This text discusses the derivation of the Navier-Stokes Equations, where they come from and how/when they can be simplified (e.g. incompressible or ideal flows).

The Navier-Stokes Equations is the name that take the conservation equations of three magnitudes related to fluids, namely:

**Mass****Momentum****Energy**

In their general form, govern the motion of fluids, so they constitute the backbone of Fluid Mechanics. The Navier-Stokes Equations are a set of five partial-differential equations (PDE) given that momentum is a vector variable, so it has three components (X, Y and Z).

Unfortunately, the analytical solution to these equations has not been found so far. In plain words, when you manipulate them, at some point, you end up with more unknown variables than equations.

This imposes a series of very restrictive issues regarding their manipulation which are essential to understand computational fluid dynamics (CFD). Actually, finding strategies to overcome such limitations is what CFD is all about.

As stated in the introduction section (which readers are advised to read to become familiar with key concepts to understand this section), the key to derive the Navier-Stokes equations is **conservation**. We will see how such complex equations actually derive from this widely accepted statement:

Mass, momentum and energy cannot be created nor destroyed, only transformed.

We will now establish the conservation law to mass, momentum and energy.

Before that, one previous remarks: we will refer the Navier-Stokes to a control volume, so all the properties will be divided by volume $(\forall)$. E.g., mass $(m)$ will be replaced by density $(\rho)$, momentum $(m \vec U)$ will be momentum per unit volume ($\rho \vec U)$, etc.

## Mass conservation

The first of the Navier-Stokes Equations is also known as Continuity Equation and stands on the universal principle of Newtonian physics:

Mass is conserved

Mass, or the property we will focus on: density, is a scalar property. Therefore, we must apply the conservation law applied to scalar quantities:

$ \begin{equation*}

\frac{\partial \xi}{\partial t} +

\vec \nabla \cdot (\vec U \xi) =

\vec \nabla \cdot (\lambda \vec \nabla \xi) +

\vec \nabla \cdot \vec Q_S +

Q_V

\end{equation*}$

Now, let us consider the following: we agreed that mass cannot be created, destroyed nor transformed, so the source/sink terms ($Q_S$ and $Q_V$) will always be zero. Additionally, mass diffusive transport does not occur, leaving the conservation law as follows:

$ \begin{equation*}

\frac{\partial \rho}{\partial t} +

\vec \nabla \cdot (\vec U \rho) =

0

\end{equation*}$

This is the general form of the continuity equation. However, when flow can be considered incompressible, which will be most of cases, density becomes constant and so can be removed from derivatives. This makes our lives much easier, leading to the expression that we will be using most of times, the incompressible form of the first of the Navier-Stokes Equations:

$\boxed{\begin{equation*}

\vec \nabla \cdot \vec U =

0

\end{equation*}}$

## Momentum conservation

The second of the Navier-Stokes Equations is nothing but another way to formulate Newton’s Second Law $(F=ma)$ and stands on the well-accepted principle:

Momentum is conserved

Momentum ($\vec U m), or the property we will focus on: $\rho \vec U$, is a vector property, as velocity has three components $(u_x, u_y, u_z)$. Therefore, in this case, we must apply the conservation law for vector quantities:

$ \begin{equation*}

\frac{\partial \vec v}{\partial t} +

\vec \nabla \cdot (\vec U \otimes \vec v) =

\vec \nabla \cdot (\lambda \vec \nabla \times \vec v) +

\vec \nabla \cdot \mathbf{Q_S} +

\vec Q_V

\end{equation*}$

We can infer from Newton’s Second Law that all time variation of momentum must be caused by a force. Therefore, in this case, we know that there will be non-zero source/sink terms

##### >>> External forces

External forces are those acting in proportion to a given fluid body volume and so they will take the place of volume sources/sinks $(\vec Q_V)$ in the conservation law. Examples of type of forces, often referred to as “body forces”, are gravity or electromagnetic forces.

The only term that will normally appear in most cases, if any, is gravity, although it can be neglected where hydrostatic and buoyancy forces are negligible. We will use the symbol $\vec f_b$ for body forces per unit mass. Therefore:

$\vec Q_V = \vec f_b$

##### >>> Internal forces

Internal forces are those acting in proportion to a given fluid body surface and so they will take the place of surface sources/sinks $(\mathbf{Q_S})$ in the conservation law. The only example of this type of forces, often referred to as “surface forces”, are internal stresses and their treatment is rather complex compared to external forces.

The internal stress tensor $(\mathbf{\sigma})$ can be decomposed into:

$\mathbf{\sigma} = -p\mathbf{I} + \mathbf{\tau}$

Where $p$ is pressure, which is assumed isotropic, i.e. acting perpendicularly to the fluid body surfaces, $\mathbf{I}$ is the identity tensor and $\mathbf{\tau}$ is the viscous shear stress tensor, constituted by all the viscous stresses acting in all possible directions parallel to the fluid body surface.

Under the hypothesis of Newtonian fluid, which all this discussion relies on, and for a one-dimensional scenario, the relationship between viscous shear stress and velocity gradient in the direction perpendicular to the wall (say $L$) is:

$\tau = \mu \frac{dU}{dL}$

When transposed to a Cartesian system of three dimensions, the former expression becomes the viscous shear stress tensor and its definition turns more complex:

$\begin{equation*}

\mathbf{\tau} = \mu \left[ (\frac{\partial U_j}{x_i} + \frac{\partial U_i}{x_j}) – \frac{2}{3} (\vec \nabla \cdot \vec U) \mathbf{I} \right]

\end{equation*}$

However, for the sake of simplicity, we will continue the discussion with the compact form stated above. Therefore:

$\mathbf{Q_S} = \mathbf{\sigma} = -p\mathbf{I} + \mathbf{\tau}$

Replacing the previous terms in the conservation law, and given that momentum cannot be in principle transported by diffusion, we obtain the second of the Navier-Stokes Equations:

$ \begin{equation*}

\frac{\partial (\rho \vec U)}{\partial t} +

\vec \nabla \cdot (\rho \vec U \otimes \vec U) =

-\vec \nabla p + \vec \nabla \cdot \mathbf{\tau} +

\rho \vec f_b

\end{equation*}$

This is the general form of the momentum conservation equation. However, when flow can be considered incompressible, which will be most of cases, as occurs in the mass conservation equation, density becomes constant and so can be removed from derivatives.

Also, the viscous terms in $\mathbf{\tau}$ appear as second derivatives, so they can be simplified using the Laplacian operator $(\vec \nabla^2)$. This makes our lives much easier, leading to the expression that we will be using most of times, the incompressible form of the second of the Navier-Stokes Equations:

$ \boxed{\begin{equation*}

\rho \frac{\partial \vec U}{\partial t} +

\rho (\vec \nabla \cdot \vec U) \vec U =

-\vec \nabla p + \mu \vec \nabla^2 \vec U +

\rho \vec f_b

\end{equation*}}$

## Energy conservation

The same reasoning can be applied to obtain the third of the Navier-Stokes Equations, a.k.a. energy conservation equation. However, making the assumptions we will normally will, i.e. isothermal incompressible flow, this equation will become trivial and so it will not be derived here.

For a more in-depth discussion on the Navier-Stokes Equations and their derivation, the reader is encouraged to read Charles Hirsch’s book [1] or watch this video from the very recommendable Youtube channel *Numberphile* [2]:

**Closure problem**

At this point, the reader will probably be aware that the Navier-Stokes have an issue. Consider this: if we take the first and second of the Navier-Stokes Equations, we have five unknown variables:

- Pressure $(p)$
- Density $(\rho)$
- Velocity in $x$ $(U_x)$
- Velocity in $y$ $(U_y)$
- Velocity in $z$ $(U_z)$

But only four equations. The mass conservation equation and the three momentum conservation equations (one for each direction: $x$, $y$ and $z$).

This leads to a mathematical closure problem, which, unfortunately, cannot be solved by coupling the energy conservation equation (the third of the Navier-Stokes Equations), as that would make things worse by introducing new unknown variables (e.g. temperature) to the mix.

That is why we do not know the solution to the Navier-Stokes Equations. To be more accurate, we do not even know if a solution actually exists. In fact, the Clay Mathematics Institute made this one of its turbulence, so this seems like a good time to jump the section devoted to what many call one of the last unresolved problem of Physics.

**References**

[1] Hirsch, C. (2007). Numerical computation of internal and external flows: The fundamentals of computational fluid dynamics. Elsevier.

[2] Numberphile (Youtube channel): https://www.youtube.com/watch?v=ERBVFcutl3M