From: eritchie@aol.com (ERitchie) Newsgroups: sci.math Subject: An equation in need is an equation indeed Date: 8 Nov 1994 21:50:13 -0500 Solving this equation will make you a better lover, because misery loves company, and this equation is making me miserable! dx/dt=z dy/dt=-tz dz/dt=ty-x solve for x,y,z in terms of t If you can solve this I love you for your brains, not your body. ERitchie@aol.com ============================================================================== Newsgroups: sci.math,sci.math.symbolic From: pmontgom@cwi.nl (Peter L. Montgomery) Subject: Re: An equation in need is an equation indeed Date: Sun, 13 Nov 1994 07:12:01 GMT In article <39pdd5$hpo@newsbf01.news.aol.com> eritchie@aol.com (ERitchie) writes: >Solving this equation will make you a better lover, because misery loves >company, and this equation is making me miserable! >dx/dt=z >dy/dt=-tz >dz/dt=ty-x >solve for x,y,z in terms of t. It is not hard to get a series solution, if we assume inf a(i) * t^i x(i) = sum ------------ . i=0 i! Use the first equation to solve for z, the third to solve for y, and the second to get a recurrence relation. a(n) = -a(n-2) - (n-2)*(n-4)*a(n-4) (n >= 4) We also need a2 = -a0, but find no restrictions on a0, a1, a3. This gives three linearly independent solutions. I don't recognize the coefficient pattern, except that the terms involving a0 and a1 are eventually divisible by 5 and the term involving a3 is eventually divisible by 49. The Maple program a2 := -a0; for j from 4 to 20 do a.j := -a.(j-2) - (j-2)*(j-4)*a.(j-4); od; x := sum('a.i*t^i/i!', i = 0..20); z := diff(x, t): y := (x + diff(z, t))/t: eq := diff(y, t) + t*z: eq1 := collect(expand(eq*t^2), t); quit; gives |\^/| Maple V Release 2 (Stichting Mathematisch Centrum) ._|\| |/|_. Copyright (c) 1981-1992 by the University of Waterloo. \ MAPLE / All rights reserved. MAPLE is a registered trademark of <____ ____> Waterloo Maple Software. | Type ? for help. > a2 := -a0; a2 := - a0 > for j from 4 to 20 do > a.j := -a.(j-2) - (j-2)*(j-4)*a.(j-4); > od; a4 := a0 a5 := - a3 - 3 a1 a6 := 7 a0 a7 := - 14 a3 + 3 a1 a8 := - 31 a0 a9 := 49 a3 + 102 a1 a10 := - 305 a0 a11 := 833 a3 - 291 a1 a12 := 2785 a0 a13 := - 5684 a3 - 9807 a1 a14 := 33815 a0 a15 := - 113435 a3 + 51420 a1 a16 := - 501695 a0 a17 := 1221815 a3 + 1860945 a1 a18 := - 7072865 a0 a19 := 27704110 a3 - 14973045 a1 a20 := 151561025 a0 -- Peter L. Montgomery pmontgom@cwi.nl Mathematically gifted, unemployed. Interested in program optimization, computer arithmetic, cryptography, compilers, computational mathematics. 17 years industrial experience. ============================================================================== Date: Tue, 15 Nov 94 14:50:17 CST From: rusin (Dave Rusin) To: eritchie@aol.com Subject: your differential equations After I posted a response to your problem it occured to me that one could say more. At any moment t the flow lines indicating the directions any solution would have to be pointing consist of a bunch of parallel circles on the sphere. At time t=0 they are all parallel to the xz plane (y=0); as time progresses, the axis running through them tilts through the xy plane (z=0) having slope y/x = exactly t at time t, so the tilting is rapid at first but then gets slower and only as t-->oo we get that the circles of flow are parallel to the yz plane (x=0). In particular, when t is very large, the flow lines are close to stationary, and are circles almost parallel to the yz plane; thus these circles are roughly what the trajectories will look like. Note though that you will be going faster and faster around the circles, since the differential equations are roughly x'=0, y'=-tz, z'=ty, giving x=0, y=-sin(t^2/2), z=cos(t^2/2), say. You can reduce the amount of handwaving in this analysis a bit. Let T=t^2/2 and view x, y, z as functions of T. Using the chain rule, z=dx/dt= (dx/dT)(dT/dt) = t.(dx/dT). Treating y and z likewise leads to x'= e z y'= -z z'= y - e x where e=1/sqrt(2T) is small. It is in taking e=0 in these equations (and using the presumed presence on the sphere, x^2+y^2+z^2=1) that we get the general solution: for some constants a and b, x=a, y=b sin(T-c), z=b cos(T-c) where a^2+b^2=1. I wish I could prove for you that one of these curves will in fact be the limit of the particle's path as t --> oo, but e does not get small fast enough for me to be able to write down a proof. The fact that z oscillates must also be used. I'm pretty sure the limiting behaviour is as I described but without plotting a solution I wouldn't want to stake my reputation on it. So my expectation is that given any initial starting point, a particle subject to the DE's you wrote would travel around on a sphere centered at the origin. At first, the motion may be chaotic, but fairly fast the particle would find a vertical circle which it would hop around, going around at a steadily increasing speed, wiggling around the circle but with amplitudes diminishing as time goes on. So, how does this problem arise? Are there any other things you'd like to know about this solution? dave ============================================================================== Date: Tue, 15 Nov 1994 22:09:24 -0500 From: ERitchie@aol.com To: rusin@math.niu.edu Subject: Equation still in need Dave, Thank you for your input. This one is really a brain buster. As to how does this problem arise, all I can tell you is what you might easily guess.... Balls that are rolling. Thanks again. Ed Ritchie ============================================================================== Date: Wed, 16 Nov 94 11:40:02 CST From: rusin (Dave Rusin) To: ERitchie@aol.com, rusin@math.niu.edu Subject: Re: Equation still in need By the way I did see how to prove that the vertical circles are the limit curves as t--> \infty. dave