Linear field in $\mathbb{R^4}$ such that $\gamma\subset \omega(\gamma)$

Question

I was wondering if there is a example of a Linear field $F:\mathbb{R^4}\to\mathbb{R^4}$(i.e. F(x)=Ax, where A is a 4x4 matrix) such that $F$ admit a non-singular and non-periodic orbit $\gamma$, with the propriety that $\gamma\subset \omega(\gamma)$.

Answer 1

In fact it is possible. The idea is to take two rotations with frequencies which are incommensurable. Then the orbit is not periodic, but in the forward limit you get everything on the torus we are walking on (the quasiperiodic motion lies dense on the torus).

Take $f_1 , f_2 \in \mathbb{R}$ linearly independent over $\mathbb{Z}$. Then consider

$$ A = \begin{pmatrix} 0 & f_1 &0 & 0 \\ -f_1 & 0 &0 &0 \\ 0& 0& 0 & f_2 \\ 0 & 0 & -f_2 &0 \end{pmatrix}.$$

Then the flow of the associated ODE is given by for $x= (x_1, x_2)\in \mathbb{R}^4$

$$ e^{tA} x = \begin{pmatrix} \cos(t f_1) & - \sin(t f_1) & 0 &0 \\ \sin(t f_1) & \cos(t f_1) & 0 &0 \\ 0 & 0& \cos(t f_2) & - \sin(t f_1) \\ 0 &0 & \sin(t f_2) & \cos(t f_2) \end{pmatrix} x $$

Thus, we get that the orbit $\gamma$ of $x$ is contained in a torus depending on $x$. More precisely

$$\gamma \subseteq \{ (y_1, y_2) \in \mathbb{R}^4 \ : \ \vert y_1 \vert = \vert x_1 \vert, \ \vert y_2 \vert = \vert x_2 \vert \} $$

Next we prove

$$\omega(\gamma) = \{ (y_1, y_2) \in \mathbb{R}^4 \ : \ \vert y_1 \vert = \vert x_1 \vert, \ \vert y_2 \vert = \vert x_2 \vert \}.$$

Pick $(y_1, y_2) \in \mathbb{R}^4$ such that $\vert y_1 \vert = \vert x_1 \vert, \ \vert y_2 \vert = \vert x_2 \vert.$ There exists $t_0$ such that

$$ y_1 = \begin{pmatrix} \cos(t_0 f_1) & - \sin(t_0 f_1) \\ \sin(t_0 f_1) & \cos(t_0 f_1) \end{pmatrix} x_1. $$

Now we set $T=2\pi/f_1$, then we get for all $n\in \mathbb{N}$

$$ e^{(t+nT)A}x = \begin{pmatrix} y_1 \\ \begin{pmatrix} \cos((t_0+nT) f_2) & - \sin((t_0+nT) f_2) \\ \sin((t_0+nT) f_2) & \cos((t_0+nT) f_2) \end{pmatrix} x_2 \end{pmatrix}.$$

As $f_1, f_2$ are linearly independent over $\mathbb{Z}$, we have that

$$ \left(\begin{pmatrix} \cos((t_0+nT) f_2) & - \sin((t_0+nT) f_2) \\ \sin((t_0+nT) f_2) & \cos((t_0+nT) f_2) \end{pmatrix}x_2 \right)_{n\in \mathbb{N}}$$

lies dense in $\{ z \in \mathbb{R}^2 \ : \ \vert z \vert = \vert x_2 \vert \}$ and hence, find a sequence $(n_k)_{k\in \mathbb{N}} \subseteq \mathbb{N}$ such that

$$ \lim_{k\rightarrow \infty} e^{(t_0+n_kT)A}x =y.$$

I.e. $y\in \omega(\gamma)$. As our flow preserves the norm, we get that

$$ \omega(\gamma) = \{ (y_1, y_2) \in \mathbb{R}^4 \ : \ \vert y_1 \vert = \vert x_1 \vert, \ \vert y_2 \vert = \vert x_2 \vert \} $$

and thus $\gamma \subseteq \omega(\gamma)$. Note that the orbit is not periodic if $\vert x_1 \vert \neq 0$ and $\vert x_2 \vert \neq 0$, as the frequencies $f_1,f_2$ are linearly independent.