Melnikov method

Remarks

The integral converges because $f(q_0(t)) \to f(p) = 0$ exponentially fast as $t \to \pm\infty$. The function $M$ is $T$-periodic in $t_0$. For small $\varepsilon$, the perturbed system has a hyperbolic periodic orbit near $p$. The distance between its stable and unstable manifolds, measured along the normal to the unperturbed loop at $q_0(0)$, is $\varepsilon M(t_0) / \norm{f(q_0(0))} + O(\varepsilon^2)$, with a suitable sign convention. The basic theorem says: if $M$ has a simple zero, then for small $\varepsilon \ne 0$ these manifolds intersect transversally; if $M$ has no zero, they do not intersect.

Example: for $H = \frac{1}{2} y^2 - \frac{1}{2} x^2 + \frac{1}{4} x^4$ and $g = (0, -\delta y + \gamma \cos(\omega t))$, the system is the damped, forced Duffing equation. Here $q_0(t) = (\sqrt{2} \operatorname{sech} t, -\sqrt{2} \operatorname{sech} t \tanh t)$, and a direct computation gives $M(t_0) = -\frac{4}{3} \delta + c\, \gamma \sin(\omega t_0)$ with a constant $c$ that depends only on $\omega$.

Non-example: if $g(u, t) = (0, t)$, then $g$ is not periodic in $t$. The perturbed system is then not periodic in time, there is no Poincaré map of period $T$, and the theorem above does not apply.