Zika Virus Model with the Caputo–Fabrizio Fractional Derivative

Abstract

In this article, we examine a deterministic Zika virus model that takes into account the vector and sexual transmission route, in the absence of disease-induced deaths, symmetrically observing the impact of human knowledge and vector control. In order to construct the model, we suppose that the Zika virus is first spread to humans through mosquito bites, and then to their sexual partner. In this article, we conduct analytical studies which often begin by proving the existence and uniqueness of solutions for the Zika virus model using the fractional derivative from the Caputo–Fabrizio derivative. Then, the uniqueness of the solution is investigated. After that, we also identify under which circumstances and symmetry the model provides a unique solution.

1. Introduction

Fractional calculus, which generalizes the concept of integer-order differentiation and symmetrical integration to non-integer (fractional) orders, has gained considerable attention in recent years. This is due to its ability to more accurately model real-world phenomena exhibiting memory and hereditary properties. Fractional-order models are better at symmetrically representing complicated dynamics than traditional integer-order models, particularly the dynamics present in biological and epidemiological systems, where previous states have an impact on future dynamics. In the context of infectious disease modeling, fractional calculus allows us to incorporate memory effects that reflect prolonged incubation, immunity, or latency periods, which are often oversimplified in traditional models. By using fractional-order calculus to create a deterministic model of Zika virus transmission, this work improves the model’s ability to depict the virus’s spread via sexual and vector-borne routes. This study offers a more thorough knowledge of the transmission dynamics by employing a fractional-order method, which could result in more successful prediction and intervention techniques. Adding fractional calculus also presents special mathematical problems, which make sure that solutions exist, are positive, and are unique [1,2]. By thoroughly proving the existence and uniqueness of positive solutions inside the fractional-order framework, this article tackles these problems and advances the theoretical underpinnings of fractional epidemiological models, as discussed in [3,4]. The Zika virus, a mosquito-borne flavivirus, gained global attention due to its rapid spread and potentially severe health consequences, especially during outbreaks in various regions. Zika transmission primarily occurs through the bite of infected Aedes mosquitoes (Aedes aegypti and Aedes albopictus), which act as vectors for the virus. However, another significant mode of transmission has emerged—sexual transmission, adding complexity to the epidemiology of the disease [5,6]. The Zika virus garnered international concern during the 2015–2016 outbreak, particularly due to its association with congenital anomalies such as microcephaly in newborns. Understanding and modeling the dynamics of Zika transmission is crucial for implementing effective control strategies. This study focuses on symmetrically integrating both vector and sexual transmission routes into a comprehensive model, considering the influence of human awareness and vector control measures [7,8]. To study the Zika virus transmission dynamics, the investigators have incorporated the following factors which can be easily seen in many sources which are mentioned in the references.

1.1. Vector Transmission

• Aedes mosquitoes are the primary vectors responsible for Zika transmission. Understanding their population dynamics and behavior is essential for predicting and controlling the spread of the virus.
• Environmental factors, such as temperature and rainfall, impact mosquito breeding and survival, influencing the intensity of Zika transmission.

1.2. Sexual Transmission

• Emerging evidence suggests that the Zika virus can be sexually transmitted between humans. The duration of viral persistence in bodily fluids and the prevalence of sexual transmission contribute to the overall risk of infection.

1.3. Human Awareness

• Public awareness plays a pivotal role in controlling Zika transmission. Educating communities about the risks, preventive measures, and early detection is vital for minimizing the impact of the disease.
• The level of awareness affects individual behaviors, including the use of protective measures and seeking medical advice, thus influencing the overall transmission dynamics.

1.4. Vector Control

• Implementing vector control measures, such as insecticide spraying, larval source reduction, and the use of bed nets can significantly reduce mosquito populations and interrupt the transmission cycle.
• The effectiveness of vector control strategies is influenced by community engagement, resource allocation, and the adaptation of interventions to local contexts.

In the absence of disease-induced mortality, this model aims to explore the interplay between human awareness, vector control measures, and the dual transmission routes of Zika. By integrating these factors into a comprehensive model, we seek to provide insights that can guide public health policies and interventions for mitigating the impact of Zika outbreaks [9]. This research contributes to the ongoing efforts to understand and manage the complex dynamics of emerging vector-borne diseases in a changing global landscape [10,11,12]. In this work, we have proven the existence, uniqueness, positivity, boundedness, and well-posedness of the solutions. Lastly, by contrasting the model predictions with the reported Zika-infected data of a specific region, we have estimated and validated the parameters of the model under consideration. Using the normalized forward sensitivity index, we examined the sensitivity analysis to determine the model’s robustness to the parameter values employed. Additionally, we conducted a numerical analysis of the influence of sexual transmission on both the basic reproduction number and the epidemic growth rate. To the best of our knowledge, no Zika model has yet taken into account the effects of vector transmission, sexual transmission, vector control, and human awareness. Therefore, in order to present a Zika model and plan to investigate its dynamic behavior, we are taking into account these few crucial factors in our research [13,14,15].

This paper is organized as follows: Section 1 is introductory. In Section 2, mathematical modeling is discussed. In Section 3, we briefly give the basic definitions of fractional derivative. In Section 4, we give a general description of the homotopy perturbation transform method. Section 5 involves an analysis of the existence and uniqueness of solutions via the Caputo–Fabrizio derivative, and lastly, a conclusion is drawn.

2. Mathematical Modeling

This paper focuses on the concept of generalization of fractional calculus. Although this idea was developed shortly after the classical one, it has recently captured the interest of many researchers. The notions of fractional calculus are especially effective and more useful when analyzing and capturing the behavior of phenomena related to complexity, memory repercussions, hereditary possessions, and other crucial and exciting features. With the help of the Mittag-Leffler function, which is regarded as the queen of fractional calculus, we examine the recently defined fractional operator known as the Caputo–Fabrizio derivative [1,2,3]. Many scholars have been using this derivative extensively in recent years to study a wide range of real-world models. In this current study, four compartments, described as susceptible human $S_h\left(t\right)$, exposed human $E_h\left(t\right)$, infected human $I_h\left(t\right)$, and recovered human $R_h\left(t\right)$ should be created from the entire human population $N_h\left(t\right)$. Here, we have taken into consideration that a person who recovers from a Zika virus infection develops permanent immunity to it. Since the Zika virus is exclusively propagated by female mosquitoes, the entire population of female mosquitoes $N_v\left(t\right)$ is separated into three groups: susceptible mosquitoes $S_v\left(t\right)$, exposed mosquitoes $E_v\left(t\right)$, and infected mosquitoes $I_v\left(t\right)$. Because of its brief lifespan, recovery from Zika infection is not taken into account.

The following elements are taken into account when creating the model:

(i)

Steady rate of recruitment for the vector population and vulnerable humans.

(ii)

Rate of natural death in human and vector populations.

(iii)

When infected vectors bite vulnerable humans, the infection is spread horizontally from the vector to the host.

(iv)

When a susceptible person engages in sexual contact with an infected person, the infection is sexually transmitted to the susceptible person.

(v)

When susceptible vectors bite an infected person, the infection is spread horizontally from the sick person to the vector.

(vi)

Human awareness has been divided into two stages. The following preventative measures can be taken by susceptible humans at a steady rate: (a) utilizing mosquito nets and repellents; (b) avoiding sexual contact or wearing a condom during the Zika outbreak.

(vii)

Vector control can be achieved by removing standing water and using pesticide spray.

Let $m$ be the population’s natural death rate and $\pi$ be the constant rate at which susceptible humans are recruited. Let us say that susceptible individuals become infected through effective contact with an infected vector at a rate of ${\lambda }_1={\displaystyle \frac{b_2{\alpha }_{1}I\_v }{N_h}},{\mathsf{\lambda }}_2={\displaystyle \frac{\mathrm{c}{\mathsf{\alpha }}_2{\mathrm{I}}_{\mathrm{h}}}{N_h}}.$ This means that the susceptible human becomes aware of the infection at a constant rate and enters the recovered class Rh. Therefore, ${\lambda }_h={\lambda }_1+{\lambda }_2$ is the total infection strength of humans. In this case, we suppose that the susceptible mosquitoes contract the virus from the infected person at a rate ${\lambda }_v{S}_v$ where ${\lambda }_v={\displaystyle \frac{b_2{\alpha }_3{I}_h}{N_h}}$. Under the above assumptions, both the human population and vector population of the proposed Zika virus can be represented by the following system [1], and the explanation of the parameters and values found in Equation (1) are shown in

Table 1. An explanation of the parameters and values found in Equation (1) [1]:.

Notation	Description of Parameters
$N_h$	The whole population of humans
$S_h$	Human population (susceptible)
$E_h$	Human population (expected)
$I_h$	Human population (infected)
$R_h$	Human population (recovered)
$N_{\nu }$	The entire population of vectors
$S_{\nu }$	Population of susceptible vectors
$E_{\nu }$	Population of exposed vectors
$I_{\nu }$	Population of infected vectors
π, π1	Human and mosquito recruitment rates, correspondingly
µ, µ1	Human and mosquito natural mortality rates, respectively
b2	Rate of mosquito bites
A1	Probability of transmission per bite of an infected mosquito on a vulnerable individual
α3	The likelihood of infection spreading when vulnerable people are bitten by infected people
c	Rate of sexual contact between an infected person and a vulnerable person
α	The likelihood of transmission between susceptible and diseased humans during sexual interaction
Σ	Rate of progression after being exposed to an infected person
Γ	The rate at which sick humans recover
A	The host population’s awareness rate
Σ1	Rate of progression following exposure to an infected mosquito
B	Steady pace of successful mosquito control

:

$$\left\{\begin{array}{c}{\displaystyle \frac{d{S}_h}{dt}}=\pi - \left({\lambda }_1+{\lambda }_2\right)S_h - \left(\mu +a\right)S_h \\ {\displaystyle \frac{d{E}_h}{dt}}= \left({\lambda }_1 + {\lambda }_2\right)S_h + \left(\sigma +\mu \right)E_h \\ {\displaystyle \frac{d{I}_h}{dt}}=\sigma E_h-\left(\gamma +\mu \right)I_h \\ {\displaystyle \frac{d{R}_h}{dt }}=\gamma I_h-\mu R_h+a{S}_h \\ {\displaystyle \frac{d{S}_v}{dt}}={\pi }_1-{\lambda }_v{S}_v-\left({\mu }_1+b \right)S_v \\ {\displaystyle \frac{d{\mathsf{{\rm E}}}_{\nu }}{dt}}= {\lambda }_{\nu }{S}_{\nu }-{(\sigma }_1+{\mu }_1+b)E_{\nu } \\ {\displaystyle \frac{d{I}_v}{dt}}={\sigma }_1{E}_v-\left({\mu }_1 + b\right)I_v \end{array}\right.$$

(1)

We examine Equation (1)’s fractional order using the Caputo–Fabrizio derivative, which looks like this:

$$\left\{\begin{array}{l}{}_0{}^{CF}{D}_t^{\alpha }{S}_h=\pi - \left({\lambda }_1+{\lambda }_2\right)S_h - \left(\mu +a\right)S_h \\ {}_0{}^{CF}{D}_t^{\alpha }{E}_h= \left({\lambda }_1 + {\lambda }_2\right)S_h + \left(\sigma +\mu \right)E_h \\ {}_0{}^{CF}{D}_t^{\alpha }{I}_h=\sigma E_h-\left(\gamma +\mu \right)I_h \\ {}_0{}^{CF}{D}_t^{\alpha }{R}_h=\gamma I_h-\mu R_h+a{S}_h \\ {}_0{}^{CF}{D}_t^{\alpha }{S}_v={\pi }_1-{\lambda }_v{S}_v-\left({\mu }_1+b \right)S_v \\ {}_0{}^{CF}{D}_t^{\alpha }{E}_v= {\lambda }_{\nu }{S}_{\nu }-{(\sigma }_1+{\mu }_1+b)E_{\nu } \\ {}_0{}^{CF}{D}_t^{\alpha }{I}_v={\sigma }_1{E}_v-\left({\mu }_1 + b\right)I_v\end{array}\right.$$

(2)

3. Preliminaries

In this section, we present some fundamental notions of fractional calculus and the Laplace transform, which are essential in the present framework.

Definition 1.

The Caputo fractional derivative operator of order $\alpha \ge 0$  and $n\in N\cup \left\{0\right\}$  is defined as [10]

$${}_0{}^C{D}_t^{\alpha }h\left(t\right)={\displaystyle \frac{1}{\Gamma (n-\alpha )}}{\int }_0^t{\left(t-\xi \right)}^{\left(n-\tau -1\right)}{\displaystyle \frac{d^n}{d{t}^n}}h\left(\xi \right)d\xi ,$$

(3)

 where  ${}_0{}^C{D}_t^{\alpha }$  is a Caputo–Fabrizio derivative with respect to t.

Definition 2

Let $h\in K^1\left(a,b\right), b>a, 0<h<1.$ Then, the time-fractional Caputo–Fabrizio fractional differential operator is defined by [10].

$${}_0{}^{CF}{D}_t^{\alpha }h\left(t\right)={\displaystyle \frac{M\left(\alpha \right)}{(1-\alpha )}}{\int }_a^t\exp \left[-{\displaystyle \frac{\alpha \left(1-\xi \right)}{1-\alpha }}\right]h^{\prime }\left(\xi \right)d\xi , t\ge 0,0<\alpha <1,$$

(4)

 where  $M\left(\alpha \right)$ is a normalisation function which depends on$\alpha$ and satisfies

$$M\left(0\right)=M\left(1\right)=1.$$

Definition 3.

The CF fractional integral operator of order $0<\gamma <1$ is given by [10]

$${}_0{}^{CF}{D}_t^{\alpha }h\left(t\right)={\displaystyle \frac{2\left(1-\alpha \right)}{\left(2-\alpha \right)M\left(\alpha \right)}}u\left(t\right)+{\displaystyle \frac{2\alpha }{\left(2-\alpha \right)M\left(\alpha \right)}}{\int }_0^{t}h\left(\xi \right)d\xi , t\ge 0,$$

(5)

 where  ${}_0{}^{CF}{D}_t^{\alpha }h\left(t\right)=0,$ if  $h$ is a constant function.

Definition 4.

The Laplace transform (LT) for the CF fractional operator of order $0<\tau \le 1$ for $m\in N$ is given as [10]

$$\begin{array}{c}L\left[{}_0{}^{CF}{D}_t^{\left(m+\alpha \right)}h\left(t\right)\right]\left(s\right)={\displaystyle \frac{1}{1-\alpha }}L\left[h^{\left(m+1\right)}\left(t\right)\right]L\left[\exp \left({\displaystyle \frac{-\alpha }{\left(1-\alpha \right)}}t\right)\right]\\ ={\displaystyle \frac{s^{(m+1)}L\left[h\left(t\right)\right]-s^{m}h\left(0\right)-s^{\left(m-1\right)}{h}^{\prime }\left(0\right)\dots \dots ..-h^{\left(m\right)}\left(0\right)}{s+\alpha (1-s)}}\end{array}$$

(6)

In particular, we have

$$L\left({}_0{}^{CF}{D}_t^{\gamma }h\left(t\right)\right)\left(s\right)={\displaystyle \frac{sL\left(h\left(t\right)\right)-h\left(0\right)}{s+\alpha (1-s)}}, m=0$$

$$L\left({}_0{}^{CF}{D}_t^{\alpha +1}h\left(t\right)\right)\left(s\right)={\displaystyle \frac{s^{2}h\left(f\left(t\right)\right)-sh\left(0\right)-h^{\prime }\left(0\right)}{s+\alpha (1-s)}}, m=1$$

4. General Description of FHPTM Using Caputo–Fabrizio Type Operator

Consider the following nonlinear partial differential equation in the sense of Caputo–Fabrizio [10,11]:

$${}_0{}^{CF}{D}_t^{\left(m+\alpha \right)}u\left(x,t\right)+\rho u\left(x,t\right)+\sigma u\left(x,t\right)=k\left(x,t\right), m-1<\alpha +n\le m,$$

(7)

along with the following initial conditions:

$${\displaystyle \frac{{\partial }^{j}h(x.0)}{\partial t^j}}=f_j\left(k\right), j=\mathrm{0,1},2\dots \dots .., m-1$$

(8)

on applying the Laplace transform on both Equations (7) and (8), we get:

$$L\left[u\left(x,t\right)\right]=\eta \left(x,s\right)-\left({\displaystyle \frac{s+\alpha \left(1-s\right)}{s^{\left(m+1\right)}}}\right)L\left[\rho u\left(x,t\right)+\sigma u\left(x,t\right)\right]$$

(9)

where

$$\eta \left(x,s\right)={\displaystyle \frac{1}{s^{\left(m+1\right)}}}\left[s^m{f}_0\left(x\right)+s^{m-1}{f}_1\left(x\right)+\dots \dots +f_m\left(x\right)\right]+{\displaystyle \frac{s+\alpha \left(1-s\right)}{s^{\left(m+1\right)}}}\tilde{t}u\left(x,s\right)$$

(10)

Taking the inverse Laplace transformation the Equation (9) yields:

$$u\left(x,t\right)=\eta \left(x,s\right)-L^{-1}\left[\left({\displaystyle \frac{s+\alpha \left(1-s\right)}{s^{\left(m+1\right)}}}\right)L\left[\rho u\left(x,t\right)+\sigma u\left(x,t\right)\right]\right]$$

(11)

where$\eta \left(x,s\right)$ arises from the source term. Now, we apply the FHPTM to obtain the solution of Equation (11), starting with the hypothesis that $u\left(x,t\right)$ expressed below is a solution of this equation.

$$u\left(x,t\right)={\sum }_{n=0}^{\infty }{z}^m{u}_m\left(x,t\right),$$

(12)

where $u_m\left(x,t\right)$ are known functions, the nonlinear term can be decomposed as

$$\sigma u\left(x,t\right)={\sum }_{n=0}^{\infty }{z}^m{H}_m\left(x,t\right),$$

(13)

The polynomials $H_m\left(x,t\right)$ are given by [13]

$$H_n\left(u_0, { u}_1, u_2 , \dots \dots , u_m\right)={\displaystyle \frac{1}{m! }}{\displaystyle \frac{{\partial }^m}{\partial p^m}}{\left[\sigma \left({\sum }_{i=0 }^{\infty }{z}^i{u}_i\right)\right]}_{z=0 }, m=\mathrm{0,1},2,\dots \dots .,$$

(14)

substituting Equations (12) and (13) into (11) we get

$${\sum }_{n=0}^{\infty }{u}_m\left(x,t\right)=\eta \left(x,s\right)-z{L}^{-1 }\left[\left({\displaystyle \frac{s+\alpha \left(1-s\right)}{s^{\left(m+1\right)}}}\right)L\left[\rho {\sum }_{m=0}^{\infty }{z}^m{u}_m\left(x,t\right),+\sigma {\sum }_{m=0}^{\infty }{z}^m{H}_m\right]\right],$$

(15)

Comparing the coefficients of z⁰, ${ z}^1$, ${ z}^2$ and $z^3$ we get:

$$z^0 :u_0\left(x,t\right)= \eta \left(x,s\right)$$

$$z^1 :u_1\left(x,t\right)=-L^{-1 }\left[\left({\displaystyle \frac{s+\alpha \left(1-s\right)}{s^{\left(m+1\right)}}}\right)L\left[\rho u_0(x,t)+H_0\left(u\right)\right]\right]$$

$$z^2 :u_2\left(x,t\right)=-L^{-1 }\left[\left({\displaystyle \frac{s+\alpha \left(1-s\right)}{s^{\left(m+1\right)}}}\right)L\left[\rho u_1(x,t)+H_1\left(u\right)\right]\right]$$

$$z^3 :u_3\left(x,t\right)=-L^{-1 }\left[\left({\displaystyle \frac{s+\alpha \left(1-s\right)}{s^{\left(m+1\right)}}}\right)L\left[\rho u_2(x,t)+H_2\left(u\right)\right]\right]$$

$$z^{m+1} :u_{m+1}\left(x,t\right)=-L^{-1 }\left[\left({\displaystyle \frac{s+\alpha \left(1-s\right)}{s^{\left(m+1\right)}}}\right)L\left[\rho u_{m+1 }\left(x,t\right)+H_{m+1}\left(u\right)\right]\right]$$

(16)

with the help of HPTM the series solutions are

$$u(x,t)={\sum }_{m=0}^{\mathrm{\infty }}{u}_m(x,t).$$

5. Analysis of the Existence of Solutions

In this section, we will provide the existence of the solution for the Zika virus model by [2]. Next, we shall also demonstrate the affirmative solution’s uniqueness. The following we get when we use the fractional integral in Equation (2).

$$\left\{\begin{array}{c}{S}_h\left(t\right)-S_0\left(t\right)={\displaystyle \frac{2\left(1-\nu \right)}{2M \left(\nu \right)-vM\left(\nu \right)}}\left[ \pi -\left({\lambda }_1-{\lambda }_2\right)s_h-\left(\mu +\alpha \right)s_h\right]\\ +{\displaystyle \frac{2v}{2M \left(\nu \right)-vM\left(\nu \right)}}{\int }_0^t\left[\pi -\left({\lambda }_1-{\lambda }_2\right)s_h-\left(\mu +\alpha \right)s_h\right]dy. \\ E_h\left(t\right)-E_0\left(t\right)=={\displaystyle \frac{2\left(1-\nu \right)}{2M \left(\nu \right)-vM\left(\nu \right)}}\left[\left({\lambda }_1+{\lambda }_2\right)s_h-\left(\sigma +\mu \right)E_h\right] \\ +{\displaystyle \frac{2v}{2M \left(\nu \right)-vM\left(\nu \right)}}{\int }_0^t\left[\left({\lambda }_1+{\lambda }_2\right)s_h-\left(\sigma +\mu \right)E_h\right]dy. \\ I_h\left(t\right)-I_0\left(t\right)={\displaystyle \frac{2\left(1-\nu \right)}{2M \left(\nu \right)-vM\left(\nu \right)}}\left[ \sigma E_h-\left(\gamma +\mu \right)I_h\right]\\ +{\displaystyle \frac{2v}{2M \left(\nu \right)-vM\left(\nu \right)}}{\int }_0^t\left[\sigma E_h-\left(\gamma +\mu \right)I_h\right] dy. \\ R_h\left(t\right)-R_o\left(t\right)= {\displaystyle \frac{2\left(1-\nu \right)}{2M \left(\nu \right)-vM\left(\nu \right)}}\left[\gamma I_{h }-\mu R_h+a{S}_h\right]\\ +{\displaystyle \frac{2v}{2M \left(\nu \right)-vM\left(\nu \right)}}{\int }_0^t\begin{array}{c}\left[\gamma I_h-\mu R_h+a{S}_h\right] dy.\\ \end{array}\\ S_v\left(t\right)-S_o\left(t\right)={\displaystyle \frac{2\left(1-\nu \right)}{2M \left(\nu \right)-vM\left(\nu \right)}}\left[{\pi }_1-{\lambda }_v{S}_v-\left({\mu }_1+b\right)S_v\right]\\ +{\displaystyle \frac{2v}{2M \left(\nu \right)-vM\left(\nu \right)}}{\int }_0^t\left[{\pi }_1-{\lambda }_v{S}_v-\left({\mu }_1+b\right)S_v\right] dy. \\ E_v\left(t\right)-E_o\left(t\right)={\displaystyle \frac{2\left(1-\nu \right)}{2M \left(\nu \right)-vM\left(\nu \right)}}\left[{\lambda }_v{S}_v-\left({\sigma }_1+{\mu }_1+b\right)E_v\right]\\ +{\displaystyle \frac{2v}{2M \left(\nu \right)-vM\left(\nu \right)}}{\int }_0^t\left[{\lambda }_v{S}_v-\left({\sigma }_1+{\mu }_1+b\right)E_v\right] dy. \\ I_v\left(t\right)-I_0\left(t\right)={\displaystyle \frac{2\left(1-\nu \right)}{2M \left(\nu \right)-vM\left(\nu \right)}}\left[ {\sigma }_1{E}_v-\left({\mu }_1+b\right)I_v\right] \\ +{\displaystyle \frac{2v}{2M \left(\nu \right)-vM\left(\nu \right)}}{\int }_0^t\left[{\sigma }_1{E}_v-\left({\mu }_1+b\right)I_v\right] dy. \end{array}\right.$$

(17)

for simplicity, we choose our kernels

$$k\left(t,S_h\left(t\right)\right), k\left(t,E_h\left(t\right)\right), k\left(t,I_h\left(t\right)\right), k\left(t,R_h\left(t\right)\right), k\left(t,S_v\left(t\right)\right),k\left(t,E_v\left(t\right)\right), k\left(t,I_v\left(t\right)\right)$$

as follows:

$$\left\{\begin{array}{c}k\left(t,S_h\left(t\right)\right)= \pi - \left({\lambda }_1+{\lambda }_2\right)S_h - \left(\mu +a\right)S_h \\ k\left(t,E_h\left(t\right)\right)= \left({\lambda }_1\mathrm{ }+\mathrm{ }{\lambda }_2\right)S_h + \left(\sigma +\mu \right)E_h \\ k\left(t,I_h\left(t\right)\right)=\sigma E_h-\left(\gamma +\mu \right)I_h \\ k(t,R_h\left(t\right))=\gamma I_h-\mu R_h+a{S}_h \\ k\left(t,S_v\left(t\right)\right)={\pi }_1-{\lambda }_v{S}_v-\left({\mu }_1+b\mathrm{ }\right)S_v \\ k\left(t,E_v\left(t\right)\right)= {\lambda }_{\nu }{S}_{\nu }-{(\sigma }_1+{\mu }_1+b)E_{\nu } \\ k\left(t,I_v\left(t\right)\right)={\sigma }_1{E}_v-\left({\mu }_1\mathrm{ }+\mathrm{ }b\right)I_v \end{array}\right.$$

(18)

We must first be able to recognize an operator. We shall then demonstrate that this operator is compact. So that $T:H\to H$ is the operator. Next, we obtained

$$\left\{\begin{array}{c}T\left(S_h\left(t\right)\right)={\displaystyle \frac{2\left(1-\nu \right)}{2M\left(\nu \right)-vM\left(\nu \right)}}k\left(t,S_h\left(t\right)\right)+{\displaystyle \frac{2v}{2M\left(\nu \right)-vM\left(\nu \right)}} {\int }_0^{t}k\left(y,S_h\left(y\right)\right)dy \\ T\left(E_h\left(t\right)\right)= {\displaystyle \frac{2\left(1-\nu \right)}{2M\left(\nu \right)-vM\left(\nu \right)}}k\left(t,E_h\left(t\right)\right)+{\displaystyle \frac{2v}{2M\left(\nu \right)-vM\left(\nu \right)}} {\int }_0^{t}k\left(y,E_h\left(y\right)\right)dy\\ T\left(I_h\left(t\right)\right)={\displaystyle \frac{2\left(1-\nu \right)}{2M\left(\nu \right)-vM\left(\nu \right)}}k\left(t,I_h\left(t\right)\right)+{\displaystyle \frac{2v}{2M\left(\nu \right)-vM\left(\nu \right)}} {\int }_0^{t}k\left(y,I_h\left(y\right)\right)dy\\ T\left(R_h\left(t\right)\right)={\displaystyle \frac{2\left(1-\nu \right)}{2M\left(\nu \right)-vM\left(\nu \right)}}k\left(t,R_h\left(t\right)\right)+{\displaystyle \frac{2v}{2M\left(\nu \right)-vM\left(\nu \right)}} {\int }_0^{t}k\left(y,R_h\left(y\right)\right)dy \\ T\left(S_v\left(t\right)\right)= {\displaystyle \frac{2\left(1-\nu \right)}{2M\left(\nu \right)-vM\left(\nu \right)}}k\left(t,S_v\left(t\right)\right)+{\displaystyle \frac{2v}{2M\left(\nu \right)-vM\left(\nu \right)}} {\int }_0^{t}k\left(y,S_v\left(y\right)\right)dy\\ T\left(E_v\left(t\right)\right)={\displaystyle \frac{2\left(1-\nu \right)}{2M\left(\nu \right)-vM\left(\nu \right)}}k\left(t,S_v\left(t\right)\right)+{\displaystyle \frac{2v}{2M\left(\nu \right)-vM\left(\nu \right)}} {\int }_0^{t}k\left(y,S_v\left(y\right)\right)dy \\ T\left(I_v\left(t\right)\right)={\displaystyle \frac{2\left(1-\nu \right)}{2M\left(\nu \right)-vM\left(\nu \right)}}k\left(t,I_v\left(t\right)\right)+{\displaystyle \frac{2v}{2M\left(\nu \right)-vM\left(\nu \right)}} {\int }_0^{t}k\left(y,I_v\left(y\right)\right)dy\end{array}\right.$$

(19)

Lemma 1.

$T:H\to H$ is a fully continuous mapping.

Proof: 

Let $M\subset H$ be bounded. There exist constants $l,m$ > 0, such that $‖S_h‖,<l, ‖ E_h‖<l,$

$$‖I_h‖<l, ‖R_h‖<l, ‖S_v‖<l, ‖E_v‖<l, \mathrm{a}\mathrm{n}\mathrm{d} ‖I_v‖<l.$$

$$\left\{\begin{array}{c} {\mathrm{L}}_1=\underset{0\le t\le 1}{\max }k(t,S_h\left(t\right))\\ {\mathrm{L}}_2=\underset{0\le t\le 1}{\max }k\left(t,E_h\left(t\right)\right)\\ {\mathrm{L}}_3=\underset{0\le t\le 1}{\max }k\left(t,I_h\left(t\right)\right)\\ {\mathrm{L}}_4=\underset{0\le t\le 1}{\max }k\left(t,R_h\left(t\right)\right)\\ {\mathrm{L}}_5=\underset{0\le t\le 1}{\max }k\left(t,S_v\left(t\right)\right)\\ {\mathrm{L}}_6=\underset{0\le t\le 1}{\max }k\left(t,E_v\left(t\right)\right)\\ {\mathrm{L}}_7=\\ \underset{0\le t\le 1}{\max }k\left(t,I_v\left(t\right)\right)\end{array}\right. \forall S_h, E_h, I_h, R_h, S_v, E_v and I_v\in M$$

we have

$$\begin{array}{c}\left|T\left(S_h\right(t)\right|=\left|{\displaystyle \frac{2\left(1-\nu \right)}{2M\left(\nu \right)-vM\left(\nu \right)}}k\left(t,S_h\left(t\right)\right)+{\displaystyle \frac{2v}{2M\left(\nu \right)-vM\left(\nu \right)}}\underset{0}{\overset{t}{\int }}k\left(y,S_h\left(y\right)\right)dy\right|\\ \le \left|{\displaystyle \frac{2\left(1-\nu \right)}{2M\left(\nu \right)-vM\left(\nu \right)}}\right|k\left(t,S_h\left(t\right)\right)+\left|{\displaystyle \frac{2v}{2M\left(\nu \right)-vM\left(\nu \right)}}\right|\left.\underset{0}{\overset{t}{\int }}k\left(y,S_h\left(y\right)\right)dy\right|,\\ \le \left[{\displaystyle \frac{2\left(1-\nu \right)}{2M\left(\nu \right)-vM\left(\nu \right)}}+{\displaystyle \frac{2v}{2M\left(\nu \right)-vM\left(\nu \right)}}{c}_1\right]\left|k\left(t,S_h\left(t\right)\right)\right|,\\ \le \left[{\displaystyle \frac{2\left(1-\nu \right)}{2M\left(\nu \right)-vM\left(\nu \right)}}+{\displaystyle \frac{2v}{2M\left(\nu \right)-vM\left(\nu \right)}}{c}_1\right]\left|L_1,\right.\end{array}$$

$$‖T\left(S_{h }\right)‖\le {\displaystyle \frac{2L_1}{2M \left(\nu \right)-vM\left(\nu \right)}}\left[1-\nu +\nu c_1 \right].$$

(20)

Similarly,

$$\begin{array}{c}\left|T\left(E_h\right(t)\right|=\left|{\displaystyle \frac{2\left(1-\nu \right)}{2M\left(\nu \right)-vM\left(\nu \right)}}k\left(t,E_h\left(t\right)\right)+{\displaystyle \frac{2v}{2M\left(\nu \right)-vM\left(\nu \right)}}\underset{0}{\overset{t}{\int }}k\left(y,E_h\left(y\right)\right)dy\right|,\\ \le \left|{\displaystyle \frac{2\left(1-\nu \right)}{2M\left(\nu \right)-vM\left(\nu \right)}}\right|k\left(t,E_h\left(t\right)\right)+\left|{\displaystyle \frac{2v}{2M\left(\nu \right)-vM\left(\nu \right)}}\right|\left.\underset{0}{\overset{t}{\int }}k\left(y,E_h\left(y\right)\right)dy\right|,\\ \le \left[{\displaystyle \frac{2\left(1-\nu \right)}{2M\left(\nu \right)-vM\left(\nu \right)}}+{\displaystyle \frac{2v}{2M\left(\nu \right)-vM\left(\nu \right)}}{c}_2\right]\left|k\left(t,E_h\left(t\right)\right)\right|,\le \left[{\displaystyle \frac{2\left(1-\nu \right)}{2M\left(\nu \right)-vM\left(\nu \right)}}+{\displaystyle \frac{2v}{2M\left(\nu \right)-vM\left(\nu \right)}}{c}_2\right]\left|L_2,\right.\end{array}$$

$$‖T\left(E_h\right)‖\le {\displaystyle \frac{2L_2}{2M \left(\nu \right)-vM\left(\nu \right)}}\left[1-\nu +\nu c_2 \right].$$

(21)

$$\begin{array}{c}\left|T\left(I_h\right(t)\right|=\left|{\displaystyle \frac{2\left(1-\nu \right)}{2M\left(\nu \right)-vM\left(\nu \right)}}k\left(t,I_h\left(t\right)\right)+{\displaystyle \frac{2v}{2M\left(\nu \right)-vM\left(\nu \right)}}\underset{0}{\overset{t}{\int }}k\left(y,I_h\left(y\right)\right)dy\right|,\\ \le \left|{\displaystyle \frac{2\left(1-\nu \right)}{2M\left(\nu \right)-vM\left(\nu \right)}}\right|k\left(t,I_h\left(t\right)\right)+\left|{\displaystyle \frac{2v}{2M\left(\nu \right)-vM\left(\nu \right)}}\right|\left.\underset{0}{\overset{t}{\int }}k\left(y,I_h\left(y\right)\right)dy\right|,\\ \le \left[{\displaystyle \frac{2\left(1-\nu \right)}{2M\left(\nu \right)-vM\left(\nu \right)}}+{\displaystyle \frac{2v}{2M\left(\nu \right)-vM\left(\nu \right)}}{c}_3\right]\left|k\left(t,I_h\left(t\right)\right)\right|,\\ \le \left[{\displaystyle \frac{2\left(1-\nu \right)}{2M\left(\nu \right)-vM\left(\nu \right)}}+{\displaystyle \frac{2v}{2M\left(\nu \right)-vM\left(\nu \right)}}{c}_3\right]\left|L_3,\right.\end{array}$$

$$‖T\left(I_{h }\right)‖\le {\displaystyle \frac{2L_3}{2M \left(\nu \right)-vM\left(\nu \right)}}\left[1-\nu +\nu c_3 \right].$$

(22)

$$\begin{array}{c}\left|T\left(R_h\right(t)\right|=\left|{\displaystyle \frac{2\left(1-\nu \right)}{2M\left(\nu \right)-vM\left(\nu \right)}}k\left(t,R_h\left(t\right)\right)+{\displaystyle \frac{2v}{2M\left(\nu \right)-vM\left(\nu \right)}}\underset{0}{\overset{t}{\int }}k\left(y,R_h\left(y\right)\right)dy\right|,\\ \le \left|{\displaystyle \frac{2\left(1-\nu \right)}{2M\left(\nu \right)-vM\left(\nu \right)}}\right|k\left(t,R_h\left(t\right)\right)+\left|{\displaystyle \frac{2v}{2M\left(\nu \right)-vM\left(\nu \right)}}\right|\left.\underset{0}{\overset{t}{\int }}k\left(y,R_h\left(y\right)\right)dy\right|,\\ \le \left[{\displaystyle \frac{2\left(1-\nu \right)}{2M\left(\nu \right)-vM\left(\nu \right)}}+{\displaystyle \frac{2v}{2M\left(\nu \right)-vM\left(\nu \right)}}{c}_4\right]\left|k\left(t,R_h\left(t\right)\right)\right|,\\ \le \left[{\displaystyle \frac{2\left(1-\nu \right)}{2M\left(\nu \right)-vM\left(\nu \right)}}+{\displaystyle \frac{2v}{2M\left(\nu \right)-vM\left(\nu \right)}}{c}_4\right]\left|L_4,\right.\end{array}$$

$$‖T\left(R_{h }\right)‖\le {\displaystyle \frac{2L_4}{2M \left(\nu \right)-vM\left(\nu \right)}}\left[1-\nu +\nu c_4 \right].$$

(23)

$$\begin{array}{c}\left|T\left(S_{\nu }\right(t)\right|=\left|{\displaystyle \frac{2\left(1-\nu \right)}{2M\left(\nu \right)-vM\left(\nu \right)}}k\left(t,S_{\nu }\left(t\right)\right)+{\displaystyle \frac{2v}{2M\left(\nu \right)-vM\left(\nu \right)}}\underset{0}{\overset{t}{\int }}k\left(y,S_{\nu }\left(y\right)\right)dy\right|,\\ \le \left|{\displaystyle \frac{2\left(1-\nu \right)}{2M\left(\nu \right)-vM\left(\nu \right)}}\right|k\left(t,S_{\nu }\left(t\right)\right)+\left|{\displaystyle \frac{2v}{2M\left(\nu \right)-vM\left(\nu \right)}}\right|\left.\underset{0}{\overset{t}{\int }}k\left(y,S_{\nu }\left(y\right)\right)dy\right|,\\ \le \left[{\displaystyle \frac{2\left(1-\nu \right)}{2M\left(\nu \right)-vM\left(\nu \right)}}+{\displaystyle \frac{2v}{2M\left(\nu \right)-vM\left(\nu \right)}}{c}_5\right]\left|k\left(t,S_{\nu }\left(t\right)\right)\right|,\\ \le \left[{\displaystyle \frac{2\left(1-\nu \right)}{2M\left(\nu \right)-vM\left(\nu \right)}}+{\displaystyle \frac{2v}{2M\left(\nu \right)-vM\left(\nu \right)}}{c}_5\right]\left|L_5,\right.\end{array}$$

$$‖T\left(S_{\nu }\right)‖\le {\displaystyle \frac{2L_5}{2M \left(\nu \right)-vM\left(\nu \right)}}\left[1-\nu +\nu c_5 \right].$$

(24)

$$\begin{array}{c}\left|T\left(E_{\nu }\right(t)\right|=\left|{\displaystyle \frac{2\left(1-\nu \right)}{2M\left(\nu \right)-vM\left(\nu \right)}}k\left(t,E_{\nu }\left(t\right)\right)+{\displaystyle \frac{2v}{2M\left(\nu \right)-vM\left(\nu \right)}}\underset{0}{\overset{t}{\int }}k\left(y,E_{\nu }\left(y\right)\right)dy\right|,\\ \le \left|{\displaystyle \frac{2\left(1-\nu \right)}{2M\left(\nu \right)-vM\left(\nu \right)}}\right|k\left(t,E_{\nu }\left(t\right)\right)+\left|{\displaystyle \frac{2v}{2M\left(\nu \right)-vM\left(\nu \right)}}\right|\left.\underset{0}{\overset{t}{\int }}k\left(y,E_{\nu }\left(y\right)\right)dy\right|,\\ \le \left[{\displaystyle \frac{2\left(1-\nu \right)}{2M\left(\nu \right)-vM\left(\nu \right)}}+{\displaystyle \frac{2v}{2M\left(\nu \right)-vM\left(\nu \right)}}{c}_6\right]\left|k\left(t,E_{\nu }\left(t\right)\right)\right|,\\ \le \left[{\displaystyle \frac{2\left(1-\nu \right)}{2M\left(\nu \right)-vM\left(\nu \right)}}+{\displaystyle \frac{2v}{2M\left(\nu \right)-vM\left(\nu \right)}}{c}_6\right]\left|L_6,\right.\end{array}$$

$$‖T\left(E_{\nu }\right)‖\le {\displaystyle \frac{2L_6}{2M \left(\nu \right)-vM\left(\nu \right)}}\left[1-\nu +\nu c_6 \right].$$

(25)

Similarly,

$$\begin{array}{c}\left|T\left(I_{\nu }\right(t)\right|=\left|{\displaystyle \frac{2\left(1-\nu \right)}{2M\left(\nu \right)-vM\left(\nu \right)}}k\left(t,I_{\nu }\left(t\right)\right)+{\displaystyle \frac{2v}{2M\left(\nu \right)-vM\left(\nu \right)}}\underset{0}{\overset{t}{\int }}k\left(y,I_{\nu }\left(y\right)\right)dy\right|,\\ \le \left|{\displaystyle \frac{2\left(1-\nu \right)}{2M\left(\nu \right)-vM\left(\nu \right)}}\right|k\left(t,I_{\nu }\left(t\right)\right)+\left|{\displaystyle \frac{2v}{2M\left(\nu \right)-vM\left(\nu \right)}}\right|\left.\underset{0}{\overset{t}{\int }}k\left(y,I_{\nu }\left(y\right)\right)dy\right|,\\ \le \left[{\displaystyle \frac{2\left(1-\nu \right)}{2M\left(\nu \right)-vM\left(\nu \right)}}+{\displaystyle \frac{2v}{2M\left(\nu \right)-vM\left(\nu \right)}}{c}_7\right]\left|k\left(t,I_{\nu }\left(t\right)\right)\right|,\\ \le \left[{\displaystyle \frac{2\left(1-\nu \right)}{2M\left(\nu \right)-vM\left(\nu \right)}}+{\displaystyle \frac{2v}{2M\left(\nu \right)-vM\left(\nu \right)}}{c}_7\right]\left|L_7,\right.\end{array}$$

$$‖T\left(I_{\nu }\right)‖\le {\displaystyle \frac{2L_7}{2M \left(\nu \right)-vM\left(\nu \right)}}\left[1-\nu +\nu c_7 \right].$$

(26)

Hence, $T\left(M\right)$ is bounded.

Now, in the following part, we will consider $t_1<t_2$ and $S_h\left(t\right),E_h\left(t\right),I_h\left(t\right),R_h\left(t\right),S_v\left(t\right),E_v\left(t\right)$, and

$I_v\left(t\right)$ ∊ M and then, for a given $\epsilon >0, if \left|t_2-t_1\right|<\delta .$ We have

$$\begin{array}{c}‖T\left(S_h\right(t_2\left)\right)-T\left(S_h\right(t_1)‖\le \left|{\displaystyle \frac{2-2\nu }{2M\left(\nu \right)-\nu M\left(\nu \right)}}\left(k\right(t_2,S_h\left(t_2\right))-k(t_1,S_h\left(t_1\right)\left)\right)\right|+|{\displaystyle \frac{2v}{2M\left(\nu \right)-vM\left(\nu \right)}}{\int }_0^{t_2}k\left(y,S_h\left(y\right)\right)dy-\\ {\displaystyle \frac{2v}{2M\left(\nu \right)-vM\left(\nu \right)}}{\int }_0^{t_1}k\left(y,S_h\left(y\right)\right)dy|,\end{array}$$

$$\le {\displaystyle \frac{2-2\nu }{2M\left(\nu \right)-\nu M\left(\nu \right)}}\left|\left(k\right(t_2,S_h\left(t_2\right))-k(t_1,S_h\left(t_1\right)\left)\right)\right|+{\displaystyle \frac{2v}{2M\left(\nu \right)-vM\left(\nu \right)}}{L}_1\left|\left(k\right(t_2,S_h\left(t_2\right))-k(t_1,S_h\left(t_1\right)\left)\right)\right|.$$

(27)

$$\begin{array}{c}‖T\left(E_h\right(t_2\left)\right)-T\left(E_h\right(t_1)‖\le \left|{\displaystyle \frac{2-2\nu }{2M\left(\nu \right)-\nu M\left(\nu \right)}}\left(k\right(t_2,E_h\left(t_2\right))-k(t_1,E_h\left(t_1\right)\left)\right)\right|+|{\displaystyle \frac{2v}{2M\left(\nu \right)-vM\left(\nu \right)}}{\int }_0^{t_2}k\left(y,E_h\left(y\right)\right)dy-\\ {\displaystyle \frac{2v}{2M\left(\nu \right)-vM\left(\nu \right)}}{\int }_0^{t_1}k\left(y,E_h\left(y\right)\right)dy|,\end{array}$$

$$\le {\displaystyle \frac{2-2\nu }{2M\left(\nu \right)-\nu M\left(\nu \right)}}\left|\left(k\right(t_2,E_h\left(t_2\right))-k(t_1,E_h\left(t_1\right)\left)\right)\right|+{\displaystyle \frac{2v}{2M\left(\nu \right)-vM\left(\nu \right)}}{L}_2\left|\left(k\right(t_2,E_h\left(t_2\right))-k(t_1,E_h\left(t_1\right)\left)\right)\right|.$$

(28)

$$\begin{array}{c}‖T\left(I_h\right(t_2\left)\right)-T\left(I_h\right(t_1)‖\le \left|{\displaystyle \frac{2-2\nu }{2M\left(\nu \right)-\nu M\left(\nu \right)}}\left(k\right(t_2,I_h\left(t_2\right))-k(t_1,I_h\left(t_1\right)\left)\right)\right|+|{\displaystyle \frac{2v}{2M\left(\nu \right)-vM\left(\nu \right)}}{\int }_0^{t_2}k\left(y,I_h\left(y\right)\right)dy-\\ {\displaystyle \frac{2v}{2M\left(\nu \right)-vM\left(\nu \right)}}{\int }_0^{t_1}k\left(y,I_h\left(y\right)\right)dy|,\end{array}$$

$$\le {\displaystyle \frac{2-2\nu }{2M\left(\nu \right)-\nu M\left(\nu \right)}}\left|\left(k\right(t_2,I_h\left(t_2\right))-k(t_1,I_h\left(t_1\right)\left)\right)\right|+{\displaystyle \frac{2v}{2M\left(\nu \right)-vM\left(\nu \right)}}{L}_3\left|\left(k\right(t_2,I_h\left(t_2\right))-k(t_1,I_h\left(t_1\right)\left)\right)\right|.$$

(29)

$$\begin{array}{c}‖T\left(R_h\right(t_2\left)\right)-T\left(R_h\right(t_1)‖\le \left|{\displaystyle \frac{2-2\nu }{2M\left(\nu \right)-\nu M\left(\nu \right)}}\left(k\right(t_2,R_h\left(t_2\right))-k(t_1,R_h\left(t_1\right)\left)\right)\right|+|{\displaystyle \frac{2v}{2M\left(\nu \right)-vM\left(\nu \right)}}{\int }_0^{t_2}k\left(y,R_h\left(y\right)\right)dy-\\ {\displaystyle \frac{2v}{2M\left(\nu \right)-vM\left(\nu \right)}}{\int }_0^{t_1}k\left(y,R_h\left(y\right)\right)dy|,\end{array}$$

$$\le {\displaystyle \frac{2-2\nu }{2M\left(\nu \right)-\nu M\left(\nu \right)}}\left|\left(k\right(t_2,R_h\left(t_2\right))-k(t_1,R_h\left(t_1\right)\left)\right)\right|+{\displaystyle \frac{2v}{2M\left(\nu \right)-vM\left(\nu \right)}}{L}_4\left|\left(k\right(t_2,R_h\left(t_2\right))-k(t_1,R_h\left(t_1\right)\left)\right)\right|.$$

(30)

$$\begin{array}{c}‖T\left(S_{\nu }\right(t_2\left)\right)-T\left(S_{\nu }\right(t_1)‖\le \left|{\displaystyle \frac{2-2\nu }{2M\left(\nu \right)-\nu M\left(\nu \right)}}\left(k\right(t_2,S_{\nu }\left(t_2\right))-k(t_1,S_{\nu }\left(t_1\right)\left)\right)\right|+|{\displaystyle \frac{2v}{2M\left(\nu \right)-vM\left(\nu \right)}}{\int }_0^{t_2}k\left(y,S_{\nu }\left(y\right)\right)dy-\\ {\displaystyle \frac{2v}{2M\left(\nu \right)-vM\left(\nu \right)}}{\int }_0^{t_1}k\left(y,S_{\nu }\left(y\right)\right)dy|,\end{array}$$

$$\le {\displaystyle \frac{2-2\nu }{2M\left(\nu \right)-\nu M\left(\nu \right)}}\left|\left(k\right(t_2,S_{\nu }\left(t_2\right))-k(t_1,S_{\nu }\left(t_1\right)\left)\right)\right|+{\displaystyle \frac{2v}{2M\left(\nu \right)-vM\left(\nu \right)}}{L}_5\left|\left(k\right(t_2,S_{\nu }\left(t_2\right))-k(t_1,S_{\nu }\left(t_1\right)\left)\right)\right|.$$

(31)

$$\begin{array}{c}‖T\left(E_{\nu }\right(t_2\left)\right)-T\left(E_{\nu }\right(t_1)‖\le \left|{\displaystyle \frac{2-2\nu }{2M\left(\nu \right)-\nu M\left(\nu \right)}}\left(k\right(t_2,E_{\nu }\left(t_2\right))-k(t_1,E_{\nu }\left(t_1\right)\left)\right)\right|+|{\displaystyle \frac{2v}{2M\left(\nu \right)-vM\left(\nu \right)}}{\int }_0^{t_2}k\left(y,E_{\nu }\left(y\right)\right)dy-\\ {\displaystyle \frac{2v}{2M\left(\nu \right)-vM\left(\nu \right)}}{\int }_0^{t_1}k\left(y,E_{\nu }\left(y\right)\right)dy|,\end{array}$$

$$\le {\displaystyle \frac{2-2\nu }{2M\left(\nu \right)-\nu M\left(\nu \right)}}\left|\left(k\right(t_2,E_{\nu }\left(t_2\right))-k(t_1,E_{\nu }\left(t_1\right)\left)\right)\right|+{\displaystyle \frac{2v}{2M\left(\nu \right)-vM\left(\nu \right)}}{L}_6\left|\left(k\right(t_2,E_{\nu }\left(t_2\right))-k(t_1,E_{\nu }\left(t_1\right)\left)\right)\right|.$$

(32)

$$\begin{array}{c}‖T\left(I_{\nu }\right(t_2\left)\right)-T\left(I_{\nu }\right(t_1)‖\le \left|{\displaystyle \frac{2-2\nu }{2M\left(\nu \right)-\nu M\left(\nu \right)}}\left(k\right(t_2,I_{\nu }\left(t_2\right))-k(t_1,I_{\nu }\left(t_1\right)\left)\right)\right|+|{\displaystyle \frac{2v}{2M\left(\nu \right)-vM\left(\nu \right)}}{\int }_0^{t_2}k\left(y,S_h\left(y\right)\right)dy-\\ {\displaystyle \frac{2v}{2M\left(\nu \right)-vM\left(\nu \right)}}{\int }_0^{t_1}k\left(y,S_h\left(y\right)\right)dy|,\end{array}$$

$$\le {\displaystyle \frac{2-2\nu }{2M\left(\nu \right)-\nu M\left(\nu \right)}}\left|\left(k\right(t_2,I_{\nu }\left(t_2\right))-k(t_1,I_{\nu }\left(t_1\right)\left)\right)\right|+{\displaystyle \frac{2v}{2M\left(\nu \right)-vM\left(\nu \right)}}{L}_7\left|\left(k\right(t_2,I_{\nu }\left(t_2\right))-k(t_1,I_{\nu }\left(t_1\right)\left)\right)\right|.$$

(33)

Now we will investigate the following:

$$\left|k\left(t_2,S_h\left(t_2\right)\right)-k\left(t_1,S_h\left(t_1\right)\right)\right|\le \left|\pi -\left({\lambda }_1-{\lambda }_2\right)\left(S_h\left(t_2\right)-S_h\left(t_1\right)\right)-\left(\mu +a\right)\left(S_h\left(t_2\right)-S_h\left(t_1\right)\right)\right|$$

$$\le \pi -(c_1-c_2)\left|{(S}_h\left(t_2\right)-S_h\left(t_1\right))\right|$$

$$\le C_1\left|t_2-t_1\right|.$$

(34)

Now putting Equation (34) and integral parts of Equation (33) in Equation (33), we get

$$\left|T\left(S_h\right(t_2\left)\right)-{T(S}_h\left(t_1\right))\right|\le {\displaystyle \frac{2v}{2M\left(\nu \right)-vM\left(\nu \right)}}{C}_1\left|t_2-t_1\right|+{\displaystyle \frac{2-2\nu }{2M\left(\nu \right)-\nu M\left(\nu \right)}}{L}_1\left|t_2-t_1\right|,$$

$$\delta ={\displaystyle \frac{ϵ}{\frac{2\nu }{2M\left(\nu \right)-\nu M\left(\nu \right)}{C}_1+\frac{2(1-\nu )}{2M\left(\nu \right)-\nu M\left(\nu \right)}{L}_1}},$$

(35)

Such that $\left|T\left(S_h\right(t_2\left)\right)-{T(S}_h\left(t_1\right))\right|\le ϵ \mathrm{i}\mathrm{s} \mathrm{s}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{s}\mathrm{f}\mathrm{i}\mathrm{e}\mathrm{d}.$

Using the same guidelines, we can obtain the following for the function

$$E_h, I_h, R_h, S_v E_{\nu } and I_v$$

$$\left\{\begin{array}{c}\delta ={\displaystyle \frac{ϵ}{\frac{2\nu }{2M\left(\nu \right)-\nu M\left(\nu \right)}{C}_2+\frac{2(1-\nu )}{2M\left(\nu \right)-\nu M\left(\nu \right)}{L}_2}}\\ \delta ={\displaystyle \frac{ϵ}{\frac{2\nu }{2M\left(\nu \right)-\nu M\left(\nu \right)}{C}_3+\frac{2(1-\nu )}{2M\left(\nu \right)-\nu M\left(\nu \right)}{L}_3}}\\ \delta ={\displaystyle \frac{ϵ}{\frac{2\nu }{2M\left(\nu \right)-\nu M\left(\nu \right)}{C}_4+\frac{2(1-\nu )}{2M\left(\nu \right)-\nu M\left(\nu \right)}{L}_4}}\\ \delta ={\displaystyle \frac{ϵ}{\frac{2\nu }{2M\left(\nu \right)-\nu M\left(\nu \right)}{C}_5+\frac{2(1-\nu )}{2M\left(\nu \right)-\nu M\left(\nu \right)}{L}_5}}\\ \delta ={\displaystyle \frac{ϵ}{\frac{2\nu }{2M\left(\nu \right)-\nu M\left(\nu \right)}{C}_6+\frac{2(1-\nu )}{2M\left(\nu \right)-\nu M\left(\nu \right)}{L}_6}}\\ \delta ={\displaystyle \frac{ϵ}{\frac{2\nu }{2M\left(\nu \right)-\nu M\left(\nu \right)}{C}_7+\frac{2(1-\nu )}{2M\left(\nu \right)-\nu M\left(\nu \right)}{L}_7}}\end{array}\right.$$

(36)

such that

$$\left\{\begin{array}{c}\left|T\left(E_h\right(t_2\left)\right)-{T(E}_h\left(t_1\right))\right|\le ϵ\\ \left|T\left(I_h\right(t_2\left)\right)-{T(I}_h\left(t_1\right))\right|\le ϵ\\ \left|T\left(R_h\right(t_2\left)\right)-{T(R}_h\left(t_1\right))\right|\le ϵ \\ \left|T\left(S_{\nu }\right(t_2\left)\right)-{T(S}_{\nu }\left(t_1\right))\right|\le ϵ\\ \left|T\left(E_{\nu }\right(t_2\left)\right)-{T(E}_{\nu }\left(t_1\right))\right|\le ϵ\\ \left|T\left(I_{\nu }\right(t_2\left)\right)-{T(I}_{\nu }\left(t_1\right))\right|\le ϵ\end{array}\right.$$

(37)

are satisfied. Therefore, $T\left(M\right)$ is equicontinuous, so that $\overline{T\left(M\right)}$ is compact via the Arzela–Ascoli theorem. □

Theorem 1.

Let $N: [S_{h1 }, S_{h2}]\times [0,\infty )\to [0,\infty ),$ then  $N(t,\bullet )$ is non- decreasing for each  $t$ in $\left[S_{h1 }, S_{h2}\right].$ There exist positive constants, $v_1$ and  $v_2$, so that $B\left(n\right)v_1 \le S\left(t,v_1\right), B\left(n\right)v_2 \ge N\left(t,v_2\right), 0 \le v_1 \left(t\right)\le v_2 \left(t\right), S_{h1} \le t \le S_{h2}$. As a result, the solution to the equation is positive.

Proof: 

For the operator of T, we simply need to take into account the fixed point.

We assume that $\mathrm{T}:\mathrm{H}\to \mathrm{H}$ is a continuous function.

Let $S_{h1}\le S_{h2},E_{h1}\le E_{h2},I_{h1}\le I_{h2},R_{h1}\le R_{h2},S_{v1}\le S_{v2},E_{v1}\le E_{v2},I_{v1}\le I_{v2}.$

$$T{S}_{h1}\left(t\right)=\mathrm{ }{\displaystyle \frac{2-2v}{2M\left(v\right)-vM\left(v\right)}}\mathrm{ }k\left(t,S_{h1}\left(t\right)\right)+\mathrm{ }{\displaystyle \frac{2v}{2M\left(v\right)-vM\left(v\right)}}{\int }_0^{t}k\left(y,S_{h1}\left(y\right)\right)dy$$

$$\le \mathrm{ }{\displaystyle \frac{2-2v}{2M\left(v\right)-vM\left(v\right)}}\mathrm{ }k\left(t,S_{h2}\left(t\right)\right)+\mathrm{ }{\displaystyle \frac{2v}{2M\left(v\right)-vM\left(v\right)}}{\int }_0^{t}k\left(y,S_{h2}\left(y\right)\right)dy$$

$$\le T{S}_{h2}\left(x,t\right).$$

(38)

$$T{E}_{h1}\left(t\right)=\mathrm{ }{\displaystyle \frac{2-2v}{2M\left(v\right)-vM\left(v\right)}}\mathrm{ }k\left(t,E_{h1}\left(t\right)\right)+\mathrm{ }{\displaystyle \frac{2v}{2M\left(v\right)-vM\left(v\right)}}{\int }_0^{t}k\left(y,E_{h1}\left(y\right)\right)dy$$

$$\le \mathrm{ }{\displaystyle \frac{2-2v}{2M\left(v\right)-vM\left(v\right)}}\mathrm{ }k\left(t,E_{h2}\left(t\right)\right)+\mathrm{ }{\displaystyle \frac{2v}{2M\left(v\right)-vM\left(v\right)}}{\int }_0^{t}k\left(y,E_{h2}\left(y\right)\right)dy$$

$$\le T{E}_{h2}\left(x,t\right).$$

(39)

$$T{I}_{h1}\left(t\right)=\mathrm{ }{\displaystyle \frac{2-2v}{2M\left(v\right)-vM\left(v\right)}}\mathrm{ }k\left(t,I_{h1}\left(t\right)\right)+\mathrm{ }{\displaystyle \frac{2v}{2M\left(v\right)-vM\left(v\right)}}{\int }_0^{t}k\left(y,I_{h1}\left(y\right)\right)dy$$

$$\le \mathrm{ }{\displaystyle \frac{2-2v}{2M\left(v\right)-vM\left(v\right)}}\mathrm{ }k\left(t,I_{h2}\left(t\right)\right)+\mathrm{ }{\displaystyle \frac{2v}{2M\left(v\right)-vM\left(v\right)}}{\int }_0^{t}k\left(y,I_{h2}\left(y\right)\right)dy$$

$$\le T{I}_{h2}\left(x,t\right).$$

(40)

$$T{R}_{h1}\left(t\right)=\mathrm{ }{\displaystyle \frac{2-2v}{2M\left(v\right)-vM\left(v\right)}}\mathrm{ }k\left(t,R_{h1}\left(t\right)\right)+\mathrm{ }{\displaystyle \frac{2v}{2M\left(v\right)-vM\left(v\right)}}{\int }_0^{t}k\left(y,R_{h1}\left(y\right)\right)dy$$

$$\le \mathrm{ }{\displaystyle \frac{2-2v}{2M\left(v\right)-vM\left(v\right)}}\mathrm{ }k\left(t,R_{h2}\left(t\right)\right)+\mathrm{ }{\displaystyle \frac{2v}{2M\left(v\right)-vM\left(v\right)}}{\int }_0^{t}k\left(y,R_{h2}\left(y\right)\right)dy$$

$$\le T{R}_{h2}\left(x,t\right).$$

(41)

$$T{S}_{v1}\left(t\right)=\mathrm{ }{\displaystyle \frac{2-2v}{2M\left(v\right)-vM\left(v\right)}}\mathrm{ }k\left(t,S_{v1}\left(t\right)\right)+\mathrm{ }{\displaystyle \frac{2v}{2M\left(v\right)-vM\left(v\right)}}{\int }_0^{t}k\left(y,S_{v1}\left(y\right)\right)dy$$

$$\le \mathrm{ }{\displaystyle \frac{2-2v}{2M\left(v\right)-vM\left(v\right)}}\mathrm{ }k\left(t,S_{v2}\left(t\right)\right)+\mathrm{ }{\displaystyle \frac{2v}{2M\left(v\right)-vM\left(v\right)}}{\int }_0^{t}k\left(y,S_{v2}\left(y\right)\right)dy$$

$$\le T{S}_{v2}\left(x,t\right)$$

(42)

$$T{E}_{v1}\left(t\right)=\mathrm{ }{\displaystyle \frac{2-2v}{2M\left(v\right)-vM\left(v\right)}}\mathrm{ }k\left(t,E_{v1}\left(t\right)\right)+\mathrm{ }{\displaystyle \frac{2v}{2M\left(v\right)-vM\left(v\right)}}{\int }_0^{t}k\left(y,E_{v1}\left(y\right)\right)dy$$

$$\le \mathrm{ }{\displaystyle \frac{2-2v}{2M\left(v\right)-vM\left(v\right)}}\mathrm{ }k\left(t,E_{v2}\left(t\right)\right)+\mathrm{ }{\displaystyle \frac{2v}{2M\left(v\right)-vM\left(v\right)}}{\int }_0^{t}k\left(y,E_{v2}\left(y\right)\right)dy$$

$$\le T{E}_{v2}\left(x,t\right).$$

(43)

$$T{I}_{v1}\left(t\right)=\mathrm{ }{\displaystyle \frac{2-2v}{2M\left(v\right)-vM\left(v\right)}}k\left(t,I_{v1}\left(t\right)\right)+{\displaystyle \frac{2v}{2M\left(v\right)-vM\left(v\right)}}{\int }_0^{t}k\left(y,I_{v1}\left(y\right)\right)dy$$

$$\le \mathrm{ }{\displaystyle \frac{2-2v}{2M\left(v\right)-vM\left(v\right)}}\mathrm{ }k\left(t,I_{v2}\left(t\right)\right)+\mathrm{ }{\displaystyle \frac{2v}{2M\left(v\right)-vM\left(v\right)}}{\int }_0^{t}k\left(y,I_{v2}\left(y\right)\right)dy$$

$$\le T{I}_{v2}\left(x,t\right).$$

(44)

The operator $T: {⟨\nu }_1, {\nu }_2⟩ ⟶ ⟨{\nu }_1, {\nu }_2⟩$ is compact and continuous via lemma l, since T is a non-decreasing operator. H is the normal cone of T in that scenario. □

Analysis of Uniqueness Solutions

In this section, we have demonstrated that the Zika model with the Caputo–Fabrizio fractional derivative has a uniqueness and a solution by applying the fixed-point theorem. This section aims to demonstrate that solutions to system (17) with the initial conditions are unique. Additionally, let us assume that we can identify seven unique connected solutions.

$$\left\{\left(S_{h1},S_{h2}\right),\left(E_{h1},E_{h2}\right),\left(I_{h1},I_{h2}\right),\left(R_{h1},R_{h2}\right),\left(S_{v1},S_{v2}\right),(E_{v1},E_{v2})\right\} \mathrm{and} (I_{v1},I_{v2}).$$

Thus, the following is how the solution’s uniqueness is presented:

$$\left|T{S_h}_1\left(t\right)-T{S_h}_2\left(t\right)\right|=\left|{\displaystyle \frac{2-2v}{2M\left(v\right)-vM\left(v\right)}}\right|k(t,{S_h}_1\left(t\right)-k\left(t,{S_h}_2\left(t\right)\right)+{\displaystyle \frac{2v}{2M\left(v\right)-vM\left(v\right)}}\left|{\int }_0^t\left(k\left(y,{S_h}_1\left(y\right)\right)-k\left(y,{S_h}_2\left(y\right)\right)\right)dy\right|$$

$$\le {\displaystyle \frac{2-2v}{2M\left(v\right)-vM\left(v\right)}}\left|k(t,{S_h}_1\left(t\right)-k\left(t,{S_h}_2\left(t\right)\right)\right|+{\displaystyle \frac{2v}{2M\left(v\right)-vM\left(v\right)}}\left|{\int }_0^t\left(k\left(y,S_1\left(y\right)\right)-k\left(y,{S_h}_2\left(y\right)\right)\right)dy\right|$$

$$\le {\displaystyle \frac{2-2v}{2M\left(v\right)-vM\left(v\right)}}{F}_1\left|{S_h}_1\left(t\right)-{S_h}_2\left(t\right)\right|+{\displaystyle \frac{2v}{2M\left(v\right)-vM\left(v\right)}}{F}_1\left|{S_h}_1\left(t\right)-{S_h}_2\left(t\right)\right|$$

(45)

Thus, we can write the above equation as (45):

$$\left|T{S_h}_1\left(t\right)-T{S_h}_2\left(t\right)\right|\le \left\{{\displaystyle \frac{2-2v}{2M\left(v\right)-vM\left(v\right)}}{F}_1+{\displaystyle \frac{2v}{2M\left(v\right)-vM\left(v\right)}}{F}_1\right\}\left|{S_h}_1\left(t\right)-{S_h}_2\left(t\right)\right|.$$

$$\left|T{E_h}_1\left(t\right)-T{E_h}_2\left(t\right)\right|=\left|{\displaystyle \frac{2-2v}{2M\left(v\right)-vM\left(v\right)}}\right|k(t,s_1\left(t\right)-k\left(t,s_2\left(t\right)\right)+{\displaystyle \frac{2v}{2M\left(v\right)-vM\left(v\right)}}\left|{\int }_0^t\left(k\left(y,s_1\left(y\right)\right)-k\left(y,s_2\left(y\right)\right)\right)dy\right|$$

$$\le {\displaystyle \frac{2-2v}{2M\left(v\right)-vM\left(v\right)}}{F}_2\left|{E_h}_1\left(t\right)-{E_h}_2\left(t\right)\right|+{\displaystyle \frac{2v}{2M\left(v\right)-vM\left(v\right)}}{F}_2\left|{E_h}_1\left(t\right)-{E_h}_2\left(t\right)\right|,$$

(46)

Thus, we can write the above Equation (46) as:

$$‖T{E_h}_1\left(t\right)-T{E_h}_2\left(t\right)‖\le \left\{{\displaystyle \frac{2-2v}{2M\left(v\right)-vM\left(v\right)}}{F}_2+{\displaystyle \frac{2v}{2M\left(v\right)-vM\left(v\right)}}{F}_2\right\}\left|{E_h}_1\left(t\right)-{E_h}_2\left(t\right)\right|.$$

$$\left|T{I_h}_1\left(t\right)-T{I_h}_2\left(t\right)\right|=\left|{\displaystyle \frac{2-2v}{2M\left(v\right)-vM\left(v\right)}}\right|k(t,s_1\left(t\right)-k\left(t,s_2\left(t\right)\right)+{\displaystyle \frac{2v}{2M\left(v\right)-vM\left(v\right)}}\left|{\int }_0^t\left(k\left(y,s_1\left(y\right)\right)-k\left(y,s_2\left(y\right)\right)\right)dy\right|$$

$$\le {\displaystyle \frac{2-2v}{2M\left(v\right)-vM\left(v\right)}}{F}_3\left|{I_h}_1\left(t\right)-{I_h}_2\left(t\right)\right|+{\displaystyle \frac{2v}{2M\left(v\right)-vM\left(v\right)}}{F}_3\left|{I_h}_1\left(t\right)-{I_h}_2\left(t\right)\right|.$$

(47)

Thus, we can write the above equation as (47):

$$‖T{E_h}_1\left(t\right)-T{E_h}_2\left(t\right)‖\le \left\{{\displaystyle \frac{2-2v}{2M\left(v\right)-vM\left(v\right)}}{F}_3+{\displaystyle \frac{2v}{2M\left(v\right)-vM\left(v\right)}}{F}_3\right\}\left|{E_h}_1\left(t\right)-{E_h}_2\left(t\right)\right|.$$

$$\left|T{R_h}_1\left(t\right)-T{R_h}_2\left(t\right)\right|=\left|{\displaystyle \frac{2-2v}{2M\left(v\right)-vM\left(v\right)}}\right|k(t,s_1\left(t\right)-k\left(t,s_2\left(t\right)\right)+{\displaystyle \frac{2v}{2M\left(v\right)-vM\left(v\right)}}\left|{\int }_0^t\left(k\left(y,s_1\left(y\right)\right)-k\left(y,s_2\left(y\right)\right)\right)dy\right|$$

$$\le {\displaystyle \frac{2-2v}{2M\left(v\right)-vM\left(v\right)}}{F}_4\left|{R_h}_1\left(t\right)-{R_h}_2\left(t\right)\right|+{\displaystyle \frac{2v}{2M\left(v\right)-vM\left(v\right)}}{F}_4\left|{R_h}_1\left(t\right)-{R_h}_2\left(t\right)\right|.$$

(48)

Thus, we can write the above Equation (48) as:

$$‖T{R_h}_1\left(t\right)-T{R_h}_2\left(t\right)‖\le \left\{{\displaystyle \frac{2-2v}{2M\left(v\right)-vM\left(v\right)}}{F}_4+{\displaystyle \frac{2v}{2M\left(v\right)-vM\left(v\right)}}{F}_4\right\}\left|{R_h}_1\left(t\right)-{R_h}_2\left(t\right)\right|.$$

$$\left|T{S_v}_1\left(t\right)-T{S_v}_2\left(t\right)\right|=\left|{\displaystyle \frac{2-2v}{2M\left(v\right)-vM\left(v\right)}}\right|k(t,s_1\left(t\right)-k\left(t,s_2\left(t\right)\right)+{\displaystyle \frac{2v}{2M\left(v\right)-vM\left(v\right)}}\left|{\int }_0^t\left(k\left(y,s_1\left(y\right)\right)-k\left(y,s_2\left(y\right)\right)\right)dy\right|$$

$$\le {\displaystyle \frac{2-2v}{2M\left(v\right)-vM\left(v\right)}}{F}_5\left|{S_v}_1\left(t\right)-{S_v}_2\left(t\right)\right|+{\displaystyle \frac{2v}{2M\left(v\right)-vM\left(v\right)}}{F}_5\left|{S_v}_1\left(t\right)-{S_v}_2\left(t\right)\right|.$$

(49)

Thus, we can write the above Equation (49) as:

$$‖T{S_v}_1\left(t\right)-T{S_v}_2\left(t\right)‖\le \left\{{\displaystyle \frac{2-2v}{2M\left(v\right)-vM\left(v\right)}}{F}_5+{\displaystyle \frac{2v}{2M\left(v\right)-vM\left(v\right)}}{F}_5\right\}\left|{S_v}_1\left(t\right)-{S_v}_2\left(t\right)\right|.$$

$$\left|T{E_v}_1\left(t\right)-T{E_v}_2\left(t\right)\right|=\left|{\displaystyle \frac{2-2v}{2M\left(v\right)-vM\left(v\right)}}\right|k(t,s_1\left(t\right)-k\left(t,s_2\left(t\right)\right)+{\displaystyle \frac{2v}{2M\left(v\right)-vM\left(v\right)}}\left|{\int }_0^t\left(k\left(y,s_1\left(y\right)\right)-k\left(y,s_2\left(y\right)\right)\right)dy\right|$$

$$\le {\displaystyle \frac{2-2v}{2M\left(v\right)-vM\left(v\right)}}{F}_6\left|{E_v}_1\left(t\right)-{E_v}_2\left(t\right)\right|+{\displaystyle \frac{2v}{2M\left(v\right)-vM\left(v\right)}}{F}_6\left|{E_v}_1\left(t\right)-{E_v}_2\left(t\right)\right|.$$

(50)

Thus, we can write the above Equation (50) as:

$$‖T{E_v}_1\left(t\right)-T{E_v}_2\left(t\right)‖\le \left\{{\displaystyle \frac{2-2v}{2M\left(v\right)-vM\left(v\right)}}{F}_7+{\displaystyle \frac{2v}{2M\left(v\right)-vM\left(v\right)}}{F}_7\right\}\left|{E_v}_1\left(t\right)-{E_v}_2\left(t\right)\right|.$$

Similarly,

$$\left|T{I_v}_1\left(t\right)-T{I_v}_2\left(t\right)\right|=\left|{\displaystyle \frac{2-2v}{2M\left(v\right)-vM\left(v\right)}}\right|k(t,s_1\left(t\right)-k\left(t,s_2\left(t\right)\right)+{\displaystyle \frac{2v}{2M\left(v\right)-vM\left(v\right)}}\left|{\int }_0^t\left(k\left(y,s_1\left(y\right)\right)-k\left(y,s_2\left(y\right)\right)\right)dy\right|$$

$$\le {\displaystyle \frac{2-2v}{2M\left(v\right)-vM\left(v\right)}}{F}_7\left|{I_v}_1\left(t\right)-{I_v}_2\left(t\right)\right|+{\displaystyle \frac{2v}{2M\left(v\right)-vM\left(v\right)}}{F}_7\left|{I_v}_1\left(t\right)-{I_v}_2\left(t\right)\right|.$$

(51)

Thus, we can write the above Equation (51) as:

$$‖T{I_v}_1\left(t\right)-T{I_v}_2\left(t\right)‖\le \left\{{\displaystyle \frac{2-2v}{2M\left(v\right)-vM\left(v\right)}}{F}_8+{\displaystyle \frac{2v}{2M\left(v\right)-vM\left(v\right)}}{F}_8\right\}\left|{I_v}_1\left(t\right)-{I_v}_2\left(t\right)\right|,$$

Therefore, if the following conditions hold:

$$\left\{\begin{array}{c}\left\{{\displaystyle \frac{2-2v}{2M\left(v\right)-vM\left(v\right)}}{F}_1+{\displaystyle \frac{2v}{2M\left(v\right)-vM\left(v\right)}}{F}_1\right\}<1,\\ \left\{{\displaystyle \frac{2-2v}{2M\left(v\right)-vM\left(v\right)}}{F}_2+{\displaystyle \frac{2v}{2M\left(v\right)-vM\left(v\right)}}{F}_2\right\}<1,\\ \left\{{\displaystyle \frac{2-2v}{2M\left(v\right)-vM\left(v\right)}}{F}_3+{\displaystyle \frac{2v}{2M\left(v\right)-vM\left(v\right)}}{F}_3\right\}<1,\\ \left\{{\displaystyle \frac{2-2v}{2M\left(v\right)-vM\left(v\right)}}{F}_4+{\displaystyle \frac{2v}{2M\left(v\right)-vM\left(v\right)}}{F}_4\right\}<1,\\ \left\{{\displaystyle \frac{2-2v}{2M\left(v\right)-vM\left(v\right)}}{F}_5+{\displaystyle \frac{2v}{2M\left(v\right)-vM\left(v\right)}}{F}_5\right\}<1,\\ \left\{{\displaystyle \frac{2-2v}{2M\left(v\right)-vM\left(v\right)}}{F}_6+{\displaystyle \frac{2v}{2M\left(v\right)-vM\left(v\right)}}{F}_6\right\}<1,\\ and \\ \left\{{\displaystyle \frac{2-2v}{2M\left(v\right)-vM\left(v\right)}}{F}_7+{\displaystyle \frac{2v}{2M\left(v\right)-vM\left(v\right)}}{F}_7\right\}<1.\end{array}\right.$$

(52)

Using the fixed point theorem, we can state that the model has a unique positive solution, since mapping T is a contraction.

6. Conclusions

In this article, first we combined the novel fractional derivative with the Zika virus model. Next, we discovered the Zika virus model’s existence solution. Lastly, we examined the model’s unique positive solution and the circumstances in which it exists. With our initiative, we attempted to support the researchers who are involved in education on the Zika virus. Examining the results revealed that the fractional derivative provides crucial process information. This framework provides a more realistic tool for studying and controlling Zika outbreaks, with insights into how historical conditions influence current and future transmission. Finally, applying this model to other infectious diseases with symmetrical transmission routes can test its generalizability and robustness, potentially broadening its applicability in epidemiological research.