Boundary layer

Boundary layer

Model problem

Lesson 14

Matched asymptotic expansions

• The fluid velocity near the solid wall

• The velocity at the edge of a jet of fluid

• Condensing vapor on a cool surface

• Constructing of the outer expansion (straightforward expansion)

• Constructing of the inner expansion

• Matching them together obtaining matched asymptotic expansion

Matched asymptotic expansions

• The fluid velocity near the solid wall

• The velocity at the edge of a jet of fluid

• Condensing vapor on a cool surface

• Constructing of the outer expansion (straightforward expansion)

• Constructing of the inner expansion

• Matching them together obtaining matched asymptotic expansion

Matched asymptotic expansions

• The fluid velocity near the solid wall

• The velocity at the edge of a jet of fluid

• Condensing vapor on a cool surface

• Constructing of the outer expansion (straightforward expansion)

• Constructing of the inner expansion

• Matching them together obtaining matched asymptotic expansion

Matched asymptotic expansions

• The fluid velocity near the solid wall

• The velocity at the edge of a jet of fluid

• Condensing vapor on a cool surface

• Constructing of the outer expansion (straightforward expansion)

• Constructing of the inner expansion

• Matching them together obtaining matched asymptotic expansion

Matched asymptotic expansions

• The fluid velocity near the solid wall

• The velocity at the edge of a jet of fluid

• Condensing vapor on a cool surface

• Constructing of the outer expansion (straightforward expansion)

• Constructing of the inner expansion

• Matching them together obtaining matched asymptotic expansion

Matched asymptotic expansions

• The fluid velocity near the solid wall

• The velocity at the edge of a jet of fluid

• Condensing vapor on a cool surface

• Constructing of the outer expansion (straightforward expansion)

• Constructing of the inner expansion

• Matching them together obtaining matched asymptotic expansion

Model example

εd²f

dx² + df

dx = 2x + 1, ε 1, x ∈ (0, 1), f(0) = 1, f(1) = 4.

f(x, ε) ∼ f₀(x) + εf₁(x) + ε²f₂(x) + . . . , ε⁰ : df₀

dx = 2x + 1, f₀(0) = 1, f₀(1) = 4 εⁿ : df_n

dx = −d²f_n⁻₁

dx² for ^{n ∈ N, f}n(0) = 0, f_n(1) = 0,

the differential equation of the first order can not satisfy two boundary conditions.

Model example

The general solution of the first equation ^df_dx^{0 = 2x + 1} is

f₀ = x² + x + c. We abandon the initial condition at

x = 0 and use

f₀(1) = 4 ⇒ c = 2 and ^f0 = x² + x + 2 df₁

dx = −d²f₀

dx² = −2 ⇒ f₁ = −2(x − 1).

f^outer = x² + x + 2 + ε2(1 − x) f^outer(0) 6= 1

Model example

Let us look on the exact solution of ^εd_dx^{2f2 +} _dx^{df = 2x + 1}, This is a linear equation with constant coefficients. Its exact solution is

f = A+Be^−x/ε+x²+x(1−2ε), A = 2(1+ε), B = −(1+2ε)

because we neglect the term ^{e−x/ε = o(εN)}. The exact solution is

f^exact = 2(1 + ε) − (1 + 2ε)e^−x/ε + x² + x(1 − 2ε)

= x² + x + 2 − e^−x/ε + ε 2(1 − x) − 2e^−x/ε

The term ^e−x/ε is absent in ^fouter.

Model example

Model example

The region near ^{x = 0} is called the boundary layer.

The stretched variables and inner expansion

The gradient of ^fexact is increase rapidly in the boundary layer.

f₀^ex = x² + x + 2 − e^−x/ε, df₀^ex

dx = 2x + 1 + 1

εe^−x/ε, d²f₀^ex

dx² = 2 − 1

ε² e^−x/ε

out. BL ^f₀^{ex = O(1), df0ex}

dx = O(1), d²f₀^ex

dx² = O(1), df^ex 1 d²f^ex 1

The stretched variables and inner expansion

Let us introduce the new variable ^{s = x}_ε. Then

df

ds = df dx

dx

ds = ε df

dx, df

dx = O(1

ε) 7→ df

ds = O(1), d²f

dx² = ε² d²f

dx² , d²f

dx² = O( 1

ε² ) 7→ d²f

ds² = O(1).

New variable ^{s = x}_ε is called stretched variable. The equation ^εd_dx^{2f2 +} _dx^{df = 2x + 1} take the form

ε ε²

d²f

ds² + 1 ε

df

ds = 2εs + 1,

multiplying by we get

The stretched variables and inner expansion

d²f

ds² + df

ds = 2ε²s + ε, f(s, ε) ∼ f₀(s) + εf₁(s) + . . . , ε⁰ d²f₀ⁱⁿ

ds² + df₀ⁱⁿ

ds = 0, f₀ⁱⁿ(0) = 1 ⇒ f₀ⁱⁿ = A + (1 − A)e^−s ε¹ d²f₁ⁱⁿ

ds² + df₁ⁱⁿ

ds = 1, f₁ⁱⁿ(0) = 0 ⇒ f₁ⁱⁿ = B − Be^−s + s ε² d²f₂ⁱⁿ

ds² + df₂ⁱⁿ

ds = 2s, f₂ⁱⁿ(0) = 0 ⇒ f₂ⁱⁿ = C − Ce^−s + s² − 2s d²fⁱⁿ dfⁱⁿ

Prandtl’s matching condition

To find the constant in the equations

f₀ⁱⁿ = A + (1 − A)e^−s f₁ⁱⁿ = B − Be^−s + s

f₂ⁱⁿ = C − Ce^−s + s² − 2s

f_nⁱⁿ = D_n − D_ne^−s, n = 3, 4, . . .

we can not use the boundary condition at ^{x = 1}. We need to match the inner and outer solutions. Let us do it for the leader terms ^f₀^in(s) and ^f₀^out(x).

Prandtl’s matching condition

Let us chose two points ^{x = 5ε} and ^{x = 6ε} and equal

f₀^out(x) = x² + x + 2 with ^f₀^{in(s) = A + (1 + A)e−s}, ^{s = x/ε} Then

A = 2 + 5ε + 25ε² − e⁻⁵

1 − e⁻⁵ for ^{x = 5ε}

A = 2 + 6ε + 36ε² − e⁻⁶

1 − e⁻⁶ for ^{x = 6ε} Prandtl’s matching condition

xlim0 f₀^out(x) = lim

s f₀ⁱⁿ(s)

Prandtl’s matching condition

In our case it gives

xlim→0 x² + x + 2 = lim

s→∞

A + (1 + A)e^−s ⇒ A = 2.

So the leader terms are

Outer region ^f₀^{out = x2 + x + 2,} for ^{x = O(1),} Inner region ^f₀^{in = 2 − e−x/ε,} for ^{x = O(ε).}

Comparing with the exact solution

If ^{x = O(1)} then ^f₀^{ex = x2 + x + 2 + . . . ,}

If then ^{ex x/ε}

The composite expansion

We want to have

For ^{x = O(1) f}₀^{out = f}₀^{comp +} small terms^, For ^{x = O(ε) f}₀^{in = f}₀^{comp +} small terms^,

If we choose ^f₀^{comp = f}₀^{in + f}₀^{out − f}₀^match, where ^f₀^match is given by the Prandtl’s matching condition. Since

for ^{x = O(1) f}₀^{in = f}₀^{match +} small terms^, for ^{x = O(ε) f}₀^{out = f}₀^{match +} small terms^.

Boundary layer location

How define where the boundary layer is located.

Consider the example

εd²f

dx² +(x−2) df

dx+f = 0, ε 1, x ∈ (0, 1), f(0) = 3, f(1) = 2.

Let us suppose that boundary layer is at ^{x = 0}. Then

(x − 2)df₀^out

dx + f₀^out = 0, f(1) = 2 ⇒ f₀^out = 2 2 − x.

Use the stretched variables ^{s = x/ε} to obtain new equation

Boundary layer location

d²fⁱⁿ

ds² + (εs − 2)dfⁱⁿ

ds + εfⁱⁿ = 0.

The leading term ^f₀ⁱⁿ satisfies

d²f₀ⁱⁿ

ds² − 2df₀ⁱⁿ

ds = 0, f₀ⁱⁿ(0) = 3

with solution

f₀ⁱⁿ = 3 − A + Ae^2s.

If ^{s → ∞}, then ^{e2s → ∞} very quickly. It happens

2 in in

Boundary layer location

Let us suppose that the boundary layer is located at

x = 1. Then

(x − 2)df₀^out

dx + f₀^out = 0, f(0) = 3 ⇒ f₀^out = 6 2 − x.

The appropriate stretched variable is ^{s = (1 − x)/ε}. Then

df

dx = −1 ε

f

ds, d²f

dx² = 1 ε²

d²f ds² .

The new equation is

d²fⁱⁿ

ds² + (εs + 2)dfⁱⁿ

ds + εfⁱⁿ = 0.

Boundary layer location

The leading term ^f₀ⁱⁿ satisfies

d²f₀ⁱⁿ

ds² + 2df₀ⁱⁿ

ds = 0, f₀ⁱⁿ(0) = 2 (x = 1 7→ s = 0)

with solution

f₀ⁱⁿ = 2 − A + Ae^−s.

Prandtl’s matching condition applied to

f₀^out = 6

2 − x and ^f₀^{in = 2 − A + Ae−s}

Boundary layer location

It gives ^{A = −4} and

f₀^comp = 6

2 − x − 4e^{−(1−x)/ε}.

General linear equation

Some general consideration about boundary location of

εd²f

dx² + a(x) df

dx + b(x)f = c(x), x₁ < x < x₂

I. If ^{a(x) > 0} on the interval ^(x1, x₂) then the boundary layer will occur at ^{x = x}1. The stretching transformation is ^{s = (x − x}1)/ε. The one term inner expansion will

satisfy

d²f₀ⁱⁿ

ds² + a(x₁)df₀ⁱⁿ

ds = 0.

The solution of the equation is

General linear equation

εd²f

dx² + a(x) df

dx + b(x)f = c(x), x₁ < x < x₂

II. If ^{a(x) < 0} on the interval ^(x1, x₂) then the boundary layer will occur at ^{x = x}2. The stretching transformation is ^{s = (x}2 − x)/ε. The the inner expansion will involve term

exp

a(x₂)x₂ − x ε

.

General linear equation

εd²f

dx² + a(x) df

dx + b(x)f = c(x), x₁ < x < x₂

III. If ^a(x) change the sign on the interval ^(x1, x₂) then the boundary layer will occur at an interior point ^x0 of

(x₁, x₂) where ^a(x0) = 0. Moreover the boundary layer can occur at both ends ^x₁ and ^x₂.

The end