Lyapunov function for a non linear 3 dimensional system

Question

How it is possible to find a Lyapunov function for the following system? \begin{cases} \dot {x_1}=x_2+x_3 & \\ \dot {x_2}=-\sin x_1-x_3 & \\ \dot {x_3}=-\sin x_1+x_2 & \end{cases} I have tried with $V(x)=\frac{1}{2}x^TPx+\int_0^{x_1} \sin{(s)} ds$ but I can't find out the right elements of P in order to obtain $\dot{V}<0$. Can you help me?

Answer 1

This system is not asymptotically stable. Near the origin we have $\sin x\approx x$ so the dynamics are practically

$$ \cases{ \dot x_1 = x_2 + x_3\\ \dot x_2 = -x_1 - x_3\\ \dot x_3 = -x_1 + x_2 } $$

or

$$ \cases{ x_1\dot x_1 = x_1x_2 + x_1x_3\\ x_2\dot x_2 = -x_1x_2 - x_2x_3\\ x_3\dot x_3 = -x_1x_3 + x_2x_3 } $$

and after adding up

$$ x_1^2+x_2^2+x_3^2 = C $$

those orbits represent a center at the origin.

Answer 2

The Lyapunov function (and also the first integral) of the system is $$ V(x_1,x_2,x_3)= 1-\cos x_1+\frac12 x_2^2+\frac12 x_3^2. $$ Indeed, $$ \dot V(x_1,x_2,x_3)= \sin x_1\cdot x_1+x_2\dot x_2+x_3\dot x_3 $$ $$ = \sin x_1 \cdot (x_2+x_3)+x_2(-\sin x_1-x_3)+x_3(-\sin x_1+x_2)=0 $$ Since $V$ is positive definite and $\dot V$ is nonpositive, the origin is stable. It is not asymptotically stable because $V$ remains constant on the solutions of the system (and doesn't tend to zero).