I have the following system and set S:
$$x' = – y + x – x^3 – xy^2\\y' = y + x – y^3 – x^2y
\\\ S = \{(x, y) \in R^2 : x^2 + y^2 \leqslant \sqrt2 \}$$
I need to prove that for this system, the set S is positively invariant.
My first thought was to find a Lyapunov function.
However, all candidates I tried to use was't suitable to deduce anything.
I would be thankful for any help!
Best Answer
Your set $S$ is the closed, circular disc in $\mathbb R^2$ centered on the origin and with radius equal to $\sqrt[4]{2}$. The boundary of $S$ is the circle centered on the origin of radius $\sqrt[4]{2}$.
So, fixing a point $(x,y)$ on that circle, what you need to prove is that the vector $\langle x',y' \rangle$ is pointing inside the circle.
At the point $(x,y)$, the vector $\langle x,y \rangle$ is an outward normal of the circle. Therefore, $\langle x',y' \rangle$ points inside the circle if and only if the dot product with the outward normal is negative: $$\langle x',y' \rangle \cdot \langle x,y \rangle < 0 $$ Substituting and simplifying: \begin{align*} (-y + x - x^3 - x y^2) x + (y + x - y^3 + x^2 y) y &= x^2 + y^2 - x^4 - 2 x^2 y^2 + y^4 \\ &= \sqrt[4]{2} - (\sqrt[4]{2})^2 \\ &< 0 \end{align*} which finishes the proof.
One can express this in terms of Lyaponov functions, if one wants: $x^2 + y^2$ is a Lyaponov function, because $\langle x,y \rangle$ is equal to ($\frac{1}{2}$ times) the gradient of $x^2 + y^2$. Geometrically, the Lyaponov condition is just saying that the vector $\langle x',y'\rangle$ points inward along each circle $x^2 + y^2 = r^2$, which is exactly what my computations prove.