THE MOST POWERFUL TOOL AVAILABLE to the (APPLIED)

MATHEMATICIAN?

Robin Johnson

(Newcastle University)

Outline

background ideas

algebraic problems

differential equations

applications

Preamble

Mathematicians (but perhaps mainly applied) – and physicists & engineers – use rather specific families of skills, such as

• algebra, integration

• classical methods for solving DEs

• (also complex variables, group theory ...)

You will be familiar with those ideas & techniques that are relevant to you.

3

Typically, physical systems are represented by DEs

– but rarely standard ones;

– cannot use familiar solution-methods.

BUT not unusual for such problems to contain a small parameter e.g.

Can we take advantage of this? YES!4

celestial mechanics: small mass ratio fluid mechanics: 1/(large Reynolds No.)

Leads to the idea of asymptotic expansions (a.e.s), based on the small parameter, and to singular perturbation theory.

Not the forum for precise definitions and careful developments, but

can give an overview of the ideas

show some techniques and properties

discuss some elementary examples

indicate what can be done5

An Example (to set the scene)

Given ( ; ) 1 e xf x x 0x with and .0

Note that . 14

(0; ) 2 2 1 ...f Approximate (asymptotic) representation requires two ‘sizes’ of x :

x fixed (“= O(1)”) as 0 ; x X , X = O(1) as 0 .

6

To see this, we expand appropriately:

( ; ) 1 e xf x x

O(1):x 12

1 ~ 1 11

f x xx

, O(1) :x X X

12

(1 )( ; ) ~ 1 e 1

1 e

XX

Xf X

Note: x = 0 in first gives - wrong!12

~ 1

but X = 0 in second: - correct. 14

~ 2 (1 )7

Matching

N.B. Two expansions are required here to cover the domain – a singular perturbation problem.

The two a.e.s are directly related:

1 12 2

1 1 1 11 1

x Xx X

1 12 2

~ 1 X

1 12 2

(1 )1 e 1 1 e 1

1 e 1 e

X xX x

X x

12

~ 1 ( )x 8

The two ‘expansions of expansions’ agree precisely (to this order);

they are said to ‘match’ – a fundamental property of a.e.s with a parameter: the matching principle.

Plotted for decreasing ε

9

Graph of our example:

( ; ) 1 e xf x x

Breakdown

Another important property of a singular perturbation problem: breakdown of a.e.s.

12

1( ; ) ~ 1 e 1

1 eX

X

Xf X

E.g.

which is valid for X = O(1), and correct on X = 0.

The expansion ‘breaks down’ (‘blows up’) where two terms become the same size; here

10

- the variable used in the other a.e. !

O(1) i.e. O(1)X x (‘large’ X)

Introductory examples

Start with a simple exercise: quadratic equation

2( ; ) 1 0.f x x x

Treat the expression as a function to be expanded:

O(1) : ~ 1x f x so (approx.) root .~ 1x

Second root? Can arise only for large x.

11

Seek better approx.:1

~ 1 nn

nx a

Breakdown of the ‘a.e.’:2( ; ) 1f x x x

is where i.e. 2 O( )x x 1O( )x

so rescale: to give x X 1 2 1( ; ) ( ) ( ; ) 0.f X X X F X

O(1):X 2~F X X

so (approx.) roots , but X = 0 corresponds to a breakdown: .

0, 1X O( ) :X X x

12Roots are . ~ 1, 1x

Another algebraic example

Consider , 2 5 3( ; ) 1 0f z z z

Breakdown where : rescale2 5 3O( )z z z

to give 3 3 5 3 3( ; ) ( ; ) 0.f F

3O(1) : ~ 1z f z : (approx.) roots2~ 1, , .z

then

5 3O(1) : ~F : relevant (approx.) rootsi

13

Roots:2 i i~ 1, , , , .z

Ordinary differential equations

First example, to show ideas & methods:

22 1 2 ; 1; (1; ) 1.y y xy x x y

ODE implies that for0

( ; ) ~ ( )nn

n

y x y x

O(1),x

so2

0 0 1 1 02 1 2 ; 2 0, etc.,y y x y y xy

with 0 (1) 1; (1) 0, 1.ny y n

14Can now solve the sequence of problems.

This procedure gives

2 3 2(1 )18

( ; ) ~ 3 6( ) 4 e xy x x x x x

for x = O(1), but breaks down where3 O( )x x

so and then .O(1 )x O(1 )y

Rescale: , to give the ODE, ( ; )x X y Y X

22 2Y Y XY X and no b.c.!

15

Seek a solution 2

0( ; ) ~ ( )n

nn

Y X Y X

then so2

0 02 2XY Y X 20

1 11 2 .Y X

X X

Invoke the Matching Principle:2 3 2(1 )1

8( ; ) ~ 3 6( ) 4 e xy x x x x x

gives 312

~Y X X , and

0( ; ) ~ ( )Y X Y X

2 2 412

1~ 1 (1 ) .y x x

x

gives

Matching accomplished with the positive sign.16

Another type of ODE

A ‘boundary-layer’ problem:

26 2 , 0 1,y y xy x x

with (0; ) 2, (1; ) 3.y y

N.B. Boundary layer – a scaling – is near x = 1.

For x away from x = 1 :0

( ; ) ~ ( ),nn

ny x y x

with 0(0) 2; (0) 0, 1.ny y n 17

We obtain, for the first two terms in the a.e. :

2 2 3( ; ) ~ 2 (2 ) 2 8y x x x x

which gives on x = 1 – not correct.~ 1 9y

Rescale: with 1x X (1 ; ) ( ; ),y X Y X

then 2 26 (1 ) 2 (1 )Y Y X Y X

with (0; ) 3.Y

Write 00

( ; ) ~ ( ); (0) 3; (0) 0, 1.nn n

nY X Y X Y Y n

18

The first two terms, satisfying the given boundary condition, are

~ (3 )e 2 (1 e ) .X XY A A X B

(A and B arb. consts.)

Match :2 2 3( ; ) ~ 2 (2 ) 2 8y x x x x

gives ~ 1 (2 9),Y X

and above gives ~ 2(1 ) ;y A x B

a.e.s match with the choice 1, 9.A B 19

One further technique

Probably the most powerful & useful: the method of multiple scales.Describe idea by an example:

3(2 ) 0, 0x x x x t (a Duffing equation with damping; λ > 0, constant)

with (0; ) 1, (0; ) 0.x x

Oscillation described by a ‘fast’ scale – carrier wave, and a ‘slow’ scale – amplitude modulation.

20

An example of a modulated wave:

In this approach, we use both scales at the same time!

We introduce (fast)2

( ) ~ 1 nn

nT t t

and (slow).t

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Impose periodicity in T, and uniformity in τ(as ).

Now seek a solution

0

( ; ) ( , ; ) ~ ( , );nn

n

x t X T X T

the equation for X becomes:

2 2 32 2 ( ) 0.TT T TX X X X X X X

22

Then 0 0 0;TTX X 3

1 1 0 0 02 2 0,TT T TX X X X X

and so on (together with the initial data).

Solving gives

0 0 0( , ) ( ) cos ( )X T A T

and then periodicity of , satisfying the initial data, requires

1( , )X T

0 ( ) eA and23

0 16( ) (e 1).

This leaves

E.g. boundedness of requires2X 22 2.

23

3 2311 1 1 32 16( , ) ( ) cos ( ) e cos 3 (e 1) ,X T A T T

and so on.

Comment

Is it consistent to treat T and τ as independent variables? (They are both proportional to t !)

If a uniformly valid solution exists, then it holds for ; thus it will be valid on any line in the first quadrant of (T,τ)-space.

0, 0T

bounded (and uniformly valid) periodic (and bounded) ( ) T

T

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And ever onwards

1. These ideas go over, directly, to PDEs.

Asymptotic expansions take the same form, but now with coefficients that depend on more than one variable cf. multiple scales for ODEs.

Breakdown (scaling) occurs, typically, in one variable, as all the others remain O(1).

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2. Relevant scalings are usually deduced directly

from the governing differential equation(s).

Some Applications – a small selection

1. Gas-lubricated slider bearing

Based on Reynolds’ thin-layer equations, this describes the pressure (p) in a thin film of gas between two (non-parallel) surfaces:

3 , 0 1, (0; ) (1; ) 1h pp hp x p p

where h(x) is the gap between the surfaces, and ε is the (small) inverse bearing number.

This is a boundary-layer problem, with the boundary layer near x = 1.

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2. Restricted 3-body problem

(It was for this type of problem that Poincaré first developed his asymptotic methods.)

The ‘restricted’ problem is one for which one of the masses is far smaller than the other two.

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In a frame centred on one of the larger masses, we obtain2 2

2 3 3 3 3 2 3d d

andd dt t

x x x y x y y y

x x y x y y

(position of small mass: x, of second large mass: y).

Solution for small μ is a singular perturbation problem if

O( ) x y

i.e. the small mass is close to the second large mass.

Introduce to give x y X

23 3 3 3

(1 ) and

X y X y yX y

X y X y y

and then near to the time of close encounter.

0t t T

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Expand each and match.

3. Michaelis-Menten kinetics

This is a model for the kinetics of enzymes, describing the conversion of a substrate (x) into a product, via a substrate-enzyme complex (y) :

Equations exhibit a boundary-layer structure in y but not in x ;

d d( ) and ( )

d d

x yx x y x x y

t t

with x = 1 and y = 0 at t = 0, for . 0

a convenient approach is to use multiple scales.

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Introduce 1and ( )T t g t

and seek an asymptotic solution for

( , ; ) and ( , ; ).x X T y Y T

Problem now becomes

1 ( ) ; ( )T Tg X X X X Y g Y Y X X Y

with (0,0; ) 1, (0,0; ) 0.X Y

Obtain, for example, 00( ) ( ) d

Tg T X t t

where , the solution of0~X X 0 0ln 1 .X X T 30

4. Josephson junction

This junction, between two superconductors which are separated by a thin insulator, can produce an AC current when a DC voltage is applied – by the tunnelling effect.

An equation that models an aspect of this is

(1 cos ) sinu a u u u b

with , for the voltage u(t;ε).(0; ) (0; ) 0u u

Relevant solution is u = εU(T,τ;ε), using multiple scales.

31

Introduce

τ = εt and 2

, ~ 1 nn

nT t

and then we find that

12

~ 1 (cos )exp (1 )u b T a

for 1.a

32

Higher-order terms can be found directly, and in the process we determine each .n

5. Fluid mechanics I: water waves

The equations for the classical (1-D) inviscid water-wave problem:

( ) ;zu u uu wu p

2[ ( )] ;z zw w uw ww p

0,zu w

with & ( ) on ( , )p h w h h uh z h

and 0 on .w z b

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These are written in suitable variables, with two parameters: δ and ε.

The governing equations are essentially an elliptic system, but the surface b.c.s produce a hyperbolic problem for the surface profile, z = h.

This problem can be analysed, for example, for

0 (small amplitude waves) with δ fixed

0 (long waves) with ε fixed

0 and 0 (small amplitude, long waves)

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6. Fluid mechanics II:

The appropriate form of the Navier-Stokes equation, mass conservation, etc., is

1( );x y x xx yyuu vu p R u u 1( );x y y xx yyuv vv p R v v

0,x yu v 2 2with ( ), 0, 0 asu U x v p x y

and 0 on the body,u v

viscous boundary layer

35for 2D, incompressible, steady flow.

The classical boundary layer, of thickness , is represented schematically as:

1 2O( ) asR R

However, at the trailing edge, where there is the necessary adjustment to the wake, we have a ‘triple-deck’ structure:

all described by matched asymptotics.37

Conclusions

We have

outlined the ideas and methods that underpin the use of asymptotic

expansions with parameters;

described, in particular, their rôle in the solution of differential equations;

38

mentioned a few classical examples.

The End