# On the Analytical Solution of Fractional SIR Epidemic Model

Authors: Ahmad Qazza; Rania Saadeh
Journal: Applied Computational Intelligence and Soft Computing
 (2023)

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## Abstract

This article presents the solution of the fractional SIR epidemic model using the Laplace residual power series method. We introduce the fractional SIR model in the sense of Caputo’s derivative; it is presented by three fractional differential equations, in which the third one depends on the first coupled equations. The Laplace residual power series method (LRPSM) is implemented in this research to solve the proposed model, in which we present the solution in a form of convergent series expansion that converges rapidly to the exact one. We analyze the results and compare the obtained approximate solutions to those obtained from other methods. Figures and tables are illustrated to show the efficiency of the LRPSM in handling the proposed SIR model.


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## Body

## 1. Introduction

Epidemiology is a discipline of biology that examines the prevalence and causes health-related problems in particular groups or communities. It may be used to manage community health concerns. Frequency, distribution, and the factors that cause diseases are the elements of epidemiology. By establishing priorities among these services, the Department of Epidemiology seeks to offer the data required for the planning, execution, and assessment of services aimed at disease prevention, control, and treatment. Epidemiology is used to investigate the historical development and decline of illnesses in the population, community diagnosis, planning and evaluation, assessment of an individual’s risks and chances, identification of syndromes, and completion of the disease’s natural history [1–5].The study of fractional calculus depends on computing integrals and derivatives of noninteger orders. The importance of these studies has appeared in the various applications in physics and engineering, in which these derivatives can describe them more realistically in the fields of science. A more accurate explanation of the challenges in the actual world can be provided by the fractional application models [6–11].Through the process of mathematical modeling, one may examine, anticipate, and offer insight into issues that arise in the actual world. It is advantageous since conventional theoretical approaches are inadequate for the study of technological, ecological, economic, and other systems examined by modern research. Fractional differential equations and systems have been used to simulate a variety of real-world issues [12–18].The SIR model is one of the simplest fractional models, and many models are derived depending on its form. The model consists of three components:φ: The number of susceptible individuals. When a susceptible and an infectious individual come into “infectious contact,” the susceptible individual contracts the disease and transitions to the infectious side.ψ: The number of infectious individuals. These are individuals who have been infected and are capable of infecting susceptible individuals.ϕ: The number of removed (and immune) or deceased individuals.The study of the SIR model was first introduced in 2009 by Ahmet and Cherruault [19], then in 2011 they present new research on analytical solutions of some related models. After that, many authors have investigated the susceptible-infected recovered models of integer fractional orders [20–23].The integer-order SIR epidemic model is given by(1)φ′τ=−q1φτψτ+q3ψτ,ψ′τ=q1φτψτ−q3ψτ−q2ψτ,ϕ′τ=q2ψτ,under the initial conditions:(2)φ0=α,ψ0=β,ϕ0=λ,where φτ denotes the susceptible individuals, ψτ denotes the infected individuals, ϕτ denotes the removed individuals, and τ denotes the time. N is the total number of the studied population such that φτ+ψτ+ϕτ=N. The rate of changes between the previous and the new values are q1,q2andq3, for more details see [24, 25]. There are some constraints on the model concerning the population number N, which should be large enough, the parameters of the system should be fit, and finally, the healing does not provide immunity.This research presents the LRPSM, which is a new analytical method that combines the Laplace transform with the residual power series method; it was first introduced in [26], and it is implemented by researchers to solve several models of fractional ordinary and partial differential equations and systems. This method shows its efficiency and applicability in solving similar problems [27–35].The main aim of this article is to present an analytical series solution of the fractional SIR model. We use the LRPSM to get the solution in the form of a rapidly convergent series. We introduce the method and state the convergence analysis of the method, then we apply it to solve the proposed model. Numerical simulations of the results are discussed, and comparisons are made with the results obtained from other numerical methods. The strength of the LRPSM arises in the ability of solving similar models and presenting many terms of the series solutions with fewer calculations and without the need of linearization, discretization, or differentiation as with other numerical methods.The layout of this article is as follows: in Section2, we present some definitions and theorems related to fractional power series and the analysis of LRPSM. In Section 3, we construct the series solution of the fractional SIR model in the sense of Caputo’s derivative by LRPSM. In Section 4, we introduce some numerical simulations of our results and comparisons to other numerical methods. Finally, the conclusion section is presented in Section 5.


## 2. Fractional Power Series

In this section of the article, we introduce some basic definitions and characteristics of the Caputo fractional derivative and the Laplace transform. Also, we present theorems about the fractional Taylor’s series of expansions.Definition 1.
Caputo fractional derivative of the functionφτ of order γ is given by the following equation:(3)Dγφτ=Jτm−γφrτ,r−1<γ<r,τ≥0,φrτ,γ=r,τ≥0,where r∈ℕ and Jτγ is the Riemann–Liouville integral of the fractional order γ to the function φτ, provided the integral exists.
There are many properties of Caputo’s derivative, that might be found in [23, 24], and we mention some of them as follows:(i)
Dγc=0,c∈ℝ(ii)
Dγtβ=Γβ+1/Γβ+1−γτβ−γ,τ≥0,β>−1,γ>0(iii)
DγJτγφτ=φτ(iv)
JτγDγφτ=φτ−∑i=0r−1φi0+τi/i!,r−1<γ≤rDefinition 2.
Letφτ be a piecewise continuous function on 0,∞. If φτ is of exponential order, then φτ has the Laplace transform which is defined as follows:(4)Φs=ℒφτ=∫0∞e−stφτdτ,s>0.
The inverse Laplace transform is given by(5)φτ=ℒ−1Φs=∫α−i∞α+i∞estΦsds,α=Res>c,for some, c provided the integral exists.
In the following arguments, we list some of the most popular properties of the Laplace transform.
IfΦs=ℒφτ, Ψs=ℒψτ, and c,d are constants, then(i)
ℒcφτ+dψτ=cΦs+dΨs(ii)
ℒ−1cΦs+dΨs=cφτ+dψτ(iii)
lims⟶∞sΦs=φ0(iv)
ℒDγφτ=sγΦs−∑k=0r−1sγ−k−1φk0,r−1<α≤r(v)
ℒDnγφτ=snγΦs−∑k=0n−1sn−kγ−1Dkγφ0,0<α<1
For the proof, readers can see [2, 4, 23].
The following theorems illustrate some results about Taylor’s fractional series and the convergence analysis of the new approach.Theorem 1 (see [23]).
Suppose the functionφτ has a power series representation at τ=δ of the shape:(6)φτ=∑n=0∞αnτ−δnγ,0<r−1<γ≤r,0<τ<ξ.Ifφτ∈Cδ,δ+ξ and Dnγφτ∈Cδ,δ+ξ for n=0,1,…, then we have the following formula for the coefficient αn in equation (6), that takes the expression as follows:(7)αn=DnγφδΓnγ+1,n=0,1,2,…,⊂,where Dnγ=DγDγ⋯Dγ⏟n−times.Note that puttingγ=1 in expression (6), we get the usual Taylor power series.The next theorem illustrates a new form of Theorem1 in the Laplace space considering δ=0.Theorem 2.
The fractional power series in (6) has three possibilities for convergence:(i)
Ifτ=δ, then the series is convergent and the radius of convergent is zero(ii)
Ifτ≥δ, the series is convergent and the radius of convergent is ∞(iii)
Ifδ≤τ<δ+ζ, for some possible real number ζ, then the series converges, and if τ>δ+ζ, then the series (1) divergesTheorem 3 (see [23]).
If the fractional power series expansion of the functionΦs=ℒφτ is expressed as follows:(8)Φs=∑n=0∞αnsnγ+1,s>0,0<γ≤1,then the coefficient αn can be obtained from the following formula:(9)αn=Dnγφ0.Moreover, the inverse Laplace transform of the series expansion (8) in Theorem 3 has the form:(10)φτ=∑n=1∞Dnγφ0Γnγ+1τnγ,τ≥0,0<γ≤1.Theorem 4 (see [23]) (Convergence analysis).
LetΦs=ℒφτ be a function that has the fractional power series in equation (2). If sℒDn+1γφτ≤V, on 0<s≤b, where 0<γ≤1 and V>0, then the remainder Rns of the series representation (2) has the following bound:(11)Rns≤Vsn+1γ+1,0<s≤b.


## 3. Construction of Solutions to Fractional SIR Model

### 3.1. Fractional SIR Epidemic Model

There are several phenomena in the real world in engineering and physics, which can be reformulated by fractional initial value problems. Not all of these problems can be solved exactly, so, they are challenging researchers around the whole world. Our aim in this section is to introduce the main idea of the LRPSM in solving systems of nonlinear fractional differential equations that might be difficult to solve by usual techniques.Now, we present the fractional SIR model:(12)Dγ1φτ=−q1φτψτ+q3ψτ,Dγ2ψτ=q1φτψτ−q3ψτ−q2ψτ,Dγ3ϕτ=q2ψτ,where γi∈0,1,∀i=1,2,3, Dγi denotes the Caputo derivative, q1,q2, and q3 are real positive numbers, q1 denotes the infection rate, q2 denotes the removal rate, and q3 is the recovery rate.The given initial conditions for system (12) are as follows:(13)φ0=α,ψ0=β,ϕ0=λ.Also, we have:(14)φτ+ψτ+ϕτ=N,and the relation:(15)Dγ1φτ+Dγ2ψτ+Dγ3ϕτ=0.The relation (15) gives an extra condition that enable us to solve only two equations of three variables.


### 3.2. LRPSM for Solving Fractional SIR Model

The basic idea of LRPSM is to apply the Laplace transform on the target equations, then define the so-called Laplace residual functions. After that, multiply each equation byskα+1 by the truncated Laplace residual functions and take the limit at infinity to get the required values of the series coefficients recursively.To get the series solution of the system (12) using the proposed method, we first operate the Laplace transform on both sides of each equation in system (5), to get:(16)ℒDγ1φτ=−q1ℒφτψτ+q3ℒψτ,ℒDγ2ψτ=q1ℒφτψτ−q3ℒψτ−q2ℒψτ,ℒDγ3ϕτ=q2ℒψτ.Running Laplace transform on system (16), we get:(17)sγ1Φs−sγ1−1φ0=−q1ℒℒ−1Φsℒ−1Ψs+q3Ψs,sγ2Ψs−sγ1−1ψ0=q1ℒℒ−1Φsℒ−1Ψs−q3Ψs−q2Ψs,sγ3Gs−sγ1−1ϕ0=q2Ψs.Simplifying the equations in system (17) and substituting the initial conditions (13), we get:(18)Φs=αs−q1sγ1ℒℒ−1Φsℒ−1Ψs+q3sγ1Ψs,Ψs=βs+q1sγ2ℒℒ−1Φsℒ−1Ψs−q3sγ2Ψs−q2Ψs,Gs=λs+q2sγ3Ψs.Suppose that the solution of system (18) can be presented in the following series forms:(19)Φs=∑n=0∞αnsnγ1+1,Ψs=∑n=0∞βnsnγ2+1,Gs=∑n=0∞λnsnγ3+1.Using the property thatlims⟶∞sΦs=φ0, it is obvious that(20)lims⟶∞sΦs=φ0=α0=α,lims⟶∞sΨs=ψ0=β0=β,lims⟶∞sGs=ϕ0=λ0=λ.The series expansion in (19) can be written as follows:(21)Φs=αs+∑n=1∞αnsnγ1+1,Ψs=βs+∑n=1∞βnsnγ2+1,Gs=λs+∑n=1∞λnsnγ3+1.Now, define the Laplace residual functions of system (18) as follows:(22)ℒResΦs=Φs−αs+q1sγ1ℒℒ−1Φsℒ−1Ψs−q3sγ1Ψs,ℒResΨs=Ψs−βs−q1sγ2ℒℒ−1Φsℒ−1Ψs+q3sγ2Ψs+q2Ψs,ℒResGs=Gs−λs−q2sγ3Ψs.Thekth truncated series of (21) has the form:(23)Φks=αs+∑n=1kαnsnγ1+1,Ψks=βs+∑n=1kβnsnγ2+1,Gks=λs+∑n=1kλnsnγ3+1.and the kth Laplace residual functions as follows:(24)ℒReskΦs=Φks−αs+q1sγ1ℒℒ−1Φksℒ−1Ψks−q3sγ1Ψks,ℒReskΨs=Ψks−βs−q1sγ2ℒℒ−1Φksℒ−1Ψks+q3sγ2Ψks+q2Ψks,ℒReskGs=Gks−λs−q2sγ3Ψks.Now, the first truncated series of (23) are as follows:(25)Φ1s=αs+α1sγ1+1,Ψ1s=βs+β1sγ2+1,G1s=λs+λ1sγ3+1.Substituting the above values in the first Laplace residual functions to get the following:(26)ℒRes1Φs=α1sγ1+1+q1sγ1ℒℒ−1αs+α1sγ1+1ℒ−1βs+β1sγ2+1−q3sγ1λs+λ1sγ3+1,ℒRes1Ψs=β1sγ2+1−q1sγ2ℒℒ−1αs+α1sγ1+1ℒ−1βs+β1sγ2+1+q3sγ2βs+β1sγ2+1+q2βs+β1sγ2+1,ℒRes1Gs=λ1sγ3+1−q2sγ3βs+β1sγ2+1.Using the following facts of the LRPSM, which can be found in [23],(i)
ℒResΦs=0(ii)
limk⟶∞ℒReskΦs=ℒResΦs, for each s>0(iii)
lims⟶∞sℒResΦs=0 and lims⟶∞ℒReskΦs=0(iv)
lims⟶∞ℒReskΦs=0,k=1,2,… and 0<α≤1Now, multiplying each equation in (26) by sγ1+1,sγ2+1, and sγ3+1, respectively, and then taking the limit as s⟶∞, we get the first coefficients of the series expansion (23) as follows:(27)α1=−q1α0β0+q3β0=−q1αβ+q3β,β1=−α1−q2β0=q1αβ−q3β−q2β,λ1=q2β1.Repeating the previous steps, we can get other coefficients.In addition, if we takeγ1=γ2=γ3=ϑ, one can get the coefficients of the series solution as follows:(28)α1=βq3−αq1,α2=βαq1−q3q1β−α+q2+q3,α3=βαq1−q3q12α−β2Γ2ϑ+1−αβΓ2ϑ+1Γ2ϑ+1+βαq1−q3q1q2β−2αΓ2ϑ+1+βΓ2ϑ+1+q3βΓ2ϑ+1−2α−βΓ2ϑ+1Γ2ϑ+1+βαq1−q3q2+q32Γ2ϑ+1Γ2ϑ+1,…,β1=βαq1−q2−q3,β2=βαq12α−β+q1q3β−2α−2αq2+q2+q32,β3=q12βq32αβΓ2ϑ+1−3α2−4αβ+β2Γ2ϑ+1+αq22β−3αΓ2ϑ+1+βΓ2ϑ+1Γ2ϑ+1+αβq13α−β2Γ2ϑ+1−αβΓ2ϑ+1Γ2ϑ+1+q2+q3q1q33α−2βΓ2ϑ+1−βΓ2ϑ+13αβq2Γ2ϑ+1Γ2ϑ+1−q2+q33Γ2ϑ+1Γ2ϑ+1,…,λ1=βq2,λ2=−βq2−αq1+q2+q3,λ3=βq2αq12α−β+q1q3β−2α−2αq2+q2+q32,….Thek th coefficients of the series solutions (21) have the form:(29)αk+1=−q1∑i=0kαiβk−iΓ1+nϑΓ1+iϑΓ1+n−iϑ+q3βk,βk+1=−αk+1−q2βk,λk+1=q2βk,k=0,1,2,….Thus, thekth solution of systems (12) and (13) in Laplace space can be expressed as follows:(30)Φks=αs+βq3−αq1sϑ+1+βαq1−q3q1β−α+q2+q3s2ϑ+1+⋯+1skϑ+1−q1∑i=0kαiβn−iΓ1+nϑΓ1+iϑΓ1+n−iϑ+q3βk,Ψks=βs+βαq1−q2−q3sϑ+1+βαq12α−β+q1q3β−2α−2αq2+q2+q32s2ϑ+1+⋯+−αk+1−q2βkskϑ+1,Gks=λs+q2β1sϑ+1+q2β2s2ϑ+1+⋯+q2βkskϑ+1.Operating the inverse Laplace transform on each equation in (32), we obtain the kth solution of system (12) as follows:(31)φkτ=α+βq3−αq1Γϑ+1τϑ+βαq1−q3q1β−α+q2+q3Γ2ϑ+1τ2ϑ+⋯+τkϑΓkϑ+1−q1∑i=0kαiβn−iΓ1+nϑΓ1+iϑΓ1+n−iϑ+q3βk,ψτ=β+βαq1−q2−q3Γϑ+1τϑ+βαq12α−β+q1q3β−2α−2αq2+q2+q32Γ2ϑ+1τ2ϑ+⋯+−αk+1−q2βkΓkϑ+1τkϑ,ϕkτ=λ+q2β1Γϑ+1τϑ+q2β2Γ2ϑ+1τ2ϑ+⋯+q2βkΓkϑ+1τkϑ.Ask⟶∞, the kth truncated solutions converge to the exact solutions in the integer orders:(32)φkτ⟶φτ,ψkτ⟶ψτ,ϕkτ⟶ϕτ.Hence, we get the required solution.


## 3.1. Fractional SIR Epidemic Model

There are several phenomena in the real world in engineering and physics, which can be reformulated by fractional initial value problems. Not all of these problems can be solved exactly, so, they are challenging researchers around the whole world. Our aim in this section is to introduce the main idea of the LRPSM in solving systems of nonlinear fractional differential equations that might be difficult to solve by usual techniques.Now, we present the fractional SIR model:(12)Dγ1φτ=−q1φτψτ+q3ψτ,Dγ2ψτ=q1φτψτ−q3ψτ−q2ψτ,Dγ3ϕτ=q2ψτ,where γi∈0,1,∀i=1,2,3, Dγi denotes the Caputo derivative, q1,q2, and q3 are real positive numbers, q1 denotes the infection rate, q2 denotes the removal rate, and q3 is the recovery rate.The given initial conditions for system (12) are as follows:(13)φ0=α,ψ0=β,ϕ0=λ.Also, we have:(14)φτ+ψτ+ϕτ=N,and the relation:(15)Dγ1φτ+Dγ2ψτ+Dγ3ϕτ=0.The relation (15) gives an extra condition that enable us to solve only two equations of three variables.


## 3.2. LRPSM for Solving Fractional SIR Model

The basic idea of LRPSM is to apply the Laplace transform on the target equations, then define the so-called Laplace residual functions. After that, multiply each equation byskα+1 by the truncated Laplace residual functions and take the limit at infinity to get the required values of the series coefficients recursively.To get the series solution of the system (12) using the proposed method, we first operate the Laplace transform on both sides of each equation in system (5), to get:(16)ℒDγ1φτ=−q1ℒφτψτ+q3ℒψτ,ℒDγ2ψτ=q1ℒφτψτ−q3ℒψτ−q2ℒψτ,ℒDγ3ϕτ=q2ℒψτ.Running Laplace transform on system (16), we get:(17)sγ1Φs−sγ1−1φ0=−q1ℒℒ−1Φsℒ−1Ψs+q3Ψs,sγ2Ψs−sγ1−1ψ0=q1ℒℒ−1Φsℒ−1Ψs−q3Ψs−q2Ψs,sγ3Gs−sγ1−1ϕ0=q2Ψs.Simplifying the equations in system (17) and substituting the initial conditions (13), we get:(18)Φs=αs−q1sγ1ℒℒ−1Φsℒ−1Ψs+q3sγ1Ψs,Ψs=βs+q1sγ2ℒℒ−1Φsℒ−1Ψs−q3sγ2Ψs−q2Ψs,Gs=λs+q2sγ3Ψs.Suppose that the solution of system (18) can be presented in the following series forms:(19)Φs=∑n=0∞αnsnγ1+1,Ψs=∑n=0∞βnsnγ2+1,Gs=∑n=0∞λnsnγ3+1.Using the property thatlims⟶∞sΦs=φ0, it is obvious that(20)lims⟶∞sΦs=φ0=α0=α,lims⟶∞sΨs=ψ0=β0=β,lims⟶∞sGs=ϕ0=λ0=λ.The series expansion in (19) can be written as follows:(21)Φs=αs+∑n=1∞αnsnγ1+1,Ψs=βs+∑n=1∞βnsnγ2+1,Gs=λs+∑n=1∞λnsnγ3+1.Now, define the Laplace residual functions of system (18) as follows:(22)ℒResΦs=Φs−αs+q1sγ1ℒℒ−1Φsℒ−1Ψs−q3sγ1Ψs,ℒResΨs=Ψs−βs−q1sγ2ℒℒ−1Φsℒ−1Ψs+q3sγ2Ψs+q2Ψs,ℒResGs=Gs−λs−q2sγ3Ψs.Thekth truncated series of (21) has the form:(23)Φks=αs+∑n=1kαnsnγ1+1,Ψks=βs+∑n=1kβnsnγ2+1,Gks=λs+∑n=1kλnsnγ3+1.and the kth Laplace residual functions as follows:(24)ℒReskΦs=Φks−αs+q1sγ1ℒℒ−1Φksℒ−1Ψks−q3sγ1Ψks,ℒReskΨs=Ψks−βs−q1sγ2ℒℒ−1Φksℒ−1Ψks+q3sγ2Ψks+q2Ψks,ℒReskGs=Gks−λs−q2sγ3Ψks.Now, the first truncated series of (23) are as follows:(25)Φ1s=αs+α1sγ1+1,Ψ1s=βs+β1sγ2+1,G1s=λs+λ1sγ3+1.Substituting the above values in the first Laplace residual functions to get the following:(26)ℒRes1Φs=α1sγ1+1+q1sγ1ℒℒ−1αs+α1sγ1+1ℒ−1βs+β1sγ2+1−q3sγ1λs+λ1sγ3+1,ℒRes1Ψs=β1sγ2+1−q1sγ2ℒℒ−1αs+α1sγ1+1ℒ−1βs+β1sγ2+1+q3sγ2βs+β1sγ2+1+q2βs+β1sγ2+1,ℒRes1Gs=λ1sγ3+1−q2sγ3βs+β1sγ2+1.Using the following facts of the LRPSM, which can be found in [23],(i)
ℒResΦs=0(ii)
limk⟶∞ℒReskΦs=ℒResΦs, for each s>0(iii)
lims⟶∞sℒResΦs=0 and lims⟶∞ℒReskΦs=0(iv)
lims⟶∞ℒReskΦs=0,k=1,2,… and 0<α≤1Now, multiplying each equation in (26) by sγ1+1,sγ2+1, and sγ3+1, respectively, and then taking the limit as s⟶∞, we get the first coefficients of the series expansion (23) as follows:(27)α1=−q1α0β0+q3β0=−q1αβ+q3β,β1=−α1−q2β0=q1αβ−q3β−q2β,λ1=q2β1.Repeating the previous steps, we can get other coefficients.In addition, if we takeγ1=γ2=γ3=ϑ, one can get the coefficients of the series solution as follows:(28)α1=βq3−αq1,α2=βαq1−q3q1β−α+q2+q3,α3=βαq1−q3q12α−β2Γ2ϑ+1−αβΓ2ϑ+1Γ2ϑ+1+βαq1−q3q1q2β−2αΓ2ϑ+1+βΓ2ϑ+1+q3βΓ2ϑ+1−2α−βΓ2ϑ+1Γ2ϑ+1+βαq1−q3q2+q32Γ2ϑ+1Γ2ϑ+1,…,β1=βαq1−q2−q3,β2=βαq12α−β+q1q3β−2α−2αq2+q2+q32,β3=q12βq32αβΓ2ϑ+1−3α2−4αβ+β2Γ2ϑ+1+αq22β−3αΓ2ϑ+1+βΓ2ϑ+1Γ2ϑ+1+αβq13α−β2Γ2ϑ+1−αβΓ2ϑ+1Γ2ϑ+1+q2+q3q1q33α−2βΓ2ϑ+1−βΓ2ϑ+13αβq2Γ2ϑ+1Γ2ϑ+1−q2+q33Γ2ϑ+1Γ2ϑ+1,…,λ1=βq2,λ2=−βq2−αq1+q2+q3,λ3=βq2αq12α−β+q1q3β−2α−2αq2+q2+q32,….Thek th coefficients of the series solutions (21) have the form:(29)αk+1=−q1∑i=0kαiβk−iΓ1+nϑΓ1+iϑΓ1+n−iϑ+q3βk,βk+1=−αk+1−q2βk,λk+1=q2βk,k=0,1,2,….Thus, thekth solution of systems (12) and (13) in Laplace space can be expressed as follows:(30)Φks=αs+βq3−αq1sϑ+1+βαq1−q3q1β−α+q2+q3s2ϑ+1+⋯+1skϑ+1−q1∑i=0kαiβn−iΓ1+nϑΓ1+iϑΓ1+n−iϑ+q3βk,Ψks=βs+βαq1−q2−q3sϑ+1+βαq12α−β+q1q3β−2α−2αq2+q2+q32s2ϑ+1+⋯+−αk+1−q2βkskϑ+1,Gks=λs+q2β1sϑ+1+q2β2s2ϑ+1+⋯+q2βkskϑ+1.Operating the inverse Laplace transform on each equation in (32), we obtain the kth solution of system (12) as follows:(31)φkτ=α+βq3−αq1Γϑ+1τϑ+βαq1−q3q1β−α+q2+q3Γ2ϑ+1τ2ϑ+⋯+τkϑΓkϑ+1−q1∑i=0kαiβn−iΓ1+nϑΓ1+iϑΓ1+n−iϑ+q3βk,ψτ=β+βαq1−q2−q3Γϑ+1τϑ+βαq12α−β+q1q3β−2α−2αq2+q2+q32Γ2ϑ+1τ2ϑ+⋯+−αk+1−q2βkΓkϑ+1τkϑ,ϕkτ=λ+q2β1Γϑ+1τϑ+q2β2Γ2ϑ+1τ2ϑ+⋯+q2βkΓkϑ+1τkϑ.Ask⟶∞, the kth truncated solutions converge to the exact solutions in the integer orders:(32)φkτ⟶φτ,ψkτ⟶ψτ,ϕkτ⟶ϕτ.Hence, we get the required solution.


## 4. Numerical Simulation

This section presents the solution of a numerical example of a SIR epidemic system using the LRPSM, and it introduces numerical simulations and figures to show the effectiveness of the suggested approach. The outcomes demonstrate the dependability and strength of LRPSM in solving such problems.We consider the epidemic SIR model as follows:(33)Dϑφτ=−0.001φτψτ+0.005ψτ,Dϑψτ=0.001φτψτ−0.005ψτ−0.072ψτ,Dϑϕτ=0.072ψτ,0<ϑ≤1,subject to the initial conditions:(34)φ0=620,ψ0=10,ϕ0=70.The total number of populations isN=700.To solve system (33) by the proposed method, we apply the Laplace transform on each equation in the system, and using the initial conditions (36), we get:(35)Φs=620s−0.001sϑℒℒ−1Φsℒ−1Ψs+0.005sϑΨs,Ψs=10s+0.001sϑℒℒ−1Φsℒ−1Ψs−0.005sϑΨs−0.072sϑΨs,Gs=70s+0.072sϑΨs.Now, thekth truncated expansion of the solution of system (35) is as follows:(36)Φks=620s+∑n=1kαnsnϑ+1,Ψks=10s+∑n=1kβnsnϑ+1,Gks=70s+∑n=1kλnsnϑ+1.We define thekth Laplace residual functions as follows:(37)ℒReskΦs=Φks−620s+0.001sϑℒℒ−1Φsℒ−1Ψs−0.005sϑΨks,ℒReskΨs=Ψks−10s−0.001sϑℒℒ−1Φsℒ−1Ψs+0.005sϑΨks+0.072sϑΨks,ℒReskGs=Gks−70s−0.072sϑΨks.Multiplying each equation in (39) by skϑ+1,k=1,2,⋯ recursively, and taking the limit as s⟶∞, we get the coefficients of the series solutions as follows:(38)α1=−6.15,β1=5.43,λ1=0.72,α2=−3.27795,β2=2.88699,λ2=0.39096,α3=0.03339Γ2ϑ+1Γ2ϑ+1−1.74272,β3=1.53486−0.0334Γ2ϑ+1Γ2ϑ+1,λ3=0.20786,α4=−0.92651+0.02021Γ2ϑ+1Γ2ϑ+1+0.03555Γ3ϑ+1Γϑ+1Γ2ϑ+1,β4=0.816−0.0178Γ2ϑ+1Γ2ϑ+1−0.03556Γ3ϑ+1Γϑ+1Γ2ϑ+1,λ4=0.11051−0.00241Γ2ϑ+1Γ2ϑ+1,α5=−0.49257−0.00039Γ4ϑ+1Γ2ϑ+1Γ3ϑ+1Γ3ϑ+1+0.01074Γ2ϑ+1Γ2ϑ+1+0.02151Γ3ϑ+1Γ2ϑ+1+0.0189Γ4ϑ+1Γ3ϑ+1+0.00946Γ4ϑ+1Γ22ϑ+1,β5=0.43382+0.00039Γ4ϑ+1Γ2ϑ+1Γ3ϑ+1Γ3ϑ+1−0.00946Γ2ϑ+1Γ2ϑ+1−0.01895Γ3ϑ+1Γ2ϑ+1Γϑ+1−0.018902Γ4ϑ+1Γ3ϑ+1Γϑ+1−0.009463Γ4ϑ+1Γ22ϑ+1,λ5=0.058752−0.001282Γ2ϑ+1Γ2ϑ+1−0.00256Γ3ϑ+1Γϑ+1Γ2ϑ+1,(39)φ6τ=620−6.15τϑΓ1+ϑ−3.27795τ2ϑΓ1+2ϑ+0.03339Γ2ϑ+1Γ2ϑ+1−1.74272τ3ϑΓ1+3ϑ+−0.92651+0.02021Γ2ϑ+1Γ2ϑ+1+0.03555Γ3ϑ+1Γϑ+1Γ2ϑ+1τ4ϑΓ1+4ϑ+−0.49257−0.00039Γ4ϑ+1Γ2ϑ+1Γ3ϑ+1Γ3ϑ+1+0.01074Γ2ϑ+1Γ2ϑ+1+0.02151Γ3ϑ+1Γ2ϑ+1+0.0189Γ4ϑ+1Γ3ϑ+1+0.00946Γ4ϑ+1Γ22ϑ+1τ5ϑΓ1+5ϑ+−0.261876+0.0057256Γ4ϑ+1Γ22ϑ+1+0.010062Γ5ϑ+1Γ2ϑ+1Γ3ϑ+1+0.011439Γ3ϑ+1/Γ2ϑ+1+0.010049Γ5ϑ+1/Γ4ϑ+1+0.011436Γ4ϑ+1/Γ3ϑ+1Γα+1+0.005712Γ2ϑ+1−0.000209Γ5ϑ+1/Γ3ϑ+1−0.000412Γ3ϑ+1Γ5ϑ+1/Γ4ϑ+1Γ2ϑ+1Γ2ϑ+1+Γ2ϑ+1−0.000234Γ24ϑ+1/Γ3ϑ+1−0.000219Γ5ϑ+1Γ3ϑ+1Γ4ϑ+1τ6ϑΓ1+6ϑ+…,(40)ψ6τ=10+5.43τϑΓ1+ϑ+2.88699τ2ϑΓ1+2ϑ+1.53486−0.0334Γ2ϑ+1Γ2ϑ+1τ3ϑΓ1+3ϑ+0.816−0.0178Γ2ϑ+1Γ2ϑ+1−0.03556Γ3ϑ+1Γϑ+1Γ2ϑ+1τ4ϑΓ1+4ϑ+0.43382+0.00039Γ4ϑ+1Γ2ϑ+1Γ3ϑ+1Γ3ϑ+1−0.00946Γ2ϑ+1Γ2ϑ+1−0.01895Γ3ϑ+1Γ2ϑ+1Γϑ+1−0.018902Γ4ϑ+1Γ3ϑ+1Γϑ+1−0.009463Γ4ϑ+1Γ22ϑ+1τ5ϑΓ1+5ϑ+0.23064−0.005044Γ4ϑ+1Γ22ϑ+1−0.010062Γ5ϑ+1Γ2ϑ+1Γ3ϑ+1+−0.010075Γ3ϑ+1/Γ2ϑ+1−0.010049Γ5ϑ+1/Γ4ϑ+1−0.010075Γ4ϑ+1/Γ3ϑ+1Γα+1+−0.005031Γ2ϑ+1+0.000206Γ5ϑ+1/Γ3ϑ+1+0.000412Γ3ϑ+1Γ5ϑ+1/Γ4ϑ+1Γ2ϑ+1Γ2ϑ+1+Γ2ϑ+10.000206Γ24ϑ+1/Γ3ϑ+1+0.00022Γ5ϑ+1Γ3ϑ+1Γ4ϑ+1τ6ϑΓ1+6ϑ+…,ϕ6τ=70+0.72τϑΓ1+ϑ+0.39096τ2ϑΓ1+2ϑ+0.20786τ3ϑΓ1+3ϑ+0.11051−0.00241Γ2ϑ+1Γ2ϑ+1τ4ϑΓ1+4ϑ+0.058752−0.001282Γ2ϑ+1Γ2ϑ+1−0.00256Γ3ϑ+1Γϑ+1Γ2ϑ+1τ5ϑΓ1+5ϑ+0.031235+0.000028Γ4ϑ+1Γ2ϑ+1/Γ3ϑ+1Γ3ϑ+1−0.000681Γ2ϑ+1/Γ2ϑ+1+−0.001364Γ3ϑ+1/Γ2ϑ+1−0.001361Γ4ϑ+1/Γ3ϑ+1Γϑ+1−0.000681Γ4ϑ+1Γ22ϑ+1τ6ϑΓ1+6ϑ+⋯.The 20th terms solution of systems (33) and (34) at ϑ=1 is given by the following equation:(41)φ20τ=620−6.15τ−1.63898τ2−0.27932τ3−0.03248τ4−0.00231τ5+6.631×10−7τ6+2.85675×10−5τ7+4.90414×10−6τ8+4.94538×10−7τ9+2.5347×10−8τ10−1.6516×10−9τ11−5.91732×10−10τ12−8.16453×10−11τ13−6.84287×10−12τ14−1.83575×10−13τ15+5.1703×10−14τ16+1.11303×10−14τ17+1.28567×10−15τ18+8.57039×10−17τ19−9.08862×10−19τ20,ψ20τ=10+5.43τ+1.44349τ2+0.24468τ3+0.028078τ4+0.00191τ5−2.3538×10−5τ6−2.83254×10−5τ7−4.64921×10−6τ8−4.57345×10−7τ9−2.20542×10−8τ10+1.79596×10−9τ11+5.80956×10−10τ12+7.84277×10−11τ13+6.43953×10−12τ14+1.52665×10−13τ15−5.239×10−14τ16−1.090842×10−14τ17−1.24203×10−15τ18−8.09972×10−17τ19+1.20045×10−18τ20,ϕ20τ=70+0.72τ+0.19548τ2+0.03464τ3+0.0044τ4+0.00041τ5+2.28749×10−5τ6−2.42105×10−7τ7−2.549283×10−7τ8−3.719366×10−8τ9−3.29288×10−9τ10−1.443545×10−10τ11+1.07757×10−11τ12+3.2176×10−12τ13+4.0334×10−13τ14+3.09097×10−14τ15+6.86992×10−16τ16−2.21887×10−16τ17−4.36337×10−17τ18−4.70665×10−18τ19−2.9159×10−19τ20.The following figures illustrate the 20th solution of systems (33) and (34) with various values of ϑ=1,0.9,0.8,0.7,0.6,0.5,0.4 and 0.3.Figure1 of φ20τ,ψ20τ, and ϕ20τ shows the strength of LRPSM in solving the proposed model, which is obvious, with ϑ increasing from 0 to one, the solution is stable and coincides with the exact solution obtained in the case ϑ=1.Figure 1
The values of (a) the solution ofφτ, (b) the solution of ψτ, and (c) the solution of ϕτ, for distinct values of ϑϑ=1,0.9,0.8,0.7,0.6,0.5,0.4,0.3 using the LRPSM.
(a)(b)(c)In Figure2 below, we sketch the graph of φ20τ,ψ20τ, and ϕ20τ.Figure 2
The values ofφ20τ, ψ20τ, and ϕ20τ using LRPSM for ϑ=1.The following tables, Tables1–3 present the LRPSM solution of systems (33) and (34) at various values of ϑϑ=1,0.9,0.8,0.7,0.6,0.5,0.4,0.3. In Table 1, we propose different values of φ20τ with different values of ϑ, in Table 2, we propose different values of ψ20τ with different values of ϑ, and in Table 3, we propose different values of ϕ20τ with different values of ϑ. We notice the efficiency and strength of the method from the agreement of the values, and they all coincide and converge to the exact solution in the integer order when ϑ=1.Table 1
The value ofφ20τ for distinct values of ϑϑ=1,0.9,0.8,0.7,0.6,0.5,0.4,0.3 using the LRPSM.
τϑ = 1ϑ = 0.9ϑ = 0.8ϑ = 0.7ϑ = 0.6ϑ = 0.5ϑ = 0.4ϑ = 0.306206206206206206206206200.1619.368619.163618.894618.538618.066617.433616.568615.360.2618.702618.384617.991617.501616.886616.107615.109613.8110.3618617.596617.112616.529615.823614.961613.901612.5940.4617.259616.785616.23615.577614.807613.895612.812611.5380.5616.478615.945615.332614.626613.811612.871611.793610.5790.6615.655615.071614.411613.664612.821611.872610.818609.6860.7614.788614.161613.463612.687611.827610.884609.873608.840.8613.874613.211612.483611.688610.824609.902608.949608.0320.9612.911612.218611.469610.663609.808608.919608.04607.2511611.897611.181610.418609.612608.774607.933607.14606.494Table 2
The value ofψ20τ for distinct values of ϑϑ=1,0.9,0.8,0.7,0.6,0.5,0.4,0.3 using the LRPSM.
τϑ = 1ϑ = 0.9ϑ = 0.8ϑ = 0.7ϑ = 0.6ϑ = 0.5ϑ = 0.4ϑ = 0.3010101010101010100.110.557710.738810.976811.290511.706912.265813.028414.09340.211.145711.426111.773512.20612.748213.434814.314915.45840.311.765812.12212.54913.063113.685814.445515.37916.52970.412.419412.837813.327413.902614.581315.385116.338117.45950.513.108313.57914.119414.741515.45916.286617.235818.30360.613.834314.349414.931215.588816.331717.16718.09419.08940.714.599315.152115.767116.450517.20718.036318.925619.83290.815.405215.989616.630417.330718.090218.90119.738520.54410.916.254116.864217.523818.232618.985119.765620.538321.2299117.148117.778218.449819.158819.894520.633421.32921.8954Table 3
The value ofϕ20τ for distinct values of ϑϑ=1,0.9,0.8,0.7,0.6,0.5,0.4,0.3 using the LRPSM.
τϑ = 1ϑ = 0.9ϑ = 0.8ϑ = 0.7ϑ = 0.6ϑ = 0.5ϑ = 0.4ϑ = 0.3070707070707070700.170.07470.09870.129770.171470.226970.301570.403770.54690.270.152170.189470.235770.293470.365970.45870.576470.73110.370.234670.282170.339170.407970.491470.593770.719970.87620.470.321670.377570.443170.520370.611670.720370.849671.00260.570.413570.476570.549170.632870.729770.84270.971471.11770.670.510570.579670.657970.746770.847570.961371.088171.2250.770.612870.687270.770370.862970.965971.079371.201571.32680.870.720870.799770.886570.981871.085671.19771.312671.42440.970.834770.917471.007171.103971.207271.31571.422271.5187170.95571.040671.132371.229671.331171.433671.530771.6103In the following, we introduce Figure3, which illustrates the absolute error between the Runge–Kutta method solution and the LRPSM solution at ϑ=1.Figure 3
Absolute error between the Runge–Kutta method solution and the LRPSM solutions forϑ=1.
(a)(b)(c)The following tables (Tables4–6), present comparisons between the obtained results φ20τ,ψ20τ, and ϕ20τ from the LRPSM and the fourth-order Runge–Kutta method (RK4). These comparisons prove the strength and convergence of the presented method. We notice from Tables 4–6 the efficiency of the proposed method, since the difference between the methods is too small.Table 4
The value ofφ20τ using RK4 and LRPSM for ϑ=1.
τiRK4LRPSMRK4φτ−LRPSMφτ0620.000000000000000620.00000000000000000.1619.368327701129830619.3683276575103504.36195×10−80.2618.702153816378770618.7021537246529309.17258×10−80.3617.999682039910570617.9996818952802201.4463×10−70.4617.259032603941480617.2590324012826402.02659×10−70.5616.478239561945540616.4782392957946502.66151×10−70.6615.655248125904110615.6552477904431303.35461×10−70.7614.787912080138200614.7879116691812004.10957×10−70.8613.873991297021010613.8739908039999604.93021×10−70.9612.911149382834760612.9111488007873705.82047×10−71611.896951485213090611.8969508067699506.78443×10−7Table 5
The value ofψ20τ using RK4 and LRPSM for ϑ=1.
τiRK4LRPSMRK4ψτ−LRPSMψτ010.00000000000000010.00000000000000000.110.55768241537714010.5576824541241353.8747×10−80.211.14574266529731811.1457427467669538.14696×10−80.311.76575271857131211.7657528470139211.28443×10−70.412.41935643032271512.4193566102758181.79953×10−70.513.10827173327717113.1082719695779062.36301×10−70.613.83429275946765513.8342930572650342.97797×10−70.714.59929187039647414.5992922351633723.64767×10−70.814.59929187039647415.4052220086472544.37545×10−70.916.25411628080169316.2541167972790115.16477×10−7117.14809392979923817.1480945317206526.01921×10−7Table 6
The value ofϕ20τ using RK4 and LRPSM for ϑ=1.
τiRK4LRPSMRK4ϕτ−LRPSMϕτ070.00000000000000070.00000000000000000.170.07398988349301370.0739898883654044.87239×10−90.270.15210351832384170.1521035285799291.02561×10−80.370.23456524151806970.2345652577056971.61876×10−80.470.32161096573575870.3216109884413072.27055×10−80.570.41348870477727970.4134887346273792.98501×10−80.670.51045911462827670.5104591522918873.76636×10−80.770.61279604946533170.6127960956556394.61903×10−80.870.72078713187640670.7207871873528265.54764×10−80.970.83473433636351070.8347344019336456.55701×10−8170.95495458498766170.9549546615092207.65216×10−8


## 5. Conclusions

In this study, we introduce the fractional SIR epidemic model in the sense of Caputo’s fractional derivative. The LRPSM is used to solve the proposed system. We present a series of solutions of the model and compare our results with those obtained by the fourth-order Runge–Kutta method. In addition, we sketch the graphs of the solutions with different values ofϑ and analyze the results. Finally, we conclude that the LRPSM is efficient in solving the SIR epidemic system and similar models. Moreover, as the method is applicable and easy in treating similar problems, it could provide many terms of the series solution with less calculations and effort compared to other numerical methods. In the future, we intend to solve new models by LRPSM and make comparisons to other analytical methods. We conclude the following points from our study:(i)
LRPSM is a powerful technique for solving fractional models(ii)
LRPSM is simple and could provide many terms of the series solution without requiring discretization, linearization, or special assumptions on the conditions(iii)
The method could give exact solutions when the exact one is a polynomial



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