Find the Lyapunov function

Question

We have the dynamical system $$ \dot{x}=-x-2 y^{6}, \quad \dot{y}=3 x^{3} y-2 y^{9} $$ Can we find an Lyapunov function to prove that $(0,0)$ is an asymptotically stable equilibrium?

(I'm really new on this subject)

Answer 1

I find (by inspection) that $V(x, y) = 9x^4 + 4y^6 $ a Lyapunov function, since \begin{align}-\frac{dV(x(t), y(t))}{dt}&=36x^3(x+2y^6)+ 24y^5(-3x^3y + 2y^9) \\&= 36x^4 + 72x^3y^6 - 72x^3y^6 + 48y^{14} \\ &= 36x^4 + 48y^{14} \end{align}

Therefore the origin is globally asymptotically stable.

My strategy is that try to find polynomial functions $V$ for which $\frac{dV(x(t), y(t))}{dt}$ is a sum of square expressions. Here is an another exercise similar to your one.

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Put $V(x, y) = x^2 + \beta y^2$ and $$G(x, y):=-\frac12\frac{dV(x(t), y(t))}{dt}=x^2+2xy^6+2\beta y^{10}-3\beta x^3y^2$$ Let $B_r$ be the open ball of radius $r$ centered at origin.

[Case 1: $\beta \leq 0$]

In this case, $\beta$ is not the one you seek because of the following:

For any $r>0$, there exists $(x, y) \in B_r$ such that $G(x, y)<0$

I'll leave the proof to you. (Hint: If $\beta = 0 $, take small $y>0$ and put $x = -y^7$. Otherwise, put $x=0$ and see what happens.)

[Case 2: $\beta > 0 $]

You can use $V$ to show the origin is locally asymptotically stable, since the following holds:

For given $\beta>0$, there exists $r>0$ such that $(x, y) \in B_r \setminus \{ 0 \}$ implies $G(x, y)>0$

proof: Take $r>0$ small so that $6\beta > 4r^2$ and $3 \beta xy^2 \leq 1$ holds for all $(x, y) \in B_r$. Now let $(x, y) \in B_r \setminus \{ 0 \}$ be given. If $x=0$ or $y=0$, it is trivial that $G(x, y)>0$. Suppose both $x$ and $y$ are nonzero. If $x>0$, our choice of $r$ forces $x^2+2xy^6+2\beta y^{10} > x^2 \geq 3\beta x^3y^2$ so $G(x, y) > 0$. If $x<0$, put $-x = t$. We have to show $t^2 - 2ty^6 + 2\beta y^{10} > -3\beta t^3y^2$.

Set $f(t) = (3 \beta y^2)t^3 + t^2 - (2y^6)t + 2\beta y^{10}$ Then $f'(t) = (9\beta y^2) t^2 + 2t - 2y^6$, so $f$ attains its local extrema at $$t= \frac{ -1 \pm \sqrt{1+18 \beta y^8}}{9 \beta y^2}$$ So it suffices to show that $f(s)>0$ where $$s= \frac{ -1 + \sqrt{1+18 \beta y^8}}{9 \beta y^2}$$ Since $(9\beta y^2)s^2 =2y^6 - 2s$, we have $(3\beta y^2)s^3 = \frac23 sy^6 - \frac 23 s^2$. Hence $f(s) = \frac13 s^2 - \frac 43 y^6 s + 2\beta y^{10}$. Again, $r>0$ satisfies $6\beta > 4r^2 > 4y^2 $ so $6\beta y^{10} > 4y^{12}$. Finally, $3f(s) = s^2 - 4y^6 s + 6\beta y^{10} > s^2 - 4y^6 + 4y^{12} \geq 0$. $\blacksquare$

However, for any $\beta>0$, there exists $(x, y) \in \mathbb R^2$ such that $G(x,y) < 0 $. For example, put $x = y^3$ and letting $y \rightarrow \infty$. In other words, you cannot use $V$ to show the global asymptotical stability whatever $\beta$ is.