On the Mean Value Property for Polyharmonic Functions in the Ball

c Allerton Press, Inc., 2014. ISSN 1055-1344, Siberian Advances in Mathematics, 2014, Vol. 24, No. 3, pp. 169–182.  c V. V. Karachik, 2013, published in Matematicheskie Trudy, 2013, Vol. 16, No. 2, pp. 69–88. Original Russian Text 

On the Mean Value Property for Polyharmonic Functions in the Ball V. V. Karachik1* 1

South Ural State University, Chelyabinsk, 454080 Russia Received April 15, 2013

Abstract—We obtain the mean value property for the normal derivatives of a polyharmonic function with respect to the unit sphere. We find the values of a polyharmonic function and its Laplacians at the center of the unit ball expressed via the integrals of the normal derivatives of this function over the unit sphere. DOI: 10.3103/S1055134414030031 Keywords: polyharmonic function, mean value property, normal derivatives on the sphere.

1. INTRODUCTION Suppose that a function u(x) is harmonic in a domain Ω ⊂ Rn and   Br (x) = y ∈ Rn : |y − x| < r . The following mean value property for harmonic functions is well known [8]: If x ∈ Ω and Br (x) ⊂ Ω then every function harmonic in Ω satisfies the equality  1  u(y) dsy . u(x) =  ∂Br (x) ∂Br (x) This mean value property was generalized by Pisetti [1, 7] for a function u(x) which is k-harmonic in Ω, as follows:  k−1  r 2i Δi u(x0 ) 1   , u(y) ds = Γ(n/2) y ∂Br (x) ∂B (x) 4i i!Γ(i + n/2) r i=0

where Γ(α) is the Euler gamma-function. It is not hard to rewrite this property for a function u ∈ C k−1 (S) which is k-harmonic in the unit ball S ⊂ Rn :  k−1  Δi u(0) 1 u(x) dsx = , (1) ωn ∂S (2, 2)i (n, 2)i i=0   where ωn is the volume of the unit sphere ∂S and (a, b)k = a(a + b) · · · a + (k − 1)b is the generalized Pochhammer symbol with the agreement (a, b)0 = 1. For example, (2, 2)i = (2i)!!. In [6, Theorem 7], we proved a similar formula for calculating the integral of a homogeneous polynomial Qm (x) over the unit sphere: ⎧  m ∈ 2N − 1, ⎨0, Qm (x) dsx = Δm/2 Qm (x) ⎩ ωn , m ∈ 2N. |x|=1 m!! n · · · (n + m − 2) *

E-mail: [email protected]

169

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Introduce the operator Λ by the equality Λ=

n  i=1

xi

∂ . ∂xi

(2)

This operator plays an important role in further investigations because we established in [5] the following equality on ∂S: ∂k u = Λ[k] u, ∂ν k

(3)

where ν is the outward normal to the unit sphere ∂S and t[k] = t(t − 1) · · · (t − k + 1) is the factorial power of t of order k. Moreover, it is known (for example, see [2]) that if u is a harmonic function then the function P (Λ)u, where P (λ) is a polynomial, is also harmonic. The structure of the article is as follows: In Section 2, we prove the mean value property (4) for the normal derivatives with respect to ∂S of a polyharmonic function. In Sections 3–5, we study the arithmetic triangle H. This triangle appears in formula (7) which defines the value of a polyharmonic function at the center of the unit ball u(0) via the normal derivatives of this function with respect to the boundary S of the ball. The properties of the numbers hsk are studied in Lemmas 1–7. In Theorem 2 of Section 6, we prove formula (7) which is subsequently generalized to the values Δm u(0) in Theorem 4. 2. THE MEAN VALUE PROPERTY FOR NORMAL DERIVATIVES Generalize (1) to the case of the normal derivatives of a polyharmonic function with respect to ∂S. Let N0 ≡ N ∪ {0}. Theorem 1. For each m ∈ N0 and every function u ∈ C m (S) polyharmonic in the unit ball S, we have  ∞  ∂ mu (2k)[m] 1 ds = Δk u(0), (4) x ωn ∂S ∂ν m (2, 2)k (n, 2)k k=0

where ν is the outward normal to the unit sphere ∂S. Proof. In [4, Theorem 4], it was proved that, for every function u(x) polyharmonic in S, there holds the Almansi representation  ∞  1 |x|2k 1 (1 − α)k−1 n/2−1 α vk (αx) dα (5) u(x) = v0 (x) + 4k k! 0 (k − 1)! k=1

in which the harmonic functions v0 (x), . . . , vk (x), . . . in S are defined by the formula  ∞  (−1)s |x|2s 1 (1 − α)s−1 αs−1 n/2−1 k+s k α Δ u(αx) dα. vk (x) = Δ u(x) + 4s s! 0 (s − 1)! s=1

(6)

The summation limits in the sums above are infinite, and since u(x) is polyharmonic in S, the sum is in fact finite and vk (x)=0 starting from some k0 . It is easy to see that   Λ |x|2k u = |x|2k (2k + Λ)u, and therefore,

    Λ[2] |x|2k u = (Λ − 1) |x|2k (2k + Λ)u = |x|2k (2k − 1 + Λ)(2k + Λ)u,

and hence,

  Λ[m] |x|2k u = |x|2k (2k − m + 1 + Λ) · · · (2k − 1 + Λ)(2k + Λ)u = (2k)[m] u + Qm (Λ)u, SIBERIAN ADVANCES IN MATHEMATICS Vol. 24 No. 3 2014

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where Qm (λ) is a polynomial with Qm (0) = 0. Consequently, in S we have  ∞  (2k)[m] |x|2k 1 (1 − α)k−1 n/2−1 [m] [m] Λ u(x) = Λ v0 (x) + α vk (αx) dα 4k k! (k − 1)! 0 k=1  ∞  |x|2k 1 (1 − α)k−1 n/2−1 α Qm (Λ)vk (αx) dα. + 4k k! 0 (k − 1)! k=1

Using the mean value property for harmonic functions and the harmonicity of Λvk , it is not hard to obtain the equality ∂S Qm (Λ)vk (αx) dsx = 0. Thus, with the use of (6), we find  1  ∞  (2k)[m] 1 [m] Λ u(x) dsx = (1 − α)k−1 αn/2−1 dα vk (0) ωn ∂S 4k k!(k − 1)! 0 k=1

=

∞  (2k)[m] vk (0) Γ(k)Γ(n/2) k=1

∞ ∞   (2k)[m] vk (0) (2k)[m] Δk u(0) = = . 4k k!(k − 1)! Γ(k + n/2) (2, 2)k (n, 2)k (2, 2)k (n, 2)k k=1

k=0

In view of (3), this yields what is required in the theorem for m > 0. If m = 0 then, by (1), formula (4) holds in this case as well. Example 1. Suppose that u(x) is a harmonic function in S and u ∈ C ∞ (S). Then Theorem 1 implies that  ∂mu dsx = 0, m ≥ 1. m ∂S ∂ν For a function u ∈ C ∞ (S) biharmonic in S, we obtain from Theorem 1 that  ∂mu 2[m] Δu(0) = 0, m ≥ 3, dsx = ωn m 2n ∂S ∂ν since 2[m] = 0 for m ≥ 3 (see also Example 3). In the general case, if a function u(x) is k-harmonic in S and u ∈ C ∞ (S) then Theorem 1 implies that  k−1  ∂mu (2i)[m] Δi u(0) ds = ω = 0, m ≥ 2k − 1, x n m (2, 2)i (n, 2)i ∂S ∂ν i=0

because (2k −

2)[m]

= 0 for m ≥ 2k − 1. 3. THE ARITHMETIC TRIANGLE H

Prove that a function u ∈ C k−1 (S) which is k-harmonic in the unit ball S ⊂ Rn satisfies the equality  k−1 u  1 0 1 ∂u k−1 ∂ + · · · + hk (7) hk u + hk dsx , u(0) = ωn ∂S ∂ν ∂ν k−1 where ν is the outward normal to the sphere S and hik are some numbers. It is easy to see that h0k = 1 since, for a function u ∈ C(S) harmonic in S (k = 1), we have  h0k u(x) dsx = h0k u(0). u(0) = ωn ∂S If we choose u(x) biharmonic in S (k = 2) and u ∈ C 1 (S) then, by Almansi’s formula, we have u(x) = u0 (x) + |x|2 u1 (x), and hence,

  Λu(x) = Λu0 (x) + Λ |x|2 u1 (x) = Λu0 (x) + 2|x|2 u1 (x) + |x|2 Λu1 (x).

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Therefore, integrating the previous equalities over ∂S, we infer   1 1 u(x) dsx = (u0 (x) + u1 (x)) dsx = u0 (0) + u1 (0), ωn ∂S ωn ∂S      1 1 ∂u 1 dsx = Λu(x) dsx = Λu0 (x) + 2u1 (x) + Λu1 (x) dsx = 2u1 (0), ωn ∂S ∂ν ωn ∂S ωn ∂S and hence we easily find 1 u(0) = u0 (0) = ωn

 ∂S

1 ∂u u(x) − 2 ∂ν

 dsx .

The study of the numbers hsk for s = 0, k − 1 and k ∈ N is the contents of the study in Sections 3–5. For example, for k = 1, 2, we have h01 = 1 and h02 = 1, h12 = −1/2. Compose of bsk an arithmetic triangle H like Pascal’s triangle (the rows are numbered by the subscript k ∈ N0 ≡ N ∪ {0} and the oblique rows parallel to the left-hand side are numbered by the superscript s ∈ N0 but s < k): 1 1 −1/2 H=

1 h13 h23

.

(8)

1 h14 h24 h34 ··· In (7), put u(x) = |x|2i v(x), where v(x) is a function v ∈ C k−1 (S) harmonic in S such that v(0) = 0 and 0 ≤ i ≤ k − 1. Then Λu(x) = v(x)Λ|x|2i + |x|2i Λv(x) = v(x)Λ|x|2i + v1 (x), where v1 (x) = |x|2i Λv(x), and so   v1 (x) dsx = ∂S

Λv(x) dsx = ωn Λv(0) = 0. ∂S

Therefore, Λ[2] u(x) = Λ(Λ − 1)u(x) = v(x)Λ(Λ − 1)|x|2i + Λv(x)Λ|x|2i + (Λ − 1)v1 (x) = v(x)Λ(Λ − 1)|x|2i + v2 (x) = v(x)Λ[2] |x|2i + v2 (x), where v2 (x) = Λv(x)Λ|x|2i + (Λ − 1)v1 (x), and hence,       v2 (x) dsx = 2iΛv(x) + (Λ − 1)v1 (x) dsx = ωn 2iΛv(0) + (Λ − 1)v1 (0) = 0. ∂S

∂S

Consequently, Λ[s] u(x) = v(x)Λ[s] |x|2i + vs (x), where the functions vs (x) are such that ∂S vs (x) dsx = 0. Hence, by (7) and (3),    1 v(x) dsx (2i)[0] h0k v(x) + (2i)[1] h1k v(x) + · · · + (2i)[k−1] hk−1 0= k ωn ∂S   = v(0) (2i)[0] h0k + (2i)[1] h1k + · · · + (2i)[k−1] hk−1 , k and so for 1 ≤ i ≤ k − 1 we have = 0. (2i)[0] h0k + (2i)[1] h1k + · · · + (2i)[k−1] hk−1 k SIBERIAN ADVANCES IN MATHEMATICS Vol. 24 No. 3 2014

(9)



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2[0] h0k + · · · + 2[k−2] hk−2 = −2[k−1] hk−1 k k .. .

(10)

 Hence, the vector hk = h0k , h1k , . . . , hk−1 must satisfy the following system of equations: k

= −(2k − 2)[k−1] hk−1 (2k − 2)[0] h0k + · · · + (2k − 2)[k−2] hk−2 k k . Since a solution to this homogeneous system is unique up to a constant factor, we must choose a solution for which h0k = 1. For this vector hk = (1, h1k , . . . , hk−1 k ), there is fulfilled (7). The converse k−1 1 also holds, i.e., if hk = (1, hk , . . . , hk ) satisfies (10) then equalities (9) hold, and hence equality (7) is fulfilled for a k-harmonic function of the form u(x) = |x|2i v(x), where 0 ≤ i < k and v(x) is a function v ∈ C k−1 (S) harmonic in S. By Almansi’s formula, the assertion (7) is fulfilled for an arbitrary kharmonic function u ∈ C k−1 (S). Denote by Δk the determinant of the matrix on the left-hand side of (10) and designate the determinant of the matrix obtained from this matrix by replacing the (s + 1)-th column for 0 ≤ s ≤ k − 2 T  by the column 2[k−1] , . . . , (2k − 2)[k−1] as Δsk , i.e.,     [0] [s−1] [k−1] [s+1] [k−2]   2 . . . 2 2 2 . . . 2     [0] [s−1] [k−1] [s+1] [k−2]   4 . . . 4 4 4 . . . 4 s  . Δk =   .. .. .. .. ..   . . . . .     (2k − 2)[0] . . .(2k − 2)[s−1](2k − 2)[k−1](2k − 2)[s+1] . . .(2k − 2)[k−2]  It is easy to notice that    2[0]    4[0] Δk =  ..  .   (2k − 2)[0]

... ... .. . ···

     [k−2]  4  = W [2, 4, . . . , 2k − 2] = (2k − 4)!! · · · 2!!,  ..  .   [k−2]  (2k − 2) 2[k−2]

where W [λ0 , . . . , λk−2 ] stands for the Vandermonde determinant of order k − 1. Therefore, system (10) k−1 so that b0k = 1. is uniquely solvable for all bk−1 k . Choose bk Compute the determinants Δsk . Consider a more general case. Let λ0 , . . . , λk−2 ∈ R. Put      [0]  [0] [0]  [0]   λ0  λ0 . . . λk−2  . . . λk−2    ..  ..   ..  .. .. ..  .  . . . .  .     [s−1]   [s−1] [s−1]   λ0 λ[s−1] · · · λk−2  · · · λk−2    0  k−s−2  [s+1] [k−1]  [s+1]  (11) Δsk (λ) ≡ Δsk (λ0 , . . . , λk−2 ) = λ[k−1] λ0 · · · λk−2  = (−1) · · · λk−2   0   [s+1]  .. ..  .. [s+1]  λ0  . . · · · λk−2  .     ..  [k−2] ..  .. [k−2]   . λ0 . · · · λk−2  .     [k−2]  [k−1] [k−2]  [k−1]  λ0 λ0 · · · λk−2  · · · λk−2  and

  [i] Δk (λ) = det λj

i,j=0,k−2

Obviously, Δk (λ) = Δk−1 k (λ). SIBERIAN ADVANCES IN MATHEMATICS

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Lemma 1. If an integer s is such that 0 ≤ s ≤ k − 1 then k−2 

s[0] Δ0k (λ) + · · · + s[s] Δsk (λ) = −Δk (λ)

(s − λi ).

(12)

i=0

Proof. Let s be an integer such that 0 ≤ s ≤ k − 1. Consider the determinant of the form    [0] [0] [0]  s λ0 . . . λk−2   ..   .. ..  . . .    [s] [s] [s]  s λ0 · · · λk−2   . V (s; λ) =  [s+1]    0 λ[s+1] · · · λ 0 k−2    .. ..  ..  . . .    [k−1]   0 λ[k−1] · · · λk−2  0 Decompose it with respect to the first column. In view of (11) and the equality s[i] = 0 valid for i > s, we have V (s; λ) = (−1)k−2 s[0] Δ0k (λ)−(−1)k−3 s[1] Δ1k (λ)+ · · · +(−1)s (−1)k−s−2 s[s] Δsk (λ)   = (−1)k−2 s[0] Δ0k (λ) + · · · + s[s] Δsk (λ) . On the other hand, it is easy to observe that, by linear operations over the rows, the determinant V (s; λ) is reduced to the Vandermonde determinant W [s, λ0 , . . . , λk−2 ]. Therefore, V (s; λ) = W [s, λ0 , . . . , λk−2 ] =

k−2 

(λi − s)W [λ0 , . . . , λk−2 ]

i=0

=

k−2 

(λi − s)Δk (λ) = (−1)

k−1

i=0

k−2 

(s − λi )Δk (λ).

i=0

Consequently, k−1

(−1)

k−2 

  (s − λi )Δk (λ) = (−1)k−2 s[0] Δ0k (λ) + · · · + s[s] Δsk (λ) .

i=0

This implies (12). Make use of Lemma 1 to determine hsk . Lemma 2. The following equality holds for all k > 1 and s with 0 ≤ s ≤ k − 1:

[k−1] (−1)k−1 s [0] 0 [s] s −1 . s hk + · · · + s hk = (k − 1)! 2

(13)

Proof. In view of the above notation, Δsk (2, . . . , 2k − 2) = Δsk , and so, by Lemma 1 for λi = 2i + 2, we conclude that s

[0]

Δ0k

+ ··· + s

[s]

Δsk

= −Δk

k−2 

(s − 2i − 2) = −2

i=0

k−1

[k−1] s −1 Δk . 2

(14)

Hence, for s = 0 and k > 1, we obtain Δ0k = −Δk 2k−1 (−1)[k−1] = (−1)k (2k − 2)!!Δk . SIBERIAN ADVANCES IN MATHEMATICS Vol. 24 No. 3 2014

POLYHARMONIC FUNCTIONS IN THE BALL s k−1 Δk k Δk

Recalling (10), we find hsk = −h

175

. For s = 0,

h0k = −hk−1 k

Δ0k = (−1)k−1 hk−1 k (2k − 2)!!, Δk

and hence, for hk−1 = (−1)k−1 /(2k − 2)!!, we have h0k = 1. Thus, k hsk = (−1)k

Δsk . (2k − 2)!!Δk

Consequently, dividing both sides of (14) by (2k − 2)!!Δk = 0 and multiplying by (−1)k , we get (13). The proof of Lemma 2 implies that if k ≥ 1 then = hk−1 k

(−1)k−1 . (2k − 2)!!

(15)

Thus, we have determined the left- and right-hand sides of the arithmetic triangle (8) consisting of the numbers hsk . 4. FORMULAS FOR THE DETERMINATION OF hsk Lemma 3. For k ∈ N and an integer s such that 0 ≤ s ≤ k − 1, the numbers hsk are defined by the equality [k−1]

 s (−1)k+s  s j −1 (−1)j+1 . (16) hsk = s!(k − 1)! 2 j j=0

a

˜ s = s!hs and assume that Proof. Put h k k b = 0 for b > a. Then (13) can be rewritten as

[k−1]





 k−1 s s ˜s s (−1) s ˜0 k−2 ˜ −1 h + ··· + h = , h + ··· + (k − 1)! 2 s k k−2 k 1 k ˜0 , . . . , h ˜ k−2 ) is a solution to the algebraic system ˜ k = (h and hence the vector h k k  [k−1]  k−1 j (−1) −1 C˜ hk = (k − 1)! 2 j=0,k−2

with the matrix

  i C= j

.

i,j=0,k−2

It is easy to verify that the triangular matrix C is invertible and 

 −1 i+j i C = (−1) j

.

i,j=0,k−2

This follows from the equality

  k−2  s s+j i (−1) = δi,j . s j s=0 Thus, (−1)k−1 ˜ hk = (k − 1)!

 k−2 [k−1] 

  j i+j i −1 (−1) 2 j j=0

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˜ s = s!hs , we obtain Therefore, involving the fact that h k k hsk

 [k−1] k−2 j (−1)k+s  j+1 s −1 = (−1) . j s!(k − 1)! 2 j=0

This immediately implies (16). Check the so-obtained formula (16) for s = 0. We have h0k = (−1)k+1 (−1)[k−1] /(k − 1)! = 1 which coincides with what was found earlier. Calculate the unknown number h1k . For k > 1, from (16) we infer  [k−1]  

k+2 (−1) 1 (2k − 3)!! 1 [k−1] hk = . (17) (−1) − − =− 1− (k − 1)! 2 (2k − 2)!! Thus, we have found the value of the first row parallel to the left-hand side of triangle (8). For k = 2, we have the above-computed value h12 = −1/2. Let us try to transform (16) to a more compact form. Lemma 4. If k ∈ N and an integer s is such that 0 ≤ s ≤ k − 1 then

 s (k−1) (−1)k−1 1 √ s t−1 . (18) hk = s!(k − 1)! t |t=1 Proof. It is easy to see that [k−1] 

 

(k−1) j j j −1 − 1 ... − k + 1 = t(j/2)−1 = . 2 2 2 |t=1 Therefore, for integer k > 1 and s such that 0 ≤ s ≤ k − 1, in accordance with (16), we have the equalities [k−1]



 s s √ j−2 (k−1) (−1)k+s  (−1)k+1  j s j+1 s s−j s −1 (−1) = (−1) t hk = s!(k − 1)! 2 s!(k − 1)! j j |t=1 j=0 j=0 ⎛ ⎞(k−1)

  s s (k−1) √ j−2 (−1)k+1 ⎝ (−1)k−1 1 √ s−j s ⎠ = (−1) t = t−1 ; s!(k − 1)! s!(k − 1)! t j |t=1 j=0

|t=1

i.e., (18) holds for k and s mentioned above. 5. A RECURRENT FORMULA FOR hsk Transform formula (18) for the numbers hsk a little more. Prove the following auxiliary assertion. Lemma 5. Let fi ∈ C k (a, b) for i = 1, . . . , s. Then

  (k)  k (i ) = f 1 (t) · · · fs(is ) (t), f1 (t) · · · fs (t) i1 . . . is 1

t ∈ (a, b),

(19)

i1 +···+is =k

where i1 , . . . , is ∈ N0 . Proof. The formula is obvious for k = 0. If k = 1 then   f1 (t) · · · fs (t) = f1 (t)f2 (t) · · · fs (t) + f1 (t)f2 (t) · · · fs (t) + f1 (t)f2 (t) · · · fs (t), and (19) yields  i1 +···+is =1

 1 f (i1 ) (t) · · · f (is ) (t) i1 . . . is SIBERIAN ADVANCES IN MATHEMATICS Vol. 24 No. 3 2014

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= 1 · f1 (t)f2 (t) · · · fs (t) + 1 · f1 (t)f2 (t) · · · fs (t) + 1 · f1 (t)f2 (t) · · · fs (t), and hence, (19) is valid for k = 1. Suppose now that (19) holds for some k ∈ N. Prove its validity for k + 1. We have ⎛ ⎞

   (k+1)  (k)  k (i ) f1 (t) · · · fs (t) = f1 (t) · · · fs (t) =⎝ f 1 (t) · · · fs(is ) (t)⎠ i1 . . . is 1 i1 +···+is =k

   k (i +1) (i ) = f1 1 (t) · · · fs(is ) (t) + · · · + f1 1 (t) · · · fs(is +1) (t) i1 . . . is i1 +···+is =k   

 k k (i ) f1 1 (t) · · · fs(is ) (t) + ··· + = i1 . . . is − 1 i1 − 1 . . . is i1 +···+is −1=k

  k+1 (i ) f 1 (t) · · · fs(is ) (t), = i1 . . . is 1 

i1 +···+is =k+1



k = 0 if ij < 0. where we should assume that i1 ...i s With the use of Lemma 5, formula (18) makes it possible to calculate hsk for large s that is rather difficult to do using (16). For example,

 k−2 (k−1) 1 √ (−1)k−1 = t − 1 hk−2 k (k − 2)!(k − 1)! t |t=1  

 k−1 k−2 k−1 (−1)k + = k−2 2 (k − 2)!(k − 1)! 1...1 2 02...1

=

k + 2 (−1)k−2 (−1)k−2 (4 + k − 2) = . 4(2k − 4)!! 4 (2k − 4)!!

Lemma 6. The numbers hsk satisfy the recurrent equality 

s 1 s hsk + hs−1 , hk+1 = 1 − 2k 2k k

(20)

where hsk = 0 whenever s ≥ k and h0k = 1, k ∈ N. Proof. Given k ∈ N and s ∈ N0 , introduce the notation

 s (k−1) 1 √ s t−1 . ak = t |t=1 We first prove that the numbers ask satisfy the recurrent equality 

s s s − k ask + as−1 , ak+1 = 2 2 k

(21)

(22)

where ask = 0 whenever s ≥ k and a0k = (−1)k−1 (k − 1)!, k ∈ N. Study the triangle A which is analogous to (8) but consisting of the numbers ask for k ∈ N, s ∈ N0 , and s ≤ k − 1. If s = 0 or k = 1 then ask can be computed as well: a0k = (−1)k−1 (k − 1)! and as1 = 0. Moreover, by (21), we have a12 = 12 and a01 = 1. It is not hard to see that if the degree s is at least k then (21) implies that ask = 0. The value of a0k was calculated by formula (21) before the lemma. Make use relation (19) in Lemma 5 for s = s + 1 and √ f0 (t) = 1/t, fi (t) = t − 1. Since



 (m) 1 1 √ 1 (−1)m−1 − 1 ··· − m + 1 t(1/2)−m = t−1 = (2m − 3)!! t(1/2)−m , 2 2 2 2m SIBERIAN ADVANCES IN MATHEMATICS

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we infer

 s (k−1) 1 √ t−1 = t +s



i0 +i1 +···+is =k−1, i0 ≥0,i1 >0,...,is >0



i0 +i1 +···+is−1 =k−1, i0 ≥0,i1 >0,...,is−1 >0

 (2i0 )!!(2i1 − 3)!! · · · (2is − 3)!! s/2−k k−1 t (−1)k−s−1 2k−1 i0 i1 . . . is

  k−1 (2i0 )!!(2i1 − 3)!! · · · (2is−1 − 3)!! (s−1)/2−k √ (−1)k−s t t − 1 i0 i1 . . . is−1 2k−1



+ s(s − 1)

i0 +i1 +···+is−2 =k−1, i0 ≥0,i1 >0,...,is−2 >0

 k−1 (−1)k−s+1 i0 i1 . . . is−2

2 (2i0 )!!(2i1 − 3)!! · · · (2is−2 − 3)!! (s−2)/2−k √ t t − 1 + ··· k−1 2  2 √ √ s−1 (s−1)/2−k s s/2−k (s−2)/2−k + sak t t − 1 + s(s − 1)as−2 t t − 1 + · · · . (23) = ak t k ×

Therefore,

√

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

On the other hand, differentiating (23), we get 

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

s s s − k ask + as−1 . ak+1 = 2 2 k Recalling that ask = hsk (−1)k−1 s!(k − 1)!, we infer 

s s s k − k hsk (−1)k−1 s!(k − 1)! + hs−1 (−1)k−1 (s − 1)!(k − 1)!. hk+1 (−1) s!k! = 2 2 k Dividing both sides of this equality by the value (−1)k s!k!, we arrive at the relation (20). If s = 0 then 0 0 hk = ak (−1)k−1 (k − 1)! = 1, and if s ≥ k then hsk = 0. Notice that the equation ask+1 = (k − 2s + 1)ask + 12 as−1 similar to (22) was applied in [3] in conk structing special polynomials. Equation (20) defines an arithmetic triangle like Pascal’s, Euler’s, or Stirling’s triangles whose elements are however rational fractions. Computing hsk by (20), extend the triangle H of (8) to five rows: 1

H=

· · · hsk+1

1 1 − 2 5 1 1 − 8 8 1 11 3 − 1 − 16 16 48 7 1 93 29 − 1 − 128 128 192 384

.

(24)

···   = 1 − s/(2k) hsk − 1/(2k)hs−1 ··· k

Another view on the numbers hsk is given by SIBERIAN ADVANCES IN MATHEMATICS Vol. 24 No. 3 2014

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Lemma 7. The rows of H have the following property: For 0 ≤ s ≤ k − 1 and k ≥ 2, the numbers hsk are the coefficients of the expansion of the polynomial of degree (k − 1) Hk (λ) =

(−1)k−1 (λ − 2) · · · (λ − 2k + 2) (2k − 2)!!

in the factorial powers λ[s] : [k−1] [k−2] + hk−2 + · · · + h1k λ[1] + h0k . Hk (λ) = hk−1 k λ k λ

Proof. Consider the following polynomial of degree (k − 1) with respect to λ: [k−1] [k−2] Hk (λ) = hk−1 + hk−2 + · · · + h1k λ[1] + h0k . k λ k λ

Equality (13) of Lemma 2 implies that, for 0 ≤ s ≤ k − 1 and k ≥ 2, Hk (s)

=

[k−1] hk−1 k s

+ ··· +

h1k s[1]

+

h0k

=

hsk s[s]

+ ··· +

h0k s[0]

[k−1] (−1)k−1 s −1 = . (k − 1)! 2

Therefore, for even s = 2, . . . , 2k − 2, we have s/2 − 1 ∈ N0 , s/2 − 1 ≤ k − 2, and hence, by the properties of the factorial power, we obtain

[k−1] (−1)k−1 s  −1 = 0. Hk (s) = (k − 1)! 2 However, Hk (λ) has the same k − 1 zeros; therefore, Hk (λ) = CHk (λ). To find C, put s = 0 in this equality. We have 1 = h0k = Hk (0) = CHk (0) = C(−2) · · · (−2k + 2) = (−1)k−1 (2k − 2)!!C; consequently, Hk (λ) =

(−1)k−1 Hk (λ). (2k − 2)!!

6. THE VALUE OF A POLYHARMONIC FUNCTION AT THE CENTER OF THE BALL Reckoning with the above study, we have the following assertion. Theorem 2. Every function u polyharmonic in the unit ball S ⊂ Rn , u ∈ C k−1 (S), satisfies (7):   k−1 u 1 k−1 ∂ 0 1 ∂u + · · · + hk hk u + hk dsx , u(0) = ωn ∂S ∂ν ∂ν k−1 where the numbers hsk are found from (16) or (18), satisfy the recurrent equation (20), and are the coefficients of the expansion of the polynomial Hk−1 (λ) =

(−1)k−1 (λ − 2) · · · (λ − 2k + 2) (2k − 2)!!

(25)

in the factorial powers λ[s] : [k−1] [k−2] + hk−2 + · · · + h1k λ[1] + h0k . Hk−1 (λ) = hk−1 k λ k λ

(26)

Proof. For every function u ∈ C k−1 (S) polyharmonic in the unit ball S ⊂ Rn , the numbers hsk of (7) must satisfy the system of equations (10). By Lemma 3, the solutions to this system are representable in the form (16), and Lemma 4 implies that they can also be written down in the form (18). It was proved in Lemma 6 that hsk also satisfy the recurrent equation (20). Finally, by Lemma 7, we have (26). SIBERIAN ADVANCES IN MATHEMATICS

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Remark 1. In view of (26) and (3), formula (7) can be rewritten as  1 u(0) = Hk−1 (Λ)u(x) dsx . ωn ∂S Example 2. In accordance with the fourth row of the triangle H in (24), a 4-harmonic function u ∈ C 3 (S) satisfies the equality   11 ∂u 3 ∂2u 1 ∂3u 1 u− dsx . + − u(0) = ωn ∂S 16 ∂ν 16 ∂ν 2 48 ∂ν 3 Consider the polynomial (m)

Hk−1 (λ) = λ(λ − 2) · · · (λ − 2m + 2)(λ − 2m − 2) · · · (λ − 2k + 2). (0)

(0)

(27)

(m)

Obviously, Hk−1 (λ) = Hk−1 (λ)/Hk−1 (0) and Hk−1 (2m) = 0. Lemma 8. Let u(x) = u0 (x) + · · · + |x|2k−2 uk−1 (x) be the Almansi representation of a function u(x) that is k-harmonic in S and such that u ∈ C k−1 (S). If m ∈ N0 and m < k then  1 (m) Hk−1 (Λ)u(x) dsx . (28) um (0) = (m) ωn Hk−1 (2m) ∂S Proof. Suppose that λ ∈ R, i ∈ N0 , i < k, and v(x) is a function harmonic in S. It is easy to see that in S there holds the equality     (Λ − λ) |x|2i v(x) = |x|2i (2i − λ)v(x) + Λv(x) ; therefore,

    (m) (m) Hk−1 (Λ) |x|2i v(x) = |x|2i Hk−1 (2i)v(x) + Qk−1 (Λ)v(x) , (m)

where Qk−1 (λ) is a polynomial of degree (k − 1) depending on Hk−1 and such that Qk−1 (0) = 0. The function Qk−1 (Λ)v is also harmonic in S. Let Sr be the sphere of radius r centered at the origin. For every r ∈ (0, 1), we have Qk−1 (Λ)v ∈ C(S r ). Then   Qk−1 (Λ)v(x) dsx = Qk−1 (0) v(x) dsx = 0. ∂Sr

∂Sr

(m)

Therefore, if i = m then Hk−1 (2i) = 0, and hence,    2i  (m) (m) Hk−1 (Λ) |x| v(x) dsx = Hk−1 (2i) ∂Sr

 v(x) dsx + ∂Sr

Qk−1 (Λ)v(x) dsx = 0. ∂Sr

If i = m then we obtain similarly to the above:    2m  1 1 (m) (m) (m) H (Λ) |x| v(x) dsx = Hk−1 (2m) r v(x) dsx = Hk−1 (2m)v(0), ωnr ∂Sr k−1 ωn ∂Sr where ωnr is the volume of ∂Sr . Therefore, u(x) satisfies the equality   k−1    1 1 (m) (m) (m) H (Λ)u(x) ds = Hk−1 (Λ) |x|2i ui (x) dsx = Hk−1 (2m)um (0). x k−1 r r ωn ∂Sr ωn ∂Sr

(29)

i=0

(m)

Since u ∈ C k−1 (S), dividing (29) by Hk−1 (2m) = 0 and passing to the limit as r → 1, we obtain (28). SIBERIAN ADVANCES IN MATHEMATICS Vol. 24 No. 3 2014

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Theorem 3. Every function u ∈ C k (S) that is k-harmonic in the unit ball S satisfies the equality  (k) Hk (Λ)u(x) dsx = 0, ∂S

where

(k) Hk (λ)

= λ(λ − 2) · · · (λ − 2k + 2). (k)

Proof. It is easy to see that Hk (2i) = 0 for each i < m. Therefore, by equality (29) in Lemma 8, write for r ∈ (0, 1):   k−1    1 1 (k) (k) Hk (Λ)u(x) dsx = Hk (Λ) |x|2i ui (x) dsx = 0. r r ωn ∂Sr ωn ∂Sr i=0

Passage to the limit as r → 1 yields the desired equality. The following assertion is a generalization of the well-known property of harmonic functions  ∂u dsx = 0 ∂S ∂ν to polyharmonic functions. Corollary 1. If the numbers ai are found from the equality (k)

Hk (λ) = λ[k] + ak−1 λ[k−1] + · · · + a1 λ[1] + a0 then a function u that is polyharmonic in the unit ball S and such that u ∈ C k (S) satisfies the equality   ∂u ∂k u + · · · + ak k dsx = 0. a0 u + a1 ∂ν ∂ν ∂S To prove the corollary, it suffices to recall (3) and use Theorem 3. Theorem 2 can be generalized as follows. Theorem 4. Every function u ∈ C k−1 (S) polyharmonic in the unit ball S satisfies the equality  1 (2, 2)m (n, 2)m (m) Hk−1 (Λ)u(x) dsx , (30) Δm u(0) = ωn H (m) (2m) ∂S k−1 (m)

where the polynomial Hk−1 (λ) is found from (27) for m = 0, . . . , k − 1. Proof. Let u(x) = u0 (x) + · · · + |x|2k−2 uk−1 (x)

(31)

be the Almansi representation of a function u(x) that is k-harmonic in S and such that u ∈ C k−1 (S). Then (28) holds for m ∈ N0 and m < k. Moreover, if v is a function harmonic in S then (see [5])   Δ |x|2m v(x) = |x|2m−2 2m(2m + n − 2 + 2Λ)v(x). Therefore, for i < m we have m 

  Δ |x|2m v(x) = |x|2m−2i i

2j(2j + n − 2 + 2Λ)v(x),

j=m−i+1

  and hence, Δi |x|2m v(x) |x=0 = 0. If i = m then

m m     2j(2j + n − 2 + 2Λ)v(x) = 2j(2j + n − 2)v(x) + Pk (Λ)v(x) Δm |x|2m v(x) = j=1

j=1

= 2m!! n · · · (n + 2m − 2)v(x) + Pk (Λ)v(x) = (2, 2)m (n, 2)m v(x) + Pk (Λ)v(x), (32) SIBERIAN ADVANCES IN MATHEMATICS

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KARACHIK

where Pk (λ) is a polynomial of degree k, with Pk (0) = 0. Therefore,   Δm |x|2m v(x) |x=0 = (2, 2)m (n, 2)m v(0). It follows from (31) that if i > m then   Δi |x|2m v(x) = (2, 2)m (n, 2)m Δi−m v(x) + Δi−m Pk (Λ)v(x) = 0. Thus, applying the operator Δm for m ∈ N0 and m < k to (31), then putting x = 0 and using (28), we infer k−1    m Δm |x|2i ui (x) |x=0 = (2, 2)m (n, 2)m um (0) Δ u(x)|x=0 = i=0

1 (2, 2)m (n, 2)m = ωn H (m) (2m) k−1



(m)

∂S

Hk−1 (Λ)u(x) dsx .

Formula (30) is proved. Remark 2. It is easy to see that if a function u ∈ C k (S) is (k + 1)-harmonic in the unit ball then, by Theorem 4, the numbers ai of Corollary 1 satisfy the equality   (n, 2)k ∂u ∂k u k + · · · + ak k dsx . a0 u + a1 Δ u(0) = ωn ∂ν ∂ν ∂S Example 3. Suppose that u(x) is a 3-harmonic function in S and u ∈ C 2 (S). It is easy to see that (2)

H2 (λ) = λ(λ − 2) = −λ[1] + λ[2] , (2)

H2 (4) = 8, (2, 2)2 = 8, (n, 2)2 = n(n + 2), and hence,   n(n + 2) ∂u ∂ 2 u + 2 dsx . − Δ2 u(0) = ωn ∂ν ∂ν ∂S If u(x) is biharmonic in S then, by Corollary 1, we have     ∂u ∂2u ∂u ∂ 2 u + 2 dsx = 0 ⇒ dsx = dsx . − 2 ∂ν ∂ν ∂S ∂S ∂ν ∂S ∂ν REFERENCES 1. R. Dalmasso, “On the mean-value property of polyharmonic functions,” Studia Sci. Math. Hungar. 47, 113 (2010). 2. V. V. Karachik, “A problem for a polyharmonic equation in a ball,” Sibirsk. Mat. Zh. 32, 51 (1991) [Siberian Math. J. 32, 767 (1991)]. 3. V. V. Karachik, “On some special polynomials,” Proc. Amer. Math. Soc. 132, 1049 (2004). 4. V. V. Karachik, “On one representation of analytic functions by harmonic functions,” Mat. Tr. 10, 142 (2007) [Sib. Adv. Math. 18, 103 (2008)]. 5. V. V. Karachik, “Construction of polynomial solutions to some boundary value problems for Poisson’s equation,” Zh. Vychisl. Mat. Mat. Fiz. 51, 1674 (2011) [Comput. Math. Math. Phys. 51, 1567 (2011). 6. V. V. Karachik, “On some special polynomials and functions,” Sib. Elektron. Mat. Izv. 10, 205 (2013). 7. M. Nicolescu, Les Fonctions Polyharmoniques (Hermann, Paris, 1936). 8. E. M. Stein and G. Weiss, Introduction to Fourier Analysis on Euclidean Spaces (Princeton Univ. Press, Princeton, NJ, 1971).

SIBERIAN ADVANCES IN MATHEMATICS Vol. 24 No. 3 2014