Can anyone give an example of two vector fields $X_1$ and $X_2$ which are complete but their sum $X_1+X_2$ is not complete?
2 Answers
Let $M = \mathbb{R}$, $X_1 = (\sin^2{x}) x^2 \frac{\partial}{\partial x}$ and $X_2 = (\cos^2{x}) x^2 \frac{\partial}{\partial x}$. Then $X_i$ are complete but $X_1 + X_2 = x^2 \frac{\partial}{\partial x}$ which is not complete.
Why is $X_1$ complete? If $\alpha \colon I \rightarrow \mathbb{R}$ is an integral curve of $X_1$ satisfying $\alpha(0) = t$ for some $t \in \mathbb{R}$ then either $\alpha \equiv 0$ (if $t = \pi k$ for some $k \in \mathbb{Z}$) or $\alpha(I) \subset (\pi k, \pi (k + 1))$ (if $\pi k < t < \pi (k + 1)$ for $k \in \mathbb{Z}$) which both imply that $I = \mathbb{R}$. Similarly for $X_2$.
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Can you say it more specific? I know $x^2$ is non-complete. Thanks~ – user146507 May 11 '14 at 08:46
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Why does $\alpha(I) \subset (\pi k, \pi (k+1)$ imply that $I = \mathbb{R}$? – ImHackingXD Jan 06 '24 at 16:17
Consider the complete vector fields $X = y^2 \frac{\partial}{\partial x}$ and $Y = x^2 \frac{\partial}{\partial y}$ defined on $\mathbb R^2$. We have that $X + Y $ is not complete. In order to see that it suffices to find the solution of $$\begin{cases}\frac{dx}{dt} = y^2 \\ \frac{dy}{dt} = x^2\end{cases}$$ when $(c,c)$ moves along $x = y$.
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I think the solution is x=y=-1/t. So it is not complete since it is not defined at the origin. Thanks for your answer! – user146507 Jun 29 '16 at 21:42