A compactly supported cutoff of an incomplete vector field is complete

Apply [L1] to $[-1,1]\subset (-2,2)$ to obtain the required smooth bump. Put $f(x)=\chi (x)x^2$. It is smooth, equals $x^2$ on $[-1,1]$, and vanishes off $[-2,2]$. Hence $\mathrm{\mid }f\mathrm{\mid }\le 4$. Its derivative is continuous and zero off $[-2,2]$, so [L4] bounds $\mathrm{\mid }{f}^{\mathrm{\prime }}\mathrm{\mid }$ globally. By [L3], $f$ is Lipschitz.

Fix any initial value $x(0)=a$ and take its maximal solution $x:(\alpha ,\beta )\to \mathbb{R}$ from [L2]. Since $\mathrm{\mid }{x}^{\mathrm{\prime }}\mathrm{\mid }=\mathrm{\mid }f(x)\mathrm{\mid }\le 4$, [L3] on the interval between $0$ and $t$ gives $\mathrm{\mid }x(t)-a\mathrm{\mid }\le 4\mathrm{\mid }t\mathrm{\mid }$. If $\beta <\mathrm{\infty }$, set $t_j=\beta (1-1\mathrm{/}(j+1))$ for $j\ge 1$. These times increase to $\beta$, and the graph points lie in the compact rectangle $[0,\beta ]\times [-R,R]$, where $R=\mathrm{\mid }a\mathrm{\mid }+4\beta +1$. The ODE domain is all of ${\mathbb{R}}^2$, so [L5] extends the solution past $\beta$, contrary to maximality. Therefore $\beta =\mathrm{\infty }$. Reflecting time applies the same argument to $-f$ and gives $\alpha =-\mathrm{\infty }$. This uses an explicitly defined sequence and no choice assumption.

Every integral curve of $Y$ is thus global, so $Y$ is complete. In contrast, $x(t)=1\mathrm{/}(1-t)$ solves $x^{\mathrm{\prime }}=x^2$ with $x(0)=1$ for $t<1$ and diverges as $t\uparrow 1$, so the original field is incomplete. This verifies both the local agreement and the claimed change in completeness.