Consider an inviscid irrotational fluid in two dimensions. There are some explicit connections with symplectic geometry that I do not understand. I am not well versed in the later topic, so please forgive me if I offend anyone’s sensibilities with bad notation/dumb questions.

First, I consider a small blob of fluid with area $dx dy$. If $x=x(a,b,t)$ and $y=y(a,b,t)$ where $x(a,b,0)=a$ and $y(a,b,0)=b$ (i.e. $a$ and $b$ label the initial positions of a particle), we have

$$dx\ dy = da\ db\ (x_ay_b-x_by_a).$$

Conservation of mass requires $ J=(x_ay_b-x_by_a)$ to be time invariant. We can think about $dx\ dy$ as a simple symplectic 2-form $dx \wedge dy$ where here the sign is always oriented to be positive, and mass conservation simply corresponds to preservation of the symplectic form.

Now, for this system one also has conservation of vorticity $\omega$ (corresponding to a particle relabeling symmetry), which is equivalent to saying that $$J\omega=\dot{x}_ax_b-\dot{x}_bx_a +\dot{y}_ay_b-\dot{y}_by_a,$$ is conserved.

The miracle of this formulation is that if you find $x=x(a,b,t)$ and $y=y(a,b,t)$ that satisfy conservation of mass and vorticity, one has an $\bf{exact}$ solution to Euler’s equations.

Is there a way to get this second relationship, i.e. the form of $\omega$, from the Hamiltonian nature of the equations without having to generate the action and find the conservation law associated with the particle-relabeling symmetry? I suppose the question I'm explicitly asking is, does the inverse of the symplectic 2-form define an appropriate bracket such that with the proper Hamiltonian (which I assume I'll have to guess on the first go around?) I recover directly the vorticity conservation law $\omega$?


  1. J.Marsden & A. Weinstein, https://doi.org/10.1016/0167-2789(83)90134-3
  • $\begingroup$ This is loosely based on ideas in this paper by Marsden: sciencedirect.com/science/article/pii/0167278983901343 as well as an alternative formulation of the Euler equations by Abrashkin and Yakubovich. $\endgroup$ – Nick P Jan 17 at 6:06
  • $\begingroup$ Do you think that finding a solution for $x(a,b,t), y(a,b,t)$ can be easier than a solution for Euler? $\endgroup$ – Alex Trounev Jan 17 at 18:15
  • 1
    $\begingroup$ Rajeev. $\endgroup$ – Cosmas Zachos Jan 17 at 20:05
  • $\begingroup$ @AlexTrounev Yes, you can find entire classes of solutions rather easily. For example, if $z=x+i*y$, then $z=g(z)+f(z*)e^{i\omega t}$ for f, g analytic solve these equations. $\endgroup$ – Nick P Jan 17 at 20:57
  • $\begingroup$ Hi @Nick P: Abrashkin & Yakubovich? $\endgroup$ – Qmechanic Jan 20 at 19:15

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