I am reading the book on virus dynamics by Novak and May, and it describes a system of ODEs to describe the viral load when antivirals are used. The systems is
$\dot y = -ay\\\dot v = ky-uv$
The solutions of this system are
$y(t)= y^*e^{-at}\\v(t) = {v^*(ue^{-at}-ae^{-ut})\over (u-a)}$.
it is assumed that $u>>a$.
Then it is analysed that the virus load ($v(t)$) start to decline at rate $e^{-at}$ only after a certain shoulder phase of duration given as $\Delta t \approx 1/u$ (more precisely, $\Delta t= (-1/a) ln (1-a/u) $ ).
Can someone please explain how this $\Delta t$ is obtained?
To see it, first note that since $u\gg a>0$ (assuming they are positive constants), the term $ae^{-ut}$ will go to $0$ more quickly than $ue^{-at}$. That is why $v(t)$ can be approximated by $ue^{-at}/(u-a)v_0$ after a certain time. To compute that time, see that on such a logarithmic scale as in your figure, $ue^{-at}/(u-a)v_0$ and $e^{-at}v_0$ correspond to parallel lines. Now since we start at $v_0$, i.e. at $e^{-at}v_0|_{t=0}$ and converge to $ue^{-at}/(u-a)v_0$ then the question can be rephrased to finding the translation from line to line, i.e. solving \begin{align} e^{-a(t-\Delta t)}v_0 &= \frac{u}{u-a}e^{-at}v_0\\ a\Delta t &=\mathrm{ln}(u)-\mathrm{ln}(u-a)\\ \Delta t &= -\frac{1}{a}\big(\mathrm{ln}(u-a)-\mathrm{ln}(u)\big)\\ &= -\frac{1}{a}\mathrm{ln}((u-a)/u)\\ &= -\frac{1}{a}\mathrm{ln}(1-a/u). \end{align}