On the solvability conditions of the first boundary value problem for a system of elliptic equations that strongly degenerate at a point

Boundary Value Problems20112011:16

DOI: 10.1186/1687-2770-2011-16

Received: 11 April 2011

Accepted: 22 August 2011

Published: 22 August 2011

Abstract

A system of elliptic equations which are irregularly degenerate at an inner point is considered in this article. The equations are weakly coupled by a matrix that has multiple zero eigenvalue and corresponding to it adjoint vectors. Two statements of a well-posed Dirichlet type problem in the class of smooth functions are given and sufficient conditions on the existence and uniqueness of the solutions are obtained.

Keywords

systems of elliptic equations degenerate elliptic equations boundary value problems Dirichlet type problem

1 Introduction and statement of the problems

The first results in the area of boundary value problems for an elliptic equation with degeneracy at an inner point of the considered domain are obtained in [1]. In that study, the Dirichlet problem for a weakly (regularly) degenerating elliptic equation with the main part of Laplace's operator is studied. These results are developed in [2], where the degenerate elliptic operator is generalized and, over and above, the second boundary value problem is investigated. In [3], the existence of a weak solution to the Dirichlet problem for an elliptic equation degenerating at isolated points in the class of Hölder functions is proved. In the case of the strong (irregular) degeneracy, can new effects emerge which influence the well-posedness of the boundary value problems. For instance, in [4], it is shown that in a well-posed Dirichlet type problem the asymptotic of the solution near the degeneracy point is supposed to be known. Many more difficulties come into being in the investigation of the systems of degenerate elliptic equations. Some results for weakly related degenerate elliptic systems are obtained in [57]. Particularly, these articles deal with Dirichlet type problems for the elliptic system
a ( r ) Δ u + i = 1 n B i ( x ) u x i + C ( x ) u = 0 , x D , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equ1_HTML.gif
(1)
where r = |x|, a is a continuous function such that a(r) = o(1) as r → 0, and a(r) > 0 for r > 0, x = 0 is an inner point of domain D, Δ is Laplace's operator, B i (x) and C(x) are diagonal and square matrices, consequently, which are smooth enough in D ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq1_HTML.gif. In [5, 6], the Dirichlet problem in the class of vector functions u bounded in D0 = D\{x = 0} is solved under the assumption that elements of the matrices B i (x) tend to zero, as x → 0, fast enough. In [7], a weighted Dirichlet problem with supplementary weighted condition of the shape
lim x 0 Ψ ( x ) u ( x ) - h x r = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equ2_HTML.gif
(2)

is considered under the condition a(r) = O(r2α), α >1, as r → 0. In the same study, Ψ (x) is some matrix entries of which are decreasing as x → 0, and h is a given vector function smooth on the unit sphere. It is noteworthy that the matrix C(x) is assumed to be negatively definite in D, i.e., it does not have any zero eigenvalue. Moreover, C(0) should be a normal matrix for the weighted Dirichlet problem to be well-posed. (If coefficients B i (x) have the main influence to the asymptotic of the solutions of system (1), then the last requirement is dispensable [8, 9]). Therefore, it is important to consider the case where C(0) has multiple zero eigenvalue and corresponding to it adjoint vectors.

Hence, the present article deals with a particular case of system (1) of the shape
Δ u - q ( r ) Λ u = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equ3_HTML.gif
(3)
in the ball ∑ R = {x : |x| <R}⊂ R3 with the Dirichlet condition
u | S R = f . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equ4_HTML.gif
(4)
In this article, Λ is a real constant non-negative definite N × N matrix having the eigenvalue λ = 0, q is scalar continuous function positive for r ≠ 0 and such that
q ( r ) = O ( r - 2 α ) , α > 1 , as r 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equ5_HTML.gif
(5)

S R = ∂∑ R , f = (f1, f2, ..., f N } and u = (u1, u2, ..., u N ) are the given and unknown vector functions, respectively. (Condition (5) means with respect to system (1) that a(r) vanishes as r → 0 not faster than any power of r.) Hence, the order of system (3) is strongly degenerate at the point x = 0 because of α > 1.

Let S be a non-degenerate matrix such that
S Λ S - 1 = J Λ = diag L m 0 ( λ 0 ) L m 1 ( λ 1 ) L m p ( λ p ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equa_HTML.gif
is the canonical Jordan form of Λ with m i × m i lower blocks
L m i ( λ i ) = λ i 0 0 0 1 λ i 0 0 0 0 λ i 0 0 0 1 λ i , i = 0 , p ¯ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equb_HTML.gif
Multiplying both (3) and (4) from the left by S, we get the system
Δ v - q ( r ) J Λ v = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equ6_HTML.gif
(6)
and the Dirichlet condition
v | S R = g , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equ7_HTML.gif
(7)
where v = Su, and g = Sf. Therefore, system (6) and Dirichlet condition (7) can be split into p + 1 separate systems
Δ v i - q ( r ) L m i ( λ i ) v i = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equc_HTML.gif
and Dirichlet conditions
v i | S R = g i , i = 0 , p ¯ , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equd_HTML.gif

which correspond to the blocks of Jordan matrix JΛ, where both v i and g i are m i -dimensional vector functions. If Λ is a matrix of simple structure, then all m i = 1, i.e., (6) splits into N separate equations, obviously.

Let λ0 = 0 and, for convenience, only one eigen vector corresponds to this eigen-value of Λ. Then, Re λ i < 0 for the rest i = 1 , p ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq2_HTML.gif, since the matrix Λ is non-negatively defined. As mentioned above, the solvability of a Dirichlet type problem under the condition Re λ i < 0 is investigated in [6, 7].

The main aim of this article is to give a well-posedness of the Dirichlet type problems to the system
Δ v 0 - q ( r ) L m 0 ( 0 ) v 0 = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equ8_HTML.gif
(8)
which is in accordance with eigenvalue λ0 = 0 of Λ. In order to avoid the complicated notations, instead of (8), we consider the system
Δ v - q ( r ) L s ( 0 ) v = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equ9_HTML.gif
(9)
where v = (v1, v2, ..., v s ) and L s (0) is a s × s lower Jordan block with zero diagonal entries. It is easily seen that
Δ v 1 = 0 , Δ v k + 1 = q ( r ) v k , k = 1 , s - 1 ¯ , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Eque_HTML.gif

is the outspread form of (9).

Denote Σ R 0 = Σ R \ { x = 0 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq3_HTML.gif. Let ||v|| be the Euclidian norm of a vector v. We propose the two following statements of the Dirichlet type problem to system (9).

Problem D1. Find a solution v = ( v 1 , v 2 , , v s ) C 2 ( Σ R 0 ) C Σ R 0 S R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq4_HTML.gif of Equation 9 that satisfies Dirichlet condition
v | S R = g , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equ10_HTML.gif
(10)
and relation
r v ( x ) = o ( 1 ) , a s x 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equ11_HTML.gif
(11)

Problem D2. Find a solution v C 2 ( Σ R 0 ) C Σ R 0 S R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq5_HTML.gif of Equation 9, such that it satisfies Dirichlet condition (11) and is bounded in Σ R 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq6_HTML.gif.

2 The properties of particular solutions of Equation 8

Let H n m ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq7_HTML.gif be m th the harmonic of a homogeneous harmonic polynomial of degree n,a i.e., H n m ( λ x ) = λ n H n m ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq8_HTML.gif and Δ H n m ( x ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq9_HTML.gif. Then, r - n H n m ( x ) = H n m ( ω ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq10_HTML.gif (here ω = x /r) is the m th spherical harmonic of order n continuous on the unit sphere S1. Let c nm be any constant vector, and let Q n (r) be a matrix solution of ODEs system
l n ( w ) - q ( r ) L s ( 0 ) w = 0 , 0 < r < R , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equ12_HTML.gif
(12)
where
l n = d 2 d r 2 + 2 ( n + 1 ) r d d r http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equf_HTML.gif
and w is an unknown s-dimensional vector function. Then, the functions
v n m ( x ) = r n H n m ( ω ) Q n ( r ) c n m , n = 0 , 1 , 2 , , m = 0 , ± 1 , , ± n , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equ13_HTML.gif
(13)

represent the particular solutions of system (9).

We seek for a solution Q n (r) of system (12), which satisfies condition Q(R) = E, where E is the unit matrix. To this end, on the set of functions ψ bounded on the interval (0, R), we consider the integral operator
K n ( ψ ) ( r ) = 0 R K n ( r , t ) q ( t ) ψ ( t ) d t , K n ( r , t ) = - t 2 n + 2 2 n + 1 × r - 2 n - 1 - R - 2 n - 1 , 0 < t r , t - 2 n - 1 - R - 2 n - 1 , r t R , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equg_HTML.gif
and its integer powers
K n σ ( ψ ) ( r ) = K n K n σ - 1 ( ψ ) ( r ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equh_HTML.gif
where by definition K0(ψ)(x) ≡ ψ (x). Obviously, according to this definition
K n σ ( ψ ) ( r ) = - 1 2 n + 1 r - 2 n - 1 - R - 2 n - 1 0 r t 2 n + 2 q ( t ) K n σ - 1 ( ψ ) ( t ) d t (1) + r R t 1 - t R 2 n + 1 q ( t ) K n σ - 1 ( ψ ) ( t ) d t , σ = 1 , 2 , . (2) (3) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equ14_HTML.gif
(14)
Lemma 1. Let relation (5) hold. If n > σ ( α - 1 ) - 1 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq11_HTML.gif, then
K n σ ( ψ ) ( r ) M σ n σ r 2 σ ( 1 - α ) ) on ( 0 , R ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equ15_HTML.gif
(15)

where M σ is some constant independent of n, and K n σ ( ψ ) ( R ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq12_HTML.gif (σ = 1, 2, ...).

Proof. We prove relation (15) by induction.

Since ψ (r) is bounded on (0, R), inequality (15) holds for σ = 0 with some constant M0. It follows from relation (5) that 0 <q(r) ≤ Mr-2αr ∈ (0, R), where M is a positive constant. Then, q ( t ) | ψ ( t ) | M M 0 t - 2 α http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq13_HTML.gift ∈ (0, R). Assuming that n > α - 3 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq14_HTML.gif, we obtain that
K n ( ψ ) ( r ) M 2 n + 1 r - 2 n - 1 0 r t 2 ( n + 1 - α ) d t + r R t 1 - 2 α d t M 2 n + 1 1 2 n - 2 α + 3 + 1 2 ( α - 1 ) r 2 ( 1 - α ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equi_HTML.gif
i.e., the estimate
K n ( ψ ) ( r ) M 1 n r 2 ( 1 - α ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equj_HTML.gif

with some constant M1 independent of n holds. Thus, the validity of (15) is proved for σ = 1.

Let (15) be valid for σ = k -1 under the condition n > ( k - 1 ) ( α - 1 ) - 1 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq15_HTML.gif. Then,
t 2 n + 2 q ( t ) K n k - 1 ( ψ ) ( t ) M M k - 1 n k - 1 t 2 ( n + k ( 1 - α ) ) t ( 0 , R ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equk_HTML.gif
i.e., the first integral in expression (14) converges, if n > k ( α - 1 ) - 1 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq16_HTML.gif, and
K n k ( ψ ) ( r ) M M k - 1 n k - 1 2 n + 1 r - 2 n - 1 0 r t 2 ( n + k ( 1 - α ) ) d t + r R t 2 k ( 1 - α ) - 1 d t M M k - 1 n k - 1 2 n + 1 1 2 n + 2 k ( 1 - α ) + 1 + 1 2 k ( α - 1 ) r 2 k ( 1 - α ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equl_HTML.gif

Therefore, there exists a constant M k such that (15) holds for σ = k under the condition n > k ( α - 1 ) - 1 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq17_HTML.gif.

If n > σ ( α - 1 ) - 1 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq18_HTML.gif, then the first integral on the right-hand side of (14) converges as r ∈ (0, R), and, evidently, K n σ ( ψ ) ( R ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq19_HTML.gif. ■

It is easy to verify that
l n K n σ ( ψ ) = q ( r ) K n σ - 1 ( ψ ) ( r ) , σ = 1 , 2 , , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equ16_HTML.gif
(16)

under the conditions of Lemma 1.

Note that w1 ≡ 1 and w2 = r-2n-1are linearly independent solutions of the differential equation l n (w) = 0. Thus, if w is the solution of this equation such that w(r) = o(r-2n-1) as r → 0, then w(r) ≡ const.

Denoting, as usual, by [a] the integer part of the real number a, we introduce the integer α k = k ( α - 1 ) + 1 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq20_HTML.gif, where k is a non-negative integer. (Note that α0 = 0.)

We use below denotation K n σ - 1 ( r ) = K n σ - 1 ( ψ ) ( r ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq21_HTML.gif in the case ψ (x) ≡ 1.

Theorem 1. Let relation (5) hold. If nαs-1, then there exists a unique matrix solution Q n (r) = {q nij (r)} of Equation 12 such that
q n i j ( r ) = o ( r - 2 n - 1 ) , as r 0 , i , j = 1 , s ¯ , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equ17_HTML.gif
(17)
and
Q n ( R ) = E . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equ18_HTML.gif
(18)
Proof. Let the condition nαs-1be valid. Then, according to Lemma 1, the functions K n σ ( r ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq22_HTML.gif, σ = 0 , s - 1 ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq23_HTML.gif, are continuous on the interval (0, R). Introduce the s × s matrix Q n (r) = {q nij (r)} by the formula
q n i j ( r ) = 0 , if i < j , K n i - j ( r ) , if i j . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equ19_HTML.gif
(19)
Note that estimate (15) yields the relations
q n i j ( r ) = O r 2 ( i - j ) ( 1 - α ) , as r 0 , 1 j i s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equ20_HTML.gif
(20)
This implies the validity of condition (17), because 2n+1 > 2(s -1)(α - 1) ≥ 2(i - j)(α - 1), for i , j = 1 , s ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq24_HTML.gif. Moreover,
l n ( Q n ) = q ( r ) L s ( 0 ) Q n on ( 0 , R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equm_HTML.gif

because of (16), i.e., Q n is the matrix solution of Equation 12. Evidently, equality (18) follows from (14).

It remains to prove the uniqueness of the solution of problems (12), (17), and (18). Let Q ̃ n ( r ) = { q ̃ n i j } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq25_HTML.gif be a matrix solution of system (12) continuous on (0, R), and satisfying both conditions q ̃ n i j ( r ) = o ( r - 2 n - 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq26_HTML.gif, as r → 0, and Q ̃ n ( R ) = Θ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq27_HTML.gif, where Θ is zero matrix. Then, the equalities
l n ( q ̃ n 1 j ) = 0 , l n ( q ̃ n , k + 1 , j ) = q ( r ) q ̃ n k j ( r ) , k = 1 , s - 1 ¯ , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equn_HTML.gif

on the interval (0, R) hold. Since q ̃ n 1 j ( r ) = o ( r - 2 n - 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq28_HTML.gif as r → 0, we obtain that q ̃ n 1 j ( r ) = const http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq29_HTML.gif on the interval (0, R). Then, the condition q ̃ n 1 j ( R ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq30_HTML.gif yields the identity q ̃ n 1 j ( r ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq31_HTML.gif because of the continuity of the function q ̃ n 1 j http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq32_HTML.gif on (0, R). In such a case, the elements of the second row of matrix Q ̃ n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq33_HTML.gif satisfy the equation l n ( q ̃ n 2 j ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq34_HTML.gif. For the same reason as above, we obtain that q ̃ n 2 j ( r ) 0 ( j = 1 , s ¯ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq35_HTML.gif on (0, R). Further, continuing this process, we get that q ̃ n 3 j ( r ) 0 , . . . , q ̃ n s j ( r ) 0 ( j = 1 , s ¯ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq36_HTML.gif on (0, R). Hence, Q ̃ n ( r ) Θ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq37_HTML.gif on (0, R). This yields the uniqueness of the solution of problems (12), (17), and (18).

What is the structure of the solutions of system (12) that increase slower than r-2n-1in the case where n does not satisfy the condition nαs-1? In order to get the answer to this question, we introduce s × s matrices E k = e i j ( k ) ( k = 1 , s ¯ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq38_HTML.gif with entries e i i ( k ) = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq39_HTML.gif for s - k + 1 ≤ is, and e i j ( k ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq40_HTML.gif for all rest i and j. (Note that E s = E according to this definition.) Let us compose the matrixes
Q n ( k ) ( r ) = Q n ( r ) E k , k = 1 , s ¯ , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equo_HTML.gif
where Q n is matrix elements of which are given by (19). It is easily seen that Q n ( s ) ( r ) = Q n ( r ) E = Q n ( r ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq41_HTML.gif, and the elements q n i j ( k ) ( r ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq42_HTML.gif of rest matrixes Q n ( k ) ( r ) ( k = 1 , s - 1 ¯ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq43_HTML.gif are defined by following formula
q n i j ( k ) ( r ) = K n i - j ( r ) , if s - k + 1 j i s , 0 , if 1 j s - k and i > j . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equp_HTML.gif
If nαk-1, then the powers K n i - j ( r ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq44_HTML.gif exist for all i and j such that s - k + 1 ≤ jis, and according to (20), the relation
q n i j ( k ) ( r ) = O ( r 2 ( i - j ) ( 1 - α ) ) = o ( r - 2 n - 1 ) , as r 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equ21_HTML.gif
(21)
holds. Moreover, we obtain by direct calculation that
l n ( Q n ( k ) ) = q ( r ) L s ( 0 ) Q n ( k ) on ( 0 , R ) , Q n ( k ) ( R ) = E k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equ22_HTML.gif
(22)

for k = 1 , s - 1 ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq45_HTML.gif due to the definition of matrix Q n ( k ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq46_HTML.gif. Hence, there holds the following

Theorem 2. Let relation (5) hold, and let natural k, 1 ≤ ks - 1, be such that αk+1n α k . Then, there exists a unique matrix solution Q n ( k ) ( r ) = q n i j ( k ) ( r ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq47_HTML.gif of Equation 12 such that relation (21) holds, and boundary value condition (22) is satisfied.

The uniqueness of the matrix solution Q n ( k ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq48_HTML.gif can be proved in the same way as that of the matrix solution Q n . In this case, condition (21) is essential, just similar to condition (17) in Theorem 1.

Hence, we obtain to system (9) the following set of particular solutions (see (13)):
v n m ( k ) = r R n H n m ( ω ) Q n ( k ) ( r ) c n m for n k - 1 n < α k , k = 1 , s - 1 ¯ , v n m ( s ) = r R n H n m ( ω ) Q n ( r ) c n m for n α s - 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equq_HTML.gif

where c nm is arbitrary constant column vector.

3 Existence and uniqueness of the solutions of problems D1and D2

Let us compose the superposition
v = k = 1 s - 1 α k - 1 n < α k r R n Q n ( k ) ( r ) m n H n m ( ω ) c n m + n = α s - 1 r R n Q n ( r ) m n H n m ( ω ) c n m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equ23_HTML.gif
(23)

of the particular solutions obtained above. Note, if α k 0 = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq49_HTML.gif for some k0, 1 ≤ k0s -1, then α k = 0 for all natural kk0 - 1. (Such a situation can come to exist, if α < 2.) Therefore, all the sums α k - 1 n < α k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq50_HTML.gif in (23), in which the inequality αk-1<α k is impossible, are taken to be equal to zero.

Evidently, if the series (23) converges and its sum v is twice differentiable in the spherical layer Σ R δ = { x : δ < | x | < R } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq51_HTML.gif with arbitrarily small δ, then this series satisfies system (9) in the ball Σ R 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq52_HTML.gif. Note that
v | S R = k = 1 s - 1 α k - 1 n < α k m n H n m x R E k c n m + n = α s - 1 m n H n m x R c n m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equ24_HTML.gif
(24)

due to both (18) and (22).

Assume that the boundary vector function g = (g1, g2, ..., g s ) (see (10)) is twice differentiable on unit sphere S1. Thus, it can be expressed on the sphere S R by Laplace series [10]:
g ( x ) = n = 0 m n H n m ω a n m , x S R , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equ25_HTML.gif
(25)
which converge (component-wise) uniformly and absolutely according to the assumed smoothness of the vector function g. The coefficients a n m = ( a n m ( 1 ) , a n m ( 2 ) , . . . , a n m ( s ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq53_HTML.gif in (25) can be calculated as follows b :
a n m ( i ) = 2 n + 1 4 π R 2 ( n - m ) ! ( n + m ) ! S R h i ( φ , ϑ ) Y n m ( φ , ϑ ) d φ d ϑ , m n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equr_HTML.gif

where h i (φ, ϑ) = g i (x), for |x| = R and Y n m ( φ , ϑ ) = H n m ω http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq54_HTML.gif, φ, ϑ (0 ≤ φ ≤ 2π, 0 ≤ ϑ ≤ π) are spherical coordinates which are introduced by the rule: x1 = r sin ϑcos φ, x2 = r sin ϑsin φ, and x3 = r cos ϑ.

It is easily seen that series (24) coincides with series (25), if c nm = a nm for nαs -1, and E k c nm = a nm for αk-1n <α k , k = 1 , s - 1 ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq55_HTML.gif, i.e., if components h1, h2, ..., hs-1of vector function h satisfy the following orthogonality conditions
S R h k ( φ , ϑ ) Y n m ( φ , ϑ ) d φ d ϑ = 0 for 0 n < α s - k , k = 1 , s - 1 ¯ , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equ26_HTML.gif
(26)
on sphere S R . Let us consider series (23), in which c nm = a nm :
v = k = 1 s - 1 α k - 1 n < α k r R n Q n ( k ) ( r ) m n H n m ( ω ) a n m + n = α s - 1 r R n Q n ( r ) m n H n m ( ω ) a n m . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equ27_HTML.gif
(27)

Assume that condition (26) is fulfilled in addition to the smoothness of g. Then, v | S R = g http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq56_HTML.gif, i.e., series (27) converges (component-wise) uniformly and absolutely on the sphere S R .

We shall prove that series (27) converges uniformly and absolutely in the spherical layer Σ R δ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq57_HTML.gif with arbitrarily small δ. Note that components v i ( i = 1 , s ¯ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq58_HTML.gif of the vector function v = (v1, v2, ..., v s ) in (27) can be formally represented in the form
v 1 = w 1 ( x ) , v k = w k ( x ) + W k ( x ) , k = 2 , s ¯ , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equ28_HTML.gif
(28)
where
w k ( x ) = n = α s - k r R n m n a n m ( k ) H n m ω , k = 1 , s ¯ , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equ29_HTML.gif
(29)
W k ( x ) = l = 1 k - 1 n = α s - l r R n K n k - l ( r ) m n a n m ( l ) H n m ω , k = 2 , s ¯ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equ30_HTML.gif
(30)
The terms
r R n m n a n m ( k ) H n m ω http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equs_HTML.gif

of the series on the right-hand side of (29) are harmonic functions in ∑ R . Since these series converge uniformly on the sphere S R , they also converge uniformly in ∑ R , and their sums w k (r, ω), k = 1 , s ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq59_HTML.gif, are harmonic functions in ∑ R because of Harnack's theorem [11].

Further, according to Lemma 1 estimates
K n k - l ( r ) M k - l n k - l r 2 k - l ( 1 - α ) , l = k - 1 , s + k - 1 , ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equt_HTML.gif
hold, where nαk-land Mk-lis a constant independent of n. Consequently,
r R n K n k - l ( r ) m n H n m ω a n m ( l ) < M k - l n k - l r 2 k - l ( 1 - α ) m n H n m ω a n m ( l ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equu_HTML.gif

in Σ R δ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq60_HTML.gif for ∀nαk-l. Note that the constants M k , k = 1 , s - 1 ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq61_HTML.gif, do not depend on n as well as on δ. Evidently, they yield the uniform and absolute convergence of series (30) in Σ R δ ¯ = Σ R δ S R S δ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq62_HTML.gif.

Let G δ (x, ξ) be the Green function of the Dirichlet problem to Laplace equation in Σ R δ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq63_HTML.gif, and let w kn (x) and W kn (x) be the n th partial sum of corresponding series (29) and (30). Since
Δ W 2 n ( x ) q ( r ) w 1 n ( x ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equv_HTML.gif
in Σ R 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq64_HTML.gif, relations
W 2 n ( x ) = Σ R δ G δ ( x , ξ ) w 1 n ξ d σ ξ , W k n ( x ) = Σ R δ G δ ( x , ξ ) w ( k - 1 ) n ξ + W ( k - 1 ) n ξ d σ ξ , k = 3 , s ¯ , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equw_HTML.gif
hold, where dσ ξ is a volume element of Σ R δ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq65_HTML.gif. These yield the equalities
2 W 2 n ( x ) x i 2 = Σ R δ 2 G δ ( x , ξ ) x i 2 w 1 n ξ d σ ξ , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equ31_HTML.gif
(31)
2 W k n ( x ) x i 2 = Σ R δ 2 G δ ( x , ξ ) x i 2 w ( k - 1 ) n ξ + W ( k - 1 ) n ξ d σ ξ , k = 3 , s ¯ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equ32_HTML.gif
(32)
Owing to the uniform and absolute convergence in Σ R δ ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq66_HTML.gif of sequences {w kn (r, ω)} and {W kn (r, ω)}, as n → ∞, we obtain, from (31) and (32), coherently, that the functions w k (r, ω) and W k (r, ω), defined by (29) and (30), are twice differentiable and
2 w k ( x ) x i 2 = n = α s - k 2 x i 2 r R n m n a n m ( 1 ) H n m ω , k = 1 , s ¯ , 2 W k ( x ) x i 2 = l = 1 k - 1 n = α s - l 2 x i 2 r R n K n k - l ( r ) m n a n m ( l ) H n m ω , k = 2 , s ¯ , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equx_HTML.gif

in Σ R δ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq67_HTML.gif (i = 1,2, and 3).

Hence, the vector function v = (v1, v2, ..., v s ) with the components v i defined by (28)-(30) is from class C 2 ( Σ R 0 S R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq68_HTML.gif, and it satisfies system (9) in Σ R 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq69_HTML.gif and the Dirichlet condition v | S R = g http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq70_HTML.gif, only if orthogonality conditions (26) hold. Besides, it follows from Lemma 1 that
v k ( x ) = O ( r α s - 1 - 2 ( k - 1 ) ( α - 1 ) ) , as x 0 , k = 1 , s ¯ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equy_HTML.gif
Therefore, r||v(x)|| = o(1) as x → 0, if
α s - 1 - 2 ( s - 1 ) ( α - 1 ) > - 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equ33_HTML.gif
(33)
Note that this inequality holds, if, for instance,
1 < α < 2 s - 1 2 ( s - 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equz_HTML.gif

We prove thereby the existence of the solution of problem D1, if both α and s are related by (33).

If the coefficients a n m ( k ) ( k = 1 , s - 1 ¯ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq71_HTML.gif in (28) and (29) are such that
a n m ( k ) = 0 for 0 n < 2 s - k ( α - 1 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equaa_HTML.gif
i.e., the components h k ( k = 1 , s - 1 ¯ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq72_HTML.gifof the vector function h satisfy the orthogonality conditions
S R h k ( φ , ϑ ) Y n m ( φ , ϑ ) d φ d ϑ = 0 for 0 n < 2 ( s - k ) ( α - 1 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equ34_HTML.gif
(34)
then the solution v of system (8), given by (28)-(30), is bounded in Σ R 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq73_HTML.gif and continuous in Σ ¯ R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq74_HTML.gif. Thus, under ortogonality conditions (34), we obtain the solution v = (v1, v2, ..., v s ) of problem D2 of the shape
v k ( x ) = l = 1 k n = n s - l r R n K n k - l ( r ) m n a n m ( l ) H n m ω , k = 1 , s , ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equ35_HTML.gif
(35)
where
n k = 2 k ( α - 1 ) , if 2 k ( α - 1 ) is an integer, 2 k ( α - 1 ) + 1 in the opposite case . . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equab_HTML.gif

The uniqueness of the solutions of both the problems D1 and D2 yields the following lemma.

Lemma 2. Let v = (v1, v2, ..., v s ) be a solution of problem D1 or problem D2 with the homogeneous Dirichlet condition v | S R = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq75_HTML.gif. If relation (33) holds, then v i = 0 in Σ R 0 ( i = 1 , n ¯ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq76_HTML.gif.

Proof. Assume that v = (v1, v2, ..., v s ) is a solution of problem D1. Since Δv1 = 0 in Σ R 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq77_HTML.gif and v 1 | S R = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq78_HTML.gif, we get that v1 ≡ 0 in Σ R 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq79_HTML.gif because of the relation v1(x) = o(r-1), as x → 0, which holds because of the validity of condition (11). Then, it follows from system (9) that Δv2 = 0 in Σ R 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq80_HTML.gif. Both the conditions v 2 | S R = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq81_HTML.gif and v2(x) = o(r-1), as x → 0, yield the identity v2 ≡ 0 in Σ R 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq82_HTML.gif to (11). Continuing this process, we obtain that all the components v i 0 ( i = 1 , n ¯ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq83_HTML.gif in Σ R 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq84_HTML.gif.

If v = (v1, v2, ..., v s ) is a solution of problem D2, then it satisfies (11), too. This implies the identity v ≡ 0 in Σ R 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq85_HTML.gif, without doubt.

One can summarize the reasoning given above as follows:

Theorem 3. Let gC2(S R ), and let relation (5) hold. If orthogonality conditions (26) are fulfilled, and the parameters α and s satisfy inequality (33), then there exists a unique solution v of problem D1, which can be represented by formulas (28)-(30). If orthogonality conditions (34) hold, then there exists a unique solution v of problem D2 with the components v i of the shape (35).

Endnotes

aOne can express the spherical function H n m ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq86_HTML.gif in Cartesian coordinates x = (x1, x2, x3) by formula [12]:
H n m x = r n r 2 - x 3 2 - m 2 P n m x 3 r × Re x 1 - i x 2 , if  0 m n , Im x 1 - i x 2 , if  - n m < 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equac_HTML.gif
where P n m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq87_HTML.gif is adjoint Legendre's function, and i is the imaginary unit. b Our opinion is that spherical coordinates are more convenient than Cartesian in the calculation of the coefficients a nm of series (25). The matter is such that spherical functions Y n m ( φ , ϑ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_IEq88_HTML.gif have quite a simple expresion:
Y n m ( φ , ϑ ) = P n m ( cos ϑ ) × cos m φ , - n m 0 , sin m φ , 0 < m n . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-16/MediaObjects/13661_2011_Article_19_Equad_HTML.gif

Authors’ Affiliations

(1)
Institute of Mathematics and Informatics of Vilnius University

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© Rutkauskas; licensee Springer. 2011

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