System of Jacobi elliptic functions

Jacobi’s elliptic functions are sorta like trig functions. His functions sn and cn have names that reminiscent of sine and cosine for good reason. These functions come up in applications such as the nonlinear pendulum (i.e. when θ is too
large to assume θ is a good enough approximation to sin θ) and in conformal mapping.

I ran across an article [1] yesterday that shows how Jacobi’s three elliptic functions—sn, cn, and dn—could be defined by one dynamical system

\begin{eqnarray*} x' &=& yz \\ y' &=& -zx \\ z' &=& -k^2 xy \end{eqnarray*}

with initial conditions x(0) = 0, y(0) = 1, and z(0) = 1.

The parameter k is the modulus. (In Mathematica’s notation below, k² is the parameter. See this post on parameterization conventions.) As k decreases to 0, sn converges to sine, cn to cosine, and dn to 1. As k increases to 1, sn converges tanh, and cn and dn converge to sech. So you could think of k as a knob you turn to go from being more like circular functions (ordinary trig functions) to more like hyperbolic functions.

Since we have a dynamical system, let’s plot the solution, varying the modulus each time.The Jacobi functions are periodic (in fact they’re doubly periodic) and so the plots will be closed loops.

f[t_, m_] = {JacobiSN[t, m], JacobiCN[t, m], JacobiDN[t, m]}
ParametricPlot3D[f[t, 0.707], {t, 0, 10}, AspectRatio -> 1]

phase portrait k = 1/2

ParametricPlot3D[f[t, 0.99], {t, 0, 20}, AspectRatio -> 1]

phase portrait k = 1/2

ParametricPlot3D[f[t, 0.01], {t, 0, 20}, AspectRatio -> 1]

phase portrait k = 1/2

Note that this last plot is nearly flat because the modulus is small and so z is nearly constant. The small modulus also makes the phase portrait nearly circular because x is approximately sine and y is approximately cosine.

[1] Kenneth R. Meyer. Jacobi Elliptic Functions from a Dynamical Systems Point of View. The American Mathematical Monthly, Vol. 108, No. 8 (Oct., 2001), pp. 729-737

3 thoughts on “System of Jacobi elliptic functions

  1. As stated in the blog sn and cn coverge to sin and cos as k-> 0. That means if k = 0 you have a circle in the xy-plane located at z = 1.

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