Closed periodic curve

Question

Let $\textbf{$\gamma$}: \mathbb{R} \rightarrow \mathbb{R}^n$ be a smooth curve and let $T \in \mathbb{R}$. We say that $\textbf{$\gamma$}$ is $T$-periodic if $$\textbf{$\gamma$}(t+T)=\textbf{$\gamma$}(t) \text{ for all } t \in \mathbb{R}.$$ If $\textbf{$\gamma$}$ is not constant and is $T$-periodic for some $T\neq 0$, then $\textbf{$\gamma$}$ is said to be closed.

The period of a closed curve $\textbf{$\gamma$}$ is the smallest positive number $T$ such that $\textbf{$\gamma$}$ is $T$-periodic.

A curve $\textbf{$\gamma$}$ is said to have a self-intersection at a point $p$ of the curve if there exist parameter values $a\neq b$ such that

• $\textbf{$\gamma$}(a) = \textbf{$\gamma$}(b) = p$, and
• if $\textbf{$\gamma$}$ is closed with period $T$, then $a-b$ is not an integer multiple of $T$.

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I am looking at the following exercise:

Show that the Cayley sextic $$\textbf{$\gamma$}(t) = \left (\cos^3t \cos {3t}, \cos^3t \sin {3t}\right ), t \in \mathbb{R}$$ is a closed curve which has exactly one self-intersection. What is its period?

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I have done the following:

Since $\textbf{$\gamma$} (0)=(1,0)$ and $\textbf{$\gamma$}\left (\frac{\pi}{6}\right )=\left (0, \frac{\sqrt{3}}{2}\right )$, i.e., $\textbf{$\gamma$} (0)\neq \textbf{$\gamma$} \left (\frac{\pi}{6}\right )$ we have that $\textbf{$\gamma$}$ is not constant.

We have that $$\cos (t+2\pi )=\cos t , \ \ \cos (3(t+2\pi ))=\cos (3t) , \ \ \sin (3(t+2\pi ))=\sin (3t)$$ so then $$\textbf{$\gamma$}(t+2\pi ) = \left ((\cos (t+2\pi ))^3 \cos {[3(t+2\pi )]}, (\cos (t+2\pi ))^3 \sin {[3(t+2\pi )]}\right )=\left (\cos^3t \cos {3t}, \cos^3t \sin {3t}\right )=\textbf{$\gamma$}(t), \forall t$$

That means that the non-constant curve $\textbf{$\gamma$}$ is $2\pi$-periodic.

So, $\textbf{$\gamma$}$ is a closed curve.

Is this correct?

Is $2\pi$ the period? Is this the smallest positive number $T$ such that $\textbf{$\gamma$}$ is $T$-periodic? Or does the period change when we have $\cos 3t$ and $\sin 3t$ instead of $\cos t$ and $\sin t$ ?

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EDIT:

We have that $$\textbf{$\gamma$}\left (\frac{\pi}{3}\right )=\left (-\frac{1}{8}, 0 \right ) \\ \textbf{$\gamma$}\left (\frac{2\pi}{3}\right )=\left (-\frac{1}{8}, 0\right )$$

So, when we take $p=\left (-\frac{1}{8}, 0\right )$ we have that $\textbf{$\gamma$}\left (\frac{\pi}{3}\right )=\textbf{$\gamma$}\left (\frac{2\pi}{3}\right )=\left (-\frac{1}{8}, 0\right )$

Since the period is $\pi$, we have that $\frac{2\pi}{3}-\frac{\pi}{3}=\frac{\pi}{3}=\frac{1}{3}(\pi)$, i.e., $\frac{2\pi}{3}-\frac{\pi}{3}$ is not an integer multiple of the period.

That means that $\gamma$ has a self-intersection at $p=\left (-\frac{1}{8}, 0\right )$.

Now we want to show that this self-intersection is unique.

To do that we want to find all points $x,y$ within one period of $[0,\pi]$ such that $\gamma(x)=\gamma(y)$. That is, find all solutions of the simultaneous system $$ \begin{align*} \cos^3(x) \cos(3x)&=\cos^3(y) \cos(3y) \\ \cos^3(x) \sin(3x)&=\cos^3(y) \sin(3y). \end{align*} $$

$$\frac{\cos^3(x) \cos(3x)}{\cos^3(x) \sin(3x)}=\frac{\cos^3(y) \cos(3y)}{\cos^3(y) \sin(3y)} \Rightarrow \frac{ \cos(3x)}{\sin(3x)}=\frac{ \cos(3y)}{ \sin(3y)} \\ \Rightarrow \cot (3x)=\cot (3y) \Rightarrow 3x=3y+k\pi \Rightarrow x=y+\frac{k}{3}\pi$$

Since the period is $\pi$, the only possible values for $k$ are $k=0$, $k=1$ and $k=2$.

For $k=0$ we have $3x=3y \Rightarrow x=y$. But for $x=y$ the second condition of the definition of a self-intersection is not satisfied.

So, for $k=1,2$ we have that $3x=3y+k\pi \Rightarrow x=y+\frac{k\pi}{3}$.

The first condition of the definition is satisfied with each of these values.

As for the second definition we want that $x-y=\frac{k\pi}{3}$ is not an integer multiple of $\pi$, which is also satisfied with each of these values for $k$.

So we want somehow to show that the points that we get this relation are $\frac{\pi}{3}$ and $\frac{2\pi}{3}$, or not?

But how? I don't have any idea...

Answer 1

Just using trig:

$T = \pi$

see this by computing $\gamma(0) = (1,0) = \gamma(T)$

Solve this using the second component first so that $$0 = \textrm{sin}(3T)\textrm{cos}^3(T)$$ Then $T = n \pi/3$, Do the same thing for the first component and you must have $n$ is a multiple of 3. In fact this is sufficient for all $t$, $\gamma(t+T) = \gamma(t)$.

The norm of the vector is $|\gamma(t)| = \cos^6(t)$ so to have an intersection we must have $\cos^6(t) = \cos^6(s)$ or in other words $s = \pi - t$ in this interval of $[0,\pi]$ This condition for the intersection is satisfied by the first component easily but leads to the following for the second component: $$\cos^3(t) \sin (3t) = -\cos^3(t) \sin (3t)$$ or in other words $$0 =\cos^3(t) \sin (3t) $$ The only possible intersections are then at $t = {0, \pi/3, \pi/2, 2\pi/3}$ Let $t<s$ from above then you must have either $t =0$ or $t = \pi/3$. But $t=0$ leads to a degenerate case so $t = \pi/3$ and $s = 2\pi/3$. Which you can check with substitution directly.

So $\gamma(\pi/3) = \gamma(2\pi/3)$ is the unique intersection.

Answer 2

The period $2\pi$ is not right. Let's complexify $\gamma(t)=(x(t),y(t))$ as $x(t)+iy(t)$: $$ \gamma=\cos^3(t)e^{i3t}=(\cos(t)e^{it})^3=\Bigl(\frac{e^{it}+e^{-it}}{2}e^{it}\Bigr)^3=\Bigl(\frac{z^2+1}{2}\Bigr)^3 $$ where $z=e^{it}$. The curve depends only on $z^2$ which is $\pi$ periodic, hence, the period is at most $\pi$. It is easy to see that it is actually $\pi$ since $\zeta=\frac{z^2+1}{2}$ runs along the small circle with the center at $1/2$ and radius $1/2$, so to get the initial value $\zeta^3=\gamma=1$ again it needs to make the whole loop which corresponds to $t=\pi$.

Self-intersection is guessed right. It can also be concluded without finding it explicitly as the function $\zeta^3$ is not univalent function inside the small circle mentioned above.

Answer 3

We had the same question a few days ago (Find the self-interection.Differential Geometry). I'm extending my accepted answer there to a full solution.

Write your curve in the form $$\gamma: \quad t\mapsto z(t):=x(t)+iy(t)={1\over8}(e^{it}+e^{-it})^3\>e^{3it}\ .$$ Substituting $t:={u\over2}$ we obtain, after a simple calculation, the parametrization $$\gamma:\quad s\mapsto{1\over8}\bigl(1+e^{iu}\bigr)^3 \qquad(u\in{\mathbb R})\ .$$ This already shows that in terms of the original variable $t$ the period is at most $\pi$.

In order to find possible double points (or even a smaller period) we have to investigate in detail for which pairs $(u,v)$ with $$0\leq u<v<2\pi\tag{1}$$ the equation $$\bigl(1+e^{iu}\bigr)^3=\bigl(1+e^{iv}\bigr)^3$$ is true. This equation is satisfied iff $1+e^{iu}$ and $1+e^{iv}$ coincide up to a third root of unity. The case $1+e^{iu}=1+e^{iv}$ gives no solution within the bounds $(1)$. Therefore it remains to discuss the cases $${\rm(a)}\quad 1+e^{iu}=\omega\bigl(1+e^{iv}\bigr),\qquad{\rm(b)}\quad 1+e^{iu}=\bar\omega\bigl(1+e^{iv}\bigr)\ ,$$ where $\omega:=e^{2\pi i/3}$, which is the same as $${\rm(a)}\quad \cos{u\over2}=\omega e^{i(v-u)/2}\>\cos{v\over2},\qquad {\rm(b)}\quad \cos{u\over2}=\bar\omega e^{i(v-u)/2}\>\cos{v\over2}\ .\tag{2}$$ In case (a) we want $\omega e^{i(v-u)/2}\in{\mathbb R}$ within the bounds $(1)$. This can only be done with ${v-u\over2}={\pi\over3}$, and $(2)$(a) then implies $\cos{u\over2}=-\cos{v\over2}$. The pair $u={2\pi\over3}$, $v={4\pi\over3}$ is the only one that satisfies these conditions, and gives rise to a double point $z_*$ of our curve, namely $$z_*={1\over8}\bigl(1+e^{2\pi i/3}\bigr)^3=-{1\over8}\ .$$

In case (b) we want $\bar \omega e^{i(v-u)/2}\in{\mathbb R}$ within the bounds $(1)$. This can only be done with ${v-u\over2}={2\pi\over3}$, and $(2)$(b) then implies $\cos{u\over2}=\cos{v\over2}$. The latter equation is not satisfiable within the bounds $(1)$.