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Welcome to Calculus. I'm professor Ghrist and we're about to

Â begin Lecture 20 on O.D.E, Linearization. Differential equations open up an entire

Â world of applications of Calculus. However, there are limits as to how far

Â integration will get you. >> In this lesson, we'll explore those

Â limits of integration, and see what to do when integration doesn't work.

Â Along the way we'll learn how to linearize a differential equation.

Â >> We have seen in the past how to use Calculus to linearize functions, but

Â there are many other things that can be linearized.

Â Today's lesson will consider linearization, differential equations.

Â We're going to do so, through examining a simple model for a simple oscillator.

Â By which I mean something that goes around and around, and is periodic in

Â it's behavior. We're going to consider theta an angular

Â variable, as a function of time. And we want the simplest possible Model.

Â That is, theta simply evolves linearly over time from 0 to 2 pi, and then back

Â to the beginning. What kind of differential equation would

Â model that? Well, I think you can figure that out

Â one. D theta dt equals a, a constant.

Â This has solution by integrating theta is AT plus the constant of integration, the

Â initial condition, theta naught. But notice that because we're

Â interpreting this as an angle, everything is modulo 2 pi.

Â When you get to 2 pi you are reset to 0. What happens if instead of a single

Â oscillator we have a pair of simple oscillators?

Â The first one, let's say, has an angle theta 1 of t.

Â The second has an angle theta 2 of t. The differential equations would be

Â simply d theta one dt equals a and d theta two dt equals a.

Â Or we assume that both oscillators have the same natural frequency a that

Â determines how fast they spin. What happens however if we have these

Â oscillators coupled, if they exert a mutual influence, one on the other.

Â That is if theta 2 is a little bit ahead of theta 1, we want to add something to d

Â theta 1 dt and subtract something from d theta 2 dt.

Â Speed one up, slow the other down. This is going to be a small amount, a

Â small coupling. And it's going to have to depend on the

Â values of data two and data one. The model that we'll use is adding or

Â subtracting. A small number epsilon times the sin of

Â theta 2 minus theta 1. We use that expression to keep things

Â angular. This quantity, theta 2 minus theta 1, has

Â a name. It is called the Phase of this pair of

Â oscillators. It's the difference in angles.

Â What we're going to do is consider how the phase evolves.

Â How does it change? Well, to determine its rate of change, we

Â compute the derivative, dphi dt, from the definition, that is, simply, the

Â derivative of theta 2 minus theta 1. Now, we know what d theta 1 dt and d

Â theta 2 dt are from our model. It is a minus epsilon Sin phi, minus

Â quantity a plus epsilon Sin phi, or notice we've substituted in phi at the

Â appropriate place in our model. There's some obvious cancellation that

Â goes on and we're left with negative 2 epsilon sin phi.

Â Notice this is a differential equation that we can solve.

Â Dphi dt is negative 2 epsilon sin phi. So, let's try to solve it and see what

Â happens. If we use separation, we get dphi over

Â sin phi equals negative 2 epsilon dt. And now we integrate both sides.

Â What do we get? Well, on the right hand side, we get,

Â clearly, negative 2 epsilon t plus a constant.

Â On the left hand side, well, we need to integrate the cosecant.

Â Well, integration can be difficult. This is one of the problems in

Â differential equations. I don't remember the integral of

Â cosecant. But let's say that I look that up and I

Â get that the integral of cosecant is minus log absolute value cosecant plus

Â cotangent. Even when I have that, that does not help

Â me, hardly at all. I can write down that solution and

Â exponentiate to get cosecant phi plus cotangent phi is a constant times e to

Â the 2 epsilon of t. But what I want is to know phi of t, what

Â the phase is doing as a function of time. And I cannot solve that equation for phi.

Â What do we do when we can do the Calculus, but can't interpret the result?

Â We're going to simplify through linearization, not linearizing a

Â solution, but linearizing the equation itself.

Â If we consider dphi, dt equals negative 2 epsilon, sign phi, and we expand that

Â sign phi about phi equals 0. We can drop thehigher order terms and

Â keep only the linear term. Giving the linearized differential

Â equation, dphi dt equals negative 2 epsilon phi.

Â Now you know and I know the solution to such an equation depends on the constant,

Â in front. Therefore we obtain the linearized

Â solution given by the initial condition, phi naught, times e to the negative two

Â epsilon t. This is only an approximate solution to

Â the original differential equation, but this approximation gives us a hint.

Â It predicts that there is an exponential decay in the phase.

Â That is, if you have two oscillators with this coupling between them, then their

Â angle difference. Their phase, should be decreasing

Â exponentially to zero, this is called synchronization.

Â It says or predicts that these two oscillators will become in sync.

Â Now the wonderful thing about this is that you can observe this phenomenon in

Â simple, physical systems involving oscillatory agents.

Â With a small amount of coupling between them.

Â And you can see that they do synchronize. Once the synchronization occurs, it is

Â fixed, and it never changes. This is an example of something called an

Â equilibrium solution. More generally, an equilibrium of the

Â differential equation x dot equals f of x is a solution of the form x of t is a

Â constant. Now otherwise said, the x dt or x dot

Â equals zero. Now since x dot equals f of x, one could

Â also define an equilibrium as a root of f the right hand side, the differential

Â equation. Now if you look at solutions to x dot

Â equals f of x and plot them as a function of time, t.

Â You will occasionally find some initial condition for which you have a constant

Â solution, these are equilibria of the system.

Â If you start there, you stay there. Now these equilibria come in two forms in

Â general these are the stable equilibria and the unstable equilibria.

Â A stable equilibrium has the property that if you have an initial condition

Â that is nearby and you evolve it over time.

Â You get closer and closer to the stable equilibrium.

Â This is what we saw with the coupled oscillator with phi equals zero.

Â On the other hand there are unstable equilibria for which if you start at a

Â nearby initial condition you do not converge to the equilibrium solution over

Â time. Rather you diverge, and move away.

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Now, one way to determine stability of an equilibrium is to plot not x versus t,

Â but x dot versus x. Remember, x dot equals f of x in this

Â model. And so, by plotting f, one can see the

Â equilibria clearly. The roots of f, that is, where it crosses

Â the horizontal axis, are precisely the equilibria.

Â And looking at the graph tells you whether x dot is positive, that is x is

Â increasing or x dot is negative. That is x is decreasing and so one can

Â see the stable equilibria as things like syncs where nearby values of x tends

Â toward the equilibria. The unstable equilibria looks like

Â sources, starting nearby pushes you further and further away, in the context

Â of the system that we looked at, the phase equation.

Â Phi dot equals negative two epsilon sin phi, what did we see?

Â If we plot plhi dot versus phi, then we get a sin wav with a negative coefficient

Â out in front of it. Notice that this has a root at phi equals

Â 0. And when phi, is, less than zero, then

Â phi dot is positive, and you're increasing, phi, as a function of time.

Â To the right of zero, phi dot is negative, which means, that phi is

Â decreasing as a function of time, and this tells us that zero is a stable

Â equilibrium for this system. That means if you have no angle

Â difference between the two isolators, it will remain that way.

Â If you have a small angle difference. Well, it will disappear.

Â And as we saw from linearization, that will happen exponentially fast.

Â Notice, however, that this system also has another root.

Â Another equilibrium at phi equals pi. Which is the same as negative pi, and

Â this is unstable. Physically, this corresponds to an anti

Â synchronization, where the two oscillators are perfectly out of phase.

Â Where their angles, differ, by a value of pi.

Â And then we've seen how linearization was useful in predicting a stable equilibrium

Â for our phase equation. How does linearization work, in the

Â general case, where x dot equals f of x? Well, let's make an assumption that x has

Â an equilibrium at a. Then, what happens if we Taylor expand

Â the equation about this equilibrium? X dot equals f of x, but that is, as you

Â know, f of a, times f prime at a, times x minus a, plus some terms in big O of x

Â minus a squared. Now, because a is in equilibrium, that

Â means x dot equals 0 there, x dot is f of x, so f at a vanishes.

Â That means, we have x is f prime of a times x minus a plus higher ordered

Â terms. To linearize, we drop the higher ordered

Â terms and look at the coefficient in front of the linear term.

Â This coefficient is f prime at a. The derivative of f at the equilibrium.

Â Now what does this mean? This means that if the derivative is

Â positive at a, then what? Well to the left, x dot is negative and

Â is decreasing. On the right, x dot is positive and x is

Â increasing. That means we have an unstable

Â equilibrium, if the derivative is positive.

Â On the other hand, if the derivative is negative.

Â If f prime at a is less than 0. Then to the left x dot is positive, and x

Â is increasing to the right. X dot is negative, and x is decreasing,

Â and we have a stable equilibrium. This follows the exact pattern that we

Â saw for our simple linear ODE. The sign of the constant out in front

Â tells you whether you have exponential growth or exponential decay.

Â Whether you're unstable, or whether you're stable.

Â