Math 1270 Honors ODE I Fall, 2008 Class notes # 14. x 0 = F (x; y) y 0 = G (x; y) u 0 = au + bv = cu + dv

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1 Math 1270 Honors ODE I Fall, 2008 Class notes # 1 We have learned how to study nonlinear systems x 0 = F (x; y) y 0 = G (x; y) (1) by linearizing around equilibrium points. If (x 0 ; y 0 ) is an equilibrium point, then the linearization of (1) around (x 0 ; y 0 ) is the system u 0 = au + bv v 0 (2) = cu + dv where a = F x (x 0 ; y 0 ) ; b = F y (x 0 ; y 0 ) ; c = G x (x 0 ; y 0 ) ; and d = G y (x 0 ; y 0 ). a summary of Theorem 9.3 of the text: Here is a b Theorem 1 Assume that :has nonzero eigenvalues r c d 1 ; r 2 : If r 1 6= r 2 and (0; 0) is a saddle point, a node, or a spiral for the linearized system (2), then (x 0 ; y 0 ) is the same for the nonlinear system (1). If the eigenvalues are equal, (x 0 ; y 0 ) can be a node or a spiral point for (1). If (0; 0) is a center for the linearized system, then (x 0 ; y 0 ) can be a center or a spiral for (1) In the case of a node or a spiral, the stability is the same for each system. In the case of a center, the stability of (x 0 ; y 0 ) is unknown based on the linearized system. To understand this theorem better, we should give careful denitions of saddle points, nodes, centers, and spirals for the nonlinear system (1). As far as I could see, the book does not do this. Here I will just dene saddle points, mainly because they are the easiest to dene! But before doing so, I need to dene \trajectory" more carefully, and discuss the denition. 1 1 To understand the denition, you have to recall what a \curve" is. A curve in R 2 is dened to be a function, or mapping, whose domain is an interval I on the real line, and whose image (i.e. its range) is in R 2 : Informally, we think of a curve as the image of the function, meaning the set of all points in R 2 that are in the image of the function, rather than the function itself. That is sometimes misleading, but probably adequate for our purposes here. 1

2 Denition 2 A trajectory for the system (1) is the image of a curve dened by the mapping t! (x (t) ; y (t)), where (x (t) ; y (t)) is a solution of (1), and the domain of this mapping is the entire interval of existence of the solution. Here is an important point: Suppose that (x (t) ; y (t)) is a solution for the system (1) : Then for any constant c; (x (t c) ; y (t c)) is also a solution. This is seen by substitution. For example, suppose c = 1; and let X (t) = x (t 1) and Y (t) = y (t 1). Then X 0 (t) = x 0 (t 1) = F (x (t 1) ; y (t 1)) = F (X (t) ; Y (t)) : This works because the system (1) is autonomous. A similar equation works for Y (t), and for any c. The meaning of this is that (x (t) ; y (t)) and (x (t c) ; y (t c)) have the same trajectories. The only dierence is that t = 0 at dierent points along this trajectory. For an autonomous system, a shift of the time axis does not mean a change in a trajectory. 2 Denition 3 An equilibrium point (x 0 ; y 0 ) of (1) is called a \saddle point" if there are exactly two trajectories on which solutions tend to (x 0 ; y 0 ) as t! 1, and two other trajectories on which solutions tend to (x 0 ; y 0 ) as t! 1. Denition If (x 0 ; y 0 ) is a saddle point, suppose that 1 and 2 are the two trajectories on which solutions tend to (x 0 ; y 0 ) as t! 1. Then the set 1 [ 2 [f(x 0 ; y 0 )g is called the \stable manifold" for (1) at the point (x 0 ; y 0 ). The union of f(x 0 ; y 0 )g and the two trajectories which tend to (x 0 ; y 0 ) as t! 1 is called the \unstable manifold". It can be shown that the stable and unstable manifolds are each smooth curves. This means that there is no sudden change in direction along one of these curves as you cross the midpoint (x 0 ; y 0 ). This, plus the statements in Theorem 9.3, are essentially what is called the \stable manifold theorem" for ode's. \Manifold" can be interpreted as \curve" in two dimensions. The proof of this theorem is often given in graduate ode courses. As an example of a system with more than one saddle point, we looked at the system 2 In fact, phase planes do not make sense for nonautonomous systems, because then, two solutions could reach the same point at dierent times, and then go in dierent directions. This would cause dierent trajectories to intersect, which is not possible for autonomous systems, as long as the uniqueness theorem for initial value problems applies 2

3 x 0 = y y 0 = x + x 3 (3) and found that its equilibrium points are (1; 0) ; (0; 0) ; and ( 1; 0) : By linearization we know that we have one of the following pictures, where the right graph should have a spiral around (0; 0) ; though we don't know if it is stable or unstable. (I can't draw spirals with my math printing software.) To look at this more carefully, consider again the linearization of (3) around (1; 0) ; which is the linear system u 0 v 0 = u v : () We do not expect that the phase plane of (3) around (x 0 ; y 0 ) = (1; 0) is exactly like that of () around (u; v) = (0; 0). The nonlinear term x 3 must have some eect. It is often hard to see just what this eect is. There are several tools for trying to determine this. One method is to try to convert the system into a rst order ode, at least near the equilibrium point. This is described on page 99 of the text. The tangent vector to a trajectory (x (t) ; y (t)) is the vector (x 0 (t) ; y 0 (t)) : If x 0 (t) 6= 0 at some point, then the trajectory is not vertical there, and its slope is given by the slope of the tangent vector, namely y0 (t) : So near such a point, we can nd the phase curve x 0 (t) by solving the ode dy G (x; y) = dx F (x; y) : We will try this for equation (3). We get dy dx = x3 y x ; 3

4 and we can solve this by separation of variables. We get Z Z y dy = x 3 x dx 1 2 y2 = 1 1 x 2 x2 + c: (5) We want a curve which tends to the point (1; 0) : The reason is that we think that the point (1; 0) is a saddle, because (u; v) = (0; 0) is a saddle for the linearized system (). To get the curve whose equation is (5) to pass through (1; 0), we set y = 0; x = 1 and get c = 1: Therefore, one phase curve has the equation y 2 = 1 2 x x : (6) It can be a little trick to see what the graph of this looks like. We could write it as x 2 + y 2 = x ; and this tells us that for small x; it looks like the circle x 2 + y 2 = 1. But as x! 1; 2 the distance from (0; 0) gets larger. If we use a computer to plot the curve (6) in a region near (1; 0), we get We can add arrows based on the fact that x 0 = y; so x 0 > 0 if y > 0 and x 0 < 0 if y < 0: We see that there are solutions tending to, and away from (1; 0) ; just as in a saddle. But these trajectories are no longer straight lines, as they were for the linearized system (). However, the tangents to the trajectories at (1; 0) will the be eigenvectors for the linearized system. Keep in mind that in the picture above, there are actually ve dierent trajectories shown. One is the single point (1; 0) : The other four are the four dierent

5 trajectories which tend toward (1; 0) ; either as t! 1 or as t! 1. These never reach (1; 0) : No two trajectories can intersect. And all of these lie on the set of points which satisfy 1 2 y x2 1 x = 1. We can learn a lot more from the equation (5), by considering other values of c: We will write it as 1 2 y x2 1 x = c: The function H (x; y) = 1 2 y x2 x (7) is called an \energy function", and for the system () ; we can see that H is constant along trajectories. This helps us graph them. For instance, we see that all for the trajectories are symmetric around the x and y axes, because ( y) 2 = y 2 and ( x) 2 = x 2. Let's plot the particular case c = 1 ; which we graphed earlier, for a larger range of x and y. We get (using a computer) : You should gure out the directions of the arrows, and nd eight dierent trajectories which lie on the set of points with H = 1. By adding arrows as before, we see that ( 1; 0) is also a saddle point. This is consistent because the linearization of (3) around ( 1; 0) is also the system (). Further, there are two trajectories which connect (1; 0) and ( 1; 0). One goes from left to right with y > 0; and the other goes from right to left, with y < 0. This type of trajectory is important enough to have a name: Denition 5 A trajectory for an autonomous system x 0 = f (x) is called \heteroclinic" if solutions on this trajectory tend to one equilibrium point as t! 1 and another equilibrium point as t! By a \solution on a trajectory", we mean a solution (x (t) ; y (t)) such that for some t 0, (x (t 0 ) ; y (t 0 )) is on the curve which forms the trajectory. In this case (x (t) ; y (t)) is on the 5

6 Continuing with our analysis of (3), we consider the nature of the equilibrium point (0; 0). The linearization of (3) around (0; 0) is 0 1 u 0 = u (8) 1 0 u where u = v : This is easy to see, because we can just drop the x 3 term from the nearly linear system (3). The eigenvalues of A in this case are i. Hence, (0; 0) is a center for (8) : We ask whether this is the case for (3) : This question can be hard to answer. The table on page 508 of the text indicates that we can't decide the nature of an equilibrium point when its linearization is a center. The unlinearized system might have a center or a spiral point there. To answer the question it is usually necessary to have an energy function. Fortunately, we have one in this case, namely, the function H in (7) We see that H (0; 0) = 0; meaning that in (5) ; c = 0: To get nearby trajectories, we take small values of c, smaller than the 1 we used earlier. For example, we plot adding this to the previous plot. H (x; y) = 1 8 ; This suggests that there is indeed a center at (0; 0), since no spiraling is seen. This is easy to see mathematically. Suppose there were a spiral around (0; 0). Start at a point where the spiral intersects the positive y axis, and follow it around trajectory for every t: There are many solutions on the same trajectory, but they dier just in where their \starting point" (x (0) ; y (0)) is. They are related simply by shifting t to t c for some c. Therefore, they all have the same limits as t! 1. 6

7 one loop until again it is on the positive y axis This will be at a dierent y value, either higher or lower. But H has not changed. This is a contradiction, since H = 1 2 y2 whenever x = 0: Finally, we use the computer to ll in more of the complete phase plane, by plotting H (x; y) = c for some other values of c. One nal example, related to the homework. Consider the system x 0 = y y 0 = x + y 3 (9) The only equilibrium point is (0; 0). The linearized system is found to be u 0 = v v 0 = u: We have already seen that (0; 0) is a center for (u; v) Hence, we don't know whether (0; 0) is a center or a spiral for the (x; y) system. We can't nd a function H (x; y) which is constant for the system (9). But consider the function E (x; y) = x 2 + y 2 : Let p (t) = E (x (t) ; y (t)) = x (t) 2 + y (t) 2. p 0 (t) = 2xx 0 + 2yy 0 = 2xy + 2y = 2y 0: If (x (t) ; y (t)) is a solution of (9), then x + y 3 Thus, p (t) is increasing, which means that (x (t) ; y (t)) gets farther and farther from (0; 0) as t increases. This means that (0; 0) is an unstable spiral point for (9). 7

8 If our second equation had been y 0 = x y 3 ; we would nd that p 0 (t) = 2y 0; so (0; 0) is a stable spiral point. Yet, the linearized system would not x + y 3 = 3y 2 x y 3 = 3y 2 and these are the same at (0; 0). To summarize: If there is a function H;which is not constant for all (x; y), but which is constant along trajectories (phase curves), then there can be no spirals. If there is a function E which is increasing along trajectories, or one which is decreasing along trajectories, then there can be no centers. I hope to have time to explain this further in class. This last example is equivalent to the second order equation where w = x; w 0 = y: We can see that w 00 + w 03 + w = 0; (10) E (x; y) = x 2 + y 2 = w 2 + w 02 : One way of deriving E is to multiply (10) by w 0, to get w 0 w 00 + w 0 + w 0 w = 0: We can integrate part of this expression. Namely, Z w 0 w 00 = 1 2 w02 Z w 0 w = 1 2 w2 So we put w 0 on the other side to get 1 2 w2 + w 02 = 8 Z w 0 :

9 The left side is 1E (x (t) ; y (t)) ; or 1 p (t) in the notation from earlier. 2 2 multiply by 2 and dierentiate to get We can p 0 (t) = w 0 0; so p (t) is decreasing. (It can't be constant over any interval.) Since p (t) > 0 if the solution is nonzero, we conclude that lim t!1 w (t) = 0; verifying the stability found earlier. The technique of multiplying a second order equation by w 0 and integrating is a very useful tool. 9