On the Arithmetic Triangle Arising from the Solvability ... - Springer Link

ISSN 0001-4346, Mathematical Notes, 2014, Vol. 96, No. 2, pp. 217–227. © Pleiades Publishing, Ltd., 2014. Original Russian Text © V. V. Karachik, 2014, published in Matematicheskie Zametki, 2014, Vol. 96, No. 2, pp. 228–238.

On the Arithmetic Triangle Arising from the Solvability Conditions for the Neumann Problem V. V. Karachik* South-Ural State University, Chelyabinsk, Russia Received July 4, 2012; in final form, August 12, 2013

Abstract—We study the arithmetic triangle arising from the solvability conditions for the Neumann problem for the polyharmonic equation in the unit ball. Recurrence relations for the elements of this triangle are obtained. DOI: 10.1134/S0001434614070232 Keywords: Neumann problem, polyharmonic equation, arithmetic triangle, Vandermond determinant.

1. INTRODUCTION Consider the homogeneous Neumann problem (see, for example, [1]) for the following polyharmonic equation in the unit ball: Δk u(x) = 0,

x ∈ S,

∂iu = fi (s), ∂ν i |∂S

s ∈ ∂S,

i = 1, . . . , k,

(1)

where S = {x ∈ Rn : |x| < 1}, and ν is the unit vector of the outward normal to ∂S. A more general boundary-value problem containing polynomials of high degree in normal derivatives in the boundary conditions was studied in [2]. A theorem on the solvability of this boundary-value problem and the representation of its solution was proved there. The following statement is a particular case of the theorem from [2]. Theorem 1. For fi ∈ C k−i (∂S), i = 1, . . . , k, the Neumann problem (1) has a solution if and only if any vector bk which is the solution of the equation NkT (0)bk = 0 satisfies the condition ˆ (b1k f1 + · · · + bkk fk ) ds = 0, ∂S

where bk = (b1k , . . . , bkk ), Nk (λ) = ((λ + 2j − 2)[i] )i,j=1,...,k · C,     t[k] = t(t − 1) · · · (t − k + 1) is the factorial power and C = (−1)i ji i,j=0,...,k−1 , ji = 0 when j < i. The purpose of the present paper is to study the numbers bsk , where s = 1, . . . , k and k ∈ N, normalized so that b1k > 0 and |bkk | = 1. For example, for k = 1, 2 it is known [1], [3] that b11 = 1 and *

E-mail: [email protected]

217

218

KARACHIK

b12 = −b22 = 1. Let us arrange the numbers bsk in a form similar to the Pascal triangle [4]: 1 −1

1 B=

b13 b14

b23 b24

(2)

b33 b34

b44

......................

2. THE MATRIX Nk (0) Let us find the solutions of the homogeneous system NkT (0)bk = 0 for k > 1. Let us write out the matrix NkT (0). It is of the form NkT (0) = C T · ((2i − 2)[j] )i,j=1,...,k , where

⎛

((2i − 2)[j] )i,j=1,...,k

0 0 ⎜ ⎜ ⎜ 2[1] 2[2] =⎜ ⎜ .. .. ⎜ . . ⎝ (2k − 2)[1] (2k − 2)[2]

⎞

···

0

··· .. .

2[k]

⎟ ⎟ ⎟ ⎟. ⎟ ⎟ ⎠

.. .

· · · (2k − 2)[k]

Since the matrix C is nonsingular, we obtain a system of equations of the form = −2[k] bkk , 2[1] b1k + · · · + 2[k−1] bk−1 k .......................................... (2k −

2)[1] b1k

+ · · · + (2k −

2)[k−1] bk−1 k

= −(2k −

(3)

2)[k] bkk .

Denote by Δ0k the determinant of the matrix on the left and by Δsk the determinant of the matrix obtained from this matrix by replacing the sth column for 1 ≤ s ≤ k − 1 by the column (2[k] , . . . , (2k − 2)[k] )T , i.e.,





[1] [s−1] [k] [s+1] [k−1]



2 ... 2 2 2 ... 2





[1] [s−1] [k] [s+1] [k−1]



4 ... 4 4 4 ... 4 s

. Δk =

.. .. .. .. ..



. . . . .





(2k − 2)[1] . . . (2k − 2)[s−1] (2k − 2)[k] (2k − 2)[s+1] . . . (2k − 2)[k−1]

It is easy to see that



2[1]



4[1] 0 Δk =

..

.



(2k − 2)[1]

... ... .. . ···







1





[k−1]



4

= 2 · 4 · · · (2k − 2) 1

.. ..

. .





[k−1]

1 (2k − 2) 2[k−1]

... ... .. . ···







[k−2]

3



..

.



[k−2]

(2k − 3)

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1[k−2]

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ARITHMETIC TRIANGLE ARISING FROM SOLVABILITY CONDITIONS



1



1 = (2k − 2)!!

.

..



1

... ... .. . ···





1k−2



3k−2

= (2k − 2)!! W [1, 3, . . . , (2k − 3)]

..

.



k−2 (2k − 3)

219

= (2k − 2)!! (2k − 4)!! · · · 2!!, where W [λ1 , . . . , λk−1 ] denotes the Vandermonde determinant of order k − 1. Therefore, system (3) is uniquely solvable for any bkk . Choose a particular value of bkk = ±1 so that b1k > 0. Let us calculate the determinants Δsk . Consider a more general case. Let λ1 , . . . , λk−1 ∈ C. Denote





[1]

[1]

[1] [1]

λ1

λ1 . . . λk−1

. . . λk−1





..

.. ..

..

.. ..

.

. . . .

.





[s−1]



[s−1]

λ

λ[s−1] · · · λ[s−1]

· · · λk−1

k−1

1

1







[k] = (−1)k−s−1 [s+1] [s+1]

Δsk (λ) ≡ Δsk (λ1 , . . . , λk−1 ) = λ[k] (4)



· · · λ λ · · · λ k−1

k−1

1

1



..

. .. ..

[s+1]

. . [s+1]

λ1



· · · λk−1



[k−1] [k−1]

..

..

λ1 .. · · · λ k−1

.

. .







[k−1]

[k] [k−1]

[k]

λ1



· · · λk−1 λ1 · · · λk−1

[i]

and Δ0k (λ) = det(λj )i,j=1,...,k−1 . Lemma 2. For an integer s such that 1 ≤ s ≤ k − 1, the following equality holds: s[1] Δ1k (λ) + · · · + s[s]Δsk (λ) = −Δ0k (λ)s

k−1 

(s − λi ).

(5)

i=1

Proof. Let s be an integer such that 1 ≤ s ≤ k − 1. Consider the determinant



[1] [1] [1]



s λ · · · λ 1 k−1



.. . .. ..

. .





[s] [s] [s]

s λ1 · · · λk−1



V (s; λ) =

.

[s+1]

0 λ[s+1] · · · λk−1

1

..

.. ..

. . .



[k]

0 λ[k] · · · λk−1

1 Let us expand it along the first column. By (4) and the equality s[i] = 0, which holds for i > s, we have V (s; λ) = (−1)k−2 s[1] Δ1k (λ) − (−1)k−3 s[2] Δ2k (λ) + · · · + (−1)s−1 (−1)k−s−1 s[s]Δsk (λ) = (−1)k−2 (s[1] Δ1k (λ) + · · · + s[s]Δsk (λ)). Let us calculate V (s; λ) by another method. Let us factor out s from the first column and λi−1 from the ith column for i > 2. It is easy to see that the resulting determinant can be reduced to the Vandermonde determinant W [s, λ1 , . . . , λk−1 ] by using linear transformations over the rows. Therefore, V (s; λ) = sλ1 . . . λk−1 W [s, λ1 , . . . , λk−1 ] = s

k−1  i=1

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λi (λi − s)W [λ1 , . . . , λk−1 ]

220

KARACHIK

=s

k−1 

(λi −

s)Δ0k (λ)

k−1

= (−1)

s

i=1

k−1 

(s − λi )Δ0k (λ).

i=1

Equating the calculated values of V (s; λ), we obtain the equality (−1)k−1 s

k−1 

(s − λi )Δ0k (λ) = (−1)k−2 (s[1] Δ1k (λ) + · · · + s[s] Δsk (λ)).

i=1

Hence we immediately have (5). 3. AUXILIARY EQUALITIES FOR THE NUMBERS bsk Let us use Lemma 2 to define bsk . Lemma 3. For any integer k > 1 and s such that 1 ≤ s ≤ k − 1, the following equality holds: [k] [1] 1 [s] s k+1 k s . (6) s bk + · · · + s bk = (−1) 2 2 Proof. In view of of the above notation Δsk (2, . . . , 2k − 2) = Δsk , by Lemma 2 for λi = 2i, we have s[1] Δ1k + · · · + s[s] Δsk = −Δ0k

k−1 

(s − 2i) = −2k

i=0

[k] s Δ0k . 2

(7)

Hence, for s = 1 and k > 1, we obtain Δ1k

=

−Δ0k 2k

[k] 1 = (−1)k Δ0k (2k − 3)!!. 2

From system (3), we see that bsk = −bkk Δsk /Δ0k . For s = 1, we have b1k = −bkk

Δ1k = (−1)k+1 bkk (2k − 3)!!, Δ0k

and hence b1k > 0 for bkk = (−1)k+1 . Thus, bsk = (−1)k Δsk /Δ0k . Therefore, dividing both sides of relation (7) by Δ0k = 0 and multiplying by (−1)k , we obtain (6). It follows from the proof of Lemma 3 that b1k = (2k − 3)!!

for k > 1,

bkk

for k ≥ 1.

k+1

= (−1)

(8)

Thus, we have defined the left- and right-hand sides of the arithmetic triangle (2) consisting the numbers bsk . Lemma 4. For k ∈ N and an integer s such that 1 ≤ s ≤ k − 1, the numbers bsk satisfy the equality [k] s k  j s k+s 2 j+1 s (−1) . (9) bk = (−1) s! 2 j j=1

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ARITHMETIC TRIANGLE ARISING FROM SOLVABILITY CONDITIONS

a

221

Proof. Denote bsk = s! bsk and assume that b = 0 for b > a. Then (6) can be rewritten as

[k] s 1 s s s k−1 k+1 k s  b + ··· + b + ··· + b = (−1) 2 1 k 2 s k k−1 k and hence the vector bk = (b1k , . . . , bkk−1 ) is a solution of the system [k] j k+1 k  C bk = (−1) 2 2 j=1,...,k−1  i  with the matrix C = j i,j=1,...,k−1 . It is readily verified that the triangular matrix C is invertible and

i . C −1 = (−1)i+j j i,j=1,...,k−1 This follows from the equality (for i ≥ j) i−j i k−1   i!  (−1)k+j s s s+j i j s i (−1) (−1) = (−1) = (−1)j j! (i − j − k)! k! s j s j s=1

s=j

k=0

 (i − j)! i! = (−1)k (i − j)! j! (i − j − k)! k! i−j

=

k=0

i δi−j,0 = δi,j . j

Therefore, bk = (−1)k+1 2k

 k−1

i+j

(−1)

j=1

[k] i j 2 j i=1,...,k−1

and hence, taking into account the equality bsk = s! bsk , we obtain bsk = (−1)k+s

[k] k−1 2k  s j (−1)j+1 . s! 2 j j=1

This yields (9). Let us verify the resulting formula (9) for s = 1. We have [k] 1 k+1 k 1 = (2k − 3)!!, bk = (−1) 2 2 which coincides with the result obtained above. Let us calculate the unknown value b2k . For k > 1, using (9), we obtain [k] 1 = −b1k = −(2k − 3)!!. (10) b2k = (−1)k+2 2k 2 Thus, we have found the first row, which is parallel to the left-hand side of the triangle (2). For k = 2, we have the well-known value b22 = −1. Let us transform formula (9) to a more compact form. Lemma 5. For k ∈ N and any integer s such that 1 ≤ s ≤ k, the following relation holds: bsk = (−1)k+1 MATHEMATICAL NOTES

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2k √ (k) (( t − 1)s )|t=1 . s!

(11)

222

KARACHIK

Proof. It is easy to see that [k]



j j j j (k) − 1 ··· − k + 1 = (tj/2 )|t=1 . = 2 2 2 2 Therefore, for an integer k > 1 and any s such that 1 ≤ s ≤ k − 1, by (9), we see that the following equalities hold: [k] s s k  k  √ j (k) s k+s 2 j+1 s k+1 2 s−j s (−1) = (−1) (−1) (( t )j )|t=1 bk = (−1) s! 2 s! j j j=1 j=1

s 2k  2k √ s √ j (k) (k) = (−1)k+1 (−1)s−j = (−1)k+1 (( t − 1)s )|t=1 , ( t) s! s! j |t=1 j=1

i.e., equality (11) is valid for such k and s. Formula (11) also holds for s = k ≥ 1. Indeed, substituting s = k into (11), we obtain

(k−1) k √ k k √ k k+1 2 k (k) k+1 2 k−1 √ ( t − 1) (( t − 1) )|t=1 = (−1) bk = (−1) k! k! 2 t |t=1 k+1 k−1 √ 2 (−1) (k−1) k−1 1 √ (( t − 1)k−1 )|t=1 = −bk−1 = b1 . k−1 = (−1) t (k − 1)! Substituting s = k = 1 into (11), we obtain b11 = 1, and hence bkk = (−1)k−1 = (−1)k+1 . It was calculated above that bkk = (−1)k+1 ; therefore, formula (11) holds for s = k. In what follows, we shall rewrite the formula for the numbers bsk . But, first, let us prove the following auxiliary lemma. Lemma 6. Let f ∈ C k (a, b). Then  (f s (t))(k) = i1 +···+is =k



k f (i1 ) (t) · · · f (is ) (t), i1 · · · is

t ∈ (a, b).

(12)

Proof. For k = 0, the formula is obvious. For k = 1, we have (f s (t)) = sf s−1 (t)f  (t); using formula (12), we obtain

 1 f (i1 ) (t) · · · f (is ) (t) i1 · · · is i1 +···+is =1

= f  (t)(f (0) (t))s−1 + · · · + (f (0) (t))s−1 f  (t) = sf s−1 (t)f  (t), and hence formula (12) is valid for k = 1. Now let (12) hold for some k ∈ N. Let us prove its validity also for k + 1. We have 



k f (i1 ) (t) · · · f (is ) (t) (f s (t))(k+1) = ((f s (t))(k) ) = i1 · · · is i1 +···+is =k

 k = (f (i1 +1) (t) · · · f (is ) (t) + · · · + f (i1 ) (t) · · · f (is +1) (t)) i1 · · · is i1 +···+is =k



 k k f (i1 ) (t) · · · f (is ) (t) + ··· + = i1 − 1 · · · is i1 · · · is − 1 i1 +···+is −1=k

 k (i1 + · · · + is )f (i1 ) (t) · · · f (is ) (t) = i1 · · · is i1 +···+is =k+1

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ARITHMETIC TRIANGLE ARISING FROM SOLVABILITY CONDITIONS



=

i1 +···+is =k+1



k+1 f (i1 ) (t) · · · f (is ) (t), i1 · · · is

223

which proves the assertion. 4. RECURRENCE EQUATION For k, s ∈ N, we introduce the notation √ (k) ask = (( t − 1)s )|t=1 .

(13)

Let us study triangle A, which is similar to triangle (2), but consists of the numbers ask for k, s ∈ N and s ≤ k. If s = 0 or k = 0, then the value of ask can also be calculated: a0k = as0 = 0. In addition, by (13), we have a11 = 1/2 and a00 = 1. Lemma 7. For the numbers ask , the following recurrence equality holds:

s s − k ask + as−1 , ask+1 = 2 2 k

(14)

where ask = 0 for s > k and a0k = 0, k ∈ N, but a00 = 1. Proof. It is easy to see that if the exponent s is greater than the order k of the derivative, then it follows from (13) that ask = 0. The values of a11 and a00 can be calculated from formula (13) preceding the lemma. √ Let us use formula (12) from Lemma 6 for f (t) = t − 1. Since



√ 1 1 1 (−1)m−1 − 1 ··· − m + 1 t1/2−m = (2m − 3)!! t1/2−m , ( t − 1)(m) = 2 2 2 2m we can write √ (( t − 1)s )(k)  =

(2i1 − 3)!! · · · (2is − 3)!! s/2−k k t (−1)k−s 2k i1 · · · is i1 +···+is =k, ij >0

 (2i1 − 3)!! · · · (2is−1 − 3)!! k +s (−1)k−s+1 2k i1 · · · is−1 i1 +···+is−1 =k, ij >0 √ × t(s−1)/2−k ( t − 1)

 (2i1 − 3)!! · · · (2is−2 − 3)!! k (−1)k−s+2 + s(s − 1) 2k i1 · · · is−2 i1 +···+is−2 =k, ij >0 √ (s−2)/2−k ×t ( t − 1)2 + ··· √ √ (s−1)/2−k s−2 (s−2)/2−k t ( t − 1) + s(s − 1)a t ( t − 1)2 + · · · . = ask ts/2−k + sas−1 k k

Therefore,



√ √ (s−1)/2−k−1 t ( t − 1) + · · · . (( t − 1)s )(k+1) = ask+1 ts/2−k−1 + sas−1 k+1

On the other hand, differentiating (15), we obtain

√ √ s s s (k+1) (s−2)/2−k − k ask ts/2−k−1 + as−1 = t + ( t − 1)F (t). (( t − 1) ) 2 2 k Setting t = 1 in the previous equalities, we see that

s s s − k ask + as−1 , ak+1 = 2 2 k which proves the assertion. MATHEMATICAL NOTES

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(15)

224

KARACHIK

Note that the equation 1 s−1 a , 2 k which is similar to (14), was used in [5] to construct special polynomials. Using formulas (14), the author calculated the values of ask for several rows of triangle A that are parallel to its lateral sides and the resulting formulas yielded the values of ask . Omitting these calculations, let us prove the resulting formula. ask+1 = (k − 2s + 1)ask +

= 0, k ∈ N, and a00 = 1, the solution of the Lemma 8. Under the boundary conditions a0k = ak+1 k recurrence equation (14) is unique and has the form ask = (−1)k+s

(2k − s − 1)!s , 2k (2k − 2s)!!

(16)

where s, k ∈ N and s ≤ k. Proof. Let us prove the existence and uniqueness of the solution of Eq. (14) under the given boundary conditions 1 a11

0 a12

0 0

0

a13

0 a22

a23

0 a33

0

.........................

Step 1. In view of the boundary conditions, setting k = 0 and s = 1 in formula (14), we obtain 1 1 1 a11 = a10 + a00 = . 2 2 2 Formula (16) yields the same value. Step 2. Let us calculate the value of a1k using formula (14). Taking into account the equality a0k = 0, we obtain

1 1 2k − 3 1 (2k − 3)!! 1 1 − k + 1 a1k−1 + a0k−1 = − ak−1 = (−1)k−1 a1 ak = 2 2 2 2k−1 (2k − 3)!! (2k − 1)!! . = (−1)k−1 = (−1)k+1 k k 2 2 (2k − 1) This formula was derived for k > 1, but is also valid for k = 1. Using formula (16), we obtain the same value: (2k − 2)! (2k − 3)!! = (−1)k+1 . a1k = (−1)k+1 k 2 (2k − 2)!! 2k

Step 3. Let us calculate the value of akk using formula (14). Taking into account the equality = 0, we obtain ak+1 k

k k k k! k! k − k + 1 akk−1 + ak−1 = ak−1 = k−1 a11 = k . ak = k−1 k−1 2 2 2 2 2 This formula was obtained for k > 1, but taking into account the equality a11 = 1/2, we see that it will also be valid for k ≥ 1. By formula (16), we obtain the same value: akk =

k! (2k − k − 1)! k = k. k 2 (2k − 2k)!! 2 MATHEMATICAL NOTES

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225

Thus, the values of ask on the lateral sides of triangle A are calculated and formula (16) specifies the solution of Eq. (14) on the sides of triangle A. The structure of Eq. (14) is such that, given the values of ask on the lateral sides of triangle A, all the values of ask inside triangle A are uniquely determined by Eq. (14). Therefore, if the numbers ask defined by (16) satisfy Eq. (14) inside triangle A, then formula (16) gives the solution of Eq. (14). Let us prove this. Let ask be obtained from (16). Substituting them into the right-hand side of formula (14), we obtain

s s s − k ask + as−1 ak+1 = 2 2 k

(2k − s − 1)!s s (2k − s)!(s − 1) k+s s −k + (−1)k+s−1 k = (−1) k 2 2 (2k − 2s)!! 2 2 (2k − 2s + 2)!! (2k − s)!s ((2k − 2s + 2) + (s − 1)) = (−1)k+s−1 k+1 2 (2k − 2s + 2)!! (2k − s + 1)!s = (−1)k+s+1 k+1 , 2 (2k − 2s + 2)!! i.e., ask+1 can also be found from (16). The lemma is proved. 5. THE TRIANGLE B AND THE NEUMANN PROBLEM Let us return to triangle B. Lemma 9. The numbers bsk , s, k ∈ N, are determined by the formula

2k − s − 1 (2k − 2s + 1)!! bsk = (−1)s+1 (2k − 2s + 1) s−1

(17)

and are the unique solution of the recurrence equation bsk+1 = (2k − s)bsk − bs−1 k

(18)

under the boundary conditions b1k =

(2k − 1)!! (2k − 1)

and

bkk = (−1)k+1 ,

k ∈ N,

which agrees with (8). Proof. Using formulas (11), (13), and (16), we can write 2k (2k − s − 1)! bsk = (−1)k+1 ask = (−1)s+1 s! (2k − 2s)!!(s − 1)!

s+1 2k − s − 1 (2k − 2s + 1)!! , = (−1) (2k − 2s + 1) s−1 i.e., formula (17) specifies triangle B. On the lateral sides of triangle B, formula (17) yields the expressions

2k − 2 (2k − 1)!! 1 = (2k − 3)!!, bk = (2k − 1) 0

2k − k − 1 (2k − 2k + 1)!! = (−1)k+1 , bkk = (−1)k+1 (2k − 2k + 1) k−1 which coincide with the boundary conditions. Let us substitute the value of ask = (−1)k+1 MATHEMATICAL NOTES

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s! s b 2k k

226

KARACHIK

in Eq. (14), obtaining k+2

(−1)



s!

bs 2k+1 k+1

=

s s s! (s − 1)! s−1 − k (−1)k+1 k bsk + (−1)k+1 bk . 2 2 2 2k

After canceling (−1)k+2 s!/2k+1 , we obtain Eq. (18). The lemma is proved. Equation (18) specifies an arithmetic triangle similar to Pascal, Euler, or Stirling triangles [6]. Such triangles are studied modulo prime numbers; see, for example, [7]. Let us calculate the value of bsk for the first rows parallel to the lateral sides of triangle B, obtaining

2k − 3 (2k − 3)!! 2 = −(2k − 3)!! = −b1k , k ≥ 2, bk = − (2k − 3) 1 which corresponds to (10), bk−1 k bk−2 k



k 3!! k k = (−1) = (−1) , k−2 3 2



k+1 k−1 k + 1 5!! k−1 = (−1) 3 = (−1) , 4 k−3 5

k ≥ 2,

k

k ≥ 3.

Calculating bsk by formula (17), we expand triangle B from (2) up to six rows: 1 −1

1 −3

3 B=

−15

15 −105

105 −945

945

1

..............

−10

45 420

−1

6 −105

bsk+1 = (2k − s)bsk − bs−1 k

(19) 1

15

−1

.........

With the foregoing study taken into account, Theorem 1 can be rewritten in the following form. Theorem 10. For fi ∈ C k−i (∂S), i = 1, . . . , k, there exists a solution of the Neumann problem (1) if and only if the vector bk = (b1k , . . . , bkk ), equal to the kth row of triangle B from (19), is orthogonal to the vector ˆ

ˆ f1 (s) ds, . . . , fk (s) ds , Fk = ∂S

∂S

i.e.,

k  j+1 2k − j − 1 (−1) (2k − 2j − 1)!! fj (s) ds = 0. (bk , Fk ) = j−1 ∂S ˆ

j=1

For example, for the 4-harmonic equation, the solvability conditions for the Neumann problem are of the form ˆ (15f1 (s) − 15f2 (s) + 6f3 (s) − f4 (s)) ds = 0. ∂S

The solution of the Dirichlet problem for the 3-harmonic equation in the ball with polynomial data was given in [8]. MATHEMATICAL NOTES

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MATHEMATICAL NOTES

Vol. 96

No. 2 2014