Mathematical model and analysis of hepatitis B virus transmission dynamics

Mathematical model

A variety of mathematical models exist, such as the SIR, SIS, SIRS, and their variations; where S=Susceptible class, I=Infective class, and R=Recovered class. The model used in this study takes the form of SEIR based on ordinary differential equations, which shall be solved to obtain the disease-free equilibrium (DFE) state.

Governing equations

Table 1 lists the parameters used. Figure 1 shows a schematic presentation of the model.

Figure 1. Schematic presentation of the model.

We formulate the HB transmission model as follows:

$\frac{dM(t)}{dt}=cP-\varphi {\rm M}(t)-\beta M(t)$                                                                                                                                                           (1)

$\frac{dS(t)}{dt}=(1-c)P+\varphi M\text{(}t\text{)}+\pi R\text{(}t\text{)}-(kI\text{(}t\text{)}+\beta )S\text{(}t\text{)}$                                                                                                                    (2)

$\frac{dL\text{(}t\text{)}}{dt}=kS\text{(}t\text{)}I\text{(}t\text{)}-qL\text{(}t\text{)}-\mu L\text{(}t\text{)}-\beta L\text{(}t\text{)}$                                                                                                                                      (3)

$\frac{dI\text{(}t\text{)}}{dt}=\mu L\text{(}t\text{)}-\psi I\text{(}t\text{)}-\eta I\text{(}t\text{)}-\beta I\text{(}t\text{)}$                                                                                                                                              (4)

$\frac{dR\text{(}t\text{)}}{dt}=qL\text{(}t\text{)}-\psi I\text{(}t\text{)}-\pi R\text{(}t\text{)}-\beta R\text{(}t\text{)}$                                                                                                                                           (5)

$N\text{(}t\text{)}=M\text{(}t\text{)}+S\text{(}t\text{)}+L\text{(}t\text{)}+I\text{(}t\text{)}+R\text{(}t\text{)}$                                                                                                                                        (6)

where M = immunized individuals, S = susceptible, L = latently infected/exposed, I = infectious individuals, and R = recovered.

Equilibrium solutions

Let E(M, S, L, I, R) be the equilibrium point of the system described by (1)–(6), At the equilibrium state, we have

i.e.,

                                                                                                                                                              cP – 𝜙M – βM = 0                 (7)

                                                                                                                                                              cP – (ϕ + β)M = 0                  (8)

                                                                                                                                     (1 – c)P + ϕM + πR – kSI – βS = 0                 (9)

                                                                                                                                    (1 – c)P + ϕM + πR – (kI + β)S = 0               (10)

                                                                                                                                                       kSI – qL – μL – βL = 0                (11)

                                                                                                                                                         kSI – (q + μ + β)L = 0               (12)

                                                                                                                                                           μL – 𝜓I – ηI – βI = 0               (13)

                                                                                                                                                          μL – (𝜓 + η + β)I = 0                (14)

                                                                                                                                                         qL – 𝜓I – πR – βR = 0               (15)

                                                                                                                                                         qL + 𝜓I – (π + β)R = 0               (16)

In order to obtain the DFE state we solve equation (7)–equation (15) simultaneously.

Existence of a trivial equilibrium state (TES)

Let E_o (M_o, S_o, L_o, I_o, R_o) be TES of (1)–(6) of the model, ∃ no TES since the population cannot be extinct, so long as new babies are born into the population (i.e. cP ≠ 0 and (1 − c)P ≠ 0).

That is, E_o (M_o, S_o, L_o, I_o, R_o) ≠ (0, 0, 0, 0, 0)

DFE state

DFE state is the state of total eradication of disease. Let E^o (M^o, S^o, L^o, I^o, R^o) be the DFE state. Suppose, both I and L must be zero. That is, for DFE state:

                                                                                                                      I^o = L^o = 0                                                                   (17)

Substituting I − L = 0 into equation (7)–equation (11) and solving simultaneously we have:

From Equation (7):

                                                                                                                                      cP − (ϕ + β)M = 0

                                                                                                                                    $M^o=\frac{cP}{\varphi +\beta }(18)$

From Equation (9)

                                                                                                                 $(1-c)P+\frac{\varphi cP}{\varphi +\beta }+\pi R-\beta S=0(19)$

From Equation (11)

                                                                                                                                   qL + 𝜓I − (π + β)R = 0

                                                                                                                    ⇒ (π + β)R = 0(since L = I = 0)                                      (20)

                                                                                                                     ⇒ Either (π + β) = 0 Or R = 0                                        (21)

Since π and β are positive constants, (π + β) ≠ 0.

Therefore, R^o = 0.

If R = 0, Equation (9) becomes

or

                                                                                                                                  $S^o=\frac{\varphi +\beta -c\beta )P}{\beta (\varphi +\beta )}(22)$

Therefore the DFE state of the model is

Stability analysis of the DFE state

To determine the stability of the DFE state E^o, we examine the behavior of the model population near this equilibrium solution. Here, we determine condition(s) that must be met if the disease is to be totally eradicated.

Recall that the system of equations in this model at equilibrium state is:

                                                                                                                                                      cP – (ϕ + β)M = 0

                                                                                                                            (1 – c)P + ϕM + πR – (kI + β)S = 0

                                                                                                                                                 kSI – (q + μ + β)L = 0

                                                                                                                                                  μL – (𝜓 + β + η)I = 0                        (23)

                                                                                                                                                qL + 𝜓I – (π + β)R = 0

We now linearize the system of equations to get the Jacobian matrix J.

                                                                                                                        $J=\left[\begin{array}{c}{\omega }_1\\ \varphi \\ 0\\ 0\\ 0\end{array}\begin{array}{c}0\\ {\omega }_2\\ k{{\rm I}}^o\\ 0\\ 0\end{array}\begin{array}{c}0\\ 0\\ {\omega }_3\\ \mu \\ q\end{array}\begin{array}{c}0\\ -k{S}^o\\ k{S}^o\\ {\omega }_4\\ \psi \end{array}\begin{array}{c}0\\ \pi \\ 0\\ 0\\ {\omega }_5\end{array}\right](24)$

where

ω₁ = −(ϕ + β), ω₂ = −(kI^o + β), ω₃ = −(q + μ + β), ω₄ = −(𝜓 + β + η), ω₅ = −(π + β)

At the disease-free equilibrium, E^o (M^o, S^o, L^o, I^o, R^o), the Jacobian Matrix becomes

                                                                                                                       $J=\left[\begin{array}{c}{\omega }_6\\ \varphi \\ 0\\ 0\\ 0\end{array}\begin{array}{c}0\\ -\beta \\ k{{\rm I}}^o\\ 0\\ 0\end{array}\begin{array}{c}0\\ 0\\ {\omega }_8\\ \mu \\ q\end{array}\begin{array}{c}0\\ {\omega }_7\\ {\omega }_9\\ {\omega }_{10}\\ \psi \end{array}\begin{array}{c}0\\ \pi \\ 0\\ 0\\ {\omega }_{11}\end{array}\right](25)$

where

ω₆ = −(𝜙 + β), ω₇ = $-k\frac{(\varphi +\beta -c\beta )P}{\beta (\varphi +\beta )},{\omega }_8$= −(q + μ + β), ω₉ = $k\frac{(\varphi +\beta -c\beta )P}{\beta (\varphi +\beta )},{\omega }_{10}$ = −(𝜓 + β + η), ω₁₁ = (q + μ + β + λ) > −(π + β)

The characteristic equation |J_o − Iλ| = 0 is obtained from the Jacobian determinant with the Eigen values λ_i (i = 1, 2, 3, 4, 5)

                                                                                             = (λ² + (ϕ + 2β)λ + (ϕβ + β²))(– π – β – λ)[X] = 0                              (26)

where

From Equation (26), either

                                                                                                  (λ² + (ϕ + 2β)λ + (ϕβ + β²))(– π – β – λ) = 0                                 (27)

or

                                                                                                     $\left[\begin{array}{c}-(q+\mu +\beta )-\lambda \\ \mu \end{array}\begin{array}{c}-k\frac{(\varphi +\beta -c\beta )P}{\beta (\varphi +\beta )}\\ -(\psi +\beta +\eta )-\lambda \end{array}\right]=0(28)$

From Equation (27), we deduce

                                                                                                                                λ₁ = –(π + β)                                                     (29)

                                                                                                                                λ₂ = –β                                                             (30)

                                                                                                                                and

                                                                                                                                λ₃ = –(ϕ + β)                                                    (31)

Let

For the DFE to be asymptotically stable, trace(A) < 0 and det A > 0.

And the trace of A is

                                                                                                                                Trace(A) = –(q + μ + β + λ) – (𝜓 + β + η + λ)

Obviously, trace(A) < 0 since all the parameters q, t, β, 𝜓, β and η are positive.

For the determinant of A to be positive, we must have

or

From equation (29)–equation (31), λ₁, λ₂, λ₃ of (25) all have negative real parts. We now establish the necessary and sufficient conditions for the remaining two Eigen values of (25) to have negative real part. The remaining two Eigen values of equation (25) will have negative real part if and only if det A > 0. i.e.

The Routh-Hurwitz theorem states that the equilibrium state will be asymptotically stable if and only if all the Eigen values of the characteristic equation |J − Iλ| = 0 have negative real part. Using this theorem we see that the DFE of this model will be asymptotically stable if and only if

or

The inequality (32) gives the condition, necessary and sufficient for the DFE state of the model to be stable (asymptotically). This means that the product of total contraction and total breakdown of latent class given by $k\mu \left(\frac{(\varphi +\beta -c\beta )P}{\beta (\varphi +\beta )}\right)$ must be less than the total removal rate from both latent and infectious classes given by (q + µ + β + λ)(𝜓 + β + η).

Alternatively, the inequality (32) can also be expressed as

The inequality (33) also gives the condition necessary and sufficient for the stability of DFE state, thus sum of the rate of recovery of latently infected people, the rate at which latently infected individuals progress to active infection and the rate of natural death of individuals (in the population, i.e. total removal rate from the latent class) must have a lower bound given by