---
title: "Blasius boundary layer"
sort_title: "Blasius boundary layer"
date: 2021-05-29
categories:
- Physics
- Fluid mechanics
- Fluid dynamics
layout: "concept"
---

In fluid dynamics, the **Blasius boundary layer** is an application of
the [Prandtl equations](/know/concept/prandtl-equations/),
which govern the flow of a fluid
at large [Reynolds number](/know/concept/reynolds-number/)
$$\mathrm{Re} \gg 1$$ close to a surface.
Specifically, the Blasius layer is the solution
for a half-plane approached from the edge by a fluid.

Let the half-plane lie in the $$(x,z)$$-plane (i.e. at $$y = 0$$)
and exist for all $$x \ge 0$$, such that its edge lies on the $$z$$-axis.
A fluid with velocity field $$\va{v} = U \vu{e}_x$$
approaches the half-plane's edge head-on.
To describe the fluid's movements around the plane,
we make an ansatz for the so-called **slip-flow region**'s $$x$$-velocity $$v_x(x, y)$$:

$$\begin{aligned}
    v_x
    = U f'(s)
    \qquad \qquad
    s
    \equiv \frac{y}{\delta(x)}
\end{aligned}$$

Where $$\delta(x) \equiv \sqrt{\nu x / U}$$ is the boundary layer thickness
estimate that was used to derive the Prandtl equations.
Note that $$f'(s)$$ is the derivative of an unknown $$f(s)$$,
and that it obeys the boundary conditions $$f'(0) = 0$$ and $$f'(\infty) = 1$$,
i.e. the fluid is stationary at the half-plane's surface $$s = 0$$,
and has velocity $$U$$ far away at $$s \to \infty$$.

Inserting the ansatz into the incompressibility condition $$\nabla \cdot \va{v} = 0$$ yields:

$$\begin{aligned}
    \pdv{v_y}{y}
    = - \pdv{v_x}{x}
    = - \pdv{v_x}{s} \pdv{s}{x}
    = U \frac{\delta'}{\delta} s f''
\end{aligned}$$

Which we integrate by parts to get an expression for the $$y$$-velocity $$v_y$$, namely:

$$\begin{aligned}
    v_y
    = U \frac{\delta'}{\delta} \int s f'' \dd{y}
    = U \delta' \bigg( s f' - \int f' \dd{s} \bigg)
    = U \delta' \: (s f' - f)
\end{aligned}$$

Now, consider the main Prandtl equation,
assuming the attack velocity $$U$$ is constant.
Inserting our expressions for $$v_x$$ and $$v_y$$ into it gives:

$$\begin{aligned}
    \nu \pdvn{2}{v_x}{y}
    &= v_x \pdv{v_x}{x} + v_y \pdv{v_x}{y}
    \\
    \nu \pdvn{2}{v_x}{s} \pdvn{2}{s}{y}
    &= v_x \pdv{v_x}{s} \pdv{s}{x} + v_y \pdv{v_x}{s} \pdv{s}{y}
    \\
    \nu U \frac{1}{\delta^2} f'''
    &= - U^2 \frac{\delta'}{\delta} s f'' f' + U^2 \frac{\delta'}{\delta} f'' (s f' - f)
\end{aligned}$$

We multiply by $$\delta^2 / U$$, cancel out some terms,
and substitute $$\delta(x) \equiv \sqrt{\nu x / U}$$, leaving:

$$\begin{aligned}
    \nu f'''
    &= - U \delta' \delta f'' f
    = - U \frac{\nu}{2 U} f'' f
\end{aligned}$$

This leads us to the **Blasius equation**,
which is a nonlinear ODE for $$f(s)$$:

$$\begin{aligned}
    \boxed{
        2 f''' + f'' f = 0
    }
\end{aligned}$$

Unfortunately, this cannot be solved analytically, only numerically.
Nevertheless, the result shows a boundary layer $$\delta(x)$$
exhibiting the expected downstream thickening.



## References
1.  B. Lautrup,
    *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition,
    CRC Press.