AMS 212B Perturbation Methods Lecture 14 Copyright by Hongyun Wang, UCSC. Example: Eigenvalue problem with a turning point inside the interval

Σχετικά έγγραφα
Oscillatory integrals

ΤΜΗΜΑ ΗΛΕΚΤΡΟΛΟΓΩΝ ΜΗΧΑΝΙΚΩΝ ΚΑΙ ΜΗΧΑΝΙΚΩΝ ΥΠΟΛΟΓΙΣΤΩΝ

Solutions 3. February 2, Apply composite Simpson s rule with m = 1, 2, 4 panels to approximate the integrals:

Example Sheet 3 Solutions

Solutions_3. 1 Exercise Exercise January 26, 2017

2 Composition. Invertible Mappings

Fourier Series. MATH 211, Calculus II. J. Robert Buchanan. Spring Department of Mathematics

Section 8.3 Trigonometric Equations

3.4 SUM AND DIFFERENCE FORMULAS. NOTE: cos(α+β) cos α + cos β cos(α-β) cos α -cos β

CHAPTER 25 SOLVING EQUATIONS BY ITERATIVE METHODS

Απόκριση σε Μοναδιαία Ωστική Δύναμη (Unit Impulse) Απόκριση σε Δυνάμεις Αυθαίρετα Μεταβαλλόμενες με το Χρόνο. Απόστολος Σ.

Exercises 10. Find a fundamental matrix of the given system of equations. Also find the fundamental matrix Φ(t) satisfying Φ(0) = I. 1.

Exercises to Statistics of Material Fatigue No. 5

Section 7.6 Double and Half Angle Formulas

D Alembert s Solution to the Wave Equation

Second Order RLC Filters

CHAPTER 101 FOURIER SERIES FOR PERIODIC FUNCTIONS OF PERIOD

Solutions to Exercise Sheet 5

LAPLACE TYPE PROBLEMS FOR A DELONE LATTICE AND NON-UNIFORM DISTRIBUTIONS

Statistical Inference I Locally most powerful tests

ST5224: Advanced Statistical Theory II

Review-2 and Practice problems. sin 2 (x) cos 2 (x)(sin(x)dx) (1 cos 2 (x)) cos 2 (x)(sin(x)dx) let u = cos(x), du = sin(x)dx. = (1 u 2 )u 2 ( du)

Some definite integrals connected with Gauss s sums

If ABC is any oblique triangle with sides a, b, and c, the following equations are valid. 2bc. (a) a 2 b 2 c 2 2bc cos A or cos A b2 c 2 a 2.

EE512: Error Control Coding

Differential equations

Matrices and Determinants

Econ 2110: Fall 2008 Suggested Solutions to Problem Set 8 questions or comments to Dan Fetter 1

4.6 Autoregressive Moving Average Model ARMA(1,1)

Srednicki Chapter 55

CHAPTER (2) Electric Charges, Electric Charge Densities and Electric Field Intensity

Areas and Lengths in Polar Coordinates

Parametrized Surfaces

Integrals in cylindrical, spherical coordinates (Sect. 15.7)

Phys460.nb Solution for the t-dependent Schrodinger s equation How did we find the solution? (not required)

derivation of the Laplacian from rectangular to spherical coordinates

Homework 3 Solutions

Σχολή Εφαρμοσμένων Μαθηματικών και Φυσικών Επιστημών. Εθνικό Μετσόβιο Πολυτεχνείο. Thales Workshop, 1-3 July 2015.

INTEGRAL INEQUALITY REGARDING r-convex AND

1 String with massive end-points

Lecture 5: Numerical Integration

ANSWERSHEET (TOPIC = DIFFERENTIAL CALCULUS) COLLECTION #2. h 0 h h 0 h h 0 ( ) g k = g 0 + g 1 + g g 2009 =?

Areas and Lengths in Polar Coordinates

Math221: HW# 1 solutions

Inverse trigonometric functions & General Solution of Trigonometric Equations

b. Use the parametrization from (a) to compute the area of S a as S a ds. Be sure to substitute for ds!

Approximation of distance between locations on earth given by latitude and longitude

Concrete Mathematics Exercises from 30 September 2016

9.09. # 1. Area inside the oval limaçon r = cos θ. To graph, start with θ = 0 so r = 6. Compute dr

HOMEWORK 4 = G. In order to plot the stress versus the stretch we define a normalized stretch:

University of Illinois at Urbana-Champaign ECE 310: Digital Signal Processing

C.S. 430 Assignment 6, Sample Solutions

( )( ) La Salle College Form Six Mock Examination 2013 Mathematics Compulsory Part Paper 2 Solution

Lecture 26: Circular domains

Appendix to On the stability of a compressible axisymmetric rotating flow in a pipe. By Z. Rusak & J. H. Lee

Chapter 6: Systems of Linear Differential. be continuous functions on the interval

Practice Exam 2. Conceptual Questions. 1. State a Basic identity and then verify it. (a) Identity: Solution: One identity is csc(θ) = 1

forms This gives Remark 1. How to remember the above formulas: Substituting these into the equation we obtain with

w o = R 1 p. (1) R = p =. = 1

Notes on Tobin s. Liquidity Preference as Behavior toward Risk

Electromagnetic Waves I

Jesse Maassen and Mark Lundstrom Purdue University November 25, 2013

Oscillation of Nonlinear Delay Partial Difference Equations. LIU Guanghui [a],*

Uniform Convergence of Fourier Series Michael Taylor

Differentiation exercise show differential equation

Section 9.2 Polar Equations and Graphs

Other Test Constructions: Likelihood Ratio & Bayes Tests

The Negative Neumann Eigenvalues of Second Order Differential Equation with Two Turning Points

PARTIAL NOTES for 6.1 Trigonometric Identities

HOMEWORK#1. t E(x) = 1 λ = (b) Find the median lifetime of a randomly selected light bulb. Answer:

Reminders: linear functions

Finite Field Problems: Solutions

Problem Set 9 Solutions. θ + 1. θ 2 + cotθ ( ) sinθ e iφ is an eigenfunction of the ˆ L 2 operator. / θ 2. φ 2. sin 2 θ φ 2. ( ) = e iφ. = e iφ cosθ.

DESIGN OF MACHINERY SOLUTION MANUAL h in h 4 0.

Physics 505 Fall 2005 Practice Midterm Solutions. The midterm will be a 120 minute open book, open notes exam. Do all three problems.

CRASH COURSE IN PRECALCULUS

Bessel functions. ν + 1 ; 1 = 0 for k = 0, 1, 2,..., n 1. Γ( n + k + 1) = ( 1) n J n (z). Γ(n + k + 1) k!

Higher Derivative Gravity Theories

ΚΥΠΡΙΑΚΗ ΕΤΑΙΡΕΙΑ ΠΛΗΡΟΦΟΡΙΚΗΣ CYPRUS COMPUTER SOCIETY ΠΑΓΚΥΠΡΙΟΣ ΜΑΘΗΤΙΚΟΣ ΔΙΑΓΩΝΙΣΜΟΣ ΠΛΗΡΟΦΟΡΙΚΗΣ 19/5/2007

MATH423 String Theory Solutions 4. = 0 τ = f(s). (1) dτ ds = dxµ dτ f (s) (2) dτ 2 [f (s)] 2 + dxµ. dτ f (s) (3)

To find the relationships between the coefficients in the original equation and the roots, we have to use a different technique.

Ordinal Arithmetic: Addition, Multiplication, Exponentiation and Limit

Partial Differential Equations in Biology The boundary element method. March 26, 2013

Similarly, we may define hyperbolic functions cosh α and sinh α from the unit hyperbola

Homework 8 Model Solution Section

Trigonometric Formula Sheet

Lecture 34 Bootstrap confidence intervals

Graded Refractive-Index

Problem Set 3: Solutions

Lecture 2. Soundness and completeness of propositional logic

Overview. Transition Semantics. Configurations and the transition relation. Executions and computation

( y) Partial Differential Equations

SCHOOL OF MATHEMATICAL SCIENCES G11LMA Linear Mathematics Examination Solutions

Second Order Partial Differential Equations

Tridiagonal matrices. Gérard MEURANT. October, 2008

ECE Spring Prof. David R. Jackson ECE Dept. Notes 2

Homomorphism in Intuitionistic Fuzzy Automata

Testing for Indeterminacy: An Application to U.S. Monetary Policy. Technical Appendix

P AND P. P : actual probability. P : risk neutral probability. Realtionship: mutual absolute continuity P P. For example:

Transcript:

AMS B Perturbtion Methods Lecture 4 Copyright by Hongyun Wng, UCSC Emple: Eigenvlue problem with turning point inside the intervl y + λ y y = =, y( ) = The ODE for y() hs the form y () + λ f() y() = with f() = (-/). The WKB epnsion for > nd < : φ ( ) = ±i f ( s)ds = ±i s s 3 ds = ±i 3 5 5 = ±it +, > ±T, < where T + = 3 3 5 5 > T = 3 3 + 5 5 < φ ( ) = 4 log f A generl solution t lrge λ is y( ) = = 4 log ( /) ( /) cos λt /4 ( + )+ b sin( λt + ), > c ep λt /4 ( )+ d ep( λt ), < Suppose y(t) is the eigenfunction of lrge eigenvlue λ. y(-) = (T ) = = 3/5 - -

AMS B Perturbtion Methods c ep λ 3 3 5 + d ep λ 5 = c d 6 = ep λ 5 Using the connection formuls, we hve = d + c b = d c + c /d = Constrint from y(-) = b c /d On the other hnd, using the boundry condition t =, we obtin y() =, (T+) = = 7/5 cos λ 7 5 + b sin λ 7 5 = b = tn λ 7 5 Constrint from y() = Combining the two constrints on /b from two boundry conditions, we rrive t tn λ 7 5 = λ(7/5) = (n /4)π λ = (5/7)(n /4)π Lrge eigenvlues re λ n 5 7 n 4 π, n = integer, n lrge. Net we consider the cse where one boundry condition flls on the turning point. Emple (Skip this emple in lecture): Eigenvlue problem with turning point t boundry - -

AMS B Perturbtion Methods y + λ =, y( ) = y y = The WKB epnsion for > : ( ) = y WKB ( /) where T + = 3 3 5 5 The inner solution ner = : ( ) = α Ai λ y inn At =, we hve 3 cos λt /4 ( + )+ b sin( λt + ), > +βbi ( λ ) y() = α Ai() + β Bi() = Property of Airy functions: = Bi ( ) Ai 3 = 3 /3 Γ /3 3 (We will derive this property when we discuss symptotic epnsions of integrls). Using this property of Airy function, we hve α β = Bi Ai = 3 Recll the reltion between (α, β) nd (, b) = b = α +β π λ /6 α β π λ /6 b = α +β α /β + = α β α /β = 3 + 3 Constrint from y() = On the other hnd, using the boundry condition t =, we obtin - 3 -

AMS B Perturbtion Methods y() =, (T+) = = 7/5 cos λ 7 5 + b sin λ 7 5 = b = tn 7 5 λ Constrint from y() = Combining the two constrints on /b from two boundry conditions, we rrive t tn λ 7 5 = 3 + 3 7 5 λ = 3 tn + nπ + 3 λ = 5 3 7 tn + nπ + 3 Lrge eigenvlues re λ n 5 3 7 tn + nπ + 3, n: integer, n A more generl cse of turning point: y + λ g( )+ λh( ) y =, g = Gol: Find WKB pproimtions for > nd <, find inner solution, nd mtch them. Prt B: Find the solution ner = Ner =, we do scling u = λ α, α = /3. The eqution becomes d y λ α du + λ λ α u g λ α u + λh λ α u y = d y du + λ 3α u g( λ α u)+ λ α h( λ α u ) y u d y + u+o λ /3 du y u = = - 4 -

AMS B Perturbtion Methods The leding term eqution is the sme s in the specil cse: d y du u y ( u ) =. Thus, the (leding term) inner solution ner = is unffected by the term λh() y. y ( inn) ( ) = α Ai( u)+βbi( u), u = λ Prt A: Find the WKB epnsions for > nd <. The differentil eqution is y + λ g( )+ λh( ) Let y y =, g = ep δ n λ n= φ n Substituting into eqution yields ( λ)φ n ( ) + δ n ( λ)φ n ( ) δ n n= Clculting ϕ(): δ(λ) = λ δ φ ( λ) φ ( ) = g n= ( ) = λ g φ ( ) = ±i g( ) φ ( ) = ±i s g( s)ds = - 5-3 = λ g( ) λh( ) Tht is, ϕ() is unffected by the presence of λh() y. Clculting ϕ(): δ(λ) = λδ ( λ) φ φ ( ) φ φ φ ( ) = λ h φ ( ) = h( ) φ ( ) ( ) = h ( ) φ ( ) φ φ ( ) = ±i h( ) g( ) log φ

AMS B Perturbtion Methods φ ( ) = ±i ds h s s g s 4 log g ( ) Prt C: Mtching We consider the region of = O λ ds ds. h s = s g s s h s = g s = h 3 +O φ ( ) = 4 log g ( ) +O λ 4 = O ( λ 4 ) ( s h ( )+O( s) )ds Tht is, in the region of = O(λ -/ ), y (WKB) () is unffected by the term λh() y. Therefore, the mtching reltions re unffected by the presence of λh() y. = b = α +β π λ /6 α β π λ /6 c = d = π λ /6 β α π λ /6 Note: = ( d + c ) b = ( d c ) *) The connection formul is unffected by the term λh() y. *) In the rnge of = O(), y (WKB) () will be ffected by the term λh() y - 6 -

AMS B Perturbtion Methods Asymptotic pproimtions of integrls f t b ep λh( t ) dt, λ + f t b cos λh( t ) dt, λ + f t b sin λh( t ) dt, λ + We strt with + e λt f ( t )dt, λ + Region of dominnt contribution Emple: I = e λt cos( πt )dt, λ + Since e -λ t decys very fst for lrge λ, the region of dominnt contribution is very smll region ner t =. So we only need to pproimte cos(πt) ner =. We use the Tylor epnsion of cos(πt) t t = I = e λt cos( πt )dt = e λt π t! + π4 t 4 4!! dt To finish the emple, we need to clculte t α e λt dt. We use chnge of vrible: s = λ t, t = s/λ, dt = ds/λ t α e λt dt = λ α sα e s λ ds = λ α+ where the Gmm function is defined s z e d Γ z We obtin very useful formul s α e s ds = Γ α + λ α+ - 7 -

t α e λt dt = Γ α + λ α+ AMS B Perturbtion Methods Properties of the Gmm function Γ() = Γ(α + ) = α Γ(α) Γ(n + ) = n! n = integer Γ = π The lst property cn be derived using the probbility density of the Gussin distribution Derivtion: Γ = s e s ds Chnge of vrible s = u = e u du = πσ ep u πσ σ du σ = = π Bck to our emple Applying the formul to our emple, we hve I = e λt π t + π4 t 4! dt = Γ! 4! λ = λ π λ 3 + π4 λ 5! λ 3 Γ 3 π! Γ( 5) + π4! 4! λ 5 Integrl e λt f ( t )dt cn be treted in similr wy. Now we consider slightly more generl cse: b e λt f ( t )dt where nd/or b +. The emple below shows how to del with the sitution where the lower integrtion limit is not nd the upper integrtion limit is not infinity. - 8 -

AMS B Perturbtion Methods Emple: 3 I = e λt dt, λ + +t 3 First, we use chnge of vrible to shift the lower limit of integrtion to zero: s = t, t = s +, dt = ds 3 I = e λt λ +s dt = e = e λ e λs +t 3 + + s + + s 3 ds Net, we del with the upper limit of integrtion Clim: 3 ds I = e λ e λs ds = e λ e λs ds e λs ds + ( + s) 3 + ( + s) 3 + ( + s) 3! ###" ### $!###" ### $ I I I = T.S.T. T.S.T. stnds for Trnscendentlly Smll Term, which mens it is symptoticlly smller thn ny powers of /λ. Derivtion: We use chnge of vrible: u = s, s = u + I = e λs λ u+ = e = e λ e λu + + s + u+3 + u+3 = T.S.T. 3 ds 3 du 3 du Since we re going to hve power series of /λ in the epnsion of I, we do not need to keep ny T.S.T. term. Summry: In generl, we hve b e λt f ( t )dt = e λt f ( t )dt + T.S.T, λ + Bck to our emple To clculte I, we epnd - 9 -

AMS B Perturbtion Methods 3 = + + s = +3s +3s +! + 3 s + 3 s +! / = 3 s + 3 s + 3 3 8 s Thus, we hve I ~e λ I = e λ e λs 3 4 s + 3 3 s +! ds = e λ Γ( ) λ 3 Γ + 3 Γ( 3) 4 λ 3 = e λ λ 3 4λ + 3 6λ +O 3 λ 3 +! = 3 4 s + 3 3 s +! +O λ 3 λ 3 Emple: (Skip this emple in lecture) 4 I = e λt t +t dt ~ e λt t +t dt = e λt t + t 8 t +! dt = Γ 3 + λ 3 = λ 3 + 3 λ 4 3 λ 5 +! Γ( 4) Γ( 5) +! λ 4 8 λ 5 Emple (non-integer power): I = e λt t +t dt, λ + = e λt t +t! t dt = e λt t 3 t +t! dt = Γ ( / ) Here we hve used Γ 3/ + Γ 5/!= π λ / λ 3/ λ 5/ λ π / λ + 3 π 3/ 4 λ 5/! - -

AMS B Perturbtion Methods Γ = π Γ 3 = Γ = Γ 5 = 3 Γ 3 = 3 4 π π Emple: (Skip this emple in lecture) 4 I = e λt log( +t )dt ~ e λt t t log( +t)dt = e λt t t 4 t + t 6 3! dt = Γ 5/ λ 5/ Γ( 9/) + Γ( 3/)! λ 9/ 3 λ 3/ Γ 5 = 3 Γ = 3 4 π Γ 9 = 7 5 Γ 5 = 5 6 π Γ 3 = 9 Γ 9 = 395 64 π Now we summrize the procedure discussed bove s lemm. Wtson s lemm: Consider the integrl T = e λt t α g t I λ Suppose ) α > ) g t Ke ct Then we hve dt - -

I( λ)~ N n= n! g n AMS B Perturbtion Methods Γ( α + n+) λ α+n+ Note: Condition ) is to mke sure tht the integrl is convergent t t =. Condition ) is to mke sure tht for t >, g(t) is dominted by e λ t so tht for lrge λ the region of dominnt contribution is smll region ner t =. Proof: T = e λt t α g t I λ dt ~ e λt t α g( t )dt ~ e λt N g ( n ) t α n! n= N g ( n ) = e λt t α+n dt n! +!= n= N n= n! g n Γ( α + n+) λ α+n+ t n dt Reling the condition α >. Suppose g() = = g (k-) () = but g (k) (). Then α is restricted by α + k >. Emple: 4 I = e λt log( t +t )dt ~ e λt t t t log( +t)dt = e λt t t 4 t t + t 6 3! dt = e λt t / t 3/ + 3 t 7/! dt = Γ ( / ) Γ( 5/) + Γ( 9/)! λ / λ 5/ 3 λ 9/ - -