---
title: "Derive Navier-Stokes equation"
subject: "Mathematics III"
chapter: 2
key: "semester_3/mathematics_iii:derive-navier-stokes-equation"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/derive-navier-stokes-equation
site: BCT Engineering
---

# Derive Navier-Stokes equation

Mathematics III · Chapter 2: Line, Surface and Volume Integrals

- Asked in: 2082 Chaitra (new course), 4 marks; 2082 Kartik (new course), 4 marks; 2081 Chaitra (new course), 4 marks (asked 3 times)
- Video: [Description and Derivation of the Navier-Stokes Equations](https://www.youtube.com/watch?v=NjoMoH51UZc) — LearnMechE
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/derive-navier-stokes-equation
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-2

## Question

Derive Navier-Stoke's equation of motion for fluid flow.

## Short answer

Apply Newton's second law to a fluid element, using the material derivative for acceleration. For incompressible Newtonian fluid with constant viscosity, the result is ρ(∂V/∂t + (V·∇)V) = −∇p + μ∇²V + ρg, with continuity ∇·V = 0.

## Answer

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:

```text
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

```text
( ∂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:

```text
∂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:

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

In the x-component:

```text
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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Navier–Stokes is derived from:
   A. Conservation of energy only
   B. Newton's second law on a fluid element
   C. Bernoulli's equation only
   D. Ohm's law
   Answer: B. F = ma applied to a fluid element. (Key point: Basis)

2. Particle acceleration is:
   A. ∂v/∂t + (v·∇)v
   B. ∂v/∂t
   C. (v·∇)v only
   D. ∇²v
   Answer: A. Material derivative. (Key point: Acceleration)

3. With ∇·v = 0 and constant μ, viscous force per volume reduces to:
   A. μ∇p
   B. ρg
   C. 2μ ∇·v
   D. μ∇²v
   Answer: D. Cross terms vanish via incompressibility. (Key point: Viscous term)

4. Full equation:
   A. ρ ∂v/∂t = ∇p
   B. ρ(v·∇)v = μ∇²v
   C. ρ(∂v/∂t+(v·∇)v) = −∇p + μ∇²v + ρg
   D. ∂v/∂t = −∇p
   Answer: C. Inertia equals pressure, viscous and body forces. (Key point: Equation)

## Related in Mathematics III

- [Derive or solve wave equation](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Aderive-or-solve-wave-equation-two) — Mathematics III, chapter 4
- [Derive or solve wave equation](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Aderive-or-solve-wave-equation) — Mathematics III, chapter 4
- [Solve one dimensional heat equation](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Asolve-one-dimensional-heat-equation-two) — Mathematics III, chapter 4

## Related in other subjects

- [Navier-Stokes equations for Newtonian fluid](https://www.bctengineer.com/question.md?key=semester_4%2Ffluid_mechanics_mechanical%3Anavier-stokes-newtonian-fluid) — Fluid Mechanics, chapter 4
- [Plane Poiseuille flow from Navier-Stokes](https://www.bctengineer.com/question.md?key=semester_7%2Fapplied_computational_fluid_dynamics%3Aplane-poiseuille-flow-numerical) — Applied Computational Fluid Dynamics, chapter 2
- [Navier-Stokes equation and its terms](https://www.bctengineer.com/question.md?key=semester_7%2Fapplied_computational_fluid_dynamics%3Anavier-stokes-terms-meaning) — Applied Computational Fluid Dynamics, chapter 2
- [Navier-Stokes simplification, velocity distribution](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Anavier-stokes-simplified-couette) — Fluid Mechanics, chapter 5