# How to pick a Lyapunov Function and make asymptotically stable?

Dynamics I have , $$\dot{x_1} = x_2$$

$$\dot{x_2} = u$$ if I pick Lyapunov function $$V(x) = \frac{1}{2}*x^2_1 + \frac{1}{2}*x^2_2$$ then

$$\dot{V(x)} = x_1*\dot{x_2} +x_2*u$$

and

$$u = -x_1 -x_2$$ $$\dot{V(x)} = -x^2_2 \leq 0$$ But this in not asymptotically stable. How to make it ? Either we can change u or V(x) How to do that?

You could pick the control law as follows

$$u(x) = -x_1 - x_2 - \frac{x_1^2}{x_2},$$

but that is not well defined when $$x_2=0$$.

Instead one could make use of the fact that your proposed control law makes the system dynamics linear, with

$$\dot{x} = A\,x,$$

$$A = \begin{bmatrix}0 & 1 \\ -1 & -1\end{bmatrix}.$$

A Lyapunov function for such system can be found of the form

$$V(x) = x^\top P\,x,$$

with $$P$$ positive definite which satisfies the Lyapunov equation

$$A^\top P + P\,A = -Q,$$

with $$Q$$ positive definite. If $$A$$ is stable any positive definite $$Q$$ should also yield a corresponding positive definite $$P$$.

For example when setting $$Q$$ equal to the identity matrix yields

$$V(x) = \frac{1}{2} (3\,x_1^2+2\,x_1\,x_2+2\,x_2^2) = \frac{1}{2} (2\,x_1^2+(x_1+x_2)^2+x_2^2),$$

$$\dot{V}(x) = -x_1^2 - x_2^2.$$

• Thank you for your response. I can proceed with this control law except when x_2 becomes zero. The second procedure is helpful, I will try to proceed with that and see what happens. Thank You – Pjdeepu Oct 18 at 10:51