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# Navier-Stokes equations/Derivation

The intent of this article is to highlight the important points of the derivation of the Navier-Stokes equations as well as the application and formulation for different families of fluids.

## Basic assumptions

The Navier-Stokes equations are based on the assumption that the fluid, at the scale of interest, is a continuum, in other words is not made up of discrete particles but rather a continuous substance. Another necessary assumption is that all the of fields of interest like pressure, velocity, density, temperature and so on are differentiable, weakly at least.

The equations are derived from the basic principles of conservation of mass, momentum, and energy. For that matter, sometimes it is necessary to consider a finite arbitrary volume, called a control volume, over which these principles can be applied. This finite volume is denoted by Ω and its bounding surface $\partial \Omega$. The control volume can remain fixed in space or can move with the fluid.

## The convective derivative

Main article: convective derivative

Changes in properties of a moving fluid can be measured in two different ways. One can measure a given property by either carrying out the measurement on a fixed point in space as particles of the fluid pass by, or by following a parcel of fluid along its streamline. The derivative of a field with respect to a fixed position in space is called the spatial derivative while the derivative following a moving parcel is called the convective derivative.

The convective derivative is defined as the operator: $\frac{D}{Dt}(\star) \ \stackrel{\mathrm{def}}{=}\ \frac{\partial}{\partial t}(\star) + \mathbf{v}\cdot\nabla (\star)$

where $\mathbf{v}$ is the velocity of the fluid. The first term on the right-hand side of the equation is the ordinary Eulerian derivative (i.e. the derivative on a fixed reference frame, representing changes at a point with respect to time) whereas the second term represents changes of a quantity with respect to position (advection). This "special" derivative is in reality the ordinary derivative of a function of many variables and can be derived easily through application of the chain rule.

For example, the measurement of changes in wind velocity in the atmosphere can be obtained with the help of an anemometer in a weather station or by mounting it on a weather balloon. The anemometer in the first case is measuring the velocity of all the moving particles passing through a fixed point in space, whereas in the second case the instrument is measuring changes in velocity as it moves with the fluid.

## Conservation laws

The Navier-Stokes equation is a special case of the (general) continuity equation. It, and associated equations such as mass continuity, may be derived from conservation principles of:

This is done via the Reynolds transport theorem, an integral relation stating that the changes of some intensive property (call it L) defined over a control volume must be equal to what is lost (or gained) through the boundaries of the volume plus what is created/consumed by sources and sinks inside the control volume. This is expressed by the following integral equation: $\frac{d}{dt}\int_{\Omega} L \ d\Omega = -\int_{\partial\Omega} L\mathbf{v\cdot n} \ d\partial\Omega + \int_{\Omega} Q \ d\Omega$

where v is the velocity of the fluid and Q represents the sources and sinks in the fluid. Recall that Ω represents the control volume and $\partial \Omega$ its bounding surface.

The divergence theorem may be applied to the surface integral, changing it into a volume integral: $\frac{d}{d t} \int_{\Omega} L \ d\Omega = -\int_{\Omega} \nabla \cdot ( L\mathbf{v}) \ d\Omega + \int_{\Omega} Q \ d\Omega$

Applying Leibniz's rule to the integral on the left and then combining all of the integrals: $\int_{\Omega} \frac{\partial L}{\partial t} \ d\Omega = - \int_{\Omega}\nabla \cdot (L\mathbf{v}) \ d\Omega + \int_{\Omega} Q \ d\Omega \qquad \Rightarrow \qquad \int_{\Omega} \left( \frac{\partial L}{\partial t} + \nabla \cdot (L\mathbf{v}) + Q\ \right) d\Omega = 0$

The integral must be zero for any control volume; this can only be true if the integrand itself is zero, so that: $\frac{\partial L}{\partial t} + \nabla \cdot (L\mathbf{v}) + Q = 0$

From this valuable relation (a very generic continuity equation), three important concepts may be concisely written: conservation of mass, conservation of momentum, and conservation of energy.

### Conservation of momentum

The most elemental form of the Navier-Stokes equations is obtained when the conservation relation is applied to momentum. Writing momentum as ρvi gives: $\frac{\partial}{\partial t}(\rho v_i) + \nabla \cdot (\rho v_i \mathbf{v}) + Q_i = 0$

The index i indicates that the above equation is applied to each component of velocity (generally three of them). Noting that a body force (notated b) is a source or sink of momentum and expanding the derivatives completely: $\frac{\partial \rho}{\partial t} v_i + \rho \frac{\partial v_i}{\partial t} + \nabla(\rho v_i) \cdot \mathbf{v} + \rho v_i \nabla \cdot \mathbf{v} = b_i$ $\frac{\partial \rho}{\partial t} v_i + \rho \frac{\partial v_i}{\partial t} + \nabla(\rho) v_i \cdot \mathbf{v} + \rho \nabla(v_i) \cdot \mathbf{v} + v_i \rho \nabla \cdot \mathbf{v} = b_i$ $v_i \frac{\partial \rho}{\partial t} + \rho \frac{\partial v_i}{\partial t} + v_i \mathbf{v} \cdot \nabla \rho + \rho \mathbf{v} \cdot \nabla v_i + \rho v_i \nabla \cdot \mathbf{v} = b_i$

Rearranging and recognizing that $\mathbf{v} \cdot \nabla \rho + \rho \nabla \cdot \mathbf{v} = \nabla \cdot (\rho \mathbf{v})$: $v_i \left(\frac{\partial \rho}{\partial t} + \mathbf{v} \cdot \nabla \rho + \rho \nabla \cdot \mathbf{v}\right) + \rho \left(\frac{\partial v_i}{\partial t} + \mathbf{v} \cdot \nabla v_i\right) = b_i$ $v_i \left(\frac{\partial \rho}{\partial t} + \nabla \cdot (\rho \mathbf{v})\right) + \rho \left(\frac{\partial v_i}{\partial t} + \mathbf{v} \cdot \nabla v_i\right) = b_i$

The leftmost expression enclosed in parentheses is, by mass continuity (shown in a moment), equal to zero. Noting that what remains on the left side of the equation is the convective derivative and writing the equation as a vector equation yields: $\rho \left(\frac{\partial \mathbf{v}}{\partial t} + \mathbf{v} \cdot \nabla \mathbf{v}\right) = \mathbf{b} \qquad \Rightarrow \qquad \rho\frac{D \mathbf{v}}{D t} = \mathbf{b}$

This is simply an expression of Newton's second law (F = ma) in terms of body forces instead of point forces. Each term in any case of the Navier-Stokes equations is a body force. A shorter though less rigorous way to arrive at this result would be the application of the chain rule to acceleration: $\rho \frac{d}{d t}(\mathbf{v}(x, y, z, t)) = \mathbf{b} \qquad \Rightarrow \qquad \rho \left( \frac{\partial \mathbf{v}}{\partial t} + \frac{\partial \mathbf{v}}{\partial x}\frac{d x}{d t} + \frac{\partial \mathbf{v}}{\partial y}\frac{d y}{d t} + \frac{\partial \mathbf{v}}{\partial z}\frac{d z}{d t} \right) = \mathbf{b} \qquad \Rightarrow$ $\rho \left( \frac{\partial \mathbf{v}}{\partial t} + u \frac{\partial \mathbf{v}}{\partial x} + v \frac{\partial \mathbf{v}}{\partial y} + w \frac{\partial \mathbf{v}}{\partial z} \right) = \mathbf{b} \qquad \Rightarrow \qquad \rho \left(\frac{\partial \mathbf{v}}{\partial t} + \mathbf{v} \cdot \nabla \mathbf{v}\right) = \mathbf{b}$

where $\mathbf{v} = (u, v, w)$.

### Conservation of mass

Mass may be considered also. Taking Q = 0 (no sources or sinks of mass) and putting in density: $\frac{\partial \rho}{\partial t} + \nabla \cdot (\rho \mathbf{v}) = 0$

where ρ is the mass density (mass per unit volume), and $\mathbf{v}$ is the velocity of the fluid. This equation is called the mass continuity equation, or simply "the" continuity equation. This equation generally accompanies the Navier-Stokes equation.

In the case of an incompressible fluid, ρ is a constant and the equation reduces to: $\nabla\cdot\mathbf{v} = 0$

which is in fact a statement of the conservation of volume.

## General form of the Navier-Stokes equations

The generic body force $\mathbf{b}$ seen previously is made specific first by breaking it up into two new terms, one to describe forces resulting from stresses and one for "other" forces such as gravity. By examining the forces acting on a small cube in a fluid, it may be shown that $\rho\frac{D\mathbf{v}}{D t} = \nabla \cdot \sigma_{ij} + \mathbf{f}$

where σij is the stress tensor, and $\mathbf{f}$ accounts for other body forces present. This equation is called the Cauchy momentum equation and describes the non-relativistic momentum conservation of any continuum that conserves mass. σij is a symmetric tensor given by: $\sigma_{ij} = \begin{pmatrix} \sigma_{xx} & \tau_{xy} & \tau_{xz} \\ \tau_{yx} & \sigma_{yy} & \tau_{yz} \\ \tau_{zx} & \tau_{zy} & \sigma_{zz} \end{pmatrix}$

where the σ are normal stresses and τ shear stresses. This tensor is split up into two terms: $\sigma_{ij} = \begin{pmatrix} \sigma_{xx} & \tau_{xy} & \tau_{xz} \\ \tau_{yx} & \sigma_{yy} & \tau_{yz} \\ \tau_{zx} & \tau_{zy} & \sigma_{zz} \end{pmatrix} = -\begin{pmatrix} p&0&0\\ 0&p&0\\ 0&0&p \end{pmatrix} + \begin{pmatrix} \sigma_{xx}+p & \tau_{xy} & \tau_{xz} \\ \tau_{yx} & \sigma_{yy}+p & \tau_{yz} \\ \tau_{zx} & \tau_{zy} & \sigma_{zz}+p \end{pmatrix} = -p I + \mathbb{T}$

where I is the 3 x 3 identity matrix. The motivation for doing this is that pressure is typically a variable of interest, and also this simplifies application to specific fluid families later on since the rightmost tensor $\mathbb{T}$ in the equation above must be zero for a fluid at rest. Note that $\mathbb{T}$ is traceless. The Navier-Stokes equation may now be written in the most general form: $\rho\frac{D\mathbf{v}}{D t} = -\nabla p + \nabla \cdot\mathbb{T} + \mathbf{f}$

This equation is still incomplete. For completion, one must make hypotheses on the form of $\mathbb{T}$, that is, one needs a constitutive law for the stress tensor which can be obtained for specific fluid families; additionally, if the flow is assumed compressible an equation of state will be required, which will likely further require a conservation of energy formulation.

## Application to different fluids

The most general form of the Navier-Stokes equations is not "ready for use", the stress tensor contains too many unknowns so that more information is needed; this information is normally some knowledge of the viscous behavior of the fluid.

### Newtonian fluid

Main article: Newtonian fluid

The formulation for Newtonian fluids stems from an observation made by Newton that, for most fluids, $\tau \propto \frac{\partial u}{\partial y}$

In order to apply this to the Navier-Stokes equations, three assumptions were made by Stokes:

• The stress tensor is a linear function of the strain rates.
• The fluid is isotropic.
• For a fluid at rest, $\nabla \cdot \mathbb{T}$ must be zero (so that hydrostatic pressure results).

Applying these assumptions will lead to: $\mathbb{T}_{ij} = \mu\left(\frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i}\right) + \delta_{ij} \lambda \nabla \cdot \mathbf{v}$

δij is the Kronecker delta. μ and λ are proportionality constants associated with the assumption that stress depends on strain linearly; μ is called the first coefficient of viscosity (usually just called "viscosity") and λ is the second coefficient of viscosity (related to bulk viscosity). The value of λ, which produces a viscous effect associated with volume change, is very difficult to determine, not even its sign is known with absolute certainty. Even in compressible flows, the term involving λ is often negligible; however it can occasionally be important even in nearly incompressible flows and is a matter of controversy. When taken nonzero, the most common approximation is $\lambda \approx 2/3 \mu$.

A straightforward substitution of $\mathbb{T}_{ij}$ into the momentum conservation equation will yield the Navier-Stokes equations for a compressible Newtonian fluid: $\rho \left(\frac{\partial u}{\partial t} + u \frac{\partial u}{\partial x} + v \frac{\partial u}{\partial y} + w \frac{\partial u}{\partial z}\right) = -\frac{\partial p}{\partial x} + \frac{\partial}{\partial x}\left(2 \mu \frac{\partial u}{\partial x} + \lambda \nabla \cdot \mathbf{v}\right) + \frac{\partial}{\partial y}\left(\mu\left(\frac{\partial u}{\partial y} + \frac{\partial v}{\partial x}\right)\right) + \frac{\partial}{\partial z}\left(\mu\left(\frac{\partial u}{\partial z} + \frac{\partial w}{\partial x}\right)\right) + \rho g_x$ $\rho \left(\frac{\partial v}{\partial t} + u \frac{\partial v}{\partial x} + v \frac{\partial v}{\partial y}+ w \frac{\partial v}{\partial z}\right) = -\frac{\partial p}{\partial y} + \frac{\partial}{\partial x}\left(\mu\left(\frac{\partial v}{\partial x} + \frac{\partial u}{\partial y}\right)\right) + \frac{\partial}{\partial y}\left(2 \mu \frac{\partial v}{\partial y} + \lambda \nabla \cdot \mathbf{v}\right) + \frac{\partial}{\partial z}\left(\mu\left(\frac{\partial v}{\partial z} + \frac{\partial w}{\partial y}\right)\right) + \rho g_y$ $\rho \left(\frac{\partial w}{\partial t} + u \frac{\partial w}{\partial x} + v \frac{\partial w}{\partial y}+ w \frac{\partial w}{\partial z}\right) = -\frac{\partial p}{\partial z} + \frac{\partial}{\partial x}\left(\mu\left(\frac{\partial w}{\partial x} + \frac{\partial u}{\partial z}\right)\right) + \frac{\partial}{\partial y}\left(\mu\left(\frac{\partial w}{\partial y} + \frac{\partial v}{\partial z}\right)\right) + \frac{\partial}{\partial z}\left(2 \mu \frac{\partial w}{\partial z} + \lambda \nabla \cdot \mathbf{v}\right) + \rho g_z$

or, more compactly in vector form, $\rho \left(\frac{\partial \mathbf{v}}{\partial t} + \mathbf{v} \cdot \nabla \mathbf{v}\right) = -\nabla p + \nabla \cdot (\mu \cdot (\nabla \otimes \mathbf{v} + (\nabla \otimes \mathbf{v})^T)) + \nabla (\lambda \nabla \cdot \mathbf{v}) + \rho \mathbf{g}$

where the outer product and matrix transpose have been used. Gravity has been accounted for as "the" body force, ie $\mathbf{f} = \rho \mathbf{g}$. The associated mass continuity equation is: $\frac{\partial \rho}{\partial t} + \nabla \cdot (\rho \mathbf{v}) = 0$

In addition to this equation, an equation of state and an equation for the conservation of energy is needed. The equation of state to use depends on context (often the ideal gas law), the conservation of energy will read: $\rho \frac{D h}{D t} = \frac{D p}{D t} + \nabla \cdot (k \nabla T) + \Phi$

Here, h is the enthalpy, T is the temperature, and Φ is a function representing the dissipation of energy due to viscous effects: $\Phi = \mu \left(2\left(\frac{\partial u}{\partial x}\right)^2 + 2\left(\frac{\partial v}{\partial y}\right)^2 + 2\left(\frac{\partial w}{\partial z}\right)^2 + \left(\frac{\partial v}{\partial x} + \frac{\partial u}{\partial y}\right)^2 + \left(\frac{\partial w}{\partial y} + \frac{\partial v}{\partial z}\right)^2 + \left(\frac{\partial u}{\partial z} + \frac{\partial w}{\partial x}\right)^2\right) + \lambda (\nabla \cdot \mathbf{v})^2$

With a good equation of state and good functions for the dependence of parameters (such as viscosity) on the variables, this system of equations seems to properly model the dynamics of all known gases and most liquids.

For the special but very common case of incompressible flow, the momentum equations simplify significantly. For example, looking at the viscous terms of the x momentum equation (note that viscosity will now be a constant and the second viscosity effect will be zero): \begin{align} &\frac{\partial}{\partial x}\left(2 \mu \frac{\partial u}{\partial x} + \lambda \nabla \cdot \mathbf{v}\right) + \frac{\partial}{\partial y}\left(\mu\left(\frac{\partial u}{\partial y} + \frac{\partial v}{\partial x}\right)\right) + \frac{\partial}{\partial z}\left(\mu\left(\frac{\partial u}{\partial z} + \frac{\partial w}{\partial x}\right)\right) \\ \\ & = 2 \mu \frac{\partial^2 u}{\partial x^2} + \mu \frac{\partial^2 u}{\partial y^2} + \mu \frac{\partial^2 v}{\partial y \, \partial x} + \mu \frac{\partial^2 u}{\partial z^2} + \mu \frac{\partial^2 w}{\partial z \, \partial x} \\ \\ & = \mu \frac{\partial^2 u}{\partial x^2} + \mu \frac{\partial^2 u}{\partial y^2} + \mu \frac{\partial^2 u}{\partial z^2} + \mu \frac{\partial^2 u}{\partial x^2} + \mu \frac{\partial^2 v}{\partial y \, \partial x} + \mu \frac{\partial^2 w}{\partial z \, \partial x} \\ \\ & = \mu \nabla^2 u + \mu \frac{\partial}{\partial x} \left(\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} + \frac{\partial w}{\partial z}\right) = \mu \nabla^2 u \end{align}

The Navier-Stokes equations are almost universally dealt with for Newtonian fluids. Part of this is because, as of 2007, good models for non-Newtonian flow simply do not exist. As with Newtonian flow, formulations are inspired by examining specific cases, but unlike Newtonian flow there are no models that will work beyond such special cases. Development and implementation of good non-Newtonian models is an area of ongoing research.

### Bingham fluid

Main article: Bingham plastic

In Bingham fluids, the situation is slightly different: $\frac {\partial u} {\partial y} = \left\{ \begin{matrix} 0 &, \quad \tau < \tau_0 \\ (\tau - \tau_0)/ {\mu} &, \quad \tau \ge \tau_0 \end{matrix}\right.$

These are fluids capable of bearing some shear before they start flowing. Some common examples are toothpaste and clay.

### Power-law fluid

Main article: Power-law fluid

A power law fluid is an idealised fluid for which the shear stress, τ, is given by $\tau = K \left(\frac{\partial u}{\partial y}\right)^n$

This form is useful for approximating all sorts of general fluids, including shear thinning (such as latex paint) and shear thickening (such as corn starch water mixture).

## The stress tensor

The derivation of the equations involves the consideration of forces acting on fluid elements, so that a quantity called the stress tensor appears naturally in the Cauchy momentum equation. Since the divergence of this tensor is taken, it is customary to write out the equation fully simplified, so that the original appearance of the stress tensor is lost.

However, the stress tensor still has some important uses, especially in formulating boundary conditions at fluid interfaces. Recalling that $\sigma_{ij} = -p I + \mathbb{T}$, for a Newtonian fluid the stress tensor is: $\sigma_{ij} = -\begin{pmatrix} p&0&0\\ 0&p&0\\ 0&0&p \end{pmatrix} + \mu\left(\frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i}\right) + \delta_{ij} \lambda \nabla \cdot \mathbf{v}$

If the fluid is assumed to be incompressible, the tensor simplifies significantly: \begin{align} \sigma_{ij} &= -\begin{pmatrix} p&0&0\\ 0&p&0\\ 0&0&p \end{pmatrix} + \mu \begin{pmatrix} 2 \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} + \frac{\partial v}{\partial x} & \frac{\partial u}{\partial z} + \frac{\partial w}{\partial x} \\ \frac{\partial v}{\partial x} + \frac{\partial u}{\partial y} & 2 \frac{\partial v}{\partial y} & \frac{\partial v}{\partial z} + \frac{\partial w}{\partial y} \\ \frac{\partial w}{\partial x} + \frac{\partial u}{\partial z} & \frac{\partial w}{\partial y} + \frac{\partial v}{\partial z} & 2\frac{\partial w}{\partial z} \end{pmatrix} \\ &= -p I + \mu\left(\frac{\partial(u, v, w)}{\partial(x, y, z)} + \frac{\partial(u, v, w)}{\partial(x, y, z)}^T\right) \\ &= -p I + \mu (\nabla \otimes \mathbf{v} + (\nabla \otimes \mathbf{v})^T) = -p I + 2 \mu E\\ \end{align}

Note that the outer product yields the Jacobian of velocity with respect to position (a special case of tensor derivative). Sometimes, the product symbols are left out and it is understood that the gradient of a column vector is contextually a matrix. E is called the deviatoric stress tensor, it is a measure of strain rate.

## References

White, Frank M. (2006). Viscous Fluid Flow. New York, NY: McGraw Hill.

Surface Tension Module, by John W. M. Bush, at MIT OCW.