Notes on linearization around fixed points. Polking et al treats this for 2 x 2 systems in 10.1. We are just getting a head start on the idea. We are considering an autonomous equation of the form y' = f(y). Where f(a) = 0, y' = 0, and y(t) = a is a fixed point for the differential equation. Suppose we want to know what happens near y = a. We construct an approximation to the equation. We assume that the variable h = y-a is small. We shift the equation to be written in terms of h. This looks like h' = (y-a)' = y' - a' = y' = f(y) = f(h+a) ~ f(a) + f'(a)h + error. By Taylor's theorem, the error is small. We call the equation h' = f(a) + f'(a)h = f'(a)h the linearization about the fixed point y = a. It is of course, a theorem, that under the right circumstances, solutions near y(t) = a behave like the solutions of the linearization. Of course, if f'(a) < 0, the solutions h ~ k expt(f'(a)t) approach 0 and we can deduce that the solutions y(t) approach a as t goes to infinity. If f'(a) > 0, the solutions h(t) grow exponentially, and after awhile the linear approximation is false. Lets do an example. Lets try our old friend the logistics equation y' = ry(1 - y/K). This has fixed points y = 0 and y = K. It is always easier to linearize about y = 0. The other fixed point is y = K. Now h = y - 0 = y; The linearization is y' = ry. The solutions to the linearization are y = k exp(rt). They grow exponentially, so y = 0 is an unstable fixed point. At y = K, f'(K) = r - 2 r y/K = -r. Now h = y-K, and the linearization is: h' = -rh. We know that nearby solutions decay into the fixed point h = 0 or y = K. We also know the decay rate: y-K ~ k exp(-rt). Hence y = K is a stable fixed point. In general, when we are modeling with non-linear equations, we cannot expect to solve them. We may not even be able to graph them (try graphing something that is 10 X 10!) However, we can always look for fixed points and linearize around the fixed point for clues as to behavior. Now for some exercises: Find all the fixed points of the equation y' = y(y^2 - 9) and linearize the equation around each fixed point. Find the fixed points of the equation y' = y cos(y). Explain which ones are stable, and verify your answer by finding the linearization.