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CLASSICAL 2-ORTHOGONAL POLYNOMIALS AND DIFFERENTIAL EQUATIONS

BOUKHEMIS AMMAR AND ZEROUKI EBTISSEM

Received 16 May 2005; Revised 17 April 2006; Accepted 25 April 2006

We construct the linear differential equations of third order satisfied by the classical 2- orthogonal polynomials. We show that these differential equations have the following form:R4,n(x)Pn(3)+3(x)+R3,n(x)Pn+3(x)+R2,n(x)Pn+3(x)+R1,n(x)Pn+3(x)=0, where the coeffi- cients{Rk,n(x)}k=1,4are polynomials whose degrees are, respectively, less than or equal to 4, 3, 2, and 1. We also show that the coefficientR4,n(x) can be written asR4,n(x)= F1,n(x)S3(x), whereS3(x) is a polynomial of degree less than or equal to 3 with coefficients independent ofnand deg(F1,n(x))1. We derive these equations in some cases and we also quote some classical 2-orthogonal polynomials, which were the subject of a deep study.

Copyright © 2006 Hindawi Publishing Corporation. All rights reserved.

1. Introduction

The classical orthogonal polynomials (Hermite, Laguerre, Jacobi, and Bessel) satisfy a hypergeometric-type differential equation of second order [5]:

σ(x)y(x) +τ(x)y(x) +λny(x)=0, where degσ2, degτ1, λn= −n(n1)

2 σ=0, n0. (1.1)

These polynomials are the unique polynomial solutions of a second-order linear dif- ferential equation of hypergeometric type [14].

The aim of this work is to generalize the results obtained in the standard orthogonality to 2-orthogonality. We first look for the differential equations whose the solutions are classical 2-orthogonal polynomials and we explicit them, where it is possible.

First, we recall some basic notions of thed-orthogonality, then we study the nature of coefficients of recurrence relations satisfied by the classical 2-orthogonal polynomials se- quences. We show afterwards that these polynomials are solutions of a third-order linear differential equation with polynomial coefficients of degree less than or equal to 4, 3, 2,

Hindawi Publishing Corporation

International Journal of Mathematics and Mathematical Sciences Volume 2006, Article ID 12640, Pages1–32

DOI10.1155/IJMMS/2006/12640

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and 1, depending generally onn. The main result is that the coefficient associated with highest derivative can be written as the product of 2 polynomials of which one is of degree

3 and independent ofn. The latter, will allow us not only to enumerate some polyno- mial solutions, but also to explicit some ODEs. Of course, these equations generalize the Sturm-Liouville equations.

The cases where the polynomial solutions are 2-symmetric orthogonal are completely derived. Finally, we mention some examples of classical 2-orthogonal polynomials with some of their properties.

The final goal being naturally to search for the analog theorem of Bochner, that is, first, to enumerate all sequences of classical 2-orthogonal polynomials and afterwards, to study their properties, in particular the representation of the pair of linear forms in each case.

2. Preliminary notions

First, we recall some definitions and properties of the sequences ofd-orthogonal polyno- mials, without forgetting to mention however, that thed-orthogonal polynomialsPn(n 0) are a special case of type II multiple orthogonal polynomialsRs(n), where the sequence

s(n) (n0) of multi-indices inNd, withn=md+α, 0αd1,m0, is defined by

s(n)=

m + 1,m+ 1, ...,m+ 1

αtimes

,m,m,...,m

, (2.1)

and wherePn(x)=Rs(n)(n0) [1,21].

Note that the multiple orthogonal polynomials are narrowly related to simultaneous vectorial Pade approximation, to be more precise as Hermite-Pade approximation. In particular, the type II multiple orthogonal polynomialsRn,n =(n1,n2,...,nd), for the measures{μj}dj=1, that is, the monic polynomialRn of degree|−n| =n1+n2+···+nd

which satisfies the orthogonal conditions

k

xmRn(x)dμj=0, k=0, 1,...,nj1, j=1, 2,...,d (2.2) (resp., thed-orthogonal polynomials with respect to the vector linear form=(ᏸ0,ᏸ1, ...,d1)T) represent the common denominator of rational approximation of thedStielt- jes functions [3,17,19,21]

fj(z)=

j

j

zx, zj, j=1, 2,...,d, (2.3) that is,

Rn(z)fj(z)Qn,j(z)=Oznj1, |z| −→ ∞, j=1, 2,...,d. (2.4) Definition 2.1. Let{Pn}n0be a sequence of monic polynomials (i.e.,Pn(x)=xn+···).

Call the dual sequence of the sequence{Pn}n0, the sequence of linear forms{£n}n0

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defined by

£nPm(x)=

£n,Pm(x)=δn,m, m,n0, (2.5) where·,·denotes the duality bracket between the vector space of polynomialsᏼand its algebraic dual spaceᏼ.

Definition 2.2 [16,21]. A sequence of polynomials{Pn}n0isd-orthogonal with respect to £=0, £1,..., £d1)Tif it satisfies

£αxmPn(x)=0, nmd+α+ 1,m0,

£αxmPmd+α(x)=0, m0, 0αd1.

(2.6)

Theorem 2.3 [16,21]. Let{Pn}n0be a monic sequence of polynomials, then the following statements are equivalent.

(a) The sequence{Pn}n0isd-orthogonal with respect to £=0, £1,..., £d1)T. (b) The sequence{Pn}n0satisfies a recurrence relation of orderd+ 1 (d1):

Pm+d+1(x)=

xβm+d

Pm+d(x)d

1

ν=0

γm+dd1ννPm+d1ν(x), m0, (2.7)

with the initial data

P0(x)=1, P1(x)=xβ0, Pm(x)=

xβm1

Pm1(x)m

2

ν=0

γdm11ννPm2ν(x), 2md,

(2.8)

whereγ0m+1=0,m0. (Regularity conditions.)

Remark 2.4 [12,18]. This result generalizes the Shohat-Favard theorem.

Definition 2.5. The sequence{Pn}n0is said to bed-symmetric if it satisfies

Pn

ρkx=ρnkPn(x), n0, whereρk=exp 2ikπ

d+ 1

,k=1,...,d. (2.9)

Theorem 2.6 [9–11]. For each monicd-orthogonal sequence{Pn}n0, the following equiv- alences hold.

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(a){Pn}n0isd-symmetric.

(b){Pn}n0satisfies the recurrence relation

Pn(x)=xn, 0nd,

Pn+d+1(x)=xPn+d(x)γn0+1Pn(x), n0. (2.10) Definition 2.7 [10]. A sequence of polynomials{Pn}n0(d1) is said to be “classical” if the sequence of the derivatives is alsod-orthogonal.

Corollary 2.8 [9, 15]. When the sequence {Pn}n0 is classical d-orthogonal and d- symmetric, then the monic sequence of derivatives{Qn}n0(i.e.,Qn(x)=Pn+1(x)/(n+ 1)) satisfies the following recurrence relation:

Qn(x)=xn, 0nd,

Qn+d+1(x)=xQn+d(x)δ0n+1Qn(x) withδn+10 =0,n0. (2.11) 3. Classical 2-orthogonal polynomials

Statement of the problem. In this work, we try to answer three main questions.

(i) Which type of differential equations have as solutions classical 2-orthogonal polynomials?

(ii) Can we exhibit these differential equations?

(iii) What are these polynomials solutions?

For this, we consider a monic sequence of classical 2-orthogonal polynomials {Pn(x)}n0, such that the recurrence relations satisfied by the polynomials Pn(x) and Pn(x) (n0) are given, respectively, by

P0(x)=1, P1(x)=xβ00, P2(x)= xβ01

P1(x)γ01, Pn+3(x)=

xβ0n+2

Pn+2(x)γ0n+2Pn+1(x)δn0+1Pn(x), n0,

(3.1) with the regularity conditionδn0=0,n1, and

P1(x)=1, P2(x)=2xβ11

, P3(x)=3 2

xβ12

P2(x)γ12

, n+ 3

n+ 4Pn+4(x)= xβn1+3

Pn+3(x)γ1n+3Pn+2(x)δn1+2Pn+1(x), n0,

(3.2)

with the regularity conditionδn1+1=0,n1.

Proposition 3.1 [9]. The coefficientsβ0n1n0n1n0n, andδn1satisfy the following finite difference system:

(n+ 2)β1n+1n1(n+ 1)β0n+1+ (n1)β0n=0, n0, (3.3) (n+ 3)γ1n+2(n+ 2)γ0n+2

n+ 2

1n+1(n1)γ0n+1

n+ 1 =

βn1+1β0n+1

2

, n0, (3.4)

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(n+ 4)δn1+2(n+ 3)δn0+2

n+ 3

1n+1(n1)δn0+1

n+ 1

=γn0+2

β0n+2+β0n+11n+1

γ1n+2

0n+2β1n+2β1n+1

, n0,

(3.5)

δ0n+1

βn0β1nδn1+1

βn0+2β1n+2

+δ0n+1δ1n+1

β0n+2βn1

=γn1+1

γ0n+2γ1n+2

γn0+2

γn0+1γn1+1

, n1,

(3.6)

δ0n+2

γ0n+1γ1n+1

δn1+1

γn0+3γn1+3

=γn1+1

δn0+2δn1+2

γ0n+3

δn0+1δn1+1

, n1,

(3.7)

δn0+3

δn0+1δn1+1

=δn1+1

δn0+3δn1+3

, n1. (3.8)

Proof. From (3.1) and (3.2), we get the relation

Pn+3(x)= 1

n+ 4Pn+4(x) +β0n+3β1n+3

Pn+3+γ0n+3γ1n+3

Pn+2(x) +δn0+2δn1+2

Pn+1(x), n0.

(3.9)

Multiplying byxboth hand sides of this relation and using once again (3.2), we get the

precedent system.

Remark 3.2. We see that the determination of all the 2-orthogonal sequences goes through the resolution of the system (3.1)–(3.8). Many authors have tried to solve it, but up to now, its resolution is still giving many problems because it is nonlinear as well as the number of unknowns is relatively high (six). Nevertheless, we will analyze the cases where its resolution is complete. In fact, we have the following.

Lemma 3.3. Equation (3.8) admits the following as a unique set of solutions.

(A)δn1+1=δn0+1,n1.

(B)δ21n=(n+ρ2)/(n1 +ρ220nandδ21n+1=δ02n+1,n1.

(C)δ21n+1=(n+ρ3)/(n1 +ρ320n+1andδ12n=δ20n,n1.

(D)δ21n=(n+ρ2)/(n1 +ρ220nandδ21n+1=(n+ρ3)/(n1 +ρ302n+1,n1, where ρ2= −δ20/(δ20δ21),ρ3= −δ30/(δ30δ31), and (ρ2andρ3/ Z).

Remark 3.4. In the last case if we putρ=2=31, then we obtain the important particular case denoted byD1 and where

δn+11 = n+ 1 +ρ

n1 +ρδn+10 , n1. (3.10)

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Proof ofLemma 3.3. δn+11 =δ0n+1,n1, is a trivial solution of (3.8).

In case, where there existsn01 such thatδ1n0+1=δn00+1, then forn0=2k0(resp.,n0= 2k0+ 1),k0N, (3.8) becomes

δ2k00+3δ2k00+1δ12k0+1=δ12k0+1δ2k00+3δ2k10+3=0 (3.11) (resp.,δ20k0+402k0+2δ21k0+2)=δ21k0+220k0+4δ21k0+4)=0).

Thusδ12(k0+1)+1=δ2(0k0+1)+1,k01 (resp.,δ2(1k0+1)+2=δ02(k0+1)+2,k00), and therefore δ21n0+1=δ20n0+1,n01 (resp.,δ21n0+2=δ20n0+2,n00). Equation (3.8) can be written as

δ02n0+3

δ20n0+3δ21n0+3

δ20n0+1

δ20n0+1δ21n0+1

= −1

resp., δ02n0+4

δ2n00+4δ12n0+4

δ20n0+2

δ2n00+2δ2n10+2

= −1

(3.12)

then δ20n0+3

δ20n0+3δ21n0+3

δ30

δ30δ31

= −n0, n00 or δ21n0+1= n0+ρ3

n01 +ρ3δ20n0+1, n01,

resp.,δ21n0= n0+ρ2

n01 +ρ2δ20n0,n01

.

(3.13) Lemma 3.5. In case (A) (i.e.,δn1+1=δn0+1,n1), (3.7) admits the following four solutions.

(A1)γ0n+1=γn1+1,n1.

(A2)γ02n=γ12nandγ02n+1γ12n+1=03γ13)(δ1020)nν=1001),n1.

(A3)γ02n+1=γ12n+1andγ02nγ21n=02γ12)(1/δ01)nν=112ν+10 0), (δ00=1),n1.

(A4)γ02nγ12n=02γ12)(1/δ01)nν=1120ν+120ν) and γ02n+1γ21n+1=30γ13)(δ1002)

×n

ν=120ν20ν1),n1.

Lemma 3.6 [9]. In case (A1) (i.e.,γ0n+1=γ1n+1andδn1+1=δn0+1,n1), β20n=β00+nb1+ 3b2

, n0, β20n+1=β01+n3b1+b2

, n0,

(3.14) γ2n+10 =(2n+ 1)γ01+nb21+b22

, n0, γ02n+2=2(n+ 1)γ01+ (n+ 1)b12+nb22

, n0,

(3.15) δ2n+10 =(n+ 1)(2n+ 1)δ10+ 2nb22

b2b1

, n0, δ20n+2=(n+ 1)(2n+ 3)δ10+ 2b1b2

γ01+ (n+ 1)b12

, n0,

(3.16)

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Table 3.1

Case β0n,n0 γ0n+1,n0

(A1.1) β0n=0 γ0n+1=(n+ 1)γ10

(A1.2) β0n=2nb1 γn+10 =(n+ 1)γ01+nb21

(A1.3)

β02n+1=n γ2n+10 =(2n+ 1)

n+δ01k1

2 β02n=3n γ2n+20 =(n+ 1)2n+δ01k1

(A1.4)

β2n+10 =3n+ 2 γ2n+10 =(2n+ 1)

n+k22δ01

2 β02n=n γ2n+20 =(n+ 1)2n+k2δ01

(A1.5) See (3.14) See (3.15)

whereβ000110, andδ10are arbitrary andb1andb2are constants defined by

b1:=β01β11=1 2

β01β00

, b2:=β02β12=1 6

20β01β00

. (3.17)

Proof. From (3.6) we have (β0nβ1n)=n0+2β1n+2),n0.

In particular forn=2k, β20k+2β12k+2=

β02kβ12k= ··· =β02β21=1 6

02β01β00

=b2, (3.18)

and forn=2k+ 1, β02k+3β12k+3=

β02k+1β21k+1)= ··· =β01β11=1 2

β01β00

=b1. (3.19)

Using (3.3), we get

2(k+ 1)b1kb2

=β02k+1β2k0 , n0, (2k+ 3)b2(2k+ 1)b1=β02k+2β20k+1, n0.

(3.20)

By adding up term by term these last 2 relations and summing this last result, we obtain the first relation of (3.14). The second relation of (3.14) is obtained in the same way.

Equations (3.15) and (3.16) are obtained similarly by using, respectively, (3.4) and

(3.5).

Proposition 3.7 [9]. The case (A1) is constituted by the following five canonical classical 2-orthogonal polynomials.

(A1.1)b1=b2=0.

(A1.2)b1=b2=0.

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Table 3.2

Case δn+10 ,n0δ02=2 Initials parameters

(A1.1) δ0n+1=(n+ 1)(n+ 2)δ20

2 β00=0;γ01andδ10arbitrary (A1.2) δn+10 (n+ 1)(n+ 2)δ20

2 β00=0;b1δ01andγ10arbitrary (A1.3) δ2n+10 =(n+ 1)(2n+ 1)2n+δ10

b2=1,β00=0

δ2n+20 =k1(n+ 1)(2n+ 3) δ10andk1=δ1001=0 arbitrary (A1.4) δ2n+10 =(n+ 1)(2n+ 1)δ10 b1=1,β00=0

δ02n+2=(n+ 1)(2n+ 3)2n+k2

δ10andk2=δ10+ 2γ01+ 2=0 arbitrary (A1.5) See (3.16) β00,b1,γ01, andδ10arbitrary

(A1.3)b1=0 andb2=0.

(A1.4)b2=0 andb1=0.

(A1.5)b1=b2andb1b2=0.

Remark 3.8. In the precedent case (i.e., (A1)) the coefficients β0nn+10 , andδn+10 can be written, respectively, in the simplified forms in Tables3.1and3.2.

Proposition 3.9 [9]. There exist only four sequences of classical 2-symmetric 2-orthogonal polynomials. The coefficientsδn0+1andδn1+1(n0) are explicit inTable 4.1.

4. Main results

4.1. Differential equations. In this section, we will construct the differential equations, whose solutions are classical 2-orthogonal polynomials, afterwards, we will give the na- ture of these equations by the study of the coefficient associated with highest derivative.

Let us note that the polynomials enumerated inProposition 3.7and the 2-symmetric solution polynomials will be completely exhibited (perfectly identified).

An analysis of a particular case (already studied) is done at the end of this section, as well as the citation of some classical 2-orthogonal polynomials, which were the subject of a deep study.

First, let us note

dn+2:=δ1n+2δ0n+2, n0, Bn:=(n+ 4)βn1+3(n+ 3)β0n+3, n0, hn=δ1n+2β0n+3δ0n+2β1n+3, n0, Gn:=(n+ 4)γ1n+3(n+ 3)γn0+3, n0, Cn:=δn1+2γn0+3δn0+2γ1n+3, n0, Dn:=

(n+ 4)δn1+2(n+ 3)δn0+2

n+ 4 , n0.

(4.1) Then, we have the following result.

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Table 4.1

Case δn+10 ,n0

(A) δ0n+1=(n+ 1)(n+ 2)δ01

2

(B)

δ2n+10 =(n+ 1)(2n+ 1) 3n+ 1 +ρ2

ρ2+ 1δ10

δ02n+2= (n+ 1)(2n+ 3)n+ρ2

3n+ 1 +ρ2

3n+ 4 +ρ2

ρ2+ 1δ01

(C)

δ2n+10 =(n+ 1)(2n+ 1)n1 +ρ3

3n1 +ρ3

3n+ 2 +ρ3

ρ3+ 2δ10

δ02n+2=(n+ 1)(2n+ 3) 3n+ 2 +ρ3

ρ3+ 1)δ10

(D)

δ2n+10 = (n+ 1)(2n+ 1)n1 +ρ3 3n1 +ρ3

3n+ 2 +ρ3

3n+ 1 +ρ2

ρ2+ 1ρ3+ 2δ10

2 δ2n+20 = (n+ 1)(2n+ 3)n+ρ2

3n+ 1 +ρ2

3n+ 4 +ρ2

3n+ 2 +ρ3

ρ2+ 1ρ3+ 2δ10

2 (D1) δ0n+1= (n+ 1)(n+ 2)(n1 +ρ)

(3n1 +ρ)(3n+ 2 +ρ)(3n+ 5 +ρ)+ 2)(ρ+ 5)δ01

2

Theorem 4.1. When

Cndn+2=0, n0, (4.2)

the classical 2-orthogonal polynomialsPn+3(x) (n0) which satisfy a differential equation are solutions of a third-order linear differential equation with polynomial coefficients of the form

R4,n(x)P(3)n+3(x) +R3,n(x)Pn+3(x) +R2,n(x)Pn+3(x) +R1,n(x)Pn+3(x)=0, n0, (4.3) with

R4,n(x) :=F1,n(x)S3,n(x), R3,n(x) :=F1,n(x)V2,n(x)F1,n(x)S3,n(x), R2,n(x) :=F1,n(x)W1,n(x)F1,n(x)T2,n(x), R1,n(x) :=(n+ 3)δn1+2

dn+2

xhn1

dn+1

+ Cn1Gn

(n+ 4)Dndn+1

F1,n(x) δn1+1

dn+1+ 2

F1,n(x)

, (4.4)

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