# Problems finding a Lyapunov function

I've been trying to find the Lyapunov function for the following system,

$$\dot x= x^{2}-x-y$$, $$\dot y=x$$.

Since Ive tried all typical ones, and some variations I havent been able to conclude that the derivative of the function is $$<0$$. The main question, I guess, is how to face the problem of finding a Lyapunov function when some term is even, which is the case of $$\dot x$$.

Probably there is a clever way to find this kind of functions and for sure Im missing it so, any help would be incredibly appreciated <3

• All kinds of $ax^{2n}+bx^{2n}$(not ALL but lots haha), some $(ax+by+c)^{2}+dy^{2}+e$ and some others which I've been adapting according the derivatives I got haha :/ I dont know a better way to find them :/ Nov 11, 2019 at 5:15

The equilibrium of this system is not globally asymptotically stable:

so there is no "global" Lyapunov function. A "local" Lyapunov function can be found by linearizing the original system. The linearized system is $$\dot x= -x-y,$$ $$\dot y=x$$ or $$\left(\begin{array}{c}\dot x\\\dot y\end{array}\right)= A \left(\begin{array}{c}x\\y\end{array}\right),$$ where $$A=\left(\begin{array}{rr}-1&-1\\1&0 \end{array}\right).$$ Solving the Lyapunov equation $$A^TP+PA=-Q$$, where $$P$$ is an unknown matrix, $$Q$$ is any positive-definite matrix, one can obtain the solution $$P$$, which can be used to construct the Lyapunov function $$V(x,y)=\left(\begin{array}{cc}x&y\end{array}\right) P \left(\begin{array}{r}x\\y \end{array}\right).$$ Update begin

$$V$$ obtained in this way is not only a Lyapunov function for the linearized system $$\tag{1}\dot x=Ax,$$ but for the original system $$\tag{2}\dot x= f(x)$$ too. Indeed, the original system can be written as $$\tag{2a}\dot x=Ax+\alpha(x)\|x\|,$$ where $$\lim_{x\to 0}\alpha(x)=0$$. The directional derivative $$\dot V\Big|_{(2a)}= (Ax + \alpha (x)\|x\|)^T Px+x^T P(Ax + \alpha (x)\|x\|)$$ $$=x^T(A^T P+ PA)x + 2x^T P\alpha (x)\|x\|= -x^T Qx + 2x^T P\alpha (x)\|x\|.$$ The inequality $$x^T Qx \geq \lambda_{\min}(Q)\|x\|^2,$$ where $$\lambda_{\min}(Q)$$ is the minimum eigenvalue of $$Q$$, implies $$\dot V\Big|_{(2a)}\leq -\lambda_{\min}(Q)\|x\|^2+ \Big|2x^T P\alpha (x)\|x\|\Big| \leq -\lambda_{\min}(Q)\|x\|^2+ 2\|\alpha (x)\|\cdot\|P\|\cdot\|x\|^2$$ $$=\|x\|^2 \left( -\lambda_{\min}(Q)+2\|\alpha (x)\|\cdot\|P\| \right).$$ Now notice that $$\|\alpha (x)\|$$ can be made arbitrarily small in some neighborhood of the origin because $$\lim_{x\to 0}\alpha(x)=0$$, hence, $$\dot V\Big|_{(2a)}$$ is negative definite.

This is actually a part of the proof of the indirect Lyapunov method.

Update end

For example, if we take $$Q=\left(\begin{array}{rr}2&0\\0&2 \end{array}\right),$$ we obtain $$V(x,y)=2x^2+2xy+3y^2$$.

• Thanks so much for your clear answer, but what was the intuition behind the application of the Lyapunov equation? Why did you think in that? Nov 11, 2019 at 19:55
• @user1trill I have updated the answer
– AVK
Nov 11, 2019 at 20:46
• Amazing answer dude! Thanks so much <3 Nov 11, 2019 at 22:38
• @user1trill Nothing is wrong, this is correct. The "local" Lyapunov function does not have to have a negative derivative everywhere, the derivative should be negative only in some neighborhood of zero
– AVK
Nov 16, 2019 at 5:21
• @user1trill The proof for the general case is given in my answer and it is completely rigorous. If you want а proof for your specific system, consider $\dot V= -2(x^2+y^2)\left(1-\frac{x^2}{x^2+y^2}(2x+y)\right)$ and use the inequality $\frac{x^2}{x^2+y^2}\le 1$.
– AVK
Nov 16, 2019 at 19:16