Derivation of the Navier-Stokes Equations
Theorem
$$ \mathbf{u} = \mathbf{u} \left( t ; \mathbf{x} \right) = \left( u_{1} \left( t ; \mathbf{x} \right) , u_{2} \left( t ; \mathbf{x} \right) , u_{3} \left( t ; \mathbf{x} \right) \right) $$ In particular, suppose that in three-dimensional space, the velocity field at time $t$ and spatial coordinate $\mathbf{x} = \left( x_{1} , x_{2} , x_{3} \right)$ is represented by the velocity vector as above. Similarly, $p : \mathbb{R}^{3} \to \mathbb{R}$ represents the pressure $p = p \left( \mathbf{x} \right)$ applied at each coordinate. If $\mathbf{u}$ is the flow velocity of an incompressible Newtonian fluid, it follows the following governing equation. $$ {\frac{ \partial \mathbf{u} }{ \partial t }} + \left( \mathbf{u} \cdot \nabla \right) \mathbf{u} = - \nabla w + \nu \nabla^{2} \mathbf{u} + \mathbf{g} $$ Here, $\nabla \cdot$ is the divergence, $\nu = \mu / \rho$ is the ratio of the kinematic viscosity coefficient $\mu$ to the density $\rho$, $\nabla w = \nabla p / \rho$ is the thermodynamic work, and $\mathbf{g}$ is the gravitational acceleration.
Explanation
The Navier-Stokes equation is a governing equation obtained by assuming a Newtonian fluid for the Euler equation in fluid mechanics, with the viscous term $\nabla^{2} \mathbf{u}$ added.
The problem of whether a smooth solution to this partial differential equation always exists in three-dimensional space, and if not, of finding a counterexample, is called the Navier–Stokes existence and smoothness problem, and it remains one of the Millennium Prize Problems.
It is sometimes said that one million dollars is awarded just for solving it, but given its difficulty, the expression “just for solving it” is not appropriate. For humanity’s wisdom to deepen regarding the equations of fluid mechanics is close to becoming able to read the will of heaven. Naturally, it is extremely difficult, and even understanding its derivation requires no small amount of study.
Derivation
Euler equation in fluid mechanics: $$ {\frac{ \partial \mathbf{u} }{ \partial t }} + \left( \mathbf{u} \cdot \nabla \right) \mathbf{u} = - {\frac{ 1 }{ \rho }} \nabla p + \mathbf{g} $$
We shall start from the Euler equation. Using the material derivative $D$ and multiplying both sides by $\rho$ gives the following. $$ \rho {\frac{ D \mathbf{u} }{ D t }} = - \nabla p + \rho \mathbf{g} $$ In this state, the repulsive force due to viscosity is not taken into account, so we now wish to modify the $- \nabla p$ acting on the right-hand side.
Cauchy stress tensor: Mainly in physics, a square matrix $\mathbf{\sigma} \in \mathbb{R}^{3 \times 3}$ that has as its components the stress acting at a point in each direction and in shear, defined as follows, is called the Cauchy stress tensor. $$ \sigma = \begin{bmatrix} \sigma_{11} & \sigma_{12} & \sigma_{13} \\ \sigma_{21} & \sigma_{22} & \sigma_{23} \\ \sigma_{31} & \sigma_{32} & \sigma_{33} \end{bmatrix} $$
In the Euler equation, the Cauchy stress tensor contained only the pressure term, as in $\sigma = - p I$. The force $\mathbf{f}$ acting on a fluid parcel with volume $V$ can be expressed as the closed-surface integral $\mathbf{f} = \oint_{\partial V} \sigma \cdot d S$ over the surface $\partial V$ enclosing $V$.
Divergence theorem: For a three-dimensional vector function $\mathbf{F}$, the following holds. $$ \int_{\mathcal{V}} \nabla \cdot \mathbf{F} dV = \oint_{\mathcal{S}} \mathbf{F} \cdot d \mathbf{S} $$ Here, $\nabla \cdot \mathbf{F}$ is the divergence, $\int_{\mathcal{V}}$ is the volume integral, and $\oint_{\mathcal{S}}$ is the closed-surface integral.
Converting to a volume integral according to the divergence theorem, since $\nabla \cdot \left( - p I \right) = - \nabla p$, we obtain the following. $$ \begin{align*} \mathbf{f} =& \oint_{\partial V} \sigma \cdot d S \\ =& \int_{V} \nabla \cdot \sigma d V \\ =& \int_{V} \nabla \cdot \left( - p I \right) d V \\ =& \int_{V} - \nabla p d V \\ \implies {\frac{ d \mathbf{f} }{ d V }} =& - \nabla p \end{align*} $$ Since this can be confusing, let us briefly verify the units through dimensional analysis of $d \mathbf{f} / d V$ and $\rho \mathbf{g}$. $$ \begin{align*} {\frac{ d \mathbf{f} }{ d V }} :& {\frac{ \text{force} }{ \text{volume} }} = \left[ {\frac{ \mathrm{kg} \cdot m / s^{2} }{ \mathrm{m}^{3} }} \right] = \left[ {\frac{ \mathrm{kg} }{ m^{2} \cdot s^{2} }} \right] \\ \rho \mathbf{g} :& {\frac{ \text{mass} }{ \text{volume} }} \cdot \text{acceleration} = \left[ {\frac{ \mathrm{kg} }{ \mathrm{m}^{3} }} \right] \cdot \left[ {\frac{ m }{ s^{2} }} \right] = \left[ {\frac{ \mathrm{kg} }{ m^{2} \cdot s^{2} }} \right] \end{align*} $$
The point is that the derivation of the Euler equation is also possible via the closed-surface integral over the Cauchy stress tensor $\sigma$. More generally, the influences other than the gravity term $\rho \mathbf{g}$ can be reflected by adding the divergence of the Cauchy stress tensor, as in $d \mathbf{f} / d V = \nabla \cdot \sigma$. Now let us add the shear stress $\tau$ to $\sigma$ as follows. $$ \sigma = - p I + \tau $$
Newton’s law of viscosity: In fluid mechanics, the principle that the stress applied to an incompressible and isotropic fluid is directly proportional to the symmetrized gradient of the flow velocity is called Newton’s law of viscosity, and mathematically it is expressed as follows in terms of the stress tensor $\tau \in \mathbb{R}^{3 \times 3}$ and the Jacobian $\nabla \mathbf{u}$ of the velocity field $\mathbf{u}$. $$ \tau = \mu \left( \nabla \mathbf{u} + \left( \nabla \mathbf{u} \right)^{T} \right) $$
Since a Newtonian fluid was assumed in the fundamental premise of the equation, $\nabla \cdot \sigma$ can be written as follows according to Newton’s law of viscosity. $$ \begin{align*} \nabla \cdot \sigma =& \nabla \cdot \left( - p I + \tau \right) \\ =& \nabla \cdot \left( - p I + \mu \left( \nabla \mathbf{u} + \left( \nabla \mathbf{u} \right)^{T} \right) \right) \\ =& - \nabla p + \mu \nabla \cdot \left( \nabla \mathbf{u} + \left( \nabla \mathbf{u} \right)^{T} \right) \\ =& - \nabla p + \mu \nabla \cdot \left( \nabla \mathbf{u} \right) + \mu \nabla \cdot \left( \nabla \mathbf{u} \right)^{T} \end{align*} $$
Divergence of $\nabla \mathbf{u}$: $$ \nabla \cdot \nabla \mathbf{u} = \nabla^{2} \mathbf{u} $$ Divergence of $\left( \nabla \mathbf{u} \right)^{T}$: $$ \nabla \cdot \left( \nabla \mathbf{u} \right)^{T} = \nabla \left( \nabla \cdot \mathbf{u} \right) $$
Meanwhile, since the fluid was assumed to be incompressible, $\nabla \cdot \mathbf{u} = 0$, and the last term can be eliminated. $$ \begin{align*} \nabla \cdot \sigma =& - \nabla p + \mu \nabla^{2} \mathbf{u} + \mu \nabla \left( \nabla \cdot \mathbf{u} \right) \\ =& - \nabla p + \mu \nabla^{2} \mathbf{u} \end{align*} $$
Now, returning to the original equation, we obtain both of the following expressions. $$ \begin{align*} \rho {\frac{ D \mathbf{u} }{ D t }} =& \nabla \cdot \sigma + \rho \mathbf{g} \\ =& - \nabla p + \mu \nabla^{2} \mathbf{u} + \rho \mathbf{g} \\ \overset{\div \rho}{\implies} {\frac{ D \mathbf{u} }{ D t }} =& - \nabla w + \nu \nabla^{2} \mathbf{u} + \mathbf{g} \end{align*} $$
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