I'm practicing for an upcoming test, and this one has been giving me some problems.
Suppose we have
$$
\begin{cases}
\dot x = x^2 + 2y – 4 \\
\dot y = -2xy
\end{cases}.
$$
Let $R = \{(x,y) : |x| \leq 2,\, 0 \leq y \leq 4 – x^2\}$. I'm to prove this is invariant, i.e., if $x_0 \in R$, then
$$
\Phi(t, x_0) \in R, \hspace{1em} \forall t\in \mathbb{R}.
$$
So far, I have $y = 0$ stays on $\{(x,0) : |x| \leq 2\}$, so the bottom line boundary is invariant. But, I'm having trouble with the boundary of the parabola. If it's invariant, then obviously we are done; but it may just be a trapping region.
Any hints on where to go would be most appreciated.
Best Answer
Fix $x_0$, and consider the upper parabola near the point $(x_0,4-x_0^2)$. Differentiating, the vector $(1,-2x_0)$ points tangent to the parabola.
OTOH, your system satisfies (at that point) \begin{align*} \dot{x}&=x_0^2+2(4-x_0)^2-4=4-x_0^2 \\ \dot{y}&=-2x_0(4-x_0^2) \end{align*} That is, $$(\dot{x},\dot{y})=(4-x_0^2)(1,-2x_0)$$
So the upper parabola is also an invariant curve, tending towards $(2,0)$.
Since both the upper and bottom boundaries are invariant and the system is in continuous time, the region between is invariant too.