Thin layer of fluid surrounding a body. It can be laminar or turbulent. A laminar boundary layer is one in which the fluid velocity is well-ordered and regular throughout the boundary layer. The speed of the fluid is taken to be zero at the surface of the body. A turbulent boundary layer is one in which the fluid velocity shows a great deal of randomness throughout the layer. A turbulent boundary layer is responsible for increased drag on the body, since the mean velocity profile of the boundary layer is 'fuller' than that of a laminar boundary layer, resulting in increased shear stress at the body's surface. Laminar to turbulent transition occurs on, for example, the raised red stitches on a baseball. This drag transition is responsible for a certain portion of the behavior of knuckleballs, curve balls, sliders, etc.

Aircraft whose value is determined by the cost of operation per passenger-mile typically require a great deal of thought and planning be put into where, when, and if boundary layer transition occurs on their wings. When a boundary layer is forced to turn through an excessively high angle on the surface of the body, it detaches and forms the boundary of a wake. On an airfoil, boundary layer detachment results in a stall. A laminar boundary layer will detach more readily than a turbulent one. This is why a smooth ping pong ball curves quite a lot when struck hard through the air. It would curve less if it had raised stitches, if memory serves.

If the fluid in question is air, the boundary layer is typically the only location in a flowfield where fluid viscosity cannot be neglected. Many calculations ignore the effects of the boundary layer since it is generally very thin with respect to the size of the body it surrounds. The boundary layer is responsible for the bulk of the heat transfer which occurs at the surface of a hypersonic vehicle (like the space shuttle).

In a pipe, the boundary layer's thickness grows until the thickness of the boundary layer equals the radius of the pipe. Using this knowledge, and formulae for the speed of boundary layer growth, calculations of pipe pressure loss per linear foot are possible.

Basically, a critical concept in fluid mechanics.

Boundary Layer theory can be thought of as being equivalent to the theory of matched asymptotic expansions.


Most of the time, it's okay to throw away small things - this is true 'far away from boundary layers'. Sometimes, it's not okay to do this - this is true 'near boundary layers'.

(Sometimes it's never okay to throw away things - even exponentially small things - but that's not true here).

Simple Example

We have an equation, such as (for a simple example)

q * y'' + y' + y = 0 (*)

Where y' means the derivative of y with respect to x, and q is a very small number. Let's have some boundary conditions, too. Say, y(0)=0, y(1)=1. The equation we end up being interested in could describe the NSEs near a boundary, for example, or deviation from Stokes Flow, with Re as a natural choice for q.

Now, our instinct as human beings is to throw away small things. Let's throw away anything that looks small, leaving us with

y' + y = 0.

Hum. Alarm bells should be ringing here - we have two boundary conditions, but only a first order ODE. What we have actually assumed by throwing away our small thing is that x=ord(1). More on this in a moment, but if we've assumed x=ord(1), then it makes sense that we apply our boundary condition at x = 1. Okay, great, we know what to do. This gives us

y = A*exp(-x) = exp(1 - x)

And we're left pretty unhappy, because this can never satisfy our boundary condition at x = y.

We think for a little bit, and realise that we wrote down 'assumed' a few lines back, and scrabble back to look at this. x = ord(1)? x is roughly the size of 1? Well, what about when x isn't? What about when x is teeny?

(Considering the balances in the original equation (*), we see that 'teeny' here means that x = ord(q^1/2), so that the first term q*y'' is roughly the size of the last term y - but this leaves us with something nasty in the middle, so we go a little smaller to find x = ord(q) is a nice rescaling).

We'd like to look when x is teeny with the mathematical equivalent of a magnifying glass. Think of a new variable, r, which will be our magnifying glass. When x is small (around the size of q) we want r to be pretty big - around the size of 1. So let's write r=x/q, and rewrite (*) in terms of our new variable. I'll use capital Y instead of y to make it clear when I'm in the small bit or the big bit, respectively.

Y'' + Y' + q*Y = 0

Right! Let's throw away the small terms again!

Y'' + Y' = 0 => Y = B*exp(-r) + C.

Okay, we have two arbitrary constants, B and C. Apply our x=0 (=> r=0) boundary condition:

Y = B*(1 - exp(-r)) = B*(1 - exp(-x/q)).

And again, we're unhappy, because we wanted a full solution, not to have some dumb constant B left over. Arse.

We go and have a cup of tea and remember that we want y and Y to be, in some sense, two different bits of the same function. Well, in physical problems, we'd like almost all of our functions to join up in a nice way, right? In the middle, they should meet somewhere.

Okay. Let's look at y first, and imagine that x has gotten really quite small (but not quite small enough for it to make our assumptions invalid). Then

y -> exp(1)

As the exp(-x) has pretty much become 1. Now, let's look at Y, and let our x there get pretty big, but not big enough to make our assumptions invalid (what I'm doing here is secretly applying Van Dyke's Matching Principle. Then

Y - > B

As our exp(-x/q) has become pretty much zero. We frown and gurn and feel pretty guilty, and eventually decide that yes, B = exp(1).

So now, we have our full solution. It's not exact, but it's pretty good in each region. To recap:

Outer solution (x = ord(1))

y = exp(1-x)

Inner Solution (x = ord(q))

Y = exp(1)*(1 - exp(-x/q))

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