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# Logistic Equation

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 Sub Topics Population is the number of entities living in a certain area at a given point of time. At first it was assumed that the population will have a constant rate of growth where the growth will be directly proportional to the present population but later on it was considered unrealistic. After a point all the resources will be falling short to support the population growth. Logistic equation is a model for population growth which was published by Verhulst. A logistic equation is a model of population growth which is not based just on the present population but also on the carrying capacity of the given region based on the available resources to sustain the population growth. The logistic equation has its use in ecology, statistics, finding the carrying capacity of a region, in economical surveys and so on. This model is also applied in the chemical reactors to model the rate of change in given quantity.

## Logistic Equation of Growth

The population growth depends upon the present population N, the saturation level K which is the carrying capacity and the growth rate r. The population will grow at a fast rate in the beginning but once it reaches the saturation level the growth slows down. $\frac{dN}{dt}$ is the population after time t.

The curve of the population growth will be of S shape which implies that the population will grow fastly in the beginning but after reaching a saturation level it will be stagnant for a while and then it will again grow at a faster rate. When we take the exponential growth of population where the present population is $N_0$ which becomes N after time t, the equation will be N = $N_0e^rt$ for exponential growth rate r.

## Logistic Differential Equation

A differential equation is an equation which contains derivative of a function and relates the derivative to the function. In the equation for population growth, the logistic differential equation tells the relation between rate of change in the population, dN/dt and the present population, N.

The logistic differential equation can be written as  $\frac{dN}{dt}$= rN(1 - $\frac{N}{K}$)  which shows the population change with respect to time t.

Solving analytically we get the solution as N = $\frac{K}{1 + Ae^{rt}}$ where A = $\frac{K - N_{0}}{N_{0}}$.

## Problems

Example 1: Let the population of a state be 15 million in year 2000 which does change to 18 million in 2010. If the growth rate is said to be 0.0225, then what is the saturation level in the logistic growth?

Solution: We have r = 0.0225, t = 10, N0 = 15 million and N = 18 million. Putting all the given data in N =

$\frac{K}{1+Ae^{-rt}}$ = $\frac{K}{1+(\frac{K - N_{0}}{N_{0}})e^{-rt}}$ and solving we get K = 86.8023 million.

Example 2: In 1900 the population of an Asian country is said to be 5 million. In the year 1920 it reaches 15 million. What will be the exponential growth rate and logistic growth rate if the carrying capacity is said to be 45 million.

Solution: When we take the exponential model of growth, we get the equation as N =$N_0e^rt$.We have N = 15 million, $N_0$ = 5 million and t = 20 years. Now, putting all the values in the formula we get, 15 = 5$e^{r20}$ where r is the exponential growth rate. Solving we get $e^{r20}$ = 3. Taking log on both sides, 20r = $log3$ $\rightarrow$  r = $\frac{log3}{20}$ = 0.023856062.

For the logistic model we use the logistic differential equation $\frac{dN}{dt}$= rN(1 - $\frac{N}{K}$) whose analytical solution will be N = $\frac{K}{(1+Ae^{-rt})}$. We get,

A = $\frac{(K - N_0)}{N_0}$ = $\frac{(45 - 5)}{5}$ = 8.

Putting the value of A in the above solution we get, 15 = $\frac{45}{(1 + 8e^{-r20})}$.

Solving we get, r = $\frac{log4}{20}$ = 0.03010299957.