= Solution
Here the <autonomous planar system> is $\dot x=v$, $\dot v=-\cos x$. Its <equilibrium points> are
$$
\boxed{(x_*,v_*)=(\pi/2+n\pi,0),\qquad n\in\mathbb Z.}
$$
The <Jacobian matrix> is
$$
J_*=\begin{pmatrix}0&1\\ \sin x_*&0\end{pmatrix},
\qquad \mu^2=\sin x_*=(-1)^n.
$$
For even $n$, the <eigenvalues> are $1,-1$, so these points are unstable <saddle equilibria>. Writing $\xi=x-x_*$, their local unstable and stable directions are $v=\xi$ and $v=-\xi$ respectively. For odd $n$, the <eigenvalues> are $i,-i$.
To classify the latter points nonlinearly, differentiate the <first integral>
$$
H(x,v)=\frac{v^2}{2}+\sin x:
\qquad \dot H=v(-\cos x)+\cos x\,v=0.
$$
At $x_*=3\pi/2+2m\pi$ this <energy> has a strict local minimum. Its nearby <level sets> are closed curves, proving that the odd-$n$ points are stable <center equilibria>, with no attraction to the center. The <phase portrait> consists of closed oscillatory orbits for $-1<H<1$, <heteroclinic orbits> at $H=1$ joining consecutive <saddle equilibria>, and open rotating orbits for $H>1$. The separating curves have
$$
v=\pm\sqrt{2(1-\sin x)}.
$$
Above the horizontal axis $\dot x=v>0$, so flow is rightward; below it flow is leftward. The closed orbits are clockwise, and these same signs orient every separating and rotating orbit.
Back to article page
