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Let $T$ be the torus defined as the square $0\leq \theta_1, \theta_2\leq 2\pi$ with opposite sides identified. Let $F(\theta_1,\theta_2)=\cos\theta_1+\cos\theta_2$. Sketch the phase portrait for the system $-\text{grad}\, F$ in $T$. Sketch a three-dimensional representation of this phase portrait with $T$ represented as the surface of a doughnut.

I understand that the $-\text{grad}(F) = (\sin(\theta_1), \sin(\theta_2))$, but I'm very confused on how the torus is defined here. Why is a doughnut defined as a square?

I'm guessing in essence that we can find $\theta'_1$ and $\theta'_2$ and apply the domain to it, but the confusion on $T$ leaves me stuck?

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2 Answers 2

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It's called flat torus. Here is the animation from Wikipedia page about torus, that shows how you can map points of a square with identical sides to a torus surface in 3D.

enter image description here

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To continue the answered of Vasily Mitch, the torus is homeomorphic to $ S^1\times S^1 $, where $S^1$ is the circle.

Thus, if we consider the square $0\leq\theta_1,\theta_2\leq 2\pi$, the top boundary $\theta_2=2\pi$ and the bottom one $\theta_2=0$ should be consider the same. In a completely analogous way, the left and right boundaries $\theta_1=0$ and $\theta_2=2\pi$ are supposed to be consider the same.

Take for example, the following vector field on a $T^2$ torus,

\begin{align}\theta_1'&=1\\\theta_2'&=\omega,\quad \omega\in\mathbb{R}\backslash\mathbb{Q}\end{align}

then you going to have something of this form, where the line represents the vectors field and they have slope $\omega$.. Note that this phase portrait is an approximation, because the solution $(\theta_1(t),\theta_2(t))$ fills $T^2$ densely.

In your exercise,

\begin{align} \theta_1'&=\sin\theta_1\\ \theta_2'&=\sin\theta_2 \end{align}

with the following phase portrait in the square $0\leq\theta_1,\theta_2\leq 2\pi$, found by means of nullclines. Thus we have 4 equilibrium points: 1 sink, 2 saddle and 1 source. From this you can easily found the 3 dimensional phase portrait.

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  • $\begingroup$ How would we know where the equilibrium points are? I'm trying to find them and how to determine which one are the sink, saddle and source.. $\endgroup$
    – Math420
    Commented Jan 12, 2023 at 10:46

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