Categories: Fluid dynamics, Fluid mechanics, Physics.
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.
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.