**3. Differential Equations Associated with** *q***-Hermite Polynomials**

In this section, we introduce differential equations arising from the generating functions of *q*-Hermite polynomials. By using these differential equations, we can obtain the explicit identities for these polynomials. Many authors studied differential equations derived in the generating functions of special polynomials in order to derive explicit identities for special polynomials, see [11–20].

Let

$$\mathcal{G} := \mathcal{G}(t, [x]\_q) = e^{2[x]\_q t - t^2} = \sum\_{n=0}^{\infty} H\_{n, q}(\mathbf{x}) \frac{t^n}{n!}, \quad \mathbf{x}, t \in \mathbb{R}. \tag{23}$$

Then, we obtain the following equations using mathematical induction:

$$\begin{split} \mathcal{G}^{(1)} &= \frac{\partial}{\partial t} \mathcal{G}(t, [\mathbf{x}]\_q) = \frac{\partial}{\partial t} \Big( \epsilon^{2[\mathbf{x}]\_q t - t^2} \Big) = \epsilon^{2[\mathbf{x}]\_q t - t^2} (2[\mathbf{x}]\_q - 2t) \\ &= (2[\mathbf{x}]\_q - 2t) \mathcal{G}(t, [\mathbf{x}]\_q) \\ &= (2[\mathbf{x}]\_q) \mathcal{G}(t, [\mathbf{x}]\_q) \\ &\quad + (-2)t \mathcal{G}(t, [\mathbf{x}]\_q), \end{split} \tag{24}$$
 
$$\begin{split} \mathcal{G}^{(2)} &= \frac{\partial}{\partial t} \mathcal{G}^{(1)}(t, [\mathbf{x}]\_q) = -2 \mathcal{G}(t, [\mathbf{x}]\_q) + (2\mathbf{x} - 2t) \mathcal{G}^{(1)}(t, [\mathbf{x}]\_q) \\ &= (-2 + 4[\mathbf{x}]\_q^2) \mathcal{G}(t, [\mathbf{x}]\_q) \\ &\quad + (-8[\mathbf{x}]\_q) t \mathcal{G}(t, [\mathbf{x}]\_q) \\ &\quad + (-2)^2 t^2 \mathcal{G}(t, [\mathbf{x}]\_q), \end{split} \tag{25}$$

and

$$\begin{split} \mathcal{G}^{(3)} &= \frac{\partial}{\partial t} \mathcal{G}^{(2)}(t, [\mathbf{x}]\_q) \\ &= (-8[\mathbf{x}]\_q + 8t) \mathcal{G}(t, [\mathbf{x}]\_q) + (-2 + 4[\mathbf{x}]\_q^2 - 8[\mathbf{x}]\_q t + 4t^2) \mathcal{G}^{(1)}(t, [\mathbf{x}]\_q) \\ &= (-12[\mathbf{x}]\_q + 8[\mathbf{x}]\_q^3) \mathcal{G}(t, [\mathbf{x}]\_q) \\ &\quad + (12 - 24[\mathbf{x}]\_q^2) t \mathcal{G}(t, [\mathbf{x}]\_q) \\ &\quad + (24[\mathbf{x}]\_q) t^2 \mathcal{G}(t, [\mathbf{x}]\_q) \\ &\quad + (-2)^3 t^3 \mathcal{G}(t, [\mathbf{x}]\_q). \end{split} \tag{26}$$

If we continue this process *N*-times, we can conjecture as follows.

$$\mathcal{G}^{(N)} = \left(\frac{\partial}{\partial t}\right)^N \mathcal{G}(t, [\mathbf{x}]\_q) = \sum\_{i=0}^N a\_i(N, [\mathbf{x}]\_q) t^i \mathcal{G}(t, [\mathbf{x}]\_q), (N = 0, 1, 2, \dots). \tag{27}$$

By differentiating <sup>G</sup>(*N*) with respect to *<sup>t</sup>* in Equation (27), we find

<sup>G</sup>(*N*+1) <sup>=</sup> *<sup>∂</sup>*G(*N*) *∂t* = *N* ∑ *i*=0 *ai*(*N*, [*x*]*q*)*iti*−1G(*t*, [*x*]*q*) + *N* ∑ *i*=0 *ai*(*N*, [*x*]*q*)*t i* <sup>G</sup>(1) (*t*, [*x*]*q*) = *N* ∑ *i*=0 *ai*(*N*, [*x*]*q*)*iti*−1G(*t*, [*x*]*q*) + *N* ∑ *i*=0 *ai*(*N*, [*x*]*q*)*t i* (2[*x*]*<sup>q</sup>* − 2*t*)G(*t*, [*x*]*q*) = *N* ∑ *i*=0 *iai*(*N*, [*x*]*q*)*t <sup>i</sup>*−1G(*t*, [*x*]*q*) + *N* ∑ *i*=0 (2[*x*]*q*)*ai*(*N*, [*x*]*q*)*t i* G(*t*, [*x*]*q*) + *N* ∑ *i*=0 (−2)*ai*(*N*, [*x*]*q*)*t <sup>i</sup>*+1G(*t*, [*x*]*q*) = *N*−1 ∑ *i*=0 (*i* + 1)*ai*+1(*N*, [*x*]*q*)*t i* G(*t*, [*x*]*q*) + *N* ∑ *i*=0 (2[*x*]*q*)*ai*(*N*, [*x*]*q*)*t i* G(*t*, [*x*]*q*) + *N*+1 ∑ *i*=1 (−2)*ai*−1(*N*, [*x*]*q*)*<sup>t</sup> i* G(*t*, [*x*]*q*). (28)

Replace *N* by *N* + 1 in (27), and we obtain

$$\mathcal{G}^{(N+1)} = \sum\_{i=0}^{N+1} a\_i (N+1, [x]\_q) t^i \mathcal{G}(t, [x]\_q). \tag{29}$$

**Theorem 4.** *For N* = 0, 1, 2, . . . , *the differential equation*

$$\mathcal{G}^{(N)} = \left(\frac{\partial}{\partial t}\right)^N \mathcal{G}(t, [\mathbf{x}]\_\emptyset) = \left(\sum\_{i=0}^N a\_i(N, [\mathbf{x}]\_\emptyset) t^i\right) \mathcal{G}(t, [\mathbf{x}]\_\emptyset) \tag{30}$$

*has a solution*

$$\mathcal{G} = \mathcal{G}(t, [\mathbf{x}]\_{\mathfrak{q}}) = e^{2[\mathbf{x}]\_{\mathfrak{q}}t - t^{\bar{z}}},\tag{31}$$

*where*

$$\begin{split} a\_{0}(N\_{\prime}[\mathbf{x}]\_{q}) &= \sum\_{k=0}^{N-1} [\mathbf{x}]\_{q}^{\mathbf{i}} a\_{1}(N-1-k\_{\prime}[\mathbf{x}]\_{q}) + (2[\mathbf{x}]\_{q})^{N} \\ a\_{N-1}(N\_{\prime}[\mathbf{x}]\_{q}) &= (-2)^{N-1} N(2[\mathbf{x}]\_{q}) \\ a\_{N}(N\_{\prime}[\mathbf{x}]\_{q}) &= (-2)^{N} \\ a\_{i}(N+1, [\mathbf{x}]\_{q}) \\ &= (i+1) \sum\_{k=0}^{N} 2^{k} [\mathbf{x}]\_{q}^{\mathbf{i}} a\_{i+1}(N-k\_{\prime}[\mathbf{x}]\_{q}) + (-2) \sum\_{k=0}^{N} 2^{k} [\mathbf{x}]\_{q}^{\mathbf{i}} a\_{i-1}(N-k\_{\prime}[\mathbf{x}]\_{q}) \\ & (1 \le i \le N-2). \end{split} \tag{32}$$

**Proof.** Comparing the coefficients on both sides of (28) and (29), we obtain

$$\begin{aligned} a\_0(N+1, [x]\_q) &= a\_1(N, [x]\_q) + (2[x]\_q) a\_0(N, [x]\_q), \\ a\_N(N+1, [x]\_q) &= (2[x]\_q) a\_N(N, [x]\_q) + (-2) a\_{N-1}(N, [x]\_q), \\ a\_{N+1}(N+1, [x]\_q) &= (-2) a\_N(N, [x]\_q), \end{aligned} \tag{33}$$

$$\begin{aligned} a\_i &(N+1, [\mathbf{x}]\_q) = (i+1) a\_{i+1} (N\_\prime[\mathbf{x}]\_q) \\ &+ (2[\mathbf{x}]\_q) a\_i (N\_\prime[\mathbf{x}]\_q) + (-2) a\_{i-1} (N\_\prime[\mathbf{x}]\_q), (1 \le i \le N-1). \end{aligned} \tag{34}$$

In addition, from Equation (27), we get

<sup>G</sup>(*t*, [*x*]*q*) = <sup>G</sup>(0) (*t*, [*x*]*q*) = *a*0(0, [*x*]*q*)G(*t*, [*x*]*q*), (35)

which gives

$$a\_0(0, [x]\_\emptyset) = 1.\tag{36}$$

It is not difficult to show that

$$\begin{aligned} &(2\lfloor \mathbf{x} \rfloor\_q) \mathcal{G}(t, \lfloor \mathbf{x} \rfloor\_q) + (-2)t \mathcal{G}(t, \lfloor \mathbf{x} \rfloor\_q) \\ &= \mathcal{G}^{(1)}(t, \lfloor \mathbf{x} \rfloor\_q) \\ &= \sum\_{i=0}^{1} a\_i(1, \lfloor \mathbf{x} \rfloor\_q) \mathcal{G}(t, \lfloor \mathbf{x} \rfloor\_q) \\ &= a\_0(1, \lfloor \mathbf{x} \rfloor\_q) \mathcal{G}(t, \lfloor \mathbf{x} \rfloor\_q) + a\_1(1, \lfloor \mathbf{x} \rfloor\_q) t \mathcal{G}(t, \lfloor \mathbf{x} \rfloor\_q). \end{aligned} \tag{37}$$

By using Equation (29), we can present the following as

$$a\_0(\mathbf{1}, [\mathbf{x}]\_q) = \mathbf{2}[\mathbf{x}]\_{q^\prime} \quad a\_1(\mathbf{1}, [\mathbf{x}]\_q) = -\mathbf{2}.\tag{38}$$

From the Equation (33), we express

$$\begin{aligned} a\_0(N+1, [\mathbf{x}]\_q) &= a\_1(N, [\mathbf{x}]\_q) + (2[\mathbf{x}]\_q)a\_0(N, [\mathbf{x}]\_q), \\ a\_0(N, [\mathbf{x}]\_q) &= a\_1(N-1, [\mathbf{x}]\_q) + (2\mathbf{x})a\_0(N-1, [\mathbf{x}]\_q), \dots \\ a\_0(N+1, [\mathbf{x}]\_q) &= \sum\_{i=0}^{N} (2[\mathbf{x}]\_q)^i a\_1(N-i, [\mathbf{x}]\_q) + (2[\mathbf{x}]\_q)^{N+1}, \end{aligned} \tag{39}$$

$$\begin{aligned} a\_N(N+1, [x]\_q) &= (2[x]\_q) a\_N(N, [x]\_q) + (-2) a\_{N-1}(N, [x]\_q), \\ a\_{N-1}(N, [x]\_q) &= (2[x]\_q) a\_{N-1}(N-1, [x]\_q) + (-2) a\_{N-2}(N-1, [x]\_q), \dots \\ a\_N(N+1, [x]\_q) &= (-2)^N (N+1) (2[x]\_q), \end{aligned} \tag{40}$$

$$\begin{aligned} a\_{N+1}(N+1, [\mathbf{x}]\_q) &= (-2)a\_N(N, [\mathbf{x}]\_q), \\ a\_N(N, [\mathbf{x}]\_q) &= (-2)a\_{N-1}(N-1, [\mathbf{x}]\_q), \dots \\ a\_{N+1}(N+1, [\mathbf{x}]\_q) &= (-2)^{N+1}. \end{aligned} \tag{41}$$

Choose *i* = 1 in (34). Then, we can find

$$a\_1(N+1, [\mathbf{x}]\_q) = 2\sum\_{k=0}^{N} (2[\mathbf{x}]\_q)^k a\_2(N-k, [\mathbf{x}]\_q) + (-2)\sum\_{k=0}^{N} (2[\mathbf{x}]\_q)^k a\_0(N-k, [\mathbf{x}]\_q). \tag{42}$$

For 1 ≤ *i* ≤ *N* − 1, by containing this process, we can deduce

$$\begin{split} a\_i(N+1, [\mathbf{x}]\_q) &= (i+1) \sum\_{k=0}^N (2[\mathbf{x}]\_q)^k a\_{i+1}(N - k, [\mathbf{x}]\_q) \\ &+ (-2) \sum\_{k=0}^N (2[\mathbf{x}]\_q)^k a\_{i-1}(N - k, [\mathbf{x}]\_q). \end{split} \tag{43}$$

Here, notice that the matrix *ai*(*j*, [*x*]*q*)0≤*i*,*j*≤*N*+<sup>1</sup> is given by

$$\begin{pmatrix} 1 & 2[\mathbf{x}]\_q & -2 + 4[\mathbf{x}]\_q^2 & -12[\mathbf{x}]\_q + 8[\mathbf{x}]\_q^3 & \cdots & \cdot & \cdot \\\\ 0 & (-2) & (-2)2(2[\mathbf{x}]\_q) & 12 - 24[\mathbf{x}]\_q^2 & \cdots & \cdot & \cdot \\\\ 0 & 0 & (-2)^2 & (-2)^2 3(2[\mathbf{x}]\_q) & \cdots & \cdot & \cdot \\\\ 0 & 0 & 0 & (-2)^3 & \ddots & \cdot & \cdot \\\\ \vdots & \vdots & \vdots & \vdots & \ddots & (-2)^N (N+1)(2[\mathbf{x}]\_q) \\\\ 0 & 0 & 0 & 0 & \cdots & (-2)^{N+1} \end{pmatrix} \tag{44}$$

From (33) to (43), we investigate the desired result immediately.

**Theorem 5.** *For N* = 0, 1, 2, . . . , *we have*

$$\mathbf{H}\_{m+N,q}(\mathbf{x}) = \sum\_{i=0}^{m} \frac{\mathbf{H}\_{m-i,q}(\mathbf{x}) a\_i(N, [\mathbf{x}]\_q) m!}{(m-i)!},\tag{45}$$

*where*

$$\begin{split} a\_{0}(N\_{\prime}\lceil\mathbf{x}\rceil\_{q}) &= \sum\_{k=0}^{N-1} 2^{k} \lfloor \mathbf{x} \rfloor\_{q}^{k} a\_{1}(N-1-k, \lceil \mathbf{x} \rceil\_{q}) + (2\lfloor \mathbf{x} \rfloor\_{q})^{N}, \\ a\_{N-1}(N\_{\prime}\lceil\mathbf{x}\rceil\_{q}) &= (-2)^{N-1} N(2\lceil \mathbf{x} \rceil\_{q}), \\ a\_{N}(N\_{\prime}\lceil\mathbf{x}\rceil\_{q}) &= (-2)^{N}, \\ a\_{i}(N+1, \lceil \mathbf{x} \rceil\_{q}) \\ &= (i+1) \sum\_{k=0}^{N} 2^{k} \lfloor \mathbf{x} \rfloor\_{q}^{k} a\_{i+1}(N-k, \lceil \mathbf{x} \rceil\_{q}) + (-2) \sum\_{k=0}^{N} 2^{k} \lfloor \mathbf{x} \rfloor\_{q}^{k} a\_{i-1}(N-k, \lceil \mathbf{x} \rceil\_{q}), \\ &\quad (1 \le i \le N-2). \end{split} \tag{46}$$

**Proof.** By making the *N*-times derivative for (4) with respect to *t*, we get

$$\mathcal{E}\left(\frac{\partial}{\partial t}\right)^N \mathcal{G}(t, [x]\_q) = \left(\frac{\partial}{\partial t}\right)^N e^{2[x]\_q t - t^2} = \sum\_{m=0}^\infty \mathbf{H}\_{m+N, q}(x) \frac{t^m}{m!}.\tag{47}$$

From (46) and (47), we obtain

$$a\_0(N\_\prime[\mathbf{x}]\_q)\mathcal{G}(t,[\mathbf{x}]\_q) + \dots + a\_1(N\_\prime[\mathbf{x}]\_q)t^N\mathcal{G}(t,[\mathbf{x}]\_q) = \sum\_{m=0}^\infty \mathbf{H}\_{m+N,q}(\mathbf{x})\frac{t^m}{m!},\qquad(48)$$

which makes the required result.

**Corollary 1.** *For N* = 0, 1, 2, . . . , *if we take m* = 0 *in (45), then, the following holds*

$$\mathbf{H}\_{N\mathcal{A}}(\mathbf{x}) = a\_0 (N\_{\prime\prime}[\mathbf{x}]\_{\mathcal{G}})\_{\prime\prime}$$

*where,*

$$\begin{aligned} a\_0(N, [\mathbf{x}]\_q) &= \sum\_{k=0}^{N-1} 2^k [\mathbf{x}]\_q^k a\_1(N - 1 - k, [\mathbf{x}]\_q) + (2[\mathbf{x}]\_q)^N, \\ a\_1(N, [\mathbf{x}]\_q) & \\ &= 2 \sum\_{k=0}^{N-1} (2[\mathbf{x}]\_q)^k a\_2(N - k - 1, [\mathbf{x}]\_q) + (-2) \sum\_{k=0}^{N-1} (2[\mathbf{x}]\_q)^k a\_0(N - k - 1, [\mathbf{x}]\_q). \end{aligned} (49)$$

For *N* = 0, 1, 2, . . . , the differential equation

$$\mathcal{G}^{(N)} = \left(\frac{\partial}{\partial t}\right)^N \mathcal{G}(t, [\mathbf{x}]\_q) = \left(\sum\_{i=0}^N a\_i(N, [\mathbf{x}]\_q) t^i\right) \mathcal{G}(t, [\mathbf{x}]\_q) \tag{50}$$

has a solution

$$\mathcal{G} = \mathcal{G}(t, [\mathbf{x}]\_q) = e^{2[\mathbf{x}]\_q t - t^2}. \tag{51}$$

The following Figure 1 is the graph representation for this solution by using MATHE-MATICA.

**Figure 1.** The surface for the solution G(*t*, [*x*]*q*).

We can find the left surface of Figure 1 when we choose −1 ≤ *x* ≤ 1, *q* = 1/10, and 0 ≤ *t* ≤ 1. Additionally, we can see the right surface of Figure 1 when we choose a condition such as −1 ≤ *x* ≤ 1, *q* = 1/3, and 0 ≤ *t* ≤ 1. It particularly shows a higher-resolution density of the plots in the right surface of Figure 1.
