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Topic: Differential Equations on a Torus (Read 6750 times) |
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uMRod
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Re: Differential Equations on a Torus
« Reply #50 on: Jul 16th, 2005, 9:32am » |
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Cool dynamical system. What did you plot it with?
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Skydiver
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Re: Differential Equations on a Torus
« Reply #51 on: Jul 17th, 2005, 5:45pm » |
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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.
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Ken_Wiley
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Re: Differential Equations on a Torus
« Reply #52 on: Jul 18th, 2005, 8:02am » |
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on Jul 16th, 2005, 9:32am, uMRod wrote:| Cool dynamical system. What did you plot it with? |
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With Matlab.
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| « Last Edit: Jul 18th, 2005, 8:04am by Ken_Wiley » |
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tmiller
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Re: Differential Equations on a Torus
« Reply #53 on: Jul 18th, 2005, 8:17am » |
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It can be shown it has such solution using the Poncare/Bendiixson theorm.
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uMRod
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Re: Differential Equations on a Torus
« Reply #54 on: Jul 18th, 2005, 9:37am » |
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on Jul 18th, 2005, 8:17am, tmiller wrote:| It can be shown it has such solution using the Poncare/Bendiixson theorm. |
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Come on with it.
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Michael Dagg
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Re: Differential Equations on a Torus
« Reply #55 on: Jul 19th, 2005, 5:58pm » |
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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.
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| « Last Edit: Jul 19th, 2005, 8:43pm by Michael Dagg » |
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Michael Dagg
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Michael Dagg
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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.
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| « Last Edit: Jul 27th, 2005, 4:46pm by Michael Dagg » |
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Michael Dagg
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Skydiver
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Re: Differential Equations on a Torus
« Reply #57 on: Jul 28th, 2005, 4:25pm » |
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on Jul 27th, 2005, 3:32pm, 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).
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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?
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ConeHead
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Re: Differential Equations on a Torus
« Reply #58 on: Jul 28th, 2005, 6:27pm » |
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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.
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Michael Dagg
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Re: Differential Equations on a Torus
« Reply #59 on: Jul 28th, 2005, 9:02pm » |
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on Jul 28th, 2005, 4:25pm, 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... |
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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.
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Michael Dagg
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ConeHead
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Re: Differential Equations on a Torus
« Reply #60 on: Jul 30th, 2005, 12:12am » |
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on Jul 28th, 2005, 9:02pm, Michael_Dagg wrote:| ...l and I would love to see it. |
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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 - a5/16 and v(x',y) = x2 - y3 - a2'/3, u(x,y') = f(x,x')v(x',y) as the bore paddle, where f is a Liapunov function.
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Michael Dagg
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Re: Differential Equations on a Torus
« Reply #61 on: Jul 30th, 2005, 11:13am » |
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I am not familiar with drill bit modeling but it sounds interesting.
What is the parameter a?
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Michael Dagg
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ConeHead
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Re: Differential Equations on a Torus
« Reply #62 on: Jul 30th, 2005, 1:18pm » |
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I made a couple of typos in the equations. Here they are fixed
u(x,y') = x2 + y2 - a5/16
v(x',y) = x2 - y3 - a2/3
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.
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Michael Dagg
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Re: Differential Equations on a Torus
« Reply #63 on: Jul 31st, 2005, 3:25am » |
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What do x and y represent?
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Michael Dagg
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ConeHead
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Re: Differential Equations on a Torus
« Reply #64 on: Aug 2nd, 2005, 11:02am » |
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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.
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Michael Dagg
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Re: Differential Equations on a Torus
« Reply #65 on: Aug 3rd, 2005, 9:26am » |
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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.
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Michael Dagg
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uMRod
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Re: Differential Equations on a Torus
« Reply #66 on: Aug 4th, 2005, 8:55am » |
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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.
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autumn
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Re: Differential Equations on a Torus
« Reply #67 on: Aug 17th, 2005, 8:42pm » |
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on Jul 30th, 2005, 1:18pm, ConeHead wrote:
u(x,y') = x2 + y2 - a5/16
v(x',y) = x2 - y3 - a2/3
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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?
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Michael Dagg
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Re: Differential Equations on a Torus
« Reply #68 on: Aug 18th, 2005, 9:20am » |
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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.
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Michael Dagg
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autumn
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Re: Differential Equations on a Torus
« Reply #69 on: Aug 18th, 2005, 3:43pm » |
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What exactly do you mean by sweep - is that "a" changes during each solver step or changes over a loop of solver runs?
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Michael Dagg
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Re: Differential Equations on a Torus
« Reply #70 on: Aug 19th, 2005, 9:26am » |
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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.
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Michael Dagg
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autumn
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Re: Differential Equations on a Torus
« Reply #71 on: Aug 23rd, 2005, 9:09pm » |
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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?
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Michael Dagg
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Re: Differential Equations on a Torus
« Reply #72 on: Aug 24th, 2005, 10:56am » |
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Taylor series is an asymptotic expansion that will hug it up to an accuracy determined by the number of terms used.
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Michael Dagg
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autumn
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Re: Differential Equations on a Torus
« Reply #73 on: Aug 24th, 2005, 9:38pm » |
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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.
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