Mathematical ModellingLecture 4

The devil’s in the details (singular perturbations)

Review of regular perturbations• Asymptotic expansions:

• Assumed:

• Collecting terms:

• Consequently,

• chosen to achieve balance, starting at the lowest order in

• Leads to a hierarchy of approximations, e.g.

• Convergence in epsilon, divergent series

• Initial value problems (expansion of ODE and initial conditions)

↵0 < ↵1 < · · ·

0 < " "0, x 2 D

↵0, ↵1, · · · "

x ⇠ "

↵0x0 + "

↵1x1 + · · · ,

"

↵0f0(xi) + "

↵1f1(xi) + · · · = 0,

f0(xi) = 0, f1(xi) = 0, · · ·

x ⇡ x0

x ⇡ x0 + "x1

x ⇡ x0 + "x1 + "

2x2

...

Singular perturbationsConsider the solutions of the quadratic equation

where ε is a small parameter.

One root converges to 1/2, one diverges in the limit

Using a regular perturbation expansion, we can approximate the root at x=1/2, but not the other root.

Instead we introduce a rescaling . The problem becomes

for which both roots exist in the limit , and regular expansions can be applied.

!

!!!

"#!!

"$!

"x

2 + 2x� 1 = 0

1� 2x = "x

2

" ! 0

x = "x

x

2 + 2x� " = 0

" ! 0

Boundary layersSingular perturbation problems arise also in differential equations. Typically when ε multiplies the highest derivative.

Again the character of the problem changes for : Both boundary conditions cannot be satisfied.

Exact solution

Solution varies (rapidly) overa region of width ∼ ε“Boundary layer”

Solution has two parts: inner and outer layers

"y

00(x) + 2y0(x) + 2y(x) = 0, 0 < x < 1

y(0) = 0, y(1) = 1

" = 0

y(x) =e

r+x � e

r�x

e

r+ � e

r�, r± = (�1±

p1� 2")/"

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.5

1

1.5

2

2.5

3

x−axis

Sol

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e = 1 e = 0.1 e = 0.01

Boundary layersSingular perturbation problems arise also in differential equations. Typically when ε multiplies the highest derivative.

To obtain the outer solution we just apply the regular perturbation method:

One free parameter, can satisfy only one b.c.

"y

00(x) + 2y0(x) + 2y(x) = 0, 0 < x < 1

y(0) = 0, y(1) = 1

y ⇠ y0(x) + "y1(x) + · · ·

"(y000 (x)+"y

001 (x)+ · · · )+2(y00(x)+"y

01(x)+ · · · )+2(y0(x)+"y1(x)+ · · · ) = 0

y0(0) + "y1(0) + · · · = 0, y0(1) + "y1(1) + · · · = 1

) y0(x) = e

1�x

O(") : y

000 + 2y01 + 2y1 = 0, y1(1) = 1 ) y1(x) = (b� x/2)e1�x

) y1(x) = (1� x)e1�x

/2

O(1) : 2y00 + 2y0 = 0, y0(0) = 0, y0(1) = 1 ) y0(x) = ae

�x

Boundary layersSingular perturbation problems arise also in differential equations. Typically when ε multiplies the highest derivative.

Next we consider what happens in a neighborhood of the left boundary. We rescale in x:

We want to choose γ such that the first term remains as .

We balance with either or at the lowest order.

"y

00(x) + 2y0(x) + 2y(x) = 0, 0 < x < 1

y(0) = 0, y(1) = 1

x =x

"

�,

d

dx

=dx

dx

d

dx

=1

"

�

d

dx

,

d

2

dx

2=

1

"

2�

d

2

dx

2

Y (x) = y(x)

"1�2�Y 00 + 2"��Y 0 + 2Y = 02�1� 3�

" ! 0

1� 2� 3�

Boundary layers

Balance

"1�2�Y 00 + 2"��Y 0 + 2Y = 0

1� 2� = �� ) � = 1,O("�1), 3� ⇠ O(1)

1� 2� = �0 ) � = 1/2,O(1), 2� ⇠ O("�1/2)1� ⇠ 3� :

1� ⇠ 2� :

Y 00 + 2Y 0 + 2"Y = 0

Y = Y0 + "Y1 + · · ·

Y0(0) + "Y1(0) = 0

O(1) : Y 000 + 2Y 0

0 = 0, Y0(0) = 0 ) Y0 = A+Be�2x

) Y0(x) = A(1� e

�2x)

Y 000 + "Y 00

1 + · · ·+ 2(Y 00 + "Y 0

1 + · · · ) + 2"(Y0 + "Y1 + · · · ) = 0

Boundary layersNext, we want to match the solutions in the “overlap region”. We require the matching condition:

Since both solutions are constant outside of their respective regions, we can construct a composite solution:

limx!1

Y0 = limx!0

y0 y0(x) = e

1�x ! e

Y0(x) = A(1� e

�2x) ! A

) A = e

y ⇠ y0(x) + Y0(x)� y0(0)

= e

1�x � e

1�2x/"

2.4 Introduction to Boundary Layers 65

10!6

10!5

10!4

10!3

10!2

10!1

100

0

e

A

x!axis

So

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Outer

Inner

Figure 2.9 Graph of the inner approximation (2.57), and the outer approximation(2.50), before matching.

approximations are both constant. Given that they are approximations of thesame function then we need to require that the inner and outer expansionsare equal in this region. In more mathematical terms, the requirement wewill impose on these two expansions is

limx!1

Y0 = limx!0

y0. (2.58)

This is called the matching condition. With this we conclude A = e andthe resulting functions are plotted in Figure 2.10 for ✏ = 10�4. The overlapdomain is clearly seen in this figure.

Step 4. Composite ExpansionThe approximation of the solution we have comes in two pieces, one thatapplies near x = 0 and another that works everywhere else. Because neithercan be used over the entire interval we say that they are not uniformly validfor 0 x 1. The question we consider now is whether we can combine them

10−6

10−5

10−4

10−3

10−2

10−1

100

0

0.5

1

1.5

2

2.5

3

x−axis

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Outer Inner

Figure 2.10 Graph of the inner approximation (2.57), and the outer approximation(2.50), after matching in the particular case of when ✏ = 10�4. Note the overlapregion where the two approximations produce, approximately, the same result.

2.4 Introduction to Boundary Layers 65

10!6

10!5

10!4

10!3

10!2

10!1

100

0

e

A

x!axis

So

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on

Outer

Inner

Figure 2.9 Graph of the inner approximation (2.57), and the outer approximation(2.50), before matching.

approximations are both constant. Given that they are approximations of thesame function then we need to require that the inner and outer expansionsare equal in this region. In more mathematical terms, the requirement wewill impose on these two expansions is

limx!1

Y0 = limx!0

y0. (2.58)

This is called the matching condition. With this we conclude A = e andthe resulting functions are plotted in Figure 2.10 for ✏ = 10�4. The overlapdomain is clearly seen in this figure.

Step 4. Composite ExpansionThe approximation of the solution we have comes in two pieces, one thatapplies near x = 0 and another that works everywhere else. Because neithercan be used over the entire interval we say that they are not uniformly validfor 0 x 1. The question we consider now is whether we can combine them

10−6

10−5

10−4

10−3

10−2

10−1

100

0

0.5

1

1.5

2

2.5

3

x−axis

Sol

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Outer Inner

Figure 2.10 Graph of the inner approximation (2.57), and the outer approximation(2.50), after matching in the particular case of when ✏ = 10�4. Note the overlapregion where the two approximations produce, approximately, the same result.

Multiple boundary layersOur second example illustrates multiple boundary layers and nonconstant coefficients:

For the solution is simply , which satisfies neither b.c.

Outer solution:

Boundary layer at x=0:

Balance:

Matching condition:

" = 0

"

2y

00 + "xy

0 � y(x) = �e

x

, 0 < x < 1

y(0) = 2, y(1) = 1

y(x) = e

x

y(x) = e

x

"

2�2�Y

00 + "xY

0 � Y = e

"

�x

2�1� 3� 4�

x =x

"

�

1� ⇠ 3� :

1� ⇠ 2� : 2� 2� = 1 ) � = 1/2,O("), 3� ⇠ O(1)

2� 2� = 0 ) � = 1,O(1), 2� ⇠ O(")

O(1) Y

000 � Y0 = �1, Y0(0) = 2 ) Y0(x) = 1 +Ae

x +Be

�x ) B = (1�A)

limx!1

Y0 = limx!0

y0 limx!1

1 +Aex = 1 ) A = 0

Y

00 + "xY

0 � Y = �e

"x Y = Y0 + "Y1 + · · ·

Multiple boundary layersOur second example illustrates multiple boundary layers and nonconstant coefficients:

For the solution is simply , which satisfies neither b.c.

Outer solution:

Boundary layer at x=1:

Balance:

Matching condition:

" = 0

"

2y

00 + "xy

0 � y(x) = �e

x

, 0 < x < 1

y(0) = 2, y(1) = 1

y(x) = e

x

y(x) = e

x

2�1� 3� 4�

1� ⇠ 3� :

"

2�2�Y

00 + "

1��(1 + "

�

x)Y 0 � Y = �e

1+"

�x

2� 2� = 0 ) � = 1,O(1), 2� ⇠ O(1)

Y = Y0 + "Y1 + · · ·Y

000 + "Y

001 + · · · (1 + "x)(Y 0

0 + "Y

01 + · · · )� (Y0 + "Y1 + · · · ) = �e

1+"x

Y

00 + (1 + "x)Y 0 � Y = �e

1+"x

O(1) Y

000 + Y

00 � Y0 = �e, Y0(0) = 1 ) Y0(x) = 1+Ae

r+x+Be

r�x r± =�1±

p5

2

x =x� 1

"

�

limx!�1

Y0 = limx!1

y0 ) Y0 = e+ (1� e)er+x

Multiple boundary layersComposite solution:

y ⇠ y0(x) + Y0(x) + Y0(x)� y0(0)� y0(1)

= e

x + e

�x/" + (1� e)er+(x�1)/"2.5 Multiple Boundary Layers 71

!!

""##"~""

$"#!$"#!

""#!

"~$"!$"##!$"##!

%&!' %&!'

"~$"

Figure 2.15 Sketch of the three regions and the values of the approximations inthose regions.

region (i.e., ex ! �1) that you get the same value as when you enter theboundary layer from the outer region (i.e., x ! 1). Given that r+ > 0 andr� < 0 then limex!�1 er�ex = 1 and limex!�1 er+ex = 0. For eY0 to be ableto match with the outer solution we must set 1 � e � A = 0. With this ourfirst term approximation in this boundary layer is

eY0(ex) = e + (1� e)er+ex. (2.71)

Step 4. CompositeIn a similar manner as in the last example, it is possible to combine the threeapproximations we have derived to produce a uniform approximation. Thesituation is shown schematically in Figure 2.15. It is seen that in each regionthe two approximations not associated with that region add to 1 + e. Thismeans we simply add the three approximations together and subtract 1 + e.In other words,

y ⇠ y0(x) + Y0(x) + eY0(ex)� y0(0)� y0(1)

= ex + e�x/✏ + (1� e)er+(x�1)/✏. (2.72)

This function is a composite expansion of the solution and it is valid for0 x 1. To demonstrate its e↵ectiveness the composite approximation isplotted in Figure 2.16 along with the numerical solution for ✏ = 10�1 andfor ✏ = 10�2. The approximations are not very accurate for ✏ = 10�1, butthis is not unexpected given that ✏ is not particularly small. In contrast, for✏ = 10�2 the composite approximation is quite good over the entire interval,and it is expected to get even better for smaller values of ✏.

72 2 Perturbation Methods

2.6 Multiple Scales and Two-Timing

As the last two examples have demonstrated, the presence of a boundary layerlimits the region over which an approximation can be used. Said another way,the inner and outer approximations are not uniformly valid over the entireinterval. The tell-tale sign that this is going to happen is that when ✏ = 0the highest derivative in the problem is lost. However, the lack of uniformitycan occur in other ways and one investigated here relates to changes in thesolution as a function of time. It is easier to explain what happens by workingout a typical example. For this we use the pendulum problem. Letting ✓(t)be the angular deflection made by the pendulum, as shown in Figure 2.17,the problem is

✓00 + sin(✓) = 0, (2.73)

where✓(0) = ✏, (2.74)

and✓0(0) = 0. (2.75)

The equation of motion (2.73) comes from Newton’s second law, F = ma,where the external forcing F is gravity. It is assumed the initial angle is small,

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 11

2

x−axis

So

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! = 0.1

Numerical

Composite

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 11

2

3

x−axis

So

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! = 0.01

Numerical

Composite

Figure 2.16 Graph of the numerical solution of the boundary value problem (2.60)-(2.62) and the composite approximation of the solution (2.72). In the upper plot✏ = 10�1 and in the lower plot ✏ = 10�2.

Two time-scalesPendulum:

Regular perturbation analysis:

✓00 + sin(✓) = 0, ✓(0) = ", ✓0(0) = 0

"✓000 + "↵+1✓001 + · · ·+ "✓0 + "↵+1✓1 �1

6"3✓30 + · · · = 0

"✓0(0) + "↵+1✓1(0) + · · · = "

"✓00(0) + "↵+1✓01(0) + · · · = 0

O(") ✓000 + ✓0 = 0, ✓0(0) = 1, ✓00(0) = 0 ✓0(t) = A cos(t+B) ) ✓0 = cos t74 2 Perturbation Methods

0 20 40 60 80 100 120 140 160 180 200−1

−0.5

0

0.5

1

t−axis

Sol

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180 185 190 195 200−1

−0.5

0

0.5

1

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So

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Numerical

Asymptotic (1 term)

Figure 2.18 Graph of the numerical solution of the pendulum problem (2.60)-(2.62)and the first term in the regular perturbation approximation (2.76). Shown are thesolutions over the entire time interval, as well as a close up of the solutions neart = 200. In the calculation ✏ = 1

3 and both solutions have been divided by ✏ = 13 .

✏✓00(0) + ✏↵+1✓0

1(0) + · · · = 0. (2.80)

Proceeding in the usual manner yields the following problem.

O(✏) ✓000 + ✓0 = 0

✓0(0) = 1, ✓00(0) = 0

The general solution of the di↵erential equation is ✓0 = a cos(t) +b sin(t), where a, b are arbitrary constants. It is possible to write thissolution in the more compact form ✓0 = A cos(t + B), where A,B arearbitrary constants. As will be explained later, there is a reason forwhy the latter form is preferred in this problem. With this, and theinitial conditions, it is found that ✓0 = cos(t).

The plot of the one-term approximation, ✓ ⇠ ✏ cos(t), and the numericalsolution are shown in Figure 2.18. The asymptotic approximation describesthe solution accurately at the start, and reproduces the amplitude very wellover the entire time interval. What it has trouble with is matching the phaseand this is evident in the lower plot in Figure 2.18. One additional commentto make is the value for ✏ used in Figure 2.18 is not particularly small, sogetting an approximation that is not very accurate is no surprise. However,if a smaller value is used the same di�culty arises. The di↵erence is that the

sin(✓) ⇠ sin("(✓0 + "↵✓1 + · · · )) ⇠ "✓0 + "↵+1✓1 �1

6"3✓30 + · · ·

✓ ⇠ "(✓0 + "↵✓1 + · · · )

Two time-scalesPendulum:

Regular perturbation analysis:

✓00 + sin(✓) = 0, ✓(0) = ", ✓0(0) = 0

"✓000 + "↵+1✓001 + · · ·+ "✓0 + "↵+1✓1 �1

6"3✓30 + · · · = 0

"✓0(0) + "↵+1✓1(0) + · · · = "

"✓00(0) + "↵+1✓01(0) + · · · = 0

O("3) ✓001 + ✓1 =1

6✓30, ✓0(0) = 0, ✓00(0) = 0 ✓1 = a cos t+ b sin t� 1

16

t sin t

✓1 = � 1

16t sin t

2.6 Multiple Scales and Two-Timing 75

first term approximation works over a longer time interval but eventually thephase error seen in Figure 2.18 occurs.

In looking to correct the approximation to reduce the phase error we cal-culate the second term in the expansion. With the given ✓0 there is an ✏3✓3

0

term in (2.78). To balance this we use the ✓1 term in the expansion and thisrequires ↵ = 2. With this we have the following problem to solve.

O(✏3) ✓001 + ✓1 = 1

6✓30

✓1(0) = 0, ✓01(0) = 0

The method of undetermined coe�cients can be used to find a par-ticular solution of this equation. This requires the identity cos3(t) =14 (3 cos(t) + 3 cos(3t)), in which case the di↵erential equation becomes

✓001 + ✓1 =

124

(3 cos(t) + 3 cos(3t)). (2.81)

With this the general solution is found to be ✓1 = a cos(t) + b sin(t)�116 t sin(t), where a, b are arbitrary constants. From the initial condi-tions this reduces to ✓1 = � 1

16 t sin(t).

The plot of the two term approximation,

✓ ⇠ ✏ cos(t)� 116

✏3t sin(t), (2.82)

and the numerical solution is shown in Figure 2.19. It is clear from thisthat we have been less than successful in improving the approximation. Theculprit here is the t sin(t) term. As time increases its contribution grows, andit eventually gets as large as the first term in the expansion. Because of this itis called a secular term, and it causes the expansion not to be uniformly validfor 0 t <1. This problem would not occur if time were limited to a finite

0 20 40 60 80 100 120 140 160 180 200−5

0

5

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Numerical

Asymptotic (2 term)

Figure 2.19 Graph of the numerical solution of the pendulum problem (2.60)-(2.62)and the regular perturbation approximation (2.72). In the calculation ✏ = 1

3 and the

solution has been divided by ✏ = 13 .

✓ ⇠ " cos t� "3

6

t sin t+ · · ·

✓ ⇠ "(✓0 + "↵✓1 + · · · )

✓001 + ✓1 =

1

24

(3 cos t+ 3 cos 3t)

Secular term

Two time-scalesPendulum:

The problem with the phase is that for the nonlinear problem, the phase is not constant, but changes slowly. We have two time scales: (1) the time scale on which the oscillations occur, (2) the time scale upon which the phase slowly changes.

We construct an approximation that explicitly uses these scales:

✓00 + sin(✓) = 0, ✓(0) = ", ✓0(0) = 0

t1 = t, t2 = "�t d

dt=

dt1dt

@

@t1+

dt2dt

@

@t2=

@

@t1+ "�

@

@t2d2

dt2= (

@

@t1+ "�

@

@t2)2 =

@2

@t21+ 2"�

@

@t1

@

@t2+ "2�

@2

@t22

✓ ⇠ "(✓0(t1, t2) + "↵✓1(t1, t2) + · · · )

"@2

@t21✓0 + "↵+1 @2

@t21✓1 +2"�+1 @2

@t1@t2✓0 + · · ·+ "✓0 + "↵+1✓1 �

1

6"3✓30 + · · · = 0

"✓0(0, 0) + "↵+1✓1(0, 0) + · · · = 0

"@

@t1✓0(0, 0) + "↵+1 @

@t1✓1(0, 0) + "�+1 @

@t2✓0(0, 0) + · · · = 0

Two time-scalesPendulum: ✓00 + sin(✓) = 0, ✓(0) = ", ✓0(0) = 0

✓ ⇠ "(✓0(t1, t2) + "↵✓1(t1, t2) + · · · )

"@2

@t21✓0 + "↵+1 @2

@t21✓1 +2"�+1 @2

@t1@t2✓0 + · · ·+ "✓0 + "↵+1✓1 �

1

6"3✓30 + · · · = 0

"✓0(0, 0) + "↵+1✓1(0, 0) + · · · = 0

"@

@t1✓0(0, 0) + "↵+1 @

@t1✓1(0, 0) + "�+1 @

@t2✓0(0, 0) + · · · = 0

O(") @2

@t21✓0 + ✓0 = 0, ✓0(0, 0) = 1, @

@t1✓0(0, 0) = 0

) ✓0 = A(t2) cos(t1 +B(t2)) A(0) = 1, B(0) = 0

Two time-scalesPendulum:

Next order in is . Applying balance as for singular perturbations:

The next order term gives:

or

To avoid a secular (growing in time) term, we may choose

✓00 + sin(✓) = 0, ✓(0) = ", ✓0(0) = 0

"3"

↵+ 1 = � + 1 = 3

✓1(0, 0) = 0, @@t1

✓1(0, 0) +@@t2

✓0(0, 0) = 0

O("3) @2

@t21✓1 + ✓1 + 2 @2

@t1@t2✓0 = 1

6✓30 = 0

✓001+✓1 =

124 [3 cos(t1+B)+3 cos(3(t1+B))]+2A0

sin(t1+B)+2AB0cos(t1+B)

A0 = 0, 2AB0 = � 18 ) A = 1, B = �t2/16

✓ ⇠ " cos(t� "2t/16) + · · ·

Exercises 79

0 20 40 60 80 100 120 140 160 180 200−1

−0.5

0

0.5

1

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180 185 190 195 200−1

−0.5

0

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Multiple Scale

Figure 2.20 Graph of the numerical solution of the pendulum problem (2.60)-(2.62)and the multiple scale approximation (2.92). Shown are the solutions over the entiretime interval, as well as a close up of the solutions near t = 200. In the calculation✏ = 1

3 and the solution has been divided by ✏ = 13 .

Exercises

2.1. Assuming f ⇠ a1✏↵ + a2✏� + · · · find ↵, � (with ↵ < �), and nonzeroa1, a2, for the following:

(a) f = esin(✏).(b) f =

p1 + cos(✏).

(c) f = 1/p

sin(✏).(d) f = 1/(1� e✏).(e) f = sin(

p1 + ✏x), for 0 x 1.

(f) f = ✏ exp(p

✏ + ✏x), for 0 x 1.

2.2. Let f(✏) = sin(e✏).(a) According to Taylor’s theorem, f(✏) = f(0)+✏f 0(0)+ 1

2✏2f 00(0)+· · · . Showthat this gives (2.13).

(b) Explain why the formula used in part (a) can not be used to find anexpansion of f(✏) = sin(e

p✏). Also, show that the method used to derive

(2.13) still works, and derive the expansion.

2.3. Consider the equation

x2 + (1� 4✏)x�p

1 + 4✏ = 0.