For this model it is assumed that ther rate of dy change of the population y is proportional to the dt The Logistic Population Model product of the current population y and K − y, or Math 121 Calculus II what is the same thing, proportion to the product D Joyce, Spring 2013 y(1 − y=K). That gives us the logistic differential equation dy Summary of the exponential model. Back a = ry(1 − y=K): while ago we discussed the exponential population dt model. For that model, it is assumed that the rate Here, r is a positive constant. Note that when dy dy of change of the population y is proportional y < K , is positive, so y increases, but when dt dt to the current population. If r is the constant of y < K, the derivative is negative, so y decreases. proportionality, that's the exponential differential We can solve this differential equation by the equation method of separation of variables. First, separate the variables to get dy = ry dt 1 dy = r dt y(1 − y=K) and that has the general solution and integrate y = Aert Z 1 Z where A is the initial population y(0). Rather than dy = r dt: using the base e for exponentiation, any other con- y(1 − y=K) venient base b can be used Z Of course, r dt = rt + C, but what about the y = Abst integral on the left side of the equation? r where s = . For that, we'll need the method of partial frac- ln b 1 tions. Write as the sum of two simpler There are many assumptions of this exponen- y(1 − y=K) tial model. In particular, it is assumed that there rational functions: are unlimited resources and there is no competition within the population. 1 A B = + More reasonable models for population growth y(1 − y=K) y 1 − y=K can be devised to fit actual populations better at the expense of complicating the model. where A and B are coefficients yet to be deter- mined. Clear the denominators to get the equation The logistic model. Verhulst proposed a model, A called the logistic model, for population growth in 1 = A(1 − y=K) + By = A − y + By K 1838. It does not assume unlimited resources. In- stead, it assumes there is a carrying capacity K for from which it follows that A = 1 and B = 1=K. the population. This carrying capacity is the stable Thus, population level. If the population is above K, then the population will decrease, but if below, then it 1 1 1=K 1 1 = + = + : will increase. y(1 − y=K) y 1 − y=K y K − y 1 Therefore, Z dy Z dy Z dy = + y(1 − y=K) y K − y = ln y − ln jK − yj y = ln : y − K Now that we've integrated the left side of the equation, we can continue y ln = rt + C y − K Next, the application of a little algebra gives us K y = 1 + Ae−rt where A is a constant. The graph of this function is asymptotic to the y-axis on the left, asymptotic to the line y = K on the right, and symmetric with respect to the point where y = K=2, which is an inflection point. To its left, the graph is concave upward, but to its right, concave downward. Here's the graph of for K = 1, A = 1, and r = 1, so that 1 y = : 1 + e−t Math 121 Home Page at http://math.clarku.edu/~djoyce/ma121/ 2.