# Conditions for a trapping region.

Suppose I have an autonomous system of ordinary differential equations and I want to show that I have a trapping region: a region of phase space which trajectories can enter but can never leave. One way to do this is to show that the vector field points "inward" on the boundary of the region, which is how trapping regions are presented in many books.

Is it possible to weaken this condition and just require that the vector field not point "outward"? In other words, if the vector field is inward pointing or tangent to the boundary everywhere on the boundary, is that enough to have a trapping region?

I am assuming that the system has unique solutions, otherwise there is the following counter-example: Let $\frac{dy}{dt} = -3 y^{2/3}$. Consider the region to be the positive real line. Then at the boundary of the region the vector field is zero, but there is still the solution $y=-t^3$ that goes from inside the region to outside.

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Suppose I have an autonomous system of ordinary differential equations and I want to show that I have a trapping region: a region of phase space which trajectories can enter but can never leave.

Such a region is usually called a positively invariant set.

One way to do this is to show that the vector field points "inward" on the boundary of the region, which is how trapping regions are presented in many books. Is it possible to weaken this condition and just require that the vector field not point "outward"?

Suppose that we have an autonomous dynamical system given by

$$\dot{x} = f (x)$$

where $f : \mathbb{R}^n \to \mathbb{R}^n$. We are given a set $\Omega \subset \mathbb{R}^n$ and would like to determine whether $\Omega$ is positively invariant. Suppose that there exists a "well-behaved" function $V$ such that $\Omega$ is a sublevel set of $V$, i.e.,

$$\Omega = \{ x \in \mathbb{R}^n \mid V (x) \leq 0 \}$$

Then $\Omega$ is positively invariant if the following universally-quantified formula evaluates to True

$$\left( \forall \, x \in \mathbb{R}^n \right) \left( V (x) = 0 \implies \partial_x V (x) \, f (x) \leq 0 \right)$$

If $V$ is a polynomial function, then the formula above is decidable (in theory) via quantifier elimination (you can use software like QEPCAD or REDLOG). Note that the formula above says that if $x$ is on the boundary of $\Omega$ (level set $V (x) = 0$), then the inner product between the the gradient $\partial_x V$ and the state velocity $\dot{x}$ is nonpositive, which is a way of saying that the vector does not point outwards.

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