Chapter 6 PART B

Some Fundamental Procedures Simplification, Dimensional Analysis, Illustrated on Ordinary Differential and Scaling Equation

6.1.1 The Basic Simplification Procedure

Basic simplification procedure: 1. Identify relatively small terms Section 6.1 the smallness is gauged relative to other terms.

2. Delete small terms and solve the simplified problem. The Basic Simplification Procedure 3. Check for consistency. use the simplified solution to check the deleted are small.

~ : seems to be approximately equal to : is approximately equal to

6.1.1 The Basic Simplification Procedure 6.1.1 The Basic Simplification Procedure Example 1. The projectile problem. The acceleration on the earth surface is g Mm R2 *** 2 mg G2 gR GM F mg 2 x x t;,, V R g R R x* v** t x*: distance from earth surface Therefore, we have the IBV problem t*: time d2 x * gR 2 * V: initial velocity dx 0 * GMm *2 2 V x 0 0 F dt * * R: radius of earth (6378.1 KM) x** t 2 R x dt R x* g: gravity acceleration Assuming x* R M,m: mass of earth and ball The Simplified IBV problem now reads d2 x * dx* 0 g V x* 0 0 dt*2 dt* * * 1 *2 * dx * Solution x gt Vt * gt V 2 dt Genuine * * 2 Newton's law of universal gravitation at tVg / , the maximum xVgRmax / 2 consistency 6.1.1 The Basic Simplification Procedure 6.1.2 Two chastening examples Example 2. x+10y=21, 5x+y=7 Example 3. 0.01xy 0.1, xy 101 11 solution x = 1, y = 2 solution x90, y 1

1. Identify small term 1x+10y=21, 5x+1y=7 Simplification 0.01xy 0.1, xy 101 11 1 < 10 Approximated y~ 0.1, x ~ 11 101*0.1 0.9 2. Delete small terms 1x+10y=21, 5x+1y=7 solution Simpilified solution y~2.1, x~0.98 Consistency 0.01xy 0.1, xy 101 11 check 3. Check for consistency 1x+10y=21 0.01 0.9 0.1 0.109 0.9 101 0.1 11 1 0.98 10 2.1 21.98 0.009 0.1 0.98 << 21 Almost Perfect Consistency ! genuine consistency But it is wrong! Why?

6.1.2 Two chastening examples 6.1.2 Two chastening examples 0.9 0.1 11 Introducing a small parameter , and rewrite as x y 1 101 1 101 xy0.1, xy 101 11 the exact solution x 0 0.9 y 0.01 1 y 0 0.1 0.9 0.1 11 x y , 1/101 1 101 1 101 0.0099 0.0099 The previous simplified solutions actually are x 0.01 90

x0 0.9 y 0 0.1 as 0 x0.01 90 x 0 0.9 y0.01 1 y 0 0.1 In simplified solution, we see But in fact, it is not Our problem is whether apparent consistency genuine consistency 0.01x 0 0.01x 0.01 x x0 y y 0 as 0.01 0.09 as =0 0.9 as =0.01 y 0 y 0.01 0.01xy 0.1, xy 101 11

6.1.2 Two chastening examples 6.1.2 Two chastening examples Example 4. But if there is a tiny numerical error like fxxxx1 2 3 x 20 x20 210 x 19 fxx20210 x 19 with = 2 23 We have zeros at the 20 positive integers • the first 10 zeros are almost unaltered. • the rest 10 zeros are radically altered to five pairs of complex conjugate roots

10i 0.64, 12 ii 1.7, 14 2.5, 17 i 2.8, 20 i 19 • apparent and genuine consistency verified

x19210 x 19 or x 0 19 210 x 0 19 as = 2 23 • but some roots are still wrong 6.1.3 Conditioning and sensitivity 6.1.3 Conditioning and sensitivity ill conditioned problem ( ): What is large or what is small ? • solution is sensitive to a neglected term T.

The only secure statement: Consider Temperature: 0o C=273.15K (absolute temp.) • No apparent consistency means poor approximation. T = 1o C. T =1.1 o C. error: 10% To see this, let true approx • x the true solution = 274.15K. =274.25K. error: 0.036% • the approximate solution with term T neglected. • ( ) small, apparently consistent • ( ) small, genuinely consistent To decide larger or small, • scientific grounds If the solution is not sensitive to the neglect of T, • must be independent of the units employed smallTx xx small Tx • Where is the zero point

6.1.4 Zeros of a function 6.1.4 Zeros of a function Let ) a zero of f(x, ) , i.e. In the previous example fxx,20 210 x 19 0 f x, 0 f x , 0

fxx,020 210 0 x 19 0 f x,0 0 = 0 is a zero of the simplified equation 20 19 19 f x,0 0 fxx, 210 x xr fx, r

f(x, ) 20 19 19 x f x,0 x 210 0 x x g f x,0 g g r

6.1.4 Zeros of a function 6.1.4 Zeros of a function Define equation error fx, fx ,0 We also have full Taylor expansion 0 0 2 fx, fx 0 ,0 fxx 0 fO Genuine equation error fx00 f 0 0 gfx, fx ,0 To the first order x g fxx 0 f or g r 0 fx0 ,0 xxO 0 2 r f x x 0 Therefore the error of the solution : 0 fx0 0 O 2 f r g x hx x0 x 0 f0 f 0 f 0 Residual equation error x x x the error h depends on : f 2 rfx0 , fx 0 ,0 fx 0 ,0 O • the residual r or the genuine g x 0 • the condition f0 O 2 6.1.4 Zeros of a function 6.1.4 Zeros of a function In example 4 f x, f x,0 fxx20210 x 19 fxx20210 x 19 with = 2 23 rfx0 , f 0 19 For later zeros, h/x x h1 r 1 x h xf0 x f 0 x increases because: x 0 x g r x x >1, varies

0 gfx,0 fxx 0 r hx x0 x 0 h1 r f 0 0 x xfx x ill conditioned

0 if fx 1 Small residual r , big error in solution h

6.1.4 Zeros of a function 6.1.5 Second order differential equations Newton’s method (also Newton–Raphson method) Consider an initial value problem x(t, ) , f x 0 d To find the root of ftxxx,,,, 0; x 0 Ax , 0 B , 1, dt

If xn is an approximate root, we have Taylor expansion 2 fx fxn fx x n xx n Oxx n 0 An approximate zero ftx, 0 , x 0 , x 0 ,0 0 f x xx x n n1 n Genuine equation error fx x n g ftxxx, , , ,0 fx fx fx O 2 f xn is the residual 2 3 4 0

r Residual equation error hx x 0 0 f rftxxx, 0 , 0 , 0 , f O 2 x 5 0

6.1.5 Second order differential equations 6.1.5 Second order differential equations We see Example: non-dimensionalized projectile problem ftxxx,,,, ftx ,0,0,0, x x 2 2 xx1 0; x 0 0, x 0 1 fx2 fx 3 fx 4 f 5 O 0 2 Thus to the first order V /Rg is small. fx0 0 fx 0 0 fx 0 0 f 0 or g r 2 3 4 5 In this case 2 Define ftxx, ,0, , x 1 x 0 0 0 0 F t fff2,, 3 4 F t fff0, 0 , 0 0,0,1 h txx0 , xx 0 , xx 0 xxx0 , 0 , 0 O 2 2 3 4 0 0 0 F h r h t xxx,, r So, r Fh Fhcos Fh , Fh h rf2 xt0 F 5 0 6.1.5 Second order differential equations 6.1.5 Second order differential equations F h r Example: nonlinear pendulum 1/2sin 1/2 0; 0 0 2 x2 xtO 0 1, 0 0, 0 1

0 0 2 With (t) angular displacement. is the initial x2 max xtO displacement.

Zeroth approximation:

1/2sin 1/2 00 0 0 0

0 cos t

6.1.5 Second order differential equations 6.1.5 Second order differential equations Forcing term f t, ,0, ,1/2 sin 1/2 0

0 0 0 F t fff2, 3 , 4 1,0,1 h t 0,, 0 0

0 0 0 1/2 1/2 1/2 3 0sin cos 0 r f5 lim 0 23/2 6 3 F h r 0 0 0 6 We obtained DE on 1 0 0 cos3 t 00 0, 0 0 0 6

6.2.0 Dimensional analysis

Many of those who have taught dimensional analysis have realized that it has suffered an unfortunate fate. Section 6.2 The idea of dimensional analysis is based is very simple: physical laws do not depend on arbitrarily chosen basic units of measurement. Dimension Analysis An important conclusion: the functions that express physical laws must possess a certain fundamental property, which in mathematics is called symmetry.

Dimensional analysis, researchers have been able to obtain remarkably deep results that have sometimes changed entire branches of science. 6.2.0 Dimensional analysis 6.2.0 Dimensional analysis Proof of Pythagorean Theorem Partition the right triangle into two right triangles • c Physics: angle and side determine a right triangle. A,,, c A b A a • a, b, are functions of ,c. 1 2 • Dimension [a]=[b]=[c]=L [A]=L2 cf2 bf 2 af 2 with dimensional reasoning, a, c cf a A1 A2 A1 A2 c2 2 +b2 b, c cfb ,c f Pythagorean Theorem A, c cf2 Pythagorean Theorem

6.2.0 Dimensional analysis 6.2.0 Dimensional analysis

Dimensional Analysis refers to the physical nature of The international system of units (SI) the quantity (Dimension) and the type of unit used to specify it.

Distance has dimension L. Area has dimension L2. Volume has dimension L3. Time has dimension T. Speed has dimension L/T

6.2.0 Dimensional analysis 6.2.0 Dimensional analysis X = x U = x’ U’ Secondary dimension size =10 cm X : physical quantity Velocity [v -1 =100 mm x, x’ : measurement -2 U, U’ : unit Force [F -2 -1 T-2 Mass Density [ M L-3 ……… • The measured value of a quantity depends on the ……… unit of measurement. • Such dependency is called dimensional. • Otherwise, called dimensionless [Physical Quantity] = Ma Lb Tc 6.2.1 Nondimensionalization of a differential equation 6.2.1 Nondimensionalization of a differential equation The projectile problem: Step A. List all parameters and variables, together 2 * 2 * dx gR dx 0 * with their dimensions *2 2 * V x 0 0 dt R x* dt

Solution x*** xtgRV;,,

x*: distance from earth surface t*: time V: initial velocity R: radius of earth (6378.1 KM) g: gravity acceleration A two-step procedure for nondimensionalization

6.2.1 Nondimensionalization of a differential equation 6.2.1 Nondimensionalization of a differential equation Step B. Construct new dimensionless variables. Remarks: x* t* (i) Intrinsic reference quantities are defined to be gauges in a y given problem. L T (ii) We have different choices of Intrinsic reference quantities L R intrinsic length T RV 1 intrinsic time intrinsic length L intrinsic time T • the two new variables are dimensionless 1 R RV 1 • their numerical value is the same whatever 1 unit of measurement is used. 2 R Rg if R miles, x* then y 3 V2 g 1 Vg 1 If R x* then y

6.2.1 Nondimensionalization of a differential equation 6.2.1 Nondimensionalization of a differential equation x** t x** t 1 2 R Vt R2 gt 1. y , 1 2. z , 1 R RV R Rg 1 t RV 1 1 2 * 2 2 2 t~ Rg dx R dyVdy dx2 * Rdz 2 dz 2 *2 2 2 2 2 dt R V d R d *2 1 2g 2 dt Rg d1 d 1 gR2 gR 2 g gR2 gR 2 g * 2 2 2 R x RRy1 y 2 2 2 R x* RRz1 z

2 d y 1 y y ; d2 z 1 2 2 z z 1; d 1 y d 2 2 V 2 h 1 1 z V 2 h y0 0, y ' 0 1 max 1 max 1 gR R z0 0, z ' 0 gR R 6.2.1 Nondimensionalization of a differential equation 6.2.1 Nondimensionalization of a differential equation ** 1 2 1 x t hmax 2 V g • There exists many reference times, e.g. 3. x 2 1, 1 1 V g Vg 1 tmax Vg 1 1 T Vg T1 RV T2 Rg 3 * d2 x * V 2 g 1 dx d 2 x x g 0,1 • Transition between them, T h T dt*2 V 2 g 2 d 2 d 2 i j hmax gR2 gR 2 g 2 V 1 1 2 2 2 T RV Rg T R x* RRx1 x 1gR 2

2 V 1 1 2 T RV Vg T d x 1 x x ; 1gR 3 2 2 d 1 x 2 2 h V 1 1 V max T Rg Vg T x0 0, x ' 0 1 1 2 3 gR R gR

6.2.1 Nondimensionalization of a differential equation 6.2.1 Nondimensionalization of a differential equation f xt**, ; gRV , , 0 1. First combination xt**1 2 gRV3 4 5 dimensionless 1, 2 , 3 , 4 , 5 1,0,0, 1,0

3 5 *1 *0 0 1 0 * 1 * LT1 2 LT2 L 4 LT 1 xtgRV xR x x/ R

L1 3 4 5 T 22 3 5 1 1, 2 , 3 , 4 , 5 0,1,0, 1,1

*0 *1 0 1 1 * 1 * 1 1 3 4 5 0 xtgRV tRV t t/ RV

22 3 5 0

1, 2 , 3 , 4 , 5 0,0,1,1, 2

4 1 2 3 1,, 2 3 xt*0 *0 gRV 1 1 2 gRV 2 V2 / gR 5 22 3 3 degree of freedom

6.2.1 Nondimensionalization of a differential equation 6.2.1 Nondimensionalization of a differential equation 2. Second combination 3. Third combination , , , , 1,0,1,0 2 1, 2 , 3 , 4 , 5 1,0,0, 1,0 1 2 3 4 5 xtgRV*1 *0 0 1 0 xR * 1 x x* / R xtgRV*1 *0 1 0 2 xgV * 2 x x*/ V 2 g 1

1 1 ,,,, 0,1,, ,0 1, 2 , 3 , 4 , 5 0,1,1,0, 1 1 2 3 4 5 2 2 *0 *1 1 0 1 * 1 * 1 *0 *1 1/2 1/2 0 * 1/2 1/2 x t g R V t gV t t/ Vg xtgR V tgR tt*/ gR 1

1, 2 , 3 , 4 , 5 0,0,1,1, 2 1, 2 , 3 , 4 , 5 0,0,1,1, 2 *0 *0 1 1 2 2 2 x*0 t *0 g 1 R 1 V 2 gRV 2 V2 / gR xt gRV gRV V/ gR 6.2.1 Nondimensionalization of a differential equation 6.2.2 Nondimensionalization of a functional relationship

Via dimensionless process • The following relation obtained from dimensionless DE. *** x xtgRV;,, * 2 tM V 1 f x x ; RV gR • Now we try to deduce it without DE We see many benefits. And one can say that the time to reach maximum height is How? M f

t* V 2 M f RV1 gR

6.2.2 Nondimensionalization of a functional relationship 6.2.2 Nondimensionalization of a functional relationship (1) Assume that a dimensionless quantity of interest is a (4) To require a dimensionless parameter function of the dimensional parameters. 0, 2 0 t* M V,g,R (5) Find the general solution in terms of arbitrary R1/2 g 1/2 constants c1, c2, ... , etc. (2) Construct a product of powers of the parameters in 2c , c , c the argument of 1 1 1 (6) Substitute back into . V gR c1 2c c c gR (3) Insert the dimensions of the various quantities. V1 gR 1 1 V 2 1 2 2 LT LT L L T gR 1, withc 1 1 where, [x] is the dimension of x V 2

6.2.2 Nondimensionalization of a functional relationship 6.2.2 Nondimensionalization of a functional relationship

Finally, we have Buckingham Pi Theorem 2 t* gR V Historical Note M V,g,R • The Buckingham Pi Theorem puts the ‘method of Rg1/2 1/2 V 2 gR dimensions’ first proposed by Lord Rayleigh in his book “The Theory of Sound” (1877) on a solid theoretical gR gR basis, and is based on ideas of matrix algebra and t* R 1/2 g 1/2 RR1/2 VV 1 g 1/2 M V 2 V 2 concept of the ‘rank’ of non-square matrices which you may see in math classes. V2 gR RV 1 • Although it is credited to E. Buckingham (1914), in fact, Rg V 2 White points out that the theorem has also appeared RV 1 1/ earlier in independent publications by A. Vaschy (1892) V 2 and D. Riabouchinsky (1911). * 1 1 tM RV f RV f gR On Physically Similar Systems; Illustrations of the Use of Dimensional Equations. Buckingham, Phys. Rev. 4, 345 (1914). 6.2.2 Nondimensionalization of a functional relationship 6.2.2 Nondimensionalization of a functional relationship Consider a physical relationship: • The dimensions of group b1, b2,… bm is expressed as

a f(,, a1 a 2 ak ;,, b 1 b 2 b m ) p1 p 2 pk b1 a 1 a 2 ak • a is the quantity under investigating q1 q 2 qk • Governing parameters a1, a2,…, ak , b1, b2,… bm b2 a 1 a 2 ak • Group a1, a2,…, ak have independent dimensions. • Group b1, b2,… bm is dependent on Group a1, a2,…, ak b as1 a s 2 a sk For example m1 2 k

a1 length [a1 -1 • The dimension of a can be expressed by the first group a2 velocity [a2 a a 2 -2 m1 m 2 mk 3 energy [ 3 T a a a a -2 -1 2 0 1 2 k b1 acceleration [b1 a1] [a2] [a3] -1 -1 -2 -1 1 b2 viscosity [b2] M L T [a1] [a2] [a3]

6.2.2 Nondimensionalization of a functional relationship 6.2.2 Nondimensionalization of a functional relationship Let’s change the units, • The fundamental physical covariance principle a a,,, a a a a claims that all physical laws can be represented in 1 1 1 2 2 2 k kk a form equally valid for all observers.

m1 m 2 mk • This principle is valid for observers using different a a1 a 2 ak magnitudes of basic units. X = x U = x’ U’ mm1 2 mk x’= x =U/U’ We may rewrite a1 2 k a X : physical quantity With the same reason, a f(,,, a1 a 2 ak ;,,,) b 1 b 2 b m x, x’ : measurement p1 pk s 1 sk pp1 2 pk U, U’ : unit f(,,;1 a 1kk a 1 k b 1 ,, 1 km b ) b1 1 2k b 1 mm1 2 mk 1 2 k a ss1 2 sk mm1 2 mk f(,, a a a ;,, b b b ) bm1 2 km b 1 2k 1 2 k 1 2 m

6.2.2 Nondimensionalization of a functional relationship 6.2.2 Nondimensionalization of a functional relationship 1 1 1 In all, Let’s choose ,,, 1a 2 ak a 1 2 k a f(,, a a a ;,, b b b ) a 1 2k 1 2 m then a mm1 2 mk a1 a 2 ak

b1 bm f (1,1, 1;1 , ,m ) f (1,,1;pp1 2 p 1 ,, ss 1 2 s 1 ) a1 a 2 ak a 1 a 2 a k now introduce the dimensionless parameters a (,, ) mm1 2 mk , 1 m a1 a 2 ak b b 1 m Buckingham Pi Theorem 1 pp1 2 pk,, m ss 1 2 sk a1 a 2 ak a 1 a 2 a k 6.2.2 Nondimensionalization of a functional relationship Example: in the flow of a fluid through a long cylindrical G. I. Taylor’s 1947 Analysis pipe, the pressure drop per length reads dp/ dx f (,,;) U D

2 2 • U the mean fluid velocity dp/ dx ML T • D the diameter of the pipe U LT1 DL • the fluid density 3 1 1 • the fluid viscosity ML MLT • The dimensions of , , are independent. • The dimensions of and dp/dx can be expressed as Published U.S. Atomic Bomb 1 3 1 1 UD LT LML MLT was 18 kiloton device dp/ dx U2 D 1 1 L2 T 2 L 1 ML 3 ML 2 T 2 U 2 dp/ dx D UD

Two assumptions: ab c RE t 1. The energy (E) was released in a small space. 1 2. The shock wave was spherical. a b c a c 1 RE t LMLT2 2 TMLb 3 R Rt,, E L1 2acba 3 T 2 M ac • R: the size of the fire ball (function of t) • t: time 1 2a 3 c 0 • E: energy ba2 0 a 1/5, b 2/5, c 1/5 • : density of the surrounding air a c 0 Let’s perform a dimensional analysis: 1/5 2/5 1/5 2/5 1/5 1/5 [ -3 [ 2T-2 1 REt RctE R5 • n dimensions the atomic energy E • Pi theorem: only one dimensionless group c5 t 2

6.2.2 Nondimensionalization of a functional relationship R (E/ )1/5 t2/5

log R t + 0.2 log(E/

Blast Radius vs Time

3 0.2 log(E/

log R = 0.4058 log t + 1.5593 kg/m3

2 E 7.9x1013 J

1 -2 -1 0 1 2 log (t) 6.2.2 Nondimensionalization of a functional relationship 6.2.2 Nondimensionalization of a functional relationship

6.2.2 Nondimensionalization of a functional relationship 6.2.2 Nondimensionalization of a functional relationship

6.2.2 Nondimensionalization of a functional relationship

Section 6.3

Scaling 6.3.0 Introduction 6.3.0 Introduction

Last section shows: And alternatively, • deleting a small parameter is not as simple as it looks, Let xwu2 , , 2 , 2 e because of ill conditioning and sensitivity. • a first approximation can be obtained w,0 lim2 e 1 • But we now see it is dangerous in doing so. For 0 example, consider Now we have three limits for the first approximation ux, xex/ , 0 x 1, 0 ux,0 x , ux ,0 e ex/ , ux ,0 1 ux,0 lim xex/ x 0 or we can rewrite as Which one is correct? Let xvu , , , e v,0 lim ee 0

6.3.0 Introduction 6.3.1 Definition of scaling The projectile problem: ** x** t x t z , The answer is Scaling: 1. y , 1 2. 1 R RV R Rg • Scaling is the correct way of nondimensionalization 2 d2 y 1 d z 1 • The key is to select intrinsic reference (scales) quantities 2 2 d 2 2 d 1 y 1 z • In the dimensional equations, each term is a product of y0 0, y ' 0 1 z0 0, z ' 0 two part:

** 2 • dimensionless factor: small or larger x t V hmax 3. x , 1 • V2 g 1 Vg 1 gR R dimensionless variable: O(1) 2 • We can determine the smallness from the factor. d x 1 d 2 1 x 2 How to choose dimensionless variables? x0 0, x ' 0 1

6.3.2 Scaling the projectile problem 6.3.3 Order of magnitude

1 2 2 1 mV mghmax h max V g / 2 2 The order of magnitude of a number A 2 d x 2 2 1 x n 1 dt • is said to be 10 , if V gt20 t 2 Vg 1 2 max max x0 0, x ' 0 1 1 1 nlog10 A n . 2 2 2 ** x t V hmax x, t 1 V2 g 1 Vg 1 gR R 3 10n1 A 3 10 n

Simplified properly V2 V x** xt, t g g d2 x The order of magnitude of a function f 2 1 dt • is the order of magnitude of f the given 2 1 1 max Vg , V g : x0 0, x ' 0 1 region. order of maginitude of x , t 6.3.4 Scaling known functions 6.3.4 Scaling known functions

Consider a first order ordinary differential equation and we have du* d Uu U du du* ux** Uux , Fx**, 0, with xI dx* d Lx L dx dx* u* is velocity with scale U If U and L are appropriate scales x* is spatial variable with length scale L • is a good estimate of the first derivative.

Dimensionless variables are • U, actually are the maximum absolute values

** x u Umax ux** x, u x* I u* max L U * L U du * max du/* dx Lx* I dx* max

6.3.4 Scaling known functions 6.3.4 Scaling known functions Example 1. Find scales U and L when uxAx** sin*, x *,, A 0 du* U dx* L * U A u * max u U max max du* A du* dx* L Acos u* AL 1 max max dx* max

U A du* A dx* L L max

length scale L is an estimate of the shortest distance 1 over which the function undergoes a significant change. L

6.3.4 Scaling known functions 6.3.4 Scaling known functions Example 2. Find scales U and L when

x* U A ux* * Ax * exp ,

* with x * 0,1 , A 0, 0 1 du A dx* max L

* Uu* A du A max dx* max L

* du1 x* 1 A1 exp AL L dx * max max 6.3.4 Scaling known functions 6.3.4 Scaling known functions

Suppose that the problem is governed by For each of the derivatives, we have length scale l(i) ,

**N i** i i **du d u du Udu du U Fx,* , *N 0, with xI , dx dx **iii i i dx l(i) dx dxmax l (i)

velocity scale U 1/i U l, iN 1,2, * (i) i** i ** x d u/ dx Umax ux** u x U u max x* I L we select the length scale L as u* L max Lmin l , i 1,2, N du* /* dx (i) max Since now we have multiple derivatives dui * UU *i i i dxmax l(i) L

6.3.4 Scaling known functions 6.3.4 Scaling known functions Example 3. Find scales U and L when Example 4. Find scales U and L when **N du** dN u **du d u ** F x , , 0, with x I F x ,* , *N 0, with x I dx* dx *N dx dx uxAx* * sin *, x * , A , 0 uxMAx** sin*, x *,,0 A

U M A length scale l(i) for i-th derivatives,

1/i 1/i 1/i1/ i U A 1/i 1 U MA1 M l l(i) 1 (i) di u**/ dx i A i di u**/ dx i Ai A max max

1 M 1/N Lmin l 1 1 (i) A Llmin(i) , i 1,2, N

6.3.5 Orthodoxy 6.3.5 Orthodoxy after the process of scaling, we have two issues : Split example of orthodoxy: x* • We have seen neglecting small terms may be wrong uxAx* * * exp , with x * 0,1 , A 0, 0 1 • Another one is the Orthodoxy of each terms. the order of magnitude of a term estimates that U A, L term’s maximum magnitude. du If the absolute value of a term deviates too much uu*/A xe x e x from their maximum values, then the order of dx magnitude may be misleading. as x * 0

Harmless unorthodoxy: L • terms decay at same rate unorthodoxy orthodoxy • With oscillatory terms of as 3x * 1 large amplitude, terms of u* unit order can be deleted du* 1 3 A 3 L max * AeL1 3 e 10 du/* dx max dx* max Ae 6.3.5 Orthodoxy 6.3.5 Orthodoxy To satisfy the orthodoxy requirement, Inner regions: (within a few • split [0, 1] into outer and inner regions du* uA*, A 1 max U A, L • different length scale in each part. dx* max

u** x 1 outer regions: (more than a few from ) v , v,*, u e A du* A uA*, A max dx* max To obtain a first approximation in the two regions, U A, L 1 inner outer let 0, u** x u,* x x outer regions: ux, xex/ x A L 1 ux, e x/ ux, ux * *, xe x/ Inner regions: v, ee A

6.3.5 Orthodoxy 6.3.6 Scaling unknown functions