This one is crucial for the explanation of positivity and boundedness of model (6) as per they designate population. Population survey denoted by positivity and boundedness might be understood as an usual limitation byway for growing by means of an acceptance of inadequate resources. Here, some key results for the existence of model (6) equilibrium, positive invariance, and solution boundedness are presented.
Positive invariance
Rearrange model (6) in terms explained by
$$\begin{aligned} \dot{\psi }(t) =&G\bigl(\psi (t)\bigr), \end{aligned}$$
(7)
where \(\psi (t) = (\psi _{1}, \psi _{2}, \psi _{3}, \psi _{4},\psi _{5})^{T} := (S, L, I, C, R)^{T}\), \(\psi (0) = (S(0), L(0), I(0), C(0), R(0))^{T} \in R^{5}_{+}\)
$$ G(\psi )= \begin{pmatrix} G_{1}(\psi ) \\ G_{2}(\psi ) \\ G_{3}(\psi ) \\ G_{4}(\psi ) \\ G_{5}(\psi ) \end{pmatrix} = \begin{pmatrix} \mu \omega (1-\nu C(t)) -(\mu _{0}+\beta I(t) +\epsilon \beta C(t) + \gamma _{3} )S(t) \\ (\beta I(t)+\epsilon \beta C(t))S(t) -(\mu _{0}+\sigma )L(t) \\ \sigma L(t) -(\mu _{0} +\gamma _{1})I(t) \\ \mu \omega \nu C(t) +q\gamma _{1}I(t) -(\mu _{0}+\mu _{1}+\gamma _{2})C(t) \\ \gamma _{2}C(t) +(1-q)\gamma _{1}I(t) -\mu _{0}R(t) \end{pmatrix}. $$
The situation is stress-free in order to shape \(G_{i}(\psi )|_{\psi _{i}=0} \geq 0\), \(i = 1,\ldots,5\). According to the well-known result of Nagumo [24], any model (6) solution with an initial point \(\psi _{0} \in R^{5}_{+}\), say \(\psi (t) = \psi (t; \psi _{0})\), is such that \(\psi (t) \in R^{5}_{+}\) ∀ \(t > 0\).
Boundedness
Theorem 1
There is a positive Ψ for nonzero such that all solutions meet \(\boldsymbol{\Psi } > (S(t), L(t), I(t), C(t), R(t))\) for the long time t.
Proof
All model (6) solutions are grater than zero, now in the first compartment of (6) as
$$\begin{aligned} \frac{dS(t)}{dt} =& \mu \omega (1-\nu C) -(\mu _{0}+\beta I + \epsilon \beta C +\gamma _{3} )S \leq \mu \omega (1-\nu C)-\mu _{0} S- \gamma _{3} S. \end{aligned}$$
Thus, \(\frac{dS(t)}{dt}< 1+\frac{\mu \omega }{\mu _{0}+\gamma _{3}}\) for excessive time t, let us say \(t>t_{0}\). Set \(R_{1}(t)=(S(t)+L(t)+I(t)+C(t))\).
Differentiating \(R_{1}\) along the solutions of model (6) yields
$$\begin{aligned} \frac{dR_{1}(t)}{dt} =&-(\mu _{0}+\gamma _{3})S-\mu _{0}L-(\mu _{0}+ \gamma _{1}-q\gamma _{1})I-(\mu _{0}+\mu _{1}+\gamma _{2})C+\mu \omega \\ \leq& -hR_{1}(t)+\mu \omega , \end{aligned}$$
where \(h=\min ((\mu _{0}+\gamma _{3}),\mu _{0},(\mu _{0}+\gamma _{1}-q \gamma _{1}),(\mu _{0}+\mu _{1}+\gamma _{2}))\). Remember \(S(t)\leq \frac{\mu \omega }{\mu _{0}+\gamma _{3}}+1\) for \(t >t_{0}\). \(\boldsymbol{\Psi }_{1}\) exists, depending simply on the model parameters (6), such that \(R_{1}(t)\leq \boldsymbol{\Psi }_{1}\) for ultimately \(t>t_{0}\), \(L(t)\), \(I(t)\), and \(C(t)\) are bounded above. After the third and fourth equations of model (6) and \(R(t)\) are ultimately bounded above, let alone Ψ is the maximum. Theorem 2.1 formerly states that keep an eye on now and thus proven. This shows that model (6) is destructive.
$$\begin{aligned} \Omega =&\biggl\{ (S,L,I,C,R)|1+ \frac{\mu \omega }{\mu _{0}+\gamma _{3}} \geq S\geq 0, I\geq 0, \boldsymbol{\Psi }\geq R\biggr\} . \end{aligned}$$
Clearly, Ω is convex. □
Disease-free equilibrium and basic reproductive number
As we know that the population is infection free, we set the right-hand side of all equations to zero such as \(L = I = C = R =0\) for the purpose of determining infection-free equilibrium of the model. Directly calculating we come to the point which is a disease-free equilibrium denoted by \(E^{\mathrm{HBV}}_{0}\).
$$\begin{aligned} E^{\mathrm{HBV}}_{0}=\bigl(S^{0},L^{0},I^{0},C^{0},R^{0} \bigr)= \biggl( \frac{\mu \omega }{\mu _{0}+\gamma _{3}},0,0,0,0 \biggr). \end{aligned}$$
(8)
The basic reproduction number represented by \(R^{\mathrm{HBV}}_{0}\) is here calculated. The number of secondary infections caused by a single infection in a population that is fully susceptible is called basic number of reproductions. Infected and noninfected cells must be separated, and then the next generation matrix approach method is applied to calculate \(R^{\mathrm{HBV}}_{0}\) [25]. Infected and noninfected cells are denoted by L, I, C and S, R respectively. By picking the infection terms only, the infection-free equilibrium model (6) is presented in (8) by applying the notation in the system matrices which are used for modeling of infection terms F and V. The term F is used in the matrix to denote the right-hand side of the equation, and the rest of the terms of the equation are denoted by V respectively and are given below.
$$ F= \begin{pmatrix} (\beta I+\epsilon \beta C)S \\ 0 \\ 0 \end{pmatrix}, \qquad V= \begin{pmatrix} -(\mu _{0}+\sigma )L \\ \sigma L-(\mu _{0}+\gamma _{1})I \\ \mu \omega \nu C+q \gamma _{1} I-(\mu _{0}+\mu _{1}+\gamma _{2})C \end{pmatrix}, $$
taking the Jacobian of the above using \(E^{\mathrm{HBV}}_{0}\), we get
$$ F^{*}= \begin{pmatrix} 0 & \beta \frac{\mu \omega }{\mu _{0}+\gamma _{3}} & \epsilon \beta \frac{\mu \omega }{\mu _{0}+\gamma _{3}} \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix},\qquad V^{*}= \begin{pmatrix} -(\mu _{0}+\sigma ) & 0 & 0 \\ \sigma & -(\mu _{0}+\gamma _{1}) & 0 \\ 0 & q \gamma _{1} & \mu \omega \nu -(\mu _{0}+\mu _{1}+\gamma _{2}) \end{pmatrix}. $$
Near the equilibrium the transition and infection rates are denoted by \(F^{*}\) and \(V^{*}\). Here, the duration which is spent in each state is denoted by \({V^{*}}^{-1}\). During the pandemic the production of new infection is denoted by \(F^{*}{V^{*}}^{-1}\). The highest eigenvalue of \(F^{*}{V^{*}}^{-1}\) is the associative of the fundamental number.
$$\begin{aligned} R_{0}^{\mathrm{HBV}}=\rho \bigl(F^{*}{V^{*}}^{-1} \bigr) =&\frac{\sigma \beta \mu \omega (\epsilon q\gamma _{1}-(\mu \omega \nu -(\mu _{0}+\mu _{1}+\gamma _{2})))}{(\mu _{0}+\sigma )(\mu _{0}+\gamma _{1})(\mu _{0}+\gamma _{3})(\mu \omega \nu -(\mu _{0}+\mu _{1}+\gamma _{2}))}. \end{aligned}$$
Endemic equilibria
In setup to find the endemic equilibria \(E^{\mathrm{HBV}}_{1}\) of model (6), let us set \(S = S^{*}\), \(L = L^{*}\), \(I = I^{*}\), \(C = C^{*}\), \(R = R^{*}\) and the answering right-hand side of model (6) equals zero. The endemic equilibria denoted by \(E^{\mathrm{HBV}}_{1}\) is acquired and administered after some rearrangement
$$\begin{aligned}& {E^{\mathrm{HBV}}_{1}= \bigl(S^{*}, L^{*}, I^{*}, C^{*}, R^{*}\bigr).} \\& \textstyle\begin{cases} S^{*}= \frac{k_{2}k_{3}(k_{1}-\mu \nu \omega )}{\beta \sigma (q\gamma _{1}\epsilon +k_{1}-\mu \nu \omega )}=- \frac{S^{0}}{R^{\mathrm{HBV}}_{0}}, \\ L^{*}= \frac{-k^{2}_{2}k_{3}k_{4}(k_{1}-\mu \nu \omega )^{2}(R^{\mathrm{HBV}}_{0}+1)}{\beta \sigma (q\epsilon \gamma _{1}+k_{1}-\mu \nu \omega )(k_{1}k_{2}k_{3}-\mu \nu (k_{3}\mu _{0}+\mu _{0}\gamma _{1}\omega -\omega \sigma \gamma _{1}(1-q)))}, \\ I^{*}= \frac{-k_{2}k_{3}k_{4}(k_{1}-\mu \nu \omega )^{2}(R^{\mathrm{HBV}}_{0}+1)}{\beta (q\epsilon \gamma _{1}+k_{1}-\mu \nu \omega )(k_{1}k_{2}k_{3}-\mu \nu (k_{3}\mu _{0}+\mu _{0}\gamma _{1}\omega -\omega \sigma \gamma _{1}(1-q)))}, \\ C^{*}= \frac{-k_{2}k_{3}k_{4}(\mu \nu \omega -k_{1})(R^{\mathrm{HBV}}_{0}+1)}{\beta (q\epsilon \gamma _{1}+k_{1}-\mu \nu \omega )(k_{1}k_{2}k_{3}-\mu \nu (k_{3}\mu _{0}+\mu _{0}\gamma _{1}\omega -\omega \sigma \gamma _{1}(1-q)))}, \\ R^{*}= \frac{-k_{2}k_{3}k_{4}(\gamma _{1}(\gamma _{2}+1-q))(\mu _{0}+\mu _{1}-\mu \nu \omega )(k_{1}-\mu \nu \omega )(R^{\mathrm{HBV}}_{0}+1)}{\beta \mu _{0}(q\epsilon \gamma _{1}+k_{1}-\mu \nu \omega )(k_{1}k_{2}k_{3}-\mu \nu (k_{3}\mu _{0}+\mu _{0}\gamma _{1}\omega -\omega \sigma \gamma _{1}(1-q)))}, \end{cases}\displaystyle \end{aligned}$$
(9)
where
$$\begin{aligned}& k_{1}=\gamma _{2}+\mu _{0}+\mu _{1},\qquad k_{2}=\gamma _{1}+\mu _{0}, \qquad k_{3}=\mu _{0}+\sigma \quad \text{and} \quad k_{4}=\gamma _{3}+\mu _{0}. \end{aligned}$$
Local stability and global stability of \(E^{\mathrm{HBV}}_{0}\)
Theorem 2
Providing the condition \(R_{0}^{\mathrm{HBV}}<1\), the disease-free equilibrium point \(E_{0}^{\mathrm{HBV}}\) is locally asymptotically stable and unstable at the condition \(R_{0}^{\mathrm{HBV}}>1\).
Proof
The system’s Jacobian matrix of (6) becomes
$$ J_{1}= \begin{pmatrix} -(\mu _{0}+\gamma _{3}) & 0 & -\beta S^{0} & -\mu \omega \nu - \epsilon \beta S^{0} & 0 \\ 0 & -(\mu _{0}+\sigma ) & \beta S^{0} & \epsilon \beta S^{0} & 0 \\ 0 & \sigma & -(\mu _{0}+\gamma _{1}) & 0 & 0 \\ 0 & 0 & q\gamma _{1} & \mu \omega \nu -(\mu _{0}+\mu _{1}+\gamma _{2}) & 0 \\ 0 & 0 & (1-q)\gamma _{1} & \gamma _{2} & -\mu _{0} \end{pmatrix}. $$
By applying the \(E_{0}^{\mathrm{HBV}}\), we have the following matrix:
$$ J_{1}\bigl(E_{0}^{\mathrm{HBV}}\bigr)= \begin{pmatrix} -(\mu _{0}+\gamma _{3}) & 0 & - \frac{\beta \mu \omega }{\mu _{0}+\gamma _{3}} & - \frac{(\mu _{0}+\gamma _{3})\mu \omega \nu -\epsilon \mu \omega }{\mu _{0}+\gamma _{3}} & 0 \\ 0 & -(\mu _{0}+\sigma ) & \frac{\beta \mu \omega }{\mu _{0}+\gamma _{3}} & \frac{\epsilon \beta \mu \omega }{\mu _{0}+\gamma _{3}} & 0 \\ 0 & \sigma & -(\mu _{0}+\gamma _{1}) & 0 & 0 \\ 0 & 0 & q\gamma _{1} & \mu \omega \nu -(\mu _{0}+\mu _{1}+\gamma _{2}) & 0 \\ 0 & 0 & (1-q)\gamma _{1} & \gamma _{2} & -\mu _{0} \end{pmatrix}. $$
Definitely eigenvalues \(\lambda _{1}=-(\mu _{0}+\gamma _{3})\) and \(\lambda _{2}=-\mu _{0}\) are negative, for the remaining eigenvalues, we are going to take the following \(3\times 3\) matrix:
$$ \det \begin{pmatrix} -\mathfrak{X}_{1}-\lambda &\frac{\beta \mu \omega }{\mathfrak{X}_{2}} & \frac{\epsilon \beta \mu \omega }{\mathfrak{X}_{2}} \\ \sigma & -\mathfrak{X}_{4}-\lambda & 0 \\ 0 & q\gamma _{1} & \mathfrak{X}_{3}-\lambda \end{pmatrix} , $$
(10)
where
\(\mathfrak{X}_{1}=(\mu _{0}+\sigma )\), \(\mathfrak{X}_{2}=(\mu _{0}+\gamma _{3})\), \(\mathfrak{X}_{3}=(\mu \omega \nu -(\mu _{0}+\mu _{1}+\gamma _{2}))\), and \(\mathfrak{X}_{4}=(\mu _{0}+\gamma _{1})\).
To determine the nature of eigenvalues in (10), we present the conditions of Routh–Hurwitz for all roots characteristic polynomial to have negative parts:
$$ \left \{ \lambda ^{3}+[\mathfrak{X}_{1}+ \mathfrak{X}_{4}-\mathfrak{X}_{3}] \lambda ^{2} + \biggl[\frac{\sigma \beta \mu \omega }{\mathfrak{X}_{2}}+ \mathfrak{X}_{3} \mathfrak{X}_{4}-\mathfrak{X}_{1}(\mathfrak{X}_{3}+ \mathfrak{X}_{4})\biggr]\lambda +\mathfrak{X}_{1} \mathfrak{X}_{3} \mathfrak{X}_{4}\bigl(R^{\mathrm{HBV}}_{0}-1 \bigr)=0. \right . $$
Let
$$\begin{aligned} &\tau _{1}=(\mathfrak{X}_{1}+\mathfrak{X}_{4}- \mathfrak{X}_{3}), \\ &\tau _{2}=\biggl[\frac{\sigma \beta \mu \omega }{\mathfrak{X}_{2}}+ \mathfrak{X}_{3} \mathfrak{X}_{4}-\mathfrak{X}_{1}(\mathfrak{X}_{3}+ \mathfrak{X}_{4})\biggr], \\ &\tau _{3}=\mathfrak{X}_{1} \mathfrak{X}_{3} \mathfrak{X}_{4}\bigl(R^{\mathrm{HBV}}_{0}-1\bigr). \end{aligned}$$
Then the equation becomes
$$\begin{aligned}& \lambda ^{3}+ \tau _{1}\lambda ^{2}+ \tau _{2}\lambda +\tau _{3} = 0, \\& \tau _{1}>0\quad \text{if} \\& \mathfrak{X}_{1}+\mathfrak{X}_{4}- \mathfrak{X}_{3}>0\quad \text{if} \\& \mathfrak{X}_{1}+\mathfrak{X}_{4}> \mathfrak{X}_{3}, \\& \tau _{2}>0\quad \text{if} \\& \frac{\sigma \beta \mu \omega }{\mathfrak{X}_{2}}+\mathfrak{X}_{3} \mathfrak{X}_{4}- \mathfrak{X}_{1}(\mathfrak{X}_{3}+\mathfrak{X}_{4})>0, \end{aligned}$$
and
$$\begin{aligned}& \tau _{3}>0\quad \text{if} \\& \mathfrak{X}_{1} \mathfrak{X}_{3} \mathfrak{X}_{4} \bigl(R^{\mathrm{HBV}}_{0}-1\bigr)>0. \end{aligned}$$
Moreover, we will show that \(\tau _{1}\tau _{2}>\tau _{3}\)
$$\begin{aligned}& [\mathfrak{X}_{1}+\mathfrak{X}_{4}- \mathfrak{X}_{3}] \biggl[ \frac{\sigma \beta \mu \omega }{\mathfrak{X}_{2}}+\mathfrak{X}_{3} \mathfrak{X}_{4}-\mathfrak{X}_{1}(\mathfrak{X}_{3}+ \mathfrak{X}_{4})\biggr]> \mathfrak{X}_{1} \mathfrak{X}_{3} \mathfrak{X}_{4}\bigl(R^{\mathrm{HBV}}_{0}-1 \bigr). \end{aligned}$$
Thus, according to the Routh–Hurwitz criterion, system (6) is locally asymptotically stable if \(\tau _{1}>0\), \(\tau _{3}>0\), and \(\tau _{1}\tau _{2}>\tau _{3}\). \(E_{0}^{\mathrm{HBV}}\) is therefore locally asymptotically stable. □
Theorem 3
If \(R^{\mathrm{HBV}}_{0} < 1\), then model (6) is globally asymptotically stable at disease-free equilibrium \(E^{\mathrm{HBV}}_{0} = (S^{0},L^{0},I^{0},C^{0},R^{0})\) and unstable otherwise.
Proof
In order to demonstrate the global stability of model (6) at \(E_{0}^{\mathrm{HBV}}\), we construct the Lyapunov function
$$ V(t)=\bigl(S-S^{0}\bigr)+L+I+C. $$
(11)
Now we calculate the time derivative of equation (11) and then using model (6), we get
$$ \begin{gathered} \frac{dV}{dt}=\frac{dS}{dt}+ \frac{dL}{dt}+\frac{dI}{dt}+ \frac{dC}{dt}, \\ \frac{dV}{dt}=\mu \omega -(\mu _{0}+\gamma _{3})S-\mu _{0}(L+I+C)-(1-q) \gamma _{1}I-( \mu _{1}+\gamma _{2}), \\ \frac{dV}{dt}=-(\mu _{0}+\gamma _{3}) \bigl(S-S^{0}\bigr)-\mu _{0}(L+I+C)-(1-q) \gamma _{1}I-(\mu _{1}+\gamma _{2})C, \\ \frac{dV}{dt}=-\bigl[(\mu _{0}+\gamma _{3}) \bigl(S-S^{0}\bigr)+\mu _{0}(L+I+C)+(1-q) \gamma _{1}I+(\mu _{1}+\gamma _{2})C\bigr]< 0. \end{gathered} $$
(12)
Thus, \(\frac{dV}{dt} < 0\) if \(R_{0}^{\mathrm{HBV}}<1\). Also \(\frac{dV}{dt} = 0\) if and only if \(S = S^{0}\) and L = I = C = 0. Thus the invariant principle of LaSalle’s [26] implies that \(E_{0}^{\mathrm{HBV}}\) is asymptotically stable globally. This completes the proof. □
Local stability and global stability of \(E^{\mathrm{HBV}}_{1}\)
Theorem 4
Model (6) is locally asymptotically stable at the endemic equilibrium \(E^{\mathrm{HBV}}_{1} = (S^{*}, L^{*}, I^{*}, C^{*}, R^{*})\) if \(R^{\mathrm{HBV}}_{0}>1\), otherwise unstable.
Proof
This can be demonstrated by linearizing system (6) around \(E^{\mathrm{HBV}}_{1} = (S^{*}, L^{*}, I^{*}, C^{*}, R^{*})\), making an elementary row transformation for the Jacobian matrix at \(E^{\mathrm{HBV}}_{1}\), and obtaining the following matrix:
$$ J_{1}= \begin{pmatrix} -\mu _{0}-\beta I^{*}-\epsilon \beta C^{*}-\gamma _{3} & 0 & -\beta S^{*} & -\mu \omega \nu -\epsilon \beta S^{*} & 0 \\ 0 & -\mu _{0}-\sigma & N_{1} & \epsilon N_{1}-\mu \omega \nu N_{2} & 0 \\ 0 & 0 & -(\mu _{0}+\gamma _{1})+\frac{\sigma N_{1}}{\mu _{0}+\sigma } & \frac{\sigma (\epsilon N_{1}-\mu \omega \nu N_{2})}{\mu _{0}+\sigma } & 0 \\ 0 & 0 & 0 & N_{3}-\mu \omega \nu N_{4} & 0 \\ 0 & 0 & 0 & 0 & -\mu _{0} \end{pmatrix}, $$
where
$$\begin{aligned} \begin{aligned} N_{1}={}& \frac{\beta S^{*}(\mu _{0}+\gamma _{3})}{\mu _{0}+\beta I^{*}+\epsilon \beta C^{*}+\gamma _{3}}, \\ N_{2}={}& \frac{\beta (I^{*}+\epsilon \beta C^{*})}{\mu _{0}+\beta I^{*}+\epsilon \beta C^{*}+\gamma _{3}}, \\ N_{3}={}&\mu \omega \nu -(\mu _{0}+\mu _{1}+ \gamma _{2})\\ &{}- \frac{q\gamma _{1} \sigma \beta \epsilon S^{*}(\gamma _{3}+\mu _{0})}{\sigma \beta S^{*}(\mu _{0}+\gamma _{3})-(\gamma _{1}+\mu _{0}) (\mu _{0}+\sigma )(\mu _{0}+\beta I^{*}+\epsilon \beta C^{*}+\gamma _{3})}, \\ N_{4}={}&{-} \frac{q\gamma _{1}\sigma \beta (I^{*}+\epsilon C^{*})}{\sigma \beta S^{*}(\mu _{0}+\gamma _{3})-(\mu _{0}+\gamma _{1}) (\mu _{0}+\sigma )(\mu _{0}+\beta I^{*}+\epsilon \beta C^{*}+\gamma _{3})}. \end{aligned} \end{aligned}$$
The eigenvalues are
$$\begin{aligned} \begin{gathered} \lambda _{1} = -\mu _{0}- \beta I^{*}-\epsilon \beta C^{*}- \gamma _{3} < 0, \qquad \lambda _{2} = -\mu _{0}-\sigma < 0, \\ \lambda _{3} = -(\mu _{0}+\gamma _{1})+ \frac{\sigma N_{1}}{\mu _{0}+\sigma }, \qquad \lambda _{4} = N_{3}-\mu \omega \nu N_{4}, \\ \lambda _{5} = -\mu _{0}< 0. \end{gathered} \end{aligned}$$
Since the endemic \(E^{\mathrm{HBV}}_{1}\) equilibrium coordinates are \(I^{*}\), \(C^{*}\), and \(L^{*}\), we have
$$\begin{aligned} \begin{aligned} \mu _{0}+\sigma =\bigl(\beta I^{*}+\epsilon \beta C^{*}\bigr) \frac{S^{*}}{L^{*}}, \qquad \mu _{0}+\gamma _{1} =\sigma \frac{L^{*}}{I^{*}}. \end{aligned} \end{aligned}$$
Thus, \(\lambda _{3}<0\) if and only if
$$\frac{(\mu _{0}+\gamma _{3})}{(\mu _{0}+\beta I^{*}+\epsilon \beta C^{*}+\gamma _{3})(I^{*}+\epsilon C^{*})} < 1, $$
which is equivalent to
$$(\mu _{0}+\gamma _{3})\epsilon C^{*}+\beta \bigl(I^{*}+\epsilon C^{*}\bigr)^{2}>0. $$
It holds as long as the endemic equilibrium \(E^{\mathrm{HBV}}_{1}\) is existent. Furthermore, \(C^{*}\) and \(I^{*}\) are satisfied with
$$q\gamma _{1}=\frac{C^{*}}{I^{*}}(\mu _{0}+\mu _{1}+\gamma _{2}- \mu \omega \nu ). $$
Similar to the \(\lambda _{3}\) proof, we receive that \(\lambda _{4}<0\) if and only if
$$\bigl(I^{*}+\epsilon C^{*}\bigr)S^{*}\beta + \mu \omega \nu >0, $$
which holds as long as there is \(E^{\mathrm{HBV}}_{1}\). Therefore, all eigenvalues are negative, and we have the following conclusion on the disease endemic equilibrium \(E^{\mathrm{HBV}}_{1}\). □
Theorem 5
The endemic equilibrium state \(E^{\mathrm{HBV}}_{1} = (S^{*},L^{*}, I^{*}, C^{*}, R^{*})\) of model (6) is globally asymptotically stable if \(R^{\mathrm{HBV}}_{0} < 1\), otherwise unstable.
Proof
We defined the Lyapunov function in order to prove the global stability of model (6) at the endemic equilibrium point \(E^{\mathrm{HBV}}_{1} = (S^{*},L^{*}, I^{*}, C^{*}, R^{*})\), which is given by
$$ \Phi =\frac{1}{2}\bigl[\bigl(S-S^{*}\bigr)+ \bigl(L-L^{*}\bigr)+\bigl(I-I^{*}\bigr)+ \bigl(C-C^{*}\bigr)\bigr]^{2}. $$
(13)
After the calculation of derivatives of the above function with respect to time by using model (6), we obtain
$$\begin{aligned} \frac{d\Phi }{dt} =&\bigl[\bigl(S-S^{*}\bigr)+ \bigl(L-L^{*}\bigr)+\bigl(I-I^{*}\bigr)+ \bigl(C-C^{*}\bigr)\bigr]\\ &{}\times \bigl[\mu \omega -(\mu _{0}+ \gamma _{3})S-\mu _{0} L-(\mu _{0}+\gamma _{1}-q \gamma _{1})I-(\mu _{0}+\mu _{1}+\gamma _{2})C\bigr], \end{aligned}$$
where
\(\eta _{1}=\mu _{0}+\gamma _{3}\), \(\mathfrak{\eta }_{2}=\mu _{0}+\gamma _{1}\) and \(\mathfrak{\eta }_{3}=\mu _{0}+\mu _{1}+\gamma _{2}\).
We obtain the solution of the above equation as follows:
$$\begin{aligned}& \begin{aligned} \frac{d\Phi }{dt}={}&\bigl[\bigl(S-S^{*} \bigr)+\bigl(L-L^{*}\bigr)+\bigl(I-I^{*}\bigr)+ \bigl(C-C^{*}\bigr)\bigr]\\ &{}\times \bigl[-S^{*}R^{\mathrm{HBV}}_{0} \mathfrak{\eta }_{1}-S \mathfrak{\eta }_{1}-\mu _{0}L-(\mathfrak{\eta }_{2}-q \gamma _{1})I- \mathfrak{\eta }_{3}C\bigr], \\ \frac{d\Phi }{dt}={}&{-}\bigl[\bigl(S-S^{*}\bigr)+ \bigl(L-L^{*}\bigr)+\bigl(I-I^{*}\bigr)+ \bigl(C-C^{*}\bigr)\bigr] \\ &{}\times\bigl[S^{*}R^{\mathrm{HBV}}_{0} \mathfrak{\eta }_{1}+S\mathfrak{\eta }_{1} +\mu _{0}L+(\mathfrak{\eta }_{2}-q \gamma _{1})I+ \mathfrak{\eta }_{3}C\bigr], \\ \frac{d\Phi }{dt}={}&{-}\bigl[\bigl(S-S^{*}\bigr)+ \bigl(L-L^{*}\bigr)+\bigl(I-I^{*}\bigr)+ \bigl(C-C^{*}\bigr)\bigr]\\ &{}\times \bigl[ \mathfrak{\eta }_{1} \bigl(S^{*}R^{\mathrm{HBV}}_{0}+S\bigr) +\mu _{0}L+(\mathfrak{\eta }_{2}-q \gamma _{1})I+ \mathfrak{\eta }_{3}C\bigr]. \end{aligned} \end{aligned}$$
(14)
Hence \(\frac{d\Phi }{dt} \leq 0\) for all \((S^{*},L^{*}, I^{*}, C^{*}, R^{*})\). The equality \(\frac{d\Phi }{dt}= 0\) holds only for \(S= S^{*}\), \(L= L^{*}\), \(I= I^{*}\), \(C=C^{*}\). Then the endemic equilibrium \(E^{\mathrm{HBV}}_{1}\) is the only positively invariant set contained in \([(S, L, I, L, R), S= S^{*}, L= L^{*}, I= I^{*}, C= C^{*}]\). Therefore, the positive \(E^{\mathrm{HBV}}_{1}\) is globally asymptotically stable. □
Permanence of system (6)
Definition
Model (6) is assumed to be persistent if there are constants \(\mathbf{M_{1}}\), \(\mathbf{m_{1}}\) greater than zero such that each one positive solution \((S(t), L(t), I(t), C(t), R(t))\) of model (6) with initial conditions \(S(0) > 0\), \(L(0) > 0\), \(I(0) > 0\), \(C(0) > 0\), \(R(0) > 0\) satisfies
$$\begin{aligned}& \mathbf{M_{1}}\geq \lim_{t\longrightarrow +\infty }\operatorname{Sup} \ S(t)\geq \lim_{t\longrightarrow +\infty }\operatorname{Inf}\ S(t)\geq \mathbf{m_{1}}, \\& \mathbf{M_{1}}\geq \lim_{t\longrightarrow +\infty }\operatorname{Sup} \ L(t)\geq \lim_{t\longrightarrow +\infty }\operatorname{Inf}\ L(t)\geq \mathbf{m_{1}}, \\& \mathbf{M_{1}}\geq \lim_{t\longrightarrow +\infty }\operatorname{Sup} \ I(t)\geq \lim_{t\longrightarrow +\infty }\operatorname{Inf}\ I(t)\geq \mathbf{m_{1}}, \\& \mathbf{M_{1}}\geq \lim_{t\longrightarrow +\infty }\operatorname{Sup} \ C(t)\geq \lim_{t\longrightarrow +\infty }\operatorname{Inf}\ C(t)\geq \mathbf{m_{1}}, \\& \mathbf{M_{1}}\geq \lim_{t\longrightarrow +\infty }\operatorname{Sup} \ R(t)\geq \lim_{t\longrightarrow +\infty }\operatorname{Inf}\ R(t)\geq \mathbf{m_{1}}. \end{aligned}$$
Optimal strategy
In this section, optimal control theory of the HBV model’s application is discussed. Optimal control is the most viable tool of mathematics that enables us to design a strategy for the control of numerous kinds of infectious diseases. First, the optimal control theory is applied to the infection model of HBV. For the purpose of controlling the spread of HBV in the population or community, optimal control techniques are applied. Incorporation of three optimal control variables \(\mathfrak{B}_{1}(t)\), \(\mathfrak{B}_{2}(t)\), and \(\mathfrak{B}_{3}(t)\) is done to serve the purpose. The three different variables represent three different techniques for the control of the disease in a population, \(\mathfrak{B}_{1}(t)\) represents the vaccination of the drug for blocking the generation of new infected cells, \(\mathfrak{B}_{2}(t)\) represents the treatment that helps to decrease the production rate of the viruses in the infected individuals, and \(\mathfrak{B}_{3}(t)\) represents the campaign of awareness in public about the mentioned infection causing agent (virus). Sole meaning of our control problem focuses on the agents that tend to reduce the number of infected individuals in the population and to give rise to a number of recovered (healthy) individuals in the population. For the purpose of developing a control strategy, we bring into use the optimal control theory [27–30]. Our main area of focus here is to minimize the infection of HBV in a given population through a proper way that goes through susceptible to infection S(t), latently infected L(t), acute infection I(t), chronic C(t) and all the way leading to a protective immune population R(t).
$$\begin{aligned} \textstyle\begin{cases} \frac{dS}{dt}=\mu \omega (1-\nu C) -(\mu _{0}+\beta I + \epsilon \beta C +\gamma _{3} )S-(1-\mathfrak{B}_{1})S -(1- \mathfrak{B}_{3})S, \\ \frac{dL}{dt}=(\beta I+\epsilon \beta C)S -(\mu _{0}+\sigma )L -(1- \mathfrak{B}_{1})L, \\ \frac{dI}{dt}=\sigma L -(\mu _{0} +\gamma _{1})I -(1-\mathfrak{B}_{2})I, \\ \frac{dC}{dt}=\mu \omega \nu C +q\gamma _{1}I -(\mu _{0}+\mu _{1}+ \gamma _{2})C -(1-\mathfrak{B}_{2})C -(1-\mathfrak{B}_{3})C, \\ \frac{dR}{dt}=\gamma _{2}C +(1-q)\gamma _{1}I -\mu _{0}R +(1- \mathfrak{B}_{1})S +(1-\mathfrak{B}_{3})S +(1-\mathfrak{B}_{1})L \\ \hphantom{\frac{dR}{dt}=}{} +(1-\mathfrak{B}_{2})I +(1-\mathfrak{B}_{2})C -(1-\mathfrak{B}_{3})C \end{cases}\displaystyle \end{aligned}$$
(15)
with the initial conditions
$$ \mathbf{S(0)>0}, \qquad \mathbf{L(0)\geqslant 0}, \qquad \mathbf{I(0)\geqslant 0}, \qquad \mathbf{C(0)\geqslant 0}, \qquad \mathbf{R(0)\geqslant 0}. $$
(16)
To represent the weight constants, we use \(\mathbf{A^{*}_{1}}\), \(\mathbf{A^{*}_{2}}\), \(\mathbf{A^{*}_{3}}\), \(\mathbf{A^{*}_{4}}\), \(\mathbf{A^{*}_{5}}\), \(\mathbf{A^{*}_{6}}\), \(\mathbf{A^{*}_{7}}\), and \(\mathbf{A^{*}_{8}}\). Let an objective functional be defined to maximize the concentration of uninfected individuals as follows:
$$\begin{aligned} J\bigl(\mathfrak{B}_{1}(t),\mathfrak{B}_{2}(t), \mathfrak{B}_{3}(t)\bigr)= \int _{0}^{t} \left \{ \textstyle\begin{array}{c} A^{*}_{1}S(t)+ A^{*}_{2}L(t)+ A^{*}_{3}I(t)+ A^{*}_{4}C(t) + A^{*}_{5}R(t) \\ {}+ \frac{1}{2} A^{*}_{6}\mathfrak{B}_{1}^{2}(t)+\frac{1}{2} A^{*}_{7} \mathfrak{B}_{2}^{2}(t)+\frac{1}{2} A^{*}_{8}\mathfrak{B}_{3}^{2}(t) \end{array}\displaystyle \right \} \,dt. \end{aligned}$$
(17)
Furthermore, to minimize the objective functional, we have to find the optimal control pair \(\mathfrak{B}^{*}_{1}(t)\), \(\mathfrak{B}^{*}_{2}(t)\), and \(\mathfrak{B}^{*}_{3}(t)\) such that
$$\begin{aligned} & J\bigl\{ \mathfrak{B^{*}}_{1}(t), \mathfrak{B^{*}}_{2}(t), \mathfrak{B^{*}}_{3}(t) \bigr\} \\ &\quad = \min \bigl\{ J\bigl(\mathfrak{B}_{1}(t), \mathfrak{B}_{2}(t),\mathfrak{B}_{3}(t)\bigr), \mathfrak{B}_{1}(t), \mathfrak{B}_{2}(t), \mathfrak{B}_{3}(t) \in B \bigr\} , \end{aligned}$$
(18)
subject to system (15), where the control set is \(B=\{(\mathfrak{B}_{1}(t),\mathfrak{B}_{2}(t),\mathfrak{B}_{3}(t)) \setminus \mathfrak{B}i(t)\}\) is Lebesgue measurable on \([0,1]\), \(0 \leq \mathfrak{B}i(t) \leq 1 \), \(i=\{1,2,3\}\).
Existence of the optimal control problem
In this section, we give proof for the existence of the control problem. In order to show the existence of the optimal control problem, we take up the reference [31]. In order to prove the existence of the pair of the optimal control problems, we bring into consideration the control system having all the conditions at the initial stage at time \(t= 0\). Positive bounded solution to the state system and positive initial conditions held existence for bounded Lebesgue measurable control [32]. For the determination of optimal solution, we push back to optimal control problems (15) and (18). So for optimal control problems (15) and (18), exploration of the Lagrangian and Hamiltonian is a must. The optimal control problem in the Lagrangian is represented by the following equation:
$$\begin{aligned} \begin{gathered} L\bigl\{ S(t),L(t),I(t),C(t),R(t), \mathfrak{B}_{1}(t),\mathfrak{B}_{2}(t), \mathfrak{B}_{3}(t)\bigr\} \\ \quad = A^{*}_{1}S(t)+ A^{*}_{2}L(t) + A^{*}_{3}I(t)+ A^{*}_{4}C(t)+ A^{*}_{5}R(t) \\ \qquad {}+ \frac{1}{2} A^{*}_{6} \mathfrak{B}^{2}_{1}(t)+ \frac{1}{2} A^{*}_{7}\mathfrak{B}^{2}_{2}(t)+ \frac{1}{2} A^{*}_{8}\mathfrak{B}^{2}_{3}(t). \end{gathered} \end{aligned}$$
(19)
To seek the minimal value of the Lagrangian, we define Hamiltonian H for the optimal control problem as follows:
$$\begin{aligned} \begin{aligned} \boldsymbol{H}={}& L \bigl(S(t),L(t),I(t),C(t),R(t),\mathfrak{B}_{1}(t), \mathfrak{B}_{2}(t),\mathfrak{B}_{3}(t)\bigr) \\ &{}+\mathfrak{T}_{1}\frac{dS(t)}{dt}+\mathfrak{T}_{2} \frac{dL(t)}{dt}+ \mathfrak{T}_{3}\frac{dI(t)}{dt}+ \mathfrak{T}_{4}\frac{dC(t)}{dt}+ \mathfrak{T}_{5} \frac{dR(t)}{dt}. \end{aligned} \end{aligned}$$
(20)
Adjoint variable \(\mathfrak{T}_{1}(t)\), \(\mathfrak{T}_{2}(t)\), \(\mathfrak{T}_{3}(t)\), \(\mathfrak{T}_{4}(t)\), and \(\mathfrak{T}_{5}(t)\) and optimal control variables \(\mathfrak{B}_{1}(t)\), \(\mathfrak{B}_{2}(t)\), and \(\mathfrak{B}_{3}(t)\),
$$\begin{aligned} \mathfrak{T}^{\prime }_{1}(t) =&- \bigl\{ A^{*}_{1}-\mathfrak{T}_{1}(\mu _{0}+ \beta I+ \epsilon \beta C+\gamma _{3}- \mathfrak{B}_{1}- \mathfrak{B}_{3}+ 2)+ \mathfrak{T}_{2}(\beta I+ \epsilon \beta C) \\ &{} +\mathfrak{T}_{5}(2- \mathfrak{B}_{1}- \mathfrak{B}_{3}) \bigr\} , \\ \mathfrak{T}^{\prime }_{2}(t) =&- \bigl\{ A^{*}_{2}- \mathfrak{T}_{2}(\mu _{0}+ \mu _{1}+ \sigma - 1)+ \mathfrak{T}_{3} \sigma + \mathfrak{T}_{5}(1- \mathfrak{B}_{1}) \bigr\} , \\ \mathfrak{T}^{\prime }_{3}(t) =&- \bigl\{ A^{*}_{3}- \beta S(\mathfrak{T}_{1}- \mathfrak{T}_{2})- \mathfrak{T}_{3}(\mu _{0}+ \gamma _{1}+ \mathfrak{B}_{2}- 1) \\ &{}+ \mathfrak{T}_{4} q\gamma _{1} + \mathfrak{T}_{5}( \gamma _{1}-q \gamma _{1}- \mathfrak{B}_{2}+1) \bigr\} , \\ \mathfrak{T}^{\prime }_{4}(t) =&- \bigl\{ A^{*}_{4}- \mathfrak{T}_{1}(\mu \upsilon \omega +\epsilon \beta S)+ \mathfrak{T}_{2} \epsilon \beta S\\ &{}+ \mathfrak{T}_{4}(\mu \omega \upsilon -\mu _{0} -\mu _{1}-\gamma _{2} + \mathfrak{B}_{2} +\mathfrak{B}_{3} -2)+ \mathfrak{T}_{5}(\gamma _{2}- \mathfrak{B}_{2}+ \mathfrak{B}_{3}) \bigr\} , \\ \mathfrak{T}^{\prime }_{5}(t) =&- \bigl\{ A^{*}_{5}- \mathfrak{T}_{5} \mu _{0} \bigr\} , \\ \mathfrak{B}_{1}(t) =& \biggl\{ \frac{-\mathfrak{T}_{1} S-\mathfrak{T}_{2} L + \mathfrak{T}_{5} (L+S)}{A^{*}_{6}} \biggr\} , \\ \mathfrak{B}_{2}(t) = &\biggl\{ \frac{-\mathfrak{T}_{3} I-\mathfrak{T}_{4} C+\mathfrak{T}_{5} (I+C)}{A^{*}_{7}} \biggr\} , \\ \mathfrak{B}_{3}(t) =& \biggl\{ \frac{-\mathfrak{T}_{1} S-\mathfrak{T}_{4} C+\mathfrak{T}_{5} (S-C)}{A^{*}_{8}} \biggr\} . \end{aligned}$$
(21)
Theorem 6
For control problem (15), there exists \(\mathfrak{B}^{*}(t)=(\mathfrak{B}^{*}_{1}(t), \mathfrak{B}^{*}_{2}(t), \mathfrak{B}^{*}_{3}(t))\in B\) such that
$$ \min_{(\mathfrak{B}_{1}(t),\mathfrak{B}_{2}(t), \mathfrak{B}_{3}(t))\in B}J\bigl(\mathfrak{B}_{1}(t), \mathfrak{B}_{2}(t), \mathfrak{B}_{3}(t)\bigr)=J\bigl( \mathfrak{B}^{*}_{1}(t),\mathfrak{B}^{*}_{2}(t), \mathfrak{B}^{*}_{3}(t)\bigr). $$
Proof
We brought several techniques into use for validation of the optimal control prevalence shown in [33, 34]. Thus all the control and state variables are nonnegative. That is why this process of reducing the problem, the required convexity of objective functional is elaborated in equation (20) in \(\mathbf{\mathfrak{B}_{1}(t), \mathfrak{B}_{2}(t)}\) and \(\mathbf{\mathfrak{B}_{3}(t)}\) is gratified. The set of control variables \(\mathbf{\mathfrak{B}_{1}, \mathfrak{B}_{2}, \mathfrak{B}_{3} \in B}\) is also convex and closed by definition. This optimal system is delineated, and it provides surety about the solidity that is needed for the validation of the optimal control system. Moreover, the integrand in the objective functional \(\mathbf{A^{*}_{1}S(t)+ A^{*}_{2}L(t)+ A^{*}_{3}I(t)+ A^{*}_{4}C(t)+A^{*}_{5}R(t)+ \frac{1}{2} A^{*}_{6}\mathfrak{B}^{2}_{1}(t)+ \frac{1}{2} A^{*}_{7}\mathfrak{B}^{2}_{2}(t)+ \frac{1}{2} A^{*}_{8}\mathfrak{B}^{2}_{3}(t)}\) is convex on the control set B, which certifies the proof. Thus, for our proposed control problem, we dug out an optimal solution. We bring use of the Pontryagin maximum principle [35] in order to find solution to our proposed problem. By using this principle, the Hamiltonian is given by
$$ \begin{aligned} \boldsymbol{H}={}& L\bigl(S(t),L(t),I(t),C(t),R(t), \mathfrak{B}_{1}(t),\mathfrak{B}_{2}(t), \mathfrak{B}_{3}(t)\bigr) \\ &{}+\mathbf{\mathfrak{T}_{1}}\frac{dS(t)}{dt}+\mathbf{ \mathfrak{T}_{2}} \frac{dL(t)}{dt}+\mathbf{\mathfrak{T}_{3}} \frac{dI(t)}{dt}+ \mathbf{\mathfrak{T}_{4}}\frac{dC(t)}{dt}+ \mathbf{\mathfrak{T}_{5}} \frac{dR(t)}{dt}. \end{aligned} $$
(22)
A nontrivial vector function \(\mathfrak{T}(t) = (\mathfrak{T}_{1}(t), \mathbf{\mathfrak{T}_{2}(t)}, . . . , \mathfrak{T}_{n}(t))\) exists if we consider \((y^{*}, \mathfrak{B}^{*})\) as an optimal solution of our proposed optimal control problem such that
$$\begin{aligned}& \frac{dy}{dt}=\frac{\partial H(t, y, \mathfrak{B},\mathfrak{T})}{\partial \mathfrak{B}}, \\& 0=\frac{\partial H(t, y, \mathfrak{B},\mathfrak{T})}{\partial \mathfrak{B}}, \\& \mathfrak{T}'(t)=- \frac{\partial H(t, y, \mathfrak{B},\mathfrak{T})}{\partial \mathfrak{B}}. \end{aligned}$$
(23)
Applying the necessary conditions to the Hamiltonian, we have these mentioned above results. □
Theorem 7
Given optimal controls \(\mathfrak{B}^{*}_{1}(t)\), \(\mathfrak{B}^{*}_{2}(t)\), \(\mathfrak{B}^{*}_{3}(t)\) and solutions \(S^{*}(t)\), \(L^{*}(t)\), \(I^{*}(t)\), \(C^{*}(t)\), \(R^{*}(t)\) of the corresponding state system (15), there exist adjoint variables \(\mathfrak{T}_{m}(t), m=1,\ldots , 5\),
$$\begin{aligned} \begin{aligned} \mathfrak{T}'_{1}(t)={}&{-} \bigl\{ A^{*}_{1}-\mathfrak{T}_{1}( \mu _{0}+ \beta I+ \epsilon \beta C+\gamma _{3}- \mathfrak{B}_{1}- \mathfrak{B}_{3}+ 2)\\ &{}+ \mathfrak{T}_{2}(\beta I+ \epsilon \beta C)+ \mathfrak{T}_{5}(2- \mathfrak{B}_{1}- \mathfrak{B}_{3}) \bigr\} , \\ \mathfrak{T}'_{2}(t)={}&{-} \bigl\{ A^{*}_{2}- \mathfrak{T}_{2}( \mu _{0}+ \mu _{1}+ \sigma - 1)+ \mathfrak{T}_{3} \sigma + \mathfrak{T}_{5}(1- \mathfrak{B}_{1}) \bigr\} , \\ \mathfrak{T}'_{3}(t)={}&{-} \bigl\{ A^{*}_{3}- \beta S( \mathfrak{T}_{1}-\mathfrak{T}_{2})- \mathfrak{T}_{3}(\mu _{0}+ \gamma _{1}+ \mathfrak{B}_{2}- 1)\\ &{}+ \mathfrak{T}_{4} q\gamma _{1} + \mathfrak{T}_{5}(\gamma _{1}-q \gamma _{1}- \mathfrak{B}_{2}+1) \bigr\} , \\ \mathfrak{T}^{\prime }_{4}(t)={}&{-} \bigl\{ A^{*}_{4}- \mathfrak{T}_{1}( \mu \upsilon \omega +\epsilon \beta S)+ \mathfrak{T}_{2} \epsilon \beta S\\ &{}+\mathfrak{T}_{4}(\mu \omega \upsilon -\mu _{0} -\mu _{1}- \gamma _{2} +\mathfrak{B}_{2} +\mathfrak{B}_{3} -2)+ \mathfrak{T}_{5}( \gamma _{2}- \mathfrak{B}_{2}+ \mathfrak{B}_{3}) \bigr\} , \\ \mathfrak{T}^{\prime }_{5}(t)={}&{-} \bigl\{ A^{*}_{5}- \mathfrak{T}_{5} \mu _{0} \bigr\} , \end{aligned} \end{aligned}$$
(24)
with transversality conditions \(\mathfrak{T}_{m}(t) = 0\), \(m = 1, 2,\ldots , 5\).
Proof
If we take the values as \(S(t) = S^{*}\), \(L(t) =L^{*}\), \(I(t) =I^{*}\), \(L(t) C^{*}\), and \(R(t)= R^{*}\) and distinguish the Hamiltonian with respect to state variables \(S(t)\), \(L(t)\), \(I(t)\), \(C(t)\), and \(R(t)\), respectively, we get the following adjoint system:
$$\begin{aligned} \begin{aligned}\mathfrak{T}'_{1}(t)={}&{-} \bigl\{ A^{*}_{1}-\mathfrak{T}_{1}( \mu _{0}+ \beta I+ \epsilon \beta C+\gamma _{3}- \mathfrak{B}_{1}- \mathfrak{B}_{3}+ 2)\\ &{}+ \mathfrak{T}_{2}(\beta I+ \epsilon \beta C) + \mathfrak{T}_{5}(2- \mathfrak{B} {1}- \mathfrak{B}_{3}) \bigr\} , \\ \mathfrak{T}'_{2}(t)={}&{-} \bigl\{ A^{*}_{2}- \mathfrak{T}_{2}( \mu _{0}+ \mu _{1}+ \sigma - 1)+ \mathfrak{T}_{3} \sigma + \mathfrak{T}_{5}(1- \mathfrak{B}_{1}) \bigr\} , \\ \mathfrak{T}'_{3}(t)={}&{-} \bigl\{ A^{*}_{3}- \beta S( \mathfrak{T}_{1}-\mathfrak{T}_{2})- \mathfrak{T}_{3}(\mu _{0}+ \gamma _{1}+ \mathfrak{B}_{2}- 1)\\ &{}+ \mathfrak{T}_{4} q\gamma _{1} + \mathfrak{T}_{5}(\gamma _{1}-q \gamma _{1}- \mathfrak{B}_{2}+1) \bigr\} , \\ \mathfrak{T}^{\prime }_{4}(t)={}&{-} \bigl\{ A^{*}_{4}- \mathfrak{T}_{1}( \mu \upsilon \omega +\epsilon \beta S)+ \mathfrak{T}_{2} \epsilon \beta S\\ &{}+\mathfrak{T}_{4}(\mu \omega \upsilon -\mu _{0} -\mu _{1}- \gamma _{2} +\mathfrak{B}_{2} +\mathfrak{B}_{3} -2)+ \mathfrak{T}_{5}( \gamma _{2}- \mathfrak{B}_{2}+ \mathfrak{B}_{3}) \bigr\} , \\ \mathfrak{T}^{\prime }_{5}(t)={}&{-} \bigl\{ A^{*}_{5}- \mathfrak{T}_{5} \mu _{0} \bigr\} , \end{aligned} \end{aligned}$$
(25)
with the transversality conditions \(\mathfrak{T}_{m}(t) = 0\), \(m = 1, 2,\ldots , 5\). □
Theorem 8
The control pair (\(\mathfrak{B}^{*}_{1}(t)\), \(\mathfrak{B}^{*}_{2}(t)\), \(\mathfrak{B}^{*}_{3}(t)\)), which maximizes the objective functional J over the region B, is given by
$$\begin{aligned} \begin{aligned} \mathfrak{B}^{*}_{1}(t)&= \max \biggl\{ \min \biggl\{ \frac{-\mathfrak{T}_{1} S-\mathfrak{T}_{2} L + \mathfrak{T}_{5} (L+S)}{A^{*}_{6}},1 \biggr\} ,0 \biggr\} , \\ \mathfrak{B}^{*}_{2}(t)&=\max \biggl\{ \min \biggl\{ \frac{-\mathfrak{T}_{3} I-\mathfrak{T}_{4} C+\mathfrak{T}_{5} (I+C)}{A^{*}_{7}},1 \biggr\} ,0 \biggr\} , \\ \mathfrak{B}^{*}_{3}(t)&=\max \biggl\{ \min \biggl\{ \frac{-\mathfrak{T}_{1} S-\mathfrak{T}_{4} C+\mathfrak{T}_{5} (S-C)}{A^{*}_{8}},1 \biggr\} ,0 \biggr\} . \end{aligned} \end{aligned}$$
(26)
Proof
By using the optimality condition, we get
$$\begin{aligned} \begin{aligned} \frac{\partial H}{\partial \mathfrak{B}_{1}} &= A^{*}_{6} \mathfrak{B}_{1}(t)+ \mathfrak{T}_{1} S+\mathfrak{T}_{2} L- \mathfrak{T}_{5} (L+S), \\ \frac{\partial H}{\partial \mathfrak{B}_{2}} &= A^{*}_{7} \mathfrak{B}_{2}(t)+\mathfrak{T}_{3} I+ \mathfrak{T}_{4} C- \mathfrak{T}_{5} (I+C), \\ \frac{\partial H}{\partial \mathfrak{B}_{3}} &= A^{*}_{8} \mathfrak{B}_{3}(t)+\mathfrak{T}_{1} S+ \mathfrak{T}_{4} C- \mathfrak{T}_{5} (S-C). \end{aligned} \end{aligned}$$
(27)
Optimal control variables ∀, \(\mathfrak{B}^{*}_{1}(t)\), \(\mathfrak{B}^{*}_{2}(t)\) & \(\mathfrak{B}^{*}_{3}(t)\), solving equations (27), we have
$$\begin{aligned} \begin{aligned} \mathfrak{B}^{*}_{1}(t)= \frac{-\mathfrak{T}_{1} S-\mathfrak{T}_{2} L + \mathfrak{T}_{5} (L+S)}{A^{*}_{6}}, \\ \mathfrak{B}^{*}_{2}(t)= \frac{-\mathfrak{T}_{3} I-\mathfrak{T}_{4} C+\mathfrak{T}_{5} (I+C)}{A^{*}_{7}}, \\ \mathfrak{B}^{*}_{3}(t)= \frac{-\mathfrak{T}_{1} S-\mathfrak{T}_{4} C+\mathfrak{T}_{5} (S-C)}{A^{*}_{8}}. \end{aligned} \end{aligned}$$
(28)
The property of control space equations (28) can be written as follows:
$$\begin{aligned}& \mathfrak{B}^{*}_{1}= \textstyle\begin{cases} 0, & \text{if } \frac{-\mathfrak{T}_{1} S-\mathfrak{T}_{2} L + \mathfrak{T}_{5} (L+S)}{A^{*}_{6}} \leq 0, \\ \frac{-\mathfrak{T}_{1} S-\mathfrak{T}_{2} L + \mathfrak{T}_{5} (L+S)}{A^{*}_{6}}, & \text{if } 0< \frac{-\mathfrak{T}_{1} S-\mathfrak{T}_{2} L + \mathfrak{T}_{5} (L+S)}{A^{*}_{6}}< 1, \\ 1, & \text{if} \frac{-\mathfrak{T}_{1} S-\mathfrak{T}_{2} L + \mathfrak{T}_{5} (L+S)}{A^{*}_{6}} \geq 1. \end{cases}\displaystyle \\& \mathfrak{B}^{*}_{2}= \textstyle\begin{cases} 0, & \text{if } \frac{-\mathfrak{T}_{3} I-\mathfrak{T}_{4} C+\mathfrak{T}_{5} (I+C)}{A^{*}_{7}} \leq 0, \\ \frac{-\mathfrak{T}_{3} I-\mathfrak{T}_{4} C+\mathfrak{T}_{5} (I+C)}{A^{*}_{7}}, & \text{if } 0< \frac{-\mathfrak{T}_{3} I-\mathfrak{T}_{4} C+\mathfrak{T}_{5} (I+C)}{A^{*}_{7}}< 1, \\ 1, & \text{if} \frac{-\mathfrak{T}_{3} I-\mathfrak{T}_{4} C+\mathfrak{T}_{5} (I+C)}{A^{*}_{7}} \geq 1. \end{cases}\displaystyle \\& \mathfrak{B}^{*}_{3}= \textstyle\begin{cases} 0, & \text{if } \frac{-\mathfrak{T}_{1} S-\mathfrak{T}_{4} C+\mathfrak{T}_{5} (S-C)}{A^{*}_{8}} \leq 0, \\ \frac{-\mathfrak{T}_{1} S-\mathfrak{T}_{4} C+\mathfrak{T}_{5} (S-C)}{A^{*}_{8}}, & \text{if } 0< \frac{-\mathfrak{T}_{1} S-\mathfrak{T}_{4} C+\mathfrak{T}_{5} (S-C)}{A^{*}_{8}}< 1, \\ 1, & \text{if} \frac{-\mathfrak{T}_{1} S-\mathfrak{T}_{4} C+\mathfrak{T}_{5} (S-C)}{A^{*}_{8}} \geq 1. \end{cases}\displaystyle \end{aligned}$$
According to compact notation \(\mathfrak{B}^{*}_{1}(t)\), \(\mathfrak{B}^{*}_{2}(t)\), and \(\mathfrak{B}^{*}_{3}(t)\) can be written as
$$\begin{aligned} \begin{aligned} \mathfrak{B}^{*}_{1}(t)&= \max \biggl\{ \min \biggl\{ \frac{-\mathfrak{T}_{1} S-\mathfrak{T}_{2} L + \mathfrak{T}_{5} (L+S)}{A^{*}_{6}},1 \biggr\} ,0 \biggr\} , \\ \mathfrak{B}^{*}_{2}(t)&=\max \biggl\{ \min \biggl\{ \frac{-\mathfrak{T}_{3} I-\mathfrak{T}_{4} C+\mathfrak{T}_{5} (I+C)}{A^{*}_{7}},1 \biggr\} ,0 \biggr\} , \\ \mathfrak{B}^{*}_{3}(t)&=\max \biggl\{ \min \biggl\{ \frac{-\mathfrak{T}_{1} S-\mathfrak{T}_{4} C+\mathfrak{T}_{5} (S-C)}{A^{*}_{8}},1 \biggr\} ,0 \biggr\} . \end{aligned} \end{aligned}$$
(29)
By using equation (29), the optimality system is
$$\begin{aligned} \textstyle\begin{cases} \frac{dS^{*}}{dt}=\mu \omega (1-\nu C^{*}) -(\mu _{0}+ \beta I^{*} +\epsilon \beta C^{*} +\gamma _{3} )S^{*}\\ \hphantom{\frac{dS^{*}}{dt}=}{}- (1-\max \{\min \{ \frac{-\mathfrak{T}_{1} S-\mathfrak{T}_{2} L + \mathfrak{T}_{5} (L+S)}{A^{*}_{6}},1 \},0 \} )S^{*} \\ \hphantom{\frac{dS^{*}}{dt}=}{}- (1-\max \{\min \{ \frac{-\mathfrak{T}_{1} S-\mathfrak{T}_{4} C+\mathfrak{T}_{5} (S-C)}{A^{*}_{8}},1 \},0 \} )S^{*}, \\ \frac{dL^{*}}{dt}=(\beta I^{*}+\epsilon \beta C^{*})S^{*} -(\mu _{0}+ \sigma )L^{*} - (1-\max \{\min \{ \frac{-\mathfrak{T}_{1} S-\mathfrak{T}_{2} L + \mathfrak{T}_{5} (L+S)}{A^{*}_{6}},1 \},0 \} )L^{*}, \\ \frac{dI^{*}}{dt}=\sigma L^{*} -(\mu _{0} +\gamma _{1})I^{*} - (1-\max \{\min \{ \frac{-\mathfrak{T}_{3} I-\mathfrak{T}_{4} C+\mathfrak{T}_{5} (I+C)}{A^{*}_{7}},1 \},0 \} )I^{*}, \\ \frac{dC^{*}}{dt}=\mu \omega \nu C^{*} +q\gamma _{1}I^{*} -(\mu _{0}+ \mu _{1}+\gamma _{2})C^{*}\\ \hphantom{\frac{dC^{*}}{dt}=}{} - (1-\max \{\min \{ \frac{-\mathfrak{T}_{3} I-\mathfrak{T}_{4} C+\mathfrak{T}_{5} (I+C)}{A^{*}_{7}},1 \},0 \} )C^{*} \\ \hphantom{\frac{dC^{*}}{dt}=}{}- (1-\max \{\min \{ \frac{-\mathfrak{T}_{1} S-\mathfrak{T}_{4} C+\mathfrak{T}_{5} (S-C)}{A^{*}_{8}},1 \},0 \} )C^{*}, \\ \frac{dR^{*}}{dt}=\gamma _{2}C^{*} +(1-q)\gamma _{1}I^{*} -\mu _{0}R^{*} + (1-\max \{\min \{ \frac{-\mathfrak{T}_{1} S-\mathfrak{T}_{2} L + \mathfrak{T}_{5} (L+S)}{A^{*}_{6}},1 \},0 \} )S^{*} \\ \hphantom{\frac{dR^{*}}{dt}=}{}+ (1-\max \{\min \{ \frac{-\mathfrak{T}_{1} S-\mathfrak{T}_{4} C+\mathfrak{T}_{5} (S-C)}{A^{*}_{8}},1 \},0 \} )S^{*}\\ \hphantom{\frac{dR^{*}}{dt}=}{} + (1-\max \{\min \{ \frac{-\mathfrak{T}_{1} S-\mathfrak{T}_{2} L + \mathfrak{T}_{5} (L+S)}{A^{*}_{6}},1 \},0 \} )L^{*}\\ \hphantom{\frac{dR^{*}}{dt}=}{}+ (1-\max \{\min \{ \frac{-\mathfrak{T}_{3} I-\mathfrak{T}_{4} C+\mathfrak{T}_{5} (I+C)}{A^{*}_{7}},1 \},0 \} )I^{*} \\ \hphantom{\frac{dR^{*}}{dt}=}{} + (1-\max \{\min \{ \frac{-\mathfrak{T}_{3} I-\mathfrak{T}_{4} C+\mathfrak{T}_{5} (I+C)}{A^{*}_{7}},1 \},0 \} )C^{*} \\ \hphantom{\frac{dR^{*}}{dt}=}{} - (1-\max \{\min \{ \frac{-\mathfrak{T}_{1} S-\mathfrak{T}_{4} C+\mathfrak{T}_{5} (S-C)}{A^{*}_{8}},1 \},0 \} )C^{*}. \end{cases}\displaystyle \end{aligned}$$
(30)
We used the solution of the adjoint system \(\frac{d\mathfrak{T}_{i}}{dt}\) and optimality system (30) along with equations of optimal control, both initial and boundary conditions, to acquire the state and optimal control variables. Additionally, it can be seen that the derivatives of Lagrangian of second order is clearly positive with respect to \(\mathfrak{B}_{1}\), \(\mathfrak{B}_{2}\), and \(\mathfrak{B}_{3}\). Hence, at control variables \(\mathfrak{B}^{*}_{1}(t)\), \(\mathfrak{B}^{*}_{2}(t)\), and \(\mathfrak{B}^{*}_{3}(t)\) the optimal problem is minimum. In the consequences of which the Hamiltonian is written in the following order:
$$\begin{aligned} H^{\ast } =&A^{*}_{1}S^{\ast }(t)+A^{*}_{2}L^{\ast }(t)+A^{*}_{3}I^{\ast }(t)+A^{*}_{4}C^{\ast }(t)+A^{*}_{5}R^{\ast }(t)+ \frac{1}{2}A^{*}_{6} \mathfrak{B}^{2}_{1}(t) \\ &{} +\frac{1}{2}A^{*}_{7} \mathfrak{B}^{2}_{2}(t)+\frac{1}{2}A^{*}_{8} \mathfrak{B}^{2}_{3}(t) +\Sigma _{m=1}^{5} \mathfrak{T}_{m}(t)g_{m}\bigl(S^{\ast },L^{\ast },I^{\ast },C^{\ast },R^{*} \bigr). \end{aligned}$$
□
