# Global Dynamics and Implications of an HBV Model with Proliferating Infected Hepatocytes
## 3 Global Stability Result for \(E_{0}\)
This condition and the condition for \(\lambda_{1}<0\) contradict, so \(U_{n}\) is always unstable. If \(R_{0}>(\frac{r}{\mu}+1)R^{*}\), then \(\limsup_{t\to\infty}z(t),w(t)=\infty\). Proof. : Let \((z(t),w(t))\) be a solution of:  
\[\frac{dz}{dt} = \beta w-az+(\rho-r)z\bigg{(}1-\frac{x(1+z)}{K}\bigg{)}\geq\beta w -(a+r-\rho)z,\] \[\frac{dw}{dt} = \gamma z-\mu w-rw\bigg{(}1-\frac{x(1+z)}{K}\bigg{)}+\frac{\beta w ^{2}}{1+z}\geq\gamma z-(\mu+r)w.\]  
Let \((Z(t),W(t))\) be a solution of:  
\[\frac{dZ}{dt} = \beta W-(a+r-\rho)Z, \tag{29}\] \[\frac{dW}{dt} = \gamma Z-(\mu+r)W. \tag{30}\]  
\((Z,W)=(0,0)\) is the only steady state of (29) and (30) and is unstable when \(R_{0}>(\frac{r}{\mu}+1)R^{*}\). Since there are no other steady states, \(Z,W\) are unbounded, and (29) and (30) is a cooperative, monotone system, that is \(\lim_{t\to\infty}Z(t),W(t)=\infty\). Let \(z_{0}=Z_{0}\) and \(w_{0}=W_{0}\), then by the comparison theorem, \((z(t),w(t))\geq(Z(t),W(t))\) for \(t>0\). Therefore,  
\[\liminf_{t\to\infty}z(t)\geq\lim_{t\to\infty}Z(t)=\infty,\]  
\[\liminf_{t\to\infty}w(t)\geq\lim_{t\to\infty}W(t)=\infty.\]  
Therefore, \(\liminf_{t\to\infty}z(t),w(t)=\infty\). If \(R_{0}>(\frac{r}{\mu}+1)R^{*}\) and \(\rho<a<\mu+\rho\), then \(\lim_{t\to\infty}x(t)=0\). **Proof. ** To prove that \(E_{0}\) is globally stable, we need it to be the only steady state in \(\mathcal{R}^{+}\). Therefore, set \(\rho<a\) to ensure that \(E_{i}\) no longer exists. Since:  
\[\frac{dx}{dt} = rx(t)\bigg{(}1-\frac{x(t)(1+z(t))}{K}\bigg{)}-\frac{\beta w(t)x(t)}{ 1+z(t)}<\bigg{(}r-\frac{\beta w(t)}{1+z(t)}\bigg{)}x(t),\] it is sufficient to show:  
\[\liminf_{t\to\infty}\frac{\beta w(t)}{1+z(t)}>r. \tag{31}\]  
Let \(\theta(t)=\frac{\beta w(t)}{1+z(t)}\). Then:  
\[\det(A) = a\mu\bigg{[}\frac{R_{0}}{1+z_{n}}-\frac{(\mu+r)R^{*}}{(1+z_{n}) \mu}\bigg{]},\]0 \[\det(A) = a\mu\bigg{[}\frac{R_{0}}{1+z_{n}}-\frac{(\mu+r)R^{*}}{(1+z_{n}) \mu}\bigg{]},\]1 \[\det(A) = a\mu\bigg{[}\frac{R_{0}}{1+z_{n}}-\frac{(\mu+r)R^{*}}{(1+z_{n}) \mu}\bigg{]},\]2  
By Lemma (9), for all \(\epsilon>0\), \(\exists\ t^{*}\) s.t. \(\forall\ t\geq t^{*}\),  
\[\det(A) = a\mu\bigg{[}\frac{R_{0}}{1+z_{n}}-\frac{(\mu+r)R^{*}}{(1+z_{n}) \mu}\bigg{]},\]3  
Combining (32) and (34) for \(t\geq t^{*}\),  
\[\det(A) = a\mu\bigg{[}\frac{R_{0}}{1+z_{n}}-\frac{(\mu+r)R^{*}}{(1+z_{n}) \mu}\bigg{]},\]4 \[\det(A) = a\mu\bigg{[}\frac{R_{0}}{1+z_{n}}-\frac{(\mu+r)R^{*}}{(1+z_{n}) \mu}\bigg{]},\]5 Letting \(\Gamma(\epsilon)=(a+r-\rho)(1-\epsilon)-\mu-r\) and solving for \(\theta(t)\) yields:  
\[\det(A) = a\mu\bigg{[}\frac{R_{0}}{1+z_{n}}-\frac{(\mu+r)R^{*}}{(1+z_{n}) \mu}\bigg{]},\]6  
Since \(\rho<a<\mu+\rho\), \(\Gamma(\epsilon)<0\). Therefore, \(\lim_{t\to\infty}\Theta(t)=\frac{\beta k(1-\epsilon)}{-\Gamma(\epsilon)}\). Since \(R_{0}>(\frac{r}{\mu}+1)R^{*}\), \(\exists\ \epsilon^{*}>0\) s.t. \(\forall\ \epsilon\in(0,\epsilon^{*}]\), \(\frac{\beta k(1-\epsilon)}{\mu+r}>a+r\). Therefore,  
\[\det(A) = a\mu\bigg{[}\frac{R_{0}}{1+z_{n}}-\frac{(\mu+r)R^{*}}{(1+z_{n}) \mu}\bigg{]},\]7  
Thus, (31) is satisfied. \(\Box\)  
According to Lemma 8, Lemma 9, and Theorem 10, this model presents the following biological hypothesis: if there is a sufficiently virulent infection and a small proliferation rate of infected hepatocytes, then the liver will fail. Note that Theorem 3.3 only provides sufficient conditions for the collapse of the hepatocyte population, which can happen before \(R_{0}\) is larger than \(\left(1+\frac{r}{\mu}\right)R^{*}\), as shown in \[\det(A) = a\mu\bigg{[}\frac{R_{0}}{1+z_{n}}-\frac{(\mu+r)R^{*}}{(1+z_{n}) \mu}\bigg{]},\]8 This leaves an open mathematical question for a necessary condition for the collapse of the hepatocyte population.