Some Functions that Generalize the Askey-Wilson Polynomials

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SomeFunctionsthatGeneralizetheAskey–WilsonPolynomials

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Commun. Math Phys. 184, 173 – 202 (1997) Communications inMathematical

Physicsc© Springer-Verlag 1997

Some Functions that Generalize the Askey–WilsonPolynomials

F. Alberto Gr unbaum1,?, Luc Haine2,??

1 Department of Mathematics, University of California, Berkeley, CA 94720–3840, USA.2 Department of Mathematics, Universite Catholique de Louvain, 1348 Louvain-la-Neuve, Belgium.

Received: 7 May 1996 / Accepted: 30 August 1996

Abstract: We determine all biinfinite tridiagonal matrices for which some family ofeigenfunctions are also eigenfunctions of a second orderq-difference operator. Thesolution is described in terms of an arbitrary solution of aq-analogue of Gauss hyper-geometric equation depending on five free parameters and extends the four dimensionalfamily of solutions given by the Askey-Wilson polynomials. There is some evidencethat this bispectral problem, for an arbitrary orderq-difference operator, is intimatelyrelated with someq-deformation of the Toda lattice hierarchy and its Virasoro symme-tries. When tridiagonal matrices are replaced by the Schroedinger operator, andq = 1,this statement holds with Toda replaced by KdV. In this context, this paper determinesthe analogs of the Bessel and Airy potentials.

Table of Contents1 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1742 An Operator Identity and its Solution. . . . . . . . . . . . . . . . . . . . . . . . . . . 1763 Theq-Riccati Equation forf1(k) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1804 The OperatorB . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1835 Linearizing theq-Riccati Equation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1876 The Gauss-Askey-Wilson Equation & Proof of Theorem 1. . . . . . . . . . 1917 Solving the Gauss-Askey-Wilson Equation in Terms of Basic Hyper-

geometric Series. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1958 The Case of the Classical Orthogonal Polynomials. . . . . . . . . . . . . . . . . 1989 The Casev = 0 in theq-Riccati Equation. . . . . . . . . . . . . . . . . . . . . . . . 200References. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201

? The first author was supported in part by NSF Grant # DMS94-00097 and by AFOSR under ContractAFO F49629-92.

?? The second author is a Research Associate for FNRS.

174 F.A. Grunbaum, L. Haine

1. Introduction

In [4] Askey and Wilson introduced a family of polynomials depending on four param-etersa, b, c, d that satisfy a three term recursion relation and a second orderq-differenceequation.

As the title of [4] clearly states these polynomials generalize those of Jacobi and infact most of the polynomials used in mathematical physics and harmonic analysis canbe seen to be special or limiting cases of the Askey–Wilson polynomials. It is fair to saythat the main properties of all these polynomials are indeed closely related to the twomentioned above.

In the last few years these polynomials have become a working tool in many devel-oping areas of mathematics and physics among them quantum groups, see [8, 20, 22].A sample of other applications is given in [10, 12, 23–25].

In [14] and particularly in [1] and [15] some instances were uncovered where thesepolynomials play a role in relation to nonlinear evolution equations connected with theToda lattice and its Virasoro symmetries. We defer to a later paper for a more detailedlook into this issue.

In [3] Andrews and Askey proposed that a family of orthogonal polynomials shouldbe named “classical” exactly when the two properties mentioned above, namely a threeterm recursion relation and a second orderq-difference equation hold, i.e. we have

fn(k) = (s(k)−bn)fn−1(k)−an−1fn−2(k) for n ≥1 andf−1 =0, f0 = 1 (1.1)

and

Bfn(k) ≡ a(k)(fn(qk)−fn(k)) + b(k)(fn(k/q)−fn(k)) = θnfn(k) , (1.2)

for some appropriate choice of the “spectral parameter”s(k).When (1.2) is replaced by a differential equation ink the choice of “spectral param-

eter” is immaterial, as discussed in [14]. Within the context ofq-difference equations thechoice of spectral parameter cannot be swept under the rug. We limit ourselves to thecase

s(k) = γ(k + ε/k), (1.3)

with γ andε arbitrary parameters. The choicess(k) = k ands(k) = (k+1/k)/2 are up toscaling the only ones that deserve attention. These two familiar cases are connected withthe bigq-Jacobi and the Askey–Wilson polynomials respectively. We make an effort topush both cases in a unified fashion. At times we abuse the standard convention byreferring to the cases(k) = γ(k + ε/k) as the Askey–Wilson case.

In [14] a proof is given of the fact that the Askey–Wilson (and the bigq-Jacobi)polynomials are indeed the only ones that satisfy these two properties. In [7] Bochnerhad proved the corresponding statement when one is dealing with ordinary second orderdifferential equations (i.e.q = 1), namely the only families of polynomials satisfying athree term recursion relation and a second order differential equation are those of Jacobi,Laguerre, Hermite and Bessel.

In [15] this result of Bochner is “revisited” by doing away with the requirement thatthe functionsfn(k) should be polynomials in the spectral variablek. This is achievedby removingthe conditionf−1(k) = 0. The main result is that all the instances whenthe two properties hold are given by an arbitrary choice offiveparameters which afterscaling and translation ink can be reduced tothree. For any such choice of parametersone considers an arbitrary solution of Gauss’ hypergeometric equation (which contains

Some Functions that Generalize Askey–Wilson Polynomials 175

threefree parameters), and buildsf1(k) out of it. The rest of the family is then determinedsince we can still assumef0(k) = 1.

The main motivation for “revisiting” this result of Bochner is our interest in the“bispectral property” studied in the continuous-continuous case in [9] and then analyzedin [13–15]. A specially simple case of the “bispectral property” is given by the existenceof nontrivial simultaneous solutions to (1.1) and (1.2) above. In the context of thisproblem one sees that the restriction to polynomials is not natural and that starting withthe largest possible family of “second order” bispectral situations is crucial in obtainingthe “higher order” ones. If one had imposed in [9] an initial condition of the typeφ(0, k) = 0 — in line with the conditionf−1(k) = 0 in the present setup — one wouldhave missed completely the “Korteweg-deVries” cases, and would have only found theVirasoro cases. These considerations could also be relevant in providing new examplesin the theory of random matrices, see for instance [1, 2, 23, 26].

The purpose of this paper is to obtain the general (q 6= 1) version of the resultsobtained in [15], i.e. we remove the conditionf−1(k) = 0 in (1.1) and retain (1.2). Moreprecisely (1.1) is replaced by (2.1).

We can now state the main result in this paper.

Theorem 1. All the instances of familiesfn(k), n ∈ Z, f0(k) = 1, satisfying a threeterm recursion (1.1) with spectral parameters(k) = γ(k + ε/k) and a second orderq-difference equation (1.2) are obtained as follows:

(1) Choose arbitrarilyfive parametersa, b, c, d, z and use them to definean andbn

in the three term recursion relation by

an = An−1Cn,bn = γ(εa + 1

a ) − (An−1 + Cn−1)(1.4)

with

An =γ(q−nz − εab)(q−nz − ac)(q−nz − ad)(q1−nz − abcd)

a(q1−2nz2 − abcd)(q−2nz2 − abcd),

Cn =γa(q−nz − 1)(q1−nz − bc)(q1−nz − bd)(εq1−nz − cd)

(q2−2nz2 − abcd)(q1−2nz2 − abcd).

(2) For each such choice consider the second orderq-difference equation

A(k)(y(qk) − y(k)) + A(ε/k)(y(k/q) − y(k)) = (1− z)(abcd

qz− 1

)y(k) (1.5)

with

A(k) =(1−ka)(1−kb)(ε−kc)(ε−kd)

(ε − k2)(ε − k2q)

≡ y4k4 − y3k

3 + y2k2 − εy1k + ε2

(ε − k2)(ε − k2q),

and with anarbitrary solutiony(k) of this equation constructf1(k) via the “Rodriguestype formula”

f1(k) = s(k) − b1 + γqzy4−qz2 A(k)(k − ε

k )(y(qk)

y(k)− 1

)+

γ(z − 1)y4 − qz2

(qz(qzy3 − y1y4)

y4 − q2z2+ y4k +

εqz

k

).

(1.6)


Then, the functionsfn(k), n ∈ Z, defined by the three term recursion relation

fn(k) = (s(k) − bn)fn−1(k) − an−1fn−2(k) , n ∈ Z ,

with an, bn as in (1.4),f0(k) = 1 andf1(k) as in (1.6), are eigenfunctions of a secondorder q-difference operator

Bfn(k) := a(k)(fn(qk) − fn(k)) + b(k)(fn(k/q) − fn(k)) = θnfn(k) ,

with coefficientsa(k) andb(k) defined in terms ofa, b, c, d, z andf1(k) and eigenvaluesθn given by

θn = (1− q−n)(abcdqn−1 − z2) . (1.7)

Furthermore, different choices of the functionf1(k) (corresponding to different solutionsof (1.5)) result in second orderq-difference operators that are conjugate to each other.

Notice that the Eq. (1.5) is strikingly similar to the one satisfied by the Askey-Wilsonpolynomials, but with anarbitrary value ofz.

In the specialcase withz = 1, one can pick a solution of theq-difference equa-tion (1.5) to be a constant and then the construction givesf1(k) = γ(k + ε/k) − b1 andwe are in the case when all thefn(k), n ≥ 0, are polynomials in the spectral parameterγ(k + ε/k), i.e. the Askey-Wilson case according to [14]. For a more complete picture,see Sect. 8.

For generic values ofz in what one may want to call the “Gauss-Askey-Wilson”equation we are led to functionsfn(k) that are not polynomials, but generalize theAskey-Wilson polynomial situation as the title of this paper indicates. One can considerthe parameterz of the Gauss-Askey-Wilson equation as the extra degree of freedom thatmoves us away from the polynomial case.

The equation that we are calling the Gauss-Askey-Wilson equationconvergesasqapproaches 1, to the classical equation of Gauss (see Sect. 6). This happens for anychoice of the spectral parameters(k). For a discussion of the relation between (1.5) andthe equation that is usually referred to as theq-hypergeometric equation, [11, 17, 20],one can see [16].

A final remark is in order. The appearance of the Gauss-Askey-Wilson equation inthe case of a generalz may be confusing, but it reflects the central role played by thisequation. Its solutions do not givef1(k) directly, ratherf1(k) is obtained through theRodrigues type formula (1.6). Exactly the same situation arises in the case ofq = 1, see[15]. In the polynomial case the Askey-Wilson equation plays a second role, namely itappears in the bispectral operatorB in (1.2).

2. An Operator Identity and its Solution

Rewrite (1.1) and (1.2) as

Lf = s(k)f, f = (. . . , f−1(k), f0(k), f1(k), f2(k), . . .)t (2.1)

Bf = Θf, (2.2)

with


L =

· · ·a−2 b−1 1

a−1 b0 1a0 b1 1

a1 b2 1a2 b3 1

· · ·· · ·

(2.3)

s(k) as in (1.3) andΘ the diagonal matrixΘ = diag(. . . , θ−1, θ0, θ1, θ2, . . .).The next lemma was already derived in [14]. It provides theq-version of a lemma

used in [15] in the caseq = 1, following a basic observation in [9].

Lemma 1. Any solution of (2.1) and (2.2) satisfies the matrix identity

(L3Θ − ΘL3) + x(L2ΘL − LΘL2) + x(LΘ − ΘL) = 0 (2.4)

with

x = − 1 + q + q2

q, x = εγ2 (q − 1)2(q + 1)2

q2.

Proof. From (2.1) and (2.2) one obtains immediately that

(L3Θ − ΘL3)f + x(L2ΘL − LΘL2)f + x(LΘ − ΘL)f =

(B(s3f ) − s3Bf ) + x(sB(s2f ) − s2B(sf )) + x(B(sf ) − sBf ) =

a(k)(s(qk) − s(k))[s2(k) + s2(qk) + (1 +x)s(k)s(qk) + x]f (qk)+

b(k)(s(q−1k) − s(k))[s2(k) + s2(q−1k) + (1 +x)s(k)s(q−1k) + x]f (q−1k).

Notice that the choice ofs(k) given in (1.3) cancels the two terms in the square brackets.As k varies, thef (k) are linearly independent vectors, so that thefinite bandoperator(L3Θ−ΘL3)+x(L2ΘL−LΘL2)+ x(LΘ−ΘL) has infinite dimensional kernel; henceit must vanish identically, proving our lemma. �

We can now exploit the lemma above to derivenecessaryconditions thatL andΘshould satisfy if (2.1) and (2.2) are to hold. We restrict ourselves to the case whereall an’s are nonzero. The same requirement is needed to derive the classical result ofBochner.

In the sequel we shall denote by

[α] =qα − 1q − 1

theq-analogue ofα.In order to solve (2.4) forL andΘ, we proceed along the lines of [14]. Equating the

diagonals of (2.4) to zero, starting with the upper one, we obtain at the (n, n+ 3)th entrythe equations:

θn+2 − [3]q

θn+1 +[3]q

θn − θn−1 = 0, n ∈ Z , (2.5)


whose general solution is given by:

θn = q1−n[n]([n]u + v) + w, (2.6)

with u, v, w free parameters. Since we can shift theθn’s by an arbitrary constant, wemay always assume thatw = 0. For our purposes it is clear that only the ratiou/v (or

v/u) plays an important role in (2.4). Equating the (n, n + 2)th and (n, n + 1)th entriesto zero, we obtain

(θn+1 − θn+2)bn+2 + (θn+1 − θn−1)bn+1 + (θn−2 − θn−1)bn = 0 (2.7)

and(θn − θn+2)an+1 + (θn+1 + θn − θn−1 − θn−2)an

+(θn−3 − θn−1)an−1 + (θn − θn−1)(bn+1q − bn)(bn+1 − bnq)

q

+εγ2 (q − 1)2(q + 1)2

q2(θn − θn−1) = 0,

(2.8)

where theθn’s, n ∈ Z, are given by (2.6).Using (2.6) we see that the general solution of (2.7) depends on two free parameters

b1 andr = b2 − b1, explicitly:

bn = b1 +[n − 1]zn−1

z2n−3z2n−1(rqn−2([3]u + v) + b1(qn−2 − 1)((1− q)v + (1− qn)u)) (2.9)

withzn = v + [n]u.

Going now into Eq. (2.8) one sees, after some labor, that the general solution foran

depends on two free parameters,a0 anda1, and is given by

an =qn−1[n − 1][n]zn−1znan ãn

(q + 1)2z2n−2z22n−1z2n

+ a1qn−1[n](v + [2]u)zn−1

z2n−2z2n

− a0qn−1[n − 1](q2v − [2]u)zn

z2n−2z2n+ εγ2 (q − 1)2[n − 1][n]zn−1zn

z2n−2z2n

(2.10)

withan = −r(v + [3]u) + b1v(q − 1) + b1u(qn+1 + qn − q2 − 1)

ãn = −qn−1r(v + [3]u) + b1v(qn − qn−1 − q2 + 1)

+ b1u(1 + q − qn−1 − qn+1).

In summary,Θ is given by (2.6) andL is determined by (2.9) and (2.10). Theexpressions above make it clear that it is safer to assume thatv + [n]u is nonzero forall integersn, see however the remark at the end of the section for a discussion of thesespecial cases.

For further use, it will be crucial to observe that Eqs. (2.7) and (2.8) can both beintegratedonce.


Indeed we can write (2.7) as

[(θn+1 − θn+2)bn+2 + (θn − θn−1)bn+1]

−[(θn − θn+1)bn+1 + (θn−1 − θn−2)bn] = 0,

which is equivalent to

(θn − θn+1)bn+1 + (θn−1 − θn−2)bn = β (2.11)

with

β = −1q

[(r + b1(1 − q2))v + ((q + 1)2b1 + [3]r)u]. (2.12)

Then using (2.5), (2.6) and (2.11), (2.8) becomes

[(θn − θn+2)an+1 + (θn − θn−2)an + xn+1b2n+1 − βbn+1 + xθn]

−[(θn−1 − θn+1)an + (θn−1 − θn−3)an−1 + xnb2n − βbn + xθn−1] = 0,

with

x = εγ2 (q − 1)2(q + 1)2

q2,

xn = q1−n((q − 1)v − (1 + q2n−2)u),(2.13)

or equivalently

(θn−1 − θn+1)an + (θn−1 − θn−3)an−1 + xnb2n − βbn + xθn−1 = α, (2.14)

with

α =(rb1 +(1−q)b2

1 + [2](a0q2− a1))v + ((1+q2)b2

1 + [3]rb1 −(1 + q)2(a0 + a1))uq

.

(2.15)Observe that one can solve (2.12) and (2.15) forr anda0,

r =b1((q2 − 1)v − (q + 1)2u) − qβ

(q2 + q + 1)u + v,

a0 =(q + 1)((q + 1)u + v)a1 + q((2u − (q − 1)v)b2

1 + βb1 + α)(q + 1)(q2v − (q + 1)u)

,

(2.16)

and therefore the set of parameters (u, v, a1, b1, α, β) is equivalent to the set of parameters(u, v, a0, a1, b1, r). From now on up to the end of Sect. 5, it will be more convenient forus to work with the set of parameters (u, v, a1, b1, α, β) and all our formulas will bewritten in terms of these equivalent parameters.

Notation. Equations (2.11) and (2.14) will be the crucial ingredient in Sect. 4 in orderto establish the existence ofB in (2.2) under the condition (2.4). We shall abbreviatethem byV an = V bn = 0 with


V an = β − (θn−1 − θn−2)bn − (θn − θn+1)bn+1,

V bn = α − xθn−1 + βbn − xnb2n

− (θn−1 − θn−3)an−1 − (θn−1 − θn+1)an .

(2.17)

Whenq = 1, we showed in [15] that the vector field_an = 2anV an, _bn = V bn is acombination of the Toda lattice vector field and thesl(2) part of its Virasoro symmetries.Remark As long asv + [n]u 6= 0, for all n ∈ Z, one can re-express the solution (2.9)

and (2.10) to Eqs. (2.7) and (2.8) using as free parametersu, v, bk−1, rk = bk − bk−1,ak−2, ak−1 instead ofu, v, b1, r = b2 − b1, a0, a1 for an arbitrary choice ofk ∈ Z. Onechecks easily that in this way one obtains formulas foran andbn for which the limitsv + [2k − 1]u = 0 andv + [2k − 4]u = 0 make sense and provide the solution of (2.7)and (2.8) corresponding to these special choices.

3. Theq-Riccati Equation for f1(k)

In the last section we obtained expressions for the entries in the matricesL andΘ (interms of the free parametersu, v, a1, b1, α, β) which follow from the existence of anontrivial family of functionsfn(k) that satisfy both (2.1) and (2.2).

In this section we will obtain a further equation that the functionf1(k) has to satisfyif (2.1) and (2.2) are to hold.

Section 4 will be devoted to proving that the conditions derived in Sect. 2 coupledwith the requirement thatf1(k) should satisfy the equation derived in the present sectionare not only necessary but also sufficient for (2.1) and (2.2) to hold.

FromBf1 = θ1f1 andBf2 = θ2f2 one can solve for the coefficientsa(k) andb(k)in B exceptin the case when the determinant of the corresponding system of equationsis zero.

This determinant is seen to be zero exactly whenf1(k) satisfies the nonlinearq-difference equation of order two given by

f1(qk)f1(k/q)(s(k/q) − s(qk)) + f1(qk)f1(k)(s(qk) − s(k))+f1(k)f1(k/q)(s(k) − s(k/q)) = 0 .

(3.1)

If f1(k) is (locally) identically zero, the equationBf2 = θ2f2 reduces toa1θ2 = 0, andunder our assumption thatv + [n]u 6= 0 for all integers this meansa1 = 0. This casegives thenfi = 0 for all i ≥ 1, and we get (directly from (2.1)) forf−i a family ofpolynomials of degreei in s and we are back into familiar territory.

If we assume thatf1 is not (locally) identically zero by introducing the ratiog(k) =f1(k)/f1(k/q), Eq. (3.1) can be rewritten as

g(qk) = Q(k)/(g(k)R(k) + S(k)) (3.2)

with

Q(k) = s(k/q)−s(k), R(k) = s(qk)−s(k), S(k) = s(k/q)−s(kq).

Notice that this equation is of the same form as the equation forf1(k) given later in (3.4)with P (k) = 0. This equation will be of fundamental importance for our development,and Sect. 5 is devoted to solving it.


We now solve the simpler Eq. (3.2) by adapting the method to be described in Sect. 5.By putting g(k) = S(k)/R(k) (w(qk)/w(k) − 1), the nonlinear equation above

becomes the second order linearq-difference equation

w(q2k) − w(qk) − R(kq)Q(k)/S(kq)S(k)w(k) = 0, (3.3)

and the coefficient ofw(k) is q/(q + 1)2 whenε = 0 and is given by

(k2 − q)(k2q3 − 1)/((k2 − 1)(k2q2 − 1)(q + 1)2) in the caseε = 1.

Appropriate substitutions allow one to proceed. The general solution of (3.3) is a linearcombination of two independent solutions with coefficientsc(k) arbitrary q-periodicfunctions, i.e., satisfyingc(kq) = c(k). See [6].

In the caseε = 0 this gives for the first order equation forg the general solution

g(k) =(−q − 1)((kn2qn2 − kn2)t + kn1qn1 − kn1)

q(kn2t + kn1)

with t an arbitraryq-periodic function andn1, n2 given by

qn1 = q/(q + 1) andqn2 = 1/(q + 1),

and one concludes thatf1(k) is given by

f1(k) = a/(k + t)

with a (andt) an arbitraryq-periodic function.In the caseε = 1 the first orderq-difference equation forg has a general solution

given byg(k) = ((q2 + k2)t + kq)/(q(t(k2 + 1) +k))

with t an arbitraryq-periodic function. In this case, once again, one concludes thatf1(k)has the formc1/(s(k) − c2), wherec1, c2 are arbitraryq-periodic functions.

To complete this discussion we need to see what is the sequence of functionsfn(k)that one obtains whenf1(k) is given by the choice above. The equationsBfi = θifi,i = 1, 2, force the following quantities to vanish:

γ(q + 1)(v + (q + 1)u)(a1 − c1) and a1(v + [3]u)(b2 − c2) .

The first condition pins downc1. Since we already assumed thatf1(k) is not (locally)identically zero, the second condition pins downc2 and we get

f1(k) = a1/(s(k) − b2) .

By using (2.1) this means thatf2(k) = 0 andf3(k) = a2f1(k). FromBf3 = θ3f3 weconclude then, sinceθ1 6= θ3, thata2 = 0 and from (2.1) we get thatfi(k) = 0 for i ≥ 3.Furthermoref1, f0, f−1, . . . , f−i, . . . are given byf1(k) times a polynomial of degreei + 1. Conjugating byf1 we are back in the polynomial case.

Assume from now on that the 2× 2 system of equationsBfi = θifi, i = 1, 2, has anonzero determinant.

We proceed now to explore the conditions under which

Bfi(k) = θifi(k) for i = 1, 2, 3, . . . .


It will turn out that we can always conclude thatf1(k) has to satisfy a certain first ordernonlinearq-difference equation of the form

f1(kq) =P (k)f1(k) + Q(k)R(k)f1(k) + S(k)

(3.4)

with P, Q, R, S given by the expressions

P (k) = qα + q[β + b1(q − 1)(qv − u)]s(kq)

+q(q + 1)[u −(q−1)v]s(kq)2 + γkq(qv − u)(q2−1)(s(kq) − b1),

Q(k) = −a1(q + 1)((q + 1)u + v)(s(kq) − s(k)),

R(k) = q(q + 1)v(s(kq) − s(k)),

S(k) = qα + q[β + b1(q − 1)(qv − u)]s(k)

+q(q + 1)[u −(q−1)v]s(k)2 +εγ

k(qv − u)(q2−1)(s(k) − b1) .

(3.5)

We will devote Sect. 5 to a discussion of this equation including a method that allowsus to solve it explicitly in terms of “classical functions” for theP, Q, R, S that appearin our problem. We observe for later use that

S(ε/kq) = P (k) , Q(ε/kq) = −Q(k) and R(ε/kq) = −R(k) . (3.6)

We proceed now to explain the method that allows one to trapf1(k) into a relationas the one given in (3.4).

Recall that we can read off the coefficientsa(k) andb(k) in the operatorB in termsof f1(k) ands(k) as well as their forward and backwardq-shifted versions.

Insisting onBf3 = θ3f3 gives an expression forf1(k/q) in terms off1(k) andf1(kq),in the form

f1(k/q) = F (k, f1(k), f1(kq)) (3.7)

with F a rational function.Insisting onBf4 = θ4f4 and using the above expression to eliminatef1(k/q) we get

for f1(kq) two possible expressions in terms ofk andf1(k). The first one is the expression(3.4) mentioned above. The other possible expression forf1(kq) has the form in (3.4)exceptfor an extra multiplicative factorf1(k) in the right-hand side.

Indicate this last relation by

f1(kq) = G(k, f1(k)). (3.8)

If one replacesk by k/q this becomes

f1(k) = G(k/q, f1(k/q)). (3.9)

This expression can be combined with relation (3.7) to give

f1(k/q) = F (k, f1(k), G(k, f1(k))) (3.10)

and if this one is finally plugged into (3.9) we get the equation inf1(k) given by

f1(k) = G(k/q, F (k, f1(k), G(k, f1(k)))). (3.11)


It turns out that this equation has two possible solutions: one isf1(k) = 0 and the otherone is given by

f1(k) = a1/(s(k) − b2).

In both cases the determinant of the two-by-two system discussed above vanishes, con-tradicting our assumption. Thus we are forced to rule out the case (3.8) and to keep thecondition (3.4).

We call this Eq. (3.4) theq-Riccati equation, since it appears in our development atexactly the point where the usual Riccati equation appears in [15] (see case b) in Sect. 3).

Now something amazing happens: as long asf1(k) satisfies (3.4) all the remainingconditionsBfi = θifi, i = 5, 6, . . . are automatically met and (3.4) is not only necessarybut in fact in conjunction with the conditions derived in Sect. 2, it is sufficient to ensurethe existence of an operatorB so that (2.1) and (2.2) should hold. This will be seen inthe next section.

We close this section with the observation that Eq. (3.4) can be put in a form thatallows one to see the classical Riccati equation emerging in the limitq → 1.

One first rewrites (3.4) as

(Rf1 + S)f1(kq)−f1(k)s(kq)−s(k)

= − R

s(kq)−s(k)f2

1 +P − S

s(kq)−s(k)f1 +

Q

s(kq)−s(k). (3.12)

Using the notation

D+sf =

f (kq) − f (k)s(kq) − s(k)

(3.13)

and the definitions ofP, Q, R, S one can express (3.12) in the form

(Rf1 + S)D+sf1 = −q(q + 1)vf2

1

+ [q(u + v)(s(k) + s(kq)) + qβ + 2qb1(u − qv)]f1

− (q + 1)a1((q + 1)u + v) .

(3.14)

As q → 1, the equation becomes

(α + βs(k) + 2us2(k))df

ds= −2vf2

1 + [2(u + v)s(k) + β + 2b1(u − v)]f1

− 2a1(2u + v) .

Whens(k) = k, this equation agrees with the one obtained in [15].

4. The Operator B

Substitutingf2 + b2f1 + a1f0 for s(k)f1, the q-Riccati equation as written in (3.14)becomes

A+f1(k) = q(u + v)f2(k) + (q(u + v)b2 + qβ + 2qb1(u − qv))f1(k)

− a1((1 + q + q2)u + v)f0(k) + qs(kq)(u + v)f1(k)(4.1)

with


A+ = (R(k)f1(k) + S(k))D+s + q(q + 1)vf1(k) .

Rewriting (3.4) as

f1(k/q) =S(k/q)f1(k) − Q(k/q)−R(k/q)f1(k) + P (k/q)

,

we could of course have obtained a similar formula in terms of the operator

D−s f (k) =

f (k/q) − f (k)s(k/q) − s(k)

,

namely

A−f1(k) = q(u + v)f2(k) + (q(u + v)b2 + qβ + 2qb1(u − qv))f1(k)− a1((1 + q + q2)u + v)f0(k) + qs(k/q)(u + v)f1(k)

(4.2)

withA− = (−R(k/q)f1(k) + P (k/q))D−

s + q(q + 1)vf1(k) .

In this section we show that formulas (4.1) and (4.2) generalize into differentiationformulas for allfn(k). This will follow by induction from the integrated form (2.11) and(2.14) of the operator identity (2.4). The difference between these two differentiationformulas will give the bispectral operatorB in (2.2), therefore establishing that theoperator identity (2.4) is not only necessary but is also sufficient to guarantee a solutionof (2.1) and (2.2). LetT be the tridiagonal matrix defined by

Tnn−1 = −an[n]q1−n([n + 2]u + v),

Tnn = q(u − qv)b1 + q1−n((q − [n])v − ([2n + 1] + q2[n][n − 2])u)bn+1,

Tnn+1 = −[n − 2]q3−n([n]u + v).

(4.3)

With f = (. . . , f−1(k), f0(k) = 1, f1(k), . . .)t, we have the

Lemma 2. (Differentiation formulas)

A+f = Tf + qs(kq)Θf, (4.4)

A−f = Tf + qs(k/q)Θf (4.5)

with T as in (4.3) andΘ the diagonal matrix of the eigenvaluesθn’s as in (2.6).

Remark. Whens(k) = k, s(kq) = qs(k) and, using the three term recursion relation,formulas (4.4) and (4.5) can be rewritten as

A±f = Q±f , (4.6)

for some tridiagonal matricesQ±. Whenq → 1, these formulas reduce to a differ-entiation formula which we already used in [15] and which generalizes the standarddifferentiation formulas satisfied by the Hermite, Laguerre, Jacobi and Bessel polyno-mials. If we now express the compatibility between (4.6) andLf = kf , we obtain aq-version of a “string-like” equation


LQ+ − qQ+L = S(L) andLQ− − q−1Q−L = P (q−1L),

(4.7)

with S(k) andP (k) the polynomials of degree 2 ink defined in (3.5) withγ = 1 andε = 0. In [15] (see also [1] for a version in the context of orthogonal polynomials) wehave shown that whenq → 1 Eqs. (4.7) can be interpreted as saying that the solutions toour problem are fixed points of an arbitrary linear combination of the Toda lattice vectorfield and thesl(2) part of its Virasoro symmetries. A similar interpretation remains tobe worked out whenq 6= 1.

Before proving Lemma 2, we deduce

Theorem 2. Let

B =A+ − A−

q(s(kq) − s(k/q)). (4.8)

Then

Bf = Θf. (4.9)

More explicitly, the coefficientsa(k) andb(k) in (1.2) are given by

a(k) =R(k)f1(k) + S(k)

q(s(kq) − s(k))(s(kq) − s(k/q))(4.10)

and

b(k) =R(k/q)f1(k) − P (k/q)

q(s(k/q) − s(k))(s(kq) − s(k/q)), (4.11)

with f1(k) a solution of theq-Riccati equation (3.4). Moreover, a choice of a differentsolutionf1(k) in (3.4) leads toB conjugate toB as follows:

B = g(k)Bg(k)−1 , (4.12)

with g(k) a solution of the equation

1g(k)

D+sg(k) =

q(q + 1)v(f1 − f1)

R(k)f1 + S(k).

Proof. Formula (4.9) follows by taking the difference between (4.4) and (4.5). Theconjugation formula (4.12) follows from the explicit expressions (4.10) and (4.11) fora(k) andb(k) if one uses Eq. (3.4) satisfied byf1(k) andf1(k). �

Remark.Note that the form ofB is independent off1(k) exactly whenv = 0. Themeaning of the conjugation result above is that one can get the sameB for differentchoices off1 if one is willing to abandon the normalizationf0 = 1.


Proof of Lemma 2.We only establish (4.4) since (4.5) can be proved in a similar way.Rewrite (4.4) as

(R(k)f1(k) + S(k))fn(kq) = S(k)fn(k) + (s(kq) − s(k))(Tnn+1fn+1(k)+ Tnnfn(k) + Tnn−1fn−1(k) + qs(kq)θnfn(k)) .

(4.13)

The casen = 0 of this identity is trivially satisfied, using the definition ofR(k) in(3.5) and the casen = 1 has been established in (4.1). Assume that we have proved(4.13) for 0≤ j ≤ n, n ≥ 1, we establish it forj = n + 1.

Let us denote by (RHS)n the right-hand side of (4.13). Since

s(kq) = qs(k) +εγ

kq(1 − q2) ,

by using the three term recursion relation defining thefn’s and the definitions ofT andΘ, (RHS)n can be rewritten as

(RHS)n = qn+1(q + 1)ufn+2(k)

+ [qn+1(q+1)ubn+2 + q1−n((q2n + 1)u−(q−1)v)bn+1 + qβ]fn+1(k)

+ [q1−n((q2n + 1)u − (q − 1)v)b2n+1 + qβbn+1 + qα

+ qn+1(q + 1)uan+1 + q1−n(q + 1)(u − (q − 1)v)an]fn(k)

+ [q1−n((q2n + 1)u − (q − 1)v)bn+1

+ q1−n(q + 1)(u − (q − 1)v)bn + qβ]anfn−1(k)

+ q1−n(q + 1)(u − (q − 1)v)an−1anfn−2(k)

− εγ(q+1)kqn

[Pnn+1fn+1(k) +

(bn+1 − εγ

k

)Pnnfn(k) + anPnn−1fn−1(k)

]

with

Pnn+1 = (2q2n+1 + q2n − 2qn+1 − 2qn + q)u + (q − 1)(qn+1 + qn − q)v,

Pnn = (q + 1)(qn − 1)((qn − 1)u + (q − 1)v),

Pnn−1 = (q2n+1 − 2qn+1 − 2qn + 2q + 1)u + (q−1)(qn+1 + qn− 2q−1)v .

Sincefn+1 = (s − bn+1)fn − anfn−1, using the induction hypothesis that (4.13) holdsfor n − 1 andn, and remembering the definitions ofV an andV bn in (2.17), we obtain


(R(k)f1(k) + S(k))fn+1(kq) − (RHS)n+1 =

(s(kq)−bn+1)(RHS)n − an(RHS)n−1 − (RHS)n+1 = q(q−1)(V an+1)fn+2(k)

+[q(q − 1)(bn+2V an+1 + V bn+1) − εγk (q − 1)(q + 1)V an+1]fn+1(k)

+[q(q − 1)(an+1V an+1 + anV an + bn+1V bn+1)

− εγk (q − 1)(q + 1)V bn+1]fn(k)

+[q(q − 1)(bnV an + V bn+1) − εγk (q − 1)(q + 1)V an]anfn−1(k)

+q(q − 1)(V an)an−1anfn−2(k)

− εγkqn (q − 1)(q + 1)2(qn − 1)((q − 1)v + (qn − 1)u)

×[an(fn + bnfn−1 + an−1fn−2 − sfn−1)

+(bn+1 − εγ

kq (q + 1))

(fn+1 + bn+1fn + anfn−1 − sfn)

+fn+2 + bn+2fn+1 + an+1fn − sfn+1] .

SinceV an = V bn = 0 for all n, (see (2.11), (2.14) and (2.17)), using the three termrecursion relation satisfied by thefn(k)’s, the last expression is obviously identicallyzero, which establishes (4.13) forn + 1.

By a similar argument one shows that, assuming that (4.13) is true forn−1 ≤ j ≤ 1,n ≤ 1, it is also true forj = n − 2, which completes the proof. �

5. Linearizing the q-Riccati Equation

Whenv = 0, the Eq. (3.4) satisfied byf1 becomes a linear first order non homogeneousq-difference equation. We will consider this case in Sect. 9. Our objective in this sectionis to linearize (3.4) whenv 6= 0, by aq-analogue of the “log derivative trick” which isused to solve a standard Riccati equation. In this way we will reduce the solution ofthis equation to solving a second order linearq-difference equation which is strikinglysimilar to the celebrated second orderq-difference equation satisfied by the Askey–Wilson polynomials.

The following trick converts theq-Riccati equation (3.4) into a second order linearequation. Put

f1(k) =S(k)R(k)

(w(kq) − w(k)

w(k)

), (5.1)

thenw(k) satisfies the second order linearq-difference equation

R(k)S(k)S(kq)w(kq2) − (R(kq)P (k) + R(k)S(kq))S(k)w(kq)+ (P (k)S(k) − Q(k)R(k))R(kq)w(k) = 0.

(5.2)

From the definition ofP (k), R(k) andS(k), see (3.5), it follows that


R(kq) = R(k)s(kq2) − s(kq)s(kq) − s(k)

and(s(kq2) − s(kq))P (k) + (s(kq) − s(k))S(kq) = (s(kq2) − s(k))U (kq)

withU (k) = qα + qβs(k) + q(2u − (q − 1)v)s(k)2. (5.3)

Substituting the above identities into (5.2) one finds

(s(kq) − s(k))S(k)S(kq)w(kq2) + (s(k) − s(kq2))U (kq)S(k)w(kq)+(s(kq2) − s(kq))(P (k)S(k) − Q(k)R(k))w(k) = 0.

(5.4)

Whenq → 1, we have seen in Sect. 3 that theq-Riccati equation (3.4) becomes

S(s)df1

ds+ 2vf2

1 − (2(u + v)s + β + 2b1(u − v))f1 + 2a1(2u + v) = 0,

and (5.1) reduces to

f1 =S(s)2v

d

dslogw

with S(s) = α + βs + 2us2, leading to the second order linear equation

S(s)2 d2w

ds2+ 2(u − v)(s − b1)S(s)

dw

ds+ 4a1v(2u + v)w = 0. (5.5)

Notice that in this case, since our equation does not depend explicitly onk, there isno difference between the casesε = 0 or ε 6= 0. Whenq 6= 1, the explicit dependenceon k cannot be eliminated and the situation becomes richer. We refer the reader to ourprevious paper [15] for a detailed study of the caseq = 1. As long as the rootsp1 andp2of S(s) are distinct, (5.5) is a Fuchsian differential equation with three regular singularpoints atp1, p2 and infinity and can therefore be reduced to the standard form of theGauss hypergeometric equation by puttingw = (s − p1)r1(s − p2)r2y, with r1 (resp.r2)a root of the indicial equation atp1 (resp.p2).

Our strategy to solve (5.4) will be to mimic this approach using that theq-analogueof (1 − k)−r is provided by theq-hypergeometric series

1φ0(r; −; q, k) =∞∑n=0

(r; q)n(q; q)n

kn,

where (r; q)n denotes theq-shifted factorial

(r; q)0 = 1, (r; q)n = (1− r)(1 − rq) . . . (1 − rqn−1), n = 1, 2, . . . .

Denoting byhr(k) the above series, we recall that

hr(qk) =1 − k

1 − rkhr(k) (5.6)

from which, with |q| < 1, one derives immediately Cauchy’s formulation of theq-binomial theorem

hr(k) =(rk; q)n(k; q)n

hr(qnk) =(rk; q)∞(k; q)∞

, (5.7)


where (r; q)∞ denotes the infinite product

(r; q)∞ =∞∏n=0

(1 − rqn).

Observe from (3.6) that the Laurent polynomialP (k)S(k) − Q(k)R(k) is invariantunder the changek 7→ ε/kq and therefore it can be factorized as

P (k)S(k) − Q(k)R(k) = V (k)V (ε/kq), (5.8)

with

V (k) ≡ x4k2 − x3k + x2 − εx1

k+

ε2x0

k2.

One can easily see thatx0 = 0 only if u = 0 or (q − 1)v = u. This case can be treatedseparately with the same technique that is used below forx0 6= 0.

Denote bypi (1 ≤ i ≤ 4) the four roots ofk2V (k) and put

w(k) = kρ4∏

i=1

1φ0(ri; −; q,k

pi)y(k) . (5.9)

From the assumptionx0 6= 0 it follows thatpi 6= 0. Then, using (5.6), one sees easilythat by pickingri = pi/ki, with ki (1 ≤ i ≤ 4) the roots ofk2S(k), and choosingρ suchthatqρ = limk→0 V (k)/S(k), after changingk to k/q, Eq. (5.4) simplifies to

V (k)(s(k/q)−s(kq))(s(kq)−s(k)) y(kq) + U (k)

(s(k)−s(k/q))(s(kq)−s(k)) y(k)

+ V (ε/k)(s(k)−s(k/q))(s(k/q)−s(kq)) y(k/q) = 0.

(5.10)

Notice that (5.10) can be written as

W (k)y(kq) + T (k)y(k) + W (ε/k)y(k/q) = 0 , (5.11)

with W (k) the coefficient in front ofy(kq) andT (k) the coefficient in front ofy(k) in(5.10). We now establish the following crucial

Lemma 3. There are eight ways to perform the factorization (5.8) so that

W (k) + T (k) + W (ε/k) = x5, (5.12)

for some constantx5 (independent onk).

As a consequence (5.11) can be written as

W (k)(y(kq) − y(k)) + W (ε/k)(y(k/q) − y(k)) + x5y(k) = 0. (5.13)

The expert reader will immediately notice the striking resemblance between this equationand the second orderq-difference equation which is satisfied by the Askey-Wilsonpolynomials, see for example [19]. Indeed, in Sect. 7, we will show that solutions of thisequation can be given in terms of some basic hypergeometric series, which in generalare not polynomials.


Proof of Lemma 3. Cleaning up the denominators in (5.12), one obtains a Laurent polyno-mial running fromk−3 tok3. SinceU (k) in (5.3) is invariant under the changek y ε/k,one sees easily that this Laurent polynomial changes sign under the substitutionk y ε/k,and therefore (5.12) amounts to three independent equations, which can be solved forx4, x3, x2:

x4 = −γ2(q − 1)2(q + 1)q

x5 − qx0 − γ2q(q + 1)((q − 1)v − 2u),

x3 = −q(x1 + γ(q + 1)β),

x2 =εγ2(q−1)2(q+1)

q2x5 − ε(q−1)x0−εγ2q(q + 1)((q − 1)v − 2u) + qα.

(5.14)

Since (5.8) is a Laurent polynomial running fromk−4 to k4, which is invariantunder the substitutionk y ε/kq, this relation amounts to a system of five independentequations in the five unknownsx0, x1, x2, x3, x4. Substituting (5.14) into these equations,one discovers that two of these equations, corresponding to the coefficients ofk−3 andk−1, become proportional, and thus we only have four independent equations. One cansolve the equation given by the coefficient ofk−4 for x5:

x5 = −q2(x0 − γ2(q + 1)u)(x0 + γ2(q + 1)((q − 1)v − u))γ2(q − 1)2(q + 1)x0

. (5.15)

Substituting (5.15) into the three remaining independent equations (corresponding tok−2, k−1 andk0), one obtains that the equations given by the coefficientsk−2 andk0

become proportional, and thus we have two independent equations forx0 andx1. Theequation given by the coefficient ofk−1 can be solved forx1:

x1 = −γ(q + 1)x0[qβ(x0 + γ2((q−1)v −2u)) − γ2b1(q−1)2(u + v)(qv−u)]qx2

0 + γ4u(q + 1)2((q−1)v − u), (5.16)

and, by substituting (5.16) into the coefficient ofk0, one gets thatx0 must be a root ofthe following degree 8 polynomial:

a1γ2(q − 1)2(q + 1)2v(v + (q + 1)u)x2

0p3(x0)2

+αqx0p1(−x0)p2(qx0)p3(x0)2

−βb1γ2(q − 1)2q(qv − u)x2

0p1(−x0)p2(qx0)p4(x0)

+β2γ2q2x20p1(−x0)p1(−qx0)p2(x0)p2(qx0)

−b21γ

2(q − 1)2q(qv − u)2x20p1(x0)p1(−x0)p2(qx0)p2(−qx0)

+εp1(−x0)p2(x0)p1(−qx0)p2(qx0)p3(x0)2 = 0,

(5.17)

with


p1(x0) = x0 + γ2(q + 1)u,

p2(x0) = x0 + γ2(q + 1)((q − 1)v − u),

p3(x0) = qx20 + γ4(q + 1)2u((q − 1)v − u),

p4(x0) = qx20 − 2γ2q(q + 1)vx0 + γ4(q + 1)2u(u − (q − 1)v).

This completes the proof of the lemma. �The results of this section allow us to express the solution of theq-Riccati equa-

tion (3.4) in terms of an arbitrary solution of the linear second orderq-difference equa-tion (5.13), as explained below. From (5.9), using property (5.6), we get that

w(kq)w(k)

= qρ4∏

i=1

(1 − k/pi

1 − k/ki

)y(kq)y(k)

,

or equivalently, using the definition ofρ, pi andki,

w(kq)w(k)

=V (k)S(k)

y(kq)y(k)

.

If we substitute this last expression into (5.1) we obtain thatf1(k) is given by theRodrigues type formula

f1(k) =1

R(k)

(V (k)

y(kq)y(k)

− S(k)

), (5.18)

with y(k) an arbitrary solution of (5.13). Of course bothV (k) defined by (5.8) with theextra requirement that (5.12) should hold and Eq. (5.13) definingy(k) depend on oneof the eight choices forx0 in (5.17). However any choice forx0 will lead to the samefamily of solutions for theq-Riccati equation (3.4).

6. The Gauss-Askey-Wilson Equation & Proof of Theorem 1

In view of Eq. (5.13) to which we have reduced the solution of theq-Riccati equationsatisfied byf1(k), it is natural to try to describe the solution of our problem in terms ofthe parametersxi, 0 ≤ i ≤ 5, definingV (k) in (5.8) in such a way that (5.12) holds.This is easy to do with 0≤ i ≤ 4 butx5 requires a different treatment.

Putyi =

xi

x0, 1 ≤ i ≤ 4 ,

and define

z = − γ2(q + 1)((q − 1)v − u)x0

.

Observe from (5.15) thatz = 1 implies thatx5 = 0. A better motivation for this choicewill be given in Sect. 8, where we shall see thatz = 1 precisely pins down the case wherethe functionsfn(k), n = 0, 1, 2, . . . , are polynomials in the spectral parameters(k).

Using (5.14), (5.15), (5.16) and (5.17) one sees easily that the set ofparameters (x0, y1, y2, y3, y4, z) is equivalent to the set of parameters(u, v, a1, b1, α, β) and thus it is also equivalent to (u, v, a0, a1, b1, r). Explicitly:


u =x0y4

γ2q(q + 1)z, v =

x0(y4 − qz2)γ2q(q2 − 1)z

,

b1 =γz[q(y3 + qy1)z2 − (q + 1)(y1y4 + qy3)z + (y3 + qy1)y4]

(z2 − y4)(q2z2 − y4),

r =γ(q − 1)z

[(y3 + qy1)(qz4 + 2(q2 + q + 1)y4z

2 + qy24)

−(q + 1)2(y1y4 + qy3)z(z2 + y4)

](z2 − y4)(z2 − q2y4)(q2z2 − y4)

,

a0 =γ2(z − 1)(q2z − y4)

[(q2z2 − y4)2(ε(q2z2 + y4) − qy2z)+q2z2(qy1z − y3)(qy3z − y1y4)

](qz2 − y4)(q2z2 − y4)2(q3z2 − y4)

,

a1 =γ2(z − q)(qz − y4)

[(z2 − y4)2(ε(z2 + y4) − y2z)+z2(y1z − y3)(y3z − y1y4)

](z2 − qy4)(z2 − y4)2(qz2 − y4)

.

(6.1)

Introduce nowa, b, c, d by means of the relation

k2V (k)x0

= y4k4−y3k

3+ y2k2 − εy1k + ε2

= (1−ka)(1−kb)(ε −kc)(ε −kd) ,(6.2)

and substitute (6.1) into (2.9) and (2.10) to obtain:

an = An−1Cn,bn = γ(εa + 1

a ) − (An−1 + Cn−1)(6.3)

with

An =γ(q−nz − εab)(q−nz − ac)(q−nz − ad)(q1−nz − abcd)

a(q1−2nz2 − abcd)(q−2nz2 − abcd),

Cn =γa(q−nz − 1)(q1−nz − bc)(q1−nz − bd)(εq1−nz − cd)

(q2−2nz2 − abcd)(q1−2nz2 − abcd).

In terms of these new parameters, one finds from (5.15) that

x5

x0=

q(1 − z)(abcd − qz)γ2(q − 1)2(q + 1)z

, (6.4)

and Eq. (5.13) takes on the form

A(k)(y(qk) − y(k)) + A(ε/k)(y(k/q) − y(k)) = (1− z)

(abcd

qz− 1

)y(k) (6.5)

with

A(k) =(1 − ka)(1 − kb)(ε − kc)(ε − kd)

(ε − k2)(ε − k2q), (6.6)

and its consequence


A(ε/k) =(k − εa)(k − εb)(k − c)(k − d)

(k2 − ε)(k2 − εq).

Replacingk by kq, we can rewrite Eq. (6.5) in terms ofD+s andD+2

s , with D+s as in

(3.13). Put then

(1 − z)

(abcd

qz− 1

)= (q − 1)2t , y4 = (q − 1)y′

4 + 1,

y3 = (q − 1)y′3 + y1 ,

with yi as in (6.2). The limitq = 1 of (6.5) becomes

(s(k)2−γy1s(k) + γ2(y2−2ε))d2

ds2y(k) + (y′

4s(k) − γy′3)

d

dsy(k)−ty(k) = 0 . (6.7)

Notice that in the limity(k) can be considered as a function ofs(k). By translation andscaling ins, (6.7) can be brought to the standard form of the Gauss hypergeometricequation. Since for the special choicez = q−n, Eq. (6.5) is nothing but the celebratedequation satisfied by the Askey-Wilson polynomials, we propose to call this equationthe Gauss-Askey-Wilson equation.

Since (6.5) has the form

c2(k)y(q2k) + c1(k)y(qk) + c0(k)y(k) = 0

for appropriate polynomialsci, i = 0, 1, 2, it is known, see [6], that the existence ofsolutions of the form

y(k) = kr∑i≥0

xiki (6.8)

is controlled by the “indicial equation”

c2(0)q2r + c1(0)qr + c1(0) = 0 .

In our case this equation becomes, for any nonzeroε (which we taketo be 1),

(qr − z)

(qr+1 − abcd

z

)= 0 .

For generic values ofabcd andz the two values ofr given by this equation will not differby an integer and the general theory guarantees the existence of two convergent powerseries of the form (6.8).

For |q| > 1 one can see that the two series converge for|k| < min(|q|, |q|max(a,b,c,d) ).

If 0 < |q| < 1 one obtains two convergent power series for|k| < min(|q|−1, |q|−1 min(a, b, c, d)).

Forε = 0 we get as “indicial equation”, the expression

(qr − 1)(qr − q) = 0 .

This gives rise to a pair of solutions of the form

y(k) = 1 +∑i≥1

xiki , y(k) = k

∑i≥0

xiki ,


which converge when|q| > 1 as long as|k| < |q|max(a,b) and when 0< |q| < 1 as long

as|k| < |q|−1 min(c, d).These “power series” solutions do not make explicit, even whenz = q−n, the role

of the basic hypergeometric functions. In the next section we will see that for anyz, the solutions of the Gauss-Askey-Wilson equation can be written in terms of basichypergeometric series in a way that extends the representation of the well known Askey-Wilson polynomials.

We can now summarize the results of this section and the previous ones and give the

Proof of Theorem 1. Formula (1.4) giving the explicit form of the coefficients enteringthe three term recursion relation satisfied by the functions{fn(k)}n∈Z has just beenestablished in (6.3). We now explain how to obtain formula (1.6) which determinesthe functionf1(k) in terms of an arbitrary solution of the Gauss-Askey-Wilson equa-tion (1.5), by combining the results of this section with those of Sects. 3 and 5.

The linearization of theq-Riccati equation (3.4) satisfied byf1(k) led us to theRodrigues type formula (5.18), withy(k) an arbitrary solution of Eq. (5.13) which, whenexpressed in terms of the new parametersa, b, c, d andz introduced at the beginning ofthis section, becomes the Gauss-Askey-Wilson equation (6.5). Substituting intoV (k) asdefined in (5.8) the expressions (5.14) and (5.16) forx1, x2, x3 andx4, with x5 given by(5.15) andx0 replaced by

x0 = − γ2(q + 1)((q − 1)v − u)z

,

one obtains by a straightforward computation using the definitions ofS(k) andR(k) in(3.5) that

V (k)−S(k)R(k)

= s − b1 + (z−1)

(δ +

γu

(q−1)vk − εγ((q−1)v−u)

(q−1)vz

1k

), (6.9)

with

δ =(u + v)(uz − q2v + qv + qu)

(q + 1)v(uz2 + q2v − qv − qu)b1 +

q(uz + qv − v − u)(q2 − 1)v(uz2 + q2v − qv − qu)

β .

Replace nowS(k)/R(k) in (5.18) by the expression obtained from (6.9), expressV (k)in terms ofA(k) from (6.2) and (6.6) and use the definition (3.5) ofR(k) to get

f1(k) = s − b1 +x0

γ(q2 − 1)vA(k)

(k − ε

k

) (y(qk)y(k)

− 1

)+ (z − 1)

(δ +

γu

(q − 1)vk − εγ((q − 1)v − u)

(q − 1)vz

1k

).

From formulas (2.12) and (6.1), we can expressδ, x0/v, u/v in terms ofyi (1 ≤ i ≤ 4)andz, which leads to formula (1.6).

In Sect. 4, Theorem 2, we have established the existence of the bispectral operatorB and the fact that different choices of the functionf1(k) result in operators that areconjugate to each other. Finally, substituting the expressions foru andv in (6.1) into(2.6) gives (up to an inessential scaling factor) formula (1.7) for the eigenvaluesθn. Thiscompletes the proof of Theorem 1. �


7. Solving the Gauss-Askey-Wilson Equation in Terms of Basic HypergeometricSeries

The purpose of this section is to obtain solutions of the equation

A(k)(y(qk)−y(k)) + A(ε/k)(y(k/q)−y(k)) = (1− z)(abcd

qz− 1)y(k) (7.1)

with

A(k) =(1 − ka)(1 − kb)(ε − kc)(ε − kd)

(ε − k2)(ε − k2q),

and its consequence

A(ε/k) =(k − εa)(k − εb)(k − c)(k − d)

(k2 − ε)(k2 − εq)

in terms of basic hypergeometric series.Introduce the standardq-shifted factorials

(a; q)n =

{1 for n = 0

(a; q)n−1(1 − aqn−1) for n ≥ 1

and define

(a; q)∞ =∞∏k=0

(1 − aqk) ,

for |q| < 1. Whenever (a; q)∞ appears in a formula, we shall assume that|q| < 1. Whenproducts ofq-shifted factorials occur, we shall use more compact notations

(a1, a2, . . . , am; q)n = (a1; q)n(a2; q)n . . . (am; q)n ,(a1, a2, . . . , am; q)∞ = (a1; q)∞(a2; q)∞ . . . (am; q)∞ .

The next theorem can be extracted from [5]. For the convenience of the reader we shallpresent our “own proof” of this result, most of which was developed before we becameaware of [5]. We shall follow the notations of [11] for the definition and the notation ofthe basic hypergeometric series.

Theorem 3. Let r = 1, q/εab, q/ac or q/ad. Then the function

y(k) =(ak, aε/k; q)∞

(ark, arε/k; q)∞4φ3

z, abcdqz , ak, aε

k

; q, qabε, ac, ad

(7.2)

solves the inhomogeneous equation

A(k)(y(kq) − y(k)) + A(ε/k)(y(k/q) − y(k)) − (1 − z)(

abcdqz − 1

)y(k)

=1r

(rz, r abcdqz , ak, aε

k ; q)∞(rabε, rac, rad, rq; q)∞

,(7.3)

wherea = ra , z = rz , (7.4)

and, depending on the choice ofr above, one must pick

b = b, c = c, d = d, or b = q/εa, c = c, d = d, orb = b, c = q/a, d = d, or b = b, c = c, d = q/a .

(7.5)


Remark.Clearly if we pickr = q/εab above, we must assume thatε 6= 0. The otherchoices ofr are valid for anyε.

Proof. First we observe that it suffices to establish the caser = 1. Indeed, assuming theresult forr = 1, denoting byL(a, b, c, d, z)y(k) the left-hand side of (7.3) and putting

y(k) =(ak, aε/k; q)∞

(ark, arε/k; q)∞y(k) ,

one finds from Cauchy’s formulation of the binomial theorem (5.6) and (5.7) that

L(a, b, c, d, z)y(k) =1r

(ak, aε/k; q)∞(ark, arε/k; q)∞

L(a, b, c, d, z)y(k)

=1r

(z, abcd

qz , ak, aεk ; q

)∞

(abε, ac, ad, q; q)∞,

which coincides with (7.3), using the definitions (7.4) and (7.5) of ã, b, c, d andz cor-responding to the possible choices ofr distinct from 1. It remains to establish the caser = 1 of (7.3).

Define

zn ≡ (ak; q)n(εa

k; q)n.

We first collect some useful properties, which are easily proved by induction:

1. (1− a)(aq; q)n−1 = (a; q)n,2. zn(kq) − zn(k) = 1−qn

qk (ak2q − εa)(akq; q)n−1( εak ; q)n−1,

3. zn(k) − zn(k/q) = 1−qn

qk (ak2 − aεq)(ak; q)n−1( εaqk ; q)n−1,

4. (1− bk)(ε − ck)(ε − dk)(ak; q)n(aεk ; q)n−1 − (k − εb)(k − c)(k − d)k(ak; q)n−1

(aεk ; q)n = k(k2 − ε)(ak; q)n−1(aε/k; q)n−1[γn(k + ε

k ) + δn],

with γn = (abcdqn−1 − 1) and

δn = aqn−1(ε(1 − bc − bd) − cd) + bε + c + d − bcd.

We now look for a solution of (7.3) in the form

y(k) =∑n≥0

xnzn.

It will turn out that the unknown coefficientsxn satisfy a first order recursion relationwhich makesy(k) into a basic hypergeometric series.

Put

yp(k) =p∑

n=0

xnzn .

Using 2) and then 1) obtain


A(k)(yp(kq)−yp(k))

= A(k)(ak2q−εa)

qk

p∑n=0

xn(1−qn)(akq; q)n−1(aε

k; q)n−1

=a(1−kb)(ε−kc)(ε−kd)

(k2 − ε)qk

p∑n=0

xn(1−qn)(ak; q)n(aε

k; q)n−1.

Similarly, using 3) and then 1) obtain

A(ε/k)(yp(k/q)−yp(k))

= −A(ε/k)(ak2−aεq)

qk

p∑n=0

xn(1−qn)(ak; q)n−1(εaq

k; q)n−1

= −a(k−εb)(k−c)(k−d)k

(k2 − ε)qk

p∑n=0

xn(1−qn)(ak; q)n−1(εa

k; q)n.

Adding these two expressions and using 4) we get

A(k)(yp(qk) − yp(k)) + A(ε/k)(yp(k/q) − yp(k))

=a

(k2−ε)qk

p∑n=0

xn(1−qn)k(k2−ε)(ak; q)n−1(aε

k; q)n−1(γn(k +

ε

k) + δn)

=a

q

p∑n=0

xn(1 − qn)zn−1(γn(k +ε

k) + δn).

If we make use of the property

5. (k + ε/k)zn = 1aqn ((1 +a2εq2n)zn − zn+1),

we obtain that the left-hand side of (7.3) withy(k) replaced byyp(k) is equal to

p−1∑n=0

[(1−zqn)

(1− abcd

qzqn

)qxn

− (1−abεqn)(1−acqn)(1−adqn)(1−qqn)xn+1

]zn

qn+1

− xp(1 − qp)γpzp

qp− (1 − z)

(abcd

qz− 1

)xpzp .

(7.6)

If we determinexn inductively by

xn+1 =(1 − zqn)(1 − abcd

qz qn)q

(1 − acqn)(1 − adqn)(1 − abεqn)(1 − qqn)xn ,

with x0 = 1, the firstp terms in the expression above vanish and we find


xp =

(z, abcd

qz ; q)pqp

(abε, ac, ad, q; q)p.

Taking the limitp → ∞, by definition of the basic hypergeometric series,yp(k) tendsto the expression given in (7.2) withr = 1 and, since limp→∞ xpzp = 0, the expression(7.6) reduces to (

z, abcdqz , ak, aε

k ; q)∞

(abε, ac, ad, q; q)∞,

as desired, which concludes the proof of the theorem.�

It is clear that by taking appropriate linear combinations of the functions given in(7.2), for different choices ofr, we shall obtain a basis of solution of the Gauss–Askey–Wilson equation (7.1) expressed in terms of a basic hypergeometric series. An interestingconsequence of these explicit formulas is

Theorem 4. If in Theorem 1 one buildsf1(k) from (1.6) using any linear combination ofthe functions (7.2) giving rise to a solution of the Gauss–Askey–Wilson equation (1.5),f1(k) and the resulting familyfn(k), n ∈ Z, are functions of the variables(k) =γ(k + ε/k).

Proof. Clearly it suffices to consider the caseε = 1. Observe that any function that ismeromorphic atk = 0 and is invariant under the changek y 1/k is a function ofk+1/k.It is clear that for any solution of (1.5) which is obtained by taking a linear combinationof the functions given in (7.2), the expression forf1(k) given in (1.6) is meromorphic atk = 0, and thus it is enough to show that it is invariant under the changek y 1/k. Since,whenε = 1, a solutiony(k) of (1.5) which is a linear combination of the functions (7.2)is invariant under the changek y 1/k, it satisfies also the equation

A(k)(y(qk)

y(k)− 1

)+ A(k−1)

(y(qk−1)y(k−1)

− 1)

= (1− z)( y4

qz− 1

).

From this identity it follows that checking the invariance of the right-hand side of (1.6)under the changek y 1/k amounts to checking the identity

γ(k − 1k

)((z − 1)(y4 − qz) − qz(1 − z)

(1 − y4

qz

))= 0 ,

which is trivially satisfied. Thusf1(k) and therefore all the otherfn(k)’s are functionsof s(k), proving the theorem. �

8. The Case of the Classical Orthogonal Polynomials

In Sect. 6 we have shown that the solution of our bispectral problem can be nicelyparametrized by five parametersa, b, c, d and z which are equivalent to our originalparametrization of the solution in (2.9) and (2.10) in terms ofu/v (or v/u), a0, a1, b1andr (see formulas (6.1), (6.2) and (6.3)). In this section we show that whenz = 1, we getback the classical orthogonal polynomials in the sense of Andrews and Askey [3], so thatz is the extra parameter which moves us away from the polynomial case. In fact we geta little bit more: the classical orthogonal polynomials are part of a (nontrivial) biinfinite


sequence of functions{fn(s(k))}n∈Z which are eigenfunctions of the celebrated Askey–Wilson second orderq-difference operator; among those only thefn with n ≥ 0 arepolynomials ins(k).

To see this picky(k) to be a solution of the Gauss–Askey–Wilson equation (7.1)given by

y(k) = 4φ3

z, abcdqz , ak, aε

k

; q, q

abε, ac, ad

− r(z, abcd

qz , rabε, rac, rad, rq, ak, aε/k; q)∞(rz, rabcd

qz , abε, ac, ad, q, ark, arε/k; q)∞

× 4φ3

z, abcdqz , ak, aε

k

; q, qabε, ac, ad

,

(8.1)

corresponding to the appropriate combination of the functions in (7.2) withr = 1 andany other admissible choice ofr.

Clearly we can writey(kq)y(k)

− 1 = (z − 1)g(k) , (8.2)

and from (1.6) we obtain that

limz→1

f1(k) = s(k) − b1 ,

and therefore all thefn(k), n ≥ 2, become (monic) polynomials of degreen in thevariables(k). Also, sincea0 is divisible byz − 1 (see (6.1)), formulas (1.6) and (8.2)show that the limit

limz→1

f−1(k) = limz→1

s(k) − b1 − f1(k)a0

exists. Since limz→1 g(k) is not a polynomial,f−1(k) and consequently all thefn(k),n ≤ −2, are not polynomials. Substitutings(k)− b1 for f1(k) into (4.10) and (4.11) andusing (6.1) one computes that whenz = 1 the bispectral equation (4.9) becomes:

A(k)(fn(kq)−fn(k)) + A(ε/k)(fn(k/q)−fn(k))

= (1−q−n)(abcdqn−1−1)fn(k),(8.3)

with A(k) as in (6.6), which is nothing but the celebrated Askey–Wilson equation.To summarize, the special solutions of theq-Riccati equation corresponding to the

solutions (8.1) of the Gauss–Askey–Wilson equation whenz → 1 lead to a biinfinitesequence{fn(k)}n∈Z, withfn(k) polynomials in the variables(k) for n ≥ 0. In this casethefn(k) themselves are eigenfunctions of the Askey–Wilson second orderq-differenceoperator. Whenγ = 1

2 andε = 1, one sees from (6.3) that the three term recursionrelation satisfied by the polynomialsfn(k) (n ≥ 0) coincides with the standard relationsatisfied by the (monic) Askey-Wilson polynomials, see for example [19]. Whenγ = 1andε = 0, the change of parameters


a = q−1ac , b = q−1bd , c = c , d = −d ,

brings (6.3) to one of the standard forms of the three term recursion relation satisfied bythe (monic) bigq-Jacobi polynomials as given in [20] or [19, p. 58].

9. The Casev = 0 in the q-Riccati Equation

As we remarked at the beginning of Sect. 5, theq-Riccati equation (3.4) becomes linearwhenv = 0. We devote this section to the study of some explicit solutions that can beobtained in this case. In the simpler case ofq = 1 this is all reduced to a question ofintegrating a first order linear inhomogeneous differential equation, a rather “trivial” taskfor the cases at hand, see Sect. 5.1 in [15]. For a generalq this last step is not necessarilytrivial given our present knowledge about explicit evaluation ofq-integrals.

Equation (3.4) takes the form

f1(kq) = A(k)f1(k) + B(k) with A(k) = P (k)/S(k) andB(k) = Q(k)/S(k).

This equation can be handled in a way that is similar to the simpler caseq = 1.Let g(k) be a particular solution of the equation

g(kq) = A(k)g(k),

and notice thatf1(k) = C(k)g(k) solves the original equation ifC(k) is chosen to satisfythe first orderq-difference equation

C(qk) − C(k) = B(k)/(A(k)g(k)) ,

whose solution is determined up to the addition of an arbitraryq-periodic function.As an illustration we do the explicit integration in the caseε = 0 with the further

conditiona0 + a1 = 0, which is the simplest case considered in [15].Using (2.16) this condition becomesα = −b1(2b1u+β) andA(k) andB(k) are given

by the expressions

A(k) =kq − b1

k − b1, B(k) = − a1k(q − 1)(q + 1)2u

(k − b1)q(kqu + ku + 2b1u + β).

This results ing(k) = k − b1 .

We get that theq-difference ratio

(C(qk) − C(k))/(k(q − 1))

is given by

− a1(q + 1)2u(k − b1)q(kq − b1)(kqu + ku + 2b1u + β)

,

which can conveniently be expressed as

w3

(k − b1)(kq − b1)+

w2

k − b1+

w1

kqu + ku + 2b1u + β

with


w1 = − a1(q + 1)4u3

q(b1qu + 3b1u + β)(3b1qu + b1u + βq),

w2 =a1(q + 1)3u2

q(b1qu + 3b1u + β)(3b1qu + b1u + βq),

w3 = − a1(q + 1)2u3b1qu + b1u + βq

.

From here we obtain forC(k) the expression

C(k) = w3/(k − b1) + w2 logq k/b1 + w1 logq k/c1

with w3, w2, w1 simply related tow3, w2, w1 andc1 the root (ink) of the denominatorin the last summand above. We are using the notation logq z =

∑∞n=1 zn/(1− qn) from

[21].Finally we can use thisC(k), as observed above, to get

f1(k) = (k − b1)C(k) .

Acknowledgement.We thank R. Askey, M. Ismail, D. Masson and S. Suslov for help with the statement ofTheorem 3.

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Communicated by T. Miwa

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Some Functions that Generalize the Askey-Wilson Polynomials