orthogonal polynomials and second-order pseudo-spectral linear differential equations

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Orthogonal polynomials and second-order pseudo-spectral lineardifferential equationsFrancisco Marcellán a & Ridha Sfaxi ba Departamento de Matemáticas, Universidad Carlos III de Madrid,Avenida de la Universidad 30, 28911, Leganés, Spainb Faculté des Sciences de Gabès, Département de Mathématiques,Cité Erriadh 6072, Gabès, TunisiePublished online: 12 Nov 2009.

To cite this article: Francisco Marcelln & Ridha Sfaxi (2010) Orthogonal polynomials and second-order pseudo-spectral linear differential equations, Integral Transforms and Special Functions, 21:7,487-501, DOI: 10.1080/10652460903403240

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Integral Transforms and Special FunctionsVol. 21, No. 7, July 2010, 487–501

Orthogonal polynomials and second-order pseudo-spectrallinear differential equations

Francisco Marcellána* and Ridha Sfaxib

aDepartamento de Matemáticas, Universidad Carlos III de Madrid, Avenida de la Universidad 30, 28911Leganés, Spain; bFaculté des Sciences de Gabès, Département de Mathématiques, Cité Erriadh 6072

Gabès, Tunisie

(Received 9 July 2009 )

In this paper, we deal with monic orthogonal polynomial sequences which satisfy the second-order pseudo-spectral linear differential equation:

φ2(x)y′′(x) + φ1(x)y′(x) = χ(x, n)y(x), n ∈ N,

where φi, i = 1, 2 are polynomials with φ2 monic, and the degrees of the polynomials χ(·, n) are uniformlybounded. These polynomial sequences are semiclassical of class either s = 0 or 1. They are, up to a linearchange of variable, the classical polynomials (Hermite, Laguerre, Bessel, and Jacobi) and symmetricsemiclassical polynomials of class one. For them, we deduce the three-term recurrence relations, thestructure relations, and the second-order linear differential equations that these polynomial sequencessatisfy.

Keywords: orthogonal polynomials; three-term recurrence relations; semiclassical polynomials; firststructure relations; second-order linear differential equations

MSC: 42C05; 33C45

1. Introduction and preliminary results

Let P be the linear space of polynomials in one variable with complex coefficients and P′ its

algebraic dual space. We denote by 〈L, f 〉 the action of L ∈ P′ on f ∈ P. Let us introduce some

useful operations in P′ which will be used in the sequel. For any linear functionalL, any polynomial

q, and complex numbers b, c and a, a non-zero complex number, let DL = L′, qL, τbL, haL,

and (x − c)−1L be the linear functionals defined by duality

〈L′, f 〉 := −〈L, f ′〉, 〈qL, f 〉 := 〈L, qf 〉,〈τ−bL, f 〉 := 〈L, τbf 〉 = 〈L, f (x − b)〉, 〈haL, f 〉 := 〈L, haf 〉 = 〈L, f (ax)〉,

〈(x − c)−1L, f 〉 := 〈L, θcf 〉 = 〈L,

f (x) − f (c)

x − c〉, f ∈ P.

*Corresponding author. Email: [email protected]

ISSN 1065-2469 print/ISSN 1476-8291 online© 2010 Taylor & FrancisDOI: 10.1080/10652460903403240http://www.informaworld.com

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488 F. Marcellán and R. Sfaxi

A linear functional L is said to be quasi-definite (regular) if there exists a sequence of monicpolynomials {Bn}n∈N with deg Bn = n such that

〈L, BnBm〉 = rnδn,m, n, m ∈ N, rn �= 0, (1)

where δn,m is the Kronecker symbol. In such a case, {Bn}n∈N is said to be the monic orthogonalpolynomial sequence (MOPS) with respect to L. The orthogonality of the sequence {Bn}n∈N canbe characterized by a three-term recurrence relation (TTRR) [5] :{

Bn+2(x) = (x − βn+1)Bn+1(x) − γn+1Bn(x), n ∈ N,

B1(x) = x − β0, B0(x) = 1,(2)

where βn and γn+1 are complex numbers and γn+1 �= 0 for every n ∈ N.

Notice that if L is a quasi-definite linear functional, then the shifted linear functionalL = (ha−1 ◦ τ−b)L, where a, b are complex numbers and a �= 0, is also quasi-definite.Furthermore, if {Bn}n∈N is the MOPS with respect to L, then the shifted sequence {Bn}n∈N whereBn(x) = a−nBn(ax + b), n ∈ N, is the MOPS with respect to L [5,9].

A linear functional L is said to be semiclassical if it is quasi-definite and there existsan admissible pair of polynomials (φ, ψ), i.e., φ is monic, deg φ = t, deg ψ = p ≥ 1, with(1/p!)ψ(p)(0) /∈ N

∗ if p = t − 1, such that

D(φL) + ψL = 0. (3)

The pair (φ, ψ) is not unique [10], because (3) can be simplified if there exists a zero ξ of φ

satisfying

φ′(ξ) + ψ(ξ) = 0 and 〈L, θ2ξ (φ) + θξ (ψ)〉 = 0. (4)

The division by x − ξ, yields

(θξ (φ)L)′ + (θ2ξ (φ) + θξ (ψ))L = 0. (5)

The minimum value of max(t − 2, p − 1), among all possible admissible pairs (φ, ψ), is said tobe the class of L. The pair (φ, ψ) associated with the class s, where s ∈ N, is unique. Moreover,when L is semiclassical of class s, its corresponding MOPS is said to be semiclassical of classs. If s = 0, L is said to be classical (Hermite, Laguerre, Bessel, and Jacobi). For more details,see [8,9].

When L is semiclassical of class s and satisfies (3), L = (ha−1 ◦ τ−b)L is also semiclassical ofclass s and

(φL)′ + ψL = 0, (6)

holds with φ(x) = a−tφ(ax + b) and ψ(x) = a1−tψ(ax + b).

The semiclassical orthogonal polynomial sequences are characterized by the followingequivalent relations.

(1) First structure relation [9].

φ(x)B ′n(x) = Cn(x) − C0(x)

2Bn(x) − γnDn(x)Bn−1(x), n ∈ N, (7)

where deg Cn ≤ s + 1, deg Dn ≤ s, and

Cn+1(x) + Cn(x) = 2(x − βn)Dn(x), (8)

γn+1Dn+1(x) + φ(x) = γnDn−1(x) + (x − βn)2Dn(x) − (x − βn)Cn(x), (9)

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Integral Transforms and Special Functions 489

for every n ∈ N, where

C0(x) = −φ′(x) − ψ(x), (10)

D0(x) = −(Lθ0φ)′(x) − (Lθ0ψ)(x). (11)

Here (Lθ0f )(x) = 〈Ly, θx(f )(y)〉 = 〈Ly, (f (x) − f (y)/x − y)〉, f ∈ P, with the conven-tion γ0 = 1 and D−1(x) = B−1(x) = 0.

In the sequel, we will use the following relations:

Cn(x) − C0(x)

2= ηnC0(x) + (−1)n−1

n−1∑ν=0

(−1)ν(x − βν)Dν(x), n ∈ N, (12)

where ηn := (−1)n − 1/2, n ∈ N, and

C2n(x) − C2

0 (x)

4− γnDn−1(x)Dn(x) = φ(x)

n−1∑ν=0

Dν(x), n ∈ N. (13)

(2) Second-order linear differential equation of Maroni type [9].

φ(x)B ′′n (x) − ψ(x)B ′

n(x) = M(x, n)Bn(x) − γnD′n(x)Bn−1(x), n ∈ N, (14)

with M(·, n) = C ′n − C ′

0/2 + ∑n−1ν=0 Dν, and Cn, Dn are given by (8)–(11).

(3) Second-order linear differential equation of Laguerre-Perron type [2] (holonomic equation).

J (x, n)B ′′n (x) + K(x, n)B ′

n(x) + L(x, n)Bn(x) = 0, n ∈ N, (15)

where

J (x, n) = φ(x)Dn(x), K(x, n) = (φ′(x) + C0(x))Dn(x) − φ(x)D′n(x),

L(x, n) =(

Cn(x) − C0(x)

2

)D′

n(x) −(

n−1∑ν=0

Dν(x) + C ′n(x) − C ′

0(x)

2

)Dn(x).

Obviously, the triplet (φ, {Cn}n∈N, {Dn}n∈N) given in (7) is not unique. So, (7) can besimplified if and only if there exists a zero ξ of φ such that C0(ξ) = D0(ξ) = 0, i.e.,the two conditions given in (4). From (8)–(10), we get Cn(ξ) = Dn(ξ) = 0, n ∈ N. Thus,dividing by x − ξ in (7), we obtain θξ (φ)(x)B ′

n(x) = (θξ (Cn)(x) − θξ (C0)(x)/2)Bn(x) −γnθξ (Dn)(x)Bn−1(x), where deg θξ (Cn) ≤ s and deg θξ (Dn) ≤ s − 1, for every n ∈ N.

Furthermore, the following result will be needed.

Lemma 1 Let {Bn}n≥0 be a semiclassical polynomial sequence satisfying (7) and let s =max(deg(φ) − 2, deg(ψ) − 1). Then,

(i) If there exists a pair (k, l) ∈ N2 such that deg Dn ≤ l for every n ≥ k, then s ≤ l.

(ii) The set {n ∈ N | deg Dn = s} is infinite.

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Proof Assume the existence of a pair (k, l) ∈ N2 such that deg Dn ≤ l, n ≥ k. Then, if we take

n = k + j, j = 1, 2 in (9), we get

γk+j+1Dk+j+1 + φ = γk+jDk+j−1 + (x − βk+j )2Dk+j − (x − βk+j )Ck+j .

Subtracting the two previous expressions, we deduce

(x − βk+2)Ck+2 − (x − βk+1)Ck+1 = γk+2Dk+1 + (x − βk+2)2Dk+2 − γk+3Dk+3

− γk+1Dk − (x − βk+1)2Dk+1 + γk+2Dk+2. (16)

The analysis of the degrees of both sides in (16) yields

deg ((x − βk+2)Ck+2 − (x − βk+1)Ck+1) ≤ l + 2. (17)

Now, from (2) we can write (16) as follows:

(x − βk+2)Ck+2 − (x − βk+1)Ck+1 = x(Ck+2 − Ck+1) + βk+1Ck+1 − βk+2Ck+2. (18)

Inserting (8) into (9) with n = k + 1, we have

φ(x) = Ck+2(x) − Ck+1(x)

2(x − βk+1) − γk+2Dk+2(x) + γk+1Dk(x). (19)

Notice that deg φ ≤ l + 2. Otherwise, if deg φ ≥ l + 3, then from (19) and the assumptiondeg Dk+1 ≤ l, we get

deg(Ck+2 − Ck+1) = deg(φ) − 1 ≥ l + 2. (20)

On the other hand, rewriting (8) with n = k + 1 in two ways: Ck+1+j (x) = (x − βk+1)Dk+1(x) −(−1)jCk+2(x) − Ck+1(x)/2, j = 0, 1, and using (20) as well as the assumption deg Dk+1 ≤ l,

we obtain

deg Ck+2 = deg Ck+1 = deg φ − 1. (21)

From (18), (20), and (21), deg((x − βk+2)Ck+2 − (x − βk+1)Ck+1) = deg φ, and together with(17), deg φ ≤ l + 2 holds. This is a contradiction. Hence, since deg φ ≤ l + 2, deg Dn ≤ l, n ≥ k,and by (9), deg Cn ≤ l + 1, n ≥ k + 1.

From (7), with n = k + 1, we have C0Bk+1 = −2φB ′k+1 + Ck+1Bk+1 − 2γk+1Dk+1Bk.

But deg φ ≤ l + 2, deg Dk+1 ≤ l and deg Ck+1 ≤ l + 1, then deg C0Bk+1 ≤ k + l + 2,and so deg C0 ≤ l + 1. Since ψ = −C0 − φ′ then deg ψ ≤ l + 1. Thus, max(deg(φ) − 2,

deg(ψ) − 1) ≤ l. Hence, (i) holds.The set {n ∈ N | deg Dn = s} is infinite. Otherwise, there exists k ∈ N such that deg Dn ≤

s − 1, n ≥ k, since deg Dn ≤ s, n ∈ N. But, according to (i) this implies s ≤ s − 1. This is acontradiction. Hence, (ii) holds. �

The aim of this manuscript is to solve the following problem raised by Belmehdi [2]. Classifyall MOPS {Bn}n≥0 satisfying the second-order pseudo-spectral linear differential equation

φ2(x)y ′′(x) + φ1(x)y ′(x) = χ(x, n)y(x), n ∈ N, (22)

where φi , i = 1, 2 are polynomials with φ2 monic, and the degrees of the polynomials χ(·, n) areuniformly bounded.The differential equation (22) and the more general (15) having an MOPS as solution are veryimportant in the theory of orthogonal polynomials. Such equations are used as a basic tool in the

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electrostatic interpretation of the zeros of the polynomials Bn(x) (see [4,7] and the referencestherein).

When max(deg(φ2) − 2, deg(φ1) − 1) = 0, then the polynomial solutions of (22) are the clas-sical orthogonal polynomials. This result has been obtained in 1889 by Routh [11] (see also [3]).However, when max(deg(φ2) − 2, deg(φ1) − 1) ≥ 1, the classification is not completed and onlysome particular cases have been studied in the literature. For the of an example, we can mentionthe generalized Gegenbauer sequence analysed in [2] by Belmehdi. In addition, the same authorshows that (15) is pseudo-spectral, if and only if Dn(x) is the product of a function of n and apolynomial in x, i.e., Dn(x) = nD(x), n ∈ N. In the present paper, we obtain all MOPS whichare solutions of (22).

The structure of this paper is as follows. In Section 2, we prove (Theorem 3) that the MOPSwhich are solutions of (22) are semiclassical of class either s = 0 or 1. They also satisfy apseudo-spectral second-order linear differential equation of Laguerre–Perron type, with Dn(x) = n(x − d)s . In Section 3, we establish the system that satisfy the corresponding elements Cn, Dn

in the pseudo-spectral cases. Finally, we describe all the resulting cases.

2. Main results

Let {Bn}n∈N be an MOPS with respect to a linear functional L satisfying the TTRR (2) as well asthe second-order pseudo-spectral linear differential equation

φ2(x)B ′′n (x) + φ1(x)B ′

n(x) = χ(x, n)Bn(x), n ≥ 0, (23)

where φν , ν = 1, 2 and χ(·, n) are polynomials, φ2 is monic, the coefficients of χ(·, n) dependon n, and there exists σ ∈ N such that

deg χ(·, n) ≤ σ, for every n ∈ N. (24)

Setting

κ = max(σ, μ), (25)

where μ = max(deg(φ2) − 2, deg(φ1) − 1), then

Proposition 2 The MOPS {Bn}n∈N is semiclassical of class less than or equal to κ. Furthermore,the corresponding quasi-definite linear functional L satisfies

D(φ2L) + ψ2L = 0, (26)

where ψ2 = 1/2(A0B1 − φ1 − A1), deg ψ2 ≥ 1, and Ai = ∑σ+iν=0(〈L, χ(·, ν)BiBν〉/〈L, B2

ν 〉)Bν,

i = 0, 1.

Proof From (1) 〈(φ2BiL)′′ − (φ1BiL)′, Bn〉 = 0, n ≥ σ + 1 + i, i = 0, 1. Then, for i = 0, 1there exists Ai ∈ P, deg Ai ≤ σ + i such that

(φ2BiL)′′ − (φ1BiL)′ = AiL, (27)

where Ai = ∑σ+iν=0〈L, B2

ν 〉−1〈L, χ(·, ν)BiBν〉Bν , i = 0, 1. For i = 1, (27) becomesB1((φ2L)′′ − (φ1L)′) + 2(φ2L)′ − φ1L = A1L, and by inserting (27) with i = 0, we getB1A0L + 2(φ2L)′ − φ1L = A1L. This yields (26), where ψ2 = 1/2(A0B1(x) − φ1 − A1).Clearly, deg ψ2 ≤ max(σ + 1, deg φ1), since deg Ai ≤ σ + i, i = 0, 1. Besides, deg ψ2 ≥ 1,

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since φ2 is a non-zero polynomial and L is quasi-definite. Using (25), we obtain max(deg(φ2) − 2,deg(ψ2) − 1) ≤ κ. Thus, {Bn}n≥0 is semiclassical of class less than or equal to κ . �

In the sequel, we will assume that s is the class of {Bn}n∈N, and L satisfies

D(φL) + ψL = 0, (28)

where (φ, ψ) ∈ P2, with φ monic, deg ψ ≥ 1, and s = max(deg(φ) − 2, deg(ψ) − 1). From

Proposition 2, we deduce that

s ≤ κ. (29)

Of course, (26) can be simplified and then we can assume that there exists a monic polynomial �

such that

φ2(x) = �(x)φ(x), ψ2(x) = �(x)ψ(x) − �′(x)φ(x). (30)

Now, we can give the main result of this section.

Theorem 3 The sequence of monic orthogonal polynomials which are solutions of a second-order pseudo-spectral linear differential equation is semiclassical of class at most one and satisfiesa second-order pseudo-spectral linear differential equations of the Laguerre–Perron type.

First, we need the three following technical lemmas.

Lemma 4 Let {Bn}n∈N be defined as above and satisfying (7). Then, there exists a monicpolynomial D, deg D = s, and k ∈ N, k ≤ κ + s + 1, such that

Dn(x) = nD(x), n ≥ k, (31)

(φ1(x) + �(x)ψ(x))D + �(x)φ(x)D′(x) = 0. (32)

Proof Multiplying both sides of (14) by �(x) and using (30), we get

φ2B′′n − �ψB ′

n = �M(·, n)Bn − γn�D′nBn−1.

The elimination of φ2B′′n between (23) and the previous equation yields

(φ1 + �ψ)B ′n = (χ(·, n) − M(·, n)�)Bn + γn�D′

nBn−1, n ∈ N. (33)

Multiplying both sides of (33) by φ and inserting (7),

�2(·, n)Bn = �1(·, n)Bn−1, n ∈ N, (34)

where

�1(·, n) = γn((φ1 + ψ�)Dn + �φD′n), deg �1(·, n) ≤ σ + s + 1,

�2(·, n) = Cn − C0

2(φ1 + ψ�) − (χ(·, n) − M(·, n)�)φ, deg �2(·, n) ≤ σ + s + 2.

But, as a consequence of the TTRR, Bn and Bn−1 are coprime polynomials for each n ≥ 1. So, by(34) there exists k ∈ N, k ≤ κ + s + 1, such that �1(·, n) = �2(·, n) = 0, n ≥ k. In particular,this implies

(φ1(x) + �(x)ψ(x))Dn(x) + �(x)φ(x)D′n(x) = 0, n ≥ k, (35)

and then Dn(x) = nD(x), n ≥ k. Hence (31) holds.

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From (31), (35) becomes n((φ1 + �ψ)D + �φD′) = 0, n ≥ k. But, by Lemma 1, (ii), withDn = nD, n ≥ k, we deduce that deg D = s and the set {n ∈ N | n ≥ k and n �= 0} is infinite.Hence, (32) follows immediately. �

Lemma 5 If deg D ≥ 1, then for any zero d of D(x), we have

(i) Cn(d) = (−1)n−kCk(d), n ≥ k.(ii) Cn(d) �= 0, n ≥ k.

(iii) βn = φ(d)/Ck(d)(−1)n−k + d , n ≥ k + 1.

Proof Assume that d is a zero of D(x). Then, (8) with Dn(x) = nD(x), n ≥ k and evaluating atx = d, yields Cn+1(d) + Cn(d) = 0, n ≥ k. Hence, (i) follows as a straightforward consequence.

From (i) and (9), we get

φ(d) = (−1)n−k+1(d − βn)Ck(d), n ≥ k + 1. (36)

Notice that Ck(d) �= 0. Otherwise, Cn(d) = 0, n ≥ k, and φ(d) = 0. Then, by (13) with x = d

and n ≥ k, we obtain C0(d) = 0.Let us consider the following recurrence property. For each i ∈ N and i ≤ k, we have Dn−i (d) =

0 and Cn−i (d) = 0, n ≥ k. For i = 0, it is easy to check that the recurrence property is true.Assumethat it holds until order i, i ≤ k − 1, and let us show that it remains valid for order i + 1. From(9) with n = k − i and x = d , and the recurrence hypothesis, we get γk−iDk−i−1(d) = 0. Sinceγn �= 0, n ≥ 1, it follows that Dk−i−1(d) = 0. Moreover, (13) with n = k − i − 1 and x = d,leads to Ck−i−1(d) = 0. Then, Dn−i−1(d) = 0 and Cn−i−1(d) = 0, n ≥ k. Thus, the recurrenceproperty holds.

So, φ(d) = Cn(d) = Dn(d) = 0, n ≥ 0, and then we can divide by x − d in (7) and in ananalogue way in (28). This is a contradiction. Hence, (ii) holds.

The statement (iii) follows in a straightforward way from (36) and (ii). �

Lemma 6 We have D(x) = (x − d)s, where s ∈ {0, 1}.

Proof If di , i = 1, 2 are two distinct zeros of D(x), then by Lemma 5, (iii), βn = φ(di)/

Ck(di)(−1)n−k + di , n ≥ k + 1, i = 1, 2. Handling in the previous relation, (φ(d1)/Ck(d1) −φ(d2)/Ck(d2))(−1)n−k = d2 − d1, n ≥ k + 1. This means that φ(d1)/Ck(d1) − φ(d2)/Ck(d2) =0 and d1 = d2. But, d1 �= d2 by hypothesis. Thus, D(x) = (x − d)s .

Notice that s ∈ {0, 1}. Otherwise, if s ≥ 2, then D(d) = D′(d) = 0. Differentiating both sidesof (8) with Dn = nD, n ≥ k, and evaluating at x = d, we get C ′

n+1(d) + C ′n(d) = 0, n ≥ k, i.e.,

C ′n(d) = (−1)n−kC ′

k(d), n ≥ k. (37)

In the same way, differentiating both sides of (9) with Dn = nD, n ≥ k, and evaluating atx = d, we get −φ′(d) − (d − βn)C

′n(d) − Cn(d) = 0, n ≥ k + 1. By (37) and Lemma 5, (iii),

we obtain (−1)n−kCk(d) + φ′(d) − φ(d)C ′k(d)/Ck(d) = 0, n ≥ k + 1. This means that φ′(d) −

φ(d)C ′k(d)/Ck(d) = 0 and Ck(d) = 0 hold. Thus, we get a contradiction and, as a consequence,

our statement follows. �

Now, we prove the main result of this section.

Proof of Theorem 3 According to Lemma 6, we will analyse two cases.

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• s = 0. Then, Dn(x) = n ∈ C, where n ∈ N. From (14), {Bn}n≥0 satisfies:

φ(x)B ′′n (x) − ψ(x)B ′

n(x) = λnBn(x), n ∈ N, (38)

where deg φ ≤ 2, deg ψ = 1, and λn = n((n − 1)φ′′(0)/2 − ψ ′(0)).• s = 1. Then, deg Dn ≤ 1, where n ∈ N, and

Dn(x) = n(x − d) + υn, n ∈ N, (39)

where υn ∈ C, n ∈ N, and υn = 0, n ≥ k. Let us prove that υn = 0, 0 ≤ n < k. If we multiplyboth sides of (23) by x − d and we use (30) and (32), we obtain �((x − d)(φB ′′

n − ψB ′n) −

φB ′n) = (x − d)χ(·, n)Bn. Then, (14) and the fact that D′

n(x) = n, n ∈ N, yield

�φB ′n(x) = Cn(x) − C0(x)

2Bn(x) − γnDn(x)Bn−1(x), n ∈ N, (40)

where C0 = − (φ′ + ψ)�, Cn − C0/2 = (x − d)(�M(·, n) − χ(·, n)), and Dn = n(x − d)�.

From (7), we can divide by � in (40). In this case, � divides Cn, n ∈ N.We can write Cn = �Cn,n ∈ N. So, if we set C0 = C0, it follows that

φ(x)B ′n(x) = Cn(x) − C0(x)

2Bn(x) − γn n(x − d)Bn−1(x), n ∈ N. (41)

CancellingφB ′n(x) from (7) and (41), and using (39), we easily obtain (Cn(x) − Cn(x)/2)Bn(x) =

−γnυnBn−1(x), where n ∈ N. Thus υn = 0 and Cn = Cn, n ∈ N, since γn �= 0, n ≥ 1, deg Bn = n,

and B−1(x) = 0. Consequently, Dn(x) = n(x − d), n ∈ N, and then {Bn}n∈N satisfies a second-order pseudo-spectral linear differential equation of the Laguerre–Perron type [2]. �

3. Orthogonal polynomials as solutions of second-order pseudo-spectral differentialequation of the Laguerre-Perron type

Let {Bn}n∈N be a semiclassical MOPS of class s which is a solution of a second-order pseudo-spectral linear differential equation of the Laguerre–Perron type with Dn(x) = nD(x), n ∈ N.

Then (15) becomes

J (x)B ′′n (x) + K(x)B ′

n(x) + L(x, n)Bn(x) = 0, n ∈ N, (42)

where

J (x) = φ(x)D(x), K(x) = (φ′(x) + C0(x))D(x) − φ(x)D′(x),

L(x, n) =(

Cn(x) − C0(x)

2

)D′(x) −

(n−1∑ν=0

νD(x) + C ′n(x) − C ′

0(x)

2

)D(x).

The first structure relation (7) reads

φ(x)B ′n(x) = Cn(x) − C0(x)

2Bn(x) − γn nD(x)Bn−1(x), n ∈ N, (43)

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where ⎧⎪⎨⎪⎩

Cn+1(x) + Cn(x) = 2 n(x − βn)D(x), C0(x) = −φ′(x) − ψ(x)

[γn+1 n+1 − γn n−1 − n(x − βn)2]D(x) = −φ(x) − (x − βn)Cn(x)

D0(x) = −(Lθ0ψ)(x) − (Lθ0φ)′(x), ( −1 = 0).

(44)

In addition, from (12) and (13),

Cn(x) − C0(x)

2= ηnC0(x) + (anx − bn)D(x), n ∈ N, with (45)

an = (−1)n−1n−1∑ν=0

(−1)ν ν, bn = (−1)n−1n−1∑ν=0

(−1)ν νβν, (46)

C2n(x) − C2

0 (x)

4− γn n n−1D

2(x) = φ(x)D(x)

n−1∑ν=0

ν, n ∈ N. (47)

3.1. The associated system in the pseudo-spectral case

From (44), we have

φ(x) = [ n(x − βn)2 − n+1γn+1 + γn n−1]D(x) − (x − βn)Cn(x), (48)

φ(x) = [ 0(x − β0)2 − 1γ1]D(x) − (x − β0)C0(x). (49)

Cancelling φ(x) in (48) and (49), and using (45), we obtain

E(x, n)D(x) = F(x, n)C0(x), n ≥ 1, where (50)

E(x, n) = �n,2x2 + �n,1x + �n,0, n ≥ 1, with (51)

�n,0 = 1

2[− nβ

2n + 0β

20 + n+1γn+1 − γn n−1 − 1γ1 + 2bnβn],

�n,1 = −β0 0 + ( n − an)βn − bn, �n,2 = 0 − n

2+ an, and

F(x, n) = −ηnx + (−1)nβn − β0

2. (52)

For n = 1, (50) reads

C∗0 (x)D(x) = D∗(x)C0(x), (53)

where C∗0 (x) := E(x, 1) and D∗(x) := F(x, 1).

Multiplying both sides of (50) by D∗(x) and using (53), we get

D∗(x)E(x, n) = C∗0 (x)F (x, n), n ≥ 1. (54)

Thus, the identification of the coefficients yields the following system:⎧⎪⎪⎨⎪⎪⎩

�n,2 = 1 − 3 0

2ηn, �n,1 = ηnR + (−1)nβn − β0

2�1,2, n ≥ 1,

�n,0 = −ηnS − (−1)nβn − β0

2R, [−(β1 + β0)ηn + (−1)nβn − β0]S = 0, n ≥ 1,

where R = −(β1 + β0/2)�1,2 − �1,1, and S = −(β1 + β0/2)R + �1,0.

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Using (46) and (51), the first equation of the previous system reads as follows n+1 + n =2 n + 1 − 0, n ∈ N, i.e., n = ( 1 − 0)n + 0, n ∈ N. Then, from (46) we get

an = 1 − 0

2n + 1 − 3 0

2ηn, n ∈ N. (55)

As a consequence, the previous system is equivalent to the following one:

n = ( 1 − 0)n + 0, n ∈ N (56)

bn = − 1 + 0

4β0 +

[ 1 − 0

2n + 1 + 0

4

]βn − Rηn, n ≥ 1 (57)

n+1γn+1 − γn n−1 =[ 0 − 1

2βn − R + 1 + 0

2β0

]βn

− 2Sηn + γ1 1 + β0(R − β0 0), n ≥ 1 (58)

[−(β1 + β0)ηn + (−1)nβn − β0]S = 0, n ≥ 1, (59)

where R = ( 1 + 5 0/4)β0 + ( 0 − 3 1/4)β1 and S = 1/2( 2γ2 − ( 0 + 1)(γ1 +(β0 − β1/2)2)).

3.2. Classification and resolution of the system

First, we give the following result:

Lemma 7 Let {Bn}n∈N be an MOPS which is a solution of (42). Then it is semiclassical of classat most one and satisfies the following first structure relation:

φ(x)B ′n(x) = C∗

n(x) − C∗0 (x)

2Bn(x) − γn nD

∗(x)Bn−1(x), n ∈ N, with (60)

φ∗(x) = ( 0(x − β0)2 − 1γ1)D

∗(x) − (x − β0)C∗0 (x),

C∗n(x) − C∗

0 (x)

2= ηnC

∗0 (x) + (anx − bn)D

∗(x), n ∈ N, (61)

C∗0 (x) =

(3 0 − 1

2x − R

)D∗(x) + S, D∗(x) = x − β1 + β0

2.

Proof If we multiply both sides of (45), (resp. (49)), by D∗(x) and we take (53) into account,we obtain

Cn(x) − C0(x)

2D∗(x) = C∗

n(x) − C∗0 (x)

2D(x), n ≥ 1, (62)

φ(x)D∗(x) = φ∗(x)D(x). (63)

Multiplying both sides of (43) by D∗(x) and using (62) and (63), then (60) follows. Notice thatthe quasi-definite linear functional L satisfies

D(φ∗L) + ψ∗

L = 0, (64)

where

φ∗(x) = ( 0(x − β0)2 − 1γ1)D

∗(x) − (x − β0)C∗0 (x),

ψ∗(x) = −φ∗′(x) − C∗

0 (x) = − 1B2(x).

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Since max(deg(φ∗) − 2, deg(ψ∗) − 1) = 1, then {Bn}n∈N is semiclassical of class at most one.�

As a consequence, {Bn}n∈N satisfies

J ∗(x)B ′′n (x) + K∗(x)B ′

n(x) + L∗(x, n)Bn(x) = 0, n ∈ N, (65)

where

J ∗(x) = φ∗(x)D∗(x), K∗(x) = (φ∗(x)′ + C∗

0 (x))D∗(x) − φ∗(x),

L∗(x, n) = C∗n(x) − C∗

0 (x)

2−

(n−1∑ν=0

νD∗(x) + C∗

n(x)′ − C∗0 (x)′

2

)D∗(x).

From Lemma 7, we get φ∗(β0 + β1/2) = (β0 − β1/2)C∗0 (β0 + β1/2). Therefore,

C∗0

(β0 + β1

2

)= S, φ∗

(β0 + β1

2

)= β0 − β1

2S. (66)

Proposition 8 Let {Bn}n∈N be an MOPS satisfying the TTRR (2) that is a solution of (42), andlet S = 1/2[ 2γ2 − ( 0 + 1)(γ1 + (β0 − β1/2)2)].(i) If S = 0, then {Bn}n∈N is a classical MOPS.

(ii) If S �= 0, then {Bn}n∈N is a semiclassical MOPS of class one.

Proof From Lemma 7, (60) can be simplified since the class of {Bn}n∈N is less than or equal toone. The simplification in (60) is possible, only by dividing by D∗(x). This requires that S = 0.

In this case, C∗0 (β0 + β1/2) = φ∗(β0 + β1/2) = 0 and C∗

n(β0 + β1/2) = 0, n ∈ N, by (66) and(61), respectively. These are the conditions that we need in order to divide by D∗(x). �

Now, we can give the description of all MOPS {Bn}n∈N solutions of a second-order pseudo-spectral linear differential equation of the Laguerre–Perron type.

I.S = 0, i.e., 2γ2 = ( 0 + 1)[γ1 + (β0 − β1/2)2]. In this case,

C∗0 (x) = D∗(x)C0(x),

φ∗(x) = φ(x)D∗(x),

C∗n(x) − C∗

0 (x)

2= Cn(x) − C0(x)

2D∗(x), n ≥ 1,

where

φ(x) = 1 − 0

2(x − β0)

2 + (β0 − β1)(3 1 − 0)

4(x − β0) − γ1 1,

Cn(x) − C0(x)

2=

(3 0 − 1

2ηn + an

)x − Rηn − bn, n ≥ 1, C0(x) = 3 0 − 1

2x − R.

Division by D∗(x) in (60) leads to

φ(x)B ′n+1(x) = Cn+1(x) − C0(x)

2Bn+1(x) − γn+1 n+1Bn(x), n ∈ N.

Thus, {Bn}n∈N is classical, the linear functional L satisfies

D(φL) + ψL = 0,

where φ(x) = ( 1 − 0/2)(x − β0)2 + ((β0 − β1)(3 1 − 0)/4)(x − β0) − γ1 1, and ψ(x) =

−φ′(x) − C0(x) = −( 0 + 1/2)B1(x). Notice that, when 1 �= 0, φ(x) = ( 1 − 0/2)

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[(x − d)2 − (1/4)μ], with d = β0 − (β0 − β1)(3 1 − 0)/4( 1 − 0), and μ = ((β0 − β1)2

(3 1 − 0)2/4)( 1 − 0)

2 + 8 1γ1/ 1 − 0.

The second-order linear differential equation satisfied by Bn is reduced to

J (x)B ′′n (x) + K(x)B ′

n(x) + L(x, n)Bn(x) = 0, n ∈ N,

with J (x) = φ(x), K(x) = −ψ(x), and L(x, n) = −n(n − 1)( 1 − 0)/2 − ( 1 + 0)/2n.According to the number of zeros of the polynomial φ(x)we recover, by using a suitable shifting,

the well-known four canonical cases of classical orthogonal polynomials [8,9] (Tables 1–4).II. S �= 0, i.e., 2γ2 �= ( 0 + 1)[γ1 + (β0 − β1/2)2]. In this case, {Bn}n∈N is semiclassical of

class one.

Lemma 9 When S �= 0, the system (56)–(59) becomes

βn = β0, n ∈ N, (67)

n+1γn+1 − n−1γn = −2ηnS + γ1 1, n ≥ 1, (68)

Table 1. Hermite polynomials.

(I1) Hermite: R = S = 0,

φ(x) = 1, ψ(x) = 2x, βn = 0, γn+1 = n + 1

2, n ∈ N,

Cn(x) = −2x, n = −2, an = 2ηn, bn = 0, n ∈ N,

B ′n(x) = nBn−1(x), n ∈ N,

B ′′n (x) − ψ(x)B ′

n(x) = −2nBn(x), n ∈ N.

Table 2. Laguerre polynomials.

(I2) Laguerre: R = −α, S = 0, α �= −n, n ≥ 1,

φ(x) = x, ψ(x) = x − α − 1, βn = 2n + α + 1, γn+1 = (n + 1)(n + α + 1), n ∈ N,

Cn(x) = −x + 2n + α, n = −1, an = ηn, bn = −n + αηn, n ∈ N,

φ(x)B ′n(x) = nBn(x) + n(n + α)Bn−1(x), n ∈ N,

φ(x)B ′′n (x) − ψ(x)B ′

n(x) = −nBn(x), n ∈ N.

Table 3. Bessel polynomials.

(I3) Bessel: R = −2, S = 0, α �= −n

2, n ∈ N,

φ(x) = x2, ψ(x) = −2(αx + 1), β0 = − 1

α, βn+1 = 1 − α

(n + α)(n + α + 1), n ∈ N,

γ1 = − 1

α2(2α + 1), γn+1 = − (n + 1)(n + 2α − 1)

(2n + 2α − 1)(n + α)2(2n + 2α + 1), n ≥ 1,

C0(x) = 2(α − 1)x + 2, Cn(x) = 2(n + α − 1)x + 2(α − 1)

(n + α − 1), n ≥ 1,

n = 2n + 2α − 1, an = n − 2(α + 1)ηn, bn = n

n + α − 1+ 2ηn, n ∈ N,

φ(x)B ′n(x) = n

(x − 1

n + α − 1

)Bn(x) − (2n + 2α − 1)γnBn−1(x), n ∈ N,

φ(x)B ′′n (x) − ψ(x)B ′

n(x) = n(n + 2α − 1)Bn(x), n ∈ N.

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Table 4. Jacobi polynomials.

(I4) Jacobi: R = α − β, S = 0, α, β �= −n, α + β �= −n − 1, n ≥ 1,

φ(x) = x2 − 1, ψ(x) = −(α + β + 2)x + α − β,

β0 = α − β

α + β + 2, βn+1 = α2 − β2

(2n + α + β + 2)(2n + α + β + 4), n ∈ N,

γ1 = 4(α + 1)(β + 1)

(α + β + 2)2(α + β + 3), γn+1 = 4(n + 1)(n + α + β + 1)(n + α + 1)(n + β + 1)

(2n + α + β + 1)(2n + α + β + 2)2(2n + α + β + 3), n ≥ 1,

C0(x) = (α + β)x + β − α, Cn(x) = (2n + α + β)x + β2 − α2

2n + α + β, n ≥ 1,

n = 2n + α + β + 1, an = n − (α + β)ηn, bn = (β − α)

(ηn + n

2n + α + β

), n ∈ N,

φ(x)B ′n(x) = n

(x + α − β

2n + α + β

)Bn(x) − (2n + 2α + β + 1)γnBn−1(x), n ∈ N,

φ(x)B ′′n (x) − ψ(x)B ′

n(x) = n(n + α + β + 1)Bn(x), n ∈ N.

where n = ( 1 − 0)n + 0, n ∈ N, S = 1/2[ 2γ2 − ( 0 + 1)γ1], and R = (3 0 − 1/2)β0.

Proof From (59) and the assumption S �= 0, we obtain

βn = β0 − β1

2(−1)n + β0 + β1

2, n ∈ N. (69)

Clearly,∑n−1

ν=0(−1)ν = −ηn and∑n−1

ν=0 ν(−1)ν = (2n − 1)(−1)n + 1/4, where n ∈ N. From(46), (56), and (69), bn = (−1)n−1(( 1 − 0)(β0 − β1)/4n(n − 1) + (β0 − β1/2) 0n − (β0 +β1)( 1 − 0)/8) + (β0 + β1)( 1 − 0)/8(2n − 1) − (β0 + β1/2) 0ηn,n ∈ N. On the other hand,bn = −( 1 + 0/4)β0 + (( 1 − 0/2)n + ( 1 + 0/4))((β0 − β1/2)(−1)n + (β0 + β1/2)) −Rηn, n ∈ N, due to (57) and (69). By identification of the two expressions of bn, (β0 − β1) i = 0,i = 0, 1. Then, β1 = β0. Otherwise, n = 0, n ∈ N, due to (56). So, S = 0 which yields acontradiction. Since β1 = β0, then by (68) we obtain (67).

Finally, (68) follows from (58) and (67). �

By shifting, we can take β0 = 0. Thus, βn = 0, n ∈ N, and then {Bn}n∈N will be a symmet-ric polynomial sequence, i.e., if Bn(−x) = (−1)nBn(x), n ∈ N. In this case, φ∗(x) = x(( 1 − 0/2)x2 − 1γ1 − S), and ψ∗(x) = − 1(x

2 − γ1).

In [1], (see also [6]), a description of all symmetric semiclassical polynomial sequences of classone is done. There are three canonical choices for φ∗ (Tables 5–7)

φ∗(x) = x, φ∗(x) = x(x2 − 1), φ∗(x) = x3.

Table 5. Generalized Hermite polynomials.

(II1) The generalized Hermite: R = 0, S = 2μ, μ �= 0 and μ �= − 2n + 1

2, n ∈ N,

φ∗(x) = x, ψ∗(x) = 2x2 − (1 + 2μ), βn = 0, γn+1 = 1

2(n + 1 − 2μηn+1), n ∈ N,

C∗n(x) = −2(x2 − (−1)nμ), n ∈ N, D∗(x) = x, n = −2, an = 2ηn, bn = 0, n ∈ N,

φ∗(x)B ′n(x) = 2μηnBn(x) + 2γnxBn−1(x), n ∈ N,

xφ∗(x)B ′′n (x) + 2x(−x2 + μ)B ′

n(x) = −2(nx2 + μηn)Bn(x), n ∈ N.

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Table 6. Symmetric generalized Gegenbauer polynomials.

(II2) The symmetric generalized Gegenbauer: R = 0, S = −(1 + 2β),

α �= −n, α + β �= −n, n ≥ 1 and β �= −(1/2),

φ∗(x) = x(x2 − 1), ψ∗(x) = −2(α + β + 2)x2 + 2(β + 1), βn = 0, n ∈ N,

γn+1 = [2n + 2 + (2β + 1)(1 + (−1)n)][2n + 2 + 4α + (2β + 1)(1 + (−1)n)]16(n + α + β + 1)(n + α + β + 2)

, n ∈ N,

C∗n(x) = (2n + 2α + 2β + 1)x2 − (2β + 1)(−1)n, n ∈ N, D∗(x) = x,

n = 2(n + α + β + 1), an = n − (2α + 2β + 1)ηn, bn = 0, n ∈ N,

φ∗(x)B ′n(x) = (nx2 − (2β + 1)ηn)Bn(x) − 2(n + α + β)γnxBn−1(x), n ∈ N,

xφ∗(x)B ′′n (x) + x((2α + 2β + 3)x2 − 2β − 1)B ′

n(x)

= (n(2α + 2β + 2 + n)x2 + (2β + 1)ηn)Bn(x), n ∈ N.

Table 7. Symmetric generalized Bessel polynomials.

(II3) The symmetric generalized Bessel: R = 0, S = (1/2), ν �= −n, n ∈ N,

φ∗(x) = x3, ψ∗(x) = −2(ν + 1)x2 − 1

2, βn = 0, γn+1 = 1 − 2ν − (−1)n(2n + 2ν + 1)

16(n + ν)(n + ν + 1), n ∈ N,

C∗n(x) = (2n + 2ν − 1)x2 + (−1)n

2, n ∈ N, D∗(x) = x,

n = 2(n + ν), an = n − 1 − 2ν, bn = 0, n ∈ N,

φ∗(x)B ′n(x) =

(nx2 + 1 + (−1)n

4

)Bn(x) − 2(n + ν)γnxBn−1(x), n ∈ N,

xφ∗(x)B ′′n (x) + x

((2ν + 1)x2 + 1

2

)B ′

n(x) =(

n(n + 2ν)x2 − 1

2ηn

)Bn(x), n ∈ N.

Acknowledgements

The first author (F.M.) was supported by Dirección General de Investigación (Ministerio de Ciencia e Innovación) ofSpain under grant MTM 2009-12740-C03-02. The work of second author 1 was supported by Faculté des Sciences deGabès and Faculté des Sciences de Sfax, Tunisia.

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orthogonal polynomials and second-order pseudo-spectral linear differential equations

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