$r$-extension of Dunkl operator in one variable and Bessel functions of vector index

0.19 MB·English

by  Ahmed Fitouhi

r
-extension of Dunkl operator in one variable and
Bessel functions of vector index
Ahmed Fitouhi ∗ & Lazhar Dhaouadi †
2
1
0 Abstract
2
Inthis workwe presentanoperatorDµ constructedwith the help ofthe cyclic groupset
p oftherth rootsofunity. Thisoperatorconstituteanr-extensionoftheDunkloperatorinone
e
S variable because when r =2 it reduces to the classicalone and admits as eigenfunctions the
Besselfunctions of vector index earlydeeply studied by Klyuchantsev. This paper is argued
5
byspecificexamplesandcontainsthebeginningofsomeelementsofharmonicanalysisrelated
2
to this operator.

]
A
Keywords : Dunkl operator, Bessel functions, Fourier transform, transmutation operator.
F
. 2000 AMS Mathematics Subject Classification—Primary 33D15,47A05.
h
t
a
m 1 Introduction
[

2 At the beginning of the last decade of the twentieth century C. Dunkl [3, 4] in a series of articles
v using reflection groups introduced a differential-difference operator now commonly called Dunkl
7
operator andbecameagreat center of interest andinspirationinmany areas of pureandapplied
7
2 mathematics. Thisoperatorhasgeneratedarichharmonicanalysisdevelopedbyseveralauthors.
5 andinvolvesacombinationofBesselfunctionsofindexαaseigenfunctions. Soexploitingspecific
.
9 properties of these well-known special functions great analysis and an armada of applications
0
was born.
2
1
: Having knowledge of the progress of this topic in many scientific areas, we are always asked
v
i and highly intrigued by his extension in higher r-order which involves Bessel functions of index
X
vector j which are eigenfunctions of ∆ , µ = (α ,...,α ) a differential operator of order
µ µ 1 r−1
r
a r. These last functions are one of the generalized Bessel functions mentioned by Watson in his
venerable book [12] and greatly studied with applications by many authors (see [1, 6, 7, 8, 9])
and the references therein . The cyclic group C = ωr = 1, r = 1,2,3,.... plays a central
r
{ }
role in the definition of our r-extension particulary for definethe r-even and the r-odd functions
of order l = 1,...r 1 and leads to decompose the space of functions in direct sum of invariant
−
subspaceswith appropriateprojectors T , k = 1,...r 1 usefultoconstructD ther-extension
k µ
−
of the Dunkl operator having as fundamental property Dr = ∆ on a particular subspace F .
µ µ k
In addition to the construction process of the operator D , we give a particular interest in
µ
many cases especially in all the paragraphs discussed namely representation integral, associated
Riemann-Liouville transform, transmutation and r-extension Dunkl transform.
∗Facult´e des sciences deTunis, 1060 Tunis, Tunisia. E-mail : ahmed.fi[email protected]
†IPEIB, 7021 Zarzouna, Bizerte, Tunisia. E-mail : [email protected]
1
2 The operator D
µ
2iπ
Throughout this paper r is an integer great than 1, ω = e r and we put
C = 1,ω,ω2,...ωr−1
r
{ }
the cyclic group of order r. We denote by F the space of complex valued functions and we
consider on this space the following actions
s g(x) = ωkg(ωx), k = 0,1,2,....
k
Putting F the subspace of F invariant by s ; namely
k k
g F s g = g. (1)
k k
∈ ⇔
Now we introduce the collection of the projector operators defined by the relations
r−1
1
T = sn; k = 0,1,2,... (2)
k r k
n=0
X
which is slightly different from that introduced by [10].
Very often in mathematical literature [8, 9], the function T g is called the r-even part of g and
0
the functions T g, r = 1,2,...r 1 are called r-odd of order k of g.
k
−
Taking account of the fact that sr = id, one can easily show that the following properties hold:
k
1. The operators T and s , k = 0,1,2... commute in the the sense
k k
T s = s T
k k k k
2. The subspace F can be also characterized as:
k
g F T g = g.
k k
∈ ⇔
3. We have
k = l T T = 0.
k l
6 ⇔
Hence we have T2 = T so that the operators T are projectors and a simple computation leads
k k k
to that
F = F .... F .
1 r−1
⊕ ⊕
Forclarityandsincetheirimportanceandtheirinterferenceinthedemonstrationswesummarize
here the following properties which are useful in the sequel.
1. The derivative operator d maps the space F into F .
dx k k+1
2
2. The multiplication operator by 1 satisfies
x
1 1
s = s
k k+1
x x
and it maps the space F into F . Moreover
k k+1
1 1
T = T
k k+1
x x
3. For a real, we denote by L the class of operators given by
a
d a
L (f)= x−a (xaf)=f′+ f. (3)
a
dx x
Firstly we have x−bL xb = L and L maps F into F .
a a+b a k k+1
Taking into account the concepts and properties presented earlier now it seems appropriate to
introduce the following definitions.
Definition 1 Let µ = (α ,α ,...,α ) be a vector of Rr. We define the Bessel operator of
0 1 r−1
order r associated to the index vector µ by
∆ = L .... L
µ ar−1 ◦ ◦ a0
where we have put
a = rα +k k = 0,1,...,r 1, (4)
k k
−
and L is the operator given by (3).
a
Definition 2 We define the r-extension of Dunkl operator by
r−1
d 1
D = + a T
µ k k
dx x
r=0
X
where the coefficients a and the operators T area given respectively by (4) and (2).
k k
Before any thing let us justify the appellation by the following proposition
Proposition 1 For f F , k = 0...(r 1), we have
k
∈ −
Dr(f) =∆ (f).
µ µ
Proof. This result is first a consequence of the fact that D maps the F in F because the
µ k k+1
operators d and T have the same properties. Since if g F then D g = L g we deduce then
dx k ∈ k µ ak
D g = L g F , hence
µ ak ∈ k+1
D2g = L L g F .
µ ak+1 ak ∈ k+2
By induction and the fact that F = F we find
k+r k
if g F D g = ∆ g.
k µ µ
∈ ⇒
3
(cid:4)
Having defined the operators ∆ and D , it quite natural to seek their eigenfunctions . For this
µ µ
we introduce the Bessel functions of vector index
µ = (α ,α ,...,α ) Rr Zr
0 1 r−1 −
∈ \
by
∞ 1 xnr
j (x) = ( 1)n ,
µ − (α +1) (α +1) ...(α ) rnr
0 n 1 n r−1 n
n=0
X
where (β) = Γ(β+1).
n Γ(β)
The knowledgeable reader should note that this function differs from that studied in [9] by the
number of components of the index vector.
The above series is entire and taking account of
∆ xrn = rr(α +n)(α +n)...(α +n)x(n−1)r,
µ 0 1 r−1
one can state :
Proposition 2 For a complex λ we have
∆ j (λx) = λrj (λx);
µ µ µ
−
with j (0) = 1.
µ
Now we put
θ = eiπr
and we consider the r-extension function which call it also r-Dunkl kernel
1 1 1
E (x) = j (x)+ D j (x)+...+ Dr−2j (x)+ Dr−1j (x). (5)
µ µ θ µ µ θr−2 µ µ θr−1 µ µ
Proposition 3 For complex λ the function x E (λx) is an eigenfunction for the generalized
µ
7→
Dunkl operator D with θλ as eigenvalue:
µ
D E (λx) = θλE (λx).
µ µ µ
Proof. Indeed, to be convinced it suffices to make the following computations:
1 1 1
D E = D j + D2j +...+ Dr−1j + Drj
µ µ µ µ θ µ µ θr−2 µ µ θr−1 µ µ
1 1 θr
= D j + D2j +...+ Dr−1j + j
µ µ θ µ µ θr−2 µ µ θr−1 µ
1 1
= θ j + D j + + Dr−1j
µ θ µ µ ··· θr−1 µ µ
(cid:20) (cid:21)
= θE .
µ
4
so the result follows. (cid:4)
For the rest of this work we need to compute the action of the r-extension of Dunkl operator
D on the Bessel functions of vector index j .
µ µ
Indeed using the fact that D xnr = L xnr = r(α +n)xnr−1, we obtain
µ a0 0
∞ 1 xnr−1
D j (x) = r ( 1)n (α +n) .
µ µ − (α +1) ...(α +1) 0 rnr
0 n r−1 n
n=0
X
Then we distinguish two cases :
Case 1 : when α = 0 .
0
6
We have
r
D j (x) = α j ,
µ µ 0 µ−1
x
remark that we have adopt the convention : for µ = (α ,α ,...,α ) we put
0 1 r−1
µ 1 = (α 1,α ,...,α ).
0 1 r−1
− −
Case 2 : when α = 0.
0
With a slice change of computation we have
1 x r−1
D j (x) = j .
µ µ µ+1
−(α +1)...(α +1) r
1 r−1
(cid:16) (cid:17)
We have also adopt the convention : for µ = (α ,α ,...,α ) we put
0 1 r−1
µ+1= (α ,α +1,...,α +1).
0 1 r−1
As mentioned in abstract we present in the following three explicit examples which illustrate
the operators ∆ and D .
µ µ
Example 1 : r = 2,ω = 1,θ = i,µ = (0,α).
−
Then
a = 0,a = 2α+1.
0 1
As a consequence, we have
d2 2α+1 d
∆ = L L = + ,
µ 2α+1 0 dx2 x dx
whichisexactlythewellknownBesseloperatoradmittingaseigenfunctionthenormalizedBessel
function
∞ 1 x2n
j (x) = j (x) = ( 1)n .
α µ − n!(α+1) 22n
n
n=0
X
5
Now since
g(x)+s g(x) g(x) g( x)
1
T g(x) = = − −
1
2 2
we discover that the operator D is the classical Dunkl operator in one variable:
µ
d α+1
D = D = + T ,
µ α 1
dx x
and using the conventional notation introduced before namely
µ+1= (0,α+1)
and the fact that θ = i, we find that the eigenfunctions E coincide with the classical one.
µ
Example 2 : r = 3,ω = ei23π,θ = ei3π,µ = 0,α 1, 2 .
− 3 −3
(cid:0) (cid:1)
Taking account of the relation (4) between a and α we deduce that
k k
a = 0, a = 3α, a = 0.
0 1 2
So
d3 3α d2 3α d
D = L L L = + .
µ 0 3α 0 dx3 − x dx3 x2 dx
The previous operator was greatly studied in [8]. Its eigenfunction is given by
∞ 1 x3n
j (x) = ( 1)n .
µ − n! α+ 2 1 33n
n=0 3 n 3 n
X
The correspondent Dunkl operator is: (cid:0) (cid:1) (cid:0) (cid:1)
d 3α
D = + T
µ 1
dx x
with
g(x)+ωg(ωx)+ω2g(ω2x)
T g(x) = .
1
3
We have the relation
d 1 x 2
D j (x) = j (x) = j (x),
µ µ dx µ −(α+ 2)(1) 3 µ+1
3 3
(cid:16) (cid:17)
with µ+1 = (0,α+ 2,1). So
3 3
∞ 1 x3n
j (x) = ( 1)n .
µ+1 − n!(α+ 5) (4) 33n
n=0 3 n 3 n
X
Since D j F , we obtain
µ µ 1
∈
d 3α
D2j (x) = D j (x)+ T D j (x).
µ µ dx µ µ x 1 µ µ
Direct computations give
x4
D2j (x) = j (x) xj ;
µ µ 4(3α+2)(3α+5) µ+2 − µ+1
6
with µ+2 = (0,µ+ 5,4) and
3 3
∞ 1 x3n
j (x) = ( 1)n .
µ+2 − n! α+ 8 7 33n
n=0 3 n 3 n
X
(cid:0) (cid:1) (cid:0) (cid:1)
Finally taking account of the above results we state that the eigenfunction of the correspondent
Dunkl operator is then
Eµ(x) = jµ(x)+e−i3πDµjµ(x)+e−2i3πDµ2jµ(x)
e−i3π e−2i3π
E (x) = j (x) j (x)+ x4j (x).
µ µ −"32(α+ 2)(1)# µ+1 4(3α+2)(3α+5) µ+2
3 3
Remark 1 : It is easy to see that the following commutation holds:
L T = T L
a k k+1 a
and since T = T this leads that the operators ∆ and T commute:
k+r k µ k
∆ T = T ∆ .
µ k k µ
Note that if g is a function such that ∆ g = g and since we have the unique decomposition
µ
−
g = T g+...+T g.
0 r−1
One can interpret the component T g as the unique solution of the previous equation restraint
k
to the subspace F .
k
Example 3 : µ = (0, 1,..., r−1), θ = eirπ.
−r − r
In this situation as α = k we have a = 0 , hence ∆ = ( d )r and D = d . It is clear that
k −r k µ dx µ dx
the function e (x) = eθx satisfies the equation
θ
∆ e = e
µ θ θ
−
The components T eθx, k = 0,1,...,r 1 are called the r-trigonometric functions [8]. We have
k
−
in particular
cos (x) = T eθx = 1 r−1eθωkx = ∞ ( 1)n 1 xnr
r 0 r − (1) ( 1 +1) ...( r−1 +1) rnr
k=0 n=0 n −r n − r n
X X
This last function is the unique eigenfunction of ∆ = ( d )r with take the value 1 at x = 0. We
µ dx
notice that the Dunkl kernel can be written
1 1
E (x) = cos (x)+ D cos (x)+...+ Dr−1cos (x)
µ r θ µ r θr−1 µ r
= (T +...+T )eθx = eθx.
0 r−1
7
3 Integral representations
In this section we attempt to show that the functions j and E have some useful integral
µ µ
representations. For the first function the reader can found tis integral representation already
shown by klyuchantsev [9] and which is recalled in the proof of Theorem 1.
The following lemma is basic and it is a consequence of properties of Euler functions.
Lemma 1 We have
1 Γ(x)Γ(y)
r (1 ur)y−1urx−1du =
− Γ(x+y)
Z0
provided the integral converges.
We deduce then the following identities
r 1unr(1 ur)αi+ri−1ur−(i+1)du = Γ −ri +1+n Γ αi + ri .
− Γ(α +1+n)
Z0 (cid:0) i (cid:1) (cid:0) (cid:1)
r Γ(αi +1) 1 Γ −ri +1 unr(1 ur)αi+ri−1ur−(i+1)du = Γ(αi +1)
Γ αi + ri Γ −ri +1 Z0 Γ −(cid:0)ri +1+(cid:1)n − Γ(αi +1+n)
So we(cid:0)have (cid:1) (cid:0) (cid:1) (cid:0) (cid:1)
r Γ(αi+1) 1 1 unr(1 ur)αi+ri−1ur−(i+1)du = 1
Γ αi + ri Γ −ri +1 Z0 −ri +1 n − (αi +1)n
(cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1)
and in general case
1 1 1 (xu ...u )nr
... ( 1)n 0 r−1 w (u )...w (u )du ...du
− (1) ... r−1 +1 rnr 0 0 r−1 r−1 0 r−1
Z0 Z0 n − r n
1 xnr
= ( 1)n (cid:0) (cid:1).
− (α +1) ...(α +1) rnr
0 n r−1 n
Let
r−1
wµ(u) = (1−uri)αi+ri−1uri−(i+1) (6)
i=0
Y
r−1
Γ(α +1)
i
c = r (7)
µ Γ α + i Γ i +1
i=0 i r −r
Y
and (cid:0) (cid:1) (cid:0) (cid:1)
u = u ...u
r 0 r−1
du = du ...du
0 r−1
and using the function cos presented in Example 3, we have
r
8
Theorem 1 The Bessel function of vector index possess the following integral representation
j (x) = c cos (xu )w (u)du
µ µ r r µ
Z[0,1]r
where c and w are given respectively by (7) and (6).
µ µ
Note that in this representation we can remove the components u associated with indices i such
i
that a = 0. In this case the Mehler representation takes the following form
i
j (x) = c cos (xu )w (u)du
µ µ′ r r′ µ′
Z[0,1]r′
with r′ is the number of index i such that a = 0 and µ′ contains only the associate α .
i i
6
Theorem 2 the r-Dunkl kernel possess the following integral representation
r−1
1
E (x) = c T + T L ...L e (xu )w (u)du.
µ µZ[0,1]r 0 k=1 θk k ak−1 a0! θ r µ
X
Proof. This is a consequence of definition of E (x) and the identity
µ
1 1
Dkj (x) = c DkT e (xu )w (u)du
θk µ µ µZ[0,1]r θk µ 0 θ r µ
1
= c L ...L T e (xu )w (u)du
µZ[0,1]r θk ak−1 a0 0 θ r µ
1
= c T L ...L e (xu )w (u)du,
µZ[0,1]r θk k ak−1 a0 θ r µ
which prove the result. (cid:4)
Example 4 : r = 2,w = 1,θ = i, µ = (0,α)
−
Since a = 0 then we can remove the index i= 0 in the representation of the function j which
0 µ
gives
jµ(x) = jα(x) = √2πΓΓ(αα++11) 1cos(xu)(1−u2)α−21du.
2 Z0
On the other hand (cid:0) (cid:1)
Eµ(x) = Eα(x) = √2πΓΓ(αα++11) 1 T0eixu + 1iT1ddxeixu (1−u2)α−12du
2 Z0 (cid:20) (cid:21)
= √2πΓΓ(αα++11) (cid:0)1 T0e(cid:1)ixu+uT1eixu (1−u2)α−12du
2 Z0
= √2πΓΓ(cid:0)(αα++11)(cid:1) 1(cid:2)T0eixu+T1x1(xu(cid:3))eixu (1−u2)α−21du
2 Z0 (cid:20) (cid:21)
(cid:0) (cid:1)
9
To find the classical form we can write
Eα(x) = √2πΓΓ(αα++11) 1 eixu+2e−ixu +ueixu−2e−ixu (1−u2)α−12du
2 Z0 (cid:20) (cid:21)
= 1 Γ(cid:0)(α+1)(cid:1) 1 eixu(1+u)(1 u2)α−12du.
√πΓ α+ 1 −
2 Z−1
(cid:0) (cid:1)
Example 5 : r = 3,w = ei23π = j,θ = eiπ3,µ = 0,v 1, 2 .
− 3 −3
(cid:0) (cid:1)
We have
Γ v+ 2 1
j (x)= j (x) = 3 3 cos (xu)(1 u3)v−1udu.
µ v Γ(v)Γ 2 3 −
(cid:0) 3(cid:1) Z0
Then (cid:0) (cid:1)
Γ v+ 2 1 1 d 1 d 3v d
E (x) = 3 3 T eθxu+ T eθxu+ T + eθxu (1 u3)v−1udu
v Γ(v)Γ 2 0 θ 1dx θ2 2 dx x dx −
(cid:0) 3(cid:1) Z0 (cid:20) (cid:18) (cid:19) (cid:21)
which can be writte(cid:0)n (cid:1)in the following form
Γ v+ 2
E (x) = 3 3
v Γ(v)Γ 2 ×
(cid:0) 3(cid:1)
1 1 1 1 3v 1
T (xu)eθ(cid:0)xu(cid:1)+T (xu)2eθxu+T (xu)3eθxu+ T (xu)2eθxu (1 u3)v−1du.
0x 1x2 2x3 θ 2x3 −
Z0 (cid:20) (cid:21)
4 Riemann–Liouville transform
Considering the r-Riemannn Liouville operators of the form
1
R g(x) = g(xt)(1 tr)α−1dt.
α
−
Z0
The integral representation in Theorem 1 of the Bessel function of vector index can berewritten
as follows
r−1
1
jµ(x) = cµZ[0,1]rcosr(xur)wµ(u)du = cµi=0(cid:18)xr−(i+1)Rαi+ri−1xr−(i+1)(cid:19)cosr(x). (8)
Y
Now we study the inverse of R . We begin by state:
α
Theorem 3 For k integer and 0 < α < 1 we have
r2 1 d k+1 x
R−1 g(x) = xr−1 g(u)(xr ur)−αu(k+α)rdu.
k+α Γ(k+1)Γ(α)Γ(1 α) rxr−1dx −
− (cid:18) (cid:19) Z0
Proof. The operator R can be take the form
α
1 x
R g(x) = g(u)[xr ur]α−1du
α
x1+r(α−1) −
Z0
10