On the $D_{\omega }$-Classical Orthogonal Polynomials

Abstract

We investigate the $D_{\omega }$-classical orthogonal polynomials, where $D_{\omega }$ is the weighted difference operator. So, we address the problem of finding the sequence of orthogonal polynomials such that their $D_{\omega }$-derivatives is also orthogonal polynomials. To solve this problem we adopt a different approach to those employed in this topic. We first begin by determining the coefficients involved in their recurrence relations, and then providing an exhaustive list of all solutions. When $\omega =0$, we rediscover the classical orthogonal polynomials of Hermite, Laguerre, Bessel and Jacobi. For $\omega =1$, we encounter the families of discrete classical orthogonal polynomials as particular cases.

1 Introduction and Preliminary Results

The orthogonal polynomials are characterized by the fact that they satisfy a second-order recurrence relation. They said to be classical if their derivatives also form a sequence of orthogonal polynomials [11]. Hahn generalized the classical orthogonal polynomials by generalizing their characteristic properties (see [2, 6] for more details). For this, he considered the linear operator [12]

$$\begin{aligned} \left( H_{q,\omega }f\right) (x):=\frac{f(qx+\omega )-f(x)}{(q-1)x+\omega }, \end{aligned}$$

(1.1)

for all polynomial f, with q and $\omega$ are two fixed complex numbers.

Hahn showed that there is no loss of generality in assuming $\omega$ to be zero so that in what follows q may be thought of as 1 or different to 1. For $q\ne 1$ and $\omega =0$, we obtain the q-difference operator (also known as the Jackson’s q-operator) which we write $\left( {{\mathscr {D}}}_{q}f\right) (x):=\left( H_{q,0}f\right) (x)$. When $q=1$ with $\omega \ne 0$ we get the discrete operator $\left( D_{\omega }f\right) (x):=\left( H_{1,\omega }f\right) (x)$, that is,

$$\begin{aligned} \left( D_{\omega }f\right) (x)=\frac{f(x+\omega )-f(x)}{\omega }. \end{aligned}$$

(1.2)

Unless otherwise stated we assume that $\omega \in {\mathbb {C}}$.

The limiting case $\omega \rightarrow 0$ (resp. $q\rightarrow 1$) of $D_{\omega }$ (resp. ${{\mathscr {D}}}_{q}$) gives rise to the derivative operator $D=d/dx$, giving $(Df)(x):=f'(x)$. Because it is always possible to take such a limit, this point is not really important at this time. It will be dealt with a little bit later when necessary. Obviously, if we take $\omega =1$ in (1.2), we recover the finite (or forward) difference operator $\Delta f(x)= f(x+1)-f(x)$.

Motivated by the several properties common to all of the classical orthogonal polynomials, Hahn [12] posed and solved five (equivalent) problems that are related to the operator ${{\mathscr {D}}}_{q}$ and obtained that all possible solutions lead to the same orthogonal polynomial sequences (OPS) which are the so-called classical q-orthogonal polynomials. Later on, the study of such polynomials has known an increasing interest (see for instance [14] and the references therein).

The first problem studied by Hahn is the following:

Find all OPS $\{P_n\}_{n\geqslant 0}$ such that $\{{{\mathscr {D}}}_{q}P_n\}_{n\geqslant 0}$ is also an OPS.

For more details about solutions of these problems we refer the reader to [2, 6, 11, 13].

In [8] Douak and Maroni considered the problem of finding all OPS such that their D-derivatives are also OPS. Instead of basing the study of this problem on the various properties of orthogonal polynomials, the authors have rather founded their exposures in a purely algebraic point of view, focusing primarily on the explicit calculation of recurrence coefficients. The identified polynomials are none other than the classical orthogonal polynomials of Hermite, Laguerre, Jacobi and Bessel with the usual restrictions on the parameters.

Referring back to the operator $D_\omega$, we will pose the analogous problem:

$\mathbf {(P)}$ Find all OPS $\{P_n\}_{n\geqslant 0}$ such that $\{D_{\omega }P_n\}_{n\geqslant 0}$ are also OPS.

When $D_\omega$ is replaced by the operator $\Delta$, the analogous problem was solved by Lancaster [15] by considering a second order difference equation of the Sturm–Liouville type (see [2] for more details). He obtained that the eigenfunctions of the related difference operator are the discrete orthogonal polynomials. Later Lesky in [16, 17] stated the same conclusion.

Note that in this field a general method of studying the classical orthogonal polynomials of a discrete variable as solutions of a second-order difference equation of hypergeometric type was considered by Nikiforov et al. [23]. On the other hand, by introducing a new matrix approach using the basic algebraic properties of the matrices, Verde-Star [27] solved similar problems by looking for sequences of orthogonal polynomials possessing Hahn’s property.

This work is mainly intended to constructing the $D_{\omega }$-classical orthogonal polynomials by proceeding as in [8]. Such approach is rather new and, of course, different to those previously used in several studies dedicated to this topic (see for instance [1, 10, 11, 23] and the references therein). After determining the recurrence coefficients explicitly, we proceed to the identification of the resulting polynomials. Under some restrictions on the parameters, we establish that these polynomials can be reduced to one of the well-known families of discrete classical orthogonal polynomials.

Similar method was also used within the d-orthogonality context ($d\geqslant 1$) to provide many extensions of the classical orthogonal polynomials (see, e.g., [7,8,9, 18] and the references therein). The same approach used in the Dunkl context, namely, when the ordinary derivative D (or $D_\omega$) is replaced by the Dunkl operator, also gives interesting results, as will be published elsewhere.

In an earlier survey, Abdelkarim and Maroni [1] investigated the problem $\mathbf {(P)}$ according to a functional approach. The authors had established various equivalent properties characterizing the resulting polynomials. Particularly, they showed that those polynomials satisfy the so-called functional Rodrigues’s formula (1.14). Based on this last characterization, up to linear transformation of the variable, they found that there are four classes of $D_\omega$-classical orthogonal sequences satisfying Rodrigues’s formula, including the Charlier, Meixner, Krawchuk and Hahn polynomials as special cases of them.

Let ${\mathscr {P}}$ be the vector space of polynomials of one variable with complex coefficients and let ${\mathscr {P}}'$ be its algebraic dual. We denote by $\bigl <.,\,.\bigr>$ the duality brackets between ${\mathscr {P}}'$ and ${\mathscr {P}}$. By $\{P_n\}_{n\geqslant 0}$ we denote a polynomials sequence (PS), $\deg P_n=n$, and $\{u_n\}_{n\geqslant 0}$ its associated dual sequence (basis) defined by $\bigl < u_n , P_m \bigr > = \delta _{n m} ;\, n , m\geqslant 0,$ where $\delta _{n m}$ is the Kronecker’s delta symbol.

The first element $u_0$ of the dual sequence is said to be the canonical form associated to the PS $\{P_n\}_{n\geqslant 0}$. Throughout this article, we will always consider the sequence of monic polynomials, i.e., the leading coefficient of each polynomial $P_n$ is one ($P_n(x)=x^n+\cdots$).

Given a form $u\in {\mathscr {P}}'$. The sequence of complex numbers $(u)_n,\ n=0, 1, 2, \ldots ,$ defined by $(u)_n := \left\langle u , x^n \right\rangle$ denotes the moments of u with respect to the sequence $\{x^n\}_{n \geqslant 0}$. The form u is called regular (or quasi-definite) if we can associate with it a PS $\{P_n\}_{n\geqslant 0}$ such that

$$\begin{aligned} \left\langle u,P_nP_m\right\rangle =k_n\delta _{n,m},\, n, m\geqslant 0\ ; \ k_n\ne 0,\ n\geqslant 0. \end{aligned}$$

In this case $\{P_n\}_{n\geqslant 0}$ is an orthogonal polynomials sequence (OPS) with respect to (w.r.t.) u. As an immediate consequence of the regularity of u, we have $(u)_0 \ne 0$ and $u=\lambda u_0$ with $\lambda \ne 0$. Furthermore, the elements of the dual sequence $\{u_n\}_{n\geqslant 0}$ are such that

$$\begin{aligned} u_n=\left( \left\langle u_0,P^2_n\right\rangle \right) ^{-1}P_nu_0,\ \ n=0, 1, 2, \ldots . \end{aligned}$$

(1.3)

So, in all what follows, we consider the orthogonality of any PS w.r.t. its canonical form $u_0$.

First, let us introduce the two operators $h_a$ and $\tau _b$ defined for all $f\in {\mathscr {P}}$ by

$$\begin{aligned} (h_a f)(x)= f(ax) \ \ \text{ and }\ \ (\tau _b f)(x)= f(x-b), \quad a\in {\mathbb {C}}^{*}:={\mathbb {C}}\backslash \{0\}, \ b\in {\mathbb {C}}.\nonumber \\ \end{aligned}$$

(1.4)

On the other hand, for any functional u, we can write by transposition

$$\begin{aligned}{} & {} \left\langle \tau _{-b}u,f(x)\right\rangle :=\left\langle u,\tau _{b}f(x)\right\rangle =\left\langle u,f(x-b)\right\rangle , f\in {{\mathscr {P}}},\end{aligned}$$

(1.5)

$$\begin{aligned}{} & {} \left\langle h_{a}u,f(x)\right\rangle :=\left\langle u,h_{a}f(x)\right\rangle =\left\langle u,f(ax)\right\rangle , f\in {{\mathscr {P}}}. \end{aligned}$$

(1.6)

For further formulas and other properties fulfilled by the operator $D_\omega$ see [1].

We now consider the sequence of monic polynomials $\{Q_n(x)\}n\geqslant 0$ defined by $\{Q_n(x):=(n+1)^{-1}D_\omega P_{n+1}(x)\}_{n\geqslant 0}$, with its associated dual sequence denoted by $\{v_n\}_{n\geqslant 0}$ and fulfilling

$$\begin{aligned} D_{-\omega }\left( v_n\right) =-(n+1)u_{n+1},\ n\geqslant 0, \end{aligned}$$

(1.7)

where by definition

$$\begin{aligned} \left\langle D_{-\omega }u\,,\, f\right\rangle =-\left\langle u\,,\, D_\omega f\right\rangle ,\, u\in {\mathscr {P}}',\, f\in {\mathscr {P}}. \end{aligned}$$

(1.8)

Next, in the light of the so-called Hahn property [11], we give the following definition.

Definition 1.1

The OPS $\{P_n\}_{n\geqslant 0}$ is called “$D_\omega$-classical” if the sequence of its derivatives $\{Q_n\}_{n\geqslant 0}$ is also a OPS.

Thus, $\{P_n\}_{n\geqslant 0}$ is orthogonal w.r.t. $u_0$ and satisfies the second-order recurrence relation

$$\begin{aligned} P_{n+2}(x)= & {} (x - \beta _{n+1})P_{n+1}(x)-\gamma _{n+1}P_n (x), \ n\geqslant 0, \end{aligned}$$

(1.9a)

$$\begin{aligned} P_1(x)= & {} x - \beta _0,\ P_0(x) = 1, \end{aligned}$$

(1.9b)

and $\{Q_n\}_{n\geqslant 0}$ is orthogonal w.r.t. $v_0$ and satisfies the second-order recurrence relation

$$\begin{aligned} Q_{n+2}(x)= & {} (x - \tilde{\beta }_{n+1})Q_{n+1}(x) -\tilde{\gamma }_{n+1} Q_n(x), \ n\geqslant 0,\end{aligned}$$

(1.10a)

$$\begin{aligned} Q_1(x)= & {} x - \tilde{\beta }_0,\ Q_0(x) = 1, \end{aligned}$$

(1.10b)

with the regularity conditions $\gamma _{n} \ne 0$ and $\tilde{\gamma }_{n} \ne 0$ for every $n\geqslant 1$.

Finally, we will summarize in the following proposition the important properties characterizing the $D_\omega$-classical orthogonal polynomials as stated in [1, Propositions 2.1$-$2.3].

Proposition 1.2

For any OPS $\{P_n\}_{n\geqslant 0}$, the following are equivalent statements:

• (a) The sequence $\{P_n\}_{n\geqslant 0}$ is $D_\omega$-classical.

• (b) The sequence $\{Q_n\}_{n\geqslant 0}$ is orthogonal.

• (c) There exist two polynomials $\Phi$ monic $(with\, \deg \Phi = t\leqslant 2)$ and $\Psi \ (with\, \deg \Psi =1)$, and a sequence $\{\lambda _n\}_{n\geqslant 0}$, $\lambda _n\ne 0$ for all n, such that

  $$\begin{aligned} \Phi (x)\left( D_\omega \circ D_{-\omega }P_{n+1}\right) (x) -\Psi (x)\left( D_{-\omega }P_{n+1}\right) (x) +\lambda _nP_{n+1}(x)=0,\ n\geqslant 0.\nonumber \\ \end{aligned}$$

  (1.11)

• (d) The sequences $\{Q_n\}_{n\geqslant 0}$ and $\{P_n\}_{n\geqslant 0}$ are interlinked via the differentiation formula

  $$\begin{aligned} \Phi (x)Q_n(x)=\alpha _{n+2}^2P_{n+2}(x)+\alpha _{n+1}^1P_{n+1}(x) +\alpha _{n}^0P_{n}(x),\ n\geqslant 0,\quad (\alpha _{n}^0\ne 0).\nonumber \\ \end{aligned}$$

  (1.12)

  This identity is referred to as the first structure relation of the OPS $\{P_n\}_{n\geqslant 0}$.

• (e) The form $u_0$ is $D_\omega$-classical, say, it is regular and satisfies the functional equation

  $$\begin{aligned} D_{-\omega }\left( \Phi u_0\right) +\Psi u_0=0. \end{aligned}$$

  (1.13)

• (f) There exist a monic polynomial $\Phi$, $\deg \Phi \leqslant 2$, and a sequence $\{\lambda _n\}_{n\geqslant 0}$, $\lambda _n\ne 0$ for all n, such that the canonical form $u_0$ satisfies the so-called functional Rodrigues formula

  $$\begin{aligned} P_n u_0 = \lambda _n D^n_{-\omega }\Big \{\Big (\prod _{\nu =0}^{n-1} \tau _{-\nu \omega }\Phi \Big )u_0\Big \},\ n\geqslant 0,\quad \text{ with } \prod _{\nu =0}^{-1}:=1. \end{aligned}$$

  (1.14)

It is worth noting that the structure relation (1.12) was also explained in [21, Proposition 4.5] in a more general setting, namely that of the $D_\omega$-semi-classical case. Here we will add a new characterization to those established in the preceding proposition. This will be proved once we give the lemma below. To begin with, apply the operator $D_\omega$ to (1.9a)–(1.9b), taking into account (1.10a)–(1.10b), we obtain

$$\begin{aligned} P_{n+2}(x)= & {} Q_{n+2}(x) + \tilde{\alpha }^1_{n+1}Q_{n+1}(x) +\tilde{\alpha }^0_{n} Q_{n} (x),\ n\geqslant 0, \end{aligned}$$

(1.15a)

$$\begin{aligned} P_1(x)= & {} Q_1(x) +\tilde{\alpha }_0^1,\; P_0(x) = Q_0(x) =1. \end{aligned}$$

(1.15b)

We refer to the above relation as the second structure relation of the polynomials $P_n,\,n\geqslant 0$. This will be instrumental in Sect. 2 to derive the system connecting the coefficients $\alpha _{n+2}^2$, $\alpha _{n+1}^1$ and $\alpha _n^0$, for $n\geqslant 0$, and the pair of recurrence coefficients $(\beta _n,\gamma _{n})$.

Combining (1.15a)–(1.15b) with (1.9a)–(1.9b) and then use (1.10a)–(1.10b), we infer that

$$\begin{aligned} \tilde{\alpha }_n^1=(n+1)(\beta _{n+1}-\tilde{\beta }_{n}-\omega )\ \text {and} \ \tilde{\alpha }_n^0=(n+1) \gamma _{n+2}-(n+2) \tilde{\gamma }_{n+1}, n\geqslant 0.\nonumber \\ \end{aligned}$$

(1.16)

Reminder that, for the classical orthogonal polynomials ($\omega =0$), the first structure relation was given by Al Salam and Chihara [3], and the second one was established by Maroni [20].

When $D_\omega$ is replaced by the finite difference operator $\Delta$, Garcia et al. [10] proved that the structure relations (1.12) and (1.15a)–(1.15b), as well as the functional Rodrigues formula (1.14) characterize the discrete classical polynomials of Charlier, Meixner, Krawchuk and Hahn.

We will now return to the operator $D_{\omega }$ and show that (1.15a)–(1.15b) also characterize the $D_\omega$-classical orthogonal polynomials. To do so, we need the following lemma.

Lemma 1.3

([19]) Let $\{P_n\}_{n \geqslant 0}$ be a sequence of monic polynomials and let $\{u_n\}_{n \geqslant 0}$ be its associated dual sequence. For any linear functional u and integer $m\geqslant 1$, the following statements are equivalent:

• (i) $\left\langle u \, ,\, P_{m-1} \right\rangle \not = 0 \ ;\ \left\langle u \, ,\, P_{n} \right\rangle = 0 ,\; n\geqslant m;$

• (ii) $\exists \ \lambda _\nu \in \mathbb {C} ,\ 0 \leqslant \nu \leqslant m-1 , \ \lambda _{m-1}\not = 0 ,\ \text{ such } \text{ that } u = \sum ^{m-1}_{\nu = 0} \lambda _\nu { u}_\nu .$

Proposition 1.4

Let $\{P_n\}_{n\geqslant 0}$ be an OPS satisfying (1.9a)–(1.9b). The sequence $\{P_n\}_{n\geqslant 0}$ is $D_\omega$-classical if and only if it fulfils (1.15a)–(1.15b).

Proof

The proof is similar in spirit to that of [10, Proposition 2.10]. The necessary condition has already been shown. Conversely, suppose that the OPS $\{P_n\}_{n\geqslant 0}$ fulfils (1.15a) with (1.15b).

The action of the functional $v_0$ on both sides of the aforementioned identities gives rise to

$$\begin{aligned} \left\langle v_0 \,,\, P_0 \right\rangle =1 ,\ \left\langle v_0 \,,\, P_1 \right\rangle =\tilde{\alpha }_0^1\,\ \left\langle v_0 \,,\, P_2 \right\rangle =\tilde{\alpha }_0^0 \ \ \text{ and } \ \ \left\langle v_0 \,,\, P_n \right\rangle =0,\ \ \text{ for }\ \ n\geqslant 3. \end{aligned}$$

Application of Lemma 1.3 shows that

$$\begin{aligned} v_0=\lambda _0u_0+\lambda _1u_1+\lambda _2u_2, \end{aligned}$$

(1.17)

with $\lambda _0=1$, $\lambda _1=\tilde{\alpha }_0^1=\beta _{1}-\tilde{\beta }_{0}-\omega$ and $\lambda _2=\tilde{\alpha }_0^0=\gamma _{2}-2\tilde{\gamma }_{1}$.

Now, the use of (1.3) enables us to write $u_1=\left( \left\langle u_0,P^2_1\right\rangle \right) ^{-1}P_1u_0$ and $u_2=\left( \left\langle u_0,P^2_2\right\rangle \right) ^{-1}P_2u_0$.

In (1.17) we replace $u_1$ and $u_2$ by their respective expressions given above, to deduce that there exists a polynomial $\Phi$, with $\deg \Phi \leqslant 2$, such that $v_0=\Phi u_0$.

On the other hand, setting $n=0$ in (1.7), it follows immediately that

$$\begin{aligned} D_{-\omega }\left( v_0\right) =-u_1=-\left( \left\langle u_0, P^2_1\right\rangle \right) ^{-1}P_1u_0:=-\Psi u_0. \end{aligned}$$

Combining these last results, we deduce that

$$\begin{aligned} D_{-\omega }\left( \Phi u_0\right) +\Psi u_0=0,\ \ \text{ with } \ \ \deg \Phi \leqslant 2 \ \ \text{ and } \ \ \deg \Psi =1. \end{aligned}$$

By Proposition 1.2, we easily conclude that the orthogonal polynomials sequence $\{P_n\}_{n\geqslant 0}$ is $D_\omega$-classical, and the proof is complete. $\square$

At the end of this section, let us remember the definition of the shifted polynomials denoted $\{\hat{P}_n\}_{n\geqslant 0}$ corresponding to the PS $\{P_n\}_{n\geqslant 0}$. For all $n, n=0, 1,\ldots ,$ we have

$$\begin{aligned} {\hat{P}}_n(x):=\hat{a}^{-n} P_n(\hat{a}x+\hat{b}),\ \hbox {for} \ (\hat{a}; \hat{b})\in {\mathbb {C}}^*\times {\mathbb {C}}. \end{aligned}$$

(1.18)

Since the classical character of the considered polynomials is preserved by any linear change of the variable, for the OPS $\{P_n\}_{n\geqslant 0}$ satisfying (1.9a)–(1.9b), we obtain that the polynomials ${\hat{P}}_n, {n=0, 1,\ldots }$, satisfy also the second-order recurrence relation

$$\begin{aligned} {{\hat{P}}}_{n+2}(x)= & {} (x - {\hat{\beta }}_{n+1}){{\hat{P}}}_{n+1}(x) -{\hat{\gamma }}_{n+1}{{\hat{P}}}_n (x), \ n\geqslant 0, \end{aligned}$$

(1.19a)

$$\begin{aligned} {{\hat{P}}}_1(x)= & {} x - {\hat{\beta }}_0,\ {{\hat{P}}}_0(x) = 1, \end{aligned}$$

(1.19b)

with

$$\begin{aligned} {\hat{\beta }}_{n}=\frac{{\beta }_{n}-\hat{b}}{\hat{a}},\, n\geqslant 0, \ \ \text{ and }\ \ {\hat{\gamma }}_{n+1} =\frac{{\gamma }_{n+1}}{\hat{a}^2}, \, n\geqslant 0\quad (\hat{a}\ne 0). \end{aligned}$$

(1.20)

Furthermore, from (1.5)–(1.6) we readily see that (1.18) becomes $\hat{P}_n(x) ={\hat{a}}^{-n}\big (h_{\hat{a}}\circ \tau _{-\hat{b}}P_n\big )(x)$.

In addition, if $u_0$ is $D_\omega$-classical, then Eq. (1.13) leads to

$$\begin{aligned} D_{-\frac{\omega }{\hat{a}}}\big (\hat{\Phi }\hat{u}_0\big ) +\hat{\Psi } \hat{u}_0=0, \end{aligned}$$

(1.21)

where $\hat{\Phi }(x)=\hat{a}^{-t}\Phi (\hat{a}x+\hat{b})$, $\hat{\Psi }=\hat{a}^{1-t}\Psi (\hat{a}x+\hat{b})$ and $\hat{u}_0=(h_{\hat{a}^{-1}}\circ \tau _{-\hat{b}})u_0$.

The paper is organized as follows. In the next section we pose and solve two nonlinear systems. The first and most fundamental one relates the recurrence coefficients $\beta _n$, $\gamma _{n+1}$ with $\tilde{\beta }_n$, $\tilde{\gamma }_{n+1}$. The second system combines the coefficients $\alpha _n^i,\ i=0, 1, 2\ ;\ \tilde{\alpha }^j_n,\ j=0, 1$, $\tilde{\beta }_n$ and $\tilde{\gamma }_{n+1}$ with those of the polynomial $\Phi$ (Proposition 2.2). This allows to express the coefficients $\alpha _n^i$, in terms of $\beta _n$ or $\gamma _{n+1}$. In Sect. 3, we investigate the four canonical families of $D_\omega$-classical orthogonal polynomials which we identify after assigning particular values to the free parameters. The last section is devoted to the sequences of higher order derivatives. We give, principally, the explicit expressions of their recurrence coefficients in terms of the coefficients $\big (\beta _n, \gamma _{n+1}\big )_{n\in \mathbb {N}}$.

When $\omega =0$, we rediscover the link between every higher order derivative sequences for the classical polynomials of Hermite, Laguerre, Bessel and Jacobi with each of these families.

2 Computation of the Related Coefficients

In order to compute the various related recurrence coefficients, the first step is to establish the main system connecting the recurrence coefficients $\{\beta _n\}_{n\geqslant 0} , \, \{\gamma _{n+1}\}_{n\geqslant 0}$ with $\{\tilde{\beta }_n\}_{n\geqslant 0}$, $\{\tilde{\gamma }_{n+1}\}_{n\geqslant 0}$. To do this, we proceed as follows. Substituting in (1.9a), $P_{n+2}, P_{n+1}$ and $P_n$ by their expressions provided in (1.15a) we derive a relation in terms of the polynomials $Q_k$ for $k=n+2,\ldots , n-2$. Now, in this new relation we replace $xQ_{n+1}$, $xQ_n$ and $xQ_{n-1}$ by their respective expressions derived from the recurrence relation (1.10a) obtaining an expansion depending only on the polynomials $Q_{n+2},\ldots , Q_{n-2}$.

By rearranging the terms in the resulting expansion and after a few further simplifications, the next system (valid for all $n\geqslant 1$) follows by identification

$$\begin{aligned}&(n+2) \tilde{\beta }_{n} - n \tilde{\beta }_{n-1}= (n+1)\beta _{n+1} -(n-1)\beta _{n}-\omega ;\\&2\tilde{\beta }_0 = \beta _1 +\beta _0 -\omega ,\\&(n + 3) \tilde{\gamma }_{n+1} - (n + 1) \tilde{\gamma }_n \\&\quad \,\,\, = (n + 1)\gamma _{n+2}-(n-1) \gamma _{n+1} + (n + 1) \big (\beta _{n+1} - \tilde{\beta }_n\big ) \big (\beta _{n+1} - \tilde{\beta }_n-\omega \big );\\&3 \tilde{\gamma }_1 = \gamma _2 + \gamma _1 + \big (\beta _1 -\tilde{\beta }_0\big )\big (\beta _1 -\tilde{\beta }_0-\omega \big ),\\&(n+1)\tilde{\gamma }_{n} \big (2\beta _{n+1} - \tilde{\beta }_{n} -\tilde{\beta }_{n-1}-\omega \big ) -n \gamma _{n+1}\big (\beta _{n+1} + \beta _{n} - 2 \tilde{\beta }_{n-1}-\omega \big )=0,\\&(n+2)\tilde{\gamma }_n\tilde{\gamma }_{n+1} -2(n+1) \tilde{\gamma }_n \gamma _{n+2}+n \gamma _{n+1} \gamma _{n+2}=0. \end{aligned}$$

We should mention here the important role played by the Hahn property (Definition 1.1) to establish such a system, since we just assume that these polynomials as well as their $D_\omega$-derivatives are orthogonal w.r.t. regular forms. In other words, we only rely on the fact that such sequences satisfy a second-order recurrence relation, as it is introduced in Sect. 1.

When $\omega =0$, we recover the system initiated and solved by Douak and Maroni [8] whose the solutions provide the classical OPS of Hermite, Laguerre, Jacobi and Bessel, after assigning particular values to the free parameters. We will encounter these families again in this paper.

To solve the above system, let us introduce the auxiliary coefficients $\delta _n$ and $\theta _n$ by writing

$$\begin{aligned} \tilde{\beta }_n= & {} \beta _{n+1} + \delta _n \,,\; n\geqslant 0, \end{aligned}$$

(2.1)

$$\begin{aligned} {\tilde{\gamma }_n}= & {} \frac{n}{n+1}\gamma _{n+1}\theta _n\,,\; n\geqslant 1, \quad \big (\theta _n\not =0\big ). \end{aligned}$$

(2.2)

With these considerations, the two equalities (1.16) take the form

$$\begin{aligned} \tilde{\alpha }_{n}^1=-(n+1)(\delta _n+\omega ),\, \ n\geqslant 0 \,; \ \ \tilde{\alpha }_{n}^0=(n+1)\gamma _{n+2}(1-\theta _{n+1}),\,\ n\geqslant 0.\nonumber \\ \end{aligned}$$

(2.3)

2.1 The Coefficients of the Recurrence Relations

Our main objective here is to initially compute the auxiliary coefficients $\delta _n$ and $\theta _n$, and then give the explicit expressions of the coefficients $\beta _n$ and $\gamma _n$. This in turn allows to determine the coefficients $\tilde{\beta }_n$ and $\tilde{\gamma }_n$ and write significantly better each of the coefficients $\alpha _{n}^i$ and $\tilde{\alpha }_{n}^j$.

Under the formulas (2.1)–(2.2), it is easy to see that the above system can be transformed into

$$\begin{aligned}{} & {} \beta _{n+1} - \beta _{n} = n\delta _{n-1} - (n+2) \delta _{n}-\omega , \ n\geqslant 0,\quad (\delta _{-1}=0), \end{aligned}$$

(2.4)

$$\begin{aligned}{} & {} \big [(n+3) (\theta _{n+1}-1)+1\big ]\frac{\gamma _{n+2}}{n+2} -\big [n(\theta _n -1)+1\big ]\frac{\gamma _{n+1}}{n+1}=\delta _n (\delta _n+\omega ),\ n\geqslant 1,\nonumber \\ \end{aligned}$$

(2.5)

$$\begin{aligned}{} & {} \big (3 \theta _1 - 2) {\gamma _2} - 2\gamma _1 =2\delta _0(\delta _0+\omega ),\end{aligned}$$

(2.6)

$$\begin{aligned}{} & {} \big [(n+3)(\theta _{n}-1)+1\big ]\delta _{n} -\big [(n-1)(\theta _{n}-1)+1\big ]\delta _{n-1} +2(\theta _{n}-1)\omega =0,\ n\geqslant 1,\nonumber \\ \end{aligned}$$

(2.7)

$$\begin{aligned}{} & {} \left( \theta _{n+1}-2\right) \theta _n +1=0,\ n\geqslant 1. \end{aligned}$$

(2.8)

To solve this system, we begin with the Riccati equation (2.8) whose solutions are

$$\begin{aligned} \mathbf{A.}\quad \theta _n&= 1,\ \ n\geqslant 1,\\ \mathbf{B.}\quad \theta _n&= \frac{n + \theta + 1}{n +\theta }, \ \ n\geqslant 1,\quad \theta \ne - 1, -2, \ldots . \end{aligned}$$

Hence, the first three equations must be examined in the light of these solutions.

Case A. For $\theta _n = 1$, the above system reduces to

$$\begin{aligned}{} & {} \beta _{n+1} - \beta _{n} = n\delta _{n-1} - (n+2) \delta _{n} -\omega ,\ n\geqslant 0, \end{aligned}$$

(2.9)

$$\begin{aligned}{} & {} \frac{\gamma _{n+2}}{n+2}- \frac{\gamma _{n+1}}{n+1} =\delta _n(\delta _n+\omega ),\ n\geqslant 0,\end{aligned}$$

(2.10)

$$\begin{aligned}{} & {} \delta _{n+1} - \delta _{n}=0,\ n\geqslant 0. \end{aligned}$$

(2.11)

Equation (2.11) clearly shows that $\delta _n=\delta _0,\ n\geqslant 0$, giving $\delta _n(\delta _n+\omega )=\delta _0 (\delta _0+\omega ),\ n\geqslant 0$.

Thus it is quite natural to single out the two statements $\delta _0(\delta _0+\omega )=0$ and $\delta _0(\delta _0+\omega )\ne 0$. But right now we go back to the two first equations from which we readily deduce that

$$\begin{aligned} \beta _n= & {} \beta _0-(2\delta _0+\omega )n,\ n\geqslant 0,\end{aligned}$$

(2.12)

$$\begin{aligned} \gamma _{n+1}= & {} (n+1)\big (\delta _0(\delta _0+\omega )n+\gamma _1\big ), \ n\geqslant 0. \end{aligned}$$

(2.13)

If we take $\omega =0$, we recover the recurrence coefficients of the Hermite or Laguerre polynomials as shown in [8]. This will be made more precise in the subcases $\textbf{A}_1$ and $\textbf{A}_2$ below.

Case B. For $\theta _n = {(n + \theta + 1)}/{(n +\theta )}$, the system (2.4)–(2.8) becomes

$$\begin{aligned}{} & {} \beta _{n+1} - \beta _{n} = n\delta _{n-1} - (n+2) \delta _{n}-\omega ,\ n\geqslant 0, \end{aligned}$$

(2.14)

$$\begin{aligned}{} & {} \frac{(2n+\theta +4)}{(n+\theta +1)} \frac{\gamma _{n+2}}{n+2}-\frac{(2n+\theta )}{(n+\theta )} \frac{\gamma _{n+1}}{n+1} = \delta _n (\delta _n+\omega ),\ n\geqslant 0, \end{aligned}$$

(2.15)

$$\begin{aligned}{} & {} \big (2n+\theta +3\big )\delta _{n} -\big (2n+\theta -1\big )\delta _{n-1}=-2\omega ,\ n\geqslant 1, \end{aligned}$$

(2.16)

unless, of course, $\theta$ happen to be zero for the index $n = 0$. So, we will first discuss the solution of the above system when $\theta \ne 0$. The case $\theta = 0$ is special, it will be considered separately.

When $\omega =0$, with appropriate choices of the parameters, the only orthogonal polynomials obtained as solutions of this problem are those of Bessel and Jacobi (see [8] for more details).

We now return to seeking solutions for the equations (2.14)–(2.16). Observe first that the RHS of Eq. (2.15) vanishes if and only if $\delta _n=-\omega$ or $\delta _n=0$. Each of these is possible.

It is easy to check that the former statement is dismissed, since it contradicts Equality (2.16). For the latter, if we replace $\delta _n=0$ in (2.16), we immediately see that this leads to $\omega =0$.

Straightforwardly from Eqs. (2.14) and (2.15), one has

$$\begin{aligned} \beta _n=\beta _0\ ; \ \gamma _{n+1}=\gamma _1 \frac{(\theta +2)(n+1)(n+\theta )}{(2n+\theta +2)(2n+\theta )}, \ n\geqslant 0. \end{aligned}$$

(2.17)

If we set $\theta =2\lambda$ and make a linear transformation with the choice

$\hat{a}^2=2(\lambda +1)\gamma _1\ ; \;\ \hat{b}=\beta _0$, then

$$\begin{aligned} \hat{\beta }_n=0 \ ; \ \hat{\gamma }_{n+1}=\frac{(n+1)(n+2\lambda )}{(n+\lambda +1)(n+\lambda )}, \ n\geqslant 0. \end{aligned}$$

(2.18)

We thus meet the Gegenbauer polynomials which will reappear again in Subcase $\textbf{B}_{21}$.

From now on we assume that $\delta _n(\delta _n+\omega )\ne 0,\ n\geqslant 0.$ Starting from Eq. (2.16), multiply both sides by $2n+\theta +1$, after summation we get

$$\begin{aligned} \delta _n=\frac{\delta _0(\theta +3)(\theta +1)-2\omega n(n+\theta +2)}{(2n+\theta +3)(2n+\theta +1)},\, n\geqslant 0. \end{aligned}$$

(2.19)

Use a division to obtain

$$\begin{aligned} \delta _{n}= & {} \frac{2\mu }{\big (2n+\theta +3\big )\big (2n+\theta +1\big )} -\frac{1}{2}\omega ,\, n\geqslant 0, \end{aligned}$$

(2.20a)

$$\begin{aligned} \delta _{n}= & {} \mu \big (\vartheta _{n}-\vartheta _{n+1}\big ) -\frac{1}{2}\omega ,\, n\geqslant 0, \end{aligned}$$

(2.20b)

where we have written $\mu :=\frac{1}{4}(2\delta _0+\omega )(\theta +3)(\theta +1)$ and $\vartheta _n =\left( 2n+\theta +1\right) ^{-1},\ n\geqslant 0$.

Thanks to the identity (2.20b), we can write

$$\begin{aligned} \left( 2n+\theta +2\right) \delta _n(\delta _n+\omega ) =\mu ^2\big ({\vartheta _{n}^2}-{\vartheta _{n+1}^2}\big ) -\frac{1}{4}{\omega ^2}(2n+\theta +2),\ n\geqslant 0.\nonumber \\ \end{aligned}$$

(2.21)

The objective of course is to incorporate this new expression into (2.15) to derive the coefficients $\gamma _n,\ n\geqslant 1$, which will in fact be processed in a next step. We first calculate the coefficients $\beta _n$. For this, observe that the RHS of (2.14) may be rewritten using Eq. (2.16) in the form

$$\begin{aligned} n\delta _{n-1} - (n+2)\delta _{n}-\omega =\frac{1}{2}(\theta -1)(\delta _{n} - \delta _{n-1}),\ n\geqslant 1. \end{aligned}$$

(2.22)

It is easily seen that, if $\theta$ assumes the value 1, Eq. (2.22) provides $\delta _n=-\frac{1}{2}\omega ,\, n\geqslant 0$.

As a straightforward consequence of this last result, one sees immediately that (2.14), (2.15) respectively provides

$$\begin{aligned} \beta _{n}= & {} \beta _0,\ n\geqslant 0, \end{aligned}$$

(2.23)

$$\begin{aligned} \gamma _{n+1}= & {} -\frac{1}{4}\frac{(n+1)^2\left( \omega ^2 n(n+2)-12\gamma _1\right) }{(2n+3)(2n+1)},\ n\geqslant 0. \end{aligned}$$

(2.24)

When $\omega =0$, under the transformation $\hat{a}^2=3\gamma _1, \hat{b}=\beta _0$, we meet the Legendre polynomials.

We now turn to the case $\theta \ne 1$. Using the identity (2.22), Eq. (2.14) gives rise to

$$\begin{aligned} \beta _{n+1}-\beta _n= & {} \frac{1}{2}(\theta -1) (\delta _n-\delta _{n-1}),\ n\geqslant 1, \end{aligned}$$

(2.25)

$$\begin{aligned} \beta _1-\beta _0= & {} -(2\delta _0+\omega ), \end{aligned}$$

(2.26)

with $\delta _n$ is given by (2.19). From this, it may be concluded that

$$\begin{aligned} \beta _{n}=\beta _0-\frac{(2\delta _0+\omega )(\theta +3)n (n+\theta )}{(2n+\theta +1)(2n+\theta -1)},\ n\geqslant 0. \end{aligned}$$

(2.27)

We can now proceed to compute the coefficients $\gamma _{n+1}, n\geqslant 0$. To do so, multiply both sides of Eq. (2.15) by $2n+\theta +2$ and set

$$\begin{aligned} \Theta _{n+1}=\frac{\left( 2n+\theta +2\right) (2n+\theta )}{(n+\theta )} \frac{\gamma _{n+1}}{n+1},\ n\geqslant 0. \end{aligned}$$

(2.28)

Then, taking into consideration (2.21), we easily check that (2.15) takes the form

$$\begin{aligned} \Theta _{n+2}- \Theta _{n+1} =\mu ^2 \big ({\vartheta _{n}^2}-{\vartheta _{n+1}^2}\big ) -\frac{1}{4}{\omega ^2}(2n+\theta +2), \, n\geqslant 0. \end{aligned}$$

By summation, we deduce that

$$\begin{aligned} \Theta _{n+1}=\Theta _1+\mu ^2\big (\vartheta _0^2-\vartheta _n^2\big ) -\frac{1}{4}{\omega ^2}n(n+\theta +1),\ n\geqslant 0. \end{aligned}$$

(2.29)

Substituting (2.28) into (2.29) yields

$$\begin{aligned} \gamma _{n+1}=-\frac{(n + 1)(n + \theta ) \Big \{\big [\frac{1}{4}\omega ^2n\left( n+\theta + 1\right) -\left( \mu ^2\vartheta _0^2 + (\theta + 2) \gamma _1\right) \big ](2n+\theta +1)^2 + \mu ^2\Big \}}{(2n+\theta +2) (2n+\theta +1)^2(2n+\theta )}.\nonumber \\ \end{aligned}$$

(2.30)

It is possible to write the expression between braces in the numerator of (2.30) in the form

$$\begin{aligned} \big (\omega n(n+\theta +1)+\varrho n+\rho _1\big ) \big (\omega n(n+\theta +1)-\varrho n+\rho _2\big ), \end{aligned}$$

where the three parameters $\rho _1$, $\rho _2$ and $\varrho$ are such that

$$\begin{aligned}{} & {} (\theta +1)\varrho ^2+(\rho _2-\rho _1)\varrho =0, \end{aligned}$$

(2.31)

$$\begin{aligned} {}{} & {} {}\varrho ^2-(\rho _2+\rho _1)\omega =\big ((\theta +3)\delta _0 +(\theta +2)\omega \big )\big ((\theta +3)\delta _0+\omega \big )+4(\theta +2)\gamma _1,\nonumber \\ \end{aligned}$$

(2.32)

$$\begin{aligned}{} & {} \rho _2\rho _1=-(\theta +1)^2(\theta +2)\gamma _1. \end{aligned}$$

(2.33)

The roots of the quadratic equation (2.31) are $\varrho =0$ and $\varrho ={(\rho _1-\rho _2)}/{(\theta +1)}$ for $\rho _2\ne \rho _1$, with the root $\varrho =0$ being double, if $\rho _2=\rho _1$. The last equation clearly shows that $\rho _2\rho _1\ne 0$.

All these parameters will be well specified when dealing with the canonical families. But, in any way, we have to consider the following two cases.

1.:

For $\varrho =0$, we obtain

$$\begin{aligned} \gamma _{n+1}=-\frac{(n+1)(n+\theta )\big (\omega n(n+\theta +1)+\rho _1\big )\big (\omega n(n+\theta +1)+\rho _2\big )}{(2n+\theta +2)(2n+\theta +1)^2(2n+\theta )},\ n\geqslant 0.\nonumber \\ \end{aligned}$$

(2.34)

In the particular case $\rho _2=\rho _1:=\rho$, (2.34) simplifies to

$$\begin{aligned} \gamma _{n+1}=-\frac{(n+1)(n+\theta )\big (\omega n(n+\theta +1)+\rho \big )^2}{(2n+\theta +2)(2n+\theta +1)^2(2n+\theta )},\ n\geqslant 0. \end{aligned}$$

(2.35)

2.:

For the general case $\varrho \ne 0$, we have

$$\begin{aligned} \gamma _{n+1}=-\frac{(n + 1)(n+\theta )\big (\omega n(n+\theta + 1)+\varrho n+\rho _1\big )\big (\omega n(n+\theta + 1)-\varrho n+\rho _2\big )}{(2n+\theta +2)(2n+\theta +1)^2(2n+\theta )},\, n\geqslant 0.\nonumber \\ \end{aligned}$$

(2.36)

We now turn to the special case $\theta =0$. Substituting this in (2.4)–(2.7) yields

$$\begin{aligned}&\beta _{n+1} - \beta _{n} = n\delta _{n-1} -(n+2) \delta _{n}-\omega ,\ n\geqslant 0,\\&\gamma _{n+2}- \gamma _{n+1} = \frac{1}{2}(n+1)\delta _n(\delta _n+\omega ),\ n\geqslant 1,\\&{\gamma _2} - \frac{1}{2}\gamma _1 = \frac{1}{2}\delta _0(\delta _0+\omega ),\\&(2n+3)\delta _{n} - (2n-1)\delta _{n-1}=-2\omega ,\ n\geqslant 1. \end{aligned}$$

The same reasoning applies to this case gives

$$\begin{aligned} \delta _{n}&= \frac{3\delta _0-2\omega n(n+2)}{(2n+3)(2n+1)},\ n\geqslant 0,\\ \beta _{n}&=\beta _0-\frac{3(2\delta _0+\omega )n^2}{(2n+1)(2n-1)},\ n\geqslant 0,\\ \gamma _{n+1}&=-\frac{\big (\omega n(n+1)+\tau n+\tau _1\big )\big (\omega n(n+1)-\tau n+\tau _2\big )}{4(2n +1)^2},\, n\geqslant 1, \end{aligned}$$

where $\tau _1$, $\tau _2$ and $\tau$ are such that

$$\begin{aligned}&\tau ^2+(\tau _2-\tau _1)\tau =0,\\ {}&\tau ^2-(\tau _2+\tau _1)\omega =(3\delta _0+2w) (3\delta _0+\omega )+8\gamma _1,\\ {}&\tau _2\tau _1=-2\gamma _1. \end{aligned}$$

When $\omega =0$, if moreover $\delta _0=0$, which we may assume, it follows that

$$\begin{aligned} \delta _{n}= 0, \ n\geqslant 0, \ \beta _{n}=\beta _0, \ n\geqslant 0, \ \gamma _{n+1}= \frac{1}{2}\gamma _1,\, n\geqslant 1. \end{aligned}$$

Thus, choosing $\beta _0=0$ and $\gamma _1=\frac{1}{2}$, we meet the Tchebychev polynomials of the first kind.

After having finished solving the first system, we now proceed to the determination of the coefficients $\tilde{\alpha }_{n}^j$ and $\alpha _{n}^j$ in terms of $\beta _{n}$ and $\gamma _{n+1},\, n\geqslant 0$. This will be done in the next subsection.

2.2 The Coefficients of the Structure Relations

Proposition 2.1

Let $\Phi$ be the polynomial arising in Proposition 1.2. We let the degree of $\Phi$ to be two 2 and write $\Phi (x)=a_2x^2+a_1x+a_0$. Then the coefficients implicated in the two structure relations (1.12) and (1.15a)–(1.15b) are interlinked through the following system

$$\begin{aligned}{} & {} \alpha _{n+2}^2=a_2,\ n\geqslant 0, \end{aligned}$$

(2.37)

$$\begin{aligned}{} & {} \alpha _{n+1}^1 +a_2\tilde{\alpha }_{n+1}^1=a_2 (\tilde{\beta }_{n+1}+\tilde{\beta }_n)+a_1,\ n\geqslant 0,\end{aligned}$$

(2.38)

$$\begin{aligned}{} & {} \alpha _{n+1}^1\tilde{\alpha }_{n}^1 +a_2\tilde{\alpha }_{n}^0 +\alpha _{n}^0=a_2(\tilde{\gamma }_{n+1}+\tilde{\gamma }_{n} +\tilde{\beta }_n^2)+a_1\tilde{\beta }_n+a_0, \ n\geqslant 0,\nonumber \\\end{aligned}$$

(2.39)

$$\begin{aligned}{} & {} \alpha _{n+1}^1\tilde{\alpha }_{n-1}^0+\alpha _{n}^0\tilde{\alpha }_{n-1}^1 =a_2\tilde{\gamma }_{n}(\tilde{\beta }_{n}+\tilde{\beta }_{n-1}) +a_1\tilde{\gamma }_{n},\ n\geqslant 1,\end{aligned}$$

(2.40)

$$\begin{aligned}{} & {} \alpha _{n+1}^0\tilde{\alpha }_{n-1}^0=a_2\tilde{\gamma }_{n+1} \tilde{\gamma }_{n}, \ n\geqslant 1, \end{aligned}$$

(2.41)

where we have adopted the convention that $\tilde{\gamma }_0:=0$ so that (2.39) remains valid for $n = 0$.

Proof

As for the preceding system, we give only the main ideas of the proof. First, comparison of coefficients in (1.12) shows that $\alpha _{n+2}^2=a_2$. Now, use the recurrence relation (1.10a) twice to write the product $\Phi (x) Q_{n}$, which is the LHS of (1.12), in terms of the polynomials $Q_{n+2},\ldots , Q_{n-2}$. Then replace in the RHS of (1.12) $P_{n+2}, P_{n+1}$ and $P_{n}$ by their expressions provided in (1.15a) to obtain another expansion in terms of the polynomials $Q_{n+2},\ldots , Q_{n-2}$. After some simplifications, and by identification the equations (2.38)–(2.41) follow. $\square$

As far as the author knows, the technique used here to find explicit expressions for the coefficients involving in (1.15a) with (1.15b) is new, and the results obtained still unknown. So, the solution of the above system brings an answer to this question. But before doing so, recall that when the form $u_0$ is $D_\omega$-classical, namely, it is regular and satisfies (1.13), the polynomials $\Phi$ and $\Psi$ necessarily satisfy (see [1, p. 7] for further details):

$$\begin{aligned}{} & {} \kappa \Phi (x)=(1-\theta _1)x^2-\bigl ((1-\theta _1)(\beta _1 +\beta _0)+\delta _0+\omega \bigr )x + \bigl ((1-\theta _1)\beta _1 \nonumber \\{} & {} \qquad \qquad \quad +\delta _0+\omega \bigr )\beta _0+\theta _1\gamma _1, \end{aligned}$$

(2.42)

$$\begin{aligned}{} & {} \kappa \Psi (x)= P_1(x)=x-\beta _0. \end{aligned}$$

(2.43)

Note that the expression in the RHS of (2.42) is slightly transformed in accordance with the relations (2.1) and (2.2). The coefficient $\kappa$ is to be chosen later so that $\Phi (x)$ is being monic.

Since we are proceeding following the two situations Case A and Case B, we see that the leading coefficient $a_2$ of $\Phi (x)$ assumes the value 0, when $\theta _1=1$, or is such that $\kappa a_2= -(\theta +1)^{-1}$, when $\theta _1=(\theta +2)/(\theta +1)$. To get $a_2=1$ in the latter case, we choose $\kappa = -(\theta +1)^{-1}$ which, in turn, allows to write the polynomial $\Phi (x)$ in one of the four standard forms

$$\begin{aligned} \Phi (x)=1,\ \Phi (x)=x,\ \Phi (x)= x^2\ \text{ and }\ \Phi (x)=(x+1)(x-c),\ c\in {\mathbb {C}}\backslash \{-1\}. \end{aligned}$$

The classification achieved according to the degree of $\Phi$ as in [1] is of course exhaustive and is equivalent to that based on the values of $\theta _n$. It is this second alternative that we will retain in the next section to go over the diverse families of $D_\omega$-classical orthogonal polynomials or some relevant cases. But before doing so, let us discuss the solutions of the system (2.37)–(2.41) when $a_2$ takes one of the values 0 or 1, with use of (2.3).

I. For $a_2=0$, since $\alpha _{n}^0\ne 0$ for each n, Eq. (2.41) readily gives $\tilde{\alpha }_{n}^0=0, n\geqslant 0$, so, due to (2.3), we necessarily have $\theta _{n}=1$ for all $n\geqslant 1$. It turns out that $\delta _n=\delta _0, n\geqslant 0$, and so $\tilde{\alpha }_{n}^1=-(n+1)(\delta _0+\omega ), n\geqslant 0$. In this case, it is easily seen that the coefficients $\beta _n$ and $\gamma _{n+1}$ are given by (2.12)–(2.13).

This actually happens in the case A when the polynomial $\Phi$ takes the form

$$\begin{aligned} \kappa \Phi (x) =-(\delta _0+\omega )x+(\delta _0+\omega )\beta _0+\gamma _1. \end{aligned}$$

(2.44)

Using the fact that the polynomial $\Phi (x)$ is monic, we are brought to consider the following two situations according with the degree of this polynomial.

(i):

$\Phi (x)$ is constant. In this case we have $\delta _0+\omega =0$ which leads to $a_2=a_1=0$ and $a_0=1$. It follows that $\kappa =\gamma _1$, $\Phi (x)=1$ and $\Psi (x)=\gamma _1^{-1}P_1(x)$. Therefore $\tilde{\alpha }_{n}^1=0,\ n\geqslant 0$, and the above system readily gives $\alpha _{n+2}^2=\alpha _{n+1}^1=0\ ; \ \alpha _{n}^0=1, \, n\geqslant 0.$ Thus, $Q_n=P_n,\, n\geqslant 0$, and so the two structure relations coincide.

(ii):

$\Phi (x)$ is linear. By setting $\kappa =-(\delta _0+\omega )\ne 0$ and $(\delta _0+\omega )\beta _0+\gamma _1=0$, we conclude that $a_2=a_0=0$ and $a_1=1$ and so $\Phi (x)=x$ and $\Psi (x)=\gamma _1^{-1}\beta _0P_1(x)$. We thus have

$$\begin{aligned} \alpha _{n+2}^2=0; \ \alpha _{n+1}^1=1\ \text{ and } \ \alpha _{n}^0=\beta _0-{\delta }_0{n},\, n\geqslant 0. \end{aligned}$$

Accordingly, the two structure relations may be written as follows

$$\begin{aligned} xQ_n(x)&=P_{n+1}(x)+(\beta _0-\delta _0 n)P_{n}(x),\, n\geqslant 0,\\ P_n(x)&=Q_n(x)-(\delta _0+\omega )n Q_{n-1}(x),\, n\geqslant 0,\ (Q_{-1}:=0). \end{aligned}$$

For $n=1$ in the first relation, we recover the equality $(\delta _0+\omega )\beta _0+\gamma _1=0$, that is, $\kappa \beta _0=\gamma _1$. This interconnection between $\beta _0$ and $\gamma _1$ will prove useful in Subcase $\textbf{A}_{2}$.

II. For $a_2\ne 0$, $\Phi (x)$ is then quadratic and so $\kappa =1-\theta _1=-(\theta +1)^{-1}$, since we take $a_2$ to be 1.

Changing n into $n+1$ in (2.41) yields $\alpha _{n+2}^0\tilde{\alpha }_{n}^0\ne 0, \, n\geqslant 0$, from which we see that the coefficients $\displaystyle \tilde{\alpha }_{n}^0$ are not identically 0 for all n, and hence, due to (2.3), we conclude that $\theta _{n}\ne 1,\ n\geqslant 1$. This in fact shows that the case B is the one to be naturally considered here. We thus have

$$\begin{aligned} \tilde{\alpha }_{n}^1=-(n+1)(\delta _n+\omega )\ \ \text{ and }\ \ \tilde{\alpha }_{n}^0=-\frac{n+1}{n+\theta +1}\gamma _{n+2}, \ n\geqslant 0. \end{aligned}$$

Additionally, the coefficients $\beta _n$ are given by (2.31), while the $\gamma _{n}$ are generated either by (2.34) or by (2.36). Moreover, taking into account the position of the zeros of the polynomial $\Phi$, we have to consider again two subcases .

(i):

$\Phi (x)=x^2$. Since $a_2=1$ and $a_1= a_0=0$, (2.42) leads to $\beta _1+\beta _0=(\theta +1)(\delta _0+\omega )$ and $\beta _0^2=-(\theta +2)\gamma _1$. Analogously to the former case, taking into consideration (2.3), the system (2.37)–(2.41) readily provides

$$\begin{aligned} \alpha _{n+2}^2&=1\ ;\ \alpha _{n+1}^1=\beta _{n+1}+\beta _n+n(\delta _{n-1}+\omega )\ \ \text { and }\\ \alpha _{n}^0&=-\frac{n+\theta +1}{n+1}\gamma _{n+1},\, n\geqslant 0. \end{aligned}$$

(ii):

$\Phi (x)=(x+1)(x-c),\ c\ne -1$. We have $a_2=1$, $a_1=1-c$ and $a_0=-c$. Therefore, (2.42) gives rise to $\beta _1+\beta _0=(\theta +1)(\delta _0+\omega )+c-1$ and $(\beta _0+1)(\beta _0-c)=-(\theta +2)\gamma _1$. In the same manner we can see that

$$\begin{aligned} \alpha _{n+2}^2&=1\ ; \ \alpha _{n+1}^1=\beta _{n+1}+\beta _n+n(\delta _{n-1}+\omega )-c+1\ \, \text{ and } \\\alpha _{n}^0&=-\frac{n+\theta +1}{n+1}\gamma _{n+1}, \, n\geqslant 0. \end{aligned}$$

We could now proceed to present an exhaustive classification of the resulting polynomials. The sequences identified are, of course, consistent with those found in [1] and are pointed out with some of their special cases. They will be regarded as the canonical (or representatives) families. To provide these results, depending on certain parameters, we review all situations occurring in either of the cases A and B as we will see in the next section.

3 The Canonical Families of $D_\omega$-Classical Polynomials

In the sequel we are only interested in regular OPS, the finite sequences are not considered here. Under some restrictions on the parameters, we rediscover the well-known families of discrete classical orthogonal polynomials or some particular cases. This is achieved on account of the specific conditions observed in Sect. 2.2. For each situation, we summarise the relevant properties of the corresponding family of polynomials. We often use the linear transformation (1.18) with (1.20) to provide the desired results.

$\triangleright$ Case A. We first investigate the two main subcases, namely, $\delta _0+\omega =0$ and $\delta _0+\omega \ne 0$. Then, we consider the particular subcase when $\delta _0$ assumes the value 0.

$\textbf{A}_1:$ $\delta _0+\omega =0$. From (2.12)–(2.13), taking into account (2.1)–(2.2), we immediately obtain $\tilde{\beta }_n=\beta _n=\beta _0+\omega n\ ; \ \tilde{\gamma }_{n+1}=\gamma _{n+1}=\gamma _1(n+1),\ n\geqslant 0$. It follows that $Q_n=P_n,\, n\geqslant 0$, and hence the PS $\{P_n\}_{n\geqslant 0}$ belongs to the class of the so-called $D_\omega$-Appell sequences.

${\textbf {A}}_{\textbf {1a}}.$ With the choice $\hat{a}^2=2\gamma _1$ and $\hat{b}=\beta _0$, we easily get $\hat{\beta }_n=\frac{\omega }{\hat{a}}n\ ;\ \hat{\gamma }_{n+1}=\frac{1}{2}(n+1),\ n\geqslant 0$. Now, replacing $\omega$ by $\hat{a}\omega$, we obtain $\hat{\beta }_n={\omega }n\ ;\ \hat{\gamma }_{n+1}=\frac{1}{2}(n+1),\ n\geqslant 0$.

When $\omega =0$, and so $\delta _0=0$, we meet the Hermite polynomials.

${\textbf {A}}_{\textbf {1b}}.$ The choice $\hat{a}=\omega$ and $\hat{b}=\beta _0-\omega a$ with $\gamma _1=a\omega ^2$ gives $\hat{\beta }_n=a+n\ ; \;\ \hat{\gamma }_{n+1}=a(n+1),\ n\geqslant 0$. We thus encounter the Charlier polynomials [5].

${\textbf {A}}_{\textbf {2}}:$ $\delta _0+\omega \ne 0$. We can assume without loss of generality that $\delta _0(\delta _0+\omega )=1\Leftrightarrow \omega =\delta _0^{-1}-\delta _0$. Use of (2.12)–(2.13) then shows that $\; {\beta }_n=\beta _0-(\delta _0^{-1}+\delta _0)n\ ; \ {\gamma }_{n+1}=(n+1)\big (n+\gamma _1\big ),\ n\geqslant 0,$ where the parameters $\beta _0$ and $\gamma _1$ are related via $\beta _0=-\delta _0^{-1}\gamma _1$ as in the statement I-(ii) above.

By setting $\gamma _1:=\alpha +1$, we can write $\beta _0=-\delta _0^{-1}(\alpha +1)$.

If we take now $\omega =0$ which yields $\delta _0^2=1$, we recover the Laguerre polynomials for $\delta _0=-1$, and the shifted Laguerre polynomials for $\delta _0=1$.

For $\delta _0=-1$, we have $\ {\beta }_n=2n+\alpha +1\ ; \ {\gamma }_{n+1}=(n+1)\big (n+\alpha +1\big ),\ n\geqslant 0.$

From now on, we assume that $\delta _0\ne -1$ and set $\gamma _1:=\alpha +1$ again. We wish to examine two interesting situations already investigated in [1] with slight differences in notation.

The first one occurs for $\delta _0:=-e^{-\varphi }, \varphi \ne 0$, so that $\omega =-2\sinh \varphi$. It follows that

$$\begin{aligned} {\beta }_n=e^{\varphi }(\alpha +1)+2n \cosh {\varphi }\ ; \ {\gamma }_{n+1}=(n+1)(n+\alpha +1),\ n\geqslant 0. \end{aligned}$$

Afterwards, choose $\hat{a}=\omega \ ;\ \hat{b}=0$ and put $c:=e^{2\varphi }$, to get

$$\begin{aligned} \hat{\beta }_n={\frac{c}{1-c}}(\alpha +1)+\frac{1+c}{1-c}n\ ; \ \hat{\gamma }_{n+1}=\frac{c}{(1-c)^2}(n+1)(n+\alpha +1),\ n\geqslant 0. \end{aligned}$$

When the parameter $c\in \mathbb {R}\backslash \{0,1\}$, we obtain the Meixner polynomials of the first kind [22]. The Krawtchouk polynomials are a special case of the Meixner polynomials of the first kind.

The second situation appears for $\delta _0:=e^{i\phi },\ 0<\phi <\pi$ by taking $2\lambda =\gamma _1:=\alpha +1$. After making a linear transformation via the changes $\hat{a}=i\omega \ ;\ \hat{b}=-\lambda \omega$, we obtain that

$$\begin{aligned} \hat{\beta }_n=-(n+\lambda )\cot \phi \ ; \ \hat{\gamma }_{n+1} =\frac{1}{4}\frac{(n+1)(n+2\lambda )}{\sin ^2\phi },\ n\geqslant 0. \end{aligned}$$

From this, we conclude that the resulting polynomials are those of Meixner–Pollaczek [25].

${\textbf {A}}_{\textbf {3}}:$ For $\delta _0=0$, one sees that (2.12)–(2.13) yields $\beta _n=\beta _0-\omega n\ ;\ \gamma _{n+1}=\gamma _1(n+1),\ n\geqslant 0$. From this, taking into consideration (2.1)–(2.2), we check at once that $\tilde{\beta }_n=\beta _n-\omega$ and $\tilde{\gamma }_{n+1}=\gamma _{n+1},\ n\geqslant 0$.

In accordance with (1.18), it is easily seen that $Q_n(x)=\tau _{-\omega }P_n(x)=P_n(x+\omega )$ for all $n\geqslant 0$.

Table 1 Description of the OPS in Case A

It is worth pointing out the following two subcases for which we can proceed analogously to the generation of $\textbf{A}_{1a}$ and $\textbf{A}_{1b}$. This case could therefore be regarded as a subcase of $\textbf{A}_1$.

${\textbf {A}}_{\textbf {3a}}$.:

Similarly to $\textbf{A}_{1a}$, the choice $\hat{a}^2=2\gamma _1\ ;\ \hat{b}=\beta _0$ yields $\hat{\beta }_n=-\frac{\omega }{\hat{a}}n\ ;\ \hat{\gamma }_{n+1}=\frac{1}{2}(n+1),\ n\geqslant 0$. Replacing $\omega$ by $-\hat{a}\omega$, we get $\hat{\beta }_n={\omega }n\ ;\ \hat{\gamma }_{n+1}=\frac{1}{2}(n+1), \ n\geqslant 0$, which for $\omega =0$ gives rise to the Hermite polynomials.

${\textbf {A}}_{\textbf {3b}}$.:

If we take $\hat{a}=-\omega \ ;\ \hat{b}=\beta _0+\omega a$ with $\gamma _1=a\omega ^2$, we get the Charlier polynomials again. Note that, with $\hat{a}=i\omega \ ;\ \hat{b}=\beta _0+i\omega b$ and $\gamma _1=a\omega ^2$, with $a\ne 0$ and b an arbitrary constant, a specific case relative to this situation have been mentioned in [1], where $\hat{\beta }_n=b+in\ ; \ \hat{\gamma }_{n+1}=-a(n+1),\ n\geqslant 0$.

Note that the resulting polynomials encountered here are already identified in Subcases $\textbf{A}_{1a}$ and $\textbf{A}_{1b}$. All the polynomials obtained in this first case are summarized below in Table 1.

$\triangleright$ Case B. Two main situations will be also investigated with some of their special subcases. The remarks referred to in the statements II-(i) and II-(ii) above must of course be taken into account to choose more precise certain parameters involved in the recurrence coefficients.

In what follows, unless otherwise stated, we assume that $\theta \ne 1$.

$\textbf{B}_1$: $\theta =2\alpha -1$. From (2.27) and (2.30), if we choose $\beta _0=\frac{1}{2}\alpha \omega -\mu _1$, we get

$$\begin{aligned} \beta _{n}= & {} \frac{1}{2}\alpha \omega -\frac{\alpha (\alpha -1) \mu _1}{(n+\alpha )(n+\alpha -1)},\ n\geqslant 0, \end{aligned}$$

(3.1)

$$\begin{aligned} \gamma _{n+1}= & {} -\frac{(n+1)(n+2\alpha -1)\big (\omega (n+\alpha )^2 -2\alpha \mu _1\big )^2}{(2n+2\alpha +1)(2n+2\alpha )^2(2n+2\alpha -1)}, \ n\geqslant 0, \end{aligned}$$

(3.2)

where we have set $\mu _1:=-\frac{1}{2}(\alpha +1)(2\delta _0+\omega )$.

Note that (3.1) is valid for $n= 0$, except that it becomes worthless if concurrently $\alpha =1$.

For $\alpha =\frac{1}{2}$, and so $\theta =0$, the coefficients $\beta _{n}$ and $\gamma _{n+1}$ coincide with those previously established in Sect. 2.1.

$\textbf{B}_{11}.$ For $\mu _1\ne 0$, the choice $\hat{a}=\alpha \mu _1=1\ ;\ \hat{b}=0$, leads to

$$\begin{aligned} \hat{\beta }_{n}= & {} \frac{1}{2}\alpha \omega +\frac{1-\alpha }{(n+\alpha ) (n+\alpha -1)},\ n\geqslant 0, \end{aligned}$$

(3.3)

$$\begin{aligned} \hat{\gamma }_{n+1}= & {} -\frac{(n+1)(n+2\alpha -1) \big (\omega (n+\alpha )^2-2\big )^2}{4(2n+2\alpha +1) (n+\alpha )^2(2n+2\alpha -1)},\ n\geqslant 0. \end{aligned}$$

(3.4)

When $\omega =0$, we clearly encounter the Bessel polynomials.

$\textbf{B}_{12}.$ For $\omega \ne 0$ with the choice $\hat{a}=i\omega \ ; \ \hat{b}=\frac{1}{2}\alpha \omega$, another specific case was discovered in [1]. If on top of that we take $\mu _1=0$ and $\alpha >0$, the corresponding orthogonal polynomials are symmetric and their associated form is positive definite.

$\textbf{B}_2:$ $\theta =\nu -1$. Let $c\in {\mathbb {C}}-\{-1\}$, thus from and (2.36), we obtain

$$\begin{aligned} \beta _{n}= & {} \frac{1}{4}\nu \omega -\frac{1}{2}(1-c) +\frac{\nu (\nu -2)\mu _2}{(2n+\nu )(2n+\nu -2)},\ n\geqslant 0, \end{aligned}$$

(3.5)

$$\begin{aligned} \gamma _{n+1}= & {} -\frac{(n+1)(n+\nu -1)}{(2n+2\nu +1) (2n+2\nu )^2(2n+2\nu -1)}\times \nonumber \\ {}{} & {} \big [\omega n(n+\nu )+(1+c)n+\nu (\beta _0+1)\big ] \big [\omega n(n+\nu )-(1+c)n+\nu (\beta _0-c)\big ],\ n\geqslant 0, \nonumber \\ \end{aligned}$$

(3.6)

where we have put $\mu _2:=\frac{1}{4}(\nu +2)(2\delta _0+\omega )$ and chosen $\beta _0:=\frac{1}{4}\nu \omega -\frac{1}{2}(1-c)+\mu _2$.

$\textbf{B}_{21}.$ For $\omega =0$ and $c=1$, we may write $\nu (1+\beta _0)=2(\alpha +1)$ and $\nu (1-\beta _0)=2(\beta +1)$, so that

$$\begin{aligned} \nu =\alpha +\beta +2\ \ \text{ and }\ \ \beta _0=\frac{\alpha -\beta }{\alpha +\beta +2}. \end{aligned}$$

Hence, we rediscover the Jacobi polynomials whose coefficients are

$$\begin{aligned} \beta _{n}&=\frac{\alpha ^2-\beta ^2}{(2n+\alpha +\beta +2)(2n+\alpha +\beta )}, \ n\geqslant 0,\\ \gamma _{n+1}&=4\frac{(n+1)(n+\alpha +\beta +1) (n+\alpha +1)(n+\beta +1)}{(2n+\alpha +\beta +3) (2n+\alpha +\beta +2)^2(2n+\alpha +\beta +1)},\ n\geqslant 0. \end{aligned}$$

For $\alpha =\beta =\lambda -\frac{1}{2}$, we meet again the Gegenbauer polynomials.

$\textbf{B}_{22}.$ When $\omega \ne 0$, we first write the two expressions between square brackets in (3.6) as follows

$$\begin{aligned} \omega n(n+\nu )+(1+c)n+\nu (\beta _0+1)&=\big [\omega (n+\beta +1) +(1+c)\big ](n+\alpha +1),\\ \omega n(n+\nu )-(1+c)n+\nu (\beta _0-c)&=\big [\omega (n+\alpha +1) -(1+c)\big ](n+\beta +1). \end{aligned}$$

If we set $\eta :=\alpha -{(1+c)}/{\omega }$ and apply a linear transformation with $\hat{a}=i\omega \; ;\;\hat{b}=\frac{1}{4}\nu \omega -\frac{1}{2}(1-c)$, we may rewrite (3.5)–(3.6) as

$$\begin{aligned}{} & {} \hat{\beta }_{n}= \frac{1}{2}i(\alpha ^2-\beta ^2) \frac{\eta -\frac{1}{2}(\alpha +\beta )}{(2n+\alpha +\beta +2) (2n+\alpha +\beta )},\ n\geqslant 0, \end{aligned}$$

(3.7)

$$\begin{aligned}{} & {} \hat{\gamma }_{n+1}= \frac{(n+1)(n+\alpha +\beta +1) (n+\alpha +\beta +1-\eta )(n+\eta +1)(n+\alpha +1)(n+\beta +1)}{(2n+\alpha +\beta +3)(2n+\alpha +\beta +2)^2(2n+\alpha +\beta +1)}, \ n\geqslant 0.\nonumber \\ \end{aligned}$$

(3.8)

Observe that if $\alpha ^2-\beta ^2=0$ or $\eta =\frac{1}{2}(\alpha +\beta )$, the corresponding polynomials are symmetric. In consequence, it is worth while to discuss the following subcases mentioned in [1].

$\textbf{B}_{22a}$. For $\alpha -\beta =0$, we obtain

$$\begin{aligned} \hat{\gamma }_{n+1}=\frac{1}{4}\frac{(n+1)(n+2\alpha +1) (n+2\alpha +1-\eta )(n+\eta +1)}{(2n+2\alpha +3)(2n+2\alpha +1)},\ n\geqslant 0. \end{aligned}$$

(3.9)

When $-1<\eta <2\alpha +1$ or when $\alpha \in \mathbb {R}$ and $\eta +\bar{\eta }=2\alpha ,\ \alpha +1>0$, the obtained polynomials are orthogonal with respect to a positive definite form.

• For $\alpha =0$, the resulting polynomials are related to Pasternak polynomials [24] having the coefficients

  $$\begin{aligned} \hat{\gamma }_{n+1}=\frac{1}{4}\frac{(n+1)^2(n+\eta +1) (n-\eta +1)}{(2n+3)(2n+1)},\ n\geqslant 0. \end{aligned}$$

  (3.10)

  .

• For $\eta =0$, the resulting polynomials are the Touchard ones [26] which are themselves particular cases of the continuous Hahn polynomials [4]. In this case we have

  $$\begin{aligned} \hat{\gamma }_{n+1}=\frac{1}{4}\frac{(n+1)^2(n+\alpha +1) (n-\alpha +1)}{(2n+3)(2n+1)},\ n\geqslant 0. \end{aligned}$$

  (3.11)

$\textbf{B}_{22b}$. For $\alpha +\beta =0$, we get

$$\begin{aligned} \hat{\gamma }_{n+1}=\frac{1}{4}\frac{(n+\alpha +1)(n-\alpha +1) (n+\eta +1)(n-\eta +1)}{(2n+3)(2n+1)},\ n\geqslant 0. \end{aligned}$$

(3.12)

Again, observe that the associated form to these orthogonal polynomials is positive definite when $-1<\eta <2\alpha +1$ or when $\alpha \in \mathbb {R}$ and $\eta +\bar{\eta }=2\alpha ,\ \alpha +1>0$.

• When $\alpha =0$, this leads to the Pasternak polynomials.

• When $\eta =0$, we meet again the Touchard polynomials.

$\textbf{B}_{22c}$. For $\eta =\frac{1}{2}(\alpha +\beta )$, we get

$$\begin{aligned} \hat{\gamma }_{n+1}=\frac{1}{4}\frac{(n+1)(n+\alpha +\beta +1)(n+\alpha +1) (n+\beta +1)}{(2n+\alpha +\beta +3)(2n+\alpha +\beta +1)},\ n\geqslant 0. \end{aligned}$$

(3.13)

For $\alpha +1>0$ and $\beta +1>0$, we have $\hat{\gamma }_{n+1}>0$, for all $n\geqslant 0$. The form associated to these polynomials is then positive definite.

On the other hand, if both $\alpha$ and $\beta$ are assumed to be real numbers, and so $\eta +\bar{\eta }=\alpha +\beta$, it is possible to find two numbers a and b, such that

$$\begin{aligned} a+\bar{a}=\alpha +1,\, b+\bar{b}=\beta +1,\, a+\bar{b}=\eta +1 \ \text{ and }\ \bar{a}+b=\alpha +\beta +1-\eta . \end{aligned}$$

In addition, under the conditions $\Re {a}>0$ and $\Re {b}>0$, it was shown in [1, p.18] that the resulting polynomials are orthogonal w.r.t. a positive definite form and coincide with the continuous Hahn polynomials. For more details we refer the reader to the aforementioned paper. The description of the main results obtained in the second case are summarized in Table 2.

Table 2 Description of the OPS in Case B

4 The Recurrence Coefficients of the Higher Order Derivatives

Let k be a positive integer and let $\{P_n\}_{n\geqslant 0}$ be a $D_\omega$-classical OPS. The sequence of the normalized higher order $D_\omega$-derivatives, denoted as $\{P^{[k]}_n\}_{n\geqslant 0}$, is recursively defined by

$$\begin{aligned} P^{[k]}_n(x):=\frac{1}{n+1} D_\omega P^{[k-1]}_{n+1}(x), \ \ k\geqslant 1, \end{aligned}$$

(4.1a)

or, equivalently,

$$\begin{aligned} P^{[k]}_n(x):=\frac{1}{(n+1)_k} D^k_\omega P_{n+k}(x), \ \ k\geqslant 1, \end{aligned}$$

(4.1b)

where $(\mu )_n=\mu (\mu +1)\cdots (\mu +n-1), \, (\mu )_0=1, \ \mu \in {\mathbb {C}}, \, n\in \mathbb {N}$, is the Pochhammer symbol. Application of Definition 1.1, with the special notations $P^{[1]}_n:=Q_n$ and $P^{[0]}_n:=P_n$, allows one to write $\beta ^{[1]}_n:=\tilde{\beta }_n$, $\gamma ^{[1]}_{n}:=\tilde{\gamma }_{n}$ and $\beta ^{[0]}_n:= \beta _n$, $\gamma ^{[0]}_{n}:=\gamma _{n}$, respectively.

The following corollary, which is in fact an immediate consequence of Proposition 1.2, plays an important role in establishing our results.

Corollary 4.1

([1]) If the OPS $\{P_n\}_{n\geqslant 0}$ is $D_\omega$-classical, then the sequence $\{P^{[k]}_n\}_{n\geqslant 0}$ is also $D_\omega$-classical OPS for any $k\geqslant 1$.

By an application of this corollary, if we denote by $\big (\beta ^{[k]}_n, \gamma ^{[k]}_{n+1}\big )_{n\in \mathbb {N}}$ the recurrence coefficients corresponding to the OPS $\{P^{[k]}_n\}_{n\geqslant 0}$, with $k\geqslant 1$, then

$$\begin{aligned} P^{[k]}_{n+2}(x)= & {} (x - \beta ^{[k]}_{n+1})P^{[k]}_{n+1}(x) -\gamma ^{[k]}_{n+1}P^{[k]}_n (x), \ n\geqslant 0, \end{aligned}$$

(4.2a)

$$\begin{aligned} P^{[k]}_1(x)= & {} x - \beta ^{[k]}_0,\ P^{[k]}_0(x) = 1. \end{aligned}$$

(4.2b)

Our objective here is to express the coefficients $\beta ^{[k]}_n$ and $\gamma ^{[k]}_{n+1}$ in terms of the corresponding coefficients of the OPS $\{P_n\}_{n\geqslant 0}$, namely, $\beta _n$ and $\gamma _{n+1}$ obtained either in Case A or in Case B.

This will be stated in the next proposition.

Proposition 4.2

Let $\{P_n\}_{n\geqslant 0}$ be a $D_\omega$-classical OPS. Then, for every $k\in \mathbb {N}$, we have

$$\begin{aligned} \beta ^{[k]}_{n}= & {} \beta _{n+k}+k\delta _{n+k-1}, \ \ n\geqslant 0, \end{aligned}$$

(4.3)

$$\begin{aligned} \gamma ^{[k]}_n= & {} \frac{n}{n+k}\gamma _{n+k} \big (k(\theta _{n+k-1}-1)+1\big ), \ \ n\geqslant 1, \end{aligned}$$

(4.4)

where $\delta _n$ and $\theta _n$ are solutions for the Eqs. (2.7) and (2.8), respectively.

Proof

From Corollary 4.1 it follows that each sequence $\{P^{[k]}_n\}_{n\geqslant 0}$, $k\geqslant 1$, is also $D_\omega$-classical. Accordingly, both of the OPS $\{P^{[k]}_n\}_{n\geqslant 0}$ and $\{P^{[k+1]}_n\}_{n\geqslant 0}$ are characterized by the fact that they satisfy a second structure relation of type (1.15a)–(1.15b). For $\{P^{[k]}_n\}_{n\geqslant 0}$, this is given by

$$\begin{aligned} P^{[k-1]}_{n+2}= & {} P^{[k]}_{n+2} + \alpha ^{k,1}_{n+1}P^{[k]}_{n+1} +\alpha ^{k,0} _{n}P^{[k]}_{n}, \ n\geqslant 0, \end{aligned}$$

(4.5a)

$$\begin{aligned} P^{[k-1]}_1= & {} P^{[k]}_1 +\alpha ^{k,1}_0, \ \ P^{[k-1]}_0 = P^{[k]}_0 = 1, \end{aligned}$$

(4.5b)

where

$$\begin{aligned} \alpha ^{k,1}_{n}= & {} (n+1)\left( \beta ^{[k-1]}_{n+1}-\beta ^{[k]}_{n} -\omega \right) \ \ \text { and }\ \ \nonumber \\\alpha ^{k,0} _{n}= & {} (n+1)\gamma ^{[k-1]}_{n+2}-(n+2)\gamma ^{[k]}_{n+1}. \end{aligned}$$

(4.6)

Likewise, for $\{P^{[k+1]}_n\}_{n\geqslant 0}$, an equivalent relation may be written as

$$\begin{aligned} P^{[k]}_{n+2}= & {} P^{[k+1]}_{n+2} + \alpha ^{k+1,1}_{n+1}P^{[k+1]}_{n+1} + \alpha ^{k+1,0}_{n}P^{[k+1]}_{n}, \ n\geqslant 0, \end{aligned}$$

(4.7a)

$$\begin{aligned} P^{[k]}_1= & {} P^{[k+1]}_1 +\alpha ^{k+1,1}_0,\ \ P^{[k]}_0 = P^{[k+1]}_0 = 1, \end{aligned}$$

(4.7b)

with

$$\begin{aligned} \alpha ^{k+1,1}_{n}= & {} (n+1)\left( \beta ^{[k]}_{n+1}-\beta ^{[k+1]}_{n} -\omega \right) \ \text { and }\ \nonumber \\\alpha ^{k+1,0}_{n}= & {} (n+1)\gamma ^{[k]}_{n+2}-(n+2)\gamma ^{[k+1]}_{n+1}. \end{aligned}$$

(4.8)

To prove the equalities (4.3) and (4.4), we can proceed by induction on k. For this, let $\textrm{P}(k)$ be the proposition that these two equalities hold. Observe first that the assertion $\textrm{P}(1)$ is trivial. Let us check that $\textrm{P}(2)$ is true. Setting $k=1$ in (4.5a)–(4.5b) we get

$$\begin{aligned} P_{n+2}= & {} P^{[1]}_{n+2} + \alpha ^{1,1}_{n+1}P^{[1]}_{n+1} +\alpha ^{1,0} _{n}P^{[1]}_{n}, \ n\geqslant 0, \end{aligned}$$

(4.9a)

$$\begin{aligned} P_1= & {} P^{[1]}_1 +\alpha ^{1,1}_0,\ \ P_0 =P^{[1]}_0 = 1, \end{aligned}$$

(4.9b)

where $\alpha ^{1,1}_{n}:=\tilde{\alpha }^{1}_{n}$ and $\alpha ^{1,0}_{n}:=\tilde{\alpha }^{0}_{n}$, since $P^{[1]}_n:=Q_n$ and $P^{[0]}_n:=P_n$. Roughly speaking, in this special case, the formulas (4.9a)–(4.9b) and (1.15a)–(1.15b) coincide.

Similarly, if we take $k=2$ in (4.5a)–(4.5b), we have

$$\begin{aligned} P^{[1]}_{n+2}= & {} P^{[2]}_{n+2} + \alpha ^{2,1}_{n+1}P^{[2]}_{n+1} +\alpha ^{2,0} _{n}P^{[2]}_{n}, \ n\geqslant 0, \end{aligned}$$

(4.10a)

$$\begin{aligned} P^{[1]}_1= & {} P^{[2]}_1 +\alpha ^{2,1}_0, \ \ P^{[1]}_0 = P^{[2]}_0 = 1. \end{aligned}$$

(4.10b)

Replace n by $n+1$ in (4.9a) and then apply the operator $D_\omega$ yields

$$\begin{aligned} (n+3)P^{[1]}_{n+2} = (n+3)P^{[2]}_{n+2} +(n+2)\alpha ^{1,1}_{n+2}P^{[2]}_{n+1} + (n+1) \alpha ^{1,0}_{n+1}P^{[2]}_{n},\ n\geqslant 0.\nonumber \\ \end{aligned}$$

(4.11)

Multiply both sides of (4.10a) by $(n+3)$ and compare this with (4.11) readily gives

$$\begin{aligned} (n+3)\alpha ^{2,1}_{n+1}= (n+2)\alpha ^{1,1}_{n+2}, \end{aligned}$$

(4.12a)

$$\begin{aligned} (n+3)\alpha ^{2,0}_{n}= (n+1)\alpha ^{1,0}_{n+1}. \end{aligned}$$

(4.12b)

By (4.6), for $k=2$ and $k=1$, it is easy to check that (4.12a) and (4.12b), respectively, lead to

$$\begin{aligned} (n+3)(n+2)\big [\beta ^{[1]}_{n+2}-\beta ^{[2]}_{n+1}-\omega \big ]&=(n+3)(n+2)\big [\beta _{n+3}-\beta ^{[1]}_{n+2}-\omega \big ],\\ (n+3)\big [(n+1)\gamma ^{[1]}_{n+2}-(n+2)\gamma ^{[2]}_{n+1}\big ]&=(n+1)\big [(n+2)\gamma _{n+3}- (n+3)\gamma ^{[1]}_{n+2}\big ]. \end{aligned}$$

Thanks to (2.1)–(2.2), we, respectively, deduce that

$$\begin{aligned} \beta ^{[2]}_{n}&=\beta _{n+2}+2\delta _{n+1},\ \ n\geqslant 0,\\ \gamma ^{[2]}_n&= \frac{n}{n+2}\gamma _{n+2} \big (2\theta _{n+1}-1\big ), \ \ n\geqslant 1, \end{aligned}$$

which is precisely the desired conclusion, that is, $\textrm{P}(2)$ is true.

Now, we must show that the conditional statement $\textrm{P}(k)\rightarrow \textrm{P}(k+1)$ is true for all positive integers k. We can proceed analogously to the proof of $\textrm{P}(2)$. Starting from (4.5a), replacing n by $n+1$ and then apply the operator $D_\omega$ we find

$$\begin{aligned} (n\!+\!3)P^{[k]}_{n\!+\!2}\!=\! (n\!+\!3)P^{[k\!+\!1]}_{n\!+\!2} \!+\!(n\!+\!2)\alpha ^{k,1}_{n\!+\!2}P^{[k\!+\!1]}_{n\!+\!1} \!+\! (n\!+\!1) \alpha ^{k,0}_{n\!+\!1}P^{[k\!+\!1]}_{n},\ n\geqslant 0.\nonumber \\ \end{aligned}$$

(4.13)

Multiply both sides of (4.7a) by $(n+3)$ and compare this with (4.13) readily gives

$$\begin{aligned} (n+3)\alpha ^{k+1,1}_{n+1}= (n+2)\alpha ^{k,1}_{n+2}, \end{aligned}$$

(4.14a)

$$\begin{aligned} (n+3)\alpha ^{k+1,0}_{n}= (n+1)\alpha ^{k,0}_{n+1}. \end{aligned}$$

(4.14b)

On account of (4.6)–(4.8), we have

$$\begin{aligned} (n+3)(n+2)\big [\beta ^{[k]}_{n+2}-\beta ^{[k+1]}_{n+1}-\omega \big ]&=(n+3)(n+2)\big [\beta ^{[k-1]}_{n+3}-\beta ^{[k]}_{n+2}-\omega \big ],\\ (n+3)\big [(n+1)\gamma ^{[k]}_{n+2}-(n+2)\gamma ^{[k+1]}_{n+1}\big ]&=(n+1)\big [(n+2)\gamma ^{[k-1]}_{n+3}- (n+3)\gamma ^{[k]}_{n+2}\big ]. \end{aligned}$$

From this, we deduce that

$$\begin{aligned} \beta ^{[k+1]}_{n+1}= & {} 2\beta ^{[k]}_{n+2}-\beta ^{[k-1]}_{n+3}, \end{aligned}$$

(4.15)

$$\begin{aligned} \gamma ^{[k+1]}_{n+1}= & {} 2\frac{n+1}{n+2}\gamma ^{[k]}_{n+2} -\frac{n+1}{n+3}\gamma ^{[k-1]}_{n+3}. \end{aligned}$$

(4.16)

On the other hand, according to the induction hypothesis we may write

$$\begin{aligned} \beta ^{[k]}_{n+2}&=\beta _{n+k+2}+k\delta _{n+k+1}, \quad \beta ^{[k-1]}_{n+3}=\beta _{n+k+2}+(k-1)\delta _{n+k+1},\\ \gamma ^{[k]}_{n+2}&= \frac{n+ 2}{n+ k+ 2} \gamma _{n+k+2}\big (k(\theta _{n+k+1} - 1) + 1\big ),\\ \gamma ^{[k-1]}_{n+3}&= \frac{n+ 3}{n+ k+ 2} \gamma _{n+k+2}\big ((k- 1)(\theta _{n+k+1}- 1)+ 1\big ). \end{aligned}$$

Substituting these into (4.15)–(4.16), and then changing n into $n-1$, it follows that

$$\begin{aligned} \beta ^{[k+1]}_{n}&=\beta _{n+k+1}+(k+1)\delta _{n+k},\ \ n\geqslant 0,\\ \gamma ^{[k+1]}_n&= \frac{n}{n+k+1}\gamma _{n+k+1}\big ((k+1) (\theta _{n+k}-1)+1\big ), \ \ n\geqslant 1. \end{aligned}$$

These last equalities show that $\textrm{P}(k+1)$ is also true, which completes the proof. $\square$

Remark

Due to (4.3) and (4.4), from (4.6), it is seen that the two coefficients $\alpha ^{k,1}_{n}$ and $\alpha ^{k,0}_{n}$ involving in the structure relation (4.5a)–(4.5b) can be rewritten as

$$\begin{aligned} \alpha ^{k,1}_{n}\!=\! -(n\!+\!1)\left( \delta _{n\!+\!k\!-\!1}\!+\!\omega \right) \, ;\, \alpha ^{k,0} _{n}\!=\! \frac{(n\!+\!1)(n\!+\!2)}{n\!+\!k\!+\!1}\gamma _{n\!+\!k\!+\!1} \left( 1\!-\!\theta _{n\!+\!k}\right) ,\ n\geqslant 0.\nonumber \\ \end{aligned}$$

(4.17)

For $k=1$, this reduces to (2.3) providing the coefficients of (1.15a)–(1.15b).

Application.

When $\omega =0$, the identities (4.5a)–(4.5b) consist of the structure relation characterizing the higher order derivatives sequence of the classical orthogonal polynomials. In this case, a direct application of the preceding proposition enables us to express the recurrence coefficients of the sequence $\{P^{[k]}_n\}_{n\geqslant 0}$ in terms of the recurrence coefficients for each of the four classical families.

To this purpose, if we denote by $\{\hat{H}_n\}_{n\geqslant 0}$, $\{\hat{L}_n(.;\alpha )\}_{n\geqslant 0}$, $\{\hat{B}_n(.;\alpha )\}_{n\geqslant 0}$ and $\{\hat{J}_n(.;\alpha ,\beta )\}_{n\geqslant 0}$, respectively, the (monic) Hermite, Laguerre, Bessel and Jacobi polynomials, and by $\hat{H}^{[k]}_n(x)$, $\hat{L}^{[k]}_n(x;\alpha )$, $\hat{B}^{[k]}_n(x;\alpha )$ and $\hat{J}^{[k]}_n(x;\alpha ,\beta )$ their corresponding sequences of derivatives of order k, then application of Formulas (4.3) and (4.4) successively give:

Case A. For $\ \theta _n=1,\, n\geqslant 1$, and $\delta _n=\delta _0,\, n\geqslant 1$, we have

$$\begin{aligned} \hat{\beta }^{[k]}_{n}= \hat{\beta }_{n+k}+k\delta _{0}, \ \ n\geqslant 0, \end{aligned}$$

(4.18)

$$\begin{aligned} \hat{\gamma }^{[k]}_n= \frac{n}{n+k}\hat{\gamma }_{n+k}, \ \ n\geqslant 1. \end{aligned}$$

(4.19)

Hermite case: When $\delta _0=0$, $\hat{\gamma }_1=\frac{1}{2}$, this yields

$$\begin{aligned} \hat{\beta }^{[k]}_n=0, n\geqslant 0, \ \ \text{ and }\ \ \hat{\gamma }^{[k]}_{n+1}=\frac{1}{2}(n+1), n\geqslant 0. \end{aligned}$$

Laguerre case: When $\delta _0=-1$, $\hat{\gamma }_1=\hat{\beta }_0=\alpha +1$, we get

$$\begin{aligned} \hat{\beta }^{[k]}_n=2n+\alpha +k+1, n\geqslant 0, \ \ \text{ and } \ \ \hat{\gamma }^{[k]}_{n+1}=(n+1)(n+\alpha +k+1), n\geqslant 0. \end{aligned}$$

Case B. For $\theta _n=\frac{n+\theta +1}{n+\theta },\, n\geqslant 1, \ \text{ and }\ \delta _n=\frac{\delta _0(\theta +3) (\theta +1)}{\big (2n+\theta +3\big )\big (2n+\theta +1\big )},\, n\geqslant 0$, we deduce that

$$\begin{aligned} \hat{\beta }^{[k]}_{n}= & {} \hat{\beta }_{n+k}+\frac{k\delta _0 (\theta +3)(\theta +1) }{\big (2(n+k)+\theta +1\big ) \big (2(n+k)+\theta -1\big )},\ \ n\geqslant 0,\end{aligned}$$

(4.20)

$$\begin{aligned} \hat{\gamma }^{[k]}_n= & {} \frac{n(n+\theta +2k-1)}{(n+k) (n+\theta +k-1)}\hat{\gamma }_{n+k}, \ \ n\geqslant 1. \end{aligned}$$

(4.21)

Bessel case: When $\theta =2\alpha -1$, $\delta _0=-1/(\alpha +1)\alpha$, we obtain

$$\begin{aligned} \hat{\beta }^{[k]}_{n}&=\frac{1-(\alpha +k)}{(n+\alpha +k) (n+\alpha +k-1)},\ n\geqslant 0, \\ \hat{\gamma }^{[k]}_{n+1}&=-\frac{(n+1)(n+2(\alpha +k)-1)}{(2n+2(\alpha +k)+1)(n+\alpha +k)^2(2n+2(\alpha +k)-1)},\ n\geqslant 0. \end{aligned}$$

Jacobi case: When $\theta =\alpha +\beta +1$, $\delta _0=2(\alpha -\beta )/(\alpha +\beta +4)(\alpha +\beta +2)$, we get

$$\begin{aligned} \hat{\beta }^{[k]}_{n}&=\frac{(\alpha -\beta )(\alpha +\beta +2k)}{(2n+\alpha +\beta +2k+2)(2n+\alpha +\beta +2k)},\ n\geqslant 0,\\ \hat{\gamma }^{[k]}_{n+1}&=4\frac{(n+1)(n+\alpha +\beta +2k+1) (n+\alpha +k+1)(n+\beta +k+1)}{(2n+\alpha +\beta +2k+3) (2n+\alpha +\beta +2k+2)^2(2n+\alpha +\beta +2k+1)},\ n\geqslant 0. \end{aligned}$$

We thus rediscover the well known relations $\hat{H}^{[k]}_n(x)=\hat{H}_n(x)$, $\hat{L}^{[k]}_n(x;\alpha )=\hat{L}_n(x;\alpha +k)$, $\hat{B}^{[k]}_n(x;\alpha )=\hat{B}_n(x;\alpha +k)$ and $\hat{J}^{[k]}_n(x;\alpha ,\beta )=\hat{J}_n(x;\alpha +k,\beta +k)$.

The results presented above are of course known and clearly show that the sequences of higher order derivative for the classical orthogonal polynomials belonging to the same class, provided that the parameters $\alpha$ and $\beta$ take values in the range of regularity.

Corollary 4.1 asserts that the $D_\omega$-classical character remain invariants under a differentiation of any $D_\omega$-classical polynomials. However, except for the case $\omega =0$, application of the kth order operator $D^k_\omega$ does not lead to a shift on the parameters of the original sequence.

5 Conclusion

We studied the $D_\omega$-classical orthogonal polynomials using a new method in this domain. The results obtained in Sect. 3 are expected, where four representatives families are pointed out with some of their special cases. The recurrence coefficients of the resulting orthogonal polynomials are explicitly determined. Proposition 1.4 established a new characterization of these polynomials via a structure relation, and Proposition 4.2 provided relations connecting the recurrence coefficients of each sequence of polynomials with those of its higher order derivatives. For $\omega =0$, the classical orthogonal polynomials are rediscovered.

We conclude, of course, by asking whether our approach provides a rather easy answer to the first Hahn’s problem, that is, when $D_\omega$ is replaced by ${{\mathscr {D}}}_{q}$?