1 The pendulum equation

Transcription

1 Math 270 Honors ODE I Fall, 2008 Class notes # 5 A longer than usual homework assignment is at the end. The pendulum equation We now come to a particularly important example, the equation for an oscillating pendulum. We discussed several derivations of this equation earlier in the course. There are two cases, the \undamped" pendulum and the \damped" pendulum. The dierence is that the damped pendulum has a rst derivative term that causes the energy to decrease. We start o with the undamped case. In the text the equation is given, on page 537, as 00 + g sin = 0; L where is the angle the pendulum makes with the vertical (which changes with time), g is the gravitational constant and L is the length of the pendulum. Note that the mass of the pendulum does not appear. I will assume that g =, which is unlikely but will simplify the equations. We L switch to a system, setting = x; 0 = y. This gives x 0 = y y 0 = sin x () The equilibrium points are easily seen to be (0; 0) ; (; 0) ; (2; 0) ; :::; ( ; 0) ; ( 2; 0) ; ::: So there are innitely many of them. But we will see that there are only two types of behavior. The linearized system at (x 0 ; y 0 ) is u 0 = v v 0 = cos (x 0 ) u

2 At (0; 0) we get u 0 = v v 0 = u which give circles, and (0; 0) is a center. At (; 0) we get u 0 = v v 0 = u and this gives a saddle. As we move along the x-axis, we alternate between centers and saddles for the linearized system. To learn more, we look for an energy function. One way is to write () as a rst order equation, valid as long as x 0 6= 0. We get Z dy dx = ydy = sin x y Z sin x dx 2 y2 = cos x + c for some scalar c. Therefore, H (x; y) = 2 y2 cos x, and the phase curves are the graphs of the equation H (x; y) = c. To check that this is correct, we can calculate d H (x (t) ; y (t)) when (x (t) ; y (t)) dt solves (): d dt 2 y2 cos x = 2yy 0 + (sin x) x 0 = 2y ( sin x) + (sin x) y = 0: (2) Thus, H (x (t) ; y (t)) is constant. We say that H is \constant along trajectories." We use H to plot the phase plane. First, we see that since H is constant, the centers for the linearized system remain centers in the full x; y system. By Theorem 9.3, the saddle points remain saddle points. This process of \dierentiating along a trajectory" is very important. Suppose that H is a function of x and y; and we have an ode system x 0 = F (x; y) ; y 0 = G (x; y) : Then we let This is what we computed in equation (2). _H (x; y) = H x (x; y) F (x; y) + H y (x; y) G (x; y) : 2

3 Now let's evaluate H at the equilibrium points. H (0; 0) = ; H (; 0) = ; H ( ; 0) = ;and so H = at 2k and + at (2k + ), for each integer k. Notice that is the minimum possible value of H; but there is no maximum possible value, because of the term 2 y2. To get the complete phase plane we must consider the graph of the equation H (x; y) = c for every allowable value of the constant. c: First suppose c = 0. Then 2 y2 cos x = 0 Two specic points on the trajectory are 0; p 2 :and ; 0. For 0 < x < ; dy = sin x < 0; and so y decreases. But 2 2 dx y H is symmetric around the x and y axes (why?), and so we can ll in a complete loop around (0; 0). Also, H has period 2 in x; and this allows us to reproduce this loop around (2; 0) ; ( 2; 0) ; and so forth. plot of 2 y2 cos x = 0 Notice next that if c < then there are no points, because H (x; y) can't be any smaller than : And if c = then we must have y = 0; cos x = ; so x = 0; 2; 2; etc. Since ( ; 0) and (; 0) are saddle points, a trajectory with H (x; y) = connects these two points. This is then a heteroclinic orbit, as was seen in a previous example. Again, this repeats below the y axis, and every 2 in x. Now we combine these on one plot: plot of 2 y2 cos x = 3

4 plot of 2 y2 cos x = c for c = 0; Finally, what happens if c >? Then, we see that y is never zero, since 2 y2 = c + cos x and jcos xj : Here is a computer plot of 2 y2 cos x = 2; added to the previous picture. plot of 2 y2 cos x = 0; ; 2 The diagrams above have no arrows. But these are easy to supply. For y > 0 the direction is to the right, while for y < 0 it is to the left. Next, we add damping. This time we will add a linear damping term, giving the equation x 00 + kx 0 + sin x = 0; where k > 0: As usual, write as a system: x 0 = y y 0 = sin x ky: The equilibrium points are the same as before. First we linearize around (0; 0). I will denote the Jacobian matrix, Fx G x 4 F y G y

5 by DF: Then we have DF (x;y) = 0 cos x k 0 so at (0; 0) our matrix A = DF (0;0) is : The eigenvalues turn out to be k k + p 2 2 k 2 4 and k 2 2p k 2 4: We will assume that the damping is small, so that jkj < 2: In this case we get two complex eigenvalues with negative real part. The equilibrium point (0; 0) is asymptotically stable, with spiraling in orbits, at least locally. : To determine the global picture we again let H (x; y) = 2 y2 we nd that _H (x; y) = y ( sin x ky) + sin x (y) = ky 2 : cos x: This time This is similar to the previous example, x 00 + x 03 + x 3 = 0: 2 The function q (t) = y 2 (t)2 cos (x (t)) is decreasing and bounded below and so q (t) tends to a limit as t tends to innity. However this does not mean that (x; y) necessarily approaches (0; 0) : Every other equilibrium point along the x-axis { all those of the form (k; 0) where k is an even integer, also is a stable spiral, so nearby solutions wind onto it. Notice that the energy function H approaches its minimum value of. To check this we can use the computer to draw some solutions. I have switched from Maple to xpp In this example also, we could imagine that possibly the inwardly spiraling curves in this picture don't spiral all the way to zero, but instead spiral onto a curve, maybe a circle or ellipse, or something close to these. Again one can use the fact that _H (x; y) < 0 if y 6= 0 to show that this does not happen. 2 So, for consistent notation, I should have called this function E: But if we don't know ahead of time how it will come out ( _ H = 0 or _ E 0), then we don't know what to call it. 5

6 A more complete picture of the phase plane for the damped pendulum equation appears on page 50 of the text.. 2 Homework. In problem of the assignment due last week you considered the system x 0 = x ( x y) y 0 = 2 y 2 y 3 2 x : In doing this problem you can refer to your work from last time. (a) Find the nullclines. (Recall from class that the x-nullcline is the set where x 0 = 0; and the y-nullcline is the set where y 0 = 0. Thus, the y axis is part of the x-nullcline.) (b) Show that the x and y axes are \invariant", meaning that if x (t 0 ) = 0 for some t 0 ; then x (t) = 0 for all t; and similarly for y: (I don't mean to assume that they are both zero at the same time.). Draw arrows showing the trajectories on these axes. (c) In the region x > 0; y > 0; draw arrows showing the directions of trajectories as they cross the nullclines. (To indicate all changes of directions which may occur along the nullclines, at least four arrows will be needed.) (d) Draw the linearized (u; v) phase planes at (0; 0) and at 2 ; 2. (Requires two separate graphs.) (e) Explain why there is no \energy function" H for this system. (f) Referring to part (c), for one arrow on each part of each nullcline in x > 0; y > 0; draw the complete trajectory, showing where the solution goes as t! and as t!. (Some solutions will be unbounded. Just indicate the general direction in which they head (forward or backward) as they tend to innity.) (g) Complete the picture by adding the complete stable and unstable manifolds at 2 ; 2. 6

7 2.. Consider the system x 0 = x ( y) y 0 = y (x ) : (a) Find the nullclines and equilibrium points. (b). Find the linearization around each equilibrium point. Draw the linearized (u; v) phase planes. (c) Draw vectors which are tangent to trajectories at several points along each nullcline, being sure to show any changes of direction which may occur at dierent points on the nullcline. (d) Convert the system into a rst order equation and use this to nd an energy function H. (e). Draw the complete phase plane, consistent with all of the above information. You can check your answer with a computer plot using pplane, but it should be clear that you can explain the entire picture without relying on this. These two problems are models of what are called \competing species" and \predator-prey" systems. The predator-prey system is also called a \Lotka-Volterra" equation. 7