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.

## Incompressible fluids

In a fluid moving according to the velocity field $\va{v}(\va{r}, t)$, 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} m = \va{w} \rho \dd{V} = \va{f^*} \dd{V} \end{aligned}

Where $m$ and $\dd{V}$ are the particle’s mass and volume, and $\rho$ is the fluid density, which we assume to be constant in space and time in this case. Now, 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 $\partial V$. Using the divergence theorem, we then find:

\begin{aligned} \va{F} = \va{F}_g + \va{F}_p = \int_V \rho \va{g} \dd{V} - \oint_{\partial V} 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} \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 $\partial V$. Then, by the divergence theorem:

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

Since $V$ is arbitrary, the integrand must vanish by itself, leading to the continuity relation:

\begin{aligned} \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}

## Compressible fluids

If the fluid is compressible, the condition $\nabla \cdot \va{v} = 0$ no longer holds, so to update the equations we demand that mass is conserved: the mass evolution of a volume $V$ is equal to the mass flow through its boundary $\partial V$. Applying the divergence theorem again:

\begin{aligned} 0 = \dv{}{t}\int_V \rho \dd{V} + \oint_{\partial V} \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. The new continuity equation is therefore:

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

When the fluid gets compressed in a certain location, thermodynamics states that the pressure, temperature and/or entropy must increase there. For simplicity, let us assume an isothermal and isentropic fluid, such that only $p$ is affected by compression, and the fundamental thermodynamic relation reduces to $\dd{E} = - p \dd{V}$.

Then the pressure is given by a thermodynamic equation of state $p(\rho, T)$, which depends on the system being studied (e.g. the ideal gas law $p = \rho R T$). However, the quantity in control of the dynamics is not $p$, but the internal energy $E$. Dividing the fundamental thermodynamic relation by $m \: \mathrm{D}t$, where $m$ is the mass of $\dd{V}$:

\begin{aligned} \frac{\mathrm{D} e}{\mathrm{D} t} = - p \frac{\mathrm{D} v}{\mathrm{D} t} \end{aligned}

With $e$ and $v$ the specific (i.e. per unit mass) internal energy and volume. Using that $\rho = 1 / v$, and substituting the above continuity relation:

\begin{aligned} \frac{\mathrm{D} e}{\mathrm{D} t} = - p \frac{\mathrm{D}}{\mathrm{D} t} \Big( \frac{1}{\rho} \Big) = \frac{p}{\rho^2} \frac{\mathrm{D} \rho}{\mathrm{D} t} = - \frac{p}{\rho} \nabla \cdot \va{v} \end{aligned}

It makes sense to see a factor $-\nabla \cdot \va{v}$ here: an incoming flow increases $e$. This gives us the time-evolution of $e$ due to compression, but its initial value is another equation of state $e(\rho, T)$.

Putting it all 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 \quad \frac{\mathrm{D} \rho}{\mathrm{D} t} = - \rho \nabla \cdot \va{v} \qquad \quad \frac{\mathrm{D} e}{\mathrm{D} t} = - \frac{p}{\rho} \nabla \cdot \va{v} } \end{aligned}

What happens if the fluid is actually incompressible, so $\nabla \cdot \va{v} = 0$ holds again? Clearly:

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

So $e$ is constant, which is in fact equivalent to saying that $\nabla \cdot \va{v} = 0$. The equation for $\rho$ enforces conservation of mass for inhomogeneous fluids, i.e. fluids that are “lumpy”, but where the size of the lumps is conserved by incompressibility.

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