OU blog

Personal Blogs

Valentin Fadeev

Descending to chaos

Tuesday, 4 Jan 2011, 05:54
Visible to anyone in the world
Edited by Valentin Fadeev, Sunday, 18 Sept 2011, 23:24

The first (unguided) steps in dynamical systems. The task is to prove that the phase paths of the following system:

x double dot plus epsilon times absolute value of x times s times g times n reverse solidus full stop x dot above plus x equals zero

are isochronous spirals, that is, every circuit of every path around the origin takes the same time. It looks a bit scary, because it involves not one but two "bad" functions. Thle easiest way to deal with these is to break the picture in quadrants and examine each part separately.

equation left hand side y dot above equals right hand side negative epsilon times absolute value of x times s times g times n times y minus x

x dot above equals y

equation left hand side d times y divided by d times x equals right hand side negative epsilon times absolute value of x times s times g times n times y divided by y minus x divided by y

First quadrant: absolute value of x equals x , s times g times n times y equals one

open one plus epsilon close times x times d times x plus y times d times y equals zero

open one plus epsilon close times x squared plus y squared equals cap c

Second quadrant: equation left hand side absolute value of x equals right hand side negative x , s times g times n times y equals one

open one minus epsilon close times x times d times x plus y times d times y equals zero

open one minus epsilon close times x squared plus y squared equals cap c

Third quadrant: equation left hand side absolute value of x equals right hand side negative x , equation left hand side s times g times n times y equals right hand side negative one , the same solution as for the
first quadrant.

Fourth quadrant: absolute value of x equals x , equation left hand side s times g times n times y equals right hand side negative one , the same solution as for the
second quadrant.

Introduce polar coordinates.

I,III:

open one plus epsilon close times rho sub one squared times cosine squared of phi plus rho sub one squared times sine squared of phi equals cap c

equation sequence rho sub one squared equals cap c divided by open one plus epsilon close times cosine squared of phi plus sine squared of phi equals cap c divided by one plus epsilon times cosine squared of phi

II,IV:

open one minus epsilon close times rho sub two squared times cosine squared of phi plus rho sub two squared times sine squared of phi equals cap c

equation sequence rho sub two squared equals cap c divided by open one minus epsilon close times cosine squared of phi plus sine squared of phi equals cap c divided by one minus epsilon times cosine squared of phi

Let cap c be a closed circuit enclosing the origin. The elapsed time
is given by the following formula:

equation sequence t equals integral over open cap c close d t equals integral over open cap c close one divided by d times x divided by d times t d x equals integral over open cap c close d times x divided by x dot above equals integral over open cap c close d times x divided by y

As the functions are periodic, it is sufficient to give proof for one loop. Due to the axial symmetry of the trajectory we can calculate transit time only in the first and the second quadrants and then double the result. If we integrate counter-clockwise, then we would follow the trajectory in a reversed direction. Therefore, we need to put the negative signe before the integrals.(oh, how long it took me to realize this..) In polar coordinates this expression has the following form:

equation sequence t sub one equals negative integral over zero under pi divided by two rho sub one super prime times cosine of phi minus rho sub one times sine of phi divided by rho sub one times sine of phi d phi equals integral over zero under pi divided by two open one minus rho sub one super prime times cosine of phi divided by rho sub one times sine of phi close d phi

equation left hand side t sub two equals right hand side integral over pi divided by two under pi open one minus rho sub two super prime times cosine of phi divided by rho sub two times sine of phi close d phi

t equals two times open integral over zero under pi divided by two open one minus rho sub one super prime times cosine of phi divided by rho sub one times sine of phi close d phi plus integral over pi divided by two under pi open one minus rho sub two super prime times cosine of phi divided by rho sub two times sine of phi close d phi close

Differentiating the expression for rho sub one by phi we obtain:

equation left hand side two times rho sub one times rho sub one super prime equals right hand side negative cap c times epsilon times two times cosine of phi of negative sine of phi divided by open one plus epsilon times cosine squared of phi close squared

equation left hand side rho sub one times rho sub one super prime equals right hand side cap c times epsilon times sine of phi times cosine of phi divided by open one plus epsilon times cosine squared of phi close squared

equation sequence rho sub one super prime divided by rho sub one equals rho sub one times rho sub one super prime divided by rho sub one squared equals cap c times epsilon times sine of phi times cosine of phi divided by open one plus epsilon times cosine squared of phi close squared times one plus epsilon times cosine squared of phi divided by cap c equals epsilon times sine of phi times cosine of phi divided by one plus epsilon times cosine squared of phi

A-ha, that's where we get the independence of the final result on the path choice: the arbitrary constant gets cancelled out.

equation sequence one minus rho sub one super prime times cosine of phi divided by rho sub one times sine of phi equals one minus epsilon times sine of phi times cosine squared of phi divided by open one plus epsilon times cosine squared of phi close times sine of phi equals one minus epsilon times cosine squared of phi divided by one plus epsilon times cosine squared of phi equals one divided by one plus epsilon times cosine squared of phi

equation left hand side cap i sub one equals right hand side integral over zero under pi divided by two d times phi divided by one plus epsilon times cosine squared of phi

Let u equals tangent of phi , then equation left hand side d times phi equals right hand side d times u divided by one plus u squared , cosine squared of phi equals one divided by one plus u squared

equation sequence cap i sub one equals integral over zero under normal infinity d times u divided by open one plus epsilon divided by one plus u squared close times open one plus u squared close equals integral over zero under normal infinity d times u divided by sum with, 3 , summands one plus epsilon plus u squared equals one divided by one plus epsilon times integral over zero under normal infinity d times u divided by one plus open u divided by Square root of one plus epsilon close squared equals equals one divided by Square root of one plus epsilon times integral over zero under normal infinity d of u divided by Square root of one plus epsilon divided by one plus open u divided by Square root of one plus epsilon close squared equals one divided by Square root of one plus epsilon times arc tangent of u divided by Square root of one plus epsilon vertical line sub zero super normal infinity equals pi divided by two times one divided by Square root of one plus epsilon

equation left hand side two times rho sub two times rho sub two super prime equals right hand side negative cap c times epsilon of negative two times cosine of phi times open negative sine of phi close divided by open one minus epsilon times cosine squared of phi close squared

equation left hand side rho sub two times rho sub two super prime equals right hand side negative cap c times epsilon times sine of phi times cosine of phi divided by open one minus epsilon times cosine squared of phi close squared

equation sequence rho sub two super prime divided by rho sub two equals rho sub two times rho sub two super prime divided by rho sub two squared equals negative cap c times epsilon times sine of phi times cosine of phi divided by open one minus epsilon times cosine squared of phi close squared times one minus epsilon times cosine squared of phi divided by cap c equals negative epsilon times sine of phi times cosine of phi divided by one minus epsilon times cosine squared of phi

equation sequence one minus rho sub two super prime times cosine of phi divided by rho sub two times sine of phi equals one plus epsilon times sine of phi times cosine squared of phi divided by open one minus epsilon times cosine squared of phi close times sine of phi equals one divided by one minus epsilon times cosine squared of phi

equation sequence cap i sub two equals integral over pi divided by two under pi d times phi divided by one minus epsilon times cosine squared of phi equals integral over pi divided by two under pi d times open phi minus pi divided by two close divided by one minus epsilon times sine squared of phi minus pi divided by two equals integral over zero under pi divided by two d times psi divided by one minus epsilon times sine squared of psi d psi equals equals integral over zero under normal infinity d times u divided by open one plus u squared close times open one minus epsilon times open one minus one divided by one plus u squared close close equals integral over zero under normal infinity d times u divided by one plus open one minus epsilon close times u squared equals one divided by Square root of one minus epsilon times integral over zero under normal infinity d of u times Square root of one minus epsilon divided by one plus open u times Square root of one minus epsilon close squared equals equals one divided by Square root of one minus epsilon times arc tangent of u times Square root of one minus epsilon vertical line sub zero super normal infinity equals pi divided by two times one divided by Square root of one minus epsilon

I made some false starts on cap i sub two getting a negative answer which is impossible for a strictly positive integrand. Still not quite sure where I went wrong. Normally these things happen when a singularity sneaks in inside the domain of integration after variable change. Anyway, I decided to cheat and shift the scale.

Hence the total transit time equals:

equation sequence t equals two times open pi divided by two times one divided by Square root of one plus epsilon plus pi divided by two times one divided by Square root of one minus epsilon close equals pi times open one divided by Square root of one plus epsilon plus one divided by Square root of one minus epsilon close

Tags: m821, dynamical systems
Permalink Add your comment
Share post