uniform and nonuniform expansions
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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
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