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The text I am reading has an example for flow in a section titled "Flows and Lie derivatives." Below is the example:

Let $M = \mathbb{R}^2$, and let $X((x,y)) = -y \partial/\partial x + x \partial/\partial y$ be a vector field in $M$. It is easy to verify that $$\sigma(t,(x,y)) = (x \cos t - y \sin t, x \sin t + y \cos t)$$ is a flow generated by $X$. The flow through (x,y) is a circle whose centre is at the origin. Clearly, $\sigma(t,(x,y)) = (x,y)$, if $t = 2n\pi$, $n \in \mathbb{Z}$. If $(x,y) = (0,0)$, the flows stays at $(0,0)$.

Could someone explain how the given flow is found from $X((x,y)) = -y \partial/\partial x + x \partial/\partial y$?

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1 Answer 1

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The components of the vector field are $X_1(x, y) = -y$ and $X_2(x, y) = x$. These components define a system of ODEs for a curve: $$ \dot{\gamma}(t) = X(\gamma(t)), $$ or more explicitely $$ \begin{align} \dot{\gamma_1}(t) &= X_1(\gamma(t)) = -\gamma_2(t), \\ \dot{\gamma_2}(t) &= X_2(\gamma(t)) = \gamma_1(t). \end{align}$$ The solution $\gamma(t)$ is the flow generated by the vector field. The function $\sigma$ solves the given system of ODES, therefore it is a flow. For further information take a look at this script from the ETH Zurich

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