Melnikov energy-balance method

For a weak perturbation of a planar Hamiltonian system, integrate the first-order orbital derivative of the Hamiltonian along an unperturbed homoclinic orbit. This measures the first-order separation of stable and unstable manifolds on a transverse section. A simple zero as a parameter varies permits a nearby homoclinic connection by the implicit function theorem. The resulting leading balance is not a claim that the unperturbed separatrix remains an exact orbit after perturbation.

For $\dot{q}=u_0(q)+ϵu_1(q,t)$ with planar Hamiltonian flow $u_0=(-{\partial }_y{\psi }_0,{\partial }_x{\psi }_0)$, let $q_0$ be a heteroclinic orbit between two hyperbolic fixed points. Its first-order displacement satisfies ${\dot{q}}_1=D{u}_0{q}_1+u_1$. The covector $w=\nabla {\psi }_0(q_0)$ satisfies $\dot{w}=-(D{u}_0{)}^{T}w$, obtained by differentiating $\nabla {\psi }_0\cdot u_0=0$. Thus $(w\cdot q_1{)}^{\prime }=w\cdot u_1$. Integrating the stable correction from the future and the unstable correction from the past gives the displayed function. At a regular section the first-order normal separation is $ϵM/|\nabla {\psi }_0|$. A simple zero yields a transverse intersection of the perturbed manifolds, by the implicit function theorem.

Perturb the first integral of the quadratic fold-Hopf amplitude flow by $u^{\prime }=-2uv+\epsilon u({\lambda }_1+v^2)$, keeping $v^{\prime }={\lambda }_2+v^2+u^2$. Then $F^{\prime }=-\epsilon u({\lambda }_1+v^2)({\lambda }_2+u^2+v^2)$. On the interior heteroclinic orbit for ${\lambda }_2=-a^2$, $u=\sqrt{3(a^2-v^2)}$ and $v^{\prime }=2(a^2-v^2)$, so the first-order change is $-\epsilon \sqrt{3}{\int }_{-a}^a({\lambda }_1+v^2)\sqrt{a^2-v^2}dv=-\epsilon \pi \sqrt{3}{a}^2({\lambda }_1+a^2/4)/2$. The boundary connection contributes zero. The simple zero gives the displayed leading persistence curve, rather than an exact all-orders parameter equality.

For the imperfect soft Duffing-van der Pol oscillator at positive small $\lambda$ and $|\epsilon |\ll \sqrt{\lambda }$, the symmetric connection of the Hamiltonian system is $v=\sqrt{\lambda }\tanh (\sqrt{\lambda /2}t)$. Its damping integral is $4{\lambda }^{3/2}(\kappa -\lambda /5)/(3\sqrt{2})$. The two saddle equilibrium barrier heights differ by $-4\epsilon {\lambda }^{3/2}/3$ to first order. Equating the energy change to this difference gives the two oppositely directed connection curves in the display. These are weak-damping asymptotic curves, not exact global formulas at order-one parameters.

// Destination: astrophysical-fluid-dynamics.bigb

Perturb the Hamiltonian blow-up of a reflection-symmetric double-zero point by $u^{\prime }=-sv+v^3/2+\epsilon (\mu -v^2/2)u+O({\epsilon }^2)$ and $v^{\prime }=2u+\epsilon (\mu -v^2/2)v+O({\epsilon }^2)$. Then $H^{\prime }=\epsilon (\mu -v^2/2)(2u^2+s{v}^2-v^4/2)+O({\epsilon }^2)$. On the positive heteroclinic branch $dt=dv/[\sqrt{2}(s-v^2/2)]$, so its first energy change is $\epsilon M/\sqrt{2}$, where $M={\int }_{-\sqrt{2s}}^{\sqrt{2s}}(\mu -v^2/2)(s+v^2/2)dv$. A simple zero gives cancellation of the leading stable/unstable manifold splitting and continues to a nearby heteroclinic bifurcation. Since $\partial M/\partial \mu >0$, there is a unique first-order balance for fixed $s>0$.

For $u^{\prime \prime }=u^2-\kappa -\epsilon (\lambda +u)u^{\prime }$ with $\kappa >0$, the Hamiltonian homoclinic orbit has $H=2{\kappa }^{3/2}/3$. Its energy balance is $\lambda \int v^{2}dt+\int u{v}^{2}dt=0$. Along this orbit the ratio of the second integral to the first is $-5\sqrt{\kappa }/7$, giving the displayed simple zero and hence the leading homoclinic parameter curve.

The conservative equation $u^{\prime \prime }=u^2-a^2$ has a saddle loop with $v_{+}(u)=\sqrt{2/3}(a-u)\sqrt{u+2a}$ for $-2a\le u\le a$. Its homoclinic orbit integrals are $I_0=2{\int }_{-2a}^a{v}_{+}(u)du$ and $I_1=2{\int }_{-2a}^{a}u{v}_{+}(u)du$. Substitution $s=u+2a$ evaluates them as displayed. For a weak perturbation $\epsilon (\beta +u)u^{\prime }$, integrating the energy derivative gives $\beta I_0+I_1=0$, hence $\beta =5a/7$. The nonzero derivative $I_0$ makes this a simple first-order splitting zero.