Euler's equation for an incompressible perfect fluid

Using a fundamental principle in physics, we now derive the equation of motion of an incompressible perfect fluid in $\mbox{${ \mathbb{ R}}$}$ⁿ. Our interest lies in the cases where n equals 2 or 3.

Thus, imagine $\mbox{${ \mathbb{ R}}$}$ⁿ filled with some fluid material. Fluid particles are identified with points in this space. Initially, say at time t = 0, the fluid particle x $\in$ $\mbox{${ \mathbb{ R}}$}$ⁿ has some velocity u(x). As time passes by, the particle that was originally at x, will occupy a new position in space, which we denote by $\eta$(t)(x). This function is not arbitrary and its dynamics may be determined by a fundamental principle in physics.

A fluid is said to be perfect if the frictional resistance between fluid particles is zero. Often, such fluids are also called ideal, inviscic or non-viscous. Assuming this to be the case for our fluid, suppose that the initial density of the fluid is given by a function $\rho$(x). Such a function will evolve in time to $\rho$(t,$\eta$(t)(x)), the density of the fluid at time t and position $\eta$(t)(x).

Let $\Omega$ be an arbitray but fixed domain $\Omega$ in $\mbox{${ \mathbb{ R}}$}$ⁿ. Then,

$$\int_{\Omega}^{}$$$$\rho$$(t, y)dy

is the total mass in $\Omega$ at time t, and the its rate of change in time is given by

$${\frac{d}{dt}}$$$$\int_{\Omega}^{}$$$$\rho$$(t, y)dy = $$\int_{\Omega}^{}$$$${\frac{\partial \rho}{ \partial t}}$$(t, y)dy .

In the absence of sources or sinks of fluid particles, the law of conservation of mass says that this change has to be opposite to the total flux of mass through the boundary of $\Omega$. Notice that the velocity of the fluid is given by the function

v = $$\dot{\eta}$$o$$\eta^{-1}_{}$$ .

In other words, at time t and position y, the velocity of the fluid is given by the time derivative of $\eta$ at x, where y = $\eta$(t)(x). Thus, if d$\sigma$ is the surface area element of $\partial$$\Omega$ and n is the exterior normal, the flux of fluid through the boundary of $\Omega$ is given by

$$\int_{\partial \Omega}^{}$$$$\rho$$v ^. n d$$\sigma$$ ,

and the law of conservation of mass implies that

$$\int_{\Omega}^{}$$$${\frac{\partial \rho}{ \partial t}}$$(t, y)dy = - $$\int_{\partial \Omega}^{}$$$$\rho$$v ^. n d$$\sigma$$ .

By virtue of the divergence theorem, we may rewrite the right side of this expression as

- $$\int_{\partial \Omega}^{}$$$$\rho$$v ^. n d$$\sigma$$ = - $$\int_{\Omega}^{}$$div($$\rho$$v)dy ,

and therefore,

$$\int_{\Omega}^{}$$$$\left(\vphantom{ \frac{\partial \rho}{\partial t}+ {\rm div}(\rho v)}\right.$$$${\frac{\partial \rho}{ \partial t}}$$ + div($$\rho$$v)$$\left.\vphantom{ \frac{\partial \rho}{\partial t}+ {\rm div}(\rho v)}\right)$$ dy = 0 .

Since the domain $\Omega$ was chosen arbitrarily, the integral expression above can only hold provided that the integrand is identically zero:

 

$${\frac{\partial \rho}{ \partial t}} \rho$$ (5.1)

The fluid is said to be incompressible if the density function is constant. In that case, by (1), we conclude that the velocity field v must be a vector of zero divergence. But this condition implies that the Jacobian matrix Jac($\eta$(t)) have constant determinant, and since $\eta$(0) is the identity, we must have detJac($\eta$(t)) = 1. Conversely, we may define an incompressible fluid as a fluid that evolves according to a diffeomorphims $\eta$(t) such that detJac($\eta$(t)) = 1. This will imply that v = $\dot{\eta}$o$\eta^{-1}_{}$ is divergence-free, and if the initial density is assumed to be constant, by (1) the density will have to be constant for all t. Thus, these two definitions of incompressibility are equivalent.

From now on, we assume that our fluid is inviscid and incompressible, and for simplicity, we assume that its density is identically one. The time evolution of this fluid is described by a t-parameter family of diffeomorphism

$$\eta$$(t) : $$\mbox{${\mathbb{ R} }$}$$ⁿ $$\mapsto$$ $$\mbox{${\mathbb{ R} }$}$$ⁿ ,

such that $\eta$(0) is the identity. The incompressibility assumption is expressed mathematically in the fact that v = $\dot{\eta}$o$\eta^{-1}_{}$ is a divergence-free vector field, and therefore, the family of diffeomorphism $\eta$(t) is infinitesimaly volume preserving, that is to say,

detJac($$\eta$$(t)) = 1 .

Such a family is completely determined by v, since

$$\eta$$(t)(x) = x + $$\int_{0}^{t}$$v(s,$$\eta$$(s)(x))ds .

We describe the dynamics of the fluid by obtaining an equation that governs the time evolution of v.

We warn about a minor difficulty that has already been present above: sometimes we do calculations with respect to the space coordinate y = $\eta$(t)(x), coordinate that is moving with the fluid. In the literature, this is known as the Euler system. The diffeomorphism $\eta$(t) defines a transformation between this and the coordinate system x, at time t = 0, which is known as the Lagrangean system. Mathematically, this is just a simple change of variables, and any equation in one system can be translated into the other by the change rule and some (often tedious) calculation.

A single particle fluid will have kinetic energy given by ${\frac{1}{2}}$$\langle$v, v$\rangle$ at time t. The total kinetic energy at time t is obtained by summing up the contributions from all particles, or integrating over space:

$${\textstyle\frac{1}{2}}$$$$\int_{{\mbox{${\mathbb{ R} }$}}^n}^{}$$$$\langle$$v(t, y), v(t, y)$$\rangle$$dy .

Since there is no potential energy binding particles of a fluid, or dissipation due to viscosity, the Lagragian, a function of the one-parameter family of diffeomorphism $\eta$, is given by

L($$\eta$$) = $${\textstyle\frac{1}{2}}$$$$\int_{0}^{T}$$$$\int_{{\mbox{${\mathbb{ R} }$}}^n}^{}$$$$\langle$$v(s, y), v(s, y)$$\rangle$$dy ds .

The principle of least action says that the fluid will move in time along a trajectory $\eta$(t) that minimizes this Lagrangian among all one-parameter family of volume-preserving diffeomorphisms defined on the interval [0, T] that start at the identity at t = 0.

What are the critical points of the Lagrangian function L? This question is the same as the question we face in Calculus courses, except for the fact that now the domain of the function L under analysis is itself a set of functions. But the answer should be the same: $\eta$ is a critical point of L iff given any variation of $\eta_{s}^{}$ of $\eta$, that is to say, a family of diffeomorphism $\eta_{s}^{}$ parametriced by s such that $\eta_{0}^{}$ = $\eta$, then

$${\frac{d}{ds}}$$L($$\eta_{s}^{}$$)$$\mid_{s=0}^{}$$ = 0 .

This statement if completely analogous to the fact that for a real valued function f of a single variable, x is a critical point iff the function f ($\gamma_{x}^{}$(s)) has derivative zero at s = 0. Here $\gamma_{x}^{}$(s) is the variation of x defined by $\gamma_{x}^{}$(s) = x + s.

Despite this clear analogy with the notion of critical points of real valued functions of one variable, we must be rather careful when finding critical points $\eta$ = $\eta$(t) of the Lagrangian L. The variations $\eta_{s}^{}$ of $\eta$ that we are permitted to consider must be in the domain of definition of L, and therefore, for each s, $\eta_{s}^{}$ must be a one parameter family of volume preserving diffeomorphism defined on [0, T]. This implies that the vector field $\zeta$ : $\eta$(t)($\mbox{${ \mathbb{ R}}$}$ⁿ) $\mapsto$ $\mbox{${ \mathbb{ R}}$}$ⁿ defined by

$$\zeta$$($$\eta$$(t)(x)) = $${\frac{\partial}{\partial s}}$$$$\eta_{s}^{}$$(t)(x)$$\mid_{s=0}^{}$$

must have zero divergence. Furthermore, we take this variation to leave unchanged the diffeomorphisms $\eta$(0) and $\eta$(T) by requiring that $\partial_{s}^{}$$\eta_{s}^{}$(t) be zero when t = 0 or t = T. Therefore:

$${\frac{d}{ds}}$$($$\eta_{s}^{}$$)$$\mid_{s=0}^{}$$	=	$$\int_{0}^{T}$$$$\int_{{\mbox{${\mathbb{ R} }$}}^n}^{}$$$$\langle$$$$\partial_{s}^{}$$$$\partial_{t}^{}$$$$\eta_{s}^{}$$(t)(x)$$\mid_{s=0}^{}$$, v(t,$$\eta$$(t)(x))$$\rangle$$ dxdt 1mm
	=	$$\int_{0}^{T}$$$$\int_{{\mbox{${\mathbb{ R} }$}}^n}^{}$$$$\langle$$$$\partial_{t}^{}$$$$\partial_{s}^{}$$$$\eta_{s}^{}$$(t)(x)$$\mid_{s=0}^{}$$, v(t,$$\eta$$(t)(x))$$\rangle$$ dxdt 1mm
	=	$$\int_{0}^{T}$$$$\int_{{\mbox{${\mathbb{ R} }$}}^n}^{}$$$$\langle$$$$\partial_{t}^{}$$$$\zeta$$($$\eta$$(t)(x)), v(t,$$\eta$$(t)(x))$$\rangle$$dxdt  .

We now integrate by parts in t; observe that the boundary terms will vanish since $\zeta$($\eta$(0)(x)) = $\zeta$($\eta$(T)(x)) = 0. Hence:

$${\frac{d}{ds}}$$($$\eta_{s}^{}$$)$$\mid_{s=0}^{}$$	=	$$\int_{0}^{T}$$$$\int_{{\mbox{${\mathbb{ R} }$}}^n}^{}$$$$\langle$$$$\zeta$$($$\eta$$(t)(x)),$$\partial_{t}^{}$$(v(t,$$\eta$$(t)(x)))$$\rangle$$dxdt 1mm
	=	$$\int_{0}^{T}$$$$\int_{{\mbox{${\mathbb{ R} }$}}^n}^{}$$$$\langle$$$$\zeta$$,$$\partial_{t}^{}$$v+v . $$\nabla$$v$$\rangle$$ dydt .

If $\eta$ is a critical point of L, the expression above must be identically zero. But given any divergence-free vector $\zeta$, we may find a variation $\eta_{s}^{}$ of $\eta$ such that $\partial_{s}^{}$$\eta_{s}^{}$ = $\zeta$. Therefore, the vanishing of this expression says that the vector $\partial_{t}^{}$v + v ^. $\nabla$v must be perpendicular to any divergence-free vector $\zeta$. Since any vector field decomposes uniquely as the sum of a divergence-free vector field and a gradient field, this condition implies that the divergence-free component of $\partial_{t}^{}$v + v ^. $\nabla$v is zero, and so

 

$$\partial_{t}^{} \nabla \nabla$$ (5.2)

for some real valued function

p : $$\eta$$(t)($$\mbox{${\mathbb{ R} }$}$$ⁿ) $$\mapsto$$ $$\mbox{${\mathbb{ R} }$}$$ .

Equation (2) rules the dynamics of perfect incompressible fluids. The function p is the pressure. It may appear as an unknown to the problem, but p it is in fact determined by v. Indeed, by incompressibility we have that divv = 0, and taking a time derivative we see that div$\partial_{t}^{}$v = 0. Hence, calculating the divergence of both sides of (2), we see that

divv ^. $$\nabla$$v = trace(Dv)² = - $$\Delta$$p .

Here, Dv is the matrix of derivatives of v. Hence, the laplacian of p, and hence p, is determined by v, and equation (2) is a non-linear partial differential equation for v.

Etymologically, it makes more sense to call a fluid incompressible if the volume does not change infinitesimally at any given point in space. Mathematically that is just the condition that

det(Jac$$\eta$$(t)) = c ,

a constant. Therefore, the velocity field v = $\eta$o$\eta^{-1}_{}$ is divergence free, and if $\eta$(0 is the identity, this constant is necessarily equal to one. If the fluid is assumed to have density $\rho$, not necessarily constant, the equation of motion will be

$$\rho$$($$\partial_{t}^{}$$v + v . $$\nabla$$v)	=	- $$\nabla$$p , 1mm
$$\partial_{t}^{}$$$$\rho$$ + v . $$\nabla$$$$\rho$$	=	0 .

We proceed using the more restrictive definition of incompressibility given earlier.