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12URAL MATHEMATICAL JOURNAL, Vol. 6, No. 2, 2020, pp. 15-24
DOI: 10.15826/umj.2020.2.002
HAHN'S PROBLEM WITH RESPECT TO SOME PERTURBATIONS OF THE RAISING OPERATOR X - c
Baghdadi Aloui
University of Gabes, Higher Institute of Industrial Systems of Gabes Salah Eddine Elayoubi Str., 6033 Gabes, Tunisia
Jihad Souissi
University of Gabes, Faculty of Sciences of Gabes Erriadh Str., 6072 Gabes, Tunisia
Abstract: In this paper, we study the Hahn's problem with respect to some raising operators perturbed of the operator X — c, where c is an arbitrary complex number. More precisely, the two following characterizations hold: up to a normalization, the q-Hermite (resp. Charlier) polynomial is the only Ha>q-classical (resp. S^-classical) orthogonal polynomial, where Ha,q := X + aHq and S\ := (X + 1) — At— 1.
Keywords: Orthogonal polynomials, Linear functional, O-classical polynomials, Raising operators, q-Hermite polynomials, Charlier polynomials.
1. Introduction
Let O be a linear operator acting on the space of polynomials which sends polynomials of degree n to polynomials of degree n + n0, where n0 is a fixed integer (n > 0 if n0 > 0 and n > |n0| if n0 < 0). We call a sequence {Pn}n>0 of orthogonal polynomials O-classical if {OPn}n>0 is also orthogonal.
In particular, if O = D, the standard derivative, we recover the know family of classical orthogonal polynomials (Hermite, Laguerre, Bessel and Jacobi). This characterization is called Hahn's characterization (see [11, 18]) of the classical orthogonal polynomials. If O = Hq, where
we recover the so-called Hq-classical polynomials (for more details, see [12]). We can also cite [14], where the authors described the all Dw-classical orthogonal polynomials, with
Duf{x) .= T-f(x^~f(x\ T_u,f(x) = f(x + U).
The literature on these topics is extremely vast. See further examples in [1-5, 7, 8, 11, 12, 14].
In this paper we consider some raising operators related to the operator X. It is easy to see that the orthogonality is not preserved by X, then we can consider and study some perturbed operators. Here we consider the following two operators (c = 0 or c = 1):
Ha,q := X + aHq (1.1)
Sa := (X + 1) - Ar—1, (1.2)
and we study the same problem, called Hahn's problem. More precisely, we find all orthogonal polynomial sequences {Pn}n>0 such that {OPn}n>0, O = Ha,q or SA, are also orthogonal. As a result, we conclude that the q-Hermite polynomial sequence is the only Ha,q-classical sequence and the Charlier polynomial sequence is the only SA-classical sequence.
The structure of the paper is the following. In Section 2, a basic background about forms of orthogonal polynomials is given. In Section 3, we show that, up to a dilatation, the q-Hermite (resp. Charlier) polynomial is the only Ha,q-classical (resp. SA-classical) orthogonal polynomial. In Section 4, we give a conclusion and describe some prospects.
2. Preliminaries
Let P be the linear space of polynomials in one variable with complex coefficients and P' be its dual space, whose elements are forms. We denote by (u,p) the action of u € P' on p € P. In particular, we denote by (u)n := (u, xn), n > 0, the moments of u. Let us define the following operations in P'. For any form u, any polynomial f, and any (a, b, c) € C\{0} x C2, let Du = u', fu, (x — c)-1u, T-bu and hau be the forms defined by duality, [16]:
(fu,p) := (u, fp), (u',p) := —(u,p'), (fu)' = f'u + fu', (hau,p) := (u,p(ax)), (r-bu,p) := (u,p(x — b)), v-i , / p(x) — p(c) \
((x-c)-y p) :=(u, PW-M), p
\ x — c /
x — c
A form u is called normalized if it satisfies (u)o = 1. We assume that the forms used in this paper are normalized.
Let {Pn}n>0 be a sequence of monic polynomials (MPS) with deg Pn = n and let {un}n>0 be its dual sequence, un € P', defined by (un,Pm) = n, m > 0. Notice that u0 is said to be the canonical functional associated with the MPS {Pn}n>0. The sequence {Pn}n>0 is called symmetric when Pra(—x) = (—1)nPra(x), n > 0.
Let us recall the following result [17].
Lemma 1. For any u € P' and any integer m > 1, the following statements are equivalent:
(i) (u, Pm-i) = 0, (u, P„) = 0, n > m.
(ii) 3AV € C, 0 < v < m — 1, Am-1 = 0 such that u = ^m—1 uv. As a consequence, the dual sequence {ui1] }n>0 of {Pi^1] }n>0 where
Pi1!(x) := (n + 1)-1P,;+1(x), n > 0,
is given by
Duj!1 = — (n + 1)ura+1, n > 0. Similarly, the dual sequence {un}n>0 of {pPn}ra>0, where
P„(x) := a-nP„(ax + b)
with (a, b) € C\{0} x C, is given by
un = an(ha-i o r-b)ura, n > 0. The form u is called regular if we can associate with it a sequence {Pn}ra>0 such that (u, PnPm) = rra5ra;m, n, m > 0, r„ =0, n > 0.
The sequence {Pn}n>0 is then called a monic orthogonal polynomial sequence (MOPS) with respect to u. Note that u = (u)0u0, with (u)0 = 0. When u is regular, let F be a polynomial such that Fu = 0. Then F = 0, [16].
Proposition 1 [16]. Let {Pn}n>0 be a MPS with degPn = n, n > 0, and let {un}n>0 be its dual sequence. The following statements are equivalent.
(i) {Pn}n>0 is orthogonal with respect to u0.
(ii) u„ = (u0,Pn)-1P«u0, n > 0.
(iii) {Pn}n>0 satisfies the three-term recurrence relation
P0(x) = 1, Pi(x)= x - ß0, ( )
(2. 1 )
Pn+2(x) = (x - ßn+l)Pn+l(x) - Yn+lPn(x), n > 0, where = (u0, xPn)(u0, Pi)-1, n > 0 and 7„+1 = (u0, P?2+1)(u0, P2)-1 = 0, n > 0.
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 o T-b)u0, and satisfies [15]
( P0(x) = 1, P(x) = x - ß0,
\ pPn+2(x) = (x - /5„+1)ipre+1(x) - Pn+1pn(x), n > 0,
where = a-1(ß„ - b) and Yn+1 = a-2Y„+1.
A MOPS {pn}n>0 is called D-classical, if {Dpn}n>0 is also orthogonal (Hermite, Laguerre, Bessel or Jacobi), [10, 11]. Moreover, if {pn}n>0 is orthogonal with respect to u0, then there exists a monic polynomial 0 with deg 0 < 2 and a polynomial ^ with deg ^ = 1 such that u0 satisfies a Pearson's equation (PE) [15]
D(0u0) + ^u0 = 0.
Any shift leaves invariant the D-classical character. Indeed, the shifted linear functional u = (ha-i o r-b)u fulfills the equation
(Ï u)' + = 0,
where (see [15, 16])
Ï (x) = a-i$(ax + b) and $ (x) = a1-i$(ax + b).
3. Hahn's problem with respect to some perturbations of the raising
operator X - c
Clearly, the orthogonality is not preserved by the operator X - c, which is given by
(X - c)(f (x)) = (x - c)f (x), f € P.
Our goal, in this section is to describe all O-classical orthogonal polynomials. More precisely, we find all orthogonal polynomial sequences {Pn}n>0 such that {OPn}n>0 are also orthogonal, where O = Ha,q or O = are the operators defined by (1.1) and (1.2). This operators are two perturbations of the operator X - c where c = 0 and c = 1.
3.1. Orthogonal polynomials via raising operator X — aHq
Let us introduce the following lemma. Lemma 2 [12]. The following properties hold
Hq(fg)(x) = f (x)(Hqg)(x)+ g(x)(Hqf)(x) + (q — 1)x(Hqf)(x)(Hqg)(x), f,g € P,
hq-1 f ) Hq U
where
Hq(fu) = (hq-1 f)HqU + q-1(Hq-1 f)u, f €P, U € P'.
HJ(x) = hq^i1)fx{X\ and hqf(x)=f(qx).
Now, recall the operator
Ha,q : P —^ P,
f Ha,q (f ):= xf + aHq (f ).
Definition 1. We call a sequence {Pn}n>0 of orthogonal polynomials Ha,q-classical if there exists a sequence {Qn}n>0 of orthogonal polynomials such that Ha,qPn = Qn+1, n > 0.
For any MPS {Pn}n>0 we define the MPS {Qn}n>0, given by
Qn+1(x) := Ha,qPn(x), n > 0,
or equivalently
Qn+1 (x) := xPn(x) + a(HqP,n)(x), n > 0, (3.1)
with initial value Q0(x) = 1.
Our next goal is to describe all the Ha,q-classical polynomial sequences. Note that, we need a = 0 to ensure that {Qn}n>0 is an orthogonal sequence. Indeed, if we suppose that a = 0, the relation (3.1) becomes, for x = 0, Qn+1(0) =0, n > 0, which contradicts the orthogonality of {Q }n>0.
Clearly, the operator Ha,q raises the degree of any polynomial. Such operator is called raising operator [9, 13, 19]. By transposition of the operator Ha,q, we get
fHaq = X - aHq. (3.2)
Denote by {un}n>0 and {vn}n>0 the dual basis in P' corresponding to {Pn}n>0 and {Qn}n>0, respectively. Then, according to Lemma 1 and (3.2), the relation
xVn+1 - aHq(vn+1) = u,n, n > 0, (3.3)
holds. Assume that {Pn}n>0 and {Qn}n>0 are MOPS satisfying
( P0(x) = 1, P1(x)= x - £0,
] Pn+2(x) = (x - £n+1)Pn+1(x) - Yn+1Pn(x), Yn+1 =0, n > 0,
\ Qû(x) = 1, Q1(x) = x — p0,
\ Qn+2 (x) = (x — pn+1)Qn+1(x) — Qn+1Qn(x), Qn+1 =0, n > 0. Next, a first result will be deduced as a consequence of the relations (3.1), (3.4) and (3.5).
(3.4)
(3.5)
Proposition 2. The sequences {Pn}n>0 and {Qn}n>0 satisfy the following finite type relation
Pn(x) + (q - 1)xHq(P„)(x) = qnQn(x), n > 0. Proof. Using (3.4), we obtain
Hq(P„+2)(x) = Hq ((x - £n+l)Pn+l) (x) - Yn+lHq(Pn)(x), n > 0. According to the Lemma 2, we obtain for n > 0
Hq(Pn+2)(X) = (X - ^„+l)Hq(P„+l)(x) + P„+l(x) + (q - 1)xHq(P„+l)(x) - Yn+lHq(P„)(x), or equivalently
xPn+2(x) + a(HqPn+2)(x) = Qn+3(x), n > 0,
which gives us for n > 0
(x-^„+l)xP„+l(x) +a(qx-^„+l)(HqPn+l)(x)- Yn+l (xPn(x)+a(HqPn)(x^+aP„+l(x) =Qn+3(x).
We use (3.1) and the last equation becomes for n > 0
(x - ^n+l)Qn+2(x) + a(q - 1)x(HqPn+l)(x) - Yn+lQn+l(x) + aPn+l(x) = Qn+3(x). (3.6)
Inserting (3.5) in (3.6), we obtain
aPn+l(x) + a(q - 1)x(HqPn+l)(x) = (^n+l-Pn+2)Qn+2(x) + (Yn+l - £n+2)Qn+l(x), n > 0.
In fact, this result is valid for n + 1 replaced by n. More precisely, we have for all n > 0
aPn(x) + a(q - 1)x(HqPn)(x) = - Pn+l)Qn+l(x) + (Yn - £>n+l)Qn(x),
with the convention y0 = 0. By comparing the degrees in the previous equation, we get = pn+l, n > 0 and aqn = Yn - £n+b n > 0. Hence the desired result is proven. □
Note that, for n = 0 the relation (3.1) gives p0 = 0, for n = 1 the Proposition 2 gives
(x - yS0) + (q - 1)x = qx - p0 = qx,
then = pl = 0. Now we establish, in the next lemma, an algebraic relation between the forms u0 and v0.
Lemma 3. The forms u0 and v0 satisfy the following relation
vo - (q - 1)Hq(xvo) = uo. (3.7)
P r o o f. According to Proposition 2 we obtain
(vo - (q - 1)Hq(xvo),Pn) = 0, n > 1. (3.8)
On the other hand,
(vo - (q - 1)Hq(xvo), Po) = 1,
since {Qn}n>0 is orthogonal with respect to the form v0, where v0 is supposed normalized. According to Lemma 1 and using (3.8), we obtain the desired result. □
Based on the last lemma, we can state the following theorem.
Theorem 1. The form v0 satisfies the following Pearson's equation
(Hqvo) - -xvo = 0, (3.9)
a
and then the scaled q-Hermite polynomial sequence is the only Ha,q-classical sequence. Proof. According to Proposition 1 (ii), the relation (3.3) can be written as follows
xQn+1(x)V0 - aHq(Qn+1V0) = XnPn(x)u0, n > 0, (3.10)
where
An := (vc,Qn+1)(u0,Pn)-1, n > 0.
Making n = 0 in (3.10), we get
x2v0 — aHq (xv0) = —au0, (Q1(x) = x, g1 = -a). Substituting this relation in (3.7), we obtain
qHq(xv o) — — (x2 + a)vo = 0.
Note that we have qHq(xv0) = x(Hqv0) + v0, then
(Hqvo) - ~xvq = 0, (3.11)
which gives
1
((Hqvo) - -xv0)n+1 = 0, n > 0,
and then
(V0)n+2 = -a[n]q(V0)n, n > 0.
Moreover, (v0)i = p\ = 0, hence (v0)2n+i = 0, n > 0. We can conclude that {Qn}n>0 is symmetric. Using the Proposition 2, we obtain
Qn (x) = q-nPn (qx), n > 0.
Then we also conclude that {Pn}n>0 is symmetric. Moreover, the relation (3.11) corresponds to a Pearson's equation of q-Hermite linear functional, hence Qn(x) is the q-Hermite polynomial. In addition, we have Qn(x) = q-nPn(qx), n > 0, then Pn(x) is the scaled q-Hermite polynomial. □
3.2. Orthogonal polynomials via raising operator (X + 1) — Ar-1 In this part, we use the following lemma. Lemma 4 [1]. The following properties hold
Dw(fg)(x) = f (x)(Dwg)(x) + g(x)(Dwf)(x) + w(Dwf)(x)(Dwg)(x), f,g &P, D-w(fu)= g(D-wu) + (D-wg)(Twu), f eP, u eP, Tb o Dw = Dw o Tb in P and P', b e C,
where
Duf(x) := , u / 0 and t-Uf(x) = f(x + u).
Recall the operator
SA : P —► P,
f SA(f) = (x + !)(/) - At_i/.
Definition 2. We call a sequence {Pn}n>0 of orthogonal polynomials SA-classical if there exists a sequence {Qn}n>0 of orthogonal polynomials such that SAPn = Qn+l, n > 0.
For any MPS {Pn}n>0 we define the MPS {Qn}n>0, given by
Qn+l(x) := SAPn(x), n > 0, (3.12)
or equivalently
Qn+l(x) := (x + 1)Pn(x) - APn(x + 1), n > 0, (3.13)
with initial value Q0(x) = 1.
Our next goal is to describe all the SA-classical polynomial sequences. Note that, we need A = 0 to ensure that {Qn}n>0 is an orthogonal sequence. Indeed, if we suppose that A = 0, the relation (3.13) becomes, for x = -1, Qn+l(-1) =0, n > 0, which contradicts the orthogonality
of {Qn}n>0.
Clearly, the operator SA raises the degree of any polynomial. Such operator is called a raising operator [9, 13, 19]. By transposition of the operator SA, we get
*SA = (X + 1) - ATi. (3.14)
Denote by {un}n>0 and {vn}n>0 the dual basis in P' corresponding to {Pn}n>0 and {Qn}n>0, respectively. Then, according to Lemma 1 and (3.14), the relation
(x + 1)Vn+l - ATlVn+l = Un, n > 0,
holds. Assume that {Pn}n>0 and {Qn}n>0 are MOPS satisfying
Po(x) = 1, Pi(x)= x - ft,
Pn+2(x) = (x - ^n+i)Pn+i(x) - Yn+iPn(x), Yn+1 =0, n > 0,
(3.15)
(3.16)
\ Qo(x) = 1, Qi(x) = x - po,
\ Qn+2(x) = (x - Pn+i)Qn+i(x) - £n+iQn(x), = 0, n > 0.
Next, a first result will be deduced as a consequence of the relations (3.13), (3.15) and (3.16).
Proposition 3. The sequences {Pn}n>0 and {Qn}n>0 satisfy the following finite type relation
Qn(x) = r_iPra(x), n > 0,
with
Pn+i = ^n, n > 0, ^n+i = Yn + A, n > 0,
and with the convention y0 = 0.
Proof. Multiplying (3.15) by x + 1, we obtain
(x + 1)Pn+2(x) = (x - ^n+1)(x + 1)Pn+1 (x) - Yn+1(x + 1)Pn(x), n > 0.
Applying Ar-1 to the (3.15) and taking the difference between the two resulting equations, we obtain
(x + 1)Pn+2(x) - A(T-1Pn+2)(x) = (x - £n+1)((x + 1)Pn+1 (x) - A(T-1Pn+1)(x)) -Yn+1 ((x + 1)Pn(x) - A(r-1 Pn)(x)) - APn+1(x + 1).
Substituting (3.13) in the last equation, we get
Qn+3(x) = (x - ^n+1)Qn+2(x) - Yn+1Qn+1(x) - APn+1(x + 1), n > 0.
Using the three-term recurrence relation (3.16), we get
APn+1(x + 1) = (pn+2 - Pn+1)Qn+2(x) + (gn+2 - Yn+1)Qn+1(x), n > 0.
In fact, this result is valid for n + 1 replaced by n. Then, by comparing the degrees in the previous equation, we get pn+1 = and gn+1 = Yn + A, n > 0, and Qn(x) = T-1Pn(x), n > 0, with the convention y0 = 0. □
The following result is a straightforward consequence of Proposition 3. Lemma 5. The forms u0 and v0 satisfy the following relation
According to Lemma 5, and based on some characterizations of Charlier polynomials [1], we can state the following theorem.
Theorem 2. The Charlier polynomial sequence {C^(x)}n>0 where A > 0, is the only S\-classical orthogonal sequence. More precisely, we have for n > 0/
Proof. Assume that {Pn}n>0 is a monic S^-classical orthogonal sequence. Then there exists a monic orthogonal sequence {Qn}n>0 satisfying (3.13), which gives by transposition the following
T1V0 = U0.
Pn(x) = Cra(x)i
Qn(x) = C^(x + 1).
(3.17)
(3.18)
system
(V0, (x + 1)Pn(x) - APn(x + 1)) = (V0, Qn+1 (x)) =0, n > 0.
But the left hand side reads as
((x + 1)v0 - AT1V0, Pn(x)) = 0, n > 0.
In other words,
(x + 1)v0 - AT1V0 = 0.
Applying the operator t-1, we obtain
(x + 2)t-1V0 - Av0 = 0.
Equivalently,
(x + 1)t-1V0 + T-1V0 - (x + 1)v0 + (x + 1)v0 - Av0 = 0,
which also gives
(x + 1)[t_ivo - vo] + T_iVo + (x + 1)vo - Avo = 0,
or equivalently
(x + l)Divo + T_ivo + (x + l)vo - Avo = 0. By using Lemma 4, the last relation becomes
Di(x(Tivo)) + (x - A)(Tivo) = 0,
which means that vo = T_iC(A), where C(A) is the Charlier form with A > 0. In addition, using the Proposition 3, we obtain that Pn(x) = C^(x) are the monic Charlier polynomials and then
Qn(x) = C£(x + l), n > 0.
□
4. Conclusion and prospects
We described Hahn's problem for some perturbed raising operators of the operator X - c using the Pearson equation, which is satisfied by the corresponding linear functionals. Indeed, we have proved that the q-Hermite (resp. Charlier) polynomial is the only Ha,q-classical (resp. SA-classical) orthogonal polynomial, where Ha,q := X + aHq and SA := (X + 1) - At-1. Now, using (3.17), (3.18) and (3.12), we obtain
SaC,A(x) = C,A+i(x + 1), n > 0, which gives, by induction, the following formula
SAm)CA(x) = CA+m(x + m), n > 0, (4.1)
where SAm) = SAm) o...oSAm). Making n = 0 in (4.1) we get
SAm) (1) = Cm(x + m), m > 0.
For prospects, we can replace the operator Hq in Subsection 3.1 by the Dunkl operator (TM := D + 2pH-l, see [6]) and study the same problem. Indeed, we have [6]
= »>0, (42)
where H^(x) is the monic generalized Hermite polynomial and where Y^(n) is defined by , 22mm!r(m + p +1/2) . _ 22m+lm!r(m + p + 1/2)
>(2m) = —wrm—' and >(2m+1) = —wrm—•
In view of (4.2), we can say that {H™}n>0 is an O-classical polynomial sequence, since it fulfills Hahn's property relatively to the raising operator
O := X - -T,t, 2 M
i.e., it is an orthogonal polynomial sequence whose sequence of O-derivatives is also orthogonal.
Acknowledgements
The authors would like to thank the referees for their corrections and many valuable suggestions.
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