For which values of $\alpha$ is the disk $B = \{(x, y) \mid x^2+y^2 \leq 1\}$ positively invariant?

Question

Given the following dynamical system

$$\begin{aligned} \dot x &= f(x,y) = -x + \alpha y \\ \dot y &= g(x,y) = -y\end{aligned}$$

for which values of $\alpha$ is the disk $B = \{(x, y)\mid x^2+y^2 \leq 1\}$ positively invariant?

Now what I have done is that I have taken the orbital derivative of $x^2 + y^2$. This gives

$$\frac{dV}{dt} = 2x\dot x + 2y \dot y = 2x(-x+\alpha y) - 2y^2$$

Now I say at the edge of this disk we have $x^2 + y^2 = 1$ so I now get

$$\frac{dV}{dt}= -2 +2x\alpha y$$

and I want $\frac{dV}{dt} < 0$ so from this I conclude that $\alpha < \frac{1}{xy}$.

Now here I am just looking for some clarification. If I have done this wrong can someone please point me in the correct direction. I am not very confident in my answer due to the fact I have not got a actual numerate answer.

Answer 1

Your reasoning is O.K., however I don't think that it will lead to useful results. I would use the inequality $$ 2 x y \le x^2 + y^2. $$ Assume $\alpha \ge 0$. So you have $$ \frac{dV}{dt} = -2 x^2 + 2 \alpha x y - 2 y^2 \le (- 2 + \alpha) (x^2 + y^2), $$ which is $\le - 2 + \alpha$ on the boundary $x^2 + y^2 = 1$ of the disk $B$.

We have thus obtain a sufficient condition, $0 \le \alpha < 2$, for the disk $B$ to be positively invariant.

Regarding necessary conditions, let $\alpha > 2$, and consider the field vector at $(\sqrt{2}/2, \sqrt{2}/2)$. Its $x$-coordinate is bigger than $\sqrt{2}/2$, and its $y$-coordinate is $-\sqrt{2}/2$. So the vector points out of $B$, consequently, the orbit passing through $(\sqrt{2}/2, \sqrt{2}/2)$ leaves the disk $B$.

Answer 2

Using polar coordinates, $x = \cos (\theta)$ and $y = \sin (\theta)$, we obtain

$$\dot V = -2 + 2 \alpha x y = -2 + 2 \alpha \cos(\theta) \sin (\theta) = -2 + \alpha \sin(2 \theta)$$

which is nonpositive for $\alpha \in [-2,2]$.