Finding Lyapunov function to show origin is stable

Question

Consider the system \begin{cases} \dot{x}=-2y+yz \\ \dot{y}=x-xz \\ \dot{z}=xy \end{cases} I need to find a Lyapunov function of the form $V(x,y,z)=ax^2+bx^2+cz^2$ to show that the origin $(0,0,0)$ is Lyapunov stable. How do I choose the constants $a,b,c$?
I need $V(0,0,0)=0$, which is true and $V(x,y,z)>0$ is any neighbourhood of the origin, which is true as long as $a,b,c>0$.
Then $\dot{V}(x,y,z)=2ax\dot{x}+2by\dot{y}+2cz\dot{z}=-4axy+2axyz+2bxy-2bxyz+2cxyz$.
I can take $b=c$ to get $\dot{V}(x,y,z)=-4axy+2axyz+2bxy$.
What should I do next?

Answer 1

Instead of imposing $b=c$ just collect the similar terms in the expression $\nabla V\cdot X$ you have just computed so that you just need to solve the non-singular linear system: \begin{cases} -4a+2b=0 \\ 2a-2b+2c=0. \end{cases} At this point you will get infinitely many Lyapunov functions indexed by $a\in\mathbb{R}$ for instance and you can conclude.