Mathematical analysis of an extended cellular model of the Hepatitis C Virus infection with non-cytolytic process

HCV is a virus that attacks the cells of the liver and causes inflammation of the latter. This virus is present in the blood of an infected person and is, according to the WHO, mandatory declaration. It can live for about 5 to 7 weeks in the open air. In the long term, there may be very serious consequences such as cirrhosis and in some cases, liver cancer . This virus can remain for decades in the body without any apparent symptoms. According to [14, njou] six main genotypes HCV were identified whereas several subtypes play an important role in the severity of the disease and its response to treatment.

HCV infection is a major public health problem. Worldwide, the number of chronic HCV carriers is estimated at about 170 million, or $3\%$ of the world’s population, and the incidence is between 3 and 4 million infections per year according to the statistics published by the World Health Organization in 2014. The actual incidence is uncertain because the distinction between acute and chronic forms is difficult to make. HCV was only identified in 1989 with the advent of modern molecular cloning techniques.
In Cameroon, about 10,000 people die each year from hepatitis (A, B, C, D, …). The country is one of the 17 most affected worldwide, with a prevalence rate of about $13\%$, that is 2.5 million people infected.

According to WHO’s global report on hepatitis C published in April 2017 the first phase of HCV infection is said Acute: It can cause jaundice, but remains asymptomatic in the majority of cases (70 to 80 $\%$), hence the risk of going unnoticed. It is estimated that 20 to 30 $\%$ people infected will spontaneously clear the virus within the first six months after initial contact. If the virus persists, hepatitis progresses to chronicity. The liver reacts to HCV aggression by an inflammatory reaction, one of the components of which is fibrogenesis. Hepatic fibrosis is the main complication of chronic hepatitis C. Hepatitis C is likely to evolve at the chronic phase in about 25 $\%$ of cases to cirrhosis within a period of 5 to 20 years. In case of cirrhosis, the incidence of hepatocellular carcinoma is high: on the order of 1 to 4 $\%$ per year [14].(See figure 1).

Proof. We will use the local Cauchy-Lipschitz theorem to proof this. Since the system of equations (4) is autonomous, it is enough to show that the function

$f$ : $\mathbb{R}^{3}$ $\longrightarrow$ $\mathbb{R}^{3}$ $(T,I,V)$ $\longmapsto$ $\Big{(}f_{1}(T,I,V);f_{2}(T,I,V);f_{3}(T,I,V)\Big{)}$

is locally Lipschitzian with:

$$f_{1}(T,I,V)=s+r_{T}T\left(1-\frac{T+I}{T_{max}}\right)-d_{T}T-(1-\eta)\beta VT+qI.$$

(6)

$$f_{2}(T,I,V)=r_{I}I\left(1-\frac{T+I}{T_{max}}\right)+(1-\eta)\beta VT-d_{I}I-qI.$$

(7)

$$f_{3}(T,I,V)=(1-\varepsilon)pI-cV.$$

(8)

According to L. Perko [18], it is also enough to prove that $f$ is a class $\mathcal{C}^{1}$ function.
The jacobian matrix of $f$ at $(T,I,V)$ is:

$$Df(T,I,V)=\begin{pmatrix}\frac{\partial f_{1}}{\partial T}(T,I,V)&\frac{\partial f_{1}}{\partial I}(T,I,V)&\frac{\partial f_{1}}{\partial V}(T,I,V)\\ \\ \frac{\partial f_{2}}{\partial T}(T,I,V)&\frac{\partial f_{2}}{\partial I}(T,I,V)&\frac{\partial f_{2}}{\partial V}(T,I,V)\\ \\ \frac{\partial f_{3}}{\partial T}(T,I,V)&\frac{\partial f_{3}}{\partial I}(T,I,V)&\frac{\partial f_{3}}{\partial V}(T,I,V)\end{pmatrix},$$

i.e.

$$Df(T,I,V)=\begin{pmatrix}r_{T}\left(1-\frac{2T+I}{T_{max}}\right)-d_{T}+&-\frac{r_{T}T}{T_{max}}+q&-(1-\eta)\beta T\\ -(1-\eta)\beta V&&\\ \\ -\frac{r_{I}I}{T_{max}}+(1-\eta)\beta V&r_{I}\left(1-\frac{T+2I}{T_{max}}\right)-d_{I}-q&(1-\eta)\beta T\\ \\ 0&(1-\varepsilon)p&-c\end{pmatrix}.$$

Each component of this matrix being continuous, they are locally bounded for all $(T,I,V)\in\mathbb{R}^{3}$. Therefore $f$ possesses continuous and bounded partial derivatives on any compact of $\mathbb{R}$. Thus $f$ is locally Lipschitzian with respect to $(T,I,V)$. By the Cauchy-Lipschitz theorem, there is a local solution defined on $[t_{0};t_{1}[$ This completes the proof of this theorem 2.1.   

Proof. We are going to prove by contradiction. so suppose there is $t\in[t_{0},t_{1}[$ such that $T(t)=0$ or $I(t)=0$ or $V(t)=0$.
Let $x=(x_{1},x_{2},x_{3})=(T,I,V)$
Let also $t_{*}$ be the smallest of all $t$ in the interval $[t_{0},t_{1}[$ such that $x_{i}(t)>0,\\ \forall t\in[t_{0},t_{*}[,\forall i\in\left\{1,2,3\right\}$ and $x_{i}(t_{*})=0$ for a certain $i$.
Then each of the equations of the system (4) can be written $\dot{x}_{i}=-h_{i}(x)+g_{i}(x)$ where $g_{i}$ is a non negative function and $h_{i}$ any function.
Thus,

	$\displaystyle\quad\dot{T}$	$\displaystyle=$	$\displaystyle\frac{dT}{dt},$
		$\displaystyle=$	$\displaystyle-T\left(-r_{T}\left(1-\frac{T+I}{T_{max}}\right)+d_{T}+(1-\eta)\beta V\right)+s+qI,$
		$\displaystyle=$	$\displaystyle-Th_{1}(T,I,V)+g_{1}(T,I,V).$

with

$$h_{1}(T,I,V)=\left(-r_{T}\left(1-\frac{T+I}{T_{max}}\right)+d_{T}+(1-\eta)\beta V\right)$$

and

$$g_{1}(T,I,V)=s+qI;$$

similarly;

	$\displaystyle\quad\dot{I}$	$\displaystyle=$	$\displaystyle\frac{dI}{dt},$
		$\displaystyle=$	$\displaystyle-I\left(-r_{I}\left(1-\frac{T+I}{T_{max}}\right)+d_{I}+q\right)+(1-\eta)\beta VT,$
		$\displaystyle=$	$\displaystyle-Ih_{2}(T,I,V)+g_{2}(T,I,V).$

with

$$h_{2}(T,I,V)=\left(-r_{I}\left(1-\frac{T+I}{T_{max}}\right)+d_{I}+q\right)$$

and

$$g_{2}(T,I,V)=(1-\eta)\beta VT$$

and

	$\displaystyle V$	$\displaystyle=$	$\displaystyle\frac{dV}{dt},$
		$\displaystyle=$	$\displaystyle-Vc+(1-\varepsilon)pI,$
		$\displaystyle=$	$\displaystyle-Vh_{3}(T,I,V)+g_{3}(T,I,V).$

with

$$h_{3}(T,I,V)=c$$

and

$$g_{3}(T,I,V)=(1-\varepsilon)pI.$$

Without loss the generality, suppose that $x_{1}(t_{*})=0$.
As hypothesized, $g_{1}(T,I,V)$ is positive on $[t_{0},t_{*}]$, it follows that

$$\dot{T}\geq-Th_{1}(T,I,V);$$

from where

$$\frac{d}{dt}(\log T)\geq-h_{1}(T,I,V).$$

Yet $(T,I,V)$ is a solution of (4). $T$, $I$, $V$ are class class $C^{1}$ functions. They are continuous on $[t_{0},t_{*}]$ and therefore are bounded on $[t_{0},t_{*}]$. Thus $h_{1}(T,I,V)$ is bounded on $[t_{0},t_{*}]$.
There exists a constant $k>0$ such that

$$\frac{d}{dt}(\log T)\geq-h_{1}(T,I,V)>-k.$$

By integrating this previous expression on $[t_{0},t_{*}]$, we get

$$\log T(t_{*})-\log T(t_{0})\geq-k(t_{*}-t_{0}).$$

Where

$$T(t_{*})\geq T(t_{0})e^{-k(t_{*}-t_{0})}>0.$$

This is a contradiction because $T(t_{*})=0$.
Similarly, assuming that $x_{2}(t_{*})=0$, as hypothesized, $g_{2}(T,I,V)$ is positive on $[t_{0},t_{*}]$, it follows that

$$\dot{I}\geq-Ih_{2}(T,I,V).$$

Thus,

$$\frac{d}{dt}(\log I)\geq-h_{1}(T,I,V).$$

Yet $(T,I,V)$ is a solution of system (4); $T$, $I$, $V$ are class $C^{1}$ functions. They are therefore continuous on $[t_{0},t_{*}]$ and consequently are bounded on $[t_{0},t_{*}]$.
Therefore $h_{2}(T,I,V)$ is bounded on $[t_{0},t_{*}]$.
Thus, there exists a constant $\lambda>0$ such that

$$\frac{d}{dt}(\log I)\geq-h_{2}(T,I,V)>-\lambda.$$

Integrating, the last expression on $[t_{0},t_{*}]$, yields

$$\log I(t_{*})-\log I(t_{0})\geq-\lambda(t_{*}-t_{0}).$$

Therefore

$$I(t_{*})\geq I(t_{0})e^{-\lambda(t_{*}-t_{0})}>0.$$

This is a contradiction because $I(t_{*})=0$.
As far as that goes, assuming that $x_{3}(t_{*})=0$.
By hypothesis, $g_{3}(T,I,V)$ is positive on $[t_{0},t_{*}]$, we obtain $V\geq-Vc.$ i.e

$$\frac{d}{dt}(\log I)\geq-c.$$

Hence integrating on $[t_{0},t_{*}]$ we obtain

$$V(t_{*})\geq V(t_{0})e^{-k(t_{*}-t_{0})}>0.$$

This is a contraction.
Conclusion: $T$, $I$, $V$ are positive on $[t_{0},t_{1}[$.   
It will now be shown, with the help of the continuation criterion the existence of global solutions of problem (4), (5).

Proof. To prove this it is enough to show that all variables are bounded on an arbitrary finite interval $[t_{0};t)$. Using the positivity, by the theorem 2.2, of the solutions it is enough to show that all variables are bounded above.
Taking the sum of equations (1), (2) shows that :

$$\frac{d}{dt}(T+I)\leq\lambda$$

and hence that $T(t)+I(t)\leq T_{0}+I_{0}+\lambda(t-t_{0})$. Thus $T$ and $I$ are bounded on any finite interval. The third equation, i.e.

$$\frac{dV}{dt}=(1-\varepsilon)pI-cV,$$

then shows that $V(t)$ cannot grow faster than linearly and is also bounded on any finite interval. This completes the proof of this theorem.   

Proof. Summing equations (1) and (2), we get:

	$\displaystyle\frac{d}{dt}(T+I)$	$\displaystyle=$	$\displaystyle s+\left(1-\frac{T+I}{T_{max}}\right)\left(r_{T}T-r_{I}I\right)-d_{T}T-d_{I}I,$
		$\displaystyle=$	$\displaystyle s+r_{T}T-r_{I}I-d_{T}T-d_{I}I-\frac{T+I}{T_{max}},$
		$\displaystyle=$	$\displaystyle s+\left(r_{T}-d_{T}\right)T+\left(r_{I}-d_{I}\right)I-\frac{T+I}{T_{max}}(r_{T}T+r_{I}I),$
		$\displaystyle\leqslant$	$\displaystyle s+\left(r_{T}-d_{T}\right)(T+I)-\frac{T+I}{T_{max}}(r_{T}T+r_{I}I)\quad\ \hbox{since}\quad\ r_{T}-d_{T}\geq r_{I}-d_{I},$
	$\displaystyle\mbox{thus}\;\frac{d}{dt}(T+I)$	$\displaystyle\leqslant$	$\displaystyle s+\left(r_{T}-d_{T}\right)(T+I)-\frac{r_{I}}{T_{max}}(T+I)^{2}\quad\ \hbox{since}\quad r_{I}\leqslant r_{T}.$

Let $N_{1}=T+I$, $a=s>0$, $b=\left(r_{T}-d_{T}\right)>0$, $d=-\frac{r_{I}}{T_{max}}<0$ and let us solve the following equation

$$\frac{dN_{1}}{dt}=a+bN_{1}+dN_{1}^{2}$$

(9)

Coupled to equation (9) the initial condition :

$$N_{1}(t_{0})=N_{1}^{0}.$$

(10)

The resolution of the problem (9), (10) gives for all $t\in[t_{0},+\infty[$,

$$N_{1}(t)=-\frac{1}{2d}\biggl{[}\tanh\biggl{(}\frac{1}{2}\sqrt{-4ad+b^{2}}-\frac{1}{2}t_{0}\sqrt{-4ad+b^{2}}-\arctan\left(\frac{2N_{1}^{0}+b}{\sqrt{-4ad+b^{2}}}\right)\biggr{)}\sqrt{-4ad+b^{2}}\biggr{]}-\frac{b}{2d}.$$

As for all $x\in\mathbb{R},-1\leqslant\tanh x\leqslant 1$, it follows that :

$$N_{1}(t)\leqslant-\frac{1}{2d}\left(\sqrt{-4ad+b^{2}}+b\right)$$

i.e.

$$N_{1}(t)\leqslant\frac{T_{max}}{2r_{I}}\left(\sqrt{(r_{T}-d_{T})^{2}+\frac{4sr_{I}}{T_{max}}}+r_{T}-d_{T}\right).$$

Let

$$\tilde{T}_{0}=\frac{T_{max}}{2r_{I}}\left(\sqrt{(r_{T}-d_{T})^{2}+\frac{4sr_{I}}{T_{max}}}+r_{T}-d_{T}\right),$$

we obtain :

$$N_{1}(t)\leqslant\tilde{T}_{0}.$$

Therefore

$$T+I\leqslant\tilde{T}_{0}.$$

Since T and I are positive $I\leqslant T+I$ and $T\leqslant T+I$, so it follows that $T(t)\leqslant\tilde{T}_{0}$ and $I(t)\leqslant\tilde{T}_{0}$.
From (3), we have:

	$\displaystyle\frac{dV}{dt}$	$\displaystyle\leqslant$	$\displaystyle(1-\varepsilon)p(T+I)-cV,$
		$\displaystyle\leqslant$	$\displaystyle(1-\varepsilon)p\tilde{T_{0}}-cV\quad\ \hbox{since}\quad\ T+I\leqslant\tilde{T_{0}}.$

According to Gronwall inequality,

	$\displaystyle V(t)$	$\displaystyle\leqslant$	$\displaystyle V(t_{0})e^{-c(t-t_{0})}+\int_{t_{0}}^{t}(1-\varepsilon)p\tilde{T_{0}}e^{\int_{u}^{t}-cds}du,$
		$\displaystyle\leqslant$	$\displaystyle V_{0}e^{-c(t-t_{0})}+(1-\varepsilon)p\tilde{T_{0}}\int_{t_{0}}^{t}e^{-c(t-u)}du,$

$\displaystyle V(t)$

$\displaystyle\leqslant$

$\displaystyle V_{0}e^{-c(t-t_{0})}+(1-\varepsilon)p\tilde{T_{0}}\frac{e^{-c(t-t)}-e^{-c(t-t_{0})}}{c},$

		$\displaystyle\leqslant$	$\displaystyle V_{0}e^{-c(t-t_{0})}+(1-\varepsilon)p\tilde{T_{0}}\frac{1-e^{-c(t-t_{0})}}{c},$
		$\displaystyle\leqslant$	$\displaystyle\lambda_{0}\Big{(}e^{-c(t-t_{0})}+1-e^{-c(t-t_{0})}\Big{)}$
	$\displaystyle V(t)$	$\displaystyle\leqslant$	$\displaystyle\lambda_{0}.$

with

$$\lambda_{0}=\max\Big{\{}V_{0},\frac{(1-\varepsilon)}{c}p\tilde{T_{0}}\Big{\}}.$$

This completes the proof of theorem 2.4.   

Proof. When there is no viral infection, the uninfected hepatocytes dynamic is determined by :

$$\frac{dT}{dt}=s+r_{T}T\left(1-\frac{T}{T_{max}}\right)-d_{T}T.$$

(11)

The quantity $T^{0}$ of the free-virus equilibrium point $E^{0}$ is solution of the equation $\frac{dT}{dt}=0$.
Hence, let us solve the following equation :

$$-\frac{r_{T}}{T_{max}}T^{2}+(r_{T}-d_{T})T+s=0.$$

Its discriminant is given by :

$$\Delta=(r_{T}-d_{T})^{2}+4\frac{r_{T}}{T_{max}}s,$$

which yields :

$$T=\frac{d_{T}-r_{T}-\sqrt{(r_{T}-d_{T})^{2}+4\frac{r_{T}}{T_{max}}s}}{2\frac{r_{T}}{T_{max}}}.$$

Thus, in the absence of viral infection, the amount of susceptible cells or uninfected hepatocytes attend to a positive constant level $T^{0}$, which is :

$$T^{0}=\frac{T_{max}}{2r_{T}}\left(r_{T}-d_{T}+\sqrt{(r_{T}-d_{T})^{2}+4\frac{r_{T}}{T_{max}}s}\right).$$

This completes the proof.   

We are going to use Van Den Driessche and Watmough method [9, 21] for calculating the basic reproduction ratio $\mathcal{R}_{0}$ of the model (4).
Let us first present briefly the method.
Considering population whose population are grouped into $n$ homogeneous compartments $X=(X_{1},X_{2},...,X_{n})$ where $X_{i}\geq 0$ is the number of individuals in compartment $i$. For clarity we sort the compartments $X_{i}$, $i=1,...,n$ so that the first $m$ $(m\leqslant n)$ compartments correspond to infected individuals. The distinction between infected and uninfected compartments must be determined from the epidemiological interpretation of the model and cannot be deduced from the structure of the equations alone.
We define $X_{s}$ to be the set of all disease free states. That is :

$$X_{s}=\left\{X\geq 0/X_{1}=X_{2}=....=X_{m}=0\right\}.$$

Let $F_{i}(X)$ be the rate of appearance of new infections in compartment $i$ that is the infected individuals coming from other compartments and enter into $i$.
$V_{i}^{+}$ be the rate of transfer of individuals into compartment $i$ by all other means (displacement, healing, aging).
$V_{i}^{-}$ be the rate of transfer of individuals out of compartment $i$ (mortality, change of statut).
It is assumed that each function is continuously differentiable at least twice in each variable.
Figure 3 below shows the variations of the number of individuals in compartment $i$ in a population.

The variation of the number of individuals in compartment $i$ is given by :

$$\frac{dX_{i}}{dt}=F_{i}(X)-V_{i}(X)$$

where

$$V_{i}(X)=V_{i}^{-}(X)-V_{i}^{+}(X).$$

Due to the nature of the epidemiological model, we have the following properties :

1. $P_{1})$

   If $X_{i}\geq 0$ then $F_{i}(X)\geq 0$, $V_{i}^{-}(X)\geq 0$ and $V_{i}^{+}(X)\geq 0$.
   Since each function represents a directed transfer of individuals.

2. $P_{2})$

   If $X_{i}=0$ then $V_{i}^{-}(X)=0$.
   Indeed, If a compartment is empty, then there can be no transfer of individuals out of the compartment by death, infection, nor any other means : it is the essential property of a compartmental model.

3. $P_{3})$

   Pour $i>m$, $F_{i}(X)=0$.
   In fact, the compartments with an index greater than $m$ are ”uninfected”. By definition, it can not appear in these compartments infected individual.

4. $P_{4})$

   Si $X\in X_{s}$ alors $F_{i}(X)=0$ et pour $i\leqslant m$, $V_{i}^{+}(X)=0$.
   Indeed, to ensure that the disease free subspace is invariant, we assume that if the population is free of disease then the population will remain free of disease. That is, there is no (density independent) immigration of infectives. This is Lavoisier’s principle. There is no spontaneous generation.

Let $F=(F_{i})_{i=1,...,n}$ and $V=V^{+}-V^{-}=(V_{i}^{+}-V_{i}^{-})_{i=1,.....,n}$.
Let also $x^{0}$ the uninfected equilibrium point of the corresponding model.
Let $DF(x^{0})$ and $DV(x^{0})$ denote the jacobian matrices of F and V respectively at the point $x^{0}$.
It follows that $DF(x^{0})$ is a positive matrix and $DV(x^{0})$ a Metzler matrix (matrix whose the extra diagonal terms are greater or equal than zero ). Thus we have the following equivalent definition of $\mathcal{R}_{0}$ :