Define $A=\{(x,y)\in \mathbb{R}^2;\ x\leq 0\ \text{or } y \leq x^2 \}$ ( a representation of $A$ can be found here.)
Let $X:A\to \mathbb{R}^2$ be a smooth function such that $X(\partial A) = (1,0)$ and $\tau:\mathbb{R}\to \mathbb{R}$ a diffeomorphism satisfying $\tau'>0$ and $\tau(0) = 0$.
My Question: Is it possible to extend $X$ to a fuction $\tilde{X}:\mathbb{R}^2\to \mathbb{R}^2$, such that the orbit of $\tilde{X}$ startig at the point $(0,y)$ (with $y>0$) also passes through the point$(\tau(y),\tau(y)^2)$.
Just for the records, I am searching for a smooth planar vector field $\tilde{X}$, such that $\left. \tilde{X}\right|_{A} = X$ and $\forall$ $y>0$ the solution $\varphi_y$ of the ODE \begin{align*}\dot{x} &= \tilde {X}(x) \\ x(0) &= (0,y).\\ \end{align*} satisfy the condition, $\exists$ $t_y \in \mathbb{R}$ such that $\varphi_y(t_y) = \left(\tau(y),\tau(y)^2 \right)$.
If the result is true just in the neighborhood of the origin it is good enough for me. I don't have any ideas on how to tackle this problem. Can anyone help?
Yes it is possible. But my "proof" is just a drawing : you draw a smooth curve $\gamma_1$ from $(0,1)$ to $(\tau(1),\tau(1)^2$ such that $\gamma_1(0,1)=\gamma_1(\tau(1),\tau(1)^2)=(1,0)$ and you take a deformation of this to get a path $\gamma_y$ from $(0,y)$ to $(\tau(y),\tau(y)^2)$ as you want it to be. Finally contruct $\tilde{X}$ from the $\gamma_y$'s.