Derive Navier-Stokes equation

The Navier–Stokes equation is Newton's second law (F = ma) applied to a small element of a viscous fluid.

Assumptions and notation

Velocity V = (u, v, w) (capital V, so it is not confused with the y-component v), density ρ, pressure p, dynamic viscosity μ, body force per unit mass g. The fluid is Newtonian and incompressible: ∇·V = 0 (continuity), with μ constant.

Acceleration

For a fluid particle the acceleration is the material derivative:

DV/Dt = ∂V/∂t + (V·∇)V

So mass × acceleration for the element dx dy dz is ρ (DV/Dt) dx dy dz.

Forces on the element (x-direction)

• Body force: ρ g_x dx dy dz.
• Surface forces. Let σxx = −p + τxx be the normal stress and τyx, τzx the shear stresses on the faces. The net force in x is

( ∂σxx/∂x + ∂τyx/∂y + ∂τzx/∂z ) dx dy dz

• Newton's law of viscosity: τxx = 2μ ∂u/∂x, τyx = μ(∂u/∂y + ∂v/∂x), τzx = μ(∂u/∂z + ∂w/∂x).

Substituting:

∂τxx/∂x + ∂τyx/∂y + ∂τzx/∂z
 = μ[ 2u_xx + u_yy + v_xy + u_zz + w_xz ]
 = μ[ ∇²u + ∂/∂x(∇·V) ] = μ ∇²u      (since ∇·V = 0)

Since σxx = −p + τxx, the net surface force per unit volume in the x-direction is −∂p/∂x + μ∇²u.

Equation of motion

Dividing by the volume element and writing the three components together:

ρ ( ∂V/∂t + (V·∇)V ) = −∇p + μ ∇²V + ρ g        with ∇·V = 0

In the x-component:

ρ( ∂u/∂t + u ∂u/∂x + v ∂u/∂y + w ∂u/∂z )
   = −∂p/∂x + μ( ∂²u/∂x² + ∂²u/∂y² + ∂²u/∂z² ) + ρ g_x

and similarly for y and z. This is the Navier–Stokes equation. The terms are: inertia (left), pressure gradient, viscous diffusion and body force. For μ = 0 it reduces to Euler's equation for an ideal fluid.