Math 555: Differential Equations

2.5 Autonomous Equations

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2.5.1 Autonomous Equations

Autonomous equations are first-order equations where the independent variable does not explicitly appear. They are of the form

$$ \dfrac{dy}{dt} = f(y). $$
This class of differential equations is especially important in describing population dynamics, the growth or decline of a given species. We will explore a few applications of this idea to various fields of study.

In particular, we are going to analyze the differential equation geometrically in order to make qualitative determinations about the solutions to the differential equation. The information that we will be gathering will not require us to solve the differential equation, which is useful when the solution is difficult to explicitly obtain.

We will also be looking at stability and instability in the solutions to the differential equations. In applications, importance of these ideas is paramount because they describe the long-term tendencies of the solutions to the equation, whether it trends to a predictable state or grows or decays out of control.

2.5.2 Exponential Growth

The simplest form of autonomous equation is one the describes exponential growth (or decay)

$$ \dfrac{dy}{dt} = ry, $$
where $r$ is a growth (or decay) rate depending on the positivity (or negativity) of $r$. We have solved several equations of this form previously. Applying the initial condition $y(0) = y_0$, the solution is

$$ y(t) = y_0e^{rt}. $$
For $r > 0$, we have completely unrestricted growth for any initial value:

Applications of this equation are myriad: bacterial growth, continuously compounded interest, radioactive decay, etc. The issue is that there is no restriction on the growth or decay after large periods of time; the rate is fixed. In order to model things in a more sophisticated way, we require a modification to the differential equation.

2.5.3 Logistic Growth

In many applications, the growth rate $r$ will actually be a function of the population size $y$, so we replace $r$ with $h(y)$:

$$ \dfrac{dy}{dt} = h(y)y. $$
Our goal is to make a model that behaves like the previous example of exponential growth for $r>0$ when $y$ (the population size) is small, but experiences decline in population as $y$ gets larger. This means we want $h(y) > 0$ for small $y$ and $h(y) < 0$ for larger $y$. The simplest form for $h(y)$ that achieves this is $h(y) = r - ay$, where $r$ and $a$ are both positive constants. This transforms our differential equation into the form

$$ \dfrac{dy}{dt} = (r-ay)y. $$
The is commonly referred to as the logistic equation . It is also written as

$$ \dfrac{dy}{dt} = r\left(1 - \frac{y}{K}\right)y, $$
where $K = \frac{r}{a}$. In this formulation, $r$ is the intrinsic growth rate , the growth rate with no limiting factors. $K$'s meaning will become clear as we investigate this equation further.

Qualitative Analysis

We are now ready to investigate the qualitative properties of the solutions to the logistic equation. Our analysis will focus on three properties:

1. What do the simplest types of solutions look like? When do they occur?
2. What geometric information can we determine from the equation?
3. What is the limiting behavior of the solution(s)?

To answer to question 1, we notice that the expression

$$ f(y) = r\left(1 - \frac{y}{K}\right)y $$
is zero when $y$ is either $0$ or $K$. We know that the derivative of a constant is zero, so the simplest solutions must be constants.

For question 2, we want to closely examine the ODE and sketch it as $f(y)=y'$ versus $y$:

This graph is parabolic, and we can glean information about how the solution $y$ behaves from both when this figure is positive and negative and when it is increasing or decreasing. It is positive for $0 < y< K$ and negative otherwise. The magenta arrows are indicating that $y$ is increasing (since $f(y) = y' > 0$) for $0 < y < K$ and decreasing for $y > K$. In this situation, $y$ is referred to as the phase line , an idea we will utilize again shortly. Also important is that this graph is increasing for $y<\frac{K}{2}$ and decreasing for $y > \frac{K}{2}$, which tells us about the second derivative of $y$. In particular, all solutions must have an inflection point when $y = \frac{K}{2}$.

Lastly, we consider question 3 and the limiting behavior of the solutions. Here the phase line is plotted vertically (parallel to the $y$-axis). Several integral curves are sketched alongside the constant solutions $y = 0$ and $y=K$. Any solutions greater than $K$ decrease towards the value $K$ and solutions less than $K$ increase towards it, with an inflection point at $\frac{K}{2}$. This value $K$ is called the saturation level or carrying capacity of the environment for the species in question. It is the "natural" population level supported by the environmental conditions modeled by the growth rate $h(y)$ and is an example of a stable equilibrium point .

This behavior is the fundamental difference between unrestricted exponential growth and logistic growth. At any population level $y$ in exponential growth, when $t\rightarrow\infty$ we have unlimited and explosive population growth. For logistic growth, all solutions tend to the environmental carrying capacity $K$. This small restriction $h(y)$ on the system's growth, no matter how tiny, provides a check against the system that results in the population tending towards $K$.

Solving the ODE:

Our logistic equation

$$ \dfrac{dy}{dt} = r\left(1 - \frac{y}{K}\right)y $$
is separable into the expression

$$ \dfrac{dy}{(1-y/K)y} = r\,dt. $$
Integrating the left hand side requires a partial fraction decomposition

$$ \dfrac{1}{(1-y/K)y} = \dfrac{1}{y} + \dfrac{1}{K\left(1 - y/K\right)} $$

Decomposition Details

$$ \begin{aligned} \dfrac{1}{(1-y/K)y} &= \dfrac{A}{y} + \dfrac{B}{1 - y/K} \\ \\ &= \dfrac{A(1-y/K) + By}{y(1-y/K)} \end{aligned}$$
This implies that $ A(1-y/K) + By = 1 $. To find $A$ and $B$ easily, we can plug in particular values for $y$ (since the decomposition must hold for all $y$). We select values that make the factors times $A$ and $B$ equal to zero. Choosing $y=0$ tells us that $A = 1$, and $y=K$ implies that $B = 1/K$.

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Integrating both sides of our ODE yields

$$ \ln|y| - \ln|1-y/K| = rt + C. $$
We combine the left hand side using logarithm properties and then apply the exponential function to both sides, setting $C = e^c$:
$$ \dfrac{y}{1-y/K} = Ce^{rt}. $$
Applying the initial value $y(0) = y_0$, we see that $C = \frac{y_0}{1-y_0/K}$. After solving explicitly for $y$, we have the formula for the solution to the logistic equation

$$ y(t) = \dfrac{y_0 K}{y_0 + (K-y_0)e^{-rt}}. $$

2.5.4 Critical Thresholds

By modifying the logistic equation slightly, we have an equation for a critical threshold .

$$ \dfrac{dy}{dt} = -r\left(1 - \frac{y}{T}\right)y $$
The growth rate $r$ and threshold level $T$ are positive constants. Here the minus sign is "critical," so to speak. It changes the fundamental nature of the equation in a way made apparent by the same type of analysis we performed on the logistic equation.

This is essentially the reverse of the logistic equation. The derivative is negative for $0 < y < T$, and positive for $y > T$. This means that we will have sharp decay with an inflection point when $y = \frac{T}{2}$ for initial values $y_0 \in (0,T)$ and exponential like growth for $y_0 > T$. Plotting some integral curves makes this apparent.

The solution $y = T$ is an unstable equilibrium point , because solutions tend away from it for any $y$ values smaller or larger. $y=0$ is an example of a stable equilibrium point , since solutions close to 0 tend towards it.

Solution:

As a result to the similarities to the logistic equation, the solution is extremely similar in form

$$ y(t) = \dfrac{y_0 T}{y_0 + (T-y_0)e^{rt}}. $$
For these systems, the population thrives or dies. A good example of this would be a virus. If the initial "viral load" an individual is exposed to is below the threshold value the virus cannot replicate at a sufficiently high rate to overcome the body's immune response. However, if the initial amount of virus is high enough, the virus replicates rapidly and threatens the infected individual.

2.5.5 Logistic Equations with Thresholds

The threshold model serves as a good counterpart to logistic growth, but there are some systems that feature both behaviors. It is common to have systems with a threshold where populations lower than that value die out but are subject to logistic growth above the threshold value. To model these systems, we combine the factors from our previous two examples into the following ODE:

$$ \dfrac{dy}{dt} = -r\left(1 - \frac{y}{T}\right)\left(1 - \frac{y}{K}\right)y $$
where $r > 0$ and $0 < T < K$. We apply our analysis of the phase line again to determine the qualitative properties of the solutions:

For $y < T$, the derivative is negative and we will see population decline to 0. If $T < y < K$, $y'$ is positive and the solution grows towards the value $K$. For $y > K$, the growth rate is again negative and the population declines to $K$. Graphing the integral curves alongside the phase line illustrates the full behavior of the solutions for various initial values.

2.5.6 Qualitative Analysis Demonstration

Example 2.5.1

Use the qualitative analysis demonstrated above to determine the behavior of the autonomous equation

$$ \dfrac{dy}{dt} = e^y - 1,\qquad y_0\in (-\infty,\infty) $$
and then find the general solution.

Check Your Work

$$ y(t) = \ln\left(\dfrac{1}{1 - Ce^t}\right) $$

Video Solution

Example 2.5.2

Use the qualitative analysis demonstrated above to determine the behavior of the autonomous equation

$$ \dfrac{dy}{dt} = y(y-1)(y-2),\qquad y_0\geq 0. $$

Video Solution

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