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Solving the logistic differential equation

The logistic differential equation is an autonomous differential equation, so we can use separation of variables to find the general solution, as we just did in [link] .

Step 1: Setting the right-hand side equal to zero leads to P = 0 and P = K as constant solutions. The first solution indicates that when there are no organisms present, the population will never grow. The second solution indicates that when the population starts at the carrying capacity, it will never change.

Step 2: Rewrite the differential equation in the form

d P d t = r P ( K P ) K .

Then multiply both sides by d t and divide both sides by P ( K P ) . This leads to

d P P ( K P ) = r K d t .

Multiply both sides of the equation by K and integrate:

K P ( K P ) d P = r d t .

The left-hand side of this equation can be integrated using partial fraction decomposition. We leave it to you to verify that

K P ( K P ) = 1 P + 1 K P .

Then the equation becomes

1 P + 1 K P d P = r d t ln | P | ln | K P | = r t + C ln | P K P | = r t + C .

Now exponentiate both sides of the equation to eliminate the natural logarithm:

e ln | P K P | = e r t + C | P K P | = e C e r t .

We define C 1 = e c so that the equation becomes

P K P = C 1 e r t .

To solve this equation for P ( t ) , first multiply both sides by K P and collect the terms containing P on the left-hand side of the equation:

P = C 1 e r t ( K P ) P = C 1 K e r t C 1 P e r t P + C 1 P e r t = C 1 K e r t .

Next, factor P from the left-hand side and divide both sides by the other factor:

P ( 1 + C 1 e r t ) = C 1 K e r t P ( t ) = C 1 K e r t 1 + C 1 e r t .

The last step is to determine the value of C 1 . The easiest way to do this is to substitute t = 0 and P 0 in place of P in [link] and solve for C 1 :

P K P = C 1 e r t P 0 K P 0 = C 1 e r ( 0 ) C 1 = P 0 K P 0 .

Finally, substitute the expression for C 1 into [link] :

P ( t ) = C 1 K e r t 1 + C 1 e r t = P 0 K P 0 K e r t 1 + P 0 K P 0 e r t

Now multiply the numerator and denominator of the right-hand side by ( K P 0 ) and simplify:

P ( t ) = P 0 K P 0 K e r t 1 + P 0 K P 0 e r t = P 0 K P 0 K e r t 1 + P 0 K P 0 e r t · K P 0 K P 0 = P 0 K e r t ( K P 0 ) + P 0 e r t .

We state this result as a theorem.

Solution of the logistic differential equation

Consider the logistic differential equation subject to an initial population of P 0 with carrying capacity K and growth rate r . The solution to the corresponding initial-value problem is given by

P ( t ) = P 0 K e r t ( K P 0 ) + P 0 e r t .

Now that we have the solution to the initial-value problem, we can choose values for P 0 , r , and K and study the solution curve. For example, in [link] we used the values r = 0.2311 , K = 1,072,764 , and an initial population of 900,000 deer. This leads to the solution

P ( t ) = P 0 K e r t ( K P 0 ) + P 0 e r t = 900,000 ( 1,072,764 ) e 0.2311 t ( 1,072,764 900,000 ) + 900,000 e 0.2311 t = 900,000 ( 1,072,764 ) e 0.2311 t 172,764 + 900,000 e 0.2311 t .

Dividing top and bottom by 900,000 gives

P ( t ) = 1,072,764 e 0.2311 t 0.19196 + e 0.2311 t .

This is the same as the original solution. The graph of this solution is shown again in blue in [link] , superimposed over the graph of the exponential growth model with initial population 900,000 and growth rate 0.2311 (appearing in green). The red dashed line represents the carrying capacity, and is a horizontal asymptote for the solution to the logistic equation.

A graph showing exponential and logistic growth for the same initial population of 900,000 organisms and growth rate of 23.11%. Both begin in quadrant two close to the x axis as increasing concave up curves. The exponential growth curve continues to grow, passing P = 1,072,764 while still in quadrant two. The logistic growth curve changes concavity, crosses the x axis at P_0 = 900,000, and asymptotically approaches P = 1,072,764.
A comparison of exponential versus logistic growth for the same initial population of 900,000 organisms and growth rate of 23.11 % .

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Source:  OpenStax, Calculus volume 2. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11965/1.2
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