Boundary Inverse problem for star-shaped graph with different densities strings-edges Текст научной статьи по специальности «Математика»

MSC 45C05, 35G16 DOI: 10.14529/mmpl80301

BOUNDARY INVERSE PROBLEM FOR STAR-SHAPED GRAPH WITH DIFFERENT DENSITIES STRINGS-EDGES

A.M. Akhtyamov1, Kh.R. Mamedov2, E.N. Yilmazoglu2 1 Bashkir State University, Ufa, Russian Federation 2Mersin University, Mersin, Turkey

E-mail: AkhtyamovAM@mail.ru, hanlar@mersin.edu.tr, nur_9465@hotmail.com

The paper is devoted to the mathematical modelling of star-shaped geometric graphs with n rib-strings of different density and the solution of the boundary inverse spectral problem for Sturm-Liouville differential operators on these graphs. Earlier it was shown that if strings have the same length and densities, then the stiffness coefficients of springs at the ends of graph strings are not uniquely recovered from natural frequencies. They are found up to permutations of their places. We show, that if the strings have different densities, then the stiffness coefficients of springs on the ends of graph strings are uniquely recovered from all natural frequencies. Counterexamples are shown that for the unique recovery of the stiffness coefficients of springs on n dead ends of the graph, it is not enough to use n natural frequencies. Examples are also given showing that it is sufficient to use n + 1 natural frequencies for the uniqueness of the recovery of the stiffness coefficients of springs at the n ends of the strings. Those, the uniqueness or non-uniqueness of the restoration of the stiffness coefficients of springs at the ends of the strings depends on whether the string densities are identical or different.

Keywords: natural frequencies; star-shaped graph; inverse problems; strings; densities; boundary conditions.

Introduction

The main results on direct and inverse spectral problems for Sturm-Liouville operators on an interval are presented in [1-11] and other works. Differential operators on graphs (networks, trees) often appear in natural sciences and engineering (see [12-22] and the references therein). Most of the results in this direction are devoted to direct problems of studying properties of the spectrum and the root functions for operators on graphs. Inverse spectral problems, because of their nonlinearity, are more difficult to investigate, and nowadays there exists only a small number of papers in this area. In particular, inverse spectral problems of recovering the coefficients of differential operators on trees (i.e on graphs without cycles) were solved. Boundary inverse eigenvalue problems geometric graphs were considered in [23-27]. In these papers it is shown that if strings have the same length and densities, then the stiffness coefficients of springs at the ends of strings are not uniquely recovered from natural frequencies. They are found up to permutations of their places. We show, that if the strings have different densities, then the stiffness coefficients of springs on the ends of strings are uniquely recovered from natural frequencies. Those, the uniqueness or non-uniqueness of the restoration of the stiffness coefficients of springs at the ends of the strings depend on whether the string densities are identical or different.

1. Eigenvalues for Sturm-Liouville Differential Operators on the Star Graph

Let r = 71 x 72 x • • • x Yn be a star-shaped graph which consists of n finite rays Yj = {xj E (0,lj)},j = 1, ...,n, with the origin of each ray identified with the single vertex of the graph. We consider on r the equation

dPy . (x ■)

LVj =--+ q(xj)y■(x.^ = P. X y■(x.^ = P. f y■(x.Xj E Yj, W

y

yi(0) = ... = yn(0), (2)

y{(0) + ... + y'n(0) = 0. (3)

We further require the boundary conditions on the end-points x. = lj\

yj (lj) + Hj y■ (I. ) = 0, j = 1, ••• ,n. (4)

Here pi, p2, • • •, pn are different numbers, ^d f are spectral parameters, Hj E C. Each potential qj (x) E L(y.) (j = 1, • • • ,n).

Let Sj(x,f), Cj(x,f), j = 1, ...,n + 1, be the solutions of (1) on the Yj which satisfy the initial conditions Sj (0, f) = 1 — Cj (0, f) = 0, 1 — sj (0, f) = Cj (0, f) = 0 (j = 1, • • • ,n). The general solution on every interval can be represented as

yj (x,f) = aj (f)cj (x,f)+ f3j (f)sj (x,f), j = 1, ••• ,n. (5)

n

ai(f) = ••• = an(f). (6)

We denote a(f) = a1(f) = • • • = an(f).

Boundary conditions on the ends of finite intervals (4) put the following conditions on a(f),fi(f), • • • ,f2(f)

a(f) [cj (lj ,f) + Hj cj (lj ,f)] + fj (f) [sj (lj ,f) + Hj sj (lj ,f)] = 0, j = 1, ■■■ (7) We define (i = 1, • • • ,n)

Oj (f) = sj (lj, f) + Hj sj (lj, f), (8)

Kj (f) = cj (lj, f) + Hj cj (lj, f). (9)

Then the condition (8) can be rewritten as

a(f)Kj (f) + fj (f)oj (f) = 0, j = 1, ••• ,n. (10)

It is important to mention, that (f) and kj (f) cannot vanish simultaneously: multiply (8) by c. (lj, ff) and subtract (9) multip lied by sj (lj ,f). As the result

Oj (f)cj (lj ,f) — Kj (f)sj (lj ,f) = sj (lj ,f)cj (lj ,f) — cj (lj ,f)sj (lj ,f) = 1 (ii)

g Bulletin of the South Ural State University. Ser. Mathematical Modelling, Programming

& Computer Software (Bulletin SUSU MMCS), 2018, vol. 11, no. 3, pp. 5-17

The second Kirchhoff condition (3) requires

f3i(v) + ••• + /3n{v) = 0. (12)

The first series of eigenvalues. Consider the case

ai(^) • a2(v) • ■■■ • = (13)

Then it follows from (7) that

cj (j ,7) + Hj Cj (j ,7) 'sj (j ,7) + HjSj (lj ,7)'

j (7) = —a(^) J (j'7) , H sJ (j „) • (14)

Then it follows from (12) that

, n cj (lj ,7) + Hj Cj (lj ,7) , ч

,n -J j =0, I5

U sj (lj ,7) + Hj Sj (j ,7) K '

where a(7) is arbitrary.

If a(7) = 0, then yj = 0 and we don't have eigenvalues. So eigenvalues are the roots of the equation

^ c'j (lj ,7) + Hjcj (lj ,7) , ,

sj (l3,7) + Hj Sj (lj ,7) " 1 j

The second series of eigenvalues. Consider the case

ax(7) • 02(7) • ■■■ • On(7) = 0. (17)

Denote

J = {j : Oj (7) = 0} . (18)

Then Kj(7) = 0, for J E J, and, to fulfil the (7)

a(7) = 0. (19)

So the general solution in this case has to be in the form

yj (x,7) = j (7)sj (x,7), J = 1, ••• ,n. (20)

But Oj(7) = 0,j E J-, and as the result, to satisfy (7), it has to be fij = 0, and as the result the general solution in such case is

yj(x,7) = fij(7)sj(x,7), J E J, (21)

yj(x,7) = 0,JE J, (22)

where fij (7) are arbitrary.

If fjj = 0 j E J), then yj = 0 and we don't have eigenvalues. So eigenvalues are the roots of the equation

sj(x,7) = 0, J E J. (23)

2. Uniqueness of Solution for Boundary Inverse Problem

The boundary inverse problem consists of determining Hj from eigenvalues. We note

that the eigenvalues of the second series cannot be used to solve the inverse problem, since

Hj

Therefore, we assume that we know the eigenvalues from the first series. Let L denote the Sturm-Liouville eigenvalue problem (1) - (4).

In that follows, the problem of type L with different coefficients in the equation and with different parameters in the boundary forms is denoted by L. Additionally, if a certain

L

counterpart in problem L.

L

problem L counting their algebraic multiplicities, then the coefficients of the boundary conditions of L and L are also equal to each oth er; i.e., Hj = Hj for j = 1, 2,.. .n.

Proof. If Ai are eigenvalues of the first series for problem L, then from (16) it follows that Ai are the roots of the entire function

n n

A(A) = (lk,j) + Hk ck (lk,j)) • J] (s'j (j j) + Hj sj (lj,j)). (24) k=1 j=1; j=k

It can be seen that A(A) is an entire function of order 1. Moreover, according to the

L LH

multiplicities are equal to each other. Therefore, the Hadamard factorization theorem implies that A(A) = C A (A), where C is a nonzero constant. It follows that

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A(A) — C A (A) = £ (dk (lk,j) + Hk ck (lk,j)) • EI (sj (lj,j) + Hj Sj (lj ,j)) — k=1 j=i; j=k nn

-C •£ (Ck(lk,j) + Hk Ck (lk,j)) • n (sj (lj,j) + Hj Sj (lj,j)). k=1 j=i; j=k

We have the asymptotic formulas

cj(xj,J) = cOs(JXj) + ± Uj(Xj) sm(^Xj) + O (, sj(xj,J) = ± sm(^xj) - ±2 uj(xj) cOs(JXj) + O (,

Cj (xj , j) = — j sin(jXj) + Uj (xj) COs(jXj ) + O ,

sj (xj, j) = cos(j Xj) + ± Uj (xj) sin(jXj) + O ^ , j = 1, 2,... ,n,

where Uj (xj) = 2 J'X q(tj) dtj for sufficiently 1 arge j E R [3]. Hence it follows that the

cj sj

Hj = HHj

□

L

recovered from the infinite set of eigenvalues. However, computer calculations show that

enough finite number of eigenvalues is sufficient to restore n parameters Hj of L. In this case, for the reconstruction we need to use not n, but n + 1 eigenvalues of problem L.

nn

Hj

3. Examples and Counterexamples

Example 1. (Counterexample 1) Let Г = ji x 72 x 73 be a star-shaped graph which consists of 3 finite rays 7j = {xj E (0,1)}, lj = 1, j = 1, 2, 3, with the origin of each ray

Г

Lyj = — d Ух3) = Pj 72 yj (xj), xj E 7j, (26)

y

yl(0) = y2(0) = Уз(0), (27)

y'i (0)+ У2 (0)+ y3 (0) = 0. (28)

xj = lj = 1

yj (1) + H3 y3 (1) = 0,j = 1, ••• ,n. (29)

Here 7 is a spectral parameter, pi = 1, p2 = 2, p3 = 3, Hj E C.

The solutions of (26) on 7j which satisfies initial соndition sj(0,7) = 1 — cj(0,7) = 0, 1 — sj (0,7) = dj (0,7) = 0 (j = 1, 2, 3) are functions cj (xj ,7) = cos(7xj) and sj (xj ,7) =

I sin(7xjY j = 1,22,3-

Let us take the first tree eigenvalues of the first series of eigenvalues for problem (26) - (29) 71 = 0, 48450, 72 = 1, 4165, 73 = 2, 2721. Using these values, from (16) we arrive at the system of three equations:

(0, 88491 H\ — 0, 22566)(0, 85069H + 0,56613)(0, 68327H3 + 0,11703) + + (0, 96133H + 0, 88491)(0, 56613H — 0, 79876)(0, 68327H3 + 0,11703) + + (0, 96133H + 0, 88491)(0, 85069H2 + 0, 56613)(0,11703H3 — 1, 4435)) = 0, (0,15372H — 1, 3996)(0,10724H2 — 0, 95274)(—0, 21055H3 — 0, 44664) + + (0, 69759H + 0,15372)(—0, 95274H2 — 0, 86063)(—0, 21055H3 — 0, 44664)+ (30) + (0, 69759H + 0,15372)(0,10724H — 0,95274)(—0, 44664H + 3, 8020) = 0, (—0, 64523H — 1, 73588)(—0, 21696H2 — 0,16736)(0, 074565H + 0, 86121) + + (0, 33625H — 0, 64523)(—0,16736H2 + 4, 4801)(0, 074565H3 + 0, 86120) + + (0, 33625H — 0, 64523)(—0, 21696H2 — 0,16736)(0, 86120H3 — 3, 4645)) = 0,

which solution will be six sets of (Hi, H2, H3):

{Hi = 0, 081922H = 3, 4849,H3 = 6, 0394}, {Hi = 0, 31023,H2 = 1, 0390,H3 = 31, 632}, {Hi = 1, 0000,H2 = 2, 0000,H3 = 3, 0000}, {Hi = 1, 2269H = 0, 90496H = 5, 56392},

{Hi = —0, 45154+0,17493i,H2 = —3, 8954+5,6196i,H = —0, 55241 — 2, 2121i}, {H1 = —0, 45154 — 0,17493iH = —3,8954 — 5, 6196iH = ~0,55241 + 2, 2121i}.

Вестник ЮУрГУ. Серия «Математическое моделирование и программирование» (Вестник ЮУрГУ ММП). 2018. Т. 11, № 3. С. 5-17

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We substitute the fourth eigenvalue = 3, 2689 of the problem (26) - (29) in (16). As the result, we obtain the equation

(—0, 99191 Hi + 0, 41503)(0,038526 H2 + 0, 96776)—0,038005H — 0,92795) +

+(—0, 038840Hi — 0, 99191)(0, 96776 H2 — 1,6467)(-0, 038005H3 — 0, 92795)+ (32)

+—0, 038840Hi — 0, 99191)(0, 038526H2 + 0, 96776)—0, 92795H3 + 3, 65503) = 0.

Let us form a new system of equations from the first two equations of system (30) and equation (32). The solution to this new system of equations is the other 6 sets of solutions:

{Hi = 1, 4493, H2 =1, 9584, H3 = 2, 5961},

{Hi = 0, 84897+ 1, 2383 i, H2 = 0, 81210 — 0, 30203 i, H3 = 4, 3564 — 1,0462 i}, {Hi = 1, 0000, H2 = 2, 0000, H3 = 3, 0000},

{Hi = —0, 44254 + 0, 06691 i,H = 14, 573 + 6, 2654 i,H = —2, 9975— 1, 7914 i}, {Hi = —0, 44254 — 0, 066907 i, H2 = 14, 573 — 6, 2654 i,H3 = —2, 9975 + 1, 7914 i}, {H1 = 0, 84897 —1, 2383 i,H2 = 0, 81210 + 0, 30203 i,H3 = 4, 3565 + 1, 0462 i}.

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n Hj

nL

Hj

boundary conditions (4) can be uniquely determined by using n + 1 eigenvalues of problem L

Example 2. Let us take the first four eigenvalues of the first series of eigenvalues for problem (26) - (29) ji = 0, 48450, j2 = 1, 4165, j3 = 2, 2721, j4 = 3, 2689. The solution of the system of four equations (30), (32) is the intersection of sets (31) and (33). This intersection consists of a unique solution of system:

{Hi = 1, 0000, H2 = 2,0000, H3 = 3, 0000}. (34)

Example 3. (Counterexample 2) Let r = ji x j2 x y3 be a star-shaped graph which consists of 3 finite rays Yj = {xj E (0,1)},j = 1, 2, 3, with the origin of each ray identified with the single vertex of the graph. We consider on r equation

' (x)

LVj =--dj + qj(x) yj(x) = Pj j yj(x), Xj E Yj,

defined for functions y satisfying the natural Kirchhoff boundary conditions on vertex:

yi(0) = y2(0) = y3(0), (36)

y'i (0)+ y2 (0)+ y'3 (0) = 0. (37)

We further require the boundary conditions on end-points Xj = lj = 1:

yj (1) + Hj y3 (1) = 0, j = 1, ••• ,n. (38)

Here j is a spectral parameter, qi(x) = x, q2(x) = x + 2, q3(x) = x + 3 Pi = 1 P2 = 2, P3 = 3 H3 E C

| () Bulletin of the South Ural State University. Ser. Mathematical Modelling, Programming

& Computer Software (Bulletin SUSU MMCS), 2018, vol. 11, no. 3, pp. 5-17

The solutions of (35) on which satisfies initial condition Sj(0,ß) = 1 — Cj(0,ß) = 0, 1 — sj(0,ß) = Cj(0,ß) = 0 (j = 1, 2, 3) are functions that are expressed in terms of Airy

functions:

( , Bi(1, -ß2) • Ai(-ß2 + x)

Ai(1, -ß2) • Bi(-ßh + x)

C2(x,ß) = -

(Bi(-ß2) • Ai(1, -ß2) - Bi(1, -ß2) • Ai(-ß2)) '

Bi(1,-2 ß2 + 2) • Ai(-2 ß2 + x + 2)

Bi(-2 ß2 + 2) • Ai(1, -2 ß2 + 2) - Bi(1, -2 ß2 + 2) • Ai(-2 ß2 + 2) + Ai(1, -2 ß2 + 2) • Bi(-2 ß2 + x + 2)

+ Bi(-2 ß2 + 2) • Ai(1, -2 ß2 + 2) - Bi(1, -2 ß2 + 2) • Ai(-2 ß2 + 2) '

Bi(1'-3 ß2 + 3) • Ai(-3 ß2 + x + 3)

C3(x'ß) = - ^BMx-^ß^+iY^ÄM-iß^+i) +

Ai(1' -3 ß2 + 3) • Bi(-3 ß2 + x + 3)

+ Bi(-3 ß2 + 3) • Ai(1, -3 ß2 + 3) - Bi(1' -3 ß2 + 3) • Ai(-3 ß2 + 3) '

Bi(-ß2) • Ai(-ß2 + x)

Bi(x'ß)

S2(x'ß)

s3(x'ß)

Bi(— 72) • Ai(1, — 72) — Bi(1, — 72) • Ai(— 72) _Ai(— 72) • Bi(— 72 + x)_

Bi(—2 72 + 2) • Ai(—2 72 + x + 2)

Bi^L,—^T^+^^^Ai^—2 72 + 2) Ai(—272 + 2) • Bi(—272 + x + 2)

Bi—2^2^27?+2)^Щ—727?+2:),

Bi(—3 72 + 3) • Ai(—3 72 + x + 3)

Bi(—372ТЗУ^А1(Т,—372ГзУ—Bi(1,—372ТЗУ^А1(—'

Ai(—372 + 3) • Bi(—372 + x + 3) Bi(—3 72 + 3) • Ai(1, —3 72 + 3) — Bi(1, —3 72 + 3) • Ai(—3 72 + 3)'

Let us take the first tree eigenvalues of the first series of eigenvalues for problem (35) -(38) 7i = 1, 7745, 72 = 2, 243^0, 73 = 3,4376. Using these values, from (16) we arrive at the system of three equations:

(—0,12050Hi — 1, 6315)(0, 47686H2 — 0, 31384)(0, 26508H3 — 0, 72008) + + (0, 61336Hi + 0, 0059937)(—0, 42522Щ — 1, 8172)(0, 26508H3 — 0, 72008) + + (0, 61336Hi + 0, 0059937)(0, 47686H — 0, 31384)(—0, 80651H — 1, 5816) = 0, (—0,58101Hi — 1, 8116)(0,13876H2 — 0,88951)(—0, 076473H — 0, 94666) + + (0, 39850Hi — 0, 47863)(—0, 95980H2 — 1, 0540)(—0, 076473H3 — 0, 94666)+ (39) + (0, 39850Hi — 0, 47863)(0,13876H2 — 0, 88951)(—0, 98500H3 + 0, 88323) = 0, (—0,99573Hi + 0, 73943)— 0, 21611H2 — 0,11628)(—0,10434H3 + 0, 80048) + + (—0, 065638Hi — 0, 95554)(—0,11388H + 4, 5660)(—0,10434H + 0, 80048) +

+ (—0, 065638Hi — 0, 95554)(—0, 21611H2 — 0,11628)(0, 81475H + 3, 3335) = 0,

«

»

(42)

which solution will be six sets of (Hi,H2, H3):

{Hi = 3, 0000, H2 = 2, 0000, H3 = 1, 0000}, {Hi = 3, 3145, H2 = 0, 29642, H3 = 4, 6856},

{Hi = 0, 80324+6, 5631 i, H2 = 0,86403 — 1, 4760 i, H3 = 0, 33506 + 1, 3100 i}, . .

{Hi =—0, 81785 + 0, 49296 i, H2 =—0, 42624+4, 92103 i, H3 =—0,12302 — 4, 2522 i}, ^ {Hi =—0, 81785 — 0, 49295 i, H2 =—0, 42624 — 4,9210 i, H3 =—0,12302 + 4, 2522 i}, {H 1 = 0, 80324 — 6, 5631 i, H2 = 0, 86403 + 1,4760 i,H = 0, 33506~1, 3100 i}.

We substitute the fourth eigenvalue = 4,9887 of problem (35) - (38) in (16). As the result, we obtain equation

(0, 22829Hi + 4, 8126)(0, 081214H2 + 0, 82564)(0, 099045H3 — 0, 54720) + + (—0,19737Hi + 0, 21965)(0,83356H2 — 3, 8390)(0, 099045H3 — 0, 54720)+ (41)

+ (—0,19737Hi + 0, 21965)(0,081213H2 + 0, 82564)(—0, 55175H3 — 7, 0481) = 0.

Let us form the new system of equations from the first two equations of system (39) and equation (41). The solution to this new system of equations is the other 6 sets of solutions:

{Hi = 3, 0000, H2 = 2, 0000, H3 = 1, 0000},

{Hi = 1, 5805 + 2, 4752 i, H2 =—0, 22203+0, 67498 i,H = 1, 0034 — 3, 6902 i}, {Hi =—0, 059508 + 0, 67472 i, H2 =—6,1889 — 6,1889 i, H3 = 0, 93238 — 2,1799 i}, {Hi =—0, 46343, H2 = 54, 238, H3 = 11, 361}, {Hi =—0, 059508 — 0, 67473 i,H2 =—6,1889+0,090512 i,H:i = 0, 93238 + 2,1799 i}, {H1 = 1, 5804~2, 475)1 i, H2 =—6,1889+0, 090512 i, H3 = 1, 0034+3,6901 i}.

Thus, both systems do not have a unique solution.

Example 4. Let us take the first four eigenvalues of problem (35) - (38) ji = 1, 7745, j2 = 2, 24?^0, j3 = 3, 4376, = 4, 9887. The solution of the system of four equations (39), (41) is the intersection of sets (40) and (42). This intersection consists of a unique solution:

{Hi = 3, 0000, H2 = 2,0000, H3 = 1, 0000}. (43)

We note that in the case of an arbitrary infinitely differentiable functions qj (xj), in calculating the characteristic determinant (24), instead of the exact values of the linearly independent solutions Cj(x,j) and sj(x,j) of equation (1), the main parts of the Taylor series for these solutions are used with respect to the two variables x and j. Moreover, the coefficients of the boundary conditions (for the inverse problem) and the eigenvalues (for the direct problem) are found with a small error, and the accuracy of the calculation is sufficiently high [28].

Acknowledgements. The reported research was funded by Russian Foundation for Basic Research, the government of the region of the Republic of Bashkortostan (projects 18-51-06002-Az_a 18-01-00250_a, 17-41-020230-r_a), and the Science Development Fund under the President of the Republic of Azerbaijan (project on the 1st Azerbaijan-Russian International Grant Competition (EIF-BGM-4-RFTF-1/2017)).

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Received July 4, 2018

УДК 517.984.54 DOI: 10.14529/mmpl80301

ГРАНИЧНАЯ ОБРАТНАЯ ЗАДАЧА ДЛЯ ЗВЕЗДООБРАЗНОГО ГРАФА

СО СТРУНАМИ-РЕБРАМИ РАЗЛИЧНОЙ ПЛОТНОСТИ

A.M. Ахтямов1, Х.Р. Мамедов2, Э.Н. Йылмазоглу2

iБашкирский государственный университет, г. Уфа, Российская Федерация 2

Работа посвящена математическому моделированию звездообразных геометрических графов с n струнами-ребрами различной плотности и решению граничной обратной спектральной задачи для дифференциальных операторов Штурма - Диувилля

на этих графах. Ранее было показано, что, если струны имеют одинаковую длину и плотность, то коэффициенты жесткости пружин на концах струн восстанавливаются по собственным частотам неоднозначно. Они находятся с точностью до перестановок их местами. В настоящей статье показано, что, если струны имеют разную плотность, то коэффициенты жесткости пружин на концах струн восстанавливаются по всем собственным частотам однозначно. Приведены контрпримеры, показывающие, что для однозначного восстановления коэффициентов жесткостей пружинок на n тупиковых вершинах графа недостаточно использования п собственных частот. Приводятся также примеры, показывающие, что для однозначного восстановления коэффициентов жест-костей пружин на n концах струн достаточно использовать n +1 собственную частоту. Таким образом, однозначность или неоднозначность восстановления коэффициентов жесткостей пружин на концах струн зависят от того, являются ли плотности струн одинаковыми или различными.

Ключевые слова: собственные частоты; звезднообразный граф; обратные задачи; струны; плотности; краевые условия.

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Работа проводилась при финансовой поддержке РФФИ и Правительства Республики Башкортостан (проекты 18-51-06002-Аз_а, 18-01-00250-а, 17-41-020230-р_а), а также Фонда развития науки при Президенте Азербайджанской Республики (проект 1-го Азербайджанско-Российского международного конкурса грантов (EIF-BGM-4-RFTF-1/2017)).

Азамат Мухтарович Ахтямов, доктор физико-математических наук, профессор, кафедра математического моделирования, Башкирский государственный университет (г. Уфа, Российская Федерация), AkhtyamovAM@mail.ru.

Ханлар Рашидоглу Мамедов, доктор физико-математических наук, профессор, руководитель департамента математики, Мерсинский университет (г. Мерсин, Турция), hanlar@mersin.edu.tr.

Эмине Hyp Иылмазоглу, докторант, Мерсинский университет (г. Мерсин, Турция), nur_9465@hotmail.com.

Поступила в редакцию 4 июля 2018 г.