Научная статья на тему 'ON THE CAUCHY PROBLEM FOR THE BIHARMONIC EQUATION'

ON THE CAUCHY PROBLEM FOR THE BIHARMONIC EQUATION Текст научной статьи по специальности «Математика»

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Ключевые слова
BIHARMONIC EQUATIONS / CAUCHY PROBLEM / ILL-POSED PROBLEMS / CARLEMAN FUNCTION / REGULARIZED SOLUTIONS / REGULARIZATION / CONTINUATION FORMULAS

Аннотация научной статьи по математике, автор научной работы — Shodiev Dilshod S.

The work is devoted to the study of continuation and stability estimation of the solution of the Cauchy problem for the biharmonic equation in the domain G from its known values on the smooth part of the boundary ∂G. The problem under consideration belongs to the problems of mathematical physics in which there is no continuous dependence of solutions on the initial data. In this work, using the Carleman function, not only the biharmonic function itself, but also its derivatives are restored from the Cauchy data on a part of the boundary of the region. The stability estimates for the solution of the Cauchy problem in the classical sense are obtained.

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Текст научной работы на тему «ON THE CAUCHY PROBLEM FOR THE BIHARMONIC EQUATION»

DOI: 10.17516/1997-1397-2022-15-2-201-215 УДК 517.946

On the Cauchy Problem for the Biharmonic Equation

Dilshod S. Shodiev*

Samarkand State University Samarkand, Uzbekistan

Received 10.11.2021, received in revised form 30.12.2021, accepted 10.02.2021 Abstract. The work is devoted to the study of continuation and stability estimation of the solution of the Cauchy problem for the biharmonic equation in the domain G from its known values on the smooth part of the boundary dG. The problem under consideration belongs to the problems of mathematical physics in which there is no continuous dependence of solutions on the initial data. In this work, using the Carleman function, not only the biharmonic function itself, but also its derivatives are restored from the Cauchy data on a part of the boundary of the region. The stability estimates for the solution of the Cauchy problem in the classical sense are obtained.

Keywords: biharmonic equations, Cauchy problem, ill-posed problems, Carleman function, regularized solutions, regularization, continuation formulas.

Citation: D.S. Shodiev, On the Cauchy Problem for the Biharmonic Equation, J. Sib. Fed. Univ. Math. Phys., 2022, 15(2), 201-215. DOI: 10.17516/1997-1397-2022-15-2-201-215.

Introducnion

Let x = (xi,x2),y = (yi,y2) & R2 and G is a bounded simply connected domain in R2 with boundary dG, consisting of compact part T = {yi & R : ai < yi < bi} and a smooth arc of the curve S : y2 = h(yi) lying in the half-plane S : y2 = h(yi). G = G U dG, dG = S U T. In the domain G, consider the equation

A2U(y)=0, y G G, (1)

d

dy2 dyl

Problem definition. It is required to find the biharmonic function U (y) = U (yi,y2) G C 4(G) n C3(G), for which the values on the part S of the boundary dG are known, i.e.

where A = + Laplace operator.

U (yi,V2)\s = fl(y), AU (yi,V2)\s = Î2(y) dU (yi,y2)

dn

= fM, a-(A"

s dn

(2)

= h(y),

S d

here fj (y) £ C4-j (S),j = 1, 2, 3,4 are given functions, and — — operator of differentiation along the outward normal to dG.

The considered problem (1)-(2) refers to ill-posed problems of mathematical physics. The true nature of such problems was clarified for the first time in the work of A. N. Tikhonov [4],

*[email protected] © Siberian Federal University. All rights reserved

and he pointed out the practical importance of unstable problems, and also showed that if the class of possible solutions is reduced to a compact set, then the stability of the solution follows from the existence and uniqueness.

Formulas that make it possible to find a solution to an elliptic equation in the case when the Cauchy data are known only on a part of the boundary of the domain are called Carleman-type formulas. In [2] Carleman established a formula giving a solution to the Cauchy-Riemann equations in a domain of a special form. Developing his idea, G. M. Goluzin and V. I. Krylov [3] derived a formula for determining the values of analytic functions from data known only on the border on the border section, already for arbitrary domains. They found a formula for restoring a solution from its values on the boundary set of positive Lebesgue measure, and also proposed a new version of the extension formula. The monograph by L. A. Aizenberg [1] is devoted to one-dimensional and multidimensional generalizations of the Carleman formula. A formula of the Carleman type, which uses the fundamental solution of a differential equation with special properties (the Carleman function), was obtained by M. M. Lavrent'ev [7,8]. In these works, the definition of the Carleman function is given for the case when the Cauchy data are given approximately, and the a scheme of regularization of the Cauchy problem for the Laplace equation is also proposed. Using this method, Sh.Ya.Yarmukhamedov [9,10] constructed Carleman functions for a wide class of elliptic operators defined in spatial domains of a special form, when part of the boundary of the domain is a hypersurface or a conical surface. It should de noted that the Carleman function proposed by Sh.Yarmukhamedov was also studied by M. Ikehata [11].

The Carleman matrix for the Cauchy-Riemann equation in the case when S is an arbitrary set of positive measure was constructed in [13]. In [14] in classical domains, Carleman's formulas are given that restore the values of a function inside a domain from its values given on a set of positive measure on the skeleton.

The Cauchy problem for linear elliptic differential operators has numerous applications in physics, electrodynamics, fluid mechanics (see [8,12,15]). It is known that if the Carleman function is constructed, then using Green's formula one can write the regularized solution explicitly. This implies that the efficiency of constructing the Carleman function is equivalent to constructing a regularized solution to the Cauchy problem.

In [16], a method is proposed for the regularization of the solution of the Cauchy problem for the Laplace equation by introducing a biharmonic operator with a small parameter, and it is shown that if a solution to the original problem exists, then the difference between the spectral expansions of the solutions of the original and regularized equations tends to zero as the parameter tends regularization to zero in the space of square-summable functions. In recent years, many numerical methods have been presented solving the Cauchy problem for elliptic equations. In the paper [18] L. Marin investigated the iterative method of fundamental solutions algorithms together with the Tikhonov regularization method.

An estimate for the conditional stability of a boundary value problem for a fourth-order elliptic type equation in rectangular domains was obtained in [19].

In [20], using the Carleman function, not only the harmonic function itself, but also its derivatives for the Laplace equation are reconstructed from the Cauchy data on a part of the boundary of the domain.

Note that when solving applied problems, one should find the approximate values of the

dU (x)

solution U(x) and its derivative —-, x £ G, i = 1, 2.

dxi

In this paper, we construct a family of functions U(x,a,fkS) = UaS (x) and

dUaS (x) dxi

dU (x,a,fks )

dxi

, k = 1, 2, 3,4; i = -, 2 depending on a parameter a and prove that with a special

dUaS (x)

choice of parameter a = a(S) the family UaS (x) and —- at S ^ 0 converges at each point

dxi

dU (x)

x G G to the solution U (x) and its derivative —-, respectively. The family of functions

dxi

U(x,a,fkS) and dUfkS), i = 1,2 with indicated properties is said to be a regularized dxi

solution by M. M. Lavrent'ev [7]. If, under the indicated conditions, instead of the Cauchy data, their continuous approximations with a given deviation in the uniform metric are given, then an explicit regularization formula is proposed. In this case, it is assumed that the solution is bounded on the part T of the boundary.

The proof of these results is based on the construction in an explicit form of the fundamental solution of the biharmonic equation depending on a positive parameter, disappearing along with its derivatives as the parameter tends to infinity on T when the pole of the fundamental solution lies in the half-plane y2 > 0.

1. Construction of the Carleman function

Let us define the function (x,y) (from [10]) as follows

— 2ne<X2 (x,y)

Im

w — x2

udu

\/u2 + a2

Separating the imaginary part of the function (x,y), we have

$<7 (x,y) = 2ne

1 i 2 , 2 2\ = _e-7(a +X2-V2 )

e 7u cos2ay2\/u2 + a2udu

u2 + r2

e 7u (y2 — x2) sm2ay2\fu2+~a2 udu

u2 + r2

\/u2 + a2

(3)

(4)

where y' = (yi, 0), x' = (x\, 0), r = \y — x\, a = \y' — x'\, a > 0, a > 0, w = i^/u2 + a2 + y2, u > 0.

It the paper [10], one has proved that the function (x,y) defined by the equalities (3) with a > 0 is presentable in the from

(x,y) = F (r) + G< (x,y),

(5)

where F(r) = — ln -, Ga(x,y) is harmonic function with respect to y in R2, including y = x. 2n r

It follows that the function (x, y) for any a > 0 in y is a fundamental solution of the Laplace equation. The fundamental solution (x,y) with the indicated property is said to be the Carleman function for the half-space [7].

Therefore, for the function U(y) = U(yl,y2) £ C4(G) n C3(G) and any x G G the following integral Green formula holds true [17]:

2

OJ

e

0

0

OJ

0

U (x)

+

1

Ida

'dG

U ( d (AL(x,y)) dU(y) U(y)-a--AL(x, y)

dn

dL(x,y) , , AU (y)—dn--L(x,y)

dn d (AU(y)) dn

dSy+

dS y, x G G,

(6)

where L(x, y) = r2- is the fundamental solution to the equation (1). r

Since (x,y) is represented in the form (5), then in the integral representation (6) L(x,y) replacing the function La(x,y) = r2<fra(x,y), we have

U(x) =

Jd

+

dG

IdG

U ( d (ALa (x,y)) AL ( ) dU(y)

U (y)---ALc (x,y) dn

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d (AU(y))

dn

\jj( sdLc (x,y) T I \ AU (y)-dn--L° (x,y)

dn

dSy+

dSy, x e G.

(7)

2. The formula of continuation and regularization by M. M. Lavrent'ev

We denote

Ua (x)

dSy+

+

fi(y)d (AL;(x,y)) - fs(y)AL„(xy) f2(y) dLcQ^ y - fa(y)La(x,y) dSy, x & G.

(8)

The main result of this paper is contained in the following theorem.

Theorem 1. Let the function U(y) = U(y\, y2) £ C4(G) fl C3(G) on the part S of boundary dG satisfy the conditions (2), and on the part T of boundary dG the inequality be fulfilled

\U (y)| +

U(y)

n

+ \AU (y)\ +

AU(y)

n

< M, y & T, M > 0.

Then, for any x £ G and a > 0, the estimates hold true

\U(x) - Uc(x)\ < <£>(a, x2)Me-cx2, dU (x) dUc (x)

dx,

dx,

< Vi(a,x2)Me-ax2, i = 1, 2,

where

. „ 23\fan (3\fan ,— ,— \

p(a,x2) = 4 + ( + 20V*n +8x/anajx2 +

/ I— /— \ 2 9\fon 3 9Jan

+ (2Jan + 4sJono) x4 + —-x2 + —-,

v 12 2 2 ax2

1 165^0 4^na2

yi(a,x2) = 10+- +

a y/a

+ 44a +

+ 3 +

2^0

2 a/0 A/0

J2

17 9^/na

+

no

x2 + [Y + -2y/o ' vo

+ 4a x2+

3 (66Jn 4Jna 1 \ 1 20 Jn +9ax32 + + + t^ + 8)— + + 16a-

a a 2a x2 ax22

(9)

(10) (11)

(12)

(13)

S

S

2 у/ a yja

an + - +

1 Ъ^/Л 4^.

+

2 4^ VO

X2 +

+ (29y/0n + b&^fOna) x2 + x2> + 10\[Onax\ +

60^/Л

(14)

2

Proof. Let us prove the inequality (10). Denote by Ia(x) the difference

Ia (x) = U (x) — Ua (x).

dn

Further, from (7) and (8), we have I a (x)

+T

From this and the inequality (9) we obtain

\Ia (x)\ = +

U ( . д (ALj(x,y)) ( . dU(y)

U (y)-5--alj (x,y)

dn

■ ( dLj(x,y) T ( . d (AU(y))

AU (y)-Б--Lj (x,y)

dn

dSy+

dn

(15)

dSy, x G G.

U ( d (ALj (x,y)) AL ( ) dU (y)

U (y)—on--(x)

dn

dSy+

( dLa(x,y) L ( ) d (AU(y))

AU (y)-Б--LJ (x,y)

dLj (x,y)

dn ' dn

where

r d (AT (x y))

+ \ALa (x,y)\ +

To show that the estimate (10) is valid, we prove the following

2

Na(x) < a,x2)e-ax2, a > 0.

dSy

< MNV (x),

Nv (x)= f

Jt

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dn

+ \ LJ (x,y)\

dSy = Ji + J2 + J3 + J4 •

(16)

According to (4), we have

La(x, y) = r4a (x, y) = тЧ J-e-v(a2+x2-y2)

e au2 cos2ay2\/u2 + a2udu

0 u2 + r2

e-jU (y2 — x2) sm2ay2\/u2+a2 udu

u2 + r2

Hence, setting y2 =0 we get

Lv(x, y) = ((yi — xi)2 + x2W -^e-™2J

\/u2 + a2

~ e-j(u2+a2)udu

u2 + (yi — xi)2 + x22 I

Now we estimate the following integral

Ji = \Lj(x, y)\ dSy < ■Jt Ja 1

(yi — xi)2 + x2 _Jx2 f ~ \u\ e-j(u2+(yi-xi)2)d

2n

lo u2 + (yi — xi)2 +

2+Z2 > dyi <

e Jx2.

2^0

Here, in the estimation, we used the inequality

(yi — xi)2 + x2 ' + (yi — xi)2 + x2

2

2

00

0

b

u

and introduced polar coordinate systems. Considering

dLa (x,y) dLa (x,y) dLa (x,y) .

-5- = -5-cos Y +--a-Sln Y,

dn dyi dy2

get

= -f (x,y)] = 2(y2 - x2)*. (xy) + r2 ,

dy2 dy2 dy2

here cos y, sln y are the coordinates of the unit outward normal n at the point y of the boundary dG.

Further, setting y2 = 0, we estimate the following integral

J = /

JT

dLa (x,y)

dn

r- bl dSy ^ / J ai Ixi e-ax22

f" \u\\(yi - xi )2 + x

Jo u2 + (yi -

[" \u\\(yi - xi)2 + x2

f" M e-a(v2+(vi-xi)2) L u2 + (y1 - Xi)2 + x2 dU+

X2 -x I '' \u\ |(yi - X1)~ + x2\ e J{y2 + {V1 xi)2) d +

e I 2 / \2 2 du+

n «- H2 —I— Itl-i _ T" ■> \2 —I— nf2

+ aX.e-°xir l"l du\ dyi

n Jo u2 + (yi - Xi)2 + x2

In \/an

3| ^-ax2

V 4^/a 1 ^V^ ' 2 2 In what follows, we need the following expressions

ALa(x,y) = A (r2^a(x,y)) = JT^ [r2$a(x,y)\ + —2^ [r2$a(x,y)\ =

dy1 dy2

m , \ N d^a (x,y) .d&u (x,y)

= (x, y) + 4(yi - xi) a ' + 4(y2 - x2) a ■

y1 y2

d(AL£M = £ [a (r2(x,y))] =

8^.0 + i(yi - x,) + i(,J2 - x-2) ,

dy2 dyi dy2 dy.

2

where

d (ALa (x,y)) d (ALa (x,y)) , d(ALa (x,y)) .

— cos y +--ä-Sln Y.

dn dyi dy2

In these expressions, flat y2 = 0, estimating, we get

r rbl f 2 2 rUAe-a(u2+(yi-Xl)2)

Mv, - xi)2 ax2 i" |u| e

2

+__^_e-ax2 _!_!_du+

n Jo u2 + (yi - xi)2 + x22

+ 4(yi - xi)2 e-ax2 r" + x e-ax22 r"

n Jo u2 + (yi - xi)2 + x2 n Jo (u2 + (yi - x,)2 + x2)2

+ ^e- x2 f uu+y-V-^d«U < (f + >**() e-ax2■

J

4 =

d (ALj(x,y))

dn

dSy ^

(20y/On + 8y0na) x2 + 4\j0nax2 + A^/Onx^ +

0x2

In estimating the integrals, we used the inequality

\u\ \yi — xi\

+ (yi — xi)2 + x2

< 1.

Taking into account the obtained estimates, we have

d (ALj(x,y))

dn <

+ \ALj (x,y)\ +

dLj (x,y)

dn

+ \lj (x,y)\

dSy ^

Aa + f -A^- + 20VOn + 8у/0Па) x2+

( ,— ,— \ 2 9 Jan 3 9 Jan + (2Jan + 4v/ana)x2 + ^— x3 +

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2

0x2

(17)

From (17) follows the proof of the inequality (10).

Let us prove the inequality (11). Differentiating the equalities (7) and (8) by xi, i = -, 2 we

get

dU(x)

OX,

'за

+

isa

d

U (y) ^ d

AU y ш

d (ALj (x,y))

dn

dLj (x,y)

dn

d (ALj(x,y)) dU(y) dxi dn

dLj(x,y) d (AU(y))

dSy+

dx,

dn

dSy

dUj (x) = Г

dxi Js

+S

d

U y d

d

AU (y) dx

d (ALj(x,y)) dn dLj (x,y)

d (ALj(x,y)) d (AU(y))

dn

dxi dn

dLj(x,y) d (AU(y))

dSy+

dxi

dn

dSy .

Denote by Iia (x) the difference of the derivatives

Iij (x) =

dU (x) 3Ua (x)

dxi

dxi

+

AU (y) iL

d

U y ъ

dLj (x,y) dn

d (ALj(x,y))

dn

d (ALj(x,y)) dU(y)

dxi

dn

dSy+

dLj(x,y) d (AU(y))

dxi

dn

dSy .

From this and the inequality (9) it follows

' d (ALa (x,y))

where

Nij (x)

\Iij (x) = +

d

U y Ö,

dn

d (ALj(x,y)) dU(y)

dxi

dn

dSy+

d

AU (y) dx

dLj (x,y)

dn

dLj(x,y) d (AU(y))

dxi

d

dxi

d (ALj(x,y))

dn

+

d (ALj(x,y))

dxi

+

dn d

dxi

dSy

< MNij (x),

dLj (x,y)

dn

+

dLj (x,y)

dxi

dSy .

To show that the estimate (11) is valid, we prove the following inequality

Nia(x) < Vi(a,x2)e-ax2, a> 0.

2

jx

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2

e

2

u

2

e-jx2

For i = 1, we have

Nij (x)

We denote

d

dx1

d (ALj(x,y)) dn

+

d (ALj(x,y))

dx1

+

d

dx1

dLj (x,y)

dn

+

dLj (x,y)

dx1

dLj (x,y) d r 2 л

Zi = — = ^^ фJ(x,y^ =r

dxi

2 d^j (x,y) dxi

— 2(yi — xi)^j (x,y),

dSy .

(19)

Z2 =

d dLj (x,y) d

dxi dn dxi

dLj (x,y) dLj (x,y) .

cos Y +--7:-sm Y

dyi

dy2

d dLj (x,y) d ' d

dxi dn dxi dy2

d^Axy) d^j (x, y) 2 d2^j (x,y)

2(y2 — x2 )-5--2(yi — xi )----+ r

dxi

dy2

dxidy2

= = га ^ xm =

(x,y) d ^j (x,y) д2фJ (x,y) д2фJ (x,y) 4----4----h 4(yi — xi)——---+ 4(y2 — x2)-

dxi

dyi

dxi dyi

dxidy2

(20)

(21)

Z4 =

d

dxi

dxi

d (ALj(x,y)) dn

d

dxi

d (ALj(x,y)) d (ALj(x,y)) .

---cos Y +--^-sm Y

dyi dy2

d (ALj(x,y))

dn

I'd (ALj(x,y))l = 8d2^j(x,y) — 4d(x,y) + dxi[ dy2 \ dxidy2 dyidy2

(x,y)

(x,y)

+4(yi — xi)Ttrêxë +4(y2 — x2) .2

When

d

dxi

dLj (x,y) dn

and

d

dxi

dxi dyi dy2 d (ALj(x,y)) dn

(22)

dxidy^

cos y, sin y are the coordinates of the unit

outward normal n at the point y of the boundary dG.

In (19), (20), (21), (22), setting y2 =0 and estimating the resulting integrals, we have:

Nij(x) < 10+- +

1 a Wa

1 1Ъу/п 165^/On

+

+ 16a +

4sjna2 4sjni

+

+

^44a + -0= + 2Van + x2 + (-27 + + 4^0По + 4a^j x22 +

+ 9ox3 + ( ^ + 4^ + a + s) JL + ) e-Jxx2.

2 a 2a x2 ax22

(23)

The inequality (18) is proved for i = -. Now let us prove the inequality (18) for i = 2. Taking into account (15) we have

N2j (x) = f

JT

d

dx2

d (ALj(x,y)) dn

+

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d (ALj(x,y))

dx2

+

d

dx2

dLj (x,y)

dn

+

dLj (x,y)

dx2

dSy .

We introduce the following designation

*i = = ^ [r2*°(x,y)] = r2— 2(y2 — x^j(x,y), (24)

dx2 dx2 dx2

2

a

d

3

$2.

d dLa (x,y) d

dx2 n dx2

La(x, y) La(x, y)

cos y +--t.-sin y

yi

y2

d dLa (x,y) d ' d

x2 n x2 dy2

-2$a(x,y) + 2(y2 - x2)-7--2(y2 - x2)-^--+ r

x2

y2

x2 y2

(25)

$3 = d(ALa(^,y)) = ¿^ [A (r2$a(x, y))] = 4(yi - x1)d$a (x,y) - 4d$aa (x,y) + 4(y2 - x2 f$a (x y)

(26)

$4 =

d

x2

d

x2

yi x2 d (ALa(x,y))'

y2

x2 y2

n

d (ALa(x,y))

d

x2

d (ALa(x,y)) , d(ALa(x,y)) .

- cos y +--~-sin y

n

d

x2

yi

d (ALa(x,y))

y2

y2

8 d2$a (x,y) 4 d2$a (x,y) +

d3$a (x,y)

+4(yi - xi ^ a.. a.. +4(y2 - x2) Q_ Qii2

x2 y2 d3$a (x,y)

y22

(27)

dx2dyidy2 dx2dy%

In (24), (25), (26), (27), setting y2 =0 and estimating the resulting integrals, we have

N2° (x) H^ + H ^ + ^ + ^ +2) x2 +

+ [29^ +

58^0

x22 +

2

x32 +

lO^o

lx4

'x2 + vox2)

e-ax2.

(28)

The inequality (18) is proved for i = 2.

From (23) and (28) follows the proof of the inequality (11). Theorem 1 is proved.

Corollary 1. With each x € G, the equality holds true lim Ua (x) = U(x)

Let us denote

Km dUa(x) = dm, i = 1, 2.

dxi

d xi

G F

, x2) € G, a > x2 ^ e, a = max h(xi), O < e < a| ■

It is easy to see that the set GE c G is compact.

Corollary 2. If x € Ge, then the family of functions {Ua (x)} and formly for a ^ m, i.e.:

dUa (x) dxi

converges uni-

rr . . rr. . dUa(x) dU(x) Ua (x) ^ U (x), a y ^ y, i = 1, 2.

xi

xi

It should be noted that the sets ne = G\GE present the boundary lever of this problem, as in the theory of singular perturbations, where there is no uniform convergence.

2

a

2

3. An estimate of the stability of the solution to the Cauchy problem

Consider the set

(dAU (y) \

E ={U € C4(G) n C3(G): \U(y)\ + I^nn^}

+ \AU (y)\ +

We put

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I dn j

<M, M>O, y € T >.

max h(yi) = a, max W1 + ( —— t yyi> ' t \l \dyi

b.

Theorem 2. Let the function U(y) € E, satisfy the equations (1) and on the part S of the boundary of the domain G the inequality

\U (y)\ +

\ dU(y) \ \ dn J

+ \AU (y)\ +

f dAU(y)) I dn J

< S, y € S.

Then, for any x € G and a > 0 , the following estimate holds

2 2

\U(x)\ < ^(a,x2) Ma25a2,

(29)

(30)

where ^ (a, x2) = max (^ (a, x2), ^ (a, x2)),

3b \9ab^J~an o 97ab(a — x2) ab^Jan 3ab 2

^ (o,x2) =--+

00 21b Von

+ 3Oa2 b +

2 + 2 + ~2a +4a"b^+

+ _(a - x2)+ ^ +2b(a - x2) + 16abo +

, , j—/ n 5ahJ on(a — x2) 2, , „ + 2Oab + 8aboy/on(a - x2)+------- + 8a2 bo (a - x2)+

+

2b

4ab^Jon o(a - x2) o(a - x2)2

+ 4a3boy/0n + '2O7absfon +

16ab^Jon

(a - x2)2

+

+5bV~0n + 48a2bo(a - x2)2 + 16a3bo2 + 2a3bo + 182a2bo + 8a2bo^On + 128abo(a - x2)+

+42b +

24bsJ0n

o( a - x2 ) +4Oab(a - x2)3 +

+ 4bo(a - x2)2 + 16a2bo2(a - x2)2 + 16abo(a - x2)3 + 4ab^on(a - x2)2 +

3 , 8abV°n + 1Qa2 bo2 + 8a2

a - x2

(a - x2) + 4b^on(a - x2)+

2bsJ0n

+16a2bo\/on(a - x2) +--(a - x2)2 + 16a3bo2(a - x2)+

o

+8abo\/on(a - x2)2 + 4b(a - x2)2,

^ (a,x2) is determined by the formula (12). Proof. From Green's integral formula we have

U(x) =

+i

+

+

U( ) d (ALa(x,y)) AL ( . dU(y)

U (y)-dn--(x )

' d(ALa (x,y)) AT , ,

U (y)-5--ALa (x,y)

n

dU (y)

dSV +

n n

' dLa (x,y) T , JdAU(y)

(y)-dn--(x,y)-

dLa (x,y) , ,

AU (y)-dn--La (x,y)

dSV +

n

dAU (y)

n

dSV +

dSV .

2

o

S

From the condition (2) and the inequality (29) we obtain \U(x)\ <

d (ALJ (x,y))

fi(y)-ö--+ f3(y)ALj (x,y)

+

dn

U ( )d (ALj(x,y)) AL ( . dU(y)

U (y)-5--ALJ (x,y)

dSy +

+

dn

f2(y) dLj(Xny) + f4(y)Lj (x,y) ( dLj (x,y) L ( ) d (AU(y))

AU (y)-Б--LJ (x,y)

dn dSy +

dSy +

+M

dn

d (ALj(x,y))

dn

dSy

dn

dSy + \ALj(x,y)\ dSy +

JT JT

< S\Uj(x)\ + dLj (x,y)

dn

dSy +

+ J \Lj(xy)\ dS^j < S \Uj(x)\ + M^(a,x2)e-jx2.

The estimate used here M

+

d (ALj(x,y))

dSy + / \ALj(x,y)\dSy+ JT

T dLdx,y) dSy + T \Lj (x,y)\ dSy^j < M^(a,x2)e-jx2,

proved in Theorem 1. Next, estimate \Ua(x)

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\Uj(x)\ <

d (ALj(x,y))

dSy + / \ALj(x,y)\dSy +

J s J S

+ \Lj (x,y)\dSy = Ai + Ä2 + A3 + A4.

J S

dLj (x,y)

dn

dSy +

Estimating these integrals, we get

Ai = J^ \Lj(x,y)\dSy < ( — +

A + abV^\ e-j(x2-a2 )

y 2a a

(32)

A2

dLj (x,y)

dn

dSy ^

5b '2ab\fän 2, ____ .

— +-A-+ 2a2b +---—(a — x2) +

2a a

13ab ~2

2ab(a — x2) + ab^fan

+

3ab

2a

3ab 2 ,— b^jan

+ —--+ a b^an +----(a — x2)

4a

-J(x2-a2)

A3 = I \ALj(x,y)\ dSy <

■J S

17b jan

4a

+ 20ab + a?b^J an + 8aba\/an(a — x2) +

5abJan 2, / \ „,, -, +--(a — x2)8a ba(a — x2) + 2b(a — x2)

-J(x2-a2)

A4

d (ALj(x,y))

dn

dSy ^

, ч 2, bJan 4abJan

16aba + 42ab(a — x2) + 28a2b --+ —--

a a(a — x2)

+

+

2b

a(a — x2)2

+ + 207ah^fan +

16ab^Jan

(a — x2)2

+ bb^/an + 48a ba(a — x2)2 + 16a3ba2+

S

a

S

e

S

+2a3ba + 182a2ba + 8a2baVan + 128aba(a — x2) + 42b + + 4ba(a — x2)2+

a (a — x2)

+16a2ba2(a — x2)2 + 16aba(a — x2)3 + 4abVan(a — x2)2 + 40ab(a — x2)3 + +

a — x2

+16a2ba2 + 2a2bV~an +--^ (a — x2) + 4b^an(a — x2) + 16a2ba^an(a — x2) +

a

+ 16ab^ + (a— x2)2 +16a3ba2(a— x2)+ 8abaV0n(a— x2)2 +4b(a— x2)2] e-a(x2-a2).

aa

When evaluating the integrals, polar coordinates were introduced and the inequalities were

used _

4ay2\/ u2 + a2

sln '2oy2\J u2 + a2

<

1 + 2ay2\Ju2 + a2 '

2 \ x\

since sinx\ ^ -:—r, x > 0.

1 ' 1 + \x\'

Adding the estimates obtained, we have

\Ua(x)\ < 4(a,x2)e-a(x2-a2),

here

, . „ 3b 19ab\fan 2, 97ab(a — x2) abJan 3ab 2, ,—

0 (a,x2) = 3- +-y-+ 30a b +-^-2) + + — + 4a2b^+

a a 2 2 2a

bJan 21bJan .— 5abJ~an(a — x2) 2

+----(a — x2) +-----+ 20ab + 8abay/an(a — x2) +---+ 8a ba(a — x2)+

4a 4a a

, , „ , 4ab\fan 2b 3, ,— , ,— 16ab\fan

+2b(a — x2) + 16aba +--.—--r + —-^ + 4a3baVan + 207absjan + --+

a(a — x2) a (a — x2)2 (a — x2)2

+5bV~an + 48a2ba(a — x2)2 + 16a3ba2 + 2a3ba + 182a2ba + 8a2 ba^On + 128aba(a — x2)+

42b + + 4ba(a — x2)2 + +16a2ba2 (a — x2)2 + 16aba(a — x2)3 + 4abVan(a — x2)2 +

a (a — x2)

No 8ab\/an 9, 9 8a2bJ an, . , ,—. .

40ab(a — x2)3 +-----+ 16a2ba2 + +----(a — x2) + 4bsjan(a — x2)+

a — x2 a

16a2baVan(a — x2) +--^ (a — x2 )2 + 16a3ba2 (a — x2) + 8abaVan(a — x2)2 + 4b(a — x2)2.

a

From the integral formula (32) and the condition (9) we obtain

\U(x)\ < 5ea(a2-x2)0(a,x2) + Mv(a,x2)e-ax2 = *(a,x2) Me-ax2 + 5ea(a2-x2)^j . (33) The best estimate for the function \U(x)\ is obtained in the case, when

Me-ax2 = 5ea(a2-x2)

or 1 M

a = - ln -r. (34)

a2 5

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Substituting the expression for a from the equality (34) into (33) we obtain the proof of the inequality (30). Theorem 2 is proved. □

Set

Ujs(x) =

J S

+

fis (y)d (ALdf,y)) — f3S (y)ALj (xy)

f2S (y) dL,Jdnx y — f4S (y)LJ (x,y)

dSy +

(35)

dSy .

Theorem 3. Let the function U(y) £ E on S satisfy the conditions (2) and instead of the functions fi(y) their approximations fig(y), i = -, 2, 3, 4 with a given deviation 6 > 0, i.e.

max \fi(y) — fig(y) \ < 6. Then, for any x £ G and a > 0, the following estimate holds:

22

\U(x) — UjS(x) \ < ^(a,x2)Mi-S.

(36)

(37)

Proof. From (31) and (35) we get

\U(x) — Ujs(x)! < \Ij(x)! + S f

■J S

d (ALj(x,y))

n

+ \ALj (x,y)\ +

dL j (x,y)

n

+ \Lj(x,y)\> dSy.

From Theorems 1 and 2 we obtain

\U(x) \ < ^(a,x2) (Me-ax2 + 6eaa2-ax2)

- M

and choosing a ^ 1^ —, we obtain the proof of Theorem 3. a2 6

Corollary 3. For each x £ G the equality

lim Uag (x) = U(x).

g^o

Corollary 4. If x £ Ge, then the family of functions {Uag (x)},

Uag (x) ^ U(x)

converges uniformly as 6 ^ 0.

dU (x) .

xi

Similarly, one can obtain stability estimates for are true:

Corollary 5. For each x £ G , the equality

dUa s (x) dU (x) , lim—--= —-, i = -, 2.

s^o dxi dxi

Corollary 6. If x £ Ge, then the family of functions

Uag(x) ^ U[x^-UM ^ WW, i =-, 2

, i = 1, 2, and the following corollaries

xi

xi

converge uniformly at S ^ 0.

S

References

10

11

12

13

14

15

16

L.A.Aizenberg, Carleman's formulas in complex analysis, Moscow, Nauka, 1990 (in Russian). T.Carleman, Les Functions quasi analytiques, Paris, 1926.

G.M.Goluzin, V.I.Krylov, Generalized Carleman's formula and its application to analytic continuation of functions, Mat. sb., 40(1933), 144-149 (in Russian).

A.N.Tikhonov, On the stability of inverse problems, DAN SSSR, 39(1943), no. 5, 147-160 (in Russian).

A.N.Tikhonov, V.Ya.Arsenin, Methods for solving ill-posed problems, Moscow, Nauka, 1995 (in Russian).

A.N.Tikhonov, A.A.Samarskiy, Equations of mathematical physics, Moscow, Nauka, 1974 (in Russian).

M.M.Lavrent'ev, On the Cauchy problem for the Laplace equation, Izv. AN SSSR, 20(1956), no. 6, 819-842 (in Russian).

M.M.Lavrent'ev, On some ill-posed problems of mathematical physics, Ed. SO AN SSSR, Novosibirsk, 1962 (in Russian).

Sh.Yarmukhamedov, On the harmonic continuation of differentiable functions given on a piece of the boundary, Siberian Math. J, 43(2002), no. 1, 183-193. DOI: 10.1023/A:1013849310522

Sh.Yarmukhamedov, Representation of a harmonic function in the form of potentials and the Cauchy problem, Mathematical Notes, 83(2008), no. 5, 693-706. DOI: 10.1134/S0001434608050131

M.Ikehata, Inverse conductivity problem in the infinite slab, Inverse Problems, 17(2001), no. 3, 437-454. DOI: 10.1088/0266-5611/17/3/305

S.I.Kabanikhin, Inverse and ill-posed problems. Novosibirsk: Siberian Scientific Publishing House, 2009 (in Russian).

L.A.Aizenberg, N.N.Tarkhanov, Abstract Carleman formula, Dokl. Math., 37(1988), no. 1, 235-238.

A.M.Kytmanov, T.N.Nikitina, Analogs of the Carleman formula for classical domains, Mat. Notes, 45(1989), no. 3, 243-248. DOI: 10.1007/BF01158560

A.N.Polkovnikov, A.A.Shlapunov, Construction of Carleman formulas by using mixed problems with parameter-dependent boundary conditions, Siberian Mathematical Journal, 58(2017), no. 4, 676-686. DOI: 10.17377/smzh.2017.58.414

V.V.Tikhomirov, N.N.Ochilov, Regularization of the Cauchy problem for the Laplace equation, Differential Equations, 50(2014), no. 8, 1128-1132.

I.N.Vekua, New methods for solving elliptic equations. OGIZ State publishing house of technology-theoretical literature, Moscow, 1948 (in Russian).

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4

5

6

7

8

9

[18] L.Marin, Relaxation procedures for an iterative MFS algorithm for two-dimensional steady-state isotropic heat conduction Cauchy problems, Eng. Anal. Bound. Elem., 35(2011), 415-429. DOI: 10.1134/S0012266114080114

[19] K.S.Fayazov, I.O.Khazhiev, Estimation of conditional stability and approximate solution of a boundary value problem for a fourth-order partial differential equation, Mathematical notes of NEFU, 22(85)(2015), no. 1, 78-88 (in Russian).

[20] A.B Khasanov, F.R.Tursunov, On the Cauchy problem for the Laplace equation, Ufa Math. J., 11(2019), no. 4, 92-107. DOI: 10.13108/2019-11-4-91

О задаче Коши для бигармонического уравнения

Дильшод С. Шодиев

Самаркандский государственный университет Самарканд, Узбекистан

Аннотация. Работа посвящена исследованию продолжения и оценки устойчивости решения задачи Коши для бигармонического уравнения в области G по его известным значениям на гладкой части границы dG. Рассматриваемая задача относится к задачам математической физики, в которых отсутствует непрерывная зависимость решений от начальных данных. В данной работе с помощью функции Карлемана восстанавливается не только сама бигармоническая функция, но и ее производные по данным Коши на части границы области. Получены оценки устойчивости решения задачи Коши в классическом смысле.

Ключевые слова: бигармонические уравнения, задача Коши, некорректные задачи, функция Карлемана, регуляризованные решения, регуляризация, формулы продолжения.

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