Physics 4550, Fall 2003 – Dynamical Systems 1

Notes on Dynamical Systems

Dynamics is the study of change. The primary ingredients of a dynamical system are its state andits rule of change (also sometimes called the dynamic). Dynamical systems can be continuous in time,with the rule being a differential equation, or discrete in time, with the rule being a difference equation.The dynamical rule can be linear or nonlinear. The rule can be deterministic or stochastic (meaningcontaining some element of randomness).

Physical dynamics refers to systems with rules of change that are derived from the equations ofmotion (Newton’s or otherwise). In physical dynamics, the state of the system is usually a set of positionsand velocities for each of the system’s particles. In physical dynamics, one is typically interested inpredicting future states given all forces acting and some starting information.

1. Examples of Simple Physical Dynamical Systems

Example #1: Consider a small mass connected to a massless, rigid rod being pushed through a containerof viscous fluid. For simplicity, suppose that the mass in the fluid isneutrally buoyant. That is, ignore gravity. Suppose that the rod exerts aforce of constant magnitude on the mass. The situation is shown in thefigure to the right. (The ends of the container are sealed and the rod fits snugly in a small hole on the leftface.) As the mass moves through the fluid it experiences a frictional force opposing its motion themagnitude of which is

†

bv, where

†

b is a coefficient describing the “gooiness” of the fluid and the cross-section of the mass and

†

v is the mass’s instantaneous speed. Suppose the mass moves only horizontallyto the right and the rod pushes to the right. In a cartesian coordinate system with x horizontal and positiveto the right, Newton’s 2nd Law for the mass is

†

max = -bvx + Frod .

Of course, the acceleration is the second derivative of position with respect to time. The custom incontinuous time dynamical systems is to express the dynamical rule as a first derivative of the statevariables—position and velocity, for physical dynamics. Thus, in this form, this problem is expressible astwo first-order differential equations:

†

˙ x = vx

˙ v x = -bm

vx +Frod

m

(1)

Throughout these notes, we use the notation that a “dot” means “derivative with respect to time.” Notethat the second equation can be solved directly for

†

vx as a function of time. The result can then beplugged into the first equation to get

†

x(t) .

The differential equations (1) are linear in that x and vx and their derivatives appear raised to the“1” power only. Two solutions of a linear equation can be added to obtain a third solution. A generalsolution to the velocity differential equation can then be expressed as a sum of two parts—one that takescare of the forcing term and a second that satisfies the “homogeneous” equation—what’s left withoutforcing. The two parts are called “particular” and “homogeneous” solutions, respectively. For the second

equation in (1), the homogeneous equation is

†

˙ v xH = -bm

vxH , which has a solution of the form

†

vxH = C exp(-bt / m), where C is some constant that depends on initial conditions. As time goes on the

Physics 4550, Fall 2003 – Dynamical Systems 2

homogeneous part of the general solution vanishes, leaving only the particular (force dependent) part. Thelate-time behavior of this equation always reduces to a particular form in which forcing and friction are inbalance—no matter what the initial state of motion is. In the language of dynamical systems, this late-time behavior is called an attractor of the dynamics; all initial states are “attracted” to it.

A general solution to this problem is

†

vx = vT + (v0 - vT )exp(-bt / m)

x = x0 + vT t + (m b)(v0 - vT )(1- exp(-bt / m))

where

†

v0 is the mass’s initial velocity,

†

x0 is its initial position, and vT , the particular (or attractor)

solution, is

†

vT =F0

b. (We’ve replaced

†

Frod by the constant force

†

F0 .) You can see from these equations

that as

†

t Æ •, vx Æ vT and x Æ vT t , irrespective of the mass’s initial state of motion. The attractingvalue vT is also sometimes called the terminal velocity: In the attracting state, the power input,

†

F0vT ,

equals the power dissipated,

†

bvT2 .

Example #2: Consider the same problem as above, only now let

†

Frod = F0 cos(wDt) . This is exactly thesame problem when

†

wD is zero. The equations of motion for the mass on the rod become

†

˙ x = vx

˙ v x = -bm

vx +F0 cos(wDt)

m

It is natural to expect that the mass will eventually also oscillate horizontally with the same frequency,

†

wD , as that of the driving force. That is, because of friction, the motion of the mass will eventually attainan attracting state. The particular or attractor value of

†

vx —let’s call it

†

vxA to remind ourselves we’re nottalking about a general velocity—cannot be written simply as

†

vmax cos(wDt) . This is because (1) the lefthand side (LHS) of the velocity differential equation would not be zero (as it was in the previous examplewhen the particular solution was substituted)—it would be a sine, and (2) the right hand side (RHS)would contain only cosines. The LHS would then at most equal the RHS only at special times, not, as isrequired, at all times. A correct particular solution can be obtained by assuming that

†

vxA is a sum ofcosines and sines. This can be done compactly by setting

†

vxA = vmax cos(wDt +f ) .

The particular solution in the previous example depended on F0 and b. To ensure that

†

vxA is a solution,plug it into the velocity differential equation above and determine what

†

vmax and f must be to make theequation true. The algebra is a bit tedious but the results can be expressed as

†

tan(f) = -mwD

b

vmax2 =

F02

(mwD )2 + b2

(2)

Physics 4550, Fall 2003 – Dynamical Systems 3

Note that when

†

wD is set to zero (remember, this condition is the same as the previous example),

†

vmax = F0 b = vT (as we expect). In addition,

†

tan(f) is zero, and so therefore is

†

f . Thus, as we expect,when the driving force is constant (as in the previous example) the velocity and the driving force areexactly in phase. They aren’t, however, if

†

wD is not zero. In fact, the larger

†

wD , the more negative is

†

tan(f) and the closer

†

f is to

†

-p / 2. Thus, at high driving frequencies the velocity lags behind the drivingforce by 90°.

Since the rate at which energy input from the rod is dissipated into “heat” by friction is

proportional to

†

vmax2 , dissipation depends on driving

frequency. In general, a “spectrum” is a classification ofbehavior as a function of constituent frequencies. Apower spectrum is power dissipated as a function of theinput driving frequency. For the problem consideredhere, (log-log) plots of dissipated power as a function of

†

wD are shown to the right. The upper graphcorresponds to a relatively small mass, while the lowerone is for a higher value. As long as

†

(mwD )2 << b2—which is true for larger values of

†

wD ifm is smaller—the dissipated power is approximatelyconstant, which means that irrespective of drivefrequency, the (velocity) response of the system isroughly the same. This limit is like the case of zerodriving frequency. Because the input force and thevelocity are in phase in this limit, the input force is always pushing the mass in the direction of itsdisplacement. Thus, the work done by the input force is always positive.

On the other hand, when

†

(mwD )2 >> b2 , the (velocity) response becomes much smaller. Thesystem response drops off sharply at high frequencies—like 1/frequency2. (The straight lines in the figurehave slope = -2.) In this limit the input force and velocity are 90° out of phase. Thus averaged over acomplete drive cycle, the input force does essentially zero work. This system is an example of what iscalled a low band pass filter. A simple electrical analog of this system is an AC driven LR circuit.

Example #3: Now, imagine that the rod is replaced by an ideal, Hooke’s Law spring with force constant k.In a suitable coordinate system, the equations of motion become

†

˙ x = vx

˙ v x = -kx -bm

vx +F0 cos(wDt)

m

Something new enters, namely the velocity differential equation now depends explicitly on the position ofthe mass. We can try the attractor velocity

†

vxA = vmax cos(wDt +f ) as in the previous example, but then we

also have to use the attractor position

†

xA = -vmax

wD

sin(wDt +f ) . Substitution of both into the velocity

differential equation produces

log(driving frequency)lo

g(po

wer

)

Physics 4550, Fall 2003 – Dynamical Systems 4

†

tan(f) = -m(wD

2 -w02 )

bwD

vmax2 =

F02wD

2

(m[wD2 -w0

2])2 + (bwD )2

(3)

where

†

w0 =km

is the so-called “natural frequency” of

the undamped, undriven oscillator. Interestingly, fromthe first equation of (3), when the driving frequency ismuch less than the natural frequency

†

tan(f) is positive.Consequently, the velocity leads the input force andthus the work done by it over a drive cycle is small (asis the dissipation). At frequencies much higher than thenatural frequency, the velocity lags the input force andso its work (and the associated dissipation) is againsmall. When the driving frequency is close to thenatural frequency, however, the velocity is close tobeing in phase with the driving force and the work done(and the dissipation) are large. This phenomenon iscalled resonance. The situation is shown to the right.The upper graph corresponds to small mass, the lower to large. A small mass does not cause much flexurein the spring, so is like the rigid rod example above.

Example #4: The physical pendulum is a mass at the end of a rigid,massless rod. (A rod can push or pull, whereas a string can only pull.You’ll see why I want that to happen in a bit.) The figure to the rightshows a typical pendulum, free to swing under the influence ofgravity with no friction. We assume that the pendulum moves in the(x, y) plane only. The cartesian components (x, y) of the mass’s

position are related through the constraint equation

†

x2 + y2 = l2. Inthe absence of friction and external driving, the forces on the massare due to gravity (pointing down and with magnitude mg) and therod (pointing along the rod and with magnitude T). While mg is aconstant, T varies as the mass moves. It turns out that the variation inT is not important for our purposes. The dynamical rules for thisproblem are:

†

˙ x = vx

˙ v x = -Tx m

˙ y = vy

˙ v y = -g + Ty m

The quantities Tx and Ty are, respectively, the x- and y-components of the rod force.

It’s easier to analyze this problem in polar coordinates

†

(r,q) (as in the figure). The constraintequation implies that r = l. We then have

l

x

y T

mg

q

log(driving frequency)

log(

pow

er)

Physics 4550, Fall 2003 – Dynamical Systems 5

†

˙ q = w

atangential = l ˙ w = -g sin(q)

{acentripetal = lw2 = -g cos(q )+ T m}

The first two equations are sufficient to determine the angular velocity. In principle, the latter could thenbe inserted into the third equation to determine T, if desired. As long as

†

q remains small,

†

sin(q ) ª q , andthe two important dynamical equations are

†

˙ q = w

˙ w = - gq l

These are equivalent to a simple harmonic oscillator with natural frequency given by

†

w02 = g / l . This is

just Galileo’s simple pendulum—perfectly periodic with a frequency that is independent of the amplitudeof the motion.

When the angle is not small, the dynamics becomes nonlinear because

†

q appears in the sinefunction of the second differential equation. If we add friction and periodic driving, we get

†

˙ q = w

˙ w = -w02 sin(q)- (b m)w + (F0 ml)cos(wDt)

Because of the nonlinearity in (6), there are no known closed-form (e.g., sines, cosines) solutions. Thebest we can do is to integrate the equations numerically. Numerical integration often requires picking“small” step sizes. But, what is meant by small? Is 1 year small? (It is compared with a century, but notwith a second.) 1 second? (Ditto: 1 yr vs. 1 ns.) 1 nanosecond? (Ditto: 1 s vs. 1 ps.) To resolve suchquestions, one makes the equations to be integrated dimensionless. This requires defining units offundamental measures and is often a bit of an art. For the problem we are considering, one way ofproceeding is to measure frequency in units of

†

w0 and time in units of

†

1 w0 . We’re going to remember weare doing this without writing frequency and time as different symbols! We are also going to introducethe dimensionless quantities

†

b = b mw0 and

†

g = F0 mg . Then,

†

˙ q = w

˙ w = -sin(q )- bw +g cos(wDt)(4)

Differential equations whose solutions wiggle (or abruptly change slope) require extra care whenintegrating numerically. The method of choice is called Fourth Order Runge-Kutta with Variable StepSize.

(Don’t worry about its details. You always have to look them up when you need them. The Excel sheetpendulum.xls integrates the above equations using the RK4 algorithm.)