Dirichlet and Neumann Boundary Value Problems for the Polyharmonic Equation in the Unit Ball

: In the previous author’s works, a representation of the solution of the Dirichlet boundary value problem for the biharmonic equation in terms of Green’s function is found, and then it is shown that this representation for a ball can be written in the form of the well-known Almansi formula with explicitly deﬁned harmonic components. In this paper, this idea is extended to the Dirichlet boundary value problem for the polyharmonic equation, but without invoking the Green’s function. It turned out to ﬁnd an explicit representation of the harmonic components of the m -harmonic function, which is a solution to the Dirichlet boundary value problem, in terms of m solutions to the Dirichlet boundary value problems for the Laplace equation in the unit ball. Then, using this representation, an explicit formula for the harmonic components of the solution to the Neumann boundary value problem for the polyharmonic equation in the unit ball is obtained. Examples are given that illustrate all stages of constructing solutions to the problems under consideration.


Introduction
Many works are devoted to the construction of Green's function in an explicit form for various classical boundary value problems. In [1], using the harmonic Green's function of the Dirichlet problem, the Green's functions of the Dirichlet, Neumann, and Robin biharmonic problems in a two-dimensional disk are constructed. In [2], an explicit representation of Green's function for the Robin problem for the Poisson equation is given, and in [3,4] an explicit form of Green's function for the biharmonic and three-harmonic equations in the unit ball are obtained. An explicit form of the Green function in a sector for the biharmonic and triharmonic equations is given in [5,6]. The Green's function of the Neumann problem for the Poisson equation in the half-space R n + is constructed in [7], and the Green's function for the Robin problem in the disk is studied in [8]. Explicit forms of Green's function G m (x, ξ) of the Dirichlet problem for the polyharmonic equation [9] in the unit ball S = {x ∈ R n : |x| < 1} are constructed in various ways in [10][11][12][13], and so forth. In the paper [12], an explicit representation of Green's function G m (x, ξ) depending on the parity of n and the positivity of 2m − n is also obtained, and in the paper [13], a representation of the polynomial solution of the Dirichlet problem with polynomial boundary data and on the polynomial right-hand side, which is not easy to obtain explicitly knowing the function G m (x, ξ), is presented. The solvability of various Neumann-type problems and their generalizations in the unit ball for the biharmonic and polyharmonic equation are analyzed in [14][15][16]. In [17], for the boundary value problems for the polyharmonic equation with normal derivatives in the boundary conditions, the sufficient condition for these problems to be Fredholm is obtained, and a formula for their index is given.
Of the recent works on the construction of Green's function for various problems, we note the papers [18][19][20][21][22], and on the application of Green's function to the problems of Mechanics and Physics, the works [22][23][24][25][26].
In the paper [27], integral representations of solutions and Green's functions of the Navier and Riquier-Neumann problems for the biharmonic equation in the unit ball are obtained. The following statement is proved in ( [28] [Theorem 2]). Theorem 1. Let ϕ 0 ∈ C 2+ε (∂S), ϕ 1 ∈ C 1+ε (∂S), and f ∈ C 1 (S), then a solution of the Dirichlet problem for the biharmonic equation ∆ 2 u(x) = f (x) for n > 4 or n = 3 can be represented as where G 2 (x, ξ) is the Green's function of the Dirichlet problem for the biharmonic equation in the unit ball [3], and ω n is the area of the unit sphere ∂S.
Further, in the paper [28] using the Green's function G 2 (x, ξ) from [3], the first two surface integrals in the formula (1) are calculated. It turned out that the sum of the values of these integrals can be written in the form where the harmonic functions u 0 (x) and u 1 (x) are solutions to the Dirichlet boundary value problems with boundary values ϕ 0 (s) and ϕ 1 (s) on ∂S, and the differential operator Λ has the form Λu = n ∑ k=1 x k u x k . Therefore, the formula (2) can be considered as Green's formula for the Dirichlet problem for the homogeneous biharmonic equation. Since the functions u 0 (x) and u 1 (x) can be represented in the integral form through the well-known Green's function G(x, ξ) of the Dirichlet problem for the Poisson equation (see, for example, [29]), then (2) can again be written in integral form. Note that in the papers [15,30], representations of solutions of the Dirichlet and Neumann boundary value problems for biharmonic and polyharmonic equations are also investigated. Since representation (2) does not formally use the Green's function G 2 (x, ξ), the idea arose to obtain a formula similar to (2), but without invoking Green's function, to represent the solution of the Dirichlet boundary value problem for the homogeneous m-harmonic equation in the unit ball where ν is the outward normal to the unit sphere ∂S. The implementation of this idea is the aim of the present work. Based on the auxiliary Lemmas 2-4 and Theorem 2, the main result is obtained in Theorem 4. In addition, in Theorem 6, using Lemma 5 and Theorem 4, a solution to the following Neumann boundary value problem is constructed. The required smoothness of the functions ϕ k (s), and ψ k (s), k = 0, . . . , m − 1 are given below.

Auxiliary Statements
Similarly to the representation of the solution to the Dirichlet problem in the biharmonic case (2), we take an m-harmonic function in S satisfying the boundary value conditions (4) in the form where p k (x) ∈ C m−1 (S), k = 0, . . . , m − 1 is a set of harmonic functions in S. The above required smoothness of the functions p k (x) is necessary, since the function u(x) must satisfy the boundary conditions (4). Let P k (t) = ∑ k s=0 c s t s be some polynomial. We define the "factorial polynomial" corresponding to the polynomial P k (t) by the equality x k u x k has an important boundary property and the function Λp(x) is harmonic if the function p(x) is harmonic.
To apply the operator Λ to a product of functions, the following lemma is needed.

Lemma 1 ([13]
). Let p, q ∈ C k (S), then the operator Λ satisfies the equality Let us prove the following main relation.
Let a polynomial P(λ) be written in factorial monomials Consider the following operation of "factorial" differentiation of polynomials It is easy to see that this differentiation reduces the degree of a polynomial by 1 and which means that Therefore, (12) implies and hence, we have the equality Moreover, the definition (13) of the derivative implies the following equality and therefore, coefficients of the polynomial P(λ) from (12) can be calculated by the formula The equality (15) can be easily proved by the induction. For k = 1, it coincides with the definition of the derivative. If it is true for the (k − 1)th derivative, then and hence the induction step is proved. Now let us look at the last transformation from (10). (10) is invertible in the form

Lemma 2. The last transformation from
By virtue of the binomial theorem, we have where m ∈ N. Hence, it follows that in the case i > j, we have c i,j = 0. For i = j from (16), we find Consider the following polynomials in λ where Using the polynomials h m,i (λ), in accordance with the formula (12), we can define the coefficients h (j) m,i by the following equalities By virtue of (14), we have h The second transformation from (10) can be reversed in the form Proof. Denote and prove that It is easy to see that for i = 1, . . . , m − 1 Consider the following polynomials of degree m − 1 Using the notation of the formula (18), we write and therefore, (20) can be written as It is not hard to see that for j = 0, . . . , m − 1, the equalities hold. Since g i (λ) is a polynomial of degree m − 1 and the equalities ( The lemma is proved.

Remark 1. In the paper ([32] [Theorem 4])
, it is established that the following connection between the kth (0 ≤ k ≤ m − 1) row of the matrix H m and the coefficients in the representation of the value ∆ k u(0) (u(x) is m-harmonic in S function) through the values of its normal derivatives in the form

Remark 2.
The last row of the matrix H m gives the solvability condition of the Neumann boundary value problem for the (m − 1)-harmonic equation with boundary conditions: Therefore, according to (17), Then, we obtain whence by virtue of (14) λ [3] , and therefore, the first row of the matrix H 4 has the form (1, − 11 16 , 3 16 , − 1 48 ). Similarly, we can find According to (22), for the four-harmonic function u(x), the first row of the matrix H 4 gives the equality Moreover, according to (23), the fourth row of the matrix H 4 gives the solvability condition of the Neumann boundary value problem for the three-harmonic equation Lemma 4. The first transformation from (10) can be reversed in the form Proof. By Lemma 11 from [13], we can write Thus, if P(x) = (p 0 (x), . . . , p m−1 (x)) T and R(x) = (r 0 (x), . . . , r m−1 (x)) T , then R(x) = CP(x) ⇒ P(x) = C −1 R(x). This proves the lemma.

Inverting the Main Relation
The following statement follows from Lemmas 2-4 and Theorem 2.
. . , m − 1 be some system of a harmonic in S functions and a harmonic in S functions p s (x), defined by the equalities where with convention H 0 (λ) = 1, and the jth-order derivative H (j) s (λ) of the polynomial H s (λ) is taken in the sense of definition (13). Then the m-harmonic function satisfies the boundary conditions of the Dirichlet boundary value problem (11).
Proof. Let us use Theorem 2 and consider the equalities (10). If we consistently apply Lemmas 2-4, then the equalities (10) can be inverted in the form where the numbers h (i) m,k are the elements of the matrix H m from (19). Therefore, by Theorem 2, for such polynomials p s (x), the function u(x) from (28) satisfies the Dirichlet boundary conditions (11).
Simplify the equalities (29). It is not hard to see that where ψ(i) denotes the internal sum over j in (29). Consider the following polynomial where the polynomials h m,k (λ) are defined in (17). Using (24), the polynomial h m,k (λ) can be written in the form Here, as in (17), it is denoted that h m (λ) = λ(λ − 2) · · · (λ − 2m + 2). Therefore, we have Next, we use the simple equality which is obvious for m = 0 and can be easily proved by the induction on m > 0: Thus, from (31) for m = m − 1 − s and µ = λ − 2s, we get Hence, for s = 0, we find H 0 (λ) = 1. If s = 1, . . . , m − 1, then using the value of h m,k (λ) from (17) we obtain H s (λ) = 1 (2s)!! λ(λ − 2) · · · (λ − 2s + 2), which is the same as the polynomial defined in (27). Further, we expand the polynomial H s (λ) in monomial powers λ [i] as in (14) H By the definition of H s (λ) and using (18) Substituting the found value of the sum into the equality (30) and using the definition of ψ(i), we get Reversing the order of summations and using the expansion of the polynomial H (j) s (λ) in the form (14), we have where s = 0, . . . , m − 1, which is the same as (26). Let us check the smoothness of the functions p s (x). In Theorem 2, it is required that p 0 (x), . . . , p m−1 (x) ∈ C m−1 (S). Since deg H (j) s (λ) = s − j, then for such smoothness from the formula (26), for example for s = m − 1, it follows that q j (x) ∈ C 2m−2−j (S). For s < m − 1, the smoothness conditions for q j (x) are weaker than those indicated. Hence, in Theorem 2, we can take k j = 2m − 2 − j, j = 0, . . . , m − 1. The theorem is proved. (28) is similar to the well-known Almansi formula [34,35], but in (28), the harmonic components are known.

Dirichlet Boundary Value Problem
The assertions proved above allow us to formulate the main result on the Dirichlet boundary value problem.
Proof. Let q i (x) ∈ C 2m−2−i (S), i = 0, . . . , m − 1 be a system of harmonic functions in S that are solutions to problems (34). Then, by Theorem 3 and using the properties of Green's function G m (x, ξ), the function where the harmonic functions p k (x) for k = 0, . . . , m − 1 are found from the equalities the polynomial H k (λ) has the form H k (λ) = 1 (2k)!! λ(λ − 2) · · · (λ − 2k + 2), k ∈ N, and the jth-order derivative H for i = 0, . . . , m − 1. We transform the first term in the resulting solution (35). It is easy to see that using the notation (33), we can write where it is taken into account that the jth-order derivative of H k (µ) for k < j is equal to zero. Finally, by virtue of paper ( [36] [Lemma 2.7]), in order for the harmonic in S functions q i (x) to have the smoothness q i (x) ∈ C 2m−2−i (S), it is sufficient to require that ϕ k ∈ C 2m−2−k+ε (∂S), k = 0, . . . , m − 1 for some ε > 0. Therefore, the m-harmonic function u(x) from (32) is a solution to the Dirichlet boundary value problem (3) and (4). The theorem is proved.

Example 2.
Let us find the solution to the Dirichlet boundary value problem for the three-harmonic equation in the unit ball. In this case m = 3. First, find the polynomials H 0 (λ), H 1 (λ) and H 2 (λ) by the formula (27). We have [2] − λ [1] ).
If we recall the equality λ [k] (m) = k [m] λ [k−m] , then it is not hard to find Hence, by the formula (32), using ([4] [Theorem 2]), we can write where G 3 (x, ξ) is the Green's function [4] of the Dirichlet boundary value problem for m = 3.
The possibility of recursive construction of the solution (32) to the Dirichlet boundary value problem (3) and (4) for the homogeneous polyharmonic equation is given in the following theorem.
Proof. By Theorem 4, solutions of the Dirichlet boundary value problems u m+1 (x) and u m (x) can be written as (32). Transform the difference of these solutions This equality proves the theorem. To do this, we need to use the "factorial" representation of the polynomial H 3 (λ) given in Example 1. Its coefficients are located in the last row of the matrix H 4 from (25): [3] .

Neumann Boundary Value Problem
Now consider the Neumann boundary value problem (5) and (6).

Theorem 6.
Let the harmonic functions q k (x) be solutions of the Dirichlet boundary value problems where ψ k ∈ C 2m−1−k+ε (∂S), k = 0, . . . , m − 1, ε > 0. Then, for the existence of a solution to the Neumann boundary value problem (5) and (6) ∆ m u(x) = 0, x ∈ S, it is necessary and sufficient to fulfill the condition where h . The solution of the Neumann problem can be written as where C is an arbitrary constant, and K (j) m−1 (λ; |x| 2 − 1) is determined in (33).
Proof. Let us make, in the Neumann boundary value problem, (5) and (6) be the change of the variable v = Λu. Then, since ∆Λu(x) = (Λ + 2)∆u(x), whence follows ∆ m Λu(x) = (Λ + 2m)∆ m u(x) = 0, we get the following boundary value problem for v(x) By Theorem 4, in which we formally replace the operator Λ by the operator Λ − 1, the solution to this problem can be written in the form where the harmonic functions q j (x) are solutions to the problems (38). Denote It is easy to see that since Λ(q j (0)) = 0 and K (j) m−1 (λ; |x| 2 − 1) is a polynomial in λ, then the following equalities hold true: where v 0 (x) is the m-harmonic polynomial. Now, it is necessary to solve the equation Note that for any m-harmonic in S function p(x), the equality (Λp)(0) = 0 is true, and therefore, v 1 (0) = 0. Find the value v 0 (0). It is not hard to see that which proves the induction step. Here, in the second line, the induction hypothesis is used. The lemma is proved.
If, in (42), we use the just-proved lemma, the equality H m (λ) = h m (λ)/(2m)!!, and the condition (39), then we get The converse is also true, that is, v(0) = 0 ⇒ v 0 (0) = 0 ⇒ (39). In [15], it is proved that the equation Λu = v is solvable in m-harmonic functions only if v(0) = 0, but this condition is satisfied. The solution to the equation Λu = v can be written as (40) u(x) = 1 0 v(tx) dt t + C. With smoothness imposed on the boundary functions ψ k ∈ C 2m−1−k+ε (∂S), k = 0, . . . , m − 1, similarly to the Theorem 4, we have q k ∈ C 2m−1−k (∂S), and hence, since in the formula (32) the differential operator applied to the function q k has the order m − 1 − k, then v = Λu ∈ C m (S), and by (40) u ∈ C m (S). Thus, the m-harmonic function (40) is a solution to the Neumann boundary value problem (5)- (6). If a solution to the Neumann boundary value problem u(x) exists, then the function v(x) from (41) must satisfy the equality Λu = v, which is possible only if v(0) = 0, whence follows equality (39). The solution u(x) is unique up to a constant. The theorem is proved. where h m+1,m (λ) are elements of the last row of the matrix H m+1 . If we notice that h m+1,m (λ) = h m (λ)/(2m)!! and take into account that for the harmonic functions q j (x), the equality 1 ω n ∂S ψ j ds ξ = q j (0) holds, then the condition (23) for the m-harmonic equation takes the form This condition is equivalent to the condition (39). The lemma is proved.

Conclusions
It is well-known (see [37]) that, if a function u is m-harmonic in a star domain, then it can be represented by Almansi's expansion u(x) = ∑ m−1 k=0 p k (x)|x| 2k , where p k (x), k = 0, . . . , m − 1 are harmonic functions. In the present work, we managed to slightly transform this expansion and find the functions p k (x) in an explicit form for solving the Dirichlet and Neumann problems in the unit ball. As far as we know, such a representation of solutions of boundary value problems for elliptic equations of arbitrary order 2m is obtained for the first time.
The representations of solutions of the Dirichlet and Neumann problems obtained in Theorems 4 and 6 have advantages over formulas like (1) using the Green's function of the problem, in their relative simplicity (no need to calculate singular integrals of Green's function) and transparency (the dependence of the solution to the problem on each of the boundary functions is visible).
The proposed method can be used to represent solutions of other types of boundary value problems for the polyharmonic equation, and can also be used when applying the Adomian decomposition method for nonlinear boundary value problems. The presented method is also useful for constructing solutions to model boundary value problems for computational purposes. As an extension of the results obtained to other differential equations, we can mark an equation of the form λ 1 u x 1 x 1 + . . . + λ n u x n x n = 0, since an analogue of the Almansi formula is known for it [35].