The Logistic Equation and Integration by Partial Fractions

Unlimited Population Growth
In our previous discussion of differential equations and population growth, we encountered the equation for exponential growth, and its initial condition,
$\frac{dy}{dt} = ~k~y$

$y(0)=y_o$
We also saw that the procedure of separation of variables and simple integration leads to the result that we can predict the value of at any future time:
$y(t) = y_o~ e^{kt}$
This result (which we had already seen in a first semester of calculus) says that the function grows without any bound - indeed that the growth gets frighteningly rapid quite quickly.

The above equation had been proposed to describe population growth by Malthus. (You may want to refresh your memory about the way we treated this equation, and other differential equations for growth and decay last semester.) The constant is then a per-capita birth rate, and the term reflects the total number of births per unit time. This equation thus describes how a population grows if its per-capita birth rate is constant, and mortality is neglected. (Actually, we can include mortality too, in the same type of equation, as shown below).

For your consideration:

(1) Suppose that the per capita mortality is $\mu$ and that the per capita birth rate is $\beta$ . Show that by combining the terms we obtain an equation just like the one for exponential growth, but with $k= \beta - \mu$ .

(2) Explain what happens in the case that $k= \beta - \mu > 0$ .

(3) Explain what happens in the case that $k= \beta - \mu < 0$ .

Because it describes an unlimited growth, this equation has been viewed as unrealistic: the argument goes that any population that gets to huge density will respond to overcrowding by a reduction in the per capita birth rate.

Limited Population Growth
The Logistic Equation has been proposed to overcome this deficiency of the above model. We have already seen The Logistic Equation once before, and you may want to review what we discussed last semester. In this equation, we will define the variables:

$N(t) = ~{\rm population ~density}$
$K = ~{\rm carrying~ capacity~ of~ the~ environment}$
$r = ~{\rm reproductive~ parameter}$

Then the proposed Logistic Equation is:

$\frac{dN}{dt} = r N \frac{(K - N)}{K}$

Like the equation for exponential growth, this is a differential equation. It is a statement about the rate of change of a population. This equation can be understood in several ways: (1) as an equation in which the rate of reproduction is no longer constant but declines with the size (or density ) of the population; the term could be viewed as that density-dependent rate of reproduction. A second approach, if we expand out the equation
$\frac{dN}{dt} = r N - \frac{r}{K} N^2$
suggests a process in which the birth rate is still constant , but in which there is a a mortality term, $\frac{r}{K} N^2$ that dominates when the population is high. Either approach is a valid interpretation of this new differential equation governing the change in the population.

We will consider the initial condition
$N(0)=N_o$
This is the starting level of the population.

For your consideration:

(1) What happens if the population is such that ? Does it change?

(2) What happens if ? If ?

(3) Are there any other values for which no change would occur?

(4) Can you suggest why the constant is called the carrying capacity?

Our task, below is to understand the solution to this differential equation using the procedure of integration. However, before we do so, it is profitable to write this equation in its simplest possible form.

Rescaling the equation
We can make this equation "look simpler" by defining a new variable,
$y = \frac{N}{K}$
The new variable is a ratio of the population size to the carrying capacity, and when we write the logistic equation in terms of this variable (by substituting and cancelling common factors), we get
$\frac{dy}{dt} = r y (1-y)$
The initial value for the new value will be called
$y(0) = y_o$

For your consideration:

(1) Show that this is the result of the substitution !

(2) Why is this a more desirable equation to study ?

(3) What are the values of for which ?

We will now show the procedure for solving this equation.

Integrating the Logistic Equation
We would like to find a function that satisfies the above Logistic differential equation. The procedure is as follows:
$\frac{dy}{dt}~=~ r ~y~(1-y)$
Divide both sides by the y-dependent expression:
$\frac{1}{ ~y~(1-y)} \frac{dy}{dt}~=~ r$
Integrate both sides with respect to time from the initial to the final .
$\int_0^T \frac{1}{ ~y~(1-y)} \frac{dy}{dt}~ dt=~ \int_0^T~r dt$
Thus,
$\int_{y_o}^{y(T)} \frac{1}{ ~y~(1-y)} ~dy=~ \int_0^T~r dt$
To go further, we need to integrate the function shown on the right hand side, but for this, we will use a TRICK: this trick is called Partial Fractions and consists of unraveling the quotient into its simpler fractional pieces.

Integrating with Partial Fractions
The idea is similar to undoing the process of finding a common denominator.

Suppose that the fraction in the integral could be expressed as
$\frac{1}{ y(1-y)} =\frac{A}{y}+ \frac{B}{(1-y)}$
where the constants are to be found. Then, if we were to express the left hand side using a common denominator, we would get:
$\frac{1}{ y(1-y)} =\frac{A~(1-y)}{y(1-y)}+ \frac{B~y}{y(1-y)}$
In order for the two sides to be equal,the numerators much match so that
$1 =A~(1-y)+ B~y$
So
$1 = A + y (B - A)$
Since this is to be true for every value of y, the constant terms must match as must the coefficients of . The constant term on the right hand side is 1 and there is no term with y in it, so the coefficient of y is zero. It may be helpful to think of this statement as the polynomial equation
$1 + 0 y = A + y (B - A)$
and to remember that two polynomials are the same if and only if the coefficients of like terms are the same. Thus, equating two sets of coefficients (for the constant and the y term) leads to the following two equations:
$1 = A ~~~~~{\rm and} ~~0 = (B-A)$
This means that , so that we have now arrived at the desired representation of the quotient as a sum of simpler fractions, that is:
$\frac{1}{ y(1-y)} =\frac{1}{y}+ \frac{1}{(1-y)}$
The usefulness of this idea resides in the fact that these simpler fractions are easily integrated. Returning to the integral, we have
$\int_{y_o}^{y(T)} \frac{1}{ ~y~(1-y)} ~ dy=~ \int_0^T~r dt$

$\int_0^T [\frac{1}{y} +\frac{1} {(1-y)}] ~ dy=~ r T$
Each of the above fractions is now easy to integrate, with the result:
$[\ln|y| -\ln|1-y|] |_{y_o}^{y(T)}=~ r T$
Now some algebraic simplification is needed. First evaluate at the endpoints and use the fact that $\ln(A)-\ln(B)=\ln(A/B)$
$\ln(\frac{|y|}{|1-y|}) |_{y_o}^{y(T)}=~ r T$
$\ln(\frac{|y(T)|}{|y_o|}\frac{|1-y_o|}{|1-y(T)|})=~ r T$
Now re-express the result in terms of the exponential function, i.e.
$\frac{|y(T)|}{|y_o|}\frac{|1-y_o|}{|1-y(T)|}=~ e^{r T}$
Group the constant terms together and bring them to the other side:
$\frac{y(T)}{(1-y(T))}=~\frac{(y_o)}{(1-y_o)} e^{r T}$
Isolate on one side of the equation. This takes a couple of steps, as shown below:
$y(T)=(1-y(T))~\frac{(y_o)}{(1-y_o)} e^{r T}$
$y(T) (1 +\frac{(y_o)}{(1-y_o)} e^{r T}) =~\frac{(y_o)}{(1-y_o)} e^{r T}$
Divide both sides by the factor next to the term, and finally we have a description of the dependence of on .
$y(T) =\frac{~\frac{(y_o)}{(1-y_o)} e^{r T}}{(1 +\frac{(y_o)}{(1-y_o)} e^{r T})}$
It looks fairly messy. However, we can simplify this expression to arrive at
$y(T) =\frac{y_o}{(1-y_o) e^{-r T} + y_o}$
Ofcourse, this is true for any final time, T, so in particular, for we have:
$y(t) =\frac{y_o}{(1-y_o) e^{-r t} + y_o}$
We have arrived (after a lengthy process) at the solution of the Logistic Equation. Below, you are asked to understand what this expression is saying.

For your consideration:

(1) Show that the above simplification is correct, i.e. do the algebra involved in obtaining the simplest expression for the solution.

(2) Explain the behaviour of the above solution of the differential equation.

(3) What is the value of for ?

(4) What is the value of for $t \to \infty$ ?

(5) Connect the behaviour for this variable with the behaviour of the original variable,

In the demonstration below, we show the behaviour of the solution to the logistic equation with K=2, starting from a variety of initial conditions. Experiment with the initial conditions to see how the behaviour changes. (This demo is labled with the variable , but it actually represents the solution of
$\frac{dy}{dt} = r y (2-y)$

Another Example of Integration with Partial Fractions

You are asked to find the integral of:
$\int ~ \frac{1}{(x^2+5x+6)} ~dy$
We can use the same idea, i.e partial fractions, by first factoring the denominator of this equation, obtaining:
$\int ~ \frac{1}{(x+2)(x+3)} ~dy$
We now need to rewrite the quotient as a sum of two simpler fractions, so suppose
$\frac{1}{(x+2)(x+3)} = \frac{A}{(x+2)} + \frac{B}{(x+3)}$
The common denominator of the right hand side implies that
$\frac{1}{(x+2)(x+3)} = \frac{A(x+3)+B(x+2)}{(x+2)(x+3)}$
Thus
$1 = A(x+3)+B(x+2)$
Isolating the constant terms and the terms multiplying x, we find that

$1 + 0 x = (3A+2B) + (A+ B)x$
Thus

$1 = (3A+2B), ~~~~~~ (A+ B)=0$
We can solve this system of equations to arrive at
$A=1, ~~~B=-1$
(Make sure you can do this too !) and thus conclude that

$\int \frac{1}{(x+2)(x+3)}~dx = \int ( \frac{1}{(x+2)}- \frac{1}{(x+3)} ) ~dx$
These integrals are now simple to do by substitution, and we obtain the result:

$\int ( \frac{1}{(x+2)} - \frac{1}{(x+3)}) ~dx = \ln|x+2| -\ln|x+3| + C = \ln| \frac{x+2}{x+3}| +C$