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

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


$$u = -x_1 -x_2$$ $$ \dot{V(x)} = -x^2_2 \leq 0 $$ 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

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

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

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  • $\begingroup$ 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 $\endgroup$ – Pjdeepu Oct 18 at 10:51

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