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I'm very lost with the following problem:

Consider $f=(f_1,f_2,f_3)\in\mathcal{C}(\mathbb{R}^3,\mathbb{R}^3)$ such that $f(0,0,0)=(0,0,0)$ and \begin{equation} x_1f_1+x_2f_2+x_3f_3<0 \tag{*} \end{equation} for all $(x_1,x_2,x_2)\in\mathbb{R}^3\backslash\{ (0,0,0)\}$. Prove that the $(0,0,0)$ is the unique equilibrium point of $f$, and its basin of attraction is all $\mathbb{R}^3$.


I have started my course of dynamical systems recently and I find this problem in my practice list of exercises. My problem is that I don't understand very well what does it means that and equilibrium point of a function. I find in my notes the definition of equilibrium point, but for a differential equation $x'=f(x)$. It means to say that I need to find the equilibria of the system $$ \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}'=\begin{bmatrix} f_1 \\ f_2 \\ f_3 \end{bmatrix} \ ? $$

If so, how does the condition $(*)$ help me? And finally, how it is obtained his basin of attraction from this?

I appreciate your help a lot.

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    Suppose you have $V(x)= x_1^2+x_2^2+x_3^2$ and $f(x) = -(x_1^{2n_1-1}x_2^{2m_1}x_3^{2p_1},x_1^{2n_2}x_2^{2m_2-1}x_3^{2p_2},x_1^{2n_3}x_2^{2m_3}x_3^{2p_3-1})$ then $V(x) > 0,\dot V(x) = \nabla V(x)f(x) < 0$ and the dynamical system $\dot x = f(x)$ has as attraction basin all $\mathbb{R}^3$ – Cesareo Dec 24 '22 at 10:16
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    See the Lyapunov stability theory. – Cesareo Dec 24 '22 at 10:25
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    @Cesareo thank you so much for your idea! I found, while reviewed my notes, that in later chapters we will see the Lyapunov stability. I'm going to take a look and see if I can figure it out :-) – hackerman Dec 24 '22 at 16:54

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You are correct in your interpretation of the question, although arguably it would be more accurate to talk about equilibrium points of flows or system of ODES and singularities of vector fields.

As to how to use the condition $(\ast)$, here is a geometric argument: put for $p=(x,y,z)$, $I(p)=p$. This is the vector field that attaches to the point $p$ the vector $p$. Let's also denote by $F(p)=(F_1(p),F_2(p),F_3(p))$ the given vector field. The $(\ast)$ condition says that the angle (in the counterclockwise direction) between $F(p)$ and $I(p)$ is larger than $\dfrac{\pi}{2}$ and less than $\dfrac{3\pi}{2}$. See https://www.desmos.com/calculator/9ljt9qnmki for a humble interactive graph. Since $I(p)$ points away from the origin, this means that any trajectory must come closer to the origin.

(To make this argument rigorous (at different levels) one can use the relationship between inner products and angles or differentiate the function $t\mapsto |\phi_t(p)|^2$, where $t\mapsto \phi_t(p)$ is the unique solution of the system with initial condition $p$.)

Alp Uzman
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    Thank you very much!! I loved the animation. It has clarified and illustrated a lot the situation to me. I finally decided to differenciate the function that you pourposed. – hackerman Dec 24 '22 at 23:30
  • @hackerman I'm glad it was useful. – Alp Uzman Dec 25 '22 at 01:50