Consider the vector field $V = x^2 \frac{\partial}{\partial x}$ on $\mathbb{R}$.
I want to find the flow of $V$ and determine whether it is complete on $\mathbb{R}$.
Given a manifold $M$, I know a flow is a map $\theta: \mathcal{D} \rightarrow M$ where $\mathcal{D} \subset \mathbb{R} \times M$ that satisfies $\theta(0,p) = p$ and $\theta(t,\theta(s,p)) = \theta(t+s, p)$. This is the defintion from Lee's book but I am not sure how this fits with the rest of my work.
My attempt using the method outlined a different stackexchange post:
First, I consider a candidate for the flow $\gamma(t) = (x(t))$. Then, I know it must satisfy $\gamma'(t) = V(\gamma(t))$ which follows that $$x'(t) \frac{\partial}{\partial x} = ((x(t))^2 \frac{\partial}{\partial x}$$ so solving the first-order nonlinear ODE $$x'(t) = (x(t))^2$$ we get that $$x(t) = \frac{1}{c-t}$$ for some constant $c\in \mathbb{R}$.
How do I make a flow with this $\gamma$? It seems like the domain of $\gamma$ doesn't match with the definition of a flow I want to use from Lee's book.
For the second part of the problem, I am guessing that this is not a global flow because there is a singularity when $c=t$.
Best Answer
If you found $x(t) = 1/(c-t)$, then $x(0) = x_0$ gives us that $x_0 = 1/c$, so that $c = 1/x_0$, and multiplying both numerator and denominator of $1/((1/x_0)-t)$ by $x_0$ yields $$x(t) = \frac{x_0}{1-tx_0}.$$Renaming $x_0\mapsto x$, this tells us that $$\theta(t,x) = \frac{x}{1-tx}.$$This vector field is not complete because the maximal connected domain of $\theta$ containing $\{0\}\times \Bbb R$ is not $\Bbb R \times \Bbb R$, but instead the region consisting of the points $(t,x)$ with $-\infty < t < 1/x$ for $x>0$, $1/x < t < \infty$ for $x<0$, and $(t,0)$ for $t\in \Bbb R$ (draw a picture, it is instructive).