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Topic: Differential Equations on a Torus (Read 6743 times) |
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Kyle Strom
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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.
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Michael Dagg
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Re: Differential Equations on a Torus
« Reply #1 on: Apr 23rd, 2005, 4:04pm » |
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on Apr 23rd, 2005, 8:13am, 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. |
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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
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Thomas Wang
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Re: Differential Equations on a Torus
« Reply #2 on: Apr 23rd, 2005, 10:44pm » |
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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?
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Michael Dagg
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Re: Differential Equations on a Torus
« Reply #4 on: Apr 24th, 2005, 9:21am » |
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on Apr 23rd, 2005, 10:44pm, 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.
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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
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Icarus
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Re: Differential Equations on a Torus
« Reply #5 on: Apr 24th, 2005, 3:30pm » |
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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].
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Michael Dagg
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Re: Differential Equations on a Torus
« Reply #6 on: Apr 24th, 2005, 5:48pm » |
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on Apr 24th, 2005, 3:30pm, Icarus wrote:| One way to define it is as S1 x S1, where S1 is the circle |
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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
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Ken Wiley
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Re: Differential Equations on a Torus
« Reply #7 on: Apr 24th, 2005, 7:52pm » |
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on Apr 24th, 2005, 5:48pm, Michael Dagg wrote:
that is, the cross product of two circles is a torus.
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That is interesting. How does one compute that cross product?
Ken
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Icarus
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Re: Differential Equations on a Torus
« Reply #8 on: Apr 24th, 2005, 7:54pm » |
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on Apr 24th, 2005, 5:48pm, 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. |
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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.
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| « Last Edit: Apr 26th, 2005, 4:45pm by Icarus » |
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Icarus
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Re: Differential Equations on a Torus
« Reply #9 on: Apr 24th, 2005, 7:56pm » |
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on Apr 24th, 2005, 7:52pm, Ken Wiley wrote:
That is interesting. How does one compute that cross product? |
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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}
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Deedlit
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Re: Differential Equations on a Torus
« Reply #10 on: Apr 25th, 2005, 7:11pm » |
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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!
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Icarus
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Re: Differential Equations on a Torus
« Reply #11 on: Apr 26th, 2005, 4:53pm » |
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on Apr 25th, 2005, 7:11pm, 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! |
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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."
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"Pi goes on and on and on ...
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I wonder: Which is larger
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Deedlit
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Re: Differential Equations on a Torus
« Reply #12 on: Apr 26th, 2005, 8:53pm » |
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on Apr 26th, 2005, 4:53pm, Icarus wrote:
And then you also have to explain to them that toruses are NOT curved, despite what they look like!
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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.
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Don't bother to tell a geometer that his tire is flat - he will reply, "Of course it is; all tires are flat." |
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Good one. 
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Michael Dagg
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Re: Differential Equations on a Torus
« Reply #13 on: Apr 27th, 2005, 12:51am » |
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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?
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| « Last Edit: Apr 27th, 2005, 1:03am by Michael Dagg » |
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Regards,
Michael Dagg
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K Strom
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Re: Differential Equations on a Torus
« Reply #14 on: Apr 27th, 2005, 11:23am » |
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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?
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K Strom
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Re: Differential Equations on a Torus
« Reply #15 on: Apr 27th, 2005, 11:38am » |
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Also, I forgot to remark about the interesting genealogy. I know the names Markov and Chebyshev and I think Tamarkin.
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Michael Dagg
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Re: Differential Equations on a Torus
« Reply #16 on: Apr 27th, 2005, 1:39pm » |
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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.
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| « Last Edit: Apr 27th, 2005, 1:41pm by Michael Dagg » |
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Michael Dagg
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K Strom
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Re: Differential Equations on a Torus
« Reply #17 on: Apr 28th, 2005, 7:33am » |
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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?
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Michael Dagg
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Re: Differential Equations on a Torus
« Reply #18 on: May 1st, 2005, 8:18am » |
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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.
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Michael Dagg
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SWF
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Re: Differential Equations on a Torus
« Reply #19 on: May 1st, 2005, 8:35pm » |
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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).
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K Strom
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Re: Differential Equations on a Torus
« Reply #20 on: May 2nd, 2005, 7:48am » |
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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.
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Another Aggie
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Re: Differential Equations on a Torus
« Reply #21 on: May 12th, 2005, 4:23pm » |
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Oh, way the way, what is a very interesting abstract group that is surprisingly also a manifold?
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Michael Dagg
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Re: Differential Equations on a Torus
« Reply #22 on: May 13th, 2005, 7:19am » |
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on May 12th, 2005, 4:23pm, Another Aggie wrote:Oh, way the way, what is a very interesting abstract group that is surprisingly also a manifold?
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One of the more interesting that comes to mind is a Lie group.
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Michael Dagg
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skydiver
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Re: Differential Equations on a Torus
« Reply #23 on: May 13th, 2005, 1:39pm » |
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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
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Michael Dagg
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Re: Differential Equations on a Torus
« Reply #24 on: May 13th, 2005, 7:56pm » |
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on May 13th, 2005, 1:39pm, skydiver wrote:
z(x2 + 2xy - y2)dx + z(2xy + y2 - x2)dy - (x2 + y2)(x + y)dz = 0 |
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I bet your instructor was disappointed.
The solution is a space curve in R3 and with some work you can produce it.
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| « Last Edit: May 13th, 2005, 8:50pm by Michael Dagg » |
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Regards,
Michael Dagg
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