Exact Solutions of Navier-Stokes Equations

TL;DR

The Navier-Stokes equations are notoriously difficult to solve, but a handful of exact analytical solutions exist for specific geometries and boundary conditions. These solutions provide crucial benchmarks for numerical methods and deep physical insight into fluid behavior. You'll learn the most important exact solutions and how to derive them systematically.

1. The Mental Model

Think of exact solutions as the "solved puzzles" in fluid mechanics—special cases where the complexity of the Navier-Stokes equations collapses into something manageable. Most real flows are too chaotic for exact solutions, but certain symmetric geometries with clever boundary conditions let us cut through the mathematical chaos. These solutions are your North Star: they show what's actually possible when the math cooperates.

2. The Core Material

Why Exact Solutions Matter

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You're dealing with one of the most famous unsolved problems in mathematics. The Clay Institute offers $1 million for proving that smooth solutions to the 3D Navier-Stokes equations exist globally—or showing they don't. But for specific cases, we can find exact solutions that satisfy the full nonlinear equations.

The Navier-Stokes equations in vector form are:
$$\frac{\partial \mathbf{u}}{\partial t} + (\mathbf{u} \cdot abla)\mathbf{u} = -\frac{1}{\rho} abla p + u abla^2 \mathbf{u}$$

where $\mathbf{u}$ is velocity, $p$ is pressure, $\rho$ is density, and $
u$ is kinematic viscosity.

The key insight: exact solutions exist when symmetry eliminates or simplifies the troublesome nonlinear term $(\mathbf{u} \cdot
abla)\mathbf{u}$.

Couette Flow: The Foundation

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Couette flow occurs between two parallel plates, where one moves with constant velocity $U$ and the other is stationary. This is your first exact solution because the geometry forces the flow to be purely one-dimensional.

Assume steady flow with $u = u(y)$, $v = 0$, and $\frac{\partial}{\partial x} = 0$. The nonlinear convection term vanishes because there's no velocity component perpendicular to the shear direction.

The momentum equation simplifies to:
$$0 = u \frac{d^2u}{dy^2}$$

With boundary conditions $u(0) = 0$ and $u(h) = U$, integration gives:
$$u(y) = \frac{U}{h}y$$

This linear velocity profile is exact—no approximations involved. The shear stress is constant: $\tau = \mu \frac{du}{dy} = \frac{\mu U}{h}$.

Poiseuille Flow: Pressure-Driven Solutions

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Poiseuille flow describes fluid motion in pipes or channels driven by pressure gradients. Unlike Couette flow, here you have a body force (pressure gradient) balanced by viscous forces.

Plane Poiseuille Flow

For flow between parallel plates separated by distance $2h$, driven by pressure gradient $-\frac{dp}{dx}$:

$$0 = -\frac{1}{\rho}\frac{dp}{dx} + u \frac{d^2u}{dy^2}$$

Since the pressure gradient is constant, $\frac{dp}{dx} = -\frac{\Delta p}{L}$ where $L$ is the channel length.

With no-slip boundary conditions $u(\pm h) = 0$, the solution is:
$$u(y) = \frac{1}{2\mu}\frac{dp}{dx}(y^2 - h^2) = \frac{\Delta p}{2\mu L}(h^2 - y^2)$$

This parabolic profile is fundamental—you'll see it everywhere in microfluidics and lubrication theory.

Circular Poiseuille Flow

In a circular pipe of radius $R$, cylindrical symmetry gives $u = u(r)$ only. The equation becomes:
$$\frac{1}{r}\frac{d}{dr}\left(r\frac{du}{dr}\right) = \frac{1}{\mu}\frac{dp}{dx}$$

Integration with $u(R) = 0$ yields:
$$u(r) = -\frac{1}{4\mu}\frac{dp}{dx}(R^2 - r^2)$$

The maximum velocity occurs at the centerline: $u_{max} = \frac{R^2}{4\mu}\left(-\frac{dp}{dx}\right)$.

Stokes' First and Second Problems

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These classical solutions show how viscosity diffuses momentum from moving boundaries.

Stokes' First Problem

An infinite flat plate suddenly starts moving with velocity $U$ parallel to itself. The fluid above is initially at rest. This is an unsteady problem where viscous effects penetrate into the fluid over time.

The governing equation is:
$$\frac{\partial u}{\partial t} = u \frac{\partial^2 u}{\partial y^2}$$

with conditions $u(0,t) = U$ and $u(\infty,t) = 0$, plus $u(y,0) = 0$.

The exact solution involves the complementary error function:
$$u(y,t) = U \text{erfc}\left(\frac{y}{2\sqrt{ u t}}\right)$$

The boundary layer thickness grows as $\delta \sim \sqrt{
u t}$—this square-root dependence is characteristic of viscous diffusion.

graph TD
    A["Initial State: Fluid at Rest"] --> B["Plate Suddenly Moves with Velocity U"]
    B --> C["Viscous Effects Penetrate Upward"]
    C --> D["Boundary Layer Grows as √(νt)"]
    D --> E["Steady State: Linear Profile"]

    F["Couette Flow"] --> G["Linear Velocity Profile"]
    F --> H["Constant Shear Stress"]

    I["Poiseuille Flow"] --> J["Parabolic Velocity Profile"]
    I --> K["Pressure Gradient Balanced by Viscosity"]

    style A fill:#f9f,stroke:#333,stroke-width:2px
    style E fill:#9f9,stroke:#333,stroke-width:2px
    style G fill:#9f9,stroke:#333,stroke-width:2px
    style J fill:#9f9,stroke:#333,stroke-width:2px

Advanced Exact Solutions

Rotating Flows

For flow between concentric cylinders where the inner cylinder rotates, you get another exact solution. If the inner cylinder (radius $R_1$) rotates with angular velocity $\Omega$ and the outer cylinder (radius $R_2$) is fixed, the azimuthal velocity is:

$$u_\theta(r) = \Omega R_1^2 \frac{R_2^2 - r^2}{r(R_2^2 - R_1^2)}$$

Hiemenz Flow

This describes two-dimensional stagnation point flow against a wall. The exact solution is:
$$u = ax f'(\eta), \quad v = -\sqrt{a u} f(\eta)$$
where $\eta = y\sqrt{a/
u}$ and $f$ satisfies the Hiemenz equation: $f''' + ff'' - (f')^2 + 1 = 0$.

3. Worked Example

Let's derive the exact solution for pressure-driven flow in a rectangular channel—a 3D extension of plane Poiseuille flow.

Problem Setup: Consider steady, fully developed flow in a rectangular duct with height $2a$ and width $2b$. The pressure gradient is $-dp/dx = G$ (constant).

Step 1: Recognize the symmetry. Due to the rectangular geometry, we expect $u = u(y,z)$ only, with $v = w = 0$.

Step 2: The Navier-Stokes equation reduces to:
$$G = \mu\left(\frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial z^2}\right)$$

Step 3: Apply boundary conditions. No-slip at all walls: $u(\pm a, z) = 0$ and $u(y, \pm b) = 0$.

Step 4: Use separation of variables. Let $u(y,z) = u_0(y,z) + u_1(y,z)$ where $u_0$ handles the pressure gradient and $u_1$ satisfies the homogeneous boundary conditions.

First, solve for the 2D Poiseuille profile: $u_0(y) = \frac{G}{2\mu}(a^2 - y^2)$.

Step 5: The correction term $u_1$ must satisfy Laplace's equation with modified boundary conditions. Using Fourier series:

$$u_1(y,z) = -\frac{32Ga^2}{\mu\pi^3} \sum_{n=1,3,5...} \frac{1}{n^3} \frac{\cosh(n\pi z/2a)}{\cosh(n\pi b/2a)} \cos\left(\frac{n\pi y}{2a}\right)$$

Step 6: The complete solution is $u = u_0 + u_1$. Near the center $(y=0, z=0)$:
$$u(0,0) = \frac{Ga^2}{2\mu}\left[1 - \frac{32}{\pi^3}\sum_{n=1,3,5...}\frac{1}{n^3}\frac{1}{\cosh(n\pi b/2a)}\right]$$

For $b >> a$ (wide channel), this approaches the 2D result: $u_{max} \approx \frac{Ga^2}{2\mu}$.

4. Key Takeaways

4.1 Most Important Concepts

• Symmetry kills nonlinearity: Exact solutions exist when geometry eliminates or simplifies the convection term $(\mathbf{u} \cdot
abla)\mathbf{u}$.

• Linear profiles from shear flows: When viscous forces balance applied stresses without pressure gradients, you get linear velocity profiles.

• Parabolic profiles from pressure gradients: When pressure forces balance viscosity, the twice-integrated force balance produces parabolic velocity distributions.

• Boundary layer growth: Unsteady viscous effects penetrate distances proportional to $\sqrt{
u t}$, reflecting the diffusive nature of momentum transport.

• No-slip boundary conditions: All exact solutions must satisfy zero velocity at solid boundaries—this constraint often determines the functional form.

• Separation of variables works: Many rectangular and cylindrical geometries yield to separation techniques, producing infinite series solutions.

• Dimensional analysis guides solutions: The characteristic velocity, length, and time scales embedded in exact solutions reveal the underlying physics.

4.2 Common Misconceptions

• "Exact solutions are just approximations": Wrong—these satisfy the full nonlinear Navier-Stokes equations exactly, with no linearization or perturbation assumptions.

• "All flows have parabolic profiles": Only pressure-driven flows between parallel boundaries give parabolas; shear-driven flows are linear, and complex geometries produce other shapes.

• "Viscosity always dominates in exact solutions": Actually, exact solutions balance all terms properly—it's the geometry that simplifies the equations, not the assumption of low Reynolds number.

• "Stokes flow is the same as exact solutions": Stokes flow assumes $Re \ll 1$ and drops inertial terms; exact solutions keep all terms but exploit symmetry to handle the nonlinearity.

4.3 Compare & Contrast

5. Now Try It

Derive the exact solution for steady flow down an inclined plane of infinite width. The fluid layer has thickness $h$, density $\rho$, viscosity $\mu$, and flows down a plane inclined at angle $\theta$ to the horizontal. Gravity acts vertically downward with acceleration $g$.

Set up your coordinate system with $x$ along the incline, $y$ perpendicular to it. Write the momentum equation, apply appropriate boundary conditions (free surface at $y = h$, no-slip at $y = 0$), and solve for $u(y)$. Calculate the maximum velocity and the volumetric flow rate per unit width.

Success looks like: You should obtain a parabolic velocity profile with maximum velocity at the free surface, and your flow rate should have dimensions of $[L^2/T]$.