Monogenic functions in the biharmonic boundary value problem

Research Article Received 10 June 2015

Published online 26 November 2015 in Wiley Online Library

(wileyonlinelibrary.com) DOI: 10.1002/mma.3741 MOS subject classification: 30G35; 31A30

Monogenic functions in the biharmonic boundary value problem Serhii V. Gryshchuk and Sergiy A. Plaksa *† Communicated by V. Didenko We consider a commutative algebra B over the field of complex numbers with a basis fe1 , e2 g satisfying the conditions .e21 C e22 /2 D 0, e21 C e22 ¤ 0. Let D be a bounded domain in the Cartesian plane xOy and D D fxe1 C ye2 : .x, y/ 2 Dg. Components of every monogenic function ˆ.xe1 C ye2 / D U1 .x, y/ e1 C U2 .x, y/ ie1 C U3 .x, y/ e2 C U4 .x, y/ ie2 having the classic derivative in D are biharmonic functions in D, that is, 2 Uj .x, y/ D 0 for j D 1, 2, 3, 4. We consider a Schwarz-type boundary value problem for monogenic functions in a simply connected domain D . This problem is associated with the following biharmonic problem: to find a biharmonic function V.x, y/ in the domain D when boundary values of its partial derivatives @V=@x, @V=@y are given on the boundary @D. Using a hypercomplex analog of the Cauchy-type integral, we reduce the mentioned Schwarz-type boundary value problem to a system of integral equations on the real axes and establish sufficient conditions under which this system has the Fredholm property. Copyright © 2015 John Wiley & Sons, Ltd. Keywords: biharmonic equation; biharmonic boundary value problem; biharmonic algebra; biharmonic plane; monogenic function; Schwarz-type boundary value problem; biharmonic Cauchy-type integral; Fredholm integral equations

1. Introduction Let D be a bounded domain in the Cartesian plane xOy, and let its boundary @D be a closed smooth Jordan curve. Let R be the set of real numbers. Consider some boundary value problems for biharmonic functions WW D ! R, which have continuous partial derivatives up to the fourth order inclusively and satisfy the biharmonic equation in the domain D 2 W.x, y/ 

@4 W.x, y/ @4 W.x, y/ @4 W.x, y/ C2 C D 0. 4 2 2 @x @x @y @y4

The principal biharmonic problem (cf., e.g., [1, p. 194] and [2, p. 13]) consists of finding a function WW D ! R, which is continuous together with partial derivatives of the first order in the closure D of the domain D and is biharmonic in D, when its values and values of its outward normal derivative are given on the boundary @D W.x0 , y0 / D !1 .s/,

@W .x0 , y0 / D !2 .s/ @n

8 .x0 , y0 / 2 @D ,

(1.1)

where s is an arc coordinate of the point .x0 , y0 / 2 @D. In the case where !1 is a continuously differentiable function, the principal biharmonic problem is equivalent to the following biharmonic problem (cf., e.g., [1, p. 194] and [2, p. 13]) on finding a biharmonic function V : D ! R with the following boundary conditions:

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Institute of Mathematics, National Academy of Sciences of Ukraine, Tereshchenkivska Str. 3, 01601, Kiev-4, Ukraine * Correspondence to: Sergiy A. Plaksa, Institute of Mathematics, National Academy of Sciences of Ukraine, Tereshchenkivska Str. 3, 01601, Kiev-4, Ukraine. † E-mail: [email protected]

S. V. GRYSHCHUK AND S. A. PLAKSA lim

@V.x,y/ @x

D !3 .s/ ,

lim

@V.x,y/ @y

D !4 .s/

.x,y/!.x0 ,y0 /, .x,y/2D .x,y/!.x 0 ,y0 /, .x,y/2D Z

8 .x0 , y0 / 2 @D ,

(1.2)

.!3 .s/ cos †.s, x/ C !4 .s/ cos †.s, y// ds D 0 . @D

Here, given boundary functions !3 , !4 have relations with given functions !1 , !2 of problem (1.1), viz. !3 .s/ D !10 .s/ cos †.s, x/ C !2 .s/ cos †.n, x/ , !4 .s/ D !10 .s/ cos †.s, y/ C !2 .s/ cos †.n, y/ , where s and n denote unit vectors of the tangent and the outward normal to the boundary @D, respectively, and †., / denotes an angle between an appropriate vector (s or n) and the positive direction of coordinate axis (x or y) indicated in the parenthesis. Furthermore, solutions of problems (1.1) and (1.2) are related by the equality V.x, y/ D W.x, y/ C c , where c 2 R . A technique of using analytic functions of the complex variable for solving the biharmonic problem is based on an expression of biharmonic functions by the Goursat formula. This expression allows to reduce the biharmonic problem to a certain boundary value problem for a pair of analytic functions. Further, expressing analytic functions via the Cauchy-type integrals, one can obtain a system of integro-differential equations in the general case. In the case where the boundary @D is a Lyapunov curve, the mentioned system can be reduced to a system of Fredholm equations. Such a scheme is developed (cf., e.g., [3–6]) for solving the main problems of the plane elasticity theory with using a special biharmonic function that is called the Airy stress function. Another methods for reducing boundary value problems of the plane elasticity theory to integral equations are developed in [2, 7–11]. In this paper, for solving the biharmonic problem, we develop a method that is based on the relation between biharmonic functions and monogenic functions taking values in a commutative algebra. We use an expression of monogenic function by a hypercomplex analog of the Cauchy-type integral. Considering a Schwarz-type boundary value problem for monogenic functions that is associated with the biharmonic problem, we develop a scheme of its reduction to a system of Fredholm equations in the case where the boundary of domain belongs to a class being wider than the class of Lyapunov curves.

2. Monogenic functions in a biharmonic algebra associated with the biharmonic equation Kovalev and Mel’nichenko [12] considered an associative commutative two-dimensional algebra B over the field of complex numbers C with the following multiplication table for basic elements e1 , e2 e21 D e1 ,

e2 e1 D e2 ,

e22 D e1 C 2ie2 ,

(2.1)

where i is the imaginary complex unit. Elements e1 , e2 satisfy the relations .e21 C e22 /2 D 0,

e21 C e22 ¤ 0,

(2.2)

thereby, all functions ˆ./ of the variable  D x e1 C y e2 , which have continuous derivatives up to the fourth order inclusively, satisfy the equalities 2 ˆ./ D ˆ.4/ ./ .e21 C e22 /2 D 0 . (2.3) Therefore, components Uj W D ! R, j D 1, 4 of the expression ˆ./ D U1 .x, y/ e1 C U2 .x, y/ ie1 C U3 .x, y/ e2 C U4 .x, y/ ie2

(2.4)

are biharmonic functions. Similar to Kovalev and Mel’nichenko [12], by the biharmonic plane, we call a linear span  :D f D x e1 C y e2 : x, y 2 Rg of the elements e1 , e2 satisfying (2.1). With a domain D of the Cartesian plane xOy, we associate the congruent domain D :D f D xe1 C ye2 : .x, y/ 2 Dg in the biharmonic plane . We say that a function ˆW D ! B is monogenic in a domain D if at every point  2 D , there exists the derivative of the function ˆ ˆ0 ./ :D

lim

h!0, h2

.ˆ. C h/  ˆ.// h1 .

It is proved in [12] that a function ˆW D ! B is monogenic in a domain D if and only if components Uj W D ! R, j D 1, 4 of the expression (2.4) are differentiable in the domain D and the following analog of the Cauchy–Riemann conditions is satisfied

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@ˆ./ @ˆ./ D e2 @y @x Copyright © 2015 John Wiley & Sons, Ltd.

8  D xe1 C e2 y 2 D .

(2.5) Math. Meth. Appl. Sci. 2016, 39 2939–2952

S. V. GRYSHCHUK AND S. A. PLAKSA It is established in [13] that every monogenic function ˆW D ! B has derivatives ˆ.n/ ./ of all orders n in the domain D and, therefore, it satisfies the equalities (2.3). At the same time, every biharmonic in D function U.x, y/ is the first component U1  U in the expression (2.4) of a certain monogenic function ˆ : D ! B, and moreover, all such functions ˆ are found in [13] in an explicit form.

3. Boundary value problem for monogenic functions that is associated with the biharmonic problem Let ˆ1 be monogenic in D function having the sought for function V.x, y/ of problem (1.2) as the first component ˆ1 ./ D V.x, y/ e1 C V2 .x, y/ ie1 C V3 .x, y/ e2 C V4 .x, y/ ie2 . It follows from condition (2.5) for ˆ D ˆ1 that @V3 .x, y/=@x D @V.x, y/=@y. Therefore, ˆ01 ./ D

@V.x, y/ @V2 .x, y/ @V.x, y/ @V4 .x, y/ e1 C ie1 C e2 C ie2 , @x @x @y @x

(3.1)

and as consequence, we conclude that the biharmonic problem with boundary condition (1.2) is reduced to the boundary value problem on finding a monogenic in D function ˆ  ˆ01 when values of two components U1 D @V.x, y/=@x and U3 D @V.x, y/=@y of the expression (2.4) are given on the boundary @D of the domain D . As in [14], by the (1-3)-problem, we shall call the problem on finding a monogenic function ˆW D ! B when values of components U1 , U3 of expression (2.4) are given on the boundary @D U1 .x0 , y0 / D u1 .0 / ,

U3 .x0 , y0 / D u3 .0 /

8 0 :D x0 e1 C y0 e2 2 @D ,

where u1 .0 /  !3 .s/ and u3 .0 /  !4 .s/. Problems of such a type on finding a monogenic function with given boundary values of two of its components were posed by Kovalev [15] who called them by biharmonic Schwarz problems, because their formulations are analogous in a certain sense to the classic Schwarz problem on finding an analytic function of the complex variable when values of its real part are given on the boundary of domain. Kovalev [15] stated a sketch of reduction of biharmonic Schwarz problems to integro-differential equations with using conformal mappings and expressions of monogenic functions via analytic functions of the complex variable. In [14], we investigated the (1-3)-problem for cases where D is either an upper half plane or a unit disk in the biharmonic plane. Its solutions were found in explicit forms with using of some integrals analogous to the classic Schwarz integral. In [16], a certain scheme was proposed for reducing the (1-3)-problem in a simply connected domain with sufficiently smooth boundary to a suitable boundary value problem in a disk with using power series and conformal mappings in the complex plane. Hypercomplex methods for investigating the biharmonic equation were also developed in the papers [17–22].

4. Biharmonic Cauchy-type integral Consider the biharmonic Cauchy-type integral ˆ./ D

Z

1 2i

'. /.  /1 d

(4.1)

@D

with a continuous density 'W @D !p B. The integral (4.1) is a monogenic function in both domains D and  n D . We use the Euclidian norm kak :D jz1 j2 C jz2 j2 in the algebra B, where a D z1 e1 C z2 e2 and z1 , z2 2 C. We use also the modulus of continuity of a function 'W @D ! B !.', "/ :D

k'.1 /  '.2 /k .

sup 1 ,2 2@D , k1 2 k"

Consider a singular integral that is understood in the sense of its Cauchy principal value Z Z '. /.  0 /1 d :D lim '. /.  0 /1 d , "!0 @D n@D" .0 /

@D

where 0 2 @D , @D" .0 / :D f 2 @D : k  0 k  "g. The following theorem can be proved in a similar way as an appropriate theorem in the complex plane (cf., e.g., [23, 24]). It presents sufficient conditions for the existence of limiting values ˆC .0 / :D

lim

!0 , 2D

ˆ./,

ˆ .0 / :D

lim

ˆ./

!0 , 2nD

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of the biharmonic Cauchy-type integral in any point 0 2 @D .

S. V. GRYSHCHUK AND S. A. PLAKSA Theorem 4.2 Let the modulus of continuity of a function 'W @D ! B satisfy the Dini condition Z1

!.', / d < 1. 

(4.2)

0

Then the integral (4.1) has limiting values ˆ˙ .0 / in any point 0 2 @D that are represented by the Sokhotski–Plemelj formulas Z 1 1 '. /.  0 /1 d , ˆC .0 / D '.0 / C 2 2i @D

1 1 ˆ .0 / D  '.0 / C 2 2i 

Z

(4.3) '. /.  0 /1 d .

@D

5. Scheme for reducing the (1-3)-problem to a system of integral equations Let the functions u1 W @D ! R, u3 W @D ! R satisfy conditions of the type (4.2). We shall find solutions of the (1-3)-problem in the class of functions represented in the form Z 1 ˆ./ D .'1 . /e1 C '3 . /e2 / .  /1 d 8  2 D , 2i

(5.1)

@D

where the functions '1 W @D ! R and '3 W @D ! R satisfy conditions of the type (4.2). Then, by Theorem 4.2, the following equality is valid for any 0 2 @D : Z 1 1 .'1 . /e1 C '3 . /e2 / .  0 /1 d . ˆC .0 / :D .'1 .0 /e1 C '3 .0 /e2 / C 2 2i

(5.2)

@D

By Dz , we denote the domain in C, which is congruent to the domain D, that is, Dz :D fz D x C iy 2 C : .x, y/ 2 Dg. We shall use a conformal mapping z D  .t/ of the upper half plane ft 2 C : Im t > 0g onto the domain Dz . Denote 1 .t/ :D Re  .t/, 2 .t/ :D Im  .t/. In as much as the mentioned conformal mapping is continued to a homeomorphism between the closures of corresponding domains, the function .s/ Q :D 1 .s/e1 C 2 .s/e2 8s 2 R (5.3) generates a homeomorphic mapping of the extended real axis R :D R [ f1g onto the curve @D . Introducing the function g.s/ :D g1 .s/e1 C g3 .s/e2 8s 2 R,

(5.4)

where gj .s/ :D 'j ..s// Q for j 2 f1, 3g, we rewrite the equality (5.2) in the form 1 1 ˆ .0 / D g.t/ C 2 2i C

Z1

1 g.s/ .Q .s/  .t// Q Q 0 .s/ ds

8t 2 R,

(5.5)

1

where the integral is understood in the sense of its Cauchy principal value (cf., e.g., [23]) and a correspondence between the points 0 2 @D n f.1/g Q and t 2 R is given by the equality 0 D Q .t/. To transform the expression under an integral sign in equality (5.5), we use the equalities i .2 .s/  2 .t// 1 C ,  .s/   .t/ 2 . .s/   .t//2 i 0 .s/ Q 0 .s/ D  0 .s/  2  , 2

1 D ..s/ Q  .t// Q

where  :D 2e1 C 2ie2

(5.6)

2

is a nilpotent element of the algebra B because  D 0. Thus, we transform equality (5.5) into the form 1 1 ˆ .0 / D g.t/ C 2 2i C

Z1 1

1 g.s/k.t, s/ ds C 2i

Z1 g.s/ 1

1 C st ds , .s  t/.s2 C 1/

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where Copyright © 2015 John Wiley & Sons, Ltd.

Math. Meth. Appl. Sci. 2016, 39 2939–2952

S. V. GRYSHCHUK AND S. A. PLAKSA k.t, s/ D k1 .t, s/e1 C i k2 .t, s/ , k1 .t, s/ :D k2 .t, s/ :D

(5.7)

1 C st  0 .s/  ,  .s/   .t/ .s  t/.s2 C 1/

 0 .s/ .2 .s/  2 .t// 2

2 . .s/   .t//



(5.8)

20 .s/ . 2 . .s/   .t//

(5.9)

We use the notations Uj Œa :D aj , j D 1, 4, where aj 2 R is the coefficient in the decomposition of element a D a1 e1 C a2 ie1 C a3 e2 C a4 ie2 2 B with respect to the basis fe1 , e2 g.  In order to single out U1 ˆC .0 / , U3 ˆC .0 / , we use the equalities (5.4), (5.6), and (5.7) and obtain the decomposition of the following expression with respect to the basis fe1 , e2 g g.s/k.t, s/ D .g1 .s/e1 C g3 .s/e2 / .k1 .t, s/e1 C i.2e1 C 2ie2 / k2 .t, s// D D .g1 .s/ .k1 .t, s/ C 2ik2 .t, s//  2g3 .s/k2 .t, s// e1 C .g3 .s/ .k1 .t, s/  2ik2 .t, s//  2g1 .s/k2 .t, s// e2 .     Now, we single out U1 ˆC .0 / , U3 ˆC .0 / and obtain the following system of integral equations for finding the functions g1 and g3   U1 ˆC .0 /    U3 ˆC .0 / 

1 2

1 2

g1 .t/ C

g3 .t/ 

1 2

1 

Z1 1 Z1

1 g1 .s/ .Im k1 .t, s/ C 2Re k2 .t, s// ds  

1 g1 .s/Im k2 .t, s/ ds C 2

1

Z1

Z1

g3 .s/Im k2 .t, s/ ds D uQ 1 .t/,

1

(5.10)

g3 .s/ .Im k1 .t, s/  2Re k2 .t, s// ds D uQ 3 .t/

8t 2 R,

1

where uQ j .t/ :D uj ..t//, Q j 2 f1, 3g. In the succeeding text, we shall state conditions that are sufficient for compactness of integral operators on the left-hand sides of equations of system (5.10).

6. Auxiliary statements For a function ' : ! C, which is continuous on the curve  C, a modulus of continuity is defined by the equality ! .', "/ :D

sup

j'.t1 /  '.t2 /j .

t1 ,t2 2 , jt1 t2 j"

  ti for all Consider the conformal mapping  .T/ of the unit disk fT 2 C : jTj < 1g onto the domain Dz such that  .t/ D  tCi t 2 ft 2 C : Im t > 0g. Denote 1 .T/ :D Re  .T/, 2 .T/ :D Im  .T/. Assume that the conformal mapping  .T/ has the continuous contour derivative  0 .T/ on the unit circle :D fT 2 C : jTj D 1g and  0 .T/ ¤ 0 for all T 2 . Then there exist constants c1 and c2 such that the following inequalities are valid: ˇ ˇ ˇ  .S/   .T/ ˇ ˇ  c2 . 0 < c1  ˇˇ ˇ ST

(6.1)

In this case, for all S, T1 , T2 2 such that 0 < jS  T1 j < jS  T2 j, the following estimates are also valid: ˇ ˇ 0 ˇ j .S/  j .T1 / j .S/  j .T2 / ˇ ˇ  c ! . , jS  T2 j/ jT1  T2 j , ˇ  ˇ ˇ S  T1 S  T2 jS  T2 j

j D 1, 2,

(6.2)

where the constant c does not depend on S, T1 , T2 . Estimates of the same type for the function  .T/ are corollaries of the estimates (6.2). They are adduced in [25]. It follows from (6.2) that the inequalities ˇ ˇ ˇ 0 ˇ ˇ .S/  j .S/  j .T/ ˇ  c ! . 0 , jS  Tj/ , ˇ j ˇ ST

j D 1, 2,

(6.3)

t2R

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are fulfilled for all S, T 2 , S ¤ T, where the constant c does not depend on S and T. Then an inequality of the same type for the function  .T/ is certainly fulfilled. Let C.R / denote the Banach space of functions gW R ! C that are continuous on the extended real axis R with the norm kgkC.R / :D sup jg.t/j.

S. V. GRYSHCHUK AND S. A. PLAKSA Lemma 6.4 Let g 2 C. R / and the conformal mapping  .T/ has the non-vanishing continuous contour derivative  0 .T/ on circle , and its modulus of continuity satisfies the Dini condition Z1

! . 0 , / d < 1. 

(6.4)

0

Then, for 0 < " < 1=4 and t 2 R, the following estimates are true: ˇ 1 ˇ ˇ Z ˇ Z1 Z2 ˇ ˇ ! . 0 , / ˇ ˇ d , g.s/ kj .t C ", s/ ds  g.s/ kj .t, s/ ds ˇ  c kgkC.R / ˇ . C / ˇ ˇ 1

1

j D 1, 2 ,

(6.5)

0

where :D "=.t2 C 1/ and the constant c does not depend on t and ". Proof Let us prove inequality (6.5) for j D 1. Denote S :D

si sCi

, T :D

ti tCi

, d.S, T/ :D

.S/.T/ ST

ST D

2i.s  t/ , .s C i/.t C i/

 0 .s/ D

2i  0 .S/ , .s C i/2

. Taking into account the equalities

(6.6)

we transform the expression 2i  0 .S/  0 .s/ .sCi/2 D D  .s/   .t/ .S  T/d.S, T/



1 C st 1 Ci 2 .s  t/.s2 C 1/ s C1



 0 .S/ d.S, T/

and represent the function k1 .t, s/ in the form k1 .t, s/ D m1 .t, s/ C i m2 .t, s/,

(6.7)

where m1 .t, s/ :D

 0 .S/  d.S, T/ 1 C st , .s  t/.s2 C 1/ d.S, T/

m2 .t, s/ :D

 0 .S/ . d.S, T/.s2 C 1/

Further, representing the integral Z1 IŒg, k1 .t/ :D

g.s/k1 .t, s/ ds

8t 2 R

1

by the sum of two integrals IŒg, k1 .t/ D IŒg, m1 .t/ C iIŒg, m2 .t/

8 t 2 R,

(6.8)

we shall obtain estimates of the type (6.5) for each of integrals IŒg, m1 .t/, IŒg, m2 .t/. For the integral IŒg, m1 .t/, we have tC2" Z

jIŒg, m1 .t C "/  IŒg, m1 .t/j  0 B C@

jg.s/j jm1 .t, s/j ds C t2"

t2" Z

1

Z1

C

tC2" Z

1

jg.s/j jm1 .t C ", s/j dsC t2"

C A jg.s/j jm1 .t, s/  m1 .t C ", s/j ds D: J1 C J2 C J3 .

tC2"

Taking into account the relations (6.1), (6.3), and (6.6), we obtain tC2" Z

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J1  c kgkC.R / t2"

! . 0 , jS  Tj/ .1 C jsjjtj/ ds  c kgkC.R / p p 2 2 2 jS  Tj s C1 t C1 s C1

Copyright © 2015 John Wiley & Sons, Ltd.

tC2" Z

! . 0 , jS  Tj/ ds  jS  Tj s2 C 1

t2"

Math. Meth. Appl. Sci. 2016, 39 2939–2952

S. V. GRYSHCHUK AND S. A. PLAKSA Z8  c kgkC.R /

! . 0 , / d  c kgkC.R / 

0

Z2

! . 0 , / d. . C /

0

Here and in the succeeding text in the proof, by c, we denote constants whose values are independent of t and " but, generally speaking, may be different even within a single chain of inequalities. The integral J2 is similarly estimated. tC"i To estimate the integral J3 , take into consideration the point T1 :D tC"Ci . Using the equality  0 .S/  d.S, T/ " C .s  t/.s  t  "/ d.S, T/ 0 1 C s.t C "/ . .S/  d.S, T// .d.S, T1 /  d.S, T// 1 C s.t C "/ d.S, T1 /  d.S, T/ C C , .s  t  "/.s2 C 1/ d.S, T/d.S, T1 / .s  t  "/.s2 C 1/ d.S, T1 /

m1 .t, s/  m1 .t C ", s/ D

we estimate J3 by the sum of three integrals 0 B J3  " @

t2" Z

Z1

C

1

0

1 C A jg.s/j

0

j .S/  d.S, T/j B ds C @ j d.S, T/jjs  tjjs  t  "j

1

tC2"

0

t2" Z

0



t2" Z

j .S/  d.S, T/j .1 C j sjj t C "j/ ds B C@ j d.S, T/jj d.S, T1 /jjs  t  "j s2 C 1

Z1

C

1

tC2"

Z1

C

1 C A jg.s/j j d.S, T1 /  d.S, T/j 

tC2"

1

3

X j d.S, T1 /  d.S, T/j .1 C jsjj t C "j/ ds C D: J3, j . A jg.s/j 2 j d.S, T1 /jjs  t  "j s C1 jD1

Taking into account the inequalities js  tj  2js  t  "j  3js  tj for all s 2 .1, t  2" [ Œt C 2", C1/ and the relations (6.1), (6.3), and (6.6), we obtain 0 B J3,1  c kgkC.R / " @

t2" Z

C

1

Z2  c kgkC.R /

Z1

0

1

0 " B C ! . , jS  Tj/ ds D c kgkC.R / 2 @ A js  tj2 t C1

t2" Z

C

1

tC2"

! . 0 , / d  c kgkC.R / 2

Z2

Z1

1 0 C ! . , jS  Tj/ ds  A jS  Tj2 s2 C 1

tC2"

! . 0 , / d . . C /

0

Using inequalities (6.2) and properties of a modulus of continuity (cf., e.g., [26, p. 176]), we obtain the following inequalities similarly to the estimation of J3,1 : 0 B J3,2  c kgkC.R / @

t2" Z

Z1

C

1

1 C jd.S, T1 /  d.S, T/j .1 C j sjj t C "j/ ds  A js  t  "j s2 C 1

tC2"

0

t2" Z

B  c kgkC.R / jT1  Tj @

C

1

0

B  c kgkC.R / @

tC2"

t2" Z

B  c kgkC.R / jT1  Tj @ 0

Z1

C

1 t2" Z

Z1

C

1

Z1

tC2"

1 0 C !. , jS  Tj/ .1 C j sjj t C "j/ ds  A jS  Tj js  tj s2 C 1

1 0 ds C !. , jS  Tj/ .1 C j sjj t C "j/  q A jS  Tj2 ps2 C 1 t2 C 1 s2 C 1

1

1

0 C !. , jS  Tj/ ds  c kgkC.R / A jS  Tj2 s2 C 1

tC2"

Z2

! . 0 , / d . . C /

0

The integral J3,3 is similarly estimated. An estimate of the type (6.5) for IŒg, m1 .t/ follows from the obtained estimates. By the scheme used earlier for estimating the integral IŒg, m1 .t/, we obtain the estimate jIŒg, m2 .t C "/  IŒg, m2 .t/j  c kgkC.R / ,

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whence an estimate of the type (6.5) follows for IŒg, m2 .t/. Thus, inequality (6.5) is proved for j D 1.  .S/ .T/ Let us prove inequality (6.5) for j D 2. Denote dj .S, T/ :D j ST j for j D 1, 2. Similar to the expression (6.7), we represent the function k2 .t, s/ in the form

S. V. GRYSHCHUK AND S. A. PLAKSA k2 .t, s/ D n1 .t, s/ C in2 .t, s/, where

(6.9)

  10 .S/ d2 .S, T/  20 .S/ C 20 .S/ 10 .S/  d1 .S, T/ 1 C st , n1 .t, s/ :D .s  t/.s2 C 1/ 2 .d.S, T//2    0 .S/ d2 .S, T/  20 .S/ C 20 .S/ 10 .S/  d1 .S, T/ n2 .t, s/ :D 1 . 2 .d.S, T//2 .s2 C 1/

Now, estimates of the type (6.5) for IŒg, n1 .t/, IŒg, n2 .t/ are established similarly to analogous estimates for IŒg, m1 .t/, IŒg, m2 .t/, respectively. Consider the notations  0 .S/  d.S, 1/  0 .S/ s C i D: m1 .1, s/ C im2 .1, s/, s2 C 1 d.S, 1/ .s2 C 1/ d.S, 1/    s k2 .1, s/ :D   0 .S/ d2 .S, 1/  20 .S/ C 20 .S/  0 .S/  d.S, 1/ C 2 2 2.s C 1/ .d.S, 1//  0 1  .S/ d2 .S, 1/  20 .S/ d.S, 1/ D: n1 .1, s/ C in2 .1, s/. Ci 2.s2 C 1/ .d.S, 1//2 k1 .1, s/ :D 

Lemma 6.11 Let g 2 C. R / and the conformal mapping  .T/ have the non-vanishing continuous contour derivative  0 .T/ on circle , and its modulus of continuity satisfies condition (6.4). Then for 0 < " < 1=4 and t 2 R such that j t j > 1=", the following estimates are true: ˇ 1 ˇ ˇ Z ˇ Z1 Z2 ˇ ˇ ! . 0 , / ˇ ˇ d , g.s/ kj .t, s/ ds  g.s/ kj .1, s/ ds ˇ  c kgkC.R / " ˇ . C "/ ˇ ˇ 1

1

j D 1, 2 ,

(6.10)

0

where the constant c does not depend on t and ". Proof Consider the case t > 1=" (the case t < 1=" is considered by analogy). In order to prove estimate (6.10) for j D 1, we shall use expression (6.8) of the integral IŒg, k1 .t/ and obtain estimates of the type (6.10) for each of the integrals IŒg, m1 .t/, IŒg, m2 .t/. For IŒg, m1 .t/, we have ˇ 0 t=2 11 ˇ 1 ˇ ˇ Z Z1 Z Z ˇ B ˇ C ˇ@ ˇ g.s/ m .t, s/ ds  g.s/ m .1, s/ ds C A jg.s/j .jm1 .t, s/j C jm1 .1, s/j/ dsC 1 1 ˇ ˇ ˇ ˇ 1

C

1

1

3t=2

3t=2 Z Zt=2 jg.s/j .jm1 .t, s/j C jm1 .1, s/j/ ds C jg.s/j jm1 .t, s/  m1 .1, s/j ds D: I1 C I2 C I3 . t=2

t=2

The integrals I1 and I2 are estimated using (6.3) 00

0  c kgkC.R /

Z1

0

1

t ds B C 0 C@ A ! . , jS  Tj/ 2 s C1

t=2 Z

Z1

1

1 ds

C C 0 A A ! . , jS  1j/ p 2 s C1 1 1 3t=2 3t=2 0 0 0 1 1 1 t=2 t=2 Z Z1 Z Z1 0 j sj ds C B B B C t ds C ! . , jS  1j/ 0 C C @! . , 6"/ @ A 2 C@ A A s jS  1j .s2 C 1/3=2

BB I1  c kgkC.R / @@

 c kgkC.R /

t=2 Z

C

1

3t=2

Z4"

0

@! . 0 , 6"/ C

2946

0

Copyright © 2015 John Wiley & Sons, Ltd.

1

1

C

3t=2

! . , / A d  c kgkC.R / " 

Z2

! . 0 , / d . C "/

0

Math. Meth. Appl. Sci. 2016, 39 2939–2952

S. V. GRYSHCHUK AND S. A. PLAKSA 3t=2 Z

I2  c kgkC.R /

 ds ! . 0 , jS  Tj/ 1 C jsjj tj  C ! . 0 , jS  1j/ s 2 p p 2 2 jS  Tj s C1 s C1 t C1

t=2

 c kgkC.R /

1 0 2" Z Z2 0 ! . , / ! . 0 , /  0 @ d C ! . , 4"/A  c kgkC.R / " d .  . C "/ 0

0

Here and in the succeeding text in the proof, by c, we denote constants whose values are independent of t and " but, generally speaking, may be different even within a single chain of inequalities. For estimating the integral I3 , we use the equality 

 0 1 s  .S/  d.S, T/ s  0 .S/  d.S, 1/  2 C 2 D st s C1 d.S, T/ s C1 d.S, 1/ 1  0 .S/  d.S, T/ s d.S, T/  d.S, 1/  0 .S/ C D 2 s C1 d.S, 1/ d.S, T/ st d.S, T/

m1 .t, s/  m1 .1, s/ D

and inequalities (6.2) and (6.3) and properties of a modulus of continuity (cf., e.g., [26, p. 176]). Thus, we obtain 0 I3  c kgkC.R /

B @jT  1j

Zt=2

! . 0 , jS  1j/ jsj ds C jS  1j s2 C 1

t=2

Zt=2  c kgkC.R / jT  1j

Zt=2

t=2

1 ! . 0 , jS  Tj/ ds C p p A 2 2 jS  Tj s C1 t C1

! . 0 , jS  1j/ j sj ds  c kgkC.R / " jS  1j2 .s2 C 1/3=2

t=2

Z2 "

! . 0 , / d  c kgkC.R / " 2

Z2

! . 0 , / d . . C "/

0

An estimate of the type (6.10) for IŒg, m1 .t/ follows from the obtained inequalities. An estimate of the type (6.10) for IŒg, m2 .t/ is similarly established. Thus, inequality (6.10) is proved for j D 1. To prove estimate (6.10) for j D 2, we use the representation (6.9) of the function k2 .t, s/ and obtain estimates of the type (6.10) for each of the integrals IŒg, n1 .t/, IŒg, n2 .t/ similar to the estimation of IŒg, m1 .t/. The next statement follows obviously from Lemmas 6.4 and 6.11. Theorem 6.13 Let the conformal mapping  .T/ have the non-vanishing continuous contour derivative  0 .T/ on circle , and its modulus of continuity satisfies condition (6.4). Let the function k.t, s/ be defined by relation (5.7), where the functions k1 .t, s/, k2 .t, s/ are defined by equalities (5.8) and (5.9), respectively. Then, the operator Z1 g.s/ k.t, s/ ds

IŒg :D 1

is compact in the space C. R /.

7. Equivalence conditions of the (1-3)-problem to a system of Fredholm integral equations From the aforementioned discussion, one can see that finding a solution of the (1-3)-problem in the form (5.1) with functions '1 W @D ! R, '3 W @D ! R satisfying conditions of the type (4.2) is reduced to solve the system (5.10). Under conditions of Theorem 6.13, integral operators, which are generated by the left parts of system (5.10), are compact in the space C. R /, that is, system (5.10) is a system of Fredholm integral equations. For any function g 2 C. R /, we use the local centered (with respect to the infinitely remote point) modulus of continuity !R,1 .g, "/ :D

jg.t/  g.1/j .

sup t2R : jtj1="

Let D.R/ denote the class of functions g 2 C.R / whose moduli of continuity satisfy the Dini conditions Z1

Copyright © 2015 John Wiley & Sons, Ltd.

Z1

!R,1 .g, / d < 1 . 

(7.1)

0

Math. Meth. Appl. Sci. 2016, 39 2939–2952

2947

0

!R .g, / d < 1, 

S. V. GRYSHCHUK AND S. A. PLAKSA Because the functions '1 W @D ! R and '3 W @D ! R in expression (5.1) of a solution of the (1-3)-problem have to satisfy conditions of the type (4.2), it is necessary to require that corresponding functions g1 , g3 satisfying system (5.10) should belong to the class D.R/. In the next theorem, we state a condition on the conformal mapping  .T/, under which all solutions of system (5.10) satisfy the mentioned requirement. Theorem 7.2 Let the functions u1 W @D ! R, u3 W @D ! R satisfy conditions of the type (4.2). Let the conformal mapping  .T/ have the nonvanishing continuous contour derivative  0 .T/ on circle , and its modulus of continuity satisfies the condition Z2

! . 0 , / 3 ln d < 1.  

(7.2)

0

Then, all functions g1 , g3 2 C. R / satisfying the system of Fredholm integral equations (5.10) belong to the class D.R/, and the corresponding functions '1 , '3 in (5.1) satisfy conditions of the type (4.2). Proof Let us rewrite system (5.10) in the matrix form g1 .t/ g3 .t/ where IŒg, k .t/ :D

1 R

! D

2Qu1 .t/ 2Qu3 .t/

! 

U1 U3

 

1 i

IŒg, k .t/

!

1 i

IŒg, k .t/



8t 2 R,

(7.3)

g.s/k.t, s/ ds and function g is defined by equality (5.4).

1

In as much as the derivative  0 .T/ is continuous on and the functions u1 , u3 satisfy conditions of the type (4.2), the right-hand sides e u3 of equations (5.10) belong to the class D.R/. Using Lemmas 6.4 and 6.11, it is easy to establish that moduli of continuity of the u1 , e function IŒg, k .t/ satisfy conditions of the type (7.1) due to condition (7.2). Therefore, in view of (7.3), the functions g1 , g3 belong to the class D.R/ also. Finally, by Lemma 3.3 in [27], we conclude that the functions '1 , '3 satisfy conditions of the type (4.2). Let us make some remarks concerning the representation of a solution of the (1-3)-problem by formula (5.1). Suppose that a solution ˆ of the (1-3)-problem is continuously extended to the boundary @D . By the (2-4)-problem conjugated with the (1-3)-problem or, briefly, the conjugated (2-4)-problem, we shall call a problem on finding a continuous function ˆ W  n D ! B, which is monogenic in the domain  n D and vanishes at the infinity with the following boundary conditions: U2 Œˆ ./ D U2 Œˆ./ ,

U4 Œˆ ./ D U4 Œˆ./

8 2 @D .

Note that if a solution of the (1-3)-problem is expressed by formula (5.1), where the functions '1 , '3 satisfy conditions of the type (4.2), then by Theorem 4.2, the integral (4.1) with '. / D '1 . /e1 C '3 . /e2 , being a solution of the (1-3)-problem, can be continuously extended to the boundary @D from each of the domains D ,  n D and is also a solution of the conjugated (2-4)-problem due to formulae (4.3). Furthermore, by virtue of assumptions that a solution ˆ of the (1-3)-problem is continuously extended to the boundary @D and the conjugated (2-4)-problem has a solution ˆ , it follows that the function ˆ is represented in the form (5.1). Indeed, by the Cauchy integral formula and the Cauchy theorem for monogenic functions in the biharmonic plane (cf., e.g, Theorem 3.2 in [28]), the following equalities are true Z Z 1 1 ˆ. /.  /1 d , 0 D ˆ . /.  /1 d 8 2 D , ˆ./ D 2i 2i @D

@D

which implies equality (5.1) with 'j . / D Uj Œˆ. /  Uj Œˆ . / D uj . /  Uj Œˆ . / ,

j 2 f1, 3g.

The made remarks can be amplified by the following theorem. Theorem 7.5 Let the functions u1 : @D ! R, u3 : @D ! R satisfy conditions of the type (4.2). Let the conformal mapping  .T/ have the nonvanishing continuous contour derivative  0 .T/ on circle , and its modulus of continuity satisfies condition (7.2). Then the following assertions are equivalent.

2948

(i) The system of Fredholm integral equations (5.10) is solvable in the space C.R/. (ii) There exists a solution of the (1-3)-problem of the form (5.1), where the functions '1 , '3 satisfy conditions of the type (4.2). (iii) A solution ˆ of the (1-3)-problem is continuously extended to the boundary @D . For this function ˆ, the conjugated (2-4)-problem is solvable, and moduli of continuity of components U1 Œˆ , U3 Œˆ of its solution ˆ satisfy conditions of the type (4.2). Copyright © 2015 John Wiley & Sons, Ltd.

Math. Meth. Appl. Sci. 2016, 39 2939–2952

S. V. GRYSHCHUK AND S. A. PLAKSA Proof Continuing the reasonings adduced before Theorem 7.5, we conclude that in the case where the functions u1 , u3 satisfy conditions of the type (4.2), the functions '1 , '3 satisfy the same conditions if and only if moduli of continuity of components U1 Œˆ , U3 Œˆ of a solution ˆ of the conjugated (2-4)-problem satisfy conditions of the type (4.2). Thus, assertions (ii) and (iii) are equivalent. The equivalence of assertions (i) and (ii) is a consequence of Theorem 7.2. Rewrite the integral equations of system (5.10) in expanded form as follows:

1 1 g1 .t/ C Im 2 2 

1 Im 2

Z1

1 Z1

1 1 g3 .t/ C Im 2 2 1 Im 2

1 Z1

Z1

1

g3 .s/

1 Z1



1  0 .s/ ds C Re g1 .s/  .s/   .t/ 2

 0 .s/.2 .s/  2 .t// 1 Re g1 .s/ 1 ds  2 . .s/   .t// 2

 2 .t// 1 Im ds C . .s/   .t//2 2

g3 .s/

1  .s/ ds  Re g3 .s/  .s/   .t/ 2

 1 .t// ds D uQ 1 .t/, . .s/   .t//2

1

g1 .s/

 0 .s/.2 .s/  2 .t// 1 Re g3 .s/ 1 ds C . .s/   .t//2 2

 2 .t// 1 Im ds C . .s/   .t//2 2

1

Z1 g3 .s/

1

Z1

10 .s/.2 .s/

g1 .s/

20 .s/.1 .s/  1 .t// ds . .s/   .t//2

20 .s/.1 .s/

1

Z1

g1 .s/

1

Z1

10 .s/.2 .s/

0

Z1

20 .s/.1 .s/

1

 1 .t// ds D uQ 3 .t/ . .s/   .t//2

20 .s/.1 .s/  1 .t// ds . .s/   .t//2 8 t 2 R.

Then, the homogeneous system transposed with system (5.10) is of the form

0 1 Z1 Z1 Z1 0 h .s/ ds 1 .2 .s/  2 .t// 20 .t/ .1 .s/  1 .t// 1 0 A  1 .t/ Re h1 .t/  Im @ .t/ Re h1 .s/ ds C h1 .s/ dsC   .s/   .t/  . .s/   .t//2  . .s/   .t//2 1

1

1

Z1

Z1

.2 .s/  2 .t//  0 .t/ .1 .s/  1 .t// 10 .t/ Im h3 .s/ ds  2 Im h3 .s/ ds D 0, 2  . .s/   .t//  . .s/   .t//2 1 1 0 1 Z1 Z1 Z1 h .s/ ds 10 .t/ 1 .2 .s/  2 .t// 20 .t/ .1 .s/  1 .t// 3 0 @ A C Re Re h3 .s/ ds  h3 .s/ dsC h3 .t/  Im  .t/ 2   .s/   .t/  . .s/   .t//  . .s/   .t//2 C

1

C

10 .t/ 

Z1 h1 .s/

Im 1

1

20 .t/

(7.4)

1

.2 .s/  2 .t// Im ds  . .s/   .t//2 

Z1 h1 .s/

1

.1 .s/  1 .t// ds D 0 . .s/   .t//2

8t 2 R.

(7.5)

Lemma 7.8 The pair of functions .h1 , h2 / :D .10 , 20 / satisfies the system of integral equations (7.4) and (7.5). Proof Substituting h1 D 10 , h2 D 20 into equation (7.4) and applying the conformal mapping z D  .t/ of the upper half plane ft 2 C : Im t > 0g onto the domain Dz of complex plane, we pass in (7.4) to integrate along the boundary @Dz of domain Dz . In such a way, using the denotations v1 :D 1 .s/, v2 :D 2 .s/, v :D  .s/  v1 C iv2 , x :D 1 .t/, y :D 2 .t/, z  x C iy, we obtain 0 1 B 10 .t/  Im @ 0 .t/ 

Z

@Dz

C

10 .t/ Im 

Z

@Dz

1 dv1 C 10 .t/ Re A vz 

Z

v2  y 20 .t/ Re dv C 1 .v  z/2 

@Dz

v2  y  0 .t/ dv2  2 Im 2 .v  z/ 

Z

Z

v1  x dv1 C .v  z/2

@Dz

v1  x dv2 D 0. .v  z/2

(7.6)

@Dz

Copyright © 2015 John Wiley & Sons, Ltd.

Math. Meth. Appl. Sci. 2016, 39 2939–2952

2949

Using polar coordinates .r, / with the relation v  z D: r expfig, it is easy to show that equality (7.6) becomes identical. In a similar way, we conclude that functions h1 D 10 , h2 D 20 satisfy equation (7.5).

S. V. GRYSHCHUK AND S. A. PLAKSA It follows from Lemma 7.8 that the condition Z1

 uQ 1 .s/10 .s/ C uQ 3 .s/20 .s/ ds D 0

1

is necessary for the solvability of system (5.10). Passing in this equality to the integration along the boundary @D , we obtain the equivalent condition Z u1 .xe1 C ye2 / dx C u3 .xe1 C ye2 / dy D 0 (7.7) @D

for given functions u1 , u3 . It is proved in [14] that condition (7.7) is also sufficient for the solvability of the (1-3)-problem for a disk. The next theorem contains assumptions, under which condition (7.7) is necessary and sufficient for the solvability of the system of integral equation (5.10) in the Cartesian product C. R /  C. R / and, therefore, for the existence of a solution of the (1-3)-problem in the form (5.1). Theorem 7.11 Assume that the conformal mapping  .T/ has the non-vanishing continuous contour derivative  0 .T/ on circle , and its modulus of continuity satisfies condition (7.2). Also, assume that all solutions .g1 , g3 / 2 C. R /  C. R / of the homogeneous system (5.10) (with uQ j  0 for j 2 f1, 3g) are differentiable on R, and the integral Z

1 2i

' 0 . /.  /1 d

@D

is bounded in both domains D and  n D ; here, ' 0 is the contour derivative of the function '. / :D g1 .s/e1 C g3 .s/e2 , where  D .s/ Q for all s 2 R. Then the following assertions are true. (i) The number of linearly independent solutions of the homogeneous system (5.10) is equal to 1. (ii) The non-homogeneous system (5.10) is solvable if and only if condition (7.7) is satisfied. Proof It follows from Lemma 7.8 that the transposed system of equations (7.4) and (7.5) has at least one nontrivial solution. Therefore, by the Fredholm theory, the homogeneous system (5.10) has at least one nontrivial solution g1 D g01 , g3 D g03 . Consider the function '0 . / :D g01 .s/e1 C g03 .s/e2 ,

 D .s/ Q ,

8s 2 R

(7.8)

corresponding to this solution. Then the function, which is defined for all  2 D by the formula ˆ0 ./ :D

1 2i

Z

'0 . /.  /1 d ,

(7.9)

@D

is a solution of the homogeneous (1-3)-problem (with u1 D u3  0). By virtue of differentiability of the functions g01 , g03 on R, the function '0 has the contour derivative '00 . / for all  2 @D and the following equality holds: ˆ00 ./ :D

1 2i

Z

'00 . /.  /1 d

8  2 D .

(7.10)

@D

Let ˆ1 be monogenic in D function such that ˆ01 ./ D ˆ0 ./ for all  2 D . Then, the function V.x, y/ :D U1 Œˆ1 ./ is a solution of the homogeneous biharmonic problem (1.2) (with !3 D !4  0). Taking into account the equalities V.x, y/ :D

    @2 V.x, y/ @2 V.x, y/ C D U1 ˆ001 ./.e21 C e22 / D U1 ˆ00 ./.e21 C e22 / 2 @x @y2

2950

and boundedness of function (7.10) in the domain D , we conclude that the Laplacian V.x, y/ is bounded in the domain D. By the uniqueness theorem (see Section 4 in [29]), solutions of the homogeneous biharmonic problem (1.2) in the class of functions V, for which V.x, y/ is bounded in D, are only constants: V D const. Therefore, taking into account the fact that functions (7.9) and (3.1) are equal, we obtain the equalities U1 Œˆ0 ./ D U3 Œˆ0 ./  0 in the domain D . Copyright © 2015 John Wiley & Sons, Ltd.

Math. Meth. Appl. Sci. 2016, 39 2939–2952

S. V. GRYSHCHUK AND S. A. PLAKSA All monogenic functions ˆ0 satisfying the condition U1 Œˆ0 ./  0 are described in [13, Lemma 3]. Taking into account the identity U3 Œˆ0 ./  0 also, we obtain the equality ˆ0 ./ D ik C i .n1 e1 C n2 e2 /

8  2 D ,

(7.11)

where k, n1 , and n2 are real numbers. Let us show that the constant k is not equal to zero in equality (7.11). Assume the contrary: k D 0. Then function (7.9), which is considered for all  2  n D , is a solution of the conjugated (2-4)-problem with boundary data C U2 Œˆ 0 ./ D U2 Œˆ0 ./  n1 ,

C U4 Œˆ 0 ./ D U4 Œˆ0 ./  n2

8 2 @D .

(7.12)

Using arguments analogous to those used in the proof of equality (7.11), we establish that a function ˆ0 , which is monogenic in  n D and satisfies the boundary conditions (7.12), is of the form ˆ0 ./ D k1  C m1 e1 C m2 e2 C in1 e1 C in2 e2

8  2  n D ,

where k1 , m1 , and m2 are real numbers. In as much as this function is a solution of the conjugated (2-4)-problem and vanishes at the infinity, we obtain k1 D m1 D m2 D n1 D n2 D 0. Therefore, as a corollary of the Sokhotski–Plemelj formula (4.3) for function (7.9), we obtain the equalities  '0 . / D ˆC 0 . /  ˆ0 . / D 0

8  2 @D .

(7.13)

As a result of (7.13), the identity '0 . /  0 contradicts to the non-triviality of at least one of the functions g01 , g03 in definition (7.8) of the function '0 . Hence, the assumption k D 0 is not true. Therefore, equality (7.11) holds with k ¤ 0. Let g1 D g11 , g3 D g13 be a nontrivial solution of the homogeneous system (5.10) and g11 , g13 be different from g01 , g03 . Then, for the Q for all s 2 R, the following equality of type (7.11) holds: function ' 1 . / :D g11 .s/e1 C g13 .s/e2 , where  D .s/ Z 1 Q C i .nQ 1 e1 C nQ 2 e2 / ' 1 . /.  /1 d D ik 8  2 D , (7.14) 2i @D

Q nQ 1 , and nQ 2 are real numbers. where k, Define a function 'Q by the equality '. Q / D ' 1 . / 

kQ '0 . / k

8  2 @D .

Taking into account equalities (7.11) and (7.14), we obtain 1 2i

Z '. Q /.  /

1

d D ie1

nQ 1  n1

kQ k

! C ie2

nQ 2  n2

kQ k

! 8 2 D .

@D

Further, in such a way as for the assumption k D 0 in equality (7.11), we obtain the identity '. Q /  0, which implies the equalities Q g1m . / D kk g0m . / for all  2 @D and m 2 f1, 3g. Assertion (i) is proved. Because of assertion (i), by the Fredholm theory, the number of linearly independent solutions of the transposed system of equations (7.4) and (7.5) is also equal to 1. Such a nontrivial solution is described in Lemma 7.8. Therefore, condition (7.7) is not only necessary but also sufficient for the solvability of system (5.10). Results of this paper have been announced at a preprint of Arxiv (arXiv: 1505.02518v1 [math.AP]).

References

Copyright © 2015 John Wiley & Sons, Ltd.

Math. Meth. Appl. Sci. 2016, 39 2939–2952

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