# How does logistic growth occur?

Oct 16, 2014

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 $M$, i.e.,

$\frac{\mathrm{dP}}{\mathrm{dt}} = k P \left(M - P\right)$, where $k$ is a constant,

with initial population $P \left(0\right) = {P}_{0}$.

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

by separating variables and integrating,

$R i g h t a r r o w \int \frac{1}{P \left(M - P\right)} \mathrm{dP} = \int k \mathrm{dt}$

by Partial Fraction Decomposition,

$R i g h t a r r o w \frac{1}{M} \int \left(\frac{1}{P} + \frac{1}{M - P}\right) \mathrm{dP} = \int k \mathrm{dt}$

by multiplying by $M$,

$R i g h t a r r o w \int \left(\frac{1}{P} + \frac{1}{M - P}\right) \mathrm{dP} = \int k M \mathrm{dt}$

$R i g h t a r r o w \ln | P | - \ln | M - P | = k M t + {C}_{1}$

$R i g h t a r r o w \ln | \frac{P}{M - P} | = k M t + {C}_{1}$

$R i g h t a r r o w | \frac{P}{M - P} | = {e}^{k M t + {C}_{1}} = {e}^{k M t} \cdot {e}^{{C}_{1}}$

$R i g h t a r r o w \frac{P}{M - P} = \pm {e}^{{C}_{1}} {e}^{k M t} = C {e}^{k M t}$

Since $P \left(0\right) = {P}_{0}$,

${P}_{0} / \left\{M - {P}_{0}\right\} = C {e}^{k M \left(0\right)} = C$

So, the equation becomes

$\frac{P}{M - P} = {P}_{0} / \left\{M - {P}_{0}\right\} {e}^{k M t}$

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

$R i g h t a r r o w P \left(t\right) = \frac{M}{1 + \left(\frac{M}{P} _ 0 - 1\right) {e}^{- k M t}}$.

I hope that this was helpful.