How does logistic growth occur?

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How does logistic growth occur?

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Answer 1 · 2014-10-16T22:20:00

Logistic growth of population occurs when the rate of its growth is proportional to the product of the population and the difference between the population and its carrying capacity $M$ $M$ , i.e.,

$\frac{d P}{d t} = k P (M - P)$ ${dP}/{dt}=kP(M-P)$ , where $k$ $k$ is a constant,

with initial population $P (0) = P_{0}$ $P(0)=P_0$ .

As you can see above, the population grows faster as the population gets larger; however, as the population gets closer to its carrying capacity $M$ $M$ , the growth slows down.

by separating variables and integrating,

$\Rightarrow \int \frac{1}{P (M - P)} d P = \int k d t$ $Rightarrow int 1/{P(M-P)}dP=int kdt$

by Partial Fraction Decomposition,

$\Rightarrow \frac{1}{M} \int (\frac{1}{P} + \frac{1}{M - P}) d P = \int k d t$ $Rightarrow 1/M int(1/P+1/{M-P})dP=int kdt$

by multiplying by $M$ $M$ ,

$\Rightarrow \int (\frac{1}{P} + \frac{1}{M - P}) d P = \int k M d t$ $Rightarrow int(1/P+1/{M-P})dP=int kMdt$

$\Rightarrow ln | P | - ln | M - P | = k M t + C_{1}$ $Rightarrow ln|P|-ln|M-P|=kMt+C_1$

$\Rightarrow ln ∣ ∣ ∣ \frac{P}{M - P} ∣ ∣ ∣ = k M t + C_{1}$ $Rightarrow ln|P/{M-P}|=kMt+C_1$

$\Rightarrow ∣ ∣ ∣ \frac{P}{M - P} ∣ ∣ ∣ = e_{1}^{k M t + C_{1}} = e^{k M t} \cdot e_{1}^{C_{1}}$ $Rightarrow |P/{M-P}|=e^{kMt+C_1}=e^{kMt}cdot e^{C_1}$

$\Rightarrow \frac{P}{M - P} = \pm e_{1}^{C_{1}} e^{k M t} = C e^{k M t}$ $Rightarrow P/{M-P}=pm e^{C_1}e^{kMt}=Ce^{kMt}$

Since $P (0) = P_{0}$ $P(0)=P_0$ ,

$\frac{P_{0}}{M - P_{0}} = C e^{k M (0)} = C$ $P_0/{M-P_0}=Ce^{kM(0)}=C$

So, the equation becomes

$\frac{P}{M - P} = \frac{P_{0}}{M - P_{0}} e^{k M t}$ $P/{M-P}=P_0/{M-P_0}e^{kMt}$

by solving for $P$ $P$ , we have the logistic equation

$\Rightarrow P (t) = \frac{M}{1 + (\frac{M}{P_{0}} - 1) e^{- k M t}}$ $Rightarrow P(t)=M/{1+(M/P_0-1)e^{-kMt}}$ .

I hope that this was helpful.

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How does logistic growth occur?

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