Understanding a dynamical system for virus populations

The equations are related to discrete approximations of the infection process of the virus.

1. 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 $$n\ln(R_0)+\ln(N) < 1$$
   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 $$n > \frac{-\ln(N)}{\ln(R_0)} = \frac{\ln(N)}{\ln \frac{1}{R_0}}$$

2. 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 $$r_0^2 + (u+a)r_0 + au (1-R_0) = 0.$$

3. 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 $$ r_0 + a(1-R_0) = 0 \Rightarrow r_0 = a(R_0-1).$$

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):

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