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I am interested in periodic orbits of mechanical systems of second-order dynamics with no damping, i.e. governed by an equation of the type \begin{equation}(1)\quad \ddot x + f(x)=0 \end{equation} for $x\in\mathbb R^n$ where $f$ is (smooth) nonlinear function. In particular, I am trying to understand why they apparently always form a two-dimensional manifold in the phase space (often referred to in mechanics as "Nonlinear normal mode").

Centre manifold theorem Let's apply the centre manifold theorem to $(1)$ in the neighbourhood of $0$. To this end, we write the linearisation of $(1)$ in the first-order form $$ \dot X = \begin{bmatrix} 0 & I_n \\ -J_f(0) & 0 \end{bmatrix} X + \begin{bmatrix}0 \\ -f(0) \end{bmatrix}:=AX+b\quad \text{with}\quad X=\begin{bmatrix} x \\ \dot x\end{bmatrix}$$ since $f(x)\approx f(0) + J_f(0)x$. The centre manifold theorem states that the eigenvalues of the matrix $A$ determines 3 (generalised) subspaces $E_s$, $E_c$ and $E_u$ corresponding to the eigenspaces of the eigenvalues of negative real part (stable), zero real part (centre) or positive real part (unstable), respectively.

The eigenvalues of $A$ are the roots of $\det(A-XI_{2n})=\det(X^2 I_n + J_f(0))$. So if $\lambda$ is an eigenvalue of $A$, then $-\lambda$ is an eigenvalue too.

Let's take a concrete example.

Concrete example in the linear case Consider \begin{align} \ddot x_1 + \omega_1^2 x_1 = 0 \\ \ddot x_2 + \omega_2^2 x_2 = 0 \end{align} Then $$ J_f(0) = \begin{bmatrix} \omega_1^2 & 0 \\ 0 & \omega_2^2 \end{bmatrix}$$ and the roots of $\det(X^2 I_2 + J_f(0))=(X^2+\omega_1^2)(X^2+\omega_2^2)$ are $\pm i \omega_1$ and $\pm i \omega_2$. They go by pair of complex conjugates, and their real parts is zero. Hence they are two centre manifolds of dimension 2 (here, planes because the equations are linear). In this well-known case, it is known that solutions are periodic $x_i=A_i\cos(\omega_i t)+B_i\sin(\omega_i t)$: all the trajectories of the centre manifold are periodic.

Adding a nonlinearity Now if we add a nonlinearity to the previous linear example, the centre manifold theorem implies that there will be two centre manifolds, tangent to the two planes and of same dimension, i.e. 2.

Back to the general nonlinear equation The centre manifold theorem implies that a solution of $(1)$ is either on the stable, the unstable or a centre manifold. Periodic solutions can only lie on the centre manifold, otherwise they would converge to 0 or diverge. But centre manifold can, a priori, contain stable, unstable or periodic orbits. My question is, why, or under what conditions, is the manifold of periodic solutions exactly of dimension 2? (according to the numerous examples in the literature) Why isn't it a simple curve on the two-dimensional centre manifold, for example? A hint could be that it is tangent to the centre manifold of the linearised system which is a plane made only of periodic orbits. But the proof is not clear to me.

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  • $\begingroup$ Could you clarify your notation a little bit? I'm slightly confused about $J_x f = x $, so I want to understand what you've meant. Thank you. $\endgroup$ – Evgeny Dec 29 '15 at 7:05
  • $\begingroup$ @Evgeny Thank you for your comment; I completely rewrote the question hoping it's more clear now. $\endgroup$ – anderstood Dec 29 '15 at 17:30
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Disclaimer: I apologize in advance is this is off-topic, But I couldn't resist noticing that your system is a Hamiltonian one, and there is quite a bit of litterature on periodic orbits of Hamiltonian systems.

The differential equation $$\ddot x +f(x) = 0$$ may be written as a system of equation

$$\dot u = v,\quad \dot v = f(u).$$

Notice that this is a Hamiltonian system which satisfies

$$\frac{\partial u}{\partial t} = -\frac{\partial H}{\partial V},\quad \frac{\partial v}{\partial t} = \frac{\partial H}{\partial u},\quad$$

where the Hamiltonian is

$$H(u,v) = -\frac {v^2}{2} - \int f(u).$$

Added: A couple known results

  • Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be a $C^1$ function and let $x_0 \in \mathbb R$ be a point at which $f (x_0) = 0$ and $f'(x_0)>0$. Then the second order differential equation

    $$ \ddot x + f (x) = 0$$

    has a nonconstant periodic solution.

  • Any critical point of a $C^2$ Hamiltonian system is not asymptotically stable
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  • $\begingroup$ Indeed, but there is apparently no general results periodic orbits of Hamiltonian systems. For example, Th. 2.4 of this paper says there is, under some conditions, no solutions (dimension 0). However, in this article the authors state that: $\endgroup$ – anderstood Dec 28 '15 at 17:19
  • $\begingroup$ In a Hamiltonian system periodic orbits are not usually isolated but form one-parametric families. But I believe it is possible to answer my question without going so far in maths, and anyway, it does not give a clear explanation under which condition the manifold is of dimension 2 (= orbits form a one-parametric family). $\endgroup$ – anderstood Dec 28 '15 at 17:21
  • $\begingroup$ @anderstood You are correct about having no general results. I will add to my answer one or two easy results. They may or may not help. $\endgroup$ – GaussTheBauss Dec 28 '15 at 17:30

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