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        riddles >> general problem-solving / chatting / whatever >> Differential Equations on a Torus
        
(Message started by: Kyle Strom on Apr 23^rd, 2005, 8:13am)

Title: Differential Equations on a Torus
Post by Kyle Strom on Apr 23^rd, 2005, 8:13am

I am wondering if anyone answer this:

Do differential equations on a torus imply that the solutions to the DE lie on the surface of the 3-manifold torus or if just the initial conditions are chosen on the torus?

Thanks.

Title: Re: Differential Equations on a Torus
Post by Michael Dagg on Apr 23^rd, 2005, 4:04pm

on 04/23/05 at 08:13:03, Kyle Strom wrote:

Do differential equations on a torus imply that the solutions to the DE lie on the surface of the 3-manifold torus or if just the initial conditions are chosen on the torus?

Thanks.

The answer is both. The study of differential equations on a torus is about studying differential equations whose solutions lie entirely on the surface of a torus. Since this is the case the initial conditions must also lie on the torus.

Your use of the term 3-manifold may confuse someone. Although a differential equation whose solutions lie on the surface of a torus is a 3d-differential in rectangular form, the surface of a torus is a 2-manifold. The solid torus is a 3-manifold with a boundary.

Regards,
MD

Title: Re: Differential Equations on a Torus
Post by Thomas Wang on Apr 23^rd, 2005, 10:44pm

That answer is very interesting from my point of view as I have not really studied domains of DE's of the such but what is being said here is that the solutions variables lie on a torus and that is fully acceptable. But, what I have a problem with is that, and I will quote it, "the surface of a torus is a 2-manifold."

It can't be a 2-manifold as the surface of a torus is 3-dimensional - hence, (x,y,z) and has the popular Euclidean metric.

Feedback?

Title: Re: Differential Equations on a Torus
Post by towr on Apr 24^th, 2005, 2:20am

The surface of the torus is locally topologically the same as the open unit disc (i.e. open ball in R²), so it's 2-manifold. At least if I read http://mathworld.wolfram.com/Manifold.html correctly.

Title: Re: Differential Equations on a Torus
Post by Michael Dagg on Apr 24^th, 2005, 9:21am

on 04/23/05 at 22:44:51, Thomas Wang wrote:

It can't be a 2-manifold as the surface of a torus is 3-dimensional - hence, (x,y,z) and has the popular Euclidean metric.
Feedback?

Your reasoning is inconsistent with the definition of a manifold. Simply because a space or an object has a certain dimension that does not mean that if it is also a manifold that the dimensions are equal.

I do understanding your reasoning - here we have the surface of a torus whose points are the triple (x,y,z) in R3 but it is a 2-dimensional manifold. It is definitely natural to ask how this can be.

Note that a circle is a 1-manifold.

Regards,
MD

Title: Re: Differential Equations on a Torus
Post by Icarus on Apr 24^th, 2005, 3:30pm

Differential geometry - where manifolds are defined - is about the intrinsic properties of the object in question, not the extrinsic properties. That is, it is concerned with properties of the object that relate only to itself, and not with those that relate the object to any surrounding space.

If you move a point in the torus, you can only move it in two independent directions - the third direction moves the point out of the torus into the surrounding space. Because there are only two independent directions, the torus is only a 2-manifold, and the location of the point on the torus only takes two coordinates (theta, phi) to describe. The third coordinate from its embedding into R³ is immaterial from the point of view of differential geometry.

In fact, from the point of view of differential geometry of the torus itself, the embedding in R³ does not even exist. The torus is definable without any mention of it being an object in R³. One way to define it is as S¹ x S¹, where S¹ is the circle, defined as R/~, where ~ is the equivalence relation x ~ y if x - [x] = y - [y].

Title: Re: Differential Equations on a Torus
Post by Michael Dagg on Apr 24^th, 2005, 5:48pm

on 04/24/05 at 15:30:47, Icarus wrote:

One way to define it is as S¹ x S¹, where S¹ is the circle

That definition is certainly conventional and tends to fuel deeper thought - that is, the cross product of two circles is a torus.

Normally one would not use rectangular coordinates to study differential equations on a torus but you could if you wanted to do a fair amount of unnecessary work. It is common place for people to think in terms of rectangular coordindates and often results in inconsistencies in dimension, measure, and metric.

However, the distinction that must be made is that the surface of a torus is a 2-manifold whereas the solid torus is a 3-manifold with a boundary. This also holds for a sphere.

Regards,
MD

Title: Re: Differential Equations on a Torus
Post by Ken Wiley on Apr 24^th, 2005, 7:52pm

on 04/24/05 at 17:48:59, Michael Dagg wrote:

that is, the cross product of two circles is a torus.

That is interesting. How does one compute that cross product?

Ken

Title: Re: Differential Equations on a Torus
Post by Icarus on Apr 24^th, 2005, 7:54pm

on 04/24/05 at 17:48:59, Michael Dagg wrote:

Normally one would not use rectangular coordinates to study differential equations on a torus but you could if you wanted to do a fair amount of unnecessary work. It is common place for people to think in terms of rectangular coordindates and often results in inconsistencies in dimension, measure, and metric.

Indeed, but I did not want to try and describe such things as coordinate patching or similar abstract methods of defining a manifold. My purpose was to demonstrate that the torus should not be thought of as a subset of 3-space, but as an object with an existance of its own separate from any imbedding. And the S¹ x S¹ definition is the conceptually easiest way of demonstrating this that I could think of.

Title: Re: Differential Equations on a Torus
Post by Icarus on Apr 24^th, 2005, 7:56pm

on 04/24/05 at 19:52:13, Ken Wiley wrote:

That is interesting. How does one compute that cross product?

You don't compute it. This is a set cross-product, not a vector cross-product: A x B = {(x, y) | x in A and y in B}

Title: Re: Differential Equations on a Torus
Post by Deedlit on Apr 25^th, 2005, 7:11pm

A representation of the torus that everyone is probably familiar with is the playing field in a video game like Asteroids. (I suppose the younger generation might not be familiar with this game!) Basically, if you go off the top of the screen you come back out the bottom, and if you go off the left you come back out the right. To see how this is a torus, imagine the screen is a malleable piece of paper, and glue the top and bottom of the screen together. This forms a cylinder, and you can how the top/bottom wrap-around is part of the geometry now. Next, bend the cylinder so that the left and right ends meet, forming a doughnut. So the Asteroids field is essentially a doughnut, and three dimensional space was not required.

If we look at a vertical or horizontal line of the Asteroids screen, it bends into a circle. So you can see how the torus is the product of two circles.

Incidentally, you see this sort of conceptual difficulty when people talk about physics - a lot of people can't understand how space-time can have a curved geometry, unless it's sitting inside a larger, uncurved space!

Title: Re: Differential Equations on a Torus
Post by Icarus on Apr 26^th, 2005, 4:53pm

on 04/25/05 at 19:11:50, Deedlit wrote:

Incidentally, you see this sort of conceptual difficulty when people talk about physics - a lot of people can't understand how space-time can have a curved geometry, unless it's sitting inside a larger, uncurved space!

And then you also have to explain to them that toruses are NOT curved, despite what they look like!

There is an old joke that a topologist is someone who can't tell the difference between his doughnut and his coffee cup.

My reply is that a differential geometer can't tell the difference between either one and the table top.

Don't bother to tell a geometer that his tire is flat - he will reply, "Of course it is; all tires are flat."

Title: Re: Differential Equations on a Torus
Post by Deedlit on Apr 26^th, 2005, 8:53pm

on 04/26/05 at 16:53:25, Icarus wrote:

And then you also have to explain to them that toruses are NOT curved, despite what they look like!

Hmmm... since doughnuts are often referred to as tori, I figured tori can be curved or noncurved - i.e. being a torus is a topological property, not dependent on geometry. But maybe (R/Z)² is the "default" torus or something.

Quote:

Don't bother to tell a geometer that his tire is flat - he will reply, "Of course it is; all tires are flat."

Good one. :D

Title: Re: Differential Equations on a Torus
Post by Michael_Dagg on Apr 27^th, 2005, 12:51am

All lovely, and with great exchange and respect, however, I ask the original proposer upon what basis did you ask about this subject - beyond simply asking if someone can answer the question?

Title: Re: Differential Equations on a Torus
Post by K Strom on Apr 27^th, 2005, 11:23am

Thanks for all the answers.

My question was based on a class project where we were given 5 topics and we had to research the topics and write up a description of them.

One thing though is that I could not find any examples of DE on a torus on the Internet any place and there seems to be little stuff about it in the library as well.

Can you give one?

Title: Re: Differential Equations on a Torus
Post by K Strom on Apr 27^th, 2005, 11:38am

Also, I forgot to remark about the interesting genealogy. I know the names Markov and Chebyshev and I think Tamarkin.

Title: Re: Differential Equations on a Torus
Post by Michael_Dagg on Apr 27^th, 2005, 1:39pm

The simple pendulum is a differential equation on a torus:

w" + g/L sin w = 0, where w" = d²w/dt²

While the differential equation is interesting associating it with a torus is not. In fact, it is also a differential equation on a sphere of radius L (or on any planar crossection as well).

There is a chapter in Theory of ODEs, Coddington & Levinson, about ODEs on a torus. Arnaud Denjoy did a fair amount of work in this area.

Tamarkin, as you may note, is Russian and McFadden is Irish and is known for his work on summability (nicknamed Dr. Power Series). Several of those genealogy mirrors are not completely up-to-date and if you can't find what you are looking for on one try another one - a good place to look for abstracts too.

Title: Re: Differential Equations on a Torus
Post by K Strom on Apr 28^th, 2005, 7:33am

Again thank you for the answers.

The book that you mention is in our library and I will get it today as I have a few more weeks until I need to turn in the paper. I come up empty with finding anything wrote by Denjoy in English.

The w in the model is an angle function of time?

Title: Re: Differential Equations on a Torus
Post by Michael_Dagg on May 1^st, 2005, 8:18am

Yes, w is an angle function of time. I used w because this board complained when I tried to use theta.

When you are done, I would be pleased to read this topic in your paper.

I would suggest that you add a more interesting example and then mention the simple pendulum at the end:

Consider constructing the parametric equations of a space curve in R3 that is analogous to a helix on the surface of a cylinder but instead transverses the surface of a torus. This construction would be the general solution. Now construct one or more differential equations by differentiating the solution and use the fewest number of differentials as possible. You may want to revisit parametric differentiation.

Title: Re: Differential Equations on a Torus
Post by SWF on May 1^st, 2005, 8:35pm

Are you interested in actually solving equations on a torus, or just using the edge matching boundary condition on a rectangle - like the Asteroids example given by Deedlit? I think the edge matching condition is more common in practical applications. I am thinking in terms of partial differential equations such as the Laplace equation or Schrodinger equation. Also, a three dimensional version of the matching boundary condition might be more common than two dimensional: for example, to solve Schrodinger equation for an infinite crystal with a simple tetragonal atom arrangement: you just need to solve for one atom inside a block and with that boundary condition on the surface.

The Laplace equation is a common and relatively simple PDE with practical applications that shouldn't be too hard to come up with example solutions. Say you want to find the voltage distribtion in a conducting plate in an infinite array of + and - point charges arranged in a checkerboard pattern (+ charges at the center of black squares, - at center of white squares). Instead of working with the infinite number of points, you just need to worry about a square of 4 point charges with that repeating boundary condition (although in this case an infinite summation solution is probably easier). Due to curvature on a torus, this is not identical to solving for voltage distribution for 4 point charges on a hollow toroidal shell. That problem can be solved solved as well, but is more complicated.

You can use a toroidal coordinate system, and that can even be used for three dimensional problems. I only used toroidal coordinates once: to find gravitational field around a torus shaped planet. (Why? Because somebody posed it as an unsolved riddle).

Title: Re: Differential Equations on a Torus
Post by K Strom on May 2^nd, 2005, 7:48am

That is too advanced for my class to understand including myself. It would be just a bunch of gibberish.

The simple example MD gave is sufficient for my purposes and also offers some graphical insight.

Thanks

K.S.

Title: Re: Differential Equations on a Torus
Post by Another Aggie on May 12^th, 2005, 4:23pm

Oh, way the way, what is a very interesting abstract group that is surprisingly also a manifold?

Title: Re: Differential Equations on a Torus
Post by Michael_Dagg on May 13^th, 2005, 7:19am

on 05/12/05 at 16:23:09, Another Aggie wrote:

Oh, way the way, what is a very interesting abstract group that is surprisingly also a manifold?

One of the more interesting that comes to mind is a Lie group.

Title: Re: Differential Equations on a Torus
Post by skydiver on May 13^th, 2005, 1:39pm

Since this area is about differential equations, here one that no one in our class could produce a solution to and Mathematica couldn't do it either.

z(x² + 2xy - y²)dx + z(2xy + y² - x²)dy - (x² + y²)(x + y)dz = 0

Title: Re: Differential Equations on a Torus
Post by Michael_Dagg on May 13^th, 2005, 7:56pm

on 05/13/05 at 13:39:56, skydiver wrote:

z(x² + 2xy - y²)dx + z(2xy + y² - x²)dy - (x² + y²)(x + y)dz = 0

I bet your instructor was disappointed.

The solution is a space curve in R3 and with some work you can produce it.

Title: Re: Differential Equations on a Torus
Post by Deedlit on May 14^th, 2005, 12:00am

on 05/13/05 at 19:56:43, Michael_Dagg wrote:

I bet your instructor was disappointed.

The solution is a space curve in R3 and with some work you can produce it.

Hmmm... it looks to me like a family of surfaces. At each point, the tangent space is restricted to a plane.

Title: Re: Differential Equations on a Torus
Post by Deedlit on May 14^th, 2005, 12:13am

on 05/13/05 at 07:19:53, Michael_Dagg wrote:

One of the more interesting that comes to mind is a Lie group.

Lie groups are by far the most well-known and studied of such groups, so I bet that's what he had in mind.

For everyone else, a Lie group is a group with the structure of an analytic manifold, such that the group operations are analytic. (that is, xy and x^-1 are analytic - but we can just assume the first, and the second comes for free.) The most common examples are linear transformations of vector spaces, like GL_n(C), Sp_n(R), etc. The study if Lie groups is surprisingly deep and connected. A related neat category are the Lie algebras - the tangent spaces to a Lie group at the identity.

Another interesting "hybrid" class of groups are the topological groups - groups with a topological structure, such that the group operations are continuous.

Title: Re: Differential Equations on a Torus
Post by Michael_Dagg on May 14^th, 2005, 8:51am

on 05/14/05 at 00:00:11, Deedlit wrote:

Hmmm... it looks to me like a family of surfaces. At each point, the tangent space is restricted to a plane.

p(x,y,z) = (x^2 + y^2)/(z(x + y)) = c

or

x^2 + y^2 - c(x + y)z = 0

or

z = (c^-1)(x^2 + y^2)/(x + y)

Title: Re: Differential Equations on a Torus
Post by skydiver on May 14^th, 2005, 11:42am

on 05/14/05 at 08:51:08, Michael_Dagg wrote:

p(x,y,z) = (x^2 + y^2)/(z(x + y)) = c

That solution works! I would be very appreciative to get the details on how you solved it. I see that you can look at it as either a family of surfaces or just a single curve since every curve that makes the surface is also a solution.

Title: Re: Differential Equations on a Torus
Post by Michael_Dagg on May 15^th, 2005, 2:06pm

You will appreciate a method of solution much more if you derive it yourself. It is not that difficult. You are looking for a solution of the form p(x,y,z) =c.

Hint: Try to eliminate one of the differentials to reduce it to the form M(u,v)du + N(u,v)dv = 0 and then integrate it.

What book did you use for this class?

Title: Re: Differential Equations on a Torus
Post by skydiver on May 16^th, 2005, 8:26am

The book we used is Diff Equations by Nagle, Saff, Snyder. The problem did not come from the book. It was on the final.

I tried endlessly to reduce it to an equation that contained only two diffs. Most of the time I would get an equation more complex than it and if I remember I did manage to eliminate z a fews ways but could never eliminate dz.

Title: Re: Differential Equations on a Torus
Post by Ken Wiley on May 16^th, 2005, 4:15pm

Humm... I tried a few things and too and the equation just got bigger instead of getting reduced. I suppose the magic is in the right substituion. If the two polys on the dx and dy terms were factorable then perhaps one or more factors could be divided out but they aren't over the real numbers. I am not sure that I can see how to get z to vanish since it is a multiplier for the dx and dy terms.

Title: Re: Differential Equations on a Torus
Post by StonedAgain on May 16^th, 2005, 5:18pm

on 05/16/05 at 16:15:55, Ken Wiley wrote:

I suppose the magic is in the right substituion.

I agree and it is like they pop up with a substitution out of nowhere and it works.

Title: Re: Differential Equations on a Torus
Post by Michael_Dagg on May 16^th, 2005, 11:01pm

You form substitutions based on what you are given, which of course are expressions upon which they are to be applied.

As for magic substitutions, they are only magic for those who know not.

The N,S,S diferential equations text is fairly widely used and is a good one to study from. The latter two authors publish a complex analysis text that is in my opinon a very good book if you want to learn complex analysis right after the calculus sequence. It was actually wrote for that purpose but it does not require hardly any linear algebra or differential equations.

Regarding the differential equations problem at present, you can eliminate both z and dz from the TDE if you substitute x = uz and y = vz. You will arrive at a TDE that is of the form I mentioned, Mdx + Ndy = 0, such that you can integrate for the solution in terms of u and v, but note that since you introduced the substitution you must convert it back to the original several variables equation in x,y,z.

Title: Re: Differential Equations on a Torus
Post by skydiver on May 19^th, 2005, 8:12am

Still doesn't reduce with x = uz and y = vz. What am doing wrong?

Title: Re: Differential Equations on a Torus
Post by Michael_Dagg on May 20^th, 2005, 3:05pm

on 05/19/05 at 08:12:25, skydiver wrote:

Still doesn't reduce with x = uz and y = vz. What am doing wrong?

There is no way for us to know what you are doing wrong since you did not write anything.

Check your algebra and confirm you differentiated the substitutions correctly. That is, given x = uz and y = vz, then dx = udz + zdu and dy = vdz + zdv.

Title: Re: Differential Equations on a Torus
Post by Michael_Dagg on Jun 17^th, 2005, 6:54pm

Skydiver I take it that you have given up on this differential equation?

Title: Re: Differential Equations on a Torus
Post by Skydiver on Jul 3^rd, 2005, 8:44am

I did manage to finally eliminate z and dz and came up with a DE in terms of u and v that was easy to integrate. It wasn't all that easy though - a lot of algebra was needed for the elimination.

Title: Re: Differential Equations on a Torus
Post by Michael_Dagg on Jul 7^th, 2005, 8:47pm

What does the reduction look like?

Title: Re: Differential Equations on a Torus
Post by Skydiver on Jul 9^th, 2005, 9:13am

Here is it abbreviated

M(u,v) dx + N(u,v) dy = 0

with

M(u,v) = a/b
N(u,v) = g/b

a = u² + 2uv - v²
b = (u² + v²)(u + v)
g = v² + 2uv - u²

Integrating gives

log((u² + v²)/(u + v))
(u² + v²)/(u + v) = C

and when you substitute u = x/z and v = y/z you get

(x² + y²)/(xz + yz) = C

Title: Re: Differential Equations on a Torus
Post by Michael_Dagg on Jul 10^th, 2005, 8:40am

Nice work.

Title: Re: Differential Equations on a Torus
Post by uMROD on Jul 10^th, 2005, 4:01pm

That is an interesting derivation of solution. I was not surprised to see those u,v polynoms in the integrals since there were x,y polynoms in parts of the DE.

Is is possible to pluck out the potential and kinetic energy component of the pendulum:

ma² Q" + mga sin(Q) = 0, where m=mass, g=gravity, and a = length of swing, and Q is the angle of swing.

Title: Re: Differential Equations on a Torus
Post by Ken Wiley on Jul 11^th, 2005, 9:23am

on 07/10/05 at 16:01:09, uMROD wrote:

Is is possible to pluck out the potential and kinetic energy component of the pendulum:

I am not completely sure, but I think they have to be given numerically since the pendulum solution cannot be given in terms of elementary functions. This is one of the things I vividly remember about the pendulum. I recall from physics that if you can get one of the energies then most of the time you can compute the other.

Title: Re: Differential Equations on a Torus
Post by Michael_Dagg on Jul 11^th, 2005, 11:00am

You can deduce expressions for both energies.

Since Q" = Q' (dQ'/dQ),

ma^2 Q" + mga sin(Q) = ma^2 Q' dQ'/dQ + mga sin(Q) = 0.

Integrating with respect to Q we get

ma^2 int[Q' dQ'] + mga int[sin(Q) dQ] = c

and therefore

1/2 ma^2 Q'^2 - mga cos(Q) = c.

The last equation expresses the conservation of energy of the pendulum. The first term is the kinetic energy and the second term is the potential energy. By choosing different values of c we can study any possible motion and compute one energy or the other. As you can see, the sum of the kinetic energy and the potential energy is constant.

Also note that the reduced form (first order equation) is, so to speak, explicitly more nonlinear that the second order form and it too cannot be expressed in terms of elementary functions. The phase portrait of the system plotted in the (Q,Q') plane reveals the behavior of the system and from studying it you can deduce a linear approximation to any given trajectory.

Title: Re: Differential Equations on a Torus
Post by Ken Wiley on Jul 12^th, 2005, 6:38am

That reduction had me stumped since I sat for quite sometime looking at Q" = Q' (dQ'/dQ) before I decided to try to come up with it, more tricks with differential relationships.

It is interesting that the kinetic energy graph is a parabola but not surprising.

Title: Re: Differential Equations on a Torus
Post by Skydiver on Jul 12^th, 2005, 5:41pm

I plugged the right side of Q" into Matlab but it did not give me Q" back until I specifed a variable to which Q is differentiated with respect to. When it did, it displayed dQ²/dt² which is the same thing.

Title: Re: Differential Equations on a Torus
Post by ConeHead on Jul 12^th, 2005, 9:03pm

Show that the system has a periodic solution:

x' = x - y - (x² + 3/2 y²)x
y' = x + y - (x² + 1/2 y²)y

Title: Re: Differential Equations on a Torus
Post by SWF on Jul 12^th, 2005, 10:32pm

on 07/11/05 at 11:00:04, Michael_Dagg wrote:

Since Q" = Q' (dQ'/dQ),
...
Integrating with respect to Q we get

You can also multiply the equation by Q' and integrate on time to get the same thing.

Same thing can be done to Newton's 2nd law:

F = m * d²y/d t²

Multiply by dy/dt and integrate to get change in kinetic energy equals work done.

Title: Re: Differential Equations on a Torus
Post by Michael_Dagg on Jul 15^th, 2005, 8:13am

on 07/12/05 at 21:03:52, ConeHead wrote:

Show that the system has a periodic solution:

This problem falls in the intermediate level of study. It can be shown that the given system has a periodic solution by various means. But more importantly would be a solution that most everyone would understand and could likely spark a trail of nice discussions.

The first step would be to plot the phase portrait of the system, say, within the bounds (-5,5) x (-5,5) and post the graphic.

Title: Re: Differential Equations on a Torus
Post by Ken_Wiley on Jul 15^th, 2005, 5:20pm

Let me kick it off, here's the plot

Re:

x' = x - y - (x2 + 3/2 y2)x
y' = x + y - (x2 + 1/2 y2)y

Title: Re: Differential Equations on a Torus
Post by uMRod on Jul 16^th, 2005, 9:32am

Cool dynamical system. What did you plot it with?

Title: Re: Differential Equations on a Torus
Post by Skydiver on Jul 17^th, 2005, 5:45pm

Hey, the dynamics of a washing machine!

Nice looking Harley handle bars but better yet the behavior of my new girl friend - kinda always converges to the same points.

Title: Re: Differential Equations on a Torus
Post by Ken_Wiley on Jul 18^th, 2005, 8:02am

on 07/16/05 at 09:32:53, uMRod wrote:

Cool dynamical system. What did you plot it with?

With Matlab.

Title: Re: Differential Equations on a Torus
Post by tmiller on Jul 18^th, 2005, 8:17am

It can be shown it has such solution using the Poncare/Bendiixson theorm.

Title: Re: Differential Equations on a Torus
Post by uMRod on Jul 18^th, 2005, 9:37am

on 07/18/05 at 08:17:05, tmiller wrote:

It can be shown it has such solution using the Poncare/Bendiixson theorm.

Come on with it.

Title: Re: Differential Equations on a Torus
Post by Michael_Dagg on Jul 19^th, 2005, 5:58pm

Ah! Good plot.

I assume the blue dot is either one of the equilibria or the zero solution.

Going fast forward, not exactly an agitating washing machine because no matter what direction you run time, the vector field doesn't reverse - hence, an autonomous system. Of course, the orbits will reverse otherwise.

The Poincare-Bendixson theorem is not a singular existence theorem but instead a classification theorem which does incidentally guarantee at least one possibility that there is a periodic solution (and among others).

If you try it this way you will eventually realize that some preliminary work as well as some post work is required if you wish to use the P-B theorem.

Title: Re: Differential Equations on a Torus
Post by Michael_Dagg on Jul 27^th, 2005, 3:32pm

Given

x' = x - y - (x^2 + (3/2) y^2)x
y' = x + y - (x^2 + (1/2) y^2)y.

*********************

Note that o is the dot product and _ denotes a vector.

From looking at the phase portrait of this system it is intuitive to consider some narrow annular region about the origin containing such solution. Therefore, we are looking for two circles centred on the origin with the required properties.

Referring to the figure, let N_ = <x,y> be the normal pointing outward at P from the circle radius r and let X_ = <X,Y> point in the direction of the path through P and tangential to the trajectory through P. Note that cos(B) = N_ o X_ /[|N_||X_|] and therefore N_ o X_ is positive or negative according to whether X_ is pointing outwards or inwards.

Noting that

X_ = <X,Y> = <x', y'> = <x - y - (x^2 + (3/2) y^2)x, x + y - (x^2 + (1/2) y^2)y>,

we have

N_ o X_ = xX + yY
= xx' + yy'
= x(x - y - (x^2 + (3/2) y^2)x) + y(x + y - (x^2 + (1/2) y^2)y)
= x^2 + y^2 - x^4 - (1/2) y^4 - (5/2) x^2 y^2
= r^2 - r^4 + (1/2) y^2 (y^2 - x^2)
= r^2 - r^4 (1 + (1/4) cos(2A) - (1/4) cos(2A)^2).

The last equation gives us enough information to conclude that there is a periodic solution. When r = 1/2, N_ o X_ is positive for all A and so all paths are directed outwards on this circle and when r = 2 it is negative, with all paths directed inwards. Therefore, somewhere between r = 1/2 and r = 2 there is at least one closed path or a periodic solution.

We could stop right here but the annulus r1,r2 = 1/2,2 is not the narrowest region.

We can pin it down by noting that the equation

r^2 - r^4 (1 + (1/4) cos(2A) - (1/4) cos(2A)^2) = 0

has a solution (r,A). Multiplying the last equation by 4/r^4 and rewriting it in the form

4(1/r^2 - 1) = cos(2A) - cos(2A)^2,

we note that the range of the right-hand side is (-2,1/4), and solutions will exist for values of the left-hand side which lie in this range.

That is, when

4/sqrt(17) <= r <= sqrt(2)

or 0.97 <= r <= 1.41.

Title: Re: Differential Equations on a Torus
Post by Skydiver on Jul 28^th, 2005, 4:25pm

on 07/27/05 at 15:32:37, Michael_Dagg wrote:

= r^2 - r^4 + (1/2) y^2 (y^2 - x^2)
= r^2 - r^4 (1 + (1/4) cos(2A) - (1/4) cos(2A)^2).

How r^2 comes is evident, but how is r^4 worked into those last two equations - especially the last one since now r^4 is being multiplied by another equation?

Title: Re: Differential Equations on a Torus
Post by ConeHead on Jul 28^th, 2005, 6:27pm

Wow. The fact is that I manufactured that differential system from another one that I had been working with and really never expected that kind of solution to my problem statement, since it turns out that the system moderately nonlinear.

I am very intrigued by your solution because it does not use any out-of-reach mathematics.

There is a way to construct a similar solution using Liapunov functions but there is less advantage in doing so - at least in my opinion.

Title: Re: Differential Equations on a Torus
Post by Michael_Dagg on Jul 28^th, 2005, 9:02pm

on 07/28/05 at 16:25:40, Skydiver wrote:

How r^2 comes is evident, but how is r^4 worked into those last two equations - especially the last one since now r^4 is being multiplied by another equation...

Wow. The fact is that I manufactured that differential system from another one ...There is a way to construct a similar solution using Liapunov functions...

Let me respond to both posts.

As you note, the circle r^2 = x^2 + y^2 appears explicitly therein but r^4 = (r^2)^2 = (x^2 + y^2)^2 or more specifically -r^4 requires some algebraic manipulation in equation #5. Equation #6 requires algebraic manipulation in addition to simple trigonometric substitutions in conjunction with familiarity with trigonometric properties of the unit circle or any other circle. Remember it is plausible to manufacture expressions within expressions as long as what you manufacture sums up to zero. There is a multiplicative analogy of this as well.

In fact, if you don't routinely think of manufactured algebraic varieties your manipulation skills will suffer.

As far as a Liapunov solution there is no loss in generality if you wish to include one since the PDE requirement is trivial and I would love to see it.

Title: Re: Differential Equations on a Torus
Post by ConeHead on Jul 30^th, 2005, 12:12am

on 07/28/05 at 21:02:14, Michael_Dagg wrote:

...l and I would love to see it.

I wish I had the background to write all the details. Most of the stuff I have done with the idea is about drill bit tracing using u(x,y') = x² + y² - a^5/16 and v(x',y) = x² - y₃ - a^2'/3, u(x,y') = f(x,x')v(x',y) as the bore paddle, where f is a Liapunov function.

Title: Re: Differential Equations on a Torus
Post by Michael_Dagg on Jul 30^th, 2005, 11:13am

I am not familiar with drill bit modeling but it sounds interesting.

What is the parameter a?

Title: Re: Differential Equations on a Torus
Post by ConeHead on Jul 30^th, 2005, 1:18pm

I made a couple of typos in the equations. Here they are fixed

u(x,y') = x² + y² - a^5/16
v(x',y) = x² - y³ - a^2/3
u(x,y')_x = f(x,y')v(x',y)

The parameter a is a psi value a = |Curl Paddle(r,w)| and is generally in the range 300 <= a <= 1000, where r is the radius of the paddle and w is its angular speed.

A blank shank is spun by a fluid driven paddle vise at a real high w and then punched into the router die to trace the bit - kinda like a torque converter in an automatic transmission driving the feeler clutch.

By perturbing f(x,x') with an e (y=y+e, y in f) within about 0 < e <= 1 we can make the bit to high precision and very fast.

You can also model high quality razor blades by perturbing Liapunov functions.

Title: Re: Differential Equations on a Torus
Post by Michael_Dagg on Jul 31^st, 2005, 3:25am

What do x and y represent?

Title: Re: Differential Equations on a Torus
Post by ConeHead on Aug 2^nd, 2005, 11:02am

The points (x,y) represent the arc of the trace cut that is lathed by the router die. If you look at a standard drill that is for steel applications you will notice a uniform arc within the cutter trace. The deeper the arc, and depending on the radius of the shank, more psi is need to punch the drill into the router die to produce a sharp edge. The time interval for the (x,y) is in terms of msec starting around the equilibria that is determined by the parameter a. Anything beyond a few msecs (x,y) have no meaning.

This is only one part of the complete model and is somewhat idealized at that.

Title: Re: Differential Equations on a Torus
Post by Michael_Dagg on Aug 3^rd, 2005, 9:26am

That sounds too easy in addition to the fact that the system given in terms of u and v has the same behavior with or without the parameter a, for in one case the (x,y) do not blow up so quickly.

Title: Re: Differential Equations on a Torus
Post by uMRod on Aug 4^th, 2005, 8:55am

From looking at plot with and one without, (a) has only a translation & scaling effect on that system and in the without case the points get large rapidly, from just t=0,x=1,y=1 to t=10 - boom.

Title: Re: Differential Equations on a Torus
Post by autumn on Aug 17^th, 2005, 8:42pm

on 07/30/05 at 13:18:53, ConeHead wrote:

u(x,y') = x² + y² - a^5/16
v(x',y) = x² - y³ - a^2/3

In practical sense, my question is why is there a need to include the power/root operations on the constant "a" in that system when you can simply pre-compute what they would be before hand?

Title: Re: Differential Equations on a Torus
Post by Michael_Dagg on Aug 18^th, 2005, 9:20am

Autumn you are correct: in computational practice the resulting values would be predetermined to avoid exponentiation on (a) each time a solver evaluated the system and, if necessary, provided up to the limit of precision being used.

On the other hand, if (a) is a sweep parameter then in certain software you might have to include exponentiation on (a) if you can't predetermine the values using code before a solver is ran on the system.

Title: Re: Differential Equations on a Torus
Post by autumn on Aug 18^th, 2005, 3:43pm

What exactly do you mean by sweep - is that "a" changes during each solver step or changes over a loop of solver runs?

Title: Re: Differential Equations on a Torus
Post by Michael_Dagg on Aug 19^th, 2005, 9:26am

It is the latter case you mentioned. Sweeping (a) during each solver step effectively changes the system during each step and will likely cause the solver to go haywire.

Title: Re: Differential Equations on a Torus
Post by autumn on Aug 23^rd, 2005, 9:09pm

I have a question that has a little different beat from the norm. Given say a simple DE like y' = f(x),with initial conditions, and an exact soution Y(x) (presumably uknown), is there way to construct an asymptotic expansion that "hugs" Y(x) up to some accuracy up to n terms?

Title: Re: Differential Equations on a Torus
Post by Michael_Dagg on Aug 24^th, 2005, 10:56am

Taylor series is an asymptotic expansion that will hug it up to an accuracy determined by the number of terms used.

Title: Re: Differential Equations on a Torus
Post by autumn on Aug 24^th, 2005, 9:38pm

Learned a couple of things here.

But, first I want to mention that this thread shows up on Google searches as well as others but this one much more.

Anyway, I thought that asymptotic expansions were all "divergent." But that is not true, as I know now as any convergent Taylor series is an asymptotic expansion. I think I just said something silly, "convergent Taylor series", as if it is a Taylor series it "is" convergent.

My idea with an asymptotic expansion was not like Taylor but more like an infinite telescoping series that can hug up on a solution.

But, construction of a Taylor series is much simplier than constructing a so to speak "series solution" based on the techniques in most elementary DE books because it is much easier and straight forward to differentiate f(x) j-th times for the terms in Tayor than to try to solve the often very wacky indicial equations associated with the series solution.