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Factorization method for difference equations of hypergeometric type on nonuniform lattices

Álvarez Nodarse, Renato; Costas Santos, Roberto Santiago

Abstract

We study the factorization of the hypergeometric-type difference equation of Nikiforov and Uvarov on nonuniform lattices. An explicit form of the raising and lowering operators is derived and some relevant examples are given.

Full text

arXiv:1003.5279v1 [math.QA] 27 Mar 2010 Factorization method for difference equations of hypergeometric type on nonuniform lattices ∗ R. ´ Alvarez-Nodarse†‡ and R. S. Costas-Santos† †Departamento de An´alisis Matem´atico. Universidad de Sevilla. Apdo. 1160, E-41080 Sevilla ‡Instituto Carlos I de F´ısica Te´orica y Computacional, Universidad de Granada, E-18071 Granada, Spain 24th May 2001 Abstract We study the factorization of the hypergeometric-type difference equation of Nikiforov and Uvarov on nonuniform lattices. An explicit form of the raising and lowering operators is derived and some relevant examples are given. 1 Introduction In this paper we will deal with the so-called factorization method (FM) of the hypergeometric-type difference equations on nonuniform lattices. The FM was already used by Darboux [14] and Schr¨odinger [27, 28] to obtain the solutions of differential equations, and also by Infeld and Hull [17] for finding analytical solutions of certain classes of second order differential equations. Later on, Miller extended it to difference equations [18] and q-differences –in the Hahn sense– [19]. For more recent works see e.g. [5, 6, 11, 29, 30] and references therein. The classical FM was based on the existence of a so-called raising and lowering operators for the corresponding equation that allows to find the explicit solutions in a very easy way. Going further, Atakishiyev and coauthors [5, 9, 6] have found the dynamical symmetry algebra related with the FM and the differential or difference equations. Of special interest was the paper by Smirnov [15] in which the equivalence of the FM and the Nikiforov et all theory [25] was shown, furthermore this paper pointed out that the aforementioned equivalence remains valid also for the nonuniform lattices that was shown later on in [20, 29]. In particular, in [29] a detailed study of the FM and its equivalence with the Nikiforov et al. approach to difference equations [25] have been established. Also, in [12], a special nonuniform lattice was considered. In fact, in [12] the author constructed the FM for the Askey–Wilson polynomials using basically the difference equation for the polynomials. In the present paper we will continue the research of the nonuniform lattice case. In fact, following the idea by Bangerezako [12] for the Askey–Wilson polynomials and Lorente [22] for the classical continuous and discrete cases, we will obtain the FM for the general polynomial solutions of the hypergeometric difference equation on the general quadratic nonuniform lattice x(s) = ∗Submitted to J. Phys. A: Math. Gen. 1 c1qs+c2q−s+c3. We will use, as it is already suggested in [9, 15], not the polynomial solutions but the corresponding normalized functions which is more natural and useful. In such a way, the method proposed here is the generalization of [12] and [22] to the aforementioned nonuniform lattice. The structure of the paper is as follows. In Section 2 we present some well-known results on orthogonal polynomials on nonuniform lattices [10, 25, 26], in section 3 we introduce the normalized functions and obtain some of their properties such as the lowering and raising operator that allow us, in Section 4, to obtain the factorization for the second order difference equation satisfied by such functions. Finally, in Section 5, some relevant examples are worked out. 2 Some basic properties of the q-polynomials Here, we will summarize some of the properties of the q-polynomials useful for the rest of the work. For further information see e.g. [25]. We will deal here with the second order difference equation of the hypergeometric type σ(s)∆ ∆x(s−1 2)∇y(s) ∇x(s)+τ(s)∆y(s) ∆x(s)+λy(s) = 0, σ(s) = ˜σ(x(s)) −1 2˜τ(x(s))∆x(s−1 2), τ(s) = ˜τ(x(s)), (1) where ∇f(s) = f(s)−f(s−1) and ∆f(s) = f(s+ 1) −f(s) denote the backward and forward finite difference derivatives, respectively, ˜σ(x(s)) and ˜τ(x(s)) are polynomials in x(s) of degree at most 2 and 1, respectively, and λis a constant. In the following, we will use the following notation for the coefficients in the power expansions in x(s) of ˜σ(s) and ˜τ(s) ˜σ(s)≡˜σ[x(s)] = ˜σ′′ 2x2(s) + ˜σ′(0)x(s) + ˜σ(0),˜τ(s)≡˜τ[x(s)] = ˜τ′x(s) + ˜τ(0).(2) An important property of the above equation is that the k-order difference derivative of a solution y(s) of (1), defined by yk(s)q=∆ ∆xk−1(s) ∆ ∆xk−2(s)... ∆ ∆x(s)y(s)≡∆(k)y(s), also satisfies a difference equation of the hypergeometric type σ(s)∆ ∆xk(s−1 2)∇yk(s)q ∇xk(s)+τk(s)∆yk(s)q ∆xk(s)+µkyk(s)q= 0,(3) where xk(s) = x(s+k 2) and [25, page 62, Eq. (3.1.29)] τk(s) = σ(s+k)−σ(s) + τ(s+k)∆x(s+k−1 2) ∆xk−1(s), µk=λ+ k−1 X m=0 ∆τm(s) ∆xm(s).(4) It is important to notice that the above difference equations have polynomial solutions of the hypergeometric type iff x(s) is a function of the form [10, 26] x(s) = c1(q)qs+c2(q)q−s+c3(q) = c1(q)[qs+q−s−µ] + c3(q),(5) 2 where c1,c2,c3and qµ=c1 c2are constants which, in general, depend on q[25, 26]. For the above lattice, a straightforward calculation shows that τk(s) is a polynomial of first degree in xk(s) of the form (see e.g. [10]) τk(s) = ˜τ′ kxk(s) + ˜τk(0),˜τ′ k= [2k]q ˜σ′′ 2+αq(2k)˜τ′, ˜τk(0) = c3eσ′′ 2(2[k]q−[2k]q) + eσ′(0)[k]q+c3τ′(αq(k)−αq(2k)) + ˜τ(0)αq(k), (6) where the q-numbers [k]qand αq(k) are defined by [k]q=qk 2−q−k 2 q1 2−q−1 2 , αq(k) = qk 2+q−k 2 2,(7) and [n]q! are the q-factorials [n]q! = [1]q[2]q···[n]q. Both difference equations (1) and (3) can be rewritten in the symmetric form ∆ ∆x(s−1 2)σ(s)ρ(s)∇y(s) ∇x(s)+λnρ(s)y(s) = 0, and ∆ ∆xk(s−1 2)σ(s)ρk(s)∇yk(s) ∇xk(s)+µkρk(s)yk(s) = 0, where ρ(s) and ρk(s) are the weight functions satisfying the Pearson-type difference equations △ ∆x(s−1 2)[σ(s)ρ(s)] = τ(s)ρ(s),△ ∆xk(s−1 2)[σ(s)ρk(s)] = τk(s)ρk(s),(8) respectively. In [25] it is shown that the polynomial solutions of (3) (and so the polynomial solutions of (1)) are determined by the q-analogue of the Rodrigues formula on the nonuniform lattices ∆ ∆xk−1(s)··· ∆ ∆x(s)Pn(x(s))q≡∆(k)Pn(x(s))q=An,kBn ρk(s)∇(n) kρn(s),(9) where ∇(n) kf(s) = ∇ ∇xk+1(s) ∇ ∇xk+2(s)··· ∇ ∇xn(s)f(s). An,k =[n]q! [n−k]q! k−1 Y m=0 αq(n+m−1)eτ′+ [n+m−1]qeσ′′ 2(10) Thus [25, page 66, Eq. (3.2.19)] Pn(x(s))q=Bn ρ(s)∇(n)ρn(s),∇(n)≡∇ ∇x1(s) ∇ ∇x2(s)··· ∇ ∇xn(s),(11) where ρn(s) = ρ(s+n) n Y k=1 σ(s+k) and λn=−[n]qαq(n−1)eτ′+ [n−1]qeσ′′ 2.(12) 3 In this paper we will deal with orthogonal q-polynomials and functions. It can be proven [25], by using the difference equation of hypergeometric-type (1), that if the boundary condition σ(s)ρ(s)xk(s−1 2)s=a,b = 0,∀k≥0,(13) holds, then the polynomials Pn(s)qare orthogonal, i.e., b−1 X s=a Pn(x(s))qPm(x(s))qρ(s)∆x(s−1 2) = δnmd2 n, s =a, a + 1,...,b−1,(14) where ρ(s) is a solution of the Pearson-type equation (8). In the special case of the linear exponential lattice x(s) = qsthe above relation can be written in terms of the Jackson q-integral (see e.g. [16, 21]) Rz2 z1f(t)dqt, defined by Zz2 z1 f(t)dqt=Zz2 0 f(t)dqt−Zz1 0 f(t)dqt, where Zz 0 f(t)dqt=z(1 −q) ∞ X k=0 f(zqk)qk,0< q < 1, as follows: Zqb qa Pn(t)qPm(t)qω(t)dqt=δnmq1/2d2 n, t =qs, ω(t)≡ω(qt) = ρ(t).(15) Notice that the above boundary condition (13) is valid for k= 0. Moreover, if we assume that ais finite, then (13) is fulfilled at s=aproviding that σ(a) = 0 [25, §3.3, page 70]. In the following we will assume that this condition holds. The squared norm in (14) is given by [25, Chapter 3, Section 3.7.2, pag. 104] d2 n= (−1)nAn,nB2 n b−n−1 X s=a ρn(s)∆xn(s−1 2). There is also a so-called continuous orthogonality. In fact, if there exist a contour Γ such that ZΓ ∆[ρ(z)σ(z)xk(z−1 2)] dz = 0,∀k≥0,(16) then [25] ZΓ Pn(x(z))qPm(x(z))qρ(z)∆x(z−1 2)dz = 0, n 6=m. A simple consequence of the orthogonality is the following three term recurrence relation: x(s)Pn(x(s))q=αnPn+1(x(s))q+βnPn(x(s))q+γnPn−1(x(s))q,(17) where αn,βnand γnare constants. If Pn(s)q=anxn(s) + bnxn−1(s) + ··· ,then using (17) we find αn=an an+1 , βn=bn an −bn+1 an+1 , γn=an−1 an d2 n d2 n−1 .(18) To obtain the explicit values of αn,βnwe will use the following lemma, –interesting in its own right– that can be proven by induction: 4 Lemma 2.1 ∆(k)xn(s) = [n]q! [n−k]q!xn−k k(s)+c3n[n−1]q! [n−k−1]q!−(n−k)[n]q! [n−k]q!xn−k−1 k(s)+···. In the case k=n−1, it becomes ∆(n−1)xn(s) = [n]q!xn−1(s) + c3[n−1]q! (n−[n]q).(19) Now, using the Rodrigues formula (9) for k=n−1, ∆(n−1)Pn(x(s))q=An,n−1Bn ρn−1(s)∇(n) n−1ρn(s) = An,n−1Bn ρn−1(s) ∇ ∇xn(s)ρn(s), as well as the identities ρn(s) = ρn−1(s+ 1)σ(s+ 1), xn(s) = xn−1(s+1 2) and the Pearson equation (8) for ρn−1(s), we find ∆(n−1)Pn(x(s))q=An,n−1Bnτn−1(s). Thus an=An,n−1Bn˜τ′ n−1 [n]q!=Bn n−1 Y k=0 αq(n+k−1)˜τ′+ [n+k−1]q ˜σ′′ 2, and bn an =[n]q˜τn−1(0) ˜τ′ n−1 +c3([n]q−n). So αn=Bn Bn+1 αq(n−1)˜τ′+[n−1]q˜σ′′ 2 (αq(2n−1)˜τ′+[2n−1]q˜σ′′ 2)(αq(2n)˜τ′+[2n]q˜σ′′ 2)=−Bn Bn+1 λn [n]q [2n]q λ2n [2n+ 1]q λ2n+1 and βn=[n]q˜τn−1(0) ˜τ′ n−1 −[n+ 1]q˜τn(0) ˜τ′ n +c3([n]q+ 1 −[n+ 1]q). Using the Rodrigues formula the following difference-recurrent relation follows [1, 25] σ(s)∇Pn(x(s))q ∇x(s)=λn [n]qτ′ nτn(s)Pn(x(s))q−Bn Bn+1 Pn+1(x(s))q, where τn(s) is given by (6), where the identity ˜τ′ n=−λ2n+1 [2n+ 1]q has been used. Then, using the explicit expression for the coefficient αn, we find σ(s)∇Pn(x(s))q ∇x(s)=λn [n]q τn(s) τ′ n Pn(x(s))q−αnλ2n [2n]q Pn+1(x(s))q.(20) This equation defines a raising operator in terms of the backward difference in the sense that we can obtain the polynomial Pn+1 of degree n+ 1 from the lower degree polynomial Pn. From the above equation and using the identity ∇= ∆ − ∇∆, the second order difference equation and the three terms recurrence relation we find [1] lowering-type operator: [σ(s) + τ(s)∆x(s−1 2)]∆Pn(x(s))q ∆x(s)=γnλ2n [2n]q Pn−1(x(s))q+ λn [n]q τn(s) τ′ n −λn∆x(s−1 2)−λ2n [2n]q (x(s)−βn)Pn(x(s))q. (21) 5 The most general polynomial solution of the q-hypergeometric equation (1) corresponds to the case σ(s) = A 4 Y i=1 [s−si]q=Cq−2s 4 Y i=1 (qs−qsi), A, C, not vanishing constants (22) and has the form [26] Pn(s)q=Dn4φ3q−n, q2µ+n−1+P4 i=1si, qs1−s, qs1+s+µ qs1+s2+µ, qs1+s3+µ, qs1+s4+µ;q , q,(23) where Dnis a normalizing constant and the basic hypergeometric series pφqare defined by [21] rφpa1,...,ar b1,...,bp;q , z= ∞ X k=0 (a1;q)k···(ar;q)k (b1;q)k···(bp;q)k zk (q;q)kh(−1)kqk 2(k−1)ip−r+1 , and (a;q)k= k−1 Y m=0 (1 −aqm),(24) is the q-analogue of the Pochhammer symbol. Instances of such polynomials are the Askey–Wilson polynomials, the q-Racah polynomials and big q-Jacobi polynomials among others [21, 26]. 3 The orthonormal functions on nonuniform lattices In this section we will introduce a set of orthonormal functions which are orthogonal with respect to the unit weight [9, 15] ϕn(s) = pρ(s)/d2 nPn(x(s))q,(25) e.g. for the case of discrete orthogonality we have b−1 X si=a ϕn(si)ϕm(si)∆x(si−1 2) = δnm. Next, we will establish several important properties of such functions which generalize, to the nonuniform lattices, the ones given in [22]. In the following we will use the notation Θ(s) = σ(s) + τ(s)∆x(s−1 2). First of all, inserting (25) into (1), (17), (20), (21) we obtain that they satisfy the following difference equation: pΘ(s)σ(s+ 1) 1 ∆x(s)ϕn(s+ 1) + pΘ(s−1))σ(s)1 ∇x(s)ϕn(s−1)− Θ(s) ∆x(s)+σ(s) ∇x(s)ϕn(s) + λn∆x(s−1 2)ϕn(s) = 0, (26) the three term recurrence relation: αn dn+1 dn ϕn+1(s) + γn dn−1 dn ϕn−1(s) + (βn−x(s))ϕn(s) = 0,(27) 6 the raising-type formula: L+(s, n)ϕn(s) = αn λ2n [2n]q dn+1 dn ϕn+1(s),(28) and the lowering-type formula: L−(s, n)ϕn(s) = γn λ2n [2n]q dn−1 dn ϕn−1(s),(29) where the raising-type operator L+(s, n) and the lowering-type operator L−(s, n) are given by L+(s, n)≡λn [n]q τn(s) τ′ n −σ(s) ∇x(s)I+pΘ(s−1)σ(s)1 ∇x(s)E−,(30) and L−(s, n)≡−λn [n]q τn(s) τ′ n +λn∆x(s−1 2) + λ2n [2n]q (x(s)−βn)−Θ(s) ∆x(s)I +pΘ(s)σ(s+ 1) 1 ∆x(s)E+, (31) respectively. In the above formulas E−f(s) = f(s−1), E+f(s) = f(s+ 1) and Iis the identity operator. Notice that the last two formulas have a remarkable property of giving all the solutions ϕn(s). In fact, from (31) setting n= 0 and taking into account that ϕ−1(s)≡ 0 we can obtain ϕ0(s). Then, substituting the obtained function in (30), we can find all the functions ϕ1(s),. . . , ϕn(s),. . . . Proposition 3.1 The raising and lowering operators (30) and (31) are mutually adjoint. Proof: The proof is straightforward. In fact using the boundary condition and after some calculations we obtain, in the case of discrete orthogonality, the expression b−1 X si=a ϕn+1(si)[2n]q λ2n L+(si, n)ϕn(si)∆x(si−1 2) = b−1 X si=a[2n+ 2]q λ2n+2 L−(si, n + 1)ϕn+1(si)ϕn(si)∆x(si−1 2) = αn dn+1 dn . The other cases can be done in an analogous way. Proposition 3.2 The operator corresponding to the eigenvalue λnin (26) is self adjoint. Proof: Again we will prove the result in the case of discrete orthogonality. Using the orthogonality conditions σ(a)ρ(a) = σ(b)ρ(b) = 0 (which is a consequence of (13)), we can write b−1 X si=a ϕn(si)pΘ(si−1)σ(si)1 ∇x(si)ϕl(si−1)∆x(si−1 2) = b−2 X s′ i=a−1 ϕn(s′ i+ 1)qΘ(s′ i)σ(s′ i+ 1) 1 ∇x(s′ i+ 1)ϕl(s′ i)∆x(s′ i+1 2) 7 = b−1 X si=a ϕn(si+ 1)pΘ(si)σ(si+ 1) 1 ∇x(si+ 1)ϕl(si)∆x(si+1 2)+ ϕn(a)pΘ(a−1)σ(a)1 ∇x(a)ϕl(a−1)∆x(a−1 2)− ϕn(b)pΘ(b−1)σ(b)1 ∇x(b)ϕl(b−1)∆x(b−1 2), where in the last two sums we first take the operations ∆ and ∇, and then substitute the corresponding value: e.g. ∆x(a) = x(a+ 1) −x(a). Now, we use the fact that ϕn(s) = pρ(s)/d2 nPn(x(s))q, as well as the boundary conditions σ(a)ρ(a) = σ(b)ρ(b) = 0, so pΘ(a−1)σ(a)ϕn(a)ϕl(a−1) = pΘ(b−1)σ(b)ϕn(b)ϕl(b−1) = 0. The other terms can be transformed in a similar way. All these yield the expression b−1 X si=a ϕl(si)pΘ(si)σ(si+ 1) 1 ∆x(si)ϕn(si+ 1)∆x(si+1 2)+ pΘ(si−1)σ(si)1 ∇x(si)ϕn(si−1)∆x(si−1 2)= = b−1 X si=a ϕn(si)pΘ(si)σ(si+ 1) 1 ∆x(si)ϕl(si+ 1)∆x(si+1 2)+ pΘ(si−1)σ(si)1 ∇x(si)ϕl(si−1)∆x(si−1 2), from where the proposition easily follows. 4 Factorization of difference equation of hypergeometric type on the nonuniform lattice We will define from (26) the following operator H(s, n)≡pΘ(s−1)σ(s)1 ∇x(s)E−+pΘ(s)σ(s+ 1) 1 ∆x(s)E+− Θ(s) ∆x(s)+σ(s) ∇x(s)−λn∆x(s−1 2)I. Clearly, the orthonormal functions satisfy H(s, n)ϕn(s) = 0. Let us rewrite the raising and lowering operators in the following way L+(s, n) = u(s, n)I+pΘ(s−1)σ(s)1 ∇x(s)E−, L−(s, n) = v(s, n)I+pΘ(s)σ(s+ 1) 1 ∆x(s)E+, 8 where, as before, Θ(s) = σ(s) + τ(s)∆x(s−1 2), and u(s, n) = λn [n]q τn(s) τ′ n −σ(s) ∇x(s), v(s, n) = −λn [n]q τn(s) τ′ n +λn∆x(s−1 2) + λ2n [2n]q (x(s)−βn)−Θ(s) ∆x(s). Proposition 4.1 The functions u(s, n)and v(s, n)satisfy u(s+1, n) = v(s, n+1), or, equivalently u(s+ 1, n −1) = v(s, n). The proof of the above proposition is straightforward but cumbersome. We will include it in appendix A. If we now calculate L−(s, n + 1)L+(s, n) = v(s, n + 1)u(s, n) + Θ(s)σ(s+ 1) 1 ∆x(s)2 + +u(s+ 1, n)pΘ(s−1)σ(s)1 ∇x(s)E−+pΘ(s)σ(s+ 1) 1 ∆x(s)E+, and substitute the values for u(s, n), v(s, n) and H(s, n) we get L−(s, n + 1)L+(s, n) = h∓(n)I+u(s+ 1, n)H(s, n), where the function h∓(n) = λn [n]q τn(s+ 1) τ′ n −σ(s+ 1) ∇x(s+ 1)λn [n]q τn(s) τ′ n −λn∆x(s−1 2)+ λn [n]q τn(s+ 1) τ′ n Θ(s) ∆x(s), is independent of s. In fact, applying the last equality to the orthonormal function ϕn(s) and taking into account (28) and (29), h∓(n) = λ2n [2n]q λ2n+2 [2n+ 2]q αnγn+1. Similarly, L+(s, n −1)L−(s, n) = h±(n)I+u(s, n −1)H(s, n), where h±(n) = −λn [n]q τn(s−1) τ′ n +λ2n [2n]q (x(s−1) −βn) + λn∆x(s−3 2)× −λn [n]q τn(s) τ′ n +λ2n [2n]q (x(s)−βn) + σ(s) ∇x(s)− −λn [n]q τn(s) τ′ n +λ2n [2n]q (x(s)−βn)Θ(s−1) ∆x(s−1), is independent of s. Furthermore, applying the last expression to the functions ϕn(s), and taking into account (28) and (29), we obtain h±(n) = λ2n−2 [2n−2]q λ2n [2n]q αn−1γn. Remark: Notice that h±(n+ 1) = h∓(n). All the above results lead us to our main theorem: 9 and whose eigenvalues are λn= 4q−n+1(1 −qn)(1 −abcdqn−1). In addition, we have eσ′′ =−4(q−1)2(1 + abcd)q−1/2, eσ′(0) = (q−1)2(a+b+c+d+abc +abd +acd +bcd)q−1/2, eσ(0) = (q−1)2(1 −ab −ac −ad −bc −bd −cd +abcd)q−1/2, τ′ n= 4q−n(q−1)(1 −abcdq2n), τn(0) = 2(q−1)(−a−b−c−d+ (abc +abd +acd +bcd)qn)q−n/2. Defining now the normalized functions (see (15)) pω(x)/d2 npn(x;a, b, c, d), the corresponding Hamiltonian H(s, n) is H(s, n) = 2q3/2 [2s−1]q G(s, a, b, c, d)E−+2q3/2 [2s+ 1]q G(s+ 1, a, b, c, d)E++ 2 q−2s+1/2Q4 i=1(1 −qsi+s) [2s+ 1]q +q−2s+1/2Q4 i=1(qs−qsi) [2s−1]q + q−n+1κ2 q(1 −qn)(1 −abcdqn−1)[2s]q!I where G(s, a, b, c, d) = v u u t4 Y i=1 (1 −2qsiq−1/2x(s−1/2) + q−1q2si), We now define u(s, n) = Dnxn(s) + DnEn+q−2s+1/2(qs−a)(qs−b)(qs−c)(qs−d) [2s−1]q where Dn=−4q−n/2+1/2(q−1)(1 −abcdqn−1). En=(−a−b−c−d+ (abc +abd +acd +bcd)qn)qn/2 2(1 −abcdq2n). Taking into account that v(s, n) = u(s+ 1, n −1), we find L+(s, n) = u(s, n)I+2q3/2 [2s−1]q G(s, a, b, c, d)E−, L−(s, n) = v(s, n)I+2q3/2 [2s+ 1]q G(s+ 1, a, b, c, d)E+, where E−f(s) = f(s−1) and E+f(s) = f(s+ 1). Thus, L−(s, n + 1)L+(s, n) = D2nD2n+2γn+1I+v(s, n + 1)H(s, n), and L+(s, n −1)L−(s, n) = D2n−2D2nγnI+u(s, n −1)H(s, n), 16 which is the factorization formula for the Askey–Wilson functions. To conclude this paper let us consider the special case of Askey–Wilson polynomials when a=b=c=d= 0, i.e., the continuous q-Hermite polynomials Hn(x|q) = 2−neinθ2φ0 q−n,0q;qne−2iθ —!, x = cos θ. These polynomials are closely related with the q-harmonic oscilator model introduced by Biedenharn [13] and Macfarlane [23], as it was pointed out in [8], where the factorization for the continuous q-Hermite polynomials were considered first. If we substitute a=b=c=d= 0 in the above formulas, we obtain the factorization for the q-Hermite functions ϕn(x) = sh(x, 1)h(x, −1)h(x, q1/2)h(x, −q1/2)(qn+1;q)∞ 2πκq(1 −x2)Hn(x|q). In fact, since for continuous q-Hermite polynomials σ(s) = −κ2 qq2s+1/2, τ(s) = 4(q−1)x(s), λn= 4q−n+1(1 −qn), and the coefficients for the three-term recurrence relation are αn= 1, βn= 0, γn= (1 −qn)/4, then we obtain H(s, n) = 2q3/2 [2s−1]q E−+2q3/2 [2s+ 1]q E++2 q−2s+1/2 [2s+ 1]q +q2s+1/2 [2s−1]q −q−n+1κ2 q(1−qn)[2s]q!I, L+(s, n) = −4q−n/2+1/2(q−1)x(s+n/2) + q2s+1/2 [2s−1]q!I+2q3/2 [2s−1]q E−, L−(s, n) = −4q−n/2+1(q−1)x(s+n/2 + 1/2) + q2s+5/2 [2s+ 1]q!I+2q3/2 [2s+ 1]q E− and h±(n) = 4κ2 qq−2n+1(1 −qn). Acknowledgements The authors thank N. Atakishiyev and Yu. F. Smirnov for interesting discussions and remarks that allowed us to improve this paper substantially, as well as the referees for their remarks. The work has been partially supported by the Ministerio de Ciencias y Tecnolog´ıa of Spain under the grant BFM-2000-0206-C04-02, the Junta de Andaluc´ıa under grant FQM-262 and the European proyect INTAS-2000-272. Appendix A Here, for the sake of completeness, we will prove Proposition 4.1, by showing that u(s+ 1, n)− v(s, n + 1) = 0. To do that, we start with computing the difference u(s+ 1, n)−v(s, n + 1) = λn [n]q τn(s+ 1) τ′ n −∆σ(s) ∆x(s)+ λn+1 [n+ 1]q τn+1(s) τ′ n+1 −λn+1∆xs−1 2−λ2n+2 [2n+ 2]q (x(s)−βn+1) + τ(s)∆xs−1 2 ∆x(s). 17 Now we use the expansion τn(s+ 1) = τ′ nxn(s+ 1) + τn(0). Since ∆(x2(s)) ∆x(s)=x2(s+ 1) −x2(s) x(s+ 1) −x(s)=x(s+ 1) + x(s) = C1qs(q+ 1) + C2q−s(q−1+ 1) + 2C3= (C1qs+1 2+C2q−s−1 2)[2]q+ 2C3= [2]qx1(s) + (2 −[2]q)C3, x(s)∆x(s−1 2) = x(s)(C1qs−1 2(q−1) + C2q−s+1 2(q−1−1)) = x(s)(C1qs−C2q−s)kq= (C2 1q2s−C2 2q−2s)kq+C3(C1qs−C2q−s)kq, where kq=q1 2−q−1 2, ∆ ∆x(s)x(s)∆x(s−1 2)= (C2 1q2s+1 +C2 2q−2s−1)[2]q+C3(C1qs+1 2+C2q−s−1 2) C1qs+1 2−C2q−s−1 2!kq, and ∆ ∆x(s)∆x(s−1 2)=∆ ∆x(s)(C1qs−C2q−s)kq=C1qs+1 2+C2q−s−1 2 C1qs+1 2−C2q−s−1 2 kq. Then ∆σ(s) ∆x(s)=∆ ∆x(s)eσ(s)−1 2eτ(s)∆x(s−1 2)= =∆ ∆x(s)eσ′′ 2x2(s) + eσ′(0)x(s) + eσ(0) −1 2(τ′x(s) + τ(0)) ∆x(s−1 2)= eσ′′ 2([2]qx1(s) + (2 −[2]q)C3) + eσ′(0) −1 2τ(0) C1qs+1 2+C2q−s−1 2 C1qs+1 2−C2q−s−1 2!kq− 1 2τ′ [2]q(C2 1q2s+1 +C2 2q−2s−1) + C3(C1qs+1 2+C2q−s−1 2) C1qs+1 2−C2q−s−1 2!kq. This yields for u(s+ 1, n)−v(s, n + 1) the expression =λn [n]q xn(s+ 1) + λn [n]q τn(0) τ′ n−eσ′′ 2[2]qx1(s) + C3 2(2 −[2]q)eσ′′ +eσ′(0)− eτ′ 2[2]q(C2 1q2s+1 +C2 2q−2s−1) C1qs+1 2−C2q−s−1 2 +C3x1(s)−C2 3 C1qs+1 2−C2q−s−1 2kq− τ(0) 2x1(s)−C3 C1qs+1 2−C2q−s−1 2kq+λn+1 [n+ 1]q τn+1(s) τ′ n+1 −λn+1∆xs−1 2− λ2n+2 [2n+ 2]qC1qs+C2q−s+C3−[n+ 1]qτn(0) τ′ n + [n+ 2]qτn+1(0) τ′ n+1 −C3(1 + [n+ 1]q−[n+ 2]q)+τ(s)∆xs−1 2 ∆x(s). Next, we expand ∆xn(s) and eσ′′ 2[2]x1(s), make some straightforward calculations and use the identities: λn [n]q τn(0) τ′ n + [n+ 1]q λ2n+2 [2n+ 2]q τn(0) τ′ n =λn [n]q + [n+ 1]q λ2n+2 [2n+ 2]qτn(0) τ′ n =−[n+ 2]qτn(0), λn+1 [n+ 1]q τn+1(s) τ′ n+1 −[n+ 2]q λ2n+2 [2n+ 2]q τn+1(0) τ′ n+1 = [n+ 1]qτn+1(0) + λn+1 [n+ 1]q xn+1(s), 18 as well as λn [n]q (C1qs+1+ n 2+C2q−s−1−n 2)−eσ′′ 2[2]q(C1qs+1 2+C2q−s−1 2)−λ2n+2 [2n+ 2]q (C1qs+C2q−s)− λn+1(C1qs−C2q−s)kq+1 2τ′(C1qs+1 2+C2q−s−1 2)(q+q−1) = C1qsτ′ 2(qn+1 2+q1 2)+ C1qseσ′′ 2(q1 2−q−1 2)(qn+1 2−q1 2) + C2q−sτ′ 2(q−n−1 2+q−1 2) + C2q−seσ′′ 2(q1 2−q−1 2)(−q−n−1 2+q−1 2) = =−λn+1 [n+ 1]q (C1qs+n+1 2+C2q−s−n+1 2), we find =−λn+1 [n+ 1]qC1qs+n+1 2+C2q−s−n+1 2+C3 λn [n]q −[n+ 2]qτn(0) −C3eσ′′ −eσ′(0) + 1 2τ′C3kq+ 1 2τ(0)kq+[n+1]qτn+1(0)+ λn+1 [n+1]q (C1qs+n+1 2+C2q−s−n+1 2+C3)+ λ2n+2 [2n+2]q C3([n+1]q−[n+2]q). 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