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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