A Novel Technique to Control the Accuracy of a Nonlinear Fractional Order Model of COVID-19: Application of the CESTAC Method and the CADNA Library

Abstract

In this paper, a nonlinear fractional order model of COVID-19 is approximated. For this aim, at first we apply the Caputo–Fabrizio fractional derivative to model the usual form of the phenomenon. In order to show the existence of a solution, the Banach fixed point theorem and the Picard–Lindelof approach are used. Additionally, the stability analysis is discussed using the fixed point theorem. The model is approximated based on Indian data and using the homotopy analysis transform method (HATM), which is among the most famous, flexible and applicable semi-analytical methods. After that, the CESTAC (Controle et Estimation Stochastique des Arrondis de Calculs) method and the CADNA (Control of Accuracy and Debugging for Numerical Applications) library, which are based on discrete stochastic arithmetic (DSA), are applied to validate the numerical results of the HATM. Additionally, the stopping condition in the numerical algorithm is based on two successive approximations and the main theorem of the CESTAC method can aid us analytically to apply the new terminations criterion instead of the usual absolute error that we use in the floating-point arithmetic (FPA). Finding the optimal approximations and the optimal iteration of the HATM to solve the nonlinear fractional order model of COVID-19 are the main novelties of this study.

1. Introduction

Corona virus infection (COVID-19) is an infectious disease caused by the newly discovered Corona virus. Most people with COVID-19 disease experience mild to moderate symptoms and recover without special treatment. The virus that causes COVID-19 is mainly transmitted through particles produced when a person coughs, sneezes, or exhales. These particles do not stay suspended in the air due to their weight and fall quickly on the ground or surfaces. If you are close to a person with COVID-19, you may become infected by inhaling the virus or touching the infected surface and then touching your eyes, nose, or mouth. According to WHO reports until 2 March 2021, more than 115 million infected, more than 2.5 million dead and more than 90 million recovered people from more than 200 countries have been identified [1].

Thus, providing mathematical models to control and predict the behavior of this pandemic is urgent [2]. Recently, many mathematical models related to COVID-19 have been studied [3,4]. In [5], an analysis model of diagnosis and treatment for the COVID-19 pandemic based on medical information fusion was discussed, in [6] a model of the COVID-19 epidemic and predicting its future situation in Ethiopia was studied and in [7] a fractional order COVID-19 model with real-life data application was analyzed. The COVID-19 epidemic and clinical risk factors of patients under an epidemiological Markov model were discussed in [8] and the incorporating social determinants of health into modeling of COVID-19 and other infectious diseases was illustrated in [9].

Recently, due to their importance and more accurate results, fractional order models are among the most studied problems in the world. In [10], the authors focused on a Caputo–Fabrizio fractional model of COVID-19. In [11], a fractional order model of HIV infection was discussed, in [12] the application of the Atangana–Baleanu fractional derivative in mathematical biology was studied and in [13], a fractional order HIV model with Caputo and constant proportional Caputo operators was illustrated. An application of the Caputo–Fabrizio fractional derivative to solve a fractional order model of an energy supply–demand system can be found in [14]. For other applications of the fractional order models and also their usual forms please see [15,16,17,18,19,20].

The homotopy analysis method (HAM) presented by Liao [21,22,23] is one of the most flexible and accurate methods for solving various linear and nonlinear problems. This method has been applied for solving many problems, such as solving integral equations [24,25], solving singular problems [26,27] and a modified non-linear epidemiological model of computer viruses [28].

We should note that all of the mentioned researches are based on the FPA, and in order to show the accuracy of the methods, we apply the traditional absolute error or residual error in the form of

$$|g\left(x\right)-g_m\left(x\right)|<\epsilon ,or|g_m\left(x\right)-g_{m-1}\left(x\right)|<\epsilon ,$$

(1)

where the exact and approximate solutions are denoted by $g\left(x\right)$ and $g_m\left(x\right)$, respectively and $\epsilon$ is a small positive value. Thus, in order to apply Condition (1), not only should we know the exact solution $g\left(x\right)$ but we should also have the optimal value of $\epsilon$, which are disadvantages of methods based on the FPA. Concerning $\epsilon$, for small values we will have extra iterations without improving the accuracy of the results and for large values we will have only one or two iterations without providing accurate results. Thus, we apply the CESTAC method and the CADNA library, which are based on the DSA. This method was presented by Laporte and Vignes [29] for the first time and after that other researchers improved the method and library [30]. Using this method we do not need to have the exact solutions and $\epsilon$ and the termination criterion will be in the following form:

$$|g_m\left(x\right)-g_{m-1}\left(x\right)|=@.0,$$

(2)

where $g_m\left(x\right)$ and $g_{m-1}\left(x\right)$ are two successive approximations and $@.0$ denotes the informatical zero. Applying Condition (2), we do not need to have the exact solution and it is important that it does not depend on a positive small value such as $\epsilon$. Additionally, in this method, instead of applying the usual softwares, we apply the CADNA library. This library should be applied on a Linux operating system and all codes should be written using C, C++, FORTRAN or ADA codes [31,32,33]. The informatical zero sign can be produced only using the DSA and the CADNA library. This shows that the number of common significant digits for two successive approximations are almost equal to the number of common significant digits of the exact and approximate solutions [33,34,35]. Thus, we will be able to apply Condition (2) instead of (1). Recently, we focused on validating the numerical results of various problems using the CESTAC method. In [36], the results of the Adomian decomposition method for solving Volterra integral equation with a discontinuous kernel was studied and in [37,38] the CESTAC method was used to validate the results on the reverse osmosis system. Dynamical control on the homotopy perturbation method and the Taylor-collocation method to solve Volterra integral equations with piecewise smooth kernels was discussed in [39,40]. In [41], we illustrated the Sinc-collocation method for solving various kinds of crisp and fuzzy integral equations and finally, we applied the CADNA library to control the accuracy of the load leveling problem in [42].

In this paper, a nonlinear fractional order model of COVID-19 will be discussed based on the Caputo–Fabrizio fractional derivative. The Picard–Lindelof approach and the Banach fixed point theorem are used to show the existence of the solution of the model. Additionally, the stability of the model is illustrated applying the fixed point theorem. The HATM is applied to solve the model and the numerical results are validated applying the CESTAC method and the CADNA library. Additionally, the main theorem of the CESTAC method is proven, which guarantees the accuracy of the numerical results obtained using Condition (2) instead of (1). Using this method, the optimal approximations and the optimal iteration to solve the model using the HATM are obtained, which are the main novelties of this study.

2. Preliminaries

In this section, we present some definitions and details of fractional derivatives [43,44].

Definition 1

([43]). Let $g\in C^n$. The ω-th order Caputo fractional derivative can be defined via integrable differentiations as follows:

$${}^C{\mathfrak{D}}^{\omega }g\left(t\right)=\frac{1}{\Gamma (n-\omega )}{\int }_0^t\frac{g^{\left(n\right)}\left(s\right)}{{(t-s)}^{\omega -n+1}}ds,n=\left[\omega \right]+1,$$

where $g^{\left(n\right)}\in L^1$.

Definition 2

([43]). The ω-th order Caputo–Fabrizio derivative for $b>a,g\in H^1(a,b)$, and $\omega \in (0,1)$ is defined in the following form:

$${}^{CF}{\mathfrak{D}}^{\omega }g\left(t\right)=\frac{M\left(\omega \right)}{1-\omega }{\int }_a^{t}exp\left(\frac{-\omega }{1-\omega }(t-s)\right)g^{\prime }\left(s\right)ds,t\ge 0,$$

where $M\left(\omega \right)$ depending on ω denotes a normalization function and $M\left(0\right)=M\left(1\right)=1$. For a function g that is not in $H^1(a,b)$, this derivative can be presented for $g\in L^1(–\infty ,b)$ as follows:

$${}^{CF}{\mathfrak{D}}^{\omega }g\left(t\right)=\frac{\omega M\left(\omega \right)}{1-\omega }{\int }_{-\infty }^b(g\left(t\right)-g\left(s\right))exp\left(\frac{-\omega }{1-\omega }(t-s)\right)ds,0<\omega <1.$$

In addition, the $\omega +n$-th order fractional derivative ${}^{CF}{\mathfrak{D}}^{\omega +n}$ for $n\ge 1$ and $\omega \in (0,1)$ can be defined as ${}^{CF}{\mathfrak{D}}^{\omega +n}g\left(t\right):{=}^{CF}{\mathfrak{D}}^{\omega }\left({\mathfrak{D}}^{n}g\left(t\right)\right)$.

Definition 3

([44]). For $0<\omega \le 1$ and $M\left(\omega \right)=1$, the Laplace transform for the Caputo–Fabrizio derivative can be presented in the following form:

$$L{[}^{CF}{\mathfrak{D}}^{\omega +n}g\left(t\right)]\left(s\right)=\frac{s^{n+1}L\left[g\left(t\right)\right]-s^{n}g\left(0\right)-s^{n-1}{g}^{\prime }\left(0\right)-\dots -f^{\left(n\right)}\left(0\right)}{s+\omega (1-s)}.$$

Definition 4

([44]). The ω-order Riemann–Liouville fractional integral for $Re\left(\omega \right)>0$ is specified as:

$$I^{\omega }g\left(t\right)=\frac{1}{\Gamma \left(\omega \right)}{\int }_0^t{(t-s)}^{\omega -1}g\left(s\right)ds.$$

Definition 5

([44]). The fractional integral of Caputo–Fabrizio is defined by

$${}^{CF}{I}^{\omega }g\left(t\right)=\frac{2(1-\omega )}{(2-\omega )M\left(\omega \right)}g\left(t\right)+\frac{2\left(\omega \right)}{(2-\omega )M\left(\omega \right)}{\int }_0^{t}g\left(s\right)ds,0<\omega <1.$$

Additionally, the left and right fractional integrals of (${}_a^{CF}{\mathfrak{D}}^{\omega }$) are defined [45] respectively by:

$$\begin{array}{c}{\displaystyle {(}^{CF}{I}_a^{\omega }g)\left(t\right)=\frac{1-\omega }{B\left(\omega \right)}g\left(t\right)+\frac{\omega }{B\left(\omega \right)}{\int }_a^{t}g\left(s\right)ds,}\hfill \\ \\ {\displaystyle {(}^{CF}{I}_b^{\omega }g)\left(t\right)=\frac{1-\omega }{B\left(\omega \right)}g\left(t\right)+\frac{\omega }{B\left(\omega \right)}{\int }_t^{b}g\left(s\right)ds.}\hfill \end{array}$$

We have the following definitions for the Sumudu transform, x which is based on the classical Fourier integral [46,47,48].

Definition 6

([47]). Let $A=\{G:\exists \lambda ,k_1,k_2\ge 0,|G\left(t\right)|<\lambda exp\left(\frac{t}{k_j}\right),t\in {(–1)}^j\times [0,\infty )\}$. Then $ST\left[g\right(t);u]=G\left(u\right)$ denotes the Sumudu transform of a function $g\left(t\right)\in A$ that is defined as

$$G\left(u\right)=ST[g\left(t\right);u]=\frac{1}{u}{\int }_0^{\infty }exp(-t/u)g\left(t\right)dt,u\in (-k_1,k_2),$$

for all $t\ge 0$. Additionally, $g\left(t\right)=S{T}^{–1}\left[G\left(u\right)\right]$ is the inverse Sumudu transform of $G\left(u\right)$.

Definition 7

([46]). For $(m–1<\omega \le m)$ we can define the Sumudu transform for the Caputo derivative as

$$ST{[}^c{\mathfrak{D}}_t^{\omega }g\left(t\right);u]=u^{-\omega }\left[G\left(u\right)-\sum _{i=0}^m{u}^{\omega -i}{{[}^c{\mathfrak{D}}^{\omega -i}g\left(t\right)]}_{t=0}\right].$$

Definition 8

([49]). Assume that G is a function and it has a Caputo–Fabrizio fractional derivative. We can define the Sumudu transform of G with the Caputo–Fabrizio fractional derivative as

$$ST{(}_0^{CF}{\mathfrak{D}}_t^{\omega })\left(G\left(t\right)\right)=\frac{M\left(\omega \right)}{1-\omega +\omega u}[ST\left(G\left(t\right)\right)-G\left(0\right)].$$

Definition 9.

[50] For a metric space $(X,d)$, a map $g:X\to X$ is called a Picard operator whenever there exists $x^{*}\in X$ such that $Fix\left(g\right)=\left\{x^{*}\right\}$ and the sequence ${\left(g^n\left(x_0\right)\right)}_{n\in N}$ converges to $x^{*}$ for all $x_0\in X$.

Definition 10.

Consider the Banach space $(Y,\parallel .\parallel )$, a self-map G on Y, and the recursive relation $P_{n+1}=\varphi (G,P_n)$. Let $\Omega \left(G\right)$ be the fixed point set of G, which includes $\Omega \left(G\right)\ne ⌀$ and ${lim}_{n\to \infty }{P}_n=p\in \Omega \left(G\right)$. Suppose that $\left\{g_n\right\}\subset \Omega$ and $e_n=g_{n+1}–\varphi (G,g_n)$, if ${lim}_{n\to \infty }{e}_n=0$ implies that ${lim}_{n\to \infty }{g}_n=p$, then the recursive procedure $P_{n+1}=\varphi (G,P_n)$ is G-stable. Suppose that our sequence $g_n$ has an upper boundary. If Picard’s iteration $P_{n+1}=G{P}_n$ is satisfied in all these conditions, then $P_{n+1}=G{P}_n$ is G-stable.

Theorem 1

([50]). Assume that $(Y,\parallel .\parallel )$ is a Banach space and G is a self-map of Y such that

$$\parallel G_x-G_y\parallel \le R\parallel x-G_x\parallel +r\parallel x-y\parallel ,$$

for all $x,y\in Y$, where $R\ge 0$ and $0\le r<1$. Then G is Picard G-stable.

3. Nonlinear Fractional Order Model of COVID-19

Consider the following non-linear bio-mathematical model [51,52]:

$$\left\{\begin{array}{c}{\displaystyle \frac{dS}{dt}=-\beta IS,}\hfill \\ \\ {\displaystyle \frac{dE}{dt}=\beta IS-\epsilon E,}\hfill \\ \\ {\displaystyle \frac{dI}{dt}=\epsilon E-\gamma I-dI-qI,}\hfill \\ \\ {\displaystyle \frac{dQ}{dt}=qI-q_{t}Q-dQ,}\hfill \\ \\ {\displaystyle \frac{dR}{dt}=\gamma I+q_{t}Q,}\hfill \\ \\ {\displaystyle \frac{dD}{dt}=dI+dQ,}\hfill \end{array}\right.$$

(3)

where $S,E,I,Q,R$ and D are susceptible, exposed, infected, quarantined, recovered and dead populations, respectively, and we denote by $N=S+E+I+Q+R+D$ the total size of the population. Susceptible individuals become infected at a rate, $\beta$, by infectious individuals, and the rate can be obtained using $\beta =R_0\gamma$ where $R_0$ is the basic reproduction number and $\gamma =\frac{1}{infectiousperiod}$. $\epsilon =\frac{1}{incubationperiod}$. We should note that D is a fraction of $I+Q$. The death rate is denoted by d, the fraction of active cases quarantined is denoted by q and the time period of quarantine is indicated by $q_t$. The graphical form of the model is shown in Figure 1 [51]. The values and parameters of this model are presented in

Table 1. List of parameters and functions [51].

Parameters	Values
Incubation Period $\frac{1}{\epsilon }$	Mean= 5.1 days
Infectious Period $\frac{1}{\gamma }$	Mean= 7 days
Basic Reproduction Number $R_0$	Mean= 2.28
Mean Death Rate	$0.43\%$
Active Cases Quarantined q	0.01
Period Quarantined $q_t$	14 days
Quarantined Cases $S_0$	20
Recovered Cases $R_0$	12
Deaths $D_0$	5
Exposed Cases $E_0$	20
Infected Cases $I_0$	15
Quarantined Cases $Q_0$	15

, based on Indian data [51].

Model (3) can be written using the $\omega$-th order Caputo–Fabrizio fractional derivative as follows:

$$\left\{\begin{array}{c}{\displaystyle {}_0^{CF}{\mathfrak{D}}_t^{\omega }S=-\beta IS,}\hfill \\ \\ {\displaystyle {}_0^{CF}{\mathfrak{D}}_t^{\omega }E=\beta IS-\epsilon E,}\hfill \\ \\ {\displaystyle {}_0^{CF}{\mathfrak{D}}_t^{\omega }I=\epsilon E-\gamma I-dI-qI,}\hfill \\ \\ {\displaystyle {}_0^{CF}{\mathfrak{D}}_t^{\omega }Q=qI-q_{t}Q-dQ,}\hfill \\ \\ {\displaystyle {}_0^{CF}{\mathfrak{D}}_t^{\omega }R=\gamma I+q_{t}Q,}\hfill \\ \\ {\displaystyle {}_0^{CF}{\mathfrak{D}}_t^{\omega }D=dI+dQ.}\hfill \end{array}\right.$$

(4)

4. Existence of Solution

Applying Definition 5 to both sides of Equation (4), we obtain:

$$\left\{\begin{array}{cc}{\displaystyle S\left(t\right)-u_1\left(t\right)}\hfill & {\displaystyle =\frac{2(1-\omega )}{(2-\omega )M\left(\omega \right)}\{-\beta I\left(t\right)S\left(t\right)\}+\frac{2\omega }{(2-\omega )M\left(\omega \right)}{\int }_0^t[-\beta I\left(s\right)S\left(s\right)]ds,}\hfill \\ \\ {\displaystyle E\left(t\right)-u_2\left(t\right)}\hfill & {\displaystyle =\frac{2(1-\omega )}{(2-\omega )M\left(\omega \right)}\{\beta I\left(t\right)S\left(t\right)-\epsilon E\left(t\right)\}+\frac{2\omega }{(2-\omega )M\left(\omega \right)}{\int }_0^t[\beta I\left(s\right)S\left(s\right)-\epsilon E\left(s\right)]ds,}\hfill \\ \\ {\displaystyle I\left(t\right)-u_3\left(t\right)}\hfill & {\displaystyle =\frac{2(1-\omega )}{(2-\omega )M\left(\omega \right)}\{\epsilon E\left(t\right)-\gamma I\left(t\right)-dI\left(t\right)-qI\left(t\right)\}+\frac{2\omega }{(2-\omega )M\left(\omega \right)}{\int }_0^t[\epsilon E\left(s\right)-\gamma I\left(s\right)-dI\left(s\right)-qI\left(s\right)]ds,}\hfill \\ \\ {\displaystyle Q\left(t\right)-u_4\left(t\right)}\hfill & {\displaystyle =\frac{2(1-\omega )}{(2-\omega )M\left(\omega \right)}\{qI\left(t\right)-qtQ\left(t\right)-dQ\left(t\right)\}+\frac{2\omega }{(2-\omega )M\left(\omega \right)}{\int }_0^t[qI\left(s\right)-qtQ\left(s\right)-dQ\left(s\right)]ds,}\hfill \\ \\ {\displaystyle R\left(t\right)-u_5\left(t\right)}\hfill & {\displaystyle =\frac{2(1-\omega )}{(2-\omega )M\left(\omega \right)}\{\gamma I\left(t\right)+qtQ\left(t\right)\}+\frac{2\omega }{(2-\omega )M\left(\omega \right)}{\int }_0^t[\gamma I\left(s\right)+qtQ\left(s\right)]ds,}\hfill \\ \\ {\displaystyle D\left(t\right)-u_6\left(t\right)}\hfill & {\displaystyle =\frac{2(1-\omega )}{(2-\omega )M\left(\omega \right)}\{dI\left(t\right)+dQ\left(t\right)\}+\frac{2\omega }{(2-\omega )M\left(\omega \right)}{\int }_0^t[dI\left(s\right)+dQ\left(s\right)]ds.}\hfill \end{array}\right.$$

(5)

By assuming $S_0\left(t\right)=u_1\left(t\right),E_0\left(t\right)=u_2\left(t\right),I_0\left(t\right)=u_3\left(t\right),Q_0\left(t\right)=u_4\left(t\right),R_0\left(t\right)=u_5\left(t\right)$ and $D_0\left(t\right)=u_6\left(t\right),$ these equations can be written in the following form:

$$\left\{\begin{array}{cc}{\displaystyle S_{n+1}\left(t\right)}\hfill & {\displaystyle =\frac{2(1-\omega )}{(2-\omega )M\left(\omega \right)}\{-\beta I_n\left(t\right)S_n\left(t\right)\}+\frac{2\omega }{(2-\omega )M\left(\omega \right)}{\int }_0^t[-\beta I_n\left(s\right)S_n\left(s\right)]ds,}\hfill \\ \\ {\displaystyle E_{n+1}\left(t\right)}\hfill & {\displaystyle =\frac{2(1-\omega )}{(2-\omega )M\left(\omega \right)}\{\beta I_n\left(t\right)S_n\left(t\right)-\epsilon E_n\left(t\right)\}+\frac{2\omega }{(2-\omega )M\left(\omega \right)}{\int }_0^t[\beta I_n\left(s\right)S_n\left(s\right)-\epsilon E_n\left(s\right)]ds,}\hfill \\ \\ {\displaystyle I_{n+1}\left(t\right)}\hfill & {\displaystyle =\frac{2(1-\omega )}{(2-\omega )M\left(\omega \right)}\{\epsilon E_n\left(t\right)-\gamma I_n\left(t\right)-d{I}_n\left(t\right)-q{I}_n\left(t\right)\}+\frac{2\omega }{(2-\omega )M\left(\omega \right)}{\int }_0^t[\epsilon E_n\left(s\right)-\gamma I_n\left(s\right)-d{I}_n\left(s\right)-q{I}_n\left(s\right)]ds,}\hfill \\ \\ {\displaystyle Q_{n+1}\left(t\right)}\hfill & {\displaystyle =\frac{2(1-\omega )}{(2-\omega )M\left(\omega \right)}\{q{I}_n\left(t\right)-qt{Q}_n\left(t\right)-d{Q}_n\left(t\right)\}+\frac{2\omega }{(2-\omega )M\left(\omega \right)}{\int }_0^t[q{I}_n\left(s\right)-qt{Q}_n\left(s\right)-d{Q}_n\left(s\right)]ds,}\hfill \\ \\ {\displaystyle R_{n+1}\left(t\right)}\hfill & {\displaystyle =\frac{2(1-\omega )}{(2-\omega )M\left(\omega \right)}\{\gamma I_n\left(t\right)+qt{Q}_n\left(t\right)\}+\frac{2\omega }{(2-\omega )M\left(\omega \right)}{\int }_0^t[\gamma I_n\left(s\right)+qt{Q}_n\left(s\right)]ds,}\hfill \\ \\ {\displaystyle D_{n+1}\left(t\right)}\hfill & {\displaystyle =\frac{2(1-\omega )}{(2-\omega )M\left(\omega \right)}\{d{I}_n\left(t\right)+d{Q}_n\left(t\right)\}+\frac{2\omega }{(2-\omega )M\left(\omega \right)}{\int }_0^t[d{I}_n\left(s\right)+d{Q}_n\left(s\right)]ds,}\hfill \end{array}\right.$$

(6)

and by letting n approach infinity in Equation (6), the approximate solutions can be obtained as:

$$\begin{array}{cc}\underset{n\to \infty }{lim}{S}_n\left(t\right)=S\left(t\right),\hfill & \underset{n\to \infty }{lim}{Q}_n\left(t\right)=Q\left(t\right),\hfill \\ \\ \underset{n\to \infty }{lim}{E}_n\left(t\right)=E\left(t\right),\hfill & \underset{n\to \infty }{lim}{R}_n\left(t\right)=R\left(t\right),\hfill \\ \\ \underset{n\to \infty }{lim}{I}_n\left(t\right)=I\left(t\right),\hfill & \underset{n\to \infty }{lim}{D}_n\left(t\right)=D\left(t\right).\hfill \end{array}$$

(7)

Now we can apply the Picard–Lindelof approach and the Banach fixed point theorem to show the existence of a solution to Equation (4). First we define the following operators as:

$$\left\{\begin{array}{c}{\displaystyle u_1(t,S)=-\beta IS,}\hfill \\ \\ {\displaystyle u_2(t,E)=\beta IS-\epsilon E,}\hfill \\ \\ {\displaystyle u_3(t,I)=\epsilon E-\gamma I-dI-qI,}\hfill \\ \\ {\displaystyle u_4(t,Q)=qI-q_{t}Q-dQ,}\hfill \\ \\ {\displaystyle u_5(t,R)=\gamma I+q_{t}Q,}\hfill \\ \\ {\displaystyle u_6(t,D)=dI+dQ.}\hfill \end{array}\right.$$

(8)

Let

$$\begin{array}{ccc}{\displaystyle {\ell }_1=\underset{C[a,{\rho }_1]}{sup}\parallel u_1(t,S)\parallel ,}\hfill & {\displaystyle {\ell }_2=\underset{C[a,{\rho }_2]}{sup}\parallel u_2(t,E)\parallel ,}\hfill & {\displaystyle {\ell }_3=\underset{C[a,{\rho }_3]}{sup}\parallel u_3(t,I)\parallel ,}\hfill \\ \\ {\displaystyle {\ell }_4=\underset{C[a,{\rho }_4]}{sup}\parallel u_4(t,Q)\parallel ,}\hfill & {\displaystyle {\ell }_5=\underset{C[a,{\rho }_5]}{sup}\parallel u_5(t,R)\parallel ,}\hfill & {\displaystyle {\ell }_6=\underset{C[a,{\rho }_6]}{sup}\parallel u_6(t,D)\parallel ,}\hfill \end{array}$$

(9)

where

$$\left\{\begin{array}{c}{\displaystyle C[a,{\rho }_1]=|t-a,t+a|\times |S-{\rho }_1,S+{\rho }_1|=B\times C_1,}\hfill \\ \\ {\displaystyle C[a,{\rho }_2]=|t-a,t+a|\times |E-{\rho }_2,E+{\rho }_2|=B\times C_2,}\hfill \\ \\ {\displaystyle C[a,{\rho }_3]=|t-a,t+a|\times |I-{\rho }_3,I+{\rho }_3|=B\times C_3,}\hfill \\ \\ {\displaystyle C[a,{\rho }_4]=|t-a,t+a|\times |Q-{\rho }_4,Q+{\rho }_4|=B\times C_4,}\hfill \\ \\ {\displaystyle C[a,{\rho }_5]=|t-a,t+a|\times |R-{\rho }_5,R+{\rho }_5|=B\times C_5,}\hfill \\ \\ {\displaystyle C[a,{\rho }_6]=|t-a,t+a|\times |D-{\rho }_6,D+{\rho }_6|=B\times C_6.}\hfill \end{array}\right.$$

(10)

Assuming a uniform norm

$${\parallel Z\left(t\right)\parallel }_{\infty }=\underset{t\in [t-a,t+a]}{sup}\left|Z\left(t\right)\right|,$$

(11)

on $C[a,{\rho }_i],i=1,\dots ,5$, and defining the Picard operator as

$$O:C(B,C_1,\dots ,C_6)\to C(B,C_1,\dots ,C_6),$$

(12)

we can write

$$O\left(Z\left(t\right)\right)=Z_0\left(t\right)+\frac{2(1-\omega )}{(2-\omega )M\left(\omega \right)}G(t,Z\left(t\right))+\frac{2\omega }{(2-\omega )M\left(\omega \right)}{\int }_0^{t}G(s,Z\left(s\right))ds,$$

(13)

where $Z\left(t\right)=\left\{S\right(t),E(t),I(t),Q(t),R(t),D(t\left)\right\}$ and $Z_0\left(t\right)=\{S\left(0\right),E\left(0\right),I\left(0\right),Q\left(0\right),R\left(0\right),$$D\left(0\right)\}$ and

$$G(t,Z\left(t\right))=\{u_1(t,S),u_2(t,E),u_3(t,I),u_4(t,Q),u_5(t,R),u_6(t,D)\},$$

(14)

and thus

$${\parallel Z\left(t\right)\parallel }_{\infty }\le max\{{\rho }_1,\dots ,{\rho }_6\}=C.$$

(15)

Assume that $L=max\{L_1,\dots ,L_6\}$ and there is $t_0$ such that $t_0\ge t$ and

$$\begin{array}{cc}{\displaystyle \parallel O\left(Z\left(t\right)\right)-Z_0\left(t\right)\parallel }\hfill & {\displaystyle =\parallel \frac{2(1-\omega )}{(2-\omega )M\left(\omega \right)}G(t,Z\left(t\right))+\frac{2\omega }{(2-v)M\left(v\right)}{\int }_0^{t}G(s,Z\left(s\right))ds\parallel }\hfill \\ \\ \hfill & {\displaystyle \le \frac{2(1-\omega )}{(2-\omega )M\left(\omega \right)}\parallel G(t,Z\left(t\right))\parallel +\frac{2\omega }{(2-\omega )M\left(\omega \right)}{\int }_0^t\parallel G(s,Z\left(s\right))\parallel ds}\hfill \\ \\ \hfill & {\displaystyle \le \left(\frac{2(1-\omega )}{(2-\omega )M\left(\omega \right)}+\frac{2\omega t}{(2-\omega )M\left(\omega \right)}\right)L}\hfill \end{array}$$

(16)

$${\displaystyle \le \left(\frac{2(1-\omega )}{(2-\omega )M\left(\omega \right)}+\frac{2\omega t_0}{(2-v)M\left(v\right)}\right)L\le \mu L\le C,}$$

(17)

where

$$\mu <\frac{C}{L}.$$

(18)

Now, we can write

$$\parallel O{Z}_1-O{Z}_2\parallel =\underset{t\in B}{sup}|Z_1\left(t\right)-Z_2\left(t\right)|,$$

(19)

and using definition of the Picard operator we obtain

$$\begin{array}{cc}{\displaystyle \parallel O{Z}_1-O{Z}_2\parallel }\hfill & {\displaystyle =\parallel \frac{2(1-\omega )}{(2-\omega )M\left(\omega \right)}\{G(t,Z_1\left(t\right))-G(t,Z_2\left(t\right))\}}\hfill \\ \\ \hfill & {\displaystyle +\frac{2\omega }{(2-\omega )M\left(\omega \right)}{\int }_0^t\{G(s,Z_1\left(s\right))-G(s,Z_2\left(s\right))\}ds\parallel }\hfill \\ \\ \hfill & {\displaystyle \le \frac{2(1-\omega )}{(2-\omega )M\left(\omega \right)}\parallel G(t,Z_1\left(t\right))-G(t,Z_2\left(t\right))\parallel }\hfill \\ \\ \hfill & {\displaystyle +\frac{2\omega }{(2-\omega )M\left(\omega \right)}{\int }_0^t\parallel G(s,Z_1\left(s\right))-G(s,Z_2\left(s\right))\parallel ds}\hfill \\ \\ \hfill & {\displaystyle \le \frac{2(1-\omega )}{(2-\omega )M\left(\omega \right)}\lambda \parallel Z_1\left(t\right)-Z_2\left(t\right)\parallel }\hfill \\ \\ \hfill & {\displaystyle +\frac{2\omega \lambda }{(2-\omega )M\left(\omega \right)}{\int }_0^t\parallel Z_1\left(s\right)-Z_2\left(s\right)\parallel ds}\hfill \\ \\ \hfill & {\displaystyle \le \left(\frac{2(1-\omega )\lambda }{(2-\omega )M\left(\omega \right)}+\frac{2\omega \lambda t_0}{(2-\omega )M\left(\omega \right)}\right)\parallel Z_1\left(t\right)-Z_2\left(t\right)\parallel }\hfill \\ \\ \hfill & {\displaystyle \le \mu \lambda \parallel Z_1\left(t\right)-Z_2\left(t\right)\parallel ,}\hfill \end{array}$$

(20)

where $\lambda <1$. We know that G is a contraction so $\mu \lambda <1$ and O is a contraction, showing that the proof is complete.

5. Special Solution via Iteration Approach

In this section, using the Sumudu transform, we can provide a special solution. Applying this transformation on both sides of (4) we obtain:

$$\left\{\begin{array}{c}{\displaystyle ST{(}_0^{CF}{\mathfrak{D}}_t^{\omega }S\left(t\right))=ST(-\beta I\left(t\right)S\left(t\right)),}\hfill \\ \\ {\displaystyle ST{(}_0^{CF}{\mathfrak{D}}_t^{\omega }E\left(t\right))=ST(\beta I\left(t\right)S\left(t\right)-\epsilon E\left(t\right)),}\hfill \\ \\ {\displaystyle ST{(}_0^{CF}{\mathfrak{D}}_t^{\omega }I\left(t\right))=ST(\epsilon E\left(t\right)-\gamma I\left(t\right)-dI\left(t\right)-qI\left(t\right)),}\hfill \\ \\ {\displaystyle ST{(}_0^{CF}{\mathfrak{D}}_t^{\omega }Q\left(t\right))=ST(qI\left(t\right)-q_{t}Q\left(t\right)-dQ\left(t\right)),}\hfill \\ \\ {\displaystyle ST{(}_0^{CF}{\mathfrak{D}}_t^{\omega }R\left(t\right))=ST(\gamma I\left(t\right)+q_{t}Q\left(t\right)),}\hfill \\ \\ {\displaystyle ST{(}_0^{CF}{\mathfrak{D}}_t^{\omega }D\left(t\right))=ST(dI\left(t\right)+dQ\left(t\right)),}\hfill \end{array}\right.$$

(21)

and based on the definition of the Sumudu transform we can write

$$\left\{\begin{array}{c}{\displaystyle \frac{M\left(\omega \right)}{1-\omega +\omega u}(ST\left(S\left(t\right)\right)-S\left(0\right))=ST(-\beta I\left(t\right)S\left(t\right)),}\hfill \\ \\ {\displaystyle \frac{M\left(\omega \right)}{1-\omega +\omega u}(ST\left(E\left(t\right)\right)-E\left(0\right))=ST(\beta I\left(t\right)S\left(t\right)-\epsilon E\left(t\right)),}\hfill \\ \\ {\displaystyle \frac{M\left(\omega \right)}{1-\omega +\omega u}(ST\left(I\left(t\right)\right)-I\left(0\right))=ST(\epsilon E\left(t\right)-\gamma I\left(t\right)-dI\left(t\right)-qI\left(t\right)),}\hfill \\ \\ {\displaystyle \frac{M\left(\omega \right)}{1-\omega +\omega u}(ST\left(Q\left(t\right)\right)-Q\left(0\right))=ST(qI\left(t\right)-q_{t}Q\left(t\right)-dQ\left(t\right)),}\hfill \\ \\ {\displaystyle \frac{M\left(\omega \right)}{1-\omega +\omega u}(ST\left(R\left(t\right)\right)-R\left(0\right))=ST(\gamma I\left(t\right)+q_{t}Q\left(t\right)),}\hfill \\ \\ {\displaystyle \frac{M\left(\omega \right)}{1-\omega +\omega u}(ST\left(D\left(t\right)\right)-D\left(0\right))=ST(dI\left(t\right)+dQ\left(t\right)),}\hfill \end{array}\right.$$

(22)

and

$$\left\{\begin{array}{c}{\displaystyle ST\left(S\left(t\right)\right)=S\left(0\right)+\frac{1-\omega +\omega u}{M\left(\omega \right)}ST(-\beta I\left(t\right)S\left(t\right)),}\hfill \\ \\ {\displaystyle ST\left(E\left(t\right)\right)=E\left(0\right)+\frac{1-\omega +\omega u}{M\left(\omega \right)}ST(\beta I\left(t\right)S\left(t\right)-\epsilon E\left(t\right)),}\hfill \\ \\ {\displaystyle ST\left(I\left(t\right)\right)=I\left(0\right)+\frac{1-\omega +\omega u}{M\left(\omega \right)}ST(\epsilon E\left(t\right)-\gamma I\left(t\right)-dI\left(t\right)-qI\left(t\right)),}\hfill \\ \\ {\displaystyle ST\left(Q\left(t\right)\right)=Q\left(0\right)+\frac{1-\omega +\omega u}{M\left(\omega \right)}ST(qI\left(t\right)-q_{t}Q\left(t\right)-dQ\left(t\right)),}\hfill \\ \\ {\displaystyle ST\left(R\left(t\right)\right)=R\left(0\right)+\frac{1-\omega +\omega u}{M\left(\omega \right)}ST(\gamma I\left(t\right)+q_{t}Q\left(t\right)),}\hfill \\ \\ {\displaystyle ST\left(D\left(t\right)\right)=D\left(0\right)+\frac{1-\omega +\omega u}{M\left(\omega \right)}ST(dI\left(t\right)+dQ\left(t\right)).}\hfill \end{array}\right.$$

(23)

Finally, the following recursive relation can be obtained as

$$\left\{\begin{array}{c}{\displaystyle S_{n+1}\left(t\right)=S_n\left(t\right)+S{T}^{-1}\left\{\frac{1-\omega +\omega u}{M\left(\omega \right)}ST(-\beta I_n\left(t\right)S_n\left(t\right))\right\},}\hfill \\ \\ {\displaystyle E_{n+1}\left(t\right)=E_n\left(t\right)+S{T}^{-1}\left\{\frac{1-\omega +\omega u}{M\left(\omega \right)}ST(\beta I_n\left(t\right)S_n\left(t\right)-\epsilon E_n\left(t\right))\right\},}\hfill \\ \\ {\displaystyle I_{n+1}\left(t\right)=I_n\left(t\right)+S{T}^{-1}\left\{\frac{1-\omega +\omega u}{M\left(\omega \right)}ST(\epsilon E_n\left(t\right)-\gamma I_n\left(t\right)-d{I}_n\left(t\right)-q{I}_n\left(t\right))\right\},}\hfill \\ \\ {\displaystyle Q_{n+1}\left(t\right))=Q_n\left(t\right)+S{T}^{-1}\left\{\frac{1-\omega +\omega u}{M\left(\omega \right)}ST(q{I}_n\left(t\right)-q_t{Q}_n\left(t\right)-d{Q}_n\left(t\right))\right\},}\hfill \\ \\ {\displaystyle R_{n+1}\left(t\right)=R_n\left(t\right)+S{T}^{-1}\left\{\frac{1-\omega +\omega u}{M\left(\omega \right)}ST(\gamma I_n\left(t\right)+q_t{Q}_n\left(t\right))\right\},}\hfill \\ \\ {\displaystyle D_{n+1}\left(t\right)=D_n\left(t\right)+S{T}^{-1}\left\{\frac{1-\omega +\omega u}{M\left(\omega \right)}ST(d{I}_n\left(t\right)+d{Q}_n\left(t\right))\right\},}\hfill \end{array}\right.$$

(24)

where the approximate solutions can be found using the following relations

$$\begin{array}{c}\underset{n\to \infty }{lim}{S}_n\left(t\right)=S\left(t\right),\hfill \\ \\ \underset{n\to \infty }{lim}{E}_n\left(t\right)=E\left(t\right),\hfill \\ \\ \underset{n\to \infty }{lim}{I}_n\left(t\right)=I\left(t\right),\hfill \\ \\ \underset{n\to \infty }{lim}{Q}_n\left(t\right)=Q\left(t\right),\hfill \\ \\ \underset{n\to \infty }{lim}{R}_n\left(t\right)=R\left(t\right),\hfill \\ \\ \underset{n\to \infty }{lim}{D}_n\left(t\right)=D\left(t\right).\hfill \end{array}$$

(25)

Fixed Point Theorem for Stability Analysis of the Iteration Method

Theorem 2.

Let F be a self-map that is defined in the following form: -5.0cm0cm

$$\left\{\begin{array}{c}{\displaystyle F\left(S_n\left(t\right)\right)=S_{n+1}\left(t\right)=S_n\left(t\right)+S{T}^{-1}\left\{\frac{1-\omega +\omega u}{M\left(\omega \right)}ST(-\beta I_n\left(t\right)S_n\left(t\right))\right\},}\hfill \\ \\ {\displaystyle F\left(E_n\left(t\right)\right)=E_{n+1}\left(t\right)=E_n\left(t\right)+S{T}^{-1}\left\{\frac{1-\omega +\omega u}{M\left(\omega \right)}ST(\beta I_n\left(t\right)S_n\left(t\right)-\epsilon E_n\left(t\right))\right\},}\hfill \\ \\ {\displaystyle F\left(I_n\left(t\right)\right)=I_{n+1}\left(t\right)=I_n\left(t\right)+S{T}^{-1}\left\{\frac{1-\omega +\omega u}{M\left(\omega \right)}ST(\epsilon E_n\left(t\right)-\gamma I_n\left(t\right)-d{I}_n\left(t\right)-q{I}_n\left(t\right))\right\},}\hfill \\ \\ {\displaystyle F\left(Q_n\left(t\right)\right)=Q_{n+1}\left(t\right))=Q_n\left(t\right)+S{T}^{-1}\left\{\frac{1-\omega +\omega u}{M\left(\omega \right)}ST(q{I}_n\left(t\right)-q_t{Q}_n\left(t\right)-d{Q}_n\left(t\right))\right\},}\hfill \\ \\ {\displaystyle F\left(R_n\left(t\right)\right)=R_{n+1}\left(t\right)=R_n\left(t\right)+S{T}^{-1}\left\{\frac{1-\omega +\omega u}{M\left(\omega \right)}ST(\gamma I_n\left(t\right)+q_t{Q}_n\left(t\right))\right\},}\hfill \\ \\ {\displaystyle F\left(D_n\left(t\right)\right)=D_{n+1}\left(t\right)=D_n\left(t\right)+S{T}^{-1}\left\{\frac{1-\omega +\omega u}{M\left(\omega \right)}ST(d{I}_n\left(t\right)+d{Q}_n\left(t\right))\right\}.}\hfill \end{array}\right.$$

(26)

Formula (26) is F-stable in $L^1(a,b)$ if

$$\left\{\begin{array}{c}{\displaystyle \left\{1-\beta P_3{k}_1\left(\eta \right)-\beta P_1{k}_2\left(\eta \right)\right\}<1,}\hfill \\ \\ {\displaystyle \left\{1+\beta P_3{k}_3\left(\eta \right)\beta P_1{k}_4\left(\eta \right)-\epsilon k_5\left(\eta \right)\right\}<1,}\hfill \\ \\ {\displaystyle \left\{1+\epsilon k_6\left(\eta \right)-(\gamma +d+q)k_7\left(\eta \right)\right\}<1,}\hfill \\ \\ {\displaystyle \left\{1+q{k}_8\left(\eta \right)-(q_t+d)k_9\left(\eta \right)\right\}<1,}\hfill \\ \\ {\displaystyle \left\{1+\gamma k_{10}\left(\eta \right)+q_t{k}_{11}\left(\eta \right)\right\}<1,}\hfill \\ \\ {\displaystyle \left\{1+d{k}_{12}\left(\eta \right)+d{k}_{13}\left(\eta \right)\right\}<1.}\hfill \end{array}\right.$$

(27)

Proof. 

In order to prove the theorem, first the following relation should be computed for $(n,m)\in N\times N$ to show that F has a fixed point as

$$\left\{\begin{array}{c}{\displaystyle F\left(S_n\left(t\right)\right)-F\left(S_m\left(t\right)\right)}\hfill \\ \\ =S_n\left(t\right)-S_m\left(t\right)+S{T}^{-1}\left\{\frac{1-\omega +\omega u}{M\left(\omega \right)}ST(-\beta I_n\left(t\right)S_n\left(t\right))\right\}\hfill \\ \\ -S{T}^{-1}\left\{\frac{1-\omega +\omega u}{M\left(\omega \right)}ST(-\beta I_m\left(t\right)S_m\left(t\right))\right\},\hfill \end{array}\right.$$

(28)

and taking the norm on both sides of Equation (28) we obtain:

$$\left\{\begin{array}{c}{\displaystyle \parallel F\left(S_n\right)-F\left(S_m\right)\parallel }\hfill \\ \\ =\parallel S_n-S_m+S{T}^{-1}\left\{\frac{1-\omega +\omega u}{M\left(\omega \right)}ST(-\beta (I_n{S}_n-I_m{S}_m))\right\}\parallel ,\hfill \\ \\ \le \parallel S_n-S_m\parallel +S{T}^{-1}\left\{\frac{1-\omega +\omega u}{M\left(\omega \right)}ST(\parallel -\beta I_n(S_n-S_m)\parallel +\parallel -\beta S_m(I_n-I_m)\parallel )\right\}.\hfill \end{array}\right.$$

(29)

Because of the same role of both solutions, we obtain:

$$\left\{\begin{array}{c}\parallel S_n\left(t\right)-S_m\left(t\right)\parallel \cong \parallel E_n\left(t\right)-E_m\left(t\right)\parallel \cong \parallel I_n\left(t\right)-I_m\left(t\right)\parallel \hfill \\ \\ \parallel Q_n\left(t\right)-Q_m\left(t\right)\parallel \cong \parallel R_n\left(t\right)-R_m\left(t\right)\parallel \cong \parallel D_n\left(t\right)-D_m\left(t\right)\parallel ,\hfill \end{array}\right.$$

(30)

and applying (29) and (30) we can write -4.6cm0cm

$$\left\{\begin{array}{c}{\displaystyle \parallel F\left(S_n\left(t\right)\right)-F\left(S_m\left(t\right)\right)\parallel }\hfill \\ \\ \le \parallel S_n\left(t\right)-S_m\left(t\right)\parallel +S{T}^{-1}\left\{\frac{1-\omega +\omega u}{M\left(\omega \right)}ST(\parallel -\beta I_n\left(t\right)(S_n\left(t\right)-S_m\left(t\right))\parallel +\parallel -\beta S_m\left(t\right)(S_n\left(t\right)-S_m\left(t\right))\parallel )\right\}.\hfill \end{array}\right.$$

(31)

We know that $S_n,E_n,I_n,Q_n,R_n,D_n$ are bounded because they are convergent sequences. Thus for all t, there are values $P_1,P_2,P_3,P_4,P_5,P_6$ such that

$$\parallel S_m\parallel <P_1,\parallel E_n\parallel <P_2,\parallel I_n\parallel <P_3,\parallel Q_n\parallel <P_4,\parallel R_n\parallel <P_5,\parallel D_n\parallel <P_6,$$

(32)

where $(m,n)\in N\times N$. Now, based on Equations (31) and (32) we obtain:

$$\left\{\begin{array}{c}{\displaystyle \parallel F\left(S_n\right)-F\left(S_m\right)\parallel \le \left\{1-\beta P_3{k}_1\left(\eta \right)-\beta P_1{k}_2\left(\eta \right)\right\}\parallel S_n-S_m\parallel ,}\hfill \end{array}\right.$$

(33)

where $k_i$ are obtained from $S{T}^{-1}\left\{\frac{1-\omega +\omega u}{M\left(\omega \right)}ST\left(.\right)\right\}$. By repeating the process we have

$$\left\{\begin{array}{c}{\displaystyle \parallel F\left(E_n\right)-F\left(E_m\right)\parallel \le \left\{1+\beta P_3{k}_3\left(\eta \right)\beta P_1{k}_4\left(\eta \right)-\epsilon k_5\left(\eta \right)\right\}\parallel E_n-E_m\parallel ,}\hfill \\ \\ {\displaystyle \parallel F\left(I_n\right)-F\left(I_m\right)\parallel \le \left\{1+\epsilon k_6\left(\eta \right)-(\gamma +d+q)k_7\left(\eta \right)\right\}\parallel I_n-I_m\parallel ,}\hfill \\ \\ {\displaystyle \parallel F\left(Q_n\right)-F\left(Q_m\right)\parallel \le \left\{1+q{k}_8\left(\eta \right)-(q_t+d)k_9\left(\eta \right)\right\}\parallel Q_n-Q_m\parallel ,}\hfill \\ \\ {\displaystyle \parallel F\left(R_n\right)-F\left(R_m\right)\parallel \le \left\{1+\gamma k_{10}\left(\eta \right)+q_t{k}_{11}\left(\eta \right)\right\}\parallel R_n-R_m\parallel ,}\hfill \\ \\ {\displaystyle \parallel F\left(D_n\right)-F\left(D_m\right)\parallel \le \left\{1+d{k}_{12}\left(\eta \right)+d{k}_{13}\left(\eta \right)\right\}\parallel D_n-D_m\parallel ,}\hfill \end{array}\right.$$

(34)

where

$$\left\{\begin{array}{c}{\displaystyle \left\{1-\beta P_3{k}_1\left(\eta \right)-\beta P_1{k}_2\left(\eta \right)\right\}<1,}\hfill \\ \\ {\displaystyle \left\{1+\beta P_3{k}_3\left(\eta \right)\beta P_1{k}_4\left(\eta \right)-\epsilon k_5\left(\eta \right)\right\}<1,}\hfill \\ \\ {\displaystyle \left\{1+\epsilon k_6\left(\eta \right)-(\gamma +d+q)k_7\left(\eta \right)\right\}<1,}\hfill \\ \\ {\displaystyle \left\{1+q{k}_8\left(\eta \right)-(q_t+d)k_9\left(\eta \right)\right\}<1,}\hfill \\ \\ {\displaystyle \left\{1+\gamma k_{10}\left(\eta \right)+q_t{k}_{11}\left(\eta \right)\right\}<1,}\hfill \\ \\ {\displaystyle \left\{1+d{k}_{12}\left(\eta \right)+d{k}_{13}\left(\eta \right)\right\}<1.}\hfill \end{array}\right.$$

(35)

Then, the F self-mapping has a fixed point. In addition, we show that F satisfies the conditions in Theorem 1. By assuming that (33) and (34) hold, we have $R=(0,0,0,0,0)$ and

$$r=\left\{\begin{array}{c}{\displaystyle \left\{1-\beta P_3{k}_1\left(\eta \right)-\beta P_1{k}_2\left(\eta \right)\right\},}\hfill \\ \\ {\displaystyle \left\{1+\beta P_3{k}_3\left(\eta \right)\beta P_1{k}_4\left(\eta \right)-\epsilon k_5\left(\eta \right)\right\},}\hfill \\ \\ {\displaystyle \left\{1+\epsilon k_6\left(\eta \right)-(\gamma +d+q)k_7\left(\eta \right)\right\},}\hfill \\ \\ {\displaystyle \left\{1+q{k}_8\left(\eta \right)-(q_t+d)k_9\left(\eta \right)\right\},}\hfill \\ \\ {\displaystyle \left\{1+\gamma k_{10}\left(\eta \right)+q_t{k}_{11}\left(\eta \right)\right\},}\hfill \\ \\ {\displaystyle \left\{1+d{k}_{12}\left(\eta \right)+d{k}_{13}\left(\eta \right)\right\}.}\hfill \end{array}\right.$$

(36)

□

Then the conditions of Theorem 1 are satisfied and the proof is concluded.

6. Application of the HATM to Solve the Model

In order to apply the HATM for solving the fractional order model (4), applying the Laplace transformation we obtain:

$$\left\{\begin{array}{c}{\displaystyle \mathcal{L}{[}_0^{CF}{D}_t^{\omega }S]=\mathcal{L}[-\beta IS],}\hfill \\ \\ {\displaystyle \mathcal{L}{[}_0^{CF}{D}_t^{\omega }E]=\mathcal{L}[\beta IS-\epsilon E],}\hfill \\ \\ {\displaystyle \mathcal{L}{[}_0^{CF}{D}_t^{\omega }I]=\mathcal{L}[\epsilon E-\gamma I-dI-qI],}\hfill \\ \\ {\displaystyle \mathcal{L}{[}_0^{CF}{D}_t^{\omega }Q]=\mathcal{L}[qI-q_{t}Q-dQ],}\hfill \\ \\ {\displaystyle \mathcal{L}{[}_0^{CF}{D}_t^{\omega }R]=\mathcal{L}[\gamma I+q_{t}Q],}\hfill \\ \\ {\displaystyle \mathcal{L}{[}_0^{CF}{D}_t^{\omega }D]=\mathcal{L}[dI+dQ].}\hfill \end{array}\right.$$

(37)

Additionally, we can write

$$\left\{\begin{array}{c}{\displaystyle \frac{s\mathcal{L}\left(S\right)-S\left(0\right)}{s+\omega (1-s)}=\mathcal{L}[-\beta IS],}\hfill \\ \\ {\displaystyle \frac{s\mathcal{L}\left(E\right)-E\left(0\right)}{s+\omega (1-s)}=\mathcal{L}[\beta IS-\epsilon E],}\hfill \\ \\ {\displaystyle \frac{s\mathcal{L}\left(I\right)-I\left(0\right)}{s+\omega (1-s)}=\mathcal{L}[\epsilon E-\gamma I-dI-qI],}\hfill \\ \\ {\displaystyle \frac{s\mathcal{L}\left(Q\right)-Q\left(0\right)}{s+\omega (1-s)}=\mathcal{L}[qI-q_{t}Q-dQ],}\hfill \\ \\ {\displaystyle \frac{s\mathcal{L}\left(R\right)-R\left(0\right)}{s+\omega (1-s)}=\mathcal{L}[\gamma I+q_{t}Q],}\hfill \\ \\ {\displaystyle \frac{s\mathcal{L}\left(D\right)-D\left(0\right)}{s+\omega (1-s)}=\mathcal{L}[dI+dQ],}\hfill \end{array}\right.$$

(38)

and we have

$$\left\{\begin{array}{c}{\displaystyle \mathcal{L}\left(S\right)-\frac{S_0}{s}-\frac{s+\omega (1-s)}{s}\mathcal{L}[-\beta IS]=0,}\hfill \\ \\ {\displaystyle \mathcal{L}\left(E\right)-\frac{E_0}{s}-\frac{s+\omega (1-s)}{s}\mathcal{L}[\beta IS-\epsilon E]=0,}\hfill \\ \\ {\displaystyle \mathcal{L}\left(I\right)-\frac{I_0}{s}-\frac{s+\omega (1-s)}{s}\mathcal{L}[\epsilon E-\gamma I-dI-qI]=0,}\hfill \\ \\ {\displaystyle \mathcal{L}\left(Q\right)-\frac{Q_0}{s}-\frac{s+\omega (1-s)}{s}\mathcal{L}[qI-q_{t}Q-dQ]=0,}\hfill \\ \\ {\displaystyle \mathcal{L}\left(R\right)-\frac{R_0}{s}-\frac{s+\omega (1-s)}{s}\mathcal{L}[\gamma I+q_{t}Q]=0,}\hfill \\ \\ {\displaystyle \mathcal{L}\left(D\right)-\frac{D_0}{s}-\frac{s+\omega (1-s)}{s}\mathcal{L}[dI+dQ]=0.}\hfill \end{array}\right.$$

(39)

We define the nonlinear operators as follows:

$$\left\{\begin{array}{c}{\displaystyle N_1({\phi }_1(t;p),\dots ,{\phi }_6(t;p))=\mathcal{L}\left(S\right)-\frac{S_0}{s}-\frac{s+\omega (1-s)}{s}\mathcal{L}[-\beta {\phi }_3(t;p){\phi }_1(t;p)],}\hfill \\ \\ {\displaystyle N_2({\phi }_1(t;p),\dots ,{\phi }_6(t;p))=\mathcal{L}\left(E\right)-\frac{E_0}{s}-\frac{s+\omega (1-s)}{s}\mathcal{L}[\beta {\phi }_3(t;p){\phi }_1(t;p)-\epsilon {\phi }_2(t;p)],}\hfill \\ \\ {\displaystyle N_3({\phi }_1(t;p),\dots ,{\phi }_6(t;p))=\mathcal{L}\left(I\right)-\frac{I_0}{s}-\frac{s+\omega (1-s)}{s}\mathcal{L}[\epsilon {\phi }_2(t;p)-\gamma {\phi }_3(t;p)-d{\phi }_3(t;p)-q{\phi }_3(t;p)],}\hfill \\ \\ {\displaystyle N_4({\phi }_1(t;p),\dots ,{\phi }_6(t;p))=\mathcal{L}\left(Q\right)-\frac{Q_0}{s}-\frac{s+\omega (1-s)}{s}\mathcal{L}[q{\phi }_3(t;p)-q_t{\phi }_4(t;p)-d{\phi }_4(t;p)],}\hfill \\ \\ {\displaystyle N_5({\phi }_1(t;p),\dots ,{\phi }_6(t;p))=\mathcal{L}\left(R\right)-\frac{R_0}{s}-\frac{s+\omega (1-s)}{s}\mathcal{L}[\gamma {\phi }_3(t;p)+q_t{\phi }_4(t;p)],}\hfill \\ \\ {\displaystyle N_6({\phi }_1(t;p),\dots ,{\phi }_6(t;p))=\mathcal{L}\left(D\right)-\frac{D_0}{s}-\frac{s+\omega (1-s)}{s}\mathcal{L}[d{\phi }_3(t;p)+d{\phi }_4(t;p)].}\hfill \end{array}\right.$$

(40)

Based on the traditional HAM, the zero-order deformation equation can be defined as

$$\left\{\begin{array}{c}{\displaystyle (1-p)\mathcal{L}[{\phi }_1(t;p)-S_0\left(t\right)]=p\hslash H\left(t\right)({\phi }_1(t;p),\dots ,{\phi }_6(t;p)),}\hfill \\ \\ {\displaystyle (1-p)\mathcal{L}[{\phi }_2(t;p)-E_0\left(t\right)]=p\hslash H\left(t\right)({\phi }_1(t;p),\dots ,{\phi }_6(t;p)),}\hfill \\ \\ {\displaystyle (1-p)\mathcal{L}[{\phi }_3(t;p)-I_0\left(t\right)]=p\hslash H\left(t\right)({\phi }_1(t;p),\dots ,{\phi }_6(t;p)),}\hfill \\ \\ {\displaystyle (1-p)\mathcal{L}[{\phi }_4(t;p)-Q_0\left(t\right)]=p\hslash H\left(t\right)({\phi }_1(t;p),\dots ,{\phi }_6(t;p)),}\hfill \\ \\ {\displaystyle (1-p)\mathcal{L}[{\phi }_5(t;p)-R_0\left(t\right)]=p\hslash H\left(t\right)({\phi }_1(t;p),\dots ,{\phi }_6(t;p)),}\hfill \\ \\ {\displaystyle (1-p)\mathcal{L}[{\phi }_6(t;p)-D_0\left(t\right)]=p\hslash H\left(t\right)({\phi }_1(t;p),\dots ,{\phi }_6(t;p)),}\hfill \end{array}\right.$$

(41)

where $p\in [0,1],\hslash ,H\left(t\right)$ and $\mathcal{L}$ are the embedding parameter, auxiliary convergence control parameter, auxiliary function and the linear operator, respectively. By assuming $S_0\left(t\right),E_0\left(t\right),I_0\left(t\right),Q_0\left(t\right),R_0\left(t\right)$ and $D_0\left(t\right)$ as initial guesses and increasing p from 0 to 1 as follows:

$$\left\{\begin{array}{cc}{\phi }_1(t;0)=S_0\left(t\right),\hfill & {\phi }_1(t;1)=S\left(t\right),\hfill \\ \\ {\phi }_2(t;0)=E_0\left(t\right),\hfill & {\phi }_2(t;1)=E\left(t\right),\hfill \\ \\ {\phi }_3(t;0)=I_0\left(t\right),\hfill & {\phi }_3(t;1)=I\left(t\right),\hfill \\ \\ {\phi }_4(t;0)=Q_0\left(t\right),\hfill & {\phi }_4(t;1)=Q\left(t\right),\hfill \\ \\ {\phi }_5(t;0)=R_0\left(t\right),\hfill & {\phi }_5(t;1)=R\left(t\right),\hfill \\ \\ {\phi }_6(t;0)=D_0\left(t\right),\hfill & {\phi }_6(t;1)=D\left(t\right),\hfill \end{array}\right.$$

(42)

the solutions of the problem can be found using the initial guesses in the form of a Taylor series

$$\left\{\begin{array}{c}{\phi }_1(t;p)=S_0+\sum _{m=1}^{\infty }{S}_m\left(t\right)p^m,\hfill \\ \\ {\phi }_2(t;p)=E_0+\sum _{m=1}^{\infty }{E}_m\left(t\right)p^m,\hfill \\ \\ {\phi }_3(t;p)=I_0+\sum _{m=1}^{\infty }{I}_m\left(t\right)p^m,\hfill \\ \\ {\phi }_4(t;p)=Q_0+\sum _{m=1}^{\infty }{Q}_m\left(t\right)p^m,\hfill \\ \\ {\phi }_5(t;p)=R_0+\sum _{m=1}^{\infty }{R}_m\left(t\right)p^m,\hfill \\ \\ {\phi }_6(t;p)=D_0+\sum _{m=1}^{\infty }{D}_m\left(t\right)p^m,\hfill \end{array}\right.$$

(43)

where $S_m\left(t\right)=\frac{1}{m!}\frac{{\partial }^m{\phi }_1(t;p)}{\partial p^m}{|}_{p=0},E_m\left(t\right)=\frac{1}{m!}\frac{{\partial }^m{\phi }_2(t;p)}{\partial p^m}{|}_{p=0},I_m\left(t\right)=\frac{1}{m!}\frac{{\partial }^m{\phi }_3(t;p)}{\partial p^m}{|}_{p=0},Q_m\left(t\right)=\frac{1}{m!}\frac{{\partial }^m{\phi }_4(t;p)}{\partial p^m}{|}_{p=0},R_m\left(t\right)=\frac{1}{m!}\frac{{\partial }^m{\phi }_5(t;p)}{\partial p^m}{|}_{p=0}$ and $D_m\left(t\right)=\frac{1}{m!}\frac{{\partial }^m{\phi }_6(t;p)}{\partial p^m}{|}_{p=0}$. Thus, the series (43) will be convergent to the exact solution by choosing the suitable values of auxiliary parameters and functions as

$$\left\{\begin{array}{c}S\left(t\right)=S_0+\sum _{m=1}^{\infty }{S}_m\left(t\right),\hfill \\ \\ E\left(t\right)=E_0+\sum _{m=1}^{\infty }{E}_m\left(t\right),\hfill \\ \\ I\left(t\right)=I_0+\sum _{m=1}^{\infty }{I}_m\left(t\right),\hfill \\ \\ Q\left(t\right)=Q_0+\sum _{m=1}^{\infty }{Q}_m\left(t\right),\hfill \\ \\ R\left(t\right)=R_0+\sum _{m=1}^{\infty }{R}_m\left(t\right),\hfill \\ \\ D\left(t\right)=D_0+\sum _{m=1}^{\infty }{D}_m\left(t\right).\hfill \end{array}\right.$$

(44)

The m-th order deformation equation can be written as follows:

$$\left\{\begin{array}{c}\mathcal{L}[S_m\left(t\right)-{\chi }_m{S}_{m-1}\left(t\right)]=\hslash H{R}_{1,m}\left(S_{m-1}\right),\hfill \\ \\ \mathcal{L}[E_m\left(t\right)-{\chi }_m{E}_{m-1}\left(t\right)]=\hslash H{R}_{2,m}\left(E_{m-1}\right),\hfill \\ \\ \mathcal{L}[I_m\left(t\right)-{\chi }_m{I}_{m-1}\left(t\right)]=\hslash H{R}_{3,m}\left(I_{m-1}\right),\hfill \\ \\ \mathcal{L}[Q_m\left(t\right)-{\chi }_m{Q}_{m-1}\left(t\right)]=\hslash H{R}_{4,m}\left(Q_{m-1}\right),\hfill \\ \\ \mathcal{L}[R_m\left(t\right)-{\chi }_m{R}_{m-1}\left(t\right)]=\hslash H{R}_{5,m}\left(R_{m-1}\right),\hfill \\ \\ \mathcal{L}[D_m\left(t\right)-{\chi }_m{D}_{m-1}\left(t\right)]=\hslash H{R}_{5,m}\left(D_{m-1}\right),\hfill \end{array}\right.$$

(45)

where

$$\left\{\begin{array}{c}{\displaystyle R_{1,m}({\overrightarrow{S}}_{m-1}\left(t\right),{\overrightarrow{E}}_{m-1}\left(t\right),{\overrightarrow{I}}_{m-1}\left(t\right),{\overrightarrow{Q}}_{m-1}\left(t\right),{\overrightarrow{R}}_{m-1}\left(t\right),{\overrightarrow{D}}_{m-1}\left(t\right))}\hfill \\ \\ {\displaystyle =\mathcal{L}\left(S\right)-\frac{S_0}{s}-\frac{s+\omega (1-s)}{s}\mathcal{L}[-\beta I_{m-1}{S}_{m-1}],}\hfill \\ \\ {\displaystyle R_{2,m}({\overrightarrow{S}}_{m-1}\left(t\right),{\overrightarrow{E}}_{m-1}\left(t\right),{\overrightarrow{I}}_{m-1}\left(t\right),{\overrightarrow{Q}}_{m-1}\left(t\right),{\overrightarrow{R}}_{m-1}\left(t\right),{\overrightarrow{D}}_{m-1}\left(t\right))}\hfill \\ \\ {\displaystyle =\mathcal{L}\left(E\right)-\frac{E_0}{s}-\frac{s+\omega (1-s)}{s}\mathcal{L}[\beta I_{m-1}{S}_{m-1}-\epsilon E_{m-1}],}\hfill \\ \\ {\displaystyle R_{3,m}({\overrightarrow{S}}_{m-1}\left(t\right),{\overrightarrow{E}}_{m-1}\left(t\right),{\overrightarrow{I}}_{m-1}\left(t\right),{\overrightarrow{Q}}_{m-1}\left(t\right),{\overrightarrow{R}}_{m-1}\left(t\right),{\overrightarrow{D}}_{m-1}\left(t\right))}\hfill \\ \\ {\displaystyle =\mathcal{L}\left(I\right)-\frac{I_0}{s}-\frac{s+\omega (1-s)}{s}\mathcal{L}[\epsilon E_{m-1}-\gamma I_{m-1}-d{I}_{m-1}-q{I}_{m-1}],}\hfill \\ \\ {\displaystyle R_{4,m}({\overrightarrow{S}}_{m-1}\left(t\right),{\overrightarrow{E}}_{m-1}\left(t\right),{\overrightarrow{I}}_{m-1}\left(t\right),{\overrightarrow{Q}}_{m-1}\left(t\right),{\overrightarrow{R}}_{m-1}\left(t\right),{\overrightarrow{D}}_{m-1}\left(t\right))}\hfill \\ \\ {\displaystyle =\mathcal{L}\left(Q\right)-\frac{Q_0}{s}-\frac{s+\omega (1-s)}{s}\mathcal{L}[q{I}_{m-1}-q_t{Q}_{m-1}-d{Q}_{m-1}],}\hfill \\ \\ {\displaystyle R_{5,m}({\overrightarrow{S}}_{m-1}\left(t\right),{\overrightarrow{E}}_{m-1}\left(t\right),{\overrightarrow{I}}_{m-1}\left(t\right),{\overrightarrow{Q}}_{m-1}\left(t\right),{\overrightarrow{R}}_{m-1}\left(t\right),{\overrightarrow{D}}_{m-1}\left(t\right))}\hfill \\ \\ {\displaystyle =\mathcal{L}\left(R\right)-\frac{R_0}{s}-\frac{s+\omega (1-s)}{s}\mathcal{L}[\gamma I_{m-1}+q_t{Q}_{m-1}],}\hfill \\ \\ {\displaystyle R_{6,m}({\overrightarrow{S}}_{m-1}\left(t\right),{\overrightarrow{E}}_{m-1}\left(t\right),{\overrightarrow{I}}_{m-1}\left(t\right),{\overrightarrow{Q}}_{m-1}\left(t\right),{\overrightarrow{R}}_{m-1}\left(t\right),{\overrightarrow{D}}_{m-1}\left(t\right))}\hfill \\ \\ {\displaystyle =\mathcal{L}\left(D\right)-\frac{D_0}{s}-\frac{s+\omega (1-s)}{s}\mathcal{L}[d{I}_{m-1}+d{Q}_{m-1}],}\hfill \end{array}\right.$$

(46)

and using the inverse Laplace transforms we can write

$$\left\{\begin{array}{c}{S}_m\left(t\right)={\chi }_m{S}_{m-1}\left(t\right)+\hslash H{\mathcal{L}}^{-1}\left[R_{1,m}\left(S_{m-1}\right)\right],\hfill \\ \\ E_m\left(t\right)={\chi }_m{E}_{m-1}\left(t\right)+\hslash H{\mathcal{L}}^{-1}\left[R_{2,m}\left(E_{m-1}\right)\right],\hfill \\ \\ I_m\left(t\right)={\chi }_m{I}_{m-1}\left(t\right)+\hslash H{\mathcal{L}}^{-1}\left[R_{3,m}\left(I_{m-1}\right)\right],\hfill \\ \\ Q_m\left(t\right)={\chi }_m{Q}_{m-1}\left(t\right)+\hslash H{\mathcal{L}}^{-1}\left[R_{4,m}\left(Q_{m-1}\right)\right],\hfill \\ \\ R_m\left(t\right)={\chi }_m{R}_{m-1}\left(t\right)+\hslash H{\mathcal{L}}^{-1}\left[R_{5,m}\left(R_{m-1}\right)\right],\hfill \\ \\ D_m\left(t\right)={\chi }_m{D}_{m-1}\left(t\right)+\hslash H{\mathcal{L}}^{-1}\left[R_{6,m}\left(D_{m-1}\right)\right],\hfill \end{array}\right.$$

(47)

where we will apply this relation to find the successive iterations of the HATM.

7. CESTAC Method with CADNA Library

Because of some advantages of the DSA in comparison to the FPA, we apply the mathematical methods based on the DSA instead of the methods based on the FPA. Thus, the CESTAC method and the CADNA library should be applied to validate the numerical results [35,53].

By collecting all representable values that are produced by the computer in set B, we can write $V^{*}\in B$ for $v^{*}\in \mathbb{R}$ with $\alpha$ mantissa bits of the binary FPA as

$$V^{*}=v^{*}-\psi 2^{E-\alpha }\eta ,$$

(48)

where the sign, missing segment of the mantissa and the binary exponent of the result are denoted by $\psi ,2^{-\alpha }\eta$ and E, respectively [31,32,33,34]. In order to find the results with single and double precisions, the value $\alpha$ can be changed to 24 and 53, respectively. Let $\eta$ be a random variable uniformly distributed on $[-1,1]$, constructing perturbation on the last mantissa bit of $v^{*}$; the mean $\left(\mu \right)$ and the standard deviation $\left(\sigma \right)$ values can be produced for results of $V^{*}$. If we repeat the process r times, then r samples of $V^{*}$ can be found as $\Phi =\left\{V_1^{*},V_2^{*},\dots ,V_r^{*}\right\}$ and we should make perturbation on the bit of mantissa. For finding ${\sigma }^2$ we need to calculate the mean value as ${\displaystyle {\tilde{V}}^{*}=\frac{{\sum }_{i=1}^r{V}_i^{*}}{r}}$. Thus we will have equality between the mean value and the exact $v^{*}$. After that, the number of common significant digits between $V^{*}$ and ${\tilde{V}}^{*}$, can be found by

$${\displaystyle {\mathcal{C}}_{{\tilde{V}}^{*},V^{*}}={log}_{10}\frac{\sqrt{r}\left|{\tilde{V}}^{*}\right|}{{\tau }_{\delta }\sigma },}$$

where ${\tau }_{\delta }$ is the value of T distribution as the confidence interval is $1-\delta$, with $r-1$ degrees of freedom [29,35,53]. This process will be stopped if ${\tilde{V}}^{*}=0,$ or ${\mathcal{C}}_{{\tilde{V}}^{*},V^{*}}\le 0$.

When we want to apply the CESTAC method, we do not need to use the method directly. For using the method we should apply the CADNA library. This library can implement the algorithm automatically. The CADNA library should be used on a Linux operating system and all codes should be written using C, C++, FORTRAN or ADA codes. Thus, in this method we do not need to apply the usual mathematical softwares such as Mathematica, Maple and MATLAB. Applying the CESTAC method and DSA we have some advantages in comparison with the methods based on the FPA. In order to apply the termination criterion (1), which is based on the FPA, we need to have the exact solution, but in the DSA we do not need the exact solution and stopping Condition (2) is based on two successive approximations. In the FPA, we do not know the optimal $\epsilon$ and in the DSA we do not have the value $\epsilon$ definitely. In the FPA, the extra iterations can be produced without improving the accuracy, but in the DSA we can find the optimal number of iterations. In the FPA, the algorithm can be stopped in the first step without producing the accurate results but in the DSA, the optimal approximation can be identified. In the CESTAC method, we can produce $@.0$, which shows that the number of common significant digits between two successive approximations are zero, but in the FPA we cannot produce this sign.

Definition 11

([32]). For two real numbers ${\tilde{a}}_1$ and ${\tilde{a}}_2$, the number of common significant digits can be defined as

$$\left\{\begin{array}{c}{\displaystyle {\mathcal{C}}_{{\tilde{a}}_1,{\tilde{a}}_2}={log}_{10}\left|\frac{{\tilde{a}}_1+{\tilde{a}}_2}{2({\tilde{a}}_1-{\tilde{a}}_2)}\right|={log}_{10}\left|\frac{{\tilde{a}}_1}{{\tilde{a}}_1-{\tilde{a}}_2}-\frac{1}{2}\right|,{\tilde{a}}_1\ne {\tilde{a}}_2,}\hfill \\ \\ {\mathcal{C}}_{{\tilde{a}}_1,{\tilde{a}}_1}=+\infty .\hfill \end{array}\right.$$

(49)

Theorem 3.

Let

$$\begin{array}{ccc}{S}_m\left(t\right)=\sum _{j=0}^m{S}_j\left(t\right),\hfill & E_m\left(t\right)=\sum _{j=0}^m{E}_j\left(t\right),\hfill & I_m\left(t\right)=\sum _{j=0}^m{I}_j\left(t\right),\hfill \\ \\ Q_m\left(t\right)=\sum _{j=0}^m{Q}_j\left(t\right),\hfill & R_m\left(t\right)=\sum _{j=0}^m{R}_j\left(t\right),\hfill & D_m\left(t\right)=\sum _{j=0}^m{D}_j\left(t\right),\hfill \end{array}$$

be the approximate solution of the mathematical model of COVID-19 (3) that is produced by the HATM; then

$$\begin{array}{cc}{\mathcal{C}}_{S_m,S_{m+1}}={\mathcal{C}}_{S_m,S}+\mathcal{O}\left(\frac{1}{m}\right),\hfill & {\mathcal{C}}_{Q_m,Q_{m+1}}={\mathcal{C}}_{Q_m,Q}+\mathcal{O}\left(\frac{1}{m}\right),\hfill \\ \\ {\mathcal{C}}_{E_m,E_{m+1}}={\mathcal{C}}_{E_m,E}+\mathcal{O}\left(\frac{1}{m}\right),\hfill & {\mathcal{C}}_{R_m,R_{m+1}}={\mathcal{C}}_{R_m,R}+\mathcal{O}\left(\frac{1}{m}\right),\hfill \\ \\ {\mathcal{C}}_{I_m,I_{m+1}}={\mathcal{C}}_{I_m,I}+\mathcal{O}\left(\frac{1}{m}\right),\hfill & {\mathcal{C}}_{D_m,D_{m+1}}={\mathcal{C}}_{D_m,D}+\mathcal{O}\left(\frac{1}{m}\right).\hfill \end{array}$$

(50)

Proof. 

Using the mentioned definition we can write

$$\begin{array}{cc}{\displaystyle {\mathcal{C}}_{S_m,S_{m+1}}-{\mathcal{C}}_{S_m,S}}\hfill & {\displaystyle ={log}_{10}\left|\frac{S_m+S_{m+1}}{2(S_m-S_{m+1})}\right|-{log}_{10}\left|\frac{S_m+S}{2(S_m-S)}\right|}\hfill \\ \\ \hfill & {\displaystyle ={log}_{10}\left|\frac{S_m+S_{m+1}}{2\left(S_{m+1}^{\hslash }\right)}\right|-{log}_{10}\left|\frac{S_m+S}{2(S_m-S)}\right|}\hfill \\ \\ \hfill & {\displaystyle ={log}_{10}\left|\frac{S_m+S_{m+1}}{S_m+S}\right|+{log}_{10}\left|\frac{S_m-S}{S_{m+1}^{\hslash }}\right|.}\hfill \end{array}$$

(51)

It is obvious that in the first term of Equation (51), by increasing the number of iterations m, the approximate and exact solutions are almost equal. Thus we can neglect that term and we have

$$\begin{array}{cc}{\displaystyle {log}_{10}\left|\frac{S_m-S}{S_{m+1}}\right|}\hfill & {\displaystyle ={log}_{10}\left|\frac{{\sum }_{i=0}^m{S}_i-{\sum }_{i=0}^{\infty }{S}_i}{S_{m+1}}\right|}\hfill \\ \\ & {\displaystyle ={log}_{10}\left|\frac{{\sum }_{i=m+1}^{\infty }{S}_i}{S_{m+1}}\right|}\hfill \\ \\ & {\displaystyle ={log}_{10}\left|1+\frac{{\sum }_{i=m+2}^{\infty }\hslash {\mathcal{L}}^{-1}\left[H\left(r\right){\Re }_i\left({\psi }_{i-1}\right)\right]}{\hslash {\mathcal{L}}^{-1}\left[H\left(r\right){\Re }_{m+1}\left({\psi }_m\right)\right]}\right|.}\hfill \end{array}$$

According to the traditional HAM [21,22,23], we can write

$$\begin{array}{c}{\displaystyle \frac{{\sum }_{i=m+2}^{\infty }\hslash {\mathcal{L}}^{-1}\left[H\left(r\right){\Re }_i\left({\psi }_{i-1}\right)\right]}{\hslash {\mathcal{L}}^{-1}\left[H\left(r\right){\Re }_{m+1}\left({\psi }_m\right)\right]}=\frac{{\sum }_{i=m+2}^{\infty }\hslash {\mathcal{L}}^{-1}\left[H\left(r\right)\frac{1}{(i-1)!}\frac{{\partial }^{i-1}N\left[\Psi (r;v)\right]}{\partial v^{i-1}}{|}_{v=0}\right]}{\hslash {\mathcal{L}}^{-1}\left[H\left(r\right)\frac{1}{m!}\frac{{\partial }^{m}N\left[\Psi (r;v)\right]}{\partial v^m}{|}_{v=0}\right]}}\hfill \\ \\ {\displaystyle =\frac{\hslash {\mathcal{L}}^{-1}\left[H\left(r\right)\frac{1}{(m+1)!}\frac{{\partial }^{m+1}N\left[\Psi (r;v)\right]}{\partial v^{m+1}}{|}_{v=0}\right]}{\hslash {\mathcal{L}}^{-1}\left[H\left(r\right)\frac{1}{m!}\frac{{\partial }^{m}N\left[\Psi (r;v)\right]}{\partial v^m}{|}_{v=0}\right]}+\frac{\hslash {\mathcal{L}}^{-1}\left[H\left(r\right)\frac{1}{(m+2)!}\frac{{\partial }^{m+2}N\left[\Psi (r;v)\right]}{\partial v^{m+2}}{|}_{v=0}\right]}{\hslash {\mathcal{L}}^{-1}\left[H\left(r\right)\frac{1}{m!}\frac{{\partial }^{m}N\left[\Psi (r;v)\right]}{\partial v^m}{|}_{v=0}\right]}+\cdots }\hfill \\ \\ {\displaystyle =\mathcal{O}\left(\frac{1}{m}\right).}\hfill \end{array}$$

Therefore,

$$\begin{array}{cc}{\displaystyle {log}_{10}\left|\frac{S_m-S}{S_{m+1}}\right|={log}_{10}\left|1+\mathcal{O}\left(\frac{1}{m}\right)\right|,}\hfill & {\displaystyle {log}_{10}\left|\frac{Q_m-Q}{Q_{m+1}}\right|={log}_{10}\left|1+\mathcal{O}\left(\frac{1}{m}\right)\right|,}\hfill \\ \\ {\displaystyle {log}_{10}\left|\frac{E_m-E}{E_{m+1}}\right|={log}_{10}\left|1+\mathcal{O}\left(\frac{1}{m}\right)\right|,}\hfill & {\displaystyle {log}_{10}\left|\frac{R_m-R}{R_{m+1}}\right|={log}_{10}\left|1+\mathcal{O}\left(\frac{1}{m}\right)\right|,}\hfill \\ \\ {\displaystyle {log}_{10}\left|\frac{I_m-I}{I_{m+1}}\right|={log}_{10}\left|1+\mathcal{O}\left(\frac{1}{m}\right)\right|,}\hfill & {\displaystyle {log}_{10}\left|\frac{D_m-D}{D_{m+1}}\right|={log}_{10}\left|1+\mathcal{O}\left(\frac{1}{m}\right)\right|.}\hfill \end{array}$$

It is clear that $\mathcal{O}\left(\frac{1}{m}\right)<<1$, and thus the right-hand sides of the above relations decrease as m increases. □

8. Numerical Results

In this section, the numerical results of the HATM for solving the nonlinear fractional order model (4) are presented. As known in the HATM, we have some auxiliary functions and parameters that can help us find more accurate solutions with high-speed of convergence. In this method, the obtained solutions are based on t and ℏ such that the parameter ℏ can help identify and control the convergence region. To this aim, by plotting some ℏ-curves we can find these regions. In Figure 2, the ℏ-curves are presented for $t=0.2$ and $m=15$. The convergence region for functions $S,E,I,Q,R$ and D is $-1\le \hslash \le -0.6,$, which is the parallel part of the ℏ-curve with the axis x. Additionally, by increasing the number of iterations of the HATM, we can plot more accurate ℏ-curves. In Figure 3, the ℏ-curves are plotted for $m=25$ and we obtain:

$$\begin{array}{cc}-0.7\le {\hslash }_S\le -0.5,& -0.55\le {\hslash }_Q\le -0.45,\\ \\ -0.7\le {\hslash }_E\le -0.5,& -0.65\le {\hslash }_R\le -0.45,\\ \\ -0.7\le {\hslash }_I\le -0.5,& -0.7\le {\hslash }_D\le -0.5.\end{array}$$

In

Table 2. Numerical results based on the DSA and using the CESTAC method for $t=0.2,\rho =1$ and $\hslash =-0.8$.

m	Approximate Solutions	Difference between Two Iterations
1	$4.365721$	$4.365721$
	$35.00683$	$35.00683$
	$15.39427$	$15.39427$
	$18.58639$	$18.58639$
	$45.94279$	$45.94279$
	$5.020640$	$5.020640$
2	$7.081621$	$2.71589$
	$31.93003$	$3.07679$
	$15.70567$	$0.31139$
	$12.3250$	$30.9114$
	$15.1206$	$30.82215$
	$5.013390$	$0.7249\times {10}^{-2}$
3	$7.329236$	$0.24761$
	$31.62606$	$0.30397$
	$15.75156$	$0.45893\times {10}^{-1}$
	$2.0715$	$14.3966$
	$29.53187$	$14.4112$
	$5.01834$	$0.495\times {10}^{-2}$
4	$7.33754$	$0.831\times {10}^{-2}$
	$31.61051$	$0.155\times {10}^{-1}$
	$15.75638$	$0.482\times {10}^{-2}$
	$1.0558$	$3.1274$
	$26.40971$	$3.1221$
	$5.017514$	$0.834\times {10}^{-3}$
5	$7.32929$	$0.825\times {10}^{-2}$
	$31.61790$	$0.738\times {10}^{-2}$
	$15.75678$	$0.39\times {10}^{-3}$
	$1.00038$	$0.5541\times {10}^{-1}$
	$26.46585$	$0.561\times {10}^{-1}$
	$5.01755$	$0.4\times {10}^{-4}$
6	$7.32617$	$0.311\times {10}^{-2}$
	$31.62090$	$0.2998352\times {10}^{-2}$
	$15.75674$	$0.3\times {10}^{-4}$
	$0.91942$	$0.8095\times {10}^{-1}$
	$26.5475$	$0.8165\times {10}^{-1}$
	$5.01759$	$0.3\times {10}^{-4}$
7	$7.32556$	$0.6\times {10}^{-3}$
	$31.62149$	$0.59\times {10}^{-3}$
	$15.75674$	$@.0$
	91948	$0.6\times {10}^{-4}$
	$26.54801$	$0.50\times {10}^{-3}$
	$5.01759$	$@.0$

, the numerical results obtained based on the DSA and using the CESTAC method are shown for $t=0.2,\rho =1$ and $\hslash =-0.8$. However, it is important to describe why we need to apply the DSA instead of the FPA. In the methods based on the FPA, generally we apply the traditional absolute error or residual error to show the accuracy of the method, but in these cases we need to know the exact solution and also in order to stop the numerical algorithm we need to apply a suitable termination criterion. Thus, instead of applying the methods based on the FPA, we propose methods based on the DSA. In this case, we apply the CESTAC method and the CADNA library to validate the numerical results. In the CESTAC method we use the difference between two successive approximations instead of the traditional absolute error. Additionally, we do not need to have positive small value such as $\epsilon$ and an important note is that we do not need to have the exact solution. Using this method, we can find the optimal number of iteration and the optimal approximation of the HATM for solving the mentioned model. According to Table 4, the optimal iteration is $m_{opt}=7$ and the optimal approximations are:

$$\begin{array}{cc}{S}_{opt}=0.732556E+001,& Q_{opt}=0.91948E+000,\\ \\ E_{opt}=0.3162149E+002,& R_{opt}=0.2654801E+002,\\ \\ I_{opt}=0.1575674E+002,& D_{opt}=0.501759E+001.\end{array}$$

Sign $@.0$ in step 7 shows that the number of common significant digits between two successive approximations is zero. Thus, we can apply the difference between two successive approximations instead of traditional absolute error. Additionally, 24 numerical instabilities are reported by the CADNA library. In

Table 3. Numerical results based on the FPA and Condition (1) for $t=0.2,\rho =1,\hslash =-0.8$ and $\epsilon ={10}^{-4}$.

m	Approximate Solutions	Difference between Two Iterations
1	$4.365721$	$4.365721$
	$35.00683$	$35.00683$
	$15.39427$	$15.39427$
	$18.58639$	$18.58639$
	$45.94279$	$45.94279$
	$5.020640$	$5.020640$
2	$7.081621$	$2.71589$
	$31.93003$	$3.07679$
	$15.70567$	$0.31139$
	$12.3250$	$30.9114$
	$15.1206$	$30.82215$
	$5.013390$	$0.7249\times {10}^{-2}$
3	$7.329236$	$0.24761$
	$31.62606$	30397
	$15.75156$	$45893\times {10}^{-1}$
	$2.0715$	$14.3966$
	$29.53187$	$14.4112$
	$5.01834$	$0.495\times {10}^{-2}$
4	$7.33754$	$0.831\times {10}^{-2}$
	$31.61051$	$0.155\times {10}^{-1}$
	$15.75638$	$0.482\times {10}^{-2}$
	$1.0558$	$3.1274$
	$26.40971$	$3.1221$
	$5.017514$	$0.834\times {10}^{-3}$
5	$0.732929$	$0.825\times {10}^{-2}$
	$31.61790$	$0.738\times {10}^{-2}$
	$15.75678$	$0.39\times {10}^{-3}$
	$1.00038$	$0.5541\times {10}^{-1}$
	$26.46585$	$0.561\times {10}^{-1}$
	$5.01755$	$0.4\times {10}^{-4}$

, the numerical results of the FPA based on Condition (1) for $t=0.2$ and $\epsilon ={10}^{-4}$ are presented. It is obvious that the stopping condition depends on the value of $\epsilon$. In

Table 4. Number of iterations based on the FPA and Condition (1) for various $\epsilon$ and $t=0.2$, $\rho =1$, $\hslash =-0.8$.

$\mathit{\epsilon }$	Small Values	${10}^{-5}$	${10}^{-4}$	${10}^{-1}$	0.5	Large Values
m	$>>9$	9	5	3	1	1

, the numbers of iterations for different values of $\epsilon$ are obtained. As we described before, the algorithm is stopped at the first steps for large values of $\epsilon$. Additionally, for small values of $\epsilon$ we have large number of iterations. In

Table 5. Numerical results based on the DSA and using the CESTAC method for $t=0.2,\rho =0.9$ and $\hslash =-0.6$.

m	Approximate Solutions	Difference between Two Iterations
1	0.6416038	0.6416040
	35.75721	35.75721
	14.30519	14.30519
	21.33839	21.33840
	47.63999	47.63999
	5.294120	5.294120
2	8.290508	1.87447
	30.37535	5.38186
	14.48619	0.18099
	34.01543	12.67704
	5.92429	41.757
	5.18415	0.10996
3	7.42754	0.86296
	31.0735	0.69816
	14.44400	$0.4218\times {10}^{-1}$
	34.8834	$0.86797\times {10}^{-2}$
	60.98040	$0.17375\times {10}^{-1}$
	5.373754	0.18960
⋮	⋮	⋮
9	7.61575	$0.32\times {10}^{-3}$
	30.7691	$0.1\times {10}^{-3}$
	14.42713	$0.19\times {10}^{-3}$
	19.1588	$0.43\times {10}^{-3}$
	45.76194	$0.2\times {10}^{-4}$
	5.33333	$0.6258\times {10}^{-1}$
10	7.61582	$0.7\times {10}^{-4}$
	30.76886	@.0
	14.4270	$0.7\times {10}^{-4}$
	19.16541	$0.6\times {10}^{-4}$
	45.76194	@.0
	5.28554	$0.4779\times {10}^{-1}$

, the numerical results are presented for $t=0.2,\rho =0.9$ and $\hslash =-0.6$ based on the CESTAC method. Using this table we can find the optimal iteration $m_{opt}=10$ and the following optimal approximations:

$$\begin{array}{cc}{S}_{opt}=0.761582E+001,& Q_{opt}=0.1916541E+002,\\ \\ E_{opt}=0.3076886E+002,& R_{opt}=0.4576194E+002,\\ \\ I_{opt}=0.144270E+002,& D_{opt}=0.528554E+001.\end{array}$$

In Figure 4, the approximate solution of the model using the HATM is demonstrated. We can see that by decreasing the number of susceptible people (red line), the quarantined people (dotted line) decreases, the infected people (dashed line) slowly decreases, the rate of exposed people (orange line) increases and after that slowly decreases, the death rate (green line) slowly increases, and the number of recovered people increases and then slowly increases. Thus, we have acceptable and logical results for the model.

9. Conclusions

Given the importance of modeling and controlling the COVID-19 pandemic, we focused on a nonlinear fractional order model of COVID-19. We applied the Caputo–Fabrizio fractional derivative to present a novel fractional order model. Some theorems were proven to show the existence of a solution and stability analysis. The HATM was used to solve the model numerically and the CESTAC method and the CADNA library were applied to validate the results. In order to show the accuracy of the results, instead of applying the absolute error we used the difference between two successive approximations. The main theorem of the CESTAC method enabled us to use the new termination criterion. This method is based on the DSA and using this method we can find the optimal iteration and the optimal approximation of the HATM for solving the fractional order model. A comparative study between the FPA and the DSA was presented to show the abilities of the CESTAC method and the CADNA library.