Phase portrait of $\ddot{x} + \sin(x)=0$ near the origin.

Question

I am trying to sketch the phase portrait of the second order ODE describing a pendulum near the origin:

$$\ddot{x}+\sin(x)=0.$$

I write this as the first order system:

$$ \begin{aligned} \dot{x} &= y \\ \dot{y} &= -\sin(x) \end{aligned} $$

The Jacobian at the origin is given by:

$$ J(0,0) = \begin{pmatrix} 0 & 1\\ -1 & 0 \end{pmatrix} $$

thus giving us eigenvalues $\pm i$ and resulting in a non-hyperbolic fixed point. The system has energy

$$ E = \frac{1}{2}y^{2}+1-\cos(x) $$

and so it is a gradient field. A well-known result is that gradient fields do not have cycles. However, searching online, I came across the plot:

https://www.researchgate.net/figure/Potential-well-of-the-simple-pendulum-and-the-phase-portrait-in-the-absence-of-friction_fig1_258272363

I am not sure where I went wrong. Any help would be greatly appreciated.

Answer 1

No, it is not a gradient field. In particular, it is not the gradient of the energy $E$. On the contrary: the energy $E$ is conserved. Thus the trajectories are the curves of constant $E$.