On solvability of the Neumann boundary value problem for a non-homogeneous polyharmonic equation in a ball

  • Batirkhan K Turmetov1Email author and

    Affiliated with

    • Ravshan R Ashurov2

      Affiliated with

      Boundary Value Problems20132013:162

      DOI: 10.1186/1687-2770-2013-162

      Received: 23 April 2013

      Accepted: 19 June 2013

      Published: 5 July 2013

      Abstract

      In this work the Neumann boundary value problem for a non-homogeneous polyharmonic equation is studied in a unit ball. Necessary and sufficient conditions for solvability of this problem are found. To do this we first reduce the Neumann problem to the Dirichlet problem for a different non-homogeneous polyharmonic equation and then use the Green function of the Dirichlet problem.

      MSC:35J40, 35J30, 35A01.

      Keywords

      non-homogeneous polyharmonic equation the Neumann problem the necessary and sufficient conditions for solvability

      1 Introduction

      Let Ω = { x R n : | x | < 1 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq1_HTML.gif be a unit ball, Ω = { x R n : | x | = 1 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq2_HTML.gif be a unit sphere and m be a positive integer.

      Consider on the domain Ω the following Neumann boundary value problem:
      ( Δ ) m u ( x ) = g ( x ) , x Ω , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equ1_HTML.gif
      (1)
      k u ν k ( x ) = φ k ( x ) , k = 1 , 2 , , m , x Ω , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equ2_HTML.gif
      (2)

      where ν is the unit outer normal vector to sphere Ω, g ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq3_HTML.gif and φ k ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq4_HTML.gif are given functions; we always suppose that these functions are sufficiently smooth, and from here on we do not pay any attention to their smoothness.

      A function u ( x ) C 2 m ( Ω ) C m + 1 ( Ω ¯ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq5_HTML.gif is called a solution of problem (1), (2) if it satisfies (1), (2) in a classical sense.

      It is well known (see, for example, [1]) that even in case m = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq6_HTML.gif the considered problem (1), (2) does not have solutions for arbitrary (even, as we supposed, smooth) functions g ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq3_HTML.gif and φ k ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq4_HTML.gif; in the case of the Poisson equation, the necessary and sufficient solvability condition for the Neumann problem is
      Ω g ( x ) d x = Ω φ 1 ( x ) d S x . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equa_HTML.gif
      In the paper [2] by Kanguzhin and Koshanov, in particular, it is shown that in case m = 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq7_HTML.gif the necessary and sufficient condition for solvability of problem (1), (2) has the form
      Ω { φ 1 ( x ) φ 2 ( x ) + Ω d 2 , n ( 4 n ) × [ ( 2 n ) | x y | n ( 1 ( x , y ) ) 2 + | x y | 2 n ( x , y ) ] g ( y ) d y } d S x = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equb_HTML.gif

      where ( x , y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq8_HTML.gif is the scalar product in R n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq9_HTML.gif and d 2 , n = Γ ( n 2 2 ) 16 π n 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq10_HTML.gif.

      The authors of the paper [3] presented this condition in a different form, which could be easily verified,
      Ω 1 | x | 2 2 g ( x ) d x = Ω [ φ 2 ( x ) φ 1 ( x ) ] d S x . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equc_HTML.gif

      In the above paper [2] the authors found a solvability condition for Neumann problem (1), (2) for arbitrary m as well (see [2], Theorem 4.2). This condition follows from equality to zero of the determinant of an m × m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq11_HTML.gif matrix, one column of which consists of integrals Ω [ φ k ( x ) k / ν k ( ε m , n g ( x ) ) ] d S x http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq12_HTML.gif, ε m , n = d m , n | x | 2 m n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq13_HTML.gif and d m , n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq14_HTML.gif is a constant. Note that the equation which one has as a result is very difficult to verify.

      The main goal of the present paper is to find a solvability condition for problem (1), (2) in a more simple form. It should be noted that in our study of problem (1), (2) the Green function of the Dirichlet problem for equation (1) is essentially used. In the paper [4] a similar method was used in the solution of the boundary value problem for the Poisson equation with the boundary operator of fractional-order.

      The paper is organized as follows. In the next section we study the properties of some integro-differential operators, which we then use throughout the paper. In Section 3 we investigate the Dirichlet problem for a polyharmonic equation, making use of the explicit form of the Green function found in [57]. Then, in the following section, reducing Neumann problem (1), (2) to the considered Dirichlet problem, we give the necessary and sufficient solvability condition for problem (1), (2) with homogeneous boundary conditions. In the same way we consider in Section 5 the Neumann boundary value problem for the homogeneous equation with non-homogeneous boundary conditions. Finally, in Section 6 we study problem (1), (2) in the general case. To present the necessary and sufficient conditions for solvability, here we apply the Almansi formula for constructing solutions to the Dirichlet problem.

      2 Properties of some integro-differential operators

      Let u ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq15_HTML.gif be a sufficiently smooth function in Ω. Consider the following operators:
      Γ c [ u ] ( x ) = ( r r + c ) u ( x ) , Γ c 1 [ u ] ( x ) = 0 1 t c 1 u ( t x ) d t , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equ3_HTML.gif
      (3)

      where r = | x | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq16_HTML.gif and c 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq17_HTML.gif is a constant. Note that the operator Γ 0 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq18_HTML.gif is not defined for functions u ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq15_HTML.gif with u ( 0 ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq19_HTML.gif.

      Note that in the class of harmonic functions in a ball the properties of operators Γ c 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq20_HTML.gif and Γ c 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq21_HTML.gif with c > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq22_HTML.gif have previously been studied in the paper [8].

      Lemma 1 Let u ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq15_HTML.gif be a smooth function. Then for any x Ω http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq23_HTML.gif one has
      1. (1)
        if c > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq22_HTML.gif, then
        Γ c [ Γ c 1 [ u ] ] ( x ) = Γ c 1 [ Γ c [ u ] ] ( x ) = u ( x ) ; http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equ4_HTML.gif
        (4)
         
      2. (2)
        if c = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq24_HTML.gif, then
        Γ 0 1 [ Γ 0 [ u ] ] ( x ) = u ( x ) u ( 0 ) ; http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equ5_HTML.gif
        (5)
         
      3. (3)
        if c = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq24_HTML.gif and u ( 0 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq25_HTML.gif, then
        Γ 0 [ Γ 0 1 [ u ] ] ( x ) = u ( x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equ6_HTML.gif
        (6)
         
      Proof Let c 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq17_HTML.gif. Then
      0 1 d d t [ t c u ( t x ) ] d t = 0 1 t c 1 [ c u ( t x ) + t d d t u ( t x ) ] d t = 0 1 t c 1 Γ c [ u ] ( t x ) d t . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equd_HTML.gif
      Therefore, if c > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq22_HTML.gif, then
      u ( x ) = 0 1 t c 1 Γ c [ u ] ( t x ) d t = Γ c 1 [ Γ c [ u ] ] ( x ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Eque_HTML.gif
      and if c = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq24_HTML.gif, then
      u ( x ) u ( 0 ) = 0 1 t 1 Γ 0 [ u ] ( t x ) d t . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equ7_HTML.gif
      (7)

      Hence, equality (5) and the second equality of (4) are proved.

      As we noted above, if u ( 0 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq25_HTML.gif, then the expression Γ 0 1 [ u ] ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq26_HTML.gif is defined. Now apply the operator Γ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq27_HTML.gif to this expression. Then
      Γ 0 [ Γ 0 1 [ u ] ] ( x ) = 0 1 t 1 Γ 0 [ u ] ( t x ) d t . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equf_HTML.gif

      But due to equality (7) and the condition u ( 0 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq25_HTML.gif, the last expression is equal to u ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq15_HTML.gif. Hence, equality (6) is proved. The first equality in (4) can be proved in the same way. □

      The following statement can be proved by a direct calculation.

      Lemma 2 Let u ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq15_HTML.gif be a smooth function. Then for any x Ω http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq23_HTML.gif one has
      1. (1)
        if c 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq17_HTML.gif, then
        Δ Γ c [ u ] ( x ) = Γ c + 2 [ Δ u ] ( x ) ; http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equg_HTML.gif
         
      2. (2)
        if c > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq22_HTML.gif, then
        Δ Γ c 1 [ u ] ( x ) = Γ c + 2 1 [ Δ u ] ( x ) ; http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equh_HTML.gif
         
      3. (3)
        if c = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq24_HTML.gif and u ( 0 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq25_HTML.gif, then
        Δ Γ 0 1 [ u ] ( x ) = Γ 2 1 [ Δ u ] ( x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equi_HTML.gif
         
      Corollary 1 Let u ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq15_HTML.gif be a smooth function. Then for any x Ω http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq23_HTML.gif one has
      1. (1)
        if c 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq17_HTML.gif, then
        Δ m Γ c [ u ] ( x ) = Γ c + 2 m [ Δ m u ] ( x ) ; http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equ8_HTML.gif
        (8)
         
      2. (2)
        if c > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq22_HTML.gif, or c = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq24_HTML.gif and u ( 0 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq25_HTML.gif, then
        Δ m Γ c 1 [ u ] ( x ) = Γ c + 2 m 1 [ Δ m u ] ( x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equ9_HTML.gif
        (9)
         

      3 Some properties of the solutions of the Dirichlet problem

      Let v ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq28_HTML.gif be a solution of the Dirichlet problem
      { ( Δ ) m v ( x ) = g 1 ( x ) , x Ω , k 1 v ν k 1 ( x ) = 0 , k = 1 , 2 , , m , x Ω . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equ10_HTML.gif
      (10)
      It is known (see, for example, [57]) that if g 1 ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq29_HTML.gif is a sufficiently smooth function, then the solution of problem (10) exists, it is unique and has the form
      v ( x ) = Ω G m , n ( x , y ) g 1 ( y ) d y , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equ11_HTML.gif
      (11)

      where G m , n ( x , y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq30_HTML.gif is the Green function of Dirichlet problem (10).

      We make use of the following explicit form of the Green function [5]:

      if n is odd, or even and n > 2 m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq31_HTML.gif, then
      G m , n ( x , y ) = d m , n [ | x y | 2 m n | x | y | y | y | | 2 m n k = 1 m 1 ( 1 ) k k ! ( m n 2 ) ( m n 2 k + 1 ) × | x | y | y | y | | 2 m n 2 k ( 1 | x | 2 ) k ( 1 | y | 2 ) k ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equj_HTML.gif
      where
      d m , n = ( 1 ) m Γ ( n 2 m ) π n 2 4 m ( m 1 ) ! ; http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equk_HTML.gif
      if n is even and n 2 m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq32_HTML.gif, then
      G m , n ( x , y ) = d m , n [ | x y | 2 m n ( ln | x y | 2 ln | x | y | y | y | | 2 ) + k = 1 m 1 | x | y | y | y | | 2 m n 2 k ( 1 | x | 2 ) k ( 1 | y | 2 ) k j = max ( k n / 2 , 0 ) min ( m n / 2 , k 1 ) ( 1 ) j k j C m n / 2 j ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equl_HTML.gif
      where
      d m , n = ( 1 ) m n / 2 2 π n 2 4 m ( m n / 2 ) ! ( m 1 ) ! . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equm_HTML.gif
      Lemma 3 Let g 1 ( x ) = Γ 2 m [ g ] ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq33_HTML.gif in Dirichlet problem (10), and let v ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq28_HTML.gif be the unique solution of this problem. Then v ( 0 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq34_HTML.gif if and only if
      Ω ( 1 | x | 2 ) m 1 g ( x ) d x = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equ12_HTML.gif
      (12)
      Proof Let v ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq28_HTML.gif be the solution of problem (10). Then it has the form (11). To use the explicit form of the Green function, we shall deal only with the case n is odd or even and n > 2 m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq31_HTML.gif, the other cases being exactly similar. So, if v ( 0 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq34_HTML.gif, then from (11) one has
      Ω [ | y | 2 m n 1 k = 1 m 1 ( 1 ) k k ! ( m n 2 ) ( m n 2 k + 1 ) ( 1 | y | 2 ) k ] g 1 ( y ) d y = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equn_HTML.gif
      If we denote ρ = | y | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq35_HTML.gif and ξ = y | y | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq36_HTML.gif, then the last integral can be rewritten as
      | ξ | = 1 0 1 ρ n 1 [ ρ 2 m n 1 k = 1 m 1 ( 1 ) k k ! ( m n 2 ) ( m n 2 k + 1 ) ( 1 ρ 2 ) k ] × g 1 ( ρ , ξ ) d ρ d ξ = | ξ | = 1 I ( ξ ) d ξ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equo_HTML.gif
      Now we consider the inner integral I ( ξ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq37_HTML.gif. Noting that g 1 ( ρ , ξ ) = ( ρ ρ + 2 m ) g ( ρ , ξ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq38_HTML.gif, we introduce the following two integrals:
      I 1 ( ξ ) = 0 1 ρ n 1 [ 2 m ρ 2 m n 2 m 2 m k = 1 m 1 ( 1 ) k k ! ( m n 2 ) ( m n 2 k + 1 ) I 1 ( ξ ) = × ( 1 ρ 2 ) k ] g ( ρ , ξ ) d ρ , I 2 ( ξ ) = 0 1 ρ n [ ρ 2 m n 1 k = 1 m 1 ( 1 ) k k ! ( m n 2 ) ( m n 2 k + 1 ) ( 1 ρ 2 ) k ] I 2 ( ξ ) = × ρ g ( ρ , ξ ) d ρ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equp_HTML.gif

      Obviously, I ( ξ ) = I 1 ( ξ ) + I 2 ( ξ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq39_HTML.gif.

      Integrating by part in the integral I 2 ( ξ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq40_HTML.gif, we obtain
      I 2 ( ξ ) = 0 1 ρ n 1 [ n ρ 2 m n + n + n [ k = 1 m 1 ( 1 ) k k ! ( m n 2 ) ( m n 2 k + 1 ) ( 1 ρ 2 ) k ] ( 2 m n ) ρ 2 m n 2 ρ 2 [ k = 1 m 1 ( 1 ) k k ! k ( m n 2 ) ( m n 2 k + 1 ) ( 1 ρ 2 ) k 1 ] ] g ( ρ , ξ ) d ρ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equq_HTML.gif
      Therefore
      I 1 ( ξ ) + I 2 ( ξ ) = 0 1 ρ n 1 S m 1 ( ρ ) g ( ρ , ξ ) d ρ , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equr_HTML.gif
      where
      S j ( ρ ) = n 2 m 2 ρ 2 k = 1 j ( 1 ) k ( k 1 ) ! ( m n 2 ) ( m n 2 k + 1 ) ( 1 ρ 2 ) k 1 + ( n 2 m ) k = 1 j ( 1 ) k k ! ( m n 2 ) ( m n 2 k + 1 ) ( 1 ρ 2 ) k , 1 j m 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equs_HTML.gif
      It is not hard to prove by induction that
      S j ( ρ ) = ( n 2 m ) ( n 2 ( m j ) ) 2 j ( 1 ρ 2 ) j j ! . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equ13_HTML.gif
      (13)
      Indeed, if j = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq41_HTML.gif, then
      S 1 ( ρ ) = n 2 m 2 ρ 2 ( 1 ) ( m n 2 ) + ( n 2 m ) ( 1 ) ( m n 2 ) ( 1 ρ 2 ) = ( n 2 m ) ( 1 ρ 2 ) + ( n 2 m ) n 2 m 2 ( 1 ρ 2 ) = ( n 2 m ) ( n 2 ( m 1 ) ) 2 ( 1 ρ 2 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equt_HTML.gif
      Now let us suppose that (13) holds true for some j and prove it for j + 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq42_HTML.gif. We have
      S j + 1 ( ρ ) = ( n 2 m ) ( n 2 ( m j ) ) 2 j ( 1 ρ 2 ) j j ! 2 j ! n 2 m 2 n 2 ( m j ) 2 ( 1 ρ 2 ) j ρ 2 + n 2 m ( j + 1 ) ! n 2 m 2 n 2 ( m j ) 2 ( 1 ρ 2 ) j + 1 = ( n 2 m ) ( n 2 ( m j ) ) 2 j ( 1 ρ 2 ) j + 1 j ! + n 2 m 2 ( j + 1 ) ( n 2 m ) ( n 2 ( m j ) ) 2 j ( 1 ρ 2 ) j + 1 j ! = ( n 2 m ) ( n 2 ( m j ) ) 2 j n 2 ( m j 1 ) 2 ( 1 ρ 2 ) j + 1 ( j + 1 ) ! . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equu_HTML.gif
      Thus equality (13) holds true for any j = 1 , 2 , , m 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq43_HTML.gif. In particular,
      S m 1 ( ρ ) = ( n 2 m ) ( n 2 ) 2 m 1 ( 1 ρ 2 ) m 1 ( m 1 ) ! = a n , m ( 1 ρ 2 ) m 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equv_HTML.gif
      where
      a n , m = ( n 2 m ) ( n 2 ) 2 m 1 ( m 1 ) ! . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equw_HTML.gif
      Therefore
      0 = | ξ | = 1 I ( ξ ) d ξ = a n , m | ξ | = 1 0 1 ρ n 1 ( 1 ρ 2 ) m 1 g ( ρ , ξ ) d ρ d ξ , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equx_HTML.gif
      and going back to the Cartesian coordinate system, we have
      Ω ( 1 | x | 2 ) m 1 g ( x ) d x = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equy_HTML.gif

       □

      4 The Neumann problem with homogeneous boundary conditions

      In this section we study problem (1), (2) with homogeneous boundary conditions.

      Theorem 1 Let g ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq3_HTML.gif be sufficiently smooth. Then the necessary and sufficient solvability condition for Neumann problem (1), (2) has the form (12).

      If a solution exists, then it is unique up to a constant and can be represented as
      u ( x ) = C + Γ 0 1 [ v ] ( x ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equz_HTML.gif

      where v ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq28_HTML.gif is the solution of Dirichlet problem (10) with the right-hand side g 1 ( x ) = Γ 2 m [ g ] ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq33_HTML.gif, which satisfies the additional condition v ( 0 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq34_HTML.gif.

      Proof Let a solution of problem (1), (2) exist and let u ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq15_HTML.gif be this solution. We apply an operator Γ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq27_HTML.gif to a function u ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq15_HTML.gif and denote v ( x ) = Γ 0 [ u ] ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq44_HTML.gif. Now we obtain the conditions for the function v ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq28_HTML.gif.

      Obviously, v ( 0 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq34_HTML.gif. If we apply the operator ( Δ ) m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq45_HTML.gif to v ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq28_HTML.gif, then by virtue of (8) we have
      ( Δ ) m v ( x ) = Γ 2 m [ ( Δ ) m u ] ( x ) = Γ 2 m [ g ] ( x ) = g 1 ( x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equaa_HTML.gif
      Further, since
      r u r ( x ) = u ν ( x ) , x Ω , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equab_HTML.gif
      then
      v ( x ) = u ν ( x ) = 0 , x Ω . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equac_HTML.gif
      It is not hard to verify that for any k = 1 , 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq46_HTML.gif and all x Ω http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq47_HTML.gif one has [9]
      k v ν k ( x ) = r r ( r r 1 ) ( r r k + 1 ) v ( x ) = r k k v r k ( x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equad_HTML.gif
      Therefore from homogeneous conditions (2) we finally have
      k v ν k ( x ) = 0 , k = 1 , 2 , , m 1 , x Ω . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equae_HTML.gif
      Thus, if u ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq15_HTML.gif is a solution of problem (1), (2) with homogeneous boundary conditions, then the function v ( x ) = Γ 0 [ u ] ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq44_HTML.gif is the solution of Dirichlet problem (10) with the right-hand side
      g 1 ( x ) = ( r r + 2 m ) g ( x ) = Γ 2 m [ g ] ( x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equaf_HTML.gif

      Moreover, the function v ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq28_HTML.gif satisfies the condition v ( 0 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq34_HTML.gif and according to Lemma 3, the necessary condition for this is (12). Hence, if a solution of problem (1), (2) exists, then it is necessary for condition (12) to be satisfied.

      Now we prove that if condition (12) is satisfied, then the solution of problem (1), (2) with homogeneous boundary conditions exists.

      Indeed, if (12) is satisfied, then according to Lemma 3 the solution v ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq28_HTML.gif of Dirichlet problem (10) with g 1 ( x ) = Γ 2 m [ g ] ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq33_HTML.gif exists and v ( 0 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq34_HTML.gif.

      Therefore, we may apply the operator Γ 0 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq18_HTML.gif to v ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq28_HTML.gif and consider the function u ( x ) = C + Γ 0 1 [ v ] ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq48_HTML.gif. It is not hard to show that this function is the solution of problem (1), (2).

      Indeed, by virtue of (9) one has
      ( Δ ) m u ( x ) = Γ 2 m 1 [ ( Δ ) m v ] ( x ) = Γ 2 m 1 [ Γ 2 m [ g ] ] ( x ) = g ( x ) , x Ω . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equag_HTML.gif

      Since Γ 0 [ u ] ( x ) = Γ 0 [ Γ 0 1 [ v ] ] ( x ) = v ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq49_HTML.gif, then one can show as above that the function u ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq15_HTML.gif satisfies all homogeneous boundary conditions. □

      5 The Neumann problem for the homogeneous equation

      In the present section we consider Neumann problem (1), (2) with g ( x ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq50_HTML.gif.

      Let A be the following matrix
      ( 1 1 1 1 0 2 [ 1 ] 4 [ 1 ] ( 2 m 2 ) [ 1 ] 0 2 [ 2 ] 4 [ 2 ] ( 2 m 2 ) [ 2 ] 0 2 [ m 1 ] 4 [ m 1 ] ( 2 m 2 ) [ m 1 ] ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equ14_HTML.gif
      (14)

      where j [ k ] = j ( j 1 ) ( j k + 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq51_HTML.gif, j [ 1 ] = j http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq52_HTML.gif. Note that j [ k ] = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq53_HTML.gif if k > j http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq54_HTML.gif. Denote by Δ j http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq55_HTML.gif, j = 1 , 2 , , m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq56_HTML.gif, the determinant of the matrix obtained from A by deleting the elements of the first column and the j th row. Obviously, | A | = det A = Δ 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq57_HTML.gif.

      Let w ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq58_HTML.gif be a solution of the following Dirichlet problem with sufficiently smooth boundary functions f k ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq59_HTML.gif:
      { ( Δ ) m w ( x ) = 0 , x Ω , k 1 w ν k 1 ( x ) = f k ( x ) , k = 1 , 2 , , m , x Ω . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equ15_HTML.gif
      (15)
      Theorem 2 Let g ( x ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq50_HTML.gif and φ k ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq4_HTML.gif, k = 1 , 2 , , m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq60_HTML.gif, be sufficiently smooth functions. Then the necessary and sufficient solvability condition for Neumann problem (1), (2) has the form
      k = 1 m Ω ( 1 ) k + 1 Δ k [ φ k ( x ) + ( k 1 ) φ k 1 ( x ) ] d S x = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equ16_HTML.gif
      (16)

      where φ 0 ( x ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq61_HTML.gif.

      If a solution exists, then it is unique up to a constant and can be represented as
      u ( x ) = C + Γ 0 1 [ w ] ( x ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equah_HTML.gif

      where w ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq58_HTML.gif is the solution of Dirichlet problem (15) with boundary functions f 1 ( x ) = φ 1 ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq62_HTML.gif, f k ( x ) = φ k ( x ) + ( k 1 ) φ k 1 ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq63_HTML.gif, k = 2 , 3 , , m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq64_HTML.gif, which satisfies the additional condition w ( 0 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq65_HTML.gif.

      Proof Let a solution of problem (1), (2) exist and let u ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq15_HTML.gif be this solution. We apply an operator Γ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq27_HTML.gif to a function u ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq15_HTML.gif and denote w ( x ) = Γ 0 [ u ] ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq66_HTML.gif. Now we prove that the function w ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq58_HTML.gif is the solution of Dirichlet problem (15) with the additional condition w ( 0 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq65_HTML.gif.

      From the properties of the operator Γ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq27_HTML.gif we have ( Δ ) m w ( x ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq67_HTML.gif, x Ω http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq23_HTML.gif. By virtue of the following formula [9]
      j u ν j = ( r r j + 1 ) ( r r 1 ) r u r , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equ17_HTML.gif
      (17)
      one has for x Ω http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq47_HTML.gif
      w ( x ) = φ 1 ( x ) , ( r r j + 1 ) ( r r 1 ) w ( x ) = φ j ( x ) , j = 2 , 3 , , m . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equai_HTML.gif
      We rewrite these conditions in a more convenient form. To do this we first consider the last two of them:
      ( r r m + 2 ) ( r r 1 ) ( r r ( m 1 ) ) w ( x ) = φ m ( x ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equ18_HTML.gif
      (18)
      ( r r m + 2 ) ( r r 1 ) w ( x ) = φ m 1 ( x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equ19_HTML.gif
      (19)
      We multiply expression (19) by m 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq68_HTML.gif and sum to (18). Then, making use of (17), we obtain
      ( r r m + 2 ) ( r r 1 ) r w ( x ) r = m 1 w ( x ) ν m 1 = φ m ( x ) + ( m 1 ) φ m 1 ( x ) = f m ( x ) , x Ω . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equaj_HTML.gif
      Further, by repeating this argument for all 1 j m 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq69_HTML.gif, we get
      j w ( x ) ν j = φ j + 1 ( x ) + j φ j ( x ) = f j + 1 ( x ) , x Ω . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equak_HTML.gif

      Thus, if u ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq15_HTML.gif is the solution of Neumann problem (1), (2), then the function w ( x ) = Γ 0 [ u ] ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq66_HTML.gif will be the solution of Dirichlet problem (15) with the additional condition w ( 0 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq65_HTML.gif. Note that, under the conditions of Theorem 2, the solution of problem (15) exists and it is unique (see, for example, [10]).

      Next we find the conditions to the boundary functions φ k ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq4_HTML.gif, which guarantee the equality w ( 0 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq65_HTML.gif. Making use of the Almansi formula (see, for example, [11], p.188) we write the solution of problem (15) as
      w ( x ) = w 0 ( x ) + r 2 w 1 ( x ) + r 4 w 2 ( x ) + + r 2 m 2 w m 1 ( x ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equ20_HTML.gif
      (20)

      where w j ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq70_HTML.gif are harmonic functions in the ball Ω. Obviously, w ( 0 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq65_HTML.gif if and only if w 0 ( 0 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq71_HTML.gif.

      Substituting function (20) into the boundary condition of (15) and integrating over the sphere, taking into account the equalities
      Ω k w j ( x ) ν k d S x = { ω n w j ( 0 ) , k = 0 , 0 , k 0 , ω n  is the area of the unit sphere, http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equal_HTML.gif
      we get the system of equations
      w 0 ( 0 ) + w 1 ( 0 ) + w 2 ( 0 ) + + w m 1 ( 0 ) = b 1 , 0 w 0 ( 0 ) + 2 w 1 ( 0 ) + 4 w 2 ( 0 ) + + ( 2 m 2 ) w m 1 ( 0 ) = b 2 , 0 w 0 ( 0 ) + 2 w 1 ( 0 ) + 4 3 w 2 ( 0 ) + + ( 2 m 2 ) ( 2 m 3 ) w m 1 ( 0 ) = b 3 , 0 w 0 ( 0 ) + 2 [ m 1 ] w 1 ( 0 ) + 4 [ m 1 ] w 2 ( 0 ) + + ( 2 m 2 ) [ m 1 ] w m 1 ( 0 ) = b m , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equam_HTML.gif
      where
      b k = 1 ω n Ω [ φ k ( x ) + ( k 1 ) φ k 1 ( x ) ] d S x , k = 1 , 2 , , m , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equan_HTML.gif

      and φ 0 ( x ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq72_HTML.gif. The matrix of this system is matrix A, defined by (14). As we noted above, | A | = det A = Δ 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq57_HTML.gif. By reducing to the Vandermonde determinant, it is not hard to find the value of this determinant; one has Δ 1 = 2 m 1 ( m 1 ) ! ( 2 m 4 ) ! ! 4 ! ! 2 ! ! http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq73_HTML.gif.

      Making use of Cramer’s rule, we find w 0 ( 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq74_HTML.gif from the above system of equations: w 0 ( 0 ) = | B | | A | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq75_HTML.gif, where | B | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq76_HTML.gif is the determinant of the following matrix
      ( b 1 1 1 1 b 2 2 [ 1 ] 4 [ 1 ] ( 2 m 2 ) [ 1 ] b 3 2 [ 2 ] 4 [ 2 ] ( 2 m 2 ) [ 2 ] b m 2 [ m 1 ] 4 [ m 1 ] ( 2 m 2 ) [ m 1 ] ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equao_HTML.gif
      Obviously,
      | B | = k = 1 m ( 1 ) k + 1 b k Δ k , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equap_HTML.gif

      where determinants Δ k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq77_HTML.gif, k = 1 , 2 , , m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq60_HTML.gif, are defined above. Therefore, the equality w 0 ( 0 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq71_HTML.gif holds if and only if k = 1 m ( 1 ) k + 1 b k Δ k = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq78_HTML.gif. But by the definition of b k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq79_HTML.gif, this condition is equivalent to (16).

      Thus, if the solution of the considered Neumann problem exists, then necessarily condition (16) holds.

      We now prove the converse, i.e., if (16) holds, then the solution of the Neumann problem exists.

      Let w ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq58_HTML.gif be the solution of Dirichlet problem (15). If condition (16) holds, then w ( 0 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq65_HTML.gif and we may consider the function
      u ( x ) = C + Γ 0 1 [ w ] ( x ) = C + 0 1 s 1 w ( s r θ ) d s , x = r θ , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equaq_HTML.gif

      and prove that this function is in fact the solution of the Neumann problem.

      Indeed, after changing of variable s r = ξ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq80_HTML.gif, the last integral can be written as
      u ( x ) = C + 0 r w ( ξ θ ) ξ d ξ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equar_HTML.gif
      Therefore,
      r u ( x ) r = r r ( 0 r w ( ξ θ ) ξ d ξ ) = w ( x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equ21_HTML.gif
      (21)
      In the subsequent discussions, we use formulas (17) and (21) and assume that x Ω http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq47_HTML.gif. So, we have
      u ( x ) ν = r u ( x ) r = w ( x ) = f 1 ( x ) = φ 1 ( x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equas_HTML.gif
      Further, for the second derivative one has
      2 u ( x ) ν 2 = ( r r 1 ) r u ( x ) r = ( r r 1 ) w ( x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equat_HTML.gif
      Then
      2 u ( x ) ν 2 + u ( x ) ν = ( r r 1 ) r u ( x ) r + r u ( x ) r = r w ( x ) r = w ( x ) ν = f 2 ( x ) = φ 2 ( x ) + φ 1 ( x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equau_HTML.gif
      Hence
      2 u ( x ) ν 2 = φ 2 ( x ) + φ 1 ( x ) u ( x ) ν = φ 2 ( x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equav_HTML.gif
      Using the same argument, we have for any j = 2 , 3 , , m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq81_HTML.gif
      j u ( x ) ν j = ( r r j + 1 ) ( r r 1 ) r u ( x ) r = ( r r j + 1 ) ( r r 1 ) w ( x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equaw_HTML.gif
      Consequently,
      j u ( x ) ν j + ( j 1 ) j 1 u ( x ) ν j 1 = ( r r ( j 1 ) ) ( r r ( j 2 ) ) ( r r 1 ) w ( x ) + ( j 1 ) ( r r ( j 2 ) ) ( r r 1 ) w ( x ) = j 1 w ( x ) ν j 1 = f j ( x ) = φ j ( x ) + ( j 1 ) φ j 1 ( x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equax_HTML.gif
      Therefore, finally we have by induction
      j u ( x ) ν j = φ j ( x ) + ( j 1 ) φ j 1 ( x ) ( j 1 ) j 1 u ( x ) ν j 1 = φ j ( x ) + ( j 1 ) φ j 1 ( x ) ( j 1 ) φ j 1 ( x ) = φ j ( x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equay_HTML.gif

       □

      6 The Neumann problem in the general case

      In this final section we consider Neumann problem (1), (2) in the case when both the equation and the conditions are non-homogeneous.

      Let the solution of problem (1), (2) exist and denote this solution by u ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq15_HTML.gif. Apply the operator Γ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq27_HTML.gif to u ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq15_HTML.gif and put z ( x ) = Γ 0 [ u ] ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq82_HTML.gif. Then the function z ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq83_HTML.gif is a solution of the following Dirichlet problem:
      { ( Δ ) m z ( x ) = g 1 ( x ) , x Ω , k 1 z ν k 1 ( x ) = φ k ( x ) + ( k 1 ) φ k 1 ( x ) , k = 1 , 2 , , m , x Ω , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equ22_HTML.gif
      (22)

      where g 1 ( x ) = Γ 2 m [ g ] ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq33_HTML.gif, φ 0 ( x ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq72_HTML.gif. Moreover, by definition z ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq83_HTML.gif satisfies the additional condition z ( 0 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq84_HTML.gif. Since functions g 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq85_HTML.gif and φ k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq86_HTML.gif are sufficiently smooth, then the solution of problem (22) exists and it is unique.

      Next we find the conditions to functions g ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq3_HTML.gif and φ k ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq4_HTML.gif, which guarantee the equality z ( 0 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq84_HTML.gif. To do this we present z ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq83_HTML.gif as
      z ( x ) = v ( x ) + w ( x ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equaz_HTML.gif

      where v ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq28_HTML.gif and w ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq58_HTML.gif are the considered above solutions of the corresponding Dirichlet problems (10) and (15). Obviously, z ( 0 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq84_HTML.gif if and only if v ( 0 ) + w ( 0 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq87_HTML.gif.

      We represent the functions v ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq28_HTML.gif and w ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq58_HTML.gif, according to (11) and (20), in the form
      v ( x ) = Ω G m , n ( x , y ) g 1 ( y ) d y , w ( x ) = w 0 ( x ) + r 2 w 1 ( x ) + r 4 w 2 ( x ) + + r 2 m 2 w m 1 ( x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equba_HTML.gif
      Then
      v ( 0 ) = Ω G m , n ( 0 , y ) g 1 ( y ) d y = C n , m Ω ( 1 | y | 2 ) m 1 g ( y ) d y , w ( 0 ) = w 0 ( 0 ) = 1 ω n | A | k = 1 m Ω ( 1 ) k Δ k [ φ k ( x ) + ( k 1 ) φ k 1 ( x ) ] d S x , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equbb_HTML.gif
      where
      C n , m = d n , m a n , m = ( 1 ) m Γ ( n 2 m ) π n 2 4 m ( m 1 ) ! ( n 2 m ) ( n 2 ) 2 m 1 ( m 1 ) ! . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equbc_HTML.gif
      It is not hard to see that using the formula ω n = 2 π n / 2 Γ ( n / 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq88_HTML.gif we can simplify C n , m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq89_HTML.gif and obtain
      C n , m = 1 ω n ( 1 ) m ( 2 m 1 ( m 1 ) ! ) 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equbd_HTML.gif
      Since | A | = Δ 1 = 2 m 1 ( m 1 ) ! ( 2 m 4 ) ! ! 4 ! ! 2 ! ! http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq90_HTML.gif, the condition v ( 0 ) + w ( 0 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq87_HTML.gif has the form
      Ω ( 1 | x | 2 ) m 1 g ( x ) d x = 2 m 1 ( m 1 ) ! ( 2 m 4 ) ! ! 4 ! ! 2 ! ! k = 1 m Ω ( 1 ) k Δ k [ φ k ( x ) + ( k 1 ) φ k 1 ( x ) ] d S x , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equ23_HTML.gif
      (23)

      where φ 0 ( x ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq61_HTML.gif.

      Thus, we proved the following statement on the necessary and sufficient solvability condition for the general Neumann boundary value problem.

      Theorem 3 Let φ k ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq4_HTML.gif, k = 1 , 2 , , m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq60_HTML.gif and g ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq3_HTML.gif be sufficiently smooth. Then the necessary and sufficient solvability condition for Neumann boundary value problem (1), (2) has the form (23).

      If a solution exists, then it is unique up to a constant and can be represented as
      u ( x ) = C + Γ 0 1 [ z ] ( x ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Eqube_HTML.gif

      where z ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq83_HTML.gif is the solution of Dirichlet problem (22) with the right-hand side g 1 ( x ) = Γ 2 m [ g ] ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq33_HTML.gif, which satisfies the additional condition z ( 0 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq84_HTML.gif.

      Example 1 Let us consider the biharmonic equation, i.e., m = 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq7_HTML.gif.

      In this case
      A = ( 1 1 0 2 ) , Δ 1 = 2 , Δ 2 = 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equbf_HTML.gif
      Then
      k = 1 2 Ω ( 1 ) k Δ k [ φ k ( x ) + ( k 1 ) φ k 1 ( x ) ] d S x = Ω [ φ 1 ( x ) φ 2 ( x ) ] d S x . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equbg_HTML.gif
      Therefore the solvability condition has the form
      Ω ( 1 | x | 2 ) g ( x ) d x = 2 Ω [ φ 2 ( x ) φ 1 ( x ) ] d S x , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equbh_HTML.gif

      i.e., this condition coincides with the result of the paper [3].

      Example 2 Let us consider the so-called three-harmonic equation, i.e., m = 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq91_HTML.gif.

      In this case
      A = ( 1 1 1 0 2 4 0 2 12 ) , Δ 1 = 16 , Δ 2 = 10 , Δ 3 = 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equbi_HTML.gif
      Then
      k = 1 3 Ω ( 1 ) k Δ k [ φ k ( x ) + ( k 1 ) φ k 1 ( x ) ] d S x = 2 Ω [ 3 φ 1 ( x ) 3 φ 2 ( x ) + φ 3 ( x ) ] d S x . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equbj_HTML.gif
      Therefore the solvability condition has the form
      Ω ( 1 | x | 2 ) 2 g ( x ) d x = 8 Ω [ 3 φ 1 ( x ) 3 φ 2 ( x ) + φ 3 ( x ) ] d S x . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_Equbk_HTML.gif

      Remark 1 Let 0 < λ < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq92_HTML.gif. Obviously, if g ( x ) C λ + 1 ( Ω ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq93_HTML.gif, then g 1 ( x ) = Γ 2 m [ g ] ( x ) C λ ( Ω ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq94_HTML.gif. Therefore, if we suppose that g ( x ) C λ + 1 ( Ω ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq93_HTML.gif and φ k ( x ) C λ + m + 1 k ( Ω ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq95_HTML.gif, k = 1 , 2 , , m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-162/MediaObjects/13661_2013_Article_428_IEq60_HTML.gif, then all the considered boundary value problems have solutions and these solutions are unique (see, for example, [2]).

      Declarations

      Acknowledgements

      This work has been supported by the MON Republic of Kazakhstan under Research Grant 0830/GF2 and the Ministry of Higher and Secondary Special Education of Uzbekistan under Research Grant F4-FA-F010.

      Authors’ Affiliations

      (1)
      Department of Mathematics, Akhmet Yasawi International Kazakh-Turkish University
      (2)
      Institute of Mathematics National University of Uzbekistan

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      © Turmetov and Ashurov; licensee Springer. 2013

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