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1URAL MATHEMATICAL JOURNAL, Vol. 9, No. 2, 2023, pp. 109-120
DOI: 10.15826/umj.2023.2.009
A NEW CHARACTERIZATION OF SYMMETRIC DUNKL AND q-DUNKL-CLASSICAL ORTHOGONAL POLYNOMIALS
Yahia Habbachi
Faculty of Sciences, University of Gabes, Erriadh City 6072, Zrig, Gabes, Tunisia
Yahiahabbachi@hotmail.com
Abstract: In this paper, we consider the following L-difference equation
$(x)LPn+i(x) = (£nx + tfn)Pn+l(x) + AnPn(x), n > 0,
where $ is a monic polynomial (even), deg$ < 2, £n, i9n, An, n > 0, are complex numbers and L is either the Dunkl operator T^ or the the q-Dunkl operator T(gq). We show that if L = T^, then the only symmetric orthogonal polynomials satisfying the previous equation are, up a dilation, the generalized Hermite polynomials and the generalized Gegenbauer polynomials and if L = T(g q), then the q2-analogue of generalized Hermite and the q2-analogue of generalized Gegenbauer polynomials are, up a dilation, the only orthogonal polynomials sequences satisfying the L-difference equation.
Keywords: Orthogonal polynomials, Dunkl operator, q-Dunkl operator.
1. Introduction
The classical orthogonal polynomials (Hermite, Laguerre, Bessel, and Jacobi) have a lot of useful characterizations: they satisfy a Hahn's property, that the sequence of their monic derivatives is again orthogonal (see [1, 8, 14, 16]), they are characterized as the polynomial eigenfunctions of a second order homogeneous linear differential (or difference) hypergeometric operator with polynomial coefficients [4, 15, 16], their corresponding linear functionals satisfy a distribution equation of Pearson type (see [11, 13, 15]).
Another characterization was established by Al-Salam and Chihara in [1], in particular they showed that the sequences Hermite, Laguerre and Jacobi are the only monic orthogonal polynomial sequences {Pn}n>0 that satisfy an equation of the form:
n(x)Pn +i(x) = (a-nx + bn)Pn+i + CnPn(x), n > 0, (1.1)
where n is a monic polynomial, deg n < 2.
Recently, Datta and J. Griffin [9] studied the q-analogue of (1.1). More precisely they studied a q-difference equation of the form:
n(x)DqPn+i(x) = (anx + bn)Pn+1 + CnPn(x), n > 0, (1.2)
where n is a monic polynomial, deg n < 2 and Dq is the Hahn operator defined by
Dqf (x) = (f (qx) - f (x))/(q - 1)x, f &V.
In particular they showed that the only orthogonal polynomials satisfying (1.2) are the Al-Salam-Carlitz I, the little and big q-Laguerre, the little and big q-Jacobi and the q-Bessel polynomials. The aim of this paper is to study the equation of the form:
$(x)LPn+i(x) = (inx + tfn)Pn+i(x) + AnPn(x), n > 0,
(1.3)
where $ is a monic polynomial (even), deg$ < 2 and L € , T^)}.
This paper is organized as follows. In Section 2, we introduce the basic background and some preliminary results that will be used in what follows. In Section 3, we show that the only symmetric orthogonal polynomials satisfying (1.3), are, up a dilation, the generalized Hermite polynomials and the generalized Gegenbauer polynomials if L = TM and the q2-analogue of generalized Hermite polynomials and the q2-analogue of generalized Gegenbauer polynomials if L = T^).
2. Preliminaries and notations
Let P be the vector space of polynomials with coefficients in C and let P' be its dual. We denote by (u, f) the action of u € P' on f € P. In particular, we denote by (u)n = (u, xn), n > 0, the moments of u. For any form u, any polynomial f and any a € C \ {0}, let fu and hau, be the forms defined by duality:
(fu, p) = (u, fp), (h„u, p) = (u, h„p), p € P,
where hap(x) = p(ax).
Let {Pn}n>0 be a sequence of monic polynomials (MPS, in short) with deg Pn = n, n > 0. The dual sequence associated with {Pn}n>0 is the sequence {un}n>0, un € P' such that (un, Pm) = ¿n>m, n, m > 0, where ¿n>m is the Kronecker symbol [14].
The linear functional u is called regular if there exists a MPS {Pn}n>0 such that (see [8, p. 7]):
(u, PmPn) = rra5ra;m, n, m > 0, r„ = 0, n > 0.
Then the sequence {Pn}n>0 is said to be orthogonal with respect to u. In this case, we have
Un = ((u0,Pn))-1PnU0, n > 0.
Moreover, u = Au0, where (u)0 = A = 0 [17].
In what follows all regular linear functionals u will be taken normalized i.e., (u)0 = 1. Therefore, u = u0.
A polynomial set {Pn}n>0 is called symmetric if
Pn(-x) = (-1)nPn(x), n > 0.
According to Favard's theorem [8], a sequence of monic orthogonal polynomials {Pn(x)}n>0 (MOPS, in short) satisfies a three-term recurrence relation
P0(x) = 1, Pi(x) = x, (21)
Pn+2(x) = (x - ^n+i)Pn+i(x) - Yn+iPn(x), n > 0, Yn+1 = 0, n > 0. (.)
with
_ {uo,xP*) _ (uo,Pw2+1)
— "/-D2T' — "1-D2\~> n - U-
(u0,P2) (u0,p2)
A dilatation preserves the property of orthogonality. Indeed, the sequence {Pn(x)}n>0 defined by
Pn(x) = a-nPn(ax), n > 0, a € C \ {0}, satisfies the recurrence relation [16]
fp0(x) = 1, Pi(x) = x - /50, | Pn+2(x) = (x - /3n+i)pPn+i(x) - 7n+i-Pn(x), n > 0,
where
= 7n+i = -n>0. (2.2)
a a2
Moreover, if {Pn}n>0 is a MOPS with respect to the regular form u0, then {Pn}n>0 is a MOPS with respect to the regular form u0 = ha-i u0.
Theorem 1 [8]. Let {Pn}n>0 be a MOPS satisfying (2.1) and orthogonal with respect to a linear functional u. The following statements are equivalent:
(i) the sequence {Pn}n>0 is symmetric;
(ii) (u)2n+i = 0, n > 0;
(iii) ^n = 0, n > 0.
Next, we introduce the Dunkl operator defined on P by [10, 18]
(Tflf)(x)=f'(x) + ^H.1f(x),
where
For the Dunkl operator, we have the property [6]
T*(fg)(x) = (Tf )(x)g(x) + f (x)(TMg)(x) - 4^x(H-if)(x)(H-ig)(x), f,g € P. In particular,
TM(xPn+i) = (1 + 2^(-1)n+i)Pn+i(x) + x(TMPn+i)(x), n > 0. (2.3)
We define the operator TM from P' to P' as follows:
(TMu,f) = -<u,Tf), f € P, u € P'.
In particular,
(TMu)n = -^n(u)n-i, n > 0, with the convention (u)-i = 0, where
^n = n + ^(1 - (-1)n), n > 0.
We introduce also the q-Dunkl operator T(#,q) defined on P by [2, 5, 7]
Remark 1. Note that when q ^ 1, we again meet the Dunkl operator. From the last definition, it is easy to prove that
T(*,q) (fg) = (T(,,q)f )g + (hq f )(T(,,q)g) + 0(h_i f - hq f )H_ig, f, g € P.
In particular,
= + 9 - d(q + 1)1 ~ ^ pn+1(x) + ^(T^i^X.r), n > 0. (2.4)
We define the operator from P' to P' as follows:
(T(,,q)u,f) = -(u,T(,,q)f), f € P, u € P'.
In particular, where (u)-i = 0 and
(T(0,q))n = -^n,q(u)n-1, n > 0,
l-(-l) 2
here [n]q, n > 0, denotes the basic q-number defined by
\n
0n,q = N« + 0-n > 0, (2.5)
1 — qn
[n]q = —-^ = l + q + ... + qn-\ 77, >1, [0]g = 0. 1—q
According to the definitions of T/ and T^), we have
T/i(xn) = ^nXn-1, T(,,q)(xn) = 0n,qXn-1
3. The main results
In this section, we will look for all symmetric MOPS satisfying (1.3). We distinguish two cases. The first case is when L = T/ and the second one is when L = T^).
3.1. First case: when L = T/
Theorem 2. The only symmetric MOPS satisfying a T^-difference equation of the form
$(x)T/Pn+i(x) = (£nx + tfn)Pn+i(x) + AnPn(x), n > 0, (3.1)
where $ is a monic polynomial (even), deg$ < 2, are, up a dilation, the generalized Hermite polynomials and the generalized Gegenbauer polynomials.
Proof. Let {Pn}n>0 be a symmetric MOPS satisfying (3.1). Since $ is a monic, even and deg$ < 2, then we distinguish two cases: $(x) = 1 and $(x) = x2 + c.
Case 1. $(x) = 1, then (3.1) becomes
T/Pn+l(x) = (CnX + tfn)Pn+l(x) + AnPn(x), n > 0.
By comparing the degrees in the last equation (in xn+2 and xn+1), we obtain £n = $n = 0, n > 0 and then
T/Pn+1(x)= AnPn(x), n > 0. (3.2)
Identifying coefficients in the monomials of degree n in the last equation, we obtain
An = ^n+1, n > 0. (3.3)
On the other hand, applying the operator T/ to (2.1) with /Sn+1 = 0 and using (2.3), we get
T/Pn+2(x) = (1 + 2^(—1)n+1)Pn+1(x) + x(T/Pn+1)(x) — Yn+1(T/Pn)(x), n > 0.
Substituting (3.2) and (3.3) in the last equation and taking into account the fact that
1 + 2^(—1)n+1 = ^n+2 — ^n+1,
we get
^n+iPn+i(x) = ^n+ixPn(x) - ^nYn+iPn-i(x), n > 0. From (2.1), the last equation is equivalent to
^n+17n Pn_ i(x) = ^n7n+iPn_i(x), n > 0,
hence,
^n+iYn = ^nYn+1, n > 1.
Therefore,
Yi ^ i
7,2+1 = —Mn+1, n > 1. Since the last relation remains valid for n = 0, then we have
Yi ^ n
Jn+1 = -Vn+1, n > 0.
Using (2.2), where a2 = 2y1 we obtain
Pn = 0, = n> 0.
So, we meet the recurrence coefficients for the generalized Hermite polynomial sequence (see [8]). Case 2. $(x) = x2 + c, then (3.1) becomes
(x2 + c)TMPn+i(x) = (£nx + ^n)Pn+i(x) + AnPn(x), n > 0. (3.4)
Identifying the coefficients of higher degree in both sides of (3.4), we obtain £n = ^n+1, n > 0. Therefore, (3.4) becomes
(x2 + c)TMPn+i(x) = (^n+ix + ^n)Pn+i(x) + AnPn(x), n > 0. (3.5)
Applying the operator TM to (2.1) with /Sn+1 = 0 and using (2.3) and the fact that
1 + 2^(-1)n+1 = ^n+2 - ¿Un+1,
we get
TMPn+2(x) = (^n+2 - ^n+i)Pn+i(x) + x(TMPn+i)(x) - Yn+i(TMPn)(x), n > 0. Multiplying the previous equation by x2 + c and using (3.5), we get
(^n+2x + $n+i)Pn+2(x) + An+iPn+i(x) = (^n+2 - ^n+i)(x2 + c)Pn+i(x) + (^n+ix2 + $nx)Pn+i(x) + AnxPn(x) - Yn+1 ((^nx + ^n_i)Pn(x) + An_iPn_i(x)) , n > 1,
or, equivalently,
(tfn+1 - ^n)xPn+i(x) - c(^n+2 - ^n+i)Pn+1 + An+iPn+i(x) (3 6)
= AnxPn(x) + Yn+1 ((^n+2x + ^n+i)Pn(x) - (^„x + ^n_i)Pn(x) - An_iPn_i(x)) , n > 1.
Comparing the degrees in the last equation, we obtain $n+1 = $n, n > 1. But, from (3.5) and the fact that {Pn}n>0 is symmetric, where n = 0 and n = 1, we get, respectively,
V0 = 0, A0 = c(1+2^),
vi =0, Ai = 2(yi + c). ( . )
Thus,
= 0, n > 0.
Therefore, (3.6) becomes
c(^n+1 — ^n+2)Pn+1(x) + An+1Pn+1(x) = AnxPn(x) + 7n+^(^n+2 — ^n)xPn(x) — An-1Pn-1(x^ , n > 1.
Taking into account (2.1), we get
(An+1 + c(^n+1 — ^n+2)) xPn(x) — Yn( An+1 + c(^n+1 — ^n+2)) Pn-1(x) = (An + (^n+2 — ^n)7n+0 xPn(x) — An-17n+1^n-1(x), n > 1.
Then,
An+1 + c(^n+1 — ^n+2) = An + (^n+2 — ^n)7n+1, n > 1, (3.8)
(An+1 + c(^n+i - ^n+2))Yn = An-iYn+1, n > 1. (3.9)
Since ^n+2 — = 2, then, substitution of (3.8) in (3.9) gives
(An + 2Yn+i)Yn = An-iYn+i, n > 1.
Therefore,
A n A n- 1
— 2, n > 1.
Yn+i Yn
So,
Ao — 2nYi f .
\n = -Yra+1) n > !• (3-10)
Yi
It is clear that (3.10) remains valid for n = 0. Then, we have
Ao — 2nYi
A„ =-Y«+i) n > 0. (3.11)
Yi
Substitution of (3.11) in (3.8) gives
Ao — 2(n — 1)yi , , , . . 1
— -:---An + C{^n+2 — Mra+lj) n ^ J--
Ao — 2nYi
By virtue of fourth equality in (3.7), we obtain that the previous equation remains valid for n = 0. Hence,
A»+i = , —+ c(nn+2 - Hn+i), n> 0. (3.12)
Ao — 2nYi
We will distinguish two situations: c = 0 and c = 0.
• If c = 0, then from (3.7) we have A0 = 0. Therefore, An = 0, n > 0. Consequently, according to (3.11) and the fourth equality in (3.7), Yn+i = 0, n > 0. This contradicts the orthogonality of {Pn}n>0.
• If c = 0, using a dilatation, we can take c = —1. Putting
1 + 2^
7i
3 + 2^ + 2a'
then (3.12) becomes
2n + 2a + 2^ + lx Xn+1 2n + 2a + 2/x + 3 + ßn+1 ~ ^ " " (3"13)
From (3.13), we can easily prove by induction that
ß„+i{ß„+i + 2a)
An —--i fl U.
n 2n + 2a + 2^ + 1 '
Thus, (3.11) gives
__Hn+iiHn+1 + 2a)__>0
7ra+1 ~ (2n + 2a + 2fj, + l)(2n + 2a + 2fj, + 3) ' " ~
So, we meet the recurrence coefficients for the generalized Gegenbauer polynomial (see [3, 8]). □ Remark 2. Notice that when ^ = 0 in (3.1), we again meet (1.1) for the symmetric case.
3.2. Second case: when L =
Theorem 3. The only symmetric MOPS satisfying a T(0,q)-difference equation of the form :
$(x)T(e>,)Pra+i(x) = (£„x + tf„)P„+i(x) + A„P„(x), n > 0, (3.14)
where $ is a monic polynomial (even), deg$ < 2, are, up a dilation, the q2-analogue of generalized Hermite polynomials and the q2-analogue of generalized Gegenbauer polynomials.
Proof. Let {Pn}n>0 be a symmetric MOPS satisfying (3.1). As in proof of Theorem 2, we distinguish two cases: $(x) = 1 and $(x) = x2 + c.
Case 1. $(x) = 1, then (3.14) becomes
T№q)Pra+i(x) = (£„x + tfn)Pn+i(x) + A„P„(x), n > 0. (3.15)
By comparing the degrees in (3.15), we obtain = = 0, n > 0. Then,
T(0)q)Pn+i(x) = A„P„(x), n > 0.
The comparison of the coefficients of xn in the previous equation leads to An = 0n+i>q, n > 0. Therefore,
T(0;q)P„+i(x) = 0ra+i,qPn(x), n > 0. (3.16)
Now, applying T(0,q) to (2.1) with /Sn+i = 0 and using (2.4), we get
( 1 - (-1)n+i \
T{e<q]Pn+2(x) = [1 + 6-6(q + 1)-^-JPn+i(x)
+qx(T(0,q)Pn+i)(x) - Y„+i(T(0;q)P„)(x), n > 0. Substituting (3.16) in the last equation, we get
0n+2,qPn+i(x) = (l + 6-6(q + l)i^^)p„+1(x) +q^n+i,q xPn(x) - 7n+i^n,qPn-i(x), n > 0.
Using the fact that
xPn - Pn+1 + Yn— 1 ,
we obtain
1 — (—1)n+1 \ 0n+2,q - 1 -6+ 6{q + 1)-^--qOn+i,q\Pn+i{x)
= q^n+1,qYnPn—1(x) — 0n,qYn+1^n— 1(x), n > 0. After easy calculations from (2.5), we have
1 — (—1)n+1
0n+2,q ~ 1 " 0 + d(q + 1)-^--qen+hq = 0, n>0. (3.17)
Therefore,
(q0n+1,qYn — ^n,qYn+1)Pn—1 (x) =0, n > 0.
Hence,
q^n+1,qYn = Yn+1, n > 1. Then, we can deduce by induction that
Yi n
Jn+1 = Y^Q 1n en+l,q, n > 1.
Moreover, the previous identity remains valid for n = 0, thus
Yi n
Tn+i = tty; n > o.
1 + U
Then, according to (2.2), with the choice
a2 = q(q + 1)- 71
1 + U and putting
- 1 + 6 I
we obtain
q(q + 1)
which are the recurrence coefficients for the q2-analogue of generalized Hermite polynomial [12], with
- 1 + 0 I
Case 2: $(x) = x2 + c, then in this case (3.14) becomes
(x2 + c)T(MPn+1 (x) = (Cnx + $n)Pn+1(x) + AnPn(x), n > 0. (3.18)
By comparing terms of higher degree in the previous equation, we obtain
£n = 0n+1,q, n > 0.
Then, equation (3.18) becomes
(x2 + c)T(0,q)Pn+1(x) = (0n+1,qx + tfn)Pn+1(x) + AnPn(x), n > 0. (3.19)
Applying the operator T(0 q) to (2.1) with ^n+1 = 0 and using (2.4), we get
Multiplying the above equation by x2 + c and substituting (3.19) into the result, we get
(Un+2,q X + $n+i)Pn+2(x) + An+iPn+i(x) = (Un+2,q — qUn+i,q)(x2 + c)Pn+i(x) +q(Un+i,qX2 + ^nX)Pn+i(x) + qAnXPn(x) — Yn+i ((Un,qX + ^n-i)Pn(x) + An-iPn-i(x)) , n > 1.
Substituting of (2.1) in the previous equation, we get
(tfn+i — q$n)xPn+i(x) + (An+i — c(Un+2,q — qUn+i,q)) Pn+i(x) = qAnXPn(x) + Yn+i (((Un+2,q — Un,q)x + $n+i — $n-i)Pn(x) — An-iPn-i(x)) , n > 1.
The comparison of the coefficients of xn+2 in the previous equation gives $n+i = q$n, n > 1 and putting n = 0 and n = 1 in (3.19), we get respectively
+qx(T(MPra+i)(x) - 7„+i(T(ö;q)Pra)(x), n > 0.
By (3.17), the last equation becomes
T(0,q)Pn+2(x) = (0n+2,q - q0n+i,g)Pn+l(x) +qx(T(ö>q)Pra+i)(x) - Y„+i(T(ö;q)Pra)(x), n > 0.
vo = 0, Ao = c(1 + 0), Vi =0, Al = (1 + q)(Yi + c).
(3.20)
Hence, $n = 0, n > 0.
Therefore, the last equation becomes
n
(An+i — c(0n+2,q — q0n+i,q )) Pn+i(x)
= qAnxPn(x) + 7n+i((0n+2,q - 0n,q)xP„(x) - An-iPn-i(x)) , n > 1.
Using the fact that Pn+1 = xPn(x) — YnPn—1, the above equation is equivalent to
(An+i - c(0n+2,q - q0n+i,q))xPn(x) - 7n(An+i - c(0n+2,q - q0n+i,q))Pn-i(x) = (qAn + (0n+2,q - 0n,q)7n+OxPn(x) - An-iYn+iPn-i(x), n > 1.
Then, we deduce
An+ i - c(0n+2,q - q0n+i,q) = qAn + (0n+2,q - 0n,q)7n+1, n > b
(3.21)
(3.22)
Un+2,q — Un,q = (1 + 9)9n,
then the substitution of (3.21) in (3.22) gives
(qAn + (1 + q)qnYn+i)Yn = An-1Yn+i, n > 1,
therefore,
We can easily deduce by induction that
qnK = ^ - q(q + l)[n]q2^Jjn+i, n> 1. It is clear that the previous identity remains valid for n = 0. Then, we have
qn\n = " <?(<? + DM^n+i, n > 0. (3.23)
Now, we will determine An. By (3.23), we have
Yra+i = q-, r 1 A„, n > 0. (3.24)
Ao - q(q + 1)7i[n],2
Therefore, (3.21) becomes
\ gA0-(g + l)7i(Ng2 -i), . (a a , . -,
An+1 — -\--,--Tl-+ C\Vn+2,q ~ qt>n+l,q), 11 > L.
Ao - q(q + 1)Yi[n]q2
By virtue of (3.20), we obtain that the previous equation remains valid for n = 0. Then,
q\0 - (g + l)7i(Ng3 - 1)
A0 -q{q + l)7i W«3
A„+1 =-r-:—, -Xn + c{0n+2,q - qOn+hq), n > 0. (3.25)
We will distinguish two situations: c = 0 and c = 0.
• If c = 0, then from (3.20) A0 = 0. Therefore, An = 0, n > 0. Consequently, according to (3.24) and the fourth equality in (3.20), y„+1 = 0, n > 0. This contradicts the orthogonality of {Pn}n>0.
• If c = 0, using a suitable dilatation, we can suppose that c = -1. Putting
71 = l + e + q^+Dia + lY (3"26)
Equation (3.25) becomes
-Wl — q —,-—TT,-1 -IN I a-An - \Vn+2,q ~ QVn+l,q), « ^ U. (O-Zi
q(q + 1)(a + 1) + 02n+i,q
Therefore, from (3.27), we can prove by induction that
tfn+i,q(q(q + 1)(a + 1) + 0n-i,,(1 + ^(1 - q)(1 - (-1)n)/2))
A„ =----;-7-7-7T-T-n > 0. (3.28)
q(q + l)(a + 1) + 02n-i,q
By virtue of (3.24), (3.26) and (3.28), we get
tfn+i,q(q(q + 1)(a + 1) + 0n-i,,(1 + 0(1 - q)(1 - (-1)n)/2))
7«+i = qn—-7-+-TT-t-^-, n > 0.
(g(q + 1)(a + 1) + 02n-i>gJ + 1)(a + 1) + 02w+i>gJ
So, we meet the recurrence coefficients for the q2-anlogue of generalized Gegenbauer polynomial
sna'ß'q2), with 1
^IWTT)-1
(see [12]). □
Remark 3. Notice that when q — 1, we recover the result in Theorem 2 and when 9 = 0 in (3.14), we again meet (1.2) for symmetric case.
4. Conclusion
To conclude this paper, we will present two tables in which we give the only symmetric MOPS verifying the L-difference (1.3).
Polynomial Cn tin A«
(a a1) Generalized Hermite H,{ 1 0 0 ßn+1, n > 0
Generalized Gegenbauer sl"'l3'q ^ x2 - 1 ßn+l 0 ßn+i(ßn+i + 2a) , ÎI u 2 n + 2a + 2/x + 1
Table 1: Case when L = T,
m
Polynomial Cn tin An
<72-analogue of generalized Hermite H^'q2) 1 0 0 On+l,q, n > 0
<72-analogue of generalized Gegenbauer ^ x2 - 1 0 q{q+l){a+l)+e2n-l,q ' n > 0.
Table 2: Case when L =
Remark 4• In this paper, we have studied only the symmetric case. The question for non-symmetric case remains open.
Acknowledgements
The author thanks Professor Francisco Marcellan for many enlightening discussions and the anonymous referees for their careful reading of the manuscript and corrections.
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