Uniform and nonuniform expansions

Lesson 7

Uniform and nonuniform expansions – p. 1/23

Uniform and nonuniform expansions

If

f(x, ε) =N

∑

n=0

an(x)δn(ε) + RN (x, ε)

is an asymptotic expansion then

RN (x, ε) = O(δN+1(ε)) as ε → 0.

Or we can say

|RN (x, ε)| ≤ K|δN+1(ε)|.

Uniform and nonuniform expansions – p. 2/23

Example of uniform expansion

11−ε sin x ∼ 1 + ε sin x + ε2 sin2 x + ε3 sin3 x + . . . with

RN =∞

∑

n=N+1

εn sinn x so limε→0

RN

εN+1= sinN+1 x

and we can estimate

|RN (x, ε)| ≤ K|εN+1|

with any K > 1.

Uniform and nonuniform expansions – p. 3/23

Example of nonuniform expansion

11−ε sin x ∼ 1 + εx + ε2x2 + ε3x3 + . . . with

RN =∞

∑

n=N+1

εnxn so limε→0

RN

εN+1= xN+1

Since x is not bounded we can not find K such that

|RN (x, ε)| ≤ K|εN+1|.

Uniform and nonuniform expansions – p. 4/23

Region of nonuniformity

The principal idea of the asymptotic expansion that thesubsequent term in the expansion smaller theprevious one an+1(x)δn+1(ε) = o

(

an(x)δn(ε))

.

11−ε sin x ∼ 1 + εx + ε2x2 + ε3x3 + . . . When twosubsequent term has the same order?

εn+1xn+1 = O(εnxn) ⇒ εx = O(1)

or x = O(1/ε). If εn+1xn+1 = o(εnxn) ⇒ εx = o(1)

or, for instance, x = O( 1√ε)

Uniform and nonuniform expansions – p. 5/23

Region of nonuniformity

The principal idea of the asymptotic expansion that thesubsequent term in the expansion smaller theprevious one an+1(x)δn+1(ε) = o

(

an(x)δn(ε))

.

11−ε sin x ∼ 1 + εx + ε2x2 + ε3x3 + . . . When twosubsequent term has the same order?

εn+1xn+1 = O(εnxn) ⇒ εx = O(1)

or x = O(1/ε). If εn+1xn+1 = o(εnxn) ⇒ εx = o(1)

or, for instance, x = O( 1√ε)

Uniform and nonuniform expansions – p. 5/23

Region of nonuniformity

• 1 + εex + ε2e2x + ε3e3x + . . .

εex = O(1) ⇒ ex = O(1/ε)

or x = O(− ln ε) as ε → 0 and x is large.

• 1 + εx + ε2

x2 + ε3

x3 + . . .If

ε

x= O(1) ⇒ x = O(ε) as ε → 0

for small values of x.

Uniform and nonuniform expansions – p. 6/23

Region of nonuniformity

• 1 + εex + ε2e2x + ε3e3x + . . .

εex = O(1) ⇒ ex = O(1/ε)

or x = O(− ln ε) as ε → 0 and x is large.

• 1 + εx + ε2

x2 + ε3

x3 + . . .If

ε

x= O(1) ⇒ x = O(ε) as ε → 0

for small values of x.

Uniform and nonuniform expansions – p. 6/23

Region of nonuniformity (moreexamples)

sin(x + ε) = sin x cos ε + cos x sin ε

= sin x(

1 −ε2

2+ O(ε4)

)

+ cos x(

ε −ε3

6+ O(ε5)

)

= sin x + ε cos x −ε2

2sin x −

ε3

6cos x + O(ε4)

sin(x(1 + ε)) = sin x cos εx + cos x sin εx

= sin x(

1 −ε2x2

2+ O(ε4x4)

)

+ cos x(

εx −ε3x3

6+ O(ε5x5)

)

= sin x + εx cos x −ε2x2

2sin x −

ε3x3

6cos x + O(ε4x4)

Uniform and nonuniform expansions – p. 7/23

Region of nonuniformity

sin(x + ε) = sin x + ε cos x −ε2

2sin x −

ε3

6cos x + O(ε4)

is the uniform expansion and

sin(x(1+ε)) = sin x+εx cos x−ε2x2

2sin x−

ε3x3

6cos x+O(ε4x4)

is nonuniform because of appearance of x incoefficients.

Uniform and nonuniform expansions – p. 8/23

Region of nonuniformity

• The expansion f(x, ε) ∼∑∞

n=0 fn(x)δn(ε) as ε → 0

is uniform if coefficients fn(x) are bounded.

• Is it true that if fn(x) are unbounded then theexpantion is nonuniform?

• The expansion is uniform x + εx + ε2x + ε3x + . . .

• For the nonuniformity is requiredfn+1(x)δn+1(ε) = O

(

fn(x)δn(ε))

⇒

fn+1(x) = fn(x)O(

δn(ε)δn+1(ε)

)

The coefficient fn+1(x)

increase faster then fn(x) since δn(ε)δn+1(ε)

increase.

Uniform and nonuniform expansions – p. 9/23

Region of nonuniformity

• The expansion f(x, ε) ∼∑∞

n=0 fn(x)δn(ε) as ε → 0

is uniform if coefficients fn(x) are bounded.• Is it true that if fn(x) are unbounded then the

expantion is nonuniform?

• The expansion is uniform x + εx + ε2x + ε3x + . . .

• For the nonuniformity is requiredfn+1(x)δn+1(ε) = O

(

fn(x)δn(ε))

⇒

fn+1(x) = fn(x)O(

δn(ε)δn+1(ε)

)

The coefficient fn+1(x)

increase faster then fn(x) since δn(ε)δn+1(ε)

increase.

Uniform and nonuniform expansions – p. 9/23

Region of nonuniformity

• The expansion f(x, ε) ∼∑∞

n=0 fn(x)δn(ε) as ε → 0

is uniform if coefficients fn(x) are bounded.• Is it true that if fn(x) are unbounded then the

expantion is nonuniform?• The expansion is uniform x + εx + ε2x + ε3x + . . .

• For the nonuniformity is requiredfn+1(x)δn+1(ε) = O

(

fn(x)δn(ε))

⇒

fn+1(x) = fn(x)O(

δn(ε)δn+1(ε)

)

The coefficient fn+1(x)

increase faster then fn(x) since δn(ε)δn+1(ε)

increase.

Uniform and nonuniform expansions – p. 9/23

Region of nonuniformity

• The expansion f(x, ε) ∼∑∞

n=0 fn(x)δn(ε) as ε → 0

is uniform if coefficients fn(x) are bounded.• Is it true that if fn(x) are unbounded then the

expantion is nonuniform?• The expansion is uniform x + εx + ε2x + ε3x + . . .

• For the nonuniformity is requiredfn+1(x)δn+1(ε) = O

(

fn(x)δn(ε))

⇒

fn+1(x) = fn(x)O(

δn(ε)δn+1(ε)

)

The coefficient fn+1(x)

increase faster then fn(x) since δn(ε)δn+1(ε)

increase.

Uniform and nonuniform expansions – p. 9/23

Sources of nonuniformity

• Infinite domains which allow long term effects ofsmall perturbations to accumulate

• Singularities in governing equations which lead tolocalized regions of rapid change

Uniform and nonuniform expansions – p. 10/23

Sources of nonuniformity

• Infinite domains which allow long term effects ofsmall perturbations to accumulate

• Singularities in governing equations which lead tolocalized regions of rapid change

Uniform and nonuniform expansions – p. 10/23

Infinite domains

Consider nonlinear oscillator equation d2udt2 + u + εu3 = 0

for t > 0 with initial conditions u(0) = a, dudt (0) = b. We

use the standard expansion

u(t, ε) = u0(t) + εu1(t) + . . .

Substituting to the equation, we obtain

(d2u0

dt2+ u0

)

+ ε(d2u1

dt2+ u1 + u3

0

)

+ O(ε2) = 0

u0(0) + εu1(0) + . . . = a,du0

dt(0) + ε

du1

dt(0) + . . . = b

Uniform and nonuniform expansions – p. 11/23

Infinite domains

We need to solve the following equations

ε0 :d2u0

dt2+ u0 = 0, u0(0) = a,

du0

dt(0) = b ⇒ u0 = a cos t,

ε1 :d2u1

dt2+ u1 + u3

0 = 0, u1(0) = 0,du1

dt(0) = 0 ⇒

d2u1

dt2+ u1 = −a3 cos3 t = −

a3

4cos 3t −

3a3

4cos t.

Uniform and nonuniform expansions – p. 12/23

Infinite domains

d2u1

dt2+ u1 = −

a3

4cos 3t −

3a3

4cos t

A cos t + B sin t is a homogeneous solution,

α cos 3t + β sin 3t is a particular solution correspondingto cos 3t ⇒ α = a3

32 , β = 0,

δt cos t + γt sin t is a particular solution corresponding tocos t since cos t has the same form as a generalsolution. We have δ = 0, γ = − 3a3

8 .

Uniform and nonuniform expansions – p. 13/23

Infinite domains

The general solution isu1 = A cos t + B sin t + a3

32 cos 3t − 3a3

8 t sin t,u1(0) = du1

dt (0) = 0.

A = −a3

32, B = 0 ⇒ u1 =

a3

32(cos 3t − cos t) −

3a3

8t sin t,

u ∼ a cos t + ε(a3

32(cos 3t − cos t) −

3a3

8t sin t

)

+ . . .

The term t sin t is called secular term. It is an oscillatoryterm of increasing amplitude. It leads to thenonuniformity. cos t = O(εt sin t) ⇒ t = O(1/ε) asε → 0.

Uniform and nonuniform expansions – p. 14/23

Small parameter multiplying thehighest derivative

Consider the equation ε dydx + y = e−x for x > 0, ε � 1

with initial conditions y(0) = 2. We use the standardexpansion

y(x, ε) = y0(x) + εy1(x) + ε2y2(x) + . . .

Substituting to the equation, we obtain

ε(dy0

dx+ ε

dy1

dx+ . . .

)

+ y0 + εy1 + ε2y2 + O(ε3) = e−x

y0(0) + εy1(0) + ε2y2(0) + . . . = 2

Uniform and nonuniform expansions – p. 15/23

Small parameter multiplying thehighest derivative

εdydx + y = e−x

ε0 : y0 = e−x, y0(0) = 2,

ε1 : y1 = −dy0

dx= e−x, y1(0) = 0,

ε2 : y2 = −dy1

dx= e−x, y2(0) = 0

We obtain the expansion y ∼ e−x + εe−x + ε2e−x + . . .,but the boundary conditions can not be satisfied. Theunperturbed equation (ε = 0) is an algebraic equation.The nature of an differential equation and algebraicequation is very different.

Uniform and nonuniform expansions – p. 16/23

Small parameter multiplying thehighest derivative

Let us compare with the exact solution ofεdy

dx + y = e−x: yex = 1−2ε1−ε e−x/ε + e−x

1−ε

yex = (1− ε− ε2 + . . .)e−x/ε + (1 + ε + ε2 + . . .)e−x = I + II.

The expansion generates the second term II, but failsto create I.

e−0/ε = 1

ande−x/ε decays rapidly for positive x

Uniform and nonuniform expansions – p. 17/23

Small parameter multiplying thehighest derivative

The first term behaves as e−x/ε = o(εn) as ε → 0, ∀n, ifx = O(1) and e−x/ε = O(1) as ε → 0 if x = O(ε). Theregion near x = 0 is called boundary layer.

Uniform and nonuniform expansions – p. 18/23

Other example

εd2ydx2 + dy

dx + y = 0, x ∈ (0, 1), ε � 1, y(0) = 0, y(1) = 1.

y = e1−x − e−x/εe1+x + O(ε) is the exact solution.

If x � 0 the behavior of the solution define e1−x,because e−x/ε is negligible. For x ∼ 0 e−x/ε ↗ 1 rapidly

Uniform and nonuniform expansions – p. 19/23

Other example

εd2ydx2 + dy

dx + y = 0,

y(x, ε) = y0(x) + εy1(x) + . . .

ε0 :dy0

dx+ y0 = 0, y0(0) = 0, y0(1) = 1

ε1 :dy1

dx+ y1 = −

d2y0

dx2, y1(0) = 0, y1(1) = 0,

Both of boundary conditions can not be satisfied. If wesatisfy y0(1) = 1 then y0 = e1−x is a good approximationaway of boundary layer. If we satisfy y0(0) = 0 we gety0 = 0, that approximate the solution only at x = 0.

Uniform and nonuniform expansions – p. 20/23

Once more example

U0∂T

∂X= α(

∂2T

∂X2+

∂2T

∂Y 2)

x =X

L, y =

Y

H, θ =

T − Tw

T0 − Tw⇒ ε

∂2θ

∂x2+

∂2θ

∂y2− Pe

∂θ

∂x= 0

with ε = H2

L2 and Pe = U0H2

αL .Uniform and nonuniform expansions – p. 21/23

Once more example

We have ε ∂2θ∂x2 + ∂2θ

∂y2 − Pe ∂θ∂x = 0 with boundary

conditions

θ = 0 on y = 0, 1, & x = 1 (outlet), θ = 1 on x = 0 (inlet)

Using θ(x, y, ε) ∼ θ0(x, y) + εθ1(x, y) + . . . we get

∂2θ0

∂y2− Pe

∂θ0

∂x= 0

All initial conditions can not be satisfied, since theinitial equation is of the elliptic type, but the equationfor θ0 is of the parabolic type. The boundary layerappears on x = 1.

Uniform and nonuniform expansions – p. 22/23

The end

Uniform and nonuniform expansions – p. 23/23