Self-intersections of a Lissajous curve

Question

Hei,

I want to find the self intersections of a Lissajous curve, for instance:

$$x(t)=\sin 2t$$

$$y(t)=\sin 3t$$

I have been trying for a couple of hours but I really don't get how to compute all the solutions. Basically I was trying to write the relation such that

$$\sin 2u=\sin 2v \text{ and }\sin 3u=\sin 3v.$$

Thanks for the support!

Answer 1

Start with a rational parametrization, using the tangent half angle formula:

$$c=\cos u=\frac{1-a^2}{1+a^2}\qquad s=\sin u=\frac{2a}{1+a^2}$$

Use Wikipedia to find formulas for double and triple angle.

$$\sin 2u=2sc=\frac{(1-a^2)2a}{(1+a^2)^2} \qquad \sin3u=3s-4s^3=\frac{6a(1+a^2)^2-32a^3}{(1+a^2)^3}$$

Now assume a second parameter $b$ instead of $a$ and write your equations, cross-multiplying in the process:

$$(1-a^2)2a(1+b^2)^2=(1-b^2)2b(1+a^2)^2\\ (6a(1+a^2)^2-32a^3)(1+b^2)^3=(6b(1+b^2)^2-32b^3)(1+a^2)^3$$

Now you have a polynomial system of two equations in two variables. You can cancel the trivial solution $a=b$ from both of them (e.g. by writing them as LHS − RHS = $0$, then factorizing and removing the factor $a-b$). Then using a computer algebra system of your choice, or some elimination technique like resultants, you find 6 algebraic solutions ignoring duplicates. Note that the tangent half-angle substitution omits one point, namely $a=\infty$ corresponding to $t=\pi$. You need to check that case separately, and will indeed find one more solution there. Turning parameters back into points you get these 7 points of intersection:

\begin{align*} x &= \tfrac{1}{2} \, \sqrt{3} & y &= 0 & \{a,b\}&=\{ -\sqrt{3}, \sqrt{\tfrac{1}{3}} \}\\ x &= -\tfrac{1}{2} \, \sqrt{3} & y &= 0 & \{a,b\}&=\{ \sqrt{3}, -\sqrt{\tfrac{1}{3}} \}\\ x &= \tfrac12 & y &= \tfrac12\sqrt2 & \{a,b\}&=\{ -\sqrt{6} + \tfrac{1}{2} \, \sqrt{8 \, \sqrt{6} + 20} - 2, \sqrt{6} - \tfrac{1}{2} \, \sqrt{-8 \, \sqrt{6} + 20} - 2 \}\\ x &= \tfrac{1}{2} & y &= -\tfrac{1}{2} \, \sqrt{2} & \{a,b\}&=\{ -\sqrt{6} - \tfrac{1}{2} \, \sqrt{8 \, \sqrt{6} + 20} - 2, \sqrt{6} + \tfrac{1}{2} \, \sqrt{-8 \, \sqrt{6} + 20} - 2 \}\\ x &= -\tfrac{1}{2} & y &= -\tfrac{1}{2} \, \sqrt{2} & \{a,b\}&=\{ -\sqrt{6} + \tfrac{1}{2} \, \sqrt{-8 \, \sqrt{6} + 20} + 2, \sqrt{6} - \tfrac{1}{2} \, \sqrt{8 \, \sqrt{6} + 20} + 2 \}\\ x &= -\tfrac{1}{2} & y &= \tfrac{1}{2} \, \sqrt{2} & \{a,b\}&=\{ -\sqrt{6} - \tfrac{1}{2} \, \sqrt{-8 \, \sqrt{6} + 20} + 2, \sqrt{6} + \tfrac{1}{2} \, \sqrt{8 \, \sqrt{6} + 20} + 2 \}\\ x &= 0 & y &= 0 & \{a,b\}&=\{0,\infty\} \end{align*}

There are also complex solutions for $\{a,b\}=\{-i,+i\}$ but I assume these are not interesting to you.

Answer 2

To find the self-intersections of the Lissajous curve $(\sin(2\pi t),\sin(3\pi t)),t\in[0,2)$,
let $u>v$ be two values in $[0,2)$ that are mapped to the same point on the curve.
So the point $(u,v)$ lies in the triangular region ${\Large◢}:\cases{u>v\\u\in[0,2)\\v\in[0,2)}$.
From $\sin(2\pi u)=\sin(2\pi v)$ we get$$2u\equiv2v\text{ or }1-2u\pmod{2}$$so the point $(u,v)$ lies on one of the blue lines.
From $\sin(3\pi u)=\sin(3\pi v)$ we get$$3u\equiv3v\text{ or }1-3v\pmod{2}$$so the point $(u,v)$ lies on one of the red lines.
We find that there are 7 intersections of blue and red lines in the region $\Large◢$.
So there are 7 self-intersections of the Lissajous curve $(\sin(2\pi t),\sin(3\pi t)),t\in[0,2)$.