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.

Solution:

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

\(dx=dz/y^3=dz(1-z^4)^{-\frac{3}{4}}\)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

\(p=\int_{0}^{1}dw[w(1-w)]^{-\frac{3}{4}}\)The true nerd will recognize the above integral is in the form of a Beta integral and is thus equal to

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