Showing that a nonlinear system is positively invariant on a subset of $\mathbb{R}^2$

Question

I have the following nonlinear system: \begin{align} x_1'=& x_1-x_2-x_1^2\\ x_2'=& x_1+x_2-x_2^3 \end{align} By examining the direction of flow on the boundary, I want to show that the region defined by $-2\leq x_1\leq 2$, $-2\leq x_2\leq2$ is positively invariant. My general idea for this problem was to examine the direction of flow at each corner of the square. That is, I calculated: $x_1'$ and $x_2'$ at the points: $(-2, -2), (-2, 2), (2, -2), (2, 2)$. The corresponding vectors were \begin{align} (-2, -2) \rightarrow& \begin{pmatrix}8\\4\end{pmatrix} \\ (-2, 2) \rightarrow& \begin{pmatrix}4\\-8\end{pmatrix} \\ (2, -2) \rightarrow& \begin{pmatrix}-4\\8\end{pmatrix} \\ (2, 2) \rightarrow& \begin{pmatrix}-8\\-4\end{pmatrix} \end{align} so, from each corner of the square the vectors are all pointing inwards. Is it sufficient to then say that this square region is positively invariant? How could I make this more clear?

Answer 1

No, checking only corners is not sufficient. Set the orientation on boundary and check the sign of dotproduct between vector field along the boundary and normals' field of the boundary.

ADDITIONS

1. The check that I've suggested is equivalent to pointing inward or outward along all boundary of domain. The inward and outward can be expressed in terms of dotproduct between normal field along boundary and vector field. If it is strictly positive or negative — then vector field points inward/outward along the boundary.

2. Setting orientation usually means that you parametrize curve such way that normal vector points inward the domain that is bounded by curve. That's done for consistency of this check

3. Positive invariance of the domain could be proven the following way. Suppose that there is a trajectory that goes out of domain. Of course, it must intersect the boundary. If trajectory goes out, its tangent vector points out of domain (another sign of dotproduct between normal vector at boundary). But we've shown that sign doesn't change.

EXAMPLE

So, I'll apply this concept to your example. Boundary has four components: $(A)$ which is straight segment from $(2, 2)$ to $(-2, 2)$, segment $(B)$ from $(-2, 2)$ to $(-2, -2)$, segment $(C)$ from $(-2, -2)$ to $(2, -2)$, segment $(D)$ from $(2, -2)$ to $(2, 2)$. We choose the normals such way that it points inside the square: $\vec{n_A} = (0, -1)$, $\vec{n_B} = (1, 0)$, $\vec{n_C} = (0, 1)$,$\vec{n_D} = (-1, 0)$. So, let's calculate the dotproduct of vector field with normal field along segment $(A)$. As we see, $x_2 \equiv 2$ along $(A)$ and dotproduct gives us $0*x'_1 + (-1)*x'_2 = x^3_2 - x_2 - x_1 = 6-x_1$ (after substituting $x_2$ with 2). Since $x_1$ changes from $2$ to $-2$ along $(A)$, the value of dotproduct is strictly positive. This means that vector field points "approximately" in the same direction as a normal field, i.e. inside the domain. Analogous calculations can be made along other parts of boundary. If they're all strictly positive too — vector field is pointed inward along all boundary.