My Physics 501 pageSpecial Topics: Mathematical Models of Physical
Problems IFall 2018
University of Wisconsin, Milwaukee
Nasser M. Abbasi
Fall 2018 Compiled on November 3, 2019 at 11:35am [public]
Contents
1 Introduction 31.1 syllabus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Topics covered . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
2 exams 112.1 Practice exam from 2017 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
2.1.1 Problem 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112.1.2 Problem 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.1.3 Problem 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152.1.4 Problem 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162.1.5 Problem 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
2.2 exam 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212.2.1 exam 1 prep sheet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212.2.2 exam 1 writeup . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
2.3 final exam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 302.3.1 exam 1 prep sheet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 302.3.2 my final exam cheat sheet . . . . . . . . . . . . . . . . . . . . . . . . 31
3 HWs 333.1 HW 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
3.1.1 Problem 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 333.1.2 Problem 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 353.1.3 Problem 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 363.1.4 key solution to HW 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
3.2 HW 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 413.2.1 Problem 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 413.2.2 Problem 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 433.2.3 Problem 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 453.2.4 Problem 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 483.2.5 Problem 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 493.2.6 key solution to HW 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
3.3 HW 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 543.3.1 Problem 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 543.3.2 Problem 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 593.3.3 Problem 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 643.3.4 key solution to HW 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . 66
3.4 HW 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 703.4.1 Problem 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 703.4.2 Problem 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 733.4.3 Problem 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 743.4.4 Problem 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 783.4.5 key solution to HW 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . 82
3.5 HW 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 863.5.1 Problem 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 863.5.2 Problem 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 963.5.3 Problem 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 993.5.4 key solution to HW 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . 102
3.6 HW 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1083.6.1 Problem 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1083.6.2 Problem 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1123.6.3 Problem 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114
2
3
3.6.4 key solution to HW 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . 1193.7 HW 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124
3.7.1 Problem 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1243.7.2 Problem 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1293.7.3 key solution to HW 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . 133
4
Chapter 1
Introduction
1.1 syllabus
1.2 Topics covered
This is list of lectures and topics covered in each
5
6
Table 1.1: Topics covered
# Date Topics
1 Wed. Sept 4, 20181. Infinite series, geometric series.
2. conditions for convergence,
3. harmonic series, alternating series.
4. tests for convergence such as ratio test, integral test.
2 Monday Sept 10, 2018
1. Talked about series solution to (1โ๐ฅ2)๐ฆโณโ2๐ฅ๐ฆโฒ+๐(๐+1)๐ฆ =0.
2. Binomial series (1 + ๐ฅ)๐ = 1 + ๐๐ฅ + ๐(๐โ1)๐ฅ2
2! + โฆ
3. series of ๐, sin(๐ฅ), cos(๐ฅ).4. Introducing Bernoulli numbers.
5. Show that alternating series is convergent but not abso-lutely.
6. Leibniz condition for convergence and its proof.
7. Showed that sum of 1 โ 1/2 + 1/3 โ 1/4 + โฆ is ln 2.8. Working with absolutely convergent series.
3 Wed Sept 12, 2018
1. Familiar series, ๐, sin ๐ฅ, cos ๐ฅ, ln(1 + ๐ฅ), arctan(๐ฅ)2. How to get Bernulli numbers. More on Bernulli numbers
but I really did not understand these well and how touse them. hopefully they will not be on the exam.
3. Started Complex analysis. Basic introduction. Propertiesof complex numbers and mapping.
4 Monday Sept 17, 2018
1. complex functions ๐ข(๐ฅ, ๐ฆ) + ๐๐ฃ(๐ฅ, ๐ฆ)
2. continuity in complex domain.
3. Derivative in complex domain and how direction is im-portant.
4. Cauchy-Riemman equation to test for analytical function.
5. Harmonic functions. Exponential function in complexdomain.
6. Multivalued functions, such as log ๐ง.7. How to obtain inverse trig function and solve ๐ค =
arcsin(๐ง)
Continued on next page
7
Table 1.1 โ continued from previous page
Lecture # Date Topics
5 Wed Sept 19, 2018
1. derivative in complex plane. Definition of analytic func-tion.
2. log(๐ง) and โ(๐ง) in complex plane and multivalued.Branch points and branch cuts.
3. Integration over contour. Parameterization โซ๐ถ๐(๐ง) ๐๐ง =
โซ๐๐๐(๐ง(๐ก))๐งโฒ(๐ก) ๐๐ก
4. Cauchy-Goursat theorem: โฎ๐(๐ง) = 0 for analyticalfunctions. Proof using Cauchy-Riemman equations andGreen theorem.
5. Cauchy integral formula 2๐๐๐(๐ง0) = โฎ๐(๐ง)๐งโ๐ง0
๐๐ง
6. Like in real, in complex domain, Continuity Does NotImply Di๏ฟฝerentiability.
7. More on analytic functions and multivalued functions.Principal value.
8. Power functions ๐ง๐ = ๐๐ ln ๐ง
9. Complex integration.
6 Monday Sept 24, 2018
1. Proof of Cauchy integral formula.
2. Maximum moduli of analytic functions. If ๐(๐ง) is analyticin ๐ท and not constant, then it has no maximum valueinside ๐ท. The maximum of ๐(๐ง) is on the boundary.
3. Taylor series for complex functions and Laurent series.
7 Wed Sept 26, 2018
1. If number of terms in principal part of Laurent series isinfinite, then it is essential singularity.
2. Proof of Laurent theorem.
3. properties of power series. Uniqueness.
4. Residues, types of singularities. How to find residues andexamples
8 Monday Oct 1, 2018
1. Residue theorem โฎ๐ถ๐(๐ง) ๐๐ง = 2๐๐โ residues inside C
2. examples using Residue theorem.
3. Analytic continuation. Examples.
4. ฮ(๐ง) function. Defined for โ(๐ง) > 0. Using analytical con-tinuation to extend it to negative complex plane. Eulerrepresentation and Weistrass represenation.
Continued on next page
8
Table 1.1 โ continued from previous page
Lecture # Date Topics
9 Wed Oct 3, 2018
1. More on Euler represenation of ฮ(๐ง) and how to use itfor extending definition ฮ(๐ง) = โซ
โ
0๐โ๐ก๐ก๐งโ1 ๐๐ก for negative ๐ง
using ฮ(๐ง) = ฮ(๐ง+1)๐ง for โ1 < ๐ง.
2. Euler reflection formula
ฮ(๐ฅ)ฮ(1 โ ๐ฅ) = ๏ฟฝโ
0
๐ก๐ฅโ1
1 + ๐ก๐๐ก =
๐sin(๐๐ฅ)
3. proof of Euler reflection formula using contour integra-tion.
4. Some useful formulas for ฮ(๐ง)
5. Method for integrations, some tricks to obtain definiteintegrations.
10 Monday Oct 8, 2018 No class.
11 Wed Oct 10, 2018 No class.
12 Monday Oct 15, 2018
1. More on method of integration. Starting Contour inte-gration.
2. How to decide that โซ๐ถ๐ ๐(๐ง) = 0 on the upper half plane.
Using Jordan inquality.
3. More examples of integrals on real line using contourintegration.
13 Wed Oct 17, 2018
1. More contour integrations.
2. Starting approximation expansion of integrals. Exampleusing error function erf(๐ฅ) = 2
โ๐โซ๐ฅ0๐โ๐ก2 ๐๐ก by applying
Taylor series.
3. Large ๐ฅ expansion by repeated integration by parts.
4. Starting Asymprtotic series. Definition. Example on find-ing ๐(๐ฅ) for erf(๐ฅ) for large ๐ฅ. When to truncate.
5. Saddle point methods of integration to approximate in-tegral for large ๐ฅ.
14 Monday Oct 22, 2018
1. More saddle point integration.
2. Saddle point methods of integration to approximate in-tegral for large ๐ฅ. Method of steepest decsent. Exampleto find ฮ(๐ฅ + 1) = โซ
โ
0๐ก๐ฅ๐โ๐ก ๐๐ก = โ2๐๐ฅ๐ฅ๐ฅ๐โ๐ฅ
3. extend saddle point method to complex plane. Findingcorrect angle. Long example.
Continued on next page
9
Table 1.1 โ continued from previous page
Lecture # Date Topics
15 Wed Oct 24, 2018
1. More on saddle point in complex plane. Angles. Exampleapplied on โซฮ(1 + ๐ง) = โซ
โ
0expโ๐ก + ๐ง ln ๐ก ๐๐ก
2. how to determine coe๏ฟฝcients of asymptotic series expan-sion.
3. Starting new topic. Fourier series. Definitions.
4. proprties of Fourier series. Examples how to find ๐ด๐, ๐ต๐.
16 Friday Oct 26, 2018 Make up lecture.
1. More on Fourier series. Examples. Fourier series usingthe complex formula.
2. Parseval identity.
3. Fourier Transform derivation.
17 Monday Oct 29, 2018
1. Fourier transform pairs.
2. How to find inverse fourier transform. Generalization tohigher dimensions.
3. Properties of Fourier transform.
4. convolution.
5. Example on driven harmonic oscillator.
6. Statring ODEโs. Order and degree of ODE.
18 Wed Oct 31, 2018
1. More on first order ODEโs. Separable, exact. How to findintegrating factor.
2. Bernulli ODE ๐ฆโฒ + ๐(๐ฅ)๐ฆ = ๐(๐ฅ)๐ฆ๐
3. Homogeneous functions. defintion. order of.
4. isobaric ODEโs.
19 Monday Nov 5, 2018
1. How to find integating factor for exact ODE.
2. Finished example on isobaric first order ODE.
๐ฅ๐ฆ2(3๐ฆ ๐๐ฅ + ๐ฅ ๐๐ฆ) โ (2๐ฆ ๐๐ฅ โ ๐ฅ ๐๐ฆ) = 0
3. Higher order ODEโs. How to solve. How to find partic-ular solution. Undetermined coe๏ฟฝcients. What to do ifforcing function has same form as one of the solutionsto homogeneous solutions.
4. How to use power series to solve nonlinear ode ๐ฆโณ = ๐ฅโ๐ฆ2
20 Wed Nov 7, 2018 First exam
Continued on next page
10
Table 1.1 โ continued from previous page
Lecture # Date Topics
21 Monday Nov 12, 2018
1. More on higher order ODEโs. Series solutions.
2. ordinary point. Regular singular point. Example Legen-dre ODE (1 โ ๐ฅ2)๐ฆโณ โ 2๐ฅ๐ฆโฒ + ๐(๐ + 1)๐ฆ = 0.
3. Example for regular singular point, Bessel ODE ๐ฅ2๐ฆโณ +๐ฅ๐ฆโฒ + (๐ฅ2 โ ๐2)๐ฆ = 0 Use ๐ฆ = ๐ฅ2โโ
๐=0 ๐๐๐ฅ๐
22 Wed Nov 14, 2018
1. Continue Bessel ODE ๐ฅ2๐ฆโณ + ๐ฅ๐ฆโฒ + (๐ฅ2 โ ๐2)๐ฆ = 0 solvingusing ๐ฆ = ๐ฅ2โโ
๐=0 ๐๐๐ฅ๐. How to find second independent
solution.
23 Monday Nov 19, 2018
1. Started on Sturm Lioville, Hermetian operators
2. setting Bessel ODE in Sturm Lioville form
3. more on Hermitian operator.
4. Wronskian to check for linear independece of solutions.
24 Wed Nov 21, 2018 Thanks Giving.
25 Monday Nov 26, 2018
1. finding second solution to Bessel ODE for ๐ integer us-ing the Wronskian. ๐(๐ฅ) = ๐ถ
๐(๐ฅ) =โ2 sin๐๐
๐
2. Generating functions to find way to generate Besself func-tions.
26 Wed Nov 28, 2018
1. Using Generating functions
2. Bessel functions of half integer order, spherical Besselfunctions
3. Legendre polynomials, recusrive relations.
4. orthonomalization.
5. physical applications
27 Monday December 3, 2018
1. Second solution to Legendre using Wronskian
2. Spherical harmonics
3. Normalization of eigenfunctions
4. Degenerncy, using Gram-Schmidt to find other L.I. solu-tions.
5. Expanding function using complete set of basis functions,example using Fourier series
6. Inhomogeneous problems, starting Green function
Continued on next page
11
Table 1.1 โ continued from previous page
Lecture # Date Topics
28 Wed December 5, 2018
1. Green function. Solution to the ODE with point source.
2. Example using vibrating string ๐ฆโณ + ๐2๐ฆ = 0. Find Greenfunction. Two Methods. Use second method.
3. Started on PDE.
29 Friday December 7, 2018 (Make up lecture)
1. more PDEโs. Solve wave PDE in 1D
2. separation of variables. Solve Wave PDE in 3D in spher-ical coordiates.
30 monday December 10, 2018
1. Solving wave PDE in 3D in spherical coordinates. Nor-mal modes.
2. Solving wave PDE in 3D in cylindrical coordinates.
3. Inhomogeneous B.C. on heat PDE. Break it into 2 parts.
31 Wed December 12, 2018 Last lecture.
1. Finish Inhomogeneous B.C. on heat PDE. Break it into2 parts. Final solution, using Fourier series.
2. Last problem. INtegral transform method. Solving heatpde on infinite line using Fourier transform.
12
Chapter 2
exams
2.1 Practice exam from 2017
2.1.1 Problem 1
i) Find Laurent series for ๐ (๐ง) = 1
๏ฟฝ๐ง2+1๏ฟฝ3 around isolated singular pole ๐ง = ๐. What is the
order of the pole? ii) Use residues to evaluate the integral โซโ
0๐๐ฅ
๏ฟฝ๐ฅ2+1๏ฟฝ3
solution
๐ง2 + 1 = 0 gives ๐ง = ยฑ๐. Hence there is a pole at ๐ง = ๐ of order 3 and also a pole at ๐ง = โ๐of order 3. Hence ๐ (๐ง) = (๐ง โ ๐)3 ๐ (๐ง) is analytic at ๐ง = ๐ and therefore it has a Taylor seriesexpansion around ๐ง = ๐ given by
๐ (๐ง) =โ๏ฟฝ๐=0
๐๐ (๐ง โ ๐)๐
(๐ง โ ๐)3 ๐ (๐ง) =โ๏ฟฝ๐=0
๐๐ (๐ง โ ๐)๐ (1)
Where ๐๐ =๐๐๐๐ง๐ ๐(๐ง)
๐! ๏ฟฝ๐ง=๐
. But
๐ (๐ง) = (๐ง โ ๐)3 ๐ (๐ง)
= (๐ง โ ๐)31
๏ฟฝ๐ง2 + 1๏ฟฝ3
= (๐ง โ ๐)31
((๐ง โ ๐) (๐ง + ๐))3
= (๐ง โ ๐)31
(๐ง โ ๐)3 (๐ง + ๐)3
=1
(๐ง + ๐)3
To find ๐๐ then ๐๐ =1๐!
๐๐
๐๐ง๐1
(๐ง+๐)3is evaluated for few ๐ terms. Since order is 3, at least 5 terms
are needed to see the residue and the first term in the analytical part of the series (๐ > 0).Starting with ๐ = 0
๐0 =1
(๐ง + ๐)3๏ฟฝ๐ง=๐
=1
(2๐)3=
1โ8๐
=18๐
For ๐ = 1
๐1 =๐๐๐ง
1(๐ง + ๐)3
๏ฟฝ๐ง=๐
=โ3
(๐ง + ๐)4๏ฟฝ๐ง=๐
=โ3(2๐)4
=โ316
For ๐ = 2
๐2 =12๐๐๐ง
โ3(๐ง + ๐)4
๏ฟฝ๐ง=๐
=12โ3 (โ4)(๐ง + ๐)5
๏ฟฝ๐ง=๐
=12โ3 (โ4)(2๐)5
=12โ3 (โ4)25๐
=632๐
=316๐
= โ3๐16
13
14
For ๐ = 3
๐3 =13!
๐๐๐งโ3 (โ4)(๐ง + ๐)5
๏ฟฝ๐ง=๐
=16โ3 (โ4) (โ5)(๐ง + ๐)6
๏ฟฝ๐ง=๐
=16โ3 (โ4) (โ5)
(2๐)6=16โ3 (โ4) (โ5)
โ26=532
For ๐ = 4
๐4 =14!
๐๐๐งโ3 (โ4) (โ5)(๐ง + ๐)6
๏ฟฝ๐ง=๐
=124
โ3 (โ4) (โ5) (โ6)(๐ง + ๐)7
๏ฟฝ๐ง=๐
=124โ3 (โ4) (โ5) (โ6)
(2๐)7=1243 (4) (5) (6)
โ๐27=15128
๐
Substituting all these back into (1) gives
(๐ง โ ๐)3 ๐ (๐ง) =โ๏ฟฝ๐=0
๐๐ (๐ง โ ๐)๐
= ๐0 + ๐1 (๐ง โ ๐) + ๐2 (๐ง โ ๐)2 + ๐3 (๐ง โ ๐)
3 + ๐4 (๐ง โ ๐)4 +โฏ
Therefore
๐ (๐ง) =1
(๐ง โ ๐)3๏ฟฝ๐0 + ๐1 (๐ง โ ๐) + ๐2 (๐ง โ ๐)
2 + ๐3 (๐ง โ ๐)3 + ๐4 (๐ง โ ๐)
4 +โฏ๏ฟฝ
=1
(๐ง โ ๐)3๏ฟฝ18๐ +
โ316(๐ง โ ๐) โ
3๐16(๐ง โ ๐)2 +
532(๐ง โ ๐)3 +
15128
๐ (๐ง โ ๐)4 +โฏ๏ฟฝ
=18
๐(๐ง โ ๐)3
โ316
1(๐ง โ ๐)2
โ316
๐(๐ง โ ๐)
+532+15128
๐ (๐ง โ ๐) โโฏ (1A)
The residue is the coe๏ฟฝcient of the term with 1๐งโ๐ factor. Hence residue is โ 3๐
16 . The order
is 3 since that is the highest power in 1๐งโ๐ .
The above method always works, but it means having to evaluate derivatives a number oftimes. For a pole of high order, it means evaluating the derivative for as many times asthe pole order and more to reach the analytical part. Another method is to expand thefunction using binomial expansion
(1 + ๐ฅ)๐ = 1 + ๐๐ง +๐ ๏ฟฝ๐ โ 1๏ฟฝ
2!๐ฅ2 +
๐ ๏ฟฝ๐ โ 1๏ฟฝ ๏ฟฝ๐ โ 2๏ฟฝ3!
๐ฅ3 +โฏ (2)
The above is valid for real ๐ , which can be negative or positive, but only for |๐ฅ| < 1. Thisis now applied to expand
๐ (๐ง) =1
๏ฟฝ๐ง2 + 1๏ฟฝ3
=1
(๐ง โ ๐)3 (๐ง + ๐)3
Let ๐ง โ ๐ = ๐, or ๐ง = ๐ + ๐ and the above becomes
๐ (๐ง) =1
๐3 (๐ + 2๐)3
=1๐3
1
(2๐)3 ๏ฟฝ1 + ๐2๐๏ฟฝ3
= ๏ฟฝ๐๐318๏ฟฝ
1
๏ฟฝ1 + ๐2๐๏ฟฝ3 (3)
Now the binomial expansion can be used on 1
๏ฟฝ1+ ๐2๐ ๏ฟฝ
3 term above, which is valid for ๏ฟฝ ๐2๐ ๏ฟฝ < 1,
which gives
1
๏ฟฝ1 + ๐2๐๏ฟฝ3 = 1 โ (3)
๐2๐+(โ3) (โ4)
2! ๏ฟฝ๐2๐๏ฟฝ
2
โ(โ3) (โ4) (โ5)
3! ๏ฟฝ๐2๐๏ฟฝ
3
+(โ3) (โ4) (โ5) (โ6)
4! ๏ฟฝ๐2๐๏ฟฝ
4
+โฏ
= 1 +32๐๐ + 6
๐2
4๐2+ 10
๐3
23๐3+ 15
๐4
24๐4+โฏ
= 1 +32๐๐ โ
32๐2 โ
108๐๐3 +
1516๐4 +โฏ
15
Therefore (3) becomes
๐ (๐ง) =๐8๐3 ๏ฟฝ
1 +32๐๐ โ
32๐2 โ
108๐๐3 +
1516๐4 +โฏ๏ฟฝ
= ๏ฟฝ๐8๐3
โ316
1๐2โ316๐๐+1064+
15(16) (8)
๐๐ +โฏ๏ฟฝ
But ๐ง = ๐ + ๐ or ๐ = ๐ง โ ๐, and the above becomes
๐ (๐ง) = ๏ฟฝ๐
8 (๐ง โ ๐)3โ316
1(๐ง โ ๐)2
โ316
๐(๐ง โ ๐)
+532+15128
๐ (๐ง โ ๐) +โฏ๏ฟฝ (4)
Which is valid for |๐ง โ ๐| < 1. In other words, inside a disk of radius 2, centered around๐ง = ๐.
Comparing (4) with (1A), shows they are the same as expected. Which is the bettermethod? After working both, I think the second method is faster, but requires carefultransformation, the first method is more direct but requires more computations.
ii) Let โซโ
0๐๐ฅ
๏ฟฝ๐ฅ2+1๏ฟฝ3 = ๐ผ, hence, because
1
๏ฟฝ๐ฅ2+1๏ฟฝ3 is even, then
๐ผ =12 ๏ฟฝ
โ
โโ
๐๐ฅ
๏ฟฝ๐ฅ2 + 1๏ฟฝ3
=12
lim๐ โโ
๏ฟฝ๐
โ๐
๐๐ฅ
๏ฟฝ๐ฅ2 + 1๏ฟฝ3
=12
lim๐ โโ
โโโโโโโโ๏ฟฝ
๐
โ๐
๐๐ฅ
๏ฟฝ๐ฅ2 + 1๏ฟฝ3 +โฎ
๐ถ
๐๐ง
๏ฟฝ๐ง2 + 1๏ฟฝ3
โโโโโโโโ
The above is valid as long as one can show โฎ๐ถ
๐๐ง
๏ฟฝ๐ง2+1๏ฟฝ3 โ 0 as ๐ โ โ. The contour ๐ถ is
from ๐ to โ๐ over semicircle, going anticlock wise. The radius of the circle is ๐ . Since theabove integration now includes ๐ง = ๐, then by residual theorem, the above is just โ 3๐
16 . Theresidue was found in the first part. In other words
12
2๐๐๏ฟฝโ 3๐16 ๏ฟฝ
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝlim๐ โโ
โโโโโโโโ๏ฟฝ
๐
โ๐
๐๐ฅ
๏ฟฝ๐ฅ2 + 1๏ฟฝ3 +โฎ
๐ถ
๐๐ง
๏ฟฝ๐ง2 + 1๏ฟฝ3
โโโโโโโโ =
12 ๏ฟฝ2๐๐ ๏ฟฝโ
3๐16๏ฟฝ๏ฟฝ
Letting ๐ง = ๐ ๐๐๐ and taking ๐ โ โ, then โฎ๐ถ
๐๐ง
๏ฟฝ๐ง2+1๏ฟฝ3 โ 0 and the above simplifies to
12
lim๐ โโ
๏ฟฝ๐
โ๐
๐๐ฅ
๏ฟฝ๐ฅ2 + 1๏ฟฝ3 =
12 ๏ฟฝ2๐
316๏ฟฝ
12
lim๐ โโ
๏ฟฝ๐
โ๐
๐๐ฅ
๏ฟฝ๐ฅ2 + 1๏ฟฝ3 = ๐
316
lim๐ โโ
๏ฟฝ๐
0
๐๐ฅ
๏ฟฝ๐ฅ2 + 1๏ฟฝ3 =
3๐16
Therefore
๏ฟฝโ
0
๐๐ฅ
๏ฟฝ๐ฅ2 + 1๏ฟฝ3 =
3๐16
2.1.2 Problem 2
Expand ๐ (๐ฅ) = ๐ฅ๐ฟ as Fourier series for 0 < ๐ฅ < ๐
๐ฟ and ๐ (๐ฅ) = โ ๐ฅ๐ฟ for โ๐๐ฟ < ๐ฅ < 0.
solution:
This function is even. For example, for ๐ฟ = 2, it looks like this
16
Hence the Fourier series will not have sin terms.
๐ (๐ฅ) =๐02+
โ๏ฟฝ๐=1
๐๐ cos ๏ฟฝ2๐๐๐๐ฅ๏ฟฝ +
โ๏ฟฝ๐=1
๐๐ sin ๏ฟฝ2๐๐๐๐ฅ๏ฟฝ
=๐02+
โ๏ฟฝ๐=1
๐๐ cos ๏ฟฝ2๐๐๐๐ฅ๏ฟฝ
Where in the above ๐ is the period of the function. In this problem ๐ = 2๐๐ฟ , hence the above
becomes
๐ (๐ฅ) =๐02+
โ๏ฟฝ๐=1
๐๐ cos
โโโโโโโ2๐๐2๐๐ฟ
๐ฅ
โโโโโโโ
=๐02+
โ๏ฟฝ๐=1
๐๐ cos (๐๐ฟ๐ฅ) (1)
Where
๐0 =1๐2
๏ฟฝ๐2
โ๐2
๐ (๐ฅ) ๐๐ฅ =22๐๐ฟ
๏ฟฝ๐๐ฟ
โ๐๐ฟ
๐ (๐ฅ) ๐๐ฅ =๐ฟ๐ ๏ฟฝ
๐๐ฟ
โ๐๐ฟ
๐ (๐ฅ) ๐๐ฅ
=๐ฟ๐(2)๏ฟฝ
๐๐ฟ
0
๐ฅ๐ฟ๐๐ฅ
=2๐ฟ๐1๐ฟ ๏ฟฝ
๐ฅ2
2 ๏ฟฝ
๐๐ฟ
0
=1๐ ๏ฟฝ
๐2
๐ฟ2 ๏ฟฝ
=๐๐ฟ2
And
๐๐ =1๐2
๏ฟฝ๐๐ฟ
โ๐๐ฟ
๐ (๐ฅ) cos (๐๐ฟ๐ฅ) ๐๐ฅ
=22๐๐ฟ
(2)๏ฟฝ๐๐ฟ
0๐ (๐ฅ) cos (๐๐ฟ๐ฅ) ๐๐ฅ
=2๐ฟ๐ ๏ฟฝ
๐๐ฟ
0
๐ฅ๐ฟ
cos (๐๐ฟ๐ฅ) ๐๐ฅ
=2๐ ๏ฟฝ
๐๐ฟ
0๐ฅ cos (๐๐ฟ๐ฅ) ๐๐ฅ
Using integration by parts โซ๐ข๐๐ฃ = ๐ข๐ฃ โ โซ๐ฃ๐๐ข. Let ๐ข = ๐ฅ, ๐๐ข = 1 and let ๐๐ฃ = cos (๐๐ฟ๐ฅ) , ๐ฃ =
17
sin(๐๐ฟ๐ฅ)๐๐ฟ , therefore the above becomes
๐๐ =2๐
โโโโโโโ๏ฟฝ๐ฅ
sin (๐๐ฟ๐ฅ)๐๐ฟ ๏ฟฝ
๐๐ฟ
0โ๏ฟฝ
๐๐ฟ
0
sin (๐๐ฟ๐ฅ)๐๐ฟ
๐๐ฅ
โโโโโโโ
=2๐
โโโโโโโ
โกโขโขโขโขโขโฃ๐๐ฟ
sin ๏ฟฝ๐๐ฟ๐๐ฟ ๏ฟฝ๐๐ฟ
โ 0
โคโฅโฅโฅโฅโฅโฆ โ
1๐๐ฟ ๏ฟฝ
๐๐ฟ
0sin (๐๐ฟ๐ฅ) ๐๐ฅ
โโโโโโโ
=2๐
โโโโโโ
1๐๐ฟ ๏ฟฝ
๐๐ฟ
0sin (๐๐ฟ๐ฅ) ๐๐ฅ
โโโโโ
=โ2๐๐๐ฟ
โโโโโ๏ฟฝ
๐๐ฟ
0sin (๐๐ฟ๐ฅ) ๐๐ฅ
โโโโโ
=โ2๐๐๐ฟ ๏ฟฝ
โcos (๐๐ฟ๐ฅ)
๐๐ฟ ๏ฟฝ
๐๐ฟ
0
=2
๐๐2๐ฟ2(cos (๐๐ฟ๐ฅ))
๐๐ฟ0
=2
๐๐2๐ฟ2๏ฟฝcos ๏ฟฝ๐๐ฟ๐
๐ฟ๏ฟฝ โ 1๏ฟฝ
=2
๐๐2๐ฟ2(cos (๐๐) โ 1)
=2
๐๐2๐ฟ2๏ฟฝ(โ1)๐ โ 1๏ฟฝ
=โ2 + 2 (โ1)๐
๐๐2๐ฟ2Therefore from (1) the Fourier series is
๐ (๐ฅ) =๐02+
โ๏ฟฝ๐=1
๐๐ cos (๐๐ฟ๐ฅ)
=๐2๐ฟ2
+โ๏ฟฝ๐=1
2๐๐2๐ฟ2
๏ฟฝ(โ1)๐ โ 1๏ฟฝ cos (๐๐ฟ๐ฅ)
The convergence is of order ๐2, so it is fast. Only few terms are needed to obtain very goodapproximation.
2.1.3 Problem 3
(i) Solve ๐ฅ๐ฆโฒ + 3๐ฅ + ๐ฆ = 0. (ii) Solve ๐ฆโฒโฒ โ 2๐ฆโฒ + ๐ฆ = ๐๐ฅ
Solution
(i). This is linear first order ODE.
๐ฆโฒ + 3 +๐ฆ๐ฅ= 0 ๐ฅ โ 0
๐ฆโฒ +๐ฆ๐ฅ= โ3
Integrating factor is ๐ = ๐โซ1๐ฅ๐๐ฅ = ๐ln ๐ฅ = ๐ฅ. Multiplying both sides of the above by ๐, the left
side becomes complete di๏ฟฝerential and it simplifies to๐๐๐ฅ๏ฟฝ๐๐ฆ๏ฟฝ = โ3๐
๐๐๐ฅ๏ฟฝ๐ฅ๐ฆ๏ฟฝ = โ3๐ฅ
๐ ๏ฟฝ๐ฅ๐ฆ๏ฟฝ = โ3๐ฅ๐๐ฅ
Integrating gives
๐ฅ๐ฆ = โ32๐ฅ2 + ๐ถ
Hence the solution is
๐ฆ = โ32๐ฅ +๐ถ๐ฅ ๐ฅ โ 0
(ii) ๐ฆโฒโฒ โ 2๐ฆโฒ + ๐ฆ = ๐๐ฅ is linear second order with constant coe๏ฟฝcients. The solution to the
18
homogeneous part ๐ฆโฒโฒโ2๐ฆโฒ+๐ฆ = 0 can be found by first finding the roots of the characteristicequation ๐ 2 โ 2๐ + 1 = 0 , hence ๐ = โ๐
2๐ ยฑ12๐โ๐
2 โ 4๐๐ or, ๐ = 22 ยฑ
12โ4 โ 4 = 1. One double root.
Therefore
๐ฆโ (๐ฅ) = ๐ถ1๐๐ฅ + ๐ถ2๐ฅ๐๐ฅ
To find the particular solution, the method of undetermined coe๏ฟฝcients is used. Since theforcing function is ๐๐ฅ, then a guess ๐ฆ๐ = ๐๐๐ฅ. But ๐๐ฅ is a basis solution. Hence ๐ฆ๐ = ๐๐ฅ๐๐ฅ isnow selected. But also ๐ฅ๐๐ฅ is basis solution. Then ๐ฆ๐ = ๐๐ฅ2๐๐ฅ is finally selected. Substitutingthis into the original ODE in order to solve for ๐, gives
๐ฆโฒโฒ๐ โ 2๐ฆโฒ๐ + ๐ฆ๐ = ๐๐ฅ
But ๐ฆโฒ๐ = 2๐๐ฅ๐๐ฅ + ๐๐ฅ2๐๐ฅ and ๐ฆโฒโฒ๐ = 2๐๐๐ฅ + 2๐๐ฅ๐๐ฅ + 2๐๐ฅ๐๐ฅ + ๐๐ฅ2๐๐ฅ. Hence the above becomes
๏ฟฝ2๐๐๐ฅ + 2๐๐ฅ๐๐ฅ + 2๐๐ฅ๐๐ฅ + ๐๐ฅ2๐๐ฅ๏ฟฝ โ 2 ๏ฟฝ2๐๐ฅ๐๐ฅ + ๐๐ฅ2๐๐ฅ๏ฟฝ + ๐๐ฅ2๐๐ฅ = ๐๐ฅ
๏ฟฝ2๐ + 2๐๐ฅ + 2๐๐ฅ + ๐๐ฅ2๏ฟฝ โ 2 ๏ฟฝ2๐๐ฅ + ๐๐ฅ2๏ฟฝ + ๐๐ฅ2 = 12๐ + 4๐๐ฅ + ๐๐ฅ2 โ 4๐๐ฅ โ 2๐๐ฅ2 + ๐๐ฅ2 = 1
2๐ = 1
๐ =12
Therefore ๐ฆ๐ =12๐ฅ2๐๐ฅ, and the complete general solution is
๐ฆ (๐ฅ) = ๐ฆโ (๐ฅ) + ๐ฆ๐ (๐ฅ)
Therefore
๐ฆ (๐ฅ) = ๐ถ1๐๐ฅ + ๐ถ2๐ฅ๐๐ฅ +12๐ฅ2๐๐ฅ
2.1.4 Problem 4
Figure 2.1: Problem 4 Statement
solution
Part (1)
๐ฅ2๐ฆโฒโฒ + 2๐ฅ๐ฆโฒ + ๐ฅ2๐ฆ = 0
๐ฆโฒโฒ +2๐ฅ๐ฆโฒ + ๐ฆ = 0
๐ฅ is a regular singular point. Becuase lim๐ฅโ0 (๐ฅ โ 0)2๐ฅ = 2. Since limit exist, then regular
singular point.
19
Part (2)
Let
๐ฆ =โ๏ฟฝ๐=0
๐๐๐ฅ๐
๐ฆโฒ =โ๏ฟฝ๐=0
๐๐๐๐ฅ๐โ1
๐ฆโฒโฒ =โ๏ฟฝ๐=0
๐ (๐ โ 1) ๐๐๐ฅ๐โ2
The ODE becomes
๐ฅ2โ๏ฟฝ๐=0
๐ (๐ โ 1) ๐๐๐ฅ๐โ2 + 2๐ฅโ๏ฟฝ๐=0
๐๐๐๐ฅ๐โ1 + ๐ฅ2โ๏ฟฝ๐=0
๐๐๐ฅ๐ = 0
โ๏ฟฝ๐=0
๐ (๐ โ 1) ๐๐๐ฅ๐ +โ๏ฟฝ๐=0
2๐๐๐๐ฅ๐ +โ๏ฟฝ๐=0
๐๐๐ฅ๐+2 = 0
โ๏ฟฝ๐=0
๐ (๐ โ 1) ๐๐๐ฅ๐ +โ๏ฟฝ๐=0
2๐๐๐๐ฅ๐ +โ๏ฟฝ๐=2
๐๐โ2๐ฅ๐ = 0
Thereforeโ๏ฟฝ๐=0
(๐ (๐ โ 1) + 2๐) ๐๐๐ฅ๐ +โ๏ฟฝ๐=2
๐๐โ2๐ฅ๐ = 0
We start from ๐ = 1 since ๐ = 0 is used to find the indicial equation. For ๐ = 1
2๐1 = 0๐1 = 0
For ๐ โฅ 2(๐ (๐ โ 1) + 2๐) ๐๐ + ๐๐โ2 = 0
๐๐ (๐ (๐ โ 1) + 2๐) = โ๐๐โ2
๐๐ =โ๐๐โ2
๐ (๐ โ 1) + 2๐For ๐ = 2
๐2 =โ๐0
2 (2 โ 1) + 4
= โ16๐0
For ๐ = 3
๐3 =โ๐1
3 (2) + 6Since ๐1 = 0 then ๐3 = 0. All odd terms are therefore zero.
For ๐ = 4
๐๐ =โ๐2
(4) (3) + 8= โ
120๐2 = โ
120 ๏ฟฝ
โ16๐0๏ฟฝ =
1120
๐0
Therefore
๐ฆ1 (๐ฅ) =โ๏ฟฝ๐=0
๐๐๐ฅ๐
= ๐0 + ๐2๐ฅ2 + ๐4๐ฅ4 +โฏ
= ๐0 โ16๐0๐ฅ2 +
160๐0๐ฅ4 โโฏ
= ๐0 ๏ฟฝ1 โ16๐ฅ2 +
11200
๐ฅ4 โโฏ๏ฟฝ
Part (3)
setting ๐0 = 1 as problem says, and since sin ๐ฅ = ๐ฅ โ ๐ฅ3
6 +๐ฅ5
120 โโฏ then the above is
๐ฆ1 (๐ฅ) =sin ๐ฅ๐ฅ
20
Part (4)
๐(๐ฅ) = ๏ฟฝ๐ฆ1 ๐ฆ2๐ฆโฒ1 ๐ฆโฒ2
๏ฟฝ = ๏ฟฝsin ๐ฅ๐ฅ ๐ฆ2
๐ฅ cos ๐ฅโsin ๐ฅ๐ฅ2 ๐ฆโฒ2
๏ฟฝ = ๐ฆโฒ2sin ๐ฅ๐ฅ
โ ๐ฆ2๐ฅ cos ๐ฅ โ sin ๐ฅ
๐ฅ2
But from Abelโs theorem, ๐(๐ฅ) = ๐ถ๐โซโ๐(๐ฅ)๐๐ฅ where ๐ (๐ฅ) is the coe๏ฟฝcient in the ODE
๐ฆโฒโฒ + ๐ (๐ฅ) ๐ฆโฒ + ๐ (๐ฅ) ๐ฆ = 0. Since the ODE is ๐ฆโฒโฒ โ 2๐ฅ + ๐ฆ = 0 then ๐ =
โ2๐ฅ and ๐(๐ฅ) = ๐โซ
โ2๐ฅ ๐๐ฅ =
๐โ2 ln ๐ฅ = ๐ฅโ2. Hence
๐ฆโฒ2sin ๐ฅ๐ฅ
โ ๐ฆ2๐ฅ cos ๐ฅ โ sin ๐ฅ
๐ฅ2=๐ถ๐ฅ2
๐ฅ sin (๐ฅ) ๐ฆโฒ2 โ ๐ฆ2 (๐ฅ cos ๐ฅ โ sin ๐ฅ) = ๐ถ
๐ฆโฒ2 โ ๐ฆ2 ๏ฟฝcos ๐ฅsin ๐ฅ โ
1๐ฅ๏ฟฝ=
๐ถ๐ฅ sin ๐ฅ
Integrating factor is ๐ = ๐โซโcos ๐ฅsin ๐ฅ +
1๐ฅ๐๐ฅ = ๐โซ
โ ๐๐๐ฅ (sin ๐ฅ)
sin ๐ฅ ๐๐ฅ๐โซ1๐ฅ๐๐ฅ = ๐โซ
โ1sin ๐ฅ๐(sin ๐ฅ) + ๐ln ๐ฅ = ๐โ ln(sin ๐ฅ)๐ฅ =
๐ฅsin ๐ฅ . Multiplying both sides by this integrating factor gives
๐๐๐ฅ๏ฟฝ๐ฆ2
๐ฅsin ๐ฅ
๏ฟฝ =๐ถ
sin2 ๐ฅIntegrating gives
๐ฆ2๐ฅ
sin ๐ฅ =๐ถ cos (๐ฅ)โ sin (๐ฅ) + ๐ถ2
๐ฆ2 (๐ฅ) = โ๐ถcos ๐ฅ๐ฅ
+ ๐ถ2sin ๐ฅ๐ฅ
= ๐ถ1cos ๐ฅ๐ฅ
+ ๐ถ2sin ๐ฅ๐ฅ
But ๐ถ2 sin ๐ฅ is the second solution, so we only keep ๐ฆ2 (๐ฅ) = ๐ถ1cos ๐ฅ๐ฅ . Hence the final solution
is
๐ฆ (๐ฅ) = ๐ถ1cos ๐ฅ๐ฅ
+ ๐ถ2sin ๐ฅ๐ฅ
2.1.5 Problem 5
Find solution to ๐2
๐๐ฅ2๐บ (๐ฅ, ๐ฅ0)+๐2๐บ (๐ฅ, ๐ฅ0) = ๐ฟ (๐ฅ โ ๐ฅ0) subject to boundary conditions ๐บ (0, ๐ฅ0) =
๐บ (๐ฟ, ๐ฅ0) = 0
solution
First we obtain the solution to the homogeneous ODE ๐ฆโฒโฒ + ๐2๐ฆ = 0 with ๐ฆ (0) = 0, ๐ฆ (๐) = 0which has the two solutions ๐ฆ1 (๐ฅ) = cos ๐๐ฅ, ๐ฆ2 (๐ฅ) = sin (๐๐ฅ).
Therefore the solution to the Green function is
๐บ (๐ฅ, ๐ฅ0) =
โงโชโชโจโชโชโฉ๐ด1๐ฆ1 (๐ฅ) + ๐ด2๐ฆ2 (๐ฅ) 0 < ๐ฅ < ๐ฅ0๐ต1๐ฆ1 (๐ฅ) + ๐ต2๐ฆ2 (๐ฅ) ๐ฅ < ๐ฅ0 < ๐
=
โงโชโชโจโชโชโฉ๐ด1 cos (๐๐ฅ) + ๐ด2 sin (๐๐ฅ) 0 < ๐ฅ < ๐ฅ0๐ต1 cos (๐๐ฅ) + ๐ต2 sin (๐๐ฅ) ๐ฅ < ๐ฅ0 < ๐
Where ๐ด๐, ๐ต๐ are constants to be found. From the condition ๐บ (0, ๐ฅ0) = 0 then the firstsolution above gives ๐ด1 cos (0) = 0 โ ๐ด1 = 0. And from ๐บ (๐ฟ, ๐ฅ0) = 0 the second solutionabove gives ๐ต1 cos (๐๐ฟ) + ๐ต2 sin (๐๐ฟ๐) = 0 or ๐ต1 = โ
๐ต2 sin(๐๐ฟ)cos(๐๐ฟ๐) , hence the solution now becomes
21
๐บ (๐ฅ, ๐ฅ0) =
โงโชโชโจโชโชโฉ
๐ด2 sin (๐๐ฅ) 0 < ๐ฅ < ๐ฅ0๐ต1 cos (๐๐ฅ) + ๐ต2 sin (๐๐ฅ) ๐ฅ < ๐ฅ0 < ๐ฟ
=
โงโชโชโจโชโชโฉ
๐ด2 sin (๐๐ฅ) 0 < ๐ฅ < ๐ฅ0โ๐ต2 sin(๐๐ฟ)
cos(๐๐ฟ) cos (๐๐ฅ) + ๐ต2 sin (๐๐ฅ) ๐ฅ < ๐ฅ0 < ๐ฟ
=
โงโชโชโจโชโชโฉ
๐ด2 sin (๐๐ฅ) 0 < ๐ฅ < ๐ฅ0๐ต2
cos(๐๐ฟ) (sin (๐๐ฅ) cos (๐๐ฟ) โ sin (๐๐ฟ) cos (๐๐ฅ)) ๐ฅ < ๐ฅ0 < ๐ฟ
Using sin (๐ โ ๐) = sin ๐ cos ๐โcos ๐ sin ๐, then sin (๐๐ฅ) cos (๐๐ฟ)โsin (๐๐ฟ) cos (๐๐ฅ) = sin (๐๐ฅ โ ๐๐ฟ) =sin (๐ (๐ฅ โ ๐ฟ)) and the above becomes
๐บ (๐ฅ, ๐ฅ0) =
โงโชโชโจโชโชโฉ
๐ด2 sin (๐๐ฅ) 0 < ๐ฅ < ๐ฅ0๐ต2
cos(๐๐ฟ) sin (๐ (๐ฅ โ ๐ฟ)) ๐ฅ < ๐ฅ0 < ๐ฟ(1A)
Now from continuity condition ๐บ (๐ฅ, ๐ฅ0)๐ฅ=๐ฅ0โ๐ = ๐บ (๐ฅ, ๐ฅ0)๐ฅ=๐ฅ0+๐ i.e. at ๐ฅ = ๐ฅ0, then (from nowon, we switch to ๐ฅ0).
๐ด2 sin (๐๐ฅ0) =๐ต2
cos (๐๐ฟ) sin (๐ (๐ฅ0 โ ๐ฟ)) (1)
Now we find the derivative of ๐บ at ๐ฅ = ๐ฅ0 gives
๐๐๐ฅ๐บ (๐ฅ, ๐ฅ0)๏ฟฝ
๐ฅ=๐ฅ0=
โงโชโชโจโชโชโฉ
๐๐ด2 cos (๐๐ฅ0) 0 < ๐ฅ < ๐ฅ0๐๐ต2
cos(๐๐ฟ) cos (๐ (๐ฅ0 โ ๐ฟ)) ๐ฅ < ๐ฅ0 < ๐ฟ
And from jump discontinuty in derivative of ๐บ at ๐ฅ = ๐ฅ0 would obtain, since ๐บโฒ๐ฅ>๐ฅ0+๐ โ๐บโฒ๐ฅ<๐ฅ0โ๐ =
โ1๐(๐ฅ) , then
๐๐ต2cos (๐๐) cos (๐ (๐ฅ0 โ ๐ฟ)) โ ๐๐ด2 cos (๐๐ฅ0) =
โ1๐ (๐ฅ)
But since the ODE is ๐ฆโฒโฒ + ๐2๐ฆ = 0 then in SL form this is โ ๏ฟฝ๐ฆโฒ๏ฟฝโฒ+ ๐2๐ฆ = 0, and comparing
to โ ๏ฟฝ๐๐ฆโฒ๏ฟฝโฒ+ ๐2๐ฆ = 0 we see that ๐ = โ1. Then above becomes
๐๐ต2cos (๐๐ฟ) cos (๐ (๐ฅ0 โ ๐ฟ)) โ ๐๐ด2 cos (๐๐ฅ0) = 1 (2)
From (1,2) we solve for ๐ด1, ๐ต2. From (1)
๐ด2 =๐ต2
cos (๐๐) sin (๐๐ฅ0)sin (๐ (๐ฅ0 โ ๐ฟ)) (3)
Substituting into (2) gives
๐๐ต2cos (๐๐ฟ) cos (๐ (๐ฅ0 โ ๐ฟ)) โ ๐ ๏ฟฝ
๐ต2cos (๐๐ฟ) sin (๐๐ฅ0)
sin (๐ (๐ฅ0 โ ๐ฟ))๏ฟฝ cos (๐๐ฅ0) = 1
๐๐ต2 cos (๐ (๐ฅ0 โ ๐ฟ)) โ ๐๐ต2 sin (๐ (๐ฅ0 โ ๐ฟ))cos (๐๐ฅ0)sin (๐๐ฅ0)
= cos (๐๐ฟ)
๐๐ต2 (sin (๐๐ฅ0) cos (๐ (๐ฅ0 โ ๐ฟ)) โ sin (๐ (๐ฅ0 โ ๐ฟ)) cos (๐๐ฅ0)) = cos (๐๐ฟ) sin (๐๐ฅ0) (4)
Using sin (๐ โ ๐) = sin ๐ cos ๐ โ cos ๐ sin ๐, thensin (๐๐ฅ0) cos (๐ (๐ฅ0 โ ๐ฟ)) โ sin (๐ (๐ฅ0 โ ๐ฟ)) cos (๐๐ฅ0) = sin (๐๐ฅ0 โ ๐ (๐ฅ0 โ ๐ฟ))
= sin (๐๐ฟ)
22
Then (3) becomes
๐๐ต2 sin (๐๐ฟ) = cos (๐๐ฟ) sin (๐๐ฅ0)
๐ต2 =cos (๐๐ฟ) sin (๐๐ฅ0)
๐ sin (๐๐ฟ)
Substituing the above in (3) gives
๐ด2 =cos (๐๐ฟ) sin (๐๐ฅ0)
๐ sin (๐๐ฟ) cos (๐๐ฟ) sin (๐๐ฅ0)sin (๐ (๐ฅ0 โ ๐ฟ))
=sin (๐ (๐ฅ0 โ ๐ฟ))๐ sin (๐๐ฟ)
Using ๐ด2, ๐ต2 found above in (1A) gives
๐บ (๐ฅ, ๐ฅ0) =
โงโชโชโจโชโชโฉ
sin(๐(๐ฅ0โ๐ฟ))๐ sin(๐๐ฟ) sin (๐๐ฅ) 0 < ๐ฅ < ๐ฅ0
cos(๐๐ฟ) sin(๐๐ฅ0)๐ sin(๐๐ฟ) cos(๐๐) sin (๐ (๐ฅ โ ๐ฟ)) ๐ฅ < ๐ฅ0 < ๐ฟ
=1
๐ sin (๐๐ฟ)
โงโชโชโจโชโชโฉ
sin (๐ (๐ฅ0 โ ๐ฟ)) sin (๐๐ฅ) 0 < ๐ฅ < ๐ฅ0sin (๐๐ฅ0) sin (๐ (๐ฅ โ ๐ฟ)) ๐ฅ < ๐ฅ0 < ๐ฟ
The following approach seems faster.
second solution
Instead of starting from
๐บ (๐ฅ, ๐ฅ0) =
โงโชโชโจโชโชโฉ๐ด1๐ฆ1 (๐ฅ) + ๐ด2๐ฆ2 (๐ฅ) 0 < ๐ฅ < ๐ฅ0๐ต1๐ฆ1 (๐ฅ) + ๐ต2๐ฆ2 (๐ฅ) ๐ฅ < ๐ฅ0 < ๐
We first find the eigenfunction ฮฆ๐ (๐ฅ) that solves ๐ฆโฒโฒ + ๐2๐ฆ = 0 which satisfies the boundaryconditions ๐ฆ (0) = 0, ๐ฆ (๐ฟ) = 0. Then write
๐บ (๐ฅ, ๐ฅ0) =
โงโชโชโจโชโชโฉ
๐ดฮฆ๐ (๐ฅ) 0 < ๐ฅ < ๐ฅ0๐ตฮฆ๐ (๐ฅ โ ๐ฟ) ๐ฅ < ๐ฅ0 < ๐
So now we have only 2 unknowns to find, ๐ด,๐ต using the continutity and jump conditionson ๐บ. Let see how this works on this same problem. The solution to ๐ฆโฒโฒ + ๐2๐ฆ = 0 is๐ฆ (๐ฅ) = ๐ด cos ๐๐ฅ + ๐ต sin ๐๐ฅ. At ๐ฆ (0) = 0 implies ๐ด = 0, so the solution becomes ๐ฆ (๐ฅ) = ๐ต sin ๐๐ฅand at ๐ฅ (๐ฟ) = 0 this gives 0 = ๐ต sin (๐๐ฟ), which implies ๐๐ฟ = ๐๐ or ๐ = ๐๐
๐ฟ . Hence the solution
is ฮฆ๐ (๐ฅ) = sin ๏ฟฝ๐๐๐ฟ ๐ฅ๏ฟฝ. Therefore we set up the Green function as
๐บ (๐ฅ, ๐ฅ0) =
โงโชโชโจโชโชโฉ
๐ด sin ๏ฟฝ๐๐๐ฟ ๐ฅ๏ฟฝ 0 < ๐ฅ < ๐ฅ0๐ต sin ๏ฟฝ๐๐๐ฟ (๐ฅ โ ๐ฟ)๏ฟฝ ๐ฅ < ๐ฅ0 < ๐
Or by letting ๐๐ =๐๐๐ฟ
๐บ (๐ฅ, ๐ฅ0) =
โงโชโชโจโชโชโฉ
๐ด sin (๐๐๐ฅ) 0 < ๐ฅ < ๐ฅ0๐ต sin (๐๐ (๐ฅ โ ๐ฟ)) ๐ฅ < ๐ฅ0 < ๐ฟ
(1)
Now continutity says
๐ด sin (๐๐๐ฅ0) = ๐ต sin (๐๐ (๐ฅ0 โ ๐ฟ)) (2)
23
Taking derivative of (1) at ๐ฅ = ๐ฅ0 gives
๐บโฒ (๐ฅ, ๐ฅ0) =
โงโชโชโจโชโชโฉ
๐ด๐๐ cos (๐๐๐ฅ0) 0 < ๐ฅ < ๐ฅ0๐ต๐๐ cos (๐๐ (๐ฅ0 โ ๐ฟ)) ๐ฅ < ๐ฅ0 < ๐ฟ
Then jump discontinutiy gives
๐ต๐๐ cos (๐๐ (๐ฅ0 โ ๐ฟ)) โ ๐ด๐๐ cos (๐๐๐ฅ0) = 1 (3)
Solving (2,3) for ๐ด,๐ต gives, as we did earlier
๐ด =sin (๐ (๐ฅ0 โ ๐ฟ))๐ sin (๐๐ฟ)
๐ต =sin (๐๐ฅ0)๐ sin (๐๐ฟ)
Using these in (1) gives
๐บ (๐ฅ, ๐ฅ0) =1
๐ sin (๐๐ฟ)
โงโชโชโจโชโชโฉ
sin (๐ (๐ฅ0 โ ๐ฟ)) sin (๐๐๐ฅ) 0 < ๐ฅ < ๐ฅ0sin (๐๐ฅ0) sin (๐๐ (๐ฅ โ ๐ฟ)) ๐ฅ < ๐ฅ0 < ๐ฟ๐
Which is the same result obtained earlier.
2.2 exam 1
2.2.1 exam 1 prep sheet
Preparation sheet for the test: November 7 in class
The test will cover notes and examples on infinite series, complex analysis, andevaluation of integrals (not including the saddle point integration) and problemsets 1-4.
1. Things to know without thought:
(a) Sum of a geometric series.(b) Definitions of convergence and absolute convergences of infinite series.(c) Ratio test for convergence.(d) Series expansions for ex, sin x, cos x, and ln(1 + x) (for |x| < 1).(e) Cauchy-Riemann conditions and definition of analytic functions.(f) The meaning of log z, Logz, and zc
(g) Cauchy-Goursat Theorem and Cauchy Integral formula.(h) The form of Taylor and Laurent series.(i) The meaning of analytic continuation.(j) The residue theorem.(k) The definition of Cauchy principle value.(l) The definition of an asymptotic series.
24
2.2.2 exam 1 writeup
Problem 1
Using a well known sum, find a closed for expression for the following series
๐ (๐ง) = 1 + 2๐ง + 3๐ง2 + 4๐ง3 + 5๐ง4 +โฏ
Using the ratio test, find for what values of ๐ง this series converges.
Solution
Method 1
Assume that the closed form is
(1 โ ๐ง)๐ = 1 + 2๐ง + 3๐ง2 + 4๐ง3 + 5๐ง4 +โฏ
For some unknown ๐. Now ๐ will be solved for. Using Binomial series definition (1 โ ๐ง)๐ =1 โ ๐๐ง + (๐)(๐โ1)
2! ๐ง2 โ ๐(๐โ1)(๐โ2)3! ๐ง3 +โฏ. in the LHS above gives
1 โ ๐๐ง +๐ (๐ โ 1)2!
๐ง2 โ๐ (๐ โ 1) (๐ โ 2)
3!๐ง3 +โฏ = 1 + 2๐ง + 3๐ง2 + 4๐ง3 + 5๐ง4 +โฏ
By comparing coe๏ฟฝcients of ๐ง in the left side and on the right side shows that ๐ = โ2 fromthe coe๏ฟฝcient of ๐ง term. Verifying this on the coe๏ฟฝcient of ๐ง2 shows it is correct since itgives (โ2)(โ3)
2 = 3. Therefore
๐ = โ2
The closed form is therefore1
(1 โ ๐ง)2= 1 + 2๐ง + 3๐ง2 + 4๐ง3 + 5๐ง4 +โฏ
Method 2
Starting with Binomial series expansion given by1
1 โ ๐ง= 1 + ๐ง + ๐ง2 + ๐ง3 + ๐ง4 +โฏ
Taking derivative w.r.t. ๐ง on both sides of the above results in
๐๐๐ง ๏ฟฝ
11 โ ๐ง๏ฟฝ
=๐๐๐ง๏ฟฝ1 + ๐ง + ๐ง2 + ๐ง3 + ๐ง4 +โฏ๏ฟฝ
โ (1 โ ๐ง)โ2 (โ1) = 0 + 1 + 2๐ง + 3๐ง2 + 4๐ง3 +โฏ1
(1 โ ๐ง)2= 1 + 2๐ง + 3๐ง2 + 4๐ง3 +โฏ
Therefore the closed form expression is1
(1 โ ๐ง)2= 1 + 2๐ง + 3๐ง2 + 4๐ง3 +โฏ
Which is the same as method 1.
The series general term of the series is
1 + 2๐ง + 3๐ง2 + 4๐ง3 +โฏ =โ๏ฟฝ๐=0
(๐ + 1) ๐ง๐
Applying the ratio test
๐ฟ = lim๐โโ
๏ฟฝ๐๐+1๐๐
๏ฟฝ
= lim๐โโ
๏ฟฝ(๐ + 2) ๐ง๐+1
(๐ + 1) ๐ง๐ ๏ฟฝ
= lim๐โโ
๏ฟฝ(๐ + 2) ๐ง๐ + 1
๏ฟฝ
= ๐ง lim๐โโ
๏ฟฝ๐ + 2๐ + 1
๏ฟฝ
= ๐ง lim๐โโ
๏ฟฝ๏ฟฝ1 + 2
๐
1 + 1๐
๏ฟฝ๏ฟฝ
25
But lim๐โโ ๏ฟฝ1+ 2
๐
1+ 1๐
๏ฟฝ = 1 and the above limit becomes
๐ฟ = ๐ง
By the ratio test, the series converges when |๐ฟ| < 1. Therefore 1+2๐ง+3๐ง2+4๐ง3+โฏ convergesabsolutely when |๐ง| < 1. An absolutely convergent series is also a convergent series. Hencethe series converges for |๐ง| < 1.
Problem 2
Find the Laurent series for the function
๐ (๐ง) =1
๏ฟฝ๐ง2 + 4๏ฟฝ3
About the isolated singular pole ๐ง = 2๐. What is the order of this pole? What is the residueat this pole?
Solution
The poles are at ๐ง2 = 4 or ๐ง = ยฑ2๐. The expansion of ๐ (๐ง) is around the isolated pole at๐ง = 2๐. This pole has order 3. The region where this expansion is valid is inside a diskcentered at 2๐ (but not including the point ๐ง = 2๐ itself) and up to the nearest pole whichis located at โ2๐. Therefore the disk will have radius 4.
2i
โ2i
R = 4
Region where Laurent seriesexpansion around z = 2i is valid
<z
=z
Let
๐ข = ๐ง โ 2๐๐ง = ๐ข + 2๐
26
Substituting this expression for ๐ง back in ๐ (๐ง) gives
๐ (๐ง) =1
๏ฟฝ(๐ข + 2๐)2 + 4๏ฟฝ3
=1
๏ฟฝ๐ข2 โ 4 + 4๐ข๐ + 4๏ฟฝ3
=1
๏ฟฝ๐ข2 + 4๐ข๐๏ฟฝ3
=1
(๐ข (๐ข + 4๐))3
=1๐ข3
1(๐ข + 4๐)3
=1๐ข3
1
๏ฟฝ4๐ ๏ฟฝ ๐ข4๐ + 1๏ฟฝ๏ฟฝ3
=1๐ข3
1
(4๐)3 ๏ฟฝ ๐ข4๐ + 1๏ฟฝ3
=1
โ๐64๐ข31
๏ฟฝ ๐ข4๐ + 1๏ฟฝ
3
= ๏ฟฝ๐
64๐ข3 ๏ฟฝ1
๏ฟฝ ๐ข4๐ + 1๏ฟฝ
3 (1)
Expanding the term 1
๏ฟฝ1+ ๐ข4๐ ๏ฟฝ
3 using Binomial series, which is valid for ๏ฟฝ ๐ข4๐ ๏ฟฝ < 1 or |๐ข| < 4 gives
1
๏ฟฝ1 + ๐ข4๐๏ฟฝ3 = 1 + (โ3)
๐ข4๐+(โ3) (โ4)
2!๏ฟฝ๐ข4๐๏ฟฝ2+(โ3) (โ4) (โ5)
3!๏ฟฝ๐ข4๐๏ฟฝ3+(โ3) (โ4) (โ5) (โ6)
4!๏ฟฝ๐ข4๐๏ฟฝ4+โฏ
= 1 โ 3๐ข4๐+3 โ 42!
๐ข2
16๐2โ3 โ 4 โ 53!
๐ข3
64๐3+3 โ 4 โ 5 โ 6
4!๐ข4
256๐4+โฏ
= 1 + 3๐๐ข4โ3 โ 42!
๐ข2
16โ3 โ 4 โ 53!
๐ข3
64 (โ๐)+3 โ 4 โ 5 โ 6
4!๐ข4
256+โฏ
= 1 + 3๐๐ข4โ3 โ 42!
๐ข2
16โ ๐3 โ 4 โ 53!
๐ข3
64+3 โ 4 โ 5 โ 6
4!๐ข4
256+โฏ (2)
Substituting (2) into (1) and simplifying gives
๐ (๐ง) = ๏ฟฝ๐
64๐ข3 ๏ฟฝ ๏ฟฝ1 + 3๐
๐ข4โ3 โ 42!
๐ข2
16โ ๐3 โ 4 โ 53!
๐ข3
64+3 โ 4 โ 5 โ 6
4!๐ข4
256+โฏ๏ฟฝ
=๐
64๐ข3+
๐64๐ข3
๏ฟฝ3๐๐ข4๏ฟฝ โ
๐64๐ข3 ๏ฟฝ
3 โ 42!
๐ข2
16๏ฟฝโ
๐64๐ข3 ๏ฟฝ
๐3 โ 4 โ 53!
๐ข3
64๏ฟฝ+
๐64๐ข3 ๏ฟฝ
3 โ 4 โ 5 โ 64!
๐ข4
256๏ฟฝ+โฏ
=๐
64๐ข3โ
164๐ข2
34โ
๐64๐ข ๏ฟฝ
3 โ 42!
116๏ฟฝ
+164 ๏ฟฝ
3 โ 4 โ 53!
164๏ฟฝ
+๐64 ๏ฟฝ
3 โ 4 โ 5 โ 64!
๐ข256๏ฟฝ
+โฏ
=๐
64๐ข3โ
3256๐ข2
โ ๐3512
1๐ข+
52048
+ ๐15
16 384๐ข +โฏ
Replacing ๐ข back by ๐ง โ 2๐ in the above results in
๐ (๐ง) =๐64
1(๐ง โ 2๐)3
โ3256
1(๐ง โ 2๐)2
โ3๐512
1(๐ง โ 2๐)
+5
2048+
15๐16 384
(๐ง โ 2๐) +โฏ (3)
This expansion is valid for |๐ง โ 2๐| < 4. The above shows that the residue is
โ 3๐512
Which is the coe๏ฟฝcient of the 1(๐งโ2๐) term in (3).
27
Problem 3
Use residues to evaluate the following integral
๐ผ = ๏ฟฝโ
0
๐๐ฅ๐ฅ4 + 6๐ฅ2 + 9
Solution
The integrand is an even function. Therefore the integral โซโ
โโ๐๐ฅ
๐ฅ4+6๐ฅ2+9is evaluated instead
and then the required integral ๐ผ will be half the value obtained. The poles of 1๐ฅ4+6๐ฅ2+9
are
the zeros of the denominator. Factoring the denominator as ๏ฟฝ๐ฅ2 + 3๏ฟฝ ๏ฟฝ๐ฅ2 + 3๏ฟฝ = 0, shows theroots are ๐ฅ = ยฑ๐โ3 from the first factor and ๐ฅ = ยฑ๐โ3 from the second factor.
Since the upper half plane will be used, the pole located there is +๐โ3 and it is of order two.
Now that pole locations are known, the following contour is used to evaluate โซโ
โโ๐๐ฅ
๐ฅ4+6๐ฅ2+9as shown in the plot below
โR +R
R
CR
<z
=z
iโ3
pole of order 2
โฎ๐ถ
๐ (๐ง) ๐๐ง = lim๐ โโ
๏ฟฝ๐ถ๐ (๐ง) ๐๐ง + lim
๐ โโ๏ฟฝ
+๐
โ๐ ๐ (๐ฅ) ๐๐ฅ
= lim๐ โโ
๏ฟฝ๐ถ
๐๐ง๐ง4 + 6๐ง2 + 9
+ lim๐ โโ
๏ฟฝ+๐
โ๐
๐๐ฅ๐ฅ4 + 6๐ฅ2 + 9
๐๐ฅ (2)
Where the integral โซ+๐
โ๐ is Cauchy principal integral. Since the contour ๐ถ is closed and
because ๐ (๐ง) is analytic on and inside ๐ถ except for the isolated singularity inside at ๐ง = ๐โ3,then by Cauchy integral formula โฎ
๐ถ
๐ (๐ง) ๐๐ง = 2๐๐โResidue. Where the sum of residues is
over all poles inside ๐ถ. Therefore (2) can becomes
๏ฟฝ+โ
โโ
๐๐ฅ๐ฅ4 + 6๐ฅ2 + 9
๐๐ฅ = 2๐๐๏ฟฝResidueโ lim๐ โโ
๏ฟฝ๐ถ๐ (๐ง) ๐๐ง (3)
But
๏ฟฝ๏ฟฝ๐ถ๐ (๐ง) ๐๐ง๏ฟฝ
maxโค ๐๐ฟ
= ๏ฟฝ๐ (๐ง)๏ฟฝmax
๐๐ (4)
Using
๏ฟฝ๐ (๐ง)๏ฟฝmax
โค1
๏ฟฝ๐ง2 + 3๏ฟฝmin
๏ฟฝ๐ง2 + 3๏ฟฝmin
By inverse triangle inequality ๏ฟฝ๐ง2 + 3๏ฟฝ โฅ |๐ง|2 โ 3. But |๐ง| = ๐ on ๐ถ, therefore ๏ฟฝ๐ง2 + 3๏ฟฝ โฅ ๐ 2 โ 3and the above can now be written as
๏ฟฝ๐ (๐ง)๏ฟฝmax
โค1
๏ฟฝ๐ 2 โ 3๏ฟฝ ๏ฟฝ๐ 2 โ 3๏ฟฝ
28
Using the above in (4) gives
๏ฟฝ๏ฟฝ๐ถ๐ (๐ง) ๐๐ง๏ฟฝ
maxโค
๐๐ ๏ฟฝ๐ 2 โ 3๏ฟฝ ๏ฟฝ๐ 2 โ 3๏ฟฝ
=๐๐
๐ 4 โ 6๐ 2 + 9
=๐๐
๐ 2 โ 6 + 9๐ 2
In the limit as ๐ โ โ then ๏ฟฝโซ๐ถ๐ (๐ง) ๐๐ง๏ฟฝ
maxโ 0. Using this result in (3) it simplifies to
๏ฟฝ+โ
โโ
๐๐ฅ๐ฅ4 + 6๐ฅ2 + 9
๐๐ฅ = 2๐๐๏ฟฝResidue (5)
What is left now is to determine the residue at pole ๐ง0 = ๐โ3 which is of order 2. This isdone using
Residue (๐ง0) = lim๐งโ๐ง0
๐๐๐ง๏ฟฝ(๐ง โ ๐ง0)
2 ๐ (๐ง)๏ฟฝ
But ๐ง0 = ๐โ3 and the above becomes
Residue ๏ฟฝ๐โ3๏ฟฝ = lim๐งโ๐โ3
๐๐๐ง
โโโโโโโโ๏ฟฝ๐ง โ ๐โ3๏ฟฝ
2 1
๏ฟฝ๐ง โ ๐โ3๏ฟฝ2๏ฟฝ๐ง + ๐โ3๏ฟฝ
2
โโโโโโโโ
= lim๐งโ๐โ3
๐๐๐ง
1
๏ฟฝ๐ง + ๐โ3๏ฟฝ2
= lim๐งโ๐โ3
โ2
๏ฟฝ๐ง + ๐โ3๏ฟฝ3
=โ2
๏ฟฝ๐โ3 + ๐โ3๏ฟฝ3
=โ2
๏ฟฝ2๐โ3๏ฟฝ3
=โ2
โ (8) (3) ๐โ3
=1
12๐โ3Using the above value of the residue in (5) gives
๏ฟฝ+โ
โโ
๐๐ฅ๐ฅ4 + 6๐ฅ2 + 9
๐๐ฅ = 2๐๐ ๏ฟฝ1
12๐โ3๏ฟฝ
=๐6โ3
Therefore the integral โซโ
0๐๐ฅ
๐ฅ4+6๐ฅ3+9is half of the above result which is
๏ฟฝโ
0
๐๐ฅ๐ฅ4 + 6๐ฅ2 + 9
=๐
12โ3
Problem 4
Find two approximations for the integral ๐ฅ > 0
๐ผ (๐ฅ) =12๐ ๏ฟฝ
๐2
โ๐2
๐๐ฅ cos2 ๐๐๐
One for small ๐ฅ (keeping up to linear order in ๐ฅ) and one for large values of ๐ฅ (keepingonly the leading order term).
Solution
The integrand has the form ๐๐ง. This has a known Taylor series expansion around zero
29
given by
๐๐ง = 1 + ๐ง +๐ง2
2!+โฏ
Replacing ๐ง by ๐ฅ cos2 ๐ in the above gives
๐๐ฅ cos2 ๐ = 1 + ๐ฅ cos2 ๐ +๏ฟฝ๐ฅ cos2 ๐๏ฟฝ
2
2+โฏ
The problem is asking to keep linear terms in ๐ฅ. Therefore
๐๐ฅ cos2 ๐ โ 1 + ๐ฅ cos2 ๐Replacing the integrand in the original integral by the above approximation gives
๐ผ (๐ฅ) โ12๐ ๏ฟฝ
๐2
โ๐2
๏ฟฝ1 + ๐ฅ cos2 ๐๏ฟฝ ๐๐
โ12๐
โโโโโโ๏ฟฝ
๐2
โ๐2
๐๐ + ๐ฅ๏ฟฝ๐2
โ๐2
cos2 ๐๐๐โโโโโโ
โ12๐
โโโโโโ๏ฟฝ
๐2
โ๐2
๐๐ + ๐ฅ๏ฟฝ๐2
โ๐2
12+12
cos 2๐๐๐โโโโโโ
โ12๐
โโโโโโโโ๐ + ๐ฅ ๏ฟฝ
12๐ +
12
sin 2๐2 ๏ฟฝ
๐2
โ๐2
โโโโโโโโ
โ12๐ ๏ฟฝ
๐ +๐ฅ4(2๐ + sin 2๐)
๐2โ๐2๏ฟฝ
โ12๐
๏ฟฝ๐ +๐ฅ4(2๐ + 0)๏ฟฝ
โ12๐
๏ฟฝ๐ +๐ฅ2๐๏ฟฝ
โ12๏ฟฝ1 +
๐ฅ2๏ฟฝ
For large value of ๐ฅ, The integrand is written as ๐๐(๐) where ๐ (๐) = ๐ฅ cos2 ๐. The value of ๐where ๐ (๐) is maximum is first found. Then solving for ๐ in
๐๐๐๐ฅ cos2 ๐ = 0
โ2๐ฅ cos๐ sin๐ = 0Hence solving for ๐ in
cos๐ sin๐ = 0There are two solutions to this. Either ๐ = ๐
2 or ๐ = 0. To find which is the correct choice,
the sign of ๐2
๐๐2๐ (๐) is checked for each choice.
๐2
๐๐2๐ฅ cos2 ๐ = ๐
๐๐(โ2๐ฅ cos๐ sin๐)
= โ2๐ฅ๐๐๐
(cos๐ sin๐)
= โ2๐ฅ (โ sin๐ sin๐ + cos๐ cos๐)= โ2๐ฅ ๏ฟฝโ sin2 ๐ + cos2 ๐๏ฟฝ (1)
Substituting ๐ = ๐2 in (1) and using cos ๏ฟฝ๐2 ๏ฟฝ = 0 and sin ๏ฟฝ๐2 ๏ฟฝ = 1 gives
๐2
๐๐2๐ฅ cos2 ๐๏ฟฝ
๐=๐2
= โ2๐ฅ (โ1)
= 2๐ฅ
Since the problem says that ๐ฅ > 0 then ๐2
๐๐2๐ฅ cos2 ๐๏ฟฝ๐=๐
2
> 0. Therefore this is a minimum.
30
Using the second choice ๐ = 0, then (1) becomes (after using cos (0) = 1 and sin (0) = 0)๐2
๐๐2๐ฅ cos2 ๐๏ฟฝ
๐=0
= โ2๐ฅ
And because ๐ฅ > 0 then ๐2
๐๐2๐ฅ cos2 ๐๏ฟฝ๐=0
< 0. Therefore the integrand is maximum at
๐๐๐๐๐ = 0
Now that peak ๐ is found, then ๐ (๐) is expanded in Taylor series around ๐๐๐๐๐ = 0. Since๐ (๐) = ๐ฅ cos2 ๐, then
๐ ๏ฟฝ๐๐๐๐๐๏ฟฝ = ๐ฅ
And ๐โฒ (๐) = โ2๐ฅ cos๐ sin๐. At ๐๐๐๐๐ this becomes ๐โฒ ๏ฟฝ๐๐๐๐๐๏ฟฝ = 0. The next term is thequadratic one, given by
๐โฒโฒ (๐) = โ2๐ฅ๐๐๐
(cos๐ sin๐)
= โ2๐ฅ ๏ฟฝโ sin2 ๐ + cos2 ๐๏ฟฝEvaluating the above at ๐๐๐๐๐ = 0 gives
๐โฒโฒ ๏ฟฝ๐๐๐๐๐๏ฟฝ = โ2๐ฅ
The problem says to keep leading term, so no need for more terms. Therefore the Taylorseries expansion of ๐ (๐) = ๐ฅ cos2 ๐ around ๐ = ๐๐๐๐๐ is
๐ฅ cos2 ๐ โ ๐ ๏ฟฝ๐๐๐๐๐๏ฟฝ + ๐โฒ ๏ฟฝ๐๐๐๐๐๏ฟฝ ๐ +12!๐โฒโฒ ๏ฟฝ๐๐๐๐๐๏ฟฝ ๐2
= ๐ฅ + 0 โ2๐ฅ2!๐2
= ๐ฅ โ ๐ฅ๐2
= ๐ฅ ๏ฟฝ1 โ ๐2๏ฟฝ
The integral now becomes
๐ผ (๐ฅ) =12๐ ๏ฟฝ
๐2
โ๐2
๐๐ฅ๏ฟฝ1โ๐2๏ฟฝ๐๐
โ12๐ ๏ฟฝ
๐2
โ๐2
๐๐ฅ๐โ๐ฅ๐2๐๐
=12๐
โโโโโโ๐๐ฅ๏ฟฝ
๐2
โ๐2
๐โ๐ฅ๐2๐๐โโโโโโ
Comparing โซ๐2
โ๐2๐โ๐ฅ๐2๐๐ to the Gaussian integral โซ
โ
โโ๐โ๐๐2๐๐ = ๏ฟฝ
๐๐ฅ , then the above can be
approximated as
๐ผ (๐ฅ) =๐๐ฅ
2๐๏ฟฝ๐๐ฅ
Summary of result
Small ๐ฅ approximation 12๏ฟฝ1 + ๐ฅ
2๏ฟฝ
Large ๐ฅ approximation ๐๐ฅ
2๐๏ฟฝ๐๐ฅ
Note that using the computer, the exact solution is
12๐ ๏ฟฝ
๐2
โ๐2
๐๐ฅ cos2 ๐๐๐ =12๐๐ฅ2 BesselI ๏ฟฝ0, ๐ฅ
2๏ฟฝ
Problem 5
Use the Cauchy-Riemann equations to determine where the function
๐ (๐ง) = ๐ง + ๐ง2
31
Is analytic. Evaluate โฎ๐ถ
๐ (๐ง) ๐๐ง where contour ๐ถ is on the unit circle |๐ง| = 1 in a counter-
clockwise sense.
Solution
Using ๐ง = ๐ฅ + ๐๐ฆ, the function ๐ (๐ง) becomes
๐ (๐ง) = ๐ฅ + ๐๐ฆ + ๏ฟฝ๐ฅ + ๐๐ฆ๏ฟฝ2
= ๐ฅ + ๐๐ฆ + ๏ฟฝ๐ฅ2 โ ๐ฆ2 + 2๐๐ฅ๐ฆ๏ฟฝ
= ๐ฅ + ๐๐ฆ + ๏ฟฝ๐ฅ2 โ ๐ฆ2 โ 2๐๐ฅ๐ฆ๏ฟฝ
= ๏ฟฝ๐ฅ + ๐ฅ2 โ ๐ฆ2๏ฟฝ + ๐ ๏ฟฝ๐ฆ โ 2๐ฅ๐ฆ๏ฟฝ
Writing ๐ (๐ง) = ๐ข + ๐๐ฃ, and comparing this to the above result shows that
๐ข = ๐ฅ + ๐ฅ2 โ ๐ฆ2
๐ฃ = ๐ฆ โ 2๐ฅ๐ฆ (1)
Cauchy-Riemann are given by
๐๐ข๐๐ฅ
=๐๐ฃ๐๐ฆ
โ๐๐ข๐๐ฆ
=๐๐ฃ๐๐ฅ
Using result in (1), Cauchy-Riemann are checked to see if they are satisfied or not. Thefirst equation above results in
๐๐ข๐๐ฅ
= 1 + 2๐ฅ
๐๐ฃ๐๐ฆ
= 1 โ 2๐ฅ
Therefore ๐๐ข๐๐ฅ โ
๐๐ฃ๐๐ฆ . This shows that ๐ (๐ง) is not analytic for all ๐ฅ, ๐ฆ.
Since ๐ (๐ง) is not analytic, Cauchy integral formula can not be used. Instead this can beintegrated using parameterization. Let ๐ง = ๐๐๐ (No need to use ๐๐๐๐ since ๐ = 1 in this casebecause it is the unit circle). The function ๐ (๐ง) becomes
๐ (๐ง) = ๐๐๐ + ๏ฟฝ๐๐๐๏ฟฝ2
= ๐๐๐ + ๐2๐๐
= ๐๐๐ + ๐โ2๐๐
And because ๐ง = ๐๐๐ then ๐๐ง = ๐๐๐๐๐. The integral now becomes
โฎ๐ถ
๐ (๐ง) ๐๐ง = ๏ฟฝ2๐
0๏ฟฝ๐๐๐ + ๐โ2๐๐๏ฟฝ ๐๐๐๐๐
= ๏ฟฝ2๐
0๏ฟฝ๐2๐๐ + ๐โ๐๐๏ฟฝ ๐๐
= ๏ฟฝ๐2๐๐
2๐ ๏ฟฝ2๐
0+ ๏ฟฝ
๐โ๐๐
โ๐ ๏ฟฝ2๐
0
=12๐[cos 2๐ + ๐ sin 2๐]2๐0 + ๐ [cos๐ โ ๐ sin๐]2๐0
=12๐[(cos 4๐ + ๐ sin 4๐) โ (cos 0 + ๐ sin 0)] + ๐ [(cos 2๐ โ ๐ sin 2๐) โ (cos 0 โ ๐ sin 0)]
=12๐[1 โ 1] + ๐ [1 โ 1]
Hence
โฎ๐ถ
๐ (๐ง) ๐๐ง = 0
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2.3 ๏ฟฝnal exam
2.3.1 exam 1 prep sheet
Preparation sheet for the exam: Dec 19 from 2:45 to 5:15 in class
The exam will cover notes and examples on infinite series, complex analysis, eval-uation of integrals, integral transforms, ordinary differential equations, eigenvalueproblems, and partial differential equations. The exam will also cover problem sets1-7. You are allowed to have a 8.5 by 11 sheet of sheet of paper with equations on it.
1. Things to know without thought:
(a) Sum of a geometric series.(b) Definitions of convergence and absolute convergences of infinite series.(c) Ratio test for convergence.(d) Series expansions for ex, sin x, cos x, and ln(1 + x) (for |x| < 1).(e) Cauchy-Riemann conditions and definition of analytic functions.(f) The meaning of log z, Logz, and zc
(g) Cauchy-Goursat Theorem and Cauchy Integral formula.(h) The form of Taylor and Laurent series.(i) The meaning of analytic continuation.(j) The residue theorem.(k) The definition of Cauchy principle value.(l) The definition of an asymptotic series.(m) The definition of Fourier series and transform The definition of a separablefirst order ODE.(n) The criterium to check if an ODE is exact.(o) Solution of linear first order differential equation using an integrating factor.(q) Definition and solution of a homogeneous ODE.(q) Use of the method of undetermined coefficients to solve inhomogeneous linearODEs.(r) Ordinary and regular singular points of a linear differential equation (and theform of the series solutions for these).(s) Definition of a Hermitian differential operator.(t) Sturm Lioville differential equation and related orthogonality of eigenfunctions.(u) Use of the Wronskian to check if two solutions are independent.(v) The definition of generating functions and how to use these to find recurrencerelations.(w) The completeness relation.(x) The definition of the Green function in solving inhomogeneous problems.(y) The eigenfunction expansion of the Green function.(z) The form of the wave, Laplace, and diffusion equations in 1,2, and 3 dimen-sions.(aa) The solution of the 1D wave equation.(bb) The use of separation of variables. Specifically in 3D spherical coordinatesand in 2D polar coordinates.
33
2.3.2 my ๏ฟฝnal exam cheat sheet
Solve for ฯ(r, ฮธ, ฯ, t) where โฯ2
โt2= c2โฯ (Wave PDE 3D spherical coordinates)
Let ฯ = T (t)X(r, ฮธ, ฯ). Firstseperation gives the following twoequations
T โฒโฒ(t) + c2k2T (t) = 0
โX + k2X = 0
k2 is firstseparation constant
T (t) = cos(ฯt) + sin(ฯt)solution
Helmholtz eqution
Let X = R(r)ฮ(ฮธ)ฮฆ(ฯ) and apply separation again.
ฮฆโฒโฒ +m2ฮฆ = 0
solve first, m2 is separation constant
ฮฆ (ฯ) =
{eimฯ
eโimฯ
}
(1โ x2
)ฮโฒโฒ โ 2xฮโฒ +
(l (l + 1)โ m2
1+x2
)ฮ = 0
angular equation.Second sepration. Usel(l + 1) as separationconstant.
Associated legendre equationNote: l is integer and โl โค m โค l
ฮ (x) =
{Pml (x)Qml (x)
}associated Legendrepolynomial of first kind
associated Legendrepolynomial of second kind.not used. blows up x = ยฑ1
x = cos ฮธ
r2Rโฒโฒ + 2rRโฒ + (r2k2 โ l(l + 1))R = 0
Radial equation(Bessel like)
Can be converted to standard Bessel ODEby some transformation (not shown)
R(r) =
{jl(kr)yl(kr)
}shperical besselfunction first kind
shperical bessel functionsecond kind. Not used as itblows up at r = 0
Final solution is summation of fundamental solutionโjl (kr)P
ml (cos(mฯ) + sin(mฯ)) (cosฯt+ sinฯt)
โ
ฯ = ck
Solve for ฯ(r, ฯ, t) where โฯ2
โt2= c2โฯ (Wave PDE in 2D disk, polar coordinates).
Membrane is fixed on edge of disk. Radius a.
Let ฯ = T (t)X(r, ฯ). Firstseperation gives the following twoequations
T โฒโฒ(t) + c2k2T (t) = 0
โX + k2X = 0
k2 is firstseparation constant
T (t) = cos(ฯt) + sin(ฯt)solution
Helmholtz eqution
ฯ = ckmn
Let X = R(r)ฮฆ(ฯ) and apply separation again.
ฮฆโฒโฒ +m2ฮฆ = 0
solve first, m2 is separation constant
ฮฆ (ฯ) =
{eimฯ
eโimฯ
}solution
Due to periodicity, m must be integer.
r2Rโฒโฒ + rRโฒ + (r2k2 โm2)R = 0
Radial equation
Bessel ODE
solutionR(r) =
{Jm(kr)Ym(kr)
}Bessel function order m
Bessel function second kind.Not used as blow up at r = 0
Jm(ka) = 0 from boundary conditions. This fixes k. Let Zmn be the nnt zeroof the Bessel Jm function. Therefore kmn = Zmn
aare allowed values of k.
ฯ โ Jm(kmnr) (cos(mฯ) + sin(mฯ)) This gives rise to modal shapesฯ(r, ฯ, t) =
โJm(kmnr) (cos(mฯ) + sin(mฯ)) (cos(ckmnt) + sin(ckmnt))
Nasser M. Abbasi December 14, 2018
wave 1D: โฯ2
โt2 = c2 โฯ2
โx2
ฯ(x, t) = 12 (f0(xโ ct) + f0(x+ ct)) + 1
2c
โซ x+ctxโct g0(s) ds
Where f0(x) is initial position of string and g0(x) isinitial velocity
Laplacian in 2D polar
โu = โu2
โr2 + 1rโuโr + 1
r2โu2
โฯ2
Bessel: x2yโฒโฒ + xyโฒ + (x2 โm2)y = 0, W = Cx . When m
not intger, solutions Jm(x), Jโm(x). When m integer,solutions Jm(x), Ym(x)
Legendre: (1โ x2)yโฒโฒ โ 2xyโฒ + n(n+ 1)y = 0,
W = C1โx2 . Solutions Pn(x), Qn(x). Where Q blows
at ยฑ1
Sturm-Lioville: (pyโฒ)โฒ โ qy + ฮปry = 0, operatorL[y] = โ(pyโฒ)โฒ + q. When q = 0, equation becomesL[y] = ฮปry. Inner product < ui, uj >=
โซuiuj dx
Green function. For boundary value problem yโฒโฒ + y = f (x) write it as
G (x, x0) =
{Ay1 (x) 0 < x < x0
By2 (x) x0 < x < L
Where y1 (x) is one of the solutions to yโฒโฒ + y = 0 that satisfies left BC and y2 (x) is one which satifies right BC. Then solve for A,B fromcontuinity condition Ay1 (x0) = By2 (x0) and jump discontinuty Byโฒ2 (x0)โAyโฒ1 (x0) = โ1
p(x)where p (x) is from SL form. (โ1) for the above.
Then flipe the x, x0 roles and then do y (x) =โซ x
0G (x0, x) f (x0) dx0 +
โซ LxG (x0, x) f (x0) dx0
Variation of parameters yp (x) = โy1 (x)โซ y2(x)f(x)
W (x)dx+ y2 (x)
โซ y1(x)f(x)W (x)
dx
Fourier series f (x) = A0
2 +โโn=0An cos
(2ฯT nx
)+B2 sin
(2ฯT nx
)where An = 2
T
โซ T2
โT2f (x) cos
(2ฯT nx
)and
Bn = 2T
โซ T2
โT2f (x) sin
(2ฯT nx
). Complex form is f (x) =
โโn=โโ Cne
i 2ฯT nx, Cn =โซ T
2
โT2f (x) eโi
2ฯT nxdx.
Fourier transform f (x) = 12ฯ
โซโโโ F (ฯ) eiฯxdฯ and F (ฯ) =
โซโโโ f (x) eโiฯxdx. Parsvales:
โซโโโ |f (x)|2 dx = 1
2ฯ
โซโโโ |F (ฯ)|2 dฯ
Wronskian W (x) = det[[y1, y2], [yโฒ1, yโฒ2]] =
Ceโโซp(x) dx Where p is from
yโฒโฒ + pyโฒ + qy = 0
Bernuli yโฒ + f (x) y = g (x) yn, where n 6= 0, 1. Divideby yn โ 1
yn yโฒ + f (x) y1โn = g (x) and let
v = y1โn โ vโฒ = (1โ n) yโnyโฒ. converts ODE toseparable in v (x)isobaric given y weight m and x weight 1. each term musthave the same weight if we can find m. Then let y = vxm.Find dy
dx. sub into ODE to get rid of y. Separable in v.
solve.Exact Write ode as Mdx+Ndx = 0 then check ifโMโy
= โNโx
. If so then set up 2 equations โฮฆโx
= M, โฮฆโy
= N .
Integrate the first to get ฮฆ =โซMdx+ f (y), differentiate
this w.r.t y, and compare to โฮฆโy
= N and solve for f (y).Solution is ฮฆ (x, y) = c
Sturm-Liouville To convert ayโฒโฒ + byโฒ + (c+ ฮปd)y = 0 to(p (x) dy
dx
)โฒ โ q (x) y (x) + ฮปr (x) y (x) = 0 use:
p (x) = eโซ b(x)a(x)
dxand q (x) = โp (x) c(x)
a(x)and r (x) = p(x)
a(x)d
generating functions. Bessel g (x.t) = ex2 (tโ 1
t ), then writeg (x, t) =
โโn=โโ Jn (x) tn. Find
An (x) = 12ฯi
โฎex2 (tโ 1
t )tn+1 dt = 1
ฯ
โซ ฯ0
cos (nฮธ โ x sin ฮธ) dฮธ. To
find recusive relations, do โgโt, โgโx
add/subtract we getJ โฒn (x) = Jnโ1 (x)โ n
xJn (x) and
J โฒn (x) = nxJn (x)โ Jnโ1 (x) For Legendre use
g (x, t) = 1โ1โ2xtโt2
=โโn=0 Pn (x) tn. Do โg
โt, โgโx
add/subtract we getP โฒn+1 (x) = (n+ 1)Pn (x) + xP โฒn (x)and P โฒnโ1 (x) = โnPn (x) + xP โฒn (x)Pn (x) = 1
2nn!dn
dxn
(x2 โ 1
)n. Special values,
P0 (x) = 1, P1 (x) = x, P2 (x) = 12
(3x2 โ 1
)Laplace PDE on disk:y (x) = A0 +
โrn (Cn cosnฮธ + kn sinnฮธ) where
A0 = 12ฯ
โซ 2ฯ
0f (ฮธ) dฮธ and
Cnan = 1
ฯ
โซ 2ฯ
0f (ฮธ) cosnฮธdฮธ and
knan = 1
ฯ
โซ 2ฯ
0f (ฮธ) sinnฮธdฮธ where a is radius and
f (ฮธ) is boundary condition.
log (z) = ln |z|+ i (ฮธ0 + 2nฯ)
Euler ode r2yโฒโฒ+ ryโฒ+ y = 0, let y = rฮฑ
Polynomial ODE. yโฒ (x) = f (ax+ by + c). LetV = ax+ by+ c, then dV = adx+ bdy, converts it to
separable dydv = f(v)
a+bf(v)
34
Eigenfunction expansion. Startting from L [y]โ ky = f (x), assuming y (x) =โn Cnฮฆn (x) where ฮฆn (x) eigenfunctions found by solving
L [y]โ ฮปy = 0 with homogenous B.C. Then f (x) =โn dnฮฆn (x). Subtituting in the ode and use that L [y] = ฮปy results inโ
n (ฮปn โ k)Cnฮฆn (x) =โn dnฮฆn (x). But dn = ใฮฆn (x) , f (x)ใ = 2
L
โซ L0
ฮฆn (x) f (x) dx. The 2L due to normalization. This gives
Cn = dnฮปnโk . Now that Cn is found the solution is y (x) =
โn
dnฮปnโkฮฆn (x) =
โ ใฮฆn(x),f(x)ใฮปnโk ฮฆn (x) or y (x) =
โ ฮฆn(x)ฮปnโk
โซฮฆn (xโฒ) f (xโฒ) dxโฒ
or y (x) =โซ โ ฮฆn(xโฒ)ฮฆn(xโฒ)
ฮปnโk f (xโฒ) dxโฒ. Compare to y (x) =โซ x
G (x, xโฒ) f (xโฒ) dxโฒ shows that G (x, xโฒ) =โ ฮฆn(xโฒ)ฮฆn(xโฒ)
ฮปnโk . This is calledGreen function eigenfunction expansion. This assumes ฮฆn are normalized so weight is 1.
completeness relationโซuiujr (x) dx = ฮดij . f =
โCnฮฆn (x) and r (xโฒ)
โn ฮฆn (x) ฮฆn (xโฒ) = ฮด (xโ xโฒ). If f (x) = ฮด (xโ xโฒ), then
L [G (x, xโฒ)]โ kG (x, xโฒ) = ฮด (xโ xโฒ) Operator is Hermite ifโซuL [v] dx =
โซvL [u] dx
Asymptotic series S (z) = c0 + c1z + c2
z2 + ยท ยท ยท is series expansion of f (z) gives good approximation for large z. we truncate S (z) before
it becomes divergent. n is the number of terms in Sn (z). It is optimally truncated when n โ |z|2. S (x) has the following twoimportant properties
1. lim|z|โโ zn (f (z)โ Sn (z)) = 0 for fixed n.
2. limnโโ zn (f (z)โ Sn (z)) =โ for fixed z.
S (z) โผ f (z) when S (z) is the asymptotic series expansion of f (z) for large z. common method to find S (z) is by integration by parts
Cauchy theoreom. Cauchy-Goursat: If f (z) is analytic on and inside closed contour C then
โฎC
f (z) dz = 0. But remember that ifโฎC
f (z) dz = 0 then this does not necessarily imply f (z). If f (z) is analytic on and inside C then and z0 is a point in C then
2ฯif (z0) =
โฎC
f(z)zโz0 dz and 2ฯif โฒ (z0) =
โฎC
f(z)
(zโz0)2dz and 2ฯi
n! f(n) (z0) =
โฎC
f(z)
(zโz0)n+1 dz
Laurent series f (z) =โโn=0 an (z โ z0)
n+โโn=1
bn(zโz0)n where an = 1
2ฯi
โฎf(z)
(zโz0)n+1 dz and bn = 12ฯi
โฎf(z)
(zโz0)โn+1 dz. Power series of
f (z) around z0 isโโ
0 an (z โ z0)n
where an = 1n! f
(n) (z)โฃโฃz=z0
For Laurent series, lets say singularity is at z = 0 and z = 1. To expand
about z = 0, get f (z) to look like 11โz and use geometric series for |z| < 1. To expand about z = 1, there are two choices, to the inside
and to the outside. For the outside, i.e. |z| > 1, get f (z) to have 11โ 1
z
form, this now valid for |z| > 1.
Euler Gamma functionฮ (z) =
โซโ0tzโ1eโtdt Re (z) > 0 To
extend ฮ (z) to the left half plane, i.e. fornegative values. Let us define, using theabove recursive formulaฮ (z) = ฮ(z+1)
zRe (z) > โ1
ฮ (z) = (z โ 1) ฮ (z โ 1) Re (z) > 1
ฮ (1) = 1
ฮ (n) = (nโ 1)!
ฮ (n+ 1) = n!
ฮ
(1
2
)=โฯ
ฮ (z + 1) = zฮ (z) recursive formula
ฮ (z) = ฮ (z)
ฮ
(n+
1
2
)=
1 ยท 3 ยท 5 ยท ยท ยท (2nโ 1)
2nโฯ
ฮ (x) ฮ (1โ x) =
โซ โ0
txโ1
1 + tdt 0 < x < 1
=ฯ
sin (ฯx)
If f (z) satisfies CReverywhere in that regionthen it is analytic. Letf (z) = u (x, y) + iv (x, y),then these two equationsin Cartesian coordinatesare โu
โx= โv
โyand
โ โuโy
= โvโx
Sometimes it iseasier to use the polarform of these. Letf (z) = r cos ฮธ + i sin ฮธ,then the equations becomeโuโr
= 1rโvโฮธ
and โ 1rโuโฮธ
= โvโr
Greenโs Theorem saysโซC Pdx+Qdy =
โซD
(โQโxโ โP
โy
)dA
ToโซC f (z) dz where C is some open
path, i.e. not closed, such as a straightline or a half circle arc. Useparameterization. This converts theintegral to line integration. If C isstraight line, use standard tparameterization,x (t) = (1โ t)x0 + tx1 andy (t) = (1โ t) y0 + ty1 where (x0, y0)in the line initial point and (x1, y1) isthe line end point. This works forstraight lines. Use the above andrewrite z = x+ iy asz (t) = x (t) + iy (t) and plug-in in thisz (t) in f (z) to obtain f (t), then theintegral becomesโซ
Cf (z) dz =
โซ t=1
t=0f (t) zโฒ (t) dt
Now evaluate this integral.
If pole of order n. to find residue, use
Residue (z0) = limzโz0
dnโ1
dzn(z โ z0)n
(nโ 1)!f (z)
residue, useR (z0) = limzโz0 (z โ z0) f (z)
ex = 1 + x+x2
2!+x3
3!+ ยท ยท ยท
ln (1 + x) = xโ x2
2+x3
3โ x4
4+ ยท ยท ยท |x| < 1
1
2ln
(1 + x
1โ x
)= x+
x3
3+x5
5+x7
7โ ยท ยท ยท |x| < 1
sinx = xโ x3
3!+ ยท ยท ยท
cosx = 1โ x2
2+x4
4!+ ยท ยท ยท
โซx cos ax =
cos ax
a2+x sin ax
aโซx sin ax =
sin ax
a2
x cos ax
aโซsin2 ax =
x
2โ
sin 2ax
4aโซcos2 ax =
x
2+
sin 2ax
4aโซsin ax cos ax =
sin2 ax
2aโซ1
ax+ b=
1
aln (ax+ b)โซ
x
ax+ b=x
aโ
b
a2ln (ax+ b)โซ
1โ
1โ x2= arcsin (x)โซ โ1
โ1โ x2
= arccos (x)โซ1
1 + x2= arctan (x)
sinx =1
2
eix โ eโix
2i
cosx =1
2
eix + eโix
2
sin (AยฑB) = sinA cosB ยฑ cosA sinB
cos (AยฑB) = cosA cosB โ sinA sinB
sin 2A = 2 sinA cosA
cos 2A = cos2 Aโ sin2 A = 1โ 2 sin2 A
sinA sinB =1
2(cos (AโB)โ cos (A+B))
cosA cosB =1
2(cos (A+B) + cos (A+B))
sinA cosB =1
2(sin (AโB) + sin (A+B))
F.S. (period 2L).โซ L0 sin2
(ฯLx)dx = L
2andโซ L
โL sin2(ฯLx)dx = L
โซxeax =
eax
a
(xโ 1
a
)โซx2eax =
eax
a
(x2 โ 2
ax+
2
a2
)โซ
sin bxeax =eax
(a2 + b2)(a sin bxโ b cos bx)โซ
lnx = x lnxโ x
Geometric series:โNn=0 r
n = 1 + r + r2 + r3 + ยท ยท ยท+ rN = 1โrNโ1
1โr andโโn=0 r
n = 1 + r + r2 + r3 + ยท ยท ยท = 11โr |r| < 1 andโโ
n=0 (โ1)n rn = 1โ r + r2 โ r3 + ยท ยท ยท = 11+r
|r| < 1
Binomial series: (x+ y)n =
xn +nxnโ1y+n(nโ1)
2!xnโ2y2 +
n(nโ1)(nโ2)3!
xnโ3y3 + ยท ยท ยทFrom this can generate all other special cases. For
1
(1 + x)= 1โ x+ x2 โ x3 + ยท ยท ยท |x| < 1
1
(1โ x)= 1 + x+ x2 + x3 + ยท ยท ยท |x| < 1
z = rei(ฮธ+2nฯ), hence
z12 = r
12 e
iฮธ2 e2nฯ for
n = 0, 1. Hence 2roots. n = 0 givesprinciple part
log(z) =ln |z|+ i(ฮธ0 + 2nฯ)n = 0 principle partzc = ec ln z
โซ โ0
eโax cosฮปx =a
a2 + ฮป2โซ โ0
eโax sinฮปx =ฮป
a2 + ฮป2
if eโy โค 1 does not work, use Jordan inqualityand use eโy โค ฯ
Rfor contour integration
sin ฮธ =z โ zโ1
2i
cos ฮธ =z + zโ1
2
Gaussianโซโโโ eโax
2dx =
โฯx
Chapter 3
HWs
3.1 HW 1
3.1.1 Problem 1
Part a
For what values of ๐ฅ does the following series converge. ๐ (๐ฅ) = 1 + 9๐ฅ2 +
81๐ฅ2 +
729๐ฅ3 +โฏ
answer
The general term of the series is
๐ (๐ฅ) =โ๏ฟฝ๐=0
๏ฟฝ3๐ฅ๏ฟฝ
2๐
The ratio test can be used to determine convergence. Since all the terms are positive(powers are even), then the absolute value is not needed.
๐ฟ = lim๐โโ
๏ฟฝ๐๐+1๐๐
๏ฟฝ
= lim๐โโ
๐๐+1๐๐
= lim๐โโ
๏ฟฝ3๐ฅ๏ฟฝ2(๐+1)
๏ฟฝ3๐ฅ๏ฟฝ2๐
= lim๐โโ
๏ฟฝ3๐ฅ๏ฟฝ2๐
๏ฟฝ3๐ฅ๏ฟฝ2๐ ๏ฟฝ
3๐ฅ๏ฟฝ
2
= lim๐โโ
32
๐ฅ2
=9๐ฅ2
The series converges when ๐ฟ < 1, which means 9๐ฅ2 < 1 or ๐ฅ
2 > 9. Therefore it convergencefor
|๐ฅ| > 3
Part b
Does the following series converges or diverges? โโ๐=1 ln ๏ฟฝ1 + 1
๐๏ฟฝ
answer
The terms are {ln (2) , ln ๏ฟฝ2 + 12๏ฟฝ , ln ๏ฟฝ2 + 1
3๏ฟฝ , ln ๏ฟฝ2 + 1
4๏ฟฝ ,โฏ}.
35
36
Since the terms become monotonically decreasing after some ๐ (after the second term inthis case), the integral test could be used. Let
๐ผ = ๏ฟฝ ln ๏ฟฝ1 +1๐ฅ๏ฟฝ๐๐ฅ
The above indefinite integral is first evaluated. Since ln ๏ฟฝ1 + 1๐ฅ๏ฟฝ = ln ๏ฟฝ1+๐ฅ๐ฅ ๏ฟฝ = ln (1 + ๐ฅ) โ ln (๐ฅ),
the above can be written as
๐ผ = ๏ฟฝ ln (1 + ๐ฅ) ๐๐ฅ โ๏ฟฝ ln (๐ฅ) ๐๐ฅ (1)
To evaluate the first integral in (2) โซ ln (1 + ๐ฅ) ๐๐ฅ, let ๐ข = 1 + ๐ฅ, then ๐๐ข = ๐๐ฅ, therefore
๏ฟฝ ln (1 + ๐ฅ) ๐๐ฅ = ๏ฟฝ ln (๐ข) ๐๐ข
= ๐ข ln (๐ข) โ ๐ขHence
๏ฟฝ ln (1 + ๐ฅ) ๐๐ฅ = (1 + ๐ฅ) ln (1 + ๐ฅ) โ (1 + ๐ฅ) (2)
The second integral in (1) is
๏ฟฝ ln (๐ฅ) ๐๐ฅ = ๐ฅ ln (๐ฅ) โ ๐ฅ (3)
Using (2,3) back into (1) gives
๐ผ = ((1 + ๐ฅ) ln (1 + ๐ฅ) โ (1 + ๐ฅ)) โ (๐ฅ ln (๐ฅ) โ ๐ฅ)= (1 + ๐ฅ) ln (1 + ๐ฅ) โ 1 โ ๐ฅ โ ๐ฅ ln (๐ฅ) + ๐ฅ= (1 + ๐ฅ) ln (1 + ๐ฅ) โ ๐ฅ ln (๐ฅ) โ 1 (4)
Now that the indefinite integral is evaluated, the limit is taken using
๐ = lim๐โโ
๏ฟฝ๐
ln ๏ฟฝ1 +1๐ฅ๏ฟฝ๐๐ฅ
Only upper limit is used following the book method1. Using the result found in (4), theabove becomes
๐ = lim๐โโ
[(1 + ๐ฅ) ln (1 + ๐ฅ) โ ๐ฅ ln (๐ฅ) โ 1]๐
The above becomes
๐ = lim๐โโ
[(1 + ๐ฅ) ln (1 + ๐ฅ) โ ๐ฅ ln (๐ฅ) โ 1]๐
= lim๐โโ
[(1 + ๐) ln (1 + ๐) โ ๐ ln (๐) โ 1]
= lim๐โโ
[ln (1 + ๐) + ๐ ln (1 + ๐) โ ๐ ln (๐) โ 1]
= lim๐โโ
๏ฟฝln (1 + ๐) + ๐ ln ๏ฟฝ1 + ๐๐ ๏ฟฝ โ 1๏ฟฝ
= lim๐โโ
ln (1 + ๐) + lim๐โโ
๐ ln ๏ฟฝ1 + ๐๐ ๏ฟฝ โ 1 (5)
But
lim๐โโ
๐ ln ๏ฟฝ1 + ๐๐ ๏ฟฝ = lim
๐โโ
ln ๏ฟฝ1+๐๐ ๏ฟฝ1๐
This gives indeterminate form 1/0. So using LโHospitalโs rule, by taking derivatives ofnumerator and denominator gives
lim๐โโ
ln ๏ฟฝ1+๐๐ ๏ฟฝ1๐
= lim๐โโ
๏ฟฝ1โ 1+๐๐ ๏ฟฝ
1+๐
โ 1๐2
= lim๐โโ
โ๏ฟฝ1 โ 1+๐
๐๏ฟฝ๐2
1 + ๐= lim๐โโ
โ(๐ โ 1 โ ๐)๐
1 + ๐= lim๐โโ
๐1 + ๐
= 1
1See page 131, second edition. Mathematical methods for physics and engineering. Riley, Hobson andBence.
37
Therefore lim๐โโ๐ ln ๏ฟฝ1+๐๐ ๏ฟฝ = 1. Using this result in (5) results in
๐ = lim๐โโ
ln (1 + ๐) + 1 โ 1
= lim๐โโ
ln (1 + ๐)
= โ
Therefore, by the integral test the series diverges.
3.1.2 Problem 2
Find closed form for the series ๐ (๐ฅ) = โโ๐=0 ๐
2๐ฅ2๐ by taking derivatives of variant of 11โ๐ฅ .
For what values of ๐ฅ does the series converge?
answer
๐ (๐ฅ) = ๐ฅ2 + 4๐ฅ4 + 9๐ฅ6 + 16๐ฅ8 + 25๐ฅ10 +โฏ
Observing that
๐2๐ฅ2๐ =๐ฅ2๐๐๐ฅ๏ฟฝ๐๐ฅ2๐๏ฟฝ
Therefore the sum can be written as
๐ (๐ฅ) =๐ฅ2
โ๏ฟฝ๐=0
๐๐๐ฅ๏ฟฝ๐๐ฅ2๐๏ฟฝ
=๐ฅ2๐๐๐ฅ
โ๏ฟฝ๐=0
๐๐ฅ2๐
=๐ฅ2๐๐๐ฅ๏ฟฝ๐ฅ2 + 2๐ฅ4 + 3๐ฅ6 + 4๐ฅ8 + 5๐ฅ10 +โฏ๏ฟฝ
=๐ฅ2๐๐๐ฅ๏ฟฝ๐ฅ2 ๏ฟฝ1 + 2๐ฅ2 + 3๐ฅ4 + 4๐ฅ6 + 5๐ฅ8 +โฏ๏ฟฝ๏ฟฝ (1)
To find what 1 + 2๐ฅ2 + 3๐ฅ4 + 4๐ฅ6 +โฏ sums to, we compare it to the binomial series
(1 + ๐ง)๐ผ = 1 + ๐ผ๐ง +๐ผ (๐ผ โ 1) ๐ง2
2!+๐ผ (๐ผ โ 1) (๐ผ โ 2) ๐ง3
3!+โฏ = 1 + 2๐ฅ2 + 3๐ฅ4 + 4๐ฅ6 + 5๐ฅ8 +โฏ
Hence, by setting
๐ง = โ๐ฅ2
๐ผ = โ2
shows they are the same. Therefore
๏ฟฝ1 โ ๐ฅ2๏ฟฝโ2= 1 + (โ2) ๏ฟฝโ๐ฅ2๏ฟฝ +
(โ2) (โ3) ๏ฟฝโ๐ฅ2๏ฟฝ2
2!+(โ2) (โ3) (โ4) ๏ฟฝโ๐ฅ2๏ฟฝ
3
3!+โฏ
= 1 + 2๐ฅ2 + 3๐ฅ2 + 3๐ฅ4 +โฏ
The above is valid for |๐ง| < 1 which implies ๐ฅ2 < 1 or |๐ฅ| < 1. Hence
1 + 2๐ฅ2 + 3๐ฅ4 + 4๐ฅ6 +โฏ = ๏ฟฝ1 โ ๐ฅ2๏ฟฝโ2
38
Using the above result in (1) gives
๐ (๐ฅ) =๐ฅ2๐๐๐ฅ
โโโโโโโโ
๐ฅ2
๏ฟฝ1 โ ๐ฅ2๏ฟฝ2
โโโโโโโโ
=๐ฅ2
โโโโโโโโ
4๐ฅ3
๏ฟฝ1 โ ๐ฅ2๏ฟฝ3 +
2๐ฅ
๏ฟฝ1 โ ๐ฅ2๏ฟฝ2
โโโโโโโโ
=2๐ฅ4
๏ฟฝ1 โ ๐ฅ2๏ฟฝ3 +
๐ฅ2
๏ฟฝ1 โ ๐ฅ2๏ฟฝ2
=2๐ฅ4 + ๐ฅ2 ๏ฟฝ1 โ ๐ฅ2๏ฟฝ
๏ฟฝ1 โ ๐ฅ2๏ฟฝ3
=2๐ฅ4 + ๐ฅ2 โ ๐ฅ4
๏ฟฝ1 โ ๐ฅ2๏ฟฝ3
=๐ฅ4 + ๐ฅ2
๏ฟฝ1 โ ๐ฅ2๏ฟฝ3
Therefore
๐ (๐ฅ) =๐ฅ2๏ฟฝ๐ฅ2+1๏ฟฝ
๏ฟฝ1โ๐ฅ2๏ฟฝ3
Where the above converges for |๐ฅ| < 1, from above, where we used Binomial expansionwhich is valid for |๐ฅ| < 1. This result could also be obtained by using the ratio test.
lim๐โโ
๏ฟฝ๐๐+1๐๐
๏ฟฝ = lim๐โโ
๏ฟฝ(๐ + 1)2 ๐ฅ2(๐+1)
๐2๐ฅ2๐ ๏ฟฝ
Since all powers are even, the absolute value is not needed. The above becomes
lim๐โโ
๐๐+1๐๐
= lim๐โโ
(๐ + 1)2 ๐ฅ2
๐2
= ๐ฅ2 lim๐โโ
(๐ + 1)2
๐2= ๐ฅ2
Therefore for the series to converge, we know that ๐๐+1๐๐
must be less than 1. Hence ๐ฅ2 < 1or |๐ฅ| < 1, which is the same result as above.
3.1.3 Problem 3
Part a
Find the sum of 1 + 14 โ
116 โ
164 +
1256 +
11024 โโฏ
solution
We would like to combine each two consecutive negative terms and combine each twoconsecutive positive terms in the series in order to obtain an alternating series which iseasier to work with. but to do that, we first need to check that the series is absolutelyconvergent. The |๐๐| term is 1
4๐ , therefore
39
๐ฟ = lim๐โโ
๏ฟฝ๐๐+1๐๐
๏ฟฝ
= lim๐โโ
๏ฟฝ๏ฟฝ
14๐+114๐
๏ฟฝ๏ฟฝ
= lim๐โโ
๏ฟฝ4๐
4๐+1๏ฟฝ
= ๏ฟฝ14๏ฟฝ
Since |๐ฟ| < 1 then the series is absolutely convergent so we are allowed now to group (orrearrange) terms as follows
๐ = ๏ฟฝ1 +14๏ฟฝโ ๏ฟฝ
116+164๏ฟฝ
+ ๏ฟฝ1256
+1
1024๏ฟฝโ ๏ฟฝ
14096
+1
16384๏ฟฝ+โฏ
=54โ564+
51024
โ5
16 384+โฏ
=54 ๏ฟฝ1 โ
116+
1256
โ1
4096+โฏ๏ฟฝ
=54
โ๏ฟฝ๐=0
(โ1)๐
42๐
=54
โ๏ฟฝ๐=0
(โ1)๐ ๏ฟฝ116๏ฟฝ
๐
(1)
But โโ๐=0 (โ1)
๐ ๏ฟฝ 116๏ฟฝ๐has the form โโ
๐=0 (โ1)๐ ๐๐ where ๐ = 1
16 and since |๐| < 1 then by the
binomial seriesโ๏ฟฝ๐=0
(โ1)๐ ๐๐ = 1 โ ๐ + ๐2 โ ๐3 +โฏ
=1
1 + ๐Therefore the sum in (1) becomes, using ๐ = 1
16
๐ =54
โโโโโโโ
11 + 1
16
โโโโโโโ
=54 ๏ฟฝ1617๏ฟฝ
Hence
๐ = 2017
Or
๐ โ 1.176
Part b
Find the sum of 11! +
82! +
163! +
644! +โฏ
solution
The sum can be written as
๐ =โ๏ฟฝ๐=1
๐3
๐!
=โ๏ฟฝ๐=1
๐๐2
๐!
40
But ๐๐! =
๐(๐โ1)!๐ =
1(๐โ1)! and the above reduces to
๐ =โ๏ฟฝ๐=1
๐2
(๐ โ 1)!Let ๐ โ 1 = ๐ or ๐ = ๐ + 1. The above becomes
๐ =โ๏ฟฝ๐=0
(๐ + 1)2
๐!
=โ๏ฟฝ๐=0
(๐ + 1)2
๐!
=โ๏ฟฝ๐=0
๐2 + 1 + 2๐๐!
=โ๏ฟฝ๐=0
๐2
๐!+
โ๏ฟฝ๐=0
1๐!+ 2
โ๏ฟฝ๐=0
๐๐!
(1)
Considering the first term โโ๐=0
๐2
๐! which can be written asโ๏ฟฝ๐=0
๐2
๐!=
โ๏ฟฝ๐=1
๐2
๐!
=โ๏ฟฝ๐=1
๐(๐ โ 1)!
(2)
Again, letting Let ๐ โ 1 = ๐ then โโ๐=1
๐(๐โ1)! becomes โโ
๐=0๐+1๐! . Hence (2) becomes
โ๏ฟฝ๐=0
๐2
๐!=
โ๏ฟฝ๐=0
๐ + 1๐!
=โ๏ฟฝ๐=0
๐ + 1๐!
=โ๏ฟฝ๐=0
๐๐!+
โ๏ฟฝ๐=0
1๐!
=โ๏ฟฝ๐=1
๐๐!+
โ๏ฟฝ๐=0
1๐!
But โโ๐=1
๐๐! = โ
โ๐=1
1(๐โ1)! . Letting ๐ โ 1 = ๐, this becomes โโ
๐=01๐! = โโ
๐=01๐!and the above
reduces toโ๏ฟฝ๐=0
๐2
๐!=
โ๏ฟฝ๐=0
1๐!+
โ๏ฟฝ๐=0
1๐!
= ๐ + ๐= 2๐ (3)
The above takes care of the first term in (1). Therefore (1) can now be written as
๐ =โ๏ฟฝ๐=0
๐2
๐!+
โ๏ฟฝ๐=0
1๐!+ 2
โ๏ฟฝ๐=0
๐๐!
= 2๐ + ๐ + 2โ๏ฟฝ๐=0
๐๐!
= 3๐ + 2โ๏ฟฝ๐=0
๐๐!
But โโ๐=0
๐๐! = โ
โ๐=1
๐๐! and โ
โ๐=1
๐๐! was calculated above. It can be written as โโ
๐=01๐! . The
above now becomes
๐ = 3๐ + 2 ๏ฟฝโ๏ฟฝ๐=0
1๐!๏ฟฝ
= 3๐ + 2๐
Therefore
๐ = 5๐
or
๐ โ 13.5914
41
3.1.4 key solution to HW 1
42
43
3.2 HW 2
3.2.1 Problem 1
Find all possible values for (put into ๐ฅ + ๐๐ฆ form)
1. log ๏ฟฝ1 + โ3๐๏ฟฝ
2. ๏ฟฝ1 + โ3๐๏ฟฝ2๐
Answer
44
Part 1
Let ๐ง = ๐ฅ + ๐๐ฆ, where here ๐ฅ = 1, ๐ฆ = 3, then |๐ง| = ๏ฟฝ๐ฅ2 + ๐ฆ2 = โ1 + 3 = 2 and arg (๐ง) = ๐0 =
arctan ๏ฟฝ๐ฆ๐ฅ๏ฟฝ = arctan ๏ฟฝโ31 ๏ฟฝ =๐6 = 60
0. The function log (๐ง) is infinitely multi-valued, given by
log (๐ง) = ln |๐ง| + ๐ (๐0 + 2๐๐) ๐ = 0, ยฑ1, ยฑ2,โฏ (1)
Where ๐0 is the principal argument, which is 600 in this example, which is when ๐ = 0.This is done to make log (๐ง) single valued. This makes the argument of ๐ง restricted toโ๐ < ๐0 < ๐. This makes the negative real axis the branch cut, including the origin. Tofind all values, we simply use (1) for all possible ๐ values other than ๐ = 0. Each di๏ฟฝerent ๐values gives di๏ฟฝerent branch cut. This gives, where ln |๐ง| = ln (2) in all cases, the following
log (๐ง) = ln (2) + ๐ ๏ฟฝ๐3๏ฟฝ ๐ = 0
= ln (2) + ๐ ๏ฟฝ๐3+ 2๐๏ฟฝ ๐ = 1
= ln (2) + ๐ ๏ฟฝ๐3โ 2๐๏ฟฝ ๐ = โ1
= ln (2) + ๐ ๏ฟฝ๐3+ 4๐๏ฟฝ ๐ = 2
= ln (2) + ๐ ๏ฟฝ๐3โ 4๐๏ฟฝ ๐ = โ2
โฎ
Or
log (๐ง) = 0.693 + 1.047๐= 0.693 + 7.330๐= 0.693 โ 5.236๐= 0.693 + 13.614๐= 0.693 โ 11.519๐โฎ
These are in ๏ฟฝ๐ฅ + ๐๐ฆ๏ฟฝ form. There are infinite number of values. Picking a specific branchcuts (i.e. specific ๐ value), picks one of these values. The principal value is one associatedwith ๐ = 0.
Part 2
Let ๐ง = 1 + ๐โ3, hence
๐ (๐ง) = ๐ง2๐
= exp (2๐ log (๐ง))= exp (2๐ (ln |๐ง| + ๐ (๐0 + 2๐๐))) ๐ = 0, ยฑ1, ยฑ2,โฏ
Where in this example, as in first part, ln |๐ง| = ln (2) = 0.693 and principal argument is๐0 =
๐3 = 60
0. Hence the above becomes
๐ (๐ง) = exp ๏ฟฝ2๐ ๏ฟฝln (2) + ๐ ๏ฟฝ๐3+ 2๐๐๏ฟฝ๏ฟฝ๏ฟฝ
= exp ๏ฟฝ2๐ ln (2) โ ๏ฟฝ2๐3+ 4๐๐๏ฟฝ๏ฟฝ
= exp ๏ฟฝ๐ ln 4 โ ๏ฟฝ2๐3+ 4๐๐๏ฟฝ๏ฟฝ
= exp (๐ ln 4) exp ๏ฟฝโ ๏ฟฝ2๐3+ 4๐๐๏ฟฝ๏ฟฝ
= ๐โ๏ฟฝ 2๐3 +4๐๐๏ฟฝ (cos (ln 4) + ๐ sin (ln 4))
= ๐โ๏ฟฝ 2๐3 +4๐๐๏ฟฝ cos (ln 4) + ๐๐โ๏ฟฝ
2๐3 +4๐๐๏ฟฝ sin (ln 4)
45
Which is now in the form of ๐ฅ + ๐๐ฆ. First few values are
๐ (๐ง) = ๐โ๏ฟฝ 2๐3 ๏ฟฝ cos (ln 4) + ๐๐โ๏ฟฝ
2๐3 ๏ฟฝ sin (ln 4) ๐ = 0
= ๐โ๏ฟฝ 2๐3 +4๐๏ฟฝ cos (ln 4) + ๐๐โ๏ฟฝ
2๐3 +4๐๏ฟฝ sin (ln 4) ๐ = 1
= ๐โ๏ฟฝ 2๐3 โ4๐๏ฟฝ cos (ln 4) + ๐๐โ๏ฟฝ
2๐3 โ4๐๏ฟฝ sin (ln 4) ๐ = โ1
= ๐โ๏ฟฝ 2๐3 +8๐๏ฟฝ cos (ln 4) + ๐๐โ๏ฟฝ
2๐3 +8๐๏ฟฝ sin (ln 4) ๐ = 2
= ๐โ๏ฟฝ 2๐3 โ8๐๏ฟฝ cos (ln 4) + ๐๐โ๏ฟฝ
2๐3 โ8๐๏ฟฝ sin (ln 4) ๐ = โ2
โฎ
Or
๐ (๐ง) = 0.0226 + ๐0.121= 7.878 ร 10โ8 + ๐4.222 ร 10โ7
= 6478 + ๐34713= 2.748 ร 10โ13 + ๐1.472 ร 10โ12
= 1.858 ร 109 + ๐9.954 ร 109
โฎ
3.2.2 Problem 2
Given that ๐ข ๏ฟฝ๐ฅ, ๐ฆ๏ฟฝ = 3๐ฅ2๐ฆ โ ๐ฆ3 find ๐ฃ ๏ฟฝ๐ฅ, ๐ฆ๏ฟฝ such that ๐ (๐ง) is analytic. Do the same for
๐ข ๏ฟฝ๐ฅ, ๐ฆ๏ฟฝ = ๐ฆ๐ฅ2+๐ฆ2
Solution
Part (1)
๐ข ๏ฟฝ๐ฅ, ๐ฆ๏ฟฝ = 3๐ฅ2๐ฆ โ ๐ฆ3. The function ๐ (๐ง) is analytic if it satisfies Cauchy-Riemann equations
๐๐ข๐๐ฅ
=๐๐ฃ๐๐ฆ
(1)
โ๐๐ข๐๐ฆ
=๐๐ฃ๐๐ฅ
(2)
Applying the first equation gives
6๐ฅ๐ฆ =๐๐ฃ๐๐ฆ
Hence, solving for ๐ฃ by integrating, gives
๐ฃ ๏ฟฝ๐ฅ, ๐ฆ๏ฟฝ = 3๐ฅ๐ฆ2 + ๐ (๐ฅ) (3)
Is the solution to (3) where ๐ (๐ฅ) is the constant of integration since it is a partial di๏ฟฝerentialequation. We now use equation (2) to find ๐ (๐ฅ). From (2)
โ ๏ฟฝ3๐ฅ2 โ 3๐ฆ2๏ฟฝ =๐๐ฃ๐๐ฅ
โ3๐ฅ2 + 3๐ฆ2 =๐๐ฃ๐๐ฅ
But (3) gives ๐๐ฃ๐๐ฅ = 3๐ฆ
2 + ๐โฒ (๐ฅ), hence the above becomes
โ3๐ฅ2 + 3๐ฆ2 = 3๐ฆ2 + ๐โฒ (๐ฅ)๐โฒ (๐ฅ) = โ3๐ฅ2 + 3๐ฆ2 โ 3๐ฆ2
= โ3๐ฅ2
Integrating gives
๐ (๐ฅ) = ๏ฟฝโ3๐ฅ2๐๐ฅ
= โ๐ฅ3 + ๐ถ
46
Therefore, (3) becomes
๐ฃ ๏ฟฝ๐ฅ, ๐ฆ๏ฟฝ = 3๐ฅ๐ฆ2 + ๐ (๐ฅ)
Or
๐ฃ ๏ฟฝ๐ฅ, ๐ฆ๏ฟฝ = 3๐ฅ๐ฆ2 โ ๐ฅ3 + ๐ถ
Where ๐ถ is arbitrary constant. To verify, we apply CR again. Equation (1) now gives
๐๐ข๐๐ฅ
=๐๐ฃ๐๐ฆ
6๐ฅ๐ฆ = 6๐ฆ๐ฅ
Verified. Equation (2) gives
โ๐๐ข๐๐ฆ
=๐๐ฃ๐๐ฅ
โ3๐ฅ2 + 3๐ฆ2 = โ3๐ฅ2 + 3๐ฆ2
Verified.
Part (2)
๐ข ๏ฟฝ๐ฅ, ๐ฆ๏ฟฝ = ๐ฆ๐ฅ2+๐ฆ2 . The function ๐ (๐ง) is analytic if it satisfies Cauchy-Riemann equations
๐๐ข๐๐ฅ
=๐๐ฃ๐๐ฆ
(1)
โ๐๐ข๐๐ฆ
=๐๐ฃ๐๐ฅ
(2)
Applying the first equation gives
โ2๐ฅ๐ฆ
๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ2 =
๐๐ฃ๐๐ฆ
Hence, solving for ๐ฃ by integrating, gives
๐ฃ = โ2๐ฅ๏ฟฝ๐ฆ
๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ2๐๐ฆ
=๐ฅ
๐ฅ2 + ๐ฆ2+ ๐ (๐ฅ) (3)
Is the solution to (3) where ๐ (๐ฅ) is the constant of integration since it is a partial di๏ฟฝerentialequation. equation (2) gives
โ1
๐ฅ2 + ๐ฆ2+
2๐ฆ2
๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ2 =
๐๐ฃ๐๐ฅ
But (3) gives ๐๐ฃ๐๐ฅ =
1๐ฅ2+๐ฆ2 โ
2๐ฅ2
๏ฟฝ๐ฅ2+๐ฆ2๏ฟฝ2 + ๐โฒ (๐ฅ), hence the above becomes
โ1
๐ฅ2 + ๐ฆ2+
2๐ฆ2
๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ2 =
1๐ฅ2 + ๐ฆ2
โ2๐ฅ2
๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ2 + ๐
โฒ (๐ฅ)
๐โฒ (๐ฅ) = โ2
๐ฅ2 + ๐ฆ2+2 ๏ฟฝ๐ฆ2 + ๐ฅ2๏ฟฝ
๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ2
= โ2
๐ฅ2 + ๐ฆ2+
2๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
= 0
Hence
๐ (๐ฅ) = ๐ถ
where ๐ถ is arbitrary constant. Therefore, (3) becomes
๐ฃ ๏ฟฝ๐ฅ, ๐ฆ๏ฟฝ = ๐ฅ๐ฅ2+๐ฆ2 + ๐ถ
47
To verify, CR is applied again. Equation (1) now gives
๐๐ข๐๐ฅ
=๐๐ฃ๐๐ฆ
โ2๐ฅ๐ฆ
๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ2 =
โ2๐ฅ๐ฆ
๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ2
Hence verified. Equation (2) gives
โ๐๐ข๐๐ฆ
=๐๐ฃ๐๐ฅ
โ1
๐ฅ2 + ๐ฆ2+
2๐ฆ2
๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ2 =
1๐ฅ2 + ๐ฆ2
โ2๐ฅ2
๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ2
โ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ + 2๐ฆ2
๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ2 =
๐ฅ2 + ๐ฆ2 โ 2๐ฅ2
๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ2
โ๐ฅ2 + ๐ฆ2
๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ2 =
โ๐ฅ2 + ๐ฆ2
๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ2
Verified.
3.2.3 Problem 3
Evaluate the integral (i) โฎ๐ถ
|๐ง|2 ๐๐ง and (ii) โฎ๐ถ
1๐ง2๐๐ง along two contours. These contours are
1. Line segment with initial point 1 and fixed point ๐
2. Arc of unit circle with Im (๐ง) โฅ 0 with initial point 1 and final point ๐
Solution
Part (1)
y
x
C
1
i
(x0, y0)
(x1, y1)
Figure 3.1: Integration path
First integral We start by finding the parameterization. For line segments that starts at
๏ฟฝ๐ฅ0, ๐ฆ0๏ฟฝ and ends at ๏ฟฝ๐ฅ1, ๐ฆ1๏ฟฝ, the parametrization is given by
๐ฅ (๐ก) = (1 โ ๐ก) ๐ฅ0 + ๐ก๐ฅ1๐ฆ (๐ก) = (1 โ ๐ก) ๐ฆ0 + ๐ก๐ฆ1
For 0 โค ๐ก โค 1. Hence for ๐ง = ๐ฅ + ๐๐ฆ, it becomes ๐ง (๐ก) = ๐ฅ (๐ก) + ๐๐ฆ (๐ก). In this case, ๐ฅ0 = 1, ๐ฆ0 =0, ๐ฅ1 = 0, ๐ฆ1 = 1, therefore
๐ฅ (๐ก) = (1 โ ๐ก)๐ฆ (๐ก) = ๐ก
48
Using these, ๐ง (๐ก) is found from
๐ง (๐ก) = ๐ฅ (๐ก) + ๐๐ฆ (๐ก)= (1 โ ๐ก) + ๐๐ก
And
๐งโฒ (๐ก) = โ1 + ๐
Since |๐ง|2 = ๐ฅ2 + ๐ฆ2, then in terms of ๐ก it becomes
|๐ง (๐ก)|2 = (1 โ ๐ก)2 + ๐ก2
Hence the line integral now becomes
๏ฟฝ๐ถ|๐ง|2 ๐๐ง = ๏ฟฝ
1
0|๐ง (๐ก)|2 ๐งโฒ (๐ก) ๐๐ก
= ๏ฟฝ1
0๏ฟฝ(1 โ ๐ก)2 + ๐ก2๏ฟฝ (โ1 + ๐) ๐๐ก
= (โ1 + ๐)๏ฟฝ1
0(1 โ ๐ก)2 + ๐ก2๐๐ก
= (โ1 + ๐)๏ฟฝ1
01 + ๐ก2 โ 2๐ก + ๐ก2๐๐ก
= (โ1 + ๐)๏ฟฝ1
01 + 2๐ก2 โ 2๐ก ๐๐ก
= (โ1 + ๐) ๏ฟฝ๏ฟฝ1
0๐๐ก +๏ฟฝ
1
02๐ก2๐๐ก โ๏ฟฝ
1
02๐ก ๐๐ก๏ฟฝ
= (โ1 + ๐)โโโโโโ(๐ก)
10 + 2 ๏ฟฝ
๐ก3
3 ๏ฟฝ1
0โ 2 ๏ฟฝ
๐ก2
2 ๏ฟฝ1
0
โโโโโโ
= (โ1 + ๐) ๏ฟฝ1 +23โ 2 ๏ฟฝ
12๏ฟฝ๏ฟฝ
Hence
โซ๐ถ|๐ง|2 ๐๐ง = 2
3(๐ โ 1)
second integral
Using the same parameterization above. But here the integrand is1๐ง2=
1((1 โ ๐ก) + ๐๐ก)2
Hence the integral becomes
๏ฟฝ๐ถ
1๐ง2๐๐ง = ๏ฟฝ
1
0
1((1 โ ๐ก) + ๐๐ก)2
๐งโฒ (๐ก) ๐๐ก
= (๐ โ 1)๏ฟฝ1
0
1((1 โ ๐ก) + ๐๐ก)2
๐๐ก
= (๐ โ 1) (โ๐)
Hence
โซ๐ถ1๐ง2๐๐ง = 1 + ๐
49
Part (2)
y
x1
i
rฮธ (x0, y0)
(x1, y1)
Figure 3.2: Integration path
First integral Let ๐ง = ๐๐๐๐ then ๐๐ง๐๐ = ๐๐๐๐๐. When ๐ง = 1 then ๐ = 0. When ๐ง = ๐ then ๐ = ๐
2 ,hence we can parameterize the contour integral using ๐ and it becomes
๏ฟฝ๐ถ|๐ง|2 ๐๐ง = ๏ฟฝ
๐2
0๐2 ๏ฟฝ๐๐๐๐๐๏ฟฝ ๐๐
= ๐๐3๏ฟฝ๐2
0๐๐๐๐๐
= ๐๐3 ๏ฟฝ๐๐๐
๐ ๏ฟฝ
๐2
0
= ๐3 ๏ฟฝ๐๐๐๏ฟฝ๐2
0
= ๐3 ๏ฟฝ๐๐๐2 โ ๐0๏ฟฝ
= ๐3 [๐ โ 1]
But ๐ = 1, therefore the above becomes
โซ๐ถ|๐ง|2 ๐๐ง = ๐ โ 1
second integral
Using the same parameterization above. But here the integrand now1๐ง2=
1๐2๐๐2๐
Therefore
๏ฟฝ๐ถ
1๐ง2๐๐ง = ๏ฟฝ
๐2
0
1๐2๐๐2๐
๏ฟฝ๐๐๐๐๐๏ฟฝ ๐๐
=๐๐ ๏ฟฝ
๐2
0๐โ๐๐๐๐
=๐๐ ๏ฟฝ๐โ๐๐
โ๐ ๏ฟฝ
๐2
0
=โ1๐๏ฟฝ๐โ๐๐๏ฟฝ
๐2
0
=โ1๐๏ฟฝ๐โ๐
๐2 โ 1๏ฟฝ
=โ1๐(โ๐ โ 1)
But ๐ = 1, hence
โซ๐ถ1๐ง2๐๐ง = 1 + ๐
50
3.2.4 Problem 4
Use the Cauchy integral formula
๐ (๐ง0) =12๐๐โฎ
๐ถ
๐ (๐ง)๐ง โ ๐ง0
๐๐ง
To evaluate
โฎ๐ถ
1(๐ง + 1) (๐ง + 2)
๐๐ง
Where ๐ถ is the circular contour |๐ง + 1| = ๐ with ๐ < 1. Note that if ๐ > 1 then a di๏ฟฝerentresult is found. Why canโt the Cauchy integral formula above be used for ๐ > 1?
Solution
The disk |๐ง + 1| = ๐ is centered at ๐ง = โ1 with ๐ < 1. The function
๐ (๐ง) =1
(๐ง + 1) (๐ง + 2)has pole at ๐ง = โ1 and at ๐ง = โ2.
R
x
y
โ1โ2
C
Figure 3.3: Showing location of pole
In the Cauchy integral formula, the function ๐ (๐ง) is analytic on ๐ถ and inside ๐ถ. Hence,to use Cauchy integral formula, we need to convert ๐ (๐ง) = 1
(๐ง+1)(๐ง+2) to look like ๐(๐ง)๐งโ๐ง0
where๐ (๐ง) is analytic inside ๐ถ. This is done as follows
1(๐ง + 1) (๐ง + 2)
=1
(๐ง+2)
๐ง โ (โ1)
=๐ (๐ง)
๐ง โ (โ1)
Where now ๐ (๐ง) = 1(๐ง+2) . This has pole at ๐ง = โ2. Since this pole is outside ๐ถ then ๐ (๐ง) is
analytic on and inside ๐ถ and can be used for the purpose of using Cauchy integral formula,which now can be written as
โฎ๐ถ
1(๐ง + 1) (๐ง + 2)
๐๐ง = โฎ๐ถ
1(๐ง+2)
๐ง โ (โ1)๐๐ง
= โฎ๐ถ
๐ (๐ง)๐ง โ (โ1)
๐๐ง
= (2๐๐) ๐ (โ1)
Therefore, we just need to evaluate ๐ (โ1) which is seen as 1. Hence
โฎ๐ถ
1(๐ง+1)(๐ง+2)๐๐ง = 2๐๐ (1)
To verify, we can solve this again using the residue theorem
โฎ๐ถ
๐ (๐ง) ๐๐ง = 2๐๐ ๏ฟฝsum of residues of ๐ (๐ง) inside ๐ถ๏ฟฝ
51
But ๐ (๐ง) = 1(๐ง+1)(๐ง+2) has only one pole inside ๐ถ, which is at ๐ง = โ1. Therefore the above
becomes
โฎ๐ถ
1(๐ง + 1) (๐ง + 2)
= 2๐๐ ๏ฟฝresidue of ๐ (๐ง) at โ 1๏ฟฝ (2)
To find residue at โ1, we can use one of the short cuts to do that. Where we write 1(๐ง+1)(๐ง+2) =
ฮฆ(๐ง)๐ง+1 where ฮฆ (๐ง) is analytic at ๐ง = โ1 and ฮฆ (โ1) โ 0. Therefore we see that ฮฆ (๐ง) = 1
๐ง+2 . Hence
residue of 1(๐ง+1)(๐ง+2) = ฮฆ (๐ง0) =
1(โ1)+2 = 1. Equation (2) becomes
โฎ๐ถ
1(๐ง + 1) (๐ง + 2)
= 2๐๐
Which is same result obtained in (1) by using Cauchy integral formula directly.
To answer last part, when ๐ > 1, then now both poles ๐ง = โ1 and = โ2, are inside ๐ถ.Therefore, we canโt split 1
(๐ง+1)(๐ง+2) into one part that is analytic (the ๐ (๐ง) in the above), in
order to obtain expression ๐(๐ง)๐งโ๐ง0
in order to apply Cauchy integral formula directly. Thereforewhen ๐ > 1 we should use
โฎ๐ถ
๐ (๐ง) ๐๐ง = 2๐๐ ๏ฟฝsum of residues of ๐ (๐ง) inside ๐ถ๏ฟฝ
3.2.5 Problem 5
Evaluate the integral
โฎ๐ถ
๐๐ง2 ๏ฟฝ1๐ง2โ1๐ง3 ๏ฟฝ
๐๐ง
Where he contour is the unit circle around origin (counter clockwise direction).
Solution
โฎ๐ถ
๐๐ง2 ๏ฟฝ1๐ง2โ1๐ง3 ๏ฟฝ
๐๐ง = โฎ๐ถ
๐๐ง2 ๏ฟฝ๐ง โ 1๐ง3 ๏ฟฝ ๐๐ง
= โฎ๐ถ
๐ (๐ง)(๐ง โ ๐ง0)
3๐๐ง
Where ๐ง0 = 0 and where
๐ (๐ง) = ๐๐ง2 (๐ง โ 1)
But ๐ (๐ง) is analytic on ๐ถ and inside, since ๐๐ง2 is analytic everywhere and ๐ง โ 1 has no poles.Hence we can use Cauchy integral formula for pole of higher order given by
โฎ๐ถ
๐ (๐ง)(๐ง โ ๐ง0)
๐+1๐๐ง =2๐๐๐!๐(๐) (๐ง0)
Where ๐ = 2 in this case. Therefore, since ๐ง0 = 0 the above reduces to
โฎ๐ถ
๐ (๐ง)๐ง3
๐๐ง =2๐๐2๐โฒโฒ (0) (1)
Now we just need to find ๐โฒโฒ (๐ง) and evaluate the result at ๐ง0 = 0
๐โฒ (๐ง) = 2๐ง๐๐ง2 (๐ง โ 1) + ๐๐ง2
๐โฒโฒ (๐ง) = 2๐๐ง2 (๐ง โ 1) + 2๐ง ๏ฟฝ2๐ง๐๐ง2 (๐ง โ 1) + ๐๐ง2๏ฟฝ + 2๐ง๐๐ง2
Hence
๐โฒโฒ (0) = โ2
Therefore (1) becomes
โฎ๐ถ
๐๐ง2 (๐ง โ 1)๐ง3
๐๐ง = โ2๐๐ (2)
To verify, we will do the same integration by converting it to line integration using param-
52
eterization on ๐. Let ๐ง (๐) = ๐๐๐๐, but ๐ = 1, therefore ๐ง (๐) = ๐๐๐, ๐๐ง = ๐๐๐๐๐๐. Therefore theintegral becomes
โฎ๐ถ
๐๐ง2 ๏ฟฝ๐ง โ 1๐ง3 ๏ฟฝ ๐๐ง = ๏ฟฝ
2๐
0๐๐2๐๐ ๏ฟฝ
๐๐๐ โ 1๐3๐๐ ๏ฟฝ ๐๐๐๐๐๐
= ๐๏ฟฝ2๐
0๐๐2๐๐ ๏ฟฝ
๐๐๐ โ 1๐2๐๐ ๏ฟฝ ๐๐
This is a hard integral to solve by hand. Using computer algebra software, it also gave โ2๐๐.This verified the result. Clearly using the Cauchy integral formula to solve this problemwas much simpler that using parameterization.
3.2.6 key solution to HW 2
53
54
55
56
3.3 HW 3
3.3.1 Problem 1
Part (a)
Use Cauchy-Riemann equations to determine if |๐ง| analytic function of the complex variable๐ง.
57
Solution
๐ (๐ง) = |๐ง|
Let ๐ง = ๐ฅ + ๐๐ฆ, then
๐ (๐ง) = ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ12
= ๐ข + ๐๐ฃ
Hence
๐ข = ๏ฟฝ๐ฅ2 + ๐ฆ2
๐ฃ = 0
Cauchy-Riemann equations are
๐๐ข๐๐ฅ
=๐๐ฃ๐๐ฆ
(1)
โ๐๐ข๐๐ฆ
=๐๐ฃ๐๐ฅ
(2)
First equation above gives ๐๐ฃ๐๐ฆ = 0 and
๐๐ข๐๐ฅ =
12
2๐ฅ
๏ฟฝ๐ฅ2+๐ฆ2, which shows that ๐๐ฃ
๐๐ฆ โ ๐๐ข๐๐ฅ . Therefore
|๐ง| is not analytic.
Part (b)
Use Cauchy-Riemann equations to determine if Re (๐ง) analytic function of the complexvariable ๐ง.
Solution
๐ (๐ง) = Re (๐ง)Let ๐ง = ๐ฅ + ๐๐ฆ, then
๐ (๐ง) = ๐ฅ= ๐ข + ๐๐ฃ
Hence
๐ข = ๐ฅ๐ฃ = 0
Cauchy-Riemann equations are
๐๐ข๐๐ฅ
=๐๐ฃ๐๐ฆ
(1)
โ๐๐ข๐๐ฆ
=๐๐ฃ๐๐ฅ
(2)
First equation above gives ๐๐ฃ๐๐ฆ = 0 and
๐๐ข๐๐ฅ = 1, which shows that ๐๐ฃ
๐๐ฆ โ ๐๐ข๐๐ฅ . Therefore Re (๐ง) is
not analytic.
Part (c)
Use Cauchy-Riemann equations to determine if ๐sin ๐ง analytic function of the complexvariable ๐ง.
Solution
๐ (๐ง) = ๐sin ๐ง is analytic since we can show that exp (๐ง) is analytic by applying Cauchy-Riemann (C-R), and also show that sin (๐ง) is analytic using C-R. Theory of analytic func-tions it says that the composition of analytic functions is also an analytic function, whichmeans ๐sin ๐ง is analytic.
But this problems seems to ask to use C-R equations directly to show this. Therefore weneed to first determine the real and complex parts (๐ข, ๐ฃ) of the function ๐sin ๐ง. Since
sin ๐ง = ๐ง โ ๐งโ1
2๐
58
Then
๐ (๐ง) = ๐sin ๐ง
= exp ๏ฟฝ๐ง โ ๐งโ1
2๐ ๏ฟฝ
= exp ๏ฟฝ ๐ง2๐๏ฟฝ exp ๏ฟฝ
โ12๐๐ง๏ฟฝ
But ๐ง = ๐ฅ + ๐๐ฆ and the above expands to
exp (sin ๐ง) = exp ๏ฟฝ12๐๏ฟฝ๐ฅ + ๐๐ฆ๏ฟฝ๏ฟฝ exp
โโโโโโ
โ12๐ ๏ฟฝ๐ฅ + ๐๐ฆ๏ฟฝ
โโโโโโ
= exp ๏ฟฝโ๐2๐ฅ +
12๐ฆ๏ฟฝ exp
โโโโโโ๐2
1๏ฟฝ๐ฅ + ๐๐ฆ๏ฟฝ
โโโโโโ
= exp ๏ฟฝโ๐2๐ฅ +
12๐ฆ๏ฟฝ exp
โโโโโโ๐2
๐ฅ โ ๐๐ฆ๏ฟฝ๐ฅ + ๐๐ฆ๏ฟฝ ๏ฟฝ๐ฅ โ ๐๐ฆ๏ฟฝ
โโโโโโ
= exp ๏ฟฝโ๐2๐ฅ +
12๐ฆ๏ฟฝ exp ๏ฟฝ
๐2๐ฅ โ ๐๐ฆ๐ฅ2 + ๐ฆ2 ๏ฟฝ
= exp ๏ฟฝโ๐2๐ฅ +
12๐ฆ๏ฟฝ exp ๏ฟฝ
๐2 ๏ฟฝ
๐ฅ๐ฅ2 + ๐ฆ2
โ๐๐ฆ
๐ฅ2 + ๐ฆ2 ๏ฟฝ๏ฟฝ
= exp ๏ฟฝโ๐2๐ฅ +
12๐ฆ๏ฟฝ exp ๏ฟฝ
๐2
๐ฅ๐ฅ2 + ๐ฆ2
+12
๐ฆ๐ฅ2 + ๐ฆ2 ๏ฟฝ
= exp ๏ฟฝโ๐2๐ฅ๏ฟฝ exp ๏ฟฝ
12๐ฆ๏ฟฝ exp ๏ฟฝ
๐2
๐ฅ๐ฅ2 + ๐ฆ2 ๏ฟฝ
exp ๏ฟฝ12
๐ฆ๐ฅ2 + ๐ฆ2 ๏ฟฝ
Collecting terms gives
exp (sin ๐ง) = exp ๏ฟฝ12๐ฆ +
12
๐ฆ๐ฅ2 + ๐ฆ2 ๏ฟฝ
exp ๏ฟฝ๐2
๐ฅ๐ฅ2 + ๐ฆ2
โ๐2๐ฅ๏ฟฝ
= expโโโโโโ12๐ฆ ๏ฟฝ1 + ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ๏ฟฝ
๐ฅ2 + ๐ฆ2
โโโโโโ exp
โโโโโโ๐2
๐ฅ๐ฅ2 + ๐ฆ2
โ๐
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ๐ฅ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
โโโโโโ
= expโโโโโโ12๐ฆ ๏ฟฝ1 + ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ๏ฟฝ
๐ฅ2 + ๐ฆ2
โโโโโโ exp
โโโโโโ๐12๐ฅ ๏ฟฝ1 โ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ๏ฟฝ
๐ฅ2 + ๐ฆ2
โโโโโโ
= expโโโโโโ12๐ฆ ๏ฟฝ1 + ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ๏ฟฝ
๐ฅ2 + ๐ฆ2
โโโโโโ
โกโขโขโขโขโฃcos
โโโโโโ12๐ฅ ๏ฟฝ1 โ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ๏ฟฝ
๐ฅ2 + ๐ฆ2
โโโโโโ + ๐ sin
โโโโโโ12๐ฅ ๏ฟฝ1 โ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ๏ฟฝ
๐ฅ2 + ๐ฆ2
โโโโโโ
โคโฅโฅโฅโฅโฆ
= expโโโโโโ๐ฆ + ๐ฆ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
โโโโโโ cos
โโโโโโ๐ฅ โ ๐ฅ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
โโโโโโ + ๐ exp
โโโโโโ๐ฆ + ๐ฆ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
โโโโโโ sin
โโโโโโ12๐ฅ โ ๐ฅ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
โโโโโโ
Therefore, since exp (sin ๐ง) = ๐ข + ๐๐ฃ, then we see from above that
๐ข = expโโโโโโ12๐ฆ + ๐ฆ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
๐ฅ2 + ๐ฆ2
โโโโโโ cos
โโโโโโ12๐ฅ โ ๐ฅ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
๐ฅ2 + ๐ฆ2
โโโโโโ
๐ฃ = expโโโโโโ12๐ฆ + ๐ฆ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
๐ฅ2 + ๐ฆ2
โโโโโโ sin
โโโโโโ12๐ฅ โ ๐ฅ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
๐ฅ2 + ๐ฆ2
โโโโโโ
Now we need to check the Cauchy-Riemann equations on the above ๐ข, ๐ฃ functions wefound.
๐๐ข๐๐ฅ
=๐๐ฃ๐๐ฆ
(1)
โ๐๐ข๐๐ฆ
=๐๐ฃ๐๐ฅ
(2)
59
Evaluating each partial derivative gives
๐๐ข๐๐ฅ
=๐๐๐ฅ
โโโโโโ12๐ฆ + ๐ฆ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
๐ฅ2 + ๐ฆ2
โโโโโโ exp
โโโโโโ12๐ฆ + ๐ฆ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
๐ฅ2 + ๐ฆ2
โโโโโโ cos
โโโโโโ12๐ฅ โ ๐ฅ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
๐ฅ2 + ๐ฆ2
โโโโโโ
+ expโโโโโโ12๐ฆ + ๐ฆ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
๐ฅ2 + ๐ฆ2
โโโโโโ ๐๐๐ฅ
cosโโโโโโ12๐ฅ โ ๐ฅ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
๐ฅ2 + ๐ฆ2
โโโโโโ
=122๐ฆ๐ฅ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ โ ๏ฟฝ๐ฆ + ๐ฆ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ๏ฟฝ 2๐ฅ
๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ2 exp
โโโโโโ12๐ฆ ๏ฟฝ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ + 1๏ฟฝ
๐ฅ2 + ๐ฆ2
โโโโโโ cos
โโโโโโ12๐ฅ ๏ฟฝ1 โ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ๏ฟฝ
๐ฅ2 + ๐ฆ2
โโโโโโ
โ expโโโโโโ12๐ฆ + ๐ฆ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
๐ฅ2 + ๐ฆ2
โโโโโโ sin
โโโโโโ12๐ฅ โ ๐ฅ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
๐ฅ2 + ๐ฆ2
โโโโโโ ๐๐๐ฅ
โโโโโโ๐ฅ โ ๐ฅ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
โโโโโโ
=โ๐ฅ๐ฆ
๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ2 exp
โโโโโโ12๐ฆ ๏ฟฝ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ + 1๏ฟฝ
๐ฅ2 + ๐ฆ2
โโโโโโ cos
โโโโโโ๐ฅ ๏ฟฝ1 โ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ๏ฟฝ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
โโโโโโ
โ expโโโโโโ๐ฆ + ๐ฆ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
โโโโโโ sin
โโโโโโ๐ฅ โ ๐ฅ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
โโโโโโ
โโโโโโโโ๏ฟฝ1 โ 3๐ฅ2 โ ๐ฆ2๏ฟฝ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ โ ๏ฟฝ๐ฅ โ ๐ฅ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ๏ฟฝ 2๐ฅ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ2
โโโโโโโโ
=โ๐ฅ๐ฆ
๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ2 exp
โโโโโโ12๐ฆ ๏ฟฝ๐ฅ2 + ๐ฆ2 + 1๏ฟฝ
๐ฅ2 + ๐ฆ2
โโโโโโ cos
โโโโโโ12๐ฅ ๏ฟฝ1 โ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ๏ฟฝ
๐ฅ2 + ๐ฆ2
โโโโโโ
โ expโโโโโโ12๐ฆ ๏ฟฝ๐ฅ2 + ๐ฆ2 + 1๏ฟฝ
๐ฅ2 + ๐ฆ2
โโโโโโ sin
โโโโโโ๐ฅ โ ๐ฅ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
โโโโโโ
โโโโโโโโ๏ฟฝโ๐ฅ4 โ 2๐ฅ2๐ฆ2 โ ๐ฅ2 โ ๐ฆ4 + ๐ฆ2๏ฟฝ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ2
โโโโโโโโ
The above can be simplified more to become
๐๐ข๐๐ฅ
=โ1
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ2 exp
โโโโโโ๐ฆ ๏ฟฝ๐ฅ2 + ๐ฆ2 + 1๏ฟฝ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
โโโโโโ
โกโขโขโขโขโฃ2๐ฅ๐ฆ cos
๐ฅ โ ๐ฅ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ+ ๏ฟฝโ๐ฅ4 โ 2๐ฅ2๐ฆ2 โ ๐ฅ2 โ ๐ฆ4 + ๐ฆ2๏ฟฝ sin
๐ฅ โ ๐ฅ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
โคโฅโฅโฅโฅโฆ (3)
Now we evaluate ๐๐ฃ๐๐ฆ to see if it the same as above. Since ๐ฃ = exp ๏ฟฝ
12๐ฆ+๐ฆ๏ฟฝ๐ฅ2+๐ฆ2๏ฟฝ
๏ฟฝ๐ฅ2+๐ฆ2๏ฟฝ ๏ฟฝ sin ๏ฟฝ12๐ฅโ๐ฅ๏ฟฝ๐ฅ2+๐ฆ2๏ฟฝ
๏ฟฝ๐ฅ2+๐ฆ2๏ฟฝ ๏ฟฝthen
๐๐ฃ๐๐ฆ
=๐๐๐ฆ
โโโโโโ๐ฆ + ๐ฆ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
โโโโโโ exp
โโโโโโ๐ฆ + ๐ฆ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
โโโโโโ sin
โโโโโโ๐ฅ โ ๐ฅ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
โโโโโโ
+ expโโโโโโ๐ฆ + ๐ฆ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
โโโโโโ cos
โโโโโโ๐ฅ โ ๐ฅ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
โโโโโโ ๐๐๐ฆ
โโโโโโ๐ฅ โ ๐ฅ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
โโโโโโ
=
โโโโโโโโ12๏ฟฝ1 + ๐ฅ2 + 3๐ฆ2๏ฟฝ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ โ ๏ฟฝ๐ฆ + ๐ฆ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ๏ฟฝ 2๐ฆ
๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ2
โโโโโโโโ exp
โโโโโโ๐ฆ + ๐ฆ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
โโโโโโ sin
โโโโโโ๐ฅ โ ๐ฅ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
โโโโโโ
+ expโโโโโโ๐ฆ + ๐ฆ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
โโโโโโ cos
โโโโโโ๐ฅ โ ๐ฅ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
โโโโโโ
โโโโโโโโ12๏ฟฝโ2๐ฅ๐ฆ๏ฟฝ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ โ ๏ฟฝ๐ฅ โ ๐ฅ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ๏ฟฝ ๏ฟฝ2๐ฆ๏ฟฝ
๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ2
โโโโโโโโ
=
โโโโโโโโ12๐ฅ4 + 2๐ฅ2๐ฆ2 + ๐ฅ2 + ๐ฆ4 โ ๐ฆ2
๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ2
โโโโโโโโ exp
โโโโโโ๐ฆ + ๐ฆ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
โโโโโโ sin
โโโโโโ12๐ฅ โ ๐ฅ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
๐ฅ2 + ๐ฆ2
โโโโโโ
+ expโโโโโโ๐ฆ + ๐ฆ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
โโโโโโ cos
โโโโโโ๐ฅ โ ๐ฅ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
โโโโโโ
โโโโโโโโ12
โ2๐ฅ๐ฆ
๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ2
โโโโโโโโ
=
โโโโโโโโ12๐ฅ4 + 2๐ฅ2๐ฆ2 + ๐ฅ2 + ๐ฆ4 โ ๐ฆ2
๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ2
โโโโโโโโ exp
โโโโโโ๐ฆ + ๐ฆ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
โโโโโโ sin
โโโโโโ12๐ฅ โ ๐ฅ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
๐ฅ2 + ๐ฆ2
โโโโโโ
โ expโโโโโโ๐ฆ + ๐ฆ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
โโโโโโ cos
โโโโโโ๐ฅ โ ๐ฅ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
โโโโโโ
โโโโโโโโ
๐ฅ๐ฆ
๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ2
โโโโโโโโ
60
Simplifying the above more gives
๐๐ฃ๐๐ฆ
=โ1
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ2 exp
โโโโโโ๐ฆ + ๐ฆ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
โโโโโโ
โกโขโขโขโขโฃ2๐ฅ๐ฆ cos
๐ฅ โ ๐ฅ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ+ ๏ฟฝโ๐ฅ4 โ 2๐ฅ2๐ฆ2 โ ๐ฅ2 โ ๐ฆ4 + ๐ฆ2๏ฟฝ sin
๐ฅ โ ๐ฅ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
โคโฅโฅโฅโฅโฆ (4)
Comparing (3) and (4) shows they are the same expressions. Therefore the first equationis verified.
๐๐ข๐๐ฅ
=๐๐ฃ๐๐ฆ
Now we verify the second equation โ๐๐ข๐๐ฆ =๐๐ฃ๐๐ฅ . Since ๐ข = exp ๏ฟฝ
12๐ฆ+๐ฆ๏ฟฝ๐ฅ2+๐ฆ2๏ฟฝ
๏ฟฝ๐ฅ2+๐ฆ2๏ฟฝ ๏ฟฝ cos ๏ฟฝ12๐ฅโ๐ฅ๏ฟฝ๐ฅ2+๐ฆ2๏ฟฝ
๏ฟฝ๐ฅ2+๐ฆ2๏ฟฝ ๏ฟฝthen
๐๐ข๐๐ฆ
=๐๐๐ฆ
โโโโโโ12๐ฆ + ๐ฆ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
โโโโโโ exp
โโโโโโ๐ฆ + ๐ฆ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
โโโโโโ cos
โโโโโโ12๐ฅ โ ๐ฅ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
๐ฅ2 + ๐ฆ2
โโโโโโ
โ expโโโโโโ๐ฆ + ๐ฆ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
โโโโโโ sin
โโโโโโ12๐ฅ โ ๐ฅ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
๐ฅ2 + ๐ฆ2
โโโโโโ ๐๐๐ฆ
โโโโโโ12๐ฅ โ ๐ฅ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
๐ฅ2 + ๐ฆ2
โโโโโโ
=๏ฟฝ1 + ๐ฅ2 + 3๐ฆ2๏ฟฝ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ โ ๏ฟฝ๐ฆ + ๐ฆ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ๏ฟฝ 2๐ฆ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ2 exp
โโโโโโ๐ฆ + ๐ฆ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
โโโโโโ cos
โโโโโโ๐ฅ โ ๐ฅ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
โโโโโโ
โ expโโโโโโ๐ฆ + ๐ฆ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
โโโโโโ sin
โโโโโโ12๐ฅ โ ๐ฅ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
๐ฅ2 + ๐ฆ2
โโโโโโ ๏ฟฝโ2๐ฆ๏ฟฝ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ โ ๏ฟฝ๐ฅ โ ๐ฅ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ๏ฟฝ 2๐ฆ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ2
=๏ฟฝ๐ฅ4 + 2๐ฅ2๐ฆ2 + ๐ฅ2 + ๐ฆ4 โ ๐ฆ2๏ฟฝ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ2 exp
โโโโโโ๐ฆ + ๐ฆ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
โโโโโโ cos
โโโโโโ๐ฅ โ ๐ฅ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
โโโโโโ
โ expโโโโโโ๐ฆ + ๐ฆ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
โโโโโโ sin
โโโโโโ12๐ฅ โ ๐ฅ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
๐ฅ2 + ๐ฆ2
โโโโโโ ๏ฟฝโ2๐ฆ๏ฟฝ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ โ 2๐ฆ๐ฅ + 2๐ฆ๐ฅ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ2
=๏ฟฝ๐ฅ4 + 2๐ฅ2๐ฆ2 + ๐ฅ2 + ๐ฆ4 โ ๐ฆ2๏ฟฝ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ2 exp
โโโโโโ๐ฆ + ๐ฆ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
โโโโโโ cos
โโโโโโ12๐ฅ โ ๐ฅ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
๐ฅ2 + ๐ฆ2
โโโโโโ
+ expโโโโโโ๐ฆ + ๐ฆ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
โโโโโโ sin
โโโโโโ12๐ฅ โ ๐ฅ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
๐ฅ2 + ๐ฆ2
โโโโโโ
๐ฆ๐ฅ
๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ2
The above can simplified more to give
๐๐ข๐๐ฆ
=1
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ2 exp
โโโโโโ๐ฆ + ๐ฆ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
โโโโโโ
โกโขโขโขโขโฃ๏ฟฝ๐ฅ
4 + 2๐ฅ2๐ฆ2 + ๐ฅ2 + ๐ฆ4 โ ๐ฆ2๏ฟฝ cos๐ฅ โ ๐ฅ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ+ 2๐ฅ๐ฆ sin
๐ฅ โ ๐ฅ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
โคโฅโฅโฅโฅโฆ
Hence
โ๐๐ข๐๐ฆ
=1
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ2 exp
โโโโโโ๐ฆ + ๐ฆ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
โโโโโโ
โกโขโขโขโขโฃโ ๏ฟฝ๐ฅ
4 + 2๐ฅ2๐ฆ2 + ๐ฅ2 + ๐ฆ4 โ ๐ฆ2๏ฟฝ cos๐ฅ โ ๐ฅ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝโ 2๐ฅ๐ฆ sin
๐ฅ โ ๐ฅ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
โคโฅโฅโฅโฅโฆ (5)
61
And since ๐ฃ = exp ๏ฟฝ12๐ฆ+๐ฆ๏ฟฝ๐ฅ2+๐ฆ2๏ฟฝ
๏ฟฝ๐ฅ2+๐ฆ2๏ฟฝ ๏ฟฝ sin ๏ฟฝ12๐ฅโ๐ฅ๏ฟฝ๐ฅ2+๐ฆ2๏ฟฝ
๏ฟฝ๐ฅ2+๐ฆ2๏ฟฝ ๏ฟฝ then
๐๐ฃ๐๐ฅ
=๐๐๐ฅ
โโโโโโ12๐ฆ + ๐ฆ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
๐ฅ2 + ๐ฆ2
โโโโโโ exp
โโโโโโ๐ฆ + ๐ฆ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
โโโโโโ sin
โโโโโโ12๐ฅ โ ๐ฅ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
๐ฅ2 + ๐ฆ2
โโโโโโ
+ expโโโโโโ๐ฆ + ๐ฆ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
โโโโโโ cos
โโโโโโ12๐ฅ โ ๐ฅ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
๐ฅ2 + ๐ฆ2
โโโโโโ ๐๐๐ฅ
โโโโโโ12๐ฅ โ ๐ฅ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
๐ฅ2 + ๐ฆ2
โโโโโโ
=12
โโโโโโโโ2๐ฅ๐ฆ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ โ ๏ฟฝ๐ฆ + ๐ฆ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ๏ฟฝ 2๐ฅ
๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ2
โโโโโโโโ exp
โโโโโโ๐ฆ + ๐ฆ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
โโโโโโ sin
โโโโโโ12๐ฅ โ ๐ฅ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
๐ฅ2 + ๐ฆ2
โโโโโโ
+ expโโโโโโ๐ฆ + ๐ฆ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
โโโโโโ cos
โโโโโโ12๐ฅ โ ๐ฅ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
๐ฅ2 + ๐ฆ2
โโโโโโ
โโโโโโโโ๏ฟฝ1 โ 3๐ฅ2 โ ๐ฆ2๏ฟฝ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ โ ๏ฟฝ๐ฅ โ ๐ฅ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ๏ฟฝ 2๐ฅ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ2
โโโโโโโโ
=โ๐ฅ๐ฆ
๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ2 exp
โโโโโโ12๐ฆ + ๐ฆ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
๐ฅ2 + ๐ฆ2
โโโโโโ sin
โโโโโโ๐ฅ โ ๐ฅ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
โโโโโโ
+ expโโโโโโ๐ฆ + ๐ฆ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
โโโโโโ cos
โโโโโโ๐ฅ โ ๐ฅ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
โโโโโโ
โโโโโโโโโ๐ฅ4 โ 2๐ฅ2๐ฆ2 โ ๐ฅ2 โ ๐ฆ4 + ๐ฆ2
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ2
โโโโโโโโ
The above can simplified more to give
๐๐ฃ๐๐ฅ
=1
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ2 exp
โโโโโโ๐ฆ + ๐ฆ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
โโโโโโ
โกโขโขโขโขโฃโ ๏ฟฝ๐ฅ
4 + 2๐ฅ2๐ฆ2 + ๐ฅ2 + ๐ฆ4 โ ๐ฆ2๏ฟฝ cos๐ฅ โ ๐ฅ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝโ 2๐ฅ๐ฆ sin
๐ฅ โ ๐ฅ ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
2 ๏ฟฝ๐ฅ2 + ๐ฆ2๏ฟฝ
โคโฅโฅโฅโฅโฆ (6)
Comparing (5,6) shows they are the same, i.e.
โ๐๐ข๐๐ฆ
=๐๐ฃ๐๐ฅ
C-R equations are satisfied, and because it is clear that all partial derivatives ๐๐ฃ๐๐ฅ ,
๐๐ฃ๐๐ฆ ,
๐๐ข๐๐ฅ ,
๐๐ข๐๐ฆ
are continuous functions in ๐ฅ, ๐ฆ as they are made up of exponential and trigonometricfunctions which are continuous, then we conclude that ๐ (๐ง) = ๐sin ๐ง is analytic functioneverywhere.
3.3.2 Problem 2
Part (a)
Represent ๐ง+3๐งโ3 by its Maclaurin series and give the region of validity for the representation.
Next expand this in powers of 1๐ง to find a Laurent series. What is the range of validity of
the Laurent series?
Solution
Maclaurin series is expansion of ๐ (๐ง) around ๐ง = 0. Since ๐ (๐ง) has simple pole at ๐ง = 3,then the region of validity will be a disk centered at ๐ง = 0 up to the nearest pole, which isat ๐ง = 3. Hence |๐ง| < 3 is the region.
๐ (๐ง) =๐ง + 3๐ง โ 3
=๐ง + 3
โ3 ๏ฟฝ1 โ ๐ง3๏ฟฝ
=๐ง + 3โ3
โโโโโโ
11 โ ๐ง
3
โโโโโโ
62
Now we can expand using Binomial to obtain
๐ (๐ง) =3 + ๐งโ3 ๏ฟฝ1 +
๐ง3+ ๏ฟฝ
๐ง3๏ฟฝ2+ ๏ฟฝ
๐ง3๏ฟฝ3+โฏ๏ฟฝ
= ๏ฟฝโ1 โ๐ง3๏ฟฝ ๏ฟฝ1 +
๐ง3+ ๏ฟฝ
๐ง3๏ฟฝ2+ ๏ฟฝ
๐ง3๏ฟฝ3+โฏ๏ฟฝ
= ๏ฟฝโ1 โ๐ง3๏ฟฝ + ๏ฟฝโ1 โ
๐ง3๏ฟฝ ๏ฟฝ๐ง3๏ฟฝ + ๏ฟฝโ1 โ
๐ง3๏ฟฝ ๏ฟฝ๐ง3๏ฟฝ2+ ๏ฟฝโ1 โ
๐ง3๏ฟฝ ๏ฟฝ๐ง3๏ฟฝ3+โฏ
= โ1 โ๐ง3โ๐ง3โ ๏ฟฝ
๐ง3๏ฟฝ2โ ๏ฟฝ
๐ง3๏ฟฝ2โ ๏ฟฝ
๐ง3๏ฟฝ3โ ๏ฟฝ
๐ง3๏ฟฝ3โ ๏ฟฝ
๐ง3๏ฟฝ4+โฏ
= โ1 โ23๐ง โ
29๐ง2 โ
227๐ง3 โ
281๐ง4 โโฏ
Or
๐ (๐ง) = โ1 โโ๏ฟฝ๐=1
23๐๐ง๐
To expand in negative powers of ๐ง, or in 1๐ง , then
๐ (๐ง) =๐ง + 3
๐ง ๏ฟฝ1 โ 3๐ง๏ฟฝ
=๐ง + 3๐ง
โโโโโโโ
11 โ 3
๐ง
โโโโโโโ
For ๏ฟฝ3๐ง ๏ฟฝ < 1 or |๐ง| < 3 the above becomes
๐ (๐ง) =๐ง + 3๐ง
โโโโโโ1 +
3๐ง+ ๏ฟฝ
3๐ง๏ฟฝ
2
+ ๏ฟฝ3๐ง๏ฟฝ
3
+โฏโโโโโโ
= ๏ฟฝ1 +3๐ง๏ฟฝ
โโโโโโ1 +
3๐ง+ ๏ฟฝ
3๐ง๏ฟฝ
2
+ ๏ฟฝ3๐ง๏ฟฝ
3
+โฏโโโโโโ
= ๏ฟฝ1 +3๐ง๏ฟฝ+ ๏ฟฝ1 +
3๐ง๏ฟฝ3๐ง+ ๏ฟฝ1 +
3๐ง๏ฟฝ ๏ฟฝ
3๐ง ๏ฟฝ
2
+ ๏ฟฝ1 +3๐ง๏ฟฝ ๏ฟฝ
3๐ง ๏ฟฝ
3
+โฏ
= 1 +3๐ง+3๐ง+ ๏ฟฝ
3๐ง๏ฟฝ
2
+ ๏ฟฝ3๐ง๏ฟฝ
2
+ ๏ฟฝ3๐ง๏ฟฝ
3
+ ๏ฟฝ3๐ง๏ฟฝ
3
+ ๏ฟฝ3๐ง๏ฟฝ
4
+โฏ
= 1 +6๐ง+18๐ง2+54๐ง3+โฏ
This is valid for |๐ง| > 3. The residue is 6, which can be confirmed using
Residue (3) = lim๐งโ3
(๐ง โ 3) ๐ (๐ง)
= lim๐งโ3
(๐ง โ 3)๐ง + 3๐ง โ 3
= lim๐งโ3
(๐ง + 3)
= 6
Summary
๐ (๐ง) = ๐ง+3๐งโ3 =
โงโชโชโจโชโชโฉโ1 โ 2
3๐ง โ29๐ง2 โ 2
27๐ง3 โ 2
81๐ง4 โโฏ |๐ง| < 3
1 + 6๐ง +
18๐ง2 +
54๐ง3 +โฏ |๐ง| > 3
Part (b)
Find Laurent series for ๐ง(๐ง+1)(๐งโ3) in each of the following domains (i) |๐ง| < 1 (ii) 1 < |๐ง| < 3
(iii) |๐ง| > 3
Solution
The possible region are shown below. Since there is a pole at ๐ง = โ1 and pole at ๐ง = 3, thenthere are three di๏ฟฝerent regions. They are named ๐ด,๐ต, ๐ถ in the following diagram
63
โ1 3A
B
C
Region A contains no poles inside.Disk centered at zero up to thenearest pole at z = โ1. Hence theseries expansion will contain onlyan analytical part and no principalpart.
Region B is annulus regionbetween region which is analytic upto the next pole at z = 3. Hencethe series expansion will containboth an analytical part and aprincipal part.
Region C is |z| > 3, Hence theseries expansion will contain only aprincipal part and no analyticalpart.
<z
=z
Figure 3.4: Laurent series regions
First the expression ๐ง(๐ง+1)(๐งโ3) is expanded using partial fractions
๐ง(๐ง + 1) (๐ง โ 3)
=๐ด
(๐ง + 1)+
๐ต(๐ง โ 3)
(1)
Hence
๐ง = ๐ด (๐ง โ 3) + ๐ต (๐ง + 1)= ๐ง (๐ด + ๐ต) โ 3๐ด + ๐ต
The above gives two equations
๐ด + ๐ต = 10 = โ3๐ด + ๐ต
First equation gives ๐ด = 1 โ ๐ต. Substituting in the second equation gives 0 = โ3 (1 โ ๐ต) + ๐ตor 0 = โ3 + 4๐ต, hence ๐ต = 3
4 , which implies ๐ด = 1 โ 34 =
14 , therefore (1) becomes
๐ง(๐ง + 1) (๐ง โ 3)
=14
1(๐ง + 1)
+34
1(๐ง โ 3)
Considering each term in turn. For 14
1(๐ง+1) , we can expand this as
14
1(๐ง + 1)
=14๏ฟฝ1 โ ๐ง + ๐ง2 โ ๐ง3 + ๐ง4 +โฏ๏ฟฝ |๐ง| < 1 (2a)
14
1(๐ง + 1)
=14๐ง
1
๏ฟฝ1 + 1๐ง๏ฟฝ=14๐ง
โโโโโโ1 โ ๏ฟฝ
1๐ง๏ฟฝ+ ๏ฟฝ
1๐ง๏ฟฝ
2
โ ๏ฟฝ1๐ง๏ฟฝ
3
+ ๏ฟฝ1๐ง๏ฟฝ
4
โโฏโโโโโโ |๐ง| > 1 (2b)
And for the term 34
1(๐งโ3) , we can expand this as
34
1(๐ง โ 3)
= โ14
1๏ฟฝ1 โ ๐ง
3๏ฟฝ= โ
14 ๏ฟฝ1 +
๐ง3+ ๏ฟฝ
๐ง3๏ฟฝ2+ ๏ฟฝ
๐ง3๏ฟฝ3+ ๏ฟฝ
๐ง3๏ฟฝ4+โฏ๏ฟฝ |๐ง| < 3 (3a)
34
1(๐ง โ 3)
=34๐ง
1
๏ฟฝ1 โ 3๐ง๏ฟฝ=34๐ง
โโโโโโ1 + ๏ฟฝ
3๐ง๏ฟฝ+ ๏ฟฝ
3๐ง๏ฟฝ
2
+ ๏ฟฝ3๐ง๏ฟฝ
3
+โฏโโโโโโ |๐ง| > 3 (3b)
Now that we expanded all the terms in the two possible ways for each each, we nowconsider each region of interest, and look at the above 4 expansions, and simply pick foreach region the expansion which is valid in for that region of interest.
For (i), region ๐ด: In this region, we want |๐ง| < 1. From (2,3) we see that (2a) and (3a) are
64
valid expansions in |๐ง| < 1. Hence
๐ง(๐ง + 1) (๐ง โ 3)
=14๏ฟฝ1 โ ๐ง + ๐ง2 โ ๐ง3 + ๐ง4 +โฏ๏ฟฝ โ
14 ๏ฟฝ1 +
๐ง3+ ๏ฟฝ
๐ง3๏ฟฝ2+ ๏ฟฝ
๐ง3๏ฟฝ3+ ๏ฟฝ
๐ง3๏ฟฝ4+โฏ๏ฟฝ
=14๏ฟฝ1 โ ๐ง + ๐ง2 โ ๐ง3 + ๐ง4 โโฏ๏ฟฝ โ
14 ๏ฟฝ1 +
๐ง3+๐ง2
9+๐ง3
27+๐ง4
81+โฏ๏ฟฝ
= ๏ฟฝ14โ14๐ง +
14๐ง2 โ
14๐ง3 +
14๐ง4 โโฏ๏ฟฝ โ ๏ฟฝ
14+๐ง12+๐ง2
36+๐ง3
108+๐ง4
324+โฏ๏ฟฝ
= โ14๐ง โ
๐ง12+14๐ง2 โ
๐ง2
36โ14๐ง3 โ
๐ง3
108+14๐ง4 โ
๐ง4
324โโฏ
โ13๐ง +
29๐ง2 โ
727๐ง3 +
2081๐ง4 โโฏ
For (ii), region ๐ต: This is for 1 < |๐ง| < 3. From equations (2,3) we see that (2b) and (3a) arevalid in this region. Hence
๐ง(๐ง + 1) (๐ง โ 3)
=14๐ง
โโโโโโ1 โ ๏ฟฝ
1๐ง๏ฟฝ+ ๏ฟฝ
1๐ง๏ฟฝ
2
โ ๏ฟฝ1๐ง๏ฟฝ
3
+ ๏ฟฝ1๐ง๏ฟฝ
4
โโฏโโโโโโ โ
14 ๏ฟฝ1 +
๐ง3+ ๏ฟฝ
๐ง3๏ฟฝ2+ ๏ฟฝ
๐ง3๏ฟฝ3+ ๏ฟฝ
๐ง3๏ฟฝ4+โฏ๏ฟฝ
=14๐ง ๏ฟฝ
1 โ1๐ง+1๐ง2โ1๐ง3+1๐ง4โโฏ๏ฟฝ โ
14 ๏ฟฝ1 +
๐ง3+๐ง2
9+๐ง3
27+๐ง4
81+โฏ๏ฟฝ
= ๏ฟฝ14๐งโ
14๐ง2
+14๐ง3
โ14๐ง4
+14๐ง5
โโฏ๏ฟฝ โ ๏ฟฝ14+๐ง12+๐ง2
36+๐ง3
108+๐ง4
324+โฏ๏ฟฝ
=
principal part
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโฏ +
14๐ง5
โ14๐ง4
+14๐ง3
โ14๐ง2
+14๐งโ
analytical part
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ14โ๐ง12โ๐ง2
36โ๐ง3
108โ๐ง4
324โโฏ
The residue is 14 by looking at the above. The value for the residue can be verified as
follows. Using
๐๐ =12๐๐โฎ
๐ถ
๐ (๐ง)(๐ง โ ๐ง0)
โ๐+1๐๐ง
Where in the above ๐ง0 is the location of the pole and ๐ is the coe๏ฟฝcient of the 1๐ง๐ is the
principal part. Since we want the residue, then ๐ = 1 and the above becomes
๐1 =12๐๐โฎ
๐ถ
๐ (๐ง) ๐๐ง
In the above, the contour ๐ถ is circle somewhere inside the annulus 1 < |๐ง| < 3. It does notmatter that the radius is, as long as it is located in this range. For example, choosing radius2 will work. The above then becomes
๐1 =12๐๐โฎ
๐ถ
๐ง(๐ง + 1) (๐ง โ 3)
๐๐ง (5)
However, since ๐ (๐ง) is analytic in this region, then โฎ๐ถ
๐ (๐ง) ๐๐ง = 2๐๐โ (residues inside).
There is only one pole now inside ๐ถ, which is at ๐ง = โ1. So all what we have to do is findthe residue at ๐ง = โ1.
Residue (โ1) = lim๐งโโ1
(๐ง + 1) ๐ (๐ง)
= lim๐งโโ1
(๐ง + 1)๐ง
(๐ง + 1) (๐ง โ 3)
= lim๐งโโ1
๐ง(๐ง โ 3)
=โ1
(โ1 โ 3)
=14
65
Using this in (5) gives
๐1 =12๐๐ ๏ฟฝ
2๐๐14๏ฟฝ
=14
Which agrees with what we found in (4) above.
For (iii), region ๐ถ: This is for |๐ง| > 3. From (2,3) we see that (2b) and (3b) are validexpansions in ๐ง > 3, Hence
๐ง(๐ง + 1) (๐ง โ 3)
=14๐ง
โโโโโโ1 โ ๏ฟฝ
1๐ง๏ฟฝ+ ๏ฟฝ
1๐ง๏ฟฝ
2
โ ๏ฟฝ1๐ง๏ฟฝ
3
+ ๏ฟฝ1๐ง๏ฟฝ
4
โโฏโโโโโโ +
34๐ง
โโโโโโ1 + ๏ฟฝ
3๐ง๏ฟฝ+ ๏ฟฝ
3๐ง๏ฟฝ
2
+ ๏ฟฝ3๐ง๏ฟฝ
3
+โฏโโโโโโ
= ๏ฟฝ14๐งโ
14๐ง2
+14๐ง3
โ14๐ง4
+14๐ง5
โโฏ๏ฟฝ +34๐ง ๏ฟฝ
1 +3๐ง+9๐ง2+27๐ง3+โฏ๏ฟฝ
= ๏ฟฝ14๐งโ
14๐ง2
+14๐ง3
โ14๐ง4
+14๐ง5
โโฏ๏ฟฝ + ๏ฟฝ34๐ง+
94๐ง2
+274๐ง3
+814๐ง4
+โฏ๏ฟฝ
= โฏ+20๐ง4+7๐ง3+2๐ง2+1๐ง
This is as expected contains only a principal part and no analytical part. The residue is 1.This above value for the residue can be verified as follows. Using
๐๐ =12๐๐โฎ
๐ถ
๐ (๐ง)(๐ง โ ๐ง0)
โ๐+1๐๐ง
Where in the above ๐ง0 is the location of the pole and ๐ is the coe๏ฟฝcient of the 1๐ง๐ is the
principal part. Since we want the residue, then ๐ = 1 and the above becomes
๐1 =12๐๐โฎ
๐ถ
๐ (๐ง) ๐๐ง
In the above, the contour ๐ถ is circle somewhere in |๐ง| > 3. It does not matter that the radiusis. The above integral then becomes
๐1 =12๐๐โฎ
๐ถ
๐ง(๐ง + 1) (๐ง โ 3)
๐๐ง (7)
However, since ๐ (๐ง) is analytic in |๐ง| > 3, then โฎ๐ถ
๐ (๐ง) ๐๐ง = 2๐๐โ (residues inside). There
are now two poles inside ๐ถ, one at ๐ง = โ1 and one at ๐ง = 3. So all what we have to do isfind the residues at each. We found earlier that Residue (โ1) = 1
4 . Now
Residue (3) = lim๐งโ3
(๐ง โ 3) ๐ (๐ง)
= lim๐งโ3
(๐ง โ 3)๐ง
(๐ง + 1) (๐ง โ 3)
= lim๐งโ3
๐ง(๐ง + 1)
=34
Therefore the sum of residues is 1. Using this result in (7) gives
๐1 =12๐๐ ๏ฟฝ
2๐๐ ๏ฟฝ14+34๏ฟฝ๏ฟฝ
= 1
Which agrees with what result from (6) above.
Summary of results
๐ (๐ง) = ๐ง(๐ง+1)(๐งโ3) =
โงโชโชโชโชโจโชโชโชโชโฉ
โ13๐ง +29๐ง2 โ 7
27๐ง3 + 20
81๐ง4 โโฏ |๐ง| < 1
โฏ + 14๐ง5 โ
14๐ง4
+ 14๐ง3 โ
14๐ง2 +
14๐ง โ
14 โ
๐ง12 โ
๐ง2
36 โ๐ง3
108 โ๐ง4
324 โโฏ 1 < |๐ง| < 3โฏ + 20
๐ง4+ 7
๐ง3 +2๐ง2 +
1๐ง |๐ง| > 3
66
3.3.3 Problem 3
Part (a)
Use residue theorem to evaluate โฎ๐ถ
๐โ2๐ง
๐ง2 ๐๐ง on contour ๐ถ which is circle |๐ง| = 1 in positive
sense.
Solution
For ๐ (๐ง) which is analytic on and inside ๐ถ, the Cauchy integral formula says
โฎ๐ถ
๐ (๐ง) ๐๐ง = 2๐๐๏ฟฝ๐
Residue ๏ฟฝ๐ง = ๐ง๐๏ฟฝ (1)
Where the sum is over all residues located inside ๐ถ. for ๐ (๐ง) = ๐โ2๐ง
๐ง2 there is a simple pole at๐ง = 0 of order 2. To find the residue, we use the formula for pole or order ๐ given by
Residue (๐ง0) = lim๐งโ๐ง0
๐๐โ1
๐๐ง๐โ1(๐ง โ ๐ง0)
๐
(๐ โ 1)!๐ (๐ง)
Hence for ๐ = 2 and ๐ง0 = 0 the above becomes
Residue (0) = lim๐งโ0
๐๐๐ง๐ง2๐ (๐ง)
= lim๐งโ0
๐๐๐ง๐ง2๐โ2๐ง
๐ง2
= lim๐งโ0
๐๐๐ง๐โ2๐ง
= lim๐งโ0
๏ฟฝโ2๐โ2๐ง๏ฟฝ
= โ2
Therefore (1) becomes
โฎ๐ถ
๐โ2๐ง
๐ง2๐๐ง = 2๐๐ (โ2)
= โ4๐๐
Part (b)
Use residue theorem to evaluate โฎ๐ถ
๐ง๐1๐ง ๐๐ง on contour ๐ถ which is circle |๐ง| = 1 in positive
sense.
Solution
The singularity is at ๐ง = 0, but we can not use the simple pole residue finding method here,
since this is an essential singularity now due to the ๐1๐ง term. To find the residue, we expand
๐ (๐ง) around ๐ง = 0 in Laurent series and look for the coe๏ฟฝcient of 1๐ง term.
๐ (๐ง) = ๐ง๐1๐ง
= ๐ง ๏ฟฝ1 +1๐ง+121๐ง2+13!1๐ง3+โฏ๏ฟฝ
= ๐ง + 1 +121๐ง+13!1๐ง2+โฏ
Hence residue is 12 . Therefore
โฎ๐ถ
๐ง๐1๐ง ๐๐ง = 2๐๐ ๏ฟฝ
12๏ฟฝ
= ๐๐
67
Part (c)
Use residue theorem to evaluate โฎ๐ถ
๐ง+2๐ง2โ ๐ง
2๐๐ง on contour ๐ถ which is circle |๐ง| = 1 in positive
sense.
Solution
๐ (๐ง) =๐ง + 2๐ง2 โ ๐ง
2
=๐ง + 2
๐ง ๏ฟฝ๐ง โ 12๏ฟฝ
Hence there is a simple pole at ๐ง = 0 and simple pole at ๐ง = 12
Residue (0) = lim๐งโ0
(๐ง) ๐ (๐ง)
= lim๐งโ0
๐ง๐ง + 2
๐ง ๏ฟฝ๐ง โ 12๏ฟฝ
= lim๐งโ0
๐ง + 2
๏ฟฝ๐ง โ 12๏ฟฝ
=2โ12
= โ4
And
Residue ๏ฟฝ12๏ฟฝ= lim๐งโ 1
2
๏ฟฝ๐ง โ12๏ฟฝ๐ (๐ง)
= lim๐งโ 1
2
๏ฟฝ๐ง โ12๏ฟฝ
๐ง + 2
๐ง ๏ฟฝ๐ง โ 12๏ฟฝ
= lim๐งโ 1
2
๐ง + 2๐ง
=12 + 212
= 5
Therefore
โฎ๐ถ
๐ง + 2๐ง2 โ ๐ง
2
๐๐ง = 2๐๐ (5 โ 4)
= 2๐๐
68
3.3.4 key solution to HW 3
69
70
71
72
3.4 HW 4
3.4.1 Problem 1
Using series expansion evaluate the integral ๐ผ = โซ1
0ln ๏ฟฝ1+๐ฅ1โ๐ฅ
๏ฟฝ ๐๐ฅ๐ฅ
Solution
73
We first need to find the Taylor series for ln ๏ฟฝ1+๐ฅ1โ๐ฅ๏ฟฝ expanded around ๐ฅ = 0. Since
ln ๏ฟฝ1 + ๐ฅ1 โ ๐ฅ๏ฟฝ
= ln ๏ฟฝ(1 + ๐ฅ) ๏ฟฝ1
1 โ ๐ฅ๏ฟฝ๏ฟฝ
= ln (1 + ๐ฅ) + ln ๏ฟฝ1
1 โ ๐ฅ๏ฟฝ
= ln (1 + ๐ฅ) โ ln (1 โ ๐ฅ) (1)
Looking at ln (1 + ๐ฅ), where now ๐ (๐ฅ) = ln (1 + ๐ฅ), then we see that ๐โฒ (๐ฅ) = 11+๐ฅ , ๐
โฒโฒ (๐ฅ) =โ1
(1+๐ฅ)2, ๐โฒโฒโฒ (๐ฅ) = 2
(1+๐ฅ)3, ๐(4) (๐ฅ) = โ 2โ 3
(1+๐ฅ)4,โฏ, therefore
ln (1 + ๐ฅ) = ๐ (0) + ๐ฅ๐โฒ (0) + ๐ฅ2
2๐โฒโฒ (0) +
๐ฅ3
3!๐โฒโฒโฒ (0) +
๐ฅ4
4!๐(4) (0) +โฏ
= 0 + ๐ฅ โ๐ฅ2
2+๐ฅ3
3โ๐ฅ4
4+โฏ (2)
Similarly for ln (1 โ ๐ฅ), where now ๐โฒ (๐ฅ) = โ11โ๐ฅ , ๐
โฒโฒ (๐ฅ) = โ1(1โ๐ฅ)2
, ๐โฒโฒโฒ (๐ฅ) = โ2(1โ๐ฅ)3
, ๐(4) (๐ฅ) =
โ 2โ 3(1โ๐ฅ)4
,โฏ, therefore
ln (1 โ ๐ฅ) = ๐ (0) + ๐ฅ๐โฒ (0) + ๐ฅ2
2๐โฒโฒ (0) +
๐ฅ3
3!๐โฒโฒโฒ (0) +
๐ฅ4
4!๐(4) (0) +โฏ
= 0 โ ๐ฅ โ๐ฅ2
2โ๐ฅ3
3โ๐ฅ4
4+โฏ (3)
Using (2,3) in (1) gives the series expansion for ln ๏ฟฝ1+๐ฅ1โ๐ฅ๏ฟฝ as
ln ๏ฟฝ1 + ๐ฅ1 โ ๐ฅ๏ฟฝ
= ๏ฟฝ๐ฅ โ๐ฅ2
2+๐ฅ3
3โ๐ฅ4
4+โฏ๏ฟฝ โ ๏ฟฝโ๐ฅ โ
๐ฅ2
2โ๐ฅ3
3โ๐ฅ4
4+โฏ๏ฟฝ
= 2๐ฅ +23๐ฅ3 +
25๐ฅ5 +
27๐ฅ7 +โฏ (4)
Using (4) in the integral given results in
๐ผ = ๏ฟฝ1
0๏ฟฝ2๐ฅ +
23๐ฅ3 +
25๐ฅ5 +
27๐ฅ7 +โฏ๏ฟฝ
๐๐ฅ๐ฅ
= ๏ฟฝ1
0๏ฟฝ2 +
23๐ฅ2 +
25๐ฅ4 +
27๐ฅ6 +โฏ๏ฟฝ๐๐ฅ
= ๏ฟฝ2๐ฅ +23๐ฅ3
3+25๐ฅ5
5+27๐ฅ7
7+โฏ๏ฟฝ
1
0
Which simplifies to
๐ผ = 2 +2313+2515+2717+โฏ
= 2 +232+252+272+292+โฏ
= 2 ๏ฟฝ1 +132+152+172+192+โฏ๏ฟฝ
= 2โ๏ฟฝ๐=0
1(2๐ + 1)2
(5)
The following are two methods to obtain closed form sum for โโ๐=0
1(2๐+1)2
. The first method
is based on writingโ๏ฟฝ๐=1
1๐2
=โ๏ฟฝ๐=1
1(2๐)2
+โ๏ฟฝ๐=0
1(2๐ + 1)2
(6)
Where the sum on the left is broken into odd and even terms on the right, as in
1 +122+132+142+152+โฏ = ๏ฟฝ
122+142+โฏ๏ฟฝ + ๏ฟฝ
112+132+152+โฏ๏ฟฝ
But, from lecture Sept. 12, 2018, we showed in class thatโ๏ฟฝ๐=1
1๐2
= ๐ (2) =๐2
6(7)
74
(This is called the Basel problem, and the above closed form sum was first given by Eulerin 1734). Now using (7) into (6) results in
โ๏ฟฝ๐=0
1(2๐ + 1)2
=โ๏ฟฝ๐=1
1๐2โ
โ๏ฟฝ๐=1
1(2๐)2
=โ๏ฟฝ๐=1
1๐2โ14
โ๏ฟฝ๐=1
1๐2
=34 ๏ฟฝ
โ๏ฟฝ๐=1
1๐2 ๏ฟฝ
=34 ๏ฟฝ๐2
6 ๏ฟฝ
=๐2
8Another way to obtained closed form sum for โโ
๐=01
(2๐+1)2is to use Fourier series. Consid-
ering the Fourier series for the following periodic function
๐ (๐ฅ) =
โงโชโชโจโชโชโฉโ๐ฅ โ๐ < ๐ฅ < 00 0 โค ๐ฅ โค ๐
Using
๐ (๐ฅ) =๐ด02+
โ๏ฟฝ๐=1
๐ด๐ cos (๐๐ฅ) +โ๏ฟฝ๐=1
๐ต๐ sin (๐๐ฅ)
Therefore
๐ด0 =1๐ ๏ฟฝ
0
โ๐โ๐ฅ๐๐ฅ =
โ1๐ ๏ฟฝ
๐ฅ2
2 ๏ฟฝ0
โ๐=โ12๐
๏ฟฝ๐ฅ2๏ฟฝ0
โ๐=โ12๐
๏ฟฝโ๐2๏ฟฝ =12๐
And
๐ด๐ =โ1๐ ๏ฟฝ
0
โ๐๐ฅ cos (๐๐ฅ) ๐๐ฅ = 1 + (โ1)๐+1
๐2
๐ต๐ =โ1๐ ๏ฟฝ
0
โ๐๐ฅ sin (๐๐ฅ) ๐๐ฅ =
(โ1)๐+1
๐๐
Hence the Fourier series for ๐ (๐ฅ) is
๐ (๐ฅ) =๐4โ1๐
โ๏ฟฝ๐=1
1 + (โ1)๐+1
๐2cos (๐๐ฅ) โ 1
๐
โ๏ฟฝ๐=0
(โ1)๐+1
๐๐ (sin ๐๐ฅ)
=๐4โ1๐
โ๏ฟฝ๐=1
1 + (โ1)๐+1
๐2cos (๐๐ฅ) โ
โ๏ฟฝ๐=0
(โ1)๐+1
๐sin (๐๐ฅ)
Evaluating the above at ๐ฅ = 0 then all the sin terms vanish and we obtain
0 =๐4โ1๐
โ๏ฟฝ๐=1
1 + (โ1)๐+1
๐2
=๐4โ2๐ ๏ฟฝ
1 +132+152+172+โฏ๏ฟฝ
=๐4โ2๐
โ๏ฟฝ๐=0
1(2๐ + 1)2
Therefore2๐
โ๏ฟฝ๐=0
1(2๐ + 1)2
=๐4
โ๏ฟฝ๐=0
1(2๐ + 1)2
=๐2
8
Now that we found closed form sum for โโ๐=0
1(2๐+1)2
, we can find the value of the integral.
75
Since ๐ผ = 2โโ๐=0
1(2๐+1)2
, then
๏ฟฝ1
0ln ๏ฟฝ
1 + ๐ฅ1 โ ๐ฅ๏ฟฝ
๐๐ฅ๐ฅ= 2 ๏ฟฝ
๐2
8 ๏ฟฝ
=๐2
4
3.4.2 Problem 2
Let ๐ผ (๐ฅ) = โซโ
0๐๐ฅ๐(๐ก)๐๐ก with ๐ (๐ก) = ๐ก โ ๐๐ก
๐ฅ , find a large ๐ฅ approximation for this integral.
Solution
๐ผ = ๏ฟฝโ
0exp ๏ฟฝ๐ฅ๐ (๐ก)๏ฟฝ ๐๐ก
= ๏ฟฝโ
0exp ๏ฟฝ๐ฅ ๏ฟฝ๐ก โ
๐๐ก
๐ฅ ๏ฟฝ๏ฟฝ๐๐ก
= ๏ฟฝโ
0exp ๏ฟฝ๐ฅ๐ก โ ๐๐ก๏ฟฝ ๐๐ก
= ๏ฟฝโ
0exp (๐น (๐ก)) ๐๐ก (1)
Where ๐น (๐ก) = ๐ฅ๐ก โ ๐๐ก. We need to find saddle point where ๐น (๐ก) is maximum. Hence๐๐๐ก๐น (๐ก) = 0
๐ฅ โ ๐๐ก = 0๐๐ก = ๐ฅ๐ก0 = ln (๐ฅ)
Where ๐ก0 is location of ๐ก where ๐น (๐ก) is maximum. We called this in class ๐ก๐๐๐๐. We nowexpand ๐น (๐ก) around ๐ก0 using Taylor series
๐น (๐ก) = ๐น (๐ก0) + ๐นโฒ (๐ก0) (๐ก โ ๐ก0) +12๐นโฒโฒ (๐ก0) (๐ก โ ๐ก0)
2 +โฏ (2)
But
๐น (๐ก0) = ๐ฅ ln (๐ฅ) โ ๐ln ๐ฅ
= ๐ฅ ln ๐ฅ โ ๐ฅAnd ๐นโฒ (๐ก) = ๐ฅ โ ๐๐ก, hence as expected ๐นโฒ (๐ก0) = 0. And ๐นโฒโฒ (๐ก) = โ๐๐ก, therefore ๐นโฒโฒ (๐ก0) = โ๐ln ๐ฅ =โ๐ฅ. We see also that ๐นโฒโฒ (๐ก0) < 0, which means the saddle point was a maximum and not aminimum (since ๐ฅ is positive). Using these in (2) gives
๐น (๐ก) โ (๐ฅ ln ๐ฅ โ ๐ฅ) + 12(โ๐ฅ) (๐ก โ ln ๐ฅ)2
= ๐ฅ ln ๐ฅ โ ๐ฅ โ 12๐ฅ (๐ก โ ln ๐ฅ)2
Substituting the above into (1) gives
๐ผ = ๏ฟฝโ
0exp ๏ฟฝ๐ฅ ln ๐ฅ โ ๐ฅ โ 1
2๐ฅ (๐ก โ ln ๐ฅ)2๏ฟฝ ๐๐ก
= ๏ฟฝโ
0exp (๐ฅ ln ๐ฅ) exp (โ๐ฅ) exp ๏ฟฝโ
12๐ฅ (๐ก โ ln ๐ฅ)2๏ฟฝ ๐๐ก
= exp (๐ฅ ln ๐ฅ) exp (โ๐ฅ)๏ฟฝโ
0exp ๏ฟฝโ
12๐ฅ (๐ก โ ln ๐ฅ)2๏ฟฝ ๐๐ก
= ๐ฅ๐ฅ๐โ๐ฅ๏ฟฝโ
0๐โ
12๐ฅ(๐กโln ๐ฅ)2๐๐ก (3)
Now, since the peak value where ๐น (๐ก) occurs is on the positive real axis, because ๐ก0 = ln (๐ฅ),therefore ๐ฅ > 1 to have a maximum, and assuming a narrow peak, then all the contribution
to the integral comes from ๐ฅ close to the peak location, so we can change โซโ
0๐โ
12๐ฅ(๐กโln ๐ฅ)2๐๐ก
76
to โซโ
โโ๐โ
12๐ฅ(๐กโln ๐ฅ)2๐๐ก without a๏ฟฝecting the final result. Therefore (3) becomes
๐ผ = ๐ฅ๐ฅ๐โ๐ฅ๏ฟฝโ
โโ๐โ
12๐ฅ(๐กโln ๐ฅ)2๐๐ก (4)
Now comparing โซโ
โโ๐โ
12๐ฅ(๐กโln ๐ฅ)2๐๐ก to the Gaussian integral โซ
โ
โโ๐โ๐(๐กโ๐)
2๐๐ก = ๏ฟฝ
๐๐ , shows that
๐ = ๐ฅ2 for our case. Hence
๏ฟฝโ
โโ๐โ
12๐ฅ(๐กโln ๐ฅ)2๐๐ก =
๏ฟฝ2๐๐ฅ
Therefore (4) becomes
๐ผ โ ๐ฅ๐ฅ๐โ๐ฅ๏ฟฝ2๐๐ฅ
For large ๐ฅ.
3.4.3 Problem 3
Evaluate the following integrals with aid of residue theorem ๐ โฅ 0. (a) โซโ
01
๐ฅ4+1๐๐ฅ (b)
โซโ0
cos(๐๐ฅ)๐ฅ2+1 ๐๐ฅ
Part (a)
Since the integrand is even, then
๐ผ =12 ๏ฟฝ
โ
โโ
1๐ฅ4 + 1
๐๐ฅ
Now we consider the following contour
โR +R
R
CR
<z
=z
Figure 3.5: contour used for problem 3
Therefore
โฎ๐ถ
๐ (๐ง) ๐๐ง = ๏ฟฝ lim๐ โโ
๏ฟฝ0
โ๐ ๐ (๐ฅ) ๐๐ฅ + lim
๏ฟฝ๏ฟฝโโ๏ฟฝ
๏ฟฝ๏ฟฝ
0๐ (๐ฅ) ๐๐ฅ๏ฟฝ + lim
๐ โโ๏ฟฝ๐ถ๐ ๐ (๐ง) ๐๐ง
Using Cauchy principal value the integral above can be written as
โฎ๐ถ
๐ (๐ง) ๐๐ง = lim๐ โโ
๏ฟฝ๐
โ๐ ๐ (๐ฅ) ๐๐ฅ + lim
๐ โโ๏ฟฝ๐ถ๐ ๐ (๐ง) ๐
= 2๐๐๏ฟฝResidue
WhereโResidue is sum of residues of 1๐ง4+1
for poles that are inside the contour ๐ถ. Thereforethe above becomes
lim๐ โโ
๏ฟฝ๐
โ๐ ๐ (๐ฅ) ๐๐ฅ = 2๐๐๏ฟฝResidueโ lim
๐ โโ๏ฟฝ๐ถ๐ ๐ (๐ง) ๐๐ง
๏ฟฝโ
โโ
1๐ฅ4 + 1
๐๐ฅ = 2๐๐๏ฟฝResidueโ lim๐ โโ
๏ฟฝ๐ถ๐
1๐ง4 + 1
๐๐ง (1)
77
Now we will show that lim๐ โโโซ๐ถ๐ 1
๐ง4+1๐๐ง = 0. Since
๏ฟฝ๏ฟฝ๐ถ๐
1๐ง4 + 1
๐๐ง๏ฟฝ โค ๐๐ฟ
= ๏ฟฝ๐ (๐ง)๏ฟฝmax
(๐๐ ) (2)
But
๐ (๐ง) =1
๏ฟฝ๐ง2 โ ๐๏ฟฝ ๏ฟฝ๐ง2 + ๐๏ฟฝ
Hence, and since ๐ง = ๐ ๐๐๐ then
๏ฟฝ๐ (๐ง)๏ฟฝmax
โค1
๏ฟฝ๐ง2 โ ๐๏ฟฝmin
๏ฟฝ๐ง2 + ๐๏ฟฝmin
But but inverse triangle inequality ๏ฟฝ๐ง2 โ ๐๏ฟฝ โฅ |๐ง|2 + 1 and ๏ฟฝ๐ง2 + ๐๏ฟฝ โฅ |๐ง|2 โ 1, and since |๐ง| = ๐ then the above becomes
๏ฟฝ๐ (๐ง)๏ฟฝmax
โค1
๏ฟฝ๐ 2 + 1๏ฟฝ ๏ฟฝ๐ 2 โ 1๏ฟฝ
=1
๐ 4 โ 1Therefore (2) becomes
๏ฟฝ๏ฟฝ๐ถ๐
1๐ง4 + 1
๐๐ง๏ฟฝ โค๐๐ ๐ 4 โ 1
Then it is clear that as ๐ โ โ the above goes to zero since lim๐ โโ๐๐ ๐ 4โ1
= lim๐ โโ
๐๐ 3
1โ 1๐ 4
=
01 = 0. Then (1) now simplifies to
๏ฟฝโ
โโ
1๐ฅ4 + 1
๐๐ฅ = 2๐๐๏ฟฝResidue (2A)
We just now need to find the residues of 1๐ง4+1
located in upper half plane. The zeros of
the denominator ๐ง4 + 1 = 0 are at ๐ง = โ114 = ๏ฟฝ๐๐๐๏ฟฝ
14 , then the first zero is at ๐๐
๐4 , and the
second zero at ๐๐๏ฟฝ๐4 +
๐2 ๏ฟฝ = ๐
๐๏ฟฝ 34๐๏ฟฝ and the third zero at ๐๐๏ฟฝ 34๐+
๐2 ๏ฟฝ = ๐
๐๏ฟฝ 54๐๏ฟฝ and the fourth zero at
๐๐๏ฟฝ 54๐+
๐2 ๏ฟฝ = ๐๐
74๐. Hence poles are at
๐ง1 = ๐๐๐4
๐ง2 = ๐๐ 34๐
๐ง3 = ๐๐ 54๐
๐ง4 = ๐๐ 74๐
Out of these only the first two are in upper half plane ๐ง1 and ๐ง1. Hence
Residue (๐ง1) = lim๐งโ๐ง1
(๐ง โ ๐ง1) ๐ (๐ง)
= lim๐งโ๐ง1
(๐ง โ ๐ง1)1
๐ง4 โ 1Applying LโHopitals
Residue (๐ง1) = lim๐งโ๐ง1
14๐ง3
=1
4 ๏ฟฝ๐๐๐4 ๏ฟฝ3
=1
4๐๐3๐4
78
Similarly for the other residue
Residue (๐ง2) = lim๐งโ๐ง2
(๐ง โ ๐ง2) ๐ (๐ง)
= lim๐งโ๐ง1
(๐ง โ ๐ง2)1
๐ง4 โ 1Applying LโHopitals
Residue (๐ง2) = lim๐งโ๐ง2
14๐ง3
=1
4 ๏ฟฝ๐๐34๐๏ฟฝ
3
=1
4๐๐9๐4
=1
4๐๐๐4
Hence (2A) becomes
๏ฟฝโ
โโ
1๐ฅ4 + 1
๐๐ฅ = 2๐๐โโโโโโ1
4๐๐3๐4
+1
4๐๐๐4
โโโโโโ
= 2๐๐โโโโโโโ24๐
โโโโโโ
=12โ
2๐
But โซโ
01
๐ฅ4+1๐๐ฅ = 1
2โซโโโ
1๐ฅ4+1
๐๐ฅ, therefore
๏ฟฝโ
0
1๐ฅ4 + 1
๐๐ฅ = โ24๐
=24โ2
๐
=๐2โ2
Part (b)
Since the integrand is even, then
๐ผ =12 ๏ฟฝ
โ
โโ
cos (๐๐ฅ)๐ฅ2 + 1
๐๐ฅ
We will evaluate โซโ
โโ๐๐๐๐ง
๐ฅ2+1๐๐ฅ and at the end take the real part of the answer. Consideringthe following contour
โR +R
R
CR
<z
=z
Figure 3.6: contour used for part b
79
Then
โฎ๐ถ
๐ (๐ง) ๐๐ง = ๏ฟฝ lim๐ โโ
๏ฟฝ0
โ๐ ๐ (๐ฅ) ๐๐ฅ + lim
๏ฟฝ๏ฟฝโโ๏ฟฝ
๏ฟฝ๏ฟฝ
0๐ (๐ฅ) ๐๐ฅ๏ฟฝ + lim
๐ โโ๏ฟฝ๐ถ๐ ๐ (๐ง) ๐๐ง
Using Cauchy principal value the integral above can be written as
โฎ๐ถ
๐ (๐ง) ๐๐ง = lim๐ โโ
๏ฟฝ๐
โ๐ ๐ (๐ฅ) ๐๐ฅ + lim
๐ โโ๏ฟฝ๐ถ๐ ๐ (๐ง) ๐
= 2๐๐๏ฟฝResidue
WhereโResidue is sum of residues of ๐๐๐๐ง
๐ฅ2+1 for poles that are inside the contour ๐ถ. Thereforethe above becomes
lim๐ โโ
๏ฟฝ๐
โ๐ ๐ (๐ฅ) ๐๐ฅ = 2๐๐๏ฟฝResidueโ lim
๐ โโ๏ฟฝ๐ถ๐ ๐ (๐ง) ๐๐ง
๏ฟฝโ
โโ
๐๐๐๐ฅ
๐ฅ2 + 1๐๐ฅ = 2๐๐๏ฟฝResidueโ lim
๐ โโ๏ฟฝ๐ถ๐
๐๐๐๐ง
๐ง2 + 1๐๐ง (1)
Now we will show that lim๐ โโโซ๐ถ๐ ๐๐๐๐ง
๐ง2+1๐๐ง = 0. Since
๏ฟฝ๏ฟฝ๐ถ๐
๐๐๐๐ง
๐ง2 + 1๐๐ง๏ฟฝ โค ๐๐ฟ
= ๏ฟฝ๐ (๐ง)๏ฟฝmax
(๐๐ ) (2)
But
๐ (๐ง) =๐๐๐๐ง
(๐ง โ ๐) (๐ง + ๐)
=๐๐๐๏ฟฝ๐ฅ+๐๐ฆ๏ฟฝ
(๐ง โ ๐) (๐ง + ๐)
=๐๐๐๐ฅโ๐๐ฆ
(๐ง โ ๐) (๐ง + ๐)
=๐๐๐๐ฅ๐โ๐๐ฆ
(๐ง โ ๐) (๐ง + ๐)Hence
๏ฟฝ๐ (๐ง)๏ฟฝmax
=๏ฟฝ๐๐๐๐ง๏ฟฝ
max|๐โ๐๐ฆ|max
|๐ง โ ๐|min |๐ง + ๐|min
=|๐โ๐๐ฆ|max
(๐ + 1) (๐ โ 1)
=|๐โ๐๐ฆ|max๐ 2 โ 1
Since ๐ > 0 and since in upper half ๐ฆ > 0 then |๐โ๐๐ฆ|max = ๏ฟฝ๐โ๐๐ ๏ฟฝmax
= 1. Jordan inequalitywas not needed here, since there is no extra ๐ฅ in the numerator of the integrand in thisproblem. The above now reduces to
๏ฟฝ๐ (๐ง)๏ฟฝmax
=1
๐ 2 โ 1Equation (2) becomes
๏ฟฝ๏ฟฝ๐ถ๐
๐๐๐๐ง
๐ง2 + 1๐๐ง๏ฟฝ โค
๐๐ ๐ 2 โ 1
๐ โ โ the above goes to zero since lim๐ โโ๐๐ ๐ 2โ1 = lim๐ โโ
๐๐ 2
1โ 1๐ 2
= 01 = 0. Equation (1) now
simplifies to
๏ฟฝโ
โโ
๐๐๐๐ฅ
๐ฅ4 + 1๐๐ฅ = 2๐๐๏ฟฝResidue
We just now need to find the residues of 1๐ง2+1 that are located in upper half plane. The
80
zeros of the denominator ๐ง2 + 1 = 0 are at ๐ง = ยฑ๐, hence poles are at
๐ง1 = ๐๐ง2 = โ๐
Only ๐ง1 is in upper half plane. Therefore
Residue (๐ง1) = lim๐งโ๐ง1
(๐ง โ ๐ง1) ๐ (๐ง)
= lim๐งโ๐ง1
(๐ง โ ๐ง1)๐๐๐๐ง
(๐ง โ ๐ง1) (๐ง โ ๐ง2)
= lim๐งโ๐ง1
๐๐๐๐ง
(๐ง โ ๐ง2)
=๐๐๐(๐)
(๐ + ๐)
=๐โ๐
2๐Since โซ
โ
โโ๐๐๐ฅ
๐ฅ4+1๐๐ฅ = 2๐๐โResidue then
๏ฟฝโ
โโ
๐๐๐๐ฅ
๐ฅ4 + 1๐๐ฅ = 2๐๐ ๏ฟฝ
๐โ๐
2๐ ๏ฟฝ
= ๐๐โ๐
Therefore
๏ฟฝโ
0
๐๐๐๐ฅ
๐ฅ4 + 1๐๐ฅ =
12 ๏ฟฝ
โ
โโ
๐๐๐ฅ
๐ฅ4 + 1๐๐ฅ
=๐2๐โ๐
But real part of the above is
๏ฟฝโ
0
cos (๐๐ฅ)๐ฅ4 + 1
๐๐ฅ =๐2๐โ๐
3.4.4 Problem 4
Using residues evaluate(a) โซ2๐
01
1+๐ cos๐๐๐ for |๐| < 1 (b) โซ๐
0(cos (๐))2๐ ๐๐ for ๐ integer.
Part (a)
Using contour which is anti-clockwise over the unit circle
<z
=z
C
z = eiฮธ
ฮธ
Figure 3.7: contour used for problem 4
Let ๐ง = ๐๐๐, hence ๐๐ง = ๐๐๐๐๐๐ = ๐๐๐๐ง. Using cos๐ = ๐ง+๐งโ1
2 then the integral can be written in
81
complex domain as
โฎ๐ถ
1๐๐ง๐๐ง
1 + ๐ ๐ง+๐งโ1
2
=2๐โฎ๐ถ
1๐ง๐๐ง
2 + ๐ ๏ฟฝ๐ง + 1๐ง๏ฟฝ
=2๐โฎ๐ถ
๐๐ง2๐ง + ๐๐ง2 + ๐
=2๐๐โฎ
๐ถ
๐๐ง๐ง2 + 2
๐๐ง + 1
=2๐๐โฎ
๐ถ
๐๐ง(๐ง โ ๐ง1) (๐ง โ ๐ง2)
Where ๐ง1, ๐ง2 are roots of ๐ง2 +2๐๐ง+ 1 = 0 which are found to be (using the quadratic formula)
as
๐ง1 =โ1 โ โ1 โ ๐2
๐
๐ง2 =โ1 + โ1 โ ๐2
๐Since |๐| < 1 then only ๐ง2 will be inside the unit disk for all ๐ values. Therefore
2๐๐โฎ
๐ถ
๐๐ง(๐ง โ ๐ง1) (๐ง โ ๐ง2)
= ๏ฟฝ2๐๐๏ฟฝ
2๐๐Residue (๐ง2)
=4๐๐Residue (๐ง2) (1)
Now we will find the Residue (๐ง2) where in this case ๐ (๐ง) = 1(๐งโ๐ง1)(๐งโ๐ง2)
. Hence
Residue (๐ง2) = lim๐งโ๐ง2
(๐ง โ ๐ง2) ๐ (๐ง)
= lim๐งโ๐ง2
(๐ง โ ๐ง2)1
(๐ง โ ๐ง1) (๐ง โ ๐ง2)
= lim๐งโ๐ง2
1(๐ง โ ๐ง1)
=1
๏ฟฝโ1+โ1โ๐2
๐ ๏ฟฝ โ ๏ฟฝโ1โโ1โ๐2
๐ ๏ฟฝ
=๐
2โ1 โ ๐2Using the above result in (1) gives
๏ฟฝ2๐
0
11 + ๐ cos๐๐๐ = ๏ฟฝ
4๐๐๏ฟฝ
๐2โ1 โ ๐2
=2๐
โ1 โ ๐2๐ โ 1
Using Maple, verified that the above result is correct.
Figure 3.8: Verification using Maple
Part (b)
Since integrand is even, then โซ๐
0(cos (๐))2๐ ๐๐ = 1
2โซ2๐0
(cos (๐))2๐ ๐๐. Using same contour as
in part (a), and letting ๐ง = ๐๐๐, hence ๐๐ง = ๐๐๐๐๐๐ = ๐๐๐๐ง and using cos๐ = ๐ง+๐งโ1
2 then the
82
integral can be written in complex domain as
๏ฟฝ2๐
0(cos (๐))2๐ ๐๐ = โฎ
๐ถ
โโโโโโโ๐ง + 1
๐ง2
โโโโโโโ
2๐๐๐ง๐๐ง
=1๐โฎ๐ถ
๏ฟฝ๐ง + 1๐ง๏ฟฝ2๐
22๐๐๐ง๐ง
=14๐๐โฎ
๐ถ๏ฟฝ๐ง +
1๐ง๏ฟฝ
2๐ ๐๐ง๐ง
=14๐๐โฎ
๐ถ๏ฟฝ๐ง2 + 1๐ง ๏ฟฝ
2๐ ๐๐ง๐ง
=14๐๐โฎ
๐ถ
๏ฟฝ๐ง2 + 1๏ฟฝ2๐
๐ง2๐๐๐ง๐ง
=14๐๐โฎ
๐ถ
๏ฟฝ๐ง2 + 1๏ฟฝ2๐
๐ง2๐+1๐๐ง
Considering ๐ (๐ง) =๏ฟฝ๐ง2+1๏ฟฝ
2๐
๐ง2๐+1, this has a pole at ๐ง = 0 of order ๐ = 2๐ + 1. Therefore
14๐๐โฎ
๐ถ
๏ฟฝ๐ง2 + 1๏ฟฝ2๐
๐ง2๐+1๐๐ง = ๏ฟฝ
14๐๐๏ฟฝ
2๐๐Residue (๐ง = 0) (1)
So we now need to find residue of ๐ (๐ง) at ๐ง = 0 but for pole of order ๐ = 2๐ + 1. Using theformula for finding residue for pole of order ๐ gives
Residue (๐ง0 = 0) = lim๐งโ๐ง0
๐๐โ1
๐๐ง๐โ1(๐ง โ ๐ง0)
๐ ๐ (๐ง)(๐ โ 1)!
But ๐ = 2๐ + 1, and ๐ง0 = 0, hence the above becomes
Residue (0) = lim๐งโ0
๐2๐
๐๐ง2๐๐ง2๐+1
(2๐)!๏ฟฝ๐ง2 + 1๏ฟฝ
2๐
๐ง2๐+1
=1
(2๐)!lim๐งโ0
๏ฟฝ๐2๐
๐๐ง2๐๏ฟฝ๐ง2 + 1๏ฟฝ
2๐๏ฟฝ
Equation (1) becomes
๏ฟฝ2๐
0(cos (๐))2๐ ๐๐ = ๏ฟฝ
14๐ ๏ฟฝ
2๐ ๏ฟฝ1
(2๐)!lim๐งโ0
๏ฟฝ๐2๐
๐๐ง2๐๏ฟฝ๐ง2 + 1๏ฟฝ
2๐๏ฟฝ๏ฟฝ
Therefore
๏ฟฝ๐
0(cos (๐))2๐ ๐๐ = 1
2 ๏ฟฝ14๐ ๏ฟฝ
2๐ ๏ฟฝ1
(2๐)!lim๐งโ0
๏ฟฝ๐2๐
๐๐ง2๐๏ฟฝ๐ง2 + 1๏ฟฝ
2๐๏ฟฝ๏ฟฝ
=14๐
๐(2๐)!
lim๐งโ0
๏ฟฝ๐2๐
๐๐ง2๐๏ฟฝ๐ง2 + 1๏ฟฝ
2๐๏ฟฝ
Will now try to obtained closed form solution. Trying for di๏ฟฝerent ๐ values in order to seethe pattern. From few lectures ago, we learned also that
ฮ ๏ฟฝ๐ +12๏ฟฝ=1 โ 3 โ 5 โ โฏ โ (2๐ โ 1)
2๐ โ๐
Now will generate a table to see the pattern
83
๐ 14๐
๐(2๐)! lim๐งโ0 ๏ฟฝ
๐2๐
๐๐ง2๐๏ฟฝ๐ง2 + 1๏ฟฝ
2๐๏ฟฝ result of integral ฮ ๏ฟฝ๐ + 1
2๏ฟฝ
1 14๐2! lim๐งโ0
๐2
๐๐ง2๏ฟฝ๐ง2 + 1๏ฟฝ
2 ๐2 ฮ ๏ฟฝ1 + 1
2๏ฟฝ = โ๐
2
2 142๐4! lim๐งโ0
๐4
๐๐ง4๏ฟฝ๐ง2 + 1๏ฟฝ
4 3๐8 ฮ ๏ฟฝ2 + 1
2๏ฟฝ = 3โ๐
4
3 143๐6! lim๐งโ0
๐6
๐๐ง6๏ฟฝ๐ง2 + 1๏ฟฝ
6 5๐16 ฮ ๏ฟฝ3 + 1
2๏ฟฝ = 15โ๐
8
4 144๐8! lim๐งโ0
๐8
๐๐ง8๏ฟฝ๐ง2 + 1๏ฟฝ
8 35๐128 ฮ ๏ฟฝ4 + 1
2๏ฟฝ = 105โ๐
16
5 145
๐10! lim๐งโ0
๐10
๐๐ง10๏ฟฝ๐ง2 + 1๏ฟฝ
10 63๐256 ฮ ๏ฟฝ5 + 1
2๏ฟฝ = 945โ๐
32
โฎ โฎ โฎ โฎ
Based on the above, we see that ๐ผ =โ๐ฮ๏ฟฝ๐+
12 ๏ฟฝ
๐! , which is verified as follows
๐ result of integral ฮ ๏ฟฝ๐ + 12๏ฟฝ
โ๐ฮ๏ฟฝ๐+12 ๏ฟฝ
๐!
1 ๐2 ฮ ๏ฟฝ1 + 1
2๏ฟฝ = โ๐
2
โ๐๏ฟฝโ๐2 ๏ฟฝ
1 = 12๐
2 3๐8 ฮ ๏ฟฝ2 + 1
2๏ฟฝ = 3โ๐
4
โ๐๏ฟฝ3โ๐4 ๏ฟฝ
2! = 38๐
3 5๐16 ฮ ๏ฟฝ3 + 1
2๏ฟฝ = 15โ๐
8
โ๐๏ฟฝ15โ๐8 ๏ฟฝ
3! = 15๐(6)(8) =
15๐48 = 3
16๐
4 35๐128 ฮ ๏ฟฝ4 + 1
2๏ฟฝ = 105โ๐
16
โ๐๏ฟฝ105โ๐16 ๏ฟฝ
4! = โ๐๏ฟฝ105โ๐๏ฟฝ(24)(16) = 105๐
384 = 35128๐
5 63๐256 ฮ ๏ฟฝ5 + 1
2๏ฟฝ = 945โ๐
32
โ๐๏ฟฝ945โ๐32 ๏ฟฝ
5! = 945๐(120)(32) =
945๐3840 =
63256๐
โฎ โฎ โฎ โฎ
Therefore
๏ฟฝ๐
0(cos (๐))2๐ ๐๐ =
โ๐ฮ ๏ฟฝ๐ +12๏ฟฝ
๐!
Tried to do pole/zero cancellation on the integrand ofโฎ๐ถ
๏ฟฝ๐ง2+1๏ฟฝ2๐
๐ง2๐+1๐๐ง in order to find a simpler
method than the above but was not able to. The above result was verified using thecomputer
In[ ]:= AssumingElement[n, Integers] && n > 0, IntegrateCos[x]2 n, {x, 0, ฯ};
TraditionalForm[%]
Out[ ]//TraditionalForm=
ฯ ฮn + 1
2
n !
Figure 3.9: Verification using Mathematica
84
3.4.5 key solution to HW 4
85
86
87
88
3.5 HW 5
3.5.1 Problem 1
Expand the following functions, which are periodic in 2๐๐ฟ , in Fourier series (i) ๐ (๐ฅ) = 1 โ |๐ฅ|
๐ฟfor โ๐ฟ
2 โค ๐ฅ โค ๐ฟ2 . (ii) ๐ (๐ฅ) = ๐
๐ฅ for โ๐ฟ2 โค ๐ฅ โค ๐ฟ
2
Solution
89
Part 1
The following is a plot of the function ๐ (๐ฅ) = 1 โ |๐ฅ|๐ฟ . In the plot below ๐ฟ = 1 was used for
illustration.
-0.4 -0.2 0.0 0.2 0.4
0.0
0.2
0.4
0.6
0.8
1.0
x
f(x)
Plot of function for one period
Figure 3.10: Function plot
L = 1;
f[x_] := 1 - Abs[x]/ L;
p = Plot[f[x], {x, -L/ 2, L/ 2},
AxesOrigin โ {0, 0}, Frame โ True,
FrameLabel โ {{"f(x)", None}, {"x", "Plot of function for one period"}},
BaseStyle โ 14,
GridLines โ Automatic, GridLinesStyle โ LightGray,
PlotStyle โ Red]
Export["../images/p1_plot_1.pdf", p]
Figure 3.11: Code used
The Fourier series of ๐ (๐ฅ) = is given by
๐ (๐ฅ) =๐02+
โ๏ฟฝ๐=1
๐๐ cos ๏ฟฝ2๐๐ฟ๐๐ฅ๏ฟฝ + ๐๐ sin ๏ฟฝ
2๐๐ฟ๐๐ฅ๏ฟฝ (1)
Where ๐ฟ is the period.
๐0 =2๐ฟ ๏ฟฝ
๐ฟ2
โ ๐ฟ2
๐ (๐ฅ) ๐๐ฅ
โซ๐ฟ2
โ ๐ฟ2๐ (๐ฅ) ๐๐ฅ is the area under the curve. Looking at the plot above shows the area is made
up of the lower rectangle of area 12๐ฟ and a triangle whose area is ๏ฟฝ12๐ฟ๏ฟฝ ๏ฟฝ
12๏ฟฝ. Therefore the
total area is 12๐ฟ +
14๐ฟ =
34๐ฟ. Hence
๐0 =2๐ฟ ๏ฟฝ
34๐ฟ๏ฟฝ
=32
And
๐๐ =2๐ฟ ๏ฟฝ
๐ฟ2
โ ๐ฟ2
๐ (๐ฅ) cos ๏ฟฝ2๐๐ฟ๐๐ฅ๏ฟฝ ๐๐ฅ
90
Since ๐ (๐ฅ) is an even function, the above simplifies to
๐๐ =4๐ฟ ๏ฟฝ
๐ฟ2
0๐ (๐ฅ) cos ๏ฟฝ
2๐๐ฟ๐๐ฅ๏ฟฝ ๐๐ฅ
=4๐ฟ ๏ฟฝ
๐ฟ2
0๏ฟฝ1 โ
๐ฅ๐ฟ๏ฟฝ cos ๏ฟฝ
2๐๐ฟ๐๐ฅ๏ฟฝ ๐๐ฅ
=4๐ฟ
โโโโโโ๏ฟฝ
๐ฟ2
0cos ๏ฟฝ
2๐๐ฟ๐๐ฅ๏ฟฝ ๐๐ฅ โ
1๐ฟ ๏ฟฝ
๐ฟ2
0๐ฅ cos ๏ฟฝ
2๐๐ฟ๐๐ฅ๏ฟฝ ๐๐ฅ
โโโโโโ
=4๐ฟ
โโโโโโโโ๏ฟฝsin ๏ฟฝ
2๐๐ฟ๐๐ฅ๏ฟฝ๏ฟฝ
๐ฟ2
0โ1๐ฟ ๏ฟฝ
๐ฟ2
0๐ฅ cos ๏ฟฝ
2๐๐ฟ๐๐ฅ๏ฟฝ ๐๐ฅ
โโโโโโโโ
=4๐ฟ
โโโโโโ๏ฟฝsin ๏ฟฝ
2๐๐ฟ๐ ๏ฟฝ๐ฟ2๏ฟฝโ 0๏ฟฝ๏ฟฝ โ
1๐ฟ ๏ฟฝ
๐ฟ2
0๐ฅ cos ๏ฟฝ
2๐๐ฟ๐๐ฅ๏ฟฝ ๐๐ฅ
โโโโโโ
=4๐ฟ
โโโโโโโ
0
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ[sin๐๐] โ 1๐ฟ ๏ฟฝ
๐ฟ2
0๐ฅ cos ๏ฟฝ
2๐๐ฟ๐๐ฅ๏ฟฝ ๐๐ฅ
โโโโโโโ
= โ4๐ฟ2 ๏ฟฝ
๐ฟ2
0๐ฅ cos ๏ฟฝ
2๐๐ฟ๐๐ฅ๏ฟฝ ๐๐ฅ
Using integration by parts: Let ๐ข = ๐ฅ, ๐๐ฃ = cos ๏ฟฝ2๐๐ฟ ๐๐ฅ๏ฟฝ then ๐๐ข = 1, ๐ฃ =sin๏ฟฝ 2๐๐ฟ ๐๐ฅ๏ฟฝ
2๐๐ฟ ๐
= ๐ฟ2๐๐ sin ๏ฟฝ2๐๐ฟ ๐๐ฅ๏ฟฝ.
The above integral becomes
๏ฟฝ๐ฟ2
0๐ฅ cos ๏ฟฝ
2๐๐ฟ๐๐ฅ๏ฟฝ ๐๐ฅ = ๏ฟฝ
๐ฟ2๐๐
๐ฅ sin ๏ฟฝ2๐๐ฟ๐๐ฅ๏ฟฝ๏ฟฝ
๐ฟ2
0โ
๐ฟ2๐๐ ๏ฟฝ
๐ฟ2
0sin ๏ฟฝ
2๐๐ฟ๐๐ฅ๏ฟฝ ๐๐ฅ
= ๏ฟฝ๐ฟ2๐๐ ๏ฟฝ
๐ฟ2๏ฟฝ
sin ๏ฟฝ2๐๐ฟ๐๐ฟ2๏ฟฝโ 0๏ฟฝ โ
๐ฟ2๐๐
โโโโโโโโโโโ
cos ๏ฟฝ2๐๐ฟ ๐๐ฅ๏ฟฝ2๐๐ฟ ๐
โโโโโโโโโโ
๐ฟ2
0
=๐ฟ2
4๐๐sin (๐๐) + ๐ฟ2
4๐2๐2 ๏ฟฝcos ๏ฟฝ
2๐๐ฟ๐ ๏ฟฝ๐ฟ2๏ฟฝ๏ฟฝ
โ 1๏ฟฝ
=๐ฟ2
4๐2๐2(cos (๐๐) โ 1)
Therefore
๐๐ = โ4๐ฟ2 ๏ฟฝ
๐ฟ2
4๐2๐2(cos (๐๐) โ 1)๏ฟฝ
=1
๐2๐2(1 โ cos (๐๐))
The above is zero for even ๐ and 2๐ฟ2
4๐2๐2 for odd ๐. Therefore the above simplifies to
๐๐ =2
๐2๐2๐ = 1, 3, 5,โฏ
Because ๐ (๐ฅ) is an even function, then ๐๐ = 0 for all ๐. The Fourier series from (1) nowbecomes
๐ (๐ฅ) =34+
โ๏ฟฝ
๐=1,3,5,โฏ
2๐2๐2
cos ๏ฟฝ2๐๐ฟ๐๐ฅ๏ฟฝ
To verify the above result, the Fourier series approximation given above was plotted forincreasing ๐ against the original ๐ (๐ฅ) function in order to see how the approximationimproves as ๐ increases. Using ๐ฟ = 2, the result is given below.
The original function is in the red color. The plot shows that the convergence is fast (dueto the 1
๐2 term). The convergence is uniform. After only 4 terms, the error between ๐ (๐ฅ)and its Fourier series approximation becomes very small. As expected, the error is largestat the top and at the lower corners where the original function changes more rapidly andtherefore more terms would be needed in those regions compared to the straight edgesregions of the function ๐ (๐ฅ) to get a better approximation.
91
x
f(x)
approximation for n=1
x
f(x)
approximation for n=3
x
f(x)
approximation for n=5
x
f(x)
approximation for n=7
Figure 3.12: Fourier series approximation, part 1
ClearAll[L, x, n, a]
L = 2;
a[n_] := 2/(Pi^2 n^2);
fApprox[x_, nTerms_] := 3/ 4 + Sum[a[n] Cos[2 Pi/ L n x], {n, 1, nTerms, 2}]
p = Table[
Plot[{f[x], fApprox[x, i]}, {x, -L/ 2, L/ 2},
Frame โ True,
FrameLabel โ {{"f(x)", None}, {"x", Row[{"approximation for n=", i}]}},
GridLines โ Automatic, GridLinesStyle โ LightGray,
PlotStyle โ {Red, Blue},
ImageSize โ 400,
BaseStyle โ 16],
{i, 1, 7, 2}
];
p = Grid[Partition[p, 2]]
Export["../images/p1_plot_2.pdf", p]
Figure 3.13: Code used
Part 2
The following is a plot of the function ๐ (๐ฅ) = ๐๐ฅ. In this plot, ๐ฟ = 1 was used.
92
x
f(x)
Plot of function for one period
Figure 3.14: Function plot part 2
L = 1;
f[x_] := Exp[x];
p = Plot[f[x], {x, -L/ 2, L/ 2}, AxesOrigin โ {0, 0},
Frame โ True,
FrameLabel โ {{"f(x)", None}, {"x", "Plot of function for one period"}},
BaseStyle โ 14, GridLines โ Automatic, GridLinesStyle โ LightGray, PlotStyle โ Red]
Export["../images/p1_plot_3.pdf", p]
Figure 3.15: Code used
The Fourier series of ๐ (๐ฅ) = is given by
๐ (๐ฅ) =๐02+
โ๏ฟฝ๐=1
๐๐ cos ๏ฟฝ2๐๐ฟ๐๐ฅ๏ฟฝ + ๐๐ sin ๏ฟฝ
2๐๐ฟ๐๐ฅ๏ฟฝ (1A)
Where ๐ฟ is the period and
๐0 =2๐ฟ ๏ฟฝ
๐ฟ2
โ ๐ฟ2
๐ (๐ฅ) ๐๐ฅ
=2๐ฟ ๏ฟฝ
๐ฟ2
โ ๐ฟ2
๐๐ฅ๐๐ฅ
=2๐ฟ[๐๐ฅ]
๐ฟ2โ ๐ฟ2
=2๐ฟ ๏ฟฝ๐๐ฟ2 โ ๐โ
๐ฟ2 ๏ฟฝ
=4๐ฟ
โกโขโขโขโขโขโฃ๐๐ฟ2 โ ๐โ
๐ฟ2
2
โคโฅโฅโฅโฅโฅโฆ
=4๐ฟ
sinh ๏ฟฝ๐ฟ2๏ฟฝ
And
๐๐ =2๐ฟ ๏ฟฝ
๐ฟ2
โ ๐ฟ2
๐ (๐ฅ) cos ๏ฟฝ2๐๐ฟ๐๐ฅ๏ฟฝ ๐๐ฅ
=2๐ฟ ๏ฟฝ
๐ฟ2
โ ๐ฟ2
๐๐ฅ cos ๏ฟฝ2๐๐ฟ๐๐ฅ๏ฟฝ ๐๐ฅ (1)
Integration by parts: Let ๐ข = cos ๏ฟฝ2๐๐ฟ ๐๐ฅ๏ฟฝ , ๐๐ข = โ2๐๐๐ฟ sin ๏ฟฝ2๐๐ฟ ๐๐ฅ๏ฟฝ and let ๐๐ฃ = ๐๐ฅ, ๐ฃ = ๐๐ฅ,
93
therefore
๐ผ = ๏ฟฝ๐ฟ2
โ ๐ฟ2
๐๐ฅ cos ๏ฟฝ2๐๐ฟ๐๐ฅ๏ฟฝ ๐๐ฅ
= ๏ฟฝ๐๐ฅ cos ๏ฟฝ2๐๐ฟ๐๐ฅ๏ฟฝ๏ฟฝ
๐ฟ2
โ ๐ฟ2
โ๏ฟฝ๐ฟ2
โ ๐ฟ2
โ2๐๐๐ฟ
sin ๏ฟฝ2๐๐ฟ๐๐ฅ๏ฟฝ ๐๐ฅ๐๐ฅ
= ๏ฟฝ๐๐ฟ2 cos ๏ฟฝ
2๐๐ฟ๐๐ฟ2๏ฟฝโ ๐โ
๐ฟ2 cos ๏ฟฝ
2๐๐ฟ๐ ๏ฟฝโ
๐ฟ2๏ฟฝ๏ฟฝ๏ฟฝ
+2๐๐๐ฟ ๏ฟฝ
๐ฟ2
โ ๐ฟ2
sin ๏ฟฝ2๐๐ฟ๐๐ฅ๏ฟฝ ๐๐ฅ๐๐ฅ
= ๏ฟฝ๐๐ฟ2 cos (๐๐) โ ๐โ
๐ฟ2 cos (๐๐)๏ฟฝ +
2๐๐๐ฟ ๏ฟฝ
๐ฟ2
โ ๐ฟ2
sin ๏ฟฝ2๐๐ฟ๐๐ฅ๏ฟฝ ๐๐ฅ๐๐ฅ
= cos (๐๐) ๏ฟฝ๐๐ฟ2 โ ๐โ
๐ฟ2 ๏ฟฝ +
2๐๐๐ฟ ๏ฟฝ
๐ฟ2
โ ๐ฟ2
sin ๏ฟฝ2๐๐ฟ๐๐ฅ๏ฟฝ ๐๐ฅ๐๐ฅ
= 2 cos (๐๐) sinh ๏ฟฝ๐ฟ2๏ฟฝ+2๐๐๐ฟ ๏ฟฝ
๐ฟ2
โ ๐ฟ2
sin ๏ฟฝ2๐๐ฟ๐๐ฅ๏ฟฝ ๐๐ฅ๐๐ฅ
Integration by parts again, let ๐ข = sin ๏ฟฝ2๐๐ฟ ๐๐ฅ๏ฟฝ , ๐๐ข =2๐๐๐ฟ cos ๏ฟฝ2๐๐ฟ ๐๐ฅ๏ฟฝ and ๐๐ฃ = ๐
๐ฅ, ๐ฃ = ๐๐ฅ. Theabove becomes
๐ผ = 2 cos (๐๐) sinh ๏ฟฝ๐ฟ2๏ฟฝ+2๐๐๐ฟ
โโโโโโโโ๏ฟฝ๐
๐ฅ sin ๏ฟฝ2๐๐ฟ๐๐ฅ๏ฟฝ๏ฟฝ
๐ฟ2
โ ๐ฟ2
โ๏ฟฝ๐ฟ2
โ ๐ฟ2
2๐๐๐ฟ
cos ๏ฟฝ2๐๐ฟ๐๐ฅ๏ฟฝ ๐๐ฅ๐๐ฅ
โโโโโโโโ
The term ๏ฟฝ๐๐ฅ sin ๏ฟฝ2๐๐ฟ ๐๐ฅ๏ฟฝ๏ฟฝ๐ฟ2
โ ๐ฟ2
goes to zero since it gives sin (๐๐) and ๐ is integer. The above
simplifies to
๐ผ = 2 cos (๐๐) sinh ๏ฟฝ๐ฟ2๏ฟฝ+2๐๐๐ฟ
โโโโโโโ2๐๐๐ฟ ๏ฟฝ
๐ฟ2
โ ๐ฟ2
cos ๏ฟฝ2๐๐ฟ๐๐ฅ๏ฟฝ ๐๐ฅ๐๐ฅ
โโโโโโ
= 2 cos (๐๐) sinh ๏ฟฝ๐ฟ2๏ฟฝโ4๐2๐2
๐ฟ2 ๏ฟฝ๐ฟ2
โ ๐ฟ2
cos ๏ฟฝ2๐๐ฟ๐๐ฅ๏ฟฝ ๐๐ฅ๐๐ฅ
Since โซ๐ฟ2
โ ๐ฟ2
cos ๏ฟฝ2๐๐ฟ ๐๐ฅ๏ฟฝ ๐๐ฅ๐๐ฅ = ๐ผ the above reduces to
๐ผ = 2 cos (๐๐) sinh ๏ฟฝ๐ฟ2๏ฟฝโ4๐2๐2
๐ฟ2๐ผ
๐ผ ๏ฟฝ1 +4๐2๐2
๐ฟ2 ๏ฟฝ = 2 cos (๐๐) sinh ๏ฟฝ๐ฟ2๏ฟฝ
๐ผ =2 cos (๐๐) sinh ๏ฟฝ๐ฟ2๏ฟฝ
1 + 4๐2๐2
๐ฟ2
Using the above in (1) gives
๐๐ =2๐ฟ
2 cos (๐๐) sinh ๏ฟฝ๐ฟ2๏ฟฝ
1 + 4๐2๐2
๐ฟ2
=2๐ฟ2
๐ฟ
2 cos (๐๐) sinh ๏ฟฝ๐ฟ2๏ฟฝ
๐ฟ2 + 4๐2๐2
=4๐ฟ
๐ฟ2 + 4๐2๐2cos (๐๐) sinh ๏ฟฝ
๐ฟ2๏ฟฝ
94
Next, ๐๐ is found:
๐๐ =2๐ฟ ๏ฟฝ
๐ฟ2
โ ๐ฟ2
๐ (๐ฅ) sin ๏ฟฝ2๐๐ฟ๐๐ฅ๏ฟฝ ๐๐ฅ
=2๐ฟ ๏ฟฝ
๐ฟ2
โ ๐ฟ2
๐๐ฅ sin ๏ฟฝ2๐๐ฟ๐๐ฅ๏ฟฝ ๐๐ฅ (2)
Integration by parts: Let ๐ข = sin ๏ฟฝ2๐๐ฟ ๐๐ฅ๏ฟฝ , ๐๐ข =2๐๐๐ฟ sin ๏ฟฝ2๐๐ฟ ๐๐ฅ๏ฟฝ and let ๐๐ฃ = ๐๐ฅ, ๐ฃ = ๐๐ฅ, therefore
๐ผ = ๏ฟฝ๐ฟ2
โ ๐ฟ2
๐๐ฅ sin ๏ฟฝ2๐๐ฟ๐๐ฅ๏ฟฝ ๐๐ฅ
= ๏ฟฝ๐๐ฅ sin ๏ฟฝ2๐๐ฟ๐๐ฅ๏ฟฝ๏ฟฝ
๐ฟ2
โ ๐ฟ2
โ๏ฟฝ๐ฟ2
โ ๐ฟ2
2๐๐๐ฟ
cos ๏ฟฝ2๐๐ฟ๐๐ฅ๏ฟฝ ๐๐ฅ๐๐ฅ
But ๏ฟฝ๐๐ฅ sin ๏ฟฝ2๐๐ฟ ๐๐ฅ๏ฟฝ๏ฟฝ๐ฟ2
โ ๐ฟ2
goes to zero as sin (๐๐) = 0 for integer ๐ and the above simplifies to
๐ผ = โ2๐๐๐ฟ ๏ฟฝ
๐ฟ2
โ ๐ฟ2
cos ๏ฟฝ2๐๐ฟ๐๐ฅ๏ฟฝ ๐๐ฅ๐๐ฅ
Integration by parts again: let ๐ข = cos ๏ฟฝ2๐๐ฟ ๐๐ฅ๏ฟฝ , ๐๐ข = โ2๐๐๐ฟ sin ๏ฟฝ2๐๐ฟ ๐๐ฅ๏ฟฝ and ๐๐ฃ = ๐
๐ฅ, ๐ฃ = ๐๐ฅ. Theabove becomes
๐ผ = โ2๐๐๐ฟ
โโโโโโโโ๏ฟฝ๐
๐ฅ cos ๏ฟฝ2๐๐ฟ๐๐ฅ๏ฟฝ๏ฟฝ
๐ฟ2
โ ๐ฟ2
โ๏ฟฝ๐ฟ2
โ ๐ฟ2
โ2๐๐๐ฟ
sin ๏ฟฝ2๐๐ฟ๐๐ฅ๏ฟฝ ๐๐ฅ๐๐ฅ
โโโโโโโโ
= โ2๐๐๐ฟ
โโโโโโ2 cos (๐๐) sinh ๏ฟฝ
๐ฟ2๏ฟฝ+2๐๐๐ฟ ๏ฟฝ
๐ฟ2
โ ๐ฟ2
sin ๏ฟฝ2๐๐ฟ๐๐ฅ๏ฟฝ ๐๐ฅ๐๐ฅ
โโโโโโ
But โซ๐ฟ2
โ ๐ฟ2
sin ๏ฟฝ2๐๐ฟ ๐๐ฅ๏ฟฝ ๐๐ฅ๐๐ฅ = ๐ผ and the above reduces to
๐ผ = โ2๐๐๐ฟ ๏ฟฝ2 cos (๐๐) sinh ๏ฟฝ
๐ฟ2๏ฟฝ+2๐๐๐ฟ๐ผ๏ฟฝ
๐ผ = โ4๐๐๐ฟ
cos (๐๐) sinh ๏ฟฝ๐ฟ2๏ฟฝโ4๐2๐2
๐ฟ2๐ผ
๐ผ ๏ฟฝ1 +4๐2๐2
๐ฟ2 ๏ฟฝ = โ4๐๐๐ฟ
cos (๐๐) sinh ๏ฟฝ๐ฟ2๏ฟฝ
๐ผ =โ4๐๐๐ฟ cos (๐๐) sinh ๏ฟฝ๐ฟ2๏ฟฝ
1 + 4๐2๐2
๐ฟ2
=โ4๐๐๐ฟ cos (๐๐) sinh ๏ฟฝ๐ฟ2๏ฟฝ
๐ฟ2 + 4๐2๐2Using the above in (2) gives
๐๐ =2๐ฟ
โ4๐๐๐ฟ cos (๐๐) sinh ๏ฟฝ๐ฟ2๏ฟฝ
๐ฟ2 + 4๐2๐2
=โ8๐๐
๐ฟ2 + 4๐2๐2cos (๐๐) sinh ๏ฟฝ
๐ฟ2๏ฟฝ
Therefore, from (1A) the Fourier series is
๐ (๐ฅ) =2๐ฟ
sinh ๏ฟฝ๐ฟ2๏ฟฝ+
โ๏ฟฝ๐=1
4๐ฟ๐ฟ2 + 4๐2๐2
cos (๐๐) sinh ๏ฟฝ๐ฟ2๏ฟฝ
cos ๏ฟฝ2๐๐ฟ๐๐ฅ๏ฟฝ โ
8๐๐๐ฟ2 + 4๐2๐2
cos (๐๐) sinh ๏ฟฝ๐ฟ2๏ฟฝ
sin ๏ฟฝ2๐๐ฟ๐๐ฅ๏ฟฝ
(3)
To verify the result, the above was plotted for increasing ๐ against the original ๐ (๐ฅ) functionto see how the approximation improves as ๐ increases. Using ๐ฟ = 2, the result is displayedbelow. The original function is in the red color.
95
Compared to part (1), more terms are needed here to get good approximation. Since theoriginal function is piecewise continuous when extending over multiple periods, the conver-gence is no longer a uniform convergence. At the point of discontinuity, the approximationconverges to the average value of the original function at that point. At about 20 termsthe approximation started to give good results. Due to Gibbs phenomena, at the points ofdiscontinuities, the error is largest. Here is a plot showing one period
x
f(x)
approximation for n=1
x
f(x)
approximation for n=3
x
f(x)
approximation for n=5
x
f(x)
approximation for n=7
x
f(x)
approximation for n=9
x
f(x)
approximation for n=11
x
f(x)
approximation for n=13
x
f(x)
approximation for n=15
x
f(x)
approximation for n=17
x
f(x)
approximation for n=19
Figure 3.16: Fourier series approximation, showing one period
96
ClearAll[L, x, n, a]
L = 2;
f[x_] := Exp[x];
a[n_] := 4 L/(L^2 + 4 Pi^2 n^2) Cos[Pi n] Sinh[L/ 2] ;
b[n_] := -8 Pi n/(L^2 + 4 Pi^2 n^2) Cos[Pi n] Sinh[L/ 2] ;
fApprox[x_, nTerms_] := 2/ L Sinh[L/ 2] + Sum[a[n] Cos[2 Pi/ L n x] + b[n] Sin[2 Pi/ L n x], {n, 1, nTerms, 1}]
p = Table[
Plot[{f[x], fApprox[x, i]}, {x, -L/ 2, L/ 2},
Frame โ True,
FrameLabel โ {{"f(x)", None}, {"x", Row[{"approximation for n=", i}]}},
GridLines โ Automatic, GridLinesStyle โ LightGray,
PlotStyle โ {Red, Blue}, ImageSize โ 400, BaseStyle โ 16],
{i, 1, 20, 2}
];
p = Grid[Partition[p, 2]]
Export["../images/p1_plot_4.pdf", p]
Figure 3.17: Code used
In the following plot, 3 periods are shown to make it easier to see the e๏ฟฝect of discontinuitiesand the Gibbs phenomena
97
x
f(x)
approximation for n=1
x
f(x)
approximation for n=2
x
f(x)
approximation for n=3
xf(x)
approximation for n=4
x
f(x)
approximation for n=5
x
f(x)
approximation for n=6
x
f(x)
approximation for n=7
x
f(x)
approximation for n=8
x
f(x)
approximation for n=9
x
f(x)
approximation for n=10
Figure 3.18: Fourier series approximation, showing 3 periods
98
ClearAll[L, x, n, a]
L = 2;
f[x_] := Piecewise[{
{Exp[x + L], -3/ 2 L < x < -L/ 2},
{Exp[x], -L/ 2 < x < L/ 2},
{Exp[x - L], L/ 2 < x < 3/ 2 L}}
];
a[n_] := 4 L/(L^2 + 4 Pi^2 n^2) Cos[Pi n] Sinh[L/ 2] ;
b[n_] := -8 Pi n/(L^2 + 4 Pi^2 n^2) Cos[Pi n] Sinh[L/ 2] ;
fApprox[x_, nTerms_] := 2/ L Sinh[L/ 2] + Sum[a[n] Cos[2 Pi/ L n x] + b[n] Sin[2 Pi/ L n x], {n, 1, nTerms, 1}];
p = Table[
Plot[{f[x], fApprox[x, i]}, {x, -3/ 2 L, 3/ 2 L},
Frame โ True, FrameLabel โ {{"f(x)", None}, {"x", Row[{"approximation for n=", i}]}},
GridLines โ Automatic, GridLinesStyle โ LightGray,
PlotStyle โ {Red, Blue},
ImageSize โ 400, BaseStyle โ 16],
{i, 1, 10, 1}
];
p = Grid[Partition[p, 2]]
Export["../images/p1_plot_5.pdf", p]
Figure 3.19: Code used
3.5.2 Problem 2
Find the general solution of
1. 2๐ฅ3๐ฆโฒ = 1 + ๏ฟฝ1 + 4๐ฅ2๐ฆ
2. ๐๐ฅ sin ๐ฆ โ 2๐ฆ sin ๐ฅ + ๏ฟฝ๐ฆ2 + ๐๐ฅ cos ๐ฆ + 2 cos ๐ฆ๏ฟฝ ๐ฆโฒ = 0
3. ๐ฆโฒ + ๐ฆ cos ๐ฅ = 12 sin ๐ฅ
Solution
part 1
This ODE is not separable and it is also not exact (It was checked for exactness and failedthe test). The ODE is next checked to see if it is isobaric. An ODE ๐ฆโฒ = ๐ ๏ฟฝ๐ฅ, ๐ฆ๏ฟฝ is isobaric(which is a generalization of a homogeneous ODE) if the substitution
๐ฆ (๐ฅ) = ๐ฃ (๐ฅ) ๐ฅ๐
Changes the ODE to be a separable one in ๐ฃ (๐ฅ). To determine if it isobaric, a weight ๐ isassigned to ๐ฆ and to ๐๐ฆ, and a weight of 1 is assigned to ๐ฅ and to ๐๐ฅ, then if an ๐ could befound such that each term in the ODE will have the same weight, then the ODE is isobaricand it can be made separable using the above substitution. Writing the above ODE as
2๐ฅ3๐๐ฆ = ๏ฟฝ1 + ๏ฟฝ1 + 4๐ฅ2๐ฆ๏ฟฝ ๐๐ฅ
๏ฟฝ2๐ฅ3๐๐ฆ โโ๐๐ฅโ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ1 + 4๐ฅ
2๐ฆ๐๐ฅ = 0
Adding the weights of the first term above gives 2๐ฅ3๐๐ฆ โ 3 + ๐. The next term weight is๐๐ฅ โ 1. The next term weight is ๏ฟฝ1 + 4๐ฅ2๐ฆ๐๐ฅ โ
12(2 + ๐) + 1 = 2 + ๐
2 . Therefore the weightsof each term are
{3 + ๐, 1, 2 +๐2}
Each term weight can be made the same by selecting ๐ = โ2. This value makes each termhave weight 1 and the above becomes
{1, 1, 1}
Therefore the ODE is isobaric. Using this value of ๐ the substitution ๐ฆ = ๐ฃ๐ฅ2 is now used
to make the original ODE separable๐๐ฆ๐๐ฅ
=1๐ฅ2๐๐ฃ๐๐ฅโ 2
๐ฃ๐ฅ3
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The original ODE now becomes (where each ๐ฆ is replaced by ๐ฃ๐ฅ2 ) separable as follows
2๐ฅ3 ๏ฟฝ1๐ฅ2๐๐ฃ๐๐ฅโ 2
๐ฃ๐ฅ3 ๏ฟฝ
= 1 +๏ฟฝ1 + 4๐ฅ2
๐ฃ๐ฅ2
2๐ฅ๐๐ฃ๐๐ฅโ 4๐ฃ = 1 + โ1 + 4๐ฃ
2๐ฅ๐๐ฃ๐๐ฅ
= 1 + โ1 + 4๐ฃ + 4๐ฃ
Solving this ODE for ๐ฃ (๐ฅ)๐๐ฃ
1 + โ1 + 4๐ฃ + 4๐ฃ=12๐ฅ๐๐ฅ
Integrating both sides gives
๏ฟฝ๐๐ฃ
1 + โ1 + 4๐ฃ + 4๐ฃ=12
ln |๐ฅ| + ๐ (2)
The integral above is solved by substitution. Let โ1 + 4๐ฃ = ๐ข, hence ๐๐ข๐๐ฃ =
12
4
โ1+4๐ฃ= 2
๐ข or
๐๐ฃ = 12๐ข๐๐ข. Squaring both sides of โ1 + 4๐ฃ = ๐ข (and assuming 1 + 4๐ฃ > 0) gives 1 + 4๐ฃ = ๐ข2
or ๐ฃ = ๐ข2โ14 . Therefore the LHS integral in (2) becomes
๏ฟฝ1
1 + โ1 + 4๐ฃ + 4๐ฃ๐๐ฃ =
12 ๏ฟฝ
๐ข
1 + ๐ข + 4 ๏ฟฝ๐ข2โ14๏ฟฝ๐๐ข
=12 ๏ฟฝ
๐ข๐ข + ๐ข2
๐๐ข
=12 ๏ฟฝ
11 + ๐ข
๐๐ข
=12
ln |1 + ๐ข|
Using this result in (2) gives the following (the absolute values are removed because theconstant of integration absorbs the sign).
12
ln (1 + ๐ข) = 12
ln ๐ฅ + ๐
ln (1 + ๐ข) = ln ๐ฅ + 2๐Let 2๐ = ๐ถ0 be a new constant. The above becomes
ln (1 + ๐ข) = ln ๐ฅ + ๐ถ0๐ln(1+๐ข) = ๐ln ๐ฅ+๐ถ0
1 + ๐ข = ๐๐ถ0๐ฅ1 + ๐ข = ๐ถ๐ฅ
Where ๐ถ = ๐๐ถ0 is a new constant. Therefore the solution is
๐ข (๐ฅ) = ๐ถ๐ฅ โ 1
Since ๐ข (๐ฅ) = โ1 + 4๐ฃ then the above becomes
โ1 + 4๐ฃ = ๐ถ๐ฅ โ 1
1 + 4๐ฃ = (๐ถ๐ฅ โ 1)2
๐ฃ (๐ฅ) =(๐ถ๐ฅ โ 1)2 โ 1
4But ๐ฆ = ๐ฃ
๐ฅ2 therefore the above gives the final solution as
๐ฆ (๐ฅ) =(๐ถ๐ฅ โ 1)2 โ 1
4๐ฅ2Where ๐ถ is the constant of integration.
Part 2
๐๐ฅ sin ๐ฆ โ 2๐ฆ sin ๐ฅ + ๏ฟฝ๐ฆ2 + ๐๐ฅ cos ๐ฆ + 2 cos ๐ฅ๏ฟฝ ๐ฆโฒ = 0
100
The first step is to write the ODE in standard form to check if it is an exact ODE
๐(๐ฅ, ๐ฆ)๐๐ฅ + ๐(๐ฅ, ๐ฆ)๐๐ฆ = 0
Hence
๐(๐ฅ, ๐ฆ) = ๐๐ฅ sin ๐ฆ โ 2 ๐ฆ sin ๐ฅ๐(๐ฅ, ๐ฆ) = ๐ฆ2 + ๐๐ฅ cos ๐ฆ + 2 cos ๐ฅ
Next, the ODE is determined if it is exact or not. The ODE is exact if the followingcondition is satisfied
๐๐๐๐ฆ
=๐๐๐๐ฅ
Applying the above on the given ODE results in
๐๐๐๐ฆ
= ๐๐ฅ cos ๐ฆ โ 2 sin ๐ฅ
๐๐๐๐ฅ
= ๐๐ฅ cos ๐ฆ โ 2 sin ๐ฅ
Because ๐๐๐๐ฆ = ๐๐
๐๐ฅ , then the ODE is exact. The following equations are used to solve for
the function ๐ ๏ฟฝ๐ฅ, ๐ฆ๏ฟฝ
๐๐๐๐ฅ
= ๐ = ๐๐ฅ sin ๐ฆ โ 2 ๐ฆ sin ๐ฅ (3)
๐๐๐๐ฆ
= ๐ = ๐ฆ2 + ๐๐ฅ cos ๐ฆ + 2 cos ๐ฅ (4)
Integrating (3) w.r.t ๐ฅ gives
๏ฟฝ๐๐๐๐ฅ๐๐ฅ = ๏ฟฝ๐๐ฅ sin ๐ฆ โ 2 ๐ฆ sin ๐ฅ๐๐ฅ
๐ ๏ฟฝ๐ฅ, ๐ฆ๏ฟฝ = ๐๐ฅ sin ๐ฆ + 2 ๐ฆ cos ๐ฅ + ๐(๐ฆ) (5)
Where ๐(๐ฆ) is used as the constant of integration because ๐ ๏ฟฝ๐ฅ, ๐ฆ๏ฟฝ is a function of both ๐ฅand ๐ฆ. Taking derivative of (5) w.r.t ๐ฆ gives
๐๐๐๐ฆ
= ๐๐ฅ cos ๐ฆ + 2 cos ๐ฅ + ๐โฒ(๐ฆ) (6)
But (4) says that๐๐๐๐ฆ = ๐ฆ
2 + ๐๐ฅ cos ๐ฆ + 2 cos ๐ฅ. Therefore by equating (4) and (6) then ๐โฒ ๏ฟฝ๐ฆ๏ฟฝcan be solved for:
๐ฆ2 + ๐๐ฅ cos ๐ฆ + 2 cos ๐ฅ = ๐๐ฅ cos ๐ฆ + 2 cos ๐ฅ + ๐โฒ(๐ฆ) (7)
Solving the above for ๐โฒ(๐ฆ) gives
๐โฒ(๐ฆ) = ๐ฆ2
Integrating w.r.t ๐ฆ gives ๐ ๏ฟฝ๐ฆ๏ฟฝ
๏ฟฝ๐โฒ๐๐ฆ = ๏ฟฝ๐ฆ2๐๐ฆ
๐(๐ฆ) =13๐ฆ3 + ๐ถ1
Where ๐ถ1 is constant of integration. Substituting the value of ๐(๐ฆ) back into (5) gives ๐ ๏ฟฝ๐ฅ, ๐ฆ๏ฟฝ
๐ = ๐๐ฅ sin ๐ฆ + 2 ๐ฆ cos ๐ฅ + 13๐ฆ3 + ๐ถ1
But since ๐ itself is a constant function, say ๐ = ๐ถ0 where ๐ถ0 is new constant, then bycombining ๐ถ1 and ๐ถ0 constants into a new constant ๐ถ1, the above gives the solution
๐ถ1 = ๐๐ฅ sin ๐ฆ (๐ฅ) + 2 ๐ฆ (๐ฅ) cos ๐ฅ + 13๐ฆ3 (๐ฅ)
The above is left in implicit form for simplicity.
Part 3
๐ฆโฒ + ๐ฆ cos ๐ฅ = 12
sin (2๐ฅ)
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This ODE is linear in ๐ฆ. It is solved using an integrating factor ๐ = ๐โซ cos ๐ฅ๐๐ฅ = ๐sin ๐ฅ.Multiplying both sides of the ODE by ๐ makes the left side an exact di๏ฟฝerential
๐ ๏ฟฝ๐ฆ๐๏ฟฝ =12๐ sin (2) ๐ฅ๐๐ฅ
Integrating both sides gives
๐ฆ๐ =12 ๏ฟฝ
๐ sin (2๐ฅ) ๐๐ฅ + ๐ถ
๐ฆ๐sin ๐ฅ =12 ๏ฟฝ
๐sin ๐ฅ sin (2๐ฅ) ๐๐ฅ + ๐ถ (1)
The above integral can be solved as follows. Since sin (2๐ฅ) = 2 sin ๐ฅ cos ๐ฅ therefore then
๐ผ =12 ๏ฟฝ
๐sin ๐ฅ sin (2๐ฅ) ๐๐ฅ = ๏ฟฝ๐sin ๐ฅ sin ๐ฅ cos ๐ฅ๐๐ฅ
Using the substitution ๐ง = sin ๐ฅ, then ๐๐ง = ๐๐ฅ cos ๐ฅ and the above becomes
๐ผ = ๏ฟฝ๐๐ง๐ง๐๐ง
Integrating the above by parts: โซ๐ข๐๐ฃ = ๐ข๐ฃ โ โซ๐ฃ๐๐ข. Let ๐ข = ๐ง, ๐๐ฃ = ๐๐ง โ ๐๐ข = 1, ๐ฃ = ๐๐ง, andthe above becomes
๐ผ = ๐ง๐๐ง โ๏ฟฝ๐๐ง๐๐ง
= ๐ง๐๐ง โ ๐๐ง
= ๐๐ง (๐ง โ 1)
Since ๐ง = sin ๐ฅ the above reduces to๐ผ = ๐sin ๐ฅ (sin (๐ฅ) โ 1)
Substituting this back in (1) results in
๐ฆ๐sin ๐ฅ = ๐sin ๐ฅ (sin (๐ฅ) โ 1) + ๐ถTherefore the final solution is
๐ฆ (๐ฅ) = sin (๐ฅ) โ 1 + ๐ถ๐โ sin ๐ฅ
Where ๐ถ is the constant of integration.
3.5.3 Problem 3
Find general solution of
1. ๐ฆโฒโฒโฒ โ 4๐ฆโฒโฒ โ 4๐ฆโฒ + 16 = 8 sin ๐ฅ
2. ๐2๐ฆโฒ2 = ๏ฟฝ1 + ๐ฆโฒ2๏ฟฝ3
Solution
Part 1
๐ฆโฒโฒโฒ โ 4๐ฆโฒโฒ โ 4๐ฆโฒ = 8 sin ๐ฅ โ 16This is linear nonhomogeneous ODE with constant coe๏ฟฝcients. Solving first the homo-geneous ODE ๐ฆโฒโฒโฒ โ 4๐ฆโฒโฒ โ 4๐ฆโฒ = 0. Since the term ๐ฆ is missing from the ODE then thesubstitution ๐ฆโฒ = ๐ข reduces the ODE to a second order ODE
๐ขโฒโฒ โ 4๐ขโฒ โ 4๐ข = 0 (1)
Let ๐ข = ๐๐๐ฅ. Substituting this into the above and simplifying gives the characteristic equation
๐2 โ 4๐ โ 4 = 0
102
The Roots are ๐ = โ ๐2๐ ยฑ
12๐โ๐
2 โ 4๐๐ or
๐ =42ยฑ12๏ฟฝ
16 โ 4 (โ4)
= 2 ยฑ12โ
32
= 2 ยฑ 2โ2
= 2 ๏ฟฝ1 ยฑ โ2๏ฟฝ
Hence the solution to (1) is given by linear combinations of ๐๐1๐ฅ, ๐๐2๐ฅ as
๐ขโ (๐ฅ) = ๐1๐2๏ฟฝ1+โ2๏ฟฝ๐ฅ + ๐2๐
2๏ฟฝ1โโ2๏ฟฝ๐ฅ
But since ๐ฆโฒ = ๐ข, then ๐ฆ is found by integrating the above
๐ฆโ = ๏ฟฝ๐1๐2๏ฟฝ1+โ2๏ฟฝ๐ฅ + ๐2๐
2๏ฟฝ1โโ2๏ฟฝ๐ฅ๐๐ฅ
= ๐1๐2๏ฟฝ1+โ2๏ฟฝ๐ฅ
2 ๏ฟฝ1 + โ2๏ฟฝ+ ๐2
๐๏ฟฝ2โ2โ2๏ฟฝ๐ฅ
2 ๏ฟฝ1 โ โ2๏ฟฝ+ ๐ถ3
To simplify the above, let ๐12๏ฟฝ1+โ2๏ฟฝ
= ๐ถ1,๐2
2๏ฟฝ1โโ2๏ฟฝ= ๐ถ2, where ๐ถ1, ๐ถ2 are new constants. The
above simplifies to
๐ฆโ = ๐ถ1๐2๏ฟฝ1+โ2๏ฟฝ๐ฅ + ๐ถ2๐
๏ฟฝ2โ2โ2๏ฟฝ๐ฅ + ๐ถ3The above solution is homogeneous solution to the original ODE. Next, the particularsolution is found. Since the RHS of the original ODE is sin ๐ฅ โ 16 then choosing ๐ฆ๐ to havethe form
๐ฆ๐ = ๐ด sin ๐ฅ + ๐ต cos ๐ฅ + ๐๐ฅTherefore
๐ฆโฒ๐ = ๐ + ๐ด cos ๐ฅ โ ๐ต sin ๐ฅ๐ฆโฒโฒ๐ = โ๐ด sin ๐ฅ โ ๐ต cos ๐ฅ๐ฆโฒโฒโฒ๐ = โ๐ด cos ๐ฅ + ๐ต sin ๐ฅ
Substituting these back into the original ODE ๐ฆโฒโฒโฒ โ 4๐ฆโฒโฒ โ 4๐ฆโฒ = 8 sin ๐ฅ โ 16 gives(โ๐ด cos ๐ฅ + ๐ต sin ๐ฅ) โ 4 (โ๐ด sin ๐ฅ โ ๐ต cos ๐ฅ) โ 4 (๐ + ๐ด cos ๐ฅ โ ๐ต sin ๐ฅ) = 8 sin ๐ฅ โ 16โ๐ด cos ๐ฅ + ๐ต sin ๐ฅ + 4๐ด sin ๐ฅ + 4๐ต cos ๐ฅ โ 4๐ด cos ๐ฅ + 4๐ต sin ๐ฅ โ 4๐ = 8 sin ๐ฅ โ 16
cos ๐ฅ (โ๐ด + 4๐ต โ 4๐ด) + sin ๐ฅ (๐ต + 4๐ด + 4๐ต) โ 4๐ = 8 sin ๐ฅ โ 16cos ๐ฅ (โ5๐ด + 4๐ต) + sin ๐ฅ (5๐ต + 4๐ด) โ 4๐ = 8 sin ๐ฅ โ 16
Comparing coe๏ฟฝcients gives the following equations to solve for the unknowns ๐ด,๐ต, ๐
โ4๐ = โ16โ5๐ด + 4๐ต = 05๐ต + 4๐ด = 8
The second equation gives ๐ต = 54๐ด. Using this in the third equation gives 5 ๏ฟฝ54๐ด๏ฟฝ + 4๐ด = 8,
solving gives ๐ด = 3241 . Hence ๐ต = 5
4๏ฟฝ3241๏ฟฝ = 40
41 . The first equation gives ๐ = 4. Therefore theparticular solution is
๐ฆ๐ = ๐ด sin ๐ฅ + ๐ต cos ๐ฅ + ๐๐ฅ
=3241
sin ๐ฅ + 4041
cos ๐ฅ + 4๐ฅ
Now that ๐ฆโ and ๐ฆ๐ are found, the general solution is found as
๐ฆ = ๐ฆโ + ๐ฆ๐
= ๐ถ1๐2๏ฟฝ1+โ2๏ฟฝ๐ฅ + ๐ถ2๐
๏ฟฝ2โ2โ2๏ฟฝ๐ฅ + ๐ถ3 +3241
sin ๐ฅ + 4041
cos ๐ฅ + 4๐ฅ
Where ๐ถ1, ๐ถ2 are the two constants of integration.
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Part 2
๐2๐ฆโฒ2 = ๏ฟฝ1 + ๐ฆโฒ2๏ฟฝ3
Let ๐ฆโฒ = ๐ด, the above becomes
๐2๐ด2 = ๏ฟฝ1 + ๐ด2๏ฟฝ3
= 1 + 3๐ด2 +(3) (2)2!
๐ด4 +(3) (2) (1)
3!๐ด6
= 1 + 3๐ด2 + 3๐ด4 + ๐ด6
Hence the polynomial is
๐ด6 + 3๐ด4 + ๐ด2 ๏ฟฝ3 โ ๐2๏ฟฝ + 1 = 0
Let ๐ด2 = ๐ต and the above becomes
๐ต3 + 3๐ต2 + ๐ต ๏ฟฝ3 โ ๐2๏ฟฝ + 1 = 0
With the help of the computer, the cubic roots of the above are
๐ต1 =3
๏ฟฝ๏ฟฝ14๐4 โ
127๐6 โ
12๐2 +
13
๐2
3
๏ฟฝ๏ฟฝ14๐4 โ 1
27๐6 โ 1
2๐2โ 1
๐ต2 =12๐โ3
โโโโโโโโโโโโโ
3
๏ฟฝ๏ฟฝ14๐4 โ
127๐6 โ
12๐2 โ
13
๐2
3
๏ฟฝ๏ฟฝ14๐4 โ 1
27๐6 โ 1
2๐2
โโโโโโโโโโโโโ โ12
3
๏ฟฝ๏ฟฝ14๐4 โ
127๐6 โ
12๐2 โ
16
๐2
3
๏ฟฝ๏ฟฝ14๐4 โ 1
27๐6 โ 1
2๐2โ 1
๐ต3 = โ12
3
๏ฟฝ๏ฟฝ14๐4 โ
127๐6 โ
12๐2 โ
12๐โ3
โโโโโโโโโโโโโ
3
๏ฟฝ๏ฟฝ14๐4 โ
127๐6 โ
12๐2 โ
13
๐2
3
๏ฟฝ๏ฟฝ14๐4 โ 1
27๐6 โ 1
2๐2
โโโโโโโโโโโโโ โ16
๐2
3
๏ฟฝ๏ฟฝ14๐4 โ 1
27๐6 โ 1
2๐2โ 1
Therefore ๐ด1 = ยฑโ๐ต1, ๐ด2 = ยฑโ๐ต2, ๐ด3 = ยฑโ๐ต3 or, since ๐ฆโฒ (๐ฅ) = ๐ด , then there are 6 solutions,each is a solution for one root.
๐๐ฆ1๐๐ฅ
= +โ๐ต1๐๐ฆ2๐๐ฅ
= โโ๐ต1๐๐ฆ3๐๐ฅ
= +โ๐ต2๐๐ฆ4๐๐ฅ
= โโ๐ต2๐๐ฆ5๐๐ฅ
= +๏ฟฝ๐ต3๐๐ฆ6๐๐ฅ
= โ๏ฟฝ๐ต3
But the roots ยฑ๐ต๐ are constants. Therefore each of the above can be solved by directintegration. The final solution which gives the solutions
๐ฆ1 = โ๐ต1๐ฅ + ๐ถ1๐ฆ2 = โโ๐ต1๐ฅ + ๐ถ2๐ฆ3 = โ๐ต2๐ฅ + ๐ถ3๐ฆ4 = โโ๐ต2๐ฅ + ๐ถ4๐ฆ5 = ๏ฟฝ๐ต3๐ฅ + ๐ถ5๐ฆ6 = โ๏ฟฝ๐ต3๐ฅ + ๐ถ6
Where the constants ๐ต๐ are given above.
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3.5.4 key solution to HW 5
105
106
107
108
109
110
3.6 HW 6
3.6.1 Problem 1
Consider the equation ๐ฅ๐ฆโฒโฒ + (๐ โ ๐ฅ) ๐ฆโฒ โ ๐๐ฆ = 0. Identify a regular singular point and findtwo series solutions around this point. Test the solutions for convergence.
Solution
111
Writing the ODE as
๐ฆโฒโฒ + ๐ด (๐ฅ) ๐ฆโฒ + ๐ต (๐ฅ) ๐ฆ = 0
Where
๐ด (๐ฅ) =(๐ โ ๐ฅ)๐ฅ
๐ต (๐ฅ) =โ๐๐ฅ
The above shows that ๐ฅ0 = 0 is a singularity point for both ๐ด (๐ฅ) and ๐ต (๐ฅ). Examining ๐ด (๐ฅ)and ๐ต (๐ฅ) to determine what type of singular point it is
lim๐ฅโ๐ฅ0
(๐ฅ โ ๐ฅ0) ๐ด (๐ฅ) = lim๐ฅโ0
๐ฅ(๐ โ ๐ฅ)๐ฅ
= lim๐ฅโ0
(๐ โ ๐ฅ) = ๐
Because the limit exists, then ๐ฅ0 = 0 is regular singular point for ๐ด (๐ฅ) .
lim๐ฅโ๐ฅ0
(๐ฅ โ ๐ฅ0)2 ๐ต (๐ฅ) = lim
๐ฅโ0๐ฅ2 ๏ฟฝ
โ๐๐ฅ๏ฟฝ = lim
๐ฅโ0(โ๐๐ฅ) = 0
Because the limit exists, then ๐ฅ0 = 0 is also regular singular point for ๐ต (๐ฅ).
Therefore ๐ฅ0 = 0 is a regular singular point for the ODE.
Assuming the solution is Frobenius series gives
๐ฆ (๐ฅ) = ๐ฅ๐โ๏ฟฝ๐=0
๐ถ๐ (๐ฅ โ ๐ฅ0)๐ ๐ถ0 โ 0
= ๐ฅ๐โ๏ฟฝ๐=0
๐ถ๐๐ฅ๐
=โ๏ฟฝ๐=0
๐ถ๐๐ฅ๐+๐
Therefore
๐ฆโฒ =โ๏ฟฝ๐=0
(๐ + ๐) ๐ถ๐๐ฅ๐+๐โ1
๐ฆโฒโฒ =โ๏ฟฝ๐=0
(๐ + ๐) (๐ + ๐ โ 1) ๐ถ๐๐ฅ๐+๐โ2
Substituting the above in the original ODE ๐ฅ๐ฆโฒโฒ + (๐ โ ๐ฅ) ๐ฆโฒ โ ๐๐ฆ = 0 gives
๐ฅโ๏ฟฝ๐=0
(๐ + ๐) (๐ + ๐ โ 1) ๐ถ๐๐ฅ๐+๐โ2 + (๐ โ ๐ฅ)โ๏ฟฝ๐=0
(๐ + ๐) ๐ถ๐๐ฅ๐+๐โ1 โ ๐โ๏ฟฝ๐=0
๐ถ๐๐ฅ๐+๐ = 0
โ๏ฟฝ๐=0
(๐ + ๐) (๐ + ๐ โ 1) ๐๐๐ฅ๐+๐โ1 + ๐โ๏ฟฝ๐=0
(๐ + ๐) ๐ถ๐๐ฅ๐+๐โ1 โ ๐ฅโ๏ฟฝ๐=0
(๐ + ๐) ๐ถ๐๐ฅ๐+๐โ1 โโ๏ฟฝ๐=0
๐๐ถ๐๐ฅ๐+๐ = 0
โ๏ฟฝ๐=0
(๐ + ๐) (๐ + ๐ โ 1) ๐ถ๐๐ฅ๐+๐โ1 +โ๏ฟฝ๐=0
๐ (๐ + ๐) ๐ถ๐๐ฅ๐+๐โ1 โโ๏ฟฝ๐=0
(๐ + ๐) ๐ถ๐๐ฅ๐+๐ โโ๏ฟฝ๐=0
๐๐ถ๐๐ฅ๐+๐ = 0
โ๏ฟฝ๐=0
((๐ + ๐) (๐ + ๐ โ 1) + ๐ (๐ + ๐)) ๐ถ๐๐ฅ๐+๐โ1 โโ๏ฟฝ๐=0
((๐ + ๐) + ๐) ๐ถ๐๐ฅ๐+๐ = 0
Since all powers of ๐ฅ have to be the same, adjusting indices and exponents gives (wherein the second sum above, the outside index ๐ is increased by 1 and ๐ inside the sum isdecreased by 1)
โ๏ฟฝ๐=0
((๐ + ๐) (๐ + ๐ โ 1) + ๐ (๐ + ๐)) ๐ถ๐๐ฅ๐+๐โ1 โโ๏ฟฝ๐=1
((๐ โ 1 + ๐) + ๐) ๐ถ๐โ1๐ฅ๐+๐โ1 = 0 (1)
Setting ๐ = 0 gives the indicial equation, which only comes from the first sum above as thesecond sum starts from ๐ = 1.
((๐) (๐ โ 1) + ๐๐) ๐ถ0 = 0
Since ๐ถ0 โ 0 then(๐) (๐ โ 1) + ๐๐ = 0
๐2 โ ๐ + ๐๐ = 0๐ (๐ + ๐ โ 1) = 0
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The roots are
๐1 = 1 โ ๐๐2 = 0
Assuming that ๐2 โ ๐1 is not an integer, in other words, assuming 1 โ ๐ is not an integer(problem did not say), then In this case, two linearly independent solutions can be con-structed directly. The first is associated with ๐1 = 1 โ ๐ and the second is associated with๐2 = 0. These solutions are
๐ฆ1 (๐ฅ) =โ๏ฟฝ๐=0
๐ถ๐๐ฅ๐+1โ๐ ๐ถ0 โ 0
๐ฆ2 (๐ฅ) =โ๏ฟฝ๐=0
๐ท๐๐ฅ๐ ๐ท0 โ 0
The coe๏ฟฝcients are not the same in each solution. For the first one ๐ถ๐ is used and for thesecond ๐ท๐ is used.
The solution ๐ฆ1 (๐ฅ) associated with ๐1 = 1 โ ๐ is now found. From (1), and replacing ๐ by1 โ ๐ givesโ๏ฟฝ๐=0
((๐ + 1 โ ๐) (๐ + 1 โ ๐ โ 1) + ๐ (๐ + 1 โ ๐)) ๐ถ๐๐ฅ๐+1โ๐โ1 โโ๏ฟฝ๐=1
((๐ โ 1 + 1 โ ๐) + ๐) ๐ถ๐โ1๐ฅ๐+1โ๐โ1 = 0
โ๏ฟฝ๐=0
((๐ + 1 โ ๐) (๐ โ ๐) + ๐ (๐ + 1 โ ๐)) ๐ถ๐๐ฅ๐โ๐ โโ๏ฟฝ๐=1
((๐ โ ๐) + ๐) ๐ถ๐โ1๐ฅ๐โ๐ = 0
โ๏ฟฝ๐=0
๐ (๐ โ ๐ + 1) ๐ถ๐๐ฅ๐โ๐ โโ๏ฟฝ๐=1
((๐ โ ๐) + ๐) ๐ถ๐โ1๐ฅ๐โ๐ = 0
For ๐ > 0 the above gives the recursive relation (๐ = 0 is not used, since it was used to find๐). For ๐ > 0 the last equation above gives
๐ (๐ โ ๐ + 1) ๐ถ๐ โ ((๐ โ ๐) + ๐) ๐ถ๐โ1 = 0
๐ถ๐ =((๐ โ ๐) + ๐)๐ (๐ โ ๐ + 1)
๐ถ๐โ1
Few terms are generated to see the pattern. For ๐ = 1
๐ถ1 =(1 โ ๐ + ๐)1 (1 โ ๐ + 1)
๐ถ0 =(1 โ ๐ + ๐)(2 โ ๐)
๐ถ0
For ๐ = 2
๐ถ2 =(2 โ ๐ + ๐)2 (2 โ ๐ + 1)
๐ถ1
=(2 โ ๐ + ๐)2 (3 โ ๐)
(1 โ ๐ + ๐)(2 โ ๐)
๐ถ0
For ๐ = 3
๐ถ3 =(3 โ ๐ + ๐)3 (3 โ ๐ + 1)
๐ถ2
=(3 โ ๐ + ๐)3 (4 โ ๐)
(2 โ ๐ + ๐)2 (3 โ ๐)
(1 โ ๐ + ๐)(2 โ ๐)
๐ถ0
And so on. The pattern for general term is
๐ถ๐ =((๐ โ ๐) + ๐)๐ (๐ โ ๐ + 1)
โ โ โ (3 โ ๐ + ๐)3 (3 โ ๐ + 1)
(2 โ ๐ + ๐)2 (2 โ ๐ + 1)
(1 โ ๐ + ๐)1 (1 โ ๐ + 1)
๐ถ0
=๐๏ฟฝ๐=1
((๐ โ ๐) + ๐)๐ (๐ โ ๐ + 1)
Therefore the solution associated with ๐1 = 1 โ ๐ is
๐ฆ1 (๐ฅ) =โ๏ฟฝ๐=0
๐ถ๐๐ฅ๐+๐
=โ๏ฟฝ๐=0
๐ถ๐๐ฅ๐+1โ๐
= ๐ถ0๐ฅ1โ๐ + ๐ถ1๐ฅ2โ๐ + ๐ถ2๐ฅ3โ๐ +โฏ
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Using results found above, and looking at few terms gives the first solution as
๐ฆ1 (๐ฅ) = ๐ถ0๐ฅ1โ๐ ๏ฟฝ1 +(1 โ ๐ + ๐)(2 โ ๐)
๐ฅ +12(2 โ ๐ + ๐)(3 โ ๐)
(1 โ ๐ + ๐)(2 โ ๐)
๐ฅ2 +16(3 โ ๐ + ๐)(4 โ ๐)
(2 โ ๐ + ๐)(3 โ ๐)
(1 โ ๐ + ๐)(2 โ ๐)
๐ฅ3 +โฏ๏ฟฝ
The second solution associated with ๐2 = 0 is now found. As above, using (1) but with ๐ท๐instead of ๐ถ๐ for coe๏ฟฝcients and replacing ๐ by zero gives
โ๏ฟฝ๐=0
(๐ (๐ โ 1) + ๐๐)๐ท๐๐ฅ๐โ1 โโ๏ฟฝ๐=1
((๐ โ 1) + ๐)๐ท๐โ1๐ฅ๐โ1 = 0
For ๐ > 0 the above gives the recursive relation for the second solution
(๐ (๐ โ 1) + ๐๐)๐ท๐ โ ((๐ โ 1) + ๐)๐ท๐โ1 = 0
๐ท๐ =๐ โ 1 + ๐
๐ (๐ โ 1) + ๐๐๐ท๐โ1
=๐ โ 1 + ๐๐๐ โ ๐ + ๐2
๐ท๐โ1
Few terms are now generated to see the pattern. For ๐ = 1
๐ท1 =๐๐๐ท0
For ๐ = 2
๐ท2 =1 + ๐
2๐ โ 2 + 4๐ท1
=1 + ๐2 (๐ + 1)
๐๐๐ท0
For ๐ = 3
๐ท3 =3 โ 1 + ๐3๐ โ 3 + 9
๐ท2
=2 + ๐3 (๐ + 2)
1 + ๐2 (๐ + 1)
๐๐๐ท0
And so on. Hence the solution ๐ฆ2 (๐ฅ) is
๐ฆ2 (๐ฅ) =โ๏ฟฝ๐=0
๐ท๐๐ฅ๐
= ๐ท0 + ๐ท1๐ฅ + ๐ท2๐ฅ2 +โฏ
Using result found above gives the second solution as
๐ฆ2 (๐ฅ) = ๐ท0 ๏ฟฝ1 +๐๐๐ฅ +
12(1 + ๐) ๐๐ (๐ + 1)
๐ฅ2 +16๐ (1 + ๐) (2 + ๐)๐ (๐ + 2) (๐ + 1)
๐ฅ3 +โฏ๏ฟฝ
The final solution is therefore the sum of the two solutions
๐ฆ (๐ฅ) = ๐ถ0๐ฅ1โ๐ ๏ฟฝ1 +(1 โ ๐ + ๐)(2 โ ๐)
๐ฅ +12(2 โ ๐ + ๐)(3 โ ๐)
(1 โ ๐ + ๐)(2 โ ๐)
๐ฅ2 +16(3 โ ๐ + ๐)(4 โ ๐)
(2 โ ๐ + ๐)(3 โ ๐)
(1 โ ๐ + ๐)(2 โ ๐)
๐ฅ3 +โฏ๏ฟฝ
(2)
+ ๐ท0 ๏ฟฝ1 +๐๐๐ฅ +
12(1 + ๐) ๐๐ (๐ + 1)
๐ฅ2 +16๐ (1 + ๐) (2 + ๐)๐ (๐ + 2) (๐ + 1)
๐ฅ3 +โฏ๏ฟฝ
Where ๐ถ0, ๐ท0 are the two constant of integration.
Testing for convergence. For ๐ฆ1 (๐ฅ) solution, the general term from above was
๐ถ๐๐ฅ๐ =((๐ โ ๐) + ๐)๐ (๐ โ ๐ + 1)
๐ถ๐โ1๐ฅ๐
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Hence by ratio test
๐ฟ = lim๐โโ
๏ฟฝ๐ถ๐๐ฅ๐
๐ถ๐โ1๐ฅ๐โ1๏ฟฝ
= lim๐โโ
๏ฟฝ๏ฟฝ
((๐โ๐)+๐)๐(๐โ๐+1)๐ถ๐โ1๐ฅ
๐
๐ถ๐โ1๐ฅ๐โ1๏ฟฝ๏ฟฝ
= lim๐โโ
๏ฟฝ((๐ โ ๐) + ๐) ๐ฅ(๐ (๐ โ ๐ + 1))
๏ฟฝ
= |๐ฅ| lim๐โโ
๏ฟฝ๐ โ ๐ + ๐๐2 โ ๐๐ + ๐
๏ฟฝ
= |๐ฅ| lim๐โโ
๏ฟฝ๏ฟฝ
1๐ โ
๐๐2 +
๐๐2
1 โ ๐๐ +
1๐
๏ฟฝ๏ฟฝ
= |๐ฅ| ๏ฟฝ01๏ฟฝ
= 0
Therefore the series ๐ฆ1 (๐ฅ) converges for all ๐ฅ.
Testing for convergence. For ๐ฆ2 (๐ฅ) solution, the general term is
๐ท๐๐ฅ๐ =๐ โ 1 + ๐๐๐ โ ๐ + ๐2
๐ท๐โ1๐ฅ๐
Hence by ratio test
๐ฟ = lim๐โโ
๏ฟฝ๐ท๐๐ฅ๐
๐ท๐โ1๐ฅ๐โ1๏ฟฝ
= lim๐โโ
๏ฟฝ๏ฟฝ
๐โ1+๐๐๐โ๐+๐2๐ท๐โ1๐ฅ๐
๐ท๐โ1๐ฅ๐โ1๏ฟฝ๏ฟฝ
= lim๐โโ
๏ฟฝ๐ โ 1 + ๐๐๐ โ ๐ + ๐2
๐ฅ๏ฟฝ
= |๐ฅ| lim๐โโ
๏ฟฝ๐ โ 1 + ๐๐๐ โ ๐ + ๐2
๏ฟฝ
= |๐ฅ| lim๐โโ
๏ฟฝ๏ฟฝ
1๐ โ
1๐2 +
๐๐2
๐๐ โ
1๐ + 1
๏ฟฝ๏ฟฝ
= |๐ฅ| ๏ฟฝ01๏ฟฝ
= 0
Therefore the series ๐ฆ2 (๐ฅ) also converges for all ๐ฅ. This means the solution ๐ฆ (๐ฅ) = ๐ฆ1 (๐ฅ) +๐ฆ2 (๐ฅ) found in (2) above also converges for all ๐ฅ.
3.6.2 Problem 2
The Sturm Liouville equation can be expressed as
๐ฟ [๐ข (๐ฅ)] = ๐๐ (๐ฅ) ๐ข (๐ฅ)
Where ๐ฟ is given as in class. Show ๐ฟ is Hermitian on the domain ๐ โค ๐ฅ โค ๐ with boundaryconditions ๐ข (๐) = ๐ข (๐) = 0. Find the orthogonality condition.
Solution
๐ฟ = โ ๏ฟฝ๐๐2
๐๐ฅ2+ ๐โฒ
๐๐๐ฅโ ๐๏ฟฝ
The operator ๐ฟ is Hermitian if
๏ฟฝ๐
๐๏ฟฝ๏ฟฝ๐ฟ [๐ข] ๐๐ฅ = ๏ฟฝ
๐
๐๏ฟฝ๏ฟฝ๐ฟ [๐ฃ] ๐๐ฅ
Where in the above ๐ข, ๐ฃ are any two functions defined over the domain that satisfy the
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boundary conditions given. Starting from the left integral to show it will result in the right
integral. Replacing ๐ฟ [๐ข] by โ ๏ฟฝ๐ ๐2
๐๐ฅ2 + ๐โฒ ๐๐๐ฅ โ ๐๏ฟฝ ๐ข in the LHS of the above gives
โ๏ฟฝ๐
๐๏ฟฝ๏ฟฝ ๏ฟฝ๐
๐2
๐๐ฅ2+ ๐โฒ
๐๐๐ฅโ ๐๏ฟฝ ๐ข ๐๐ฅ = โ๏ฟฝ
๐
๐๏ฟฝ๏ฟฝ ๏ฟฝ๐
๐2๐ข๐๐ฅ2
+ ๐โฒ๐๐ข๐๐ฅ
โ ๐๐ข๏ฟฝ ๐๐ฅ
= โ๏ฟฝ๐
๐๏ฟฝ๏ฟฝ๐๐2๐ข๐๐ฅ2
+ ๏ฟฝ๏ฟฝ๐โฒ๐๐ข๐๐ฅ
โ ๐๏ฟฝ๏ฟฝ๐ข ๐๐ฅ
= โ
๐ผ1๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ
๐
๐๐๏ฟฝ๏ฟฝ๐2๐ข๐๐ฅ2
๐๐ฅ โ๏ฟฝ๐
๐๏ฟฝ๏ฟฝ๐โฒ
๐๐ข๐๐ฅ๐๐ฅ +๏ฟฝ
๐
๐๐๏ฟฝ๏ฟฝ๐ข ๐๐ฅ (1)
Looking at the first integral above, which is ๐ผ1 = โซ๐
๐๏ฟฝ๐๏ฟฝ๏ฟฝ๏ฟฝ ๏ฟฝ๐
2๐ข๐๐ฅ2๏ฟฝ ๐๐ฅ. The idea is to integrate
this twice to move the second derivative from ๐ข to ๏ฟฝ๏ฟฝ. Applying โซ๐ด๐๐ต = ๐ด๐ตโโซ๐ต๐๐ด, where
๐ด โก ๐๏ฟฝ๏ฟฝ
๐๐ต โก๐2๐ข๐๐ฅ2
Hence
๐๐ด = ๐๐๏ฟฝ๏ฟฝ๐๐ฅ+ ๐โฒ๏ฟฝ๏ฟฝ
๐ต =๐๐ข๐๐ฅ
Therefore the integral ๐ผ1 in (1) becomes
๐ผ1 = ๏ฟฝ๐
๐๐๏ฟฝ๏ฟฝ
๐2
๐๐ฅ2๐ข
= ๏ฟฝ๐๏ฟฝ๏ฟฝ๐๐ข๐๐ฅ ๏ฟฝ
๐
๐โ๏ฟฝ
๐
๐
๐๐ข๐๐ฅ ๏ฟฝ
๐๐๏ฟฝ๏ฟฝ๐๐ฅ+ ๐โฒ๏ฟฝ๏ฟฝ๏ฟฝ ๐๐ฅ
But ๏ฟฝ๏ฟฝ (๐) = 0 and ๏ฟฝ๏ฟฝ (๐) = 0, hence the boundary terms above vanish and simplifies to
๐ผ1 = โ๏ฟฝ๐
๐๐๐๐ข๐๐ฅ๐๏ฟฝ๏ฟฝ๐๐ฅ+ ๐โฒ๏ฟฝ๏ฟฝ
๐๐ข๐๐ฅ๐๐ฅ
= โ๏ฟฝ๐
๐๐๐๐ข๐๐ฅ๐๏ฟฝ๏ฟฝ๐๐ฅ๐๐ฅ โ๏ฟฝ
๐
๐๐โฒ๏ฟฝ๏ฟฝ
๐๐ข๐๐ฅ๐๐ฅ (2)
Before integrating by parts a second time, putting the result of ๐ผ1 back into (1) first simplifiesthe result. Substituting (2) into (1) gives
๏ฟฝ๐
๐๏ฟฝ๏ฟฝ๐ฟ [๐ข] ๐๐ฅ = โ๐ผ1 โ๏ฟฝ
๐
๐๏ฟฝ๏ฟฝ๐โฒ
๐๐ข๐๐ฅ๐๐ฅ +๏ฟฝ
๐
๐๐๏ฟฝ๏ฟฝ๐ข ๐๐ฅ
= โ
๐ผ1๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ๏ฟฝ
๐
๐๐๐๐ข๐๐ฅ๐๏ฟฝ๏ฟฝ๐๐ฅ๐๐ฅ โ๏ฟฝ
๐
๐๐โฒ๏ฟฝ๏ฟฝ
๐๐ข๐๐ฅ๐๐ฅ๏ฟฝ โ๏ฟฝ
๐
๐๏ฟฝ๏ฟฝ๐โฒ
๐๐ข๐๐ฅ๐๐ฅ +๏ฟฝ
๐
๐๐๏ฟฝ๏ฟฝ๐ข ๐๐ฅ
= ๏ฟฝ๐
๐๐๐๐ข๐๐ฅ๐๏ฟฝ๏ฟฝ๐๐ฅ๐๐ฅ +๏ฟฝ
๐
๐๐โฒ๏ฟฝ๏ฟฝ
๐๐ข๐๐ฅ๐๐ฅ โ๏ฟฝ
๐
๐๏ฟฝ๏ฟฝ๐โฒ
๐๐ข๐๐ฅ๐๐ฅ +๏ฟฝ
๐
๐๐๏ฟฝ๏ฟฝ๐ข ๐๐ฅ
The second and third terms above cancel and the result becomes
๏ฟฝ๐
๐๏ฟฝ๏ฟฝ๐ฟ [๐ข] ๐๐ฅ =
๐ผ2๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ
๐
๐๐๐๐ข๐๐ฅ๐๏ฟฝ๏ฟฝ๐๐ฅ๐๐ฅ +๏ฟฝ
๐
๐๐๏ฟฝ๏ฟฝ๐ข ๐๐ฅ (3)
Now integration by parts is applied on the first integral above. Let ๐ผ2 = โซ๐๐
๐๐ข๐๐ฅ๏ฟฝ๐๐๏ฟฝ๏ฟฝ๐๐ฅ๏ฟฝ ๐๐ฅ.
Applying โซ๐ด๐๐ต = ๐ด๐ต โ โซ๐ต๐๐ด, where
๐ด โก ๐๐๏ฟฝ๏ฟฝ๐๐ฅ
๐๐ต โก๐๐ข๐๐ฅ
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Hence
๐๐ด = ๐๐2๏ฟฝ๏ฟฝ๐๐ฅ2
+ ๐โฒ๐๏ฟฝ๏ฟฝ๐๐ฅ
๐ต = ๐ข
Therefore the integral ๐ผ2 becomes
๐ผ2 = ๏ฟฝ๐๐๏ฟฝ๏ฟฝ๐๐ฅ๐ข๏ฟฝ๐
๐โ๏ฟฝ
๐
๐๐ข ๏ฟฝ๐
๐2๏ฟฝ๏ฟฝ๐๐ฅ2
+ ๐โฒ๐๏ฟฝ๏ฟฝ๐๐ฅ๏ฟฝ
๐๐ฅ
But ๐ข (๐) = 0, ๐ข (๐) = 0, hence the boundary term vanishes and the above simplifies to
๐ผ2 = โ๏ฟฝ๐
๐๐ข ๏ฟฝ๐
๐2๏ฟฝ๏ฟฝ๐๐ฅ2
+ ๐โฒ๐๏ฟฝ๏ฟฝ๐๐ฅ๏ฟฝ
๐๐ฅ
Substituting the above back into (3) gives
๏ฟฝ๐
๐๏ฟฝ๏ฟฝ๐ฟ [๐ข] ๐๐ฅ = โ๏ฟฝ
๐
๐๐ข ๏ฟฝ๐
๐2๏ฟฝ๏ฟฝ๐๐ฅ2
+ ๐โฒ๐๏ฟฝ๏ฟฝ๐๐ฅ๏ฟฝ
๐๐ฅ +๏ฟฝ๐
๐๐๏ฟฝ๏ฟฝ๐ข ๐๐ฅ
= โ๏ฟฝ๐
๐๐ข ๏ฟฝ๐
๐2๏ฟฝ๏ฟฝ๐๐ฅ2
+ ๐โฒ๐๏ฟฝ๏ฟฝ๐๐ฅโ ๐๏ฟฝ๏ฟฝ๏ฟฝ ๐๐ฅ
But โ ๏ฟฝ๐๐2๏ฟฝ๏ฟฝ๐๐ฅ2 + ๐
โฒ ๐๏ฟฝ๏ฟฝ๐๐ฅ โ ๐๏ฟฝ๏ฟฝ๏ฟฝ = ๐ฟ [๏ฟฝ๏ฟฝ] by definition, and the above becomes
๏ฟฝ๐
๐๏ฟฝ๏ฟฝ๐ฟ [๐ข] ๐๐ฅ = ๏ฟฝ
๐
๐๐ข๐ฟ [๏ฟฝ๏ฟฝ] ๐๐ฅ
But โซ๐
๐๐ข๐ฟ [๏ฟฝ๏ฟฝ] ๐๐ฅ = โซ
๐
๐๏ฟฝ๏ฟฝ๐ฟ [๐ฃ] ๐๐ฅ, and the above becomes
๏ฟฝ๐
๐๏ฟฝ๏ฟฝ๐ฟ [๐ข] ๐๐ฅ = ๏ฟฝ
๐
๐๏ฟฝ๏ฟฝ (๐ฟ [๐ฃ]) ๐๐ฅ
Therefore ๐ฟ is Hermitian.
3.6.3 Problem 3
1. For the equation ๐ฆโฒโฒ + 1โ๐ผ2
4๐ฅ2 ๐ฆ = 0 show that two solutions are ๐ฆ1 (๐ฅ) = ๐0๐ฅ1+๐ผ2 and
๐ฆ2 (๐ฅ) = ๐0๐ฅ1โ๐ผ2
2. For ๐ผ = 0, the two solutions are not independent. Find a second solution ๐ฆ20 bysolving ๐โฒ = 0 (๐ is the Wronskian).
3. Show that the second solution found in (2) is a limiting case of the two solutionsfrom part (1). That is
๐ฆ20 = lim๐ผโ0
๐ฆ1 โ ๐ฆ2๐ผ
Solution
Part 1
The point ๐ฅ0 = 0 is a regular singular point. This is shown as follows.
lim๐ฅโ๐ฅ0
(๐ฅ โ ๐ฅ0)2 1 โ ๐ผ2
4๐ฅ2= lim๐ฅโ0
๐ฅ21 โ ๐ผ2
4๐ฅ2
= lim๐ฅโ0
1 โ ๐ผ2
4
=1 โ ๐ผ2
4Since the limit exist, then ๐ฅ0 = 0 is a regular singular point. Assuming the solution is aFrobenius series given by
๐ฆ (๐ฅ) =โ๏ฟฝ๐=0
๐๐๐ฅ๐+๐ ๐0 โ 0
117
Therefore
๐ฆโฒ (๐ฅ) =โ๏ฟฝ๐=0
(๐ + ๐) ๐๐๐ฅ๐+๐โ1
๐ฆโฒโฒ (๐ฅ) =โ๏ฟฝ๐=0
(๐ + ๐) (๐ + ๐ โ 1) ๐๐๐ฅ๐+๐โ2
Substituting the above 2 expressions back into the original ODE gives
4๐ฅ2 ๏ฟฝโ๏ฟฝ๐=0
(๐ + ๐) (๐ + ๐ โ 1) ๐๐๐ฅ๐+๐โ2๏ฟฝ + ๏ฟฝ1 โ ๐ผ2๏ฟฝ ๏ฟฝโ๏ฟฝ๐=0
๐๐๐ฅ๐+๐๏ฟฝ = 0
โ๏ฟฝ๐=0
4 (๐ + ๐) (๐ + ๐ โ 1) ๐๐๐ฅ๐+๐ + ๏ฟฝ1 โ ๐ผ2๏ฟฝ ๏ฟฝโ๏ฟฝ๐=0
๐๐๐ฅ๐+๐๏ฟฝ = 0 (1)
Looking at ๐ = 0 first, in order to obtain the indicial equation gives
4 (๐) (๐ โ 1) ๐0 + ๏ฟฝ1 โ ๐ผ2๏ฟฝ ๐0 = 0
๐0 ๏ฟฝ4๐2 โ 4๐ + ๏ฟฝ1 โ ๐ผ2๏ฟฝ๏ฟฝ = 0
But ๐0 โ 0, therefore
๐2 โ ๐ +๏ฟฝ1 โ ๐ผ2๏ฟฝ
4= 0
The roots are ๐ = โ๐2๐ ยฑ
12๐โ๐
2 โ 4๐๐, but ๐ = 1, ๐ = โ1, ๐ =๏ฟฝ1โ๐ผ2๏ฟฝ
4 , hence the roots are
๐ =12ยฑ12๏ฟฝ
1 โ ๏ฟฝ1 โ ๐ผ2๏ฟฝ
=12ยฑ12โ๐ผ2
=12ยฑ12๐ผ
Hence ๐1 =12(1 + ๐ผ) and ๐2 =
12(1 โ ๐ผ). Each one of these roots gives a solution. The
di๏ฟฝerence is
๐2 โ ๐1 =12(1 + ๐ผ) โ
12(1 โ ๐ผ)
= ๐ผ
Therefore, to use the same solution form ๐ฆ1 (๐ฅ) = โโ๐=0 ๐๐๐ฅ
๐+๐1 and ๐ฆ2 (๐ฅ) = โโ๐=0 ๐๐๐ฅ
๐+๐2 foreach, it is assumed that ๐ผ is not an integer. In this case, the recursive relation for ๐ฆ1 (๐ฅ) is
found from (1) by using ๐ = 12(1 + ๐ผ) which results in
โ๏ฟฝ๐=0
4 ๏ฟฝ๐ +12(1 + ๐ผ)๏ฟฝ ๏ฟฝ๐ +
12(1 + ๐ผ) โ 1๏ฟฝ ๐๐๐ฅ
๐+ 12 (1+๐ผ) + ๏ฟฝ1 โ ๐ผ2๏ฟฝ ๏ฟฝ
โ๏ฟฝ๐=0
๐๐๐ฅ๐+ 1
2 (1+๐ผ)๏ฟฝ = 0
For ๐ > 0 the above becomes
4 ๏ฟฝ๐ +12(1 + ๐ผ)๏ฟฝ ๏ฟฝ๐ +
12(1 + ๐ผ) โ 1๏ฟฝ ๐๐ + ๏ฟฝ1 โ ๐ผ2๏ฟฝ ๐๐ = 0
๏ฟฝ4 ๏ฟฝ๐ +12(1 + ๐ผ)๏ฟฝ ๏ฟฝ๐ +
12(1 + ๐ผ) โ 1๏ฟฝ + ๏ฟฝ1 โ ๐ผ2๏ฟฝ๏ฟฝ ๐๐ = 0
4๐ (๐ + ๐ผ) ๐๐ = 0
The above can be true for all ๐ > 0 only when ๐๐ = 0 for ๐ > 0. Therefore the solution isonly the term with ๐0
๐ฆ1 (๐ฅ) =โ๏ฟฝ๐=0
๐๐๐ฅ๐+๐1 = ๐0๐ฅ๐1 = ๐0๐ฅ12 (1+๐ผ)
To find the second solution ๐ฆ2 (๐ฅ), the above is repeated but with
๐ฆ2 (๐ฅ) =โ๏ฟฝ๐=0
๐๐๐ฅ๐+๐2
Where the constants are not the same and by replacing ๐ in (1) by ๐2 =12(1 โ ๐ผ). This
results inโ๏ฟฝ๐=0
4 ๏ฟฝ๐ +12(1 โ ๐ผ)๏ฟฝ ๏ฟฝ๐ +
12(1 โ ๐ผ) โ 1๏ฟฝ ๐๐๐ฅ
๐+ 12 (1โ๐ผ) + ๏ฟฝ1 โ ๐ผ2๏ฟฝ ๏ฟฝ
โ๏ฟฝ๐=0
๐๐๐ฅ๐+ 1
2 (1โ๐ผ)๏ฟฝ = 0
118
For ๐ > 0
๏ฟฝ4 ๏ฟฝ๐ +12(1 โ ๐ผ)๏ฟฝ ๏ฟฝ๐ +
12(1 โ ๐ผ) โ 1๏ฟฝ + ๏ฟฝ1 โ ๐ผ2๏ฟฝ๏ฟฝ ๐๐ = 0
4๐ (๐ โ ๐ผ) ๐๐ = 0
The above is true for all ๐ > 0 only when ๐๐ = 0 for ๐ > 0. Therefore the solution is just theterm with ๐0
๐ฆ2 (๐ฅ) =โ๏ฟฝ๐=0
๐๐๐ฅ๐+๐2 = ๐0๐ฅ๐2 = ๐0๐ฅ12 (1โ๐ผ)
Therefore the two solutions are
๐ฆ1 (๐ฅ) = ๐0๐ฅ12 (1+๐ผ)
๐ฆ2 (๐ฅ) = ๐0๐ฅ12 (1โ๐ผ)
Part 2
When ๐ผ = 0 then the ODE becomes
4๐ฅ2๐ฆโฒโฒ + ๐ฆ = 0
And the two solutions found in part (1) simplify to
๐ฆ1 (๐ฅ) = ๐0โ๐ฅ๐ฆ2 (๐ฅ) = ๐0โ๐ฅ
Therefore the two solutions are not linearly independent. Let ๐ฆ20 (๐ฅ) be the second solution.The Wronskian is
๐(๐ฅ) = ๏ฟฝ๐ฆ1 ๐ฆ20๐ฆโฒ1 ๐ฆโฒ20
๏ฟฝ = ๐ฆ1๐ฆโฒ20 โ ๐ฆ20๐ฆโฒ1 (1)
Using Abelโs theorem which says that for ODE of form ๐ฆโฒโฒ+๐ (๐ฅ) ๐ฆโฒ+๐ (๐ฅ) ๐ฆ = 0, theWronskianis ๐(๐ฅ) = ๐ถ๐โโซ๐(๐ฅ)๐๐ฅ. Applying this to the given ODE above and since ๐ (๐ฅ) = 0 then theabove becomes
๐(๐ฅ) = ๐ถ
Where ๐ถ is constant. For ๐ฆ20 to be linearly independent from ๐ฆ1 ๐(๐ฅ) โ 0. Using ๐(๐ฅ) = ๐ถin (1) results in the following equation (here it is also assumed that ๐ฆ1 โ 0, or ๐ฅ โ 0, becausethe equation is divided by ๐ฆ1)
๐ฆ1๐ฆโฒ20 โ ๐ฆ20๐ฆโฒ1 = ๐ถ
๐ฆโฒ20 โ ๐ฆ20๐ฆโฒ1๐ฆ1=๐ถ๐ฆ1
Since ๐ฆ1 = โ๐ฅ and ๐ฆโฒ1 =121
โ๐ฅthe above simplifies to
๐ฆโฒ20 โ ๐ฆ20
121
โ๐ฅ
โ๐ฅ=
๐ถ
โ๐ฅ
๐ฆโฒ20 โ ๐ฆ2012๐ฅ
=๐ถ
โ๐ฅ(2)
But the above is linear first order ODE of the form ๐โฒ + ๐๐ = ๐, therefore the standardintegrating factor to use is ๐ผ = ๐โซ๐(๐ฅ)๐๐ฅ which results in
๐ผ = ๐โซโ12๐ฅ ๐๐ฅ
= ๐โ12 โซ
1๐ฅ๐๐ฅ
= ๐โ12 ln ๐ฅ
=1
โ๐ฅ
119
Multiplying both sides of (2) by this integrating factor, makes the left side of (2) an exactdi๏ฟฝerential
๐๐๐ฅ ๏ฟฝ
๐ฆ201
โ๐ฅ๏ฟฝ =
๐ถ๐ฅ
Integrating both sides gives
๐ฆ201
โ๐ฅ= ๐ถ๏ฟฝ
1๐ฅ๐๐ฅ + ๐ถ1
๐ฆ201
โ๐ฅ= 2๐ถ ln ๐ฅ + ๐ถ1
๐ฆ20 = 2๐ถ ln ๐ฅโ๐ฅ + ๐ถ1โ๐ฅOr
๐ฆ20 = ๐ถ1 ln ๐ฅโ๐ฅ + ๐ถ2โ๐ฅ (3)
The above is the second solution. Therefore the final solution is
๐ฆ (๐ฅ) = ๐ถ0๐ฆ1 (๐ฅ) + ๐ถ3๐ฆ20 (๐ฅ)
Substituting ๐ฆ1 = โ๐ฅ and ๐ฆ20 found above and combining the common term โ๐ฅ andrenaming constants gives
๐ฆ (๐ฅ) = ๐ถ1โ๐ฅ + ๐ถ2 ln ๐ฅโ๐ฅAnother method to find the second solution
This method is called the reduction of order method. It does not require finding๐(๐ฅ) first.Let the second solution be
๐ฆ20 = ๐ = ๐ฃ (๐ฅ) ๐ฆ1 (๐ฅ) (4)
Where ๐ฃ (๐ฅ) is unknown function to be determined, and ๐ฆ1 (๐ฅ) = โ๐ฅ which is the first solutionthat is already known. Therefore
๐โฒ = ๐ฃโฒ๐ฆ1 + ๐ฃ๐ฆโฒ1๐โฒโฒ = ๐ฃโฒโฒ๐ฆ1 + ๐ฃโฒ๐ฆโฒ1 + ๐ฃโฒ๐ฆโฒ1 + ๐ฃ๐ฆโฒโฒ1
= ๐ฃโฒโฒ๐ฆ1 + 2๐ฃโฒ๐ฆโฒ1 + ๐ฃ๐ฆโฒโฒ1Since ๐ is a solution to the ODE 4๐ฅ2๐ฆโฒโฒ + ๐ฆ = 0, then substituting the above equations backinto the ODE 4๐ฅ2๐ฆโฒโฒ + ๐ฆ = 0 gives
4๐ฅ2 ๏ฟฝ๐ฃโฒโฒ๐ฆ1 + 2๐ฃโฒ๐ฆโฒ1 + ๐ฃ๐ฆโฒโฒ1 ๏ฟฝ + ๐ฃ๐ฆ1 = 0
๐ฃโฒโฒ ๏ฟฝ4๐ฅ2๐ฆ1๏ฟฝ + ๐ฃโฒ ๏ฟฝ8๐ฅ2๐ฆโฒ1๏ฟฝ + ๐ฃ
โโโโโโโโ
0
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ4๐ฅ2๐ฆโฒโฒ1 + ๐ฆ1
โโโโโโโโ = 0
But 4๐ฅ2๐ฆโฒโฒ1 + ๐ฆ1 = 0 because ๐ฆ1 is a solution. The above simplifies to
๐ฃโฒโฒ ๏ฟฝ4๐ฅ2๐ฆ1๏ฟฝ + ๐ฃโฒ ๏ฟฝ8๐ฅ2๐ฆโฒ1๏ฟฝ = 0
But ๐ฆ1 = ๐ฅ12 , hence ๐ฆโฒ1 =
12๐ฅ
โ12 and the above simplifies to
๐ฃโฒโฒ ๏ฟฝ4๐ฅ2๐ฅ12 ๏ฟฝ + ๐ฃโฒ ๏ฟฝ4๐ฅ2๐ฅ
โ12 ๏ฟฝ = 0
๐ฅ52๐ฃโฒโฒ + ๐ฃโฒ๐ฅ
32 = 0
๐ฅ๐ฃโฒโฒ + ๐ฃโฒ = 0
๐ฃโฒโฒ +1๐ฅ๐ฃโฒ = 0
This ODE is now easy to solve because the ๐ฃ (๐ฅ) term is missing. Let ๐ค = ๐ฃโฒ and the above
first order ODE ๐คโฒ+ 1๐ฅ๐ค = 0. This is linear in ๐ค. Hence using integrating factor ๐ผ = ๐โซ
1๐ฅ๐๐ง = ๐ฅ,
this ODE becomes๐๐ฅ(๐ค๐ฅ) = 0
๐ค๐ฅ = ๐ถ
๐ค =๐ถ๐ฅ
120
Where ๐ถ is constant of integration. Since ๐ฃโฒ = ๐ค, then ๐ฃโฒ = ๐ถ1๐ฅ . Now ๐ฃ (๐ฅ) is found by
integrating both sides
๐ฃ = ๐ถ1 ln ๐ฅ + ๐ถ2Therefore the second solution from (4) becomes
๐ฆ20 = ๐ถ1 ln ๐ฅ๐ฆ1 + ๐ถ2๐ฆ1= ๐ถ1โ๐ฅ ln ๐ฅ + ๐ถ2โ๐ฅ (5)
Comparing the above to (3), shows it is the same solution. Both methods can be used, butreduction of order method is a more common method and it does not require finding theWronskian first, although it is not hard to find by using Abelโs theorem.
Part 3
The solutions we found in part (1) are
๐ฆ1 (๐ฅ) = ๐ถ1๐ฅ12 (1+๐ผ)
๐ฆ2 (๐ฅ) = ๐ถ2๐ฅ12 (1โ๐ผ)
Therefore
lim๐ผโ0
๐ฆ1 โ ๐ฆ2๐ผ
= lim๐ผโ0
๐ถ1๐ฅ12 (1+๐ผ) โ ๐ถ2๐ฅ
12 (1โ๐ผ)
๐ผApplying LโHopitalโs
lim๐ผโ0
๐ฆ1 โ ๐ฆ2๐ผ
= lim๐ผโ0
๐ถ1๐๐๐ผ๏ฟฝ๐ฅ
12 (1+๐ผ)๏ฟฝ โ ๐ถ2
๐๐๐ผ๏ฟฝ๐ฅ
12 (1โ๐ผ)๏ฟฝ
1(1)
But๐๐๐ผ
๏ฟฝ๐ฅ12 (1+๐ผ)๏ฟฝ =
๐๐๐ผ๐12 (1+๐ผ) ln ๐ฅ
=๐๐๐ผ๐๏ฟฝ 12 ln ๐ฅ+๐ผ ln ๐ฅ๏ฟฝ
= ln ๐ฅ๐๏ฟฝ12 ln ๐ฅ+๐ผ ln ๐ฅ๏ฟฝ
And๐๐๐ผ
๏ฟฝ๐ฅ12 (1โ๐ผ)๏ฟฝ =
๐๐๐ผ๐12 (1โ๐ผ) ln ๐ฅ
=๐๐๐ผ๐๏ฟฝ 12 ln ๐ฅโ๐ผ ln ๐ฅ๏ฟฝ
= โ ln ๐ฅ๐๏ฟฝ12 ln ๐ฅโ๐ผ ln ๐ฅ๏ฟฝ
Therefore (1) becomes
lim๐ผโ0
๐ฆ1 โ ๐ฆ2๐ผ
= lim๐ผโ0
๐ถ1 ln ๐ฅ๐๏ฟฝ12 ln ๐ฅ+๐ผ ln ๐ฅ๏ฟฝ
+ ๐ถ2 ln ๐ฅ๐๏ฟฝ12 ln ๐ฅโ๐ผ ln ๐ฅ๏ฟฝ
= ln ๐ฅ ๏ฟฝlim๐ผโ0
๐ถ1๐๏ฟฝ 12 ln ๐ฅ+๐ผ ln ๐ฅ๏ฟฝ
+ ๐ถ2๐๏ฟฝ 12 ln ๐ฅโ๐ผ ln ๐ฅ๏ฟฝ
๏ฟฝ
= ln ๐ฅ ๏ฟฝ๐ถ1๐12 ln ๐ฅ + ๐ถ2๐
12 ln ๐ฅ๏ฟฝ
= ln ๐ฅ ๏ฟฝ๐ถ1โ๐ฅ + ๐ถ2โ๐ฅ๏ฟฝ= ๐ถโ๐ฅ ln ๐ฅ
The above is the same as (3) found in part (2). Hence
๐ฆ20 (๐ฅ) = lim๐ผโ0
๐ฆ1 โ ๐ฆ2๐ผ
Which is what the problem asked to show.
121
3.6.4 key solution to HW 6
122
123
124
125
126
3.7 HW 7
3.7.1 Problem 1
Figure 3.20: Problem statement
Solution
๐ฆโฒโฒ โ 2๐ฅ๐ฆโฒ + 2๐๐ฆ = 0 โ โ < ๐ฅ < โ
Part 1
๐ (๐ฅ, ๐ก) =โ๏ฟฝ๐=0
๐ป๐ (๐ฅ)๐ก๐
๐!Di๏ฟฝerentiating w.r.t ๐ฅ, and assuming term by term di๏ฟฝerentiation is allowed, gives
๐๐ (๐ฅ, ๐ก)๐๐ฅ
=โ๏ฟฝ๐=0
๐ปโฒ๐ (๐ฅ)
๐ก๐
๐!Using ๐ปโฒ
๐ (๐ฅ) = 2๐๐ป๐โ1 (๐ฅ) in the above results in
๐๐ (๐ฅ, ๐ก)๐๐ฅ
=โ๏ฟฝ๐=0
2๐๐ป๐โ1 (๐ฅ)๐ก๐
๐!But for ๐ = 0, the first term is zero, so the sum can start from 1 and give the same result
๐๐ (๐ฅ, ๐ก)๐๐ฅ
=โ๏ฟฝ๐=1
2๐๐ป๐โ1 (๐ฅ)๐ก๐
๐!Now, decreasing the summation index by 1 and increasing the ๐ inside the sum by 1 gives
๐๐ (๐ฅ, ๐ก)๐๐ฅ
=โ๏ฟฝ๐=0
2 (๐ + 1)๐ป๐ (๐ฅ)๐ก๐+1
(๐ + 1)!
=โ๏ฟฝ๐=0
2 (๐ + 1)๐ป๐ (๐ฅ)๐ก๐+1
(๐ + 1) ๐!
=โ๏ฟฝ๐=0
2๐ป๐ (๐ฅ)๐ก๐+1
๐!
=โ๏ฟฝ๐=0
2๐ก ๏ฟฝ๐ป๐ (๐ฅ)๐ก๐
๐!๏ฟฝ
= 2๐กโ๏ฟฝ๐=0
๐ป๐ (๐ฅ)๐ก๐
๐!
127
But โโ๐=0๐ป๐ (๐ฅ)
๐ก๐
๐! = ๐ (๐ฅ, ๐ก) and the above reduces to
๐๐ (๐ฅ, ๐ก)๐๐ฅ
= 2๐ก๐ (๐ฅ, ๐ก)
The problem says it is supposed to be a first order di๏ฟฝerential equation and not a first orderpartial di๏ฟฝerential equation. Therefore, by assuming ๐ฅ to be a fixed parameter instead ofanother independent variable, the above can now be written as
๐๐๐ฅ๐ (๐ฅ, ๐ก) โ 2๐ก๐ (๐ฅ, ๐ก) = 0
Part 2
From the solution found in part (1)๐๐๐ฅ๐ (๐ฅ, ๐ก)๐ (๐ฅ, ๐ก)
= 2๐ก
๐๐ (๐ฅ, ๐ก)๐ (๐ฅ, ๐ก)
= 2๐ก๐๐ฅ
Integrating both sides gives
๏ฟฝ๐๐ (๐ฅ, ๐ก)๐ (๐ฅ, ๐ก)
= ๏ฟฝ2๐ก๐๐ฅ
ln ๏ฟฝ๐ (๐ฅ, ๐ก)๏ฟฝ = 2๐ก๐ฅ + ๐ถ๐ (๐ฅ, ๐ก) = ๐2๐ก๐ฅ+๐ถ
๐ (๐ฅ, ๐ก) = ๐ถ1๐2๐ก๐ฅ
Where ๐ถ1 = ๐๐ถ a new constant. Let ๐ (0, ๐ก) = ๐0 then the above shows that ๐ถ1 = ๐0 and theabove can now be written as
๐ (๐ฅ, ๐ก) = ๐ (0, ๐ก) ๐2๐ก๐ฅ
Part 3
Using the given definition of ๐ (๐ฅ, ๐ก) = โโ๐=0๐ป๐ (๐ฅ)
๐ก๐
๐! and when ๐ฅ = 0 then
๐ (0, ๐ก) =โ๏ฟฝ๐=0
๐ป๐ (0)๐ก๐
๐!
= ๐ป0 (0) + ๐ป1 (0) +โ๏ฟฝ๐=2
๐ป๐ (0)๐ก๐
๐!But ๐ป0 (๐ฅ) = 1, hence ๐ป0 (0) = 1 and ๐ป1 (๐ฅ) = 2๐ฅ, hence ๐ป1 (0) = 0 and the above becomes
๐ (0, ๐ก) = 1 +โ๏ฟฝ๐=2
๐ป๐ (0)๐ก๐
๐!For the remaining series, it can be written as sum of even and odd terms
๐ (0, ๐ก) = 1 +โ๏ฟฝ
๐=2,4,6,โฏ๐ป๐ (0)
๐ก๐
๐!+
โ๏ฟฝ
๐=3,5,7,โฏ๐ป๐ (0)
๐ก๐
๐!
Or, equivalently
๐ (0, ๐ก) = 1 +โ๏ฟฝ
๐=1,2,3,โฏ๐ป2๐ (0)
๐ก2๐
(2๐)!+
โ๏ฟฝ
๐=1,2,3,โฏ๐ป2๐+1 (0)
๐ก2๐+1
(2๐ + 1)!
But using the hint given that ๐ป2๐+1 (0) = 0 and ๐ป2๐ (0) =(โ1)๐(2๐)!
๐! the above simplifies to
๐ (0, ๐ก) = 1 +โ๏ฟฝ
๐=1,2,3,โฏ
(โ1)๐ (2๐)!๐!
๐ก2๐
(2๐)!
= 1 +โ๏ฟฝ
๐=1,2,3,โฏ(โ1)๐
๐ก2๐
๐!
But since (โ1)๐ ๐ก2๐
๐! = 1 when ๐ = 0, then the above sum can be made to start as zero and itsimplifies to
๐ (0, ๐ก) =โ๏ฟฝ๐=0
(โ1)๐๐ก2๐
๐!
128
Therefore the solution ๐ (๐ฅ, ๐ก) = ๐ (0, ๐ก) ๐๐ก๐ฅ found in part (2) becomes
๐ (๐ฅ, ๐ก) = ๏ฟฝโ๏ฟฝ๐=0
(โ1)๐๐ก2๐
๐! ๏ฟฝ๐2๐ก๐ฅ (1)
Now the sum โโ๐=0 (โ1)
๐ ๐ก2๐
๐! = 1 โ ๐ก2 + ๐ก4
2! โ๐ก6
3! +โฏ and comparing this sum to standard series
of ๐๐ง = 1 + ๐ง + ๐ง2
2! +๐ง3
3! +โฏ, then this shows that when ๐ง = โ๐ก2 and series for ๐โ๐ก2 becomes
๐โ๐ก2 = 1 + ๏ฟฝโ๐ก2๏ฟฝ +๏ฟฝโ๐ก2๏ฟฝ
2
2!+๏ฟฝโ๐ก2๏ฟฝ
3
3!+๏ฟฝโ๐ก2๏ฟฝ
4
4!โฏ
= 1 โ ๐ก2 +๐ก4
2!โ๐ก6
3!+๐ก8
4!โฏ
Henceโ๏ฟฝ๐=0
(โ1)๐๐ก2๐
๐!= ๐โ๐ก2
Substituting this into (1) gives
๐ (๐ฅ, ๐ก) = ๐โ๐ก2๐2๐ก๐ฅ
= ๐2๐ก๐ฅโ๐ก2
Part 4
Since ๐ (๐ฅ, ๐ก) = ๐2๐ก๐ฅโ๐ก2 from part (3), then
๐๐๐ก๐ (๐ฅ, ๐ก) = (2๐ฅ โ 2๐ก) ๐2๐ก๐ฅโ๐ก2
= (2๐ฅ โ 2๐ก) ๐ (๐ฅ, ๐ก)
But ๐ (๐ฅ, ๐ก) = โโ๐=0๐ป๐ (๐ฅ)
๐ก๐
๐! , therefore the above can be written as
๐๐๐ก๐ (๐ฅ, ๐ก) = (2๐ฅ โ 2๐ก)
โ๏ฟฝ๐=0
๐ป๐ (๐ฅ)๐ก๐
๐!
= 2๐ฅโ๏ฟฝ๐=0
๐ป๐ (๐ฅ)๐ก๐
๐!โ 2๐ก
โ๏ฟฝ๐=0
๐ป๐ (๐ฅ)๐ก๐
๐!
= 2๐ฅโ๏ฟฝ๐=0
๐ป๐ (๐ฅ)๐ก๐
๐!โ 2
โ๏ฟฝ๐=0
๐ป๐ (๐ฅ)๐ก๐+1
๐!
= 2๐ฅโ๏ฟฝ๐=0
๐ป๐ (๐ฅ)๐ก๐
๐!โ 2
โ๏ฟฝ๐=1
๐ป๐โ1 (๐ฅ)๐ก๐
(๐ โ 1)!
= 2๐ฅโ๏ฟฝ๐=0
๐ป๐ (๐ฅ)๐ก๐
๐!โ 2
โ๏ฟฝ๐=1
๐๐ป๐โ1 (๐ฅ)๐ก๐
๐ (๐ โ 1)!
= 2๐ฅโ๏ฟฝ๐=0
๐ป๐ (๐ฅ)๐ก๐
๐!โ 2
โ๏ฟฝ๐=1
๐๐ป๐โ1 (๐ฅ)๐ก๐
๐!(1)
On the other hand,๐๐๐ก๐ (๐ฅ, ๐ก) =
๐๐๐ก
โ๏ฟฝ๐=0
๐ป๐ (๐ฅ)๐ก๐
๐!
=โ๏ฟฝ๐=0
๐๐ป๐ (๐ฅ)๐ก๐โ1
๐!Since at ๐ = 0 the sum is zero, then it can be started from ๐ = 1 without changing the result
๐๐๐ก๐ (๐ฅ, ๐ก) =
โ๏ฟฝ๐=1
๐๐ป๐ (๐ฅ)๐ก๐โ1
๐!
=โ๏ฟฝ๐=0
(๐ + 1)๐ป๐+1 (๐ฅ)๐ก๐
(๐ + 1)!
=โ๏ฟฝ๐=0
(๐ + 1)๐ป๐+1 (๐ฅ)๐ก๐
(๐ + 1) ๐!
=โ๏ฟฝ๐=0
๐ป๐+1 (๐ฅ)๐ก๐
๐!(2)
129
Equating (1) and (2) givesโ๏ฟฝ๐=0
๐ป๐+1 (๐ฅ)๐ก๐
๐!= 2๐ฅ
โ๏ฟฝ๐=0
๐ป๐ (๐ฅ)๐ก๐
๐!โ 2
โ๏ฟฝ๐=1
๐๐ป๐โ1 (๐ฅ)๐ก๐
๐!
But โโ๐=1 ๐๐ป๐โ1 (๐ฅ)
๐ก๐
๐! = โโ๐=0 ๐๐ป๐โ1 (๐ฅ)
๐ก๐
๐! because at ๐ = 0 it is zero, so it does not a๏ฟฝect theresult to start the sum from zero, and now the above can be written as
โ๏ฟฝ๐=0
๐ป๐+1 (๐ฅ)๐ก๐
๐!= 2๐ฅ
โ๏ฟฝ๐=0
๐ป๐ (๐ฅ)๐ก๐
๐!โ 2
โ๏ฟฝ๐=0
๐๐ป๐โ1 (๐ฅ)๐ก๐
๐!Now since all the sums start from ๐ = 0 then the above means the same as
๐ป๐+1 (๐ฅ)๐ก๐
๐!= 2๐ฅ๐ป๐ (๐ฅ)
๐ก๐
๐!โ 2๐๐ป๐โ1 (๐ฅ)
๐ก๐
๐!Canceling ๐ก๐
๐! from each term gives
๐ป๐+1 (๐ฅ) = 2๐ฅ๐ป๐ (๐ฅ) โ 2๐๐ป๐โ1 (๐ฅ)
Which is the result required to show.
Part 5
The problem is asking to show that
๏ฟฝโ
โโ๐โ๐ฅ2๐ป๐ (๐ฅ)๐ป๐ (๐ฅ) ๐๐ฅ =
โงโชโชโจโชโชโฉ
0 ๐ โ ๐2๐๐!โ๐ ๐ = ๐
The first part below will show the case for ๐ โ ๐ and the second part part will show thecase for ๐ = ๐
case ๐ โ ๐ This is shown by using the di๏ฟฝerential equation directly. I found this methodeasier and more direct. Before starting, the ODE ๐ฆโฒโฒ โ 2๐ฅ๐ฆโฒ + 2๐๐ฆ = 0 is rewritten as
๐๐ฅ2๐๐๐ฅ๏ฟฝ๐โ๐ฅ2๐ฆโฒ๏ฟฝ + 2๐๐ฆ = 0 (1)
The above form is exactly the same as the original ODE as can be seen by expandingit. Now, Let ๐ป๐ (๐ฅ) be one solution to (1) and let ๐ป๐ (๐ฅ) be another solution to (1) whichresults in the following two ODEโs
๐๐ฅ2๐๐๐ฅ๏ฟฝ๐โ๐ฅ2๐ปโฒ
๐๏ฟฝ + 2๐๐ป๐ = 0 (1A)
๐๐ฅ2๐๐๐ฅ๏ฟฝ๐โ๐ฅ2๐ปโฒ
๐๏ฟฝ + 2๐๐ป๐ = 0 (2A)
Multiplying (1A) by ๐ป๐ and (2A) by ๐ป๐ and subtracting gives
๐ป๐ ๏ฟฝ๐๐ฅ2 ๐๐๐ฅ๏ฟฝ๐โ๐ฅ2๐ปโฒ
๐๏ฟฝ + 2๐๐ป๐๏ฟฝ โ ๐ป๐ ๏ฟฝ๐๐ฅ2 ๐๐๐ฅ๏ฟฝ๐โ๐ฅ2๐ปโฒ
๐๏ฟฝ + 2๐๐ป๐๏ฟฝ = 0
๏ฟฝ๐ป๐๐๐ฅ2 ๐๐๐ฅ๏ฟฝ๐โ๐ฅ2๐ปโฒ
๐๏ฟฝ + 2๐๐ป๐๐ป๐๏ฟฝ โ ๏ฟฝ๐ป๐๐๐ฅ2 ๐๐๐ฅ๏ฟฝ๐โ๐ฅ2๐ปโฒ
๐๏ฟฝ + 2๐๐ป๐๐ป๐๏ฟฝ = 0
๐ป๐๐๐ฅ2 ๐๐๐ฅ๏ฟฝ๐โ๐ฅ2๐ปโฒ
๐๏ฟฝ โ ๐ป๐๐๐ฅ2 ๐๐๐ฅ๏ฟฝ๐โ๐ฅ2๐ปโฒ
๐๏ฟฝ + 2 (๐ โ ๐)๐ป๐๐ป๐ = 0
๐ป๐๐๐๐ฅ๏ฟฝ๐โ๐ฅ2๐ปโฒ
๐๏ฟฝ โ ๐ป๐๐๐๐ฅ๏ฟฝ๐โ๐ฅ2๐ปโฒ
๐๏ฟฝ + 2 (๐ โ ๐)๐ป๐๐ป๐๐โ๐ฅ2 = 0 (3)
But
๐ป๐๐๐๐ฅ๏ฟฝ๐โ๐ฅ2๐ปโฒ
๐๏ฟฝ =๐๐๐ฅ๏ฟฝ๐โ๐ฅ2๐ปโฒ
๐๐ป๐๏ฟฝ โ ๐โ๐ฅ2๐ปโฒ
๐๐ปโฒ๐
And
๐ป๐๐๐๐ฅ๏ฟฝ๐โ๐ฅ2๐ปโฒ
๐๏ฟฝ =๐๐๐ฅ๏ฟฝ๐โ๐ฅ2๐ปโฒ
๐๐ป๐๏ฟฝ โ ๐โ๐ฅ2๐ปโฒ
๐๐ปโฒ๐
Therefore
๐ป๐๐๐๐ฅ๏ฟฝ๐โ๐ฅ2๐ปโฒ
๐๏ฟฝ โ ๐ป๐๐๐๐ฅ๏ฟฝ๐โ๐ฅ2๐ปโฒ
๐๏ฟฝ = ๏ฟฝ๐๐๐ฅ๏ฟฝ๐โ๐ฅ2๐ปโฒ
๐๐ป๐๏ฟฝ โ ๐โ๐ฅ2๐ปโฒ
๐๐ปโฒ๐๏ฟฝ โ ๏ฟฝ
๐๐๐ฅ๏ฟฝ๐โ๐ฅ2๐ปโฒ
๐๐ป๐๏ฟฝ โ ๐โ๐ฅ2๐ปโฒ
๐๐ปโฒ๐๏ฟฝ
=๐๐๐ฅ๏ฟฝ๐โ๐ฅ2๐ปโฒ
๐๐ป๐๏ฟฝ โ๐๐๐ฅ๏ฟฝ๐โ๐ฅ2๐ปโฒ
๐๐ป๐๏ฟฝ
=๐๐๐ฅ๏ฟฝ๐โ๐ฅ2 (๐ปโฒ
๐๐ป๐ โ ๐ปโฒ๐๐ป๐)๏ฟฝ
130
Substituting the above relation back into (3) gives๐๐๐ฅ๏ฟฝ๐โ๐ฅ2 (๐ปโฒ
๐๐ป๐ โ ๐ปโฒ๐๐ป๐)๏ฟฝ + 2 (๐ โ ๐)๐ป๐๐ป๐๐โ๐ฅ
2 = 0
Integrating gives
๏ฟฝโ
โโ
๐๐๐ฅ๏ฟฝ๐โ๐ฅ2 (๐ปโฒ
๐๐ป๐ โ ๐ปโฒ๐๐ป๐)๏ฟฝ ๐๐ฅ +๏ฟฝ
โ
โโ2 (๐ โ ๐)๐ป๐๐ป๐๐โ๐ฅ
2๐๐ฅ = 0
๏ฟฝโ
โโ๐ ๏ฟฝ๐โ๐ฅ2 (๐ปโฒ
๐๐ป๐ โ ๐ปโฒ๐๐ป๐)๏ฟฝ + 2 (๐ โ ๐)๏ฟฝ
โ
โโ๐ป๐๐ป๐๐โ๐ฅ
2๐๐ฅ = 0
๏ฟฝ๐โ๐ฅ2 (๐ปโฒ๐๐ป๐ โ ๐ปโฒ
๐๐ป๐)๏ฟฝโ
โโ+ 2 (๐ โ ๐)๏ฟฝ
โ
โโ๐ป๐๐ป๐๐โ๐ฅ
2๐๐ฅ = 0
But lim๐ฅโยฑโ ๐โ๐ฅ2 โ 0 so the first term above vanishes and the above becomes
2 (๐ โ ๐)๏ฟฝโ
โโ๐ป๐๐ป๐๐โ๐ฅ
2๐๐ฅ = 0
Since this is the case where ๐ โ ๐ then the above shows that
๏ฟฝโ
โโ๐ป๐๐ป๐๐โ๐ฅ
2๐๐ฅ = 0 ๐ โ ๐
Now the case ๐ = ๐ is proofed. When ๐ป๐ = ๐ป๐ then the integral becomes โซโ
โโ๐ป๐๐ป๐๐โ๐ฅ
2๐๐ฅ.Using the known Rodrigues formula for Hermite polynomials, given by
๐ป๐ (๐ฅ) = (โ1)๐ ๐๐ฅ2
๐๐
๐๐ฅ๐๐โ๐ฅ2
Then applying the above the above to one of the ๐ป๐ (๐ฅ) in the integral โซโ
โโ๐ป๐๐ป๐๐โ๐ฅ
2๐๐ฅ, gives
๏ฟฝโ
โโ๐ป๐๐ป๐๐โ๐ฅ
2๐๐ฅ = ๏ฟฝโ
โโ๏ฟฝ(โ1)๐ ๐๐ฅ
2 ๐๐
๐๐ฅ๐๐โ๐ฅ2๏ฟฝ๐ป๐๐โ๐ฅ
2๐๐ฅ
= (โ1)๐๏ฟฝโ
โโ๏ฟฝ๐๐
๐๐ฅ๐๐โ๐ฅ2๏ฟฝ๐ป๐๐๐ฅ
Now integration by parts is carried out. โซ๐ข๐๐ฃ = ๐ข๐ฃ โ โซ๐ฃ๐๐ข. Let ๐ข = ๐ป๐ and let ๐๐ฃ = ๐๐
๐๐ฅ๐ ๐โ๐ฅ2,
therefore ๐๐ข = ๐ปโฒ๐ (๐ฅ) = 2๐๐ป๐โ1 (๐ฅ) and ๐ฃ =
๐๐โ1
๐๐ฅ๐โ1๐โ๐ฅ2, therefore
๏ฟฝโ
โโ๐ป๐๐ป๐๐โ๐ฅ
2๐๐ฅ = (โ1)๐โโโโโโ๏ฟฝ๐ป๐ (๐ฅ)
๐๐โ1
๐๐ฅ๐โ1๐โ๐ฅ2๏ฟฝ
โ
โโโ๏ฟฝ
โ
โโ๏ฟฝ๐๐โ1
๐๐ฅ๐โ1๐โ๐ฅ2๏ฟฝ 2๐๐ป๐โ1 (๐ฅ) ๐๐ฅ
โโโโโโ
But ๏ฟฝ๐ป๐ (๐ฅ)๐๐โ1
๐๐ฅ๐โ1๐โ๐ฅ2๏ฟฝ
โ
โโโ 0 as ๐ฅ โ ยฑโ because each derivative of ๐๐โ1
๐๐ฅ๐โ1๐โ๐ฅ2 produces a
term with ๐โ๐ฅ2 which vanishes at both ends of the real line. Hence the above integral nowbecomes
๏ฟฝโ
โโ๐ป๐๐ป๐๐โ๐ฅ
2๐๐ฅ = (โ1)๐ ๏ฟฝโ2๐๏ฟฝโ
โโ๏ฟฝ๐๐โ1
๐๐ฅ๐โ1๐โ๐ฅ2๏ฟฝ๐ป๐โ1 (๐ฅ) ๐๐ฅ๏ฟฝ
Now the process is repeated, doing one more integration by parts. This results in
๏ฟฝโ
โโ๐ป๐๐ป๐๐โ๐ฅ
2๐๐ฅ = (โ1)๐ ๏ฟฝโ2๐ ๏ฟฝโ2 (๐ โ 1)๏ฟฝโ
โโ๏ฟฝ๐๐โ2
๐๐ฅ๐โ2๐โ๐ฅ2๏ฟฝ๐ป๐โ2 (๐ฅ) ๐๐ฅ๏ฟฝ๏ฟฝ
And again
๏ฟฝโ
โโ๐ป๐๐ป๐๐โ๐ฅ
2๐๐ฅ = (โ1)๐ ๏ฟฝโ2๐ ๏ฟฝโ2 (๐ โ 1) ๏ฟฝโ2 (๐ โ 2)๏ฟฝโ
โโ๏ฟฝ๐๐โ3
๐๐ฅ๐โ3๐โ๐ฅ2๏ฟฝ๐ป๐โ3 (๐ฅ) ๐๐ฅ๏ฟฝ๏ฟฝ๏ฟฝ
This process continues ๐ times. After ๐ integrations by parts, the above becomes
๏ฟฝโ
โโ๐ป๐๐ป๐๐โ๐ฅ
2๐๐ฅ = (โ1)๐ ๏ฟฝโ2๐ ๏ฟฝโ2 (๐ โ 1) ๏ฟฝโ2 (๐ โ 2) ๏ฟฝโฏ๏ฟฝ๏ฟฝโ
โโ๐โ๐ฅ2๐ป0 (๐ฅ) ๐๐ฅ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ
= (โ1)๐ (โ2)๐ ๐!๏ฟฝโ
โโ๐โ๐ฅ2๐ป0 (๐ฅ) ๐๐ฅ
= 2๐๐!๏ฟฝโ
โโ๐โ๐ฅ2๐ป0 (๐ฅ) ๐๐ฅ
But ๐ป0 (๐ฅ) = 1, therefore the above becomes
๏ฟฝโ
โโ๐ป๐๐ป๐๐โ๐ฅ
2๐๐ฅ = 2๐๐!๏ฟฝโ
โโ๐โ๐ฅ2๐๐ฅ
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But
๏ฟฝโ
โโ๐โ๐ฅ2 = 2๏ฟฝ
โ
0๐โ๐ฅ2
= 2โ๐2
= โ๐
Therefore
๏ฟฝโ
โโ๐ป๐๐ป๐๐โ๐ฅ
2๐๐ฅ = 2๐๐!โ๐
This completes the case for ๐ = ๐. Hence
๏ฟฝโ
โโ๐โ๐ฅ2๐ป๐ (๐ฅ)๐ป๐ (๐ฅ) ๐๐ฅ =
โงโชโชโจโชโชโฉ
0 ๐ โ ๐2๐๐!โ๐ ๐ = ๐
Which is what the problem asked to show.
3.7.2 Problem 2
Figure 3.21: Problem statement
Solution
Part (a)
๐ฆโฒโฒ (๐) +1๐๐ฆโฒ (๐) โ
๐2
๐2๐ฆ (๐) = 0 0 < ๐ < โ
Or
๐2๐ฆโฒโฒ (๐) + ๐๐ฆโฒ (๐) โ ๐2๐ฆ (๐) = 0
case ๐ = 0
The ode becomes ๐2๐ฆโฒโฒ (๐) + ๐๐ฆโฒ (๐) = 0. Let ๐ง = ๐ฆโฒ and it becomes ๐2๐งโฒ (๐) + ๐๐ง (๐) = 0 or
๐งโฒ (๐) + 1๐ ๐ง (๐) = 0. This is linear in ๐ง (๐). Integrating factor is ๐ผ = ๐โซ
1๐ ๐๐ = ๐. Multiplying the
ode by ๐ผ it becomes exact di๏ฟฝerential ๐๐๐(๐ง๐) = 0 or ๐ (๐ง๐) = 0, hence ๐ง = ๐1
๐ where ๐1 isconstant of integration. Therefore
๐ฆโฒ (๐) =๐1๐
Integrating again gives
๐ฆ (๐) =๐1
ln ๐ + ๐2Since lim๐โ0 the solution is bounded, then ๐1 must be zero. Therefore 0 = ๐2 and thisimplies ๐2 = 0 also. Therefore when ๐ = 0 the solution is
๐ฆ (๐) = 0
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Case ๐ โ 0
Since powers of ๐ is the same as order of derivative in each term, this is an Euler ODE. Itis solved by assuming ๐ฆ = ๐๐ผ. Hence ๐ฆโฒ = ๐ผ๐๐ผโ1, ๐ฆโฒโฒ = ๐ผ (๐ผ โ 1) ๐๐ผโ2. Substituting these intothe above ODE gives
๐2๐ผ (๐ผ โ 1) ๐๐ผโ2 + ๐๐ผ๐๐ผโ1 โ ๐2๐๐ผ = 0๐ผ (๐ผ โ 1) ๐๐ผ + ๐ผ๐๐ผ โ ๐2๐๐ผ = 0
๐๐ผ ๏ฟฝ๐ผ (๐ผ โ 1) + ๐ผ โ ๐2๏ฟฝ = 0
Assuming non-trivial solution ๐๐ผ โ 0, then the indicial equation is
๐ผ (๐ผ โ 1) + ๐ผ โ ๐2 = 0๐ผ2 = ๐2
๐ผ = ยฑ๐
Hence one solution is
๐ฆ1 (๐) = ๐๐
And second solution is
๐ฆ2 (๐) = ๐โ๐
And the general solution is linear combination of these solutions
๐ฆ (๐) = ๐1๐๐ + ๐2๐โ๐
The above shows that lim๐โ0 ๐ฆ1 (๐) = 0 and lim๐โโ ๐ฆ2 (๐) = 0.
Part (b)
Short version of the solution
To simplify the notations, ๐0 is used instead of ๐โฒ in all the following.
๐ฆโฒโฒ (๐) +1๐๐ฆโฒ (๐) โ
๐2
๐2๐ฆ (๐) =
1๐๐ฟ (๐ โ ๐0) 0 < ๐ < โ
Multiplying both sides by ๐ the above becomes
๐๐ฆโฒโฒ (๐) + ๐ฆโฒ (๐) โ๐2
๐๐ฆ (๐) = ๐ฟ (๐ โ ๐0) (1)
But the two solutions2 to the homogeneous ODE ๐๐ฆโฒโฒ (๐) + ๐ฆโฒ (๐) โ ๐2
๐ ๐ฆ (๐) = 0 were found inpart (a). These are
๐ฆ1 (๐) = ๐๐ (1A)
๐ฆ2 (๐) = ๐โ๐
The Green function is the solution to
๐๐บ (๐, ๐0) + ๐บ (๐, ๐0) โ๐2
๐๐บ (๐, ๐0) = ๐ฟ (๐ โ ๐0) (1B)
lim๐โ0
๐บ (๐, ๐0) = 0
lim๐โโ
๐บ (๐, ๐0) = 0
Which is given by (Using class notes, Lecture December 5, 2018) as
๐บ (๐, ๐0) =1๐ถ
โงโชโชโจโชโชโฉ๐ฆ1 (๐) ๐ฆ2 (๐0) 0 < ๐ < ๐0๐ฆ1 (๐0) ๐ฆ2 (๐) ๐0 < ๐ < โ
(2)
Note, I used +1๐ถ and not โ1
๐ถ as in class notes, since I am using ๐ฟ = โ ๏ฟฝ๏ฟฝ๐๐ฆโฒ๏ฟฝโฒโ ๐๐ฆ๏ฟฝ as the
operator and not ๐ฟ = + ๏ฟฝ๏ฟฝ๐๐ฆโฒ๏ฟฝโฒ+ ๐๐ฆ๏ฟฝ. Now ๐ถ is given by
๐ถ = ๐ (๐0) ๏ฟฝ๐ฆ1 (๐0) ๐ฆโฒ2 (๐0) โ ๐ฆโฒ1 (๐0) ๐ฆ2 (๐0)๏ฟฝ
2All the following is for ๐ โ 0, since for ๐ = 0, only trivial solution exist
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Where from (1A) we see that
๐ฆ1 (๐0) = ๐๐0๐ฆโฒ2 (๐0) = โ๐๐โ๐โ10
๐ฆโฒ1 (๐0) = ๐๐๐โ10
๐ฆ2 (๐0) = ๐โ๐0Therefore ๐ถ becomes
๐ถ = ๐ (๐0) ๏ฟฝโ๐๐โ๐โ10 ๐๐0 โ ๐๐๐โ10 ๐โ๐0 ๏ฟฝ
= 2๐๐โ10 ๐ (๐0)
We just need now to find ๐ (๐0). This comes from Sturm Liouville form. We need to convertthe ODE ๐2๐ฆโฒโฒ (๐) + ๐๐ฆโฒ (๐) โ ๐2๐ฆ (๐) = 0 to Sturm Liouville. Writing this ODE as ๐๐ฆโฒโฒ + ๐๐ฆโฒ +(๐ + ๐) ๐ฆ = 0 where ๐ = ๐2, ๐ = ๐, ๐ = 0, ๐ = โ๐2, therefore
๐ = ๐โซ๐๐๐๐ = ๐
โซ ๐๐2๐๐ = ๐
๐ = โ๐๐๐= 0
๐ =๐๐=๐๐2=1๐
Hence the SL form is ๏ฟฝ๐๐ฆโฒ๏ฟฝโฒโ ๐๐ฆ + ๐๐๐ฆ = 0. Hence the SL form is ๏ฟฝ๐๐ฆโฒ๏ฟฝ
โฒโ ๐๐ฆ + ๐๐๐ฆ = 0 or
๏ฟฝ๐๐ฆโฒ๏ฟฝโฒโ1๐๐2๐ฆ = 0 (2A)
Hence the operator is ๐ฟ ๏ฟฝ๐ฆ๏ฟฝ = โ ๏ฟฝ ๐๐๐ ๏ฟฝ๐๐๐๐๏ฟฝ๏ฟฝ ๏ฟฝ๐ฆ๏ฟฝ and in standard form it becomes ๐ฟ ๏ฟฝ๐ฆ๏ฟฝ+ 1
๐๐2๐ฆ = 0.
The above shows that ๐ (๐0) = ๐0. Therefore
๐ถ = 2๐
Hence Green function is now found from (2) as, for ๐ โ 0
๐บ (๐, ๐0) =12๐
โงโชโชโจโชโชโฉ๐๐๐โ๐0 0 < ๐ < ๐0๐๐0๐โ๐ ๐0 < ๐ < โ
Since ๐ (๐) in the original ODE is zero, there is nothing to convolve with. i.e. ๐ฆ (๐) =โซโ0๐บ (๐, ๐0) ๐ (๐0) ๐๐0 here is not needed since there is no ๐ (๐). Therefore the above is the
final solution.
Extended solution
This solution shows derivation of (2) above. It can be considered as an appendix. TheGreen function is the solution to
๐๐บ (๐, ๐0) + ๐บ (๐, ๐0) โ๐2
๐๐บ (๐, ๐0) = ๐ฟ (๐ โ ๐0) (1B)
lim๐โ0
๐บ (๐, ๐0) = 0
lim๐โโ
๐บ (๐, ๐0) = 0
In (1B), ๐0 is the location of the impulse and ๐ is the location of the observed response dueto this impulse. The solution to the above ODE is now broken to two regions
๐บ (๐, ๐0) =
โงโชโชโจโชโชโฉ๐ด1๐ฆ1 (๐) + ๐ด2๐ฆ2 (๐) 0 < ๐ < ๐0๐ต1๐ฆ1 (๐) + ๐ต1๐ฆ2 (๐) ๐0 < ๐ < โ
(2)
Where ๐ฆ1 (๐) , ๐ฆ2 (๐) are the solution to ๐๐ฆโฒโฒ (๐) + ๐ฆโฒ (๐) โ ๐2
๐ ๐ฆ (๐) = 0 and these were found inpart (a) to be ๐ฆ1 (๐) = ๐๐, ๐ฆ2 (๐) = ๐โ๐ and ๐ด1, ๐ด2, ๐ต1, ๐ต2 needs to be determined. Hence (2)becomes
๐บ (๐, ๐0) =
โงโชโชโจโชโชโฉ๐ด1๐๐ + ๐ด2๐โ๐ 0 < ๐ < ๐0๐ต1๐๐ + ๐ต2๐โ๐ ๐0 < ๐ < โ
(3)
The left boundary condition lim๐โ0 ๐บ (๐, ๐0) = 0 implies ๐ด2 = 0 and the right boundarycondition lim๐โโ ๐บ (๐, ๐0) = 0 implies ๐ต1 = 0. This is needed to keep the solution bounded.
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Hence (3) simplifies to
๐บ (๐, ๐0) =
โงโชโชโจโชโชโฉ๐ด1๐๐ 0 < ๐ < ๐0๐ต2๐โ๐ ๐0 < ๐ < โ
(4)
To determine the remaining two constants ๐ด1, ๐ต2, two additional conditions are needed.The first is that ๐บ (๐, ๐0) is continuous at ๐ = ๐0 which implies
๐ด1๐๐0 = ๐ต2๐โ๐0 (5)
The second condition is the jump in the derivative of ๐บ (๐, ๐0) given by๐๐๐๐บ (๐, ๐0)๏ฟฝ
๐>๐0โ๐๐๐๐บ (๐, ๐0)๏ฟฝ
๐<๐0=
โ1๐ (๐0)
Where ๐ (๐0) comes from the Sturm Liouville form of the homogeneous ODE. This wasfound above as ๐ (๐0) = ๐0. Hence the above condition becomes
๐๐๐๐บ (๐, ๐0)๏ฟฝ
๐>๐0โ๐๐๐๐บ (๐, ๐0)๏ฟฝ
๐<๐0=โ1๐0
Equation (4) shows that ๐๐๐๐บ (๐, ๐0)๏ฟฝ๐>๐0
= โ๐๐ต2๐โ๐โ10 and that ๐๐๐๐บ (๐, ๐0)๏ฟฝ๐<๐0
= ๐๐ด1๐๐โ10 . Using
these in the above gives the second equation needed
โ๐๐ต2๐โ๐โ10 โ ๐๐ด1๐๐โ10 =โ1๐0
(6)
Solving (5,6) for ๐ด1, ๐ต1: From (5) ๐ด1 = ๐ต2๐โ2๐0 . Substituting this in (6) gives
โ๐๐ต2๐โ๐โ10 โ ๐ ๏ฟฝ๐ต2๐โ2๐0 ๏ฟฝ ๐๐โ1 =โ1๐0
โ๐๐ต2๐โ๐โ1 โ ๐๐ต2๐โ๐โ1 =โ1๐0
โ2๐๐ต2๐โ๐โ10 = โ๐โ10
๐ต2 =โ๐โ10
โ2๐๐โ๐โ10
=12๐๐๐0
But since ๐ด1 = ๐ต2๐โ2๐0 , then
๐ด1 =12๐๐๐0๐โ2๐0
=12๐๐โ๐0
Therefore the solution (4), which is the Green function, becomes, for ๐ โ 0
๐บ (๐, ๐0) =
โงโชโชโจโชโชโฉ
12๐๐
โ๐0 ๐๐ 0 < ๐ < ๐0
12๐๐
๐0๐โ๐ ๐0 < ๐ < โ
(7)
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3.7.3 key solution to HW 7
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