Why is the recovery time for a logistic model $\frac{1}{r}$

Question

What I mean by Logistic Model is the case that $\frac{dN}{dt}=f(N)=rN(1-\frac{N}{K})$.

Also the steady states are $N$ when $f(N)=rN(1-\frac{N}{K})=0$

This model has two steady states, one at $0$ and the other at $K$. If we consider a small perturbation about $K$, say $N=K+n$, then we will find, by linearising $$\frac{dn}{dt}=-rn\Longrightarrow n(t)=n(0)e^{-rt}$$

Now, my course defines recovery time to be the time taken for a perturbation to decrease by a factor of $e$ about the steady state $K$ and concluded that the recovery time for logistic model is $\frac{1}{r}$.

My confusions

Question 1:

I am not entirely sure why it followed immediately the recovery time is $\frac{1}{r}$, is it because $n(t+\frac{1}{r})=n(0)e^{-r(t+\frac{1}{r})}=\frac{1}{e}n(t)?$

Question 2:

Is it pure a coincidence that $\frac{1}{r}$ also happens to be $\frac{1}{|f'(K)|}$, or are these two closely related?

Question 3:

Why the definition of recovery time given above sensible? I mean what exactly are we trying to 'recover' in that definition?

Thank you to everyone in advance!

Answer 1

1) $n(t)/e$ is a factor of $e$ smaller than $n(t)$.

2) Absolutely not. $r$ comes from linearising $f$ around $K$, a fixed point. So the behaviour of the solution will be governed by $f'(K)$.

3) You're saying if you move away from the fixed point for some reason, how long will it take to recover the steady state, i.e. how long will it take to come back into equilibrium.

Answer 2

1) Yes. 3) is because of 2). You could also take the time required for a reduction by a factor 2 or 10, but then you would get their logarithms in the formula. In practical applications you might need something like the time of reduction to 5% or 0.1%, there you have to compute the corresponding multiple of the "unit" recovery time anyway, so that an initial factor in the unit only adds complexity.