 2) that under the condition e^ ^ 0, k = 0,1,..., 2n — 1, the function u {x) satisfying the equation Lnu(x) = 0, x e Q, is harmonic in Q. Hence, u {x) satisfies the conditions of the following problem:

Au(x) = 0, x e Q, (15)

DHu(x) — {—1)kDgu(x*) = 0, x e 3Q+, (16)

/°u,w { ±) ^o-

Dqu(x) + (—1)kDqu(x*) = 0, X e BQ+. (17)

Let k = 1 and v(x) = u(x) — u(x*). Note that for any x e Q there is the equality v(x) = u(x) — u(x*) = — [u(x*) — u(x)] = — v(x*). Hence, if x e BQ+, then

D^x)^ = DqU(X) — D'o^u(x*)\di+ = 0,

and if x e BQ_, then x* e BQ+ and

Dov^l^ i_ = — [DqU(X*) — DZum^, = 0.

Therefore, the function ( x satisfies the conditions of the problem

Av(x) = 0, x e Q, D£v(x)|an = 0. (18)

Then, by the assertion of Theorem 2 (case a0 = 1, cij = 0, j = 2, 3,..., 2n — 1) we get v(x) " Cons t. As v(x) = —v(x*), we see that v(x) " 0, x e h. Hence, u(x) " u(x*), x e h. From this equality it follows that Dqu(x) " D^u(x*), x e BQ. On the other hand, from condition (16) we have D^^u(x) = —D^u(x*), x e BQ+. So, we obtain the equality D^u(x) = 0, x e BQ. Thus, the function u (x) satisfies the conditions of the problem

Au(x) = 0, x e Q, Dfou(x)

= 0. (19)

an

If 3 = 0, then problem (19) coincides with the Dirichlet problem and, therefore, u(x) " 0,x e h. If 3 > 0, then by the assertion of Theorem 2 we obtain u(x) " C. If k = 2, then for the function v(x) = u(x) — u(x*) we obtain the problem

Av(x) = 0, x e Q, D^v(x)

= 0. (20)

an

In this case, for all 0 ^3 < a we also get v(x) " 0, x e h. Then u(x) " u(x*), x e ^ and, hence, D0fu(x) " D%u(x*), x e ^. On the other hand, from condition (17) it follows that D0^u(x) = —D0^u(x*), x e BQ+, which is possible only in the case D^u(x) = 0, x e BQ. Thus, the function u ( x satisfies the conditions of the problem

Au(x) = 0, x e Q, D™u(x)|an = 0. (21)

Then, by the assertion of Theorem 2 we obtain u(x) " C. The theorem is proved. □

Let us consider the existence of a solution to Problem 2. In the case k = 1, the following assertion is valid:

Theorem 4. Let k = 1, and the coefficients of the operator Ln be such that the conditions £p ^ 0, p = 0,1,..., 2n — 1, are satisfied and g0(x), gi(x) e Cx+2 (BQ+), f(x) e Cx (H) , 0 < A < 1. Then

1) if 3 = 0, then a solution to problem 2 exists and is unique;

2) if ft > 0, then for the solvability of Problem 2 it is necessary and sufficient that the condition

/2"-l

22)

fi-a(y)dy + I 2 g0(y)dsy = 0

n ' Bn

be satisfied.

If a solution to the problem exists, it is unique up to a constant term and belongs to the class Cx+2 (Q).

Proof. Let k = 1 and u(x) be the solution to Problem 2. Let us represent this function as u(x) = v(x) + w(x), where

v(x) = 1 \u(x) — u(x*)\ , w(x) = 1 \u(x) + u(x*)\.

Let us find the problems that the functions v(x) and w(x) satisfy. By assumption, u(x) satisfies equation (1), i.e.,

2r-1

2 dAu (S^ ... S22Six) = f (x), x e Q.

As the operators I jr j2 ^ and A commute, at the point x*= S^... S2 S}x we get

2"-1 2r-1

^ ] Q>iAu (S ™ ... S'2S'^x ) = ^ | <iiAIgjri gj2^u (S^ ... S'^S'^x) = i"0 i"0

2r-1

= ^jr sj2s1 aiAu № ... S^S^x) = f(x ). i"0

Hence, for the functions v(x) and w(x) we get: Lnv(x) = 1 \Lnu(x) — Lnu(x*)\ = 2 \f(x) — f(x*)\ " f~(x), x e Q,

Lnw(x) = 1 \Lnu(x) + Lnu(x*)\ = 2\f(x) + f(x*)\ " f+(x), x e Q.

It is clear, that if f(x) e Cx+k (Q), 0 < X <_ 1, k = 0,1,..., then the functions f±(x) also belong to the class Cx+k (Q).

If x e BQ+, then, by virtue of condition (3), we have

Dgv(x)L+ = 1 [Dgu(x) - Dgu(x*)] and if x e BQ-, then x* e BQ+, and thus: Dgv(x)Idn_ = 1 [Dgu(x) - Dgu(x*)]Idn_ =

an i

= 2 9l(x),

= -2 [Dgu(x*)-Dgu(x)]Ix,edn+ = gi(x*).

2-

Similarly, taking into account condition (2) for the function w(x), we obtain

dqw(x)

an 4

D^u(x) + Dq u(x*)

an

= 2 9o(x),

dqw(x)

xeBn-

DU u(x*) + DU u(x)

x*eBnj

= 2 9o(x*).

Let us introduce the functions

2g°(x) J9o(x), x eBQ+, 2~gi{x) "19l(x), x edQ+ l#°(x*), x e BQ-, l-g-i_(x*), x e BQ_

Then, for ( x and w( x , we get the following problems:

L nV(x) = f-(x), x e Q; D;v(x)|an = gi(x),

L Xw(x) = f+(x), x e Q; D°[w(x) = g°(x).

0 an

(23)

(24)

If g1(x) e Cx+2 (BQ+) and the matching condition (6) is satisfied, then g1(x) e Cx+2 (BQ). Then, according to Theorem 2, for the solvability of problem (23) it is necessary and sufficient that the condition

fi-a(y)dy + Yu aA g1(y)dsy =

^ / an

(25)

be satisfied, where f1-a(y) = 1 [h-a(y) - ¡1-a(y*)}. If this condition is satisfied, then the solution to the problem exists, and it is unique up to

a constant term. Let us study the integrals in equality (25). In [22] it is proved that if S is an orthogonal matrix, then the equalities

f psy)dy ^ f (y)dy, J gpsy)dsy = J g(y)dsy. (26) n n an- an+

are valid.

From these equalities for the functions f--a(y) and g\(y), we get

fi-apv)dy = 1 J ih-a(y) - h-a(y*)s dy = 0, J gi(y)dsy = 0. n n an

Thus, the solvability condition (25) is satisfied and, therefore, a solution to problem (23) exists and is unique up to a constant term. As the function v(x) has the property v(x*) =-v(x), this is possible only in the case C " 0.

Now we can consider problem (24). If 3 = 0, then this problem coincides with the Dirichlet problem and, therefore, according to Theorem 1, the problem is unconditionally solvable. In the case 3 > 0, the solvability condition is written as

/2n-1

a*] I Qn(y)ds,, = 0. (27)

fi-a(y)dy + 1] aM 9o(y)dsy = 0.

V i=° ' Â

L—a

n x 7 an

where

fi—M = 2 [h—M + h—«(y*q].

If this condition is satisfied, the solution to the problem exists and is unique up to a constant term. Further, from equalities (26), for the integrals from (27) we have:

ft—a(y)dy = 1 J [ fi—a(y) + h—a(V*)} dy ^ f1—a(y)dy,

n n n

~go(y)d s y = 2

an

go(y)d s y + | go( y* )d s y

an an

= go(y)ds y.

an+

Then the solvability condition (27) can be rewritten as (22). Thus we have found the conditions under which solutions to problems (21) and (24)

exist. The function u(x) = v(x) + w{x) constructed from the solutions to these problems satisfies all the conditions of Problem 2. The theorem is proved. □

The following assertion is proved in a similar way.

Theorem 5. Let k = 2, the coefficients of the operator Ln be such that the following conditions £p ^ 0, p = 0,1 ,..., 2n — 1, are satisfied, and g°(x), gi(x) e Cx+2 (dQ+), f(x) e Cx (Ù), 0 < A < 1. Then, for the solvability of Problem 2, it is necessary and sufficient that the condition

be satisfied.

If a solution to the problem exists, then it is unique up to a constant term and belongs to the class Cx+2 (Q).

Acknowledgment. This research has been funded by the Science Committee of the Ministry of Science and Higher Education of the Republic of Kazakhstan (Grant No. AP23488086).

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Received August 22, 2024. In revised form,, October 13, 2024. Accepted October 13, 2024. Published online November 10, 2024.

B. Kh. Turmetov

Khoja Akhmet Yassawi International Kazakh-Turkish University 29 B. Sattarhanov ave., Turkistan 161200, Kazakhstan Alfraganus University

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