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I am trying to understand if the deterministic 2+1D Kuramoto-Sivashinsky equation $$ \partial_t h = -\nu \nabla^2 h - K \nabla^4 h + \frac{\lambda}{2} (\nabla h)^2, $$ where $\nu$, $K$, $\lambda$ are constants and $h=h(x,y,t)$ is a time-dependent real field in two spatial dimensions, can be seen in the light of a statistical field theory. It seems to me that this equation corresponds to a non-conservative system (similar to the Korteweg–de Vries equation, see wiki). Thus, my question is the following:

Is there a Hamiltonian or a free energy that has been written down and analysed for the deterministic Kuramoto-Sivashinsky? If not, has another similar equation been analyzed in terms of statistical field theory (phases, free energy, phase transitions, critical behavior etc.)?

Qmechanic
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jarm
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1 Answers1

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Comments to the question (v2):

  1. On one hand, the Kuramoto-Sivashinsky (KS) equation is a dissipative differential equation (DE). Each term has an even number of spatial derivatives. It's a non-linear version of the heat equation. Dissipative systems rarely have variational action formulations nor Hamiltonian formulations.

  2. On the other hand, in the Korteweg de Vries (KdV) equation, each term has an odd number of derivatives. The KdV equation is not a dissipative DE. It has both a Lagrangian and a Hamiltonian formulation. The energy is a conserved quantity.

Qmechanic
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  • Thanks for the comment! What is the definition of a dissipative DE that you are using? Note that both KdV and KS are nonlinear equations; in 1D they only differ by gradient terms. – jarm Apr 24 '16 at 08:25
  • I updated the answer (v2). – Qmechanic Apr 24 '16 at 09:13
  • Thanks for the update, but I still do not understand how you come up with the fact that KS is a dissipative DE, while KdV is not. Unfortunately, I cannot find the answer in the wiki link either. An odd number of derivatives does not seem to be the answer here either, as the heat equation (or the Schrodinger equation; both have a well-defined energy which is conserved) has a $\nabla^2$ term. Also note that the nonlinear term in the KS equation can be rewritten in the KdV form by partial differentiation. – jarm Apr 24 '16 at 19:16
  • Well, the full story is quite long. The Schrödinger equation (SE) is not dissipative because of the imaginary $i$ and the complex-valued wavefunction. The SE has a Lagrangian and a Hamiltonian formulation, while the heat equation does not. – Qmechanic Apr 24 '16 at 19:21
  • As far as I know there's no general definition of when a PDE is "dissipative", but the basic idea is that solutions of such an equation "lose energy", so an autonomous PDE derived from a Lagrangian or Hamiltonian is never dissipative. It would be nice to prove that the KS equation cannot be derived from a Lagrangian or Hamiltonian. I believe it, but I haven't seen it proved! – John Baez Oct 20 '21 at 16:30