# Show that the equilibrium point $(0,0)$ is asymptotically stable and an estimate of its basin of attraction

Consider the system \begin{aligned} \dot{x} &=-y-x^3+x^3y^2\\ \dot{y}&=x-y^3+x^2y^3\end{aligned} Show that the equilibrium point $$(0,0)$$ is asymptotically stable and an estimate of its attractiveness basin.

Clearly the point $$(0,0)$$ is a point of equilibrium and also $$D_f(x,y)=\begin{pmatrix}-3x^2+3y^4 & -1+2yx^3\\ 1+2xy^3 &-3y^2+3y^2x^2 \end{pmatrix}$$, so $$D_f(0,0)=\begin{pmatrix}0 & -1\\1 &0 \end{pmatrix}$$ and has eigenvalues ​​$$\pm i$$ So, I can not conclude anything of stability for this point, I need a Liapunov function and I do not know what it is or how to find it, could someone help me please? Thank you very much.

• Always try the simplest one first. For example, quadratic. More precisely, $V(x, y) = x^2+y^2$. Oct 30 '17 at 5:44

In addition to what has been said by @Evgeny and @MrYouMath: the set $$M=\left\{ (x,y)\in\mathbb R^2 :\; x^2+y^2<2 \right\}$$ is a positively invariant set of the considered system since $\forall (x,y)\in M$ $$\dot V=-x^4-y^4+x^2y^2(x^2+y^2)<-x^4-y^4+2x^2y^2=-(x^2-y^2)^2\leq 0;$$ it is also a subset (guaranteed estimation) of the domain of attraction.

• Nice one! This estimate is even better than $\| (x, y) \|_{\infty} < 1$ . Oct 30 '17 at 18:06
• +1: Very nice observation. This might be useful in the future. Oct 31 '17 at 14:03
• How Do I show that $M$ is positively invariant with this fact?
– P.G
Nov 20 '20 at 13:44
• @P.G Just as it is done in the proof of the Lyapunov stability theorem
– AVK
Nov 21 '20 at 7:02

As @Evgeny suggested you could use the Lyapunov function candidate

$$V(x,y)=\dfrac{1}{2}\left[x^2+y^2\right].$$

$V(x,y)$ is clearly positive definite at the origin and radially unbounded (which we would need for assessing global asymptotic stability).

The time derivative of $V(x,y)$ is given by $$\dot{V}=x\dot{x}+y\dot{y}=x(-y-x^3+x^3y^2)+y(x-y^3+x^2y^3)$$ $$\dot{V}=-x^4-y^4+x^2y^2(x^2+y^2)=-(1-y^2)x^4-(1-x^2)y^4.$$

As the lower order terms $-x^4-y^4$ are negative definite we can conclude that the equilibrium point is asymptotically stable in a region around the origin. Using the comment given by @Evgeny we can see that this expression is negative semidefinite if $(x,y)$ lie inside the unit circle $D=\{(x,y)\in \mathbb{R}^2|x^2+y^2<1\}$. This is the basin of attraction (correction due to @Artem).

We cannot say if the origin is globally asymptotically stable because $\dot{V}$ is not negative definite for all regions around the origin. This does not mean that the origin cannot be globally asymptotically stable. It just means we can only show (local) asymptotic stability of the origin with $V(x)$ as our Lyapunov function candidate.

• Actually, you can do better if regroup terms correctly: $\dot{V} = -x^4(1-y^2)-y^4(1-x^2)$. This also gives a rough estimate of basin of attraction since this expression is negative when $\vert x \vert < 1$ and $\vert y \vert < 1$. Oct 30 '17 at 13:18
• @Evgeny: You are right. I felt that I should be able to factor this expression. Oct 30 '17 at 13:19
• Also, speaking about radial unboundedness: it's not a necessary condition really, see recent discussion in comments. Oct 30 '17 at 18:09
• It is not necessary for asymptotic stability but it is necessary for global asymptotic stability (see the comment in the brackets in my answer). I think that is what the example showed. Or do you mean something else? Oct 30 '17 at 18:14
• Even for the global asymptotic stability it is not necessary. You can prove the global asymptotic stability with a Lyapunov function which is not radially unbounded. Moreover, you can transform the image of any Lyapunov function such that it will become bounded but still will prove that something is globally asymptotic stable. Oct 30 '17 at 20:29

This problem can be handled with an optimization procedure, having in mind that generally is a non convex problem. The result depends on the test Lyapunov function used so we will generalize to a quadratic Lyapunov function

$$V(p) = p^{\dagger}\cdot M\cdot p = a x^2+b x y + c y^2,\ \ \ p = (x,y)^{\dagger}$$

and

$$f(p) = \{-y - x^3 + x^3 y^2, x - y^3 + x^2 y^3\}$$ with $$a>0,c>0, a b-b^2 > 0$$ to assure positivity on $$M$$. We will assure a set involving the origin $$Q_{\dot V}$$ such that $$\dot V(Q_{\dot V}) < 0$$. The optimization process will be used to guarantee a maximal $$Q_{\dot V}$$.

After determination of $$\dot V = 2 p^{\dagger}\cdot M\cdot f(p)$$ we follow with a change of variables

$$\cases{ x = r\cos\theta\\ y = r\sin\theta }$$

so $$\dot V = \dot V(a,b,c,r,\theta)$$. The next step is to make a sweep on $$\theta$$ calculating

$$S(a,b,c, r)=\{\dot V(a,b,c,r,k\Delta\theta\},\ \ k = 0,\cdots, \frac{2\pi}{\Delta\theta}$$

and then the optimization formulation follows as

$$\max_{a,b,c,r}r\ \ \ \ \text{s. t.}\ \ \ \ a > 0, c> 0, a c -b^2 > 0, \max S(a,b,c,r) \le -\gamma$$

with $$\gamma > 0$$ a margin control number.

Follows a MATHEMATICA script which implements this procedure in the present case.

f = {-y - x^3 + x^3 y^2, x - y^3 + x^2 y^3};
V = a x^2 + 2 b x y + c y^2;
dV = Grad[V, {x, y}].f /. {x -> r Cos[t], y -> r Sin[t]};
rest = Max[Table[dV, {t, -Pi, Pi, Pi/30}]] < -0.1;
rests = Join[{rest}, {r > 0, a > 0, c > 0, a c - b^2 > 0}];
sols = NMinimize[Join[{-r}, rests], {a, b, c, r}, Method -> "DifferentialEvolution"]
rest /. sols[[2]]

dV0 = Grad[V, p].f /. sols[[2]]
V0 = V /. sols[[2]]
r0 = 2;
rmax = r /. sols[[2]];
gr0 = StreamPlot[f, {x, -r0, r0}, {y, -r0, r0}];
gr1a = ContourPlot[dV0, {x, -r0, r0}, {y, -r0, r0}, ContourShading -> None, Contours -> 80];
gr1b = ContourPlot[dV0 == 0, {x, -r0, r0}, {y, -r0, r0}, ContourStyle -> Blue];
gr2 = ContourPlot[x^2 + y^2 == rmax^2, {x, -r0, r0}, {y, -r0, r0}, ContourStyle -> {Red, Dashed}];
Show[gr0, gr1a, gr1b, gr2]


Follows a plot showing in black the level sets $$Q_{\dot V}$$ an in blue the trace of $$\dot V = 0$$. In dashed red is shown the largest circular set $$\delta = 1.42486$$ defining the maximum attraction basin for the given test Lyapunov function's family.