Section 1.6 Autonomous equations
An Autonomous equation \(y'=f(y)\) ONLY depends on the dependent variable. This simplification allows a lot of special types of analysis such as equilibrium values, stability, and what happens as t gets large.
Consider problems of the form
\begin{equation*}
\frac{dx}{dt} = f(x) ,
\end{equation*}
where the derivative of solutions depends only on \(x\) (the dependent variable). Such equations are called autonomous equations. If we think of \(t\) as time, the naming comes from the fact that the equation is independent of time.
We return to the cooling coffee problem (Example 1.3.3). Newton’s law of cooling says
\begin{equation*}
\frac{dx}{dt} = k (A-x) ,
\end{equation*}
where \(x\) is the temperature, \(t\) is time, \(k\) is some positive constant, and \(A\) is the ambient temperature. See Figure 1.11 for an example with \(k=0.3\) and \(A=5\text{.}\)
Note the solution \(x=A\) (in the figure \(x=5\)). We call these constant solutions the equilibrium solutions. The points on the \(x\)-axis where \(f(x) = 0\) are called critical points. The point \(x=A\) is a critical point. In fact, each critical point corresponds to an equilibrium solution. Note also, by looking at the graph, that the solution \(x=A\) is “stable” in that small perturbations in \(x\) do not lead to substantially different solutions as \(t\) grows. If we change the initial condition a little bit, then as \(t \to \infty\) we get \(x(t) \to A\text{.}\) We call such a critical point stable. In this simple example it turns out that all solutions in fact go to \(A\) as \(t \to \infty\text{.}\) If a critical point is not stable, we say it is unstable.
Video 1.6.2. Logistic Equation.
Exponential growth is great and all, but is it realistic that biological systems grow unbounded forever? In this video we will explore modelling with Logistic Growth to account for this difficulty.
Consider now the logistic equation
\begin{equation*}
\frac{dx}{dt} = kx(M-x) ,
\end{equation*}
for some positive \(k\) and \(M\text{.}\) This equation is commonly used to model population if we know the limiting population \(M\text{,}\) that is the maximum sustainable population. The logistic equation leads to less catastrophic predictions on world population than \(x'=kx\text{.}\) In the real world there is no such thing as negative population, but we will still consider negative \(x\) for the purposes of the math.
See Figure 1.12 for an example, \(x' = 0.1 x(5-x)\text{.}\) There are two critical points, \(x=0\) and \(x=5\text{.}\) The critical point at \(x=5\) is stable, while the critical point at \(x=0\) is unstable.
It is not necessary to find the exact solutions to talk about the long term behavior of the solutions. From the slope field above of \(x' = 0.1 x(5-x)\text{,}\) we see that
\begin{equation*}
\lim_{t\to \infty} x(t) =
\begin{cases}
5 & \text{if } \; x(0) > 0 , \\
0 & \text{if } \; x(0) = 0 , \\
\text{DNE or } {-\infty} & \text{if } \; x(0) < 0 . \\
\end{cases}
\end{equation*}
Here DNE means “does not exist.” From just looking at the slope field we cannot quite decide what happens if \(x(0) < 0\text{.}\) It could be that the solution does not exist for \(t\) all the way to \(\infty\text{.}\) Think of the equation \(x' = x^2\text{;}\) we have seen that solutions only exist for some finite period of time. Same can happen here. In our example equation above it turns out that the solution does not exist for all time, but to see that we would have to solve the equation. In any case, the solution does go to \(-\infty\text{,}\) but it may get there rather quickly.
If we are interested only in the long term behavior of the solution, we would be doing unnecessary work if we solved the equation exactly. We could draw the slope field, but it is easier to just look at the phase diagram or phase portrait, which is a simple way to visualize the behavior of autonomous equations. In this case there is one dependent variable \(x\text{.}\) We draw the \(x\)-axis, we mark all the critical points, and then we draw arrows in between. Since \(x\) is the dependent variable we draw the axis vertically, as it appears in the slope field diagrams above. If \(f(x) > 0\text{,}\) we draw an up arrow. If \(f(x) < 0\text{,}\) we draw a down arrow. To figure this out, we could just plug in some \(x\) between the critical points, \(f(x)\) will have the same sign at all \(x\) between two critical points as long \(f(x)\) is continuous. For example, \(f(6) = -0.6 < 0\text{,}\) so \(f(x) < 0\) for \(x > 5\text{,}\) and the arrow above \(x=5\) is a down arrow. Next, \(f(1) = 0.4 > 0\text{,}\) so \(f(x) > 0\) whenever \(0 < x < 5\text{,}\) and the arrow points up. Finally, \(f(-1) = -0.6 < 0\) so \(f(x) < 0\) when \(x <
0\text{,}\) and the arrow points down.
Armed with the phase diagram, it is easy to sketch the solutions approximately: As time \(t\) moves from left to right, the graph of a solution goes up if the arrow is up, and it goes down if the arrow is down.
Exercise: Try sketching a few solutions simply from looking at the phase diagram. Check with the preceding graphs if you are getting the type of curves.
Once we draw the phase diagram, we classify critical points as stable or unstable 1 .
Since any mathematical model we cook up will only be an approximation to the real world, unstable points are generally bad news.
Let us think about the logistic equation with harvesting. Suppose an alien race really likes to eat humans. They keep a planet with humans on it and harvest the humans at a rate of \(h\) million humans per year. Suppose \(x\) is the number of humans in millions on the planet and \(t\) is time in years. Let \(M\) be the limiting population when no harvesting is done. The number \(k > 0\) is a constant depending on how fast humans multiply. Our equation becomes
\begin{equation*}
\frac{dx}{dt} = kx(M-x) - h .
\end{equation*}
We expand the right-hand side and set it to zero.
\begin{equation*}
kx(M-x) - h = -kx^2+kMx - h = 0.
\end{equation*}
Solving for the critical points, let us call them \(A\) and \(B\text{,}\) we get
\begin{equation*}
A = \frac{kM + \sqrt{{(kM)}^2 - 4hk}}{2k}, \qquad
B = \frac{kM - \sqrt{{(kM)}^2 - 4hk}}{2k} .
\end{equation*}
Exercise: Sketch a phase diagram for different possibilities. Note that these possibilities are \(A > B\text{,}\) or \(A=B\text{,}\) or \(A\) and \(B\) both complex (i.e. no real solutions). Hint: Fix some simple \(k\) and \(M\) and then vary \(h\text{.}\)
For example, let \(M=8\) and \(k=0.1\text{.}\) When \(h=1\text{,}\) then \(A\) and \(B\) are distinct and positive. The slope field we get is in Figure 1.13. As long as the population starts above \(B\text{,}\) which is approximately 1.55 million, then the population will not die out. It will in fact tend towards \(A \approx
6.45\) million. If ever some catastrophe happens and the population drops below \(B\text{,}\) humans will die out, and the fast food restaurant serving them will go out of business.
When \(h = 1.6\text{,}\) then \(A=B=4\text{.}\) There is only one critical point and it is unstable. When the population starts above 4 million it will tend towards 4 million. If it ever drops below 4 million, humans will die out on the planet. This scenario is not one that we (as the human fast food proprietor) want to be in. A small perturbation of the equilibrium state and we are out of business. There is no room for error. See Figure 1.14.
Finally if we are harvesting at 2 million humans per year, there are no critical points. The population will always plummet towards zero, no matter how well stocked the planet starts. See Figure 1.15.
Example 1.6.1.
Geogebra Activity: Try to sketch by hand the phase diagram for each of the following equations, then use this Geogebra phase diagram plotter 2 to verify your answers.
\begin{equation*}
\frac{dx}{dt}=x^2-9x+4
\end{equation*}
\begin{equation*}
\frac{dx}{dt}=\cos(2x),\ \ -\pi<x<\pi
\end{equation*}
\begin{equation*}
\frac{dx}{dt}=(x-2)^2
\end{equation*}
\begin{equation*}
\frac{dx}{dt}=x^2+2
\end{equation*}
Subsection 1.6.1 Exercises
Exercise 1.6.1.
Consider \(x' = x^2\text{.}\)
- Draw the phase diagram, find the critical points, and mark them stable or unstable.
- Sketch typical solutions of the equation.
- Find \(\displaystyle \lim_{t\to \infty} x(t)\) for the solution with the initial condition \(x(0) = -1\text{.}\)
Exercise 1.6.2.
Consider \(x' = \sin x\text{.}\)
- Draw the phase diagram for \(-4\pi \leq x \leq 4\pi\text{.}\) On this interval mark the critical points stable or unstable.
- Sketch typical solutions of the equation.
- Find \(\displaystyle \lim_{t\to \infty} x(t)\) for the solution with the initial condition \(x(0) = 1\text{.}\)
Exercise 1.6.3.
Suppose \(f(x)\) is positive for \(0 < x < 1\text{,}\) it is zero when \(x=0\) and \(x=1\text{,}\) and it is negative for all other \(x\text{.}\)
- Draw the phase diagram for \(x' = f(x)\text{,}\) find the critical points, and mark them stable or unstable.
- Sketch typical solutions of the equation.
- Find \(\displaystyle \lim_{t\to \infty} x(t)\) for the solution with the initial condition \(x(0) = 0.5\text{.}\)
Exercise 1.6.4.
Start with the logistic equation \(\frac{dx}{dt} = kx(M-x)\text{.}\) Suppose we modify our harvesting. That is we will only harvest an amount proportional to current population. In other words, we harvest \(hx\) per unit of time for some \(h > 0\) (similar to earlier example with \(h\) replaced with \(hx\)).
- Construct the differential equation.
- Show that if \(kM > h\text{,}\) then the equation is still logistic.
- What happens when \(kM < h\text{?}\)
Solution.
(a) \(\frac{dx}{dt}=kx(M-x)-hx\)
(b) Rearranging \[ \frac{dx}{dt}=kx\left(\frac{kM-h}{k}-x\right) \] The term \(\frac{kM-h}{k}>0\) in this case, and we cane rename it to \(N\) making the equation \[ \frac{dx}{dt}=kx(N-x) \] With \(k\) and \(N\) positive, making this the logistic equation again.
(c) If \(kM<0\) then \(N<0\text{,}\) so the critical points are at \(x=0\) and \(x=-N\text{.}\) The latter point can be ignored since the population can only be positive. Substituting positive x values in \(kx(N-x)\) gives negative values, hence \(\displaystyle \lim_{t\to\infty} x(t)=0\text{,}\) i.e. the population collapses.
(b) Rearranging \[ \frac{dx}{dt}=kx\left(\frac{kM-h}{k}-x\right) \] The term \(\frac{kM-h}{k}>0\) in this case, and we cane rename it to \(N\) making the equation \[ \frac{dx}{dt}=kx(N-x) \] With \(k\) and \(N\) positive, making this the logistic equation again.
(c) If \(kM<0\) then \(N<0\text{,}\) so the critical points are at \(x=0\) and \(x=-N\text{.}\) The latter point can be ignored since the population can only be positive. Substituting positive x values in \(kx(N-x)\) gives negative values, hence \(\displaystyle \lim_{t\to\infty} x(t)=0\text{,}\) i.e. the population collapses.
Exercise 1.6.5.
A disease is spreading through the country. Let \(x\) be the number of people infected. Let the constant \(S\) be the number of people susceptible to infection. The infection rate \(\frac{dx}{dt}\) is proportional to the product of already infected people, \(x\text{,}\) and the number of susceptible but uninfected people, \(S-x\text{.}\)
- Write down the differential equation.
- Supposing \(x(0) > 0\text{,}\) that is, some people are infected at time \(t=0\text{,}\) what is \(\displaystyle \lim_{t\to\infty} x(t)\text{.}\)
- Does the solution to part b) agree with your intuition? Why or why not?
Answer.
(a) \(\frac{dx}{dt}=kx(S-x)\)
(b) \(\displaystyle \lim_{t\to\infty}x(t)=S\)
(c) It does, since the solution in part (b) says that if initially some population is infected, this will lead to the number of infected people (\(x(t)\)) to become equal to the whole susceptible population (\(S\)) after a long time. This is a reasonable expectation.
(b) \(\displaystyle \lim_{t\to\infty}x(t)=S\)
(c) It does, since the solution in part (b) says that if initially some population is infected, this will lead to the number of infected people (\(x(t)\)) to become equal to the whole susceptible population (\(S\)) after a long time. This is a reasonable expectation.
Exercise 1.6.6.
Let \(x'=(x-1)(x-2)x^2\text{.}\)
- Sketch the phase diagram and find critical points.
- Classify the critical points.
- If \(x(0)=0.5\text{,}\) then find \(\displaystyle \lim_{t\to\infty} x(t)\text{.}\)
Answer.
a) 0, 1, 2 are critical points. b) \(x=0\) is unstable (semistable), \(x=1\) is stable, and \(x=2\) is unstable. c) 1
Exercise 1.6.7.
Let \(x'=e^{-x}\text{.}\)
- Find and classify all critical points.
- Find \(\displaystyle \lim_{t\to\infty} x(t)\) given any initial condition.
Answer.
a) There are no critical points. b) \(\infty\)
Exercise 1.6.8.
Assume that a population of fish in a lake satisfies \(\frac{dx}{dt} = kx(M-x)\text{.}\) Now suppose that fish are continually added at \(A\) fish per unit of time.
- Find the differential equation for \(x\text{.}\)
- What is the new limiting population?
Answer.
a) \(\frac{dx}{dt} = kx(M-x)+A\) b) \(\frac{kM + \sqrt{{(kM)}^2 + 4Ak}}{2k}\)
Exercise 1.6.9.
Suppose \(\frac{dx}{dt} = (x-\alpha)(x-\beta)\) for two numbers \(\alpha <
\beta\text{.}\)
- Find the critical points, and classify them.
For b), c), d), find \(\displaystyle \lim_{t\to\infty} x(t)\) based on the phase diagram.
- \(x(0) < \alpha\text{,}\)
- \(\alpha < x(0) < \beta\text{,}\)
- \(\beta < x(0)\text{.}\)
Answer.
a) \(\alpha\) is a stable critical point, \(\beta\) is an unstable one. b) \(\alpha\text{,}\) c) \(\alpha\text{,}\) d) \(\infty\) or DNE.
Unstable points with one of the arrows pointing towards the critical point are sometimes called semistable.
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