General Letters in Mathematics Vol. 6, No. 2, June 2019, pp.84-105

e-ISSN 2519-9277, p-ISSN 2519-9269

Available online at http:// www.refaad.com

https://doi.org/10.31559/glm2019.6.2.3

Semi-classical Linear Functionals of Class Four:

The Symmetric Case

M. Zaatra

Institut Suprieur des Sciences et Techniques des Eaux de Gabs. Campus universitaire, Gabs 6072, Tunisia.

medzaatra@yahoo.fr

Abstract. In this paper, we obtain all the symmetric semi-classical linear functionals of class four taking into account the

irreducible expression of the corresponding Pearson equation. We focus our attention on their integral representations. Thus,

some linear functionals very well known in the literature, associated with perturbations of semi-classical linear functionals of

class two at most, appear as well as new linear functionals which have not been studied.

Keywords: Orthogonal polynomials, Semi-classical linear functionals, Integral representations2000 MSC No: Primary 33C45; Secondary 42C05

1 Introduction

Let P be the linear space of polynomials with complex coefficients and let P ′ be its dual. The elements of P ′ will becalled either linear functionals or linear forms. We denote by 〈u, f〉 the action of u ∈ P ′ on f ∈ P. In particular, wedenote by (u)n := 〈u, xn〉 , n ≥ 0 , the moments of u. For any linear functional u and any polynomial h let Du = u′,hu, δc, and (x − c)−1u be the linear functionals defined by 〈u′, f〉 := −〈u, f ′〉, 〈hu, f〉 := 〈u, hf〉 , 〈δc, f〉 := f(c),

and〈(x− c)−1u, f

〉:= 〈u, θcf〉 where

(θcf)(x) =

f(x)− f(c)x− c

, c ∈ C , f ∈ P.

It is straightforward to prove that for c , d ∈ C , c 6= d , f ∈ P and u ∈ P ′(see [15])

(fu)′

= f′u+ fu

′, (1)

(x− c)−1((x− c)u) = u− (u)0δc , (2)

((x− d)(x− c))−1((x− d)(x− c)u) = u+ 1c− d

((d(u)0 − (u)1)δc − (c(u)0 − (u)1)δd) . (3)

Let us define the operator σ : P → P by (σf)(x) := f(x2). Then, we define the even part σu of a linearfunctional u by 〈σu, f〉 = 〈u, σf〉.

Therefore, we have [14]f(x)σu = σ

(f(x2)u

), (4)

σu′ = 2(σ(xu)

)′. (5)

The linear functional u is said to be regular (quasi-definite) if there exists a sequence {Pn}n≥0 of polynomialswith degPn = n, n ≥ 0, such that

〈u, PnPm〉 = rnδn,m , n,m ≥ 0 , rn 6= 0 , n ≥ 0 .

UserNew Stamp

UserNew Stamp

Semi-classical Linear Functionals of Class Four:The Symmetric Case 85

We can always assume that each Pn is monic i.e. Pn(x) = xn + lower degree terms. Then the sequence {Pn}n≥0

is said to be orthogonal with respect to u (MOPS in short). It is a very well known fact that the sequence {Pn}n≥0satisfies a three term recurrence relation

(see, for instance, the monograph by T. S. Chihara (see [7])

)Pn+2(x) = (x− ξn+1)Pn+1(x)− ρn+1Pn(x) , n ≥ 0 ,P1(x) = x− ξ0 , P0(x) = 1 ,

(6)

with(ξn, ρn+1

)∈ C× C∗ , n ≥ 0 . By convention, we set ρ0 = (u)0 = 1.

A linear functional u is called symmetric if (u)2n+1 = 0, n ≥ 0. The conditions (u)2n+1 = 0, n ≥ 0, areequivalent to the fact that the corresponding MOPS, {Pn}n≥0, satisfies the recurrence relation (6) with ξn = 0, n ≥ 0(see [7]).

A regular linear functional u is said to be positive definite if 〈u, f〉 > 0 for all f ∈ P such that f(x) ≥ 0, forevery x ∈ R and f 6= 0 or, equivalently, its MOPS satisfies (6) with ξn ∈ R and ρn ∈ R∗+ for all n ≥ 1 (see [15]).

From a structural point of view, a very important family of regular linear functionals has been exhaustivelyanalyzed in the literature during the last two decades. Let us recall that a linear functional ũ is said to be semi-classical when it is regular and there exist two polynomials Φ̃ , a monic polynomial, and Ψ̃,deg Ψ̃ ≥ 1, such that(

Φ̃ũ)′

+ Ψ̃ũ = 0 . (7)

This is the Pearson equation associated with the linear functional. The class of the semi-classical linear functional ũis defined as the nonnegative number s̃ = max

(deg Ψ̃− 1,deg Φ̃− 2

)if and only if the following condition is satisfied∏

c

(|Φ̃′(c) + Ψ̃(c)|+

∣∣∣〈ũ, θcΨ̃ + θ2c Φ̃〉∣∣∣ ) > 0 , (8)where c belongs to the set of zeros of Φ̃. Notice that this condition means that the polynomials Φ̃ and Φ̃′(x) + Ψ̃(x)are coprime as well as the equation (7) is irreducible (see [15]).

The corresponding MOPS {Pn}n≥0 is said to be semi-classical of class s̃. When s̃ = 0, ũ is a classical linearfunctional (Hermite, Laguerre, Jacobi, and Bessel). See [7] and [15].

Semi-classical linear functionals associated with weight functions were considered first by J. Shohat [18] in theframework of the existence of sequences of orthogonal polynomials satisfying second order linear differential equationswith polynomial coefficients (holonomic equations). Later on, P. Maroni and coworkers have extensively studied sucha kind of linear functionals with a special emphasis on their structure properties. For instance, [15] constitutes arelevant survey on the subject. Many examples appearing in the literature, mainly related to spectral perturbationsof classical linear functionals, are semi-classical but a constructive theory of such linear functionals remains open.

Indeed, the classification of semi-classical linear functionals according to some criteria of optimal informationfrom their Pearson equation, plays a central role in the constructive theory of such linear functionals. In [5] S.Belmehdi makes use of this approach to provide a full description of all semi-classical linear functionals of classs = 1. In [1] the classification of symmetric semi-classical linear functionals of such a class is given. Recently, thesemi-classical linear functionals of class s = 2 are completely described by F. Marcellán et al. in [8, 10]. In [9], theauthors gives a complete description of all symmetric semi-classical linear of class s = 3 using such an approach.Finally, a complete study of the class of the symmetric companion u of a linear functional v in terms of the class of vis done in [4]. Notice that some examples of symmetric semi-classical linear functionals of class greater than 2 havebeen analyzed in [2]. The aim of our contribution is to cover this gap.

As a first step, it is natural to deal with the description of the symmetric semi-classical linear functionals ofclass s = 4.

86 M. Zaatra

The structure of the manuscript is as follows. In Section 2, the irreducible canonical Pearson equationsassociated with symmetric semi-classical linear functionals of class s = 4 are obtained. Thus, fifteen irreduciblecanonical cases appear. In Section 3, the integral representation of such linear functionals is given, with a specialemphasis on the positive definite case.

2. Irreducible canonical functional equations

First of all, let us recall the following result:

Proposition 2.1. [1] Let u be a symmetric semi-classical linear functional of class s, satisfying (8). If s is an evennonnegative integer number, then Φ is an even polynomial function and Ψ is an odd polynomial function. If s is anodd nonnegative integer number, then Φ is an odd polynomial function and Ψ is an even polynomial function.

In the sequel, we will assume that the linear functional ũ is symmetric and semi-classical of class s̃ = 4. Then,according to the above proposition ũ satisfies (7) with

Φ̃(x) = c̃6x6 + c̃4x

4 + c̃2x2 + c̃0 , Ψ̃(x) = ã5x

5 + ã3x3 + ã1x , |c̃6|+ |ã5| 6= 0 .

As a consequence, the moments (ũ)n of the linear functional ũ satisfy the linear difference equation

(ã5 − (2n+ 1)c̃6)(ũ)2n+6 + (ã3 − (2n+ 1)c̃4)(ũ)2n+4 + (ã1 − (2n+ 1)c̃2)(ũ)2n+2 − (2n+ 1)c̃0(ũ)2n = 0 , n ≥ 0 ,

with (ũ)0 = 1 and (ũ)2n+1 = 0 , n ≥ 0 . Then, the set of solutions is a linear space of dimension at most three.The linear functional haũ

(dilation of ũ

)is defined by

〈haũ, f〉 := 〈ũ, haf〉 := 〈ũ, f(ax)〉 , f ∈ P . (9)

The semi-classical character of a linear functional is preserved by a dilation. Indeed, the dilated linear functionalu = (ha−1)ũ , a ∈ C∗ , satisfies

(Φu)′+ Ψu = 0 , (10)

withΦ(x) = a−tΦ̃(ax) , Ψ(x) = a1−tΨ̃(ax) , t = degΦ̃ . (11)

The sequence {P̂n(x)}n≥0, where P̂n(x) = a−nPn(ax), is orthogonal with respect to u and fulfils (6) with

ξ̂n = 0, ρ̂n+1 =ρn+1a2

, n ≥ 0 . (12)

As we can see, a dilation ha does not modify the nature of a symmetric semi-classical linear functional. This processwill be applied to the Pearson equation satisfied by a symmetric semi-classical linear functional of class s = 4. In thesequel, a ∈ C∗ will denote an arbitrary complex number. A convenient choice of a , according to the expression of Φ̃,allows us to re-locate the zeros of Φ̃ in the complex plane. In this way, (9) can be reduced either to some situationswhich appear in the literature or yield new linear functionals not yet studied. Thus we get canonical distributionalequations of class four in a simple way, that becomes a pattern for the family of equations which can be reducedusing a shifting.

According to Proposition 2.1, we will analyze two situations.

A. deg Φ̃ = 6 and 1 ≤ deg Ψ̃ ≤ 5.B. deg Ψ̃ = 5 and 0 ≤ deg Φ̃ ≤ 4.

Case A. Set

Φ̃(x) =

3∏i=1

(x2 − α2i ) ,

Ψ̃ = x(ã4x4 + ã2x

2 + ã0) , |ã4|+ |ã2|+ |ã0| 6= 0 .

Semi-classical Linear Functionals of Class Four:The Symmetric Case 87

Then u satisfies (3∏i=1

(x2 − α2i a−2

)u

)′+ x

(ã4x

4 + a−2ã2x2 + a−4ã0

)u = 0 . (13)

We will discuss the following cases.

A1. Φ̃ has six simple zeros.

We choose a such that α1 = a 6= 0, α2 = ac and α3 = ad with c2, d2 /∈ {0, 1} and c2 6= d2. Thus (13) reduces to((x2 − 1)(x2 − c2)(x2 − d2)u

)′+ x(a4x

4 + a2x2 + a0)u = 0 . (14)

The rational function −Φ′+ Ψ

Φ′has six simple poles 1 ,−1 , c ,−c , d and −d. We denote by

α = −Φ′(1) + Ψ(1)

Φ′(1), β = −Φ

′(c) + Ψ(c)

Φ′(c)and γ = −Φ

′(d) + Ψ(d)

Φ′(d)

the corresponding residues. Then, after some straightforward calculation, we obtain a4a2a0

= M 2α+ 22β + 2

2γ + 2

,where

M =

−1 −1 −1c2 + d2 d2 + 1 c2 + 1−c2d2 −d2 −c2

.Notice that detM = (c2 − 1)(d2 − 1)(c2 − d2) 6= 0. This means that this change of parameters is bijective. (We havethe same kind of relation between the new parameters and the old ones in the other cases that we will study in thesequel.)

Now, changing the parameters in (14), according to the condition of irreducibility (8) we get

((x2 − 1)(x2 − c2)(x2 − d2)u

)′+ x(− 2(α+ β + γ + 3)x4 + 2

(β + γ + 2

+c2(α+ γ + 2) + d2(α+ β + 2))x2 − 2

(c2d2(α+ 1) + c2(γ + 1) + d2(β + 1)

))u = 0 ,{

|β|+ | − (2α+ 2β + 2γ + 5)(u)4 +(2β + 2γ + 3 + (2α+ 2β + 3)d2 − 2βc2

)(u)2

+2βc2(d2 + 1− c2)− (2β + 1)d2|}×{|α|+ | − (2α+ 2β + 2γ + 5)(u)4

+((2α+ 2γ + 3)c2 + (2α+ 2β + 3)d2 − 2α

)(u)2 + 2α(c

2 + d2 − 1)− (2α+ 1)c2d2|}

×{|γ|+ | − (2α+ 2β + 2γ + 5)(u)4 +

((2α+ 2γ + 3)c2 + 2β + 2γ + 3− 2γd2

)(u)2

+2αd2(c2 + 1− d2)− (2γ + 1)c2|}6= 0 .

(15)

This is the irreducible Pearson equation satisfied by a linear functional of class four when Φ in (10) has six simplezeros.

We will proceed in a similar way in the cases listed in below.

A2. Φ̃ has five different zeros and one of them is double zeros. In such a situation α1 = 0 and α2α3 6= 0.

We choose a such that α2 = a and α3 = ac with c /∈ {−1, 0, 1}. Then (13) can be written as(x2(x2 − 1)(x2 − c2)u

)′+ x(a4x

4 + a2x2 + a0)u = 0 . (16)

88 M. Zaatra

By an appropriate choice of the coefficients a4, a2, and a0, (16) becomes

(x2(x2 − 1)2(x2 − c2)u

)′+ x(− (α+ 2β + 2γ + 6)x4 +

((α+ 2β + 4)c2

+α+ 2γ + 4)x2 − (α+ 2)c2

)u = 0 ,{

− (α+ 2β + 2γ + 5)(u)4 +((α+ 2β + 3)c2 + α+ 2γ + 3

)(u)2 − (α+ 1)c2

}×{|β|+ | − (α+ 2β + 2γ + 5)(u)4 +

((α+ 2β + 3)c2 − 2β

)(u)2 + 2β(c

2 − 1)|}

×{|γ|+ | − (α+ 2β + 2γ + 5)(u)4 +

(α+ 3 + 2γ(1− c2)

)(u)2 + 2γc

2(1− c2)|}6= 0 .

(17)

This is the corresponding irreducible Pearson equation of a linear functional of class four when Φ in (10) has fivedifferent zeros and one of them is of multiplicity two.

Remark 1. If c2 = −1 and α = −2, then we obtain the result given in [3].

A3. Φ̃ has four different zeros and two of them are a double zero.

We choose a such that α1 = α2 = a and α3 = ac with c /∈ {−1, 0, 1}. Then (13) becomes((x2 − 1)2(x2 − c2)u

)′+ x(a4x

4 + a2x2 + a0)u = 0 . (18)

From a suitable choice of the coefficients a4, a2, and a0, (18) yields

((x2 − 1)2(x2 − c2)u

)′+ 2x

(− (α+ γ + 3)x4 +

((α+ 2)c2 + α+ β + 2γ + 4

)x2

−γ − 1− (α+ β + 2)c2)u = 0 ,{

|β|+ | − (2α+ 2γ + 5)(u)4 +((2α+ 3)c2 + 2β + 2γ + 3

)(u)2 + 2β − (2β + 1)c2|

}×{|γ|+ | − (2α+ 2γ + 5)(u)4 + 2

(3 + α+ β + γ(2− c2)

)(u)2 + 2γc

2(1− c2)− 2γ − 1|}6= 0 .

(19)

This is an irreducible Pearson equation of a linear functional of class four, when Φ in (10) has four different zeroswhen two of them are of multiplicity two.

A4. Φ̃ has three different zeros of multiplicity two, i.e. α1 = 0 and α2α3 6= 0.

We choose a such that α2 = α3 = a.

(13) becomes(x2(x2 − 1)2u)′ + x

(a4x

4 + a2x2 + a0

)u = 0. (20)

By a skilful choice of the coefficients a4, a2 and a0, (20) can be written as(x2(x2 − 1)2u)′ + x

(− (2α+ 2β + 7)x4 + 2(2α+ β + γ + 5)x2 − 2α− 3

)u = 0 ,{

− (α+ β + 3)(u)4 + (2α+ β + γ + 4)(u)2 − α− 1}

×{|γ|+ | − (α+ β + 3)(u)4 + (α+ γ + 2)(u)2 + γ|

}6= 0 .

(21)

This is the irreducible Pearson equation for a functional of class four, when Φ in (10) has a three different zeros ofmultiplicity two.

A5. Φ̃ has three different zeros and one of them of multiplicity four. This means that either α1 = α2 = 0 andα3 6= 0.

We choose a such that α3 = a. (13) can be written as

(x4(x2 − 1)u)′ + x(a4x

4 + a2x2 + a0

)u = 0. (22)

By an appropriate choice of the coefficients a4, a2 and a0, (22) reads as(x4(x2 − 1)u)′ + x

(− (α+ 2γ + 6)x4 + (α− 2β + 4)x2 + 2β

)u = 0 ,{

− (α+ 2γ + 5)(u)4 + (α− 2β + 3)(u)2 + 2β}

×{|γ|+ | − (α+ 2γ + 5)(u)4 − 2(β + γ)(u)2 − 2γ|

}6= 0 .

(23)

Semi-classical Linear Functionals of Class Four:The Symmetric Case 89

This is the irreducible Pearson equation for a functional of class four, when Φ in (10) has a three different zeros: twoof them are simple and the other one has multiplicity four.

A6. Φ̃ has two different zeros of multiplicity three.

We choose a such that α1 = α2 = α3 = a. Then (13) becomes

((x2 − 1)3u)′ + x(a4x

4 + a2x2 + a0

)u = 0. (24)

From a suitable choice of the coefficients a4, a2 and a0, (24) can be written as ((x2 − 1)3u)′ + 2x

(− (α+ 3)x4 + 2(α+ 3)x2 − α− 2β − 3

)u = 0 ,

|β|+ | − (2α+ 5)(u)4 + 2(α+ 3)(u)2 − 4β − 1|}6= 0 .

(25)

This is the irreducible Pearson equation for a functional of class four, when Φ in (10) has a two zeros of multiplicitythree.

A7. Φ̃ has one zero of multiplicity six, i.e. α1 = α2 = α3 = 0.

(13) becomes

(x6u)′ + x

(ã4x

4 + a−2Ψ̃′′′

(0)

6x2 + a−4Ψ̃

′(0)

)u = 0 . (26)

We can consider two subcases.

A7.1. Ψ̃′(0) 6= 0.

Let a be such that a−4ã0 = −8. Then (26) reduces to

(x6u)′ + x(a4x4 + a2x

2 − 8)u = 0 . (27)

By a skilful choice of the coefficients a4 and a2, (27) yields{(x6u)′ + x

(− (2α+ 1)x4 − 2βx2 − 8

)u = 0 ,

α(u)4 + β(u)2 + 4 6= 0 .(28)

This is the irreducible Pearson equation for a functional of class four, when Φ in (10) has a zero of multiplicity sixand this zero is a simple zero of Ψ.

A7.2. Ψ̃′(0) = 0 and Ψ̃

′′′(0) 6= 0.

If we chose a such that a−2Ψ̃′′′

(0) = −24, then we obtain

(x6u)′ + x3(a4x2 − 4)u = 0 . (29)

By an appropriate choice of the coefficient a4, (29) reads{(x6u)′ − x3

((4α− 1)x2 + 4

)u = 0 ,

(2α− 1)(u)4 + 2(u)2 6= 0 .(30)

This is the irreducible Pearson equation for a functional of class four, when Φ in (10) has a zero of multiplicity sixand it is a triple zero of Ψ.

Remark 2. (1) If zero is a root of multiplicity five of Ψ̃, the equation (26) writes

(x6u)′ + ã4x5u = 0 .

Then, we obtain(2n+ 1− ã4)(u)2n+6 = 0 , n ≥ 0 .

So after a certain rank the Hankel determinants associated with u are all hopeless, by following u is not regular.Then the case is to reject.

90 M. Zaatra

(2) From (30), we havex2u = kv , (31)

where k is a normalization term and v is the symmetric semi-classical linear form of class two defined by the Pearsonequation (see [8, 10, 17]) {

(x4v)′ − x(

(4α− 1)x2 + 4)v = 0 ,

(2α+ 1)(v)2 + 2 6= 0 .

The linear functional u defined by (31) is regular if and only if (see [12])

〈u, P (α)n+1〉〈u, xP (α)n 〉 − 〈u, P (α)n 〉〈u, xP(α)n+1〉 6= 0 , n ≥ 0 ,

where P(α)n (x) , n ≥ 0 , are the monic polynomials orthogonal with to v.

Case B. Set

Φ̃(x) = c̃4x4 + c̃2x

2 + c0 , |c̃4|+ |c̃2|+ |c̃0| 6= 0 ,

Ψ̃ = x(ã4x4 + ã2x

2 + ã0) , ã4 6= 0 .

We will analyze the following situations:

B1. Φ̃ has four simple zeros α1, −α1, α2 and −α2.

ũ and u satisfy, respectively,

( 2∏i=1

(x2 − α2i )ũ)′

+ x(ã4x4 + ã2x

2 + ã0)ũ = 0 ,

( 2∏i=1

(x2 − α2i a−2)u)′

+ x(ã4a

2x4 + ã2x2 + a−2ã0

)u = 0 . (32)

Let a be such that α1 = a and α2 = ac with c /∈ {−1, 0, 1}. Then (32) becomes((x2 − 1)(x2 − c2)u

)′+ x(a4x

4 + a2x2 + a0)u = 0 . (33)

By an appropriate choice of the coefficients a4, a2, and a0, with λ 6= 0 we get

((x2 − 1)(x2 − c2)u

)′+ 2x

(λx4 −

(λ(c2 + 1) + α+ β + 2

)x2

+(λ+ α+ 1)c2 + β + 1)u = 0 ,{

|β|+ |2λ(u)4 − (2λ+ 2α+ 2β + 3)(u)2 + 2β(1− c2) + 1|}

×{|α|+ |2λ(u)4 − (2λc2 + 2α+ 2β + 3)(u)2 + (2α+ 1)c2 − 2α|

}6= 0 .

(34)

This is the irreducible Pearson equation for a linear functional of class four, when Φ in (10) has four simple zeros.

B2. Φ̃ has three simple zeros and one of them of multiplicity two. This means α1 = 0 and α2 6= 0.

We choose a in such a way that α2 = a. Then (32) reads as

(x2(x2 − 1)u)′ + x(a4x4 + a2x2 + a−2a0)u = 0. (35)

By a suitable choice of the coefficients a4, a2 and a0, with λ 6= 0 we obtain(x2(x2 − 1)u)′ + x

(2λx4 − (2λ+ α+ 2β + 4)x2 + α+ 2

)u = 0 ,{

2λ(u)4 − (2λ+ α+ 2β + 3)(u)2 + α+ 1}

×{|β|+ |2λ(u)4 − (α+ 2β + 3)(u)2 − 2β|

}6= 0 .

(36)

Semi-classical Linear Functionals of Class Four:The Symmetric Case 91

This is the irreducible Pearson equation for a linear functional of class four, when Φ in (10) has a zero of multiplicitytwo.

B3. Φ̃ has two zeros of multiplicity two.

We choose a such that α1 = α2 = a. (32) becomes

((x2 − 1)2u)′ + x(a4x4 + a2x2 + a−2a0)u = 0. (37)

By an appropriate choice of the coefficients a4, a2 and a0 with λ 6= 0, (37) yields ((x2 − 1)2u)′ + 2x

(λx4 − (2λ+ α+ 2)x2 + λ+ α+ β + 2

)u = 0 ,

|β|+ |2λ(u)4 − (2λ+ 2α+ 3)(u)2 + 2β + 1| 6= 0 .(38)

This is the irreducible Pearson equation for a linear functional of class four, when Φ in (10) has two zeros ofmultiplicity two.

B4. Φ̃ has one zero of multiplicity four α1 = α2 = 0.

We choose a in such a way that ã4a2 = 2. Then (32) reads as

(x4u)′ + x(2x4 + a2x2 + a−2a0)u = 0. (39)

By a suitable choice of the coefficients a2 and a0, (39) can be written as (x4u)′ + 2x

(x4 − (α+ 2)x2 − β

)u = 0 ,

2(u)4 − (2α+ 3)(u)2 − 2β 6= 0 .(40)

This is the irreducible Pearson equation for a linear functional of class four, when Φ in (10) has a zero of multiplicityfour.

B5. Φ̃(x) = x2 − α21.

We need to discuss the following situations.

B5.1. Φ̃ has two simple zeros.

ũ and u satisfy, respectively, ((x2 − α21)ũ

)′+ x(ã4x

4 + ã2x2 + ã0)ũ = 0 ,(

(x2 − α21a−2)u)′

+ x(a4ã4x

4 + a2ã2x2 + ã0

)u = 0 . (41)

If we choose a such that α1 = a, then we have

((x2 − 1)u)′ + x(a4x4 + a2x2 + a0)u = 0 . (42)

By an appropriate choice of the coefficients a4, a2 and a0 with λ 6= 0, (42) reads as ((x2 − 1)u)′ + 2x(2λx4 − (2λ+ α)x2 + α− β − 1)u = 0 ,

|β|+ |4λ(u)4 − 2α(u)2 − 2β − 1| 6= 0 .(43)

This is the irreducible Pearson equation for a linear functional of class four, when Φ in (10) has two simple zeros.

B5.2. Φ̃ has one zero α1 = 0.

We choose a in such a way that ã4a4 = 4. Then (41) reads as

(x2u)′ + x(4x4 + a2x2 + a0)u = 0. (44)

By a suitable choice of the coefficients a2 and a0, (44) can be written as (x2u)′ + x

(4x4 − 2βx2 − α− 2

)u = 0 ,

4(u)4 − 2β(u)2 − α− 1 6= 0 .(45)

92 M. Zaatra

This is the irreducible Pearson equation for a linear functional of class four, when Φ in (10) has a zero of multiplicitytwo.

B6. Φ̃ is a constant.

After the displacement, we get

u′+ x(a6ã4x

4 + a4ã2x2 + a2ã0)u = 0 . (46)

If we choose a such that a6ã4 = 6, then (46) reduces to

u′+ 2x(3x4 + 2αx2 + β)u = 0 . (47)

Since Φ is non zero constant, condition (8) is satisfied and (47) is the irreducible Pearson equation for a linearfunctional of class four, when Φ in (10) is a constant.

3. Integral representation of symmetric semi-classical linear functionals of class s=4.

Let u be a symmetric semi-classical linear functional of class s satisfying (10) and let us assume (u)0 = 1.

Our aim will be to obtain an integral representation of u

〈u, f(x)〉 =∫ +∞−∞

f(x)U(x) dx , (48)

where we assume the function U is absolutely continuous on R and it decays as fast as its derivative U ′ . From (10),we get ∫ +∞

−∞

((ΦU)

′+ ΨU

)f(x)dx− Φ(x)U(x)f(x)]+∞−∞ = 0 , f ∈ P .

Hence, from the assumptions on U , the following conditions hold

Φ(x)U(x)f(x)]+∞−∞ = 0 , f ∈ P , (49)∫ +∞−∞

((ΦU)

′+ ΨU

)f(x)dx = 0 , f ∈ P . (50)

Condition (50) implies

(ΦU)′+ ΨU = wg , (51)

where w is arbitrary and g is a locally integrable function with rapid decay representing the null-linear functional(see [13]) ∫ +∞

−∞xng(x) = 0 , n ≥ 0 . (52)

Conversely, if U is a solution of (51) verifying the hypothesis above as well as∫ +∞−∞

U(x)dx 6= 0 , (53)

then (49)− (50) are fulfilled and (48) defines a linear functional u which is a solution of (10).If w = 0, then (51) becomes

U′

U= −Φ

′+ Ψ

Φ. (54)

We will consider the above fifteen canonical functional equations and, in each case, an integral representation ofthe corresponding linear functionals will be given.

Case A.

Semi-classical Linear Functionals of Class Four:The Symmetric Case 93

A1. From (15) to (54), we get

U′(x)

U(x)=

2αx

x2 − 1+

2βx

x2 − c2+

2γx

x2 − d2,

and, as a consequence,U(x) = |x2 − 1|α|x2 − c2|β |x2 − d2|γ (55)

is the solution at some intervals depending on c and d (see below).

On the other hand, if αβγ 6= 0 and α , β , γ > −1, then the conditions (8) and (49) hold in the followingsituations:

s1: 0 < c < d < 1 . In such a case, u is represented by

〈u, f〉 =∫ 1−1f(x)(1− x2)α|x2 − c2|β |x2 − d2|γ

(Aχ[−c,c](x) +Bχ[−d,d](x) + Cχ[−1,1](x)

)dx , (56)

since from an integration by parts we deduce that the linear functionals u1, u2 and u3 defined by

〈u1, f〉 =∫ c−cU(x)f(x)dx, 〈u2, f〉 =

∫ d−dU(x)f(x)dx and 〈u3, f〉 =

∫ 1−1U(x)f(x)dx,

are solutions of (15). The same result holds for any linear combination of u1, u2 and u3 with coefficients A, B and C

〈Au1 +Bu2 + Cu3, f〉 =∫ 1−1U(x)

(Aχ[−c,c] +Bχ[−d,d] + Cχ[−1,1]

)f(x)dx

where χ[a,b] denotes the characteristic function of the interval [a, b], i. e. χ[a,b](x) = 1 when x ∈ [a, b] and zerootherwise. A, B and C will be chosen in such a way that (u)0 = 1.

We will proceed in a similar way in the case below.

s2: d > c > 1. In such a case, we get

〈u, f〉 =∫ d−df(x)|1− x2|α|x2 − c2|β(d2 − x2)γ

(A1χ[−c,c](x) +B1χ[−d,d](x) + C1χ[−1,1](x)

)dx . (57)

Remark 3. The condition αβγ 6= 0 is sufficient to ensure that the condition of irreducibility (8) is satisfied (see secondequation in (15)). Indeed, for every parameter α, β, γ > −1 satisfying (15), we obtain the integral representationsgiven above.

Particular case:

If α = β = γ = 0, then (15) becomes

((x2 − 1)(x2 − c2)(x2 − d2)u

)′+ 2x

(− 3x4 + 2

(1 + +c2 + d2

)x2

−c2d2 − c2 − d2)u = 0 ,

{−5(u)4 + 3(1 + d2)(u)2 − d2} × {−5(u)4 + 3(c2 + d2)(u)2 − c2d2}×{−5(u)4 + 3(1 + c2)(u)2 − c2} 6= 0 .

(58)

Which may be written as(x2 − 1)(x2 − c2)(x2 − d2)u

′= 0 , (59)

{−5(u)4 + 3(1 + d2)(u)2 − d2} × {−5(u)4 + 3(c2 + d2)(u)2 − c2d2} × {−5(u)4 + 3(1 + c2)− c2} 6= 0 . (60)

(59) is equivalent to

u′

= λ1(δ1 − δ−1) + λ2(δc − δ−c) + λ3(δd − δ−d) , (61)

with λ1 =

3(c2+d2)(u)2−5(u)4−c2d22(c2−1)(d2−1) ,

λ2 =3(1+d2)(u)2−5(u)4−d2

2c(c2−1)(c2−d2) ,

λ3 =3(c2+1)(u)2−5(u)4−c2

2d(d2−1)(d2−c2) .

(62)

94 M. Zaatra

(60) is equivalent to λ1λ2λ3 6= 0.On the one hand, by virtue of the definition of the derivative of a linear functional, we have

(u′)2n+1 = −(2n+ 1)(u)2n , n ≥ 0 . (63)

On the other hand, from (61) we get

(u′)2n+1 = 2(λ1 + λ2c

2n+1 + λ3d2n+1) , n ≥ 0 . (64)

The conjunction of (63)) and (64) leads to

(u)2n = −2(λ1 + λ2c

2n+1 + λ3d2n+1)

2n+ 1, n ≥ 0 .

By a simple computation, we can see that (u)0 = 1 and (u)2, (u)4 are an arbitrary parameter.

Let us define

E0 = {(λ, µ) ∈ C2/ λ1λ2λ3 = 0} ,En = {(λ, µ) ∈ C2/∆n(λ, µ) = 0} , n ≥ 1 ,where λ = (u)2, µ = (u)4 and ∆n(λ, µ) is the Hankel determinant associated with u.

For any integer n, ∆n(λ, µ) is a polynomial in λ and µ, then the linear functional u satisfying (58) is regular and

of class four if and only if (λ, µ) /∈⋃n≥0

En.

Let us back to equation (61), then we have

u = λ1(H1 −H−1) + λ2(Hc −H−c) + λ3(Hd −H−d) , (65)

where Hx is the Heaviside step function:

Hx =

{1 over ]x, +∞[,0 over ]−∞, x] .

Then, u is represented by

〈u, f〉 = −λ1∫ 1−1f(x)dx− λ2

∫ c−cf(x)dx− λ3

∫ d−df(x)dx .

A2. From (17) to (54), we have

U′(x)

U(x)=α

x+

2βx

x2 − 1+

2γx

x2 − c2,

soU(x) = |x|α|1− x2|β |x2 − c2|γ . (66)

Finally, it is straightforward to check that for βγ 6= 0 and α, β, γ > −1, the conditions (8) and (49) hold in thefollowing situations:

s1: 0 < c < 1 . In such a case, u is represented by

〈u, f〉 =∫ 1−1f(x)|x|α(1− x2)β |x2 − c2|γ

(Aχ[−c,c] +Bχ[−1,1]

)dx . (67)

s2: c > 1. In such a case, we get

〈u, f〉 =∫ c−cf(x)|x|α|1− x2|β(c2 − x2)γ

(A1χ[−c,c] +B1χ[−1,1]

)dx . (68)

A3. From (19) to (54), we obtain

U′

U=

2αx

x2 − 1− 2βx

(x2 − 1)2+

2γx

x2 − c2,

Semi-classical Linear Functionals of Class Four:The Symmetric Case 95

and, as a consequence,

U(x) = |x2 − 1|α|x2 − c2|γe−β

1−x2 , (69)

is the solution at some intervals depending on c. For instance, if βγ 6= 0 , α , γ > −1 and β > 0, then the conditions(8) and (49) hold in the following situations:

s1: If c ∈]0, 1[, then

〈u, f〉 =∫ 1−1f(x)(1− x2)α|x2 − c2|γe

−β1−x2

(Aχ[−c,c] +Bχ[−1,1]

)dx . (70)

s2: If c > 1, then u is represented by

〈u, f〉 =∫ c−cf(x)|1− x2|α(c2 − x2)γe

−β1−x2

(A1χ[−c,c] +B1χ[−1,1]

)dx . (71)

Remark 4. For every parameter α, γ > −1 and β > 0 such that (19) is verified, we also obtain the integralrepresentation given above, except when β ≤ 0. For β = 0 see below.

Particular case:

If β = 0, thus for (19) we get(

(x2 − 1)2(x2 − c2)u)′

+ 2x(x2 − 1)(− (α+ γ + 3)x2 + (α+ 2)c2 + γ + 1

)u = 0 ,{

− (2α+ 2γ + 5)(u)4 +((2α+ 3)c2 + 2γ + 3

)(u)2 − c2

}×{|γ|+ | − (2α+ 2γ + 5)(u)4 + 2

(3 + α+ γ(2− c2)

)(u)2 + 2γc

2(1− c2)− 2γ − 1|}6= 0 .

Then,

(x2 − 1)u = kv̂ , (72)

where k is a normalization term and v̂ is the symmetric semi-classical linear form of class two defined by the Pearsonequation ( see [8, 10])

((x2 − 1)(x2 − c2)v̂

)′+ 2x

(− (α+ γ + 3)x2 + (α+ 2)c2 + γ + 1

)v̂ = 0 ,{

|α|+ | − (2α+ 2γ + 3)(v̂)2 + (2α+ 1)c2|}

×{|γ|+ | − (2α+ 2γ + 3)(v̂)2 + 2γ(1− c2) + 1|

}6= 0 .

Notice that the linear functional u defined by (72) is regular if and only if the MOPS Pn(x) , n ≥ 0 , satisfies (see[6]) ∣∣∣∣∣∣∣∣

Pn+1(−1 ; k) Pn(−1 ; k)

Pn+1(1 ;−k) Pn(1 ;−k)

∣∣∣∣∣∣∣∣ 6= 0 , n ≥ 0 ,where Pn(x) , n ≥ 0 , are the monic polynomials orthogonal with respect to v̂ and {Pn(. ;µ)}n≥0 is the co-recursiveMOPS of {Pn}n≥0 (see [7])

Pn(x;µ) = Pn(x)− µP (1)n−1(x) ,

where {P (1)n }n≥0 is the sequence of associated polynomials of the first kind for the sequence {Pn}n≥0 (see [15]).From (3) and (72), we obtain

u = k(x2 − 1)−1v̂ + 12

(δ1 + δ−1) . (73)

A4. From (21) to (54), we get

U′

U=

2βx

x2 − 1− 2γx

(x2 − 1)2+

2α+ 1

x,

96 M. Zaatra

therefore,

U(x) = |x|2α+1|x2 − 1|βe−γ

1−x2 . (74)

Notice that if γ > 0 and α , β > −1 , then the conditions (8) and (49) hold.Thus u is represented by

〈u, f〉 = k∫ 1−1f(x)|x|2α+1(1− x2)βe

−γ1−x2 dx . (75)

Here k is a constant such that (u)0 = 1.

Particular case:

If γ = 0, thus from (21) we have(x2(x2 − 1)2u)′ + x

(− (2α+ 2β + 7)x4 + 2(2α+ β + 5)x2 − 2α− 3

)u = 0 ,{

− (α+ β + 3)(u)4 + (2α+ β + 4)(u)2 − α− 1}

×{− (α+ β + 3)(u)4 + (α+ 2)(u)2

}6= 0 .

Henceu = (1− k1 − 2k2)G.G(β, α) + k1δ0 + k2(δ1 + δ−1) , (76)

where G.G(a, b) is the generalized Gegenbauer linear functional defined by the Pearson equation (see [7])(x(x2 − 1)G.G(a, b)

)′+ 2(− (a+ b+ 2)x2 + b+ 1

)G.G(a, b) .

The above linear functional u is regular if and only if k1 and k2 satisfies (see [2])∣∣∣∣∣∣1 + k1Kn(0, 0) k2Kn(0, 1) k2Kn(0,−1)k1Kn(1, 0) 1 + k2Kn(1, 1) k2Kn(1,−1)k1Kn(−1, 0) k2Kn(−1, 1) 1 + k2Kn(−1,−1)

∣∣∣∣∣∣ 6= 0 , n ≥ 0 ,where Kn(x, y) :=

n∑m=0

P(β,α)m (x)P

(β,α)m (y)

〈GG(β, α) , (P (β,α)m )2〉is the n-th reproducing Kernel associated with the linear functional

GG(β, α) and P (β,α)n (x) , n ≥ 0 , are the monic polynomials orthogonal with respect to GG(β, α) .

The action of the linear functionals defined in (76) over the polynomial x2n yields

(u)2n = (1− k1 − 2k2)Γ(n+ α+ 1)Γ(α+ β + 2)

Γ(n+ α+ β + 2)Γ(α+ 1)+ 2k2 , n ≥ 1 .

Finally from (76) we obtain

〈u, f〉 = (1− k1 − 2k2)Γ(α+ β + 2)

Γ(α+ 1)Γ(β + 1)

∫ 1−1|x|2α+1(1− x2)βf(x) dx+ k1f(0) + k2(f(1) + f(−1)) .

A5. From (23) to (54), we have

U′

U=

2γx

x2 − 1+

2β

x3+α

x.

Thus,

U(x) = |x|α|x2 − 1|γe−βx2 . (77)

Notice that if γ 6= 0 , α , γ > −1 and β ≥ 0, then the conditions (8) and (49) hold.Thus u is represented by

〈u, f〉 = k∫ 1−1f(x)|x|α(1− x2)γe

−βx2 dx . (78)

A6. From (25) to (54) we obtain

U′

U=

2αx

x2 − 1+

4βx

(x2 − 1)3,

Semi-classical Linear Functionals of Class Four:The Symmetric Case 97

hence,

U(x) = |x2 − 1|αe−β

(x2−1)2 . (79)

Furthermore, for β > 0 and α > −1, (8) and (49) hold.As a consequence u is represented by

〈u, f〉 = A∫ 1−1f(x)(1− x2)αe

−β(x2−1)2 dx . (80)

Particular case:

If β = 0, then (25) becomes {((x2 − 1)3u)′ − 2(α+ 3)x(x2 − 1)2u = 0 ,

−(2α+ 5)(u)4 + 2(α+ 3)(u)2 − 1 6= 0 .

Thus,u = G(α) + k(δ

′

1 − δ′

−1) , (81)

where G(α) is the Gegenbauer linear functional that satisfies (see [15])

((x2 − 1)G(α))′− 2(α+ 1)xG(α) = 0 ,

with α+ 1 6= −n and 2α+ 1 6= −n , n ≥ 0 .The above linear functional u is regular if and only if k satisfies (see [2])∣∣∣∣1− kKn(1, 1) kKn(1,−1)−kKn(−1, 1) 1 + kKn(−1,−1)

∣∣∣∣ 6= 0 , n ≥ 0 .Finally, from (81) we get

〈u, f〉 =Γ(α+ 32 )√πΓ(α+ 1)

∫ 1−1

(1− x2)αf(x) dx− k(f′(1)− f

′(−1)) .

When f(x) = x2n, the above expression becomes

(u)2n = Γ(n+1

2)

Γ(α+ 32 )√πΓ(n+ α+ 32 )

− 4nk , n ≥ 0 .

A7.1. In this case, it is not possible to choose w = 0. Indeed, from (28) and (54) (e.i. (51) with w = 0 ), we get

U′

U=

2α− 5x

+2β

x3+

8

x5.

A priori there is no a real path C such that x6U(x)f(x)|C = 0 , f ∈ P. This is analog of Bessel in classical case (see[13] for more information ).

Thus, we handle it differently with choice w 6= 0 in (51).From (28) to (51), we get

(x6U)′+ x{−(2α+ 1)x4 − 2βx2 − 8}U = wg(x) . (82)

For instance, let g(x) = x|x|s(x2) , x ∈ R, where s is the classical Stieltjes function [13]

s(x) =

0 , x ≤ 0 ,

e−x14 sin x

14 , x > 0 .

(83)

A possible solution of (82) is the even function

U(x) =

0 , x = 0 ,

w|x|2α−5e−βx2− 2x4

∫ +∞x2

t−αe2t2

+ βt s(t)dt , β ≥ 0 , x ∈ R− {0} .(84)

98 M. Zaatra

First, condition (49) is fulfilled since

|x6U(x)| ≤ |w||x|2α+1e−βx2− 2x4

∫ +∞x2

t−αe2t2

+ βt e−t14 dt = o

(e−

12 |x|

12), |x| → +∞ .

Furthermore, when x→ +∞

|U(x)| ≤ |w|x2α−5∫ +∞x2

t−αe−t14 dt = o

(e−

12x

12),

and when x→ +0

|U(x)| ≤ |w|x2α−5e−βx2− 2x4

∫ 1x2t−αe

2t2

+ βt dt+ o(1) .

Applying l’Hospital’s rule to the ratio

limx→+0

∫ 1x2t−αe

2t2

+ βt dt

x−2α+5e2x2

+ βx= limx→+0

2x

(2α− 5)x4 + 2βx2 + 8= 0 ,

we get limx→+0

U(x) = 0 = U(0).

Consequently, U ∈ L1.Condition (53) now becomes∫ +∞

−∞U(x)dx = 2w

∫ +∞0

ξ−αe2ξ2

+ βξ s(ξ)(∫ √ξ

0

x2α−5e−βx2− 2x4 dx

)dξ = wSα , (85)

with

Sα = 8

∫ +∞0

t3−4αe2t8

+ βt4 e−tϕα−1(t

2)sin t dt , (86)

ϕα(t) =

∫ t0

x2α−3e−βx2− 2x4 dx . (87)

Let us establish some results about Sα.

Lemma 3.1. We have for α ≥ 32 and β ≥ 0

1

2t2ϕα(t) ≤ ϕα+1(t) ≤ t2ϕα(t) , t ≥ 0 , (88)

t2α+2e−β

t2− 2t4

2(α+ 1)t4 + 2βt2 + 8≤ ϕα(t) ≤ 2

t2α+2e−β

t2− 2t4

(α+ 1)t4 + 2βt2 + 16, t ≥ 0 . (89)

Proof. It is easy to prove (88) from (87) and monotonicity.From (87), we have upon integration by parts

ϕα(t) =1

8t2α+2e−

β

t2− 2t4 − α+ 1

4ϕα+2(t)−

β

4ϕα+1(t) , t ≥ 0 . (90)

Now, in accordance of (88) and (90) we obtain the desired result (89).

Proposition 3.2. We have the following expression for n ≥ 1 and α ∈ C

Sα =(−1)n

22n−3

n−1∏k=0

(α+ 2k)

∫ +∞0

t3−4αeβ

t2+ 2t4 e−tϕα+2n−1(t

2)sin t dt

+ β

n−1∑k=0

(−1)k+1

22k−1

k−1∏i=0

(α+ 2i)

∫ +∞0

t3−4αeβ

t2+ 2t4 e−tϕα+2k(t

2)sin t dt (91)

with

−1∏0

= 1.

Semi-classical Linear Functionals of Class Four:The Symmetric Case 99

Proof. From (90) and using the Stieltjes representation (52) of the null-form, we get

Sα = −2α∫ +∞0

t3−4αeβ

t2+ 2t4 e−tϕα+1(t

2)sin t dt− 2β∫ +∞0

t3−4αeβ

t2+ 2t4 e−tϕα(t

2)sin t dt .

Suppose (91) for n ≥ 1 fixed. From (90) where α → α+ 2n and t → t2

ϕα+2n−1(t2) =

1

8t4α+8ne−

β

t4− 2t8 − α+ 2n

4ϕα+2n+1(t

2)− β4ϕα+2n(t

2) ,

hence easily (91) for n → n+ 1 .

Corollary 3.3. If β = 0, then S−2n = 0 , n ≥ 0 .This results is consistent with the fact the linear form u defined by (28) where β = 0 is not regular for these valuesof α.

Proposition 3.4. For α ≥ 52 , we have Sα > 0.

Proof. First, we need the following Lemma [13]

Lemma 3.5. Consider the following integral

S =

∫ +∞0

F (t)sin t dt , (92)

where we suppose F (t) ≥ 0, continuous, increasing in 0 < t ≤ t and decreasing to zero for t > t.Then,

0 < t ≤ π ,∫ π0

(F (t)− F (t+ π)

)sin t dt ≤ 0 ⇒ S > 0 . (93)

Now, denoting F (t) = Fα(t) = fα(t)e−t with

fα(t) = t3−4αe

β

t2+ 2t4 ϕα−1(t

2) .

We have from (89)

t3

2αt8 + 2βt4 + 8≤ fα(t) ≤

2t3

αt8 + 2βt4 + 16, t ≥ 0 , α ≥ 5

2. (94)

Then,t3e−t

2αt8 + 2βt4 + 8≤ Fα(t) ≤ 2

t3e−t

αt8 + 2βt4 + 16, t ≥ 0 , α ≥ 5

2. (95)

Consequently, Fα(t) > 0 for t > 0, Fα(0) = 0 and limt→+∞

Fα(t) = 0 which implies that Fα has a maximum for t = t

defined by f′

α(t) = fα(t).Hence

fα(t) =2t

3

t9

+ (4α− 3)t8 + 4βt4 + 16, (96)

since

f′

α(t) =2

t6−{4α− 3

t+

4β

t5+

16

t9

}fα(t) , t > 0 .

From the first inequality of (94) and by virtue of (96) necessarily t ≤ 3.Therefore the implication (93) is true if the following is verified∫ π

0

sin t2(π + t)3

α(π + t)8 + 2β(π + t)4 + 16e−t−πdt ≤

∫ π0

sin tt3

2αt8 + 2βt4 + 8e−tdt . (97)

100 M. Zaatra

The function t → 2t3

αt8+2βt4+16 is decreasing for t ≥ t1 =(√

240α+β2−β5α

) 14

and from α ≥ 52 we have easily t1 ≤(45

√6) 1

4

< π2 . We have successively∫ π0

sin t2(t+ π)3

α(t+ π)8 + 2β(t+ π)4 + 16e−t−πdt ≤ e−π π

3

απ8 + 2βπ4 + 16(1 + e−π) .

On the other hand∫ πt1

sin tt3

2αt8 + 2βt4 + 8e−tdt =

∫ πt1

sin tt3

αt8 + 2βt4 + 16

αt8 + 2βt4 + 16

2αt8 + 2βt4 + 8e−tdt

≥ 12

π3

απ8 + 2βπ4 + 16

∫ ππ2

sin t e−tdt

≥ 14e−π

π3

απ8 + 2βπ4 + 16(1 + e

π2 ).

Thus, (97) is fulfilled if

e−ππ3

απ8 + 2βπ4 + 16(1 + e−π) ≤

∫ t10

sin t2t3

αt8 + 2βt4 + 16e−tdt+

1

4e−π

π3(1 + eπ2 )

απ8 + 2βπ4 + 16. (98)

But, 1 + e−π < 14 (1 + eπ2 ), therefore the inequality (98) is satisfied and the proposition is proved.

Finally, for f ∈ P , α ≥ 52

〈u, f〉 = S−1α∫ +∞−∞

1

|x|5

∫ +∞x2

(x2t

)αexp(− βx2

+β

t

)exp( 2t2− 2x4

)s(t) dt f(x) dx . (99)

Remark 5. Multiplying (28) by x and using (1) we get

(x7u)′− 2x2((α+ 1)x4 + βx2 + 4)u = 0 .

Applying the operator σ for the above equation and using (4)− (5), we obtain(x4σu

)′− x((α+ 1)x2 + βx+ 4)σu = 0 . (100)

Moreover, since the linear form u is symmetric and regular, then σu is regular.

If α(u)4 + β(u)2 + 4 6= 0, then from (8), the linear functional σu is semi-classical of class s = 2.From (99), we get

〈σu, f(x)〉 =〈u, f(x2)

〉= S−1α

∫ +∞−∞

1

|x|5

∫ +∞x2

(x2t

)αexp(− βx2

+β

t

)exp( 2t2− 2x4

)s(t) dt f(x2) dx .

After a change of variables, we get

〈σu, f(x)〉 = S−1α∫ +∞0

xα−3exp(−βx− 2x2

)∫ +∞x

t−αexp( 2t2

+β

t

)s(t) dt f(x) dx . (101)

The integral representation (100) doesn’t exist in the list given in [10].

A7.2. By applying the same process as we did to obtain (99) with g(x) = x|x|3s(x2), we get after some straightforwardcalculations

〈u, f〉 = T−1α∫ +∞−∞

1

|x|3

∫ +∞x2

( t2x4

)1−αexp(2t− 2x2

)s(t) dt f(x) dx , α ≥ 7

4, (102)

with

Tα = 8

∫ +∞0

t11−8αe2t4 e−tψα− 32 (t

2)sin t dt > 0 , (103)

Semi-classical Linear Functionals of Class Four:The Symmetric Case 101

ψα(t) =

∫ t0

x4α−1e−2x2 dx . (104)

Besides Tα and ψα verify the following relation:

1

2t2ψα(t) ≤ ψα+ 12 (t) ≤ t

2ψα(t) , α ≥1

4, t ≥ 0 , (105)

t4α+2e−2t2

4 + 2(2α+ 1)t2≤ ψα(t) ≤

t4α+2e−2t2

4 + (2α+ 1)t2, α ≥ 1

4, t ≥ 0 , (106)

Tα =(−1)n

2n−3

n−1∏k=0

(2α− 2 + k)∫ +∞0

t11−8αe2t4 e−tψα−1+n−12

(t2)sin t dt , n ≥ 1 . (107)

Remark 6. By applying the same process as we did to obtain (100), we have

(x4σu)′− 2x2(αx+ 1)σu = 0 .

Moreover, if (1− 2α)(u)4 − 2(u)2 6= 0, then from (8), the linear functional σu is semi-classical of class s = 2.Finally, from (102), we get

〈σu, f(x)〉 = T−1α∫ +∞0

x2α−4e−2x

∫ +∞x

t2−2αe2t s(t) dt f(x) dx . (108)

The integral representation (108) doesn’t exist in the list given in [10].

Case B.

B1. From (34) to (54), we have

U′(x)

U(x)= −2λx+ 2αx

x2 − 1+

2βx

x2 − c2.

Thus,

U(x) = |1− x2|α|x2 − c2|βe−λx2

, (109)

is the solution in ]−∞,+∞[, [−1, 1] and [−c, c].Furthermore, for αβ 6= 0 and α, β > −1, (8) and (49) hold in the following situations:

s1 : If λ > 0, then u is represented by

〈u, f〉 =∫ +∞−∞

|1− x2|α|x2 − c2|βe−λx2(Aχ[−1,1](x) +Bχ[−c,c](x) + Cχ]−∞,+∞[(x)

)f(x) dx . (110)

s2 : If λ < 0 and 0 < c < 1, then

〈u, f〉 =∫ 1−1

(1− x2)α|x2 − c2|βe−λx2(A1χ[−1,1](x) +B1χ[−c,c](x)

)f(x) dx . (111)

s3 : If λ < 0 and c > 1, then

〈u, f〉 =∫ c−c|1− x2|α(c2 − x2)βe−λx

2(A2χ[−1,1](x) +B2χ[−c,c](x)

)f(x) dx . (112)

B2. From (36) to (54), we obtain

U′(x)

U(x)= −2λx+ α

x+

2βx

x2 − 1.

Hence,

U(x) = |x|α|x2 − 1|βe−λx2

, (113)

is the solution in ]−∞,+∞[ and [−1, 1].If β 6= 0 and α, β > −1, (8) and (49) hold in the following situations:

102 M. Zaatra

s1 : If λ > 0, then

〈u, f〉 =∫ +∞−∞

|x|α|x2 − 1|βe−λx2(Aχ[−1,1](x) +Bχ]−∞,+∞[(x)

)f(x) dx . (114)

s2 : If λ < 0, then

〈u, f〉 = A1∫ 1−1|x|α(1− x2)βe−λx

2

f(x) dx . (115)

Particular case:

If α = β = 0, then (36) becomes (x2(x2 − 1)u)′ + 2x

(λx4 − (λ+ 2)x2 + 1

)u = 0 ,{

2λ(u)4 − (2λ+ 3)(u)2 + 1}×{

2λ(u)4 − 3(u)2}6= 0 .

Thus,

u = (1− k)ṽ + kδ0 , (116)

where ṽ is symmetric semi-classical form of class two satisfies (see [8, 10])

((x2 − 1)ṽ)′+ 2x(λx2 − λ− 1)ṽ = 0 .

Notice that the linear functional u defined by (116) is regular if and only if k satisfies[11]

1 + k

n∑m=0

P 2m(0)

〈v, P 2m〉6= 0 , n ≥ 0 ,

where {Pn}n≥0 is the MOPS with respect to ṽ.In this case

〈u, f〉 = 1− k√λπ

∫ +∞−∞

e−λx2

f(x)dx+ kf(0) , λ > 0 .

B3. From (38) to (54) we get

U′(x)

U(x)= −2λx+ 2αx

x2 − 1− 2βx

(x2 − 1)2,

therefore

U(x) = |x2 − 1|αe−λx2− β

1−x2 , (117)

is the solution in ]−∞,+∞[ and [−1, 1].If β 6= 0 and α > −1, then (8) and (49) hold in the following situations:

s1 : If λ > 0 and β > 0, then

〈u, f〉 =∫ +∞−∞

|x2 − 1|αe−λx2− β

1−x2(Aχ[−1,1](x) +Bχ]−∞,+∞[(x)

)f(x) dx . (118)

s2 : If λ < 0 and β > 0, then

〈u, f〉 = A1∫ 1−1

(1− x2)αe−λx2− β

1−x2 f(x) dx . (119)

Particular case:

If β = 0 and α = 2, then from (38) we obtain ((x2 − 1)2u)′ + 2x

(λx4 − 2(λ+ 2)x2 + λ+ 4

)u = 0 ,

2λ(u)4 − (2λ+ 7)(u)2 + 1 6= 0 .

Semi-classical Linear Functionals of Class Four:The Symmetric Case 103

Then,u = (1− 2k)h√λ−1H+ k(δ1 + δ−1) , (120)

where H is the Hermite linear functional that satisfies[7]

H′+ 2xH = 0 .

The above linear functional u is regular if and only if k satisfies (see [2])∣∣∣∣1 + kKn(1, 1) kKn(1,−1)kKn(1,−1) 1 + kKn(−1,−1)∣∣∣∣ 6= 0 , n ≥ 0 .

From (120), applying both linear functionals x2n we get

(u)2n = (1− 2k)λ−nΓ(2n+ 1)

22nΓ(n+ 1)+ 2k , n ≥ 0 .

Finally, from (120) we obtain

〈u, f〉 = 1− 2k√λπ

∫ +∞−∞

e−λx2

f(x)dx+ k(f(1) + f(−1)) , λ > 0 .

B4. From (40) to (54), we have

U′(x)

U(x)= −2x+ 2α

x+

2β

x3.

Thus,

U(x) = |x|2αe−x2− β

x2 , (121)

is the solution in ]−∞,+∞[.

If β ≥ 0 and α > − 12 , then (8) and (49) hold.Thus

〈u, f〉 = k∫ +∞−∞

f(x)|x|2αe−x2− β

x2 dx . (122)

Remark 7. If β = 0, then the form defined by (122) is the generalized Hermite polynomials [7].

B5.1. From (43) to (54), we get

U′(x)

U(x)= −4λx3 + +2αx+ 2βx

x2 − 1.

SoU(x) = |x2 − 1|βe−λx

4+αx2 , (123)

is the solution in R and [−1, 1].If β 6= 0 and β > −1, (8) and (49) hold in the following situations:

s1 : If λ > 0, then

〈u, f〉 =∫ +∞−∞

|x2 − 1|βe−λx4+αx2

(Aχ[−1,1](x) +Bχ]−∞,+∞[(x)

)f(x) dx . (124)

s2 : If λ < 0, then

〈u, f〉 = A1∫ 1−1

(1− x2)βe−λx4+αx2f(x) dx . (125)

B5.2. From (45) to (54), we obtain

U′(x)

U(x)= −4x3 + 2βx+ α

x.

104 M. Zaatra

Hence,

U(x) = |x|αe−x4+βx2 , (126)

is the solution in R.

Furthermore, for α > −1, then (8) and (49) hold.Thus,

〈u, f〉 = A∫ +∞−∞

f(x)|x|αe−x4+βx2 dx . (127)

Particular case:

If α = 0, then (45) becomes (x2u)′ + 2x

(2x4 − βx2 − 1

)u = 0 ,

4λ(u)4 − 2β(u)2 − 1 6= 0 .

Thus,u = (1− k)v̆ + kδ0 , (128)

where v̆ is symmetric semi-classical form of class two satisfies (see [8, 10])

v̆′+ 2x(2x2 − β)v̆ = 0 .

In this case

〈u, f〉 = (1− k)A∫ +∞−∞

e−x4+βx2f(x)dx+ kf(0) ,

where A =

(∫ +∞−∞

e−x4+βx2dx

)−1.

B6. From (47) to (54), we get

U′(x)

U(x)= −6x5 − 4αx3 − 2βx .

Thus,

U(x) = e−x6−αx4−βx2 , (129)

is the solution in R. As a consequence u is represented by

〈u, f〉 = k∫ +∞−∞

f(x)e−x6−αx4−βx2 dx . (130)

This is an example of Freud linear functional (see [8] and [16]).

References

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[2] R. Álvarez-Nodarse, J. Arvesú and F. Marcellán, Modifications of quasi-definite linear functionals viaaddition of delta and derivatives of delta Dirac functions, Indag. Math. (N. S.) 15 (2004), 1-20.https://doi.org/10.1016/s0019-3577(04)90001-8

[3] M.J. Atia, Some symmetric sem-classical orthogonal polynomials of class 0 ≤ s ≤ 5. Integral Transforms Spec.Funct. 17(7) (2006), 469-483. https://doi.org/10.1080/10652460600725085

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Semi-classical Linear Functionals of Class Four:The Symmetric Case 105

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[7] T. S. Chihara, An introduction to orthogonal polynomials, Gordon and Breach, New York. 1978.

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[11] F. Marcellán, and P. Maroni, Sur l’adjonction d’une masse de Dirac à une forme régulière et semi-classique.Ann. Mat. Pura ed Appl. 162 (IV) (1992), 1-22. https://doi.org/10.1007/bf01759996

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Introduction