Euler equations (fluid dynamics)

Potential flow around a wing. This incompressible potential flow satisfies the Euler equations for the special case of zero vorticity.

In fluid dynamics, the Euler equations are a set of quasilinear hyperbolic equations governing adiabatic and inviscid flow. They are named after Leonhard Euler. The equations represent Cauchy equations of conservation of mass (continuity), and balance of momentum and energy, and can be seen as particular Navier–Stokes equations with zero viscosity and zero thermal conductivity.^[1] In fact, Euler equations can be obtained by linearization of some more precise continuity equations like Navier–Stokes equations in a local equilibrium state given by a Maxwellian. The Euler equations can be applied to incompressible and to compressible flow – assuming the flow velocity is a solenoidal field, or using another appropriate energy equation respectively (the simplest form for Euler equations being the conservation of the specific entropy). Historically, only the incompressible equations have been derived by Euler. However, fluid dynamics literature often refers to the full set – including the energy equation – of the more general compressible equations together as "the Euler equations".^[2]

From the mathematical point of view, Euler equations are notably hyperbolic conservation equations in the case without external field (i.e. in the limit of high Froude number). In fact, like any Cauchy equation, the Euler equations originally formulated in convective form (also called usually "Lagrangian form", but this name is not self-explanatory and historically wrong, so it will be avoided) can also be put in the "conservation form" (also called usually "Eulerian form", but also this name is not self-explanatory and is historically wrong, so it will be avoided here). The conservation form emphasizes the mathematical interpretation of the equations as conservation equations through a control volume fixed in space, and is the most important for these equations also from a numerical point of view. The convective form emphasizes changes to the state in a frame of reference moving with the fluid.

History

The Euler equations first appeared in published form in Euler's article "Principes généraux du mouvement des fluides", published in Mémoires de l'Academie des Sciences de Berlin in 1757 (in this article Euler actually published only the general form of the continuity equation and the momentum equation;^[3] the energy balance equation would be obtained a century later). They were among the first partial differential equations to be written down. At the time Euler published his work, the system of equations consisted of the momentum and continuity equations, and thus was underdetermined except in the case of an incompressible fluid. An additional equation, which was later to be called the adiabatic condition, was supplied by Pierre-Simon Laplace in 1816.

During the second half of the 19th century, it was found that the equation related to the balance of energy must at all times be kept, while the adiabatic condition is a consequence of the fundamental laws in the case of smooth solutions. With the discovery of the special theory of relativity, the concepts of energy density, momentum density, and stress were unified into the concept of the stress–energy tensor, and energy and momentum were likewise unified into a single concept, the energy–momentum vector.^[4]

Incompressible Euler equations with constant and uniform density

In convective form (i.e. the form with the convective operator made explicit in the momentum equation), the incompressible Euler equations in case of density constant in time and uniform in space are:^[5]

Incompressible Euler equations with constant and uniform density (convective or Lagrangian form) $\left\{ \begin{align} {D\bold u \over Dt} = - \nabla w +\bold{g} \\ \nabla\cdot \bold u=0 \end{align}\right.$

where:

• $\bold u$ is the flow velocity vector, with components in a N-dimensional space $u_1, u_2, \dots, u_N$,
• ${D \over Dt}$ denotes the material derivative in time,
• $\cdot$ denotes the scalar product,
• $\nabla$ is the nabla operator, here used to represent the specific thermodynamic work gradient (first equation), and the flow velocity divergence (second equation), and
• $\mathbf u \cdot \nabla$ is the convective operator,
• $w$ is the specific (with the sense of per unit mass) thermodynamic work, the internal source term.
• $\mathbf{g}$ represents body accelerations (per unit mass) acting on the continuum, for example gravity, inertial accelerations, electric field acceleration, and so on.

The first equation is the Euler momentum equation with uniform density (for this equation it could also not be constant in time). By expanding the material derivative, the equations become:

$\left\{ \begin{align} {\partial \bold u \over\partial t}+ \bold u \cdot \nabla \bold u =- \nabla w + \bold{g} \\ \nabla\cdot \bold u=0 \end{align}\right.$

In fact for a flow with uniform density $\rho_0$ the following identity holds:

$\nabla w \equiv \nabla \left(\frac p {\rho_0} \right) = \frac 1 {\rho_0} \nabla p$

where $p$ is the mechanic pressure. The second equation is the incompressible constraint, stating the flow velocity is a solenoidal field (the order of the equations is not casual, but underlines the fact that the incompressible constraint is not a degenerate form of the continuity equation, but rather of the energy equation, as it will become clear in the following). Notably, the continuity equation would be required also in this incompressible case as an additional third equation in case of density varying in time or varying in space. For example, with density uniform but varying in time, the continuity equation to be added to the above set would correspond to:

$\frac{\partial \rho}{\partial t} = 0$

So the case of constant and uniform density is the only one not requiring the continuity equation as additional equation regardless of the presence or absence of the incompressible constraint. In fact, the case of incompressible Euler equations with constant and uniform density being analyzed is a toy model featuring only two simplified equations, so it is ideal for didactical purposes even if with limited physical relevancy.

The equations above thus represent respectively conservation of mass (1 scalar equation) and momentum (1 vector equation containing $N$ scalar components, where $N$ is the physical dimension of the space of interest). In 3D for example $N=3$ and the $\bold r$ and $\bold u$ vectors are explicitly $(x_1, x_2, x_3)$ and $(u_1, u_2, u_3)$. Flow velocity and pressure are the so-called physical variables.^[1]

These equations may be expressed in subscript notation:

$\left\{ \begin{align} {\partial u_j \over\partial t}+ \sum_{i=1}^N u_j {\partial (u_i + w \hat e_i)\over\partial r_i} =0\\ \sum_{i=1}^N {\partial u_i\over\partial r_i}=0 \end{align}\right.$

where the $i$ and $j$ subscripts label the N-dimensional space components. These equations may be more succinctly expressed using Einstein notation:

$\left\{ \begin{align} \partial_t u_j+\partial_i (u_i u_j + w \delta_{ij})=0\\ \partial_i u_i=0, \end{align}\right.$

where the $i$ and $j$ subscripts label the N-dimensional space components, and $\delta$; is the Kroenecker delta. In 3D $N=3$ and the $\bold r$ and $\bold u$ vectors are explicitly $(x_1, x_2, x_3)$ and $(u_1, u_2, u_3)$, and matched indices imply a sum over those indices and $\partial_t \equiv \frac{\partial}{\partial t}$ and $\partial_i \equiv \frac{\partial}{\partial r_i}$.

Properties

Although Euler first presented these equations in 1755, many fundamental questions about them remain unanswered.

In three space dimensions it is not even known whether solutions of the equations are defined for all time or if they form singularities.^[6]

Smooth solutions of the free (in the sense of without source term: g=0) equations satisfy the conservation of specific kinetic energy:

${\partial \over\partial t} \left(\frac 1 2 u^2 \right) + \nabla \cdot (u^2 \bold u + w \bold I) = 0$

In the one dimensional case without the source term (both pressure gradient and external force), the momentum equation becomes the inviscid Burgers equation:

${\partial u \over\partial t}+ u {\partial u \over\partial x} = 0$

This is a model equation giving many insights on Euler equations.

Nondimensionalisation

In order to make the equations dimensionless, a characteristic length $r_0$, and a characteristic velocity $u_0$, need to be defined. These should be chosen such that the dimensionless variables are all of order one. The following dimensionless variables are thus obtained:

$\begin{align} u^* & \equiv \frac u {u_0}, \\[5pt] r^* & \equiv \frac r {r_0}, \\[5pt] t^* & \equiv \frac {u_0}{r_0} t, \\[5pt] p^* & \equiv \frac w {u_0^2}, \\[5pt] \nabla^* & \equiv r_0 \nabla \end{align}$

and of the field unit vector:

$\hat{\mathbf g}\equiv \frac {\mathbf g} g,$

Substitution of these inversed relations in Euler equations, defining the Froude number, yields (omitting the * at apix):

Incompressible Euler equations with constant and uniform density (nondimensional form) $\left\{\begin{align} {D \bold u \over Dt}= - \nabla w + \frac 1 {\mathrm{Fr}} \hat{\mathbf g}\\ \nabla\cdot \bold u=0 \end{align} \right.$

Euler equations in the Froude limit (no external field) are named free equations and are conservative. The limit of high Froude numbers (low external field) is thus notable and can be studied with perturbation theory.

Conservation form

The conservation form emphasizes the mathematical properties of Euler equations, and especially the contracted form is often the most convenient one for computational fluid dynamics simulations. Computationally, there are some advantages in using the conserved variables. This gives rise to a large class of numerical methods called conservative methods.^[1]

The free Euler equations are conservative, in the sense they are equivalent to a conservation equation:

$\frac{\partial \bold y}{\partial t}+ \nabla \cdot \bold F ={\bold 0},$

or simply in Einstein notation:

$\frac{\partial y_j}{\partial t}+ \frac{\partial f_{ij}}{\partial r_i}= 0_i,$

where the conservation quantity $\bold y$ in this case is a vector, and $\bold F$ is a flux matrix. This can be simply proved.

Demonstration of the conservation form 

First, the following identities hold:

$\nabla \cdot (w \bold I) = \bold I \cdot \nabla w + w \nabla \cdot \bold I = \nabla w$

$\bold u \cdot \nabla \cdot \bold u = \nabla \cdot (\bold u \otimes \bold u)$

where $\otimes$ denotes the outer product. The same identities expressed in Einstein notation are:

$\partial_i (w \delta_{ij} ) = \delta_{ij} \partial_i w + w \partial_i \delta_{ij} = \delta_{ij} \partial_i w = \partial_j w$

$u_j \partial_i u_i = \partial_i (u_i u_j)$

where I is the identity matrix with dimension N and δ_ij its general element, the Kroenecker delta.

Thanks to these vector identities, the incompressible Euler equations with constant and uniform density and without external field can be put in the so-called conservation (or Eulerian) differential form, with vector notation:

$\left\{ \begin{align} {\partial \bold u \over\partial t}+ \nabla \cdot (\bold u \otimes \bold u + w \bold I)=\bold{0}\\ {\partial 0 \over\partial t}+ \nabla\cdot \bold u=0, \end{align}\right.$

or with Einstein notation:

$\left\{ \begin{align} \partial_t u_j + \partial_i (u_i u_j + w \delta_{ij})=0\\ \partial_t 0 + \partial _j u_j=0, \end{align}\right.$

Then incompressible Euler equations with uniform density have conservation variables:

${\bold y}=\begin{pmatrix}\bold u \\ 0 \end{pmatrix}; \qquad {\bold F}=\begin{pmatrix}\bold u \otimes \bold u + w \bold I \\ \bold u \end{pmatrix}.$

Note that in the second component u is by itself a vector, with length N, so y has length N+1 and F has size N(N+1). In 3D for example y has length 4, I has size 3x3 and F has size 3x4, so the explicit forms are:

${\bold y}=\begin{pmatrix} u_1 \\ u_2 \\ u_3 \\0 \end{pmatrix}; \quad {\bold F}=\begin{pmatrix} u_1^2 + w & u_1u_2 & u_1u_3 \\ u_2 u_1 & u_2^2 + w & u_2 u_3 \\ u_3 u_1 & u_3 u_2 & u_3^2 + w \\ u_1 & u_2 & u_3 \end{pmatrix}.$

At last Euler equations can be recast into the particular equation:

Incompressible Euler equation(s) with constant and uniform density (conservation or Eulerian form) $\frac {\partial}{\partial t}\begin{pmatrix} \bold u \\ 0 \end{pmatrix} + \nabla \cdot \begin{pmatrix}\bold u \otimes \bold u + w \bold I \\ \bold u \end{pmatrix} = \begin{pmatrix}\bold g \\ 0\end{pmatrix}$

Spatial dimensions

For certain problems, especially when used to analyze compressible flow in a duct or in case the flow is cylindrically or spherically symmetric, the one-dimensional Euler equations are a useful first approximation. Generally, the Euler equations are solved by Riemann's method of characteristics. This involves finding curves in plane of independent variables (i.e., $x$ and $t$) along which partial differential equations (PDE's) degenerate into ordinary differential equations (ODE's). Numerical solutions of the Euler equations rely heavily on the method of characteristics.

Incompressible Euler equations

In convective form the incompressible Euler equations in case of density variable in space are:^[5]

Incompressible Euler equations (convective or Lagrangian form) $\left\{ \begin{align} {D \rho \over D t} = 0\\ {D \bold u \over Dt}=- \frac {\nabla p}{\rho}+\bold{g}\\ \nabla\cdot \bold u=0 \end{align}\right.$

where the additional variables are:

• $\rho$ is the fluid mass density,
• $p$ is the pressure, $p = \rho w$.

The first equation, which is the new one, is the incompressible continuity equation. In fact the general continuity equation would be:

${\partial \rho \over\partial t} + \bold u \cdot \nabla \rho + \rho \nabla \cdot \bold u = 0$

but here the last term is identically zero for the incompressibility constraint.

Conservation form

The incompressible Euler equations in the Froude limit are equivalent to a single conservation equation with conserved quantity and associated flux respectively:

${\bold y}=\begin{pmatrix}\rho \\ \rho \bold u \\0\end{pmatrix}; \qquad {\bold F}=\begin{pmatrix}\rho \bold u\\\rho \bold u \otimes \bold u + p \bold I\\\bold u\end{pmatrix}.$

Here $\bold y$ has length N+2 and $\bold F$ has size N(N+2).^[7] In general (not only in the Froude limit) Euler equations are expressible as:

$\frac {\partial}{\partial t}\begin{pmatrix}\rho \\ \rho \bold u \\0\end{pmatrix}+ \nabla \cdot \begin{pmatrix}\rho \bold u\\\rho \bold u \otimes \bold u + p \bold I\\ \bold u\end{pmatrix} = \begin{pmatrix}0 \\ \rho \bold g \\ 0 \end{pmatrix}$

Conservation variables

The variables for the equations in conservation form are not yet optimised. In fact we could define:

${\bold y}=\begin{pmatrix}\rho \\ \bold j \\0\end{pmatrix}; \qquad {\bold F}=\begin{pmatrix} \bold j \\ \frac {1} \rho \, \bold j \otimes \bold j+ p \bold I\\ \frac \bold j \rho \end{pmatrix}.$

where:

• $\bold j = \rho \bold u$ is the momentum density, a conservation variable.

Incompressible Euler equation(s) (conservation or Eulerian form) $\frac {\partial}{\partial t}\begin{pmatrix}\rho \\ \bold j \\0\end{pmatrix}+ \nabla \cdot \begin{pmatrix}\bold j \\ \frac 1 \rho \, \bold j \otimes \bold j + p \bold I\\ \frac \bold j \rho\end{pmatrix} = \begin{pmatrix}0 \\ \bold f \\ 0 \end{pmatrix}$

where:

• $\bold f = \rho \bold g$ is the force density, a conservation variable.

Euler equations

In differential convective form, the compressible (and most general) Euler equations can be written shortly with the material derivative notation:

Euler equations (convective form) $\left\{\begin{align} {D \rho\over Dt} = - \rho \nabla \cdot \bold u \\[1.2ex] \frac{D \mathbf{ u}}{D t} = - \frac {\nabla p}{\rho} + \mathbf{g} \\[1.2ex] {D e \over Dt} = -\frac p \rho \nabla \cdot \bold u \end{align} \right.$

where the additional variables here is:

• $e$ is the specific internal energy (internal energy per unit mass).

The equations above thus represent conservation of mass, momentum, and energy: the energy equation expressed in the variable internal energy allows to understand the link with the incompressible case, but it is not in the simplest form. Mass density, flow velocity and pressure are the so-called convective variables (or physical variables, or lagrangian variables), while mass density, momentum density and total energy density are the so-called conserved variables (also called eulerian, or mathematical variables).^[1] If one explicitates the material derivative the equations above are:

$\left\{\begin{align} {\partial\rho\over\partial t} + \bold u \cdot \nabla \rho + \rho \nabla \cdot \bold u = 0\\[1.2ex] \frac{\partial \mathbf{ u}}{\partial t} + \mathbf{u} \cdot \nabla \mathbf{u} + \frac {\nabla p}{\rho} = \mathbf{g} \\[1.2ex] {\partial e \over\partial t}+ \bold u \cdot \nabla e + \frac p \rho \nabla \cdot \bold u = 0\end{align} \right.$

Incompressible constraint

Coming back to the incompressible case, it now becomes apparent that the incompressible constraint typical of the former cases actually is a particular form valid for incompressible flows of the energy equation, and not of the mass equation. In particular, the incompressible constraint corresponds to the following very simple energy equation:

${D e \over D t} = 0$

Thus for an incompressible inviscid fluid the specific internal energy is constant along the flow lines, also in a time-dependent flow. The pressure in an incompressible flow acts like a Lagrange multiplier, being the multiplier of the incompressible constraint in the energy equation, and consequently in incompressible flows it has no thermodynamic meaning. In fact, thermodynamics is typical of compressible flows and degenerates in incompressible flows.^[8]

Basing on the mass conservation equation, one can put this equation in the conservation form:

${\partial \rho e \over \partial t} + \nabla \cdot (\rho e \mathbf u) = 0$

meaning that for an incompressible inviscid nonconductive flow a continuity equation holds for the internal energy.

Enthalpy conservation

Since by definition the specific enthalpy is:

$h = e + \frac p \rho$

The material derivative of the specific internal energy can be expressed as:

${D e \over Dt} = {D h \over Dt} - \frac 1 \rho \left({D p \over Dt} - \frac p \rho {D \rho \over Dt} \right)$

Then by substituting the momentum equation in this expression, one obtains:

${D e \over Dt}= {D h \over Dt} - \frac 1 \rho \left(p \nabla \cdot \mathbf u + {D p \over Dt} \right)$

And by substituting the latter in the energy equation, one obtains that the enthalpy expression for the Euler energy equation:

${D h \over Dt} = \frac 1 \rho {D p \over Dt}$

In a reference frame moving with an inviscid and nonconductive flow, the variation of enthalpy directly corresponds to a variation of pressure.

Thermodynamic systems

In thermodynamics the independent variables are the specific volume, and the specific entropy, while the specific energy is a function of state of these two variables.

Deduction of the form valid for thermodynamic systems 

Considering the first equation, variable must be changed from density to specific volume. By definition:

$v \equiv \frac 1 \rho$

Thus the following identities hold:

$\nabla \rho =\nabla \left(\frac 1 v \right) = - \frac 1 {v^2} \nabla v$

$\frac {\partial \rho}{\partial t} = \frac {\partial}{\partial t} \left(\frac 1 v \right) = - \frac 1 {v^2} \frac {\partial v}{\partial t}$

Then by substituting these expressions in the mass conservation equation:

$- \frac 1 {v^2} \nabla v - \frac 1 {v^2} \frac {\partial v}{\partial t} = - \frac 1 v \nabla \cdot u$

And by multiplication:

${\partial v \over\partial t}+\bold u \cdot \nabla v = v \nabla \cdot \bold u$

Note that this equation is the only belonging to general continuum equations, so only this equation have the same form for example also in Navier-Stokes equations.

On the other hand, the pressure in thermodynamics is the opposite of the partial derivative of the specific internal energy with respect to the specific volume:

$p (v,s) = - {\partial e (v,s) \over \partial v}$

since the internal energy in thermodynamics is a function of the two variables aforementioned, the pressure gradient contained into the momentum equation should be explicited as:

$- \nabla p (v,s) = - \frac {\partial p}{\partial v} \nabla v - \frac {\partial p}{\partial s} \nabla s = \frac {\partial^2 e}{\partial v^2} \nabla v + \frac {\partial^2 e}{\partial v \partial s}\nabla s$

It is convenient for brevity to switch the notation for the second order derivatives:

$- \nabla p (v,s) = e_{vv} \nabla v + e_{vs} \nabla s$

Finally, the energy equation:

${D e \over Dt} = - p v \nabla \cdot \bold u$

can be furtherly simplified in convective form by changing variable from specific energy to the specific entropy: in fact the first law of thermodynamics in local form can be written:

${D e \over Dt} = T {D s \over Dt} - p {D v \over Dt}$

by substituting the material derivative of the internal energy, the energy equation becomes:

$T {D s \over Dt} + \frac p {\rho^2} \left( {D \rho \over Dt} + \rho \nabla \cdot \bold u \right) = 0$

now the term between parenthesis is identically zero according to the conservation of mass, then the Euler energy equation becomes simply:

${D s \over Dt} = 0$

For a thermodynamic fluid, the compressible Euler equations are consequently best written as:

Euler equations (convective form, for a thermodynamic system) $\left\{\begin{align} {D v \over Dt} = v \nabla \cdot \bold u\\[1.2ex] \frac{D \mathbf{ u}}{D t} = v e_{vv} \nabla v + v e_{vs} \nabla s + \mathbf{g} \\[1.2ex] {D s \over Dt} =0 \end{align}\right.$

where:

• $v$ is the specific volume
• $\bold u$ is the flow velocity vector
• $s$ is the specific entropy

Note that, in the general case and not only in the incompressible case, the energy equation means that for an inviscid thermodynamic fluid the specific entropy is constant along the flow lines, also in a time-dependent flow. Basing on the mass conservation equation, one can put this equation in the conservation form:^[9]

${\partial \rho s \over \partial t} + \nabla \cdot (\rho s \mathbf u) = 0$

meaning that for an inviscid nonconductive flow a continuity equation holds for the entropy.

On the other hand, the two second-order partial derivatives of the specific internal energy in the momentum equation require the specification of the fundamental equation of state of the material considered, i.e. of the specific internal energy as function of the two variables specific volume and specific entropy:

$e = e(v,s)$

Note that the fundamental equation of state contains all the thermodynamic information about the system (Callen, 1985),^[10] exactly like the couple of a thermal equation of state together with a caloric equation of state.

Conservation form

The Euler equations in the Froude limit are equivalent to a single conservation equation with conserved quantity and associated flux respectively:

${\bold y}=\begin{pmatrix}\rho \\ \bold j \\E^t \end{pmatrix}; \qquad {\bold F}=\begin{pmatrix}\bold j\\ \frac 1 \rho \bold j \otimes \bold j + p \bold I\\(E^t +p) \frac \bold j \rho \end{pmatrix}.$

where:

• $\bold j = \rho \bold u$ is the momentum density, a conservation variable.
• $E^t = \rho e + 1/2 \rho u^2$ is the total energy density (total energy per unit volume).

Here $\bold y$ has length N+2 and $\bold F$ has size N(N+2).^[11] In general (not only in the Froude limit) Euler equations are expressible as:

Euler equation(s) (original conservation or Eulerian form) $\frac {\partial}{\partial t}\begin{pmatrix}\rho \\ \bold j \\E^t \end{pmatrix}+ \nabla \cdot \begin{pmatrix} \bold j \\\frac 1 \rho \bold j \otimes \bold j + p \bold I\\(E^t +p) \frac \bold j \rho \end{pmatrix} = \begin{pmatrix}0 \\ \bold f \\ \frac \bold j \rho \cdot \bold f \end{pmatrix}$

where:

• $\bold f = \rho \bold g$ is the force density, a conservation variable.

We remark that also the Euler equation even when conservative (no external field, Froude limit) have no Riemann invariants in general.^[12] Some further assumptions are required

However, we already mentioned that for a thermodynamic fluid the equation for the total energy density is equivalent to the conservation equation:

${\partial \over \partial t} (\rho s) + \nabla \cdot (\rho s \mathbf u) = 0$

Then the conservation equations in the case of a thermodynamic fluid are more simply expressed as:

Euler equation(s) (conservation form, for thermodynamic fluids) $\frac {\partial}{\partial t}\begin{pmatrix}\rho \\ \bold j \\S \end{pmatrix}+ \nabla \cdot \begin{pmatrix} \bold j \\\frac 1 \rho \bold j \otimes \bold j + p \bold I\\S \frac \bold j \rho \end{pmatrix} = \begin{pmatrix}0 \\ \bold f \\ 0 \end{pmatrix}$

where:

• $S = \rho s$ is the entropy density, a thermodynamic conservation variable.

Another possible form for the energy equation, being particularly useful for isobarics, is:

$\frac{\partial H^t }{\partial t}+ \nabla \cdot (H^t \mathbf u) = \mathbf u \cdot \mathbf f - \frac{\partial p}{\partial t}$

where:

• $H^t = E^t + p = \rho e + p + 1/2 \rho u^2$ is the total enthalpy density.

Quasilinear form and characteristic equations

Expanding the fluxes can be an important part of constructing numerical solvers, for example by exploiting (approximate) solutions to the Riemann problem. In regions where the state vector y varies smoothly, the equations in conservative form can be put in quasilinear form :

$\frac{\partial \bold y}{\partial t} + \bold A_i \frac{\partial \bold y}{\partial r_i} = {\bold 0}.$

where $\bold A_i$ are called the flux Jacobians defined as the matrices:

$\bold A_i (\bold y)=\frac{\partial \bold f_i (\bold y)}{\partial \bold y}.$

Obviously this jacobian does not exist in disconinuity regions (e.g. contact disconitnuities, shock waves in inviscid nonconductive flows). Note that if the flux Jacobians $\bold A_i$ are not functions of the state vector $\bold y$, the equations reveals linear.

Characteristic equations

The compressible Euler equations can be decoupled into a set of N+2 wave equations that describes sound in Eulerian continuum if they are expressed in characteristic variables instead of conserved variables.

In fact the tensor A is always diagonalizable. If the eigenvalues (the case of Euler equations) are all real the system is defined hyperbolic, and physically eignevalues represent the speeds of propagation of information.^[13] If they are all distinguished, the system is defined strictly hyperbolic (it will be proved to be the case of one-dimensional Euler equations). Furthermore, note that diagonalisation of compressible Euler equation is easier when the energy equation is expressed in the variable entropy (i.e. with equations for thermodynamic fluids) than in other energy variables. This will become clear by considering the 1D case.

If $\bold p_i$ is the right eigenvector of the matrix $\bold A$ corresponding to the eigenvalue $\lambda_i$, by building the projection matrix:

$\mathbf{P}=[ \mathbf{p}_1, \mathbf{p}_2, ... , \mathbf{p}_n]$

One can finally find the characteristic variables as:

$\mathbf{w}= \mathbf{P}^{-1}\mathbf{y},$

Since A is constant, multiplying the original 1-D equation in flux-Jacobian form with P⁻¹ yields the characteristic equations:^[14]

$\frac{\partial w_i}{\partial t} + \lambda_j \frac{\partial w_i}{\partial r_j} = 0_i$

The original equations have been decoupled into N+2 characteristic equations each describing a simple wave, with the eigenvalues being the wave speeds. The variables w_i are called the characteristic variablesand are a subset of the conservative variables. The solution of the initial value problem in terms of characteristic variables is finally very simple. In one spatial dimension it is:

$w_i (x, t) = w_i (x-\lambda_i t, 0)$

Then the solution in terms of the original conservative variables is obtained by transforming back:

$\mathbf{y}= \mathbf{P} \mathbf{w},$

this computation can be explicited as the linear combination of the eigenvectors:

$\mathbf{y}(x,t) = \sum_{i=1}^m w_i (x-\lambda_i t, 0) \mathbf p_i,$

Now it becomes apparent that the characteristic variables act as weights in the linear combination of the jacobian eigenvectors. The solution can be seen as superposition of waves, each of which is advected independently without change in shape. Each i-th wave has shape w_ip_i and speed of propagation λ_i. In the following we show a very simple example of this solution procedure.

Waves in 1D inviscid, nonconductive thermodynamic fluid

If one considers Euler equations for a thermodynamic fluid with the two further assumptions of one spatial dimension and free (no external field: g = 0) :

$\left\{\begin{align} {\partial v \over \partial t} + u {\partial v \over \partial x} - v {\partial u \over \partial x} = 0\\[1.2ex] {\partial u \over \partial t} + u {\partial u \over \partial x} - e_{vv} v {\partial v \over \partial x}- e_{vs} v {\partial s \over \partial x} = 0\\[1.2ex] {\partial s \over \partial t} + u {\partial s \over \partial x} = 0 \end{align}\right.$

If one defines the vector of variables:

${\bold y}=\begin{pmatrix}v\\ u \\s \end{pmatrix}$

recalling that $v$ is the specific volume, $u$ the flow speed, $s$ the specific entropy, the corresponding jacobian matrix is:

${\bold A}=\begin{pmatrix}u & -v & 0 \\ - e_{vv} v & u & - e_{vs} v \\ 0 & 0 & u \end{pmatrix}.$

At first one must find the eigenvalues of this matrix by solving the characteristic equation:

$\det(\mathbf A (\mathbf y)-\lambda (\mathbf y) \mathbf I))=0$

that is explicitly:

$\det\begin{bmatrix}u-\lambda & -v & 0 \\ - e_{vv} v & u-\lambda & - e_{vs} v \\ 0 & 0 & u-\lambda \end{bmatrix}=0$

This determinant is very simple: the fastest computation starts on the last row, since it has the highest number of zero elements.

$(u-\lambda) \det \begin{bmatrix}u-\lambda & -v \\ - e_{vv} v & u -\lambda \end{bmatrix}=0$

Now by computing the determinant 2x2:

$(u-\lambda) \left( (u-\lambda)^2 - e_{vv} v^2 \right) =0$

by defining the parameter:

$a(v,s) \equiv v \sqrt {e_{vv}}$

or equivalently in mechanical variables, as:

$a(\rho,p) \equiv \sqrt {\partial p \over \partial \rho}$

This parameter is always real according to the second law of thermodynamics. In fact the second law of thermodynamics can be expressed by several postulates. The most elementary of them in mathematical terms is the statement of convexity of the fundamental equation of state, i.e. the hessian matrix of the specific energy expresseed as function of specific volume and specific entropy:

$\begin{pmatrix}e_{vv} & e_{vs} \\ e_{vs} & e_{ss} \end{pmatrix}$

is defined positive. This statement corresponds to the two conditions:

$\left\{\begin{align} e_{vv}>0\\[1.2ex] e_{vv}e_{ss}-e_{vs}^2 > 0 \end{align}\right.$

The first condition is the one ensuring the parameter a is defined real.

The characteristic equation finally results:

$(u-\lambda) \left( (u-\lambda)^2 - a^2 \right) =0$

That has three real solutions:

$\lambda_1(v,u,s) = u-a(v,s) \quad \lambda_2(u)= u, \quad \lambda_3(v,u,s) = u+a(v,s)$

Then the matrix has three real eigenvalues all distinguished: the 1D Euler equations are a strictly hyperbolic system.

At this point one should determine the three eigenvectors: each one is obtained by substituting one eigenvalue in the eigenvalue equation and then solving it. By substituting the first eigenvalue λ₁ one obtains:

$\begin{pmatrix}a & -v & 0 \\ - e_{vv} v & a & - e_{vs} v \\ 0 & 0 & a \end{pmatrix} \begin{pmatrix}v_1\\ u_1 \\s_1 \end{pmatrix}=0$

Basing on the third equation that simply has solution s₁=0, the system reduces to:

$\begin{pmatrix}a & -v \\-a^2 /v& a \end{pmatrix} \begin{pmatrix}v_1\\ u_1 \end{pmatrix}=0$

The two equations are redundant as usual, then the eigenvector is defined with a multiplying constant. We choose as right eigenvector:

$\mathbf p_1=\begin{pmatrix}v\\ a \\0\end{pmatrix}$

The other two eigenvectors can be found with analogous procedure as:

$\mathbf p_2=\begin{pmatrix} e_{vs} \\ 0\\ - \left(\frac a v \right)^2 \end{pmatrix}, \qquad \mathbf p_3=\begin{pmatrix}v\\ -a \\0\end{pmatrix}$

Then the projection matrix can be built:

$\mathbf P (v,u,s)=( \mathbf{p}_1, \mathbf{p}_2, \mathbf{p}_3) =\begin{pmatrix} v & e_{vs} & v\\ a & 0 & -a \\ 0 & - \left(\frac a v \right)^2 & 0 \end{pmatrix}$

Finally it becomes apparent that the real parameter a previously defined is the speed of propagation of the information characteristic of the hyperbolic system made of Euler equations, i.e. it is the wave speed. It remains to be shown that the sound speed corresponds to the particular case of an isoentropic transformation:

$a_s \equiv \sqrt {\left({\partial p \over \partial \rho} \right)_s}$

Compressibility and sound speed

Sound speed is defined as the wavespeed of an isentropic transformation:

$a_s(\rho,p) \equiv \sqrt {\left({\partial p \over \partial \rho} \right)_s}$

by the definition of the isoentropic compressibility:

$K_s (\rho,p) \equiv \frac 1 \rho \left({\partial p \over \partial \rho} \right)_s$

the soundspeed results always the square root of ratio between the isoentropic compressibility and the density:

$a_s \equiv \sqrt {\frac {K_s} \rho}$

Ideal gas

The sound speed in an ideal gas depends only on its temperature:

$a_s (T) = \sqrt {\gamma \frac T m}$

Deduction of the form valid for ideal gases 

In an ideal gas the isoentropic transformation is described by the Poisson's law:

$d (p \rho^{-\gamma})_s =0$

where γ is the heat capacity ratio, a constant for the material. By explicitating the differentials:

$\rho^{-\gamma} (d p)_s + \gamma p \rho^{-\gamma-1} (d \rho)_s =0$

and by dividing for ρ^−γ dρ:

$\left({\partial p \over \partial \rho} \right)_s= \gamma p \rho$

Then by substitution in the general definitions for an ideal gas the isentropic compressibility is simply proportional to the pressure:

$K_s (p) = \gamma p$

and the sound speed results (Newton–Laplace law):

$a_s (\rho,p) = \sqrt {\gamma \frac p \rho}$

Notably, for an ideal gas the ideal gas law holds, that in mathematical form is simply:

$p = n T$

where n is the number density, and T is the absolute temperature, provided it is measured in energetic units (i.e. in joules) through multiplication with the Boltzmann constant. Since the mass density is proportional to the number density through the average molecular mass m of the material:

$\rho = m n$

The ideal gas law can be recast into the formula:

$\frac p \rho = \frac T m$

By substituting this ratio in the Newton–Laplace law, the expression of the sound speed into an ideal gas as function of temperature is finally achieved.

Since the specific enthalpy in an ideal gas is proportional to its temperature:

$h = c_p T = \frac {\gamma}{\gamma-1} \frac T m$

the sound speed in an ideal gas can also be made dependent only on its specific enthalpy:

$a_s (h) = \sqrt {(\gamma -1) h}$

Bernoulli's theorems for steady inviscid flow

Incompressible case and Lamb's form

The vector calculus identity of the cross product of a curl holds:

$\mathbf{v \ \times } \left( \mathbf{ \nabla \times F} \right) =\nabla_F \left( \mathbf{v \cdot F } \right) - \mathbf{v \cdot \nabla } \mathbf{ F} \ ,$

where the Feynman subscript notation $\nabla_F$ is used, which means the subscripted gradient operates only on the factor $\bold F$.

Lamb in his famous classical book Hydrodynamics (1895), still in print, used this identity to change the convective term of the flow velocity in rotational form:^[15]

$\mathbf u \cdot \nabla \mathbf u = \frac 1 2 \nabla (u^2) + (\nabla \times \mathbf u) \times \mathbf u$

the Euler momentum equation in Lamb's form becomes:

$\frac{\partial \mathbf{ u}}{\partial t} + \frac 1 2 \nabla (u^2) + (\nabla \times \mathbf u) \times \mathbf u + \frac {\nabla p}{\rho} = \mathbf{g}$

Now, basing on the other identity:

$\nabla \left( \frac {p}{\rho} \right) = \frac {\nabla p}{\rho} - \frac{p}{\rho^2} \nabla \rho$

the Euler momentum equation assumes a form that is optimal to demonstrate Bernoulli's theorem for steady flows:

$\nabla \left(\frac 1 2 u^2 + \frac p \rho \right) - \mathbf g = \frac{p}{\rho^2} \nabla \rho + \mathbf u \times (\nabla \times \mathbf u) - \frac{\partial u}{\partial t}$

In fact, in case of an external conservative field, by defining its potential φ:

$\nabla \left( \frac 1 2 u^2 + \phi + \frac p \rho \right) = \frac{p}{\rho^2} \nabla \rho + \mathbf u \times (\nabla \times \mathbf u) - \frac{\partial u}{\partial t}$

In case of a steady flow the time derivative of the flow velocity disappears, so the momentum equation becomes:

$\nabla \left( \frac 1 2 u^2 + \phi + \frac p \rho \right) = \frac{p}{\rho^2} \nabla \rho + \mathbf u \times (\nabla \times \mathbf u)$

And by projecting the momentum equation on the flow direction, i.e. along a streamline, the cross product disappears due to a vector calculus identity of the triple scalar product:

$\mathbf u \cdot \nabla \left( \frac 1 2 u^2 + \phi + \frac p \rho \right) = \frac{p}{\rho^2} \mathbf u \cdot \nabla \rho$

In the steady incompressible case the mass equation is simply:

$\mathbf u \cdot \nabla \rho = 0$

, that is the mass conservation for a steady incompressible flow states that the density along a streamline is constant. Then the Euler momentum equation in the steady incompressible case becomes:

$\mathbf u \cdot \nabla \left( \frac 1 2 u^2 + \phi + \frac p \rho \right) = 0$

The convenience of defining the total head for an inviscid liquid flow is now apparent:

$b_l \equiv \frac 1 2 u^2 + \phi + \frac p \rho$

, in fact the above equation can be simply written as:

$\mathbf u \cdot \nabla b_l = 0$

That is, the momentum balance for a steady inviscid and incompressible flow in an external conservative field states that the total head along a streamline is constant.

Compressible case

In the most general steady (compressibile) case the mass equation in conservation form is:

$\nabla \cdot \mathbf j = 0$

and the steady energy equation in conservation form is:

$\nabla \cdot \left( \left(e + \frac 1 2 u^2 + \frac p \rho \right) \mathbf j \right) = \mathbf j \cdot \mathbf g$

Thanks to the mass conservation equation and to the definition of the momentum density, the first member becomes simply:

$\nabla \cdot \left( \left(e + \frac 1 2 u^2 + \frac p \rho \right) \mathbf j \right) =\rho \mathbf u \cdot \nabla \left( e + \frac 1 2 u^2 + \frac p \rho \right)$

and for an external conservative field, the second member becomes:

$\mathbf j \cdot \mathbf g = - \rho \mathbf u \cdot \nabla \phi$

Then by dividing for the density, the energy equation becomes:

$\mathbf u \cdot \nabla \left( e + \frac p \rho + \frac 1 2 u^2 + \phi \right) = 0$

Since the external field potential is usually small compared to the other terms, it is convenient to group the latters in the total enthalpy:

$h^t \equiv e + \frac p \rho + \frac 1 2 u^2$

and the Bernoulli invariant for an inviscid gas flow is:

$b_g \equiv h^t + \phi = b_l + e$

, in fact the above equation can be always written as:

$\mathbf u \cdot \nabla b_g = 0$

That is, the energy balance for a steady inviscid flow in an external conservative field states that the sum of the total enthalpy and the external potential is constant along a streamline.

In the usual case of small potential field, simply:

$\mathbf u \cdot \nabla h^t \sim 0$

Friedman form and Crocco form

By substituting the pressure gradient with the entropy and enthalpy gradient, according to the first law of thermodynamics in the enthalpy form:

$v \nabla p = -T \nabla s + \nabla h$

in the convective form of Euler momentum equation, one arrives to:

$\frac{D\mathbf u}{Dt}=T \nabla\,s-\nabla \,h$

Friedmann deduced this equation for the particular case of a perfect gas and published it in 1922.^[16] However, this equation is general for an inviscid nonconductive fluid and no equation of state is implicit in it.

On the other hand, by substituting the enthalpy form of the first law of thermodynamics in the rotational form of Euler momentum equation, one obtains:

$\frac{\partial \mathbf{ u}}{\partial t} + \frac 1 2 \nabla (u^2) + (\nabla \times \mathbf u) \times \mathbf u + \frac {\nabla p}{\rho} = \mathbf{g}$

and by defining the specific total enthalpy:

$h^t =h + \frac 1 2 u^2$

one arrives to the Crocco–Vazsonyi form^[17] (Crocco, 1937) of the Euler momentum equation:

$\frac{\partial \mathbf{ u}}{\partial t} + (\nabla \times \mathbf u) \times \mathbf u - T \nabla s + \nabla h^t = \mathbf{g}$

In the steady case the two variables entropy and total enthalpy are particularly useful since Euler equations can be recast into the Crocco's form:

$\left\{\begin{align} \mathbf u \times \nabla \times \mathbf u + T \nabla s - \nabla h^t = \mathbf{g} \\ \mathbf u \cdot \nabla s = 0 \\ \mathbf u \cdot \nabla h^t = 0 \end{align} \right.$

Finally if the flow is also isothermal:

$T \nabla s = \nabla (T s)$

by defining the specific total Gibbs free energy:

$g^t \equiv h^t + Ts$

the Crocco's form can be reduced to:

$\left\{\begin{align} \mathbf u \times \nabla \times \mathbf u - \nabla g^t = \mathbf{g} \\ \mathbf u \cdot \nabla g^t = 0 \end{align} \right.$

From these relationships one deduces that the specific total free energy is uniform in a steady, irrotational, isothermal, isoentropic, inviscid flow.

Discontinuities

The Euler equations are quasilinear hyperbolic equations and their general solutions are waves. Under certain assumptions they can be simplified leading to Burgers equation. Much like the familiar oceanic waves, waves described by the Euler Equations 'break' and so-called shock waves are formed; this is a nonlinear effect and represents the solution becoming multi-valued. Physically this represents a breakdown of the assumptions that led to the formulation of the differential equations, and to extract further information from the equations we must go back to the more fundamental integral form. Then, weak solutions are formulated by working in 'jumps' (discontinuities) into the flow quantities – density, velocity, pressure, entropy – using the Rankine–Hugoniot equations. Physical quantities are rarely discontinuous; in real flows, these discontinuities are smoothed out by viscosity and by heat transfer. (See Navier–Stokes equations)

Shock propagation is studied – among many other fields – in aerodynamics and rocket propulsion, where sufficiently fast flows occur.

To properly compute the continuum quantities in discontinuous zones (for example shock waves or boundary layers) from the local forms^[18] (all the above forms are local forms, since the variables being described are typical of one point in the space caonsidered, i.e. they are local variables) of Euler equations through finite difference methods generally too many space points and time steps would be necessary for the memory of computers now and in the near future. In these cases it is mandatory to avoid the local forms of the conservation equations, passing some weak forms, like the finite volume one.

Rankine–Hugoniot equations

Starting from the simplest case, one consider a steady free conservation equation in conservation form in the space domain:

$\nabla \cdot \mathbf F = \mathbf 0$

where in general F is the flux matrix. By integrating this local equation over a fixed volume V_m, it becomes:

$\int_{V_m} \nabla \cdot \mathbf F dV = \mathbf 0.$

Then, basing on the divergence theorem, we can transform this integral in a boundary integral of the flux:

$\oint_{\partial V_m} \mathbf F ds = \mathbf 0.$

This global form simply states that there is no net flux of a conserved quantity passing through a region in the case steady and without source. In 1D the volume reduces to an interval, its boundary being its extrema, then the divergence theorem reduces to the fundamental theorem of calculus:

$\int_{x_m}^{x_{m+1}} \mathbf F(x') dx' = \mathbf 0,$

that is the simple finite difference equation, known as the jump relation:

$\Delta \mathbf F = \mathbf 0.$

That can be made explicit as:

$\mathbf F_{m+1} - \mathbf F_m= \mathbf 0$

where the notation employed is:

$\mathbf F_{m}=\mathbf F(x_m).$

Or, if one performs an indefinite integral:

$\mathbf F - \mathbf F_0= \mathbf 0.$

On the other hand, a transient conservation equation:

${\partial y \over \partial t} + \nabla \cdot \mathbf F = \mathbf 0$

brings to a jump relation:

$\frac {d x}{dt} \, \Delta u = \Delta \mathbf F.$

For one-dimensional Euler equations the conservation variables and the flux are the vectors:

$\mathbf y = \begin{pmatrix} 1/v \\ j \\ E^t \end{pmatrix},$

$\mathbf F = \begin{pmatrix} j \\ v j^2 + p \\ v j (E^t + p) \end{pmatrix},$

where:

• $v$ is the specific volume,
• $j$ is the mass flux.

In the one dimensional case the correspondent jump relations, called the Rankine–Hugoniot equations, are:^[19]

$\left\{\begin{align} \frac {d x}{dt} \Delta (1/v) = \Delta j \\[1.2ex] \frac {d x}{dt} \Delta j = \Delta (v j^2 +p)\\[1.2ex] \frac {d x}{dt} \Delta E^t = \Delta (j v (E^t + p))\end{align}\right. .$

In the steady one dimensional case the become simply:

$\left\{\begin{align} \Delta j = 0\\[1.2ex] \Delta (v j^2 +p) = 0\\[1.2ex] \Delta (j (\frac {E^t} \rho + \frac p \rho))= 0\end{align}\right. .$

Thanks to the mass difference equation, the energy difference equation can be simplified without any restriction:

$\left\{\begin{align} \Delta j = 0\\[1.2ex] \Delta (v j^2 + p)= 0\\[1.2ex] \Delta h^t = 0\end{align}\right.,$

where $h^t$ is the specific total enthalpy.

These are the usually expressed in the convective variables:

$\left\{\begin{align} \Delta j = 0\\[1.2ex] \Delta (\frac {u^2} v + p)= 0\\[1.2ex] \Delta \left(e + \frac 1 2 u^2 + p v \right) = 0 \end{align}\right.,$

where:

• $u$ is the flow speed
• $e$ is the specific internal energy.

Note that the energy equation is an integral form of the Bernoulli equation in the compressible case. The former mass and momentum equations by substitution lead to the Rayleigh equation:

$\frac{\Delta p}{\Delta v} = - \frac {u_0^2}{v_0}.$

Since the second member is a constant, the Rayleigh equation always describes a simple line in the pressure volume plane not depending of any equation of state, i.e. the Rayleigh line. By substituition in the Rankine–Hugoniot equations, that can be also made explicit as:

$\left\{\begin{align} \rho u = \rho_0 u_0\\[1.2ex] \rho u^2 + p= \rho_0 u_0^2 + p_0\\[1.2ex] e + \frac 1 2 u^2 + \frac p \rho = e_0 + \frac 1 2 u_0^2 + \frac {p_0}{\rho_0} \end{align}\right. .$

One can also obtain the kinetic equation and to the Hugoniot equation. The analytical passages are not shown here for brevity.

These are respectively:

$\left\{\begin{align} u^2(v,p) = u_0^2 + (p-p_0)(v_0+v)\\[1.2ex] e(v,p)= e_0 + \frac 1 2 (p+p_0)(v_0-v)\end{align}\right. .$

The Hugoniot equation, coupled with the fundamental equation of state of the material:

$e = e(v,p)$

describes in general in the pressure volume plane a curve passing by the conditions (v₀,p₀), i.e. the Hugoniot curve, whose shape strongly depends on the type of material considered.

It is also customary to define a Hugoniot function:^[20]

$\mathfrak h (v,s) \equiv e(v,s) - e_0 +\frac 1 2 (p(v,s) + p_0) (v -v_0)$

allowing to quantify deviations from the Hugoniot equation, similarly to the previous definition of the hydraulic head, useful for the deviations from the Bernoulli equation.

Finite volume form

On the other hand, by integrating a generic conservation equation:

$\frac {\partial \mathbf y}{\partial t} +\nabla \cdot \mathbf F = \mathbf s$

on a fixed volume V_m, and then basing on the divergence theorem, it becomes:

$\frac {d}{dt} \int_{V_m} \mathbf y dV + \oint_{\partial V_m} \mathbf F \cdot \hat n ds = \mathbf S .$

By integrating this equation also over a time interval:

$\int_{V_m} \mathbf y(\mathbf r, t_{n+1}) \, dV - \int_{V_m} \mathbf y(\mathbf r, t_n) \, dV+ \int_{t_n}^{t_{n+1}} \oint_{\partial V_m} \mathbf F \cdot \hat n \, ds \, dt = \mathbf 0 .$

Now by defining the node conserved quantity:

$\mathbf y_{m,n} \equiv \frac 1 {V_m} \int_{V_m} \mathbf y(\mathbf r, t_n) \, dV ,$

we deduce the finite volume form:

$\mathbf{y}_{m,n+1}=\mathbf{y}_{m,n} - \frac{1}{V_m} \int_{t_n}^{t_{n+1}} \oint_{\partial V_m} \mathbf{F} \cdot \hat{n}\, ds \, dt .$

In particular, for Euler equations, once the conserved quantities have been determined, the convective variables are deduced by back substitution:

$\left\{\begin{align} \mathbf u_{m,n} = \frac {\mathbf j_{m,n}}{\rho_{m,n}}\\[1.2ex] e_{m,n} = \frac {E^t_{m,n}}{\rho_{m,n}} -\frac 1 2 u^2_{m,n}\\[1.2ex] \end{align}\right. .$

Then the explicit finite volume expressions of the original convective variables are:^[21]

Euler equations (Finite volume form) $\left\{\begin{align} \rho_{m,n+1}=\rho_{m,n} - \frac 1 {V_m} \int_{t_n}^{t_{n+1}} \oint_{\partial V_m} \rho \mathbf u \cdot \hat n \, ds \, dt \\[1.2ex] \mathbf u_{m,n+1}=\mathbf u_{m,n} - \frac 1 {\rho_{m,n} V_m} \int_{t_n}^{t_{n+1}} \oint_{\partial V_m} (\rho \mathbf u \otimes \mathbf u - p \mathbf I) \cdot \hat n \, ds \, dt \\[1.2ex] \mathbf e_{m,n+1}=\mathbf e_{m,n} - \frac 1 2 (u^2_{m,n+1} - u^2_{m,n}) - \frac 1 {\rho_{m,n} V_m} \int_{t_n}^{t_{n+1}} \oint_{\partial V_m} (\rho e + \frac 1 2 \rho u^2 + p) \mathbf u \cdot \hat n \, ds \, dt \\[1.2ex] \end{align}\right. .$

Constraints

It has been shown that Euler equations are not a complete set of equations, but they require some additional constraints to admit a unique solution: these are the equation of state of the material considered. To be constistent with thermodynamics these equations of state should satisfy the two laws of thermodynamics. On the other hand, by definition non-equilibrium system are described by laws lying outside these laws. In the following we list some very simple equations of state and the corresponding influence on Euler equations.

Ideal polytropic gas

For an ideal polytropic gas the fundamental equation of state is:^[22]

$e (v,s) = e_0 e^{ (\gamma - 1) m (s - s_0)} \left({v_0 \over v}\right)^{\gamma -1}$

where $e$ is the specific energy, $v$ is the specific volume, $s$ is the specific entropy, $m$ is the molecular mass, $\gamma$ here is considered a constant (polytropic process), and can be shown to correspond to the heat capacity ratio. This equation can be shown to be consistent with the usual equations of state employed by thermodynamics.

Demostration of consistency with the thermodynamics of an ideal gas 

By the thermodynamic definition of temperature:

$T(e) \equiv {\partial e \over \partial s} = (\gamma - 1) m e$

Where the temperature is measured in energy units. At first, note that by combining these two equations one can deduce the ideal gas law:

$p v = m T$

or, in the usual form:

$p = n T$

where: $n \equiv \frac m v$ is the number density of the material. On the other hand the ideal gas law is less strict than the original fundamental equation of state considered.

Now consider the molar heat capacity associated to a process x:

$c_x = \left(m T {\partial s \over \partial T}\right)_x$

according to the first law of thermodynamics:

$d e(v,s)=-p dv + T \, ds$

it can be simply expressed as:

$c_x \equiv m \left({\partial e \over \partial T}\right)_x + m p \left({\partial v \over \partial T}\right)_x$

Now inverting the equation for temperature T(e) we deduce that for an ideal polytropic gas the isocoric heat capacity is a constant:

$c_v \equiv m \left({\partial e \over \partial T}\right)_v = m {d e \over dT} = \frac {1}{(\gamma -1)}$

and similarly for an ideal polytropic gas the isobaric heat capacity results constant:

$c_p \equiv m \left({\partial e \over \partial T}\right)_p + m p \left({\partial v \over \partial T}\right)_p = m {d e \over dT} + p \left({\partial v \over \partial T}\right)_p = \frac {1}{(\gamma -1)} + 1$

This brings to two important relations between heat capacities: the constant gamma actually represents the heat capacity ratio in the ideal polytropic gas:

$\frac {c_p}{c_v}= \gamma$

and one also arrives to the Meyer's relation:

$c_p = c_v+1$

The specific energy is then, by inverting the relation T(e):

$e(T) = \frac {mT} {\gamma - 1} = c_v m T$

The specific enthalpy results by substitution of the latter and of the ideal gas law:

$h(T) \equiv e(T) + (p v)(T) = c_v m T + m T = c_p m T$

From this equation one can derive the equation for pressure by its thermodynamic definition:

$p(v,e) \equiv - {\partial e \over \partial v} = (\gamma - 1) \frac e v$

By inverting it one arrives to the mechanical equation of state:

$e(v,p) = \frac {pv}{\gamma - 1}$

Then for an ideal gas the compressible Euler equations can be simply expressed in the mechanical or primitive variables specific volume, flow velocity and pressure, by taking the set of the equations for a thermodynamic system and modifying the energy equation into a pressure equation through this mechanical equation of state. At last, in convective form they result:

Euler equations for an ideal polytropic gas (convective form)[23] $\left\{\begin{align} {D v \over Dt} = v \nabla \cdot \bold u\\[1.2ex] \frac{D \mathbf{ u}}{D t} = v \nabla p + \mathbf{g} \\[1.2ex] {Dp \over Dt} = \gamma p \nabla \cdot \bold u \end{align}\right.$

and in one-dimensional quasilinear form they results:

$\frac{\partial \bold y}{\partial t} + \bold A \frac{\partial \bold y}{\partial x} = {\bold 0}.$

where the conservative vector variable is:

${\bold y}=\begin{pmatrix}v\\ u \\p \end{pmatrix}$

and the corresponding jacobian matrix is:^[24][25]

${\bold A}=\begin{pmatrix}u & -v & 0 \\ 0 & u & v \\ 0 & \gamma p & u \end{pmatrix}.$

Steady flow in material coordinates

In the case of steady flow, it is convenient to choose the Frenet–Serret frame along a streamline as the coordinate system for describing the steady momentum Euler equation:^[26]

$\boldsymbol{u}\cdot\nabla \boldsymbol{u} = - \frac{1}{\rho} \nabla p,$

where $\bold u$, $p$ and $\rho$ denote the flow velocity, the pressure and the density, respectively.

Let $\{ \bold e_s, \bold e_n, \bold e_b \}$ be a Frenet–Serret orthonormal basis which consists of a tangential unit vector, a normal unit vector, and a binormal unit vector to the streamline, respectively. Since a streamline is a curve that is tangent to the velocity vector of the flow, the left-handed side of the above equation, the convective derivative of velocity, can be described as follows:

$\begin{align} \boldsymbol{u}\cdot\nabla \boldsymbol{u} \\ &= u\frac{\partial}{\partial s}(u\boldsymbol{e}_s) &(\boldsymbol{u} = u \boldsymbol{e}_s ,~ {\partial / \partial s} \equiv \boldsymbol{e}_s\cdot\nabla)\\ &= u\frac{\partial u}{\partial s}\boldsymbol{e}_s + \frac{u^2}{R} \boldsymbol{e}_n &(\because~ \frac{\partial \boldsymbol{e}_s}{\partial s}=\frac{1}{R}\boldsymbol{e}_n), \end{align}$

where $R$ is the radius of curvature of the streamline.

Therefore, the momentum part of the Euler equations for a steady flow is found to have a simple form:

$\begin{cases} \displaystyle u\frac{\partial u}{\partial s} = -\frac{1}{\rho}\frac{\partial p}{\partial s},\\ \displaystyle {u^2 \over R} = -\frac{1}{\rho}\frac{\partial p}{\partial n} &({\partial / \partial n}\equiv\boldsymbol{e}_n\cdot\nabla),\\ \displaystyle 0 = -\frac{1}{\rho}\frac{\partial p}{\partial b} &({\partial / \partial b}\equiv\boldsymbol{e}_b\cdot\nabla). \end{cases}$

For barotropic flow $(\rho = \rho(p))$, Bernoulli's equation is derived from the first equation:

$\frac{\partial}{\partial s}\left(\frac{u^2}{2} + \int \frac{\mathrm{d}p}{\rho}\right) =0.$

The second equation expresses that, in the case the streamline is curved, there should exist a pressure gradient normal to the streamline because the centripetal acceleration of the fluid parcel is only generated by the normal pressure gradient.

The third equation expresses that pressure is constant along the binormal axis.

Streamline curvature theorem

Let $r$ be the distance from the center of curvature of the streamline, then the second equation is written as follows:

$\frac{\partial p}{\partial r} = \rho \frac{u^2}{r}~(>0),$

where ${\partial / \partial r} = -{\partial /\partial n}.$

This equation states:

In a steady flow of an inviscid fluid without external forces, the center of curvature of the streamline lies in the direction of decreasing radial pressure.

Although this relationship between the pressure field and flow curvature is very useful, it doesn't have a name in the English-language scientific literature.^[27] Japanese fluid-dynamicists call the relationship the "Streamline curvature theorem". ^[28]

This "theorem" explains clearly why there are such low pressures in the centre of vortices,^[27] which consist of concentric circles of streamlines. This also is a way to intuitively explain why airfoils generate lift forces.^[27]

Exact solutions

In incompressible flow, all potential flow solutions are also solutions of the Euler equations.^[29]

A two-dimensional parallel shear-flow.

Solutions to the Euler equations with vorticity are:

• parallel shear flows – where the flow is unidirectional, and the flow velocity only varies in the cross-flow directions, e.g. in a Cartesian coordinate system $(x,y,z)$ the flow is for instance in the $x$-direction – with the only non-zero velocity component being $u_x(y,z)$ only dependent on $y$ and $z$ and not on $x.$^[30]
• Arnold–Beltrami–Childress flow – an exact solution of the incompressible Euler equations.
• Two solutions of the three-dimensional Euler equations with cylindrical symmetry have been presented by Gibbon, Moore and Stuart in 2003.^[31] These two solutions have infinite energy; they blow up everywhere in space in finite time.

See also

• Bernoulli's theorem
• Kelvin's circulation theorem
• Cauchy equations
• Madelung equations
• Froude number
• Navier–Stokes equations
• Burgers equation

Notes

1. see Toro, p. 24
2. Anderson, John D. (1995), Computational Fluid Dynamics, The Basics With Applications. ISBN 0-07-113210-4
3. E226 – Principes generaux du mouvement des fluides
4. Christodoulou, Demetrios (October 2007). "The Euler Equations of Compressible Fluid Flow" (PDF). Bulletin of the American Mathematical Society 44 (4): 581–602. doi:10.1090/S0273-0979-07-01181-0. Retrieved June 13, 2009.
5. Hunter, J.K. An Introduction to the Incompressible Euler Equations
6. Hunter, An introduction to incompressible Euler equations, p.2
7. In 3D for example $\bold y$ has length 5, $\bold I$ has size 3x3 and $\bold F$ has size 3x5, so the explicit forms are:

   ${\bold y}=\begin{pmatrix}\rho \\ \rho u_1 \\ \rho u_2 \\ \rho u_3 \\0\end{pmatrix}; \quad {\bold F}=\begin{pmatrix}\rho u_1 & \rho u_2 & \rho u_3 \\ \rho u_1^2 + p & \rho u_1u_2 & \rho u_1u_3 \\ \rho u_1 u_2 & \rho u_2^2 + p & \rho u_2u_3 \\ \rho u_3 u_1 & \rho u_3 u_2 & \rho u_3^2 + p \\ u_1 & u_2 & u_3 \end{pmatrix}.$

8. (Italian) Quartapelle, Autieri, Fluidodinamica comprimibile, Chap. 9, p.13
9. Landau, Lifshits, Fluid Mechanics, par. 1.1, eq. 2.6 and 2.7
10. L.F. Henderson, par. 2.6 Thermodynamic properties of materials, in Handbook of Shock Waves, p. 152
11. In 3D for example y has length 5, I has size 3x3 and F has size 3x5, so the explicit forms are:

    ${\bold y}=\begin{pmatrix} j_1 \\ j_2 \\ j_3 \end{pmatrix}; \quad {\bold F}=\begin{pmatrix}j_1 & j_2 & j_3 \\ \frac{ j_1^2}\rho + p & \frac{j_1j_2}\rho& \frac{j_1j_3} \rho\\ \frac{j_1j_2}\rho & \frac {j_2^2} \rho + p & \frac{j_2j_3} \rho\\ \frac{j_3j_1}\rho & \frac{j_3j_2}{\rho} & \frac {j_3^2} \rho + p \\ (E^t+p) \frac {j_1} \rho & (E^t+p) \frac {j_2} \rho & (E^t+p) \frac{j_3} \rho \end{pmatrix}.$

12. Chorin, Marsden, A mathematical introduction to fluid mechanics, par. 3.2 Shocks, p.118
13. Toro, Rienmann solvers and numerical methods for fluid dynamics, par 2.1 Quasi-linear Equations: Basic concept, p.44
14. Toro, op. cit., par 2.3 Linear Hyperbolic System, p.52
15. (Italian)Valorani, Nasuti, Metodi di analisi delle turbomacchine, pp. 11–12
16. Friedmann A. An essay on hydrodynamics of compressible fluid (Опыт гидромеханики сжимаемой жидкости), Petrograd, 1922, 516 p., reprinted in 1934 under the editorship of Nikolai Kochin (see the first formula on page 198 of the reprint).
17. Handbook of Shock Waves, Vol I, par. 2.12 Crocco's theorem, p.177
18. Sometimes the local and the global forms are also called respectively differential and non-differential, but this is not appropriate in all cases. For example, this is appropriate for Euler equations, while it is not for Navier-Stokes equations since in their global form there are some residual spatial first-order derivative operators in all the caractheristic transport terms that in the local form contains second-order spatial derivatives.
19. Chorin, Marsden, A mathematical introduction to fluid mechanics, par. 3.2 Shocks, p. 122
20. L.F. Henderson, par. 2.96 The Bethe–Weyl theorem, in Handbook of Shock Waves, p. 167
21. (Italian) Quartapelle, Autieri, Fluidodinamica comprimibile, par. 11.10: Forma differenziale: metodo dei volumi finiti, p.161
22. (Italian) Quartapelle, Autieri, Fluidodinamica comprimibile, Appendix E, p. A-61
23. Toro, par 3.1.2 Nonconservative formulations, p.91
24. M. Zingale, Notes on the Euler equations
25. Toro, p.92
26. James A. Fay (June 1994). Introduction to Fluid Mechanics. MIT Press. ISBN 0-262-06165-1. see "4.5 Euler's Equation in Streamline Coordinates" pp. 150–152 <https://books.google.com/books?id=XGVpue4954wC&pg=150>
27. Babinsky, Holger (November 2003), "How do wings work?" (PDF), Physics Education
28. 今井 功 (IMAI, Isao) (November 1973). 『流体力学(前編)』(Fluid Dynamics 1) (in Japanese). 裳華房 (Shoukabou). ISBN 4-7853-2314-0.
29. Marchioro, C.; Pulvirenti, M. (1994). Mathematical Theory of Incompressible Nonviscous Fluids. Applied Mathematical Sciences 96. New York: Springer. p. 33. ISBN 0 387 94044 8.
30. Friedlander, S.; Serre, D., eds. (2003). Handbook of Mathematical Fluid Dynamics – Volume 2. Elsevier. p. 298. ISBN 978 0 444 51287 1.
31. Gibbon, J.D.; Moore, D.R.; Stuart, J.T. (2003). "Exact, infinite energy, blow-up solutions of the three-dimensional Euler equations". Nonlinearity 16 (5): 1823–1831. doi:10.1088/0951-7715/16/5/315.

Further reading

• Batchelor, G. K. (1967). An Introduction to Fluid Dynamics. Cambridge University Press. ISBN 0-521-66396-2.
• Thompson, Philip A. (1972). Compressible Fluid Flow. New York: McGraw-Hill. ISBN 0-07-064405-5.
• Toro, E.F. (1999). Riemann Solvers and Numerical Methods for Fluid Dynamics. Springer-Verlag. ISBN 3-540-65966-8.