**2. The Hermite Collocation Method (HCM)**

In this section, we present the Hermite collocation method to obtain approximate solutions to Equation (1) in the truncated Hermite series form

*Mathematics* **2018**, *6*, 305

$$S(t) = \sum\_{l=0}^{L} c\_{1,l} H\_l(t), \ I(t) = \sum\_{l=0}^{L} c\_{2,l} H\_l(t) \text{ and } R(t) = \sum\_{l=0}^{L} c\_{3,l} H\_l(t). \tag{3}$$

Here, *c*1,*l*, *c*2,*l*, and *c*3,*<sup>l</sup>* (*l* = 0, 1, 2, . . . , *L*) are the unknown Hermite coefficients, *L* is any positive number where *L* ≥ *m* (*m* is the number of equations in the system), and *Hl*(*t*), *l* = 0, 1, 2, ... , *L* are the Hermite polynomials. The Hermite polynomials are identified by

$$H\_l(t) = l! \sum\_{m=0}^{L} \frac{(-1)^m}{m!(l-2m)!} (2t)^{l-2m}, \ l \in \mathbb{N}, \ 0 \le t \le \infty \tag{4}$$

where *L* = *l*/2 if *l* is even and *L* = (*l* − 1)/2 if *l* is odd.

We represent Equation (1) in the form of matrices. Firstly, we write the approximate solutions of Equation (1):

$$\begin{aligned} S(t) &= H(t)\mathbf{C}\_1\\ I(t) &= H(t)\mathbf{C}\_2\\ R(t) &= H(t)\mathbf{C}\_3 \end{aligned} \tag{5}$$

where

$$\begin{aligned} H(t) &= \begin{bmatrix} H\_0(t) & H\_1(t) & \dots & H\_{L-1}(t) & H\_L(t) \end{bmatrix}, \mathbf{C}\_1 = \begin{bmatrix} \mathbf{c}\_{1,0} & \mathbf{c}\_{1,1} & \dots & \mathbf{c}\_{1,L} \end{bmatrix}^T \\\\ \mathbf{C}\_2 &= \begin{bmatrix} \mathbf{c}\_{2,0} & \mathbf{c}\_{2,1} & \dots & \mathbf{c}\_{2,L} \end{bmatrix}^T, \mathbf{C}\_3 = \begin{bmatrix} \mathbf{c}\_{3,0} & \mathbf{c}\_{3,1} & \dots & \mathbf{c}\_{3,L} \end{bmatrix}^T \end{aligned}$$

If *L* is an odd number,

$$\underbrace{\begin{bmatrix}H\_{0}(t)\\H\_{1}(t)\\ \vdots\\H\_{L-1}(t)\\ \hline H\_{L}(t)\end{bmatrix}}\_{H^{\mathsf{T}}(t)} = \underbrace{\begin{bmatrix}2^{0} & 0 & \dots & 0 & 0\\0 & 2^{1} & & \dots & 0 & 0\\ \vdots & & \vdots & & \ddots & \vdots\\ (-1)^{\left(\frac{L-5}{2}\right)} \frac{2^{0}}{W} \frac{(L-1)}{\left(\frac{L-1}{2}\right)!} & & \dots & 2^{L-1} & 0\\ 0 & (-1)^{\left(\frac{L-1}{2}\right)} \frac{2^{1}}{W} \frac{(L-1)}{\left(\frac{L-1}{2}\right)!} & 0 & \dots & 0 & 2^{L}\end{bmatrix}} \underbrace{\begin{bmatrix}1\\t\\ \vdots\\ \frac{t^{L-1}}{4}\\ \hline t^{L}\end{bmatrix}}\_{X^{\mathsf{T}}(t)}.\tag{6}$$

If *L* is an even number,

$$
\underbrace{\begin{bmatrix}H\_0(t) \\ H\_1(t) \\ \vdots \\ H\_{L-1}(t) \\ \hline H\_L(t) \end{bmatrix}}\_{H^{\mathsf{T}}(t)} = \underbrace{\begin{bmatrix} 2^0 & 0 & \dots & 0 & 0 \\ 0 & 2^1 & \dots & 0 & 0 \\ \vdots & \vdots & \ddots & & \vdots & \vdots \\ & & & & \\ 0 & (-1)^{(\frac{L-2}{2})} \frac{t^1}{1!} \frac{(L-1)}{(\frac{L-2}{2})!} & \dots & 2^{L-1} & 0 \\ (-1)^{(\frac{L-4}{2})} \frac{t^0}{0!} \frac{(L)}{(\frac{L}{2})!} & 0 & \dots & 0 & 2^L \end{bmatrix} \begin{bmatrix} 1 \\ t \\ \vdots \\ t^{L-1} \\ t^L \end{bmatrix}}\_{X^{\mathsf{T}}(t)} \tag{7}
$$

where *<sup>X</sup>*(*t*) = 1 *t t*<sup>2</sup> ... *t<sup>L</sup>* . Therefore, we can write the following equations:

$$\begin{aligned} S(t) &= X(t)F^T \mathbf{C}\_1\\ I(t) &= X(t)F^T \mathbf{C}\_3\\ R(t) &= X(t)F^T \mathbf{C}\_3. \end{aligned} \tag{8}$$

The relation between the matrix *X*(*t*) and its derivative *X*(1)(*t*) is

$$X^{(1)}(t) = X(t)B^T\tag{9}$$

where

$$B^T = \begin{bmatrix} 0 & 1 & 0 & \dots & 0 \\ 0 & 0 & 2 & \dots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \dots & L \\ 0 & 0 & 0 & \dots & 0 \end{bmatrix}.$$

From Equations (8) and (9), we obtain the following equations:

$$S^{(1)}(t) = X(t)B^T F^T \mathbb{C}\_{1\prime} \, ^t I^{(1)}(t) = X(t)B^T F^T \mathbb{C}\_{2\prime} \, R^{(1)}(t) = X(t)B^T F^T \mathbb{C}\_3. \tag{10}$$

Thus, we can construct the matrices *v*(*t*) and *v*(1)(*t*) as follows:

$$v(t) = \overline{X}\,\overline{F}C \text{ and } v^{(1)}(t) = \overline{X}\,\overline{B}\,\overline{F}C \tag{11}$$

where

$$\boldsymbol{w}(t) = \begin{bmatrix} \boldsymbol{S}(t) \\ \boldsymbol{I}(t) \\ \boldsymbol{R}(t) \end{bmatrix}, \boldsymbol{v}^{(1)}(t) = \begin{bmatrix} \boldsymbol{S}^{(1)}(t) \\ \boldsymbol{I}^{(1)}(t) \\ \boldsymbol{R}^{(1)}(t) \end{bmatrix}, \boldsymbol{\overline{X}}(t) = \begin{bmatrix} \boldsymbol{X}(t) & \boldsymbol{0} \\ \boldsymbol{0} & \boldsymbol{X}(t) & \boldsymbol{0} \\ \boldsymbol{0} & \boldsymbol{0} & \boldsymbol{X}(t) \end{bmatrix}, \boldsymbol{\overline{F}} = \begin{bmatrix} \boldsymbol{F}^{T} & \boldsymbol{0} \\ \boldsymbol{0} & \boldsymbol{F}^{T} & \boldsymbol{0} \\ \boldsymbol{0} & \boldsymbol{0} & \boldsymbol{F}^{T} \end{bmatrix}$$

and

$$
\overline{B} = \begin{bmatrix} B^T & 0 & 0 \\ 0 & B^T & 0 \\ 0 & 0 & B^T \end{bmatrix}, \\
\mathcal{C} = \begin{bmatrix} \mathcal{C}\_1 \\ \mathcal{C}\_2 \\ \mathcal{C}\_3 \end{bmatrix}.
$$

We can express Equation (1) in the matrix form

$$\left(\upsilon^{(1)}(t) - Kv(t) - Mv\_{1,2}(t)\right) = \mathbf{g} \tag{12}$$

where

$$\mathbf{J} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}, \mathbf{K} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & -\gamma & 0 \\ 0 & \gamma & 0 \end{bmatrix}, \mathcal{M} = \begin{bmatrix} -\beta \\ \beta \\ 0 \end{bmatrix}, \boldsymbol{\upsilon}\_{1,2} = \left[ \mathcal{S}(t)\boldsymbol{I}(t) \right].$$

Now let us determine the unknown coefficients *c*1,*l*, *c*2,*l*, and *c*3,*l*. We can use the collocation points defined by

$$t\_i = a + \frac{b - a}{L}i, \ i = 0, 1, \dots, L \tag{13}$$

for an interval *a* ≤ *t* ≤ *b*.

*Mathematics* **2018**, *6*, 305

By using the collocation points in Equation (12), we obtain the following system of matrix equations:

$$
\sigma^{(1)}(t\_i) - K\upsilon(t\_i) - M\upsilon\_{1,2}(t\_i) = \text{g.} \tag{14}
$$

$$\begin{aligned} V^{(1)} = \begin{bmatrix} v^{(1)}(t\_0) \\ v^{(1)}(t\_1) \\ \vdots \\ v^{(1)}(t\_L) \end{bmatrix}, \overline{\mathcal{K}} = \begin{bmatrix} \mathcal{K} & 0 & \cdots & 0 \\ 0 & \mathcal{K} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \mathcal{K} \end{bmatrix}\_{(L+1)\times(L+1)}, \upsilon = \begin{bmatrix} v(t\_0) \\ v(t\_1) \\ \vdots \\ v(t\_L) \end{bmatrix}, \mathcal{G} = \begin{bmatrix} \mathcal{g} \\ \mathcal{g} \\ \vdots \\ \mathcal{g} \end{bmatrix}\_{(L+1)\times1} \end{aligned}$$

$$
\begin{aligned}
\widetilde{V} = \begin{bmatrix} v\_{1,2}(t\_0) \\ v\_{1,2}(t\_1) \\ \vdots \\ v\_{1,2}(t\_L) \end{bmatrix}, \overline{\mathcal{M}} = \begin{bmatrix} \mathcal{M} & 0 & \cdots & 0 \\ 0 & \mathcal{M} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \mathcal{M} \end{bmatrix}\_{(L+1)\times(L+1)}.
\end{aligned}
$$

By aid of the upper matrices, Equation (1) can be written in the following matrix form:

$$
\overline{V}V^{(1)} - \overline{K}V - \overline{M}\,\overline{V} = \mathcal{G}.\tag{15}
$$

.

By putting the collocation points of Equation (13) in Equation (11), because we can write recurrence relations

$$w(t\_i) = \overline{X}(t\_i)\overline{F} \mathbf{C} \text{ and } w^{(1)}(t\_i) = \overline{X}(t\_i)\,\overline{B}\,\overline{F}\mathbf{C},$$

we can write

$$V = X\overline{\text{FC and }}V^{(1)} = X\overline{\text{BFC}}\tag{16}$$

so that

$$X = \begin{bmatrix} \overline{X}(t\_0) & \overline{X}(t\_1) & \cdots & \overline{X}(t\_L) \end{bmatrix}^T,\\ \overline{X}(t\_i) = \begin{bmatrix} X(t\_i) & 0 & 0 \\ 0 & X(t\_i) & 0 \\ 0 & 0 & X(t\_i) \end{bmatrix}.$$

Let us put the collocation points into the *v*1,2(*t*). We then obtain the matrix form

$$
\widetilde{\boldsymbol{V}} = \begin{bmatrix} \boldsymbol{v}\_{1,2}(t\_0) \\ \boldsymbol{v}\_{1,2}(t\_1) \\ \vdots \\ \boldsymbol{v}\_{1,2}(t\_L) \end{bmatrix} = \begin{bmatrix} I(t\_0) & 0 & \cdots & 0 \\ 0 & I(t\_1) & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & I(t\_L) \end{bmatrix} \begin{bmatrix} \boldsymbol{S}(t\_0) \\ \boldsymbol{S}(t\_1) \\ \vdots \\ \boldsymbol{S}(t\_L) \end{bmatrix} = \widetilde{\boldsymbol{I}} \, \widetilde{\boldsymbol{S}} \tag{17}
$$

where

$$
\overline{I} = \widetilde{X} \, \widetilde{F} \, \overline{C\_2}, \text{and } \overline{S} = \widetilde{T} \widetilde{F} \, \overline{C} \tag{18}
$$

so that

$$\begin{aligned} \, ^0X = \begin{bmatrix} X(t\_0) & 0 & \cdots & 0 \\ 0 & X(t\_1) & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & X(t\_L) \end{bmatrix}, \mathbf{\overline{C}}\_2 = \begin{bmatrix} \mathbf{C}\_2 & 0 & \cdots & 0 \\ 0 & \mathbf{C}\_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \mathbf{C}\_2 \end{bmatrix}\_{(L+1)\times(L+1)} \end{aligned}$$

$$
\widetilde{F} = \begin{bmatrix} F^T & 0 & \cdots & 0 \\ 0 & F^T & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & F^T \end{bmatrix}\_{(L+1)\times(L+1)}
$$

$$
\widetilde{X} = \begin{bmatrix} X(t\_0) \\ X(t\_1) \\ \vdots \\ X(t\_L) \end{bmatrix}, \widetilde{\mathcal{F}} = \begin{bmatrix} F^T & S & S \end{bmatrix}, \mathcal{S} = \begin{bmatrix} 0 & 0 & \cdots & 0 \\ 0 & 0 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 0 \end{bmatrix}\_{(L+1)\times(L+1)}.
$$

From Equations (16)–(18), we obtain the fundamental matrix equation

$$\{X\mathbb{B}\,\mathbb{F}-\mathbb{K}X\mathbb{F}-\overline{M}\hat{X}\,\mathbb{F}\mathbb{C}\_2\dot{X}\,\dot{\mathbb{F}}\}\mathbb{C}=\mathbb{G}.\tag{19}$$

Shortly, Equation (19) can be written as

$$\mathcal{NC} = \mathcal{G} \text{ or } [\mathcal{W}; \mathcal{G}] \tag{20}$$

.

$$
\mathcal{W} = X \mathcal{B} \, \mathcal{F} - \mathsf{K} X \mathcal{F} - \mathsf{M} \vec{X} \, \mathcal{F} \mathsf{C}\_2 \vec{X} \, \vec{F}.\tag{21}
$$

Equation (21) subtends a system of 3(*L* + 1) nonlinear algebraic equations with the unknown Hermite coefficients *c*1,*l*, *c*2,*l*, and *c*3,*l*. By placing *t* → 0 in Equation (5), the matrix forms of the initial conditions can be expressed by

$$\begin{aligned} S(t) &= H(0)\mathbf{C}\_1 = [N\_\mathcal{S}] \\ I(t) &= H(0)\mathbf{C}\_2 = [N\_I] \\ R(t) &= H(0)\mathbf{C}\_3 = [N\_R] \end{aligned} \tag{22}$$

That is, these matrix forms can be expressed by

$$\begin{aligned} lL\_1 &= S(0) = \begin{bmatrix} c\_{1,0} & c\_{1,1} & \cdots & c\_{1,L} \end{bmatrix} \\ lL\_2 &= I(0) = \begin{bmatrix} c\_{2,0} & c\_{2,1} & \cdots & c\_{2,L} \end{bmatrix} \\ lL\_3 &= R(0) = \begin{bmatrix} c\_{3,0} & c\_{3,1} & \cdots & c\_{3,L} \end{bmatrix} . \end{aligned} \tag{23}$$

When the rows in the matrices in Equation (23) are replaced with any three rows of the matrix in Equation (20), we obtain the solution to Equation (1) under initial conditions. Thereby, we get the augmented matrix

$$
\tilde{\mathcal{W}}\mathcal{C} = \tilde{\mathcal{G}},\tag{24}
$$

which is an algebraic system. To determine the coefficients, this system must be solved. The determined coefficients *ci*,0, *ci*,1, ··· , *ci*,*L*, (*i* = 1, 2, 3) are substituted into Equation (3), and we can then obtain approximate solutions.
