Uniform and nonuniform expansions
Lesson 7
Uniform and nonuniform expansions – p. 1/2
Uniform and nonuniform expansions
If
f (x, ε) =
N
X
an (x)δn (ε) + RN (x, ε)
n=0
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. 2/2
Example of uniform expansion
1
1−ε sin x
∼ 1 + ε sin x + ε2 sin2 x + ε3 sin3 x + . . . with
RN =
∞
X
n
n
ε sin x
n=N +1
so
RN
lim N +1 = sinN +1 x
ε→0 ε
and we can estimate
|RN (x, ε)| ≤ K|εN +1 |
with any K > 1.
Uniform and nonuniform expansions – p. 3/2
Example of nonuniform expansion
1
1−ε sin x
∼ 1 + εx + ε2 x2 + ε3 x3 + . . . with
RN =
∞
X
n=N +1
n n
ε x
so
RN
lim N +1 = xN +1
ε→0 ε
Since x is not bounded we can not find K such that
|RN (x, ε)| ≤ K|εN +1 |.
Uniform and nonuniform expansions – p. 4/2
Region of nonuniformity
The principal idea of the asymptotic expansion that the
subsequent term in the expansion smaller the
previous one an+1 (x)δn+1 (ε) = o an (x)δn (ε) .
Uniform and nonuniform expansions – p. 5/2
Region of nonuniformity
The principal idea of the asymptotic expansion that the
subsequent term in the expansion smaller the
previous one an+1 (x)δn+1 (ε) = o an (x)δn (ε) .
1
1−ε sin x
∼ 1 + εx + ε2 x2 + ε3 x3 + . . . When two
subsequent term has the same order?
εn+1 xn+1 = O(εn xn ) ⇒ εx = O(1)
or x = O(1/ε). If εn+1 xn+1 = o(εn xn ) ⇒ εx = o(1)
or, for instance, x = O( √1ε )
Uniform and nonuniform expansions – p. 5/2
Region of nonuniformity
• 1 + εex + ε2 e2x + ε3 e3x + . . .
εex = O(1) ⇒ ex = O(1/ε)
or x = O(− ln ε) as ε → 0 and x is large.
Uniform and nonuniform expansions – p. 6/2
Region of nonuniformity
• 1 + εex + ε2 e2x + ε3 e3x + . . .
εex = O(1) ⇒ ex = O(1/ε)
or x = O(− ln ε) as ε → 0 and x is large.
2
3
• 1 + ε + ε 2 + ε 3 + . . .If
x
x
x
ε
= O(1) ⇒ x = O(ε) as ε → 0
x
for small values of x.
Uniform and nonuniform expansions – p. 6/2
Region of nonuniformity (more
examples)
sin(x + ε) = sin x cos ε + cos x sin ε
3
ε2
ε
+ O(ε4 ) + cos x ε −
+ O(ε5 )
= sin x 1 −
2
6
ε2
ε3
= sin x + ε cos x − sin x − cos x + O(ε4 )
2
6
sin(x(1 + ε)) = sin x cos εx + cos x sin εx
3 x3
ε
ε2 x 2
+ O(ε4 x4 ) + cos x εx −
+ O(ε5 x5 )
= sin x 1 −
2
6
ε2 x 2
ε3 x 3
= sin x + εx cos x −
sin x −
cos x + O(ε4 x4 )
2
6
Uniform and nonuniform expansions – p. 7/2
Region of nonuniformity
ε3
ε2
sin(x + ε) = sin x + ε cos x − sin x − cos x + O(ε4 )
2
6
is the uniform expansion and
ε3 x 3
ε2 x 2
sin x−
cos x+O(ε4 x4 )
sin(x(1+ε)) = sin x+εx cos x−
2
6
is nonuniform because of appearance of x in
coefficients.
Uniform and nonuniform expansions – p. 8/2
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.
P∞
Uniform and nonuniform expansions – p. 9/2
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?
P∞
Uniform and nonuniform expansions – p. 9/2
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 + ε2 x + ε3 x + . . .
P∞
Uniform and nonuniform expansions – p. 9/2
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 + ε2 x + ε3 x + . . .
•
For the nonuniformity is required
P∞
fn+1 (x)δn+1 (ε) = O fn (x)δn (ε) ⇒
δn (ε)
fn+1 (x) = fn (x)O 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. 9/2
Sources of nonuniformity
•
Infinite domains which allow long term effects of
small perturbations to accumulate
Uniform and nonuniform expansions – p. 10/2
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. 10/2
Infinite domains
Consider nonlinear oscillator equation
for t > 0 with initial conditions u(0) = a,
use the standard expansion
d2 u
3 =
+
u
+
εu
2
dt
du
dt (0) = b. We
0
u(t, ε) = u0 (t) + εu1 (t) + . . .
Substituting to the equation, we obtain
d2 u
0
dt2
+ u0 + ε
d2 u
1
dt2
u0 (0) + εu1 (0) + . . . = a,
+ u1 + u30 + O(ε2 ) = 0
du0
du1
(0) + ε
(0) + . . . = b
dt
dt
Uniform and nonuniform expansions – p. 11/2
Infinite domains
We need to solve the following equations
2u
d
du0
0
0
ε :
+ u0 = 0, u0 (0) = a,
(0) = b ⇒ u0 = a cos t,
2
dt
dt
2u
d
du1
1
1
3
ε :
+ u1 + u0 = 0, u1 (0) = 0,
(0) = 0 ⇒
2
dt
dt
3
3
d 2 u1
a
3a
3
3
+
u
=
−a
cos
t = − cos 3t −
cos t.
1
2
dt
4
4
Uniform and nonuniform expansions – p. 12/2
Infinite domains
d 2 u1
a3
3a3
+ u1 = − cos 3t −
cos t
2
dt
4
4
A cos t + B sin t is a homogeneous solution,
α cos 3t + β sin 3t is a particular solution corresponding
a3
to cos 3t ⇒ α = 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
3a3
solution. We have δ = 0, γ = − 8 .
Uniform and nonuniform expansions – p. 13/2
Infinite domains
The general solution is
u1 = A cos t + B sin t +
1
u1 (0) = du
dt (0) = 0.
a3
32
cos 3t −
3a3
8 t sin t,
a3
a3
3a3
A = − , B = 0 ⇒ u1 = (cos 3t − cos t) −
t sin t,
32
32
8
a3
3a3
(cos 3t − cos t) −
t sin t + . . .
u ∼ a cos t + ε
32
8
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. 14/2
Small parameter multiplying the
highest derivative
dy
Consider the equation ε 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) + ε2 y2 (x) + . . .
Substituting to the equation, we obtain
dy1
ε
+ε
+ . . . + y0 + εy1 + ε2 y2 + O(ε3 ) = e−x
dx
dx
dy
0
y0 (0) + εy1 (0) + ε2 y2 (0) + . . . = 2
Uniform and nonuniform expansions – p. 15/2
Small parameter multiplying the
highest derivative
dy
ε dx
+ y = e−x
ε0 : y0 = e−x , y0 (0) = 2,
dy0
= e−x , y1 (0) = 0,
ε : y1 = −
dx
dy1
2
= e−x , y2 (0) = 0
ε : y2 = −
dx
1
We obtain the expansion y ∼ e−x + εe−x + ε2 e−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. 16/2
Small parameter multiplying the
highest derivative
Let us compare with the exact solution of
dy
1−2ε −x/ε
e−x
−x
ε dx + y = e : yex = 1−ε e
+ 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
and
e−x/ε decays rapidly for positive x
Uniform and nonuniform expansions – p. 17/2
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. 18/2
Other example
d2 y
ε dx2
+
dy
dx
+ y = 0, x ∈ (0, 1), ε 1, y(0) = 0, y(1) = 1.
y = e1−x − e−x/ε e1+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. 19/2
Other example
d2 y
ε dx2
+
dy
dx
+ y = 0,
y(x, ε) = y0 (x) + εy1 (x) + . . .
dy0
ε :
+ y0 = 0, y0 (0) = 0, y0 (1) = 1
dx
2y
dy
d
1
0
ε1 :
+ y1 = − 2 , y1 (0) = 0, y1 (1) = 0,
dx
dx
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. 20/2
Once more example
∂T
∂ 2T
∂ 2T
U0
+
)
= α(
2
2
∂X
∂X
∂Y
X
∂ 2θ ∂ 2θ
∂θ
Y
T − Tw
x = ,y = ,θ =
⇒ ε 2 + 2 − Pe
=0
L
H
T0 − T w
∂x
∂y
∂x
with ε =
H2
L2
and P e =
U0 H 2
αL .
Uniform and nonuniform expansions – p. 21/2
Once more example
∂2θ
ε ∂x2
We have
conditions
+
∂2θ
∂y 2
∂θ
− P e ∂x
= 0 with boundary
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
∂ 2 θ0
∂θ0
− Pe
=0
2
∂y
∂x
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. 22/2
The end
Uniform and nonuniform expansions – p. 23/2