The incompressible Navier-Stokes equations are partial differential equations for a velocity field and pressure field. In highly symmetric flows, many terms can vanish because of the physics and geometry, reducing the PDEs to an ordinary differential equation that can be solved exactly.

The important skill is not to delete terms by inspection, but to state the assumptions first and then evaluate each term.

Example geometry: parallel fully developed flow

Consider a Newtonian fluid between two large parallel planes. Choose

• $x$ along the plates and along the mean flow;
• $y$ normal to the plates;
• $z$ in the spanwise direction.

Suppose the flow is

1. steady, so fields do not change with time;
2. unidirectional, with velocity only in the $x$ direction;
3. fully developed, so the axial velocity profile no longer changes with $x$;
4. uniform in the spanwise direction.

These assumptions give

$$\boxed{\mathbf v=u(y),\hat{\mathbf x}}.$$

The statement $u=u(y)$ is powerful: the only unknown velocity variation is across the gap.

Continuity

For incompressible flow,

$$\nabla\cdot\mathbf v=0.$$

Here

$$\nabla\cdot\mathbf v =\frac{\partial u}{\partial x} +\frac{\partial v_y}{\partial y} +\frac{\partial v_z}{\partial z}.$$

Because

$$\frac{\partial u}{\partial x}=0,\qquad v_y=0,\qquad v_z=0,$$

continuity is satisfied automatically.

This is a useful consistency check: the assumed velocity field is compatible with incompressibility.

Local acceleration

Steady flow gives

$$\boxed{\frac{\partial\mathbf v}{\partial t}=\mathbf0}.$$

Advective acceleration

The nonlinear term is

$$({\mathbf v}\cdot\nabla)\mathbf v.$$

For $\mathbf v=u(y)\hat{\mathbf x}$,

$$({\mathbf v}\cdot\nabla) =u(y)\frac{\partial}{\partial x}.$$

Therefore

$$({\mathbf v}\cdot\nabla)\mathbf v =u(y)\frac{\partial}{\partial x} \left[u(y)\hat{\mathbf x}\right] =\mathbf0.$$

The fluid can have a nonzero and spatially varying speed while the advective acceleration is zero. What matters is whether a moving parcel travels through variation along its trajectory. In this fully developed parallel flow, parcels move in $x$ while $u$ varies only with $y$.

Viscous term

Because only $u(y)$ varies,

$$\nabla^2\mathbf v =\frac{d^2u}{dy^2}\hat{\mathbf x}.$$

For a horizontal flow with no body-force component in the $x$ direction, the $x$ component of Navier-Stokes becomes

$$0=-\frac{\partial p}{\partial x} +\mu\frac{d^2u}{dy^2}.$$

Thus

$$\boxed{\mu\frac{d^2u}{dy^2}=\frac{dp}{dx}}.$$

The original nonlinear vector PDE has reduced to one linear second-order ODE.

The transverse momentum equations give, for this horizontal case,

$$\frac{\partial p}{\partial y}=0, \qquad \frac{\partial p}{\partial z}=0.$$

Hence pressure varies only along $x$. Because the left side of

$$\mu u''(y)=\frac{dp}{dx}$$

depends only on $y$ while the right side depends only on $x$, both must be constant in a fully developed solution. The axial pressure gradient is therefore constant.

What remains to solve

Integrating the reduced equation twice produces two integration constants. A unique velocity profile requires two velocity boundary conditions, typically supplied by the velocities of the two solid walls.

The solution then depends on two physically distinct drivers:

• wall motion, transmitted through viscosity;
• an axial pressure gradient.

Because the reduced equation is linear, their contributions can be superposed.

A general reduction workflow

When looking for an exact Navier-Stokes solution:

1. choose coordinates adapted to the geometry;
2. state the velocity components that symmetry permits;
3. state which spatial derivatives vanish and why;
4. apply continuity as a consistency condition;
5. evaluate local acceleration, advection, pressure, viscous, and body-force terms separately;
6. keep every term that survives;
7. solve the reduced equation with physically appropriate boundary conditions.

The same method produces many canonical viscous-flow solutions. The differential operator changes with geometry—for example, cylindrical pipe flow uses the radial form of the Laplacian—but the reasoning is the same: symmetry and boundary conditions turn the general field equations into a tractable model without changing the underlying physics.