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Uniform and nonuniform expansions Lesson 7 Uniform and nonuniform expansions – p. 1/2
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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