Dynamics at the Horsetooth Volume 2A, Focused Issue: Asymptotics and Perturbations
Single and Multiple Boundary Layer Phenomena - Examples
Jennifer Maple
Department of Mathematics
Colorado State University
maplej@math.colostate.edu
Report submitted to Prof. I. Oprea for Math 676, Fall 2010
Abstract. A collection of boundary layer examples.
Keywords: boundary layer, multiple boundary layers
1
Introduction
Boundary layer theory is used to understand the flow of a slightly viscous fluid near a solid boundary.
The idea is that the effects of the friction are experienced only very near an object moving through
a fluid. This divides the domain into two regions. One of these regions, called the inner region, is
close to the solid boundary and the solution varies rapidly in the direction normal to the boundary
and the higher-order viscous terms become significant. In the other region, called the outer region,
the inviscid equations hold.
A boundary layer is a very thin region near to a boundary, in which the solution varies rapidly
in the direction normal to the boundary. Boundary layers arise in the solution of differential
equations in which the highest order derivative is multiplied by a small parameter. Prandtl’s
paradigm example of the harmonic oscillator will be used to illustrate the fundamental idea of
boundary layer theory. His example is a singular perturbation problem that can be treated by the
method of matched asymptotic expansions.
This paper is organized as follows: In Section 2, we introduce Prandtl’s model example of the
harmonic oscillator. In Section 3, we have an ordinary differential equation with two boundary
layers. Finally, Section 4 contains a look at the Blasius boundary layer example.
2
Prandtl’s model example of the harmonic oscillator
Now on to Prandtl’s model example of the harmonic oscillator as described in [2]. Consider the ideal
case of the oscillatory motion of a mass m, which is a material point and xm is its displacement.
The mass is subjected to an elastic force, cxm , and a dissipative force, kx0m . The inertial force,
mx00m , balances the other two forces. After some rescaling, the mathematical model derived from
Newton’s law F̄ = mā with appropriate initial and boundary conditions is
y 00 + y 0 + y = 0
y(0) = 0,
y(1) = 1.
The question of interest is to find the behavior of the solution as → 0.
There are two cases for the solution to 1) dependent on .
(1)
Single and Multiple Boundary Layer Phenomena - Examples
• If 6= 0, then the ODE is second order and the exact solution is
√
x
1 − 4
1 − x sinh 2
.
√
y(x, ) = exp
1
2
sinh 2
1 − 4
Jennifer Maple
(2)
• If = 0, then the ODE is first order with general solution
y(x) = A exp(−x).
(3)
A can be chosen to satisfy y(0) = 0 which leads to A = 0 and y(x) = 0. This doesn’t satisfy
the other boundary condition, y(1) = 1. However, if A is chosen to satisfy y(1) = 1, this
leads to A = e and y(x) = exp(1 − x). Again, the other boundary condition is not satisfied,
y(0) 6= 0. This situation occurs because the first order ODE is overdetermined when there
are two boundary conditions.
Use matched asymptotic expansions to find the asymptotic behavior of the solution y(x, ) to
(1) as → 0. We start by dividing the domain into two regions. The inner region, 0 ≤ x ≤ δ,
where y = y i satisfies the inner boundary condition, y(0) = 0. And, the outer region, δ ≤ x ≤ 1,
where y = y o satisfies the outer boundary condition, y(1) = 1. δ is the thickness of the boundary
layer located near x = 0.
Let’s deal with the outer problem first. The solution outside the boundary layer can often be
found as a regular perturbation expansion:
y o (x, ) = y0 (x) + y1 (x) + y2 (x)2 + · · ·
(4)
which is just a Taylor series expansion about = 0.
1. Substitute the outer expansion into the ODE.
2. Expand any nonlinear terms.
3. Collect terms of like power in .
4. Set the coefficient of each power of equal to zero. This gives us a set of ODE’s for the set
of coefficient functions whose solution can be uncoupled.
5. Substitute the outer expansion into the outer boundary condition. Expand any remaining
function of into a Taylor series in .
6. Find the particular solutions to each of the ODE’s.
7. Construct the outer solution.
The outer solution is
y o (x, ) = exp(1 − x) + O().
(5)
Now for the inner problem, we need to transform the independent variable to deal with the
boundary layer. The transformation used is a stretch transformation with the form X = xn with
n > 0.
1. Substitute the stretch transformation into the ODE, (1).
2. Select a value for the parameter n in the stretch transform. n is chosen such that we do not
lose the term containing the highest order derivative. In oder to do this, the term must be
lowest order in and not the only term which has this order.
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Vol. 2A, 2010
Single and Multiple Boundary Layer Phenomena - Examples
Jennifer Maple
3. Substitute this value of the parameter into the stretch transformed ODE.
4. Substitute a regular perturbation expansion into the transformed inner ODE.
5. Collect terms of like order in .
6. Set coefficients equal to zero.
7. Substitute the inner expansion into the inner boundary condition. Expand any remaining
function of into a Taylor series in .
8. Find the particular solutions to each of the ODE’s.
9. Construct the inner solution.
The inner solution is
y i = A(1 − exp(−X)) + O().
(6)
Next is to determine A so that the outer and inner solutions match. One way to do this is to use
the Primitive Matching Principle. This principle states that the outer limit of the inner solution,
(y i )o , equals the inner limit of the outer solution, (y o )i . Using this principle gives us that the inner
solution must be
y i = e(1 − exp(−X)) + O().
(7)
This means that for 0 ≤ x ≤ δ, y ≈ e(1 − exp(−x/)) and for δ ≤ x ≤ 1, y ≈ exp(1 − x). The
composite solution, y c = y i + y o − (y i )o , can be found by blending y i and y o together.
3
Two Boundary Layers
So far we have only looked at the situation where we have a single boundary layer, however there
are situations where more than one boundary layer occurs. From [1], one such situation is the
boundary value problem
2 y 00 + xy 0 − y = −ex , for 0 < x < 1,
(8)
with the boundary conditions y(0) = 2 and y(1) = 1. For this problem, if = 0 the ODE reduces
to y = ex which does not satisfy either boundary condition. Thus for the solution we will find two
boundary layers, one at each end of the interval, [0,1].
First, find the outer solution. Assume a regular perturbation expansion,
y ∼ y0 (x) + y1 (x) + · · · .
(9)
Substitute this expansion into the ODE and collect like powers of ,
2 (y0 (x) + y1 (x) + · · · )00 + x(y0 (x) + y1 (x) + · · · )0 − (y0 (x) + y1 (x) + · · · ) = −ex
2 y000 (x) + · · · + xy00 (x) + 2 xy10 (x) + · · · − y0 (x) − y1 (x) − · · · = −ex
−y0 (x) + (xy00 (x) − y1 (x)) + 2 (y000 (x) + xy10 (x) − y2 (x)) + · · · = −ex . (10)
This gives that y0 = ex , which fails to satisfy either boundary condition.
Second and Third, find the boundary layer solutions and match the solutions. Let’s deal with
the layer at x = 0 first. In this region we will call the solution Y (x̄). Set x̄ = x/γ . Using the chain
rule this gives
d
dx
d2
dx2
Dynamics at the Horsetooth
=
=
dx̄ d
1 d
= γ
and
dx dx̄
dx̄
1 d2
.
2γ dx̄2
3
(11)
(12)
Vol. 2A, 2010
Single and Multiple Boundary Layer Phenomena - Examples
Jennifer Maple
The transformation changes the ODE into
2
1 00
1
Y + (γ x̄) γ Y 0 − Y
2γ
2−2γ 00
Y + x̄Y 0 − Y
γ x̄
= −e
γ
= −e x̄ .
(13)
The terms need to be balanced as we did before. Since the last two terms are the same order, O(1),
we only need to consider one of them. We can not make all the terms have the same order with
respect to . This leads us to two options that satisfy the balancing conditions. The first term
must be lowest order in and not the only term which has this order.
• Make the first and second terms have the same order with the third term having a higher
order with respect to . This means that 2 − 2γ = 1 giving γ = 1/2. This means that the
first and second terms are O(), but the third term is O(1). Since the third term has lower
order in the required balance is not achieved.
• Make the first and third terms have the same order with the second term having a higher
order with respect to . This means that 2 − 2γ = 0 giving γ = 1. This means that the first
and third terms are O(1) and the third term is O(). Since the second term has higher order
in the required balance is achieved.
Now that γ = 1 is fixed, the ODE becomes
Y 00 + x̄Y 0 − Y = −ex̄ .
(14)
Assume a regular perturbation expansion in the new variables Y (x̄) ∼ Y0 (x̄) + Y1 (x̄) + · · · . Next
is to substitute the expansion into the ODE, expand all terms, and collect like orders of :
(Y0 (x̄) + Y1 (x̄) + · · · )00 + x̄(Y0 (x̄) + Y1 (x̄) + · · · )0 − (Y0 (x̄) + Y1 (x̄) + · · · ) = −ex̄
Y000 (x̄) + Y100 (x̄) + · · · + x̄Y00 (x̄) + 2 x̄Y10 (x̄) + · · · − Y0 (x̄) − Y1 (x̄) − · · · = −(1 + x̄ + · · · )
(Y000 (x̄) − Y0 (x̄)) + (Y100 (x̄) + x̄Y00 (x̄) − Y1 (x̄)) + · · · = −1 − x̄ − · · · ).
(15)
This gives a set of problems for the coefficients of the expansion, such as
Y000 − Y0 = −1
(16)
Y0 (0) = 2
(17)
which gives us Y0 = 1 + Aex̄ + (1 − A)e−x̄ . In order to match this boundary layer solution with the
outer solution we need
lim Y0 (x̄) =
x̄→∞
lim 1 + Aex̄ + (1 − A)e−x̄ =
x̄→∞
lim y0 (x)
x→0
lim exp(x)
x→0
lim 1 + Aex̄ + (1 − A)e−x̄ = 1
x̄→∞
lim Aex̄ + (1 − A)e−x̄ = 0.
x̄→∞
(18)
Since lim ex̄ = ∞, in order for (18) to be true we need A = 0. This leads to this boundary layer
x̄→∞
solution being
Y0 (x̄) = 1 + e−x̄ .
(19)
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Vol. 2A, 2010
Single and Multiple Boundary Layer Phenomena - Examples
Jennifer Maple
Next, is the layer at x = 1. Similarly, in this region we will call the solution Ỹ (x̃). Set x̃ =
thus x = 1 + γ x̃. Using the chain rule this gives
d
dx
d2
dx2
=
=
dx̃ d
1 d
= γ
and
dx dx̃
dx̃
1 d2
.
2γ dx̃2
x−1
γ ,
(20)
(21)
The transformation changes the ODE into
1
1 00
Ỹ + (1 + γ x̃) γ Ỹ 0 − Ỹ
2γ
2−2γ Ỹ 00 + 1−γ (1 + γ x̃)Y 0 − Y
2
γ x̃
= −e1+
γ
= −e1+ x̃ .
(22)
The terms need to be balanced as we did before. Since the last two terms are the same order,
O(1), we only need to consider one of them. It would be ideal if we could make all the terms have
the same order with respect to . Let’s try balancing the first and third terms, 2 − 2γ = 0 gives
us γ = 1. This gives us all the terms having the order O(1). Now that γ = 1 is fixed, the ODE
becomes
Ỹ 00 + (1 + x̃)Ỹ 0 − Ỹ = −e1+x̃ .
(23)
Next, assume a regular perturbation expansion in the new variables Ỹ (x̃) ∼ Y˜0 (x̃) + · · · . Substitute
into the ODE and boundary condition and collect like orders of :
(Y˜0 + Y˜1 + · · · )00 + (1 + x̃)(Y˜0 + Y˜1 + · · · )0 − (Y˜0 + Y˜1 + · · · ) = −eex̃
00
00
0
0
Y˜0 + Y˜1 + · · · + (1 + x̃)(Y˜0 + Y˜1 + · · · ) − Y˜0 − Y˜1 − · · · = −e(1 + x̃ + · · · )
00
00
0
0
0
Y˜0 + Y˜1 + · · · + Y˜0 + Y˜1 + · · · + x̃Y˜0 + · · · − Y˜0 − Y˜1 − · · · = −e − x̃e + · · ·
00
0
00
0
0
(Y˜0 + Y˜0 − Y˜0 ) + (Y˜1 + Y˜1 + x̃Y˜0 − Y˜1 ) + · · · = −e − x̃e + · · · .
(24)
This gives a set of problems for the coefficients of the expansion, such as
00
0
Y˜0 + Y˜0 − Y˜0 = −e
Y˜0 (0) = 1
(25)
(26)
which gives us Y˜0 = e + Aer+ x̃ + (1 − e − A)er− x̃ with r± = (−1 ±
boundary layer solution with the outer solution we need
lim Y˜0 (x̃) =
x̃→−∞
lim e + Aer+ x̃ + (1 − e − A)er− x̃ =
x̃→−∞
√
5)/2. In order to match this
lim y0 (x)
x→1
lim exp(x)
x→1
lim e + Aer+ x̃ + (1 − e − A)er− x̃ = e
x̃→−∞
lim Aer+ x̃ + (1 − e − A)er− x̃ = 0
(27)
x̃→−∞
Given that lim er+ x̃ = 0 since r+ > 0 and lim er− x̃ = ∞ since r+ < 0, in order for (27) to be
x̃→−∞
x̃→−∞
true we need 1 − e − A = 0 or 1 − e = A. This leads to this boundary layer solution being
Y˜0 (x̃) = e + (1 − e)er+ x̃ .
(28)
Fourth, derive the composite solution. This involves combining the three approximations to
produce a uniform approximation. To do this, we must determine what each approximation
contributes to each region. Given α() and β() such that 0 < α < β < 1 and α → 0 and
β → 1 as → 0:
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Vol. 2A, 2010
Single and Multiple Boundary Layer Phenomena - Examples
Jennifer Maple
• In the region [0, α],
– y ∼ Y0 ,
– y0 = 1 since y0 = ex → 1 as → 0 on [0, α],
– Y˜0 = e since Y˜0 (x̃) = e + (1 − e)er+ (x−1)/ → e as → 0 on [0, α],
• In the region (α, β),
– Y0 = 1 since Y0 = 1 + e−x/ as → 0 on (0, 1),
– y ∼ y0 ,
– Y˜0 = e since Y˜0 (x̃) = e + (1 − e)er+ (x−1)/ → e as → 0 on (0, 1), and
• In the region [β, 1],
– Y0 = 1 since Y0 = 1 + e−x/ as → 0 on [β, 1],
– y0 = e since y0 = ex → e as → 1 on [β, 1],
– y ∼ Y˜0 .
The approximations not associated with the specified region add 1 + e to the solution. Thus the
composite expansion of the solution adds all three of the approximations and subtracts 1 + e from
the total,
y ∼ y0 (x) + Y0 (x̄) + Y˜0 (x̃) − (1 + e)
= ex + 1 + e−x̄ + e + (1 − e)er+ x̃ − (1 + e)
= ex + 1 + e−x/ + e + (1 − e)er+ (x−1)/ − (1 + e)
= ex + e−x/ + (1 − e)er+ (x−1)/
(29)
and it is valid for 0 ≤ x ≤ 1. Figures 1 and 2 are graphs of the numerical solution and the composite
approximation of the solution with = .1 and .01.
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Vol. 2A, 2010
Single and Multiple Boundary Layer Phenomena - Examples
Jennifer Maple
Figure 1: Graph of the numerical solution and the composite approximation of the solution with
= .1.
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Single and Multiple Boundary Layer Phenomena - Examples
Jennifer Maple
Figure 2: Graph of the numerical solution and the composite approximation of the solution with
= .01.
4
Blasius Boundary Layer
The Blasius boundary layer problem is a time dependent viscous fluid problem which deals with
steady flow over a stationary flat plate, [1]. The plate is fixed at y = 0 for 0 < x < L. There is no
flow in the z-direction and the flow goes from left to right. This leads to the fluid equations
2
∂u
∂u
∂p
∂ u ∂2u
ρ u
+v
= −
+µ
+ 2
∂x
∂y
∂x
∂x2
∂y
2
∂v
∂v
∂p
∂ v
∂2v
ρ u
+v
= −
+µ
+
∂x
∂y
∂y
∂x2 ∂y 2
∂u ∂v
+
= 0
(30)
∂x ∂y
where the boundary conditions are
v = (0, 0, 0) on y = 0, 0 < x < L,
(31)
v = (u0 , 0, 0) for y → −∞.
(32)
To look at the boundary layer we need to the problem to be nondimensional. In order to accomplish
this we let x = Lx̄, y = Lȳ, u = u0 ū, v = u0 v̄, and p = pc p̄ where pc = ρu20 . Using the chain rule
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Vol. 2A, 2010
Single and Multiple Boundary Layer Phenomena - Examples
Jennifer Maple
this gives
d
dx
d2
dx2
d
dy
d2
dy 2
=
=
=
=
dx̄ d
1 d
=
,
dx dx̄
L dx̄
1 d2
,
L2 dx̄2
dȳ d
1 d
=
, and
dy dȳ
L dȳ
1 d2
.
L2 dȳ 2
(33)
(34)
(35)
(36)
This transformation changes the equations of motion into
u0 ū ∂u0 ū u0 v̄ ∂u0 ū
1 ∂pc p̄
1 ∂ 2 u0 ū
1 ∂ 2 u0 ū
ρ
= −
+ 2
+
+µ
L ∂ x̄
L ∂ ȳ
L ∂ x̄
L2 ∂ x̄2
L ∂ ȳ 2
2
pc ∂ p̄
u0 ∂ ū
u20 ∂ ū
u0 ∂ 2 ū u0 ∂ 2 ū
= −
+
ρ ū
+ v̄
+µ
L ∂ x̄
L ∂ ȳ
L ∂ x̄
L2 ∂ x̄2 L2 ∂ ȳ 2
ρu20
∂ ū
∂ ū
ρu20 ∂ p̄ µu0 ∂ 2 ū ∂ 2 ū
ū
+ v̄
= −
+ 2
+ 2
L
∂ x̄
∂ ȳ
L ∂ x̄
L
∂ x̄2
∂ ȳ
2
2
∂ ū
∂ ū
∂ p̄
µ
∂ ū ∂ ū
ū
+ v̄
= −
+
+ 2
∂ x̄
∂ ȳ
∂ x̄ ρLu0 ∂ x̄2
∂ ȳ
2
2
∂ ū
∂ ū
∂ p̄
∂ ū
2 ∂ ū
ū
+ v̄
= −
+
+ 2
∂ x̄
∂ ȳ
∂ x̄
∂ x̄2
∂ ȳ
ρ
u0 ū ∂u0 v̄ u0 v̄ ∂u0 v̄
+
L ∂ x̄
L ∂ ȳ
2
∂v̄
∂v̄
ρu0
ū
+ v̄
L
∂ x̄
∂ ȳ
∂v̄
∂v̄
ū
+ v̄
∂ x̄
∂ ȳ
∂v̄
∂v̄
ū
+ v̄
∂ x̄
∂ ȳ
=
=
=
=
1 ∂pc p̄
1 ∂ 2 u0 v̄
1 ∂ 2 u0 v̄
−
+µ
+ 2
L ∂ ȳ
L2 ∂ x̄2
L ∂ ȳ 2
ρu20 ∂ p̄ µu0 ∂ 2 v̄
∂ 2 v̄
−
+ 2
+
L ∂ ȳ
L
∂ x̄2 ∂ ȳ 2
2
∂ p̄
µ
∂ 2 v̄
∂ v̄
−
+
+
∂ ȳ ρLu0 ∂ x̄2 ∂ ȳ 2
2
∂ p̄
∂ 2 v̄
2 ∂ v̄
−
+
+
∂ ȳ
∂ x̄2 ∂ ȳ 2
(37)
(38)
1 ∂u0 ū
1 ∂u0 v̄
+
= 0
L ∂ x̄
L ∂ ȳ
u0 ∂ ū ∂v̄
+
= 0
L ∂ x̄ ∂ ȳ
∂ ū ∂v̄
+
= 0
∂ x̄ ∂ ȳ
(39)
µ
where 2 = ρLu
and 2 is the inverse of the Reynolds number for the flow.
0
First, let’s find the outer solution. Assume regular perturbation expansions of v and p̄:
v ∼ v0 + v1 + · · · and
(40)
p̄ ∼ p¯0 + p¯1 + · · · .
(41)
The first coefficient problem is the problem for an inviscid flow and the solution is v0 = (u0 , 0, 0)
and a constant p¯0 .
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Single and Multiple Boundary Layer Phenomena - Examples
Jennifer Maple
Now to deal with the boundary layer. Y = ȳ is the boundary layer coordinate. Capitols are
used to denote the dependent variables in the boundary layer. Using the chain rule this gives
∂
∂ ȳ
∂2
∂ ȳ 2
=
=
∂Y ∂
1 ∂
=
and
∂ ȳ ∂Y
∂Y
1 ∂2
.
2 ∂Y 2
(42)
(43)
The transformation changes (37) - (39) into
∂ Ū
V̄ ∂ Ū
+
∂ x̄
∂Y
∂ V̄
V̄ ∂ V̄
Ū
+
∂ x̄
∂Y
∂ Ū
1 ∂ V̄
+
∂ x̄
∂Y
Ū
∂ 2 Ū
∂ P̄
∂ 2 Ū
+ 2 2 +
∂ x̄
∂ x̄
∂Y 2
2
1 ∂ P̄
∂ V̄
∂ 2 V̄
= −
+ 2 2 +
∂Y
∂ x̄
∂Y 2
= −
(44)
(45)
= 0.
(46)
Ū ∼ U¯0 + · · · , V̄ ∼ (V¯0 + · · · ), and P̄ ∼ P¯0 + · · · are the expansions used by [1]. Using the
expansions in (44) - (46) gives
∂ ¯
1
∂ ¯
(U0 + U¯1 + · · · ) + (V¯0 + 2 V¯1 + · · · )
(U0 + U¯1 + · · · )
∂ x̄
∂Y
∂
∂2
∂2 ¯
= − (P¯0 + P¯1 + · · · ) + 2 2 (U¯0 + U¯1 + · · · ) +
(U0 + U¯1 + · · · )
∂ x̄
∂ x̄
∂Y 2 ¯
¯1
¯0
¯1
∂
U
∂
U
∂
U
∂
U
0
+
+ · · · + (V¯0 + V¯1 + · · · )
+
+ ···
(U¯0 + U¯1 + · · · )
∂ x̄
∂ x̄
∂Y
∂Y
2¯
2¯
∂ P¯0
∂ P¯1
∂ 2 U¯1
∂ U0
∂ 2 U¯1
2 ∂ U0
=−
−
− ··· + +
+ ··· +
+
+ ···
∂ x̄
∂ x̄
∂ x̄2
∂ x̄2
∂Y 2
∂Y 2
∂ U¯0
∂ U¯1
∂ U¯0
∂ U¯0
∂ U¯1
∂ U¯0
U¯0
+ U¯0
+ U¯1
+ · · · + V¯0
+ V¯0
+ V¯1
+ ···
∂ x̄
∂ x̄
∂ x̄
∂Y
∂Y
∂Y
∂ P¯1
∂ 2 U¯0
∂ 2 U¯1
∂ P¯0
−
− ··· +
+
+ ···
=−
∂ x̄ ∂Y 2
∂Y 2
∂ x̄¯
∂ U0
∂ U¯0
∂ U¯1
∂ U¯0
∂ U¯1
∂ U¯0
¯
¯
¯
¯
¯
¯
U0
+ V0
+ U0
+ U1
+ V0
+ V1
+ ···
∂ x̄
∂Y
∂ x̄
∂ x̄
∂Y
∂Y
∂ P¯0 ∂ 2 U¯0
∂ P¯1 ∂ 2 U¯1
= −
+
+
+
+ ···
∂ x̄
∂Y 2
∂ x̄
∂Y 2
(U¯0 + U¯1 + · · · )
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Vol. 2A, 2010
Single and Multiple Boundary Layer Phenomena - Examples
Jennifer Maple
∂
1
∂
(U¯0 + U¯1 + · · · ) (V¯0 + 2 V¯1 + · · · ) + (V¯0 + 2 V¯1 + · · · )
(V¯0 + 2 V¯1 + · · · )
∂ x̄
∂Y
1 ∂ ¯
∂2
∂2
=−
(V¯0 + 2 V¯1 + · · · )
(P0 + P¯1 + · · · ) + 2 2 (V¯0 + 2 V¯1 + · · · ) +
2
∂Y
∂
x̄
∂Y
¯
¯
¯
¯
∂ V0
∂ V0
2 ∂ V1
2 ∂ V1
¯
¯
¯
¯
(U0 + U1 + · · · ) +
+ · · · + (V0 + V1 + · · · ) +
+ ···
∂ x̄
∂ x̄
∂Y
∂Y
¯
2 ¯
2 ¯
∂ 2 V¯0
∂ P¯1
∂ 2 V¯0
1 ∂ P0
2 ∂ V1
2 ∂ V1
2
+ ··· + +
+ ···
+
+ ··· + 2 + =−
∂Y
∂Y
∂ x̄
∂ x̄2
∂Y 2
∂Y 2
∂ V¯0
∂ V¯0
1 ∂ P¯0 ∂ P¯1
∂ P¯2
∂ 2 V¯0
U¯0
+ ···
+ · · · + V¯0
+ ··· = −
−
−
− ··· + ∂Y
∂Y ∂Y
∂Y
∂Y 2
∂ x̄ ¯
1
∂ P¯0
∂ P¯1
∂ P¯2 ∂ 2 V¯0
∂ V¯0
∂ V0
¯
¯
+ ··· =
−
+ −
+ −
+ ···
U0
+ V0
+
∂ x̄
∂Y
∂Y
∂Y
∂Y
∂Y 2
∂ V¯0
∂ P¯2 ∂ 2 V¯0
∂ V¯0
∂ P¯0
∂ P¯1
2
2
¯
¯
U0
+ ···
(48)
+ V0
+ ··· = −
+ −
+ −
+
∂ x̄
∂Y
∂Y
∂Y
∂Y
∂Y 2
∂ ¯
1 ∂
(U0 + U¯1 + · · · ) +
(V¯0 + 2 V¯1 + · · · ) = 0
∂ x̄
∂Y
∂ U¯0
∂ U¯1
∂ V¯0
∂ V¯1
+
+ ··· +
+
+ ··· = 0
∂ x̄
∂Y
∂Y
∂ x̄¯
∂ U0 ∂ V¯0
∂ U¯1 ∂ V¯1
+
+
+
+ · · · = 0.
∂ x̄
∂Y
∂ x̄
∂Y
(49)
Taking the O(1) terms gives us the equations
∂ U¯0
∂ U¯0
U¯0
+ V¯0
∂ x̄
∂Y
∂ P¯0 ∂ 2 U¯0
+
,
∂ x̄
∂Y 2
∂ P¯0
0 = −
,
∂Y
∂ U¯0 ∂ V¯0
+
= 0.
∂ x̄
∂Y
= −
(50)
(51)
(52)
The no-slip condition on the plates requires that
(U¯0 , V¯0 ) = (0, 0) on Y = 0, 0 < x̄ < 1.
(53)
In order to match with the outer solution
U¯0 → 1 and P¯0 → 0 as Y → ∞, 0 < x̄ < 1
(54)
is required. We get that P¯0 = 0 from (51) and (54). Next, we define a stream function ϕ(x̄, Y )
such that
∂ϕ
∂ϕ
and V¯0 = − .
(55)
U¯0 =
∂Y
∂ x̄
Thus, (52) is satisfied. This reduces (50) to
∂ϕ ∂ ∂ϕ
∂ϕ ∂ ∂ϕ
+−
∂Y ∂ x̄ ∂Y
∂ x̄ ∂Y ∂Y
2
∂ϕ ∂ ϕ
∂ϕ ∂ 2 ϕ
+−
∂Y ∂ x̄∂Y
∂ x̄ ∂Y 2
Dynamics at the Horsetooth
11
=
=
∂ 2 ∂ϕ
,
∂Y 2 ∂Y
∂3ϕ
.
∂Y 3
(56)
Vol. 2A, 2010
Single and Multiple Boundary Layer Phenomena - Examples
Jennifer Maple
The conditions (53) and (54) transform into
∂ϕ
∂ϕ
=
= 0, on Y = 0,
∂Y
∂ x̄
(57)
and
∂ϕ
→ 1 as Y → ∞.
(58)
∂Y
The stream function problem has no analytic solution. If we take a semi-infinite plate, 0 ≤ x̄ < ∞,
instead of a finite√one, this gives rise the√Blasius boundary layer problem. Which can be reduced
by assuming ϕ = x̄f (η), where η = Y / x̄. This similarity variable tranforms (56) to
1
f 000 + f f 00 = 0, for 0 < η < ∞,
2
(59)
f (0) = f 0 (0) = 0, and f 0 (∞) = 1.
(60)
where (57) and (58) become
This tranformation gives us an ODE, (59), instead of the PDE, (56), which is easier to solve
numerically.
Once f is determined, the velocity functions can be obtained by
U¯0 = f 0 (η)
1
V¯0 = − √ (f − ηf 0 ).
2 x̄
5
(61)
(62)
Conclusion
We looked at three examples of boundary layer phenonmena. First was an overview of Prandtl’s
model example of the harmonic oscillator. Second, we went into detail for an ordinary differential
equation with two boundary layers and showed that the composite approximation and the numerical
solution are similar. Last, we looked at the Blasius boundary layer example.
Dynamics at the Horsetooth
12
Vol. 2A, 2010
Single and Multiple Boundary Layer Phenomena - Examples
Jennifer Maple
References
[1] Holmes, Introduction to the Foundations of Applied Mathematics, Springer, 2009.
[2] Oprea, I. M676 - Perturbations and asymptotic methods - The mathematical nature of the
boundary layer. Handout. 2010.
Dynamics at the Horsetooth
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Vol. 2A, 2010