I have the system
$x'=\sin y = f(x,y)$
$y'=-2x-3y = g(x,y)$
I am given the Lyapunov function $V(x,y)=15x^2+6xy+3y^2$.
Obviously, $V(0,0)=0$. Also, $V(x,y)=6x^2+9x^2+2 \cdot 3x y + y^2 + 2y^2 = 6x^2+2y^2+(3x+y)^2>0$, $ \forall (x,y) \in R^2 \setminus \{(0,0)\}$.
From Maple I know that the origin is asymptotically stable. But I can't show that
$\dot V(x,y) = \displaystyle \frac{\partial V}{\partial x} f(x,y) + \displaystyle \frac{\partial V}{\partial y} g(x,y)=(30x+6y)\sin y + (6x+6y)(-2x-3y) = ... <0$.
I would appreciate any help. Thanks.
A Lyapunov function can be found by linearizing the original system at the origin. The linearized system is $$\tag{1} \left\{\begin{array}{lll} \dot x &=& y\\ \dot y&=&-2x-3y \end{array}\right. $$ or $$ \left(\begin{array}{c} \dot x\\ \dot y \end{array}\right)=A \left(\begin{array}{c} x\\y \end{array}\right), \quad\mbox{where } A=\left(\begin{array}{rr} 0 & 1\\ -2& -3 \end{array}\right). $$ A Lyapunov function for (1) is also a Lyapunov function for the original system (why this is so, is described, for example, in this answer). The quadratic form $$ V(x,y)=\left(\begin{array}{cc}x&y\end{array}\right) P \left(\begin{array}{r}x\\y \end{array}\right) $$ is a Lyapunov function for (1) if and only if $P>0$ and the matrix $Q=A^TP+PA$ is negative definite. We have $$ P=\left(\begin{array}{cc} 15 & 3\\ 3& 3 \end{array}\right),\quad Q=A^TP+PA=\left(\begin{array}{rr} -12 & 0\\ 0& -12 \end{array}\right), $$ thus, $V(x,y)$ is a Lyapunov function for the original system.