Table Of Contentd2m
THE DISCRETE ANALOGUE OF THE OPERATOR AND
dx2m
ITS PROPERTIES
Kh.M.Shadimetov
0 Abstract
1
0 In this paper the discrete analogue Dm[β] of the differential operator d2m/dx2m is constructed
2 and its some new properties are proved.
n
a Key words and phrases: Discrete function, discrete analogue of the differential operator, Euler
J
polynomial.
4
] 1 Main results.
A
N
(m)
First S.L.Sobolev [1] studied construction and investigated properties of the operator D [β], which
. hH
h is inverse of the convolution operator with function G(m)[β] = hnG (hHβ). The function D(m)[β] of
t hH m hH
a
discrete variable, satisfying the equality
m
[ hnD(m)[β]∗G(m)[β] = δ[β]
hH hH
1
v is called by the discrete analogue of the polyharmonic operator ∆m. S.L.Sobolev suggested an algo-
6
(m)
5 rithm for finding function DhH [β] and proved several properties of this function. In one dimensional
5 d2m
0 case, i.e. the discrete analogue of the operator was constructed by Z.Zh.Zhamalov [2, 3]. But
dx2m
.
1 there the form of this function was written with m+1 unknown coefficients. In works [4, 5] these
0 d2m
0 coefficients were found, hereunder the discrete analogue of the operator was constructed com-
1 dx2m
pletely.
:
v
In this paper we give the results of works [4-7], concerning to construction of the discrete analogue
i
X d2m
D [β] of the operator , and discovery of its properties, which early were not known.
r m dx2m
a
Following statements are valid.
d2m
Theorem 1. The discrete analogue of the differential operator have following form
dx2m
m−1 (1−λ )2m+1λ|β|
k k for |β| ≥ 2,
λ E (λ )
k 2m−1 k
k=1
Dm[β] = (2mh2−m1)! 1X+mXk=−11 (1E−2mλ−k1)(2λmk+)1 for |β| = 1, (1)
m−1 (1−λ )2m+1
−22m−1+ k for β = 0,
λkE2m−1(λk)
k=1
X
where E (λ) is the Euler polynomial of degree α, λ are the roots of the Euler polynomial E (λ),
α k 2m−2
in module less than unity, i.e. |λ | < 1, h is the step of the lattice.
k
1
Property 1. The discrete analogue D [β] of the differential operator of order 2m have represen-
m
tation
m−1 |β|+m−2
(2m−1)! λ
D [β] = ∆[m][β]∗ k ,
m h2m 2 E′ (λ )
k=1 2m−2 k
X
m
2m
where ∆[m][β] = (−1)k+m δ[β−k] is symmetric difference of order 2m.
2 m+k
k=−m (cid:18) (cid:19)
X
Property 2. The operator D [β] and monomials [β]k = (hβ)k are connected as
m
0 for 0 ≤ k ≤ 2m−1,
D [β][β]k = (2)
m
(2m)! for k = 2m,
β (cid:26)
X
0 for 2m+1 ≤ k ≤ 4m−1,
D [β][β]k = h2m(4m)!B (3)
m 2m
for k = 4m.
Xβ (2m)!
Property 3. The operator D [β] and the function exp(2πihpβ) connected as
m
D [β]exp(2πihpβ) =
m
β
X
(−1)m22m(2m−1)!h−2msin2m(πhp)
,
m−2
(2m−2) (2m−2)
2 a cos2πhp(m−1−k)+a
k m−1
k=0
P
where
k
2m
a(2m−2) = (−1)j (k+1−j)2m−1
k j
j=0 (cid:18) (cid:19)
X
are the coefficients of the Euler polynomial E (λ).
2m−2
2 Lemmas.
As known, Euler polynomials E (λ) have following form
k
λ
λE (λ) = (1−λ)k+2Dk , (4)
k (1−λ)2
where
d d
D = λ , Dk = λ Dk−1.
dλ dλ
(k)
In [8] was shown, that all roots λ of the Euler polynomial E (λ) are real, negative and different:
j k
(k) (k) (k)
λ < λ < ... < λ < 0. (5)
1 2 k
2
Furthermore, the roots, equal standing from the ends of the chain (5) mutually inverse:
(k) (k)
λ ·λ = 1. (6)
j k+1−j
k
If we denote E (λ) = a(k)λs, then the coefficients a(k) of Euler polynomials, as this was shown by
k s s
s=0
Euler himself, are exprPessed by formula
s
k+2
a(k) = (−1)j (s+1−j)k+1.
s j
j=0 (cid:18) (cid:19)
X
From the definition E (λ) follow following statement.
k
Lemma 1. For polynomial E (λ) following recurrence relation is valid
k
E (λ) = (kλ+1)E (λ)+λ(1−λ)E′ (λ), (7)
k k−1 k−1
where E (λ) = 1, k = 1,2,....
0
Lemma 2. The polynomial E (λ) satisfies the identity
k
1
E (λ) = λkE , (8)
k k
λ
(cid:18) (cid:19)
(k) (k)
or otherwise a = a , s = 0,1,2,...,k.
s k−s
Proof of lemma 1. From (4) we can see, that
λ
E (λ) = λ−1(1−λ)k+1Dk−1 . (9)
k−1 (1−λ)2
Differentiating by λ the polynomial E (λ), we get
k−1
λ E (λ)
E′ (λ) = −(1−λ)kλ−2(kλ+1)Dk−1 + k .
k−1 (1−λ)2 λ(1−λ)
Hence and from (9) we obtain, that
λ
(kλ+1)E (λ)+λ(1−λ)E′ (λ) = (kλ+1)λ−1(1−λ)k+1Dk−1 −
k−1 k−1 (1−λ)2
λ
−(1−λ)k+1λ−1(kλ+1)Dk−1 +E (λ) = E (λ).
(1−λ)2 k k
So, lemma 1 is proved.
Proof of lemma 2. The lemma we will prove by induction method. When k = 1 from (4) we
find
E (λ) = λ+1.
1
(k−1) (k−1)
We suppose, that when k ≥ 1 the equality a = a , n = 0,1,...,k−1 is fulfilled. We assume,
n k−1−n
(k−1)
that a = 0 for n < 0 and n > k−1.
n
3
From (7) we have
a(k) = (s+1)a(k−1)+(k−s+1)a(k−1),
s s s−1
then, using assumption of induction, we get
a(k) = (k−s+1)a(k−1)+(s+1)a(k−1) = (k−s+1)a(k−1)+(s+1)a(k−1) = a(k),
k−s k−s k−s−1 s−1 s s
and lemma 2 is proved.
3 Proof of theorem 1
For this we will use function
x2m−1signx
G (x) = .
m
2·(2m−1)!
To this function we correspond following function of discrete argument:
(hβ)2m−1sign(hβ)
G [β] = .
m
2·(2m−1)!
Here we must find such function D [β], which satisfies the equality
m
hD [β]∗G [β] = δ[β]. (10)
m m
According to the theory of periodic generalized functions and Fourier transformation in them
instead of function D [β] it is convenient to search harrow shaped function [1]
m
↽⇁
Dm (x) = Dm[β]δ(x−hβ).
β
X
The equality (10) in the class of harrow shaped functions goes to equation
↽⇁ ↽⇁
h Dm (x)∗ Gm (x) = δ(x), (11)
where
↽⇁
Gm (x) = Gm[β]δ(x−hβ).
β
X
It is known [1], that the class of harrow shaped functions and the class of functions of discrete vari-
ables are isomorphic. So instead of function of discrete argument D [β] it is sufficiently to investigate
m
↽⇁
the function Dm (x), defining from equation (11).
Later on we need following well known formulas of Fourier transformation:
F[f(p)] = f(x)exp(2πipx)dx,
Z
F−1[f(p)] = f(x)exp(−2πipx)dx,
Z
4
F[f(x)∗ϕ(x)] = F[f(x)]·F[ϕ(x)],
F[δ(x)] = 1.
Applying to both parts of (11) Fourier transformation, we get
↽⇁ ↽⇁
F[Dm (x)]·F[h Gm (x)] = 1. (12)
↽⇁
Fourier transform of h Gm (p) is well known periodic function, given in R with period h−1
↽⇁ (−1)m 1
F[h Gm (x)] = (2π)2m |p−h−1β|2m, p 6= h−1β. (13)
β
X
This formula is obtained from the equalities
(−1)m 1
F[G (p)] = ( [1, p. 729])
m (2π)2m|p|2m
and
↽⇁
Gm (x) = Gm(x) δ(x−hβ).
β
X
Hence, taking into account (12), we get
−1
↽⇁ (−1)m 1
F[Dm (p)] = (2π)2m |p−h−1β|2m . (14)
β
X
Themainpropertiesofthisfunctioninmultidimensionalcase,appearinginconstructionofdiscrete
analogue of the polyharmonic operator, were investigated in [1].
We give some of them, which we will use later on.
↽⇁
1. Zeros of the function F[Dm (p)] are the points p = h−1β.
↽⇁
2. The function F[Dm (p)] is periodic with period h−1, real and analytic for all real p.
↽⇁
The function F[Dm (p)] can be represented in the form of Fourier series
↽⇁
F[Dm (p)] = Dˆm[β]exp(2πihβp), (15)
β
X
where
h−1
↽⇁
Dˆm[β] = F[Dm (p)]exp(−2πihβp)dp. (16)
Z
0
Applying inverse Fourier transformation to the equality (15), we get harrow shaped function
↽⇁
Dm (x) = Dˆm[β]δ(x−hβ). (17)
β
X
5
Thus, Dˆ [β] is searching function D [β] of discrete argument or discrete analogue of the operator
m m
d2m
. For finding the function Dˆ [β] calculation of the integral (16) inadvisable. We will find it by
dx2m m
following way.
By virtue of known formula
1 π2
=
(p−β)2 sin2πp
β
X
and from the formula (13) we get
↽⇁ −1 1 −h2
F[h G1 (p)] = (2π)2 (p−h−1β)2 = 4sin2πph.
β
X
Hence by differentiating we have
d ↽⇁ 2 1
dpF[h G1 (p)] = (2π)2 (p−h−1β)3.
β
X
Thus continuing further, we obtain
d2m−2 ↽⇁ (2m−1)! 1
dp2m−2F[h G1 (p)] = − (2π)2m (p−h−1β)2m =
β
X
↽⇁
= (−1)m−1(2m−1)!(2π)2m−2F[h Gm (p)].
So,
↽⇁ (−1)mh2 d2m−2 1
F[h Gm (p)] = 22mπ2m−2(2m−1)!dp2m−2 sin2πhp .
(cid:18) (cid:19)
Consider, the expression
d2m−2 1
.
dp2m−2 sin2πhp
(cid:18) (cid:19)
Using
exp(πihp)−exp(−πihp)
sinπhp = ,
2i
we have
d2m−2 −4 d2m−2 exp(2πihp)
= −4 .
dp2m−2 (exp(πihp)−exp(−πihp))2 dp2m−2 (exp(2πihp)−1)2
(cid:18) (cid:19) (cid:18) (cid:19)
We will do change of variables λ = exp(2πihp), then in view of that
d dλ d d d
= and = 2πihλ ,
dp dpdλ dp dλ
we get
d2m−2
= (2πih)2m−2D2m−2,
dp2m−2
6
where
d d
D = λ , D2m−2 = λ D2m−3.
dλ dλ
Thus,
↽⇁ h2m λ
F[h Gm (p)] = (2m−1)!D2m−2(1−λ)2.
Hence in virtue of (4) we have
↽⇁ h2m λE2m−2(λ)
F[h Gm (p)] = (2m−1)! (1−λ)2m . (18)
From (18), according to (12), we obtain
↽⇁ (2m−1)! (1−λ)2m
F[Dm (p)] = h2m λE (λ). (19)
2m−2
Now in order to obtain Fourier-series expansion, we will do following.
We divide the polynomial (1−λ)2m to the polynomial λE (λ):
2m−2
(1−λ)2m P (λ)
(2m−2) 2m−2
= λ−2m−a + , (20)
2m−2 2m−3 λE (λ)
λ a(2m−2)λs 2m−2
s
s=0
P
where P (λ) is a polynomial of degree 2m−2. It is not difficult to see, that the rational fraction
2m−2
P (λ)
2m−2
is proper fraction, i.e. degree of the polynomial P (λ) is less than degree of the
2m−2
λE (λ)
2m−2
polynomial λE (λ). Since the roots of the polynomial E (λ) are real and different, then the
2m−2 2m−2
P (λ)
2m−2
rational fraction is expanded to the sum of elementary fractions. Searching expansion has
λE (λ)
2m−2
following form
m−1 m−1
P (λ) A A A
2m−2 0 1,k 2,k
= + + , (21)
λE (λ) λ λ−λ λ−λ
2m−2 1,k 2,k
k=1 k=1
X X
where A , A , A are unknown coefficients, λ are the roots of the polynomial E (λ), in
0 1,k 2,k 1,k 2m−2
modulus less than unity, and λ are the roots of the polynomial E (λ), in modulus greater than
2,k 2m−2
unity. By (21) the equality (20) takes the form
(1−λ)2m A
(2m−2) 0
= λ−2m−a + +
2m−2 2m−3 λ
λ a(2m−2)λs
s
s=0
P
m−1 m−1
A A
1,k 2,k
+ + . (22)
λ−λ λ−λ
1,k 2,k
k=1 k=1
X X
Reducing to the common denominator and omitting it, we get
(1−λ)2m = λ2E (λ)−λ(2m+a(2m−2))E (λ)+
2m−2 2m−3 2m−2
7
m−1 m−1
A λE (λ) A λE (λ)
1,k 2m−2 2,k 2m−2
+A E (λ)+ + . (23)
0 2m−2
λ−λ λ−λ
1,k 2,k
k=1 k=1
X X
Assuming in the equality (23) consequently λ = 0, λ = λ and λ = λ , we find
1,k 2,k
1 = E (0)A ; (1−λ )2m = λ E′ (λ )A ;
2m−2 0 1,k 1,k 2m−2 1,k 1,k
(1−λ )2m = λ E′ (λ )A .
2,k 2,k 2m−2 2,k 2,k
Hence
(1−λ )2m (1−λ )2m
1,k 2,k
A = 1; A = ; A = .
0 1,k λ E′ (λ ) 2,k λ E′ (λ )
1,k 2m−2 1,k 2,k 2m−2 2,k
Using (6), we have
A = (1− λ21,k)2m = (λ2,k −1)2m .
1,k λ−1E′ ( 1 ) λ2m−1E′ ( 1 )
2,k 2m−2 λ2,k 2,k 2m−2 λ2,k
In virtue of (7) we obtain
1 1 1 1
1− E′ ( ) = E ( ),
λ λ 2m−2 λ 2m−1 λ
2,k (cid:18) 2,k(cid:19) 2,k 2,k
λ (1−λ )E′ (λ ) = E (λ ),
2,k 2,k 2m−2 2,k 2m−1 2,k
hence
E ( 1 )λ2
E′ ( 1 ) = 2m−1 λ2,k 2,k,
2m−2 λ λ −1
2,k 2,k
E (λ )
E′ (λ ) = 2m−1 2,k .
2m−2 2,k λ (1−λ )
2,k 2,k
From here application of the lemma 2 gives
−A (1−λ )2m+1
2,k 1,k
A = , A = . (24)
1,k λ2 1,k E (λ )
2,k 2m−1 1,k
Since |λ | < 1 and |λ | > 1, then
1,k 2,k
m−1 m−1
A A
1,k 2,k
and
λ−λ λ−λ
1,k 2,k
k=1 k=1
X X
can be represented as Laurent series on the circle |λ2| = 1:
m−1 m−1 m−1 ∞ β
A 1 A 1 λ
1,k 1,k 1,k
= = A , (25)
k=1 λ−λ1,k λ k=1 1− λ1λ,k λ k=1 1,kβ=0(cid:18) λ (cid:19)
X X X X
m−1 m−1 m−1 ∞ β
A A A λ
2,k 2,k 2,k
= − = − . (26)
Xk=1 λ−λ2,k Xk=1 λ2,k(1− λ2λ,k) Xk=1 λ2,k βX=0(cid:18)λ2,k(cid:19)
8
Putting (25), (26) to (22) and taking into account λ = exp(2πihp) from (19), (20), we obtain
↽⇁ (2m−1)! (2m−2)
F[Dm (p)] = h2m exp(2πihp)−2m−a2m−3 +
"
m−1 ∞
β
+exp(−2πihp)+ A λ exp(−2πihp(β +1))−
1,k 1,k
k=1 β=0
X X
∞ β
1
−A λ−1 exp(2πihpβ) .
2,k 2,k λ
β=0(cid:18) 2,k(cid:19) !#
X
↽⇁
Thus, searching Fourier series for F[Dm (p)] have following form
↽⇁
F[Dm (p)] = Dm[β]exp(2πihβp),
β
X
where
m−1
−β−1
A λ for β ≤ −2,
1,k 1,k
k=1
P m−1
1+ A1,k for β = −1,
k=1
(2m−1)! P m−1
Dm[β] = h2m −22m−1− k=1 A2,kλ−2,k1 for β = 0,
m−1 P
1− A λ−2 for β = 1,
2,k 2,k
k=1
−m−P1A2,kλ−2,kβ−1 for β ≥ 2.
k=1
P
With the help (24) the function D [β]we rewrite in the form
m
m−1 (1−λ )2m+1λ|β|
1,k 1,k
for |β| ≥ 2,
λkE2m−1(λ1,k)
Dm[β] = (2mh2−m1)! 1Xk=+1mXk=−11 (1E−2mλ−11,k(λ)21m,k+)1 for |β| = 1,
m−1 (1−λ )2m+1
−22m−1+ 1,k for β = 0.
λ1,kE2m−1(λ1,k)
Xk=1
We note, that
D [β] = D [−β].
m m
Theorem 1 is proved completely.
9
4 Proofs of properties.
Proof of property 1.
Following is takes placed
↽⇁[1]
F[∆ (p)] = −4sin2πph.
2
Indeed,
↽⇁[1] ↽⇁ ↽⇁ ↽⇁
∆ (x) = δ (x+1)−2 δ (x)+2 δ (x−1) = δ(x+h)−2δ(x)+δ(x−h).
2
By definition of Fourier transformation we have
↽⇁[1] ↽⇁[1]
F[∆ (p)] = exp(2πipx) ∆ (x)dx = −4sin2πph.
2 2
Z
Hence consequently we obtain
m times
↽⇁[m] ↽⇁[1] ↽⇁[1] ↽⇁[1]
F[∆ (p)] = F[∆ (p)∗ ∆ (p)∗...∗ ∆ (p)] = (−4)msin2mπph. (27)
2 2 2 2
z }| {
Immediately we have
(1−λ)2 m
= sin2mπhp. (28)
−4λ
(cid:20) (cid:21)
By virtue of (27) and (28) the formula (19) takes form
↽⇁ (2m−1)! λm−1 ↽⇁[m]
F[Dm (p)] = h2m E (λ)F[∆2 (p)]. (29)
2m−2
Now expanding rational fraction λm−1 to the sum of elementary fractions, we have
E2m−2(λ)
λm−1 m−1 B B
1,k 2,k
= + , (30)
E (λ) λ−λ λ−λ
2m−2 k=1 (cid:20) 1,k 2,k(cid:21)
X
where
λm−1 λm−1
1,k 2,k
B = , B = .
1,k E′ (λ ) 2,k E′ (λ )
2m−2 1,k 2m−2 2,k
Since |λ | < 1 and |λ | > 1, then expanding B1,k and B2,k to the Laurent series on the circle
1,k 2,k λ−λ1,k λ−λ2,k
|λ| = 1, we find
∞ β
B 1 B B λ
1,k 1,k 1,k 1,k
= · = , (31)
λ−λ1,k λ 1− λ1λ,k λ β=0(cid:18) λ (cid:19)
X
∞ β
B B B λ
2,k 2,k 2,k
= − = − . (32)
λ−λ2,k λ2,k(1− λ2λ,k) λ2,k βX=0(cid:18)λ2,k(cid:19)
10
