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The equilibrium equations give $y=0$ and $x-x^3=0$, so the fixed points areThe Jacobian isAt the origin its determinant is $-1$, so the origin is a saddle equilibrium. At either of the other points the determinant is two and the trace is $\epsilon (1-\alpha )$. For small positive $\epsilon$, they are unstable foci when $\alpha <1$ and stable foci when $\alpha >1$; at $\alpha =1$ the linearization is nonhyperbolic. When $\epsilon =0$, each is a center equilibrium.

For $\epsilon =0$, the system is Hamiltonian withbecause $\dot{x}=H_y$ and $\dot{y}=-H_x$. Its conservative planar phase portrait has two wells centered at $(\pm 1,0)$, a saddle at the origin, periodic level curves around each center for $-1/4<H<0$, two homoclinic loops on $H=0$, and outer periodic curves surrounding both wells for $H>0$.

For positive small $\epsilon$,On the unperturbed level $H=H_0$,The upper and lower halves contribute equally because $dt=dx/y$. Hence the energy balance for the weakly perturbed double-well oscillator isso one may take

For the right homoclinic loop, $H_0=0$, $x_1=0$, and $x_2=\sqrt{2}$. Its persistence requiresThe substitution $u=1-x^2/2$ givesThus the homoclinic balance for the weakly perturbed double-well oscillator occurs atThe left loop gives the same condition by symmetry.

For an orbit surrounding one center, the corresponding balance ratio tends to $\alpha =1$ as the orbit shrinks to the center and to $\alpha =5/4$ as it approaches the homoclinic loop. The single-well periodic orbit range for the weakly perturbed double-well oscillator is therefore

Put $\omega =\dot{\theta }$. The first-order system isWriteOn $[0,\pi ]$, $h$ is strictly increasing. Moreover,when $3\lambda <4\mu$. Thus there is one ${\theta }_{\ast }\in (0,\pi )$ with $h({\theta }_{\ast })=0$, in addition to the fixed points at $\theta =0,\pi$. Explicitly,

Let $g(\theta )=\sin \theta h(\theta )$. Linearization at $({\theta }_0,0)$ has eigenvalues satisfyingAt zero, $g^{\prime }(0)=h(0)<0$, and at $\pi$, $g^{\prime }(\pi )=-h(\pi )<0$, so both endpoints are saddles. At the interior point,so $({\theta }_{\ast },0)$ is a center. On the full angular interval there is a second center at $(-{\theta }_{\ast },0)$.

The system has the conserved energy of a conservative planar phase portrait,For $\lambda =1$, $\mu =3/2$,The phase portrait on the cylinder consists of centers at $\theta =\pm \arccos (-11/16)$ surrounded by closed periodic-energy curves, with saddles at $\theta =0$ and the identified point $\theta =\pi =-\pi$. The saddle-energy contours form the separatrices between librations in the potential wells and trajectories crossing the lower barrier.

If $3\lambda >4\mu$, then $h(0)>0$ and monotonicity gives no interior zero on $(0,\pi )$. The point $(0,0)$ becomes a center, while $(\pi ,0)$ remains a saddle. At $3\lambda =4\mu$, the two off-axis centers coalesce with the origin and the linearization there is degenerate.

When $k=0$, the system is conservative andis constant. The fixed points areThe potential $U$ has a strict minimum $U(0)=0$, strict maximaand tends to zero as $|x|\to \infty$. The conservative planar phase portrait therefore has:

The portrait is symmetric under $(x,y)\mapsto (-x,-y)$ and under time reversal $(x,y,t)\mapsto (x,-y,-t)$.