Why the Logistic Differential Equation Accurately Models Population

Question

I understand how the logistic differential equation (disregarding the Malthusian parameter)

$\frac{dN}{dt}=KN(A-N)$ where A is the carrying capacity

produces the graph that it does, however, I was wondering why this differential equation is a natural choice when modeling the growth of things with a carrying capacity i.e. how does the existence of a carrying capacity and growth based on the current "population" lead to this differential equation? Likewise, why do curiosities such as the rate of growth is at its maximum when $N=\frac{A}2$ arise?

Answer 1

I was wondering why this differential equation is a natural choice when modeling the growth of things with a carrying capacity

Because it is in a sense a simplest dynamical system with density-dependent regulation. To see this start with the general model $$ \dot N=NF(N), $$ where $F(N)$ is the per capita birth rate and assume that it is smooth enough to have Taylor's series $$ F(N)=a+bN+cN^2+\ldots $$ If one keeps only constant term then they get Maltus model $\dot N=aN$. The next step --- keeping linear terms only --- leads to the logistic equation $$ \dot N=N(a+b N). $$

On the other hand it would be too naive to expect that logistic equation actually represents some kind of biological law. Much more on this, together with data, can be found in a very interesting paper Feller, W., 1940. On the logistic law of growth and its empirical verifications in biology. Acta biotheoretica, 5(2), pp.51-66.

Answer 2

The solution to $$\frac{dN}{dt}=kN(A-N)$$ is $$ N(t) = \frac {AN(0)}{N(0)+(A-N(0)e^{-kt}}$$

The solution satisfies all the requirements of a population which starts at $N=N(0)$ and has the carrying capacity of $A.$

The difference between the carrying capacity and the population is decreasing very slowly as population approaches its carrying capacity.

The reason for the fastest growth at $N=A/2$ is that this point is where the second derivative of $N$ is zero.

That is an inflection point where the first derivative attains its maximum.

Note that $$\frac{d^2N}{dt^2}=k(A-2N)$$ which is zero at $N=A/2.$

Answer 3

Regarding the final question, you can maximize $f(N) = K N(A - N)$ with respect to $N$ by setting the derivative equal to $0$: $$ f'(N) = KA - 2KN = 0 \implies N = \frac{A}{2}. $$