Suppose we have a nonautonomous nearly integrable Hamiltonian system, periodic in $t$ with period $2\pi / \omega$

$$H_{\epsilon}(x,y,t)=H_{0}(x,y) + \epsilon H_{1}(x,y,t)$$

with $(x,y,t) \in \mathbb{R}^{2} \times \mathbb{S}^{1}$. Denote by $q_{0}(t)$ the homoclinic orbit to a fixed point of the unperturbed Hamiltonian. We know that we may write the nonatonomous three-dimensional vector field as

$$\dot{x} = \frac{\partial H_{0}}{\partial y} + \epsilon \frac{\partial H_{1}}{\partial y}$$ $$\dot{y} = -\frac{\partial H_{0}}{\partial x} - \epsilon \frac{\partial H_{1}}{\partial x}$$ $$\dot{\phi} = \omega$$

Then we have the Melnikov function $$M(t_{0}, \phi_{0}) = \int_{-\infty}^{\infty} \{ H_{0}, H_{1} \} (q_{0}(t),\omega t + \omega t_{0},0) dt$$

How would the expression for $M(t_{0},\phi_{0})$ change if we consider the augmented symplectic phase space with an extra variable $E$ conjugated to $t$? In other words we introduce

$$\tilde{H} = H_{\epsilon}(x,y,t)-E$$


$$\dot{E} = \frac{\partial \tilde{H}}{\partial t}$$

Now we will have a two dimensional normally hyperbolic invariant manifold, with the homoclinic orbit $q_{0}$ giving rise to three-dimensional stable and unstable manifolds in 4 dimensional space. So we still have a scalar Melnikov function, but how does it change? Logically it should be the same, but how does one account for the extra $\dot{E}$ vector component?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.