[Notes on Diffy Qs home] [PDF version] [Buy paperback on Amazon]

[next] [prev] [prev-tail] [tail] [up]

Note: 1 lecture, §2.2 in [EP], §2.5 in [BD]

Let us consider problems of the form

where the derivative of solutions depends only on (the dependent variable). Such equations are called autonomous equations. If we think of as time, the naming comes from the fact that the equation is independent of time.

Let us return to the cooling coﬀee problem (see Example 1.3.3). Newton’s law of cooling says

where is the temperature, is time, is some constant, and is the ambient temperature. See Figure 1.8 for an example with and .

Note the solution (in the ﬁgure ). We call these constant solutions the equilibrium solutions. The points on the axis where are called critical points. The point is a critical point. In fact, each critical point corresponds to an equilibrium solution. Note also, by looking at the graph, that the solution is “stable” in that small perturbations in do not lead to substantially diﬀerent solutions as grows. If we change the initial condition a little bit, then as we get . We call such critical points stable. In this simple example it turns out that all solutions in fact go to as . If a critical point is not stable we would say it is unstable.

Let us consider the logistic equation

for some positive and . This equation is commonly used to model population if we know the limiting population , that is the maximum sustainable population. The logistic equation leads to less catastrophic predictions on world population than . In the real world there is no such thing as negative population, but we will still consider negative for the purposes of the math.

See Figure 1.9 for an example. Note two critical points, and . The critical point at is stable. On the other hand the critical point at is unstable.

It is not really necessary to ﬁnd the exact solutions to talk about the long term behavior of the solutions. For example, from the above we can easily see that

Where DNE means “does not exist.” From just looking at the slope ﬁeld we cannot quite decide what happens if . It could be that the solution does not exist for all the way to . Think of the equation ; we have seen that solutions only exist for some ﬁnite period of time. Same can happen here. In our example equation above it will actually turn 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 , but it may get there rather quickly.

Often we are interested only in the long term behavior of the solution and we would be doing unnecessary work if we solved the equation exactly. 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 . We draw the axis, we mark all the critical points, and then we draw arrows in between. If , we draw an up arrow. If , we draw a down arrow.

Armed with the phase diagram, it is easy to sketch the solutions approximately.

Exercise 1.6.1: 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 can easily classify critical points as stable or unstable^{3}.

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 million humans per year. Suppose is the number of humans in millions on the planet and is time in years. Let be the limiting population when no harvesting is done. The number is a constant depending on how fast humans multiply. Our equation becomes

We expand the right hand side and set it to zero

Solving for the critical points, let us call them and , we get

Exercise 1.6.2: Draw the phase diagram for diﬀerent possibilities. Note that these possibilities are , or , or and both complex (i.e. no real solutions). Hint: Fix some simple and and then vary .

For example, let and . When , then and are distinct and positive. The graph we will get is given in Figure 1.10. As long as the population starts above , which is approximately 1.55 million, then the population will not die out. It will in fact tend towards million. If ever some catastrophe happens and the population drops below , humans will die out, and the fast food restaurant serving them will go out of business.

When , then . 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 1.11.

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 1.12.

Exercise 1.6.3: Take . a) Draw the phase diagram, ﬁnd the critical points, and mark them stable or unstable. b) Sketch typical solutions of the equation. c) Find for the solution with the initial condition .

Exercise 1.6.4: Take . a) Draw the phase diagram for . On this interval mark the critical points stable or unstable. b) Sketch typical solutions of the equation. c) Find for the solution with the initial condition .

Exercise 1.6.5: Suppose is positive for , it is zero when and , and it is negative for all other . a) Draw the phase diagram for , ﬁnd the critical points, and mark them stable or unstable. b) Sketch typical solutions of the equation. c) Find for the solution with the initial condition .

Exercise 1.6.6: Start with the logistic equation . Suppose we modify our harvesting. That is we will only harvest an amount proportional to current population. In other words, we harvest per unit of time for some (Similar to earlier example with replaced with ). a) Construct the diﬀerential equation. b) Show that if , then the equation is still logistic. c) What happens when ?

Exercise 1.6.7: A disease is spreading through the country. Let be the number of people infected. Let the constant be the number of people susceptible to infection. The infection rate is proportional to the product of already infected people, , and the number of susceptible but uninfected people, . a) Write down the diﬀerential equation. b) Supposing , that is, some people are infected at time , what is . c) Does the solution to part b) agree with your intuition? Why or why not?

Exercise 1.6.101: Let . a) Sketch the phase diagram and ﬁnd critical points. b) Classify the critical points. c) If then ﬁnd .

Exercise 1.6.102: Let . a) Find and classify all critical points. b) Find given any initial condition.

Exercise 1.6.103: Assume that a population of ﬁsh in a lake satisﬁes . Now suppose that ﬁsh are continually added at ﬁsh per unit of time. a) Find the diﬀerential equation for . b) What is the new limiting population?

Exercise 1.6.104: Suppose for two numbers .

a) Find the critical points, and classify them.

For b), c), d), ﬁnd based on the phase diagram.

b) , c) , d) .

^{3}The unstable points that have one of the arrows pointing towards the critical point are sometimes called semistable.