2. Differential equations associated with the 3-variable Hermite polynomials

In this section, we study differential equations arising from the generating functions of the 3 variable Hermite polynomials.

Let

$$F = F(t, x, y, z) = e^{\mathbf{x}t + yt^2 + zt^3} = \sum\_{n=0}^{\infty} H\_n(\mathbf{x}, y, z) \frac{t^n}{n!} \quad \text{x, y, z, t \in \mathbb{C}}.\tag{4}$$

Differential Equations Arising from the 3-Variable Hermite Polynomials and Computation of Their Zeros http://dx.doi.org/10.5772/intechopen.74355 85

Then, by (4), we have

Theorem 4. For any positive integer n, we have

� � <sup>¼</sup> <sup>P</sup><sup>n</sup>

<sup>H</sup>4ð Þ¼ <sup>x</sup>; <sup>y</sup>; <sup>z</sup> <sup>x</sup><sup>4</sup> <sup>þ</sup> <sup>12</sup>x<sup>2</sup><sup>y</sup> <sup>þ</sup> <sup>12</sup>y<sup>2</sup> <sup>þ</sup> <sup>24</sup>xz,

<sup>H</sup>5ð Þ¼ <sup>x</sup>; <sup>y</sup>; <sup>z</sup> <sup>x</sup><sup>5</sup> <sup>þ</sup> <sup>20</sup>x<sup>3</sup><sup>y</sup> <sup>þ</sup> <sup>60</sup>xy<sup>2</sup> <sup>þ</sup> <sup>60</sup>x<sup>2</sup><sup>z</sup> <sup>þ</sup> <sup>120</sup>yz,

l¼0

n l � �

l¼0

n l � �

<sup>H</sup>6ð Þ¼ <sup>x</sup>; <sup>y</sup>; <sup>z</sup> <sup>x</sup><sup>6</sup> <sup>þ</sup> <sup>30</sup>x<sup>4</sup><sup>y</sup> <sup>þ</sup> <sup>180</sup>x<sup>2</sup>y<sup>2</sup> <sup>þ</sup> <sup>120</sup>y<sup>3</sup> <sup>þ</sup> <sup>120</sup>x<sup>3</sup><sup>z</sup> <sup>þ</sup> <sup>720</sup>xyz <sup>þ</sup> <sup>360</sup>z<sup>2</sup>,

<sup>þ</sup> <sup>20160</sup>xy<sup>2</sup><sup>z</sup> <sup>þ</sup> <sup>10080</sup>x<sup>2</sup>z<sup>2</sup> <sup>þ</sup> <sup>20160</sup>yz<sup>2</sup>:

<sup>þ</sup> <sup>907200</sup>x<sup>2</sup>yz<sup>2</sup> <sup>þ</sup> <sup>907200</sup>y<sup>2</sup>z<sup>2</sup> <sup>þ</sup> <sup>604800</sup>xz<sup>3</sup>:

2. Differential equations associated with the 3-variable Hermite

xtþyt2þzt<sup>3</sup>

<sup>H</sup>8ð Þ¼ <sup>x</sup>; <sup>y</sup>; <sup>z</sup> <sup>x</sup><sup>8</sup> <sup>þ</sup> <sup>56</sup>x<sup>6</sup><sup>y</sup> <sup>þ</sup> <sup>840</sup>x<sup>4</sup>y<sup>2</sup> <sup>þ</sup> <sup>3360</sup>x<sup>2</sup>y<sup>3</sup> <sup>þ</sup> <sup>1680</sup>y<sup>4</sup> <sup>þ</sup> <sup>336</sup>x<sup>5</sup><sup>z</sup> <sup>þ</sup> <sup>6720</sup>x<sup>3</sup>yz

<sup>H</sup>7ð Þ¼ <sup>x</sup>; <sup>y</sup>; <sup>z</sup> <sup>x</sup><sup>7</sup> <sup>þ</sup> <sup>42</sup>x<sup>5</sup><sup>y</sup> <sup>þ</sup> <sup>420</sup>x<sup>3</sup>y<sup>2</sup> <sup>þ</sup> <sup>840</sup>xy<sup>3</sup> <sup>þ</sup> <sup>210</sup>x<sup>4</sup><sup>z</sup> <sup>þ</sup> <sup>2520</sup>x<sup>2</sup>yz <sup>þ</sup> <sup>2520</sup>y<sup>2</sup><sup>z</sup> <sup>þ</sup> <sup>2520</sup>xz<sup>2</sup>,

<sup>H</sup>9ð Þ¼ <sup>x</sup>; <sup>y</sup>; <sup>z</sup> <sup>x</sup><sup>9</sup> <sup>þ</sup> <sup>72</sup>x<sup>7</sup><sup>y</sup> <sup>þ</sup> <sup>1512</sup>x<sup>5</sup>y<sup>2</sup> <sup>þ</sup> <sup>10080</sup>x<sup>3</sup>y<sup>3</sup> <sup>þ</sup> <sup>15120</sup>xy<sup>4</sup> <sup>þ</sup> <sup>504</sup>x<sup>6</sup><sup>z</sup> <sup>þ</sup> <sup>15120</sup>x<sup>4</sup>yz <sup>þ</sup> <sup>90720</sup>x<sup>2</sup>y<sup>2</sup><sup>z</sup> <sup>þ</sup> <sup>60480</sup>y<sup>3</sup><sup>z</sup> <sup>þ</sup> <sup>30240</sup>x<sup>3</sup>z<sup>2</sup> <sup>þ</sup> <sup>181440</sup>xyz<sup>2</sup> <sup>þ</sup> <sup>60480</sup>z<sup>3</sup>, <sup>H</sup>10ð Þ¼ <sup>x</sup>; <sup>y</sup>; <sup>z</sup> <sup>x</sup><sup>10</sup> <sup>þ</sup> <sup>90</sup>x<sup>8</sup><sup>y</sup> <sup>þ</sup> <sup>2520</sup>x<sup>6</sup>y<sup>2</sup> <sup>þ</sup> <sup>25200</sup>x<sup>4</sup>y<sup>3</sup> <sup>þ</sup> <sup>75600</sup>x<sup>2</sup>y<sup>4</sup> <sup>þ</sup> <sup>30240</sup>y<sup>5</sup> <sup>þ</sup> <sup>720</sup>x<sup>7</sup><sup>z</sup> <sup>þ</sup> <sup>30240</sup>x<sup>5</sup>yz <sup>þ</sup> <sup>302400</sup>x<sup>3</sup>y<sup>2</sup><sup>z</sup> <sup>þ</sup> <sup>604800</sup>xy<sup>3</sup><sup>z</sup> <sup>þ</sup> <sup>75600</sup>x<sup>4</sup>z<sup>2</sup>

Recently, many mathematicians have studied the differential equations arising from the generating functions of special polynomials (see [7, 8, 12, 16–19]). In this paper, we study differential equations arising from the generating functions of the 3-variable Hermite polynomials. We give explicit identities for the 3-variable Hermite polynomials. In addition, we investigate the zeros of the 3-variable Hermite polynomials using numerical methods. Using computer, a realistic study for the zeros of the 3-variable Hermite polynomials is very interesting. Finally, we observe an interesting phenomenon of 'scattering' of the zeros of the 3-variable Hermite polynomials.

In this section, we study differential equations arising from the generating functions of the 3-

<sup>¼</sup> <sup>X</sup><sup>∞</sup> n¼0

Hnð Þ x; y; z

t n n!

, x, y, z, t∈ℂ: (4)

The 3-variable Hermite polynomials can be determined explicitly. A few of them are

Hl <sup>x</sup>1; <sup>y</sup>1; <sup>z</sup> � �Hn�<sup>l</sup> <sup>x</sup>2; <sup>y</sup><sup>2</sup>

� �:

� �:

Hl <sup>x</sup>1; <sup>y</sup>1; <sup>z</sup> � �Hn�<sup>l</sup> <sup>x</sup>2; <sup>y</sup>2; <sup>z</sup><sup>2</sup>

<sup>1</sup> Hn <sup>x</sup><sup>1</sup> <sup>þ</sup> <sup>x</sup>2; <sup>y</sup><sup>1</sup> <sup>þ</sup> <sup>y</sup>2; <sup>z</sup> � � <sup>¼</sup> <sup>P</sup><sup>n</sup>

84 Differential Equations - Theory and Current Research

2 Hn x<sup>1</sup> þ x2; y<sup>1</sup> þ y2; z<sup>1</sup> þ z<sup>2</sup>

H0ð Þ¼ x; y; z 1, H1ð Þ¼ x; y; z x, <sup>H</sup>2ð Þ¼ <sup>x</sup>; <sup>y</sup>; <sup>z</sup> <sup>x</sup><sup>2</sup> <sup>þ</sup> <sup>2</sup>y, <sup>H</sup>3ð Þ¼ <sup>x</sup>; <sup>y</sup>; <sup>z</sup> <sup>x</sup><sup>3</sup> <sup>þ</sup> <sup>6</sup>xy <sup>þ</sup> <sup>6</sup>z,

polynomials

Let

variable Hermite polynomials.

F ¼ F tð Þ¼ ; x; y; z e

$$\begin{split} F^{(1)} = \frac{\partial}{\partial t} F(t, \mathbf{x}, y, z) = \frac{\partial}{\partial t} \left( e^{\mathbf{x}t + y\mathbf{f}^2 + z\mathbf{f}^3} \right) = e^{\mathbf{x}t + y\mathbf{f}^2 + z\mathbf{f}^3} \left( \mathbf{x} + 2y\mathbf{f} + 3z\mathbf{f}^2 \right) \\ = \left( \mathbf{x} + 2y\mathbf{f} + 3z\mathbf{f}^2 \right) F(t, \mathbf{x}, y, z), \end{split} \tag{5}$$

$$\begin{split} F^{(2)} &= \frac{\partial}{\partial t} F^{(1)}(t, \mathbf{x}, y, z) = (2y + 6zt) F(t, \mathbf{x}, y, z) + \left(\mathbf{x} + 2yt + 3zt^2\right) F^{(1)}(t, \mathbf{x}, y, z) \\ &= \left( (\mathbf{x}^2 + 2y) + (6z + 4\mathbf{x}y)t + \left(4y^2 + 6\mathbf{x}z\right)t^2 + (12yz)t^3 + (9z^2)t^4 \right) F(t, \mathbf{x}, y, z). \end{split} \tag{6}$$

Continuing this process, we can guess that

$$F^{(N)} = \left(\frac{\partial}{\partial t}\right)^N F(t, x, y, z) = \sum\_{i=0}^{2N} a\_i(N, x, y, z) t^i F(t, x, y, z), (N = 0, 1, 2, \dots). \tag{7}$$

Differentiating (7) with respect to t, we have

$$F^{(N+1)} = \frac{\partial F^{(N)}}{\partial t} = \sum\_{i=0}^{2N} a\_i(N, \mathbf{x}, y, z) i^{i-1} F(t, \mathbf{x}, y, z) + \sum\_{i=0}^{2N} a\_i(N, \mathbf{x}, y, z) i^i F^{(1)}(t, \mathbf{x}, y, z)$$

$$= \sum\_{i=0}^{2N} a\_i(N, \mathbf{x}, y, z) i^{i-1} F(t, \mathbf{x}, y, z) + \sum\_{i=0}^{2N} a\_i(N, \mathbf{x}, y, z) i^i (x + 2yt + 3zt^2) F(t, \mathbf{x}, y, z)$$

$$= \sum\_{i=0}^{2N} \text{ia}(N, \mathbf{x}, y, z) i^{-1} F(t, \mathbf{x}, y, z) + \sum\_{i=0}^{2N} \text{xa}(N, \mathbf{x}, y, z) i^i F(t, \mathbf{x}, y, z)$$

$$+ \sum\_{i=0}^{2N} 2ya\_i(N, \mathbf{x}, y, z) i^{i+1} F(t, \mathbf{x}, y, z) + \sum\_{i=0}^{2N} 3za\_i(N, \mathbf{x}, y, z) i^{i+2} F(t, \mathbf{x}, y, z)$$

$$= \sum\_{i=0}^{2N-1} (i + 1)a\_{i+1}(N, \mathbf{x}, y, z) i^i F(t, \mathbf{x}, y, z) + \sum\_{i=0}^{2N} \text{xa}(N, \mathbf{x}, y, z) i^i F(t, \mathbf{x}, y, z)$$

$$+ \sum\_{i=1}^{2N+1} 2ya\_{i-1}(N, \mathbf{x}, y, z) i^i F(t, \mathbf{x}, y, z) + \sum\_{i=2}^{2N+2} 3za\_{i-2}(N, \mathbf{x}, y, z) i^i F(t, \mathbf{x}, y, z)$$

Hence we have

$$\begin{split} F^{(N+1)} &= \sum\_{i=0}^{2N-1} (i+1)a\_{i+1}(N,x,y,z)t^i F(t,x,y,z) \\ &\quad + \sum\_{i=0}^{2N} xa\_i(N,x,y,z)t^i F(t,x,y,z) \\ &\quad + \sum\_{i=1}^{2N+1} 2ya\_{i-1}(N,x,y,z)t^i F(t,x,y,z) \\ &\quad + \sum\_{i=2}^{2N+2} 3za\_{i-2}(N,x,y,z)t^i F(t,x,y,z). \end{split} \tag{8}$$

Now replacing N by N þ 1 in (7), we find

$$F^{(N+1)} = \sum\_{i=0}^{2N+2} a\_i(N+1, \mathbf{x}, \mathbf{y}, \mathbf{z}) t^i \mathbf{F}(t, \mathbf{x}, \mathbf{y}, \mathbf{z}). \tag{9}$$

a2Nþ<sup>2</sup>ð Þ¼ N þ 1; x; y; z 3za2Nð Þ N; x; y; z , a2Nð Þ¼ N; x; y; z 3za2N�<sup>2</sup>ð Þ N � 1; x; y; z , …

<sup>1</sup> <sup>x</sup> <sup>2</sup><sup>y</sup> <sup>þ</sup> <sup>x</sup><sup>2</sup> � <sup>⋯</sup> �

0 2y 4xy þ 6z � ⋯ �

0 3<sup>z</sup> <sup>6</sup>xz <sup>þ</sup> <sup>4</sup>y<sup>2</sup> � <sup>⋯</sup> �

0 0 12yz � ⋯ �

00 0 � ⋯ �

⋮⋮ ⋮ ⋮ ⋱ �

F tð Þ¼ ; <sup>x</sup>; <sup>y</sup>; <sup>z</sup> <sup>X</sup>

F ¼ F tð Þ¼ ; x; y; z e

<sup>a</sup>1ð Þþ <sup>N</sup> � <sup>i</sup>; <sup>x</sup>; <sup>y</sup>; <sup>z</sup> xNþ<sup>1</sup>

<sup>a</sup>2Nð Þ¼ <sup>N</sup> <sup>þ</sup> <sup>1</sup>; <sup>x</sup>; <sup>y</sup>; <sup>z</sup> xa2Nð Þþ <sup>N</sup>; <sup>x</sup>; <sup>y</sup>; <sup>z</sup> <sup>2</sup>ya2N�<sup>1</sup>ð Þþ <sup>N</sup>; <sup>x</sup>; <sup>y</sup>; <sup>z</sup> <sup>3</sup>za2N�<sup>2</sup>ð Þ <sup>N</sup>; <sup>x</sup>; <sup>y</sup>; <sup>z</sup> ,

a1ð Þ¼ N þ 1; x; y; z 2a2ð Þþ N; x; y; z xa1ð Þþ N; x; y; z 2ya0ð Þ N; x; y; z ,

a2Nþ<sup>1</sup>ð Þ¼ N þ 1; x; y; z 2ya2Nð Þþ N; x; y; z 3za2N�<sup>1</sup>ð Þ N; x; y; z ,

Nþ1 ,

00 0 0 ⋯ ð Þ 3z

N

i¼0

00 0 3ð Þz

<sup>2</sup> � <sup>⋯</sup> �

<sup>3</sup> <sup>⋯</sup> �

aið Þ N; x; y; z t

!

xtþyt2þzt<sup>3</sup> ,

,

Nþ1

i

F tð Þ ; x; y; z

1

http://dx.doi.org/10.5772/intechopen.74355

CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCA

Nþ1 :

Differential Equations Arising from the 3-Variable Hermite Polynomials and Computation of Their Zeros

(17)

87

a2Nþ<sup>2</sup>ð Þ¼ N þ 1; x; y; z ð Þ 3z

Note that, here the matrix aið Þ <sup>j</sup>; <sup>x</sup>; <sup>y</sup> <sup>0</sup> <sup>≤</sup> <sup>i</sup> <sup>≤</sup> <sup>2</sup>Nþ2, <sup>0</sup> <sup>≤</sup> <sup>j</sup> <sup>≤</sup> <sup>N</sup>þ<sup>1</sup> is given by

0

BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB@

Therefore, we obtain the following theorem.

<sup>F</sup>ð Þ <sup>N</sup> <sup>¼</sup> <sup>∂</sup>

<sup>a</sup>0ð Þ¼ <sup>N</sup> <sup>þ</sup> <sup>1</sup>; <sup>x</sup>; <sup>y</sup>; <sup>z</sup> <sup>X</sup>

a2Nþ<sup>2</sup>ð Þ¼ N þ 1; x; y; z ð Þ 3z

has a solution

where

Theorem 5. For N ¼ 0; 1; 2, …, the differential equation

∂t � �<sup>N</sup>

N

i¼0 xi

00 3ð Þz

Comparing the coefficients on both sides of (8) and (9), we obtain

$$\begin{aligned} a\_0(N+1, \mathbf{x}, y, \mathbf{z}) &= a\_1(N, \mathbf{x}, y, \mathbf{z}) + \mathbf{x}a\_0(N, \mathbf{x}, y, \mathbf{z}), \\ a\_1(N+1, \mathbf{x}, y, \mathbf{z}) &= 2a\_2(N, \mathbf{x}, y, \mathbf{z}) + \mathbf{x}a\_1(N, \mathbf{x}, y, \mathbf{z}) + 2ya\_0(N, \mathbf{x}, y, \mathbf{z}), \\ a\_{2N}(N+1, \mathbf{x}, y, \mathbf{z}) &= \mathbf{x}a\_{2N}(N, \mathbf{x}, y, \mathbf{z}) + 2ya\_{2N-1}(N, \mathbf{x}, y, \mathbf{z}) + 3za\_{2N-2}(N, \mathbf{x}, y, \mathbf{z}), \\ a\_{2N+1}(N+1, \mathbf{x}, y, \mathbf{z}) &= 2ya\_{2N}(N, \mathbf{x}, y, \mathbf{z}) + 3za\_{2N-1}(N, \mathbf{x}, y, \mathbf{z}), \\ a\_{2N+2}(N+1, \mathbf{x}, y, \mathbf{z}) &= 3za\_{2N}(N, \mathbf{x}, y, \mathbf{z}), \end{aligned} \tag{10}$$

and

$$\begin{split} a\_i(\mathcal{N}+1, \mathbf{x}, \boldsymbol{y}, \boldsymbol{z}) &= (i+1)a\_{i+1}(\mathcal{N}, \mathbf{x}, \boldsymbol{y}, \boldsymbol{z}) + \mathbf{x}a\_i(\mathcal{N}, \mathbf{x}, \boldsymbol{y}, \boldsymbol{z}) \\ &+ 2ya\_{i-1}(\mathcal{N}, \mathbf{x}, \boldsymbol{y}, \boldsymbol{z}) + 3\mathbf{z}a\_{i-2}(\mathcal{N}, \mathbf{x}, \boldsymbol{y}, \boldsymbol{z}), (2 \le i \le 2\mathcal{N}-1). \end{split} \tag{11}$$

In addition, by (7), we have

$$F(t, \mathbf{x}, y, z) = F^{(0)}(t, \mathbf{x}, y, z) = a\_0(0, \mathbf{x}, y, z) F(t, \mathbf{x}, y, z),\tag{12}$$

which gives

$$a\_0(0, \mathbf{x}, y, z) = \mathbf{1}.\tag{13}$$

It is not difficult to show that

$$\begin{aligned} &xF(t,x,y) + 2ytF(t,x,y,z) + 3zt^2F(t,x,y,z) \\ &= F^{(1)}(t,x,y,z) \\ &= \sum\_{i=0}^{2} a\_i(1,x,y,z)F(t,x,y,z) \\ &= \left(a\_0(1,x,y,z) + a\_1(1,x,y,z)t + a\_2(1,x,y,z)t^2\right)F(t,x,y,z). \end{aligned} \tag{14}$$

Thus, by (14), we also find

$$a\_0(1, \mathbf{x}, y, z) = \mathbf{x}, \quad a\_1(1, \mathbf{x}, y, z) = 2y, \quad a\_2(1, \mathbf{x}, y, z) = 3z. \tag{15}$$

From (10), we note that

$$\begin{aligned} a\_0(N+1, \mathbf{x}, y, \mathbf{z}) &= a\_1(N, \mathbf{x}, y, \mathbf{z}) + \mathbf{x} a\_0(N, \mathbf{x}, y, \mathbf{z}), \\ a\_0(N, \mathbf{x}, y, \mathbf{z}) &= a\_1(N-1, \mathbf{x}, y, \mathbf{z}) + \mathbf{x} a\_0(N-1, \mathbf{x}, y, \mathbf{z}), \dots \\ a\_0(N+1, \mathbf{x}, y, \mathbf{z}) &= \sum\_{i=0}^{N} \mathbf{x}^i a\_1(N-i, \mathbf{x}, y, \mathbf{z}) + \mathbf{x}^{N+1} \end{aligned} \tag{16}$$

and

Differential Equations Arising from the 3-Variable Hermite Polynomials and Computation of Their Zeros http://dx.doi.org/10.5772/intechopen.74355 87

$$\begin{aligned} a\_{2N+2}(N+1, \mathbf{x}, y, z) &= 3za\_{2N}(N, \mathbf{x}, y, z), \\ a\_{2N}(N, \mathbf{x}, y, z) &= 3za\_{2N-2}(N-1, \mathbf{x}, y, z), \dots \\ a\_{2N+2}(N+1, \mathbf{x}, y, z) &= \left(3z\right)^{N+1}. \end{aligned} \tag{17}$$

Note that, here the matrix aið Þ <sup>j</sup>; <sup>x</sup>; <sup>y</sup> <sup>0</sup> <sup>≤</sup> <sup>i</sup> <sup>≤</sup> <sup>2</sup>Nþ2, <sup>0</sup> <sup>≤</sup> <sup>j</sup> <sup>≤</sup> <sup>N</sup>þ<sup>1</sup> is given by

$$\begin{pmatrix} 1 & x & 2y + x^2 & \cdots & \cdots \\ 0 & 2y & 4xy + 6z & \cdots & \cdots \\ 0 & 3z & 6xz + 4y^2 & \cdots & \cdots \\ 0 & 0 & 12yz & \cdots & \cdots & \cdots \\ 0 & 0 & \cdots & \cdots & \cdots & \cdots \\ \vdots \\ 0 & 0 & \cdots & \cdots & \cdots & \cdots \\ 0 & 0 & 0 & \cdots & \cdots & \cdots \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & \cdots & \cdots \\ \end{pmatrix}$$

Therefore, we obtain the following theorem.

Theorem 5. For N ¼ 0; 1; 2, …, the differential equation

$$F^{(N)} = \left(\frac{\partial}{\partial t}\right)^N F(t, x, y, z) = \left(\sum\_{i=0}^N a\_i(N, x, y, z)t^i\right) F(t, x, y, z)$$

has a solution

$$F = F(t, x, y, z) = e^{xt + yt^2 + zt^3} \, .$$

where

Now replacing N by N þ 1 in (7), we find

86 Differential Equations - Theory and Current Research

and

which gives

In addition, by (7), we have

It is not difficult to show that

Thus, by (14), we also find

From (10), we note that

and

<sup>F</sup>ð Þ <sup>N</sup>þ<sup>1</sup> <sup>¼</sup>

Comparing the coefficients on both sides of (8) and (9), we obtain

a0ð Þ¼ N þ 1; x; y; z a1ð Þþ N; x; y; z xa0ð Þ N; x; y; z ,

a2Nþ<sup>2</sup>ð Þ¼ N þ 1; x; y; z 3za2Nð Þ N; x; y; z ,

<sup>¼</sup> <sup>F</sup>ð Þ<sup>1</sup> ð Þ <sup>t</sup>; <sup>x</sup>; <sup>y</sup>; <sup>z</sup>

<sup>¼</sup> <sup>X</sup> 2

i¼0

2 X Nþ2

i¼0

a1ð Þ¼ N þ 1; x; y; z 2a2ð Þþ N; x; y; z xa1ð Þþ N; x; y; z 2ya0ð Þ N; x; y; z ,

aið Þ¼ N þ 1; x; y; z ð Þ i þ 1 aiþ<sup>1</sup>ð Þþ N; x; y; z xaið Þ N; x; y; z

xF tð Þþ ; <sup>x</sup>; <sup>y</sup> <sup>2</sup>ytF tð Þþ ; <sup>x</sup>; <sup>y</sup>; <sup>z</sup> <sup>3</sup>zt<sup>2</sup>F tð Þ ; <sup>x</sup>; <sup>y</sup>; <sup>z</sup>

¼ a0ð Þþ 1; x; y; z a1ð1; x; y; zÞt þ a2ð1; x; y; zÞt

a0ð Þ¼ N þ 1; x; y; z a1ð Þþ N; x; y; z xa0ð Þ N; x; y; z ,

N

i¼0 xi

a0ð Þ¼ N; x; y; z a1ðN � 1; x; y; zÞ þ xa0ð Þ N � 1; x; y; z , …

<sup>2</sup> � �F tð Þ ; <sup>x</sup>; <sup>y</sup>; <sup>z</sup> :

a0ð Þ¼ 1; x; y; z x, a1ð Þ¼ 1; x; y; z 2y, a2ð Þ¼ 1; x; y; z 3z: (15)

<sup>a</sup>1ð Þþ <sup>N</sup> � <sup>i</sup>; <sup>x</sup>; <sup>y</sup>; <sup>z</sup> xNþ<sup>1</sup>

,

aið Þ 1; x; y; z F tð Þ ; x; y; z

<sup>a</sup>0ð Þ¼ <sup>N</sup> <sup>þ</sup> <sup>1</sup>; <sup>x</sup>; <sup>y</sup>; <sup>z</sup> <sup>X</sup>

a2Nþ<sup>1</sup>ð Þ¼ N þ 1; x; y; z 2ya2Nð Þþ N; x; y; z 3za2N�<sup>1</sup>ð Þ N; x; y; z ,

aið Þ N þ 1; x; y; z t

<sup>a</sup>2Nð Þ¼ <sup>N</sup> <sup>þ</sup> <sup>1</sup>; <sup>x</sup>; <sup>y</sup>; <sup>z</sup> xa2Nð Þþ <sup>N</sup>; <sup>x</sup>; <sup>y</sup>; <sup>z</sup> <sup>2</sup>ya2N�<sup>1</sup>ð Þþ <sup>N</sup>; <sup>x</sup>; <sup>y</sup>; <sup>z</sup> <sup>3</sup>za2N�<sup>2</sup>ð Þ <sup>N</sup>; <sup>x</sup>; <sup>y</sup>; <sup>z</sup> ,

i

<sup>þ</sup> <sup>2</sup>yai�<sup>1</sup>ð Þþ <sup>N</sup>; <sup>x</sup>; <sup>y</sup>; <sup>z</sup> <sup>3</sup>zai�<sup>2</sup>ð Þ <sup>N</sup>; <sup>x</sup>; <sup>y</sup>; <sup>z</sup> ,ð Þ <sup>2</sup> <sup>≤</sup> <sup>i</sup> <sup>≤</sup> <sup>2</sup><sup>N</sup> � <sup>1</sup> : (11)

F tð Þ¼ ; <sup>x</sup>; <sup>y</sup>; <sup>z</sup> <sup>F</sup>ð Þ<sup>0</sup> ð Þ¼ <sup>t</sup>; <sup>x</sup>; <sup>y</sup>; <sup>z</sup> <sup>a</sup>0ð Þ <sup>0</sup>; <sup>x</sup>; <sup>y</sup>; <sup>z</sup> F tð Þ ; <sup>x</sup>; <sup>y</sup>; <sup>z</sup> , (12)

a0ð Þ¼ 0; x; y; z 1: (13)

F tð Þ ; x; y; z : (9)

(10)

(14)

(16)

$$\begin{aligned} a\_0(N+1, \mathbf{x}, y, z) &= \sum\_{i=0}^{N} \mathbf{x}^i a\_1(N-i, \mathbf{x}, y, z) + \mathbf{x}^{N+1}, \\ a\_1(N+1, \mathbf{x}, y, z) &= 2a\_2(N, \mathbf{x}, y, z) + \mathbf{x} a\_1(N, \mathbf{x}, y, z) + 2ya\_0(N, \mathbf{x}, y, z), \\ a\_{2N}(N+1, \mathbf{x}, y, z) &= \mathbf{x} a\_{2N}(N, \mathbf{x}, y, z) + 2ya\_{2N-1}(N, \mathbf{x}, y, z) + 3za\_{2N-2}(N, \mathbf{x}, y, z), \\ a\_{2N+1}(N+1, \mathbf{x}, y, z) &= 2ya\_{2N}(N, \mathbf{x}, y, z) + 3za\_{2N-1}(N, \mathbf{x}, y, z), \\ a\_{2N+2}(N+1, \mathbf{x}, y, z) &= \left(3z\right)^{N+1}, \end{aligned}$$

and

$$\begin{aligned} a\_i(N+1, \mathbf{x}, y, z) &= (i+1)a\_{i+1}(N, \mathbf{x}, y, z) + \mathbf{x}a\_i(N, \mathbf{x}, y, z) \\ &+ 2ya\_{i-1}(N, \mathbf{x}, y, z) + 3za\_{i-2}(N, \mathbf{x}, y, z), (2 \le i \le 2N - 1). \end{aligned}$$

From (4), we note that

$$F^{(N)} = \left(\frac{\partial}{\partial t}\right)^N F(t, x, y, z) = \sum\_{k=0}^n H\_{k+N}(x, y, z) \frac{t^k}{k!}.\tag{18}$$

F ¼ F tð Þ¼ ; x; y; z e

3. Distribution of zeros of the 3-variable Hermite polynomials

y ¼ �1 � i, and z ¼ �3 � i.

Figure 1. The surface for the solution F tð Þ ; x; y; z .

Here is a plot of the surface for this solution. In Figure 1(left), we choose �2 ≤ z ≤ 2, �1 ≤ t ≤ 1, x ¼ 2, and y ¼ �4. In Figure 1(right), we choose �5 ≤ x ≤ 5, � 1 ≤ t ≤ 1, y ¼ �3, and z ¼ �1.

This section aims to demonstrate the benefit of using numerical investigation to support theoretical prediction and to discover new interesting pattern of the zeros of the 3-variable Hermite polynomials Hnð Þ x; y; z . By using computer, the 3-variable Hermite polynomials Hnð Þ x; y; z can be determined explicitly. We display the shapes of the 3-variable Hermite polynomials Hnð Þ x; y; z and investigate the zeros of the 3-variable Hermite polynomials Hnð Þ x; y; z . We investigate the beautiful zeros of the 3-variable Hermite polynomials Hnð Þ x; y; z by using a computer. We plot the zeros of the Hnð Þ x; y; z for n ¼ 20, y ¼ 1, � 1, 1 þ i, � 1 � i, z ¼ 3, � 3, 3 þ i, � 3 � i and x∈ C (Figure 2). In Figure 2(top-left), we choose n ¼ 20, y ¼ 1, and z ¼ 3. In Figure 2(top-right), we choose n ¼ 20, y ¼ �1, and z ¼ �3. In Figure 2(bottomleft), we choose n ¼ 20, y ¼ 1 þ i, and z ¼ 3 þ i. In Figure 2(bottom-right), we choose n ¼ 20,

In Figure 3(top-left), we choose n ¼ 20, x ¼ 1, and y ¼ 1. In Figure 3(top-right), we choose n ¼ 20, x ¼ �1, and y ¼ �1. In Figure 3(bottom-left), we choose n ¼ 20, x ¼ 1 þ i, and

Stacks of zeros of the 3-variable Hermite polynomials Hnð Þ x; y; z for 1 ≤ n ≤ 20 from a 3-D structure are presented (Figure 3). In Figure 4(top-left), we choose n ¼ 20, y ¼ 1, and z ¼ 3. In Figure 4 (top-right), we choose n ¼ 20, y ¼ �1, and z ¼ �3. In Figure 4(bottom-left), we choose n ¼ 20, y ¼ 1 þ i, and z ¼ 3 þ i. In Figure 4(bottom-right), we choose n ¼ 20, y ¼ �1 � i, and z ¼ �3 � i.

y ¼ 1 þ i. In Figure 3(bottom-right), we choose n ¼ 20, x ¼ �1 � i, and y ¼ �1 � i.

xtþyt2þzt<sup>3</sup> :

http://dx.doi.org/10.5772/intechopen.74355

89

Differential Equations Arising from the 3-Variable Hermite Polynomials and Computation of Their Zeros

By (4) and (18), we get

$$\begin{split} e^{-nt} \left( \frac{\partial}{\partial t} \right)^{N} F(t, x, y, z) &= \left( \sum\_{m=0}^{m} (-n)^{m} \frac{t^{m}}{m!} \right) \left( \sum\_{m=0}^{m} H\_{m+N}(x, y, z) \frac{t^{m}}{m!} \right) \\ &= \sum\_{m=0}^{\infty} \left( \sum\_{k=0}^{m} \binom{m}{k} (-n)^{m-k} H\_{N+k}(x, y, z) \right) \frac{t^{m}}{m!} . \end{split} \tag{19}$$

By the Leibniz rule and the inverse relation, we have

$$\begin{split} e^{-nt} \left( \frac{\partial}{\partial t} \right)^{N} F(t, \mathbf{x}, \mathbf{y}, \mathbf{z}) &= \sum\_{k=0}^{N} \binom{N}{k} n^{N-k} \left( \frac{\partial}{\partial t} \right)^{k} \left( e^{-nt} F(t, \mathbf{x}, \mathbf{y}, \mathbf{z}) \right) \\ &= \sum\_{m=0}^{\bullet} \left( \sum\_{k=0}^{N} \binom{N}{k} n^{N-k} H\_{m+k}(\mathbf{x} - n, \mathbf{y}, \mathbf{z}) \right) \frac{t^{m}}{m!} . \end{split} \tag{20}$$

Hence, by (19) and (20), and comparing the coefficients of <sup>t</sup> m <sup>m</sup>! gives the following theorem. Theorem 6. Let m, n, N be nonnegative integers. Then

$$\sum\_{k=0}^{m} \binom{m}{k} (-n)^{m-k} H\_{N+k}(\mathbf{x}, y, z) = \sum\_{k=0}^{N} \binom{N}{k} n^{N-k} H\_{m+k}(\mathbf{x} - n, y, z). \tag{21}$$

If we take m ¼ 0 in (21), then we have the following corollary.

Corollary 7. For N ¼ 0; 1; 2, …, we have

$$H\_N(\mathbf{x}, y, z) = \sum\_{k=0}^{N} \binom{N}{k} n^{N-k} H\_k(\mathbf{x} - n, y, z).$$

For N ¼ 0; 1; 2, …, the differential equation

$$F^{(N)} = \left(\frac{\partial}{\partial t}\right)^N F(t, x, y, z) = \left(\sum\_{i=0}^N a\_i(N, x, y, z)t^i\right) F(t, x, y, z)$$

has a solution

$$F = F(t, x, y, z) = e^{\varkappa t + y t^2 + z t^3}.$$

Here is a plot of the surface for this solution. In Figure 1(left), we choose �2 ≤ z ≤ 2, �1 ≤ t ≤ 1, x ¼ 2, and y ¼ �4. In Figure 1(right), we choose �5 ≤ x ≤ 5, � 1 ≤ t ≤ 1, y ¼ �3, and z ¼ �1.
