ODE on Circle and torus

Question

How can we define an ODE on unit Circle $S^1=x^2+y^2=1$ and torus $S^1*S^1$? Every ODE on circle must satisfies $x \dot{x}+y \dot{y}=0$. Therefor $\dot{x}=-y , \dot{y}=x$ is a system on Circle. but this system has $x^2+y^2=r$ as a general solution. Since it is the Hamiltonian and complete system as well. So what are ODE on circle and torus?

Answer 1

Any ODE of the form $\dot{x} = f(x,y), \, \dot{y} = g(x,y)$ with $xf(x,y) + yg(x,y) = 0$ has only solutions with the property $x^2 + y^2 = const.$.

So choose any smooth $f_0: \mathbb{R}^2 \to \mathbb{R}$, set $f(x,y) = f_0(x,y) y, \, g(x,y) = - f_0(x,y)x$, and you have such a system.

Answer 2

You would profit a lot from seeing an equation on the circle as one of the form $$ \theta'=f(\theta) $$ for some $1$-periodic function $f$. Taking $x=\cos \theta$ and $y=\sin\theta$ we get $$ x'=-\theta'\sin\theta\quad\text{and}\quad y'=\theta'\cos\theta, $$ which is the same as $$ x'=-f(\theta)y\quad\text{and}\quad y'=f(\theta)x. $$ You can in fact write $f(\theta)$ using $x$ and $y$ since $\tan\theta=y/x$ (the problem with $x=0$ is that $S^1$ is a manifold and in fact we need at least two charts to cover it; if you prefer, saying this in another manner: $\theta$ clearly is not globally defined although $\theta'$ is). So we can rewrite the former equations in the form $$ x'=-g(x,y)y\quad\text{and}\quad y'=g(x,y)x.\tag1 $$ Conversely, for any equation of this form we have $xx'+yy'=0$ and so it defines and equation on the circle.

Summing up, we have a differential equation on the circle if and only if it has form (1) for some function $g$.

Similarly, any equation on the torus is of the form $$ x'=f(x,y),\quad y'=g(x,y) $$ for some functions $f$ and $g$ that are $1$-periodic in both variables.