A boundary layer of thickness O ( √ ε)

Principal of least degeneracy

Higher order matching and nonlinear equations

Lesson 16

A boundary layer of thickness O ( √ ε)

Let us consider the example ^εd_dx^{2f2 + x2}_dx^df ₋ ^{f = 0} with

x ∈ (0, 1), ^{f(0) = 1}, ^{f(1) = 2}. We will look for the one

term expansion. Since ^{x2 > 1} on ^{(0, 1)} we assume that the boundary layer is at ^{x = 0}. Is something strange up to now? Thus

x²df₀^out

dx − f₀^out = 0, f₀^out(1) = 2.

Separating variables leads to

Z df₀^out f₀^out =

Z dx

x² ⇒ f₀^out = Ce^−1/x = 2e^1−1/x.

Thickness O( √ ε)

We suppose that ^{s = x/ε}, then the new equation is

d²f

ds² + ε²s² df

ds − εf = 0, f(0) = 1

then

d²f₀ⁱⁿ

ds² = 0 ⇒ f₀ⁱⁿ = As + 1.

Applying matching condition, we get

xlim→0 2e^1−1/x = lim

s→∞

As + 1.

What is wrong?

Thickness O( √ ε)

Let us try the stretched variable ^{s = x/εp}, then

df

dx = 1 ε^p

df

ds, d²f

dx² = 1 ε^2p

d²f ds² . ε^1−2p d²f

ds² + ε^ps² df

ds − f = 0.

Let us try to keep the maximum number of terms in the equation. Taking ^{p = 1/2} we keep the second derivative and ^f. Then

d²f₀ⁱⁿ

ds² − f₀ⁱⁿ = 0, f₀ⁱⁿ(0) = 1, ⇒ f₀ⁱⁿ = Ae^s + (1 − A)e^−s.

Thickness O( √ ε)

Applying Prandtl’s matching condition, we get

xlim→0 2e^1−1/x = lim

s→∞

Ae^s + (1 − A)e^−s

that implies ^{A = 2}. Thus

f₀^comp = 2e^1−1/x + e^−x/√ε.

If ^{p < 1/2} then ^f₀^{in = 0} which does not satisfy the boundary condition at ^{s = 0}.

An interior boundary layer

εd²f

dx² +xdf

dx+xf = 0 with ^x _∈ ⁽₋^{1, 1), f(}₋^{1) = e, f(1) = 2e−1.} We expect the boundary layer at ^{x = 0}, two outer

solution ^f₀^{+out = f}₀⁺ and ^f₀^{−out = f}₀⁻. We have

df₀⁺

dx + f₀⁺ = 0, f₀⁺(1) = 2e⁻¹ ⇒ f₀⁺ = 2e^−x, df₀⁻

dx + f₀⁻ = 0, f₀⁻(−1) = e ⇒ f₀⁻ = e^−x.

An interior boundary layer

The stretched variable ^{s = x/εp} leads to

ε^1−2p d²f

ds² + sdf

ds + ε^psf = 0.

Taking ^{p = 1/2}, we get

d²f₀ⁱⁿ

ds² + sdf₀ⁱⁿ

ds = 0.

We make a substitution ^{w = df}_ds⁰ⁱⁿ, then

dw + sw = 0 ⇒ w = Ae^−s2/2 ⇒

An interior boundary layer

f₀ⁱⁿ = A

Z _s

0

e^−t2/2 dt + f₀ⁱⁿ(0).

We write ^R₀^{s e−t2/2 dt = √1}₂ ^R₀^{s/√2 e−u2 du = B erf(s/√2)}. Then

f₀ⁱⁿ = B erf(s/√

2) + f₀ⁱⁿ(0).

The Prandtl’s condition gives

xlim→0⁺ 2e^−x = lim

s→+∞

B erf(s/√

2) +f₀ⁱⁿ(0) ⇒ B+f₀ⁱⁿ(0) = 2,

xlim→0⁻ e^−x = lim

s→+∞

B erf(s/√

2)+f₀ⁱⁿ(0) ⇒ −B+f₀ⁱⁿ(0) = 1.

An interior boundary layer

B = 1/2, f₀ⁱⁿ(0) = 3/2.

Summarizing

f₀⁺ = 2e^−x, x > O(√ ε) f₀ⁱⁿ = 1/2 erf(x/√

2ε) + 3/2, x = O(√ ε) f₀⁻ = e^−x, x < −O(√

ε)

The composition solution can not be produced by the standard way. We observe

An interior boundary layer

f₀ⁱⁿ

x > O(√

ε)

= 1/2 + 3/2 + sm. t., since ^erf _∼ ^1,

f₀ⁱⁿ

x < −O(√

ε)

= −1/2 + 3/2 +sm. t., since ^erf _{∼ −}^1, We conclude that

f₀^comp =

1/2 erf(x/√

2ε) + 3/2

e^−x.

An interior boundary layer

Higher order matching

We shall work with the old example

εd²f

dx² + df

dx = 2x + 1, x ∈ (0, 1), f(0) = 1, f(1) = 4, f^ex = x² + x + 2 − e^−x/ε + ε

2(1 − x) − 2e^−x/ε

Using the stretched variable ^{s = x/ε} we get

Higher order matching

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 εⁿ d²f_nⁱⁿ

ds² + df_nⁱⁿ

ds = 0, f_nⁱⁿ(0) = 0 ⇒ f_nⁱⁿ = D_n − D_ne^−s,

Higher order matching

fⁱⁿ = A + (1 − A)e^−s + ε(B − Be^−s + s) +ε²(C − Ce^−s + s² − 2s) +

X∞

n=3

εⁿD_n(1 − e^−s).

The Prandtl’s condition can not be applied, because of existence of unbounded terms ^εs and ^ε2(s2 ₋ ^2s). They represent ^x and ^x2 ₋ ^2εx. Let us write the exact

solution in terms of ^s:

Higher order matching

f^ex = x² + x + 2 − e^−x/ε + ε

2(1 − x) − 2e^−x/ε

= ε²s² + εs + 2 − e^−s + ε 2(1 − εs) − 2e^−s

= 2 − e^−s + ε(2 − 2e^−s + s) + ε²(s² − 2s).

Comparing with ^fin we get ^{A = 2}, ^{B = 2}, ^{C = 0}, ^Dn = 0,

n = 3, 4, . . .. How we can define ^{A, B, C, D}n without of exact solution?

Higher order matching

Let us look on the Prandtl’s condition from the other point of view. Let ^{x = O(ε1/2)}, then

fⁱⁿ = A+(1−A)e^−s = A+o(1), f^out = x²+x+2−e^−x/ε = 2+o(1).

We recuperated ^{A = 2}. Let us do the same, but in slightly general form. We represent ^{s = x}_ε in two

variables ^{x = tεα} then ^{s =} _ε^1t−α . Let us represent two term expansion in terms of new variable ^{t = x/εα},

α ∈ (0, 1) for ^t fixed. We get

f^out = x² + x + 2 + ε2(1 − x) = 2 + ε^αt + ε^2αt² + 2ε − 2ε^1+αt

Higher order matching

Now we use ^{s =} _ε^1t−α and fined two term expansion of

fⁱⁿ in terms of _ε^1t−α .

fⁱⁿ = A + ε(B + s) = A + ε^αt + εB.

Matching with

f^out = 2 + ε^αt + ε^2αt² + 2ε − 2ε^1+αt

we have to ask ^{ε2α = o(ε)}, or ^{α > 1/2}. We get ^{A = 2} and

B = 2.

Higher order matching

The composition solutions is

f₂^comp_term = f₂^out_term + f₂ⁱⁿ_term − f₂^match_term

where ^f₂^comp_term is given by terms of order up to ^O(ε). In our case it is ^{2 + x + 2ε}. We have

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

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

Nonlinear examples

εd²f

dx² + df

dx + f² = 0, x ∈ (0, 1), f(0) = 2, f(1) = 1/2.

df₀^out

dx + (f₀^out)² = 0, f₀^out(1) = 1/2 ⇒ f₀^out = 1 x + 1. s = x

ε ⇒ d²fⁱⁿ

ds² + dfⁱⁿ

ds + ε(fⁱⁿ)² = 0 d²f₀ⁱⁿ

ds² + df₀ⁱⁿ

ds = 0 ⇒ f₀ⁱⁿ = A + (2 − A)e^−s

Nonlinear examples

The Prandtl’s condition implies

xlim→0

1

x + 1 = lim

s→∞

A + (2 − A)e^−s ⇒ A = 1.

f^comp = 1

x + 1 + e^−x/ε.

Nonlinear examples

εd²f

dx² + 2f df

dx − 4f = 0, x ∈ (0, 1), f(0) = 0, f(1) = 4.

2f₀^outdf₀^out

dx − 4f₀^out = 0, f(1) = 4 ⇒ f₀^out = 2x + 2 s = x

ε ⇒ d²f₀ⁱⁿ

ds² +2f₀ⁱⁿdf₀ⁱⁿ

ds = 0 ⇒ d ds

df₀ⁱⁿ

ds +(f₀ⁱⁿ)²

= 0 df₀ⁱⁿ

ds + (f₀ⁱⁿ)² = c = a²

since ^f₀^{in(0) = 0} and ^lim

x 0 f₀^out = 2 and ^f₀ⁱⁿ increase

Nonlinear examples

• R _df0ⁱⁿ

a²−(f₀ⁱⁿ)² = R

ds

• −_2a¹ R _df0ⁱⁿ

f₀ⁱⁿ−a + _2a¹ R _df0ⁱⁿ

f₀ⁱⁿ+a = s + b

• − ln |f₀ⁱⁿ − a| + ln |f₀ⁱⁿ + a| = 2as + 2ab

• ^f0in−a

f₀ⁱⁿ+a = Ae^2as, ^{A =} _±^e2ab

• A = 1, ^f₀^{in = ae}_e^22asas−₊₁^{1 = a tanh as}

• lim

x→0 f₀^out = lim

s→∞

f₀ⁱⁿ ⇒ a = 2

Nonlinear examples

• R _df0ⁱⁿ

a²−(f₀ⁱⁿ)² = R

ds

• −_2a¹ R _df0ⁱⁿ

f₀ⁱⁿ−a + _2a¹ R _df0ⁱⁿ

f₀ⁱⁿ+a = s + b

• − ln |f₀ⁱⁿ − a| + ln |f₀ⁱⁿ + a| = 2as + 2ab

• ^f0in−a

f₀ⁱⁿ+a = Ae^2as, ^{A =} _±^e2ab

• A = 1, ^f₀^{in = ae}_e^22asas−₊₁^{1 = a tanh as}

• lim

x→0 f₀^out = lim

s→∞

f₀ⁱⁿ ⇒ a = 2

Nonlinear examples

• R _df0ⁱⁿ

a²−(f₀ⁱⁿ)² = R

ds

• −_2a¹ R _df0ⁱⁿ

f₀ⁱⁿ−a + _2a¹ R _df0ⁱⁿ

f₀ⁱⁿ+a = s + b

• − ln |f₀ⁱⁿ − a| + ln |f₀ⁱⁿ + a| = 2as + 2ab

• ^f0in−a

f₀ⁱⁿ+a = Ae^2as, ^{A =} _±^e2ab

• A = 1, ^f₀^{in = ae}_e^22asas−₊₁^{1 = a tanh as}

• lim

x→0 f₀^out = lim

s→∞

f₀ⁱⁿ ⇒ a = 2

Nonlinear examples

• R _df0ⁱⁿ

a²−(f₀ⁱⁿ)² = R

ds

• −_2a¹ R _df0ⁱⁿ

f₀ⁱⁿ−a + _2a¹ R _df0ⁱⁿ

f₀ⁱⁿ+a = s + b

• − ln |f₀ⁱⁿ − a| + ln |f₀ⁱⁿ + a| = 2as + 2ab

• ^f0in−a

f₀ⁱⁿ+a = Ae^2as, ^{A =} _±^e2ab

• A = 1, ^f₀^{in = ae}_e^22asas−₊₁^{1 = a tanh as}

• lim

x→0 f₀^out = lim

s→∞

f₀ⁱⁿ ⇒ a = 2

Nonlinear examples

• R _df0ⁱⁿ

a²−(f₀ⁱⁿ)² = R

ds

• −_2a¹ R _df0ⁱⁿ

f₀ⁱⁿ−a + _2a¹ R _df0ⁱⁿ

f₀ⁱⁿ+a = s + b

• − ln |f₀ⁱⁿ − a| + ln |f₀ⁱⁿ + a| = 2as + 2ab

• ^f0in−a

f₀ⁱⁿ+a = Ae^2as, ^{A =} _±^e2ab

• A = 1, ^f₀^{in = ae}_e^22asas−₊₁^{1 = a tanh as}

• lim

x→0 f₀^out = lim

s→∞

f₀ⁱⁿ ⇒ a = 2

Nonlinear examples

• R _df0ⁱⁿ

a²−(f₀ⁱⁿ)² = R

ds

• −_2a¹ R _df0ⁱⁿ

f₀ⁱⁿ−a + _2a¹ R _df0ⁱⁿ

f₀ⁱⁿ+a = s + b

• − ln |f₀ⁱⁿ − a| + ln |f₀ⁱⁿ + a| = 2as + 2ab

• ^f0in−a

f₀ⁱⁿ+a = Ae^2as, ^{A =} _±^e2ab

• A = 1, ^f₀^{in = ae}_e^22asas−₊₁^{1 = a tanh as}

• lim

x→0 f₀^out = lim

s→∞

f₀ⁱⁿ ⇒ a = 2

Nonlinear examples

f₀ⁱⁿ = a tanh 2s

f^comp = 2x + 2 tanh(2x/ε).

The end