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(THIS POST HAS BEEN EDITED)

Ok.. So I am looking for a flow $\phi$ on the closed unit disc, where for every $x$ in $\overline{B(0,1)}$:

$\omega(x,\phi)$=$\partial B(0,1)$ and $\alpha(x,\phi)=\{0\}$

So I have tried doing this manually and every case I propposed did not fulfill the flow propperty.

My idea was, see $\phi_t(z)=\psi_t(z)\cdot \chi_t(z)$ where $\chi_t(z)=ze^{2\pi it}$ and $\psi: \mathbb{R} \times B(0,1) \to \mathbb{R}$ such that $\psi_t(z)=1$ when $\mid z\mid = 1$, $\psi_t(z)=1$ when $t=0$ and:

$\psi_t(z) \longrightarrow 0 $ when $t \longrightarrow -\infty$ , $\psi_t(z) \longrightarrow \frac{1}{\mid z \mid} $ when $t \longrightarrow +\infty$

So making a continuous function who does this is no problem at all:

For example $\psi_t(z)= \left\{ \begin{array}{lcc} exp(log(\frac{1}{\mid z\mid}) (t)) & if & t \leq 0 \\ \\ exp(log(\frac{1}{\mid z\mid}) (\frac{t}{t+1}))& if & t>0 \end{array} \right.$

Which satisfies what I am looking for, but using this function on $\phi$, makes $\phi$ not a flow.

Any suggestions would be largely appreciated, thanks.

Ok, so from here I edited it, I forgot a detail about this flow:

$\omega(x,\phi)$=$\partial B(0,1)$ and $\alpha(x,\phi)=\{0\}$ $\forall x \neq 0$

$0$ must be a fixed point. Its more of a subtelty but I think it might be worth mentioning because it surely would not make sense otherwise.

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Consider the flow obtained from solving the pair of equations $r'=r(1-r)$, $\theta'=1$ written in polar coordinates. Solving explicitly we find that it induces the flow $$ \phi_t(r,\theta)=\left(\frac{e^t}{1/r-1+e^t},\theta+t\right), $$ which you can rewrite in other coordinates if you want (it is indeed a flow because it is the solution of an autonomous equation). For example, in $(x,y)$ coordinates it takes the form $$ \psi_t(x,y)=\left(\frac{xe^t}{1-\sqrt{x^2+y^2}(1-e^t)},\frac{ye^t}{1-\sqrt{x^2+y^2}(1-e^t)}\right). $$ It follows readily from the phase portrait that the flow has the properties that you want (when restricted to the closed unit disc).

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