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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?

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  • $\begingroup$ So, is your problem about constructing Lyapunov function for proving global stability or in proving global stability no matter what? Because you seem to be very focused on constructing precious analytic expression for Lyapunov function here. Added later: also, note that this equilibrium is not really globally attractive: there is a set of points (namely, stable manifold of saddle at the origin) that go to this saddle. And the whole left halfplane goes just to infinity. $\endgroup$
    – Evgeny
    Oct 26, 2016 at 9:33
  • $\begingroup$ @Evgeny I'm interested in proving "global" stability in general (not constructing a Lyapunov function per say), but what I really mean by global stability is that the basin of attraction of (1,0) contains the whole specified domain {(x,y) : 0<x<1, y≥0} $\endgroup$ Oct 26, 2016 at 23:26
  • $\begingroup$ What is your precise definition of a globally attractive invariant curve $C$? This result depends on this definition. $\endgroup$ Nov 2, 2016 at 3:51
  • $\begingroup$ @Futurologist For the general statement, for all x0 in B, you can find a t such that for any epsilon x(t) gets epsilon close to a point on C using euclidean distance. However, a discussion on the how much this definition matters would be useful. $\endgroup$ Nov 2, 2016 at 4:09

2 Answers 2

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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).

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    $\begingroup$ I don't know about OP's interest, but mine is of course about proof of this theorem :) $\endgroup$
    – Evgeny
    Nov 3, 2016 at 9:35
  • $\begingroup$ Also, it is kind of nitpicking, but for me it seems that there is a minor contradiction in your conditions. If this invariant curve is compact and it is homeomorphic to unit segment, then its endpoints must be equilibria -- there's got to be something that prevents trajectory "to leak out" through boundary of this segment. So, you can't say that for all initial conditions on this invariant curve trajectories go to the equilibrium $o$. But I think that this might be fixable. $\endgroup$
    – Evgeny
    Nov 3, 2016 at 9:47
  • $\begingroup$ @Evgeny The invariant curve $\gamma$ is not assumed to be a maximal invariant manifold of the vector field $X$. Otherwise, you cannot really define it as a graph of $y=h(x)$ (e.g. homoclinic trajectory or heteroclinic connected to another equilibrium of type repelling sink). I have only chosen a compact portion of it around the equilibrium point $o$, where it can be written as $y=h(x)$. By assumption, the vector field $X$ restricted to $\gamma$ forms a one dimensional dynamical system with asymptotically stable equilibrium $o \in \gamma$, i.e. $\gamma$ is an invariant domain for $o$. $\endgroup$ Nov 3, 2016 at 12:24
  • $\begingroup$ @Evgeny Basically the vector field $X$ restricted to $\gamma$ points always towards the equilibrium point $o$ (otherwise $o$ is unstable in two dimensional system $X$) so no leakage can occur. And that's why $\gamma$ is an invariant domain of $X |_{\gamma}$. $\endgroup$ Nov 3, 2016 at 12:30
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    $\begingroup$ Okay, let's leave the terminological issues. For me it's important: do you have the proof, or you've skipped it for some reasons? $\endgroup$
    – Evgeny
    Nov 3, 2016 at 15:48
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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.

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