# Uniform and nonuniform expansions

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Uniform and nonuniform expansions

RN (x, ε) = O(δN+1(ε)) as ε → 0.

Or we can say

Example of uniform expansion

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

RN = ∞

RN

and we can estimate

with any K > 1.

Example of nonuniform expansion

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

RN = ∞

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

(

.

1 1−ε sin x ∼ 1 + εx + ε2x2 + ε3x3 + . . . When two subsequent 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 1−ε sin x ∼ 1 + εx + ε2x2 + ε3x3 + . . . When two subsequent 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 + ε x + ε2

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 + ε x + ε2

for small values of x.

Uniform and nonuniform expansions – p. 6/23

Region of nonuniformity (more examples)

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

= sin x (

1 − ε2

2 + O(ε4)

2 sin x −

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

= sin x (

1 − ε2x2

2 + O(ε4x4)

2 sin x −

Region of nonuniformity

2 sin x −

sin(x(1+ε)) = sin x+εx cos x− ε2x2

2 sin x−

Uniform and nonuniform expansions – p. 8/23

Region of nonuniformity

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 required fn+1(x)δn+1(ε) = O

(

increase.

Region of nonuniformity

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?

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

(

increase.

Region of nonuniformity

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 required fn+1(x)δn+1(ε) = O

(

increase.

Region of nonuniformity

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 required fn+1(x)δn+1(ε) = O

(

increase.

Sources of nonuniformity

• Infinite domains which allow long term effects of small perturbations to accumulate

• Singularities in governing equations which lead to localized regions of rapid change

Uniform and nonuniform expansions – p. 10/23

Sources of nonuniformity

• Infinite domains which allow long term effects of small perturbations to accumulate

• Singularities in governing equations which lead to localized regions of rapid change

Uniform and nonuniform expansions – p. 10/23

Infinite domains

Consider nonlinear oscillator equation d2u dt2 + u + εu3 = 0

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

use the standard expansion

u(t, ε) = u0(t) + εu1(t) + . . .

(d2u0

Infinite domains

ε0 : d2u0

du0

ε1 : d2u1

dt (0) = 0 ⇒

a3

Infinite domains

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

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

32 , β = 0,

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

8 .

Infinite domains

The general solution is u1 = A cos t + B sin t + a3

32 cos 3t − 3a3

dt (0) = 0.

3a3

32 (cos 3t − cos t) −

3a3

)

+ . . .

The term t sin t is called secular term. It is an oscillatory term of increasing amplitude. It leads to the nonuniformity. cos t = O(εt sin t) ⇒ t = O(1/ε) as ε → 0.

Uniform and nonuniform expansions – p. 14/23

Small parameter multiplying the highest derivative

Consider the equation ε dy dx + y = e−x for x > 0, ε 1

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

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

Substituting to the equation, we obtain

ε (dy0

dx + ε

y0(0) + εy1(0) + ε2y2(0) + . . . = 2

εdy dx + y = e−x

ε0 : y0 = e−x, y0(0) = 2,

ε1 : y1 = − dy0

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. The unperturbed equation (ε = 0) is an algebraic equation. The nature of an differential equation and algebraic equation is very different.

Uniform and nonuniform expansions – p. 16/23

Small parameter multiplying the highest derivative

Let us compare with the exact solution of εdy

dx + y = e−x: yex = 1−2ε 1−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 fails to create I.

e−0/ε = 1

Uniform and nonuniform expansions – p. 17/23

Small parameter multiplying the highest derivative

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

Uniform and nonuniform expansions – p. 18/23

Other example

εd2y dx2 + 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

ε1 : dy1

dx + y1 = −

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

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

Uniform and nonuniform expansions – p. 20/23

Once more example

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

Once more example

∂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

∂x = 0

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

Uniform and nonuniform expansions – p. 22/23

The end

Uniform and nonuniform expansions

Example of uniform expansion

Example of nonuniform expansion

Region of nonuniformity

Region of nonuniformity

Region of nonuniformity

Region of nonuniformity

Region of nonuniformity

Sources of nonuniformity

Sources of nonuniformity

Other example

Other example

RN (x, ε) = O(δN+1(ε)) as ε → 0.

Or we can say

Example of uniform expansion

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

RN = ∞

RN

and we can estimate

with any K > 1.

Example of nonuniform expansion

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

RN = ∞

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

(

.

1 1−ε sin x ∼ 1 + εx + ε2x2 + ε3x3 + . . . When two subsequent 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 1−ε sin x ∼ 1 + εx + ε2x2 + ε3x3 + . . . When two subsequent 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 + ε x + ε2

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 + ε x + ε2

for small values of x.

Uniform and nonuniform expansions – p. 6/23

Region of nonuniformity (more examples)

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

= sin x (

1 − ε2

2 + O(ε4)

2 sin x −

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

= sin x (

1 − ε2x2

2 + O(ε4x4)

2 sin x −

Region of nonuniformity

2 sin x −

sin(x(1+ε)) = sin x+εx cos x− ε2x2

2 sin x−

Uniform and nonuniform expansions – p. 8/23

Region of nonuniformity

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 required fn+1(x)δn+1(ε) = O

(

increase.

Region of nonuniformity

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?

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

(

increase.

Region of nonuniformity

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 required fn+1(x)δn+1(ε) = O

(

increase.

Region of nonuniformity

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 required fn+1(x)δn+1(ε) = O

(

increase.

Sources of nonuniformity

• Infinite domains which allow long term effects of small perturbations to accumulate

• Singularities in governing equations which lead to localized regions of rapid change

Uniform and nonuniform expansions – p. 10/23

Sources of nonuniformity

• Infinite domains which allow long term effects of small perturbations to accumulate

• Singularities in governing equations which lead to localized regions of rapid change

Uniform and nonuniform expansions – p. 10/23

Infinite domains

Consider nonlinear oscillator equation d2u dt2 + u + εu3 = 0

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

use the standard expansion

u(t, ε) = u0(t) + εu1(t) + . . .

(d2u0

Infinite domains

ε0 : d2u0

du0

ε1 : d2u1

dt (0) = 0 ⇒

a3

Infinite domains

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

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

32 , β = 0,

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

8 .

Infinite domains

The general solution is u1 = A cos t + B sin t + a3

32 cos 3t − 3a3

dt (0) = 0.

3a3

32 (cos 3t − cos t) −

3a3

)

+ . . .

The term t sin t is called secular term. It is an oscillatory term of increasing amplitude. It leads to the nonuniformity. cos t = O(εt sin t) ⇒ t = O(1/ε) as ε → 0.

Uniform and nonuniform expansions – p. 14/23

Small parameter multiplying the highest derivative

Consider the equation ε dy dx + y = e−x for x > 0, ε 1

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

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

Substituting to the equation, we obtain

ε (dy0

dx + ε

y0(0) + εy1(0) + ε2y2(0) + . . . = 2

εdy dx + y = e−x

ε0 : y0 = e−x, y0(0) = 2,

ε1 : y1 = − dy0

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. The unperturbed equation (ε = 0) is an algebraic equation. The nature of an differential equation and algebraic equation is very different.

Uniform and nonuniform expansions – p. 16/23

Small parameter multiplying the highest derivative

Let us compare with the exact solution of εdy

dx + y = e−x: yex = 1−2ε 1−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 fails to create I.

e−0/ε = 1

Uniform and nonuniform expansions – p. 17/23

Small parameter multiplying the highest derivative

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

Uniform and nonuniform expansions – p. 18/23

Other example

εd2y dx2 + 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

ε1 : dy1

dx + y1 = −

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

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

Uniform and nonuniform expansions – p. 20/23

Once more example

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

Once more example

∂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

∂x = 0

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

Uniform and nonuniform expansions – p. 22/23

The end

Uniform and nonuniform expansions

Example of uniform expansion

Example of nonuniform expansion

Region of nonuniformity

Region of nonuniformity

Region of nonuniformity

Region of nonuniformity

Region of nonuniformity

Sources of nonuniformity

Sources of nonuniformity

Other example

Other example