# Variational Derivation of the Equation of Motion of an Ideal Fluid

## Continuous Diffeomorphisms

For ideal fluids, the configuration space is the group of volume preserving diffeomorphisms $$\text{Diff}_{\text{Vol}}(\mathcal{F})$$ on the fluid container (a region $$\mathcal{F}(t)$$ in $$\mathbb{R}^{2}$$ or $$\mathbb{R}^{3}$$ that changes with time as a result of the boundary motion). A particle located at a point $$x_{0}\in\mathcal{F}(0)$$ will travel to a point $$x_{t}=\varphi_{t}(x_{0})\in\mathcal{F}(t)$$ at time $$t$$.

$\varphi_{t}:\quad\mathcal{F}(0)\to\mathcal{F}(t)$

The kinetic energy is a mapping from the tangent space to the real numbers

$L:\quad T\text{Diff}_{\text{Vol}}\to\mathbb{R}$

where

$L(\varphi,\dot{\varphi})=\frac{1}{2}\int_{\mathcal{F}}||\dot{\varphi}\circ\varphi^{-1}||^{2}dV$

Notice that the Lagrangian satisfies the particle relabeling symmetry expressed as invariance under right composition:

$L(\varphi\circ\phi,\dot{\varphi}\circ\phi)=L(\varphi,\dot{\varphi})$

where $$\phi,\varphi\in\text{TDiff}_{Vol}$$.

The action is given by

$S(\varphi(t),\dot{\varphi}(t))=\int_{t_{0}}^{t_{1}}L\,dt$

Due to the particle relabeling symmetry the system can be described in terms of a reduced Lagrangian

$l:\quad\mathfrak{diff}_{Vol}\to\mathbb{R}\quad\xi\mapsto l(\xi)=L(e,\xi)=L(\phi,\xi\circ\phi)$

where $$\mathfrak{\xi\in diff}_{Vol}$$ is in the Lie algebra of volume preserving diffeomorphisms, and $$e$$ is the identity element of the $$\text{TDiff}_{Vol}$$ group.

To obtain the reduced action

\begin{split}\begin{aligned} S(\varphi(t),\dot{\varphi}(t)) & =\int_{t_{0}}^{t_{1}}L(\varphi(t),\dot{\varphi}(t))\,dt\\ & =\int_{t_{0}}^{t_{1}}L(e,\dot{\varphi}(t)\circ\varphi(t)^{-1})\,dt\\ & =\int_{t_{0}}^{t_{1}}l(\xi(t))\,dt\quad\text{where }\xi=\dot{\varphi}\circ\varphi^{-1}\\ & =:s(\xi(t))\quad\text{Reduced Action Function }\end{aligned}\end{split}

Variations must satisfy the Lin Constraint:

\begin{split}\begin{aligned} \delta\xi & =\delta(\dot{\varphi}\varphi^{-1})\\ & =\delta\dot{\varphi}\varphi^{-1}-\dot{\varphi}\varphi^{-1}\delta\varphi\varphi^{-1}\\ & =\dot{\eta}+[\xi,\eta],\quad\text{where }\eta=\delta\varphi\circ\varphi^{-1}\\\end{aligned}\end{split}

The variation in the reduced action is

\begin{split}\begin{aligned} \delta s & =\delta\left(\frac{1}{2}\int_{t_{1}}^{t_{2}}\int_{\mathcal{F}}v^{2}\:dtdV\right)\\ & =\int_{t_{1}}^{t_{2}}\int_{\mathcal{F}}\,\langle v^{\flat},\delta v\rangle dtdV\\ & =\int_{t_{1}}^{t_{2}}\int_{\mathcal{F}}\,\left(\langle v^{\flat},\dot{u}\rangle+\langle v^{\flat},[v,u]\rangle\right)dtdV\\\end{aligned}\end{split}
$\langle v^{\flat},[v,u]\rangle=\langle v^{\flat},L_{v}u\rangle=L_{v}\langle v^{\flat},u\rangle-\langle L_{v}v^{\flat},u\rangle$

For a divergence free vector field,

$\int_{\mathcal{F}}L_{v}f\overset{\nabla\cdot v=0}{=}\int_{\partial\mathcal{F}}(v,n)f=0$

Therefore,

$\delta s=\int_{t_{1}}^{t_{2}}\int_{\mathcal{F}}\,\left(\langle v^{\flat},\dot{u}\rangle-\langle L_{v}v^{\flat},u\rangle\right)dtdV$

Setting the variation $$\delta s=0$$ and integrating by parts:

$\int_{\mathcal{F}}dV\,\langle\dot{v}^{\flat}+L_{v}v^{\flat},u\rangle=0$

This implies that the integrand must be the gradient of a function

$\dot{v}^{\flat}+L_{v}v^{\flat}+dp=0$

which is equivalent to the Euler equation in coordinate form:

$\dot{\mathbf{v}}+(\mathbf{v}\cdot\nabla)\mathbf{v}+\nabla\mathbf{p}=0$