Understanding a dynamical system for virus populations I am reading the book: M. A. Novak, R. M. May: Virus Dynamics: Mathematical Principles of Immunology and Virology. Oxford University Press, 2000.
I came across few places that I don’t understand. 
In the model, uninfected cells react with free virus to give rise to infected cells; the rate constant is $\beta$. Infected cells produce free virions at rate $k$. Uninfected cells, free virus and infected cells die at rates $d$, $u$ and $a$, respectively. Uninfected cells are replenished at rate $\lambda$.
The schematic of the model is:
 
The model equations are [eq. (3.1) p. 18 in the link]:
$$
\begin{alignat}{1}
\dot{x} &= λ - dx - βxv,\\
\dot{y} &= βxv - ay, \\
\dot{v} &= ky - uv,
\end{alignat}
$$
where $x$, $y$, and $v$ are the populations of uninfected cells, infected cells and virions.
The basic reproductive ratio, $R_0$ is the number of newly infected cells that arise from any infected cell when almost all cells are uninfected (and the system is near its equilibrium):  
$$R_0= \frac{\beta \lambda k}{adu}$$ 
What I don’t understand is:


*

*It is mentioned that if $R_0<1$ the virus will not spread since every infected cell will produce on average less than one infected cell. If we start with $N$ infected cells, then on average, we expect roughly ${\ln N}\over {\ln(1/R_0)}$ rounds of replication before the virus population dies out. How is this ${\ln N}\over {\ln(1/R_0)}$ found?

*If $R_0>1$ then virus will initially grow exponentially, and $r_0$ is the exponential growth rate of the population. Then it is said $r_0$ is given by the larger root of the equation $r_0^2+(a+u)r_0+au (1-R_0)=0$. How did they come up with this equation?  

*Then they say in the equation $r_0^2+(a+u)r_0+au (1-R_0)=0$ if $u \gg a+r_0$ we find the approximation $r_0=a(R_0-1)$. How is this $r_0=a(R_0-1)$ obtained?
 A: The equations are related to discrete approximations of the infection process of the virus.

*

*Let $F_n$ be the number of infected cells at replication step $n$.  Then, we have that $F_{n+1} \approx R_0 F_n$, $F_0=N$.  Then we can solve this recurrence relation to get $F_n = R_0^nN$.  We want to know for what $n$ we get $F_n <1$.  So we solve $R_0^n N <1$ for $n$.  Taking logarithms we get that $$n\ln(R_0)+\ln N < 0$$
subtracting, and dividing by $\ln(R_0)$ (and remember that $\ln(R_0) < 0$ since $R_0 <1$, so we have to switch the direction of the inequality) we get $$n > \frac{-\ln N}{\ln(R_0)} = \frac{\ln N}{\ln \left(\frac{1}{R_0}\right)}$$


*Here is my best guess for this one.  Suppose that we insert a single virion into the population when it is at equilibrium.  We then have that as time progresses $v(t) = e^{r_0t}$.  Plugging this into the $\dot{v}$ equation gives that $ky=(r_0+u)e^{r_0t}$.  Differentiating the $\dot{v}$ equation again gives $$\ddot{v} + u\dot{v} = k\beta x v - a(ky).$$
Using the above expression for $ky$ and the fact that $x \approx \frac{\lambda}{d}$ we get $$\ddot{v} + u\dot{v} - \frac{k \beta \lambda}{d}v = -a(r_0+u)e^{r_0t}.$$ Plug in $v = e^{r_0t}$, which then allows you to cancel the $e^{r_0t}$ from both sides and this gives $$r_0^2+ur_0- \frac{k \beta \lambda}{d} = -ar_0-au.$$ Thus $$r_0^2 + (u+a)r_0 + au-\frac{k\beta\lambda}{d}=0$$
Then, finally, we factor $au$ out of the constant terms and use the definition of $R_0$ to get $$r_0^2 + (u+a)r_0 + au (1-R_0) = 0.$$


*Expand the equation to get $r_0^2 + a r_0 + ur_0 + au(1-R_0) = 0$.  Dividing both sides by $u$ gives $$ r_0 \left( \frac{r_0+a}{u} \right) + r_0+a(1-R_0)=0$$
Since $u \gg r_0+a$ we get that $\frac{r_0+a}{u} \ll 1$.  since the leading term in the above equation is so small, it can be neglected.  Tossing it out gives $$ r_0 + a(1-R_0) = 0 \Rightarrow r_0 = a(R_0-1).$$
A: An alternative way to solve problem 2 using some knowledge on dynamical systems:
We first insert our knowledge of being near the uninfected equilibrium (with $x=\frac{λ}{d}$) into the differential equations:
$$
\begin{alignat}{4}
\dot{y} &=~& - ay &~+~& \frac{λβ}{d} v, \\
\dot{v} &=& ky &~-~& uv,
\end{alignat}
$$
This is a linear system of differential equations, which can be rewritten as a matrix–vector multiplication (whose matrix I denote as $A$):
$$
\frac{\mathrm{d}}{\mathrm{d}t} \begin{pmatrix}y\\v\end{pmatrix}
= 
\begin{pmatrix}
-a & \frac{λβ}{d} \\
 k & -u
\end{pmatrix}
\begin{pmatrix}y\\v\end{pmatrix}
=:
A
\begin{pmatrix}y\\v\end{pmatrix}
$$
The solutions to such a differential equation are known to be of the form:
$$ α_1 e^{ρ_1 t} w_1 + α_2 e^{ρ_2 t} w_2,$$
where the $w_i$ are the eigenvectors of $A$, the $ρ_i$ are the corresponding eigenvalues and the $α_i$ are constants determined by the initial conditions.
Now, the component corresponding to the largest eigenvalue will dominate the other one and thus we get exponential growth with this eigenvalue as a growth rate.
(As $v$ and $y$ depend on each other, everything grows with the same rate.)
So, all that is left to do is to determine the characteristic polynomial for $A$ (whose roots are the eigenvalues):
$$
  p_A(r_0)
= \det(r_0-A) = \det\pmatrix{r_0+a & \frac{-λβ}{d} \\ -k & r_0+u}
= (r_0+a)(r_0+u) - \frac{kλβ}{d} \\
= r_0^2 +(a+u)r_0 + au - auR_0
= r_0^2 +(a+u)r_0 + au (1 - R_0)
$$
