Math 1280 notes,6, Limit cycles (9.7) Periodic solutions are an important aspect of di¤erential equations, since many physical phenomena occur roughly periodically. In the last section, for example, we saw how predator-prey phenomena can be periodic, and obviously, anything connected with daily, monthly (lunar month) or yearly cycles in nature is approximately periodic. We can write a general system of ode’s in vector form as x0 = f (x) ; 0

1 x1 (t) B x2 (t) C C where x (t) = B @ ::: A : Then a solution is periodic if there is a T such that xn (t) x (t + T ) = x (t)

for all t: The simplest example (aside from constant solutions) are the solutions of x00 +x = 0; i.e. c1 sin t + c2 cos t for any constants c1 ; c2 : These all have period T = 2 : Notice that for small c1 and c2 ; when the solution is close to 0; the velocity x0 is small, so the solution changes slowly. However its amplitude is also small, so the period can, and does, remain the same. On the other hand, for nonlinear systems, the period will usually change from one solution to another. For example, consider the predator prey phase plane on page 537 of the text. Those solutions which are near the equilibrium point in the center will have period about equal to that of the linearized system. But those solutions near the outside, where the trajectory comes close to the origin, will have long periods. This is because as the solution comes near to (0; 0) it slows down, but still has to move a long way (has large amplitude). (If you consider the last sentence to be a “proof”, perhaps you should major in biology, not math! But this can be proven.) As another example, consider the nonlinear pendulum equation x00 + sin x = 0: We discussed earlier how a solution which starts very close to x = ; x0 = 0 is periodic, but has a long period. This is because it is close to the solution which tends to as t ! 1 and to as t ! 1: In these last two examples, there are many periodic solutions. The phase planes have centers, surrounded by a band of periodic solutions. In section 9.7 the situation 1

is di¤erent. There is only one periodic solution and, in most of the examples, other solutions approach the periodic solution as t ! 1: In this case, the periodic solution is called a “limit cycle”. It is di¢ cult to give examples of limit cycles in which the formulas for the solutions can be given exactly. Most of the examples in 9.7 are quite arti…cial, being constructed speci…cally so that one can give the formula for the limit cycle, and perhaps for the other solutions also. These make it clear what a limit cycle is, but are not models of a speci…c physical process. Example 1 in 9.7 is x0 = x + y x (x2 + y 2 ) : (1) y 0 = x + y y (x2 + y 2 ) These look terribly complicated and impossible to solve. The key to …nding a solution is to use polar coordinates. This means that we set x = r cos ; y = r sin and derive di¤erential equations for r and : We are lead to try this because of the terms x2 + y 2 in the equations, remembering that with the polar coordinate substitution, x2 + y 2 = r2 : I will derive the ode’s for r and in a slightly di¤erent way from the text. You should read both methods. Start with x (t) = r (t) cos (t) ; y (t) = r (t) sin (t) : Di¤erentiate both sides of each equation with respect to t: Start with the …rst equation, where we use the product rule and the chain rule, to get x0 (t) = r0 (t) sin (t)

r (t) (sin (t))

0

(t) :

I’ll write this without the t’s: x0 = r0 cos

r 0 sin

From y = r sin we get y 0 = r0 sin + r 0 cos : Now substitute all of this into the system (1) : This gives r0 cos r 0 sin = r cos + r sin 0 r sin + r 0 cos = r cos + r sin

r3 cos r3 sin :

(2)

Now multiply all terms in the …rst equation by cos and all terms in the second equation by sin ; to give r0 cos2 r 0 sin cos = r cos2 + r sin cos 2 0 r sin + r 0 cos sin = r cos sin + r sin2 2

r3 cos2 : r3 sin2 :

Now add the two equations, noticing the cancellations that occur on each side. This gives r0 = r r3 : Notice the great simpli…cation that has occurred. There is only one unknown function, r; and we can solve the equation using separation of variables. However, solving it is not so instructive as drawing the phase line for this equation. There are equilibrium points at r = 0; 1; and 1: However r = 1 doesn’t make sense in the original variables (x; y) ; so we ignore it. The phase line for r 0 has a right arrow between 0 and 1 and a left arrow in the region r > 1: All solutions with r (0) > 0 tend to 1 as t ! 1: Translating this into the x; y plane, it means that if x2 + y 2 < 1 then x2 + y 2 increases, while if x2 + y 2 > 1 then x2 + y 2 decreases. We can see this easily from the ode for r if we write it as r0 = r 1

r2 :

p In either case, limt!1 x (t)2 + y (t)2 = 1: Remember that r = x2 + y 2 : To get a complete picture we have to …nd 0 as well. We …nd the ode for by solving (2) for 0 : This can be done by multiplying the …rst equation of (2) by sin and the second by cos ; and then subtracting the …rst equation from the second. This gives r 0 = r; so 0 = 1; (t) = t + c: Now look at the phase plane in Figure 9.7.1, and try to understand it in light of what we know about and r: Notice that it is easy to …nd the equation for the limit cycle, since r = 1: We can take = t: We get x (t) = cos ( t) ; y (t) = sin ( t) : Using properties of sine and cosine, we can write this as x (t) = cos t;

y (t) =

sin t:

Note that if you substitute x2 + y 2 = 1 into (1) ; you get x0 = y y 0 = x; so the solution (cos t; sin t) is to be expected. Before going to the next part of 9.7, I’ll give some homework. 3

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Homework, part 1. due Feb. 24

1. pg. 563, # 6 2. pg. 563, # 9.

2

Chapter 9.7, 2nd part.

The systems studied above are very special, and are not derived from physical applications. It is much harder to study limit cycles for general systems x0 = F (x; y) y 0 = G (x; y) : The theorems in this section are much to hard to prove in this text. They are usually proved in a basic graduate course in ode’s. The …rst two theorems, Theorems 9.7.1 and 9.7.2, give necessary conditions for there to be a periodic solution. If these conditions are not satis…ed, then there is no periodic solution and no limit cycle. Theorem 9.7.3 on the other hand gives a su¢ cient condition for there to be a periodic solution. However this condition is quite complicated, and it is a lot of work to verify it. An important example is discussed. I will not repeat these theorem here but will go over them in class. I want to discuss the example however. It is called “Vanderpol’s equation,” named after its discoverer. It is u00 + u2 1 u0 + u = 0: (3) We can write it as a system in the usual way: v0 =

u0 = v (1 u2 ) v

u

(4)

(This is not a linear system, even though I used the letters u and v: ) Here is the phase plane as found with Pplane. You can see the periodic solution. Looking carefully at the arrow, you can see that solutions starting inside the “closed”trajectory spiral out to it, while solutions starting outside spiral in. This makes it a limit cycle.

4

First we look for equilibrium points. We must have v = 0; and therefore u = 0 also, and the only equilibrium point is (0; 0) : Next we linearize (4) around (0; 0) : We can do this most quickly by dropping any terms that are not linear, which means only the u2 v term. I’ll use p and q for the variables in the linearized system: p0 = q q 0 = p + q: There is a parameter, which prompts us to wonder if there is a bifurcation as changes. The number of equilibrium points doesn’t depend on ; but perhaps the stability does. So we …nd the eigenvalues. From (3) without the u2 u0 term we can see that the eigenvalues satisfy r2 The roots are r=

r + 1 = 0: p

2

4

: 2 We will only consider small values of ; in which case r is complex, with a real part 2 : Therefore (0; 0) is a spiral point which is asymptotically stable if < 0 and unstable if > 0: Thus some sort of bifurcation takes place as crosses zero. However, it turns out that the nature of that bifurcation is rather complicated to analyze. We will content ourselves only a partial attempt to …nd an energy function. (A Lyapunov function is even harder to …nd.) 5

Let H (p; q) = u2 + v 2 : Then H_ = 2uv + 2v

u+

1

u2 v =

1

u2 v 2 :

We therefore see that if juj < 1; then the sign of H_ is the same as the sign of ; :unless v = 0: This means that when H_ 0; all solutions starting in the circle 2 2 _ u + v < 1 spiral in, while when H 0 all solutions spiral out. Notice how H gives us more “global” information than the “local” information we got from linearizing. But the two results are consistent. If juj > 1; the results are the opposite. Now we’ll stick with the case > 0: Then “large” solutions spiral in a lot of the time, while small solutions spiral out. Neither can cross the periodic solution, which a sort of balance, where the solution spends about equal amounts of time inside and outside j j = 1:You can see all this in the Pplane picture above, which was for = :5. The proof that this periodic solution actually exists is di¢ cult, and we will not present it here. One tool is theorem 9.7.3. This theorem discusses a general system x0 = F (x; y) y 0 = G (x; y) where, here, we will assume that F and G have continuous partial derivatives everywhere. Suppose R is a bounded region in the (x; y) plane which has no equilibrium points. If (x (t) ; y (t)) is a solution which remains in R for all t 0; then Theorem 9.7.3 implies that either (x (t) ; y (t)) is a periodic solution, or it tends to a periodic solution as t ! 1: In the case of Vanderpol’s equation, we choose > 0; in which case we know that solutions within x2 + y 2 = 1 spiral outward. Let R1 denote the region outside x2 +y 2 = 1: There are no equilibrium points in this region. If a solution (x (t) ; y (t)) starts in R1 then it must remain in R1 for all t 0; since it cannot enter the region x2 + y 2 1: But we cannot immediately apply Theorem 9.7.3 because it requires a bounded region R; and R1 is not bounded. It is possible to construct a bounded region R as desired, but it is quite complicated. We won’t attempt to go further in this problem. The Poincaré-Bendixson theorem has a more far reaching consequence. By showing that solutions of two dimensional systems must either be periodic, or tend to a periodic solution, or tend to an equilibrium point, it eliminates the possibility 6

of very complicated solutions, which wander about unpredictably. This means it eliminates the possibility of what mathematicians call “chaos” for a system of two …rst order ode’s. But such complicated solutions can occur for some three dimensional systems, and this is the subject of the next section, Section 9.8, the …nal section in chapter 9.

3

Homework

There is one more problem due on February 24. Section 9., 12

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