Stability of the null solution of system of differential equations

Question

I know how to do the phase portraits for linear system of equation with constant coefficeints and also to linearize the non linear systems. But If I have to check the stability of the null solutions of the followingemphasized text type : $$x'(t) = -y\cos (x)$$ $$ y'(t) = \sin(x). $$ Then I am confuse to linearize it first or should I find the lyapunov function but how to find its lyapunov function. Any idea is helpful.

Answer 1

Linearization is useless here because the matrix of the linearized system $$\left. \left( \begin{array}{rr} y\sin x&-\cos x\\ \cos x&0 \end{array}\right)\right|_{(0,0)}= \left( \begin{array}{rr} 0&-1\\ 1&0 \end{array}\right) $$ has the pure imaginary eigenvalues $\pm i$.

Suppose the Lyapunov function is of the form $$ V(x,y)=\phi(x)+\psi(y), $$ then $$ \dot V= -\phi'(x) y\cos x +\psi'(y) \sin x. $$ It is easy to observe that $\dot V$ vanishes when $$ \phi'(x)=\tan x,\quad \psi'(y)=y, $$ thus, $$ V(x,y)=\frac{y^2}2-\ln \cos x $$ is a Lyapunov function of our system and also it is its first integral.

Hence, the origin is stable (but not asymptotically stable).