MAM2046W - Second year nonlinear dynamics

Section 1.5: Reversible systems

If you make a movie of an egg dropping on the floor and then play it forwards and backwards, it’s very obvious which one is the correct direction. Eggs don’t simply coalesce from a mess on the floor, and jump back up into your hand.

Well, actually that’s not quite true. In theory, they can, but the chances of them doing this are exponentially small in the number of particles in the system. What would it take for this strange reversal to happen?

Well, you would need that the particles in the floor which are all jiggling around because they have some temperature all happen to jiggle in just the right way to push the bits of egg together at just the right moment. The particles in the egg itself would also all have to be jiggling in just the right way that the energy from them goes into chemical bonds which reform the egg and have enough left over to propel the egg up off the floor. Unlikely, I say, but technically not impossible...although in the lifetime of the universe, even if there were a million cracked eggs on a million floors in every planet in every galaxy in the observable universe, the chances of this happening even once are basically zero, so we say that this process is not reversible.

The paragraph above was really about entropy and the arrow of time, and we’re not going to go into that now, though there’s much to be said about it.

How about if I take a video of a pendulum swinging back and forth and play it one way, and then the other way? You won’t be able to tell which one is the ‘correct’ direction, because each behaviour is just as likely. We can say that a pendulum swinging is a reversible system. The dynamics with time running in one direction is equivalent to the dynamics with time running in the other direction.

Let’s say we just take the pendulum starting on the left and swing it one half oscillation to the right. Now if you played the video backward it would look different to if you played the video forward. However, if we reversed both the time, and swapped left for right, the two videos would look exactly the same.

Swapping the direction of time is the same as swapping

t<->-t

. Let’s look at this more generally, and in terms of the equations themselves.

Let’s go back to our Newton’s law for a conservative system:



=F(x)

which is equivalent to:

=F(x)

which is really

x=F(x)

The first thing that we see here is that if we let

t->-t,

then

d(-t)

, and so

ie. it doesn’t change.

So that under the transformation of

t->-t

the equation of motion remains the same.

Actually we have to be slightly more careful here because we would like to know what’s happening in the phase plane (labeling the phase plane directions

(x,v)

). So let’s go to the coupled first order system:



F(x)



Now, if we just let

t->-t,

then the system doesn’t remain the same. If we just did this then we’d end up with:



F(x)



which is a different set of equations. If however we also let

v->-v

then all of a sudden we get back to



F(x)



Which says that if we swap the direction of time, and the direction of the velocity then the system remains invariant. This makes sense in terms of the pendulum. Swap the direction of time, and the video will look the same, just with everything moving in the opposite direction.

What does this look like in terms of a trajectory? Well, taking some part of a trajectory in the phase plane, for instance:

In[]:=

Show[Plot[Sin[x]+2,{x,-π,π},AxesLabel->Evaluate[Style[#,20]&/@{"x","v"}],AxesOrigin->{0,0},PlotRange->{-3,3},PlotStyle->Blue],Graphics[Arrow[{{0,Sin[0]+2},{0.2,Sin[0.2]+2}}]]]

Out[]=

Letting

v->-v

will reflect the curve about the x-axis, but we also have to let

t->-t

which corresponds to reversing the direction of the arrow to, so this will give the red curve in the figure below:

In[]:=

Show[Plot[Sin[x]+2,{x,-π,π},AxesLabel->Evaluate[Style[#,20]&/@{"x","v"}],AxesOrigin->{0,0},PlotRange->{-3.1,3.1},PlotStyle->Blue],Plot[-(Sin[x]+2),{x,-π,π},AxesLabel->Evaluate[Style[#,20]&/@{"x","v"}],AxesOrigin->{0,0},PlotRange->{-3,3},PlotStyle->Red],Graphics[Arrow[{{0,Sin[0]+2},{0.2,Sin[0.2]+2}}]],Graphics[Arrow[{{0,-(Sin[0]+2)},{-0.2,-(Sin[-0.2]+2)}}]]]

Out[]=

The system is reversible, if the red curve is also a trajectory of the original system.

In terms of the equations, any system for which

t->-t

and

v->-v

is a symmetry is reversible.

Given a more general system:



=f(x,y)



=g(x,y)

what constraints are there on

and

such that the system is reversible under

t->-t

and

y->-y

(note that we have written

ynowinsteadofvtobeabitmoregeneral.

Well, first of all performing the transformations we get:



=f(x,-y)



=g(x,-y)

where the



has not changed sign because both

tandy

have changed.

So for this to be invariant under the change, we need that

is odd in

and

giseveniny

which will give us back our original system.

So that's it. For a two-dimensional first order system to be reversible (ie. time reversal invariant), we need that

is odd in

and

giseveniny

Remember from the last section that if you take a conservative system, where there is a linear center (ie. when you perform the linear analysis you find a center), then this center is robust in the full non-linear model (ie. the center is really there in the full system, not just the linearised system). Well, it turns out that in reversible systems, the same holds. A center which is really there in the non-linear system is called a non-linear center.

We will state a theorem which will be in terms of a system which has a linear center at the origin of the phase plane. This can be extended to any position along the x-axis of the phase plane.

All trajectories near a linear center at the origin of the phase plane for a reversible dynamical system described by continuously differentiable functions, will themselves form closed orbits.

Essentially this says that if you know that you have the following linear center:

In[]:=

Show[ListPlot[{{0,0}},PlotStyle->{Black,PointSize[0.02]}],AspectRatio->1,AxesLabel->Evaluate[Style[#,20]&/@{"x","y"}]]

Out[]=

If there system is reversible, then whatever happens near the linear center must form a closed orbit, and must be reflectively symmetric about the x-axis. For instance:

One way to think about this is that the instability of a linear center can lead to a stable or unstable spiral in general. However, if the system is reversible, whatever is happening above the x-axis must be mirrored below the x-axis. This rules out spirals around a linear center and so the orbits about it can only be closed orbits, and not spirals.

Example

Let’s look at

Finding the fixed points is equivalent to solving

Which has solutions at:

Let’s focus on the fixed point at the origin. The Jacobian of this system is

which for the origin is

OK, so we have a reversible system, with a center at the origin, so we know that we must have closed cycles around the fixed point at the origin.

For the other fixed points we have

Which are at

respectively. ie. they are both saddle points. We can calculate the eigenvalues and eigenvectors of these Jacobians and we get:

Let’s just focus on the first fixed point’s the eigenvalues and eigenvectors. At (-1,1) we have

We can look at the same thing for the fixed point at (-1,-1)

Let's draw on what we know...

We also know that there are no other fixed points in this system, so there’s not a great deal of freedom about what the trajectories could be doing. One other piece of information that we do have though is that everything is differentiable, and so we don’t expect any sharp jumps in trajectories, and so because of reversibility, whenever we pass through the x-axis it will be perpendicular to it. This just says that we know that there must be a reflection about the x-axis in the trajectories. In the absence of the smoothness condition, we could have, for instance:

But this has a sudden turn in it on the x-axis. This could only come about because of some discontinuity in the vector field or its derivatives, so anything that passes through the x-axes has to do so orthogonally.

We can figure out the direction by the flows about the fixed points:

Let’s think about just one trajectory, the one carrying on from this arrow:

What could it possibly do next given that we know that there are no more fixed points? Well we see that the arrows directly below it are pointing in the opposite direction, so it can’t meet up with them, and the arrows along the positive x-axis are all pointing down, so the only place it could seem to go such that the arrows pointing down can link up with anything would be somewhere on that positive x-axis.

Give it a go and see if you can imagine what the flow might look like.

Well, I’m just going to write down the whole flow for a few trajectories, but you don’t have to know how to do this exactly.

Pretty neat, huh?

While we couldn’t have figured out the precise trajectories of this using our techniques, we could have had a good guess at their qualitative features.

Note that there are two trajectories joining the two saddlepoints, one going to the left of the center, and the other going to the right. These are called heteroclinic trajectories or saddle connections. Note that a homoclinic trajectory starts and ends at the same saddle point and a heteroclinic trajectory links two different saddle points.

An important generalisation

Are all conservative systems reversible, and vice versa?

For the case of a reversible system which is not conservative. See exercise 2 below.

For the case of conservative systems which are not reversible, I was able to find a couple of papers which discuss this in the literature, though the systems are a lot more complicated than we cover in this course. These papers are:

Take a look if you want to know more.

Can we find a reversible system where there is a centre which is not on the x-axis?

We start with a system where the centre is on the x-axis, but not at the origin. This is given by

Taking the red arrows, there’s only really one thing that they can do:

Plotting the flow lines on top of this we get:

This looks like a hell of a mess, but if you now plot the flow lines in this new system we see:

An exercises for you

1) Given

Sketch the phase portrait.

Note that this has an attracting fixed point, and so cannot be conservative, but it can be reversible. Check whether this is true.

Assignment

The non-dimensionalised pendulum is modelled by

1) Write this as a coupled set of first order equations.

2) Is the system reversible?

3) Is the system conservative? Find the energy as a function of the phase space coordinates.

3) Calculate the fixed points of the system

4) Calculate the Jacobian at the fixed points. You should find that there are two unique Jacobians for the infinite number of fixed points. Why is this?

5) What sort of fixed points are they?

6) What does reversibility and conservation of energy tell you about the nature of one of the linear fixed points?

7) By finding the eigenvalues and eigenvectors of the Jacobian, plot the trajectories close to the three closest fixed points to the origin in phase space.

8) Use the energy function to plot some characteristic trajectories above and below the horizontal axis.

10) What dynamics do the saddles correspond to?

12) This one is tricky to visualise! You can take the cylinder that you have, and rather than plotting the angle and angular velocity directions, plot the angle around the cylinder and the energy vertically. You should end up with something that looks like this. You are NOT expected to be able to figure this out for yourself. However, I want you to spend a while pondering what this means. Ignore the gaps in the lines...this is just a function of the way that this is plotted.

13) Extension: Now imagine that you add a damping term to this system. Is it still conservative? Is it still reversible? How will the phase space look now? Try and draw the phase space just by thinking about the behaviour of a damped pendulum and without going through all of the calculation. What would the equivalent of the strange u-shaped diagram above look like for the damped trajectories?

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