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The simple plane pendulum $$\frac{d^2\theta}{dt^2} + \frac{g}{l}\sin{\theta} = 0$$

has the very perdy phase portrait

perdy phase portrait Meanwhile, a domain coloring of $\sin(z)$ in the complex plane is

even perdier

Why are these so similar?

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2 Answers 2

up vote 5 down vote accepted

The trajectories of the differential equation satisfy the implicit equation (with $v = dy/dt$) $$f(\theta,v) = \frac{v^2}{2} - \frac{g}{l} \cos \theta = A, \ A \ge -\frac{g}{l}$$

I think the white curves are the level curves of $|\sin(x+iy)|$. These are given by $$g(x,y) = \cosh (2y) - \cos(2x) = B,\ B \ge 0$$

The two are related by the change of variables $B = Al/g + 1$, $\theta = 2 x$, $v = 2 \sqrt{g/l} \sinh y$.

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The implicit equation is just conservation of energy, yes? –  AndrewG Oct 25 '12 at 0:32
    
Yes, that's right. –  Robert Israel Oct 28 '12 at 23:06

The equations of the phase curves in the phase portrait of the simple plane pendulum actually correspond to different energy conservation relations: $$ \dot{\theta}^2 - \frac{g}{l}\cos(\theta) = C_0 $$

And in the colored graph of $\sin(z)$ in the complex plane the lines are the lines of constant magnitude: $$ \|\sin(x+yi)\|^2 = C $$ which can be transformed into another form by the steps below $$ \begin{align} \|\sin(x)\cosh(y) + i\cos(x)\sinh(y)\|^2 &= C \\ \sin(x)^2\cosh(y)^2 + \cos(x)^2\sinh(y)^2 &= C \\ (\sin(x)^2 + \cos(x)^2)\frac{e^{2y}+e^{-2y}}{2} + \sin(x)^2-\cos(x)^2 &= C \\ \frac{e^{2y}+e^{-2y}}{2} -\cos(2x) &= C \end{align} $$ when $y$ is not far from $0$, $\frac{e^{2y}+e^{-2y}}{2} \approx 4y^2 = (2y)^2$,so if we replace $(x,y)$ by $(u,v)$ with $u=2x, \, v=2y$, then the equation becomes $$ v^2 -\cos(u) = C. $$ I think this is why the two plots look so similar. When $y$ goes far from $0$, their forms may no longer be such similar.

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