Variational form of Euler's incompressible fluid equations?

I am trying to derive Euler's incompressible fluid equations in terms of a variational stationary principle. Given Euler's flow equations:

$$\frac{\partial v}{\partial t} = -\nabla p$$

$$\nabla\cdot v = 0$$

Starting with a Lagrangian consisting of the kinetic energy, and the continuity constraint (divergence free velocity):

$$\mathcal{L} = \int_\Omega{\frac{1}{2}|v|^2 - p(\nabla\cdot v)}$$

Can one simply apply the Euler-Lagrange equations:

$$\frac{\text{d}}{\text{dt}}\frac{\partial \mathcal{L}}{\partial{v}} - \frac{\partial \mathcal{L}}{\partial{{x}}} = 0$$

to arrive at the Euler equations? As far as I understand this is quite possible, but I am not sure exactly how to proceed. Specifically, how do you differentiate the divergence operator with respect to $v, x$?

I am hoping they can be derived from this simple form ($\mathcal{L} = T - V$), without invoking some of the more abstract, geometrical methods, e.g.: Arnold.

The closest I have seen is Luke's variational principle, but this is not as general. References welcome.

(Question moved from math.stackexchange.com)

• Um... You sure about that first equation? First, the units don't match, $L/T^2 = M/(T^2L^2)$(?). Also, where did the advective term ($u \cdot \nabla u$) go? Or is that something to do with the Lagrangian formulation? Oct 1, 2014 at 11:27
• @user3823992: I agree, the first equation should be written as $D_tv=-(\nabla p)/\rho$ where $D_t$ is the total derivative. Oct 1, 2014 at 12:49
• In general the Lagrangian ($L=T-V$) in Euler-Lagrange equations need not be a standard one (where the potential depends only on postion $x$). In many cases there are generalised potentials (i.e depending on velocity but in such a way that E-L equations hold with a generalised lagranagian $L=T-V_g$), example: lagrangian of Electromagnetism Oct 1, 2014 at 13:16
• Oct 1, 2014 at 14:32

Here we assume that OP is mostly interested in the Eulerian fluid picture (as opposed to the Lagrangian fluid picture). Both fluid pictures are discussed in great detail in Ref. 1.

Note however that in the methods of Ref. 1, the mass density $\rho$ is a dynamical variable. The variation of $\rho$ is important in order to obtain a full set of eoms. But OP specifically asks about incompressible fluids, which means constant $\rho$ (more precisely, along pathlines, cf. the continuity equation).

In Luke's variational principle, which OP mentions, instead of varying $\rho$, one varies wrt. a free surface, which is insufficient to derive bulk eoms.

Alternatively, the Euler fluid equations for constant $\rho$ can be viewed as as a special case of Euler-Poincare (EP) equations, which has a variational formulation, cf. Ref. 2-4 and this Phys.SE post. With a caveat. Because rather than deriving the sought-for equation

$$\frac{D(\rho u)}{Dt} ~=~ -\nabla p ,$$

one essentially only derives

$$\nabla \times \frac{D(\rho u)}{Dt}~=~0,$$

i.e. the pressure $p$ does not enter into the variational principle.

References:

1. R. Salmon, Hamiltonian Fluid Mechanics, Ann. Rev Fluid. Mech. (1988) 225. The pdf file can be downloaded from the author's webpage.

2. V.A. Arnold and B.A. Khesin, Topological Methods in Hydrodynamics, 1998; $\S$7.

3. J.E. Marsden and T.S. Ratiu, Intro to Mechanics and Symmetry, 2nd Eds, 1998; Section 13.5.

4. Terence Tao, The Euler-Arnold equation, blogpost 2010.