TITLE: Understanding a dynamical system for virus populations
QUESTION [7 upvotes]: 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?

REPLY [6 votes]: 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):
$$
  p_A(r_0)
= \det(r_0-A) = \det\pmatrix{r_0+a & \frac{-λβ}{d} \\ -k & r_0+u}
= (r_0+a)(r_0+u) - \frac{kλβ}{d} \\
= r_0^2 +(a+u)r_0 + au - auR_0
= r_0^2 +(a+u)r_0 + au (1 - R_0)
$$