An exercise about stability points $\dot x =f(x)$ with $x \in \mathbb R^2$. $f(0)=0$ $f(x)=\alpha e_r(x) + \beta e_{\theta}(x)$, I am studying for a final and can't solve the following question:
Let the ODE system $\dot x =f(x)$ with $x \in \mathbb R^2$. Suppose $f(0)=0$ and $f(x)=\alpha e_r(x) + \beta e_{\theta}(x)$, where $\alpha, \beta \in  \mathbb R$ and $e_{\theta} \in \mathbb R^2$ are respectively the radial and tangent unit vectors at point $x \in \mathbb R^2 $.  In order for the  origin to be a stable equilibrium point it is necessary that:
a)  $ \alpha < 0 $and $ \beta \le \alpha$
b) $  \beta <0$ independently from $\alpha$
c) $\alpha =0$and $\beta <0 $
d) $\alpha \le 0$ independently from $\beta$
My try:
let $ x=(x, y)$
$e_r=(\cos\theta, \sin\theta) =(\frac{x}{\sqrt(x^2 + y^2)}, \frac{y}{\sqrt(x^2 + y^2)})$ and
$e_{\theta}=(-\sin\theta, \cos\theta)=(\frac{-y}{\sqrt(x^2 + y^2)}, \frac{x}{\sqrt(x^2 + y^2)})$
$f(x)=(\alpha \cos \theta -\beta \sin \theta,\alpha \sin \theta + \beta \cos \theta)$
$=(\alpha \frac{x}{\sqrt(x^2 + y^2)} -\beta \frac{y}{\sqrt(x^2 + y^2)},\alpha \frac{y}{\sqrt(x^2 + y^2)} + \beta \frac{x}{\sqrt(x^2 + y^2)})$
Now I think I have to linearize $f$
If I try to do the jacobian of this matrix, I get a matrix with term having $\sqrt(x^2 + y^2)$ at the denominator so I can't evaluate at (0,0), so I am stuck here. I wanted to impose that the eigenvalues of the jacobian are negative to find the relationship in the answer.
Can someone please help?
 A: With $e_r = \frac{x}{\|x\|}$ we get for $x\ne0$
$$\dot x = f(x) \Rightarrow x\cdot\dot x = x\cdot (\alpha e_r + \beta e_\theta) = \|x\| e_r \cdot (\alpha e_r + \beta e_\theta) = \alpha \|x\|,$$
so
$$\frac 12 \frac {d}{dt}\left(\|x\|^2\right) = \alpha \|x\|.$$
Can you take it from here?
edit
One of possible definitions of stability: A zero solution is stable if for any $\varepsilon >0$ there exists such $\delta > 0$ that $\forall x_0$ with $\|x_0\|< \delta$ we have $$\forall t >0\quad \|x(t)\|<\varepsilon.$$
In our case, for $\alpha \le 0$ we can safely take $\delta = \varepsilon$ - and that's pretty much it, because the norm $\|x(t)\|$ is decreasing in time.
If $\alpha > 0$ then $\|x(t)\| = \alpha t + \|x_0\|$ is strictly increasing, hence the zero solution can not be stable.
Note also that $\beta$ plays no role.
A: from
$f(x)=(\alpha \cos \theta -\beta \sin \theta, \alpha \sin \theta + \beta \cos \theta)$
I rewrite the equation in polar coordinates $(\rho, \theta)$, by components:
$\frac{d}{dt}(\rho cos(\theta) )= \alpha \cos \theta -\beta \sin \theta $
$\dot\rho \cos \theta-\rho\sin\theta\dot\theta = \alpha \cos \theta -\beta \sin \theta ...(1)$
$\frac{d}{dt}(\rho sin(\theta) )= \alpha \sin \theta +\beta \cos \theta $
$\dot\rho \sin \theta-\rho\cos\theta\dot\theta  = \alpha \sin \theta +\beta \cos \theta ..(2)$
from $\cos\theta(1) +\sin\theta(2)$ and $\sin\theta(1)+\cos(\theta)(2)$, I get
$\dot\rho = \alpha$
$\dot \theta \rho=\beta$
Solving the first one yields
$\rho=\alpha t+c$, so I must choose $\alpha \le 0$ in order for the trajectory not to diverge from the origin. And it does't matter what the angle is doing as long as the radial component is not diverging
