On one mathematical model described by boundary value problem for the biharmonic equation Текст научной статьи по специальности «Математика»

MSC 35J40

DOI: 10.14529/ mm p160404

ON ONE MATHEMATICAL MODEL DESCRIBED BY BOUNDARY VALUE PROBLEM FOR THE BIHARMONIC EQUATION

V. V. Karachik, South Ural State University, Chelyabinsk, Russian Federation, karachik@susu. ru,

B. T. Torebek, Institute of Mathematics and Mathematical Modelling, Almaty, Kazakhistan, [email protected]

In this paper mathematical model described by a generalized third boundary value problem for the homogeneous biharmonic equation in the unit ball with boundary operators up to the third order containing normal derivatives and Laplacian is investigated. Particular cases of the considered mathematical model are the classical models described by Dirichlet, Riquier, and Robin problems, and the Steklov spectral problem, as well as many other mathematical models generated by these boundary conditions. Two existence theorems for the solution of the problem are proved. Existence conditions are obtained in the form of orthogonality on the boundary of some linear combination of boundary functions to homogeneous harmonic polynomials of a particular order. The obtained results are illustrated by some special cases of the general problem.

Keywords: mathematical model; biharmonic equation; boundary value problems; Laplace operator.

Introduction

A significant number of mathematical models in physics and engineering lead to partial differential equations. The steady processes of various physical nature are described by the partial differential equations of elliptic type. One of the important special cases of fourth order elliptic equations is the biharmonic equation A2u(x) = f (x).

Investigation of mathematical models of problems of the plane deformation of the elasticity theory in many cases is reduced to integration of the biharmonic equation with the appropriate boundary conditions and under some uniqueness conditions for the unknown function.

Moreover investigation of many mathematical models of continuum mechanics are reduced to solving the harmonic and biharmonic equations. However the convenient analytical expressions for the solutions are obtained only for the certain domains of particular forms. Application of biharmonic problems in mathematical models of mechanics and physics can be found in the numerous scientific investigations (see, for example, [1-3]).

Multiple applications of boundary value problems for the biharmonic equation in mathematical models of mechanics and physics encourage investigation of various formulations of boundary value problems for the biharmonic equation. The class of biharmonic functions includes a class of harmonic functions and is a subclass of polyharmonic functions. One of the first important works on the biharmonic equation is the article by Almansi [4]. Following this work, there was a big interest in studying of boundary value problems for the biharmonic equation (see [5-7]).

The Dirichlet problem (see [8]) is a well known boundary value problem for biharmonic equation. In recent years other types of boundary value problems for the biharmonic

equation, such as the problems by Eiquier (see [9]), by Neumann (see [10-12]), by Robin and etc are actively studied. In the spectral theory the Steklov spectral problem is of some interest.

The present paper is organized as follows. In Section lfor the biharmonic equation in the unit ball a boundary value problem (1) - (2) with the boundary conditions of general type, called by us as a generalized third boundary value problem is formulated. Particular cases of the problem are considered in Section 4. In Section 2, namely in Theorem 2, the conditions of unconditional solvability of the problem for the homogeneous biharmonic equation are found. These conditions coincide with the conditions of uniqueness from Theorem 1. In Theorem 3 of Section 3 the case, when the conditions of Theorem 2 are not fulfilled, but a solution of the considered problem still exists, are obtained.

1. Statement of a Problem

Let S = {x E Rra : |x| < 1} be an ^dimensional unit ball in the Euclidean space Rra with the norm |x| = \Jx\ + x2 + ... + x^, and dS = {x E : |x| = 1} be an unit sphere. S

A2u = f (x), x E S, (1)

aoou + aoi u + a^Au

dS

= ^1(s), s G dS,

aii dv u + ai2Au + ai3 djv, Au

д dv '

dS

Ф2 (s), s G dS,

(2)

where Jjj is the outside normal derivative, a0j and a1j where j = 1, 2, 3 are real constants, and f (x), p1(x), ^2(x) are defined function, smoothness of which will be given below. As a solution of problem (1) - (2) we consider a biharmonic in S function u(x) from the class u E C4 (S) fl C3 (S) which satisfies on dS conditions (2).

Problem (1) - (2) generalizes the known Dirichlet problem (aoo = 0, a11 = 0, and all other coefficients are zero), Eiquier problem (a00 = 0 a12 = 0, and all other coefficients are zero), but does not generalize the Neumann boundary value problem. If a00 = 0 a12 > 0, a11 < 0, and all other coefficients vanish, then conditions (2) coincide with the Steklov conditions [13]. In [14] uniqueness of a solution for (1) - (2) was proved.

Theorem 1. Solution of problem (1) - (2) is unique if and only if the following polynomial

A(A)

aoo + aoiA 2aoi + (2n + 4A)ao2

Aaii 2aii + (2n + 4A)ai2 + A(2n + 4A)ai3

has no roots in N0 = N U {0^. If A(m) = 0, then the homogeneous problem (1) - (2) has a solution

u(x)= (C2M2 + C1 - C2) Hm (x), where Hm(x) is an arbitrary homogeneous harmonic polynomial of degree m, and the

C1 C2

(

aoo + maoi 2aoi + (2n + 4m)ao2

maii 2aii + (2n + 4m)ai2 + m(2n + 4m)ai3

)(S)

0.

(3)

2. Unconditional Existence of a Solution

One auxiliary assertion in which harmonic functions from the Almansi representation are explicitly defined can be easily proved.

Lemma 1. Let u(x) be a biharmonic function in S, then the following functions

\x\2 r1 I r1

uo(x)= u(x) tn-1Au(t2x) dt, ui(x) = - tn-1Au(t2x) dt (4)

2 J 0 2 J 0

are harmonic in S and such that the Almansi representation u(x) = u0(x) + \x\2u1(x) holds S

Theorem 2. Solution of the problem

A2u = 0, x E S, (5)

aoo u + aoi -§v u + «02^u aii dv u + auAu + ai3 -§v Au

^1(s), x E dS,

dS (6)

= p2(s), x E dS,

dS

from the class u E C3(S) for arbitrary functions E C2(dS) and p2 E C 1(dS) exist, if and only if the polynomial

A(X)

has no roots in N0.

a00 + a0iA 2a0i + (2n + 4\)a02

a11 A 2a11 + (2n + 4A)a12 + A(2n + 4A)a13

Proof. Consider the harmonic in S functions v1 (x) and v2(x) such that for them the equalities

V1 (x)\ss = P1(s), s E dS, V2(x)\ds = P2(s), s E dS

hold, i.e. v1(x) and v2(x) are solutions of the Dirichlet problem for the Laplace equation

S u(x)

holds. Let us find the functions u0(x) and u1(x) under which the biharmonic function u(x) = u0(x) + \x\2u1(x) is a solution of (5) - (6). Consider the operators L1 and L2 from the Theorem 1. Then, because of equality

A (\x\2v(x)) = (2n + 4A)v(x)

we have

L1u = L1 (u0(x) + \x\2u1(x)) = a00u + a01Au + a02(2n + 4A)u1,

L2u = L2 (u0(x) + \x\2u1(x)) = a11Au + a12(2n + 4A)u1 + a13A(2n + 4A)u1.

Taking the limit as x ^ s E dS, as in Theorem 1, we obtain

a00(u0 + u1) + a01 (Au0 + (A + 2)u1) + a02(2n + 4A) un = p1(s), , ,

an (Au0 + (A + 2)u1) + au(2n + 4A)u1 + a^A(2n + 4A)u1 = ^(s). ^ '

S

S

aoo(uo + ui) + aoi (Auo + (A + 2)ui) + ao2(2n + 4A) ui = vi(x), a11 (Au0 + (A + 2)u1) + a12(2n + 4A)u1 + a13A(2n + 4A)u1 = v2(x),

A

equalities are harmonic in S. We rewrite the resulting equations in the matrix form

/ aoo + ao1 A aoo + (A + 2)ao1 + (2n + 4A)ao2 \ í uo \ f V1 \ , , V anA (A + 2)an + (2n + 4A)(a12 + Aa^)/ V uj V V2 ) ' { '

Equation (8) is a system of differential equations in harmonic functions in S. Let us show that (8) holds also in S. By the theorem's condition we have u(x) G C3(S), then according to formulas (4) we obtain uo(x),u1(x) G C 1(S). Taking into account that Au1 = 1 Au — nu1 (see Lemma 1) we obtain Au1(x) G C 1(S). However, beeause uo(x) G C 1(S^d Au1(x) G C 1(S) equation (8) is fulfilled in S. So, from the above reasoning we can conclude that if the function u(x) = uo(x) + |x|2u1(x) G C3(S) is a solution of the problem (5) - (6), then harmonic functions uo(x) and u1(x) satisfy (8) in S.

Converse assertion also holds, i.e. if the harmonic functions uo(x) G C1 (S) and u1(x) G C 1(S), Au1(x) G C 1(S) satisfy (8) in S, then they satisfy (7), and therefore the function u(x) = uo(x) + |x|2u1(x) is a solution of (5) - (6). Since Au(x) = (2n + 4A)u1(x), then u(x) G C3(S). Thus, problem (5) - (6) and equation (8) in S are equivalent. Consider the following matrix

A(\)=( aoo + aoA aoo + (A + 2)ao1 + (2n + 4A)ao2 \ (A) V auA (A + 2)a11 + (2n + 4A)a12 + A(2n + 4A)a1^ '

depending on a parameter A G R ft is easy to see that det A(A) = A(A) and degdet A (A) = 3. Therefore, under the conditions of the theorem the matrix A (A) is nonsingular for A G No. System (8) can be written in the form

A(A)U(x) = V(x), x G S' (9)

where U = ^ ^ and V = ^ V1 ^ • We expand the polynomial det A(A)

det A(A) = A(A) = c(A — A1)(A — A2)(A — A3)'

where Ai G C, i = 1, 2, 3 are the roots of the polynomial A(A).

1) Let all the roots Ai of the polynomial A(A) be different. It is well known that in this case

1 b1 + b2 + 63

A(A) A - A1 A - A2 A - A3' where hi E R. ^te inverse matrix A-1 (A) to A(A) has the form A"1 (A) = ^^yA*(A), where

A*A) =( (A + 2)an + (2n + 4A)(a12 + Aa13) -a00 - (A + 2)a01 - (2n + 4A)a0^

y -a11A a00 + a01A J

Ai

Mi(A)v{x)= I t-Xi-lv{tx) dt

1

0

where i = 1, 2, 3. If lim t Xiw(tx) = 0, then it is easy to see that for a differentiable function w(x) the following equalities hold

(A - Ai)Mi(A)w = (A - Ai) f01 t-Xi-1w(tx)dt = -A, f01 t-Xi-1w(tx)dt+ + /0 t-XiYTi=1 x,wXi(tx)dt = -A, f01 t-Xi-1w(tx)dt + f01 t-Xiwt(tx)dt = (10)

-A, /0 t-Xi-1w(tx)dt + t-Xiw(tx)\0 + Ai /0 t-Xi-1w(tx)dt = t-Xiw(tx)\lQ = w(x).

Since the following equalities are true

1 = A(A) (Jx- + JX- + JX-) = JTX + ITX + irX =

\ X — X- X — X- X — X— J X — X— X — X- X — X-

= cb1(A - A2)(A - A3) + cb2(A - A1)(A - A3) + cb3(A - A{)(A - A2),

then for the function w E C2(S) such that limt—Xiw(tx) = 0, i = 1, 2, 3 we have

t^0

A(A) (bM1(A) + b2M2(A) + b3M3(A)) w = (ch(A - A2)(A - A3)+

+ cb2(A - A1)(A - A3) + cb3(A - A1)(A - A2))w = 1 • w. (11)

Therefore, if the function V(x) satisfies the conditions lim t—XiV(tx) = 0, where i =

t^0

1, 2, 3

U(x) = A*(A) (b1M1(A) + b2M2(A) + b:iM3(A)) V(x). (12)

Indeed, the function U(x), which is found from (12) by virtue of (11) satisfies (9) in S, i.e.

A(A)U(x) = A(A)A*(A) (bM(A) + b2M2(A) + b3M3(A)) V(x) = = A(A) (b1M1(A) + b2M2(A) + b3M3(A)) V(x) = V(x).

2) Let two roots of the polynomial A(A) be equal A1 = A2, i.e. A(A) = c(A - A1)2(A -A3). It is known that in this case -r-^ = —+ ^ b- ^ ^ v^V, and hence

3> A(X) X—X— (X—X—)- X—X— '

i=Am ( b1 + b2 + b3 ) = b1A(A) b2A(A) b:iA(A) = 1 = A(A) 1-r" + ^-A \2 -r = ^-r- + ^-A \~2 +

A - A1 (A - A1)2 A - A3) A - A1 (A - A1)2 A - A3

= cb1(A - A1)(A - A3) + cb2(A - A3) + cb3(A - A1)2. (13) Consider the operator

Mf\A)v(x) = - i ln t t—X——1v(tx) dt. 0

w(x)

(A - A1)M((2)(A)w = M1(A)w + lntt—X—w(tx)\1 = M^A)w,

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

& Computer Software (Bulletin SUSU MMCS), 2016, vol. 9, no. 4, pp. 40-52

hold and hence (A - Ai)2M1(2)(A)w = w, if limln t t Xlw(tx) = 0. Due to (13) a function

w E C2(S) such that limlnt t-Xlw(tx) = 0 satisfies the equality

t^o

(2),

Д(Л) (ЪМ^Л) + b2MY>(Л) + 6зМа(Л^ w = = (сЬ1(Л - Л1)(Л - Аз) + еЬ2(Л - Аз) + сЬз(Л - A^2) w = 1 • w.

So, if A1 = A2 and the function V(x) satisfies the conditions limlntt-XlV(tx) = 0 and

lim t Лз V(tx) = 0, then the solution of (9) can be written as

i-S- 0

i-S- 0

U(x) = A*(A) (biMi(A) + &2Mj2)(A) + &3Ma(A)) V(x). (14)

3) Let the three roots of the polynomial A(A) be equal, i.e. A(A) = c(A - Ai)3. We consider the operator

1 i' i

Ml3)(A)w(x) = - ln2t t-Xl-1v(tx) dt. 2 J o

It is easy to see that if the function w(x) is a differentiable one, then we have

1 f i

(A - Ai)Ml3)(A)w = (A - Ai)- / ln21 t-Xl-1 w(tx)dt =

2

0

A1 1 1 1 — ln2t t-Лl-1w(tx) dt + -

2 Jo 2 J 0

I ln2t t-Лl-1w(tx) dt + - ln2tt-Лlwi(tx)dt

A [1 1 A f1

ln2t t-:Kl-1w(tx)dt + -ln2tt-Лl w(tx) 10 + A1 2 0 2 0 2

- -2 I ln2tt~Xl~ w(tx)dt +2lntrAl w(tx) 10 + 2/ ln2tt" Al" w(tx)dt-1

Лl-1w(tx)dt = M^ VA)w + .l]n2tt-^, '0 2

- I lnt t-Лl-1w(tx)dt = M12)(Л)w + -1п2й-Л1 w(tx)11 = М12)(ЛХ

provided that limln2t t Л1 w (tx) = 0. By the above (Л - A^M^^w =w

i^0

the solution of (9) in this case can be written in the form

U (x) = -А*(Л)М13)(Л^ (x). (15)

V(x)

use formulas (12), (14) and (15). First, the following limits must be true lirn t-)Ki V(tx) = 0, where i = 1, 2, 3 for the different roots, or limlnfct t-Xk1 V(tx) = 0 (k = 1, 2) in the case of

U(x)

perform the boundary conditions of the problem, it is necessary that the operator A* (Л) is applicable to the function V(x) in S. Due to the structure of the operator A* (Л) it is possible if v1 E C2(S) and v2 E C1 (S). Such smoothness of the functions v1(x) and v2(x) is provided if ф1 E C2(dS) and ф2 E C 1(dS). This is accomplished. Let us return to the conditions

limtV(tx) = 0, i = 1, 2, 3, (16)

which in fact are imposed on the functions E C2(dS) and p2 E Cl(dS). Note that due to the smoothness of the functions V(x), for Ai E N0 conditions lim t—Xi V(tx) = 0

and timlnfct t—XiV(tx) = 0 are equivalent. If Ai < 0, then conditions (16) are fulfilled.

Otherwise, i.e. if at least for one i we have Ai > 0, then (16) can be not fulfilled for some functions and p2.

Ai Ai > 0

At first note that if lim t—Xiw(tx) = 0, then for the operator

t

Mi(A)v(x) = -I t-Xi-lv(tx) dt

where w(x) E Cl(Rn), the equalities similar to (10) are true

rœ pœ pœ

L~Xi— I + — + _ \ I +~Xi—1„

(A - \i)Mi(A)w = \ t-Xi-lw(tx) dt - t-Xiwt(tx) dt =\i t-Xi-1 w(tx) dt-

-t Xiw(tx)\l - Xi J t Xi lw(tx) dt = - t Xiw(tx)\l = w(x).

For example, if |a| < Xi then we have

Mi(A)xa = -xc

t\a\

-Xi -l

dt = — xc

t\a\-Xi

№ - Xi

xa

ai - Xi

(17)

where xa = xf - ... x^" h \a\ = a1 + • • • + an, i.e. the operator Mi(A) is applicable to the polynomial xa. Change formula (12). Let ni = [A,]. We represent the function V(x) in the form

ni hk k=0 j=l

ni hk

V(x) = ^ îl V?> Hk' (x) + (V(x) - £ £ Hk] (x) ) ^ V® (x) + vr> (x),

k=0 j=l

)

r{i),

where |Hm(x), m E N0, z = 1,hkj is a complete orthogonal system of homogeneous harmonic polynomials on dS [15], and vj V(x)

by this system. It is clear that if ni = [Xi] < 0 then V(\x) = 0 and thus V2] (x) = V (x).

(i)

Consider the expression

Ui(x) = biA*(A) (Mi(A)Vl{i)(x) + Mi(A)Vii)(x)^j

(18)

Since V(i (x) is a polynomial of d egree ni5 then for each its term oft he form V(j (x) we have k - Ai < ni - Ai = [Ai] - Ai < 0 (no equals sign, because the numbers Ai are not integer) and hence by virtue of (17) the operator M,(A) is applicable to the polynomial V(i)(x). Hence -Ai + n0 + 1 = -Ai + [Ai] + 1 > 0, then t—XiV^(tx) - t—Xi+n0+1VJ(i)(x) — 0 as t — 0 So, the operator M, (A) is applicable to the fu nction V2(i\x). Thus expression (18) makes sense.

By properties of the operators M,(A) and M,(A) the following equalities hold

A(A)Ui(x) = biA(A)A*(A) ( Ml(A)V(i](x) + Ml(A)V^>(x)

r(i).

oo

oo

oo

1

математическое моделирование

= biA(A) M(A)v(\x) + Mi(A)V2(i)(x)^

= cbi(A - A2)(A - A3) (V(i)(x) + V2(i)(x)) = cbi(A - A2XA - A3)V. It means that the following function

U (x) = Ui(x) + U2(x) + U3(x) (19)

satisfies the following equality

3

A(A)U = £ A(A)Ui(x) = (cbi(A - A2XA - A3) + cb2(A - Ai)(A - A3) +

i=l

+cb3(A - Ai)(A - A2)) V = - • V = V.

This means that the function U(x) from (19) is a solution of (9) in S. Therefore the function u(x) = u0(x) + |x|2u((x) is a solution of (5) - (6).

In the case of duple roots Al = A2 > 0 together with the operator Ml(A) it is necessary to consider the operator

M}2)(A)v(x) = J ln tt-Xi-lv(tx) dt,

for which in the case of a differentiable function w(x) E Cl(Rra) the equalities hold

(A - Ai)M}2)(A)w = -t-Xiw(tx)|~ + Mi(A)w = Mi(A)w(x) 2 "—"(2)

and thus (A - Ai)2Mf;(A)w = w. For example, when lal < Al we have

rtt xa f^ tla^ — X^^ ln t

M?\A)xa = xa ln t tla-Xl-i dt =—,-- ln t dtlal-Xl = xa ^-—

Ji |a| - AiJi |a| - Ai

xa r^ 1__

, x л tlal-^-1dt = —--Mr(K)xa =-2

|a|- Aj1 |a|- A1 ^ (|a| - A1)2

and hence the operator m(2)(A) is applicable to the polynomial Vl(l)(x). The solution of (9) can be represented in the form

U (x) = Ui(x) + U(2)(x) + U3(x), (20)

(2)

where the functions U((x) and U3(x) are defined in (18), and the function U( (x) is written in the form

U(2) (x) = b2A* (A) Mi2)(A)Vl(l)(x) + m(2)(A)V2(1)(x)) . In the case of triple roots A( = A2 = A3 > 0 the solution of (9) has the form

U(x) = -A*(A) (M}3)(A)Vl(l)(x) + M(3)(A)V2(l)(x)) , (21)

where M13)^)v(x) = -2 fj ln2t t Л1 1v(tx) dt, and the operator M13)^), the functions V(1)(x) and V2(1)(x) are defined as above. Thus, the solution of (9) in S is constructed in all cases of the roots of polynomial Д^). After finding the solution of (9) that are harmonic functions u0(x) and u1(x), a solution of (5) - (6) can be written in the form u(x) = u0 (x) + |x|2u1(x).

_а

Вестник ЮУрГУ. Серия «Математическое моделирование ^7

и программирование» (Вестник ЮУрГУ ММП). 2016. Т. 9, № 4. С. 40-52

оо

3. Existence of Conditional Solutions

Consider the case of problem (5) - (6) which is not investigated in Theorem 2, when for some m E No the equality A(m) = 0 holds. Problem (1) - (2) with nonhomogeneous biharmonic equation is considered in [18].

Theorem 3. Let for some m E No the equality A(m) = 0 hold. Then a solution of (5) - (6) from the class u E C3(S) exists if and only if the functions p1 E C2(dS) and p2 E C 1(dS) satisfy the equalities

IdS

Hm(x) (qi(m)^i(x) + q2(m)v2(x)) dsx = 0, (22)

where Hm(x) is an arbitrary homogeneous harmonic polynomial of degree m, and the vector ql(m)j ^ an ar^rary s°hJ'ti°n of the system of algebraic equations

(

aoo + aoim anm

aoo + (m + 2)aoi + (2n + 4m)ao2 (m + 2)an + (2n + 4m)(ai2 + mai3)

x

*( q2im0=o- <23)

Proof. Solution of problem (5) - (6) constructed in Theorem 2 under the conditions of the present theorem is not suitable. It must be a little changed. Consider equation (9). We expand the function V(x) onto two terms V(x) = V/(x) + Vm(x), where

hm

Vm(x) = Y, P^H^(x). (24)

i=1

We find conditions under which (9) has a solution in a particular case, when the right-hand side is a monomial of the form Vm(x) = Pm^Hmfa), where

r(i) <-m

_ / fdS Hm (x)tyl(x) ds^ \

II H(i

"L2(dS) \ Ids Hm(x)'2(x) dsx J

Pi = 1 ( fdsHm (x)'i(x) dsx

" Hm llL2(dS)

By the homogeneity of the operator A (A) the solution of (9) in this case may have only the form Um(x) = QmHm(x). Substituting the polynomial Um(x) = Qm Hm(x) to (9) we have

A(A)Q«H«(x) = H$(x)A(m)Q = H«(x)P«,

and consequently the following equality should be true

A(m)Qg = PW. (25)

It is known that the obtained system of algebraic equations has a solution only if the right-hand side Pm is orthogonal to zeros of the conjugated system AT(m)Q = 0, i.e. to

all vectors ( qi(m) ) from the theorem's conditions v qi(m) '

P® '( 9l(mn = ( fds H^l(x)dsA .( ^1(mM = / qi(m)H£)(x)^i(x) dsx+

m V q2(m) J \ fds H^^'Mx^sx ) V q2(m) J Jds

+ q2(m)Hm)(x)p2(x) dsa

JdS

Ids

Hm(x) (qi(m)pi(x) + q2(m)p2(x)) dsx = 0.

By virtue of (22) this condition is satisfied. Therefore a solution of (25) exists. Going through the all terms in sum (24) and taking advantage of the theorem's conditions, we conclude that a solution of (9) for V(x) = Vm(x) exists. We denote it as Um(x). After that, if the polynomial A(A) has no other roots in No, then by Theorem 2 we solve equation (9) for V(x) = V(x) and add the resulting solutions U(x) = Um(x) + U(x). The function U(x) is a solution of (9) in and hence problem (5) - (6) has a solution. If the polynomial A(A) has also other roots in N0 except number m, then we proceed with the function V(x) in a

V(x)

A(A) from N0 conditions (22) must be satisfied.

□

4. Particular Cases of the Problem

1. Riquier - Neumann problem: let a0\ = 0, ai3 = 0 (all other coefficients are equal to zero), then

A2u = 0, x E S; a0^7—u

dv

ds

^i(s), ai3 dv Au

ds

Ms), s E dS. (26)

The uniqueness conditions of this problem have the form A(A) = \2(2n + 4A)a01a13 and hence A(0) = 0. The algebraic system (3) has the form ^ ° ^O)1 ^ (^Cj

its solution can be written in the form C2 = 0, C1 is an arbitrary constant. By Theorem 2 solution of (26) is unique up to a constant u(x) = C1H0(x) = CC1.

0 0\( Q1(0)

2ao1 0 J\ q2(0)

and hence q1(0) = 0. In this case the space of zeros of system (23) is one-dimensional and is

0

For the existence of solutions of (26) consider (23) for m = 0,

0,

0,

spanned by the vector

1

. Therefore, the existence condition (22) of solutions of (26)

has the form fdS p2(x) dsx = 0. For example, let a01 = a13 = 1 and = 0 tp2 = 1. The obtained above existence condition for the problem's solution is not fulfilled. Make sure that a solution of (26) in this case does not exist. From the second boundary condition of problem (26) follows that for solution u(x) = u0(x) + \x\2u1(x) (see Lemma 1) of this problem the following equalities must be fulfilled

ids

f д

dsx = + 4K)u\(x) dsx

JdS dv

4

ids

dv 2

u\(x) dsx = 0,

which is not true. Here it was taken into account that the integral on dS of the normal derivatives of harmonic in S functions is equal to zero (see [17]). Therefore (26) has no solution.

2. For problem (5) - (6) if a02 = 0 and a11 = 0 we have

W)

aoo + aoiA 2aoi

0 (2n + 4A)ai2 + A(2n + 4A)ai3

(2n + 4A)(aoo + Aaoi)(au + Aai3).

Let us take in (2) a00 = — 2, a01 = 1, a12 = — 3, a13 = 1, i.e. consider the problem

A2u = 0, xe S ;

du

— 2u + — du

d Au

p1(s), —3Au +—-— = p2(s), s G dS

dS ' du

(27)

dS

In this case A(\) = (2n + 4\)(\ — 2)(\ — 3) and hence A(2) = 0 and A(3) = 0. For m = 2 system (3) has the form ^ ° (2n + 8)^ ( ^C ) = 0 ant^ therefore C2 =

0 C1 being an arbitrary constant, is its solution. Thus the polynomial u2(x) = C1H2(x) is a solution of the homogeneous problem (27).

If m = 3, then the system of (3) takes th e form ^^ 0 ^ C1 ^ = 0 ant^ hence

C1 = —2C2, C2 being an arbitrary constant, is its solution. Hence the polynomial u3(x) = C2(\x\2 — 3)H3(x) is also a solution of the homogeneous problem (27). For the existence of solutions of (27) consider system (23)

m — 2 0 \ f q1(m) \ =0

/ m — 2 0 \ i q1(m) N

y m (m — 3)(2n + 4m) J y q2(m) J

(m — 3)(2n + 4m) J V q2(m) for m = ^d m = 3. Let m = 2, then we have ( 0 0 | ( qi(2) | = 0, and

y 2 —(2n + 8) ) V q2(2) J

hence q1(2) — (n + 4)q2(2) = 0. In this case the space of zeros of (23) is one-dimensional

/ n + 4 \

and is spanned by the vector ( ^ 1. So, the first condition of existence of solutions to (27) has the form fQS H2(x)((n + 4)'c1(x) + p2{x)) dsx = 0.

Let m = 3, then we obtain ^ 3 0 ^ ^ q1(3) ^ = 0, and hence q1(3) = 0. In this case

the space of zeros of (23) is also one-dimensional and is spanned by the vector ^ ^ ^. So

the second condition of existence of solutions to (27) has the form fdS H3(x)p2(x) dsx = 0. Thus, (27) has a solution if

I H2(x)((n + 4)Vi(x) + w(x)) dsx = 0, I H3(x)Mx) dsx = 0, (28) JdS JdS

where H2(x) and H3(x) are homogeneous harmonic polynomials of degrees 2 and 3.

For example, let n > 2, = s1s2 and = 0. If a solution of (27) exists, then it takes the form u(x) = u0(x) + \x\2u1(x) (see Lemma 1). From the second boundary condition we have (2n + 4A)(A — 3)u1(x) = 0, whence (A — 3)u1(x) = 0 and therefore u1(x) = H3(x). By virtue of the first boundary condition the function u0(x) should be such that (tv — 2) (u0(x) + \x\2H3(x)^j = s1s2 and hence (A — 2)u0(x) = x1x2 — 3H3(x),

V ' dS

x E S. This equation has no solution in harmonic functions in S because a harmonic function on the left-hand side (A — 2)u0(x) does not contain the second order terms in its expansion at the original, but on the right-hand side such terms are present. Note y H2 (x) = x1 x2

JdS X^X^l dsx =

Bulletin of the South Ural State University. Ser. Mathematical Modelling, Programming & Computer Software (Bulletin SUSU MMCS), 2016, vol. 9, no. 4, pp. 40-52

Acknowledgements. The second named author is financially supported by a grant

from the Ministry of Science and Education of the Republic of Kazakhstan (Grant No. 0819/GF4).

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Received September 26, 2016

УДК 517.9 DOI: 10.14529/mmpl60404

ОБ ОДНОЙ МАТЕМАТИЧЕСКОЙ МОДЕЛИ, ОПИСЫВАЕМОЙ КРАЕВОЙ ЗАДАЧЕЙ ДЛЯ БИГАРМОНИЧЕСКОГО УРАВНЕНИЯ

В.В. Карачик, Б. Т. Торебек

В данной работе рассматривается математическая модель, описываемая обобщенной третьей краевой задачи для однородного бигармонического уравнения в шаре с граничными операторами до третьего порядка, содержащие нормальные производные и лапласиан. Частными случаями рассматриваемой математической модели являются классические модели, описываемые задачами Дирихле, Рикье и Робина, спектральная задача Стеклова, а также многие другие математические модели, порожденные этими граничными условиями. Доказаны две теоремы существования рассматриваемой задачи. Условия существования получены в виде ортогональности на границе некоторой линейной комбинации граничных функций однородным гармоническим многочленам заданного порядка. Полученные результаты проиллюстрированы некоторыми частными случаями общей задачи.

Ключевые слова: математическая модель; бигармоническое уравнение; граничная задача; оператор Лапласа.

Валерий Валентинович Карачик, доктор физико-математических наук, профессор, кафедра математического анализа и методики преподавания математики, Южно-Уральский государственный университет (г. Челябинск, Российская Федерация), [email protected].

Берикбол Тиллабайулы Торебек, докторант, Институт математики и математического моделирования (г. Алматы, Республика Казахстан), [email protected].

Поступила в редакцию 26 сентября 2016 г.