1
$\begingroup$

The logistic differential equation $y'=y(b-ay) \, \textrm{with}\, a\neq 0, b\neq 0$ has the non-trivial solution

$y(t) = \frac{\frac{b}{a}}{1+c\cdot e^{-bt}}\tag{1}$

$\quad\quad = \frac{b}{(a+a\cdot c\cdot e^{-bt})}\tag{2}$ where $c$ is a constant.

Why should we assume that $c$ is a positive real number?

  • 0
    There is actually no reason to assume that c>0.2012-11-12

1 Answers 1

1

There isn't. The logistic equation is commonly written in the form $ {dP\over dt}=rP\left(1-{P\over K}\right), \quad P(0)=P_0, $ and in the context of logistic population models,

  • $P$ is population
  • $t$ is time
  • $r$ is the intrinsic growth rate
  • $K$ is the carry capacity of the environment
  • $P_0$ is the initial population

Because of their physical meaning, each is taken to be positive. The solution is $ P(t)={KP_0\over P_0+(K-P_0)e^{-rt}}={K\over 1+\left({K\over P_0}-1\right)e^{-rt}}. $ This latter formulation matches the first form of your solution, just with $b:=r$ and $a:=r/K$, and $c:={K\over P_0}-1$.

There is no mathematical nor physical reason why we must have $c>0$. A negative value for $c$ would just mean that the initial population happened to be greater than the carrying capacity.