Existence and uniqueness of solution for a fractional hepatitis B model

Nnaemeka Stanley Aguegboh; Dominic Obinna Oranugo; Phineas Roy Kiogora; Mutua Felix; Onyiaji Netochukwu; Andrew Onyeka Egwu

doi:10.1515/cmb-2024-0009

Article Open Access

Existence and uniqueness of solution for a fractional hepatitis B model

Nnaemeka Stanley Aguegboh , Dominic Obinna Oranugo , Phineas Roy Kiogora , Mutua Felix , Onyiaji Netochukwu and Andrew Onyeka Egwu

Published/Copyright: March 7, 2025

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From the journal Computational and Mathematical Biophysics Volume 13 Issue 1

Abstract

Understanding the dynamics of infectious diseases using mathematical modeling is essential for developing prevention and control measures. Hepatitis B is still a major public health issue in many places, including Kenya, where the high incidence of illness presents serious difficulties. In this article, the existence and uniqueness of solutions for a fractional hepatitis B model in the Caputo sense were explored. Furthermore, using the Adams-type Predictor-Corrector method, numerical simulations of the fractional order model are carried out. The outcomes show how well the suggested control strategies work to stop the spread of hepatitis B virus.

Keywords: Hepatits B; fractional-order differential equations; parameter estimation; numerical simulations; predictor-corrector method

MSC 2010: 37A50; 60H10; 60J65; 60J70

1 Introduction

Hepatitis B, a potentially fatal liver illness, is brought on by the hepatitis B virus (HBV). It has a big impact on global health. It can lead to high death rates from cirrhosis and liver cancer, as well as chronic infection and liver disease [10]. When the virus enters the bloodstream and makes its way to the liver, an individual get hepatitis B. In the liver, the virus divides and releases a large number of new viruses into the circulation [9]. There are two stages of an HBV infection: acute and chronic. The first 6 months following a person’s exposure to the virus are referred to as the acute period. The immune system can usually eradicate the infection during this time, but in very extreme situations, it can progress to a dangerous level and cause a permanent sickness. The chronic stage is another name for this. It should be mentioned that if a person tests positive for HBsAg for longer than 6 months, it indicates that the person has a chronic condition. Often, when someone is in the chronic state, they have never experienced the acute stage. In addition, this infection may cause liver failure, liver malignancy, and liver scarring [7]. About 80% of cases of primary liver cancer had this virus as their primary cause. About 25% of infections in children progresses to chronic infections. It is believed that 93 million people in China are afflicted with HBV. When people come into contact with contaminated blood, bodily fluids from an HBV patient, or dangerous instruments during surgery, such as syringes, the infection can spread among them [14].

In recent years, there has been a lot of interest in the mathematical modeling of infectious diseases using differential equations of integer order [9,10]. However, epidemiological models and other models in science and engineering have been effectively developed and analyzed using fractional derivatives and integrals [4,11]. Fractional calculus was first introduced by Leibniz in a memo that was transcribed in 1965. Fractional order differential equations (FDEs), have gained importance recently in mathematics. This is because nonlinear systems have many possibilities for unfolding and are used in fields including engineering, control theory, and physics [2,3]. Since they allow for more degrees of freedom and incorporate memory into the model, they were added to the epidemiological model [1]. Khan et al. [6] presented a fractional hepatitis B order derivative in Atangana-Baleanu sense and showed the existence and uniqueness of the equlibrium solution using the banach fixed-point theorem. A fractional order asymptotically carriers hepatitis B model was presented by Gul et al. [5] in the Caputo sense. Both the model analysis and the model simulations were completed. The outcome demonstrated that the carriers may contribute to a rise in the HBV transmission. Salman and Yousef [13] considered a fractional order model for a HBV infection with cured infected cells. The local asymptotic stability of the equilibria was investigated. Their numerical results show that adding a fractional order derivative to the model reduces the peak of the infection, but it will take longer for the disease to totally disappear.

This article’s primary goal is to investigate the existence and uniqueness of a fractional operator solution for the hepatitis B model. The Caputo operator is the one being examined here using the Picard-Lindelof method.

2 Preliminaries

In this section, important definitions and theorems are presented.

Definition 2.1

[10] The Caputo fractional derivative of order α of a function f : ℜ + → ℜ is given by

(2.1) D t α C f ( t ) = 1 Γ ( α − n ) ∫ α t f ( n ) ( τ ) d τ ( t − τ ) α + 1 − n ( n − 1 < α ≤ n ) .

Definition 2.2

[10] The fractional integral of the Caputo fractional derivative of order α of a function f : ℜ + → ℜ is given by

(2.2) I α C f ( t ) = 1 Γ ( α ) ∫ 0 t f ( τ ) ( t − τ ) 1 − α d τ .

Theorem 2.1

[12] The following Lipschitz condition is satisfied for any two functions f 1 ( t ) and f 2 ( t )

(2.3) ‖ D t α C f 1 ( t ) − D t α C f 2 ( t ) ‖ ≤ K ‖ f 1 ( t ) − f 2 ( t ) ‖ .

3 Model formulation

The model used in this article was adopted from [10]. The model equation is given as follows:

(3.1) D t α C S ( t ) = Λ − β 1 S I 1 − β 2 S I 2 − σ S − μ S D t α C V ( t ) = σ S − μ V D t α C E ( t ) = β 1 S I 1 + β 2 S I 2 − γ E − μ E D t α C I 1 ( t ) = γ E − ε ρ I 1 − ε ( 1 − ρ ) I 1 − μ I 1 D t α C R ( t ) = ε ρ I 1 − μ R D t α C I 2 ( t ) = ε ( 1 − ρ ) I 1 − η I 2 − δ 1 I 2 − μ I 2 D t α C T ( t ) = η I 2 − δ 2 T − μ T .

With the nonnegative initial condition:

S ( 0 ) = S 0 , V ( 0 ) = V 0 , E ( 0 ) = E 0 , I 1 ( 0 ) = I 1 0 , I 2 ( 0 ) = I 2 0 , R ( 0 ) = R 0 , T ( 0 ) = T 0 .

The flow diagram of the model is as follows [10] (Figure 1 and Table 1):

Figure 1

Flow chart. Source: Created by the authors.

Table 1

Description of parameters and variables for model (3.1)

Variables	Description	Unit
S	Susceptible individuals	People
V	Vaccinated individuals	People
E	Exposed individuals	People
I 1	Acutely infected individuals	People
R	Recoveredindividuals	People
I 2	Chronically infected individuals	People
T	Individuals under treatment	People

Parameters	Description	Unit
Λ	Recruitment rate	day ‒ 1
σ	Vaccination rate	day ‒ 1
β 1	Rate of interaction between the acutely infected and susceptible populations	day ‒ 1
β 2	Interaction rate between the population that is susceptible and the chronically infected	day ‒ 1
γ	The rate at which individuals becomes acutely sick after being exposed	day ‒ 1
ε ρ	Percentage of the population with acute infection	day ‒ 1
ε ( 1 ‒ ρ )	Percentage of the population moving from the acutely infected to the chronically infected	day ‒ 1
δ 1	Death rate from infection in the class of chronically infected patients	day ‒ 1
η	Rate of progression from the treatment class to the chronically infected	day ‒ 1
δ 2	Death rate in the treatment-infected class due to infection	day ‒ 1
μ	Natural death rate	day ‒ 1

4 Existence and uniqueness of solution of model 3.1

In the section, we show that the system 3.1 has a unique solution. First, the system is rewritten in the following form:

(4.1) D t α C S ( t ) = K 1 ( t , S ( t ) ) , D t α C V ( t ) = K 2 ( t , V ( t ) ) , D t α C E ( t ) = K 3 ( t , E ( t ) ) , D t α C I 1 ( t ) = K 4 ( t , I 1 ( t ) ) , D t α C R ( t ) = K 5 ( t , R ( t ) ) , D t α C I 2 ( t ) = K 6 ( t , I 2 ( t ) ) , D t α C T ( t ) = K 7 ( t , T ( t ) ) ,

where

(4.2) K 1 ( t , S ( t ) ) = Λ − β 1 S I 1 − β 2 S I 2 − σ S − μ S K 2 ( t , V ( t ) ) = σ S − μ V K 3 ( t , E ( t ) ) = β 1 S I 1 + β 2 S I 2 − γ E − μ E K 4 ( t , I 1 ( t ) ) = γ E − ε ρ I 1 − ε ( 1 − ρ ) I 1 − μ I 1 K 5 ( t , R ( t ) ) = ε ρ I 1 − μ R K 6 ( t , I 2 ( t ) ) = ε ( 1 − ρ ) I 1 − η I 2 − δ 1 I 2 − μ I 2 K 7 ( t , T ( t ) ) = η I 2 − δ 2 T − μ T .

Taking the integral transform of system 4.1 [12], we obtain

(4.3) S ( t ) − S ( 0 ) = 1 Γ ( α ) ∫ 0 t K 1 ( τ , S ) ( t − τ ) α − 1 d τ V ( t ) − V ( 0 ) = 1 Γ ( α ) ∫ 0 t K 2 ( τ , V ) ( t − τ ) α − 1 d τ E ( t ) − E ( 0 ) = 1 Γ ( α ) ∫ 0 t K 3 ( τ , E ) ( t − τ ) α − 1 d τ I 1 ( t ) − I 1 ( 0 ) = 1 Γ ( α ) ∫ 0 t K 4 ( τ , I 1 ) ( t − τ ) α − 1 d τ R ( t ) − R ( 0 ) = 1 Γ ( α ) ∫ 0 t K 5 ( τ , R ) ( t − τ ) α − 1 d τ I 2 ( t ) − I 2 ( 0 ) = 1 Γ ( α ) ∫ 0 t K 6 ( τ , I 2 ) ( t − τ ) α − 1 d τ T ( t ) − T ( 0 ) = 1 Γ ( α ) ∫ 0 t K 7 ( τ , T ) ( t − τ ) α − 1 d τ .

The Kernels K 1 , i = 1 , 2 , 3 , 4 , 5 , 6 , 7 satisfy the Lipschitz condition as shown below.

Theorem 4.1

K 1 satisfies the Lipschitz condition and contraction if the following condition holds: 0 ≤ β 1 b 1 + β 2 b 2 + σ + μ < 1 .

Proof

For S and S 1 ,

(4.4) ‖ K 1 ( t , S ) − K 1 ( t , S 1 ) ‖ = ‖ − β 1 I 1 ( t ) ( S ( t ) − S 1 ( t ) ) − β 2 I 2 ( S ( t ) − S 1 ( t ) ) − σ ( S ( t ) − S 1 ( t ) ) − μ ( S ( t ) − S 1 ( t ) ) ‖ ≤ ( β 1 ‖ I 1 ( t ) ‖ + β 2 ‖ I 2 ( t ) ‖ + σ + μ ) ( ‖ S ( t ) − S 1 ( t ) ‖ ) .

Suppose M 1 = β 1 b 1 + β 2 b 2 + σ + μ , where I 1 ( t ) ≤ b 1 and I 2 ( t ) ≤ b 2 are bounded functions.

Therefore,

(4.5)□ ‖ K 1 ( t , S ) − K 1 ( t , S 1 ) ‖ ≤ M 1 ( ‖ S ( t ) − S 1 ( t ) ‖ ) .

For K 1 , the Lipschtiz conditions is gotten, and if 0 ≤ β 1 b 1 + β 2 b 2 + σ + μ < 1 , then K 1 is a contraction.

Theorem 4.2

K 2 satisfies the Lipschtiz condition and contraction if the following holds: 0 ≤ μ < 1 .

Proof

For V and V 1 ,

(4.6) ‖ V 1 ( t , S ) − V 1 ( t , S 1 ) ‖ = ‖ − μ ( V ( t ) − V 1 ( t ) ) ‖ ≤ μ ‖ V ( t ) − V 1 ( t ) ‖ .

Suppose K 2 = μ , then ‖ K 2 ( t , V ) − K 2 ( t , V 2 ) ‖ ≤ M 2 ( ‖ ( V ( t ) − V 1 ( t ) ) ‖ ) .

The Lipschitz condition for K 2 is obtained, and if 0 ≤ μ < 1 , then K 2 is a contraction.

In the same vein, K j , j = 3 , 4 , 5 , 6 , 7 satisfy the Lipschitz condition as follows:

‖ K 3 ( t , E ) − K 3 ( t , E 1 ) ‖ ≤ M 3 ( ‖ E ( t ) − E 1 ( t ) ‖ ) , ‖ K 4 ( t , I 1 ) − K 4 ( t , I 1 1 ) ‖ ≤ M 4 ( ‖ I 1 ( t ) − I 1 1 ( t ) ‖ ) , ‖ K 5 ( t , R ) − K 5 ( t , R 1 ) ‖ ≤ M 5 ( ‖ R ( t ) − R 1 ( t ) ‖ ) , ‖ K 6 ( t , I 2 ) − K 6 ( t , I 2 1 ) ‖ ≤ M 6 ( ‖ I 2 ( t ) − I 2 1 ( t ) ‖ ) , ‖ K 7 ( t , T ) − K 7 ( t , T 1 ) ‖ ≤ M 7 ( ‖ T ( t ) − T 1 ( t ) ‖ ) ,

where M 3 = γ + μ , M 4 = ε ρ + μ , M 5 = μ , M 6 = η + δ 1 + μ , δ 2 + μ .

For j = 3 , 4 , 5 , 6 , 7 , we obtain 0 ≤ K j < 1 , then K j are contractions. Take a look at the recursive patterns, as suggested by system (4.7):

(4.7) ϕ 1 n ( t ) = S n ( t ) − S n − 1 ( t ) = 1 Γ ( α ) ∫ 0 t ( K 1 ( τ , S n − 1 ) − K 1 ( τ , S n − 2 ) ) ( t − τ ) α − 1 d τ ϕ 2 n ( t ) = V n ( t ) − V n − 1 ( t ) = 1 Γ ( α ) ∫ 0 t ( K 2 ( τ , V n − 1 ) − K 2 ( τ , V n − 2 ) ) ( t − τ ) α − 1 d τ ϕ 3 n ( t ) = E n ( t ) − E n − 1 ( t ) = 1 Γ ( α ) ∫ 0 t ( K 3 ( τ , E n − 1 ) − K 3 ( τ , E n − 2 ) ) ( t − τ ) α − 1 d τ ϕ 4 n ( t ) = I 1 n ( t ) − I 1 n − 1 ( t ) = 1 Γ ( α ) ∫ 0 t ( K 4 ( τ , I 1 n − 1 ) − K 4 ( τ , I 1 n − 2 ) ) ( t − τ ) α − 1 d τ ϕ 5 n ( t ) = R n ( t ) − R n − 1 ( t ) = 1 Γ ( α ) ∫ 0 t ( K 5 ( τ , R n − 1 ) − K 5 ( τ , R n − 2 ) ) ( t − τ ) α − 1 d τ ϕ 6 n ( t ) = I 2 n ( t ) − I 2 n − 1 ( t ) = 1 Γ ( α ) ∫ 0 t ( K 6 ( τ , I 2 n − 1 ) − K 6 ( τ , I 2 n − 2 ) ) ( t − τ ) α − 1 d τ ϕ 7 n ( t ) = T n ( t ) − T n − 1 ( t ) = 1 Γ ( α ) ∫ 0 t ( K 7 ( τ , T n − 1 ) − K 7 ( τ , T n − 2 ) ) ( t − τ ) α − 1 d τ .

With S ( 0 ) = S 0 , V ( 0 ) = V 0 , E ( 0 ) = E 0 , I 1 ( 0 ) = I 1 0 , I 2 ( 0 ) = I 2 0 , R ( 0 ) = R 0 , T ( 0 ) = T 0 .

We calculate the norm of the aforementioned system’s initial equation throughout, and then

(4.8) ‖ ϕ 1 n ( t ) ‖ = ‖ S n ( t ) − S n − 1 ( t ) ‖ = 1 Γ ( α ) ∫ 0 t ( K 1 ( τ , S n − 1 ) − K 1 ( τ , S n − 2 ) ) ( t − τ ) α − 1 d τ ≤ 1 Γ ( α ) ∑ 0 t ‖ ( K 1 ( τ , S n − 1 ) − K 1 ( τ , S n − 2 ) ) ( t − τ ) α − 1 d τ ‖ .

Possessing Lipschitz’s condition 4.5, we have

(4.9) ‖ ϕ 1 n ( t ) ‖ ≤ 1 Γ ( α ) M 1 ∫ 0 t ‖ ϕ 1 n − 1 ( τ ) ‖ d τ

In a similar manner, we obtain

(4.10) ‖ ϕ 2 n ( t ) ‖ ≤ 1 Γ ( α ) M 2 ∫ 0 t ‖ ϕ 2 n − 1 ( τ ) ‖ d τ ‖ ϕ 3 n ( t ) ‖ ≤ 1 Γ ( α ) M 3 ∫ 0 t ‖ ϕ 3 n − 1 ( τ ) ‖ d τ ‖ ϕ 4 n ( t ) ‖ ≤ 1 Γ ( α ) M 4 ∫ 0 t ‖ ϕ 4 n − 1 ( τ ) ‖ d τ ‖ ϕ 5 n ( t ) ‖ ≤ 1 Γ ( α ) M 5 ∫ 0 t ‖ ϕ 5 n − 1 ( τ ) ‖ d τ ‖ ϕ 6 n ( t ) ‖ ≤ 1 Γ ( α ) M 6 ∫ 0 t ‖ ϕ 6 n − 1 ( τ ) ‖ d τ ‖ ϕ 7 n ( t ) ‖ ≤ 1 Γ ( α ) M 7 ∫ 0 t ‖ ϕ 7 n − 1 ( τ ) ‖ d τ .

Consequently, we may write

□ S n ( t ) = ∑ i = 1 n ϕ 1 i ( t ) , V n ( t ) = ∑ i = 1 n ϕ 2 i ( t ) , E n ( t ) = ∑ i = 1 n ϕ 3 i ( t ) , I 1 n ( t ) = ∑ i = 1 n ϕ 4 i ( t ) , R n ( t ) = ∑ i = 1 n ϕ 5 i ( t ) , I 2 n ( t ) = ∑ i = 1 n ϕ 6 i ( t ) , T n ( t ) = ∑ i = 1 n ϕ 7 i ( t ) .

Theorem 4.3

A system of solutions described by the model 3.1 exists if there exists t 1 such that ( 1 Γ ( α ) t 1 M j ) < 1 .

Proof

From 4.9 and 4.10, we have

(4.11) ‖ ϕ 1 n ( t ) ‖ ≤ ‖ S n ( 0 ) ‖ 1 Γ ( α ) t M 1 n ‖ ϕ 3 n ( t ) ‖ ≤ ‖ V n ( 0 ) ‖ 1 Γ ( α ) t M 2 n ‖ ϕ 4 n ( t ) ‖ ≤ ‖ E n ( 0 ) ‖ 1 Γ ( α ) t M 3 n ‖ ϕ 5 n ( t ) ‖ ≤ ‖ I 1 n ( 0 ) ‖ 1 Γ ( α ) t M 4 n ‖ ϕ 6 n ( t ) ‖ ≤ ‖ R n ( 0 ) ‖ 1 Γ ( α ) t M 5 n ‖ ϕ 7 n ( t ) ‖ ≤ ‖ I 2 n ( 0 ) ‖ 1 Γ ( α ) t M 6 n ‖ ϕ 7 n ( t ) ‖ ≤ ‖ T n ( 0 ) ‖ 1 Γ ( α ) t M 7 n .

Therefore, the system is continuous and has a solution. We will now demonstrate how to create a model solution using the functions mentioned above for 3.1. Let us assume the following:

(4.12) S ( t ) − S ( 0 ) = S n ( t ) − Q 1 n ( t ) V ( t ) − V ( 0 ) = V n ( t ) − Q 2 n ( t ) E ( t ) − E ( 0 ) = E n ( t ) − Q 3 n ( t ) I 1 ( t ) − I 1 ( 0 ) = I 1 n ( t ) − Q 4 n ( t ) R ( t ) − R ( 0 ) = R n ( t ) − Q 5 n ( t ) I 2 ( t ) − I 2 ( 0 ) = I 2 n ( t ) − Q 6 n ( t ) T ( t ) − T ( 0 ) = T n ( t ) − Q 7 n ( t ) .

Hence,

‖ Q 1 n ( t ) ‖ = 1 Γ ( α ) ∫ 0 t ( K 1 ( τ , S ) − K 1 ( τ , S n − 1 ) ) d τ ≤ 1 Γ ( α ) ∫ 0 t ‖ ( K 1 ( τ , S ) − K 1 ( τ , S n − 1 ) ) d τ ‖ ≤ 1 Γ ( α ) t M 1 ‖ ( S − S n − 1 ) ‖ .

Repeating the process yields

‖ Q 1 n ( t ) ‖ ≤ 1 Γ ( α ) t n + 1 K 1 n + 1 k .

‖ Q 1 n ( t ) ‖ → 0 as n → ∞ .

Similarly, we may ascertain that ‖ Q 1 n ( t ) ‖ → 0 , j = 2 , 3 , 4 , 5 , 6 , 7 as n → ∞ .

We make the assumption that there is another solution to the problem to assess the solution’s uniqueness such as S 1 ( t ) , V 1 ( t ) , E 1 ( t ) , I 1 ( t ) , R 1 ( t ) , I 1 ( t ) , T 1 ( t ) .

Then

S ( t ) − S 1 ( t ) = 1 Γ ( α ) ∫ 0 t ( K 1 ( τ , S ) − K 1 ( τ , S n − 1 ) ) d τ .

Taking norm, we have

‖ S ( t ) − S 1 ( t ) ‖ = 1 Γ ( α ) ∫ 0 t ‖ ( K 1 ( τ , S ) − K 1 ( τ , S n − 1 ) ) d τ ‖ .

From Lipschitz condition 4.5,

‖ S ( t ) − S 1 ( t ) ‖ ≤ 1 Γ ( α ) t K 1 ‖ S ( t ) − S 1 ( t ) ‖ ,

Therefore,

(4.13)□ ‖ S ( t ) − S 1 ( t ) ‖ 1 − 1 Γ ( α ) t K 1 ∣ ≤ 0 .

Theorem 4.4

The model system 3.1 has a unique solution, provided that

(4.14) 1 − 1 Γ ( α ) t K 1 ∣ > 0 .

Proof

Assuming that condition 4.13 holds,

‖ S ( t ) − S 1 ( t ) ‖ 1 − 1 Γ ( α ) t K 1 ∣ ≤ 0 .

Then ‖ S ( t ) − S 1 ( t ) ‖ = 0 . So, we have S ( t ) = S 1 ( t ) . Similarly, we can prove that for V ( t ) = V 1 ( t ) , E ( t ) = E 1 ( t ) , I 1 ( t ) = I 1 1 ( t ) , R ( t ) = R 1 ( t ) , I 2 ( t ) = I 2 1 ( t ) , T ( t ) = T 1 ( t ) .□

5 Parameter estimation/numerical simulations

It is important to estimate the parameter values to carry out numerical analysis. We consider Kenya where hepatitis B is endemic [8]. The life expectancy in Kenya is 62.88 [15]; therefore, the natural death rate is 1 62.88 × 365 day − 1 . The vaccination parameter σ lies between 0 and 1 (that is 0 ≤ σ ≤ 1 ) since it is a proportion of the susceptible class. The natural recovery period for an acutely infected individual is 30–180 days [10]; therefore, ε = 1 180 day − 1 and 0 ≤ ρ ≤ 1 . From [10], we obtained the parameter values for Λ , η , δ 1 , δ 2 . The parameters β 1 and β 2 are obtained by fitting the model to data. We do parameter estimation using the least-squares approach, which is provided by the MATLAB optimization toolbox function fminsearch.

The variable values used are S = 1,000 , V = 15 , E = 50 , I 1 = 60 , R = 10 , I 2 = 30 , and T = 5 (Table 2).

Table 2

Estimated values of parameters

Parameters	Value	Source
Λ	0.0260	[10]
σ	0.5	Estimated
β 1	0.0055	Fitted
β 2	0.000055	Fitted
γ	0.00556	Estimated
ε ρ	3.6	[10]
δ 1	0.0063	[10]
η	0.025	[10]
δ 2	0.051	[10]
μ	0.0000044	Estimated

6 Discussion and conclusion

In Kenya, the prevalence of the HBV makes its control an important public health concern and the simulation is to show that control strategies (vaccination and treatment) can be used to curb the spread of the virus.

Figure 2 shows the dynamics of the susceptible population when the vaccination parameter is 0 and when it was increased to 0.5. It can be observed that if 0.5% of the susceptible population is vaccinated, there will be a significant decrease in contact rates β 1 and β 2 between the susceptible population and the infected population. As a result, the virus is eradicated and the success of long-term vaccinations is predictable. As the vaccination parameter in Figure 2a is σ = 0 , we can therefore observe a notable drop in the susceptible class.

$Figure 2 Evolutions of the susceptible population with time for different values of α \alpha with σ = 0 \sigma =0 and σ = 0.5 \sigma =0.5 , respectively. Source: Created by the authors.$

Figure 2

Evolutions of the susceptible population with time for different values of α with σ = 0 and σ = 0.5 , respectively. Source: Created by the authors.

When the contact rate between the susceptible population and the acutely infected population was equal to zero, Figure 3a shows a slow increase in the exposed population. The implementation of the vaccination strategy is expected to result in a noteworthy reduction in β 1 , hence mitigating the transmission of the HBV. Due to an increase in the contact rate between the susceptible and the acutely infected population in Figure 3b, β 1 = 0.00055 , the exposed population has significantly increased.

$Figure 3 The density of the exposed population with time for different values of α \alpha with β 1 = 0 {\beta }_{1}=0 and β 1 = 0.00055 {\beta }_{1}=0.00055 , respectively. Source: Created by the authors.$

Figure 3

The density of the exposed population with time for different values of α with β 1 = 0 and β 1 = 0.00055 , respectively. Source: Created by the authors.

Figure 4 shows the evolution of the acutely infected population when the proportion of the recovered individuals is zero and one. Acutely sick people may recover in 3–6 months if their immune systems are extremely robust. A decrease in our resistance to HBV might result from consuming large amounts of alcohol, smoking, and using strong medications. Figure 4a illustrates the significant progression from the actually infected population to the recovered population that will occur if these items are avoided. If not, Figure 4b illustrates the outcome that would occur.

$Figure 4 The density of the acutely infected population with time for different values of α \alpha for ρ = 1 \rho =1 and ρ = 0 \rho =0 , respectively. Source: Created by the authors.$

Figure 4

The density of the acutely infected population with time for different values of α for ρ = 1 and ρ = 0 , respectively. Source: Created by the authors.

Figure 5 illustrates the dynamics of the chronically infected population when the treatment parameter is 0.025 and 0. Figure 5a shows that if the percentage of treatment among individuals with chronic HBV increases, there will be a significant decline in that class and a consequent significant reduction in HBV-related deaths. Although a person with chronic HBV will never fully recover, treatment will enable them to live longer than a person without treatment. The scenario shown in Figure 5b occurs when individuals with chronic HBV decline to receive treatment.

$Figure 5 Dynamics of the chronically infected population with time for different values of α \alpha for η = 0.025 \eta =0.025 and η = 0 \eta =0 . Source: Created by the authors.$

Figure 5

Dynamics of the chronically infected population with time for different values of α for η = 0.025 and η = 0 . Source: Created by the authors.

In summary, this article examines the existence and uniqueness of a fractional order solution to a HBV model in the Caputo sense. The simulation results also demonstrated that implementing control strategies, i.e., vaccination and treatment, is the most effective strategy to stop the spread of HBV. In addition, the simulation findings demonstrate that the fractional order model outperforms the classical case in terms of results.

Funding information: The authors state no funding involved.
Author contributions: Nnaemeka Stanley Aguegboh: Conceptualization; Formal analysis; Validation; Visualization; Writing – original draft; and Writing. Oranugo Dominic Obinna: Conceptualization; Validation; review & editing. Phineas Roy Kiogora: Conceptualization; Supervision; review & editing. Felix Mutua: Conceptualization; Validation; review & editing. Netochukwu Onyiaji: Conceptualization; Validation; review & editing. Andrew Egwu Onyeka: Conceptualization; Validation; review & editing.
Conflict of interest: The authors have no conflicts of interest to disclose.
Ethical approval: The conducted research is not related to either human or animal use.
Data availability statement: Data sharing is not applicable to this article as no datasets were generated or analysed during the current study.

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Received: 2024-04-22

Revised: 2024-07-11

Accepted: 2024-07-23

Published Online: 2025-03-07

This work is licensed under the Creative Commons Attribution 4.0 International License.

Articles in the same Issue

https://doi.org/10.1515/cmb-2024-0009

Keywords for this article

Hepatits B; fractional-order differential equations; parameter estimation; numerical simulations; predictor-corrector method

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