An exercise about stability points $\dot x =f(x)$ with $x \in \mathbb R^2$. $f(0)=0$ $f(x)=\alpha e_r(x) + \beta e_{\theta}(x)$,

Question

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?

Answer 1

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.

Answer 2

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