# 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.

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

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

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### 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.