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Navier-Stokes equation and its terms

Short answer

The incompressible Navier–Stokes equation is ρ(∂u/∂t + u·∇u) = −∇p + μ∇²u + ρg. Terms: local acceleration, nonlinear convective acceleration, pressure gradient, viscous diffusion, body force. It is difficult because the convective term is nonlinear, pressure lacks an explicit equation, and turbulence at high Re demands very fine grids or modelling.

Write the Navier-Stokes momentum equation for an incompressible Newtonian fluid with constant properties. Explain the physical meaning of each term and state the assumptions used. Why is the equation difficult to solve?

The Navier-Stokes (N-S) equations express Newton's second law (mass×acceleration=force\text{mass}\times\text{acceleration}=\text{force}) for a fluid element.

Equation (vector form)

ρ(∂V⃗∂t+(V⃗⋅∇)V⃗)=−∇p+μ∇2V⃗+ρg⃗\rho\left(\frac{\partial\vec V}{\partial t}+(\vec V\cdot\nabla)\vec V\right)=-\nabla p+\mu\nabla^2\vec V+\rho\vec g

In Cartesian form, x-direction:

ρ(∂u∂t+u∂u∂x+v∂u∂y+w∂u∂z)=−∂p∂x+μ(∂2u∂x2+∂2u∂y2+∂2u∂z2)+ρgx\rho\left(\frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}+w\frac{\partial u}{\partial z}\right)=-\frac{\partial p}{\partial x}+\mu\left(\frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2}+\frac{\partial^2u}{\partial z^2}\right)+\rho g_x

with similar equations for yy and zz. It is used together with ∇⋅V⃗=0\nabla\cdot\vec V=0.

Meaning of terms

TermNamePhysical meaning
ρ ∂V⃗/∂t\rho\,\partial\vec V/\partial tLocal (unsteady) inertiaRate of change of momentum at a fixed point
ρ(V⃗⋅∇)V⃗\rho(\vec V\cdot\nabla)\vec VConvective inertiaMomentum carried by the flow; non-linear term
−∇p-\nabla pPressure forceForce per unit volume from the pressure gradient
μ∇2V⃗\mu\nabla^2\vec VViscous (diffusion) forceMomentum diffusion by molecular friction
ρg⃗\rho\vec gBody forceGravity (buoyancy etc.) per unit volume

The left side is mass times acceleration per unit volume; the right side is the sum of forces.

Assumptions

  • Continuum, Newtonian fluid (stress proportional to strain rate).
  • Constant density and viscosity (incompressible, isothermal).
  • Stokes' hypothesis for bulk viscosity is not needed when ∇⋅V⃗=0\nabla\cdot\vec V=0.

Why difficult to solve

  • The convective term u ∂u/∂xu\,\partial u/\partial x is non-linear, so superposition fails and no general analytical solution exists.
  • Four unknowns (u,v,w,pu,v,w,p) are coupled with four equations, and pressure has no equation of its own (needs pressure-velocity coupling).
  • At high ReRe the solution is turbulent, with a wide range of scales, so very fine meshes are needed (DNS) or modelling (RANS/LES).
  • Analytical solutions exist only for simple cases such as Couette and Poiseuille flows, which are used to test CFD codes.

✓ Answer checked by Claude Opus 5.5 (high effort)

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