# Global Dynamics and Implications of an HBV Model with Proliferating Infected Hepatocytes
## 3 Global Stability Result for \(E_{0}\)  
For a sufficiently virulent infection and small proliferation rate of the infected hepatocytes, the liver will completely fail. Mathematically, this situation is represented by the global asymptotic stability of the liver failure state, \(E_{0}\). This section builds towards proving that if \(R_{0}>\left(\frac{r}{\mu}+1\right)R^{*}\) and \(\rho<a<\mu+\rho\), then \(E_{0}\) is globally stable. This technique is modified from [PERSON] et al. [36]. The \"blow-up transformation\" has been used previously to study similar complex equilibrium [46; 47; 48]. Notice that a similar argument can be used for (7)-(9) to prove that if \(R_{0}>\frac{r+\mu}{\mu}\) and \(r<a<\mu+r\), then \(E_{0}\) is globally stable. Since there is a singularity at the liver failure state, \(E_{0}\), we used a ratio dependent transformation to arrive at the global result. We used the transformation \((x,y,v)\to(x,z,w)\), where \(z=\frac{v}{x}\) and \(w=\frac{v}{x}\). This results in the following system,  
\[\frac{dx}{dt} = rx(t)\bigg{(}1-\frac{x(t)+y(t)}{K}\bigg{)}-\frac{\beta w(t)x(t)} {1+z(t)}, \tag{24}\] \[\frac{dz}{dt} = (\rho-r)z(t)\bigg{(}1-\frac{x(t)+y(t)}{K}\bigg{)}+\beta w(t)-az(t),\] (25) \[\frac{dw}{dt} = \gamma z(t)-uw(t)-rw(t)\bigg{(}1-\frac{x(t)(1+z(t))}{K}\bigg{)}+ \frac{\beta w(t)^{2}}{1+z(t)}. \tag{26}\]  
The steady states of (24)-(26) are:  
\[U_{0}=(0,0,0),\qquad\quad U_{n}=(0,z_{n},w_{n}),\qquad\quad U_{f}=(K,0,0), \qquad\quad U^{*}=(x^{*},z^{*},w^{*}),\]  
where:  
\[z_{n}=\frac{R^{*}(1+\frac{r}{\mu})-R_{0}}{R^{*}(\frac{a}{\mu}-\frac{e}{\mu}- 1)+R_{0}},\qquad\quad w_{n}=\frac{aR^{*}}{\beta}z_{n}, \tag{27}\]  
and:  
\[x^{*}=\frac{Kra}{(r-\rho)^{2}}(1-\frac{\rho}{r}R_{0})\bigg{(}\frac{R^{*}}{R_{ 0}}-1\bigg{)},\qquad\quad z^{*}=\frac{R_{0}-1}{1-\frac{\rho}{r}R_{0}},\qquad \quad w^{*}=\frac{\gamma(R_{0}-1)}{\mu(1-\frac{\rho}{r}R_{0})}. \tag{28}\]  
Notice that \(U_{n}\) is nonnegative when \(\frac{\rho}{\mu}-\frac{a}{\mu}<\frac{R_{0}}{R^{*}}-1<\frac{r}{\mu}\) and \(U^{*}\) is nonnegative when \(R^{*}<R_{0}<1\) and \(R_{0}<\frac{r}{\rho}\). The nontrivial steady states are preserved in that \(E_{f}=U_{f}\) and \(E^{*}=U^{*}\). \(E_{0}\) has been blown up into two steady states: \(U_{0}\) and \(U_{n}\). We call these the two trivial states. To find a global stability result for \(E_{0}\), we show that if \(R_{0}>\left(\frac{r}{\mu}+1\right)R^{*}\) and \(\rho<a<\mu+\rho\), all steady states of (24)-(26) are unstable and that \(\liminf\limits_{t\to\infty}x=0\) and \(\limsup\limits_{t\to\infty}z,w=\infty\). Since \(y=xz\), \(v=xw\), and (14)-(16) is bounded, this is enough to show if \(R_{0}>\left(\frac{r}{\mu}+1\right)R^{*}\) and \(\rho<a<\mu+\rho\), then \(\liminf\limits_{t\to\infty}x,y,v=0\). Lemma 8: \(U_{0}\) _and \(U_{n}\) are always unstable._  
Proof: The variational matrix of the system (24)-(26) evaluated at \(U_{0}\) is:  
\[J(x,z,w)\big{|}_{E_{0}}=\left( \begin{array}{ccc}r&0&0\\ -r&-aR^{*}&\beta\\ 0&\gamma&-(\mu+r)\end{array} \right).\]  
Since \(\lambda_{1}=r>0\), \(U_{0}\) is always unstable. The variational matrix of the system (24)-(26) evaluated at \(U_{n}\) is:  
\[J(x,z,w)|_{U_{n}}=\left( \begin{array}{ccc}r-\beta\frac{w_{n}}{1+z_{n}}&0&0\\ \frac{r-\rho}{K}z_{n}(1+z_{n})&-aR^{*}&\beta\\ \frac{r}{K}w_{n}(1+z_{n})&\gamma-\beta\frac{w_{n}^{2}}{(1+z_{n})^{2}}&-\mu-r+2 \beta\frac{w_{n}}{1+z_{n}}\end{array} \right).\]  
\(\lambda_{1}=r-\beta\frac{w_{n}}{1+z_{n}}=\mu\left(\frac{R_{0}}{R^{*}}-1\right)\), which is negative when \(R_{0}<R^{*}\). On the other hand, \(\lambda_{23}\) are given by the following matrix,  
\[A=\left( \begin{array}{ccc}-aR^{*}&\beta\\ \gamma-\frac{\beta w_{n}^{2}}{(1+z_{n})^{2}}&-\mu-r+2\beta\frac{w_{n}}{1+z_{n} }\end{array} \right),\]  
where:  
\[tr(A)=-a-2r-\mu-z_{n}(\mu-a).\]  
The trace is negative when \(-\frac{a+\mu}{2}<\lambda_{1}\). However,  
\[\det(A) = a\mu\bigg{[}\frac{R_{0}}{1+z_{n}}-\frac{(\mu+r)R^{*}}{(1+z_{n}) \mu}\bigg{]},\]  
which is positive when \(R_{0}>(1+\frac{r}{\mu})R^{*}\).