Tues Mar 246.4 Nonlinear mechanical systems

Announcements:

Warm-up Exercise:

See if you can recallhowderivethe total

energy f w TE KE t PE

forthe pendulum problem on the next pageNow insteadof the mass hanging on

a string like in

chapter3 it's attached to a massless rigid rodwhich is fixed at the oppositeend so that itsfree to rotatearound around insteadofjust swingingback and forth

Example 1) The rigid rod pendulum.

We've already considered a special case of this configuration, when the angle from vertical is near zero. Now assume that the pendulum is free to rotate through any angle .

In Chapter 3 we used conservation of energy to derive the dynamics for this (now) swinging, or possibly rotating, pendulum. There were no assumptions about the values of in that derivation of the non-linear DE (it was only when we linearized that we assumed was near zero). We began with the total energy

TE = KE PE =12mv2 mgh

=12m L t

2m g L 1 cos t

And we set TE t 0 to arrive at the differential equation

tgL

sin t = 0 .

TE KE t PE say height

Imu t mghfrom lowpoint

Slt Left s'tt L044 HI

htt L LcosottTE EmL2 0 t ti t ngL I cosOltl

ISITISCt Vlt

TE Const for system inmotionit date o mix o O myLl sin070

O me O fo t IsinO

you can alwaysset Sys intomotion

with arbitrary 0 at anygiventimeso O t sin0 0

tgL

sin t = 0

We see that the constant solutions t = must satisfy sin = 0, i.e. = n , . In other words, the mass can be at rest at the lowest possible point (if is an even multiple of ), but also at the highest possible point (if is any odd multiple of ). We expect the lowest point configuration to be a "stable" constant solution, and the other one to be "unstable".

We will study these stability questions systematically using the equivalent first order system forx t

y t=

t

twhen t represents solutions the pendulum problem. You can quickly check that this is the system

x t = y

y t =gL

sin x .

By now this equivalence between second order differential equations and first order systems of two differential equations must be second nature! Prove it below!

Notice that constant solutions of this system, x 0, y 0, equivalentlyx t

y t=

x

y equals constant

must satisfy y = 0, sin x = 0, In other words, x = n , y = 0 are the equilibrium solutions. These correspond precisely to the constant solutions (with zero velocity) of the second order pendulum differential equation, = n , = 0.

Equilibrium points and stability definitions for higher order autonomous differential equations (or systems)always have to do with constant solutions, and can always be phrased and analyzed in terms of the equivalent first order system of differential equations, and the corresponding equilibrium points.

0 t ofsinO 0

If Olt solves 2ndorderDE

f ff L forTY 5 IF in 1storders

pEs.no f7qx

fromDE

8 fqsinx

I

Here's a phase portrait for the first order pendulum system, with gL

= 1, see below.

a) Locate the equilibrium points on the picture and verify algebraically.

b) Interpret the solution trajectories in terms of pendulum motion.

c) Looking near each equilibrium point, and recalling our classifications of the origin for linear homogenous systems (spiral source, spiral sink, nodal source, nodal sink, saddle, stable center), how would you classify these equilbrium points and characterize their stability? For the equilibrium points thatlook like stable centers and have that linearization, you will use separable DE's as in the predator-prey example in yesterday's notes, to prove that they are stable centers. This is homework problem w9.2. You will verify that your conserved function is equivalent to total energy function TE = KE PE in this conservative system. HE't penduluma x y 1 hit o HEI Y motion

y sinx

c htt O n odd agroundmass attop unstable aroundsaddlepoints pasodir

ht o n even

stable b t bachIcenters fat o C it o 10.0 ht o 1299 forth

apparently

L aroundaround

n TegoO dir

7O

Example 1 Work out the Jacobian matrices and linearizations at the equilibria for the undamped rigid rod pendulum system on the previous page

x t = y

y t =gL

sin x

d

f xyGlay

I It o

Jenner.am fog IJ att Itsroots IVE

stable centerfor linearizationindeterminate for non linear

equilptJnuoa.ao L ol II re re

X t.VESaddlepoint for linearization

for non linear equilpt

1e) Stability discussion at x* , y* = n , 0 , n even. As discussed in yesterday's notes, if a conserved energy function for an autonomous first order system of differential equations has a strict local extreme value at a critical point (constant solution) for the system of DE's, then that critical point is a stable equilibrium point. That is the context of your homework problem w9.2. For the pendulum system

x t = y

y t =gL

sin x

the TE=KE+PE function is a multiple of

E x, y =12

y2 gL

1 cos x .

Here is a graph of the TE function, in the case gL

= 1 (which can always be achieved by appropriate

choice of units). And below that graph is a graph of the countour curves, which solutions of the first ordersystem follow. Thus x* , y* = n , 0 , n even, are stable, because the TE has global minimum valuesat those points.

Example 2 What happens when you add damping?

t c t gL

sin t = 0

x t = y

y t =gL

sin x c y

same equilibriumpointshit 0 h odd

remain saddles

h II O n even

are eitherspiral sinks

underdamped

or nodalsinks

overdamped

upshot today's yesterday's lectures illustratelinearization and its use in understandingnonlinear autonomous DE's systemsofDE'snear constant i e equilibriumsolutions