Deriving the Navier-Stokes Equations

Every Navier-Stokes derivation starts with the same idea: apply Newton's second law to a tiny parcel of moving fluid. Force equals mass times acceleration. That is the basic starting point.

Pick a small blob of water, air, or any other fluid. It has some mass. Forces act on it: pressure squeezes it from all sides, internal friction tugs on it, gravity pulls it down. Newton says the net force determines how the blob accelerates.

But a fluid parcel isn't a billiard ball. It deforms as it moves. It stretches, twists, distorts as it rides the flow. So "acceleration" isn't as simple as tracking a single object. We need to follow the blob through all of that.

Think of it this way: imagine sitting in a canoe on a river. Your acceleration depends on how the current changes in time at your location, and on the fact that the current is carrying you into regions where the flow is faster or slower. Both effects contribute to how your velocity changes.

Writing this balance for every point in the fluid simultaneously gives us the starting point of the derivation of the Navier-Stokes equations.

The Navier-Stokes equation derivation begins with the Cauchy momentum equation: the continuum-mechanics form of Newton's second law applied to a material volume $\Omega(t)$ moving with the fluid.

For a fluid with density $\rho$ and velocity field $u(x,t)$, conservation of linear momentum states

$$\frac{d}{dt}\int_{\Omega(t)} \rho\, u\, dV = \int_{\partial\Omega(t)} T\, n\, dS + \int_{\Omega(t)} \rho\, f\, dV,$$

where $T$ is the Cauchy stress tensor, $n$ is the outward unit normal, and $f$ represents body forces per unit mass (typically gravity).

Localizing with the Reynolds transport theorem gives the differential form

$$\rho\frac{Du}{Dt} = \nabla \cdot T + \rho f,$$

where the material derivative

$$\frac{Du}{Dt} = \partial_t u + (u \cdot \nabla)u$$

captures both the local time rate of change and the convective acceleration. This is $ma = F$ at every point in the continuum.

Three kinds of forces show up in the derivation of the Navier-Stokes equation:

1. Pressure forces. Fluid pushes on the parcel from every direction. If pressure is higher on one side than the other, the parcel gets shoved toward the low-pressure side. This imbalance is captured by the pressure gradient $-\nabla p$: a vector pointing from high pressure to low, telling the fluid which way to go.

2. Viscous forces. Adjacent layers of fluid moving at different speeds drag on each other. Faster layers pull slower neighbors along; slower layers hold faster ones back. This internal friction smooths out velocity differences. For a simple fluid like water, the strength of this friction comes down to a single number: the viscosity $\mu$.

3. External forces. Anything acting on the fluid from outside, most commonly gravity. These are body forces: they act on every bit of fluid in the volume, not just at the surface.

The Navier-Stokes equations are what you get when you write "mass times acceleration = pressure force + viscous force + external force" at every point.

The Cauchy stress tensor $T$ encodes all internal contact forces. For any fluid, it decomposes into an isotropic pressure part and a deviatoric (viscous) part:

$$T = -pI + \tau,$$

where $p$ is the mechanical pressure (defined as $-\tfrac{1}{3}\mathrm{tr}\,T$ for compressible flow) and $\tau$ is the viscous stress tensor.

Pressure gradient. Applying the divergence theorem to $-pI$ yields the force $-\nabla p$ per unit volume. This is the isotropic part of the stress divergence.

Viscous stress. The tensor $\tau$ depends on the specific constitutive law relating stress to the rate of deformation. Its divergence $\nabla \cdot \tau$ gives the viscous force per unit volume. The form of $\tau$ is still unspecified here; that comes from the Newtonian-fluid assumption in the next section.

Body forces. External forces such as gravity enter as $\rho f$ per unit volume. Substituting the stress decomposition into the Cauchy momentum equation gives

$$\rho\frac{Du}{Dt} = -\nabla p + \nabla \cdot \tau + \rho f.$$

This is the general momentum equation for any simple fluid, Newtonian or not. To close the system, we need a constitutive law specifying $\tau$.

Not all fluids behave the same way under stress. Honey resists motion differently than water. Ketchup gets runnier when you shake it. Cornstarch mixed with water gets stiffer when you hit it.

The Navier-Stokes equations make a specific assumption: the fluid is Newtonian. This means internal friction is directly proportional to how fast the fluid is being deformed. Double the rate of deformation, double the stress. It's a linear relationship.

Water and air are very well modeled as Newtonian. But this is an assumption, not a consequence of Newton's laws. The derivation requires it. Without it, you get a different class of equations entirely (non-Newtonian fluid models).

The proportionality constant is the dynamic viscosity $\mu$. It's a material property measuring how much a fluid resists shearing. Water: low viscosity. Honey: high viscosity.

There's also a second viscosity parameter $\lambda$, sometimes called the second viscosity coefficient, which matters when the fluid compresses or expands. A common further simplification, Stokes' hypothesis, sets $\lambda = -\frac{2}{3}\mu$. This is an additional assumption, not a theorem.

The constitutive law for a Newtonian fluid assumes that the viscous stress $\tau$ is a linear, isotropic function of the rate-of-strain tensor $D(u) = \tfrac{1}{2}(\nabla u + \nabla u^T)$. By the representation theorem for isotropic tensor functions, the most general such law is

$$\tau = 2\mu\, D(u) + \lambda (\nabla \cdot u)\, I,$$

equivalently written as

$$\tau = \mu(\nabla u + \nabla u^T) + \lambda(\nabla \cdot u)\, I,$$

where $\mu > 0$ is the dynamic (shear) viscosity and $\lambda$ is the second viscosity coefficient.

This is the defining assumption of a Newtonian fluid. It isn't derived from first principles; it's a constitutive hypothesis validated empirically for many common fluids.

Stokes' hypothesis further posits that $\lambda = -\frac{2}{3}\mu$, which makes the bulk viscosity $\kappa = \lambda + \frac{2}{3}\mu$ vanish. This simplification is widely used but is an independent assumption, not a consequence of thermodynamics or the Newtonian hypothesis itself. Classical kinetic theory predicts zero bulk viscosity for monatomic ideal gases under idealized assumptions; for many real gases and liquids, Stokes' hypothesis is only approximate.

The thermodynamic constraint from the Clausius-Duhem inequality requires only $\mu \geq 0$ and $3\lambda + 2\mu \geq 0$ (equivalently, $\kappa \geq 0$).

Now we put the pieces together. We have:

• Mass times acceleration on the left (the material derivative of velocity)
• Pressure, viscous friction, and gravity on the right
• The Newtonian-fluid rule connecting friction to the rate of deformation

Substituting and simplifying gives the compressible Navier-Stokes momentum equation:

$$\rho\Big(\frac{\partial u}{\partial t} + (u \cdot \nabla) u\Big) = -\nabla p + \mu\, \Delta u + (\mu + \lambda)\,\nabla(\nabla \cdot u) + \rho\, f$$

Every term has a physical meaning:

• $\rho\,\partial_t u$: how velocity at a fixed point changes over time
• $\rho(u \cdot \nabla) u$: the fluid carrying its own velocity from place to place (advection)
• $-\nabla p$: pressure pushing from high to low
• $\mu\,\Delta u$: viscosity smoothing out velocity differences
• $(\mu + \lambda)\nabla(\nabla \cdot u)$: an extra viscous term that only matters when the fluid compresses or expands
• $\rho f$: external forces like gravity

That's the full momentum equation. Pair it with conservation of mass and an equation of state (linking pressure to density), and you have the compressible Navier-Stokes system.

Substituting the Newtonian constitutive law $\tau = 2\mu D(u) + \lambda(\nabla \cdot u)I$ into the momentum equation $\rho \frac{Du}{Dt} = -\nabla p + \nabla \cdot \tau + \rho f$ and computing the divergence of $\tau$:

$$\nabla \cdot \tau = \nabla \cdot \big[\mu(\nabla u + \nabla u^T)\big] + \nabla\big[\lambda(\nabla \cdot u)\big].$$

If $\mu$ and $\lambda$ are constant (a standard assumption for many derivations), this simplifies to

$$\nabla \cdot \tau = \mu\,\Delta u + (\mu + \lambda)\,\nabla(\nabla \cdot u),$$

using the vector identity $\nabla \cdot (\nabla u^T) = \nabla(\nabla \cdot u)$. The compressible Navier-Stokes momentum equation is then

$$\rho\big(\partial_t u + (u \cdot \nabla)u\big) = -\nabla p + \mu\,\Delta u + (\mu + \lambda)\,\nabla(\nabla \cdot u) + \rho f.$$

This must be coupled with conservation of mass (the continuity equation)

$$\partial_t \rho + \nabla \cdot (\rho u) = 0$$

and an equation of state $p = p(\rho, \theta)$ (or an energy equation) to close the system. With Stokes' hypothesis $\lambda = -\frac{2}{3}\mu$, the coefficient $\mu + \lambda = \frac{1}{3}\mu$. For a detailed comparison with the incompressible system, see Incompressible vs. Compressible Navier-Stokes.

Water in a pipe. Slow air currents. Ocean circulation. These flows involve fluids whose density stays essentially constant. Making that simplification transforms the general equations into something much cleaner.

Constant density means $\rho$ doesn't change, anywhere, ever. Conservation of mass then forces $\nabla \cdot u = 0$: the fluid can't compress or expand. This is the incompressibility constraint.

This simplification removes one term entirely. The extra viscous term $(\mu + \lambda)\nabla(\nabla \cdot u)$ vanishes entirely. The second viscosity coefficient $\lambda$ drops out. It simply doesn't matter when the fluid can't compress. The momentum equation becomes:

$$\rho\Big(\frac{\partial u}{\partial t} + (u \cdot \nabla)u\Big) = -\nabla p + \mu\,\Delta u + \rho f$$

Dividing both sides by $\rho$, writing $\nu = \mu / \rho$ (the kinematic viscosity), and redefining pressure to absorb the factor $1/\rho$ gives the standard form:

$$\frac{\partial u}{\partial t} + (u \cdot \nabla)u = -\nabla p + \nu\,\Delta u + f$$

$$\nabla \cdot u = 0$$

This is the system studied in the Clay Millennium Problem. It's the version you'll encounter throughout this site and in most mathematical treatments of Navier-Stokes. For a thorough comparison with the compressible system, see Incompressible vs. Compressible Navier-Stokes.

Set $\nu = 0$ (no viscosity at all) and you get the Euler equations, a related but importantly different system.

A common incompressible specialization assumes constant density $\rho$ throughout the flow. The continuity equation $\partial_t \rho + \nabla \cdot(\rho u) = 0$ then forces $\nabla \cdot u = 0$. More generally, incompressibility is encoded by $\nabla \cdot u = 0$ (equivalently $D\rho/Dt = 0$ through continuity), which allows for variable density in principle, but the constant-density case is the standard setting for the Clay problem.

Under incompressibility, the term $(\mu + \lambda)\nabla(\nabla \cdot u)$ vanishes identically. The second viscosity coefficient $\lambda$ becomes irrelevant: neither Stokes' hypothesis nor any other assumption about $\lambda$ matters for the incompressible system. The momentum equation reduces to

$$\rho\big(\partial_t u + (u \cdot \nabla)u\big) = -\nabla p + \mu\,\Delta u + \rho f.$$

Dividing by $\rho$ and defining the kinematic viscosity $\nu = \mu/\rho$ (with $p$ redefined to absorb the factor $1/\rho$) yields the incompressible Navier-Stokes system on $\mathbb{R}^3$:

$$\partial_t u + (u \cdot \nabla)u = -\nabla p + \nu\,\Delta u + f,$$

$$\nabla \cdot u = 0, \qquad u(x,0) = u_0(x).$$

This is the system studied in the Clay Millennium Problem (Fefferman, 2000). The official problem concerns global regularity for smooth divergence-free initial data on $\mathbb{R}^3$ (or $\mathbb{T}^3$) in the standard Clay formulation (Fefferman, 2000).

The pressure $p$ isn't an independent dynamic variable: it's determined (up to a constant) by taking the divergence of the momentum equation and using $\nabla \cdot u = 0$, yielding a Poisson equation $\Delta p = -\nabla \cdot [(u \cdot \nabla)u] + \nabla \cdot f$ (or $\Delta p = -\partial_i \partial_j(u_i u_j)$ when $f$ is divergence-free or absent). This elliptic coupling is a distinctive feature of the incompressible system.

Setting $\nu = 0$ recovers the incompressible Euler equations. The relationship between these two systems is central to many open questions in mathematical fluid dynamics.

The derivation gives us the Navier-Stokes equations. It tells us what they are and why they take the form they do. Every term traces back to a physical principle or an explicit assumption.

But deriving the equations isn't the same as understanding their solutions. The derivation doesn't answer:

• Do solutions always exist for all time?
• If they start smooth, do they stay smooth?
• Can the velocity blow up to infinity in finite time?

These are questions about the mathematical behavior of the equations, not their physical origin. In three dimensions, they're still open. That gap is the Clay Millennium Problem: whether smooth 3D incompressible Navier-Stokes data always produce global smooth solutions, or whether singularities can form in finite time.

The equations date to the 19th century, and the Clay Mathematics Institute has offered a $1 million prize since 2000 for a resolution of the modern regularity problem.

The derivation establishes the Navier-Stokes equations as a well-motivated PDE system grounded in conservation of momentum and the Newtonian constitutive law. It does not address the central question of mathematical well-posedness.

Specifically, the derivation is silent on:

• Global existence: Whether smooth solutions persist for all $t > 0$ given smooth initial data.
• Regularity: Whether solutions remain in $C^\infty(\mathbb{R}^3 \times [0,\infty))$ or can develop singularities.
• Uniqueness: Whether solutions in various function-space settings are unique.

Leray's theory (1934) guarantees global existence of weak solutions in $L^2$, but uniqueness and regularity of Leray-Hopf solutions remain unproven in 3D. The gap between the available energy estimates and the scaling of the nonlinearity is the analytical heart of the difficulty.

The Clay Millennium Problem asks precisely for a proof or disproof of global regularity in $\mathbb{R}^3$. The derivation motivates the PDE system, but it doesn't resolve existence, regularity, or uniqueness in 3D.

Now that you've seen where the equations come from:

• What Are the Navier-Stokes Equations? for the equations themselves, with term-by-term explanation
• Incompressible vs. Compressible Navier-Stokes for what changes when density varies, and why the incompressible case is special
• Euler vs. Navier-Stokes for what happens when you remove viscosity entirely
• The Millennium Problem for the open question that makes these equations famous

Related pages for further study:

• What Are the Navier-Stokes Equations? for the standard forms with detailed discussion of each term
• Incompressible vs. Compressible Navier-Stokes for the full compressible system, barotropic closures, and the low-Mach-number limit
• Euler vs. Navier-Stokes for the inviscid limit, the role of $\nu$, and the vanishing-viscosity problem
• Navier-Stokes Existence and Smoothness for the precise Clay formulation, known results, and open questions