# Incomplete vector field

Is there a way I can tell if a vector field on a manifold or just $\mathbb{R}^n$ is incomplete simply by just looking at its formula. For instance on $\mathbb{R}$, $\displaystyle X= (x^2+1) \frac{\partial}{\partial x}$, is incomplete as it shoots off to infinity in finite time as evidenced by its flow $F=\tan(t-C)$. Should I be able to see that X was going to be incomplete without computing $F$?

-

This looks like a continuation of Complete non-vanishing vector field and "Completing" a vector field on a non-compact manifold $M$. The answer is: in general, this is not a simple question to answer. The global existence of a solution to ODE is harder to prove or disprove than local existence. For example, consider the vector field $$X=x_2e_1+x_3e_2+x_4e_3-x_1^3e_4 \tag1$$ on $\mathbb R^4$ where I write $e_j$ instead of $\partial/\partial x_j$. Is $X$ complete? I suspect it's not, based on the evidence in Fourth Order Nonlinear ODE and my own experiments; however, so far I failed to come up with a proof. On the other hand, $$X=x_2e_1-x_1^3e_1\tag2$$ is a complete field on $\mathbb R^2$; its trajectories are closed curves $x_1^4/4+x_2^2/2=C$.
In one dimension the analysis is easier because we can write down the solution of initial value problem $x'=f(x)$, $x(0)=x_0$, in the implicit form $$\int_{x_0}^x \frac{ds}{f(s)}=t\tag1$$ Assuming $f>0$, the formula (3) shows that the blow up occurs if and only if the improper integral $\int_{x_0}^{+\infty} \frac{dx}{f(x)}$ converges, and the value of this integral is the blow up time. (If $f$ is negative, replace the upper limit with $-\infty$). For example, equation of the form $x'=x^p$ with $p>1$ will produce blow up.
Executive summary. If $X$ is a smooth vector field and the norm of $X(p)$ is bounded by $A\,d(p,p_0)+B$ for some constants $A,B$ and a point $p_0$, then $X$ is complete. This is because the distance of solution from $p_0$ will grow at most exponentially, as do the solutions of 1-dimensional ODE $x'=Ax+B$.
If the growth of $X$ is superlinear, the field may or may not be complete: compare (1) and (2), where the latter is complete while the former is apparently not.