1550078481-Ordinary_Differential_Equations__Roberts_

482 Ordinary Differential Equations

(Verify this fact.) Since this solution is periodic with period 2n, the origin

is a neutrally stable critical point. A phase-plane graph of Y2 versus YI (the

angular velocity of the pendulum versus its angular displacement) for the ini-

tial conditions: (i) YI (0) = 0, Y2 (0) = 1.5, (ii) YI (0) = 0, Y2 (0) = 2, and

(iii) YI(O) = 0, y 2 (0) = 2.5 is shown in Figure 10.22. The given initial con-

ditions correspond to the pendulum being at rest in the downward vertical

position at time t = 0 and being struck sharply to impart an initial angular

velocity. The maximum angular displacement, max IYI(t)I, for the given three

initial conditions is 1.5, 2, and 2.5, respectively. Notice that these values are

much larger than .1- the maximum value for which we assumed the approxi-

mation of sin y by y to be valid. The reason for choosing initial conditions so

large that maxlyI(t)I is much larger than .1 is so that the linear approxima-

tion sin y ~ y is not valid. Then we can compare the solutions of the linear

system (4) with the solutions of the corresponding nonlinear system (6) which

we obtain next.

4

2

Y2 0

-2

-4

-10 -5

2.5

0

Y1

5

Figure 10.22 Phase-Plane Portrait for System (4)

Simple Pendulum with No Damping and No Forcing Function

10

Ifwe assume there is no damping (c = 0) and no forcing function (f(t) = 0)

and if we choose e = g in magnitude, then the equation of motion for a simple

pendulum is

(5) y" + siny = O; y(O) = co, y' (0) = CI.

The general solution of this initial value problem involves elliptic integrals.
Letting YI = y and Y2 = y' as before, we can rewrite (5) as

Y~ = Y2; Y1(0) =Co

(6)

y~ = -siny1; Y2(0) = c1.

Simultaneously setting Y2 = 0 and - sin y 1 = 0, we find system (6) has an
infinite number of critical points located at (nn, 0) where n is any integer.