Winter 1995 MEAM 626Singular-Perturbation Methods in Fluid Mechanics

Instructor: William Schultz O�ce: 313 Auto LabEmail: Schultz@engin Phone: 936-0351Grading: Homework 50% Time: T Th 4:30-6:00

Project 25% Room: 3166 DowFinal 25%

References

� Bender and Orszag (1978), Advanced Mathematical Methods for Scientists and Engineers,McGraw-Hill.

� Rand and Armbruster (1987), Perturbation Methods, Bifurcation Theory and Computer Al-gebra, Springer-Verlag.

� Lagerstrom, P.A. (1988), Matched Asymptotic Expansions, Springer-Verlag.

� Kervokian and Cole (1980), Perturbation Methods in Applied Mechanics, Springer-Verlag.

� Nayfeh, A. (1973), Perturbation Methods, Wiley.

� Erdelyi, A. (1956), Asymptotic Expansions, Dover.

� Van Dyke, M. (1975), Perturbation Methods in Fluid Mechanics, Parabolic Press.

� Bush (1992), Perturbation methods for Engineers and Sciences, CRC Press.

� Hinch, E.J. (1991), Perturbation Methods, Cambridge U. Press.

� Tayler, A.B. (1986), Perturbation Methods in Applied Mechanics, Clarendon Press, Oxford.

Tentative Outline

Introduction to asymptotic methodsConvergent and asymptotic sequencesGauge functionsRegular and singular perturbationsGlobal vs. local approximationsPade approximants and continued fractionsUses in Fluid MechanicsSymbolic Computer Algebra

Unsteady ProblemsPoincare'-Linstedt MethodMulti-time methods

Method of Matched Asymptotic MethodsOseen FlowPrandtl Boundary Layers

Derivation of Lower-Dimensional TheoriesLubrication Theory (Boundary Layer Theory)Fin EquationSlender Body Theory

Asymptotic Evaluation of IntegralsIntegration by PartsSaddle Point Methods

WKBJ MethodGeometric TheoryEvaluation of Orr-Sommerfeld Equation

Bifurcation TheoryUses in Computational Fluid Mechanics

1

Participants

Joe Ivanenok SchultzMichael Martin gogators@umich SchultzNolan Dickey nsd@enginE-June Chen ejchen@engin Dowling/SchultzGary Lapham lapham@engin Dowling/PerlinScott Kukuck srk@engin uidsLei Jiang leijiang@engin SchultzZheng Xu xuz@engin JacobsNatasha Flyer n yer@engin BoydHamad Rashid hrashid@engin SchultzLixin Xu wolfxlx@engin TroeschKiang Heng Ti khti@engin VorusJames Criner mckay@engin VorusSunny Kankanala 313-323-2998 TryggvasonBernard Bunner bunner@engin Ceccio/TryggvasonStephen Scorpio sscorpio@engin Beck

|||||||||||A simple example (assuming we have forgotten the quadratic formula)

u2 + 2u+ � = 0

When � = 0, two roots are recovered u = 0;�2. How does solution behave when � is small? Try Ansatzu = u0 + �u1 + �2u2 + ::: Plug into equation and equate like powers of �.

(u0 + �u1 + �2u2 + :::)2 + 2(u0 + �u1 + �2u2 + :::) + � = 0

�0 : u20 + 2u0 = 0 u01;2 = 0;�2 (1a)

�1 : (2u0 + 2)u1 = �1 u11;2 = �1=2; 1=2 (1b)

�2 : (2u0 + 2)u2 = �u21 u21;2 = �1=8; 1=8 (1c)

�3 : (2u0 + 2)u3 = f(u2; u1; u0) :::::::::::: (1d)

......

... (1e)

Notes:

� Lowest-order may or may not be nonlinear (here nonlinear) but it is simpler! Otherwise, why bother?

� Higher-orders are almost always linear

� Operator stays same while RHS gets messier and messier. We will show this more formally later.

� Sometimes a simple power series insu�cient: simple example:

u2 + 2u+ �1=2 = 0

ANOTHER METHODOLOGY (When power series work)

We seek u = u(�), where �! 0, so try a Taylor series expansion:

u = u(0) + � _u(0) + �2�u(0)=2! + ::: ;

2

where to be consistent �u=2! (We drop \(0)") must equal u2 (more generally u(n) = n!un). Then:

�0 : u2 + 2u = 0 u1;2 = 0;�2 (2a)

2u _u+ 2 _u+ 1 = 0 (2b)

�1 : (2u+ 2) _u = �1 _u1;2 = �1=2; 1=2 (2c)

2u�u+ 2 _u2 + 2�u = 0 (2d)

�2 : (2u+ 2)�u = �2 _u2 �u1;2 = �1=4; 1=4 (2e)

......

... (2f)

What we have done is really perturbed the question rather than the answer. We certainly would have gotthe same answer if we had done a binomial expansion of the exact solution:

u =�2�p4� 4�

2= �1�p1� �

Note that the expansion will be convergent for � < 1 because the nearest singularity is at the branch pointat 1.

|||||||||||A slightly less simple example:

�u2 + 2u+ 1 = 0

again let u = u0 + �u1 + �2u2 + :::

�0 : 2u0 = �1 u0 = �1=2 (3a)

�1 : 2u1 = �u20 u1 = �1=8 (3b)

�2 : 2u2 = �2u0u1 ::: (3c)

......

... (3d)

What happened to the other root? Look at the exact solution:

u =�2�p4� 4�

2�=�1�p1� �

�

using the binomial expansion:

u =�1� (1� �=2� �2=8 + :::)

�

u = �1

2� �

8+ ::: and � 2

�+1

2+�

8+ :::

One root came from in�nity! We can see that this must be so | if the �rst term is to be important as�! 0, u must be very large. Try rescaling: let v = u��, and balance with the other terms. Make sure that �is chosen such that the u2 term comes in at leading order and balances at least one other term. Otherwise,all you know is that the term is in�nite at leading order. For this case, the rescaled problem is the sameas the �rst{hence it is \regular" and can recover both roots. This is not usually the case. Another singularexample where the roots cannot be obtained from one scaling...

|||||||||||HWK no. 1: Find roots of x3 � x2 + � = 0The classical solution is in the sense of Galois (W.H. Freeman, 1985) and is given by

x1 =1

3+1

3(z+ + z�) x2 =

1

3+1

3(!z+ + !2z�)

x3 =1

3+1

3(!2z+ + !z�)

3

where

z+ = 1=z� =

�1� 27

2�+

3

2

p�3(4�� 27�2

�1=3

and

! = �1

2+

p3

2i:

The perturbation solution: ??|||||||||||

De�nitions:

� \de�ned as" or \identically equal to"= \equals"! \approaches"� \approximately equal to" as in 1=3 � :33� \asymptotically equal to" (f � g means f=g ! 1 as �! 0)<< or � \much less than" or \asymptotically smaller than" (see below)Big O(), Little o() These are often used as remainders (see below)

Often a power series in � does not work. Then we consider gauge functions|comparing how two functionsgo to 0 at a limit.

three choices f(�)!8<:

0M1

as �! �0 = 0 (say)

lim�! 0

f(�) =

8<:

0M1

We say f(�) = O(g(�)) (f is the order of g) if

lim�! �0

f(�)

g(�)=M

where M can be zero but not in�nite. Also, we say f(�) = o(g(�)) (f is higher order than g) if

lim�! �0

f(�)

g(�)= 0 or f(�)� g(�)

Examples: 1� cos � = O(�2) = o(�) = o(1)1� cos � �2=2 as �! 0x+ x2 � x2 as x!1, but not as x! 0sinx � x, but sinx = x+ o(x)x� �10 as x! 0 (sign isn't important)

be careful off � 0 No! use f ! 00 < �� 1 | better use �! 0+

|||||||||||Defn: ffn(�)g is an asymptotic sequence as � ! 0 if fn+1(�) = o(fn(t; �)) for each n. Then the

functional representation

f(t; �) =NXn=1

an(t)fn(t; �) +RN ;

is an asymptotic expansion if RN = o(fN ) as � ! 0 as opposed to a convergent expansion whenRN ! 0 as N !1. Examples of asymptotic sequences:

4

a) ��1; 1; �; �2; �3; :::b) �1=n; �2=n; �3=n; :::c) �; �2 log2 �; �2 log �; �2; ::: (may need l'Hopital's rule!)d) z; 1; 1=z; 1=z2; ::: as z !1 (earlier terms can be singular)e) (x � x0); (x � x0)

2; ::: as x! x0 (shifting origin is better)

Examples of asymptotic expansions:

a) cos � = 1 +O(�2) as �! 0

b) Jn(x) =q

2�x cos

�x� �

4 � n�2

�+O(1=x) as x!1|||||||||||

A simple asymptotic example|The Exponential integral. Consider

Ei(1=�) = f(�) =

Z 1

1=�

e�s

sds as �! 0

Integrating by parts (u = 1=s; dv = e�sds) gives

= � e�s

s

����1

1=�

�Z 1

1=�

e�s

s2= �e�1=� �

Z 1

1=�

e�s

s2ds

or integrating n+ 1 times

f(�) = �e�1=�[1� �+ 2�2 + :::+ (�1)nn!�n] + (�1)n+1(n+ 1)!

Z 1

1=�

e�s

sn+2ds

Does it converge? Cauchy ratio test gives

� =lim

n!1j(�1)nn!�n+1e�1=�j

j(�1)n�1(n� 1)!�ne�1=�j =lim

n!1 jn�j

Which is not convergent. Is it asymptotic? It obviously is an asymptotic sequence. Now show Rn = o(fn)Z 1

1=�

e�s

sn+2ds � �n+2

Z 1

1=�

e�sds = �n+2e�1=� =�fn

(�1)nn!Therefore Rn = o(fn) for arbitrary n

|||||||||||A more interesting example....consider

y(x) =

1Xn=0

xn

(n!)2:

This converges for all x, similar to ex although it doesn't grow as fast with x and converges better. If x = 10,N = 16 gives 10 signi�cant digits, if x = 104, this requires N = 150. How does this sum behave as x!1?

We can �nd the local behavior at x ! 1 by �rst �nding an equation it solves! Note y0 =P1

n=1xn�1

(n!)(n�1)!

and (xy0)0 =P1

n=1xn�1

(n�1)!2 = y. Then y(x) is solution of

xy00 + y0 = y:

This has a regular singular point at x = 0 and an irregular singular point at x ! 1. (Bessels equationsolved by method of Frobenius).

[A regular singular point is determined from the linear, homogeneous ODE written in the standardform:

NXn=0

an(x)y(n) = 0;

where aN , the coe�cient of the highest degree is unity. If all the coe�cients are regular at thepoint in question, the point is regular; if any coe�cient is singular but all anx

N�n are not, thepoint is a regular singular point. Otherwise, it is an irregular singular point.]

5

Fuchs (1866) showed that at least one solution at a regular singular point has the form (x � x0)�A(x),

(hence a pole or branch cut) where A(x) is analytic at x0 and the solution has a radius of convergence at leastas large as the nearest singularity of the coe�cients. The point at in�nity is determined by x = 1=t; d=dx =�t2 d=dt; d2=dx2 = t4 d2=dt2+2t3 d=dt; ::: and expanding about t = 0. No comprehensive theory exists, butoften all solutions exhibit an essential singularity (pole of all orders). Two more things: A series around anirregular point will usually be asymptotic and not convergent, and the behavior is often governed by

y(x) = eS(x)

as suggested by Carlini(1817), Liuville (1837), and Green (1837). This changes

y00 + p(x)y0 + q(x)y = 0

intoS00 + S02 + p(x)S0 + q(x) = 0

which is also di�cult to solve except that usually S00 � (S0)2 at the irregular point. For our example, weobtain:

xS00 + xS02 + S0 = 1

or with the assumption above;

S0(x) � [�1� (1 + 4x)1=2]=2x; x! +1:

Since x is large, S0(x) � �x�1=2 or S(x) � �2x1=2 + C(x) where C(x) � 2x1=2 as x ! 1. We can nowshow a postereori that our assumption of dropping the S00 term is OK. Resubstituting this form into theindicial equation gives:

xC 00 + xC 02 + (2x1=2 + 1)C 0 +1

2x�1=2 = 0

With the usual asymptotic simpli�cations C 0 � �14x or C(x) � �1

4 lnx. Then y(x) � cx�1=4e2px as x! +1.

The value for c cannot be obtained by a local analysis, but from an integral analysis we obtain c = 1=p4�.

Examine �g. 3.4 in B&0:

|||||||||||

HWK no. 2: Write a computer program to compute erf(x) = 2p�

R xo e

�u2du to �ve signi�cant digits for0 < x <1.

What should you expect?

1. Cannot improve the asymptotic expansion but it is generally more accurate for few terms

2. Frustrations ... accurate results for `medium' x can become very expensive

3. Taylor series (especially with alternating sign) .... soon need quadruple precision to compute successiveterms carefully.

4. Both solutions are local | dependent variable small or large.

|||||||||||

Local vs. Global solutions (Ref: Bender & Orszag)These example expansions, using the dependent variable as the small or large value, are local expansions.We may require two expansions: one each locally valid for small and large values. Global solutions, more orless uniformly valid (see below) are often preferable although the solutions are often asymptotic rather thanconvergent. Example 3, p.322 B&O:

y00 = �e�xy y(0) = y0(0) = 1

6

Exact solution: let x = � ln t (so that t = e�x) and replace y(x) by Y (t), then

�t ddt

��tdY

dt

�= �tY Y (1) = �Y 0(1) = 1

that has primitive solutions J0(2t) and Y0(2t), (Bessel's equation of order 0). Applying boundary conditionsand using the value of the Wronskian, W (J0(2t); Y0(2t)) = �1=(�t), we �nd the solution is

Y (t) = � f[(Y0(2)� Y1(2)]J0(2t)� [J0(2)� J1(2)]Y0(2t)g :when we replace t by e�x we get the �nal result.

Taylor series expansion: In the usual way, we use the i.c. the D.E. and derivatives of the D.E. to getexactly the Taylor series expansion of the result above.

Perturbation series (global solution): Stick � in front of RHS and set � = 1! Then y(�; x) =P1n=0 �

nyn(x) and yn(0) = y0n(0) = 0; n = 1; 2; 3:::. This sets up a set of simple linear inhomogeneousdi�erential equations which can be solved by Green's functions or variation of parameters. For this verysimple equation we get

y0(x) = 1 + x

yn(x) =

Z x

0

Z t

0

�e�syn�1(s)ds dt n = 1; 2; 3:::

The solution (it helps to get symbolic algebra for the higher terms!)

y1(x) = (�3� x)e�x � 2x+ 3

y2(x) = (1 +x

4)e�2x + (1 + 2x)e�x +

3

4x� 2

y3(x) = (� 7

54� x

36)e�3x � (

3

4+x

2)e�2x + (

1

2� 3

4x)e�x � 1

9x+

41

108:::

Here, these higher orders can be obtained in Maple by:y0:=1+x; int(-exp(-x)*y0,x=0..s);y1:=int(",s=0..x); int(-y1*exp(-x),x=0..s);y2:=int(",s=0..x); int(-y2*exp(-x),x=0..s);y3:=int(",s=0..x);The four-term perturbation solution is then the plot of

y0 + y1(x) + y2(x) + y3(x) =257

108� 13

36x�

�3

2� x

4

�e�x +

1� x

4e�2x �

�7

54+

x

36

�e�3x

These solutions are also convergent for all x but converge much faster. This method of solving, say y00 = f(x)yhas the following advantages:

1. The problem will converge beyond the radius of convergence of f .

2. The problem can be expanded at a singular point of f .

3. It can more easily handle a boundary value problem.

||||{

A second example with �nite radius of convergence for Taylor Series and a y0 term:

y00 = �y0 +�

1 + xy(0) = y0(0) = 0 � = 1

The MAPLE output:

7

tayl:=taylor(y(x),x=0,8);y2:=D(y)+1/(1+x):y3:=subs(D(x)=1,D(y2)): y4:=subs(D(x)=1,D(y3)):y5:=subs(D(x)=1,D(y4)): y6:=subs(D(x)=1,D(y5)): y7:=subs(D(x)=1,D(y6)):subs((D@@7)(y)(0)=subs(x=0,y7),"): subs((D@@6)(y)(0)=subs(x=0,y6),"):subs((D@@5)(y)(0)=subs(x=0,y5),"): subs((D@@4)(y)(0)=subs(x=0,y4),"):subs((D@@3)(y)(0)=subs(x=0,y3),"): subs((D@@2)(y)(0)=subs(x=0,y2),"):subs((D@@7)(y)(0)=subs(x=0,y7),"): subs((D@@6)(y)(0)=subs(x=0,y6),"):subs((D@@5)(y)(0)=subs(x=0,y5),"): subs((D@@4)(y)(0)=subs(x=0,y4),"):subs((D@@3)(y)(0)=subs(x=0,y3),"): subs((D@@2)(y)(0)=subs(x=0,y2),"):subs((D@@7)(y)=subs(x=0,y7),"): subs((D@@6)(y)=subs(x=0,y6),"):subs((D@@5)(y)=subs(x=0,y5),"): subs((D@@4)(y)=subs(x=0,y4),"):subs((D@@3)(y)=subs(x=0,y3),"): subs((D@@2)(y)=subs(x=0,y2),");

tayl := y( 0 ) +

�1

2D( y ) +

1

2

�x2 +

1

6D( y )x3 +

�1

24D( y ) +

1

12

�x4+�

1

120D( y )� 1

30

�x5 +

�1

720D( y ) +

1

36

�x6 +

�1

5040D( y )� 5

252

�x7 +O(x8 )

subs((D)(y)=0,y=0,tayl);

tayl :=1

2x2 +

1

12x4 � 1

30x5 +

1

36x6 � 5

252x7 +O(x8 )

y0:=0; int(di�(y0,x)+1/(1+x),x=0..s); y1:=int(",s=0..x);int(di�(y1,x),x=0..s); y2:=int(",s=0..x);int(di�(y2,x),x=0..s); y3:=int(",s=0..x);int(di�(y3,x),x=0..s); y4:=int(",s=0..x);

y0 := 0

ln( 1 + s )

y1 := ln( 1 + x ) + ln( 1 + x )x� x

ln( 1 + s ) + ln( 1 + s ) s� s

y2 :=1

2ln( 1 + x ) +

1

2ln( 1 + x )x2 + ln( 1 + x )x� 3

4x2 � 1

2x

1

2ln( 1 + s ) +

1

2ln( 1 + s ) s2 + ln( 1 + s ) s� 3

4s2 � 1

2s

y3 :=1

6ln( 1 + x ) +

1

2ln( 1 + x )x� 11

36x3 � 5

12x2 +

1

2ln( 1 + x )x2 � 1

6x+

1

6ln( 1 + x )x3

1

6ln( 1 + s ) +

1

2ln( 1 + s ) s� 11

36s3 � 5

12s2 +

1

2ln( 1 + s ) s2 � 1

6s+

1

6ln( 1 + s ) s3

y4 := � 1

24x+

1

6ln( 1 + x )x3 � 7

48x2 +

1

6ln( 1 + x )x� 25

288x4

+1

4ln( 1 + x )x2 � 13

72x3 +

1

24ln( 1 + x )x4 +

1

24ln( 1 + x )

ys:=y1+y2+y3;

ys :=5

3ln( 1 + x ) +

5

2ln( 1 + x )x � 5

3x+ ln( 1 + x )x2 � 7

6x2 � 11

36x3 +

1

6ln( 1 + x )x3

||||A third example for nonlinear boundary value problem is left as an optional exercise. Can you automate theMaple (or other) manipulator with \FOR" statements?

y00 = �yy0 + � y(0) = y(1) = 0 � = 1

8

It is easy to see why these global solutions are better than the Taylor series that are made of only poly-nomials. However, now we have both � and x...

|||||||||||Nonuniform expansions

A brief de�nition of nonuniformity is given below from Kervokian and Cole:Let x be �xed. � = O( ) as � ! �0 if for the range of x and a neighborhood of �0, there exists a k(x)

such that j�j � jk(x) j: If k can be made independent of x, the relation is uniform.Let x be �xed. � = o( ) as �! �0 if for the range of x and a neighborhood of �0, given \any" �(x) > 0,

then j�j � j�(x) j. If � can be made independent of x, the relation is uniform.These de�nitions revert to the former form if and � do not vanish on the interval. Simple examples of

nonuniform behavior are

f(x; �) = 1 +1

1� x�+O(�2)

f(x; �) = x+ 2�+O(�2)

This case is somewhat unusual in that it concerns the lower-order term going to zero (as x! 0), instead ofthe higher-order term \blowing up".

f(t; �) = A sin t+ �(B sin t+ C cos t+Dt sin t) +O(�2)

sinx

�= O(x) 0 < x < 1 as �! 0

even though the limit does not exist as �! 0. (It is unusual to see a gauge function with anything but thesmall parameter.) Here, k(x) = 1=x. The following example is uniformly valid:

sinx

�= O(1) 0 < x < 1

|||||||||||Some general comments:a) convergent vs. asymptotic formsb) exponentially small terms | Van Dyke (\no problem")

f(�)\ = "f(�)(1 + e�1=�);

because when a Taylors series about � = 0 of the last term on the RHS is performed we obtain

1 + e�1=�����=0

+ �

���� 1�2 e�1=�

�����=0

+ ::: = 0

c) when do nonuniformities occur?1. � in front of highest order2. boundary conditions are placed at 1 (or singular point of coordinate system)3. change of type4. things go wrong...

d) How do you pick proper gauge functions for expansion?1. balance2. super uous terms don't get you in trouble, just redundant3. when in doubt try a logarithmic term

|||||||||||More generally we have F [u; x; �] as an implicit equation (algebraic, di�erential, or integral), where u

and x represent the dependent and independent (possibly vector) variables respectively and � is a smallparameter. Then setting � to zero gives

F [u0; x; 0] = 0

which should easily give the solution u0, otherwise why bother? Di�erentiating with respect to � gives

Fu[u; x; �] _u+ F�[u; x; �] = 0

9

set � to zeroFu[u0; x; 0]u1 + F�[u0; x; 0] � F 0

uu1 + F 0� = 0

assuming this invertible we get the solution u1 = (F 0u )�1F 0

� . Similarly at the next highest order we get

Fu[u0; x; 0]u2 = �(F 0uuu

1)u1 � 2F 0u� � F 0

��

It is easy to see that the higher-order terms are always linear. In fact, the operator is always the same anddepends on the lowest-order solution.

|||||||||||A FLUID DYNAMICS EXAMPLE!

Flow in a \squashed" pipe is given by:

Dw

Dt= �pz + �52 w

with w(r = R = �R+R0 sin 2�) = 0 for example. We choose a length scale �R, velocity scale pz �R2=�. Then

52w = 1; w(r = 1 + � sin 2�) = 0; � = R0= �R

Where the cross section can look like an ellipse, ..., a lima bean, ..., a �gure 8,....Let w = w0 + �w1, then the leading order is axisymmetric

�0 : w0rr +1

rw0r =

1

r(rw0r )r = 1

with boundary conditions w0(r = 1) = 0; jw0(r = 0)j <1. This has the answer

w0 =r2 � 1

4:

Since � didn't appear in PDE, it has the same form at higher order, except homogeneous and the �derivatives appear:

�1 :1

r(rw1r )r +

1

r2w1�� = 0

Expanding the boundary conditions in a Taylor Series about r = 1:

0 = w(R; �) = w(1; �) + wr(1; �)� sin 2� + wrr(1; �)(� sin 2�)2=2! + :::

Now expand each term on the RHS with its power series expansion. Therefore at order �:

w1(1; �) = �w0r(1; �) sin 2� = �1

2sin 2�

Since solution has sin 2� dependence try as ansatz: w1 = w1(r) sin 2�:

w1rr +1

rw1r �

4

r2w1 = 0

This is equidimensional, so we try w1 = ra to give indicial equation:

a(a� 1) + a� 4 = a2 � 4 = 0

or w1 = Ar2 +Br�2, from boundedness B = 0; A = � 12 , etc.

We can check the accuracy of the no-slip boundary condition by putting the exact radius location intothe approximate solution.

w :=1

4r2 � 1

4� 1

2" r2 sin( 2 � )

subs(r=1+epsilon*sin(2*theta),w);

1

4( 1 + " sin( 2 � ) )2 � 1

4� 1

2" ( 1 + " sin( 2 � ) )2 sin( 2 � )

10

simplify(");

� 3

4"2 +

3

4"2 cos( 2 � )2 � 1

2"3 sin( 2 � ) +

1

2"3 sin( 2 � ) cos( 2 � )2

The maximum size of the error to O(�2) is then 34�

2:We could say that the �rst-order correction explains 90percent of the variation caused by the squashed pipe when����12�(1� :90)

���� <�����3

4�2����

Hence we should require � < 230 � :1. What happens with a more general boundary condition?

|||||||||||HWK no. 3 : 2D ow in Bearings .....References:DiPrima and Stuart (1972), J. Lubr. Tech:266-274(modi�ed, bipolar coordinate system, small clearance ratio)

Kulinski and Ostrach (1967), J. Appl. Mech.:16-22 (small eccentricity, some errors)

The equations of motion for steady, incompressible N-S using a streamfunction de�ned by ~v � 5 � kgives upon taking the curl of the momentum equation

Re~v � 552 =54 : (here take Re = 0)

Continuity is automatically satis�ed and pressure has been eliminated. The boundary conditions of theeccentric circular cylinders where the coordinate system is centered on the inner cylinder and the problem ismade dimensionless by setting the inner radius, the angular rotation of the inner cylinder and density equalto 1 are:

~v(r = 1) = e� ~v(r =

q 2 � �2 sin2 � + � cos �) = 0

We expand = 0 + � 1 + ::: and at lowest order, we �nd the solution for concentric cylinders:

v0 = � 0r =A

r+Br =

r � 2=r

1� 2:::

These last two examples have been regular perturbation expansions. That is, nothing \bad" happenedduring the expansion procedure. They also have contained only one small parameter. One can develop amultiple power series expansion for more than one small parameter. My preference is to relate small param-eters and end up with only one. In HWK 3, for instance, a relevant engineering example would have 1� to be a small number comparable to �. It then makes sense to de�ne 1 � � M�, where M is an orderone parameter such that M > 1. When the problem is set up in this way, the lowest-order solution placeincompatible boundary conditions on top of each other! The problem becomes singular and we will handlethis problem later.

|||||||||||SUMMATION ACCELERATION TECHNIQUES (Chpt 8 B & 0)

Taylor series are convergent inside a circle centered at the expansion point that excludes all singularitiesin the complex plane. Hence the expansion about zero of

f(z) =1

1 + z

will converge radius when jzj < 1, and is diverging or very slowly converging near z = 1 even though thereis no singularity there. The remainder after n terms is

Rn = (�z)n+1=(1 + z);

which just alternates sign at z = 1:, Rn = � 12 : We try a Shank's transform: Assume that as n!1 that

the sequence of partial sums takes the form

An = A+ �qn;

11

Then An � A if jqj < 1 and the last term is called the \transient". If we took three consecutive partial sums,we can solve for the three unknowns A;�; and q. In particular,

A =An+1An�1 �A2

n

An+1 +An�1 � 2An

which is exact if the term has only one transient, even if the transient does not die out (the sum does notconverge)! This motivates:

S(An) =An+1An�1 �A2

n

An+1 +An�1 � 2An:

Repeated use can achieve fantastic results at eliminating more than one transient. Note that when thesolution does converge (three consecutive terms are same), the transformation gives an indeterminate result.

High precision may be required. As a simple example, take f(z) = 1=(1 + z) =PN

n=0 f(n) zn

n! =PN

n=0 anzn.

Here, An are the partial sums for z = 1.

n f (n) an An s(An)0 (1 + z)�1 1 1 ...1 �(1 + z)�2 -1 0 1/22 2(1 + z)�3 1 1 1/2

3 �6(1 + z)�4 -1 0...

For z = 2, the An are increasing in magnitude with n, however the �rst transformation gives the expectedresult 1=3. Like the example above, a second transformation gives an indeterminant result since this problemhas exactly one transient.

|||||||||||

Richardson Extrapolation

Shank's transformation does not improve the convergence ofP1

0 n�2 = �2

6 because Rn � 1=n as n ! 1.When the nth partial sum is of the form

An =

nXk=0

ak � Q0 +Q1n�1 +Q2n

�2 + :::;

it is appropriate for Richardson extrapolation.

An =

nXk=0

ak � Q0 +Q1n�1 +Q2n

�2 + :::+QNn�N ;

An+1 =

n+1Xk=0

ak � Q0 +Q1(n+ 1)�1 +Q2(n+ 1)�2 + :::+QN(n+ 1)�N ;

...

An+N =n+NXk=0

ak � Q0 +QN (n+N)�N +Q2(n+N)�2 + :::+QN (n+N)�N ;

The result for Q0 is

Q0 =

NXk=0

An+k(n+ k)N (�1)k+Nk!(N � k)!

The partial sums of this example series can be expressed as an integral

An =

nX0

k�2 =�2

6�Z 1

0

te�nt

et � 1dt � �2

6� 1

n+

1

2n2+ � � �

12

It is not necessary to know this equation using this approach to in�nity but that the form exists. See Table8.4. B&O. A summary is given below:

n N = 0 N = 1 N = 2 N = 61 1.000 1.5000 1.6250 1.6449351855 1.464 1.6303 1.6441 1.64493406010 1.550 1.6407 1.6448 1.64493406625 1.606 1.6442 1.64492 1.644934067

|||||||||||

Pad�e Approximants

Convergence problems are due to singularities. If we could remove them, we would get faster convergenceand a larger radius of convergence. How do we estimate, say, simple poles to remove them? Try

PNM (z) =

PNn=0Anz

nPMm=0Bmzm

; B0 = 1:

Usually we choose N � M(N = M is called a diagonal approximation). Actually we take N = M + J , J�xed and �nd convergence of the Pad�e approximants as M ! 1. We choose the coe�cients so that the�rst (N +M +1) terms of the T.S.E of PN

M match those of the original expansion. See the example 1, p.383B.&O. for P 0

1 .Systematic way see 8.4 B.&O. with relationship to continued fractions and generalized Shank's Transforma-tions.

A generalized Pade summation chooses the coe�cients from two distinct expansions:

f(z) �1Xn=0

an(z � z0)n; z ! z0

f(z) �1Xn=0

bn(z � z1)n; z ! z1

pick J terms from �rst K from second such that J +K =M +N + 1 Although its not explicitly stated inB&O (8.3.8b) the n exponent can be replaced by �n for large z asymptotic expansions: Some examples:

1) Pad�e Approximant of 1=(1 + z) is exact.

2) P 11 approximation of erf(z). We simply make sure that the �rst \two" terms of the Taylor series

expansion match those of the Pad�e approximant and that the �rst two terms of the large z asymptoticexpansion also match.

P 11 =

A0 +A1z

1 +B1z

P 11 =

A0�+A1

�+B1=A1

B1+

�A0

B1� A1

B21

�1

z+ ::: z !1

for small z; P = A0 + (A1 � aA0B1)z � (B1A1 + :::)z2 + :::, while

erf(z) = 0 +2p�(z � z3=3 + :::) z ! 0

erfc(z) = (1� 1

2z2+ :::)=(

p�zez

2

) z = 1=�!1

13

We can �t four conditions for P 12 , the �rst four terms for the small z expansion to give

P 12 =

2zp�(1 + z2=3)

:

or the �rst two at z = 0 and the requirement that P 11 = 1 as z ! infty to give

P 11 =

2p�z

1 + 2p�z:

If we want to study a uniform expansion which shows the asymptotic approach to in�nity, we must modifythe expansion to avoid the exponentially small terms. (See! Exponentially small terms has already got us

into trouble!) This can be done by expanding P 11 � ez

2

erfc(z) that now can be expanded for large z. Then

erf(z) � 1� P 11 e�z2

What are the e�cient uses of Pad�e? From the Maple worksheet, you can see that diagonal Pad�e approximantscan be obtained using the \convert" command using the `ratpoly' subcommand and continued fractions (aclose cousin) from the `confrac' subcommand (the `polynomial' subcommand eliminates the remainder term).We see that the Pad�e approximant is local and gives the identical results as the continued fraction...the thingyou've probably seen in A&S and have never used. This truncated continued fraction is often written as

1 +x

1 +x

�2 + x

�3 + 1

2x

= 1 +x

1+

x

�2+x

�3+x

2

How do we obtain this and how do we evaluate it | other than the brute force method? We would like tostop when we converge. The standard local continuing fraction is obtained like the Pad�e approximant. Onestep at a time. There are two big advantages:

1. The coe�cients of the continued fractions do not change with truncation

2. There is a straightforward way to �nd the convergence of such functions

If you have a normal Pad�e sequence, that is P 00 (z); P

01 (z); P

11 (z); P

12 (z); ::: all exist and are distinct, the

continued fraction can be obtained easily (p. 396 B&O). Otherwise a system of equations may need to besolved. A procedure to evaluate the function is found in pages 19 and 22 of A&S. Let

fn =An

Bn= b0 +

a1b1+

+a2b2+

+ :::+anbn

The fn can be found recursively from

An = bnAn�1 + anAn�2

Bn = bnBn�1 + anBn�2

where A�1 = 1; A0 = b0; B�1 = 0; B0 = 1: An example, compute

tan�1(x) =x

1++x2

3++4x2

5++9x2

7+:::

to �ve digits for x = :2. Here a1 = x; an = (n� 1)2x2 for n > 1; b0 = 0; bn = 2n� 1:

A0

B0=

0

0= 0

A1

B1=:2

1= 0:2

A2

B2=

:6

3:04= 0:197368

14

A3

B3=

3:032

15:36= 0:197396

some questions....Is Pad�e always superior to Shank's transformation? Yes...How can one determine the radius of convergence? (see section 8.5 B&O...tough going))What about other continued functions?? c0

pc1zpc2z:::

|||||||||||

Kummer's Transformation (A&S 3.6.26, K&K 4.8-8)

We seek the slowly converging sum1Xk=0

ak = s

If we can compare to a known series,

1Xk=0

ck = c such thatlim

k !1akck

= � 6= 0;

then the following series will converge must faster

s = �c+

1Xk=0

�1� �

ckak

�ak

|||||||||||

Euler's Transformation (A&S 3.6.27, K&K 4.8-7)

For the alternating series1Xk=0

(�1)kak = a0 � a1 + a2 + ::: = s

can be accelerated using1Xk=0

(�1)k�ka02k+1

where �ka0 =

kXm=0

(�1)m�

km

�ak�m

|||||||||||

METHOD OF MATCHED ASYMPTOTIC EXPANSIONS

references: B&O Chpt. 9, Kervorkian and Cole,van Dyke, O'Malley (Sing. Pert. Methods)Eckhaus (Matched Asymptotic expansions and singular perturbations)Bender and Orszag (chpt.9)

Example 1: Di�erential Equation: Linear 2nd Order, BVP:

�u00 + 2bu0 + cu = 0

where u(0) = � and u(1) = �: For simplicity, Let b = c = � = � = 1:Why should we study this as a perturbed problem? Assume we can solve two terms at a time but not

three! Actually we could solve almost any 1st order NL equation, so this could really be useful. What arethe problems? It is obvious that a regular perturbation expansion cannot satisfy both boundary conditions.

15

The �u00 term is almost negligible everywhere except at a boundary layer where u00 is large to counteract the�.

First, the exact solution is given as exponentials of the roots from the characteristic equation �D2+2D+c = 0 giving:

u = A1e(�1+

p1��c)x=� +A2e

(�1�p1��c)x=�

which has the expansion u = A1e[�1+(1��c=2+:::)]x=� + A2e

[�1�(1��c=2+:::)]x=�: The �rst term is regular thesecond is singular. Where is the boundary layer{at x = 1 or 0?, depends on sign of the second term (assum-ing � > 0). Usually can tell from physics, but for this problem the boundary layer is at x = 0. Sometimesthe \boundary layer" location may be obvious as in the initial layer problem: �u00 + 2bu0 + cu = 0; whereu(0) = � and u0(0) = �: However, it then is not obvious which initial condition the outer solution shouldsatisfy. The procedure to solve this problem is as follows:

1. Outer expansion, perform regular asymptotic expansion{usually a power series expansion in �. Thisresult will be nonuniformly valid because it cannot satisfy one of the boundary conditions: (perhaps)uo =

Puoj(x)�

j .

2. Stretch the boundary-layer variable by � = x=�(�) where �! 0 as �! 0. Balance terms to make surehighest derivative term survives. (We have assumed that the nonuniform behavior or boundary layer willform at x = 0. A simple coordinate translation will su�ce if it is elsewhere.)

3. Rescale equation and �nd inner expansion: (perhaps) uI =PUj(�)�

j .

4. Find overlap interval:

valid region of

outer solution

|----------->

<----------|

valid region of

inner solution

The existence of this overlap region is hard to prove. My experiences are that it is there{don't worryabout it.

5. Express outer expansion in inner variable � or vice versa:

uo =X

uj(�)�j :

6. Match at x = O(�(�)). The idea of the overlap region is that the outer region is valid for all x > �(�),where �(�)! 0 as �! 0 but not as fast as �(�) goes to zero, i.e., � = o(�). Then the matching condition is

lim�! 0

hXuj(�)�

j �X

Uj(�)�ji= 0

x = O(�(�)) If all is well, the matching proceeds in orders, essentially the lowest order outer solution asx! 0 matches the lowest order inner solution as � !1.

7. Build a composite solution by adding inner and outer parts and subtract out the common part (thatpart which was matched).

Here is a maple session for this problem:e:=epsilon:eq:=e*di�(y(x),x$2) +2*di�(y(x),x)+y(x)=0;

eq := "

�@2

@x2y(x )

�+ 2

�@

@xy(x )

�+ y(x ) = 0

16

dsolve(eq, y(x));

y(x ) = C1 e

�(�1+

p1�") x"

�+ C2 e

�� (1+

p1�") x"

�

ans:=dsolve(epsilon*di�(y(x),x$2) +2*di�(y(x),x)+y(x), y(0)=1,y(1)=1,y(x));

ans := y(x ) = �

e

�� 1+

p1�"

"

�� 1

!e

�(�1+

p1�") x"

�

e

��1+

p1�"

"

�� e

�� 1+

p1�"

"

� +

�1 + e

��1+

p1�"

"

�!e

�� (1+

p1�") x"

�

e

��1+

p1�"

"

�� e

�� 1+

p1�"

"

�

taylor(rhs(ans),epsilon,2);Error, (in series/exp) unable to compute series#the problem is singular!#Form an outer solution for the regular part of pblm{away from boundary layereqo:=subs(y(x)=yo[0](x)+e*yo[1](x),eq);

eqo := "

�@2

@x2(yo0(x ) + " yo1(x ))

�+ 2

�@

@x(yo0(x ) + " yo1(x ))

�+ yo0(x ) + " yo1(x ) = 0

eqo0:=subs(e=0,eqo);anso:=dsolve(eqo0,yo[0](1)=1,yo[0](x));

eqo0 := 2

�@

@xyo0(x )

�+ yo0(x ) = 0

anso := yo0(x ) =e(� 1=2 x )

e(�1=2 )

eqo1:=subs(e=0,di�(eqo,e));

eqo1 :=

�@2

@x2yo0(x )

�+ 2

�@

@xyo1(x )

�+ yo1(x ) = 0

eqo1:=simplify(subs(yo[0](x)=rhs(ans),eqo1));

eqo1 :=1

4e(� 1=2 x+1=2 ) + 2

�@

@xyo1(x )

�+ yo1(x ) = 0

dsolve(eqo1,yo[1](1)=0,yo[1](x));# the second correction is

yo1(x ) = �1

8e(� 1=2 x ) e( 1=2 ) x+

1

8e(� 1=2 x ) e( 1=2 )

eqi:=subs(y(x)=Y(xi),eq);

eqi := "

�@2

@x2Y( � )

�+ 2

�@

@xY( � )

�+Y( � ) = 0

eqi:=e(1-2*a)*di�(Y(xi),xi$2)+e(-a)*di�(Y(xi),xi)+Y(xi);#Balancing terms

eqi := "(1�2a

�@2

@�2Y( � )

�+ "(�a )

�@

@�Y( � )

�+Y( � )

simplify(subs(a=1/2,eqi)=0); This doesn't work�@2

@�2 Y( � )� p

"+�

@@� Y( � )

�+Y( � )

p"

p"

= 0

17

simplify(subs(a=1,eqi)=0); �@2

@�2 Y( � )�+�

@@� Y( � )

�+Y( � ) "

"= 0

eqi:="*e;Y(xi):=Yi[0](xi)+e*Yi[1](xi);

eqi :=

�@2

@�2Y( � )

�+

�@

@�Y( � )

�+Y( � ) " = 0

Y( � ) := Yi0( � ) + "Yi1( � )

eqi0:=subs(e=0,eqi);

eqi0 :=

�@2

@�2Yi0( � )

�+

�@

@�Yi0( � )

�= 0

ansi:=dsolve(eqi0,Yi[0](0)=1,Yi[0](20)=exp(1/2),Yi[0](xi)):ansi:=subs(xi=x/e,ansi); #I had problems setting this to in�nity

ansi := Yi0

�x"

�= � �e

(�20 ) + e( 1=2 )

�1 + e(�20 )+( e( 1=2 ) � 1 ) e(�

x" )

�1 + e(�20 )

cmp:=rhs(anso)+subs(e=.05,rhs(ansi))-exp(1/2);

cmp :=e(� 1=2 x )

e(�1=2 )� �e(�20 ) + e( 1=2 )

�1 + e(�20 )+( e( 1=2 ) � 1 ) e(�20:00000000 x )

�1 + e(�20 )� e( 1=2 )

|||||||||||

Questions:1) What happens if boundary conditions are reversed, i.e, u(1) = 0; u(0 = 1)?2) What if middle term is \�" instead. Easy way to see{call z = x� 1.3) What if middle term disappeared? Maybe it only disappears in one place such as (x � 1=2)u0. We willuse WKBJ later for the internal layers that develop.4) Note the exact solution has exponentially small terms. How could an expansion procedure �x this?

Example 2: p. 421, B&O: A �rst-order problem with boundary layer

(x� �y)y0 + ay = e�x y(1) = 1=e

You can see the problem in that at x = 0, there is only one term to balance e�x, so yy0 must be large. Seetext for details.

Example 3: �y00 + 2y0 + ey = 0 y(0) = y(1) = 0Nonlinear equations can be easily handled if the �rst-order outer solution is integrable. Here, y0(x) = ln 2

1+xand the stretch is againY = y=�:

HWK no. 4: Find the leading-order composite solution for

�y00 + (x2 + 1)y0 +�x3y = 0 y(0) = y(1) = 1

\extra HWK"

1) �w0000 � w00 = p(x); w(0) = w0(0) = w(1) = w0(1) = 0 � = EITL2

2) m�y + b _y + ky = 0; m! 0; y(0)� 1 = _(y) = 0 initial-layer problem

3) �y00 + y0 � xy = 0 y(0) = 0; y(1) =pe

Compare with the expansion of exact solution in terms of I1=3 and K1=3

18

|||||||||||

Poincar�e-Linstedt Method

Singularities can occur from unbounded domains. An example is the initial value problem for vibratingpendulums, 0 < t <1:

�00 + �2 sin � = 0 �2 = g=l

No dissipation, so can solve by phase plane analysis...multiply by �0 and integrate

�0�00 + �0�2 sin � = 0�1

2�02�0� �2[cos �]0 = 0

1

2�02 � �2 cos � = C

where C is the constant energy determined from i.c. This �nal equation can be integrated once more to getthe �nal result in terms of elliptic integrals.

The phase plane has the form:

Note that the equation has steady solutions �0 = 0 at � = n� | n even is stable, n odd unstable.

We scale time to get make � = 1 to obtain:

�00 + sin � = 0

with i.c. say � = �, �0 = 0.

We try a standard expansion in �. We quickly �nd that only odd powers are required:

�(t; �) = �1�1(t) + �3�3(t) + :::

That is expected|the solution to be an odd function of the initial displacement.at \lowest" order: O(�1):

�001 + �1 = 0 �1 = 1; �01 = 0:

which has the solution � = cos t. So far so good.at \next" order: O(�3):

�003 + �3 = �31=3! =1

24[3 cos t+ cos3t] �3 = 0; �03 = 0:

which has the solution � = A cos t+B sin t+C cos 3t+Dt cos t+Et sin t. The constants C;D;E are obtainedfrom method of undetermined coe�cients, A and B are used to satisfy the i.c. The presence of E (D turnsout to be zero) is troublesome because it blows up at t ! 1. This term is trying to locally change thefrequency. It works OK for small time!

19

We need to eliminate secular terms that cause this behavior. We allow the frequency to be a functionof amplitude (energy) by � = !t, d=dt = !d=d� , let X(�) = �(t) gives:

!2X 00 + sinX = 0 X = �; X 0 = 0:

We took out a constant term by scaling and put in a variable one! [We could also rescale to put � in operator].We will �nd that ! = !0 + !2�

2 + ::: = 1+ !2�2 + :::. An even function for !. Note that !0 = 1=n would

also give 2� periodic solutions. We instead examine the primary frequency, the lowest value of n that is stillperiodic.

The lowest order remains the same except t is replaced by !t. At O(�3):

X 003 +X3 = X3

1=3! =1

24[3 cos � + cos 3� ] + 2!2 cos � X3 = 0; X 0

3 = 0:

we simply set the coe�cients of cos � equal to zero to avoid secular terms. Then !2 = �1=16 i:e: thefrequency slows as amplitude increases. (Note: quadratic nonlinearities are more di�cult because the nexthighest order DOES NOT produce a secularity condition. Hence, a �rst correction to linear theory mustbe produced before the correction to frequency is found. This often happens in Fluid Mechanics since thesenonlinearities are usually quadratic. The result is often called a Hopf Bifurcation (see AM623 HydrodynamicStability).)

See how MACSYMA and the program lc(); by Rand and Armbruster (This MACSYMA program is avail-able at http://alamos.math.arizona.edu/ rychlik/577-dir/RAND/program16.)can help the tedious procedureof the Poincar�e-Linstedt method. For example, the van der Pol Oscillator :

THE D.E. IS OF THE FORM: X'' + X + E * F(X,X') = 0

ENTER F(X,Y), REPRESENTING X' AS Y, Z = W*T

y*(x^2-1);

2

THE D.E. IS: X'' + X + E ( (X - 1) Y ) = 0

ENTER TRUNCATION ORDER

2;

CHOICES FOR LIMIT CYCLE AMPLITUDE:

1 ) - 2

2 ) 2

3 ) 0

ENTER CHOICE NUMBER 2;

(SIN(3 Z) - 3 SIN(Z)) E

X = 2 COS(Z) - ----------------------- + . . .

4

2

E

W = 1 - --

16

Here, we have an X 0 term, but instead of being positive (dissipation), part of it is negative so a limit cycleexists. We should note that at present, higher truncations do not work on lc().

FORCED OSCILLATORS

What about a singing wire or the Tacoma Narrows bridge? These represent an interaction between anatural frequency and periodic forcing. Let's take a di�erent example this time, the Du�ng Oscillator:

x00 + p2x� ��x3 = q cos!t

20

with I.C x(0) = a; x0(o) = b. Note that we have introduced the small parameter � into the equationrather than the initial condition by a suitable rescaling. This is important because we will deemphasizethe I.C. soon. This also means that we will expand in all integer orders.) p2 = k=m is the linear naturalfrequency, � is the nonlinear term; � > 0 is a weak spring, � < 0 is a strong spring. We look for 2�=!periodic solutions in time. Note p2 is not scaled because we will use 1=! as the time scale. The solution isno longer autonomous due to the forcing, so initial conditions cannot be simpli�ed by a change of the timeorigin. We rescale � = !t, set u(�) = x(t), and look for 2� periodic solutions in � .

!2u00 + p2u� ��u3 = q cos �

u(0) = a u0(0) = b=!

Here ! is known so it is not expanded. We will perform a T.S.E this time:

u(�; �) = u(0; �) + _u(0; �)�+ �u(0; �)�2=2 + ::: = u0 + u1�+ u2�2=2 + :::

set � = 0 to get lowest order equation:

!2u000 + p2u0 = q cos �

that has the solution u0 =q

p2�!2 cos � + A cos p! � + B sin p

! � . We assume i.c. are such that A = B = 0. Alittle dissipation will cause the solution to forget its i.c., otherwise, solution is not 2� periodic. Then WLOG,we can assume that q > 0. The solution is OK if p2 � !2 is not small (nonresonance case). To get higherorders, we di�erentiate with respect to �:

!2 _u00 + p2 _u� 3��u2 _u� �u3 = 0

then set � = 0:

!2 _u00 + p2 _u = �u(0; �)3 = �q3

[p2 � !2]31

4[3 cos � + cos 3� ]

This equation is now forced by the nonlinearity, has homogeneous i.c., and has the solution

u1 =3�q3

4(p2 � !2)4cos � +

�q3

4(p2 � !2)3(p2 � 9!2)cos 3�

Now the superharmonic blows up as well....

Near-resonance case. Suppose !2 is near p2, then let

! = p+ !1�+ !2�2=2 + :::

(we again expand ! | not because we don't know it, but so the � (amplitude) can be determined from !iand p. This is often called \detuning"). The lowest-order equation

p2[u000 + u0] = q cos �

has no periodic solutions. We now must insist that the forcing is small for the solution to be only weaklynonlinear. Then, q = q0�, and at leading order

p2[u000 + u0] = 0

leads to u0 = a cos � + b sin � , where a and b are to be determined. Di�erentiating with respect to �

!2 _u00 + 2! _!u00 + p2 _u� 3��u2 _u� �u3 = q0 cos �

setting � = 0 gives the �rst-order equation:

p2[ _u00 + _u] = �2p _!u00 + �u3 + q0 cos �

now substitute u(� = 0) = u0 = a cos � + b sin � ... lots of algebra ...

p2[ _u00 + _u] =

�2p!1a+

3�

4a(a2 + b2) + q0

�cos � +

�2p!1b+

3�

4b(a2 + b2)

�sin � + [:::] cos 3� + [:::] sin 3�

21

We now have two secularity conditions to enforce. A simple solution to this is b = 0 and a[2p!1+3�4 a

2]+q0 =0. We can substitute for !1 at this order by:

!2 = (p+ !1�+ :::)2 = p2 + 2p!1�+ :::

or 2p!1 = (!2 � p2)=�+ ::: which put in the secularity condition gives

a

�(!2 � p2)=�+

3�

4a2�+ q=� = 0

In spite of all these �'s, these terms do balance. The resonance �gures then look like:

matching to the nonresonant case is challenging... referencesStoker, \Nonlinear Vibrations"Minorsky, \Nonlinear Oscillations"Nayfeh (1979), \Nonlinear Oscillations", Wiley.

Fluid mechanics examples:Taylor, G.I. (1954), \An experimental study of standing waves", Proc. R. Soc. Lond. A 218:44-59.Henderson, D.M. and J.W. Miles (1991), \Faraday waves in 2:1 internal resonance", J. Fluid Mech.

222:449-470.|||||||||||

Multiple Scale (Time) Methods

So far we have had only essentially one time behavior, with the time dependence modi�ed a smallamount to accommodate the nonlinear dependence on frequency. What happens when something happensfast (oscillation) and something slow (approach to limit cycle)? We will be able to see which periodic solutionscomputed above are stable or unstable.

Simple example: linear oscillator with small damping...

u00 + 2�u0 + u = 0 u(0) = u0(0)� 1 = 0

that has the exact solution u = e��tp1��2 sin

p1� �2t. We try a straightforward perturbation expansion.

u(t; �) = u(t; 0) + _u(t; 0)�+ �u(t; 0)�2=2 + ::: = u0 + u1�+ u2�2=2 + :::

Lowest order gives u0 = sin t. Next order gives

u001 + u1 = �2u00 = �2 cos t u1(0) = u01(0) = 0

which has the solution u1 = A cos t + B sin t + Ct cos t + Dt sin t. C = 0 and D = �1 from method ofundetermined coe�cients. This time, the secular term is trying to change the amplitude (at this order) andthe frequency (at higher order).

How do we �x this? The period and the time constant of the envelope are nearly independent of eachother. The frequency is determined by the mass and spring, the envelope by the damper and spring. Theperiod is O(1), the time constant is O(��1). We rescale and pretend they are independent!

22

fast time ~t = tslow time � = �t

We now look for an asymptotic expansion as �! 0 keeping t (drop tilde) and � �xed.

u = u0(t; �) + u1(t; �)� + :::

which we wish to remain bounded. Because the independent variables are independent,

u0 =du

dt= ut + u�

d�

dt+ ::: = u0t + [u0� + u1t ]�+ :::

u00 = utt + 2ut��+ u���2 = u0tt + [2u0t� + u1tt]�+ :::

Now substitute into the di�erential equation. The lowest-order equation is the same but the solution isu0 = A cos t + B sin t; where A and B are functions of � . The initial conditions at lowest order giveA(� = 0) = B(� = 0)� 1 = 0. At next order we eliminate secular terms to form the amplitude equations:

A� +A = 0 B� +B = 0

With the i.c., this gives the lowest-order solution u0 = e�� sin t.||||||||||{

Multiple Time Methods for limit cycles

The problems we have been considering belong to the broad class of oscillators:

y00 + y + �f(y; y0; t) = 0

Note that forcing (inhomogeneous term) can be included if it is small (in the last term). These problems areoften solved with a combination of Poincar�e-Linstedt methods and multiple scales, i.e., t+ = t(1 + !2�

2 +:::); ~t = �t. The amplitude equations can generally be shown to be integrable. See Kervokian and Cole foran excellent discussion. We examine the van der Pol's oscillator again:

u00 + u� �(1� u2)u0 = 0

(Oscillator with triode). It is easy to see that a limit cycle will develop from heuristic arguments. This isof the form that can be used in Rand's program twovar(), (http://alamos.math.arizona.edu/ rychlik/577-dir/RAND/program16)which does not go as far as to �nd !2 which can be determined by lc() independently,

twovar();

DO YOU WANT TO ENTER NEW DATA (Y/N)

y;

NUMBER OF D.E.'S

1;

THE 1 D.E.'S WILL BE IN THE FORM:

X[I]'' + W[I]^2 X[I] = E F[I](X[1],...,X[ 1 ],T)

ENTER SYMBOL FOR X[ 1 ]

u;

ENTER W[ 1 ]

1;

ENTER F[ 1 ]

(1-u^2)*diff(u,t);

THE D.E.'S ARE ENTERED AS:

2 dU

U '' + U = E (1 - U ) --

dT

THE METHOD ASSUMES A SOLUTION IN THE FORM:

X[I] = X0[I] + E X1[I]

WHERE X0[I] = A[I](ETA) COS W[I] XI + B[I](ETA) SIN W[I] XI

23

WHERE XI = T AND ETA = E T

REMOVAL OF SECULAR TERMS IN THE X1[I] EQS. GIVES:

2 3

A B A

d 1 1 1

2 (---- (A )) + ----- + -- - A = 0

dETA 1 4 4 1

3 2

B A B

d 1 1 1

- 2 (---- (B )) - -- - ----- + B = 0

dETA 1 4 4 1

DO YOU WANT TO TRANSFORM TO POLAR COORDINATES (Y/N)

y;

3

R R

d 1 1 d

[[---- (R ) = -- - --, ---- (THETA ) = 0]]

dETA 1 2 8 dETA 1

DO YOU WANT TO SEARCH FOR RESONANT PARAMETER VALUES (Y/N)

n;

We can see that a nontrivial limit cycle is A1 = 2; B1 = 0, If we de�ne A1 = R cos �; B1 = R sin � we obtainthe polar equations above, that are separable (here ~t = �)

R(~t) =2R(0)e~t=2q

(e~t � 1)R(0)2 + 4

||||||||||{

Check the stability near resonance of the Du�ng Equation:

!2u00 + p2u� ��u3 = �q0 cos � (4)

we have already scaled the small forcing amplitude and the forcing frequency is unity. Since ! is near p weuse

! = p+ !1�+ !2�2=2 + :::

and reintroduce t as the slow time t = �� . At leading order we obtain:

u0�� + u0 = 0 (5)

which has the solutionu0 = a(t) cos � + b(t) sin � (6)

Taking the derivative with respect to �

!2 _u�� + 2! _!u�� + p2 _u� 3��u2 _u� �u3 = q0 cos � � 2p2ut�

collecting the secular terms on the RHS we obtain the same as before with the addition of two terms:

p2[ _u�� + _u] = [�2p2b0 + 2p!1a+3�

4a(a2 + b2) + q0] cos �

+[2p2a0 + 2p!1b+3�

4b(a2 + b2)] sin � + [:::] cos 3� + [:::] sin 3�

which as before has the periodic solution: �a[2p!1 +3�4 �a

2] + q0 = 0, and �b = 0, where the bar indicates thesteady amplitude (or basic state). Now we allow small disturbances into the original equations such that

a = �a + a and b = b. The terms with only basic-state terms will drop out{we force them to. We drop all

24

higher-order terms that have more than one-hatted quantity. Then we end up with homogeneous equations(and boundary conditions) with constant coe�cients. We seek a normal mode solution of the form�

a

b

�=

�~a~b

�e�t:

We now have a second-order eigenequation (characteristic equation) for �: To show stability, both roots of� must have negative real part.

HWK no.5: a) Show which periodic solutions shown on the previous �gure are stable and unstable. b)Use the P-L method for the quadratic nonlinear problem

x00 + p2x� ��x2 = 0

||||||||||{Comparing the MMAE and MS Methods Consider the initial layer-problem

�u00 + u0 + u = 0 u(0) = u0(0)� 1=� = 0

If we balance we �nd ~t = t=�, giving after renaming u(t) � y(~t)

y00 + y0 + �y = 0 y(0) = y0(0)� 1 = 0

If we seek the solution y = y0(~t; t) + :::: we get the same result. In �rst case we have regular and fast(stretched) time, in the second regular and slow.

|||||||||||Multiple Time Methods for PDE{The NL Schroedinger Eqn.For weakly nonlinear waves in deep water we expand the velocity potential � for small amplitude a: Thenorm of �1 is 1.

� = a�1 + a2�2 + a3�3 + � � � : (7)

We introduce the slow variables

� = a2t; � = a(x� cgt); �y = ay; (8)

where cg is the group velocity that will be determined from the solution. The lowest-order equation givesthe solution

�1 = (AE +AcE�1)ey; (9)

where E � expfi(x� t)g satis�es the Laplace equation and the deep-water boundary condition. and A(�; �)describes a slow variation of amplitude with a mean value of 1

2 at � = 0, since a is the initial amplitude ofthis mode. Note space and time have been made dimensionless by setting ! = k = 1

The Bernoulli equation combined with the kinematic condition at lowest order gives the dispersion rela-tionship, phase velocity cp = !=k as a function of k. Since our equation is dimensionless, that only sets thescales here. The Bernoulli equation at second order gives the group velocity cg = d!=dk. The third-orderamplitude equation for E0 giving �01 is not required for deep water (see C.C. Mei). At third-order, theamplitude equation with E1 gives:

A� +�2

2p2� �[�+ i(

p2 � �)]

�2 �p2�+ 1A+

i

8(4T 2 � 8T + 1)A�� = �i9T

2 � 15T + 8

4(1� 3T )jA j2 A: (10)

Where T is a dimensionless surface tension equal to one for capillary waves and 0 for gravity waves and � is asurfactant dissipation coe�cient (see Joo, et al.). This equation is usually called a nonlinear Schroedinger(Schro�dinger) equation. It is also often called a Ginzburg-Landau Equation since A ius complex. Notethere is a nonuniformity when T = 1=3. At this value we have second-harmonic resonance (Wilton's ripples)which requires the E and E2 term at leading order for the \carrier wave". We now consider the stability ofa uniform wave-train (@=@� = 0) for gravity waves (T = � = 0) and the dissipation term is scaled di�erentlyfor the much weaker dissipation without surfactant so the second term is 2�2A This gives the result

A0(�) = a0 exp

��2�2� + i

2�2ja0j24

(e�4�2� � 1)

�; (11)

25

where � is the dissipation rate. In the inviscid limit (� = 0) we recover Stokes waves A = ao exp[ijaoj2� ].We perturb this uniform wave-train by

A = A0(�) [1 +B(�; �)] : (12)

Substituting into the amplitude (NLS) equation gives

iB� � 1

8B�� = 2 ja0 j2 e�4�2� (B +Bc) (13)

after the quadratic terms in B are neglected. For a pair of sideband modes, we seek a solution in the form

B = B1(�)eil� +B2(�)e

�il� ; (14)

where B1 and B2 are complex coe�cients, and l is a real wavenumber.Although the disturbance can grow for small values of the slow time variable � , it will eventually be

damped out due to viscous dissipation. We obtain a pair of linear homogeneous equations for B1 and B2:

B01 �il2

8B1 + 2i ja0 j2 e�4�2� (B1 +Bc

2) = 0 (15a)

B02 �il2

8B2 + 2i ja0 j2 e�4�2� (Bc

1 +B2) = 0; (15b)

where a prime denotes di�erentiation with respect to � , and a0 =12 : The transformation

B1 = e�2�2� �B(�) (16)

yields�B00 � � �B = 0; (17)

where

� =l2

64(32 ja0 j2 e�4�2� � l2) +

i

2�2l2 + 4�4: (18)

In the inviscid limit (�! 0), jA0 j is constant, to give

� =l2

64(32 ja0 j2 �l2) (19)

and �B = B1. The stability of the ow is then determined by the sign of �: � > 0 indicates instability,whereas � < 0 gives neutral stability, in agreement with the inviscid instability condition of Benjamin & Feir(1967):

0 < l2 < 32 ja0 j2 : (20)

The details of the instability for inviscid ow were given by Stuart & DiPrima (1978).

REFERENCES

Benjamin, T. B. and Feir, J. E. 1967 The disintegration of wave trains on deep water. J. Fluid. Mech.,27, 417-430. (Initial description of B-F instability).

Davey, A. and Stewartson, K. 1974 On three-dimensional packets of surface waves. Proc. R. Soc. Lon-

don A, 338, 101-110. (Best description of amplitude equations).

Djordjevic, V. D. and Redekopp, L. G. 1977 On two-dimensional packets of capillary-gravity waves. J. FluidMech., 79, 703-714. (Show resonance in amplitude equations for some T ).

Hasimoto, H. and Ono, H. 1972 Non-linear modulation of gravity waves. J. Phys. Soc. Japan, 33, 805.(Credited with being �rst NL.).

Joo, S. W, A.F. Messiter and W.W. Schultz 1991 Evolution of weakly nonlinear water waves in the presence

26

of viscosity and surfactant. J. Fluid Mech. 229, 135-158. (First to FORMALLY add dissipation).

Lake, B. M., Yuen, H. C., Rungaldier, H., and Ferguson, W. E. 1977 Nonlinear deep-water waves: the-ory and experiment. Part 2. Evolution of a continuous wave train. J. Fluid Mech., 83, 49-74.

Mei, C.C. 1983 The Applied Dynamics of Ocean Surface Waves, Wiley.(Fixed up the deep-water limit).

Peregrine, D. H. 1983Water waves, nonlinear Schr�odinger equations and their solutions. J. Austral. Math. Soc.

B, 25, 16-43. Good summary.

Phillips, O. M. 1966 Dynamics of the upper ocean. Cambridge Univ. Press. (Good summary).

Stuart, J. T. and DiPrima, R. C. 1978 The Eckhaus and Benjamin-Feir resonance mechanisms. Proc. R. Soc.London A, 362, 27-41. (Best description of B-F instability).

Whitham, G. B. 1967 Nonlinear dispersion of water waves. J. Fluid Mech., 27, 399-412. (Early work{di�cult).

Zakharov, V. E. 1968 Stability of periodic waves of �nite-amplitude on the surface of deep uid. J. Appl. Mech.

Tech. Phys., 9, 190-194. (Carrier wave changes frequency and amplitude).

|||||||||||Reduction to lower dimension

What happens in the earlier bearing homework problem when Ri ! Ro?In boundary-layer theory we stretched a variable in one dimension. If we have a PDE instead of an ODE we(usually) expand only one-dimension...

Consider the following two-dimensional, steady, incompressible ow.

We scale x � L; y � H; � = H=L, where H is some representative (say average) h. This scaling impliesthat @=@y � @=@x: To satisfy continuity in a non-trivial way we scale u � U; y � �U; � = H=L.The scaled equations using a viscous pressure scale become:

(�U2=L)(uux + vuy) = �(P=L)px + (�U=H2)(�2uxx + uyy)

(��U2=L)(uvx + vvy) = �(P=H)py + (�U�=H2)(�2vxx + vyy)

ux + vy = 0

u(0)� 1 = v(1) = u(h) = v(h) = 0

simplifying(�UH2=L�)(uux + vuy) = �(PH2=LU�)px + �2uxx + uyy

27

Note that we must choose P to balance uyy, otherwise the velocity is linear and we cannot satisfy continuity.On choosing the pressure scale P and Re appropriately.

�2Re(uux + vuy) = �px + �2uxx + uyy

similarly�2Re(uux + vuy) = �1=�2py + �2vxx + vyy

We have chosen Re so that only even powers of � appear. We of course hope that the newly de�nedRe = O(1). Hence, we expand

u = u0 + u2�2 + :::

v = v0 + v2�2 + :::

p = p0 + p2�2 + :::

at lowest order, the y�momentum equation says that p0 = p0(x), hence we can integrate the x�momentumequation that becomes an ODE to give

u0 = p0xy2

2+A(x)y +B(x)

Because A;B are functions of x, they can satisfy the local vertical boundary conditions.

u0 = p0xy2

2��U

h+ p0x

h

2

�y + U

Once u0 is known we can post-process to �nd p0 using

Q0 =

Z h

0

dy = p0xh3

6��U

h+ p0x

h

2

�h2

2+ Uh+ C;

which when integrated gives

p0 = 6

Zdx

h2� 12Q

Zdx

h3;

which is the Reynold's Lubrication equation.This technique is quite di�erent and generally more applicable than perturbing the boundary about

parallel surfaces. It is singular in a way. Think of the bearing problem. In the limit of gap thickness goingto zero, at lowest order we try to apply two di�erent boundary conditions on the same place! It is betterto perhaps use a curvilinear coordinate system x = r � Ri y = Ri�. Then to lowest order you developthe equation in Cartesian coordinates! The restriction is on slope not on small de ections. This is oftenviolated...example is using shallow water equations about an obstacle.

We proceed to higher orders. We can see that the lowest-order lubrication approximation breaks downwhen �2p2 is no longer � p0. See:

Schumack, M., W.W. Schultz, J.-B. Chung and E. Kannatey-Asibu (1991), \Analysis of uid ow undera grinding wheel", J. Eng. for Industry 113:190-197.They use streamfunction, also �nd correction due to inertia, surface roughness, etc.

What have we lost? The ability to apply boundary conditions at the ends. These are boundary layersthat must be \restretched" to recover the original equations of motion{numerical. Remember that we saidthat we expect @=@y � @=@x: But p0 = p0(x)! This often turns out that way, as in shallow water equations,viscous jets...

Schultz, W.W. and S.H. Davis, \One-dimensional liquid �bers", J. Rheol. 26:331-345, 1982.

|||||||||||HWK no. 6: The �n equation, simple linear theory. Find limitations on slope and Bi = hL=k; L = A=Pfrom the requirement �2T2 � T0. It is possible to have both heat transfer from the side (Bi 6= 0) and a sloping

28

(Y = ax+ b) simultaneously, but for simplicity you may want to do them one at a time.

The two-dimensional steady equations of conduction and convection for a symmetric �n of slowly-varyingheight Y are:

�2Txx + Tyy = 0 (21a)

�(��2YxTx + Ty)(1 + �2Y 2x )�1=2 = Bi T on y = �Y (x) (21b)

where the ambient temperature is 0, we use the average temperature at the base of the �n (x = 0) as thetemperature scale, we use the nondimensional height at the base and the length of the �n to scale the y andx directions, respectively. The slenderness parameter � � 1 is the ratio of these scales. The Biot numberis Bi = hY (0)=k, where h is the heat transfer coe�cient and k is the thermal conductivity. Note that Binumber must be rescaled in order for the problem to be one-dimensional to leading order. In other words,the convection should not be too strong compared to the conduction. Since � appears in even orders, itbehhoves us to introduce an �2 into the modi�ed Biot number.

|||||||||||

Slender Body Theory

Ref: Van Dyke and ??

Another example where di�erent length scales are chosen is in external ow about a slender body. Con-sider potential ow about a symmetrical body

�� = 0

�y=(U + �x) = �T 0(x) on y = ��T5�! 0 as x2 + y2 !1

where � in the potential for the disturbance to uniform ow U . How fast the disturbance potential goes tozero at in�nity determines if there is any circulation or net out ow from the body. Here, because we havea closed, symmetric body at 0 angle of attack, we expect no circulation and a balance of sources and sinks.Note that both �T and �T 0 are O(�) and hence we must expand in all powers of �. Near the body we stretch� = y=� and let

� � �I = ��1(x; �) + �2�2(x; �) + :::

Then the �rst two orders give�1�� = �2�� = 0

with higher orders given by�n�� = ��(n�2)xx n = 3; 4; 5; :::

The boundary conditions on � = T are

�1� = �2� � T 0U = �n� � T 0�(n�2)x = 0

These have solutions�1 = B1(x)

29

�2 = UT 0� +B2(x)

�3 = �1

2�2B001 (x) + �(TB01)

0 +B3(x)

�4 = �1

6U�3T 000 � 1

2�2B002 (x) + �(

1

2T 2T 00U + TB02)

0 +B4(x)

Combining and writing in terms of y:

~� = �[B1 + yT 0U � 1

2B001 y

2 � 1

6T 000Uy3 +O(y4)] + �2[B2 + (TB01)

0y � 1

2B002 y

2 +O(y3)] +O(�3):

Now far from the body� � �O = ��1(x; y) + �2�2(x; y) + :::

As seen from the outer ow, the body collapses onto a line, so we replace it by a line of singularities. UsingCauchy's theorem in the form

f(zo) =1

2�i

If(z)

z � zodz

wn = �nx � i�ny =1

2�i

Iwn(�; 0)

� � zd�

where w is the complex velocity and z = x+ iy. Integration is clockwise because ow is exterior. Then

wn =1

2�i

"Z l

0

wn(�; 0+)

� � zd� +

Z 0

l

wn(�; 0�)

� � zd�

#

=1

�

Z l

0

�ny (�; 0+)

z � �d�

since by symmetry�nx(�; 0

+) = �nx(�; 0�)

and�ny (�; O

+) = ��ny (�; O�):

These solutions satisfy the outer di�erential equations as well as the in�nite boundary conditions. The innerexpansion at y = 0+ (but still � !1) gives

~�y(x; 0+) = �T 0U + �2(TB01)

0 + �3(1

2T 2T 00U + TB02)

0 +O(�4)

Then from matching�1y (x; 0

+) = T 0U

�2y (x; 0+) = (TB01)

0

�3y (x; 0+) = (

1

2T 2T 00U + TB02)

0

But, it also gives~�nx(x; 0

+) = B0n

or upon integrating~�n(x; 0

+) = Bn;

where the constant of integration is ignored. And combining with the outer integral equation result withevaluation only at the surface gives

B01(z) =1

�

limy ! 0+

Z l

0

T 0(�)

z � �d�

B02(z) =1

�

limy ! 0+

Z l

0

T (�)B01(�)

z � �d�

30

:::

where we use the principal part and we note that the integrand is real. This end result is called von Karman'smethod although he did not construct the solution so formally.

Van Dyke uses a di�erent method. He only poses an outer solution and expands his result about y = 0.Hence using a power series expansion for the disturbance potential and a Taylor series expansion about y = 0gives (we have set U = 1):

�y(x;��T ) = ��1y (x;�0) + �2[�2y (x;�0)� T�1yy (x;�0)]

+�3[�3y (x;�0)� T�2yy (x;�0) +T 2

2�1yyy (x;�0)] + :::

and similarly for �x.We thus have a solution provided T 0; T 00; ::: are O(1). Near a blunt nose this is not the case. Suppose at

x = 0 x � y2, thenT =

p2ax(1 + bx+ :::)

T 0 =pa=2x(1 + 3bx+ :::)

In this region we must stretch both x and y, and match to the solution we have just derived. The procedurecan become complicated! From experience with construction of solutions with simple sources, i.e. RankineBodies, we know that the answer is to have no sources near the leading edge!

Example: Flow about ellipse

T (x) =p1� x2

Has a major axis of 2 and a minor axis of 2 � At lowest order we get (drop hats)

�1x(x; 0+) =

1

�

Z TE

LE

�1y (�; 0+)

x� �d� =

1

�

Z TE

LE

T 0(�; 0+)

x� �d�

= � 1

�

Z 1

�1

�p1� �2(x� �)

d� =

(1 x2 < 1

1�q

x2

x2�1 x2 > 1

Where a table of defenite integrals has been used! More generally, the analytic solution that goes withthis result on the boundary is �1x � i�1y = 1 � z=

pz2 � 1 (Jones and Cohen, 1960). This has the typical

thin airfoil result that the velocity does not vary along the foil and decays like 1=z away from the foil. Whencombined with the basic free-stream velocity, it becomes q � 1+�. It only has the right value at x = 0. Notethat the velocities at the stagnation points at x = �1 are not satis�ed. Going to higher order just makesthese nonuniformities worse. On the airfoil the speed q =

pu2 + v2 is important to �nd pressure for lift and

drag. The value of v = �y is determined from the tangency condition. At higher order we get

q=U � 1 + �� 1

2�2

x2

1� x2+ :::

which has \worse" nonuniformities at the stagnation points.

Multiplicative correction for round edges (4.8 Van Dyke)Ref: Lighthill (1951), Aero Quart. 3, 193-210.

From conformal mapping, the exact surface speed of a parabola with radius of curvature � is

q=U =

rs

s+ �2=2;

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where s is the distance from the leading edge. The ow around a Rankine body would do as well. For small�2=s (as in the outer expansion for airfoil theory) we obtain

~q=U = 1� �2

4s+ :::

This result has a simple pole at s = 0 that can \cancel" the pole at x = �1 in the outer expansion. Theratio of the last two expressions is

q

~q=

rs

s+ �2=2(1� �2

4s+ :::)�1 =

rs

s+ �2=2(1 +

�2

4s+ :::)

This multiplying factor has the right behavior at the stagnation point and approaches a constant 1 in theouter region when �! 0; s 6= 0. Multiplying it with the outer solution gives the uniformly valid solution

q=U =

s1 + x

1 + x+ �2=2

�1 + �+

�2

4

1� 2x

1� x+ :::

�

Gives a very good result as shown in �g 4.3 of Van Dyke, even for � = 0:5.|||||||||||

External Low Reynolds FlowSince the smallest parameter doesn't appear in front of the highest derivative, this singular perturbation

must come from boundary conditions at in�nity. Consider the ow about a cylinder or sphere:

The ow is steady and either two dimensional (@=@z = 0 for cylinder) or axisymmetric (@=@� = 0 for sphere).In either case we solve in terms of stream function:

u =1

r2 sin � � v = � 1

r sin � r (axisymmetric)

u =1

r � v = � r (2D)

For the cylinder, from the uniform ow we try the Ansatz = f(r) sin �:

�2 =

�d2

dr2+1

r

d

dr� 1

r2

�2

f = 0

which has the solution

f = Ar3 +Br ln r + Cr +D

r

To satisfy (r; �) � UR sin � as r ! 1 requires A = B = C � U = 0 and we have only one constant tosatisfy two boundary conditions on the cylinder surface. ...

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