Theorem: Let $Z:U\subset \mathbb{R}^2\rightarrow \mathbb{R}^2$ a $\mathcal{C}^1$ field defined in a simply connected set $U$. If $\mathrm{div} Z(x)\neq0$ for all $x\in U$, then $Z$ does not have any periodic orbit entirely contained in $U$.
Application: Let $X$ be a linear focus of $\mathbb{R}^2$ (i.e $X(x)=A\cdot x$ where the eigenvalues of $A$ have non-zero real and imaginary part). Prove there is $\delta>0$ such that every field $Y$ of $\mathbb{R}^2$ which satisfies $$ \underset{x\in\mathbb{R}^2}{\sup} \left\Vert Y(x) \right\Vert\leq \delta $$ $$ \underset{\left\Vert x \right\Vert\leq 1}{\sup} \left\Vert DY(x) \right\Vert\leq \delta $$ does not admit any periodic orbit.
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What I tried:
Since $X$ is a linear focus : $X(x_1, x_2)=P\underset{A}{\underbrace{\left(\begin{array}{cc}a+ib & 0 \\ 0 & a-ib \end{array}\right)}}P^{-1} \left(\begin{array}{c} x_1\\x_2 \end{array}\right)$, so $\mathrm{div} X(x) = 2a \neq 0$
For $\left\Vert x \right\Vert\leq 1,\, \left| \mathrm{div} Y(x) \right|=\left| \mathrm{TrD} Y(x) \right|\leq \left\Vert \mathrm{Tr} \right\Vert\left\Vert \mathrm{D}Y(x) \right\Vert \leq \left\Vert \mathrm{Tr} \right\Vert\delta<\left| 2a \right|$ for a good choice of $\delta$. Here lies my first issue: I don't want the norm of the transpose ($ \left\Vert \mathrm{Tr} \right\Vert $) to be involved, but I don't know how to use another constant.
So we are sure there is no periodic orbit in the unit disk.
Now if we are out of the unit disk ($ \left\Vert x \right\Vert > 1 $):
If $\left\Vert A \right\Vert <1$:
$\left\Vert \left( X+Y\right)(x) \right\Vert \leq \left\Vert A \right\Vert \left\Vert x \right\Vert + \delta \leq \left(\left\Vert A \right\Vert + \delta \right) \left\Vert x \right\Vert < \left\Vert x \right\Vert $ for $0<\delta<\frac{1-\left\Vert A \right\Vert}{2}$
Second issue: I would like to conclude that because of this property, any solution which started (i.e at $t=0$) outside of the unit disk eventually goes into the unit disk and never leaves it.
If $\left\Vert A \right\Vert >1$: Third issue: I can consider $\left(-X-Y \right)$ to use the previous statement. Or can I?
Fourth issue: I don't know how to handle the case $\left\Vert A \right\Vert =1$