Nonlinear system producing periodic functions

Consider the following system:

\(y’\) = \(-z^3\); \(z’\) = \(y^3\); \(y[0]=1\), \(z[0]=0\).

Show that the solutions \(y=f[x]\) and \(z=g[x]\) are periodic functions, and evaluate the period.


Clearly, \(y^3 y’\) + \(z^3 z’=0\), so that the quantity \(y^4+z^4\) is constant.  Given the boundary conditions, \(y^4+z^4=1\) for all values of \(x\).  When plotted in phase space, this is a closed curve and is therefore representative of a periodic system.

To compute the period \(p\), we begin with the obvious observation that

\(p=\int_{0}^{p}dx=4 \int_{0}^{p/4}dx\).

Note that


so that

\(p=4 \int_{0}^{1}dz(1-z^4)^{-\frac{3}{4}}\)

This integral is doable by using the substitution \(w=1-z^4\) to reveal that


The true nerd will recognize the above integral is in the form of a Beta integral and is thus equal to

\(p\) = \(\frac{\Gamma[\frac{1}{4}]^2}{\sqrt{\pi}}\) \(\approx 7.4163\)

Be Sociable, Share!

Published by

Ron Gordon

Math nerd in his early 40's who seems to have an opinion about everything and an inability to keep it to himself.