For the pendulum equation x sin x x xt find all equilibri

For the pendulum equation x = - sin x, x = x(t) find all equilibrium positions and sketch level curves of the energy near the lower and upper equilibrium positions.

Solution

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 di erence 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 L sin = 0; 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 L = 1, which is unlikely but will simplify the equations. We switch to a system, setting = x; 0 = y. This gives x 0 = y y 0 = sin x (1) The equilibrium points are easily seen to be (0; 0);(; 0);(2; 0); :::;(; 0);(2; 0); ::: So there are in nitely many of them. But we will see that there are only two types of behavior. The linearized system at (x0; y0) is u 0 = v v 0 = cos (x0) u 1 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 (1) as a rst order equation, valid as long as x 0 6= 0. We get dy dx = sin x y Z ydy = Z sin x dx 1 2 y 2 = cos x + c for some scalar c. Therefore, H (x; y) = 1 2 y 2 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 dtH (x (t); y (t)) when (x (t); y (t)) solves (1): d dt 1 2 y 2 cos x = 2yy0 + (sin x) x 0 = 2y ( sin x) + (sin x) y = 0: (2) Thus, H (x (t); y (t)) is constant.1 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.