Linear Hamiltonian systems and complex quartets

In the text Hamiltonian Systems: Chaos and quantization by Alfredo M. Ozorio De Almeida section 1.2 there is a discussion of the possible types of eigenvalues for Linear Hamiltonian systems. One possibility is a quartet of complex eigenvalues $\pm\alpha\pm i\beta$.

• How does one reconcile this possibility with the conservation of energy (or conservation of the Hamiltonian)?

• Do we not have spiral type motion in linear dynamical systems if the real part of the eigenvalues is not zero?

These complex quartets have recently appeared in my own research in atomic physics. Thank you in advance for any clarifications.

• Hamilton system is divergence-free. How can $\alpha$ be nonzero? Dec 1, 2013 at 15:01
• @ Zhang, This is exactly what I am trying to understand.
– JEM
Dec 1, 2013 at 15:13
• Well, I think I should read the section before answering it. And could you specify in which page you got the question? Dec 1, 2013 at 15:22
• The classification of the eigenvalues is made on page 14.
– JEM
Dec 1, 2013 at 16:20

Let the system be $\dot{x}=f(x), x\in R^n$

Divergence$= Tr(D_xf)$ or $Tr(A)$ if the system is linear $\dot{x}=Ax$

For $\lambda_i=\pm\alpha\pm i \beta$ Divergence free condition is not violated here. Divergence= trace of Jacobian (sum of all e.values) is still zero. The system being hamiltonian does not restrict unbounded growth/shrinkage in any one direction, all it says that the total phase space volume is conserved.

Conservation of energy or Hamiltonian is not violated either, since all that needs is $\frac{dH}{dt}=0$

In any linear Hamiltonian system, instability associated with eigenvalues $\pm\alpha\pm i\beta$ with $\alpha\ne0$ requires the Hamiltonian $H$ to be of indefinite sign. This is necessary but not sufficient. The unstable solution must lie on the level set $H=0$ (since it approaches zero in backward time) and this level set must be unbounded.

In classical mechanics, the simplest example of spiral-type instability for a Hamiltonian system (both $\alpha$ and $\beta$ nonzero) involves 2 degrees of freedom. Start with a planar oscillator, with spring constants $a$ and $b$ along the $x$ and $y$ axes respectively, and impose a rotation of plane with angular velocity $\omega$. (This classical example is mentioned in this paper, see the references there for more information concerning gyroscopic stabilization.)

The Lagrangian for this system takes the form $$L = \frac12 ((\dot x-\omega y)^2+(\dot y+\omega x)^2)-(a x^2+b y^2))$$ and the Hamiltonian is $$H = \frac12 (p_x^2+p_y)^2+ax^2+by^2)+ \omega(p_xy-p_yx) .$$ If $0<a<-b<3a$, the system is unstable without rotation, and as the rotation speed $\omega$ is increased it becomes first stable, but then unstable with oscillatory instability: The characteristic equation for solutions of the system that are proportional to $e^{\lambda t}$ reduces to the equation $$0={\rm det}( (\lambda I-\omega J)^2 +A ) =0, \quad J = \left(\begin{matrix} 0& 1\\ -1&0 \end{matrix}\right), \quad A = \left(\begin{matrix}a&0\\ 0&b\end{matrix}\right).$$ One computes that the characteristic roots satisfy $$-\lambda^2 = \omega^2+ \frac{a+b}2 \pm \left( \left(\omega+\frac{a+b}2\right)^2-(\omega^2-a)(\omega^2-b)\right)^{1/2}$$ or $$-\lambda^2 = \omega^2+ \frac{a+b}2 \pm \frac 12 \left( \left({a-b}\right)^2 + 8\omega^2(a+b)\right)^{1/2}$$ The roots are of the form $\lambda =\pm\alpha\pm i\beta$ with nonzero $\alpha$ and $\beta$ whenever $-\lambda^2$ is not real. Clearly this is the case and the system has an oscillatory instability if and only if $a+b<0$ and $\omega^2$ is large enough so that $$\Delta:=8\omega^2(a+b)+(b-a)^2<0.$$

For the value of $\omega^2$ that makes $\Delta=0$, we have $$-8(a+b)\lambda^2 = -(b-a)^2+ 4(a+b)^2.$$ One finds that the system is stable ($\lambda^2<0$) at the onset of oscillatory instability ($\Delta=0$) if $$0<a<-b<3a.$$ In this situation, as $\omega^2$ increases from 0, the system goes from i) being unstable with one real root $\lambda>0$, to ii) stable with four pure imaginary roots, to iii) unstable with four roots $\lambda =\pm\alpha\pm i\beta$ with nonzero $\alpha$ and $\beta$.

• thank you. very helpful discussion.
– JEM
Dec 8, 2013 at 23:37