**1. Introduction**

There is a special case in the Sturm–Liouville boundary value problem the called Hermite differential equation that arises when dealing with harmonic oscillator in quantum mechanics. The ordinary Hermite differential equation is defined as

$$\frac{d^2y}{dx^2} - 2x\frac{dy}{dx} + (\rho - 1)y = 0,\tag{1}$$

where *ρ* is a constant. When *ρ* = 2*n* + 1, *n* = 0, 1, 2, ... , then one of the solutions of Equation (1) becomes a polynomial. These polynomial solutions are known as Hermite polynomials *Hn*(*x*), which are defined by means of the generating function

$$e^{(2x-t)t} = \sum\_{n=0}^{\infty} H\_{\text{ll}}(x) \frac{t^n}{n!} \qquad |t| < \infty. \tag{2}$$

The numbers *Hn* := *Hn*(0) are the Hermite numbers. Hermite polynomials, first defined by Laplace, are one of the classic orthogonal polynomials and many studies have been conducted by mathematicians. These Hermite polynomials also have many mathematical applications, such as quantum mechanics, physics, and probability theory; see [1–6].

We define the *q*-numbers also referred by Jackson as follows; see [7–9]

$$[\mathbf{x}]\_q = \frac{1 - q^{\mathbf{x}}}{1 - q}, \quad 0 < q < 1,\tag{3}$$

Note that lim*q*→1[*x*]*<sup>q</sup>* = *x*. In [8], we recall that the *q*-Hermite polynomials **H***n*,*q*(*x*) defined by <sup>∞</sup>

$$\sum\_{n=0}^{\infty} \mathbf{H}\_{\mathcal{U}, \mathfrak{q}}(\mathbf{x}) \frac{t^n}{n!} = e^{2[\mathbf{x}]\_{\mathfrak{q}} t - \mathfrak{q}^2} = \mathcal{G}(t, [\mathbf{x}]\_{\mathfrak{q}}),\tag{4}$$

where 0 < *q* < 1. In the definition of *q*-Hermite polynomials, we can observe that if *q* → 1 , then **H***n*,*q*(*x*) → *Hn*(*x*).

**Citation:** Ryoo, C.-S.; Kang, J. Some Properties Involving *q*-Hermite Polynomials Arising from Differential Equations and Location of Their Zeros. *Mathematics* **2021**, *9*, 1168. https://doi.org/10.3390/math9111168

Academic Editor: Arsen Palestini

Received: 21 April 2021 Accepted: 19 May 2021 Published: 22 May 2021

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In [10], authors defined the two-variable partially degenerate Hermite polynomials **<sup>H</sup>***n*(*x*, *<sup>y</sup>*, *<sup>λ</sup>*) as <sup>∞</sup>

$$\sum\_{n=0}^{\infty} \mathbf{H}\_n(x, y, \lambda) \frac{t^n}{n!} = (1 + \lambda t)^{\frac{n}{\lambda}} e^{yt^2}, \quad \lambda \neq 0,\tag{5}$$

and we can see some useful properties of these polynomials. Representatively, we can confirm the following theorems in [10].

$$\begin{aligned} \text{(i)} \quad a^m \mathbf{H}\_m(bx, b^2y, \frac{\lambda}{a}) &= b^m \mathbf{H}\_m(ax, a^2y, \frac{\lambda}{b}) \\ \text{(ii)} \quad \mathbf{H}\_n(x\_1 + x\_2, y, \lambda) &= \sum\_{l=0}^n \binom{n}{l} (x\_2|\lambda) \mathbf{H}\_{n-l}(x\_1, y, \lambda) . \end{aligned} \tag{6}$$

The differential equations derived from the generating functions of special numbers and polynomials have been studied by many mathematicians; see [11–21].

Based on the results to date, in the present work, we can investigate the differential equations generated from the generating function of *q*-Hermite polynomials **H***n*,*q*(*x*). The rest of the paper is organized as follows. In Section 2, we obtain the basic properties of the *q*-Hermite polynomials. In Section 3, we construct the differential equations generated from the definition of *q*-Hermite polynomials:

$$\left(\frac{\partial}{\partial t}\right)^{N}\mathcal{G}(t,[\mathbf{x}]\_{q}) - a\_{0}(N,[\mathbf{x}]\_{q})\mathcal{G}(t,[\mathbf{x}]\_{q}) - \cdots - a\_{N}(N,[\mathbf{x}]\_{q})t^{N}\mathcal{G}(t,[\mathbf{x}]\_{q}) = 0. \tag{7}$$

We also consider explicit identities for **H***n*,*q*(*x*) using the coefficients of this differential equation. In Section 4, we find the zeros of the *q*-Hermite polynomials using numerical methods and observe the scattering phenomenon of the zeros of these polynomials. Finally, in Section 5, conclusions and discussions on this work are provided.

## **2. Basic Properties for the** *q***-Hermite Polynomials**

To derive various properties of **H***n*,*q*(*x*), the generating function (4) is an useful function. The following basic properties of polynomials **H***n*,*q*(*x*) are derived from (4). Hence, we choose to omit the details involved.

**Theorem 1.** *Let n be any positive integer. Then, we have*

$$\begin{aligned} (1) \quad \mathbf{H}\_{n,q}(\mathbf{x}) &= \sum\_{k=0}^{n} \binom{n}{k} 2^{n-k} [\mathbf{x}]\_q^{n-k} H\_k. \\ (2) \quad \mathbf{H}\_{n,q}(\mathbf{x}) &= n! \sum\_{k=0}^{\lfloor \frac{n}{2} \rfloor} \frac{(-1)^k 2^{n-2k} [\mathbf{x}]\_q^{n-2k}}{k!(n-2k)!}. \\ (3) \quad \mathbf{H}\_{n,q}(\mathbf{x}\_1 + \mathbf{x}\_2) &= \sum\_{k=0}^{n} \binom{n}{k} \mathbf{H}\_{k,q}(\mathbf{x}\_1) 2^{n-k} q^{\mathbf{x}\_1(n-k)} [\mathbf{x}\_2]\_q^{n-k}. \end{aligned} \tag{8}$$

*where* [*x*] *is the greatest integer not exceeding x.*

**Theorem 2.** *The q-Hermite polynomials are the solutions of equation*

$$\begin{aligned} &\left(\left(\frac{d}{d[\mathbf{x}]\_q}\right)^2 - 2[\mathbf{x}]\_q \left(\frac{d}{d[\mathbf{x}]\_q}\right) + 2n\right) \mathbf{H}\_{n,q}(\mathbf{x}) = 0, \\ &\mathbf{H}\_{n,q}(0) = \begin{cases} (-1)^k \frac{(2k)!}{k!}, & \text{if } n = 2k, \\ 0, & \text{otherwise} \end{cases} \end{aligned} \tag{9}$$

**Proof.** From Equation (4), we can note that

$$\mathcal{G}(t,[x]\_q) = \mathcal{e}^{2[x]\_{\mathfrak{q}}t - t^2} \tag{10}$$

which is satisfied as

$$\frac{\partial \mathcal{G}(t, [x]\_q)}{\partial t} - \left(2[x]\_q - 2t\right) \mathcal{G}(t, [x]\_q) = 0. \tag{11}$$

By substituting the series in (11) for G(*t*, [*x*]*q*), we find

$$\mathbf{H}\_{n+1,q}(\mathbf{x}) - 2[\mathbf{x}]\_q \mathbf{H}\_{n,q}(\mathbf{x}) + 2n \mathbf{H}\_{n-1,q}(\mathbf{x}) = 0, n = 1, 2, \dots, \tag{12}$$

which is the recurrence relation for *q*-Hermite polynomials. Another recurrence relation comes from 

$$2\left(\frac{d}{d[\mathbf{x}]\_q}\right)\mathcal{G}(t,[\mathbf{x}]\_q) - 2t\mathcal{G}(t,[\mathbf{x}]\_q) = 0. \tag{13}$$

The following equation implies

$$\mathbb{P}\left(\frac{d}{d[\mathbf{x}]\_q}\right)\mathbf{H}\_{\mathbf{n},\boldsymbol{\eta}}(\mathbf{x}) - 2n\mathbf{H}\_{n-1,\boldsymbol{\eta}}(\mathbf{x}) = 0, n = 1, 2, \dots \tag{14}$$

Remove **H***n*−1,*q*(*x*) from Equations (12) and (13) to obtain

$$\mathbf{H}\_{n+1,q}(\mathbf{x}) - 2[\mathbf{x}]\_q \mathbf{H}\_{n,q}(\mathbf{x}) + \left(\frac{d}{d[\mathbf{x}]\_q}\right) \mathbf{H}\_{n,q}(\mathbf{x}) = 0. \tag{15}$$

By differentiating the following equation and using Equations (12) and (13) again, we can obtain

$$\left(\frac{d}{d[\mathbf{x}]\_q}\right)^2 \mathbf{H}\_{\mathbb{H},\mathbb{q}}(\mathbf{x}) - 2[\mathbf{x}]\_q \left(\frac{d}{d[\mathbf{x}]\_q}\right) \mathbf{H}\_{\mathbb{H},\mathbb{q}}(\mathbf{x}) + 2n \mathbf{H}\_{\mathbb{H},\mathbb{q}}(\mathbf{x}) = 0, n = 0, 1, 2, \dots \tag{16}$$

From the above equation, we complete the proof of Theorem 2.

**Theorem 3. H***n*,*q*(*x*) *in the Equation (4) is the solution of equation*

$$\begin{aligned} \left( \frac{q-1}{q^x \log q} \frac{d^2}{dx^2} + \left( \frac{1-q}{q^x} - \frac{2(1-q^x)}{1-q} \right) \frac{d}{dx} + 2n \frac{\log q}{q-1} q^x \right) \mathbf{H}\_{n,q}(\mathbf{x}) &= 0, \\ \mathbf{H}\_{n,q}(0) = \begin{cases} (-1)^k \frac{(2k)!}{k!}, & \text{if } n=2k, \\ 0, & \text{otherwise.} \end{cases} \end{aligned} \tag{17}$$

**Proof.** We consider another form of the differential equation for **H***n*,*q*(*x*). We consider

$$\mathcal{G}(t, [\mathbf{x}]\_q) = \epsilon^{2\langle \mathbf{x} \rangle\_q t - t^2},\tag{18}$$

which satisfies

$$\frac{d\mathcal{G}(t,[\mathbf{x}]\_{\mathfrak{q}})}{d\mathbf{x}} - \frac{\log q}{q - 1} q^{\mathbf{x}} 2t \mathcal{G}(t,[\mathbf{x}]\_{\mathfrak{q}}) = 0. \tag{19}$$

Substitute the series in Equation (19) for G(*t*, [*x*]*q*), in order to find

$$\frac{d\mathbf{H}\_{n,q}(\mathbf{x})}{d\mathbf{x}} - \frac{2n\log q}{q-1} q^x \mathbf{H}\_{n-1,q}(\mathbf{x}) = 0, n = 1, 2, \dots \tag{20}$$

To use Equation (15), we note

$$\frac{d}{d\mathbf{x}}\left(\frac{q-1}{q^{\mathbf{x}}\log q}\frac{d}{d\mathbf{x}}\mathbf{H}\_{n,\emptyset}(\mathbf{x})\right) = \frac{1-q}{q^{\mathbf{x}}}\frac{d}{d\mathbf{x}}\mathbf{H}\_{n,\emptyset}(\mathbf{x}) + \frac{q-1}{q^{\mathbf{x}}\log q}\left(\frac{d}{d\mathbf{x}}\right)^{2}\mathbf{H}\_{n,\emptyset}(\mathbf{x}).\tag{21}$$

By differentiating Equation (15) and using the above Equation (21), we derive

$$\begin{aligned} &2n\frac{\log q}{q-1}q^x \mathbf{H}\_{n,q}(\mathbf{x}) + \left(\frac{1-q}{q^x} - \frac{2(1-q^x)}{1-q}\right)\frac{d\mathbf{H}\_{n,q}(\mathbf{x})}{d\mathbf{x}}\\ &+\frac{q-1}{q^x\log q}\frac{d^2\mathbf{H}\_{n,q}(\mathbf{x})}{d\mathbf{x}^2} = 0, \end{aligned} \tag{22}$$

where the equation is obtained as the required result immediately.
