Categories: Fluid dynamics, Fluid mechanics, Physics.

# Euler equations

The Euler equations are a system of partial differential equations that govern the movement of ideal fluids, i.e. fluids without viscosity. There exist several forms, depending on the surrounding assumptions about the fluid.

## Incompressible fluid

In a fluid moving according to the velocity vield $$\va{v}(\va{r}, t)$$, the acceleration felt by a particle is given by the material acceleration field $$\va{w}(\va{r}, t)$$, which is the material derivative of $$\va{v}$$:

\begin{aligned} \va{w} \equiv \frac{\mathrm{D} \va{v}}{\mathrm{D} t} = \pdv{\va{v}}{t} + (\va{v} \cdot \nabla) \va{v} \end{aligned}

This infinitesimal particle obeys Newton’s second law, which can be written as follows:

\begin{aligned} \va{w} \dd{m} = \va{w} \rho \dd{V} = \va{f^*} \dd{V} \end{aligned}

Where $$\dd{m}$$ and $$\dd{V}$$ are the particle’s mass volume, and $$\rho$$ is the fluid density, which we assume, in this case, to be constant in space and time. Then the effective force density $$\va{f^*}$$ represents the net force-per-particle. By dividing the law by $$\dd{V}$$, we find:

\begin{aligned} \rho \va{w} = \va{f^*} \end{aligned}

Next, we want to find another expression for $$\va{f^*}$$. We know that the overall force $$\va{F}$$ on an arbitrary volume $$V$$ of the fluid is the sum of the gravity body force $$\va{F}_g$$, and the pressure contact force $$\va{F}_p$$ on the enclosing surface $$S$$. Using Gauss’ theorem, we then find:

\begin{aligned} \va{F} = \va{F}_g + \va{F}_p = \int_V \rho \va{g} \dd{V} - \oint_S p \dd{\va{S}} = \int_V (\rho \va{g} - \nabla p) \dd{V} = \int_V \va{f^*} \dd{V} \end{aligned}

Where $$p(\va{r}, t)$$ is the pressure field, and $$\va{g}(\va{r}, t)$$ is the gravitational acceleration field. Combining this with Newton’s law, we find the following equation for the force density:

\begin{aligned} \va{f^*} = \rho \va{w} = \rho \va{g} - \nabla p \end{aligned}

Dividing this by $$\rho$$, we get the first of the system of Euler equations:

\begin{aligned} \boxed{ \va{w} = \frac{\mathrm{D} \va{v}}{\mathrm{D} t} = \va{g} - \frac{\nabla p}{\rho} } \end{aligned}

The last ingredient is incompressibility: the same volume must simultaneously be flowing in and out of an arbitrary enclosure $$S$$. Then, by Gauss’ theorem:

\begin{aligned} 0 = \oint_S \va{v} \cdot \dd{\va{S}} = \int_V \nabla \cdot \va{v} \dd{V} \end{aligned}

Since $$S$$ and $$V$$ are arbitrary, the integrand must vanish by itself everywhere:

\begin{aligned} \boxed{ \nabla \cdot \va{v} = 0 } \end{aligned}

Combining this with the equation for $$\va{w}$$, we get a system of two coupled differential equations: these are the Euler equations for an incompressible fluid with spatially uniform density $$\rho$$:

\begin{aligned} \boxed{ \frac{\mathrm{D} \va{v}}{\mathrm{D} t} = \va{g} - \frac{\nabla p}{\rho} \qquad \quad \nabla \cdot \va{v} = 0 } \end{aligned}

The above form is straightforward to generalize to incompressible fluids with non-uniform spatial densities $$\rho(\va{r}, t)$$. In other words, these fluids are “lumpy” (variable density), but the size of their lumps does not change (incompressibility).

To update the equations, we demand conservation of mass: the mass evolution of a volume $$V$$ is equal to the mass flow through its boundary $$S$$. Applying Gauss’ theorem again:

\begin{aligned} 0 = \dv{t} \int_V \rho \dd{V} + \oint_S \rho \va{v} \cdot \dd{\va{S}} = \int_V \dv{\rho}{t} + \nabla \cdot (\rho \va{v}) \dd{V} \end{aligned}

Since $$V$$ is arbitrary, the integrand must be zero. This leads to the following continuity equation, to which we apply a vector identity:

\begin{aligned} 0 = \dv{\rho}{t} + \nabla \cdot (\rho \va{v}) = \dv{\rho}{t} + (\va{v} \cdot \nabla) \rho + \rho (\nabla \cdot \va{v}) \end{aligned}

Thanks to incompressibility, the last term disappears, leaving us with a material derivative:

\begin{aligned} \boxed{ 0 = \frac{\mathrm{D} \rho}{\mathrm{D} t} = \dv{\rho}{t} + (\va{v} \cdot \nabla) \rho } \end{aligned}

Putting everything together, Euler’s system of equations now takes the following form:

\begin{aligned} \boxed{ \frac{\mathrm{D} \va{v}}{\mathrm{D} t} = \va{g} - \frac{\nabla p}{\rho} \qquad \nabla \cdot \va{v} = 0 \qquad \frac{\mathrm{D} \rho}{\mathrm{D} t} = 0 } \end{aligned}

Usually, however, when discussing incompressible fluids, $$\rho$$ is assumed to be spatially uniform, in which case the latter equation is trivially satisfied.

1. B. Lautrup, Physics of continuous matter: exotic and everyday phenomena in the macroscopic world, 2nd edition, CRC Press.