Suppose \(q(t)\) is a quantity that is varying over time, and \(r(t) = q'(t)\) is its association rate function. If \(q\) and \(r\) satisfy a rate equation of the form
\begin{equation*}
r(t) = f(q)
\end{equation*}
for some function \(f\text{,}\) then the rate of variation of \(q\) is entirely dependent on the amount \(q\) and not on the actual time at which that particular amount occurs. We call such a model an autonomous system, and the expression in \(q\) on the right-hand side is called the rate function. Population dynamics are a typical example of autonomous systems: if a scientist is studying bacterial growth in controlled lab environment, the rate of growth doesn’t have much to do with the reading on the clock on the wall, only on the current number of bacteria.
We can determine a lot about the general behavioural characteristics of solutions to autonomous systems without actually solving the rate equation, by applying the concepts we learned in Chapter 18.
The first thing to notice is that at quantity levels \(q = -1\) and \(q = 2\text{,}\) we have \(dq/dt = 0\text{,}\) which means that the rate of growth is zero, and neither growth nor decay can occur. So if the quantity reaches a level of \(q = -1\) or \(q = 2\text{,}\) it will remain constant and never deviate from that level afterward.
Next, we can analyze which quantity levels will result in growth and which will result in decay. As in Section 18.1, we can determine this by solving \(q'(t) \gt 0\) and \(q'(t) \lt 0\text{,}\) but this time we will solve in terms of \(q\) rather than in terms of \(t\text{.}\) Using our rate equation, we can determine when the derivative is positive or negative by determining the values of \(q\) that make the right-hand side of (✶) positive or negative. We already know that this right-hand expression is zero at \(q = -1\) and \(q = 2\text{,}\) so we just need to test around those values.
\(q + 1\)
\(q - 2\)
\(dq/dt\)
\(q\)
\(q \lt -1\)
\(-\)
\(-\)
\(+\)
increasing
\(-1 \lt q \lt 2\)
\(+\)
\(-\)
\(-\)
decreasing
\(q \gt 2\)
\(+\)
\(+\)
\(+\)
increasing
We can perform a similar analysis for concavity, but for that we need to know something about the second derivative. If we differentiate both sides of (✶), on the left we will be differentiating the derivative of \(q\text{,}\) and so we’ll get the second derivative. On the right, we could expand before differentiating so that we can just apply Pattern 12.4.3, or we can directly apply the Product Rule. But in either case we will also need to apply the Chain Rule, since we are differentiating with respect to \(t\text{,}\) not \(q\text{.}\)
We already know the quantity levels where the derivative is positive, negative, or zero, so we just need to combine that with an analysis of the new factor \(2 q - 1\text{.}\) We see that \(d^2q/dt^2\) will be zero at \(q = -1\) and \(q = 2\) (since these make \(dq/dt\) zero) and at \(q = 1/2\text{,}\) so again we test around these levels.
\(2 q - 1\)
\(dq/dt\)
\(d^2/dt^2\)
\(q\)
\(q \lt -1\)
\(-\)
\(+\)
\(-\)
concave down
\(-1 \lt q \lt 1/2\)
\(-\)
\(-\)
\(+\)
concave up
\(1/2 \lt q \lt 2\)
\(+\)
\(-\)
\(-\)
concave down
\(q \gt 2\)
\(+\)
\(+\)
\(+\)
concave up
We see now that any solution curve that passes through the quantity level \(q = 1/2\) will experience a concavity change there, so we will always see inflection points at that level. However, even though the concavity also changes at above and below \(q = -1\) and also above and below \(q = 2\text{,}\) those will not become inflection points, since any solution that reaches one of those levels would become constant after that.
It’s useful to summarize all of the information so far pictorially on a phase line. Since our behaviours are based on different ranges of \(q\)-values, we’ll use a vertical axis to lay out the different behaviour ranges.
Now we can sketch some typical solution curves that demonstrate the range of behaviours above.
The constant solutions are an important feature of our analysis in Example 20.1.1, as they can be boundaries between different types of solution behaviours, both for increasing/decreasing and concavity. But also notice that solution curves near the constant solutions tend to exhibit either an “attraction” to or a “repulsion” from the constant solution lines. We often model our systems as if they are “closed,” in the sense that no unaccounted-for outside factors can influence the system’s behaviour. But reality is typically messier than that, and outside factors can “nudge” a system out of a constant, stable state into a different type of behaviour. After such a “nudge”, sometimes the system gently trends back to that original constant state, and sometimes the system reacts violently and moves quickly away from the original constant state.
Definition20.1.4.Stable and unstable equilibriums.
For an autonomous system \(q'(t) = f(q)\text{,}\) a constant solution function \(q(t) = C\) (so that \(f(C) = 0\)) is called an equilibrium solution for the system.
If solution curves with \(q \approx C\) tend toward that equilibrium as \(t \to \infty\text{,}\) then the equilibrium is called stable.
If solution curves with \(q \approx C\) tend away from that equilibrium as \(t \to \infty\text{,}\) then the equilibrium is called unstable.
If solution curves exhibit different trending behaviours for \(q \approx C^+\) versus \(q \approx C^-\text{,}\) then the equilibrium is called semi-stable.
So, for example, the system in Example 20.1.1 has a stable equilibrium \(q(t) = -1\) and an unstable equilibrium \(q(t) = 2\text{.}\)
Remark20.1.5.
A constant function \(q(t) = C\) has \(dq/dt = 0\text{,}\) so, as noted in the definition, equilibrium solutions occur precisely at \(q\)-levels where \(f(q) = 0\text{.}\)
We can easily determine whether an equilibrium solution is stable, unstable, or semi-stable without going through the full analysis, but instead just considering the increasing/decreasing behaviour of solutions. A stable equilibrium must have decreasing behaviour just above it and increasing behaviour just below, and an unstable equilibrium must have the opposite. A half-stable equilibrium will have same type of behaviour both just above and just below, either both increasing or both decreasing.
A useful analogy for stable/unstable behaviour is to think of taking a bicycle ride along a path. The equilibrium state is to steer straight along the center of the path. If the sides of the path are curved upward, as if in a valley with steep sides, then any deviation from the center of the path will cause the bicycle to tend back towards that central equilibrium. On the other hand, if the sides of the path are curved downward, as if the path runs along the top of a berm, then any deviation from the center of the path will cause the rider to go careening down a slope away from the path. In a way, we could imagine that all of the solution curves stitch together to create a surface, and there is a third dimension, perpendicular to the plane where we have sketched our representative solutions, for another type of “concavity.” Near a stable equilibrium, this solution-curve surface is concave up (out of the page), with the equilibrium running along the floor of a depression. Near an unstable equilibrium, the solution-curve surface is concave down (into the page), with the equilibrium running along the peak of a long hill. Near a semi-stable equilibrium, the solution-curve surface “flattens” out between two regions of the same concavity, like a narrow horizontal shelf on the side of a hill.
We further remark on why the equilibrium solutions in Example 20.1.1 are not considered inflection points, even though the concavity of solutions changes from below to above each of those equilibriums. We take it as a general assumption that two different solution curves cannot intersect, since if they did then that point of intersection would act like an “initial condition” for both solutions. In an autonomous system the behaviour is completely determined by the quantity amount \(q\) and not by the independent variable \(t\text{.}\) So if two solution curves met at the same level, they would exhibit the same behaviour from the point forward, and, extrapolating backward, they should have exhibited the same behaviour to get to that point. Which means they weren’t actually different solutions to begin with.
Principle20.1.6.Principle of non-intersection.
In an autonomous system, different solution curves can never intersect.
This principle explains why we cannot have an inflection point at an equilibrium \(q\)-level: because solution curves cannot actually cross over an equilibrium solution, and so an individual solution curve cannot change its concavity there. Indeed, noting that equilibrium solutions necessarily have \(q'(t) = 0\text{,}\) in Figure 20.1.3 you will find it impossible to draw a curve that passes through the equilibrium solution at level \(q = -1\) (with a horizontal slope as it does so) that meets the increasing/decreasing and concavity behaviour requirements both above and below.
Finally, here is a procedure we can use to analyze solution behaviour for an autonomous system without actually solving it.
Procedure20.1.7.Analyzing an autonomous system.
For \(q'(t) = f(q)\text{:}\)
Determine the equilibrium solutions by solving \(f(q) = 0\) for \(q\text{.}\) Plot these values on a phase line.
Determine ranges of increase/decrease by testing for \(f(q) \gt 0\) or \(f(q) \lt 0\) between the equilibrium solutions, and between equilibrium solutions and any \(q\)-levels where \(f(q)\) is undefined.
Determine possible inflection levels by solving \(f'(q) = 0\) for \(q\text{,}\) and also noting any \(q\)-levels where \(f'(q)\) is undefined. Plot these levels on the phase line.
determine ranges of concavity by testing for \(f'(q) f(q) \gt 0\) or \(f'(q) f(q) \lt 0\) between equilibrium solutions, and equilibrium solutions and the levels identified in the previous step.
Add the information concerning ranges of increase/decrease and concavity to the phase line. Sketch some representative curves with various initial \(q\)-levels, illustrating all of the different types of behaviours possible according to the analysis.
Section20.2Exponential growth
We have already analyzed this system several times, but we will review it once again as it will serve as a “base” model that we will modify in different ways in the rest of this chapter.
Suppose \(q\) is a quantity that varies over time so that its rate of variation is always proportional to its current amount:
\begin{equation*}
\frac{dq/dt}{q} = k \text{,}
\end{equation*}
for some constant of proportionality \(k\text{.}\) We might use such a model when we study a population that has lots of food and space and no predators or disease. In this case, we should assume that the growth rate factor \(k\) is positive, since when \(q\) is positive we should observe growth, in which case \(dq/dt\) should also be positive.
Rewriting the rate equation as
\begin{equation*}
\dqdt = k q \text{,}
\end{equation*}
our rate function is \(f(q) = k q\text{.}\) The only equilibrium solution is \(q(t) = 0\text{,}\) and above that we have increasing behaviour. The second derivative is
which will also be positive when \(q\) is, so we have concave-up behaviour for \(q > 0\text{.}\)
Section20.3Logistic growth
Let’s consider a modification to our initial population model. Suppose as the population grows, competition for space and food increases and the rate of growth slows. We might identify a particular carrying capacity of the environment: where if the population starts to approach the carrying capacity, the growth rate is significantly reduced, and if the population somehow “jumped” over the carrying capacity (say, by a significant number of new individuals moving or being placed into the system’s environment), then the population would experience decline instead of growth.
To model this mathematically, we might assume that the rate of growth is still proportional to the population level, but in a proportion that varies according to how close the population level is to the carrying capacity:
Constant \(M\) is the carrying capacity; if \(q \approx M\) then \(dq/dt \approx 0\) and the population growth has stalled.
The growth rate factor \(\rho\) is also a (positive) constant, and the quantity \(k = \rho M\) is called the intrinsic growth rate; if \(q \approx 0\text{,}\) then our rate equation becomes
\begin{equation*}
\frac{dq/dt}{q} \approx k \text{,}
\end{equation*}
and we should observe approximately exponential growth as in Section 20.2.
For this model, we will demonstrate an entirely graphical analysis. Our rate equation is
\begin{equation*}
f(q) = \rho M q - \rho q^2
\end{equation*}
is a concave-down parabola as a function in \(q\text{.}\)
The equilibrium solutions are \(q\)-levels that make \(f(q) = 0\text{,}\) and from the graph we see that this occurs at \(q = 0\) and \(q = M\text{.}\) These make sense, since if the population cannot grow if there are no individuals at all, and also if the population is at the carrying capacity it will not be able to grow, but it also will not decline as the system tendency will be to try to grow but will be impeded by the space/food limits.
Assuming \(q \gt 0\text{,}\) for \(0 \lt q \lt M\) we see that the rate function is positive, which translates into a positive \(dq/dt\) hence an increasing \(q\text{.}\) But for \(q \gt M\text{,}\) the rate function is negative and so \(q\) is decreasing on that range. These behaviours are in line with our intentions when we created the model. We can also identify \(q = 0\) as an unstable equilibrium as solutions above that level increase away from it, and \(q = M\) as a stable equilibrium as solutions below increase toward it and solutions above decrease toward it.
So concavity is determined by combinations of positive/negative \(f(q)\) (where our plot in Figure 20.3.1 is above/below the horizontal axis) and positive/negative \(f'(q)\) (where our plot in Figure 20.3.1 is increasing/decreasing). By symmetry of the parabola, peak in the graph \(y = f(q)\) occurs at \(q = M/2\text{.}\)
Up to \(q = M/2\text{,}\)\(f\) is positive and increasing as a function in \(q\text{,}\) so that \(f'\) is also positive. These two positives combine to a positive \(d^2q/dt^2\) and a concave up \(q\text{.}\)
After \(q = M/2\text{,}\) the graph of \(y = f(q)\) starts to decrease but remains positive, so a positive \(f\) combined with a negative \(f'\) makes a negative \(d^2q/dt^2\) and a concave down \(q\text{.}\)
At \(q = M\text{,}\) the graph of \(y = f(q)\) becomes negative and is still decreasing, so the two negatives combine to a positive \(d^2q/dt^2\) and a concave up \(q\text{.}\)
From this analysis we note an inflection level at \(q = M/2\text{,}\) but not at \(q = M\) since that is one of our equilibrium solutions.
We now have enough information to draw a phase line and sketch some representative solution curves.
Section20.4Growth with threshold
Another way to modify our initial exponential-growth model is to assume that there is some “healthy” population threshold \(q = T\text{,}\) below which the population is not large enough to thrive and instead dwindles away. Again we create a model where the rate of variation is proportional to the population, but this time with a variable proportion that is negative when \(q\) is below the threshold and positive when \(q\) is above it.
Similarly to the logistic growth model, the factor \(k = - \rho T\) (with \(\rho\) assumed to be positive) is the intrinsic growth rate, so that for \(q \approx 0\) we have approximately exponential decay.
Rewriting the rate equation to leave only \(dq/dt\) on the left, we have rate function
The analysis of this model is very similar to the analysis of logistic growth in Section 20.3, so we will leave the details to you, the reader.
Notice that the equilibriums have “flipped” in nature from logistic growth. The equilibrium at \(q(t) = 0\) is now stable, since if we have \(q \approx 0^+\) we are below the threshold and the population will wither away. However, the equilibrium \(q(t) = T\) at the threshold is unstable equilibrium, since below the threshold we always observe decay toward \(0\text{,}\) while above the threshold the population will now thrive and grow away from the threshold.
Section20.5Logistic growth with threshold
Finally, we can combine our previous modifications into one model that takes into account both a carrying capacity, above which the environment is not able to sustain such a large population, and a threshold, below which the population is not robust enough to thrive and grow. The rate equation for this model is of the form
is our rate function, and we assume that the threshold \(T\) is less than the carrying capacity \(M\text{.}\)
(a)Graph of the rate function for logistic growth with threshold. Note that \(q\) is on the horizontal axis.
(b)Phase line and representative solution curves for logistic growth with threshold.
Figure20.5.1.Rate function and associated representative solution curves for logistic growth with threshold.
In this model, the equilibrium at \(q(t) = 0\) remains stable, since it is below the threshold. The equilibrium at the threshold itself \(q(t) = T\) is again unstable, as populations below it will decay and those above it will thrive. The equilibrium at the carrying capacity is stable, since populations approaching that level will significantly slow in their growth, and populations above that level will decay as the environment cannot supprt them.