Pathological vector fields

Question

Is it true that any smooth curve is an integral curve for some smooth vector field?
More concretely, consider the curve $\gamma:\mathbb{R}\rightarrow \mathbb{R}^2$ defined by $\gamma(t)=(t,t^3sin(1/t))$ if $t\neq 0$ and by $\gamma(0)=(0,0)$. It is a $C^1$ curve, so i was wondering if it could exist a $C^1$ vector field $v=v(x,y,t)$ such that $\gamma$ in an integral curve.
Of course i'm interested in the somehow pathological behaviour near $(0,0)$, more generally i was wondering if this kind of oscillation phenomena could happen with integral curves of vector fields (also in the $C^k$, $k\in \mathbb{N}\cup\{\infty\}$ case).
Thank you for reading.

Answer 1

I was thinking about this, and actually it seems that it is true, at least locally.
It is actually the same construction for the canonic form of flows near non singular points (for an autonomous vector field).
Say that you have a smooth curve $t\rightarrow \gamma(t)\in \mathbb{R}^2$ for $t\in I=(-\epsilon, \epsilon)$ with $\gamma(0)=(0,0)$ and $\gamma'(0,0)=(1,0)$. Define the function $F:I\times\mathbb{R}\rightarrow \mathbb{R}^2$ by $F(x,y)=\gamma(x)+(0,y)$. We have that $F(x,0)=\gamma(x)$ and $DF(0,0)=I_2$. So $F$ is a diffeomorfism between two open sets $U\subseteq I\times\mathbb{R}$ and $V\subseteq\mathbb{R}^2$ both containing $(0,0)$. Now if we take the constant vector field $v(x,y)=(1,0)$ on $U$, we note that tha curve $\alpha(t)=(t,0)$ is an integral curve for $v$, so $F\circ\alpha$ is an integral curve for the push forward vector field $F_*v$ on $V$, but $F\circ\alpha=\gamma$.