Is a fixed point, which is attracting on a manifold $M$, always globally stable if $M$ is globally attracting?

Question

Lets start with a toy example system of equations

\begin{align} \frac{dx}{dt}&=x(1-x)-xy\\ \frac{dy}{dt}&=-y \end{align}

I'd like to show $(1,0)$ has a basin of attraction that includes all of $B:=\{(x,y): 0<x<1, y\geq 0 \}$. The x-axis is obviously attracting on $B$, given $dy/dt=-y$. And on the x-axis, the point x=1 attracts all positive $x$. How can we prove that all trajectories with initial conditions in $B$ flow to $(1,0)$. And does this concept generalize to systems as follows?

The general question in 2D, given a dynamical system

\begin{align} \frac{dx}{dt}=f(x,y)\\ \frac{dy}{dt}=g(x,y) \end{align}

with a globally attracting curve $C:=\{(x,y): y=h(x)\}$, such that all points $(x,y)$ not on $C$ approach $C$. My question is, if there exists a unique fixed point $(x^*,y^*)$ on $C$ such that it attracts all points on the curve $C$, can we call that fixed point globally stable. Intuitively it seems like it has to be. I'm wondering if anyone has a proof for this proposition or a counterexample?

Answer 1

So for the example, one need only choose the Lyapunov function

\begin{equation} V(x,y)=y^2+(x+y-1)^2 \end{equation}

Clearly, $V(k,0)=0$ and $V(x,y)>0$ for all $(x,y)\neq(k,0)$ in $B$. Note

\begin{align} \dot{V}(x,y)&=2y\dot{y}+2(x+y-1)\dot{x}+2(x+y-1)\dot{y}\\ \end{align}

Case (1) $x>y-1$: the first and third therm are always negative because $\dot{y}$ is negative. The second term is negative because $x>y-1$ only includes points above the $dx/dt$ nullcline and hence, $\dot{x}>0$.

Case (2) $x<y-1$: The second term is negative because $x<y-1$ means $(x+y-1)<0$ sincle it also means $(x,y)$ is below the $dx/dt$ nullcline and hence, $\dot{x}>0$. Note that the third term is positive. However because $|x+y-1|<y$, (note $0<x \leq 1$ required to be in $B$) it is less positive than the first term is negative.

Case (3) $x=y-1$: $\dot{V}=-2y^2<0$.

Therefore $(k,0)$ attracks all trajectories on $B$. Here I exploited the fact that trajectories always flow down to the x axis, and the fact that in the x direction they flow towards the $dx/dt$ nullcline, which is why I chose $[x-x_{nullcline}(y)]^2$ in the Lyapunov function.

Unfortunately I don't think you will always be able to exploit this, so it doesn't help with the general proposition, but it might provide some insight.

Answer 2

If we have the right definitions, we can have a precise statement which is more or less what you need.

For simplicity let $z = (x,y) \in \mathbb{R}^2$ and denote by $$X(z) = f(x,y) \frac{\partial}{\partial x} + g(x,y) \frac{\partial}{\partial y}$$ the vector field corresponding to the smooth (or real analytic or polynomial) dynamical system \begin{align*} \frac{dx}{dt} &= f(x,y)\\ \frac{dy}{dt} &= g(x,y) \end{align*} which we will also write as $$\dot{z} = X(z)$$ For simplicity, we denote by $o = (o_1,o_2) \in \mathbb{R}^2$ the equilibrium point of $X(z)$, i.e. $X(o) = 0$. Furthermore, by $\phi^t(z)$ we define the phase flow of $X(z)$. That is $\phi^t(z)$ solves the initial value problem

$$\frac{d}{dt} \, \phi^t(z) = X\big(\phi^t(z)\big)$$ $$\phi^0(z) = z$$

Definition 1. Let $\gamma = \big\{(x,h(x)) \in \mathbb{R}^2 \,\, | \,\, x \in [a,b]\, \big\}$ be a compact embedded smooth (or real analytic) curve in $\mathbb{R}^2$. We call it invariant curve of the vector field $X$ whenever

$\bullet$ $\gamma$ is invariant under the flow $\phi^t(z)$, i.e. if $z \in \gamma$ then $\phi^t(z) \in \gamma$ for all $t\geq 0$, which is true if and only if the vector field $X(z)$ is tangent to $\gamma$ for each $z \in \gamma$ and points inwards at the endpoints of $\gamma$.

Definition 2. The invariant curve $\gamma$ of $X$ is called stable whenever

$\bullet$ for any open set $U$ containing $\gamma$, there exists an open set $V$ containing $\gamma$ such that if $z\in V$ then $\phi^t(z) \in U$ for all $t\geq 0$. Clearly, $V \subseteq U$.

Definition 3. The invariant curve $\gamma$ of $X$ is called attracting with basin of attraction $BA \subseteq \mathbb{R}^2$ whenever

1. $\gamma$ is stable;

2. $\gamma \subset BA$, where $BA$ is an open subset of $\mathbb{R}^2$;

3. For any open set $U$ containing $\gamma$ and any $z \in BA$ there exists $T\geq 0$ such that $\phi^t(z) \in U$ for any $t \geq T$.

4. The open set $BA$ is the maximal open set with properties 1, 2, 3 above.

Similar definitions apply to the point $o = (o_1,o_2) \in \mathbb{R}^2$.

Definition 4. The equilibrium point $o = (o_1,o_2) \in \mathbb{R}^2$ of $X$ is called stable whenever

$\bullet$ for any open set $U$ containing $o$, there exists an open set $V$ containing $o$ such that if $z \in V$ then $\phi^t(z) \in U$ for all $t\geq 0$. Clearly, $V \subseteq U$.

Definition 5. The equilibrium point $o = (o_1,o_2) \in \mathbb{R}^2$ of $X$ is called attracting with basin of attraction $BA \subseteq \mathbb{R}^2$ whenever

1. $o$ is stable;

2. $o \in BA$, where $BA$ is an open subset of $\mathbb{R}^2$;

3. For any open set $U$ containing $o$ and any $z \in BA$ there exists $T\geq 0$ such that $\phi^t(z) \in U$ for any $t \geq T$.

4. The open set $BA$ is the maximal open set with properties $1, 2, 3$ above.

For both definition of attraction, we can show that, due to its maximality, the basin of attraction $BA$ is an invariant open set for the vector field $X$, i.e. $\phi^t(z) \in BA$ for any $z \in BA$ and $t \geq 0$.

Theorem. Let $\gamma = \big\{(x,h(x)) \in \mathbb{R}^2 \,\, | \,\, x \in [a,b]\, \big\}$ be a smooth (or real analytic) invariant (compact!) curve for the smooth (real analytic or polynomial) dynamical system in $\mathbb{R}^2$ given by the vector field $X(z)$ above.

1. Let $o = (o_1,o_2)$ be the only equilibrium point of $X(z)$ lying on $\gamma$;

2. Let $\lim_{t \to \infty} \, \phi^t(z) = o$ for all $z \in \gamma$.

3. Assume that the invariant curve $\gamma$ is attracting with basin of attraction $BA \subset \mathbb{R}^2$ (open);

Then the equilibrium point $o$ is attracting with basin of attraction $BA$ and hence (asymptotically) stable.

Observe that in your example, the set $B$ is not a basin of attraction. It is not maximal. And the equilibrium point is handled only on one side of the invariant curve, while in general you should handle it from all sides (right and left up and down).