Consider the following system of ODEs : $\dot x =(|x|^2 -1)(\bar x -x)$ with $x \in \mathbb R^n, n\in \mathbb N, n\ge 1$ and $\bar x \in \mathbb R^n$, assigned. Which of the following statements is surely true?
(a) $\bar x$ is a locally attractive equilibrium point if it is close enough to the origin
(b) $\bar x$ is a stable equilibrium point if it is far enough away from the origin
(c) $\bar x$ is the only point of equilibrium in the system
(d) $\bar x$< is a globally asymptotically stable equilibrium point if it is close enough to the origin
The given correct answer was (a) (but seems to be wrong)
Can someone help me out to prove it formaly and also what is the reasoning to rule out the other ones? If it were just in n=1, I could just solve $(|x|^2 -1)(\bar x -x)=0 $ to find the equilibrium points and $(|x|^2 -1)(\bar x -x)>0 $ to find the intervals where the solution trajectory increases and $(|x|^2 -1)(\bar x -x)<0 $ the intervals where it decreases, but in n dimensions how do I deal with that ?**
So so far I only have that the equilibrium points are $x=\bar x$ and all the points in the hypersphere $|x|=1$
Edit: Even in the one-dimensional case my analysis is not making any sense with the answer that is supposed to be correct: For n=1, the equilibrium points are $\bar x, 1,-1$ Now analyzing the sign of the right-hand side of the ODE: Suppose $\bar x$ is close enough to the origin,as option (a) states, that is $ -1< \bar x < 1 $, then $(x^2-1)(\bar x- x)<0 \iff x \in (-1,\bar x)\cup(1,+\infty)$, so I get the following diagram for the trajectories over the real line:
---->>>-1<<<<$\bar x$>>>>1<<<<---- , meaning that inside $\bar x$ is unstable. What am doing wrong?
Edit 2 So with the help of MatthewH, it seems like (a) is certanly wrong at least in the n=1 case I will continue the analysis at least in the n=1 case Now option (b) seems to be the correct one because if $\bar x$ is far enough away from the origin, that is $\bar x>1$ or $\bar x<-1$. To fix the idea I consider $\bar x >1$: $(x^2-1)(\bar x- x)<0 \iff x \in (-1,1)\cup(\bar x,+\infty)$ so I get the following diagram for the trajectories over the real line:
---->>>-1<<<<1>>>>$\bar x$<<<<---- , meaning that $\bar x$ is a sink, that is asymptotically stable: attractive and stable. Therefore option b is the correct one
To rule out the other ones:
Option c is not correct because there are infinite equilibrium points
Option d is not correct : $\bar x$ can't be globally asymptotically stable equilibrium point because when it is inside $(-1,1)$, solution curves with initial value outside of that interval can't reach it
Does this seems ok? Can someone provide a solution in n dimensions?
Following the advice I give in the comments, assume $$x(t)=\lambda(t)(\bar{x}-x_0)+x_0$$ where $\lambda(t)$ is a function satisfying $\lambda(0)=0$. If you plug this expression into the DE $\dot{x}=\left(1-|x|^2\right)(x-\bar{x})$ and assume $x_0\neq \bar{x}$ you'll see how $\lambda(t)$ must satisfy the first order ODE $$\frac{dy}{dt}=\left(|y(\bar{x}-x_0)+x_0|^2-1\right)(1-y) \text{ subject to } y(0)=0$$ If $\bar{x},x_0$ are chosen to have magnitudes larger than $1$ such that the line segment connecting $\bar{x}$ with $x_0$ doesn't intersect the unit hypersphere $\{x:|x|=1\}$ then $$|y(\bar{x}-x_0)+x_0|>1 \text{ for any } y\in [0,1]$$ Since the DE in $y$ is autonomous with $y(0)=0$ it follows any solution $y(t)$ must satisfy $y(t)\mapsto 1$ as $t\mapsto +\infty$. This implies $x(t)\mapsto \bar{x}$ as $t\mapsto +\infty$.