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University Course
Engineering Physics (EP) 548Engineering analysis II
University of Wisconsin, MadisonSpring 2017
My Class Notes
Nasser M. Abbasi
Spring 2017
Contents
1 Introduction 11.1 links . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 syllabus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
2 Resources 52.1 Review of ODEโs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.2 Sturm Liouville . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172.3 Wave equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
3 HWs 393.1 HW1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 403.2 HW2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 803.3 HW3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1243.4 HW4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1783.5 HW5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1893.6 HW6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195
4 Study notes 2094.1 question asked 2/7/2017 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2104.2 note on Airy added 1/31/2017 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2114.3 note p1 added 2/3/2017 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2134.4 note added 1/31/2017 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2204.5 note p3 figure 3.4 in text (page 92) reproduced in color . . . . . . . . . . . . 2254.6 note p4 added 2/15/17 animations of the solutions to ODE from lecture 2/9/17
for small parameter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2274.7 note p5 added 2/15/17 animation figure 9.5 in text page 434 . . . . . . . . . . 2314.8 note p7 added 2/15/17, boundary layer problem solved in details . . . . . . . 2374.9 note p9. added 2/24/17, asking question about problem in book . . . . . . . . 2434.10 note p11. added 3/7/17, Showing that scaling for normalization is same for
any ๐ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2454.11 note p12. added 3/8/17, comparing exact solution to WKB solution using
Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2474.12 Convert ODE to Liouville form . . . . . . . . . . . . . . . . . . . . . . . . . . . 2524.13 note p13. added 3/15/17, solving ๐2๐ฆโณ(๐ฅ) = (๐ + ๐ฅ + ๐ฅ2)๐ฆ(๐ฅ) in Maple . . . . . . 254
5 Exams 255
iii
Contents CONTENTS
5.1 Exam 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2565.2 Exam 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284
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Chapter 1
Introduction
Local contents1.1 links . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 syllabus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1
1.1. links CHAPTER 1. INTRODUCTION
1.1 links
1. Professor Leslie M. Smith web page
2. piazza class page Needs login
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1.2. syllabus CHAPTER 1. INTRODUCTION
1.2 syllabus
NE 548
Engineering Analysis II
TR 11:00-12:15 in Van Vleck B341
Leslie Smith: [email protected], Office Hours in Van Vleck Tuesday/Thursday 12:30-2:00, http://www.math.wisc.edu/ lsmith.
Textbook 1 Required: Advanced Mathematical Methods for Scientists and Engineers,Bender and Orszag, Springer.
Textbook 2 Recommended: Applied Partial Differential Equations, Haberman, Pear-son/Prentice Hall. This text is recommended because some of you may already own it (fromMath 322). Almost any intermediate-advanced PDEs text would be suitable alternative asreference.
Pre-requisite: NE 547. If you have not taken NE 547, you should have previously takencourses equivalent to Math 319, 320 or 340, 321, 322.
Assessment: Your grade for the course will be based on two take-home midterm exams(35% each) and selected homework solutions (30%). The homework and midterms examsfor undergraduate and graduate students will be different in scope; please see the separatelearning outcomes below. Graduate students will be expected to synthesize/analyze materialat a deeper level.
Undergraduate Learning Outcomes:
โข Students will demonstrate knowledge of asymptotic methods to analyze ordinary differ-ential equations, including (but not limited to) boundary layer theory, WKB analysisand multiple-scale analysis.
โข Students will demonstrate knowledge of commonly used methods to analyze partialdifferential equations, including (but not limited to) Fourier analysis, Greenโs functionsolutions, similarity solutions, and method of characteristics.
โข Students will apply methods to idealized problems motivated by applications. Such ap-plications include heat conduction (the heat equation), quantum mechanics (Schrodingerโsequation) and plasma turbulence (dispersive wave equations).
Graduate Learning Outcomes:
โข Students will demonstrate knowledge of asymptotic methods to analyze ordinary differ-ential equations, including (but not limited to) boundary layer theory, WKB analysisand multiple-scale analysis.
โข Students will demonstrate knowledge of commonly used methods to analyze partialdifferential equations, including (but not limited to) Fourier analysis, Greenโs functionsolutions, similarity solutions, and method of characteristics.
โข Students will apply methods to idealized problems motivated by applications. Such ap-plications include heat conduction (the heat equation), quantum mechanics (Schrodingerโsequation) and plasma turbulence (dispersive wave equations).
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1.2. syllabus CHAPTER 1. INTRODUCTION
โข Students will apply methods to solve problems in realistic physical settings.
โข Students will synthesize multiple techniques to solve equations arising from applica-tions.
Grading Scale for Final Grade: 92-100 A, 89-91 AB, 82-88 B, 79-81 BC, 70-78 C, 60-69D, 59 and below F
Midterm 1: Given out Thursday March 2, 2015 and due Thursday March 9, 2017.
Midterm 2: Given out Thursday April 27, 2017 and due Thursday May 4, 2015.
Homework: Homework problems will be assigned regularly, either each week or every otherweek, paced for 6 hours out-of-class work every week. Homework groups are encouraged,but each student should separately submit solutions reflecting individual understanding ofthe material.
Piazza: There will be a Piazza course page where all course materials will be posted. Piazzais also a forum to facilitate peer-group discussions. Please take advantage of this resource tokeep up to date on class notes, homework and discussions.
Piazza Sign-Up Page: piazza.com/wisc/spring2017/neema548
Piazza Course Page: piazza.com/wisc/spring2017/neema548/home
Course Outline
Part I: Intermediate-Advanced Topics in ODEs from Bender and Orszag.
1. Review of local analysis of ODEs near ordinary points, regular singular points and irregularsingular points (BO Chapter 3, 1.5 weeks)
2. Global analysis using boundary layer theory (BO Chapter 9, 1.5 weeks).
3. Global analysis using WKB theory (BO Chapter 10, 1.5 weeks).
4. Greenโs function solutions (1 week)
5. Multiple-scale analysis (BO Chapter 11, 1.5 weeks).
Part II: Intermediate-Advanced Topics in PDEs
1. Review of Sturm-Liouville theory and eigenfunction expansions (1.5 weeks)
2. Non-homogeneous problems and Greenโs function solutions (1.5 weeks)
3. Infinite domain problems and Fourier transforms (1.5 weeks)
4. Quasilinear PDEs (1.5 weeks)
5. Dispersive wave systems (time remaining)
2
4
Chapter 2
Resources
Local contents2.1 Review of ODEโs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.2 Sturm Liouville . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172.3 Wave equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
5
2.1. Review of ODEโs CHAPTER 2. RESOURCES
2.1 Review of ODEโs
6
2.1. Review of ODEโs CHAPTER 2. RESOURCES
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2.1. Review of ODEโs CHAPTER 2. RESOURCES
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2.1. Review of ODEโs CHAPTER 2. RESOURCES
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2.1. Review of ODEโs CHAPTER 2. RESOURCES
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2.1. Review of ODEโs CHAPTER 2. RESOURCES
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2.1. Review of ODEโs CHAPTER 2. RESOURCES
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2.1. Review of ODEโs CHAPTER 2. RESOURCES
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2.1. Review of ODEโs CHAPTER 2. RESOURCES
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2.1. Review of ODEโs CHAPTER 2. RESOURCES
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2.1. Review of ODEโs CHAPTER 2. RESOURCES
16
2.2. Sturm Liouville CHAPTER 2. RESOURCES
2.2 Sturm Liouville
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2.2. Sturm Liouville CHAPTER 2. RESOURCES
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2.2. Sturm Liouville CHAPTER 2. RESOURCES
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2.2. Sturm Liouville CHAPTER 2. RESOURCES
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2.2. Sturm Liouville CHAPTER 2. RESOURCES
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2.2. Sturm Liouville CHAPTER 2. RESOURCES
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2.2. Sturm Liouville CHAPTER 2. RESOURCES
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2.2. Sturm Liouville CHAPTER 2. RESOURCES
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2.2. Sturm Liouville CHAPTER 2. RESOURCES
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2.2. Sturm Liouville CHAPTER 2. RESOURCES
26
2.2. Sturm Liouville CHAPTER 2. RESOURCES
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2.3. Wave equation CHAPTER 2. RESOURCES
2.3 Wave equation
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2.3. Wave equation CHAPTER 2. RESOURCES
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2.3. Wave equation CHAPTER 2. RESOURCES
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2.3. Wave equation CHAPTER 2. RESOURCES
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2.3. Wave equation CHAPTER 2. RESOURCES
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2.3. Wave equation CHAPTER 2. RESOURCES
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2.3. Wave equation CHAPTER 2. RESOURCES
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2.3. Wave equation CHAPTER 2. RESOURCES
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2.3. Wave equation CHAPTER 2. RESOURCES
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2.3. Wave equation CHAPTER 2. RESOURCES
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2.3. Wave equation CHAPTER 2. RESOURCES
38
Chapter 3
HWs
Local contents3.1 HW1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 403.2 HW2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 803.3 HW3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1243.4 HW4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1783.5 HW5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1893.6 HW6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195
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3.1. HW1 CHAPTER 3. HWS
3.1 HW1
3.1.1 problem 3.3 (page 138)
3.1.1.1 Part c
problem Classify all the singular points (finite and infinite) of the following
๐ฅ (1 โ ๐ฅ) ๐ฆโฒโฒ + (๐ โ (๐ + ๐ + 1) ๐ฅ) ๐ฆโฒ โ ๐๐๐ฆ = 0
Answer
Writing the DE in standard form
๐ฆโฒโฒ + ๐ (๐ฅ) ๐ฆโฒ + ๐ (๐ฅ) ๐ฆ = 0
๐ฆโฒโฒ +๐ โ (๐ + ๐ + 1) ๐ฅ
๐ฅ (1 โ ๐ฅ)๐ฆโฒ โ
๐๐๐ฅ (1 โ ๐ฅ)
๐ฆ = 0
๐ฆโฒโฒ +
๐(๐ฅ)
๏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
๐๐ฅ (1 โ ๐ฅ)
โ(๐ + ๐ + 1)(1 โ ๐ฅ) ๏ฟฝ๐ฆโฒ โ
๐(๐ฅ)
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๐๐๐ฅ (1 โ ๐ฅ)
๐ฆ = 0
๐ฅ = 0, 1 are singular points for ๐ (๐ฅ) as well as for ๐ (๐ฅ). Now we classify what type of singularityeach point is. For ๐ (๐ฅ)
lim๐ฅโ0
๐ฅ๐ (๐ฅ) = lim๐ฅโ0
๐ฅ ๏ฟฝ๐
๐ฅ (1 โ ๐ฅ)โ(๐ + ๐ + 1)(1 โ ๐ฅ) ๏ฟฝ
= lim๐ฅโ0
๏ฟฝ๐
(1 โ ๐ฅ)โ๐ฅ (๐ + ๐ + 1)(1 โ ๐ฅ) ๏ฟฝ
= ๐
Hence ๐ฅ๐ (๐ฅ) is analytic at ๐ฅ = 0. Therefore ๐ฅ = 0 is regular singularity point. Now we checkfor ๐ (๐ฅ)
lim๐ฅโ0
๐ฅ2๐ (๐ฅ) = lim๐ฅโ0
๐ฅ2โ๐๐
๐ฅ (1 โ ๐ฅ)
= lim๐ฅโ0
โ๐ฅ๐๐(1 โ ๐ฅ)
= 0
Hence ๐ฅ2๐ (๐ฅ) is also analytic at ๐ฅ = 0. Therefore ๐ฅ = 0 is regular singularity point. Now we
40
3.1. HW1 CHAPTER 3. HWS
look at ๐ฅ = 1 and classify it. For ๐ (๐ฅ)
lim๐ฅโ1
(๐ฅ โ 1) ๐ (๐ฅ) = lim๐ฅโ1
(๐ฅ โ 1) ๏ฟฝ๐
๐ฅ (1 โ ๐ฅ)โ(๐ + ๐ + 1)(1 โ ๐ฅ) ๏ฟฝ
= lim๐ฅโ1
(๐ฅ โ 1) ๏ฟฝโ๐
๐ฅ (๐ฅ โ 1)+(๐ + ๐ + 1)(๐ฅ โ 1) ๏ฟฝ
= lim๐ฅโ1
๏ฟฝโ๐๐ฅ+ (๐ + ๐ + 1)๏ฟฝ
= โ๐ + (๐ + ๐ + 1)
Hence (๐ฅ โ 1) ๐ (๐ฅ) is analytic at ๐ฅ = 1. Therefore ๐ฅ = 1 is regular singularity point. Now wecheck for ๐ (๐ฅ)
lim๐ฅโ1
(๐ฅ โ 1)2 ๐ (๐ฅ) = lim๐ฅโ1
(๐ฅ โ 1)2 ๏ฟฝโ๐๐
๐ฅ (1 โ ๐ฅ)๏ฟฝ
= lim๐ฅโ1
(๐ฅ โ 1)2 ๏ฟฝ๐๐
๐ฅ (๐ฅ โ 1)๏ฟฝ
= lim๐ฅโ1
(๐ฅ โ 1) ๏ฟฝ๐๐๐ฅ ๏ฟฝ
= 0
Hence (๐ฅ โ 1)2 ๐ (๐ฅ) is analytic also at ๐ฅ = 1. Therefore ๐ฅ = 1 is regular singularity point.Therefore ๐ฅ = 0, 1 are regular singular points for the ODE. Now we check for ๐ฅ at โ. Tocheck the type of singularity, if any, at ๐ฅ = โ, the DE is first transformed using
๐ฅ =1๐ก
(1)
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3.1. HW1 CHAPTER 3. HWS
This transformation will always results in 1 new ODE in ๐ก of this form
๐2๐ฆ๐๐ก2
+๏ฟฝโ๐ (๐ก) + 2๐ก๏ฟฝ
๐ก2+๐ (๐ก)๐ก4๐ฆ = 0 (2)
But
๐ (๐ก) = ๐ (๐ฅ)๏ฟฝ๐ฅ= 1
๐ก= ๏ฟฝ
๐ โ (๐ + ๐ + 1) ๐ฅ๐ฅ (1 โ ๐ฅ) ๏ฟฝ
๐ฅ= 1๐ก
=
โโโโโโโโโโ
๐ โ (๐ + ๐ + 1) 1๐ก1๐ก๏ฟฝ1 โ 1
๐ก๏ฟฝ
โโโโโโโโโโ
=๐๐ก2 โ ๐ก (๐ + ๐ + 1)
(๐ก โ 1)(3)
And
๐ (๐ก) = ๐ (๐ฅ)๏ฟฝ๐ฅ= 1
๐ก= ๏ฟฝโ
๐๐๐ฅ (1 โ ๐ฅ)๏ฟฝ
๐ฅ= 1๐ก
= โ๐๐
1๐ก๏ฟฝ1 โ 1
๐ก๏ฟฝ
= โ๐๐๐ก2
(๐ก โ 1)(4)
1Let ๐ฅ = 1๐ก, then
๐๐๐ฅ
=๐๐๐ก
๐๐ก๐๐ฅ
= โ๐ก2๐๐๐ก
And
๐2
๐๐ฅ2=
๐๐๐ฅ ๏ฟฝ
๐๐๐ฅ๏ฟฝ
= ๏ฟฝโ๐ก2๐๐๐ก ๏ฟฝ ๏ฟฝ
โ๐ก2๐๐๐ก ๏ฟฝ
= โ๐ก2๐๐๐ก ๏ฟฝ
โ๐ก2๐๐๐ก ๏ฟฝ
= โ๐ก2 ๏ฟฝโ2๐ก๐๐๐ก
โ ๐ก2๐2
๐๐ก2 ๏ฟฝ
= 2๐ก3๐๐๐ก
+ ๐ก4๐2
๐๐ก2
The original ODE becomes
๏ฟฝ2๐ก3๐๐๐ก
+ ๐ก4๐2
๐๐ก2 ๏ฟฝ๐ฆ + ๐ (๐ฅ)๏ฟฝ
๐ฅ= 12๏ฟฝโ๐ก2
๐๐๐ก ๏ฟฝ
๐ฆ + ๐ (๐ฅ)๏ฟฝ๐ฅ= 1
๐ก
๐ฆ = 0
๏ฟฝ2๐ก3๐ฆโฒ + ๐ก4๐ฆโฒโฒ๏ฟฝ โ ๐ก2๐ (๐ก) ๐ฆโฒ + ๐ (๐ก) ๐ฆ = 0
๐ก4๐ฆโฒโฒ + ๏ฟฝโ๐ก2๐ (๐ก) + 2๐ก3๏ฟฝ ๐ฆโฒ + ๐ (๐ก) ๐ฆ = 0
๐ฆโฒโฒ +๏ฟฝโ๐ (๐ก) + 2๐ก๏ฟฝ
๐ก2๐ฆโฒ +
๐ (๐ก)๐ก4
๐ฆ = 0
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3.1. HW1 CHAPTER 3. HWS
Substituting equations (3,4) into (2) gives
๐ฆโฒโฒ +๏ฟฝโ ๏ฟฝ ๐๐ก
2โ๐ก(๐+๐+1)(๐กโ1)
๏ฟฝ + 2๐ก๏ฟฝ
๐ก2+๏ฟฝโ ๐๐๐ก2
(๐กโ1)๏ฟฝ
๐ก4๐ฆ = 0
๐ฆโฒโฒ +๏ฟฝ2๐ก (๐ก โ 1) โ ๐ก2๐ + (๐ + ๐ + 1) ๐ก๏ฟฝ
๐ก2 (๐ก โ 1)๐ฆโฒ โ
๐๐๐ก2 (๐ก โ 1)
๐ฆ = 0
Expanding
๐ฆโฒโฒ +
๐(๐ก)
๏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๏ฟฝ2๐ก โ 1 โ ๐ก๐ + ๐ + ๐
๐ก (๐ก โ 1) ๏ฟฝ๐ฆโฒ โ
๐(๐ก)
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๐๐๐ก2 (๐ก โ 1)
๐ฆ = 0
We see that ๐ก = 0 (this means ๐ฅ = โ) is singular point for both ๐ (๐ฅ) , ๐ (๐ฅ). Now we checkwhat type it is. For ๐ (๐ก)
lim๐กโ0
๐ก๐ (๐ก) = lim๐กโ0
๐ก ๏ฟฝ2๐กโ
๐(๐ก โ 1)
+(๐ + ๐ + 1)๐ก (๐ก โ 1) ๏ฟฝ
= lim๐กโ0
๏ฟฝ2 โ๐ก๐
(๐ก โ 1)+(๐ + ๐ + 1)(๐ก โ 1) ๏ฟฝ
= 1 โ ๐ โ ๐
๐ก๐ (๐ก) is therefore analytic at ๐ก = 0. Hence ๐ก = 0 is regular singular point. Now we checkfor ๐ (๐ก)
lim๐กโ0
๐ก2๐ (๐ก) = lim๐กโ0
๐ก2 ๏ฟฝโ๐๐
๐ก2 (๐ก โ 1)๏ฟฝ
= lim๐กโ0
๏ฟฝโ๐๐(๐ก โ 1)๏ฟฝ
= ๐๐
๐ก2๐ (๐ก) is therefore analytic at ๐ก = 0. Hence ๐ก = 0 is regular singular point for ๐ (๐ก). Therefore๐ก = 0 is regular singular point which mean that ๐ฅ โ โ is a regular singular point for theODE.
Summary
Singular points are ๐ฅ = 0, 1. Both are regular singular points. Also ๐ฅ = โ is regular singularpoint.
3.1.1.2 Part (d)
Problem Classify all the singular points (finite and infinite) of the following
๐ฅ๐ฆโฒโฒ + (๐ โ ๐ฅ) ๐ฆโฒ โ ๐๐ฆ = 0
solution
Writing the ODE in standard for
๐ฆโฒโฒ +(๐ โ ๐ฅ)๐ฅ
๐ฆโฒ โ๐๐ฅ๐ฆ = 0
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3.1. HW1 CHAPTER 3. HWS
We see that ๐ฅ = 0 is singularity point for both ๐ (๐ฅ) and ๐ (๐ฅ). Now we check its type. For ๐ (๐ฅ)
lim๐ฅโ0
๐ฅ๐ (๐ฅ) = lim๐ฅโ0
๐ฅ(๐ โ ๐ฅ)๐ฅ
= ๐
Hence ๐ฅ๐ (๐ฅ) is analytic at ๐ฅ = 0. Therefore ๐ฅ = 0 is regular singularity point for ๐ (๐ฅ). For ๐ (๐ฅ)
lim๐ฅโ0
๐ฅ2๐ (๐ฅ) = lim๐ฅโ0
๐ฅ2 ๏ฟฝโ๐๐ฅ๏ฟฝ
= lim๐ฅโ0
(โ๐๐ฅ)
= 0
Hence ๐ฅ2๐ (๐ฅ) is analytic at ๐ฅ = 0. Therefore ๐ฅ = 0 is regular singularity point for ๐ (๐ฅ).
Now we check for ๐ฅ at โ. To check the type of singularity, if any, at ๐ฅ = โ, the DE is firsttransformed using
๐ฅ =1๐ก
(1)
This results in (as was done in above part)
๐2๐ฆ๐๐ก2
+๏ฟฝโ๐ (๐ก) + 2๐ก๏ฟฝ
๐ก2+๐ (๐ก)๐ก4๐ฆ = 0
Where
๐ (๐ก) =(๐ โ ๐ฅ)๐ฅ
๏ฟฝ๐ฅ= 1
๐ก
=๏ฟฝ๐ โ 1
๐ก๏ฟฝ
1๐ก
= (๐๐ก โ 1)
And ๐ (๐ก) = โ ๐๐ฅ = โ๐๐ก Hence the new ODE is
๐2๐ฆ๐๐ก2
+(โ (๐๐ก โ 1) + 2๐ก)
๐ก2โ๐๐ก๐ก4๐ฆ = 0
๐2๐ฆ๐๐ก2
+โ๐๐ก + 1 + 2๐ก
๐ก2โ๐๐ก3๐ฆ = 0
Therefore ๐ก = 0 (or ๐ฅ = โ) is singular point. Now we will find the singularity type
lim๐กโ0
๐ก๐ (๐ก) = lim๐กโ0
๐ก ๏ฟฝโ๐๐ก + 1 + 2๐ก
๐ก2 ๏ฟฝ
= lim๐กโ0
๏ฟฝโ๐๐ก + 1 + 2๐ก
๐ก ๏ฟฝ
= โ
Hence ๐ก๐ (๐ก) is not analytic, since the limit do not exist, which means ๐ก = 0 is an irregular
44
3.1. HW1 CHAPTER 3. HWS
singular point for ๐ (๐ก) . We can stop here, but will also check for ๐ (๐ก)
lim๐กโ0
๐ก2๐ (๐ก) = lim๐กโ0
๐ก2 ๏ฟฝโ๐๐ก3๏ฟฝ
= lim๐กโ0
๏ฟฝโ๐๐ก๏ฟฝ
= โ
Therefore, ๐ก = 0 is irregular singular point, which means ๐ฅ = โ is an irregular singular point.
Summary
๐ฅ = 0 is regular singular point. ๐ฅ = โ is an irregular singular point.
3.1.2 problem 3.4
3.1.2.1 part d
problem Classify ๐ฅ = 0 and ๐ฅ = โ of the following
๐ฅ2๐ฆโฒโฒ = ๐ฆ๐1๐ฅ
Answer
In standard form
๐ฆโฒโฒ โ๐1๐ฅ
๐ฅ2๐ฆ = 0
Hence ๐ (๐ฅ) = 0, ๐ (๐ฅ) = โ ๐1๐ฅ
๐ฅ2 . The singularity is ๐ฅ = 0. We need to check on ๐ (๐ฅ) only.
lim๐ฅโ0
๐ฅ2๐ (๐ฅ) = lim๐ฅโ0
๐ฅ2โโโโโโโโ๐1๐ฅ
๐ฅ2
โโโโโโโ
= lim๐ฅโ0
๏ฟฝโ๐1๐ฅ ๏ฟฝ
The above is not analytic, since lim๐ฅโ0+ ๏ฟฝโ๐1๐ฅ ๏ฟฝ = โ while lim๐ฅโ0โ ๏ฟฝโ๐
1๐ฅ ๏ฟฝ = 0. This means ๐
1๐ฅ is
not di๏ฟฝerentiable at ๐ฅ = 0. Here is plot ๐1๐ฅ near ๐ฅ = 0
45
3.1. HW1 CHAPTER 3. HWS
-0.4 -0.2 0.0 0.2 0.4
0
200
400
600
x
f(x)
Plot of exp(1/x)
Therefore ๐ฅ = 0 is an irregular singular point. We now convert the ODE using ๐ฅ = 1๐ก in order
to check what happens at ๐ฅ = โ. This results in (as was done in above part)
๐2๐ฆ๐๐ก2
+๐ (๐ก)๐ก4๐ฆ = 0
But
๐ (๐ก) = ๐ (๐ฅ)๏ฟฝ๐ฅ= 1
๐ก
=
โโโโโโโโ๐1๐ฅ
๐ฅ2
โโโโโโโ ๐ฅ= 1
๐ก
= โ๐ก2๐๐ก
Hence the ODE becomes
๐ฆโฒโฒ โโ๐ก2๐๐ก
๐ก4๐ฆ = 0
๐ฆโฒโฒ +๐๐ก
๐ก2๐ฆ = 0
We now check ๐ (๐ก).
lim๐กโ0
๐ก2๐ (๐ก) = lim๐กโ0
๐ก2๐๐ก
๐ก2= lim
๐กโ0๐๐ก
= 1
This is analytic. Hence ๐ก = 0 is regular singular point, which means ๐ฅ = โ is regular singularpoint.
Summary ๐ฅ = 0 is irregular singular point, ๐ฅ = โ is regular singular point.
46
3.1. HW1 CHAPTER 3. HWS
3.1.2.2 Part e
problem
Classify ๐ฅ = 0 and ๐ฅ = โ of the following
(tan ๐ฅ) ๐ฆโฒ = ๐ฆAnswer
๐ฆโฒ โ1
tan ๐ฅ๐ฆ = 0
The function tan ๐ฅ looks like
-ฯ - ฯ
20 ฯ
2ฯ
-6
-4
-2
0
2
4
6
x
tan(x)
tan(x)
Therefore, tan (๐ฅ) is not analytic at ๐ฅ = ๏ฟฝ๐ โ 12๏ฟฝ ๐ for ๐ โ โค. Hence the function 1
tan(๐ฅ) is not
analytic at ๐ฅ = ๐๐ as seen in this plot
-ฯ - ฯ
20 ฯ
2ฯ
-6
-4
-2
0
2
4
6
x
1/tan(x)
1/tan(x)
47
3.1. HW1 CHAPTER 3. HWS
Hence singular points are ๐ฅ = {โฏ , โ2๐, โ๐, 0, ๐, 2๐,โฏ}. Looking at ๐ฅ = 0
lim๐ฅโ0
๐ฅ๐ (๐ฅ) = lim๐ฅโ0
๏ฟฝ๐ฅ1
tan (๐ฅ)๏ฟฝ
= lim๐ฅโ0
โโโโโโโ
๐๐ฅ๐๐ฅ
๐ tan(๐ฅ)๐๐ฅ
โโโโโโโ
= lim๐ฅโ0
1sec2 ๐ฅ
= lim๐ฅโ0
cos2 ๐ฅ
= 1
Therefore the point ๐ฅ = 0 is regular singular point. To classify ๐ฅ = โ, we use ๐ฅ = 1๐ก substitu-
tion. ๐๐๐ฅ =
๐๐๐ก
๐๐ก๐๐ฅ = โ๐ก
2 ๐๐๐ก and the ODE becomes
๏ฟฝโ๐ก2๐๐๐ก๏ฟฝ
๐ฆ โ1
tan ๏ฟฝ1๐ก ๏ฟฝ๐ฆ = 0
โ๐ก2๐ฆโฒ โ1
tan ๏ฟฝ1๐ก ๏ฟฝ๐ฆ = 0
๐ฆโฒ +1
๐ก2 tan ๏ฟฝ1๐ก ๏ฟฝ๐ฆ = 0
Hence
๐ (๐ก) =1
๐ก2 tan ๏ฟฝ1๐ก ๏ฟฝ
This function has singularity at ๐ก = 0 and at ๐ก = 1๐๐ for ๐ โ โค. We just need to consider ๐ก = 0
48
3.1. HW1 CHAPTER 3. HWS
since this maps to ๐ฅ = โ. Hence
lim๐กโ0
๐ก๐ (๐ก) = lim๐กโ0
โโโโโโโโโโ๐ก
1
๐ก2 tan ๏ฟฝ1๐ก ๏ฟฝ
โโโโโโโโโโ
= lim๐กโ0
โโโโโโโโโโ
1
๐ก tan ๏ฟฝ1๐ก ๏ฟฝ
โโโโโโโโโโ
= lim๐กโ0
โโโโโโโโโโ
1๐ก
tan ๏ฟฝ1๐ก ๏ฟฝ
โโโโโโโโโโ
โLโhopitals
lim๐กโ0
โโโโโโโโโโโโ
โ 1๐ก2
โsec2๏ฟฝ 1๐ก ๏ฟฝ
๐ก2
โโโโโโโโโโโโ
= lim๐กโ0
โโโโโโโโโโ
1
sec2 ๏ฟฝ1๐ก ๏ฟฝ
โโโโโโโโโโ
= lim๐กโ0
๏ฟฝcos2 ๏ฟฝ1๐ก ๏ฟฝ๏ฟฝ
The following is a plot of cos2 ๏ฟฝ1๐ก ๏ฟฝ as ๐ก goes to zero.
- ฯ
20 ฯ
2
0.0
0.2
0.4
0.6
0.8
1.0
t
f(t)
cos21
t
This is the same as asking for lim๐ฅโโ cos2 (๐ฅ) which does not exist, since cos (๐ฅ) keepsoscillating, hence it has no limit. Therefore, we conclude that ๐ก๐ (๐ก) is not analytic at ๐ก = 0,hence ๐ก is irregular singular point, which means ๐ฅ = โ is an irregular singular point.
Summary
๐ฅ = 0 is regular singular point and ๐ฅ = โ is an irregular singular point
49
3.1. HW1 CHAPTER 3. HWS
3.1.3 Problem 3.6
3.1.3.1 part b
Problem Find the Taylor series expansion about ๐ฅ = 0 of the solution to the initial valueproblem
๐ฆโฒโฒ โ 2๐ฅ๐ฆโฒ + 8๐ฆ = 0๐ฆ (0) = 0๐ฆโฒ (0) = 4
solution
Since ๐ฅ = 0 is ordinary point, then we can use power series solution
๐ฆ (๐ฅ) =โ๏ฟฝ๐=0
๐๐๐ฅ๐
Hence
๐ฆโฒ (๐ฅ) =โ๏ฟฝ๐=0
๐๐๐๐ฅ๐โ1 =โ๏ฟฝ๐=1
๐๐๐๐ฅ๐โ1 =โ๏ฟฝ๐=0
(๐ + 1) ๐๐+1๐ฅ๐
๐ฆโฒโฒ (๐ฅ) =โ๏ฟฝ๐=0
๐ (๐ + 1) ๐๐+1๐ฅ๐โ1 =โ๏ฟฝ๐=1
๐ (๐ + 1) ๐๐+1๐ฅ๐โ1 =โ๏ฟฝ๐=0
(๐ + 1) (๐ + 2) ๐๐+2๐ฅ๐
Therefore the ODE becomesโ๏ฟฝ๐=0
(๐ + 1) (๐ + 2) ๐๐+2๐ฅ๐ โ 2๐ฅโ๏ฟฝ๐=0
(๐ + 1) ๐๐+1๐ฅ๐ + 8โ๏ฟฝ๐=0
๐๐๐ฅ๐ = 0
โ๏ฟฝ๐=0
(๐ + 1) (๐ + 2) ๐๐+2๐ฅ๐ โโ๏ฟฝ๐=0
2 (๐ + 1) ๐๐+1๐ฅ๐+1 +โ๏ฟฝ๐=0
8๐๐๐ฅ๐ = 0
โ๏ฟฝ๐=0
(๐ + 1) (๐ + 2) ๐๐+2๐ฅ๐ โโ๏ฟฝ๐=1
2๐๐๐๐ฅ๐ +โ๏ฟฝ๐=0
8๐๐๐ฅ๐ = 0
Hence, for ๐ = 0 we obtain(๐ + 1) (๐ + 2) ๐๐+2๐ฅ๐ + 8๐๐๐ฅ๐ = 0
2๐2 + 8๐0 = 0๐2 = โ4๐0
For ๐ โฅ 1(๐ + 1) (๐ + 2) ๐๐+2 โ 2๐๐๐ + 8๐๐ = 0
๐๐+2 =2๐๐๐ โ 8๐๐(๐ + 1) (๐ + 2)
=๐๐ (2๐ โ 8)
(๐ + 1) (๐ + 2)Hence for ๐ = 1
๐3 =๐1 (2 โ 8)
(1 + 1) (1 + 2)= โ๐1
50
3.1. HW1 CHAPTER 3. HWS
For ๐ = 2
๐4 =๐2 (2 (2) โ 8)(2 + 1) (2 + 2)
= โ13๐2 = โ
13(โ4๐0) =
43๐0
For ๐ = 3
๐5 =๐3 (2 (3) โ 8)(3 + 1) (3 + 2)
= โ110๐3 = โ
110(โ๐1) =
110๐1
For ๐ = 4
๐6 =๐4 (2 (4) โ 8)(4 + 1) (4 + 2)
= 0
For ๐ = 5
๐7 =๐5 (2 (5) โ 8)(5 + 1) (5 + 2)
=121๐5 =
121 ๏ฟฝ
110๐1๏ฟฝ =
1210
๐1
For ๐ = 6
๐8 =๐6 (2 (6) โ 8)(6 + 1) (6 + 2)
=114๐6 = 0
For ๐ = 7
๐9 =๐7 (2 (7) โ 8)(7 + 1) (7 + 2)
=112๐7 =
112 ๏ฟฝ
1210
๐1๏ฟฝ =1
2520๐1
Writing now few terms
๐ฆ (๐ฅ) =โ๏ฟฝ๐=0
๐๐๐ฅ๐
= ๐0๐ฅ0 + ๐1๐ฅ1 + ๐2๐ฅ2 + ๐3๐ฅ3 + ๐4๐ฅ4 + ๐5๐ฅ5 + ๐6๐ฅ6 + ๐7๐ฅ7 + ๐8๐ฅ8 + ๐9๐ฅ9 +โฏ
= ๐0 + ๐1๐ฅ + (โ4๐0) ๐ฅ2 + (โ๐1) ๐ฅ3 +43๐0๐ฅ4 +
110๐1๐ฅ5 + 0 +
1210
๐1๐ฅ7 + 0 +1
2520๐1๐ฅ9 +โฏ
= ๐0 + ๐1๐ฅ โ 4๐0๐ฅ2 โ ๐1๐ฅ3 +43๐0๐ฅ4 +
110๐1๐ฅ5 +
1210
๐1๐ฅ7 +1
2520๐1๐ฅ9 +โฏ
= ๐0 ๏ฟฝ1 โ 4๐ฅ2 +43๐ฅ4๏ฟฝ + ๐1 ๏ฟฝ๐ฅ โ ๐ฅ3 +
110๐ฅ5 +
1210
๐ฅ7 +1
2520๐ฅ9 +โฏ๏ฟฝ (1)
We notice that ๐0 terms terminates at 43๐ฅ
4 but the ๐1 terms do not terminate. Now we needto find ๐0, ๐1 from initial conditions. At ๐ฅ = 0, ๐ฆ (0) = 0. Hence from (1)
0 = ๐0Hence the solution becomes
๐ฆ (๐ฅ) = ๐1 ๏ฟฝ๐ฅ โ ๐ฅ3 +110๐ฅ5 +
1210
๐ฅ7 +1
2520๐ฅ9 +โฏ๏ฟฝ (2)
Taking derivative of (2), term by term
๐ฆโฒ (๐ฅ) = ๐1 ๏ฟฝ1 โ 3๐ฅ2 +510๐ฅ4 +
7210
๐ฅ6 +9
2520๐ฅ8 +โฏ๏ฟฝ
Using ๐ฆโฒ (0) = 4 the above becomes
4 = ๐1
51
3.1. HW1 CHAPTER 3. HWS
Hence the solution is
๐ฆ (๐ฅ) = 4 ๏ฟฝ๐ฅ โ ๐ฅ3 +110๐ฅ5 +
1210
๐ฅ7 +1
2520๐ฅ9 +โฏ๏ฟฝ
Or
๐ฆ (๐ฅ) = 4๐ฅ โ 4๐ฅ3 + 25๐ฅ
5 + 2105๐ฅ
7 + 1630๐ฅ
9 +โฏ
The above is the Taylor series of the solution expanded around ๐ฅ = 0.
3.1.4 Problem 3.7
Problem: Estimate the number of terms in the Taylor series (3.2.1) and (3.2.2) at page 68 ofthe text, that are necessary to compute
Ai(๐ฅ) and Bi(๐ฅ) correct to three decimal places at ๐ฅ = ยฑ1, ยฑ100, ยฑ10000
Answer:
Ai (๐ฅ) = 3โ23
โ๏ฟฝ๐=0
๐ฅ3๐
9๐๐!ฮ ๏ฟฝ๐ + 23๏ฟฝโ 3
โ43
โ๏ฟฝ๐=0
๐ฅ3๐+1
9๐๐!ฮ ๏ฟฝ๐ + 43๏ฟฝ
(3.2.1)
Bi (๐ฅ) = 3โ16
โ๏ฟฝ๐=0
๐ฅ3๐
9๐๐!ฮ ๏ฟฝ๐ + 23๏ฟฝโ 3
โ56
โ๏ฟฝ๐=0
๐ฅ3๐+1
9๐๐!ฮ ๏ฟฝ๐ + 43๏ฟฝ
(3.2.2)
The radius of convergence of these series extends from ๐ฅ = 0 to ยฑโ so we know this willconverge to the correct value of Ai (๐ฅ) ,Bi (๐ฅ) for all ๐ฅ, even though we might need largenumber of terms to achieve this, as will be shown below. The following is a plot of Ai (๐ฅ)and Bi (๐ฅ)
-10 -5 0 5 10
-0.4
-0.2
0.0
0.2
0.4
x
AiryAi(x)
0.35503
-10 -5 0
-0.5
0.0
0.5
1.0
1.5
2.0
x
AiryBi(x)
0.6149
52
3.1. HW1 CHAPTER 3. HWS
3.1.4.1 AiryAI series
For Ai (๐ฅ) looking at the (๐+1)๐กโ
๐๐กโ term
ฮ =
3โ23
๐ฅ3(๐+1)
9๐+1(๐+1)!ฮ๏ฟฝ๐+1+ 23 ๏ฟฝโ 3
โ43
๐ฅ3๐+2
9๐+1(๐+1)!ฮ๏ฟฝ๐+1+ 43 ๏ฟฝ
3โ23
๐ฅ3๐
9๐๐!ฮ๏ฟฝ๐+ 23 ๏ฟฝโ 3
โ43
๐ฅ3๐+1
9๐๐!ฮ๏ฟฝ๐+ 43 ๏ฟฝ
This can be simplified using ฮ (๐ + 1) = ๐ฮ (๐) giving
ฮ ๏ฟฝ๏ฟฝ๐ +23๏ฟฝ+ 1๏ฟฝ = ๏ฟฝ๐ +
23๏ฟฝฮ ๏ฟฝ๐ +
23๏ฟฝ
ฮ ๏ฟฝ๏ฟฝ๐ +43๏ฟฝ+ 1๏ฟฝ = ๏ฟฝ๐ +
43๏ฟฝฮ ๏ฟฝ๐ +
43๏ฟฝ
Hence ฮ becomes
ฮ =
3โ23 ๐ฅ3(๐+1)
9๐+1(๐+1)!๏ฟฝ๐+ 23 ๏ฟฝฮ๏ฟฝ๐+ 2
3 ๏ฟฝโ 3
โ43 ๐ฅ3๐+2
9๐+1(๐+1)!๏ฟฝ๐+ 43 ๏ฟฝฮ๏ฟฝ๐+ 4
3 ๏ฟฝ
3โ23 ๐ฅ3๐
9๐๐!ฮ๏ฟฝ๐+ 23 ๏ฟฝโ 3
โ43 ๐ฅ3๐+1
9๐๐!ฮ๏ฟฝ๐+ 43 ๏ฟฝ
Or
ฮ =
3โ23 ๐ฅ3(๐+1)
9(๐+1)๏ฟฝ๐+ 23 ๏ฟฝฮ๏ฟฝ๐+ 2
3 ๏ฟฝโ 3
โ43 ๐ฅ3๐+2
9(๐+1)๏ฟฝ๐+ 43 ๏ฟฝฮ๏ฟฝ๐+ 4
3 ๏ฟฝ
3โ23 ๐ฅ3๐
ฮ๏ฟฝ๐+ 23 ๏ฟฝโ 3
โ43 ๐ฅ3๐+1
ฮ๏ฟฝ๐+ 43 ๏ฟฝ
Or
ฮ =
3โ23 ๐ฅ3(๐+1)๏ฟฝ๐+ 4
3 ๏ฟฝฮ๏ฟฝ๐+ 43 ๏ฟฝโ3
โ43 ๐ฅ3๐+2๏ฟฝ๐+ 2
3 ๏ฟฝฮ๏ฟฝ๐+ 23 ๏ฟฝ
9(๐+1)๏ฟฝ๐+ 23 ๏ฟฝฮ๏ฟฝ๐+ 2
3 ๏ฟฝ๏ฟฝ๐+ 43 ๏ฟฝฮ๏ฟฝ๐+ 4
3 ๏ฟฝ
3โ23 ๐ฅ3๐ฮ๏ฟฝ๐+ 4
3 ๏ฟฝโ3โ43 ๐ฅ3๐+1ฮ๏ฟฝ๐+ 2
3 ๏ฟฝ
ฮ๏ฟฝ๐+ 23 ๏ฟฝฮ๏ฟฝ๐+ 4
3 ๏ฟฝ
Or
ฮ =
3โ23 ๐ฅ3(๐+1)๏ฟฝ๐+ 4
3 ๏ฟฝฮ๏ฟฝ๐+ 43 ๏ฟฝโ3
โ43 ๐ฅ3๐+2๏ฟฝ๐+ 2
3 ๏ฟฝฮ๏ฟฝ๐+ 23 ๏ฟฝ
9(๐+1)๏ฟฝ๐+ 23 ๏ฟฝ๏ฟฝ๐+ 4
3 ๏ฟฝ
3โ23 ๐ฅ3๐ฮ ๏ฟฝ๐ + 4
3๏ฟฝ โ 3
โ43 ๐ฅ3๐+1ฮ ๏ฟฝ๐ + 2
3๏ฟฝ
53
3.1. HW1 CHAPTER 3. HWS
For large๐, we can approximate ๏ฟฝ๐ + 23๏ฟฝ , ๏ฟฝ๐ + 4
3๏ฟฝ , (๐ + 1) to just ๐+1 and the above becomes
ฮ =
3โ23 ๐ฅ3(๐+1)ฮ๏ฟฝ๐+ 4
3 ๏ฟฝโ3โ43 ๐ฅ3๐+2ฮ๏ฟฝ๐+ 2
3 ๏ฟฝ
9(๐+1)2
3โ23 ๐ฅ3๐ฮ ๏ฟฝ๐ + 4
3๏ฟฝ โ 3
โ43 ๐ฅ3๐+1ฮ ๏ฟฝ๐ + 2
3๏ฟฝ
Or
ฮ =3โ23 ๐ฅ3๐๐ฅ3ฮ ๏ฟฝ๐ + 4
3๏ฟฝ โ 3
โ43 ๐ฅ3๐๐ฅ2ฮ ๏ฟฝ๐ + 2
3๏ฟฝ
9 (๐ + 1)2 ๏ฟฝ3โ23 ๐ฅ3๐ฮ ๏ฟฝ๐ + 4
3๏ฟฝ โ 3
โ43 ๐ฅ3๐๐ฅฮ ๏ฟฝ๐ + 2
3๏ฟฝ๏ฟฝ
Or
ฮ =0.488๐ฅ3๐๐ฅ2 ๏ฟฝ๐ฅฮ ๏ฟฝ๐ + 4
3๏ฟฝ โ 0.488ฮ ๏ฟฝ๐ + 2
3๏ฟฝ๏ฟฝ
(0.488) 9 (๐ + 1)2 ๐ฅ3๐ ๏ฟฝฮ ๏ฟฝ๐ + 43๏ฟฝ โ 0.488๐ฅฮ ๏ฟฝ๐ + 2
3๏ฟฝ๏ฟฝ
=๐ฅ2
9 (๐ + 1)2
โโโโโโโโโโ
๐ฅฮ ๏ฟฝ๐ + 43๏ฟฝ โ 0.488ฮ ๏ฟฝ๐ + 2
3๏ฟฝ
ฮ ๏ฟฝ๐ + 43๏ฟฝ โ 0.488๐ฅฮ ๏ฟฝ๐ + 2
3๏ฟฝ
โโโโโโโโโโ
We want to solve for ๐ s.t. ๏ฟฝ๐ฅ2
9(๐+1)2
โโโโโโ๐ฅฮ๏ฟฝ๐+ 4
3 ๏ฟฝโ0.488ฮ๏ฟฝ๐+ 23 ๏ฟฝ
ฮ๏ฟฝ๐+ 43 ๏ฟฝโ0.488๐ฅฮ๏ฟฝ๐+ 2
3 ๏ฟฝ
โโโโโโ ๏ฟฝ < 0.001.
For ๐ฅ = 1
19 (๐ + 1)2
โโโโโโโโโโ
ฮ ๏ฟฝ๐ + 43๏ฟฝ โ 0.488ฮ ๏ฟฝ๐ + 2
3๏ฟฝ
ฮ ๏ฟฝ๐ + 43๏ฟฝ โ 0.488ฮ ๏ฟฝ๐ + 2
3๏ฟฝ
โโโโโโโโโโ < 0.001
19 (๐ + 1)2
< 0.001
9 (๐ + 1)2 > 1000
(๐ + 1)2 >10009
๐ + 1 >๏ฟฝ10009
๐ > 9.541๐ = 10
For ๐ฅ = 100
๏ฟฝ๏ฟฝ
1002
9 (๐ + 1)2
โโโโโโโโโโ
100ฮ ๏ฟฝ๐ + 43๏ฟฝ โ 0.488ฮ ๏ฟฝ๐ + 2
3๏ฟฝ
ฮ ๏ฟฝ๐ + 43๏ฟฝ โ 0.488 (100) ฮ ๏ฟฝ๐ + 2
3๏ฟฝ
โโโโโโโโโโ
๏ฟฝ๏ฟฝ< 0.001
I could not simplify away the Gamma terms above any more. Is there a way? So wrote small
54
3.1. HW1 CHAPTER 3. HWS
function (in Mathematica, which can compute this) which increments ๐ and evaluate theabove, until the value became smaller than 0.001. At ๐ = 11, 500 this was achieved.
For ๐ฅ = 10000
๏ฟฝ๏ฟฝ100002
9 (๐ + 1)2
โโโโโโโโโโ
100ฮ ๏ฟฝ๐ + 43๏ฟฝ โ 0.488ฮ ๏ฟฝ๐ + 2
3๏ฟฝ
ฮ ๏ฟฝ๐ + 43๏ฟฝ โ 0.488 (10000) ฮ ๏ฟฝ๐ + 2
3๏ฟฝ
โโโโโโโโโโ
๏ฟฝ๏ฟฝ< 0.001
Using the same program, found that ๐ = 11, 500, 000 was needed to obtain the result below0.001.
For = โ1 also ๐ = 10. For ๐ฅ = โ100, ๐ = 10, 030, which is little less than ๐ฅ = +100. For๐ฅ = โ10000, ๐ = 10, 030, 000
Summary table for AiryAI(x)
๐ฅ ๐1 10โ1 10100 11, 500โ100 10, 03010000 11, 500, 000โ10000 10, 030, 000
The Mathematica function which did the estimate is the following
estimateAiryAi[x_, n_] := (x^2/(9*(n + 1)^2))*((x*Gamma[n + 4/3] -0.488*Gamma[n + 2/3])/(Gamma[n + 4/3] - 0.488*x*Gamma[n + 2/3]))estimateAiryAi[-10000, 10030000] // N-0.0009995904estimateAiryAi[100, 11500] // N0.000928983646707407estimateAiryAi[10000, 11500000] // N0.000929156800155198
55
3.1. HW1 CHAPTER 3. HWS
3.1.4.2 AiryBI series
For Bi (๐ฅ) the di๏ฟฝerence is the coe๏ฟฝcients. Hence using the result from above, and justreplace the coe๏ฟฝcients
ฮ =3โ16 ๐ฅ3๐๐ฅ3ฮ ๏ฟฝ๐ + 4
3๏ฟฝ โ 3
โ58 ๐ฅ3๐๐ฅ2ฮ ๏ฟฝ๐ + 2
3๏ฟฝ
9 (๐ + 1)2 ๏ฟฝ3โ16 ๐ฅ3๐ฮ ๏ฟฝ๐ + 4
3๏ฟฝ โ 3
โ58 ๐ฅ3๐๐ฅฮ ๏ฟฝ๐ + 2
3๏ฟฝ๏ฟฝ
=๐ฅ2 ๏ฟฝ๐ฅฮ ๏ฟฝ๐ + 4
3๏ฟฝ โ 0.604 39ฮ ๏ฟฝ๐ + 2
3๏ฟฝ๏ฟฝ
9 (๐ + 1)2 ๏ฟฝฮ ๏ฟฝ๐ + 43๏ฟฝ โ 0.604 39๐ฅฮ ๏ฟฝ๐ + 2
3๏ฟฝ๏ฟฝ
Hence for ๐ฅ = 1, using the above reduces to1
9 (๐ + 1)2< 0.001
Which is the same as AiryAI, therefore ๐ = 10. For ๐ฅ = 100, using the same small functionin Mathematica to calculate the above, here are the result.
Summary table for AiryBI(x)
๐ฅ ๐1 10โ1 10100 11, 290โ100 9, 90010000 10, 900, 000โ10000 9, 950, 000
The result between AiryAi and AiryBi are similar. AiryBi needs a slightly less number ofterms in the series to obtain same accuracy.
The Mathematica function which did the estimate for the larger ๐ value for the above tableis the following
estimateAiryBI[x_, n_] := Abs[(x^2/(9*(n + 1)^2))*((x*Gamma[n + 4/3] -0.60439*Gamma[n + 2/3])/(Gamma[n + 4/3] - 0.6439*x*Gamma[n + 2/3]))]
3.1.5 Problem 3.8
Problem How many terms in the Taylor series solution to ๐ฆโฒโฒโฒ = ๐ฅ3๐ฆ with ๐ฆ (0) = 1, ๐ฆโฒ (0) =๐ฆโฒโฒ (0) = 0 are needed to evaluate โซ
1
0๐ฆ (๐ฅ) ๐๐ฅ correct to three decimal places?
Answer
๐ฆโฒโฒโฒ โ ๐ฅ3๐ฆ = 056
3.1. HW1 CHAPTER 3. HWS
Since ๐ฅ is an ordinary point, we use
๐ฆ =โ๏ฟฝ๐=0
๐๐๐ฅ๐
๐ฆโฒ =โ๏ฟฝ๐=0
๐๐๐๐ฅ๐โ1 =โ๏ฟฝ๐=1
๐๐๐๐ฅ๐โ1 =โ๏ฟฝ๐=0
(๐ + 1) ๐๐+1๐ฅ๐
๐ฆโฒโฒ =โ๏ฟฝ๐=0
๐ (๐ + 1) ๐๐+1๐ฅ๐โ1 =โ๏ฟฝ๐=1
๐ (๐ + 1) ๐๐+1๐ฅ๐โ1 =โ๏ฟฝ๐=0
(๐ + 1) (๐ + 2) ๐๐+2๐ฅ๐
๐ฆโฒโฒโฒ =โ๏ฟฝ๐=0
๐ (๐ + 1) (๐ + 2) ๐๐+2๐ฅ๐โ1 =โ๏ฟฝ๐=1
๐ (๐ + 1) (๐ + 2) ๐๐+2๐ฅ๐โ1
=โ๏ฟฝ๐=0
(๐ + 1) (๐ + 2) (๐ + 3) ๐๐+3๐ฅ๐
Hence the ODE becomesโ๏ฟฝ๐=0
(๐ + 1) (๐ + 2) (๐ + 3) ๐๐+3๐ฅ๐ โ ๐ฅ3โ๏ฟฝ๐=0
๐๐๐ฅ๐ = 0
โ๏ฟฝ๐=0
(๐ + 1) (๐ + 2) (๐ + 3) ๐๐+3๐ฅ๐ โโ๏ฟฝ๐=0
๐๐๐ฅ๐+3 = 0
โ๏ฟฝ๐=0
(๐ + 1) (๐ + 2) (๐ + 3) ๐๐+3๐ฅ๐ โโ๏ฟฝ๐=3
๐๐โ3๐ฅ๐ = 0
For ๐ = 0(1) (2) (3) ๐3 = 0
๐3 = 0
For ๐ = 1(2) (3) (4) ๐4 = 0
๐4 = 0
For ๐ = 2
๐5 = 0
For ๐ โฅ 3, recursive equation is used
(๐ + 1) (๐ + 2) (๐ + 3) ๐๐+3 โ ๐๐โ3 = 0
๐๐+3 =๐๐โ3
(๐ + 1) (๐ + 2) (๐ + 3)Or, for ๐ = 3
๐6 =๐0
(4) (5) (6)For ๐ = 4
๐7 =๐1
(4 + 1) (4 + 2) (4 + 3)=
๐1(5) (6) (7)
For ๐ = 5
๐8 =๐2
(5 + 1) (5 + 2) (5 + 3)=
๐2(6) (7) (8)
57
3.1. HW1 CHAPTER 3. HWS
For ๐ = 6
๐9 =๐3
(6 + 1) (6 + 2) (6 + 3)= 0
For ๐ = 7
๐10 =๐4
(๐ + 1) (๐ + 2) (๐ + 3)= 0
For ๐ = 8
๐11 =๐5
(8 + 1) (8 + 2) (8 + 3)= 0
For ๐ = 9
๐12 =๐6
(9 + 1) (9 + 2) (9 + 3)=
๐6(10) (11) (12)
=๐0
(4) (5) (6) (10) (11) (12)For ๐ = 10
๐13 =๐7
(10 + 1) (10 + 2) (10 + 3)=
๐7(11) (12) (13)
=๐1
(5) (6) (7) (11) (12) (13)For ๐ = 11
๐14 =๐8
(11 + 1) (11 + 2) (11 + 3)=
๐8(12) (13) (14)
=๐2
(6) (7) (8) (12) (13) (14)And so on. Hence the series is
๐ฆ =โ๏ฟฝ๐=0
๐๐๐ฅ๐
= ๐0 + ๐1๐ฅ + ๐2๐ฅ2 + 0๐ฅ3 + 0๐ฅ4 + 0๐ฅ5 +๐0
(4) (5) (6)๐ฅ6 +
๐1(5) (6) (7)
๐ฅ7 +๐2
(6) (7) (8)๐ฅ8
+ 0 + 0 + 0 +๐0
(4) (5) (6) (10) (11) (12)๐ฅ12 +
๐1(5) (6) (7) (11) (12) (13)
๐ฅ13
+๐2
(6) (7) (8) (12) (13) (14)๐ฅ14 + 0 + 0 + 0 +โฏ
Or
๐ฆ (๐ฅ) = ๐0 + ๐1๐ฅ + ๐2๐ฅ2 +๐0
(4) (5) (6)๐ฅ6 +
๐1(5) (6) (7)
๐ฅ7 +๐2
(6) (7) (8)๐ฅ8+
๐0(4) (5) (6) (10) (11) (12)
๐ฅ12 +๐1
(5) (6) (7) (11) (12) (13)๐ฅ13 +
๐2(6) (7) (8) (12) (13) (14)
๐ฅ14 +โฏ
Or
๐ฆ (๐ฅ) = ๐0 ๏ฟฝ1 +๐ฅ6
(4) (5) (6)+
๐ฅ12
(4) (5) (6) (10) (11) (12)+โฏ๏ฟฝ
+ ๐1 ๏ฟฝ๐ฅ +๐ฅ7
(5) (6) (7)+
๐ฅ13
(5) (6) (7) (11) (12) (13)+โฏ๏ฟฝ
+ ๐2 ๏ฟฝ๐ฅ2 +๐ฅ8
(6) (7) (8)+
๐ฅ14
(6) (7) (8) (12) (13) (14)+โฏ๏ฟฝ
58
3.1. HW1 CHAPTER 3. HWS
๐ฆ (๐ฅ) = ๐0 ๏ฟฝ1 +1120
๐ฅ6 +1
158400๐ฅ12 +โฏ๏ฟฝ
+ ๐1 ๏ฟฝ๐ฅ +1210
๐ฅ7 +1
360360๐ฅ13 +โฏ๏ฟฝ
+ ๐2 ๏ฟฝ๐ฅ2 +1336
๐ฅ8 +1
733 824๐ฅ14 +โฏ๏ฟฝ
We now apply initial conditions ๐ฆ (0) = 1, ๐ฆโฒ (0) = ๐ฆโฒโฒ (0) = 0 . When ๐ฆ (0) = 1
1 = ๐0Hence solution becomes
๐ฆ (๐ฅ) = ๏ฟฝ1 +1120
๐ฅ6 +1
158400๐ฅ12 +โฏ๏ฟฝ
+ ๐1 ๏ฟฝ๐ฅ +1210
๐ฅ7 +1
360360๐ฅ13 +โฏ๏ฟฝ
+ ๐2 ๏ฟฝ๐ฅ2 +1336
๐ฅ8 +1
733 824๐ฅ14 +โฏ๏ฟฝ
Taking derivative
๐ฆโฒ (๐ฅ) = ๏ฟฝ6120
๐ฅ5 +12
158400๐ฅ11 +โฏ๏ฟฝ
+ ๐1 ๏ฟฝ1 +7210
๐ฅ6 +13
360360๐ฅ12 +โฏ๏ฟฝ
+ ๐2 ๏ฟฝ2๐ฅ +8336
๐ฅ7 +12
733 824๐ฅ13 +โฏ๏ฟฝ
Applying ๐ฆโฒ (0) = 0 gives
0 = ๐1And similarly, Applying ๐ฆโฒโฒ (0) = 0 gives ๐2 = 0. Hence the solution is
๐ฆ (๐ฅ) = ๐0 ๏ฟฝ1 +๐ฅ6
(4) (5) (6)+
๐ฅ12
(4) (5) (6) (10) (11) (12)+โฏ๏ฟฝ
= 1 +๐ฅ6
(4) (5) (6)+
๐ฅ12
(4) (5) (6) (10) (11) (12)+
๐ฅ18
(4) (5) (6) (10) (11) (12) (16) (17) (18)+โฏ
= 1 +๐ฅ6
120+
๐ฅ12
158 400+
๐ฅ18
775 526 400+โฏ
We are now ready to answer the question. We will do the integration by increasing thenumber of terms by one each time. When the absolute di๏ฟฝerence between each incrementbecomes less than 0.001 we stop. When using one term For
๏ฟฝ1
0๐ฆ (๐ฅ) ๐๐ฅ = ๏ฟฝ
1
0๐๐ฅ
= 1
59
3.1. HW1 CHAPTER 3. HWS
When using two terms
๏ฟฝ1
01 +
๐ฅ6
120๐๐ฅ = ๏ฟฝ๐ฅ +
๐ฅ7
120 (7)๏ฟฝ1
0
= 1 +1840
=841840
= 1.001190476
Di๏ฟฝerence between one term and two terms is 0.001190476. When using three terms
๏ฟฝ1
01 +
๐ฅ6
120+
๐ฅ12
158 400๐๐ฅ = ๏ฟฝ๐ฅ +
๐ฅ7
840+
๐ฅ13
2059 200๏ฟฝ1
0
= 1 +1840
+1
2059 200
=14 431 56714 414 400
= 1.001190962
Comparing the above result, with the result using two terms, we see that only two terms areneeded since the change in accuracy did not a๏ฟฝect the first three decimal points. Hence weneed only this solution with two terms only
๐ฆ (๐ฅ) = 1 +1120
๐ฅ6
3.1.6 Problem 3.24
3.1.6.1 part e
Problem Find series expansion of all the solutions to the following di๏ฟฝerential equationabout ๐ฅ = 0. Try to sum in closed form any infinite series that appear.
2๐ฅ๐ฆโฒโฒ โ ๐ฆโฒ + ๐ฅ2๐ฆ = 0
Solution
๐ฆโฒโฒ โ12๐ฅ๐ฆโฒ +
๐ฅ2๐ฆ = 0
The only singularity is in ๐ (๐ฅ) is ๐ฅ = 0. We will now check if it is removable. (i.e. regular)
lim๐ฅโ0
๐ฅ๐ (๐ฅ) = lim๐ฅโ0
๐ฅ12๐ฅ
=12
60
3.1. HW1 CHAPTER 3. HWS
Therefore ๐ฅ = 0 is regular singular point. Hence we try Frobenius series
๐ฆ (๐ฅ) =โ๏ฟฝ๐=0
๐๐๐ฅ๐+๐
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 in 2๐ฅ2๐ฆโฒโฒ โ ๐ฅ๐ฆโฒ + ๐ฅ3๐ฆ = 0 results in
2๐ฅ2โ๏ฟฝ๐=0
(๐ + ๐) (๐ + ๐ โ 1) ๐๐๐ฅ๐+๐โ2 โ ๐ฅโ๏ฟฝ๐=0
(๐ + ๐) ๐๐๐ฅ๐+๐โ1 + ๐ฅ3โ๏ฟฝ๐=0
๐๐๐ฅ๐+๐ = 0
โ๏ฟฝ๐=0
2 (๐ + ๐) (๐ + ๐ โ 1) ๐๐๐ฅ๐+๐ โโ๏ฟฝ๐=0
(๐ + ๐) ๐๐๐ฅ๐+๐ +โ๏ฟฝ๐=0
๐๐๐ฅ๐+๐+3 = 0
โ๏ฟฝ๐=0
2 (๐ + ๐) (๐ + ๐ โ 1) ๐๐๐ฅ๐+๐ โโ๏ฟฝ๐=0
(๐ + ๐) ๐๐๐ฅ๐+๐ +โ๏ฟฝ๐=3
๐๐โ3๐ฅ๐+๐ = 0 (1)
The first step is to obtain the indicial equation. As the nature of the roots will tell us howto proceed. The indicial equation is obtained from setting ๐ = 0 in (1) with the assumptionthat ๐0 โ 0. Setting ๐ = 0 in (1) gives
2 (๐ + ๐) (๐ + ๐ โ 1) ๐๐ โ (๐ + ๐) ๐๐ = 02 (๐) (๐ โ 1) ๐0 โ ๐๐0 = 0
Since ๐0 โ 0 then we obtain the indicial equation (quadratic in ๐)
2 (๐) (๐ โ 1) โ ๐ = 0๐ (2 (๐ โ 1) โ 1) = 0
๐ (2๐ โ 3) = 0
Hence roots are
๐1 = 0
๐2 =32
Since ๐1 โ ๐2 is not an integer, then we know we can now construct two linearly independentsolutions
๐ฆ1 (๐ฅ) = ๐ฅ๐1๏ฟฝ๐=0๐๐๐ฅ๐
๐ฆ2 (๐ฅ) = ๐ฅ๐2๏ฟฝ๐=0๐๐๐ฅ๐
Or
๐ฆ1 (๐ฅ) = ๏ฟฝ๐=0๐๐๐ฅ๐
๐ฆ2 (๐ฅ) = ๏ฟฝ๐=0๐๐๐ฅ
๐+ 32
61
3.1. HW1 CHAPTER 3. HWS
Notice that the coe๏ฟฝcients are not the same. Since we now know ๐1, ๐2, we will use the aboveseries solution to obtain ๐ฆ1 (๐ฅ) and ๐ฆ2 (๐ฅ).
For ๐ฆ1 (๐ฅ) where ๐1 = 0
๐ฆ1 (๐ฅ) = ๏ฟฝ๐=0๐๐๐ฅ๐
๐ฆโฒ (๐ฅ) = ๏ฟฝ๐=0๐๐๐๐ฅ๐โ1 = ๏ฟฝ
๐=1๐๐๐๐ฅ๐โ1 = ๏ฟฝ
๐=0(๐ + 1) ๐๐+1๐ฅ๐
๐ฆโฒโฒ (๐ฅ) = ๏ฟฝ๐=0๐ (๐ + 1) ๐๐+1๐ฅ๐โ1 = ๏ฟฝ
๐=1๐ (๐ + 1) ๐๐+1๐ฅ๐โ1 = ๏ฟฝ
๐=0(๐ + 1) (๐ + 2) ๐๐+2๐ฅ๐
Substituting the above in 2๐ฅ2๐ฆโฒโฒ โ ๐ฅ๐ฆโฒ + ๐ฅ3๐ฆ = 0 results in
2๐ฅ2๏ฟฝ๐=0
(๐ + 1) (๐ + 2) ๐๐+2๐ฅ๐ โ ๐ฅ๏ฟฝ๐=0
(๐ + 1) ๐๐+1๐ฅ๐ + ๐ฅ3๏ฟฝ๐=0๐๐๐ฅ๐ = 0
๏ฟฝ๐=02 (๐ + 1) (๐ + 2) ๐๐+2๐ฅ๐+2 โ๏ฟฝ
๐=0(๐ + 1) ๐๐+1๐ฅ๐+1 +๏ฟฝ
๐=0๐๐๐ฅ๐+3 = 0
๏ฟฝ๐=22 (๐ โ 1) (๐) ๐๐๐ฅ๐ โ๏ฟฝ
๐=1๐๐๐๐ฅ๐ +๏ฟฝ
๐=3๐๐โ3๐ฅ๐ = 0
For ๐ = 1 (index starts at ๐ = 1).
โ๐๐๐๐ฅ๐ = 0โ๐1 = 0๐1 = 0
For ๐ = 2
2 (๐ โ 1) (๐) ๐๐๐ฅ๐ โ ๐๐๐๐ฅ๐ = 02 (2 โ 1) (2) ๐2 โ 2๐2 = 0
2๐2 = 0๐2 = 0
For ๐ โฅ 3 we have recursive formula
2 (๐ โ 1) (๐) ๐๐ โ ๐๐๐ + ๐๐โ3 = 0
๐๐ =โ๐๐โ3
๐ (2๐ โ 3)Hence, for ๐ = 3
๐3 =โ๐0
(3) (6 โ 3)=โ๐09
For ๐ = 4
๐4 =โ๐1
๐ (2๐ โ 3)= 0
For ๐ = 5
๐5 =โ๐2
๐ (2๐ โ 3)= 0
For ๐ = 6
๐6 =โ๐3
๐ (2๐ โ 3)=
โ๐3(6) (12 โ 3)
=โ๐354
=๐0
(54) (9)=
1486
๐0
62
3.1. HW1 CHAPTER 3. HWS
And for ๐ = 7, 8 we also obtain ๐7 = 0, ๐8 = 0, but for ๐9
๐9 =โ๐6
9 (2 (9) โ 3)=โ๐6135
=โ๐0
(135) (486)= โ
165 610
๐0
And so on. Hence from ๏ฟฝ๐=0๐๐๐ฅ๐ we obtain
๐ฆ1 (๐ฅ) = ๐0 โโ๐09๐ฅ3 +
1486
๐0๐ฅ6 โ1
65 610๐0๐ฅ9 +โฏ
= ๐0 ๏ฟฝ1 โ19๐ฅ3 +
1486
๐ฅ6 โ1
65 610๐ฅ9 +โฏ๏ฟฝ
Now that we found ๐ฆ1 (๐ฅ).
For ๐ฆ2 (๐ฅ) with ๐2 =32 .
๐ฆ (๐ฅ) = ๏ฟฝ๐=0๐๐๐ฅ
๐+ 32
๐ฆโฒ (๐ฅ) = ๏ฟฝ๐=0
๏ฟฝ๐ +32๏ฟฝ๐๐๐ฅ
๐+ 12
๐ฆโฒโฒ (๐ฅ) = ๏ฟฝ๐=0
๏ฟฝ๐ +12๏ฟฝ ๏ฟฝ
๐ +32๏ฟฝ๐๐๐ฅ
๐โ 12
Substituting this into 2๐ฅ2๐ฆโฒโฒ โ ๐ฅ๐ฆโฒ + ๐ฅ3๐ฆ = 0 gives
2๐ฅ2๏ฟฝ๐=0
๏ฟฝ๐ +12๏ฟฝ ๏ฟฝ
๐ +32๏ฟฝ๐๐๐ฅ
๐โ 12 โ ๐ฅ๏ฟฝ
๐=0๏ฟฝ๐ +
32๏ฟฝ๐๐๐ฅ
๐+ 12 + ๐ฅ3๏ฟฝ
๐=0๐๐๐ฅ
๐+ 32 = 0
๏ฟฝ๐=02 ๏ฟฝ๐ +
12๏ฟฝ ๏ฟฝ
๐ +32๏ฟฝ๐๐๐ฅ
๐โ 12+2 โ๏ฟฝ
๐=0๏ฟฝ๐ +
32๏ฟฝ๐๐๐ฅ
๐+ 12+1 +๏ฟฝ
๐=0๐๐๐ฅ
๐+ 32+3 = 0
๏ฟฝ๐=02 ๏ฟฝ๐ +
12๏ฟฝ ๏ฟฝ
๐ +32๏ฟฝ๐๐๐ฅ
๐+ 32 โ๏ฟฝ
๐=0๏ฟฝ๐ +
32๏ฟฝ๐๐๐ฅ
๐+ 32 +๏ฟฝ
๐=0๐๐๐ฅ
๐+ 92 = 0
๏ฟฝ๐=02 ๏ฟฝ๐ +
12๏ฟฝ ๏ฟฝ
๐ +32๏ฟฝ๐๐๐ฅ
๐+ 32 โ๏ฟฝ
๐=0๏ฟฝ๐ +
32๏ฟฝ๐๐๐ฅ
๐+ 32 +๏ฟฝ
๐=3๐๐โ3๐ฅ
๐+ 92โ3 = 0
๏ฟฝ๐=02 ๏ฟฝ๐ +
12๏ฟฝ ๏ฟฝ
๐ +32๏ฟฝ๐๐๐ฅ
๐+ 32 โ๏ฟฝ
๐=0๏ฟฝ๐ +
32๏ฟฝ๐๐๐ฅ
๐+ 32 +๏ฟฝ
๐=3๐๐โ3๐ฅ
๐+ 32 = 0
Now that all the ๐ฅ terms have the same exponents, we can continue.
For ๐ = 0
2 ๏ฟฝ๐ +12๏ฟฝ ๏ฟฝ
๐ +32๏ฟฝ๐0๐ฅ
32 โ ๏ฟฝ๐ +
32๏ฟฝ๐0๐ฅ
32 = 0
2 ๏ฟฝ12๏ฟฝ ๏ฟฝ
32๏ฟฝ๐0๐ฅ
32 โ ๏ฟฝ
32๏ฟฝ๐0๐ฅ
32 = 0
32๐0๐ฅ
32 โ ๏ฟฝ
32๏ฟฝ๐0๐ฅ
32 = 0
0๐0 = 0
63
3.1. HW1 CHAPTER 3. HWS
Hence ๐0 is arbitrary.
For ๐ = 1
2 ๏ฟฝ๐ +12๏ฟฝ ๏ฟฝ
๐ +32๏ฟฝ๐๐๐ฅ
๐+ 32 โ ๏ฟฝ๐ +
32๏ฟฝ๐๐๐ฅ
๐+ 32 = 0
2 ๏ฟฝ1 +12๏ฟฝ ๏ฟฝ
1 +32๏ฟฝ๐1 โ ๏ฟฝ1 +
32๏ฟฝ๐1 = 0
5๐1 = 0๐1 = 0
For ๐ = 2
2 ๏ฟฝ๐ +12๏ฟฝ ๏ฟฝ
๐ +32๏ฟฝ๐๐๐ฅ
๐+ 32 โ ๏ฟฝ๐ +
32๏ฟฝ๐๐๐ฅ
๐+ 32 = 0
2 ๏ฟฝ2 +12๏ฟฝ ๏ฟฝ
2 +32๏ฟฝ๐2 โ ๏ฟฝ2 +
32๏ฟฝ๐2 = 0
14๐2 = 0๐2 = 0
For ๐ โฅ 3 we use the recursive formula
2 ๏ฟฝ๐ +12๏ฟฝ ๏ฟฝ
๐ +32๏ฟฝ๐๐ โ ๏ฟฝ๐ +
32๏ฟฝ๐๐ + ๐๐โ3 = 0
๐๐ =โ๐๐โ3
๐ (2๐ + 3)Hence for ๐ = 3
๐3 =โ๐03 (9)
=โ๐027
For ๐ = 4, ๐ = 5 we will get ๐4 = 0 and ๐5 = 0 since ๐1 = 0 and ๐2 = 0.
For ๐ = 6
๐6 =โ๐3
6 (12 + 3)=โ๐390
=๐0
27 (90)=
๐02430
For ๐ = 7, ๐ = 8 we will get ๐7 = 0 and ๐8 = 0 since ๐4 = 0 and ๐5 = 0
For ๐ = 9
๐9 =โ๐6
๐ (2๐ + 3)=
โ๐69 (18 + 3)
=โ๐6189
=โ๐0
2430 (189)=
โ๐0459270
And so on. Hence, from ๐ฆ2 (๐ฅ) = ๏ฟฝ๐=0๐๐๐ฅ
๐+ 32 the series is
๐ฆ2 (๐ฅ) = ๐0๐ฅ32 โ
๐027๐ฅ3+
32 +
๐02430
๐ฅ6+32 โ
๐0459 270
๐ฅ9+32 +โฏ
= ๐0๐ฅ32 ๏ฟฝ1 โ
๐ฅ3
27+
๐ฅ6
2430โ
๐ฅ9
459270+โฏ๏ฟฝ
64
3.1. HW1 CHAPTER 3. HWS
The final solution is
๐ฆ (๐ฅ) = ๐ฆ1 (๐ฅ) + ๐ฆ2 (๐ฅ)
Or
๐ฆ (๐ฅ) = ๐0 ๏ฟฝ1 โ19๐ฅ
3 + 1486๐ฅ
6 โ 165 610๐ฅ
9 +โฏ๏ฟฝ + ๐0๐ฅ32 ๏ฟฝ1 โ ๐ฅ3
27 +๐ฅ6
2430 โ๐ฅ9
459270 +โฏ๏ฟฝ (2)
Now comes the hard part. Finding closed form solution.
The Taylor series of cos ๏ฟฝ13๐ฅ32โ2๏ฟฝ is (using CAS)
cos ๏ฟฝ13๐ฅ32โ2๏ฟฝ โ 1 โ
19๐ฅ3 +
1486
๐ฅ6 โ1
65610๐ฅ9 +โฏ (3)
And the Taylor series for sin ๏ฟฝ13๐ฅ32โ2๏ฟฝ is (Using CAS)
sin ๏ฟฝ13๐ฅ32โ2๏ฟฝ โ
13๐ฅ32โ2 โ
181๐ฅ92โ2 +
17290
๐ฅ152 โ2 โโฏ
= ๐ฅ32 ๏ฟฝ13โ
2 โ181๐ฅ3โ2 +
17290
๐ฅ6โ2 โโฏ๏ฟฝ
=13โ
2๐ฅ32 ๏ฟฝ1 โ
127๐ฅ3 +
12430
๐ฅ6 โโฏ๏ฟฝ (4)
Comparing (3,4) with (2) we see that (2) can now be written as
๐ฆ (๐ฅ) = ๐0 cos ๏ฟฝ13๐ฅ32โ2๏ฟฝ +
๐013โ2
sin ๏ฟฝ13๐ฅ32โ2๏ฟฝ
Or letting ๐ = ๐013โ2
๐ฆ (๐ฅ) = ๐0 cos ๏ฟฝ13๐ฅ32โ2๏ฟฝ + ๐0 sin ๏ฟฝ13๐ฅ
32โ2๏ฟฝ
This is the closed form solution. The constants ๐0, ๐0 can be found from initial conditions.
3.1.6.2 part f
Problem Find series expansion of all the solutions to the following di๏ฟฝerential equationabout ๐ฅ = 0. Try to sum in closed form any infinite series that appear.
(sin ๐ฅ) ๐ฆโฒโฒ โ 2 (cos ๐ฅ) ๐ฆโฒ โ (sin ๐ฅ) ๐ฆ = 0Solution
In standard form
๐ฆโฒโฒ โ 2 ๏ฟฝcos ๐ฅsin ๐ฅ
๏ฟฝ ๐ฆโฒ โ ๐ฆ = 0
65
3.1. HW1 CHAPTER 3. HWS
Hence the singularities are in ๐ (๐ฅ) only and they occur when sin ๐ฅ = 0 or ๐ฅ = 0, ยฑ๐, ยฑ2๐,โฏbut we just need to consider ๐ฅ = 0. Let us check if the singularity is removable.
lim๐ฅโ0
๐ฅ๐ (๐ฅ) = 2 lim๐ฅโ0
๐ฅcos ๐ฅsin ๐ฅ = 2 lim
๐ฅโ0๐ฅ1 โ ๐ฅ2
2 +๐ฅ4
4! โโฏ
๐ฅ โ ๐ฅ3
3! +๐ฅ5
5! โโฏ= 2 lim
๐ฅโ0
1 โ ๐ฅ2
2 +๐ฅ4
4! โโฏ
1 โ ๐ฅ2
3! +๐ฅ4
5! โโฏ= 2
Hence the singularity is regular. So we can use Frobenius series
๐ฆ (๐ฅ) =โ๏ฟฝ๐=0
๐๐๐ฅ๐+๐
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 in sin (๐ฅ) ๐ฆโฒโฒ โ 2 cos (๐ฅ) ๐ฆโฒ โ sin (๐ฅ) ๐ฆ = 0 results in
sin (๐ฅ)โ๏ฟฝ๐=0
(๐ + ๐) (๐ + ๐ โ 1) ๐๐๐ฅ๐+๐โ2 โ 2 cos (๐ฅ)โ๏ฟฝ๐=0
(๐ + ๐) ๐๐๐ฅ๐+๐โ1 โ sin (๐ฅ)โ๏ฟฝ๐=0
๐๐๐ฅ๐+๐ = 0
Now using Taylor series for sin ๐ฅ, cos ๐ฅ expanded around 0, the above becomesโ๏ฟฝ๐=0
(โ1)๐๐ฅ2๐+1
(2๐ + 1)!
โ๏ฟฝ๐=0
(๐ + ๐) (๐ + ๐ โ 1) ๐๐๐ฅ๐+๐โ2
โ2โ๏ฟฝ๐=0
(โ1)๐๐ฅ2๐
(2๐)!
โ๏ฟฝ๐=0
(๐ + ๐) ๐๐๐ฅ๐+๐โ1
โโ๏ฟฝ๐=0
(โ1)๐๐ฅ2๐+1
(2๐ + 1)!
โ๏ฟฝ๐=0
๐๐๐ฅ๐+๐ = 0 (1)
We need now to evaluate products of power series. Using what is called Cauchy product rule,where
๐ (๐ฅ) ๐ (๐ฅ) = ๏ฟฝโ๏ฟฝ๐=0
๐๐๐ฅ๐๏ฟฝ ๏ฟฝโ๏ฟฝ๐=0
๐๐๐ฅ๐๏ฟฝ =โ๏ฟฝ๐=0
โ๏ฟฝ๐=0
๐๐๐๐๐ฅ๐+๐ (2)
Applying (2) to first term in (1) givesโ๏ฟฝ๐=0
(โ1)๐๐ฅ2๐+1
(2๐ + 1)!
โ๏ฟฝ๐=0
(๐ + ๐) (๐ + ๐ โ 1) ๐๐๐ฅ๐+๐โ2 =โ๏ฟฝ๐=0
โ๏ฟฝ๐=0
(โ1)๐ (๐ + ๐) (๐ + ๐ โ 1)(2๐ + 1)!
๐๐๐ฅ2๐+๐+๐โ1
(3)
Applying (2) to second term in (1) givesโ๏ฟฝ๐=0
(โ1)๐๐ฅ2๐
(2๐)!
โ๏ฟฝ๐=0
(๐ + ๐) ๐๐๐ฅ๐+๐โ1 =โ๏ฟฝ๐=0
โ๏ฟฝ๐=0
(โ1)๐ (๐ + ๐)(2๐)!
๐๐๐ฅ2๐+๐+๐โ1 (4)
Applying (2) to the last term in (1) givesโ๏ฟฝ๐=0
(โ1)๐๐ฅ2๐+1
(2๐ + 1)!
โ๏ฟฝ๐=0
๐๐๐ฅ๐+๐ =โ๏ฟฝ๐=0
โ๏ฟฝ๐=0
(โ1)๐ ๐๐(2๐ + 1)!
๐ฅ2๐+๐+๐+1 (5)
66
3.1. HW1 CHAPTER 3. HWS
Substituting (3,4,5) back into (1) givesโ๏ฟฝ๐=0
โ๏ฟฝ๐=0
(โ1)๐ (๐ + ๐) (๐ + ๐ โ 1)(2๐ + 1)!
๐๐๐ฅ2๐+๐+๐โ1
โ2โ๏ฟฝ๐=0
โ๏ฟฝ๐=0
(โ1)๐ (๐ + ๐)(2๐)!
๐๐๐ฅ2๐+๐+๐โ1
โโ๏ฟฝ๐=0
โ๏ฟฝ๐=0
(โ1)๐ ๐๐(2๐ + 1)!
๐ฅ2๐+๐+๐+1 = 0
We now need to make all ๐ฅ exponents the same. This givesโ๏ฟฝ๐=0
โ๏ฟฝ๐=0
(โ1)๐ (๐ + ๐) (๐ + ๐ โ 1)(2๐ + 1)!
๐๐๐ฅ2๐+๐+๐โ1
โ2โ๏ฟฝ๐=0
โ๏ฟฝ๐=0
(โ1)๐ (๐ + ๐)(2๐)!
๐๐๐ฅ2๐+๐+๐โ1
โโ๏ฟฝ๐=0
โ๏ฟฝ๐=2
(โ1)๐ ๐๐โ2(2๐ + 1)!
๐ฅ2๐+๐+๐โ1 = 0 (6)
The first step is to obtain the indicial equation. As the nature of the roots will tell us how toproceed. The indicial equation is obtained from setting ๐ = ๐ = 0 in (6) with the assumptionthat ๐0 โ 0. This results in
(โ1)๐ (๐ + ๐) (๐ + ๐ โ 1)(2๐ + 1)!
๐๐๐ฅ2๐+๐+๐โ1 โ 2(โ1)๐ (๐ + ๐)
(2๐)!๐๐๐ฅ2๐+๐+๐โ1 = 0
(๐) (๐ โ 1) ๐0 โ 2๐๐0 = 0
๐0 ๏ฟฝ๐2 โ ๐ โ 2๐๏ฟฝ = 0
Since ๐0 โ 0 then we obtain the indicial equation (quadratic in ๐)
๐2 โ 3๐ = 0๐ (๐ โ 3) = 0
Hence roots are
๐1 = 3๐2 = 0
(it is always easier to make ๐1 > ๐2). Since ๐1 โ ๐2 = 3 is now an integer, then this is case IIpart (b) in textbook, page 72. In this case, the two linearly independent solutions are
๐ฆ1 (๐ฅ) = ๐ฅ3๏ฟฝ๐=0๐๐๐ฅ๐ = ๏ฟฝ
๐=0๐๐๐ฅ๐+3
๐ฆ2 (๐ฅ) = ๐๐ฆ1 (๐ฅ) ln (๐ฅ) + ๐ฅ๐2๏ฟฝ๐=0๐๐๐ฅ๐ = ๐๐ฆ1 (๐ฅ) ln (๐ฅ) +๏ฟฝ
๐=0๐๐๐ฅ๐
67
3.1. HW1 CHAPTER 3. HWS
Where ๐ is some constant. Now we will find ๐ฆ1. From (6), where now we set ๐ = 3โ๏ฟฝ๐=0
โ๏ฟฝ๐=0
(โ1)๐ (๐ + 3) (๐ + 2)(2๐ + 1)!
๐๐๐ฅ2๐+๐++2
โ2โ๏ฟฝ๐=0
โ๏ฟฝ๐=0
(โ1)๐ (๐ + 3)(2๐)!
๐๐๐ฅ2๐+๐+2
โโ๏ฟฝ๐=0
โ๏ฟฝ๐=2
(โ1)๐ ๐๐โ2(2๐ + 1)!
๐ฅ2๐+๐+2 = 0
For ๐ = 0, ๐ = 1(โ1)๐ (๐ + 3) (๐ + 2)
(2๐ + 1)!๐๐ โ 2 ๏ฟฝ
(โ1)๐ (๐ + 3)(2๐)!
๐๐๏ฟฝ = 0
(4) (3) ๐1 โ 2 (4๐1) = 0๐1 = 0
For ๐ = 0, ๐ โฅ 2 we obtain recursive equation
(๐ + 3) (๐ + 2) ๐๐ โ 2 (๐ + 3) ๐๐ โ ๐๐โ2 = 0
๐๐ =โ๐๐โ2
(๐ + 3) (๐ + 2) โ 2 (๐ + 3)=
โ๐๐โ2๐ (๐ + 3)
Hence, for ๐ = 0, ๐ = 2
๐2 =โ๐010
For ๐ = 0, ๐ = 3
๐3 =โ๐1
๐ (๐ + 3)= 0
For ๐ = 0, ๐ = 4
๐4 =โ๐24 (7)
=๐0280
For ๐ = 0, ๐ = 5
๐5 = 0
For ๐ = 0, ๐ = 6
๐6 =โ๐4
6 (6 + 3)=
โ๐06 (6 + 3) (280)
=โ๐015120
For ๐ = 0, ๐ = 7, ๐7 = 0 and for ๐ = 0, ๐ = 8
๐8 =โ๐6
8 (8 + 3)=
11330560
๐0
68
3.1. HW1 CHAPTER 3. HWS
And so on. Since ๐ฆ1 (๐ฅ) = ๏ฟฝ๐=0๐๐๐ฅ๐+3, then the first solution is now found. It is
๐ฆ1 (๐ฅ) = ๐0๐ฅ3 + ๐1๐ฅ4 + ๐2๐ฅ5 + ๐3๐ฅ6 + ๐4๐ฅ7 + ๐5๐ฅ8 + ๐6๐ฅ9 + ๐7๐ฅ10 + ๐8๐ฅ11 +โฏ
= ๐0๐ฅ3 + 0 โ๐010๐ฅ5 + 0 +
๐0280
๐ฅ7 + 0 โ๐0
15 120๐ฅ9 + 0 +
11330560
๐0๐ฅ11 +โฏ
= ๐0๐ฅ3 ๏ฟฝ1 โ๐ฅ2
10+๐ฅ5
280โ
๐ฅ6
15120+
๐ฅ8
1330560โโฏ๏ฟฝ (7)
The second solution can now be found from
๐ฆ2 = ๐๐ฆ1 (๐ฅ) ln (๐ฅ) +๏ฟฝ๐=0๐๐๐ฅ๐
I could not find a way to convert the complete solution to closed form solution, or even findclosed form for ๐ฆ1 (๐ฅ). The computer claims that the closed form final solution is
๐ฆ (๐ฅ) = ๐ฆ1 (๐ฅ) + ๐ฆ2 (๐ฅ)
= ๐0 cos (๐ฅ) + ๐0 ๏ฟฝโโcos2 ๐ฅ โ 1 + cos ๐ฅ ln ๏ฟฝcos (๐ฅ) + โcos2 ๐ฅ โ 1๏ฟฝ๏ฟฝ
Which appears to imply that (7) is cos (๐ฅ) series. But it is not. Converting series solution toclosed form solution is hard. Is this something we are supposed to know how to do? Otherby inspection, is there a formal process to do it?
69
3.1. HW1 CHAPTER 3. HWS
3.1.7 key solution of selected problems
3.1.7.1 section 3 problem 8
Problem Statement: How many terms are needed in the Taylor series solution to yโฒโฒโฒ = x3y,
y(0) = 1, yโฒ(0) = 0, yโฒโฒ(0) = 0 are needed to evaluateโโซ0
y(x)dx correct to 3 decimal places?
Solution: Let
y =โโ
n=0
anxn
yโฒ =โโ
n=0
nanxnโ1
yโฒโฒ =โโ
n=0
nan(nโ 1)xnโ1
yโฒโฒโฒ =
โโn=0
n(nโ 1)(nโ 2)anxnโ3 = x3y = x3
โโn=0
anxn =
โโn=0
anxn+3.
Shift indices by letting n โ (nโ 6) in the RHS:
โโn=0
anxn+3 โ
โโn=6
anโ6xnโ3
โโn=0
n(nโ 1)(nโ 2)anxnโ3 = 0 ยท xโ3 + 0 ยท xโ2 + 0 ยท xโ1 + 3 ยท 2 ยท 1 ยท a3x
0 + 4 ยท 3 ยท 2 ยท a4x1
+ 5 ยท 4 ยท 3 ยท a5x2 +
โโn=6
n(nโ 1)(nโ 2)anxnโ3
From here,a3 = 0, a4 = 0, a5 = 0
xnโ3{n(nโ 1)(nโ 2)an โ anโ6} = 0 for n = 6, 7, 8, ...
The recurrence relation becomes:
an =1
n(nโ 1)(nโ 2)anโ6 n โฅ 6
Solution to D.E. becomes:
y =
โโn=0
anxn = a0 + a1 ยท x+ a2 ยท x
2 +
โโn=6
[1
n(nโ 1)(nโ 2)anโ6x
n]
y(0) = a0 = 1 ==> a0 = 1
1
70
3.1. HW1 CHAPTER 3. HWS
yโฒ(0) = a1 = 0 ==> a1 = 0
yโฒโฒ(0) = 2 ยท a2 = 0 ==> a2 = 0
y(x) = 1 +
โโn=6
1
n(nโ 1)(nโ 2)anโ6x
n
โซ 1
0
y(x)dx = [x+โโ
n=6
1
(n+ 1)n(nโ 1)(nโ 2)anโ6x
n+1]10
= 1 +
โโn=6
1
(n+ 1)(n)(nโ 1)(nโ 2)anโ6
= 1 +1
7 ยท 6 ยท 5 ยท 4(1) +
1
13 ยท 12 ยท 11 ยท 10(
1
6 ยท ยท4(1)) + ...
= 1 +1
840+
1
2, 059, 200+ ...
โซ 1
0
y(x)dx โ 1 + 0.00119 + 4.856ร 10โ7
So the 2nd non-zero term is required. All other terms are below the specified tolerance of0.001.
2
71
3.1. HW1 CHAPTER 3. HWS
3.1.7.2 section 3 problem 6
72
3.1. HW1 CHAPTER 3. HWS
73
3.1. HW1 CHAPTER 3. HWS
3.1.7.3 section 3 problem 4
74
3.1. HW1 CHAPTER 3. HWS
75
3.1. HW1 CHAPTER 3. HWS
76
3.1. HW1 CHAPTER 3. HWS
3.1.7.4 section 3 problem 24
1 Closed form of Problem 3.24e
In problem 3.24e, we find Frobenius series solutions
yฮฑ(x) =โโ
n=0
anxn+ฮฑ (1)
for ฮฑ = 0, 3
2, and recurrence relation on the coefficients:
an+3 =โan
(n+ ฮฑ + 3)(2n+ 2ฮฑ+ 3).
For each ฮฑ, a1 = a2 = 0, so only the a3k survive. Rewriting the recurrence relation:
a3k+3 =โa3k
(3k + 3 + ฮฑ)(6k + 3 + 2ฮฑ)=
โa3k9(k + 1 + ฮฑ/3)(2k + 1 + 2ฮฑ/3)
.
To identify a closed-form sum, we need to find the general form of the coefficients a3k. Towards thisend, consider the case ฮฑ = 0. Now
a3k =โ1/9
k(2k โ 1)a3kโ3 =
(โ1/9)2
k(k โ 1) ยท (2k โ 1)(2k โ 3)a3kโ6 = ยท ยท ยท =
(โ1/9)k
k! ยท (2k โ 1)!!a0.
Now note that
(2k โ 1)!! =(2k)!
(2k)!!=
(2k)!
2k ยท k!=โ k! ยท (2k โ 1)!! = 2โk(2k)!,
so
a3k =(โ1/9)k
2โk(2k)!a0 =
(โ2/9)k
(2k)!a0.
Therefore,
y0(x) = a0
โโ
k=0
(โ2/9)k
(2k)!x3k = a0
โโ
k=0
(โ1)k
(2k)!(x3/2
โ2/3)2k =: a0
โโ
k=0
โz2k
(2k)!,
where z := x3/2โ2/3. We recognize the remaining series as the Taylor series for cos z, so
y0(x) = a0 cos z(x) = a0 cos
(โ2
3x3/2
)
.
For ฮฑ = 3
2,
a3k+3 =โa3k
9(k + 3/2)(2k + 2)=
โa3k9(k + 1)(2k + 3)
.
Following the same argument as before,
a3k =(โ1/9)k
k! ยท (2k + 1)!!a0.
This is the same as before, but divided by 2k + 1, so
y3/2(x) = a0
โโ
k=0
(โ2/9)k
(2k + 1)!x3k = a0 sin
(โ2
3x3/2
)
.
Therefore, the general solution is
y(x) = y0(x) + y3/2(x) = c1 cos
(โ2
3x3/2
)
+ c2 sin
(โ2
3x3/2
)
.
1
77
3.1. HW1 CHAPTER 3. HWS
2 Closed form of Problem 3.24f
Problem 3.24f asks us to solve (sin x)yโฒโฒ โ 2(cosx)yโฒ โ (sin x)y = 0. Dividing by sin x,
yโฒโฒ โ 2(cotx)yโฒ โ y = 0,
so x = 0 is a regular singular point (cotx is not analytic at x = 0, but x cot x is). Thus, we can look fora Frobenius series solution.
Expanding sin x and cos x:
0 =
(
โโ
m=0
(โ1)m
(2m+ 1)!x2m+1
)
โโ
n=0
(n+ ฮฑ)(n+ ฮฑโ 1)anxn+ฮฑโ2 โ 2
(
โโ
m=0
(โ1)m
(2m)!x2m
)
โโ
n=0
(n + ฮฑ)anxn+ฮฑโ1
โ
(
โโ
m=0
(โ1)m
(2m+ 1)!x2m+1
)
โโ
n=0
anxn+ฮฑ.
The lowest order is xฮฑโ1:0 = ฮฑ(ฮฑโ 1)a0 โ 2ฮฑa0 = ฮฑ(ฮฑโ 3)a0,
so ฮฑ = 0, 3. We are in case II(b) on page 72.Let ฮฑ = 0. Then
0 =
(
โโ
m=0
(โ1)m
(2m+ 1)!x2m+1
)
โโ
n=0
n(nโ 1)anxnโ2 โ 2
(
โโ
m=0
(โ1)m
(2m)!x2m
)
โโ
n=0
nanxnโ1
โ
(
โโ
m=0
(โ1)m
(2m+ 1)!x2m+1
)
โโ
n=0
anxn.
We solve order-by-order:
x0 : 0 = โ2a1 =โ a1 = 0,
x1 : 0 = 2a2 โ 2 ยท 2a2 โ a0 =โ a2 = โa02,
x2k : 0 =
kโ
โ=0
(โ1)kโโ 2โ(2โ+ 1)
(2k โ 2โ+ 1)!a2โ+1 โ 2
kโ
โ=0
(โ1)kโโ 2โ+ 1
(2k โ 2โ)!a2โ+1 โ
kโ1โ
โ=0
(โ1)kโโโ1
(2k โ 2โโ 1)!a2โ+1,
x2k+1 : 0 =
k+1โ
โ=0
(โ1)kโโ+12โ(2โโ 1)
(2k โ 2โ+ 3)!a2โ โ 2
k+1โ
โ=0
(โ1)kโโ+12โ
(2k โ 2โ+ 2)!a2โ โ
kโ
โ=0
(โ1)kโโ
(2k โ 2โ+ 1)!a2โ,
where k โฅ 1. From the x2k terms, we see that a2n+1 depends only on a1, a3, ..., a2nโ1. As a1 = 0, it is asimple (strong) induction to show that each a2n+1 = 0.
We use the x2k+1 terms for k = 1 to see that
0 =
(
0a0 โ2
6a2 +
12
1a4
)
โ 2
(
0a0 โ2
2a2 +
4
1a4
)
โ(
โ1
6a0 +
1
1a2
)
=โ a4 = โ1
4
(
1
6a0 โ
2
3a2
)
.
As a2 = โ1
2a0, we see that a4 = 1
24a0. We notice a pattern that these coefficients are starting to look
like those in the Taylor series expansion of y0(x) = a0 cos x, which we now verify by plugging into theODE:
(sin x)yโฒโฒ0(x)โ 2(cosx)yโฒ0(x)โ (sin x)y0(x) = โa0 sin x cos x+ 2a0 cos x sin xโ a0 sin x cosx = 0.
2
78
3.1. HW1 CHAPTER 3. HWS
To find the other solution, we use reduction of order: look for a solution of the form y3(x) = v(x) cosx(we keep the ฮฑ = 3 subscript to remind us that we expect the Taylor series of our solution to start atthe x3-term). Plugging into the ODE:
(sin x)(vโฒโฒ(x) cosxโ 2vโฒ(x) sin xโ v(x) cosx)โ (2 cosx)(vโฒ(x) cos xโ v(x) sin x)โ (sin x)(v(x) cosx) = 0.
Simplifying,
(sin x cosx)vโฒโฒ(x)โ 2vโฒ(x) = 0 =โ vโฒ(x) = c1 tan2 x =โ v(x) = c1(tanxโ x) + c0.
We set c0 = 0, since this corresponds to y0(x). Then y3(x) = c1(tan xโ x) cosx = c1(sin xโ x cosx).The full solution is therefore
y(x) = c0 cosx+ c1(sin xโ x cosx).
3
79
3.2. HW2 CHAPTER 3. HWS
3.2 HW2
3.2.1 problem 3.27 (page 138)
Problem derive 3.4.28. Below is a screen shot from the book giving 3.4.28 at page 88, andthe context it is used in before solving the problem
Solution
For ๐๐กโ order ODE, ๐0 (๐ฅ) is given by
๐0 (๐ฅ) โผ ๐๏ฟฝ๐ฅ๐ (๐ก)
1๐ ๐๐ก
And (page 497, textbook)
๐1 (๐ฅ) โผ1 โ ๐2๐
ln (๐ (๐ฅ)) + ๐ (10.2.11)
80
3.2. HW2 CHAPTER 3. HWS
Therefore
๐ฆ (๐ฅ) โผ exp (๐0 + ๐1)
โผ exp ๏ฟฝ๐๏ฟฝ๐ฅ๐ (๐ก)
1๐ ๐๐ก +
1 โ ๐2๐
ln (๐ (๐ฅ)) + ๐๏ฟฝ
โผ ๐ [๐ (๐ฅ)]1โ๐2๐ exp ๏ฟฝ๐๏ฟฝ
๐ฅ๐ (๐ก)
1๐ ๐๐ก๏ฟฝ
Note: I have tried other methods to proof this, such as a proof by induction. But was notable to after many hours trying. The above method uses a given formula which the bookdid not indicate how it was obtained. (see key solution)
3.2.2 Problem 3.33(b) (page 140)
Problem Find leading behavior as ๐ฅ โ 0+ for ๐ฅ4๐ฆโฒโฒโฒ โ 3๐ฅ2๐ฆโฒ + 2๐ฆ = 0
Solution Let
๐ฆ (๐ฅ) = ๐๐(๐ฅ)
๐ฆโฒ = ๐โฒ๐๐
๐ฆโฒโฒ = ๐โฒโฒ๐๐ + (๐โฒ)2 ๐๐
๐ฆโฒโฒโฒ = ๐โฒโฒโฒ๐๐ + ๐โฒโฒ๐โฒ๐๐ + 2๐โฒ๐โฒโฒ๐๐ + (๐โฒ)3 ๐๐
Hence the ODE becomes
๐ฅ4 ๏ฟฝ๐โฒโฒโฒ + 3๐โฒ๐โฒโฒ + (๐โฒ)3๏ฟฝ โ 3๐ฅ2๐โฒ = โ2 (1)
Now, we define ๐ (๐ฅ) as sum of a number of leading terms, which we try to find
๐ (๐ฅ) = ๐0 (๐ฅ) + ๐1 (๐ฅ) + ๐2 (๐ฅ) +โฏ
Therefore (1) becomes (using only two terms for now ๐ = ๐0 + ๐1)
{๐0 + ๐1}โฒโฒโฒ + 3 ๏ฟฝ(๐0 + ๐1) (๐0 + ๐1)
โฒโฒ๏ฟฝ + ๏ฟฝ(๐0 + ๐1)โฒ๏ฟฝ3โ3๐ฅ2{๐0 + ๐1}
โฒ = โ2๐ฅ4
๏ฟฝ๐โฒโฒโฒ0 + ๐โฒโฒโฒ1 ๏ฟฝ + 3 ๏ฟฝ(๐0 + ๐1) ๏ฟฝ๐โฒโฒ0 + ๐โฒโฒ1 ๏ฟฝ๏ฟฝ + ๏ฟฝ๐โฒ0 + ๐โฒ1๏ฟฝ3โ3๐ฅ2๏ฟฝ๐โฒ0 + ๐โฒ1๏ฟฝ = โ
2๐ฅ4
๏ฟฝ๐โฒโฒโฒ0 + ๐โฒโฒโฒ1 ๏ฟฝ + 3 ๏ฟฝ๐โฒโฒ0 ๐โฒ0 + ๐โฒโฒ0 ๐โฒ1 + ๐โฒโฒ1 ๐โฒ0๏ฟฝ + ๏ฟฝ๏ฟฝ๐โฒ0๏ฟฝ3+ 3 ๏ฟฝ๐โฒ0๏ฟฝ
2๐โฒ1 + 3๐โฒ0 ๏ฟฝ๐โฒ1๏ฟฝ
2๏ฟฝ โ
3๐ฅ2๏ฟฝ๐โฒ0 + ๐โฒ1๏ฟฝ = โ
2๐ฅ4
(2)
Assuming that ๐โฒ0 โ ๐โฒ1, ๐โฒโฒโฒ0 โ ๐โฒโฒโฒ1 , ๏ฟฝ๐โฒ0๏ฟฝ3โ 3๏ฟฝ๐โฒ0๏ฟฝ
2๐โฒ1 then equation (2) simplifies to
๐โฒโฒโฒ0 + 3๐โฒโฒ0 ๐โฒ0 + ๏ฟฝ๐โฒ0๏ฟฝ3โ3๐ฅ2๐โฒ0 โผ โ
2๐ฅ4
Assuming ๏ฟฝ๐โฒ0๏ฟฝ3โ ๐โฒโฒโฒ0 , ๏ฟฝ๐โฒ0๏ฟฝ
3โ 3๐โฒโฒ0 ๐โฒ0, ๏ฟฝ๐โฒ0๏ฟฝ
3โ 3
๐ฅ2๐โฒ0 (which we need to verify later), then
the above becomes
๏ฟฝ๐โฒ0๏ฟฝ3โผ โ
2๐ฅ4
Verification2
2When carrying out verification, all constant multipliers and signs are automticaly simplified and removed
81
3.2. HW2 CHAPTER 3. HWS
Since ๐โฒ0 โผ ๏ฟฝโ2๐ฅ4๏ฟฝ13= 1
๐ฅ43then ๐โฒโฒ0 โผ 1
๐ฅ73and ๐โฒโฒโฒ0 โผ 1
๐ฅ103. Now we need to verify the three
assumptions made above, which we used to obtain ๐โฒ0.
๏ฟฝ๐โฒ0๏ฟฝ3โ ๐โฒโฒโฒ0
1๐ฅ4โ
1
๐ฅ103
Yes.
๏ฟฝ๐โฒ0๏ฟฝ3โ 3๐โฒโฒ0 ๐โฒ0
1๐ฅ4โ
โโโโโโ1
๐ฅ73
โโโโโโ
โโโโโโ1
๐ฅ43
โโโโโโ
1๐ฅ4โ
โโโโโโ1
๐ฅ113
โโโโโโ
Yes.
๏ฟฝ๐โฒ0๏ฟฝ3โ
3๐ฅ2๐โฒ0
1๐ฅ4โ ๏ฟฝ
1๐ฅ2 ๏ฟฝ
1
๐ฅ43
1๐ฅ4โ
1
๐ฅ103
Yes. Assumed balance ise verified. Therefore
๏ฟฝ๐โฒ0๏ฟฝ3โผ โ
2๐ฅ4
๐โฒ0 โผ ๐๐ฅโ43
Where ๐3 = โ2. Integrating
๐0 โผ ๐๏ฟฝ๐ฅโ43 ๐๐ฅ
โผ ๐๏ฟฝ๐ฅโ43๐๐ฅ
โผ โ3๐๐ฅโ13
Where we ignored the constant of integration since subdominant. To find leading behavior,we go back to equation (2) and now solve for ๐1.
๏ฟฝ๐โฒโฒโฒ0 + ๐โฒโฒโฒ1 ๏ฟฝ + 3 ๏ฟฝ๐โฒโฒ0 ๐โฒ0 + ๐โฒโฒ0 ๐โฒ1 + ๐โฒโฒ1 ๐โฒ0๏ฟฝ + ๏ฟฝ๏ฟฝ๐โฒ0๏ฟฝ3+ 3 ๏ฟฝ๐โฒ0๏ฟฝ
2๐โฒ1 + 3๐โฒ0 ๏ฟฝ๐โฒ1๏ฟฝ
2๏ฟฝ โ
3๐ฅ2๏ฟฝ๐โฒ0 + ๐โฒ1๏ฟฝ = โ
2๐ฅ4
Moving all known quantities (those which are made of ๐0 and its derivatives) to the RHS
going from one step to the next, as they do not a๏ฟฝect the final result.
82
3.2. HW2 CHAPTER 3. HWS
and simplifying, gives
๏ฟฝ๐โฒโฒโฒ1 ๏ฟฝ + 3 ๏ฟฝ๐โฒโฒ0 ๐โฒ1 + ๐โฒโฒ1 ๐โฒ0๏ฟฝ + ๏ฟฝ3 ๏ฟฝ๐โฒ0๏ฟฝ2๐โฒ1 + 3๐โฒ0 ๏ฟฝ๐โฒ1๏ฟฝ
2๏ฟฝ โ
3๐โฒ1๐ฅ2
โผ โ๐โฒโฒโฒ0 โ 3๐โฒโฒ0 ๐โฒ0 +3๐ฅ2๐โฒ0
Now we assume the following (then will verify later)
3 ๏ฟฝ๐โฒ0๏ฟฝ2๐โฒ1 โ 3๐โฒ0 ๏ฟฝ๐โฒ1๏ฟฝ
2
3 ๏ฟฝ๐โฒ0๏ฟฝ2๐โฒ1 โ ๐โฒโฒโฒ1
3 ๏ฟฝ๐โฒ0๏ฟฝ2๐โฒ1 โ ๐โฒโฒ1 ๐โฒ0
3 ๏ฟฝ๐โฒ0๏ฟฝ2๐โฒ1 โ ๐โฒโฒ0 ๐โฒ1
3 ๏ฟฝ๐โฒ0๏ฟฝ2๐โฒ1 โ ๐โฒโฒ1 ๐โฒ0
Hence
3 ๏ฟฝ๐โฒ0๏ฟฝ2๐โฒ1 โ
3๐โฒ1๐ฅ2
โผ โ๐โฒโฒโฒ0 โ 3๐โฒโฒ0 ๐โฒ0 +3๐โฒ0๐ฅ2
(3)
But
๐โฒ0 โผ ๐๐ฅโ43
๏ฟฝ๐โฒ0๏ฟฝ2โผ ๐2๐ฅ
โ83
๐โฒโฒ0 โผ โ43๐๐ฅ
โ73
๐โฒโฒโฒ0 โผ289๐๐ฅ
โ103
Hence (3) becomes
3 ๏ฟฝ๐2๐ฅโ83 ๏ฟฝ ๐โฒ1 โ
3๐โฒ1๐ฅ2
โผ289๐๐ฅ
โ103 + 3 ๏ฟฝ
43๐2๐ฅ
โ73 ๐ฅ
โ43 ๏ฟฝ +
3๐๐ฅโ43
๐ฅ2
3๐2๐ฅโ83 ๐โฒ1 โ 3๐ฅโ2๐โฒ1 โผ
289๐๐ฅ
โ103 + 4๐2๐ฅ
โ113 + 3๐๐ฅ
โ103
For small ๐ฅ, ๐ฅโ83 ๐โฒ1 โ ๐ฅโ2๐โฒ1 and ๐ฅ
โ113 โ ๐ฅ
โ103 , then the above simplifies to
3๐2๐ฅโ83 ๐โฒ1 โผ 4๐2๐ฅ
โ113
๐โฒ1 โผ43๐ฅโ1
๐1 โผ43
ln ๐ฅ
Where constant of integration was dropped, since subdominant.
83
3.2. HW2 CHAPTER 3. HWS
Verification Using ๐โฒ0 โฝ ๐ฅโ43 , ๏ฟฝ๐โฒ0๏ฟฝ
2โฝ ๐ฅ
โ83 , ๐โฒโฒ0 โฝ ๐ฅ
โ73 , ๐โฒ1 โผ
1๐ฅ , ๐
โฒโฒ1 โผ 1
๐ฅ2 , ๐โฒโฒโฒ1 โผ 1
๐ฅ3
3 ๏ฟฝ๐โฒ0๏ฟฝ2๐โฒ1 โ 3๐โฒ0 ๏ฟฝ๐โฒ1๏ฟฝ
2
๐ฅโ831๐ฅโ ๐ฅ
โ431๐ฅ2
๐ฅโ83 โ ๐ฅ
โ73
Yes.
3 ๏ฟฝ๐โฒ0๏ฟฝ2๐โฒ1 โ ๐โฒโฒโฒ1
๐ฅโ831๐ฅโ
1๐ฅ3
1
๐ฅ83
โ1๐ฅ2
Yes.
3 ๏ฟฝ๐โฒ0๏ฟฝ2๐โฒ1 โ ๐โฒโฒ1 ๐โฒ0
๐ฅโ831๐ฅโ
1๐ฅ2๐ฅโ43
๐ฅโ83 โ ๐ฅ
โ73
Yes.
3 ๏ฟฝ๐โฒ0๏ฟฝ2๐โฒ1 โ ๐โฒโฒ0 ๐โฒ1
๐ฅโ831๐ฅโ ๐ฅ
โ731๐ฅ
๐ฅโ83 โ ๐ฅ
โ73
Yes
3 ๏ฟฝ๐โฒ0๏ฟฝ2๐โฒ1 โ ๐โฒโฒ1 ๐โฒ0
๐ฅโ831๐ฅโ
1๐ฅ2๐ฅโ43
๐ฅโ83 โ ๐ฅ
โ73
Yes. All verified. Leading behavior is
๐ฆ (๐ฅ) โฝ ๐๐0(๐ฅ)+๐1(๐ฅ)
= exp ๏ฟฝ๐๐๐ฅโ13 +
43
ln ๐ฅ๏ฟฝ
= ๐ฅ43 ๐๐๐๐ฅ
โ13
I now wanted to see how Maple solution to this problem compare with the leading behaviornear ๐ฅ = 0. To obtain a solution from Maple, one have to give initial conditions a little bitremoved from ๐ฅ = 0 else no solution could be generated. So using arbitrary initial conditionsat ๐ฅ = 1
100 a solution was obtained and compared to the above leading behavior. Another
84
3.2. HW2 CHAPTER 3. HWS
problem is how to select ๐ in the above leading solution. By trial and error a constant wasselected. Here is screen shot of the result. The exact solution generated by Maple is verycomplicated, in terms of hypergeom special functions.
ode:=x^4*diff(y(x),x$3)-3*x^2*diff(y(x),x)+2*y(x);pt:=1/100:ic:=y(pt)=500,D(y)(pt)=0,(D@@2)(y)(pt)=0:sol:=dsolve({ode,ic},y(x)):leading:=(x,c)->x^(4/3)*exp(c*x^(-1/3));plot([leading(x,.1),rhs(sol)],x=pt..10,y=0..10,color=[blue,red],legend=["leading behavior","exact solution"],legendstyle=[location=top]);
3.2.3 problem 3.33(c) (page 140)
Problem Find leading behavior as ๐ฅ โ 0+ for ๐ฆโฒโฒ = โ๐ฅ๐ฆ
Solution
Let ๐ฆ (๐ฅ) = ๐๐(๐ฅ). Hence
๐ฆ (๐ฅ) = ๐๐0(๐ฅ)
๐ฆโฒ (๐ฅ) = ๐โฒ0๐๐0
๐ฆโฒโฒ = ๐โฒโฒ0 ๐๐0 + ๏ฟฝ๐โฒ0๏ฟฝ2๐๐0
= ๏ฟฝ๐โฒโฒ0 + ๏ฟฝ๐โฒ0๏ฟฝ2๏ฟฝ ๐๐0
Substituting in the ODE gives
๐โฒโฒ0 + ๏ฟฝ๐โฒ0๏ฟฝ2= โ๐ฅ (1)
85
3.2. HW2 CHAPTER 3. HWS
Assuming ๐โฒโฒ0 โผ ๏ฟฝ๐โฒ0๏ฟฝ2then (1) becomes
๐โฒโฒ0 โผ โ ๏ฟฝ๐โฒ0๏ฟฝ2
Let ๐โฒ0 = ๐ง then the above becomes ๐งโฒ = โ๐ง2. Hence ๐๐ง๐๐ฅ
1๐ง2 = โ1 or ๐๐ง
๐ง2 = โ๐๐ฅ. Integratingโ1๐ง = โ๐ฅ + ๐ or ๐ง =
1๐ฅ+๐1
. Hence ๐โฒ0 =1
๐ฅ+๐1. Integrating again gives
๐0 (๐ฅ) โผ ln |๐ฅ + ๐1| + ๐2
Verification
๐โฒ0 =1
๐ฅ+๐1, ๏ฟฝ๐โฒ0๏ฟฝ
2= 1
(๐ฅ+๐1)2 , ๐โฒโฒ0 =
โ1(๐ฅ+๐1)
2
๐โฒโฒ0 โ ๐ฅ12
1(๐ฅ + ๐1)
2 โ ๐ฅ12
Yes for ๐ฅ โ 0+.
๏ฟฝ๐โฒ0๏ฟฝ2โ ๐ฅ
12
1(๐ฅ + ๐1)
2 โ ๐ฅ12
Yes for ๐ฅ โ 0+. Verified. Controlling factor is
๐ฆ (๐ฅ) โผ ๐๐0(๐ฅ)
โผ ๐ln|๐ฅ+๐1|+๐2
โผ ๐ด๐ฅ + ๐ต
3.2.4 problem 3.35
Problem: Obtain the full asymptotic behavior for small ๐ฅ of solutions to the equation
๐ฅ2๐ฆโฒโฒ + (2๐ฅ + 1) ๐ฆโฒ โ ๐ฅ2 ๏ฟฝ๐2๐ฅ + 1๏ฟฝ ๐ฆ = 0
Solution
86
3.2. HW2 CHAPTER 3. HWS
Let ๐ฆ (๐ฅ) = ๐๐(๐ฅ). Hence
๐ฆ (๐ฅ) = ๐๐0(๐ฅ)
๐ฆโฒ (๐ฅ) = ๐โฒ0๐๐0
๐ฆโฒโฒ = ๐โฒโฒ0 ๐๐0 + ๏ฟฝ๐โฒ0๏ฟฝ2๐๐0
= ๏ฟฝ๐โฒโฒ0 + ๏ฟฝ๐โฒ0๏ฟฝ2๏ฟฝ ๐๐0
Substituting in the ODE gives
๐ฅ2 ๏ฟฝ๐โฒโฒ0 + ๏ฟฝ๐โฒ0๏ฟฝ2๏ฟฝ ๐๐0 + (2๐ฅ + 1) ๐โฒ0๐๐0 โ ๐ฅ2 ๏ฟฝ๐
2๐ฅ + 1๏ฟฝ ๐๐0(๐ฅ) = 0
๐ฅ2 ๏ฟฝ๐โฒโฒ0 + ๏ฟฝ๐โฒ0๏ฟฝ2๏ฟฝ + (2๐ฅ + 1) ๐โฒ0 โ ๐ฅ2 ๏ฟฝ๐
2๐ฅ + 1๏ฟฝ = 0
๐ฅ2 ๏ฟฝ๐โฒโฒ0 + ๏ฟฝ๐โฒ0๏ฟฝ2๏ฟฝ + (2๐ฅ + 1) ๐โฒ0 = ๐ฅ2 ๏ฟฝ๐
2๐ฅ + 1๏ฟฝ
๐โฒโฒ0 + ๏ฟฝ๐โฒ0๏ฟฝ2+(2๐ฅ + 1)๐ฅ2
๐โฒ0 = ๐2๐ฅ + 1
Assuming balance
๏ฟฝ๐โฒ0๏ฟฝ2โผ ๏ฟฝ๐
2๐ฅ + 1๏ฟฝ
๏ฟฝ๐โฒ0๏ฟฝ2โผ ๐
2๐ฅ
๐โฒ0 โผ ยฑ๐1๐ฅ
Where 1 was dropped since subdominant to ๐1๐ฅ for small ๐ฅ.
Verification Since ๏ฟฝ๐โฒ0๏ฟฝ2โผ ๐
2๐ฅ then ๐โฒ0 โผ ๐
1๐ฅ and ๐โฒโฒ0 โผ โ 1
๐ฅ2 ๐1๐ฅ , hence
๏ฟฝ๐โฒ0๏ฟฝ2โ ๐โฒโฒ0
๐2๐ฅ โ
1๐ฅ2๐1๐ฅ
๐1๐ฅ โ
1๐ฅ2
Yes, As ๐ฅ โ 0+
๏ฟฝ๐โฒ0๏ฟฝ2โ
(2๐ฅ + 1) ๐โฒ0๐ฅ2
๐2๐ฅ โ
(2๐ฅ + 1)๐ฅ2
๐1๐ฅ
๐1๐ฅ โ
(2๐ฅ + 1)๐ฅ2
๐1๐ฅ โ
2๐ฅ+1๐ฅ2
87
3.2. HW2 CHAPTER 3. HWS
Yes as ๐ฅ โ 0+. Verified. Hence both assumptions used were verified OK. Hence
๐โฒ0 โผ ยฑ๐2๐ฅ
๐0 โผ ยฑ๏ฟฝ๐1๐ฅ๐๐ฅ
Since the integral do not have closed form, we will do asymptotic expansion on the integral.
Rewriting โซ ๐1๐ฅ๐๐ฅ as โซ ๐
1๐ฅ
๏ฟฝโ๐ฅ2๏ฟฝ๏ฟฝโ๐ฅ2๏ฟฝ ๐๐ฅ. Using โซ๐ข๐๐ฃ = ๐ข๐ฃ โ โซ๐ฃ๐๐ข, where ๐ข = โ๐ฅ2, ๐๐ฃ = โ๐
1๐ฅ
๐ฅ2 , gives
๐๐ข = โ2๐ฅ and ๐ฃ = ๐1๐ฅ , hence
๏ฟฝ๐1๐ฅ๐๐ฅ = ๐ข๐ฃ โ๏ฟฝ๐ฃ๐๐ข
= โ๐ฅ2๐1๐ฅ โ๏ฟฝโ2๐ฅ๐
1๐ฅ๐๐ฅ
= โ๐ฅ2๐1๐ฅ + 2๏ฟฝ๐ฅ๐
1๐ฅ๐๐ฅ (1)
Now we apply integration by parts on โซ๐ฅ๐1๐ฅ๐๐ฅ = โซ๐ฅ ๐
1๐ฅ
โ๐ฅ3๏ฟฝโ๐ฅ3๏ฟฝ ๐๐ข = โซ ๐
1๐ฅ
โ๐ฅ2๏ฟฝโ๐ฅ3๏ฟฝ ๐๐ข, where
๐ข = โ๐ฅ3, ๐๐ฃ = ๐1๐ฅ
โ๐ฅ2 , hence ๐๐ข = โ3๐ฅ2, ๐ฃ = ๐
1๐ฅ , hence we have
๏ฟฝ๐ฅ๐1๐ฅ๐๐ฅ = ๐ข๐ฃ โ๏ฟฝ๐ฃ๐๐ข
= โ๐ฅ3๐1๐ฅ +๏ฟฝ3๐ฅ2๐
1๐ฅ๐๐ฅ
Substituting this into (1) gives
๏ฟฝ๐1๐ฅ๐๐ฅ = โ๐ฅ2๐
1๐ฅ + 2 ๏ฟฝโ๐ฅ3๐
1๐ฅ +๏ฟฝ3๐ฅ2๐
1๐ฅ๐๐ฅ๏ฟฝ
= โ๐ฅ2๐1๐ฅ โ 2๐ฅ3๐
1๐ฅ + 6๏ฟฝ3๐ฅ2๐
1๐ฅ๐๐ฅ
And so on. The series will become
๏ฟฝ๐1๐ฅ๐๐ฅ = โ๐ฅ2๐
1๐ฅ โ 2๐ฅ3๐
1๐ฅ + 6๐ฅ4๐
1๐ฅ + 24๐ฅ5๐
1๐ฅ +โฏ+ ๐!๐ฅ๐+1๐
1๐ฅ +โฏ
= โ๐1๐ฅ ๏ฟฝ๐ฅ2 + 2๐ฅ3 + 6๐ฅ4 +โฏ+ ๐!๐ฅ๐+1 +โฏ๏ฟฝ
Now as ๐ฅ โ 0+, we can decide how many terms to keep in the RHS, If we keep one term,then we can say
๐0 โผ ยฑ๏ฟฝ๐1๐ฅ๐๐ฅ
โผ ยฑ๐ฅ2๐1๐ฅ
For two terms
๐0 โผ ยฑ๏ฟฝ๐1๐ฅ๐๐ฅ โผ ยฑ๐
1๐ฅ ๏ฟฝ๐ฅ2 + 2๐ฅ3๏ฟฝ
And so on. Let us use one term for now for the rest of the solution.
๐0 โผ ยฑ๐ฅ2๐1๐ฅ
88
3.2. HW2 CHAPTER 3. HWS
To find leading behavior, let
๐ (๐ฅ) = ๐0 (๐ฅ) + ๐1 (๐ฅ)
Then ๐ฆ (๐ฅ) = ๐๐0(๐ฅ)+๐1(๐ฅ) and hence now
๐ฆโฒ (๐ฅ) = (๐0 (๐ฅ) + ๐1 (๐ฅ))โฒ ๐๐0+๐1
๐ฆโฒโฒ (๐ฅ) = ๏ฟฝ(๐0 + ๐1)โฒ๏ฟฝ2๐๐0+๐1 + (๐0 + ๐1)
โฒโฒ ๐๐0+๐1
Substituting into the given ODE gives
๐ฅ2 ๏ฟฝ๏ฟฝ(๐0 + ๐1)โฒ๏ฟฝ2+ (๐0 + ๐1)
โฒโฒ๏ฟฝ + (2๐ฅ + 1) (๐0 (๐ฅ) + ๐1 (๐ฅ))โฒ โ ๐ฅ2 ๏ฟฝ๐
2๐ฅ + 1๏ฟฝ = 0
๐ฅ2 ๏ฟฝ๏ฟฝ๐โฒ0 + ๐โฒ1๏ฟฝ2+ (๐0 + ๐1)
โฒโฒ๏ฟฝ + (2๐ฅ + 1) ๏ฟฝ๐โฒ0 (๐ฅ) + ๐โฒ1 (๐ฅ)๏ฟฝ โ ๐ฅ2 ๏ฟฝ๐2๐ฅ + 1๏ฟฝ = 0
๐ฅ2 ๏ฟฝ๏ฟฝ๐โฒ0๏ฟฝ2+ ๏ฟฝ๐โฒ1๏ฟฝ
2+ 2๐โฒ0๐โฒ1 + ๏ฟฝ๐โฒโฒ0 + ๐โฒโฒ1 ๏ฟฝ๏ฟฝ + (2๐ฅ + 1) ๐โฒ0 (๐ฅ) + (2๐ฅ + 1) ๐โฒ1 (๐ฅ) โ ๐ฅ2 ๏ฟฝ๐
2๐ฅ + 1๏ฟฝ = 0
๐ฅ2 ๏ฟฝ๐โฒ0๏ฟฝ2+ ๐ฅ2 ๏ฟฝ๐โฒ1๏ฟฝ
2+ 2๐ฅ2๐โฒ0๐โฒ1 + ๐ฅ2๐โฒโฒ0 + ๐ฅ2๐โฒโฒ1 + (2๐ฅ + 1) ๐โฒ0 (๐ฅ) + (2๐ฅ + 1) ๐โฒ1 (๐ฅ) = ๐ฅ2 ๏ฟฝ๐
2๐ฅ + 1๏ฟฝ
But ๐ฅ2 ๏ฟฝ๐โฒ0๏ฟฝ2โผ ๐ฅ2 ๏ฟฝ๐
2๐ฅ + 1๏ฟฝ since we found that ๐โฒ0 โผ ๐
1๐ฅ . Hence the above simplifies to
๐ฅ2 ๏ฟฝ๐โฒ1๏ฟฝ2+ 2๐ฅ2๐โฒ0๐โฒ1 + ๐ฅ2๐โฒโฒ0 + ๐ฅ2๐โฒโฒ1 + (2๐ฅ + 1) ๐โฒ0 (๐ฅ) + (2๐ฅ + 1) ๐โฒ1 (๐ฅ) = 0
๏ฟฝ๐โฒ1๏ฟฝ2+ 2๐โฒ0๐โฒ1 + ๐โฒโฒ0 + ๐โฒโฒ1 +
(2๐ฅ + 1)๐ฅ2
๐โฒ0 (๐ฅ) +(2๐ฅ + 1)๐ฅ2
๐โฒ1 (๐ฅ) = 0 (2)
Now looking at ๐โฒโฒ0 +(2๐ฅ+1)๐ฅ2 ๐โฒ0 (๐ฅ) terms in the above. We can simplify this since we know
๐โฒ0 = ๐1๐ฅ ,๐โฒโฒ0 = โ
1๐ฅ2 ๐
1๐ฅ This terms becomes
โ1๐ฅ2๐1๐ฅ +
(2๐ฅ + 1)๐ฅ2
๐1๐ฅ =
โ๐1๐ฅ + 2๐ฅ๐
1๐ฅ + ๐
1๐ฅ
๐ฅ2=2๐ฅ๐
1๐ฅ
๐ฅ2=2๐
1๐ฅ
๐ฅTherefore (2) becomes
๏ฟฝ๐โฒ1๏ฟฝ2+ 2๐โฒ0๐โฒ1 + ๐โฒโฒ1 +
(2๐ฅ + 1)๐ฅ2
๐โฒ1 (๐ฅ) โผโ2๐
1๐ฅ
๐ฅ๏ฟฝ๐โฒ1๏ฟฝ
2
๐โฒ0+ 2๐โฒ1 +
๐โฒโฒ1๐โฒ0
+(2๐ฅ + 1)๐ฅ2๐โฒ0
๐โฒ1 (๐ฅ) โผโ2๐
1๐ฅ
๐ฅ๐โฒ0๏ฟฝ๐โฒ1๏ฟฝ
2
๐1๐ฅ
+ 2๐โฒ1 +๐โฒโฒ1๐1๐ฅ
+(2๐ฅ + 1)
๐ฅ2๐1๐ฅ
๐โฒ1 (๐ฅ) โผโ2๐ฅ
Assuming the balance is
๐โฒ1 โผโ1๐ฅ
Hence
๐1 (๐ฅ) โผ โ ln (๐ฅ) + ๐
89
3.2. HW2 CHAPTER 3. HWS
Since ๐ subdominant as ๐ฅ โ 0+ then
๐1 (๐ฅ) โผ โ ln (๐ฅ)
Verification
๐โฒ1 โ๏ฟฝ๐โฒ1๏ฟฝ
2
๐1๐ฅ
1๐ฅโ
1๐ฅ2
1
๐1๐ฅ
๐1๐ฅ โ
1๐ฅ
Yes, for ๐ฅ โ 0+
๐โฒ1 โ๐โฒโฒ1๐1๐ฅ
1๐ฅโ
1๐ฅ2
1
๐1๐ฅ
Yes.
๐โฒ1 โ(2๐ฅ + 1)
๐ฅ2๐1๐ฅ
๐โฒ1 (๐ฅ)
1๐ฅโ
(2๐ฅ + 1)
๐ฅ2๐1๐ฅ
1๐ฅ
1๐ฅโ
1
๐ฅ3๐1๐ฅ
Yes. All assumptions verified. Hence leading behavior is
๐ฆ (๐ฅ) โผ exp (๐0 (๐ฅ) + ๐1 (๐ฅ))
โผ exp ๏ฟฝยฑ๐ฅ2๐1๐ฅ โ ln (๐ฅ)๏ฟฝ
โผ1๐ฅ๏ฟฝexp ๏ฟฝ๐ฅ2๐
1๐ฅ ๏ฟฝ + exp ๏ฟฝโ๐ฅ2๐
1๐ฅ ๏ฟฝ๏ฟฝ
For small ๐ฅ, then we ignore exp ๏ฟฝโ๐ฅ2๐1๐ฅ ๏ฟฝ since much smaller than exp ๏ฟฝ๐ฅ2๐
1๐ฅ ๏ฟฝ. Therefore
๐ฆ (๐ฅ) โผ1๐ฅ
exp ๏ฟฝ๐ฅ2๐1๐ฅ ๏ฟฝ
3.2.5 problem 3.39(h)
problem Find leading asymptotic behavior as ๐ฅ โ โ for ๐ฆโฒโฒ = ๐โ3๐ฅ๐ฆ
90
3.2. HW2 CHAPTER 3. HWS
solution Let ๐ฆ (๐ฅ) = ๐๐(๐ฅ). Hence
๐ฆ (๐ฅ) = ๐๐0(๐ฅ)
๐ฆโฒ (๐ฅ) = ๐โฒ0๐๐0
๐ฆโฒโฒ = ๐โฒโฒ0 ๐๐0 + ๏ฟฝ๐โฒ0๏ฟฝ2๐๐0
= ๏ฟฝ๐โฒโฒ0 + ๏ฟฝ๐โฒ0๏ฟฝ2๏ฟฝ ๐๐0
Substituting in the ODE gives
๏ฟฝ๐โฒโฒ0 + ๏ฟฝ๐โฒ0๏ฟฝ2๏ฟฝ ๐๐0 = ๐โ
3๐ฅ ๐๐0
๐โฒโฒ0 + ๏ฟฝ๐โฒ0๏ฟฝ2= ๐โ
3๐ฅ
Assuming ๏ฟฝ๐โฒ0๏ฟฝ2โ ๐โฒโฒ0 the above becomes
๏ฟฝ๐โฒ0๏ฟฝ2โผ ๐โ
3๐ฅ
๐โฒ0 โผ ยฑ๐โ32๐ฅ
Hence
๐0 โผ ยฑ๏ฟฝ๐โ32๐ฅ๐๐ฅ
Integration by parts. Since ๐๐๐ฅ๐
โ32๐ฅ = 3
2๐ฅ2 ๐โ 32๐ฅ , then we rewrite the integral above as
๏ฟฝ๐โ32๐ฅ๐๐ฅ = ๏ฟฝ
32๐ฅ2
๐โ32๐ฅ ๏ฟฝ
2๐ฅ2
3 ๏ฟฝ ๐๐ฅ
And now apply integration by parts. Let ๐๐ฃ = 32๐ฅ2 ๐
โ 32๐ฅ โ ๐ฃ = ๐โ
32๐ฅ , ๐ข = 2๐ฅ2
3 โ ๐๐ข = 43๐ฅ, hence
๏ฟฝ๐โ32๐ฅ๐๐ฅ = [๐ข๐ฃ] โ๏ฟฝ๐ฃ๐๐ข
=2๐ฅ2
3๐โ
32๐ฅ โ๏ฟฝ
43๐ฅ๐โ
32๐ฅ๐๐ฅ
Ignoring higher terms, then we use
๐0 โผ ยฑ2๐ฅ2
3๐โ
32๐ฅ
Verification
๏ฟฝ๐โฒ0๏ฟฝ2โ ๐โฒโฒ0
๏ฟฝ๐โ32๐ฅ ๏ฟฝ
2โ
32๐ฅ2
๐โ32๐ฅ
๐โ3๐ฅ โ
32๐ฅ2
๐โ32๐ฅ
Yes, as ๐ฅ โ โ. To find leading behavior, let
๐ (๐ฅ) = ๐0 (๐ฅ) + ๐1 (๐ฅ)
91
3.2. HW2 CHAPTER 3. HWS
Then ๐ฆ (๐ฅ) = ๐๐0(๐ฅ)+๐1(๐ฅ) and hence now
๐ฆโฒ (๐ฅ) = (๐0 (๐ฅ) + ๐1 (๐ฅ))โฒ ๐๐0+๐1
๐ฆโฒโฒ (๐ฅ) = ๏ฟฝ(๐0 + ๐1)โฒ๏ฟฝ2๐๐0+๐1 + (๐0 + ๐1)
โฒโฒ ๐๐0+๐1
Using the above, the ODE ๐ฆโฒโฒ = ๐โ3๐ฅ๐ฆ now becomes
๏ฟฝ(๐0 + ๐1)โฒ๏ฟฝ2+ (๐0 + ๐1)
โฒโฒ = ๐โ3๐ฅ
๏ฟฝ๐โฒ0 + ๐โฒ1๏ฟฝ2+ ๐โฒโฒ0 + ๐โฒโฒ1 = ๐
โ 3๐ฅ
๏ฟฝ๐โฒ0๏ฟฝ2+ ๏ฟฝ๐โฒ1๏ฟฝ
2+ 2๐โฒ0๐โฒ1 + ๐โฒโฒ0 + ๐โฒโฒ1 = ๐
โ 3๐ฅ
But ๏ฟฝ๐โฒ0๏ฟฝ2โผ ๐โ
3๐ฅ hence the above simplifies to
๏ฟฝ๐โฒ1๏ฟฝ2+ 2๐โฒ0๐โฒ1 + ๐โฒโฒ0 + ๐โฒโฒ1 = 0
Assuming ๏ฟฝ2๐โฒ0๐โฒ1๏ฟฝ โ ๐โฒโฒ1 the above becomes
๏ฟฝ๐โฒ1๏ฟฝ2+ 2๐โฒ0๐โฒ1 + ๐โฒโฒ0 = 0
Assuming 2๐โฒ0๐โฒ1 โ ๏ฟฝ๐โฒ1๏ฟฝ2
2๐โฒ0๐โฒ1 + ๐โฒโฒ0 = 0
๐โฒ1 โผ โ๐โฒโฒ02๐โฒ0
๐1 โผ โ12
ln ๏ฟฝ๐โฒ0๏ฟฝ
But ๐โฒ0 โผ ๐โ32๐ฅ , hence the above becomes
๐1 โผ โ12
ln ๏ฟฝ๐โ32๐ฅ ๏ฟฝ + ๐
โผ34๐ฅ+ ๐
Verification
๏ฟฝ2๐โฒ0๐โฒ1๏ฟฝ โ ๐โฒโฒ1
๏ฟฝ2๐โ32๐ฅโ34๐ฅ2 ๏ฟฝ
โ32๐ฅ3
32๐โ
32๐ฅ
๐ฅ2โ
32๐ฅ3
For large ๐ฅ the above simplifies to1๐ฅ2โ
1๐ฅ3
92
3.2. HW2 CHAPTER 3. HWS
Yes.
๏ฟฝ2๐โฒ0๐โฒ1๏ฟฝ โ ๏ฟฝ๐โฒ1๏ฟฝ2
๏ฟฝ2๐โ32๐ฅโ34๐ฅ2 ๏ฟฝ
โ ๏ฟฝโ34๐ฅ2 ๏ฟฝ
2
32๐โ
32๐ฅ
๐ฅ2โ
916๐ฅ4
For large ๐ฅ the above simplifies to1๐ฅ2โ
1๐ฅ4
Yes. All verified. Therefore, the leading behavior is
๐ฆ (๐ฅ) โผ exp (๐0 (๐ฅ) + ๐1 (๐ฅ))
โผ exp ๏ฟฝยฑ2๐ฅ2
3๐โ
32๐ฅ โ
12
ln ๏ฟฝ๐โ32๐ฅ ๏ฟฝ + ๐๏ฟฝ
โผ ๐๐34๐ฅ exp ๏ฟฝยฑ
2๐ฅ2
3๐โ
32๐ฅ ๏ฟฝ (1)
Check if we can use 3.4.28 to verify:
lim๐ฅโโ
|๐ฅ๐๐ (๐ฅ)| = lim๐ฅโโ
๏ฟฝ๐ฅ2๐โ3๐ฅ ๏ฟฝ โ โ
We can use it. Lets verify using 3.4.28
๐ฆ (๐ฅ) โผ ๐ [๐ (๐ฅ)]1โ๐2๐ exp ๏ฟฝ๐๏ฟฝ
๐ฅ๐ [๐ก]
1๐ ๐๐ก๏ฟฝ
Where ๐2 = 1. For ๐ = 2,๐ (๐ฅ) = ๐โ3๐ฅ , the above gives
๐ฆ (๐ฅ) โผ ๐ ๏ฟฝ๐โ3๐ฅ ๏ฟฝ
1โ24
exp
โโโโโโโ๐๏ฟฝ
๐ฅ๏ฟฝ๐โ
3๐ก ๏ฟฝ
12๐๐ก
โโโโโโโ
โผ ๐ ๏ฟฝ๐โ3๐ฅ ๏ฟฝ
โ14
exp ๏ฟฝ๐๏ฟฝ๐ฅ๐โ
32๐ก๐๐ก๏ฟฝ
โผ ๐๐34๐ฅ exp ๏ฟฝ๐๏ฟฝ
๐ฅ๐โ
32๐ก๐๐ก๏ฟฝ (2)
We see that (1,2) are the same. Verified OK. Notice that in (1), we use the approximation for
the ๐โ32๐ฅ๐๐ฅ โ 2๐ฅ2
3 ๐โ 32๐ฅ we found earlier. This was done, since there is no closed form solution
for the integral.
QED.
3.2.6 problem 3.42(a)
Problem: Extend investigation of example 1 of section 3.5 (a) Obtain the next few correctionsto the leading behavior (3.5.5) then see how including these terms improves the numerical
93
3.2. HW2 CHAPTER 3. HWS
approximation of ๐ฆ (๐ฅ) in 3.5.1.
Solution Example 1 at page 90 is ๐ฅ๐ฆโฒโฒ + ๐ฆโฒ = ๐ฆ. The leading behavior is given by 3.5.5 as(๐ฅ โ โ)
๐ฆ (๐ฅ) โผ ๐๐ฅโ14 ๐2๐ฅ
12 (3.5.5)
Where the book gives ๐ = 12๐
โ12 on page 91. And 3.5.1 is
๐ฆ (๐ฅ) =โ๏ฟฝ๐=0
๐ฅ๐
(๐!)2(3.5.1)
To see the improvement, the book method is followed. This is described at end of page 91.This is done by plotting the leading behavior as ratio to ๐ฆ (๐ฅ) as given in 3.5.1. Hence forthe above leading behavior, we need to plot
12๐
โ12 ๐ฅ
โ14 ๐2๐ฅ
12
๐ฆ (๐ฅ)We are given ๐0 (๐ฅ) , ๐1 (๐ฅ) in the problem. They are
๐0 (๐ฅ) = 2๐ฅ12
๐1 (๐ฅ) = โ14
ln ๐ฅ + ๐
Hence
๐โฒ0 (๐ฅ) = ๐ฅโ12
๐โฒโฒ0 =โ12๐ฅโ
32
๐โฒ1 (๐ฅ) = โ141๐ฅ
๐โฒโฒ1 (๐ฅ) =14๐ฅ2
(1)
We need to find ๐2 (๐ฅ) , ๐3 (๐ฅ) ,โฏ to see that this will improve the solution ๐ฆ (๐ฅ) โผ exp (๐0 + ๐1 + ๐2 +โฏ)as ๐ฅ โ ๐ฅ0 compared to just using leading behavior ๐ฆ (๐ฅ) โผ exp (๐0 + ๐1). So now we need tofind ๐2 (๐ฅ)
Let ๐ฆ (๐ฅ) = ๐๐, then the ODE becomes
๐ฅ ๏ฟฝ๐โฒโฒ + (๐โฒ)2๏ฟฝ + ๐โฒ = 1
Replacing ๐ by ๐0 (๐ฅ) + ๐1 (๐ฅ) + ๐2 (๐ฅ) in the above gives
(๐0 + ๐1 + ๐2)โฒโฒ + ๏ฟฝ(๐0 + ๐1 + ๐2)
โฒ๏ฟฝ2+1๐ฅ(๐0 + ๐1 + ๐2)
โฒ โผ1๐ฅ
๏ฟฝ๐โฒโฒ0 + ๐โฒโฒ1 + ๐โฒโฒ2 ๏ฟฝ + ๏ฟฝ๏ฟฝ๐โฒ0 + ๐โฒ1 + ๐โฒ2๏ฟฝ๏ฟฝ2+1๐ฅ๏ฟฝ๐โฒ0 + ๐โฒ1 + ๐โฒ2๏ฟฝ โผ
1๐ฅ
๏ฟฝ๐โฒโฒ0 + ๐โฒโฒ1 + ๐โฒโฒ2 ๏ฟฝ + ๏ฟฝ๏ฟฝ๐โฒ0๏ฟฝ2+ 2๐โฒ0๐โฒ1 + 2๐โฒ0๐โฒ2 + ๏ฟฝ๐โฒ1๏ฟฝ
2+ 2๐โฒ1๐โฒ2 + ๏ฟฝ๐โฒ2๏ฟฝ
2๏ฟฝ +
1๐ฅ๏ฟฝ๐โฒ0 + ๐โฒ1 + ๐โฒ2๏ฟฝ โผ
1๐ฅ
Moving all known quantities to the RHS, these are ๐โฒโฒ0 , ๐โฒโฒ1 , ๏ฟฝ๐โฒ0๏ฟฝ2, 2๐โฒ0๐โฒ1, ๐โฒ0, ๐โฒ1, ๏ฟฝ๐โฒ1๏ฟฝ
2then the
94
3.2. HW2 CHAPTER 3. HWS
above reduces to
๏ฟฝ๐โฒโฒ2 ๏ฟฝ + ๏ฟฝ+2๐โฒ0๐โฒ2 + 2๐โฒ1๐โฒ2 + ๏ฟฝ๐โฒ2๏ฟฝ2๏ฟฝ +
1๐ฅ๏ฟฝ๐โฒ2๏ฟฝ โผ
1๐ฅโ ๐โฒโฒ0 โ ๐โฒโฒ1 โ ๏ฟฝ๐โฒ0๏ฟฝ
2โ 2๐โฒ0๐โฒ1 โ
1๐ฅ๐โฒ0 โ
1๐ฅ๐โฒ1 โ ๏ฟฝ๐โฒ1๏ฟฝ
2
Replacing known terms, by using (1) into the above gives
๏ฟฝ๐โฒโฒ2 ๏ฟฝ + ๏ฟฝ2๐โฒ0๐โฒ2 + 2๐โฒ1๐โฒ2 + ๏ฟฝ๐โฒ2๏ฟฝ2๏ฟฝ +
1๐ฅ๏ฟฝ๐โฒ2๏ฟฝ โผ
1๐ฅ+12๐ฅโ
32 โ
14๐ฅ2
โ ๏ฟฝ๐ฅโ12 ๏ฟฝ
2โ 2 ๏ฟฝ๐ฅ
โ12 ๏ฟฝ ๏ฟฝโ
141๐ฅ๏ฟฝโ1๐ฅ๏ฟฝ๐ฅ
โ12 ๏ฟฝ โ
1๐ฅ ๏ฟฝโ141๐ฅ๏ฟฝโ ๏ฟฝโ
141๐ฅ๏ฟฝ
2
Simplifying gives
๏ฟฝ๐โฒโฒ2 ๏ฟฝ + ๏ฟฝ2๐โฒ0๐โฒ2 + 2๐โฒ1๐โฒ2 + ๏ฟฝ๐โฒ2๏ฟฝ2๏ฟฝ +
1๐ฅ๏ฟฝ๐โฒ2๏ฟฝ โผ
1๐ฅ+12๐ฅโ
32 โ
14๐ฅ2
โ ๐ฅโ1 +12๐ฅโ32 โ ๐ฅ
โ32 +
141๐ฅ2โ116
1๐ฅ2
Hence
๏ฟฝ๐โฒโฒ2 ๏ฟฝ + ๏ฟฝ2๐โฒ0๐โฒ2 + 2๐โฒ1๐โฒ2 + ๏ฟฝ๐โฒ2๏ฟฝ2๏ฟฝ +
1๐ฅ๏ฟฝ๐โฒ2๏ฟฝ โผ โ
116
1๐ฅ2
Lets assume now that
2๐โฒ0๐โฒ2 โผ โ116
1๐ฅ2
(2)
Therefore
๐โฒ2 โผ โ132
1๐โฒ0๐ฅ2
โผ โ132
1
๏ฟฝ๐ฅโ12 ๏ฟฝ ๐ฅ2
โผ โ132๐ฅโ32
We can now verify this before solving the ODE. We need to check that (as ๐ฅ โ โ)
2๐โฒ0๐โฒ2 โ ๐โฒโฒ22๐โฒ0๐โฒ2 โ 2๐โฒ1๐โฒ2
2๐โฒ0๐โฒ2 โ ๏ฟฝ๐โฒ2๏ฟฝ2
2๐โฒ0๐โฒ2 โ1๐ฅ๐โฒ2
Where ๐โฒโฒ2 โผ ๐ฅโ52 , Hence
2๐โฒ0๐โฒ2 โ ๐โฒโฒ2
๐ฅโ12 ๏ฟฝ๐ฅ
โ32 ๏ฟฝ โ ๐ฅ
โ52
๐ฅโ2 โ ๐ฅโ52
95
3.2. HW2 CHAPTER 3. HWS
Yes.
2๐โฒ0๐โฒ2 โ 2๐โฒ1๐โฒ2
๐ฅโ2 โ ๏ฟฝ1๐ฅ๏ฟฝ๏ฟฝ๐ฅ
โ32 ๏ฟฝ
๐ฅโ2 โ ๐ฅโ52
Yes
2๐โฒ0๐โฒ2 โ ๏ฟฝ๐โฒ2๏ฟฝ2
๐ฅโ2 โ ๏ฟฝ๐ฅโ32 ๏ฟฝ
2
๐ฅโ2 โ ๐ฅโ3
Yes
2๐โฒ0๐โฒ2 โ1๐ฅ๐โฒ2
๐ฅโ2 โ1๐ฅ๐ฅโ32
๐ฅโ2 โ ๐ฅโ52
Yes. All assumptions are verified. Therefore we can g ahead and solve for ๐2 using (2)
2๐โฒ0๐โฒ2 โผ โ116
1๐ฅ2
๐โฒ2 โผ โ132
1๐ฅ2
1๐โฒ0
โผ โ132
1๐ฅ2
1
๐ฅโ12
โผ โ132
1
๐ฅ32
Hence
๐2 โผ116
1
โ๐ฅ
The leading behavior now is
๐ฆ (๐ฅ) โผ exp (๐0 + ๐1 + ๐2)
โผ exp ๏ฟฝ2๐ฅ12 โ
14
ln ๐ฅ + ๐ + 116
1
โ๐ฅ๏ฟฝ
Now we will find ๐3. From
๐ฅ ๏ฟฝ๐โฒโฒ + (๐โฒ)2๏ฟฝ + ๐โฒ = 1
96
3.2. HW2 CHAPTER 3. HWS
Replacing ๐ by ๐0 + ๐1 + ๐2 + ๐3 in the above gives
(๐0 + ๐1 + ๐2 + ๐3)โฒโฒ + ๏ฟฝ(๐0 + ๐1 + ๐2 + ๐3)
โฒ๏ฟฝ2+1๐ฅ(๐0 + ๐1 + ๐2 + ๐3)
โฒ โผ1๐ฅ
๏ฟฝ๐โฒโฒ0 + ๐โฒโฒ1 + ๐โฒโฒ2 + ๐โฒโฒ3 ๏ฟฝ + ๏ฟฝ๏ฟฝ๐โฒ0 + ๐โฒ1 + ๐โฒ2 + ๐โฒ3๏ฟฝ๏ฟฝ2+1๐ฅ๏ฟฝ๐โฒ0 + ๐โฒ1 + ๐โฒ2 + ๐โฒ3๏ฟฝ โผ
1๐ฅ
Hence
๏ฟฝ๐โฒโฒ0 + ๐โฒโฒ1 + ๐โฒโฒ2 + ๐โฒโฒ3 ๏ฟฝ +
๏ฟฝ๏ฟฝ๐โฒ0๏ฟฝ2+ 2๐โฒ0๐โฒ1 + 2๐โฒ0๐โฒ2 + 2๐โฒ1๐โฒ2 + ๏ฟฝ๐โฒ1๏ฟฝ
2+ ๏ฟฝ๐โฒ2๏ฟฝ
2+ 2๐โฒ0๐โฒ3 + 2๐โฒ1๐โฒ3 + 2๐โฒ2๐โฒ3 + ๏ฟฝ๐โฒ3๏ฟฝ
2๏ฟฝ
+1๐ฅ๏ฟฝ๐โฒ0 + ๐โฒ1 + ๐โฒ2 + ๐โฒ3๏ฟฝ โผ
1๐ฅ
Moving all known quantities to the RHS gives
๏ฟฝ๐โฒโฒ3 ๏ฟฝ + ๏ฟฝ2๐โฒ0๐โฒ3 + 2๐โฒ1๐โฒ3 + 2๐โฒ2๐โฒ3 + ๏ฟฝ๐โฒ3๏ฟฝ2๏ฟฝ +
1๐ฅ๏ฟฝ๐โฒ3๏ฟฝ
โผ1๐ฅโ ๐โฒโฒ0 โ ๐โฒโฒ1 โ ๐โฒโฒ2 โ ๏ฟฝ๐โฒ0๏ฟฝ
2โ 2๐โฒ0๐โฒ1 โ 2๐โฒ0๐โฒ2 โ 2๐โฒ1๐โฒ2 โ ๏ฟฝ๐โฒ1๏ฟฝ
2โ ๏ฟฝ๐โฒ2๏ฟฝ
2โ1๐ฅ๐โฒ0 โ
1๐ฅ๐โฒ1 โ
1๐ฅ๐โฒ2 (3)
Now we will simplify the RHS, since it is all known. Using
๐โฒ0 (๐ฅ) = ๐ฅโ12
๏ฟฝ๐โฒ0๏ฟฝ2= ๐ฅโ1
๐โฒโฒ0 =โ12๐ฅโ
32
๐โฒ1 (๐ฅ) = โ141๐ฅ
๏ฟฝ๐โฒ1 (๐ฅ)๏ฟฝ2=116
1๐ฅ2
๐โฒโฒ1 (๐ฅ) =14๐ฅ2
๐โฒ2 (๐ฅ) = โ132
1
๐ฅ32
๏ฟฝ๐โฒ2 (๐ฅ)๏ฟฝ2=
11024๐ฅ3
๐โฒโฒ2 (๐ฅ) =364
1
๐ฅ52
2๐โฒ0๐โฒ1 = 2 ๏ฟฝ๐ฅโ12 ๏ฟฝ ๏ฟฝโ
141๐ฅ๏ฟฝ= โ
121
๐ฅ32
2๐โฒ0๐โฒ2 = 2 ๏ฟฝ๐ฅโ12 ๏ฟฝโโโโโโโ
132
1
๐ฅ32
โโโโโโ = โ
116๐ฅ2
2๐โฒ1๐โฒ2 = 2 ๏ฟฝโ141๐ฅ๏ฟฝ
โโโโโโโ
132
1
๐ฅ32
โโโโโโ =
1
64๐ฅ52
97
3.2. HW2 CHAPTER 3. HWS
Hence (3) becomes
๏ฟฝ๐โฒโฒ3 ๏ฟฝ + ๏ฟฝ2๐โฒ0๐โฒ3 + 2๐โฒ1๐โฒ3 + 2๐โฒ2๐โฒ3 + ๏ฟฝ๐โฒ3๏ฟฝ2๏ฟฝ +
1๐ฅ๏ฟฝ๐โฒ3๏ฟฝ โผ
1๐ฅ+
1
2๐ฅ32
โ14๐ฅ2
โ364
1
๐ฅ52
โ1๐ฅ+121
๐ฅ32
+1
16๐ฅ2โ
1
64๐ฅ52
โ116
1๐ฅ2โ
11024๐ฅ3
โ1
๐ฅ32
+141๐ฅ2+132
1
๐ฅ52
Simplifying gives
๏ฟฝ๐โฒโฒ3 ๏ฟฝ + ๏ฟฝ2๐โฒ0๐โฒ3 + 2๐โฒ1๐โฒ3 + 2๐โฒ2๐โฒ3 + ๏ฟฝ๐โฒ3๏ฟฝ2๏ฟฝ +
1๐ฅ๏ฟฝ๐โฒ3๏ฟฝ โผ โ
โโโโโโ
32
1024๐ฅ52
+1
1024๐ฅ3
โโโโโโ
Let us now assume that
๐โฒ0๐โฒ3 โ ๐โฒ1๐โฒ3๐โฒ0๐โฒ3 โ ๐โฒ2๐โฒ3
๐โฒ0๐โฒ3 โ ๏ฟฝ๐โฒ3๏ฟฝ2
๐โฒ0๐โฒ3 โ1๐ฅ๏ฟฝ๐โฒ3๏ฟฝ
๐โฒ0๐โฒ3 โ ๐โฒโฒ3Therefore, we end up with the balance
2๐โฒ0๐โฒ3 โผ โโโโโโโ
32
1024๐ฅ52
+1
1024๐ฅ3
โโโโโโ
๐โฒ3 โผ โ
โโโโโโโ
32
1024๐ฅ52๐โฒ0
+1
1024๐ฅ3๐โฒ0
โโโโโโโ
โผ โ
โโโโโโโโโโ
32
1024๐ฅ52 ๏ฟฝ๐ฅ
โ12 ๏ฟฝ
+1
1024๐ฅ3 ๏ฟฝ๐ฅโ12 ๏ฟฝ
โโโโโโโโโโ
โผ โโโโโโโ1
32๐ฅ2+
1
1024๐ฅ52
โโโโโโ
Hence
๐3 โผ1
1024๏ฟฝ 232๐ฅ2 +
32๐ฅ๏ฟฝ
Where constant of integration was ignored. Let us now verify the assumptions made
๐โฒ0๐โฒ3 โ ๐โฒ1๐โฒ3๐โฒ0 โ ๐โฒ1
Yes.
๐โฒ0๐โฒ3 โ ๐โฒ2๐โฒ3๐โฒ0 โ ๐โฒ2
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3.2. HW2 CHAPTER 3. HWS
Yes
๐โฒ0๐โฒ3 โ ๏ฟฝ๐โฒ3๏ฟฝ2
๐โฒ0 โ ๐โฒ3Yes.
๐โฒ0๐โฒ3 โ1๐ฅ๏ฟฝ๐โฒ3๏ฟฝ
๐โฒ0 โ1๐ฅ
๐ฅโ12 โ
1๐ฅ
Yes, as ๐ฅ โ โ, and finally
๐โฒ0๐โฒ3 โ ๐โฒโฒ3
๐ฅโ12
โโโโโโ1
32๐ฅ2+
1
1024๐ฅ52
โโโโโโ โ
โโโโโโ
5
2048๐ฅ72
+1
16๐ฅ3
โโโโโโ
๏ฟฝ32โ๐ฅ + 1๏ฟฝ1024๐ฅ3
โ128โ๐ฅ + 5
2048๐ฅ72
Yes, as ๐ฅ โ โ. All assumptions verified. The leading behavior now is
๐ฆ (๐ฅ) โผ exp (๐0 + ๐1 + ๐2 + ๐3)
โผ exp ๏ฟฝ2๐ฅ12 โ
14
ln ๐ฅ + ๐ + 116
1
โ๐ฅ+
11024 ๏ฟฝ
232๐ฅ2
+32๐ฅ ๏ฟฝ๏ฟฝ
โผ ๐๐ฅโ14 exp ๏ฟฝ2๐ฅ
12 +
116
1
โ๐ฅ+
11024 ๏ฟฝ
232๐ฅ2
+32๐ฅ ๏ฟฝ๏ฟฝ
Now we will show how adding more terms to leading behavior improved the ๐ฆ (๐ฅ) solutionfor large ๐ฅ. When plotting the solutions, we see that exp(๐0+๐1+๐2+๐3)
๐ฆ(๐ฅ) approached the ratio
1 sooner than exp(๐0+๐1+๐2)๐ฆ(๐ฅ) and this in turn approached the ratio 1 sooner than just using
exp(๐0+๐1)๐ฆ(๐ฅ) . So the e๏ฟฝect of adding more terms, is that the solution becomes more accurate
for larger range of ๐ฅ values. Below is the code used and the plot generated.
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3.2. HW2 CHAPTER 3. HWS
ClearAll[y, x];
s0[x_] :=
1
2 Pi12
Exp2 x12
y[x, 300];
s0s1[x_] :=
1
2 Pi12
Exp2 x12 -
1
4Log[x]
y[x, 300];
s0s1s2[x_] :=
1
2 Pi12
Exp2 x12 -
1
4Log[x] +
1
16
1
Sqrt[x]
y[x, 300];
s0s1s2s3[x_] :=
1
2 Pi12
Exp2 x12 -
1
4Log[x] +
1
16
1
Sqrt[x]+
1
1024
2
32 x2+
32
x
y[x, 300];
y[x_, max_] := Sum[ x^n / (Factorial[n]^2), {n, 0, max}];
LogLinearPlot[Evaluate[{s0s1[x], s0s1s2[x], s0s1s2s3[x]}], {x, 1, 30},
PlotRange โ All, Frame โ True, GridLines โ Automatic, GridLinesStyle โ LightGray,
PlotLegends โ {"exp(S0+S1)", "exp(S0+S1+S2)", "exp(S0+S1+S2+S3)"},
FrameLabel โ {{None, None}, {"x", "Showing improvement as more terms are added"}},
PlotStyle โ {Red, Blue, Black}, BaseStyle โ 14]
Out[59]=
1 2 5 10 20
0.92
0.94
0.96
0.98
1.00
x
Showing improvement as more terms are added
exp(S0+S1)
exp(S0+S1+S2)
exp(S0+S1+S2+S3)
100
3.2. HW2 CHAPTER 3. HWS
3.2.7 problem 3.49(c)
Problem Find the leading behavior as ๐ฅ โ โ of the general solution of ๐ฆโฒโฒ + ๐ฅ๐ฆ = ๐ฅ5
Solution This is non-homogenous ODE. We solve this by first finding the homogenoussolution (asymptotic solution) and then finding particular solution. Hence we start with
๐ฆโฒโฒโ + ๐ฅ๐ฆโ = 0
๐ฅ = โ is ISP point. Therefore, we assume ๐ฆโ (๐ฅ) = ๐๐(๐ฅ) and obtain
๐โฒโฒ + (๐โฒ)2 + ๐ฅ = 0 (1)
Let
๐ (๐ฅ) = ๐0 + ๐1 +โฏ
Therefore (1) becomes
๏ฟฝ๐โฒโฒ0 + ๐โฒโฒ1 +โฏ๏ฟฝ + ๏ฟฝ๐โฒ0 + ๐โฒ1 +โฏ๏ฟฝ2= โ๐ฅ
๏ฟฝ๐โฒโฒ0 + ๐โฒโฒ1 +โฏ๏ฟฝ + ๏ฟฝ๏ฟฝ๐โฒ0๏ฟฝ2+ 2๐โฒ0๐โฒ1 + ๏ฟฝ๐โฒ1๏ฟฝ
2+โฏ๏ฟฝ = โ๐ฅ (2)
Assuming ๏ฟฝ๐โฒ0๏ฟฝ2โ ๐โฒโฒ0 we obtain
๏ฟฝ๐โฒ0๏ฟฝ2โผ โ๐ฅ
๐โฒ0 โผ ๐โ๐ฅ
Where ๐ = ยฑ๐
Verification
๏ฟฝ๐โฒ0๏ฟฝ2โ ๐โฒโฒ0
๐ฅ โ121
โ๐ฅYes, as ๐ฅ โ โ. Hence
๐0 โผ32๐๐ฅ
32
Now we will find ๐1. From (2), and moving all known terms to RHS
๏ฟฝ๐โฒโฒ1 +โฏ๏ฟฝ + ๏ฟฝ2๐โฒ0๐โฒ1 + ๏ฟฝ๐โฒ1๏ฟฝ2+โฏ๏ฟฝ โผ โ๐ฅ โ ๐โฒโฒ0 โ ๏ฟฝ๐โฒ0๏ฟฝ
2(3)
Assuming
2๐โฒ0๐โฒ1 โ ๐โฒโฒ1
2๐โฒ0๐โฒ1 โ ๏ฟฝ๐โฒ1๏ฟฝ2
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3.2. HW2 CHAPTER 3. HWS
Then (3) becomes (where ๐โฒ0 โผ ๐โ๐ฅ, ๏ฟฝ๐โฒ0๏ฟฝ2โผ โ๐ฅ, ๐โฒโฒ0 โผ 1
2๐1
โ๐ฅ)
2๐โฒ0๐โฒ1 โผ โ๐ฅ โ ๐โฒโฒ0 โ ๏ฟฝ๐โฒ0๏ฟฝ2
๐โฒ1 โผโ๐ฅ โ ๐โฒโฒ0 โ ๏ฟฝ๐โฒ0๏ฟฝ
2
2๐โฒ0
โผโ๐ฅ โ 1
2๐1
โ๐ฅโ (โ๐ฅ)
2๐โ๐ฅ
โผ โ14๐ฅ
Verification (where ๐โฒโฒ1 โผ 141๐ฅ2 )
2๐โฒ0๐โฒ1 โ ๐โฒโฒ1
โ๐ฅ ๏ฟฝ14๐ฅ๏ฟฝ
โ141๐ฅ2
1
๐ฅ12
โ1๐ฅ2
Yes, as ๐ฅ โ โ
2๐โฒ0๐โฒ1 โ ๏ฟฝ๐โฒ1๏ฟฝ2
1
๐ฅ12
โ ๏ฟฝ14๐ฅ๏ฟฝ
2
1
๐ฅ12
โ1
16๐ฅ2
Yes, as ๐ฅ โ โ. All validated. We solve for ๐1
๐โฒ1 โผ โ14๐ฅ
๐1 โผ โ14
ln ๐ฅ + ๐
๐ฆโ is found. It is given by
๐ฆโ (๐ฅ) โผ exp (๐0 (๐ฅ) + ๐1 (๐ฅ))
โผ exp ๏ฟฝ32๐๐ฅ
32 โ
14
ln ๐ฅ + ๐๏ฟฝ
โผ ๐๐ฅโ14 exp ๏ฟฝ
32๐๐ฅ
32 ๏ฟฝ
Now that we have found ๐ฆโ, we go back and look at
๐ฆโฒโฒ + ๐ฅ๐ฆ = ๐ฅ5
And consider two cases (a) ๐ฆโฒโฒ โผ ๐ฅ5 (b) ๐ฅ๐ฆ โผ ๐ฅ5. The case of ๐ฆโฒโฒ โผ ๐ฅ๐ฆ was covered above.This is what we did to find ๐ฆโ (๐ฅ).
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3.2. HW2 CHAPTER 3. HWS
case (a)
๐ฆโฒโฒ๐ โผ ๐ฅ5
๐ฆโฒ๐ โผ15๐ฅ4
๐ฆ๐ โผ120๐ฅ3
Where constants of integration are ignored since subdominant for ๐ฅ โ โ. Now we check ifthis case is valid.
๐ฅ๐ฆ๐ โ ๐ฅ5
๐ฅ120๐ฅ3 โ ๐ฅ5
๐ฅ4 โ ๐ฅ5
No. Therefore case (a) did not work out. We try case (b) now
๐ฅ๐ฆ๐ โผ ๐ฅ5
๐ฆ๐ โผ ๐ฅ4
Now we check is this case is valid.
๐ฆโฒโฒ๐ โ ๐ฅ5
12๐ฅ2 โ ๐ฅ5
Yes. Therefore, we found
๐ฆ๐ โผ ๐ฅ4
Hence the complete asymptotic solution is
๐ฆ (๐ฅ) โผ ๐ฆโ (๐ฅ) + ๐ฆ๐ (๐ฅ)
โผ ๐๐ฅโ14 exp ๏ฟฝ
32๐๐ฅ
32 ๏ฟฝ + ๐ฅ4
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3.2. HW2 CHAPTER 3. HWS
3.2.8 key solution of selected problems
3.2.8.1 section 3 problem 27
104
3.2. HW2 CHAPTER 3. HWS
105
3.2. HW2 CHAPTER 3. HWS
106
3.2. HW2 CHAPTER 3. HWS
107
3.2. HW2 CHAPTER 3. HWS
3.2.8.2 section 3 problem 33 b
NEEP 548: Engineering Analysis II 2/6/2011
Problem 3.33: Bender & Orszag
Instructor: Leslie Smith
1 Problem Statement
Find the leading behavoir of the following equation as x โ 0+:
x4yโฒโฒโฒ โ 3x2yโฒ + 2y = 0 (1)
2 Solution
Need to make the substitution: y = eS(x)
yโฒ = SโฒeS(x)
yโฒโฒ = SโฒโฒeS(x) + (Sโฒ)2eS(x)
yโฒโฒโฒ = SโฒโฒโฒeS(x) + 3SโฒโฒSโฒeS(x) + (Sโฒ)3eS(x)
After substituting and diving through by y, one obtains,
Sโฒโฒโฒ + 3SโฒโฒSโฒ + (Sโฒ)3 โ3
x2Sโฒ +
2
x4โผ 0 (2)
Typically, Sโฒโฒ << (Sโฒ)2 =โ Sโฒโฒโฒ << (Sโฒ)3. Similarly, assume Sโฒโฒโฒ << (Sโฒ)3
With these assumptions, 2 becomes:
(Sโฒ)3 โ3
x2Sโฒ
โผ โ2
x4
which is still a difficult problem to solve. Therefore, also want to assume (Sโฒ)3 >> 3x2S
โฒ as x โ 0+.Then,
(Sโฒ)3 โผโ2
x4x โ 0+
Sโฒโผ wxโ4/3 x โ 0+
w = (โ2)1/3
So(x) โผ โ3wxโ1/3 x โ 0+
Sโฒโฒโผ โ
4w
3xโ7/3
Sโฒโฒโฒโผ
28w
9xโ10/3
1
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3.2. HW2 CHAPTER 3. HWS
Now, check our assumptions,
(Sโฒ)3 โผ โ2xโ12/3 >> โ3Sโฒxโ2โผ โ3wxโ10/3 x โ 0+ X
(Sโฒ)3 โผ โ2xโ12/3 >> Sโฒโฒโฒโผ
28w
9xโ10/3 x โ 0+ X
(Sโฒ)3 โผ โ2xโ12/3 >> SโฒโฒSโฒโผ
โ4w2
3xโ11/3 x โ 0+ X
Now estimate the inegrating function C(x) by letting,
S(x) = So(x) + C(x) (3)
and substitute this into 2.
Sโฒโฒโฒ
o + C โฒโฒโฒ + 3 (Sโฒโฒ
o + C โฒโฒ)(Sโฒ
o + C โฒ)๏ธธ ๏ธท๏ธท ๏ธธ
term 1
+(Sโฒ
o + C โฒ)3๏ธธ ๏ธท๏ธท ๏ธธ
term 2
โ3
x2(Sโฒ
o + C โฒ) +2
x4= 0 (4)
term 1: (Sโฒโฒ
oSโฒ
o +C โฒโฒSโฒ
o + Sโฒโฒ
oCโฒ + C โฒโฒC โฒ)
term 2: (Sโฒ
o)3 + (C โฒ)3 + 3(Sโฒ)2C โฒ + 3(Sโฒ)(C โฒ)2
From here, notice that we have already balanced the terms (Sโฒ
o)3 and โ2
x4 , so they are removedat this point. Now we make the following assumptions:
Sโฒ
o >> C โฒ, Sโฒโฒ
o >> C โฒโฒ, Sโฒโฒโฒ
o >> C โฒโฒโฒ as x โ 0+
These assumptions result in,
Sโฒโฒโฒ
o + 3Sโฒโฒ
oSโฒ
o + 3(Sโฒ
o)2C โฒ
โผ3Sโฒ
o
x2+
3C โฒ
x2x โ 0+
Insert the value of So found in the previous step,
28w
9xโ10/3
โ 4w2xโ11/3 + 3w2xโ8/3C โฒโผ 3wxโ10/3 + 3C โฒxโ6/3
In keeping with the dominant balance idea, it is clear that the middle two terms dominate, thusleading to the following simplified relation,
3w2xโ8/3C โฒโผ 4w2xโ11/3 x โ 0+
C โฒโผ
4
3xโ1 x โ 0+
C โผ4
3ln(x) x โ 0+
Then C โฒโฒโผ
โ43 xโ2, C โฒโฒโฒ
โผ83x
โ3 and once again, check assumptions made,
2
109
3.2. HW2 CHAPTER 3. HWS
Sโฒ
o โผ wxโ4/3 >> C โฒโผ
4
3xโ1 x โ 0+ X
Sโฒโฒ
o โผโ4
3xโ7/3 >> C โฒโฒ
โผโ4
3xโ2 x โ 0+ X
Sโฒโฒโฒ
o โผ28w
9xโ10/3 >> C โฒโฒโฒ
โผ8
3xโ3 x โ 0+ X
With the appearance of the ln(x) term, we have likely found the leading behavior already. However,lets find D, the third term, just to check.
Substitute y = e(So+Co+D), then divide by y,
yโฒ = [Sโฒ
o + C โฒ
o +Dโฒ]e(So+Co+D)
yโฒโฒ = [Sโฒโฒ
o + C โฒโฒ
o +Dโฒโฒ]e(So+Co+D) + [Sโฒ
o + C โฒ
o +Dโฒ]2e(So+Co+D)
yโฒโฒโฒ = [Sโฒโฒโฒ
o + C โฒโฒโฒ
o +Dโฒโฒโฒ]e(So+Co+D) + 3[Sโฒโฒ
o + C โฒโฒ
o +Dโฒโฒ][Sโฒ
o + C โฒ
o +Dโฒ]e(So+Co+D) + [Sโฒ
o + C โฒ
o +Dโฒ]3e(So+Co+D)
And...here we go...
Sโฒโฒโฒ
o +๏ฟฝ๏ฟฝC โฒโฒโฒ
o +๏ฟฝ๏ฟฝDโฒโฒโฒ+3[๏ฟฝ๏ฟฝ๏ฟฝSโฒโฒ
oSโฒ
o + Sโฒโฒ
oCโฒ
o +๏ฟฝ๏ฟฝ๏ฟฝSโฒโฒ
oDโฒ + C โฒโฒ
oSโฒ
o +๏ฟฝ๏ฟฝ๏ฟฝC โฒโฒ
oCโฒ
o +๏ฟฝ๏ฟฝ๏ฟฝC โฒโฒ
oDโฒ +๏ฟฝ๏ฟฝ๏ฟฝ
DโฒโฒSโฒ
o +๏ฟฝ๏ฟฝ๏ฟฝDโฒโฒC โฒ
o +๏ฟฝ๏ฟฝ๏ฟฝDโฒโฒDโฒ]
+๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ6Sโฒ
oCโฒ
oDโฒ + 3[๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ
(Sโฒ
o)2C โฒ
o + (Sโฒ
o)2Dโฒ + (C โฒ
o)2Sโฒ
o๏ธธ ๏ธท๏ธท ๏ธธ
Unknown balance
+๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ(C โฒ
o)2Dโฒ +๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ
(Dโฒ)2Sโฒ
o +๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ(Dโฒ)2C โฒ
o] +๏ฟฝ๏ฟฝ๏ฟฝ(Sโฒ
o)3
+๏ฟฝ๏ฟฝ๏ฟฝ(C โฒ
o)3 +๏ฟฝ๏ฟฝ๏ฟฝ(Dโฒ)3 โ
3
x2[Sโฒ
o + C โฒ
o +Dโฒ] +๏ฟฝ๏ฟฝ๏ฟฝ2
x4= 0
Assumptions: Sโฒ
o >> C โฒ
o >> Dโฒ, Sโฒโฒ
o >> C โฒโฒ
o >> Dโฒโฒ, Sโฒโฒโฒ
o >> C โฒโฒโฒ
o >> Dโฒโฒโฒ all as x โ 0+ and removepreviously balanced terms,
Sโฒโฒโฒ
o + 3Sโฒโฒ
oCโฒ
o + 3C โฒโฒ
oSโฒ
o + 3(Sโฒ
o)2Dโฒ
โผ โ3(C โฒ
o)Sโฒ
o x โ 0+
Subbing in the previously found values for So, Co,
โ3wxโ10/3 +28w
9xโ10/3
โ
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ16w
3xโ10/3
โ 4wxโ10/3 + 3w2xโ8/3Dโฒโผ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ16w
3xโ10/3
3
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3.2. HW2 CHAPTER 3. HWS
โ3w2xโ8/3Dโฒโผ
35w
9xโ10/3
Dโฒโผ
35
27wxโ2/3
D โผ35
27w(1
3)x1/3 + d
So since x1/3 โ 0 as x โ 0+, the leading order behavior is
y โผ exp[โw
3xโ1/3 +
4
3ln(x) + d] x โ 0+
4
111
3.2. HW2 CHAPTER 3. HWS
3.2.8.3 section 3 problem 33 c
112
3.2. HW2 CHAPTER 3. HWS
113
3.2. HW2 CHAPTER 3. HWS
114
3.2. HW2 CHAPTER 3. HWS
115
3.2. HW2 CHAPTER 3. HWS
116
3.2. HW2 CHAPTER 3. HWS
3.2.8.4 section 3 problem 35
Some notes on BO 3.35 (to get the leading behavior):
โข a consistent balance at lowest order seems to be
(S โฒ
o)2 โผ exp(2/x), x โ 0
โข taking the square root and integrating leads to
So โผ ยฑ
โซ x
1
exp(1/s)ds
โข then change variables s = 1/t to find
So โผ ยฑ
โซ (1/x)
1
exp(t)
(
โ1
t2
)
dt
โข integration by parts gives
So โผ โx2 exp(1/x) +ยฑao +โ
โซ (1/x)
1
2
t3exp(t)dt
โผ โx2 exp(1/x)
and subdominant terms have been dropped. [Why are these terms subdominant andwhy is this the only consistent integration by parts?]
117
3.2. HW2 CHAPTER 3. HWS
3.2.8.5 section 3 problem 42a
118
3.2. HW2 CHAPTER 3. HWS
119
3.2. HW2 CHAPTER 3. HWS
120
3.2. HW2 CHAPTER 3. HWS
121
3.2. HW2 CHAPTER 3. HWS
122
3.2. HW2 CHAPTER 3. HWS
123
3.3. HW3 CHAPTER 3. HWS
3.3 HW3
3.3.1 problem 9.3 (page 479)
problem (a) show that if ๐ (๐ฅ) < 0 for 0 โค ๐ฅ โค 1 then the solution to 9.1.7 has boundarylayer at ๐ฅ = 1. (b) Find a uniform approximation with error ๐ (๐) to the solution 9.1.7 when๐ (๐ฅ) < 0 for 0 โค ๐ฅ โค 1 (c) Show that if ๐ (๐ฅ) > 0 it is impossible to match to a boundary layerat ๐ฅ = 1
solution
3.3.1.1 Part a
Equation 9.1.7 at page 422 is
๐๐ฆโฒโฒ + ๐ (๐ฅ) ๐ฆโฒ + ๐ (๐ฅ) ๐ฆ (๐ฅ) = 0 (9.1.7)
๐ฆ (0) = ๐ด๐ฆ (1) = ๐ต
For 0 โค ๐ฅ โค 1. Now we solve for ๐ฆ๐๐ (๐ฅ), but first we introduce inner variable ๐. Weassume boundary layer is at ๐ฅ = 0, then show that this leads to inconsistency. Let ๐ = ๐ฅ
๐๐be the inner variable. We express the original ODE using this new variable. We also needto determine ๐. Since ๐๐ฆ
๐๐ฅ =๐๐ฆ๐๐
๐๐๐๐ฅ then ๐๐ฆ
๐๐ฅ =๐๐ฆ๐๐๐
โ๐. Hence ๐๐๐ฅ โก ๐
โ๐ ๐๐๐
๐2
๐๐ฅ2=๐๐๐ฅ
๐๐๐ฅ
= ๏ฟฝ๐โ๐๐๐๐๏ฟฝ ๏ฟฝ
๐โ๐๐๐๐๏ฟฝ
= ๐โ2๐๐2
๐๐2
Therefore ๐2๐ฆ๐๐ฅ2 = ๐
โ2๐ ๐2๐ฆ๐๐2 and (9.1.7) becomes
๐๐โ2๐๐2๐ฆ๐๐2
+ ๐ (๐ฅ) ๐โ๐๐๐ฆ๐๐
+ ๐ฆ = 0
๐1โ2๐๐2๐ฆ๐๐2
+ ๐ (๐ฅ) ๐โ๐๐๐ฆ๐๐
+ ๐ฆ = 0
The largest terms are ๏ฟฝ๐1โ2๐, ๐โ๐๏ฟฝ, therefore balance gives 1 โ 2๐ = โ๐ or ๐ = 1. The ODE nowbecomes
๐โ1๐2๐ฆ๐๐2
+ ๐ (๐ฅ) ๐โ1๐๐ฆ๐๐
+ ๐ฆ = 0 (1)
Assuming that
๐ฆ๐๐ (๐ฅ) =โ๏ฟฝ๐=0
๐๐๐ฆ๐ = ๐ฆ0 + ๐๐ฆ1 + ๐2๐ฆ2 +โฏ
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3.3. HW3 CHAPTER 3. HWS
And substituting the above into (1) gives
๐โ1 ๏ฟฝ๐ฆโฒโฒ0 + ๐๐ฆโฒโฒ1 +โฏ๏ฟฝ + ๐ (๐ฅ) ๐โ1 ๏ฟฝ๐ฆโฒ0 + ๐๐ฆโฒ1 +โฏ๏ฟฝ + ๏ฟฝ๐ฆ0 + ๐๐ฆ1 +โฏ๏ฟฝ = 0 (1A)
Collecting powers of ๐๏ฟฝ๐โ1๏ฟฝ terms, gives the ODE to solve for ๐ฆ๐๐0 as
๐ฆโฒโฒ0 โผ โ๐ (๐ฅ) ๐ฆโฒ0In the rapidly changing region, because the boundary layer is very thin, we can approximate๐ (๐ฅ) by ๐ (0). The above becomes
๐ฆโฒโฒ0 โผ โ๐ (0) ๐ฆโฒ0But we are told that ๐ (๐ฅ) < 0, so ๐ (0) < 0, hence โ๐ (0) is positive. Let โ๐ (0) = ๐2, to make itmore clear this is positive, then the ODE to solve is
๐ฆโฒโฒ0 โผ ๐2๐ฆโฒ0The solution to this ODE is
๐ฆ0 (๐) โผ๐ถ1๐2๐๐2๐ + ๐ถ2
Using ๐ฆ (0) = ๐ด, then the above gives ๐ด = ๐ถ1๐2 + ๐ถ2 or ๐ถ2 = ๐ด โ
๐ถ1๐2 and the ODE becomes
๐ฆ0 (๐) โผ๐ถ1๐2๐๐2๐ + ๏ฟฝ๐ด โ
๐ถ1๐2 ๏ฟฝ
โผ๐ถ1๐2๏ฟฝ๐๐2๐ โ 1๏ฟฝ + ๐ด
We see from the above solution for the inner layer, that as ๐ increases (meaning we aremoving away from ๐ฅ = 0), then the solution ๐ฆ0 (๐) and its derivative is increasing and notdecreasing since ๐ฆโฒ0 (๐) = ๐ถ1๐๐
2๐ and ๐ฆโฒโฒ0 (๐) = ๐ถ1๐2๐๐2๐.
But this contradicts what we assumed that the boundary layer is at ๐ฅ = 0 since we expectthe solution to change less rapidly as we move away from ๐ฅ = 0. Hence we conclude that if๐ (๐ฅ) < 0, then the boundary layer can not be at ๐ฅ = 0.
Let us now see what happens by taking the boundary layer to be at ๐ฅ = 1. We repeat thesame process as above, but now the inner variable as defined as
๐ =1 โ ๐ฅ๐๐
We express the original ODE using this new variable and determine ๐. Since ๐๐ฆ๐๐ฅ =
๐๐ฆ๐๐
๐๐๐๐ฅ then
๐๐ฆ๐๐ฅ =
๐๐ฆ๐๐(โ๐โ๐). Hence ๐
๐๐ฅ โก (โ๐โ๐) ๐
๐๐
๐2
๐๐ฅ2=๐๐๐ฅ
๐๐๐ฅ
= ๏ฟฝ(โ๐โ๐)๐๐๐๏ฟฝ ๏ฟฝ
(โ๐โ๐)๐๐๐๏ฟฝ
= ๐โ2๐๐2
๐๐2
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3.3. HW3 CHAPTER 3. HWS
Therefore ๐2๐ฆ๐๐ฅ2 = ๐
โ2๐ ๐2๐ฆ๐๐2 and equation (9.1.7) becomes
๐๐โ2๐๐2๐ฆ๐๐2
โ ๐ (๐ฅ) ๐โ๐๐๐ฆ๐๐
+ ๐ฆ = 0
๐1โ2๐๐2๐ฆ๐๐2
โ ๐ (๐ฅ) ๐โ๐๐๐ฆ๐๐
+ ๐ฆ = 0
The largest terms are ๏ฟฝ๐1โ2๐, ๐โ๐๏ฟฝ, therefore matching them gives 1 โ 2๐ = โ๐ or
๐ = 1
The ODE now becomes
๐โ1๐2๐ฆ๐๐2
โ ๐ (๐ฅ) ๐โ1๐๐ฆ๐๐
+ ๐ฆ = 0 (2)
Assuming that
๐ฆ๐๐ (๐ฅ) =โ๏ฟฝ๐=0
๐๐๐ฆ๐ = ๐ฆ0 + ๐๐ฆ1 + ๐2๐ฆ2 +โฏ
And substituting the above into (2) gives
๐โ1 ๏ฟฝ๐ฆโฒโฒ0 + ๐๐ฆโฒโฒ1 +โฏ๏ฟฝ โ ๐ (๐ฅ) ๐โ1 ๏ฟฝ๐ฆโฒ0 + ๐๐ฆโฒ1 +โฏ๏ฟฝ + ๏ฟฝ๐ฆ0 + ๐๐ฆ1 +โฏ๏ฟฝ = 0 (2A)
Collecting ๐๏ฟฝ๐โ1๏ฟฝ terms, gives the ODE to solve for ๐ฆ๐๐0 as
๐ฆโฒโฒ0 โผ ๐ (๐ฅ) ๐ฆโฒ0In the rapidly changing region, ๐ผ = ๐ (1), because the boundary layer is very thin, weapproximated ๐ (๐ฅ) by ๐ (1). The above becomes
๐ฆโฒโฒ0 โผ ๐ผ๐ฆโฒ0But we are told that ๐ (๐ฅ) < 0, so ๐ผ < 0, and the above becomes
๐ฆโฒโฒ0 โผ ๐ผ๐ฆโฒ0The solution to this ODE is
๐ฆ0 (๐) โผ๐ถ1๐ผ๐๐ผ๐ + ๐ถ2 (3)
Using
๐ฆ (๐ฅ = 1) = ๐ฆ (๐ = 0)= ๐ต
Then (3) gives ๐ต = ๐ถ1๐ผ + ๐ถ2 or ๐ถ2 = ๐ต โ
๐ถ1๐ผ and (3) becomes
๐ฆ0 (๐) โผ๐ถ1๐ผ๐๐ผ๐ + ๏ฟฝ๐ต โ
๐ถ1๐ผ ๏ฟฝ
โผ๐ถ1๐ผ๏ฟฝ๐๐ผ๐ โ 1๏ฟฝ + ๐ต (4)
From the above, ๐ฆโฒ0 (๐) = โ๐ถ1๐๐ผ๐ and ๐ฆโฒโฒ0 (๐) = ๐ถ1๐ผ๐๐ผ๐. We now see that as that as ๐ increases(meaning we are moving away from ๐ฅ = 1 towards the left), then the solution ๐ฆ0 (๐) is actuallychanging less rapidly. This is because ๐ผ < 0. The solution is changing less rapidly as we
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move away from the boundary layer as what we expect. Therefore, we conclude that if๐ (๐ฅ) < 0 then the boundary layer can not be at ๐ฅ = 0 and has to be at ๐ฅ = 1.
3.3.1.2 Part b
To find uniform approximation, we need now to find ๐ฆ๐๐ข๐ก (๐ฅ) and then do the matching. Sincefrom part(a) we concluded that ๐ฆ๐๐ is near ๐ฅ = 1, then we assume now that ๐ฆ๐๐ข๐ก (๐ฅ) is near๐ฅ = 0. Let
๐ฆ๐๐ข๐ก (๐ฅ) =โ๏ฟฝ๐=0
๐๐๐ฆ๐ = ๐ฆ0 + ๐๐ฆ1 + ๐2๐ฆ2 +โฏ
Substituting this into (9.1.7) gives
๐ ๏ฟฝ๐ฆโฒโฒ0 + ๐๐ฆโฒโฒ1 + ๐2๐ฆโฒโฒ2 +โฏ๏ฟฝ + ๐ (๐ฅ) ๏ฟฝ๐ฆโฒ0 + ๐๐ฆโฒ1 + ๐2๐ฆโฒ2 +โฏ๏ฟฝ + ๐ (๐ฅ) ๏ฟฝ๐ฆ0 + ๐๐ฆ1 + ๐2๐ฆ2 +โฏ๏ฟฝ = 0
Collecting terms of ๐ (1) gives the ODE
๐ (๐ฅ) ๐ฆโฒ0 + ๐ (๐ฅ) ๐ฆ0 = 0
The solution to this ODE is
๐ฆ0 (๐ฅ) = ๐ถ2๐โโซ
๐ฅ0
๐(๐ )๐(๐ )๐๐
Applying ๐ฆ (0) = ๐ด gives
๐ด = ๐ถ2๐โโซ
10
๐(๐ )๐(๐ )๐๐
= ๐ถ2๐ธ
Where ๐ธ is constant, which is the value of the definite integral ๐ธ = ๐โโซ10
๐(๐ )๐(๐ )๐๐ . Hence the
solution ๐ฆ๐๐ข๐ก (๐ฅ) can now be written as
๐ฆ๐๐ข๐ก0 (๐ฅ) =๐ด๐ธ๐โโซ
๐ฅ0
๐(๐ )๐(๐ )๐๐
We are now ready to do the matching.
lim๐โโ
๐ฆ๐๐ (๐) โผ lim๐ฅโ0
๐ฆ๐๐ข๐ก (๐ฅ)
lim๐โโ
๐ถ1๐ผ๏ฟฝ๐๐ผ๐ โ 1๏ฟฝ + ๐ต โผ lim
๐ฅโ0
๐ด๐ธ๐โโซ
๐ฅ0
๐(๐ )๐(๐ )๐๐
But since ๐ผ = ๐ (1) < 0 then the above simplifies to
โ๐ถ1๐ (1)
+ ๐ต =๐ด๐ธ
๐ถ1 = โ๐ (1) ๏ฟฝ๐ด๐ธโ ๐ต๏ฟฝ
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Hence inner solution becomes
๐ฆ๐๐0 (๐) โผโ๐ (1) ๏ฟฝ๐ด๐ธ โ ๐ต๏ฟฝ
๐ (1)๏ฟฝ๐๐(1)๐ โ 1๏ฟฝ + ๐ต
โผ ๏ฟฝ๐ต โ๐ด๐ธ ๏ฟฝ
๏ฟฝ๐๐(1)๐ โ 1๏ฟฝ + ๐ต
โผ ๐ต ๏ฟฝ๐๐(1)๐ โ 1๏ฟฝ โ๐ด๐ธ๏ฟฝ๐๐(1)๐ โ 1๏ฟฝ + ๐ต
โผ ๏ฟฝ๐ต โ๐ด๐ธ ๏ฟฝ
๏ฟฝ๐๐(1)๐ โ 1๏ฟฝ + ๐ต
The uniform solution is
๐ฆuniform (๐ฅ) โผ ๐ฆ๐๐ (๐) + ๐ฆ๐๐ข๐ก (๐ฅ) โ ๐ฆ๐๐๐ก๐โ
โผ
๐ฆ๐๐
๏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๏ฟฝ๐ต โ
๐ด๐ธ ๏ฟฝ
๏ฟฝ๐๐(1)๐ โ 1๏ฟฝ + ๐ต +
๐ฆ๐๐ข๐ก
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๐ด๐ธ๐โโซ
๐ฅ0
๐(๐ )๐(๐ )๐๐ โ
๐ด๐ธ
Or in terms of ๐ฅ only
๐ฆuniform (๐ฅ) โผ ๏ฟฝ๐ต โ๐ด๐ธ ๏ฟฝ
๏ฟฝ๐๐(1)1โ๐ฅ๐ โ 1๏ฟฝ + ๐ต +
๐ด๐ธ๐โโซ
๐ฅ0
๐(๐ )๐(๐ )๐๐ โ
๐ด๐ธ
3.3.1.3 Part c
We now assume the boundary layer is at ๐ฅ = 1 but ๐ (๐ฅ) > 0. From part (a), we found thatthe solution for ๐ฆ๐๐0 (๐) where boundary layer at ๐ฅ = 1 is
๐ฆ0 (๐) โผ๐ถ1๐ผ๏ฟฝ๐๐ผ๐ โ 1๏ฟฝ + ๐ต
But now ๐ผ = ๐ (1) > 0 and not negative as before. We also found that ๐ฆ๐๐ข๐ก0 (๐ฅ) solution was
๐ฆ๐๐ข๐ก0 (๐ฅ) =๐ด๐ธ๐โโซ
๐ฅ0
๐(๐ )๐(๐ )๐๐
Lets now try to do the matching and see what happens
lim๐โโ
๐ฆ๐๐ (๐) โผ lim๐ฅโ0
๐ฆ๐๐ข๐ก (๐ฅ)
lim๐โโ
๐ถ1๐ผ๏ฟฝ๐๐ผ๐ โ 1๏ฟฝ + ๐ต + ๐ (๐) โผ lim
๐ฅโ0
๐ด๐ธ๐โโซ
๐ฅ0
๐(๐ )๐(๐ )๐๐ + ๐ (๐)
lim๐โโ
๐ถ1 ๏ฟฝ๐๐ผ๐
๐ผโ1๐ผ๏ฟฝ
โผ๐ด๐ธโ ๐ต
Since now ๐ผ > 0, then the term on the left blows up, while the term on the right is finite.Not possible to match, unless ๐ถ1 = 0. But this means the boundary layer solution is just aconstant ๐ต and that ๐ด
๐ธ = ๐ต. So the matching does not work in general for arbitrary conditions.This means if ๐ (๐ฅ) > 0, it is not possible to match boundary layer at ๐ฅ = 1.
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3.3. HW3 CHAPTER 3. HWS
3.3.2 problem 9.4(b)
Problem Find the leading order uniform asymptotic approximation to the solution of
๐๐ฆโฒโฒ + ๏ฟฝ1 + ๐ฅ2๏ฟฝ ๐ฆโฒ โ ๐ฅ3๐ฆ (๐ฅ) = 0 (1)
๐ฆ (0) = 1๐ฆ (1) = 1
For 0 โค ๐ฅ โค 1 in the limit as ๐ โ 0.
solution
Since ๐ (๐ฅ) = ๏ฟฝ1 + ๐ฅ2๏ฟฝ is positive, we expect the boundary layer to be near ๐ฅ = 0. First we find๐ฆ๐๐ข๐ก (๐ฅ), which is near ๐ฅ = 1. Assuming
๐ฆ๐๐ข๐ก (๐ฅ) =โ๏ฟฝ๐=0
๐๐๐ฆ๐ = ๐ฆ0 + ๐๐ฆ1 + ๐2๐ฆ2 +โฏ
And substituting this into (1) gives
๐ ๏ฟฝ๐ฆโฒโฒ0 + ๐๐ฆโฒโฒ1 + ๐2๐ฆโฒโฒ2 +โฏ๏ฟฝ + ๏ฟฝ1 + ๐ฅ2๏ฟฝ ๏ฟฝ๐ฆโฒ0 + ๐๐ฆโฒ1 + ๐2๐ฆโฒ2 +โฏ๏ฟฝ โ ๐ฅ3 ๏ฟฝ๐ฆ0 + ๐๐ฆ1 + ๐2๐ฆ2 +โฏ๏ฟฝ = 0
Collecting terms in ๐ (1) gives the ODE
๏ฟฝ1 + ๐ฅ2๏ฟฝ ๐ฆโฒ0 โผ ๐ฅ3๐ฆ0
The ODE becomes ๐ฆโฒ0 โผ๐ฅ3
๏ฟฝ1+๐ฅ2๏ฟฝ๐ฆ0 with integrating factor ๐ = ๐
โซ โ๐ฅ3
๏ฟฝ1+๐ฅ2๏ฟฝ๐๐ฅ. To evaluate โซ โ๐ฅ3
๏ฟฝ1+๐ฅ2๏ฟฝ๐๐ฅ,
let ๐ข = ๐ฅ2, hence ๐๐ข๐๐ฅ = 2๐ฅ and the integral becomes
๏ฟฝโ๐ฅ3
๏ฟฝ1 + ๐ฅ2๏ฟฝ๐๐ฅ = โ๏ฟฝ
๐ข๐ฅ(1 + ๐ข)
๐๐ข2๐ฅ
= โ12 ๏ฟฝ
๐ข(1 + ๐ข)
๐๐ข
But
๏ฟฝ๐ข
(1 + ๐ข)๐๐ข = ๏ฟฝ1 โ
1(1 + ๐ข)
๐๐ข
= ๐ข โ ln (1 + ๐ข)But ๐ข = ๐ฅ2, hence
๏ฟฝโ๐ฅ3
๏ฟฝ1 + ๐ฅ2๏ฟฝ๐๐ฅ =
โ12๏ฟฝ๐ฅ2 โ ln ๏ฟฝ1 + ๐ฅ2๏ฟฝ๏ฟฝ
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3.3. HW3 CHAPTER 3. HWS
Therefore the integrating factor is ๐ = exp ๏ฟฝโ12 ๐ฅ2 + 1
2 ln ๏ฟฝ1 + ๐ฅ2๏ฟฝ๏ฟฝ. The ODE becomes
๐๐๐ฅ๏ฟฝ๐๐ฆ0๏ฟฝ = 0
๐๐ฆ0 โผ ๐
๐ฆ๐๐ข๐ก (๐ฅ) โผ ๐ exp ๏ฟฝ12๐ฅ2 โ
12
ln ๏ฟฝ1 + ๐ฅ2๏ฟฝ๏ฟฝ
โผ ๐๐12๐ฅ
2๐ln๏ฟฝ1+๐ฅ
2๏ฟฝโ12
โผ ๐๐๐ฅ22
โ1 + ๐ฅ2To find ๐, using boundary conditions ๐ฆ (1) = 1 gives
1 = ๐๐12
โ2
๐ = โ2๐โ12
Hence
๐ฆ๐๐ข๐ก0 (๐ฅ) โผ โ2๐๐ฅ2โ12
โ1 + ๐ฅ2
Now we find ๐ฆ๐๐ (๐ฅ) near ๐ฅ = 0. Let ๐ = ๐ฅ๐๐ be the inner variable. We express the original ODE
using this new variable and determine ๐. Since ๐๐ฆ๐๐ฅ =
๐๐ฆ๐๐
๐๐๐๐ฅ then ๐๐ฆ
๐๐ฅ =๐๐ฆ๐๐๐
โ๐. Hence ๐๐๐ฅ โก ๐
โ๐ ๐๐๐
๐2
๐๐ฅ2=๐๐๐ฅ
๐๐๐ฅ
= ๏ฟฝ๐โ๐๐๐๐๏ฟฝ ๏ฟฝ
๐โ๐๐๐๐๏ฟฝ
= ๐โ2๐๐2
๐๐2
Therefore ๐2๐ฆ๐๐ฅ2 = ๐
โ2๐ ๐2๐ฆ๐๐2 and ๐๐ฆ
โฒโฒ + ๏ฟฝ1 + ๐ฅ2๏ฟฝ ๐ฆโฒ โ ๐ฅ3๐ฆ (๐ฅ) = 0 becomes
๐๐โ2๐๐2๐ฆ๐๐2
+ ๏ฟฝ1 + (๐๐๐)2๏ฟฝ ๐โ๐๐๐ฆ๐๐
โ (๐๐๐)3 ๐ฆ = 0
๐1โ2๐๐2๐ฆ๐๐2
+ ๏ฟฝ1 + ๐2๐2๐๏ฟฝ ๐โ๐๐๐ฆ๐๐
โ ๐3๐3๐๐ฆ = 0
The largest terms are ๏ฟฝ๐1โ2๐, ๐โ๐๏ฟฝ, therefore matching them gives 1 โ 2๐ = โ๐ or ๐ = 1. TheODE now becomes
๐โ1๐2๐ฆ๐๐2
+ ๏ฟฝ1 + ๐2๐2๏ฟฝ ๐โ1๐๐ฆ๐๐
โ ๐3๐3๐ฆ = 0 (2)
130
3.3. HW3 CHAPTER 3. HWS
Assuming that
๐ฆ๐๐ (๐ฅ) =โ๏ฟฝ๐=0
๐๐๐ฆ๐ = ๐ฆ0 + ๐๐ฆ1 + ๐2๐ฆ2 +โฏ
And substituting the above into (2) gives
๐โ1 ๏ฟฝ๐ฆโฒโฒ0 + ๐๐ฆโฒโฒ1 +โฏ๏ฟฝ + ๏ฟฝ1 + ๐2๐2๏ฟฝ ๐โ1 ๏ฟฝ๐ฆโฒ0 + ๐๐ฆโฒ1 +โฏ๏ฟฝ โ ๐3๐3 ๏ฟฝ๐ฆ0 + ๐๐ฆ1 +โฏ๏ฟฝ = 0 (2A)
Collecting terms in ๐๏ฟฝ๐โ1๏ฟฝ gives the ODE
๐ฆโฒโฒ0 (๐) โผ โ๐ฆโฒ0 (๐)
The solution to this ODE is
๐ฆ๐๐0 (๐) โผ ๐1 + ๐2๐โ๐ (3)
Applying ๐ฆ๐๐0 (0) = 1 gives
1 = ๐1 + ๐2๐1 = 1 โ ๐2
Hence (3) becomes
๐ฆ๐๐0 (๐) โผ (1 โ ๐2) + ๐2๐โ๐
โผ 1 + ๐2 ๏ฟฝ๐โ๐ โ 1๏ฟฝ (4)
Now that we found ๐ฆ๐๐ข๐ก and ๐ฆ๐๐, we apply matching to find ๐2 in the ๐ฆ๐๐ solution.
lim๐โโ
๐ฆ๐๐0 (๐) โผ lim๐ฅโ0+
๐ฆ๐๐ข๐ก0 (๐ฅ)
lim๐โโ
1 + ๐2 ๏ฟฝ๐โ๐ โ 1๏ฟฝ โผ lim๐ฅโ0+
โ2๐๐ฅ2โ12
โ1 + ๐ฅ2
1 โ ๐2 โผ๏ฟฝ2๐
lim๐ฅโ0+
๐๐ฅ22
โ1 + ๐ฅ2
โผ๏ฟฝ2๐
lim๐ฅโ0+
๐๐ฅ22
โ1 + ๐ฅ2
=๏ฟฝ2๐
Hence
๐2 = 1 โ๏ฟฝ2๐
131
3.3. HW3 CHAPTER 3. HWS
Therefore the ๐ฆ๐๐0 (๐) becomes
๐ฆ๐๐0 (๐) โผ 1 +โโโโโโ1 โ๏ฟฝ
2๐
โโโโโโ ๏ฟฝ๐
โ๐ โ 1๏ฟฝ
โผ 1 + ๏ฟฝ๐โ๐ โ 1๏ฟฝ โ๏ฟฝ2๐๏ฟฝ๐โ๐ โ 1๏ฟฝ
โผ ๐โ๐ โ๏ฟฝ2๐๏ฟฝ๐โ๐ โ 1๏ฟฝ
โผ ๐โ๐ โ๏ฟฝ2๐๐โ๐ +
๏ฟฝ2๐
โผ ๐โ๐โโโโโโ1 โ๏ฟฝ
2๐
โโโโโโ +๏ฟฝ
2๐
โผ 0.858 + 0.142๐โ๐
Therefore, the uniform solution is
๐ฆ๐ข๐๐๐๐๐๐ โผ ๐ฆ๐๐ (๐ฅ) + ๐ฆ๐๐ข๐ก (๐ฅ) โ ๐ฆ๐๐๐ก๐โ + ๐ (๐) (4)
Where ๐ฆ๐๐๐ก๐โ is ๐ฆ๐๐ (๐ฅ) at the boundary layer matching location. (or ๐ฆ๐๐ข๐ก at same matchinglocation). Hence
๐ฆ๐๐๐ก๐โ โผ 1 โ ๐2
โผ 1 โโโโโโโ1 โ๏ฟฝ
2๐
โโโโโโ
โผ๏ฟฝ2๐
Hence (4) becomes
๐ฆ๐ข๐๐๐๐๐๐ โผ
๐ฆ๐๐๏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๏ฟฝ๐โ๐
โโโโโโ1 โ๏ฟฝ
2๐
โโโโโโ +๏ฟฝ
2๐+
๐ฆ๐๐ข๐ก๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ
โ2๐๐ฅ2โ12
โ1 + ๐ฅ2โ
๐ฆ๐๐๐ก๐โ๏ฟฝ
๏ฟฝ2๐
โผ ๐โ๐ฅ๐
โโโโโโ1 โ๏ฟฝ
2๐
โโโโโโ + โ2
๐๐ฅ2โ12
โ1 + ๐ฅ2+ ๐ (๐)
This is the leading order uniform asymptotic approximation solution. To verify the result,the numerical solution was plotted against the above solution for ๐ = {0.1, 0.05, 0.01}. We seefrom these plots that as ๐ becomes smaller, the asymptotic solution becomes more accuratewhen compared to the numerical solution. This is because the error, which is ๐ (๐) , becomessmaller. The code used to generate these plots is
132
3.3. HW3 CHAPTER 3. HWS
In[180]:= eps = 0.1;
sol = NDSolve[{1 / 100 y''[x] + (1 + x^2) y'[x] - x^3 y[x] โฉต 0, y[0] โฉต 1, y[1] โฉต 1}, y, {x, 0, 1}];
p1 = Plot[Evaluate[y[x] /. sol], {x, 0, 1}, Frame โ True,
FrameLabel โ {{"y(x)", None}, {"x", Row[{"numberical vs. asymptotic for eps =", eps}]}},
GridLines โ Automatic, GridLinesStyle โ LightGray];
mysol[x_, eps_] := Exp-x
eps 1 - Sqrt
2
Exp[1] +
Sqrt[2] Exp x2-1
2
Sqrt[1 + x^2];
p2 = Plot[mysol[x, eps], {x, 0, 1}, PlotRange โ All, PlotStyle โ Red];
Show[Legended[p1, Style["Numerical", Red]], Legended[p2, Style["Asymtotic", Blue]]]
The following are the three plots for each value of ๐
Out[6]=
0.0 0.2 0.4 0.6 0.8 1.0
0.86
0.88
0.90
0.92
0.94
0.96
0.98
1.00
x
y(x)
numberical vs. asymptotic for eps =0.1
NumericalAsymptotic
Out[12]=
0.0 0.2 0.4 0.6 0.8 1.0
0.86
0.88
0.90
0.92
0.94
0.96
0.98
1.00
x
y(x)
numberical vs. asymptotic for eps =0.05
NumericalAsymptotic
133
3.3. HW3 CHAPTER 3. HWS
Out[17]=
0.0 0.2 0.4 0.6 0.8 1.0
0.86
0.88
0.90
0.92
0.94
0.96
0.98
1.00
x
y(x)
numberical vs. asymptotic for eps =0.01
NumericalAsymptotic
To see the e๏ฟฝect on changing ๐ on only the asymptotic approximation, the following plotgives the approximation solution only as ๐ changes. We see how the approximation convergesto the numerical solution as ๐ becomes smaller.
In[236]:= Plot[{mysol[x, .1], mysol[x, .05], mysol[x, .01]}, {x, 0, 1},
PlotLegends โ {"โ=0.1", "โ=0.05", "โ=0.01"}, Frame โ True,
FrameLabel โ {{"y(x)", None}, {"x", "Asymptotic solution as eps changes"}},
BaseStyle โ 14]
Out[236]=
0.0 0.2 0.4 0.6 0.8 1.0
0.86
0.88
0.90
0.92
0.94
0.96
0.98
1.00
x
y(x)
Asymptotic solution as eps changes
โ=0.1
โ=0.05
โ=0.01
3.3.3 problem 9.6
Problem Consider initial value problem
๐ฆโฒ = ๏ฟฝ1 +๐ฅโ2
100๏ฟฝ๐ฆ2 โ 2๐ฆ + 1
With ๐ฆ (1) = 1 on the interval 0 โค ๐ฅ โค 1. (a) Formulate this problem as perturbation problemby introducing a small parameter ๐. (b) Find outer approximation correct to order ๐ witherrors of order ๐2. Where does this approximation break down? (c) Introduce inner variableand find the inner solution valid to order 1 (with errors of order ๐). By matching to the outersolution find a uniform valid solution to ๐ฆ (๐ฅ) on interval 0 โค ๐ฅ โค 1. Estimate the accuracy of
134
3.3. HW3 CHAPTER 3. HWS
this approximation. (d) Find inner solution correct to order ๐ (with errors of order ๐2) andshow that it matches to the outer solution correct to order ๐.
solution
3.3.3.1 Part a
Since 1100 is relatively small compared to all other coe๏ฟฝcients, we replace it with ๐ and the
ODE becomes
๐ฆโฒ โ ๏ฟฝ1 +๐๐ฅ2๏ฟฝ ๐ฆ2 + 2๐ฆ = 1 (1)
3.3.3.2 Part b
Assuming boundary layer is on the left side at ๐ฅ = 0. We now solve for ๐ฆ๐๐ข๐ก (๐ฅ), which is thesolution near ๐ฅ = 1.
๐ฆ๐๐ข๐ก (๐ฅ) =โ๏ฟฝ๐=0
๐๐๐ฆ๐ = ๐ฆ0 + ๐๐ฆ1 + ๐2๐ฆ2 +โฏ
Substituting this into (1) gives
๏ฟฝ๐ฆโฒ0 + ๐๐ฆโฒ1 + ๐2๐ฆโฒ2 +โฏ๏ฟฝ โ ๏ฟฝ1 +๐๐ฅ2๏ฟฝ ๏ฟฝ๐ฆ0 + ๐๐ฆ1 + ๐2๐ฆ2 +โฏ๏ฟฝ
2+ 2 ๏ฟฝ๐ฆ0 + ๐๐ฆ1 + ๐2๐ฆ2 +โฏ๏ฟฝ = 1
Expanding the above to see more clearly the terms gives
๏ฟฝ๐ฆโฒ0 + ๐๐ฆโฒ1 + ๐2๐ฆโฒ2 +โฏ๏ฟฝ โ ๏ฟฝ1 +๐๐ฅ2๏ฟฝ ๏ฟฝ๐ฆ20 + ๐ ๏ฟฝ2๐ฆ0๐ฆ1๏ฟฝ + ๐2 ๏ฟฝ2๐ฆ0๐ฆ2 + ๐ฆ21๏ฟฝ +โฏ๏ฟฝ + 2 ๏ฟฝ๐ฆ0 + ๐๐ฆ1 + ๐2๐ฆ2 +โฏ๏ฟฝ = 1
(2)
The leading order are those terms of coe๏ฟฝcient ๐ (1). This gives
๐ฆโฒ0 โ ๐ฆ20 + 2๐ฆ0 โผ 1
With boundary conditions ๐ฆ (1) = 1.๐๐ฆ0๐๐ฅ
โผ ๐ฆ20 โ 2๐ฆ0 + 1
This is separable๐๐ฆ0
๐ฆ20 โ 2๐ฆ0 + 1โผ ๐๐ฅ
๐๐ฆ0๏ฟฝ๐ฆ0 โ 1๏ฟฝ
2 โผ ๐๐ฅ
135
3.3. HW3 CHAPTER 3. HWS
For ๐ฆ0 โ 1. Integrating
๏ฟฝ๐๐ฆ0
๏ฟฝ๐ฆ0 โ 1๏ฟฝ2 โผ๏ฟฝ๐๐ฅ
โ1๐ฆ0 โ 1
โผ ๐ฅ + ๐ถ
๏ฟฝ๐ฆ0 โ 1๏ฟฝ (๐ฅ + ๐ถ) โผ โ1
๐ฆ0 โผโ1๐ฅ + ๐ถ
+ 1 (3)
To find ๐ถ, from ๐ฆ (1) = 1, we find
1 โผโ11 + ๐ถ
+ 1
1 โผ๐ถ
1 + ๐ถThis is only possible if ๐ถ = โ. Therefore from (2), we conclude that
๐ฆ0 (๐ฅ) โผ 1
The above is leading order for the outer solution. Now we repeat everything to find ๐ฆ๐๐ข๐ก1 (๐ฅ).From (2) above, we now keep all terms with ๐ (๐) which gives
๐ฆโฒ1 โ 2๐ฆ0๐ฆ1 + 2๐ฆ1 โผ1๐ฅ2๐ฆ20
But we found ๐ฆ0 (๐ฅ) โผ 1 from above, so the above ODE becomes
๐ฆโฒ1 โ 2๐ฆ1 + 2๐ฆ1 โผ1๐ฅ2
๐ฆโฒ1 โผ1๐ฅ2
Integrating gives
๐ฆ1 (๐ฅ) โผ โ1๐ฅ+ ๐ถ
The boundary condition now becomes ๐ฆ1 (1) = 0 (since we used ๐ฆ (1) = 1 earlier with ๐ฆ0).This gives
0 = โ11+ ๐ถ
๐ถ = 1
Therefore the solution becomes
๐ฆ1 (๐ฅ) โผ 1 โ 1๐ฅ
Therefore, the outer solution is
๐ฆ๐๐ข๐ก (๐ฅ) โผ ๐ฆ0 + ๐๐ฆ1
136
3.3. HW3 CHAPTER 3. HWS
Or
๐ฆ (๐ฅ) โผ 1 + ๐ ๏ฟฝ1 โ 1๐ฅ๏ฟฝ + ๐ ๏ฟฝ๐2๏ฟฝ
Since the ODE is ๐ฆโฒ โ ๏ฟฝ1 + ๐๐ฅ2๏ฟฝ ๐ฆ2 + 2๐ฆ = 1, the approximation breaks down when ๐ฅ < โ๐ or
๐ฅ < 110 . Because when ๐ฅ < โ๐, the
๐๐ฅ2 will start to become large. The term ๐
๐ฅ2 should remainsmall for the approximation to be accurate. The following are plots of the ๐ฆ0 and ๐ฆ0 + ๐๐ฆ1solutions (using ๐ = 1
100 ) showing that with two terms the approximation has improved forthe outer layer, compared to the full solution of the original ODE obtained using CAS. Butthe outer solution breaks down near ๐ฅ = 0.1 and smaller as can be seen in these plots. Hereis the solution of the original ODE obtained using CAS
In[46]:= eps =1
100;
ode = y'[x] == 1 +eps
x2 y[x]^2 - 2 y[x] + 1;
sol = y[x] /. First@DSolve[{ode, y[1] โฉต 1}, y[x], x]
Out[48]=
10 x -12 - 5 6 - 12 x2 65 + 5 6 x
2 65
- 6 - 120 x - 50 6 x + 6 x2 65 - 120 x1+
2 65 + 50 6 x1+
2 65
In[53]:= Plot[sol, {x, 0, 1}, PlotRange โ All, Frame โ True, GridLines โ Automatic, GridLinesStyle โ LightGray,
FrameLabel โ {{"y(x)", None}, {"x", "Exact solution to use to compare with"}}, BaseStyle โ 14,
PlotStyle โ Red, ImageSize โ 400]
Out[53]=
0.0 0.2 0.4 0.6 0.8 1.0
0.0
0.2
0.4
0.6
0.8
1.0
x
y(x)
Exact solution to use to compare with
In the following plot, the ๐ฆ0 and the ๐ฆ0 + ๐๐ฆ1 solutions are superimposed on same figure, toshow how the outer solution has improved when adding another term. But we also noticethat the outer solution ๐ฆ0+๐๐ฆ1 only gives good approximation to the exact solution for about๐ฅ > 0.1 and it breaks down quickly as ๐ฅ becomes smaller.
137
3.3. HW3 CHAPTER 3. HWS
In[164]:= outer1 = 1;
outer2 = 1 + eps * (1 - 1 / x);
p2 = Plot[Callout[outer1, "y0", Scaled[0.1]], {x, 0, 1}, PlotRange โ All, Frame โ True,
GridLines โ Automatic, GridLinesStyle โ LightGray,
FrameLabel โ {{"y(x)", None}, {"x", "comparing outer solution y0 with y0+ โy1"}},
BaseStyle โ 14, PlotStyle โ Red, ImageSize โ 400];
p3 = Plot[Callout[outer2, "y0+ โy1", {Scaled[0.5], Below}], {x, 0.01, 1}, AxesOrigin โ {0, 0}];
Show[p2, p3, PlotRange โ {{0.01, .5}, {0, 1.2}}]
Out[168]=
y0
y0+ โy1
0.0 0.1 0.2 0.3 0.4 0.5
0.0
0.2
0.4
0.6
0.8
1.0
1.2
x
y(x)
comparing outer solution y0 with y0+ โy1
3.3.3.3 Part c
Now we will obtain solution inside the boundary layer ๐ฆ๐๐ (๐) = ๐ฆ๐๐0 (๐) + ๐ (๐). The first stepis to always introduce new inner variable. Since the boundary layer is on the right side, then
๐ =๐ฅ๐๐
And then to express the original ODE using this new variable. We also need to determine๐ in the above expression. Since the original ODE is ๐ฆโฒ โ ๏ฟฝ1 + ๐๐ฅโ2๏ฟฝ ๐ฆ2 + 2๐ฆ = 1, then ๐๐ฆ
๐๐ฅ =๐๐ฆ๐๐
๐๐๐๐ฅ =
๐๐ฆ๐๐(๐โ๐), then the ODE now becomes
๐๐ฆ๐๐๐โ๐ โ ๏ฟฝ1 +
๐๐๐๐2 ๏ฟฝ
๐ฆ2 + 2๐ฆ = 1
๐๐ฆ๐๐๐โ๐ โ ๏ฟฝ1 +
๐1โ2๐
๐2 ๏ฟฝ ๐ฆ2 + 2๐ฆ = 1
Where in the above ๐ฆ โก ๐ฆ (๐). We see that we have ๏ฟฝ๐โ๐, ๐๏ฟฝ1โ2๐๏ฟฝ๏ฟฝ as the two biggest terms to
match. This means โ๐ = 1 โ 2๐ or
๐ = 1
Hence the above ODE becomes๐๐ฆ๐๐๐โ1 โ ๏ฟฝ1 +
๐โ1
๐2 ๏ฟฝ๐ฆ2 + 2๐ฆ = 1
138
3.3. HW3 CHAPTER 3. HWS
We are now ready to replace ๐ฆ (๐) with โโ๐=0 ๐
๐๐ฆ๐ which gives
๏ฟฝ๐ฆโฒ0 + ๐๐ฆโฒ1 + ๐2๐ฆโฒ2 +โฏ๏ฟฝ ๐โ1 โ ๏ฟฝ1 +๐โ1
๐2 ๏ฟฝ๏ฟฝ๐ฆ0 + ๐๐ฆ1 + ๐2๐ฆ2 +โฏ๏ฟฝ
2+ 2 ๏ฟฝ๐ฆ0 + ๐๐ฆ1 + ๐2๐ฆ2 +โฏ๏ฟฝ = 1
๏ฟฝ๐ฆโฒ0 + ๐๐ฆโฒ1 + ๐2๐ฆโฒ2 +โฏ๏ฟฝ ๐โ1 โ ๏ฟฝ1 +๐โ1
๐2 ๏ฟฝ๏ฟฝ๐ฆ20 + ๐ ๏ฟฝ2๐ฆ0๐ฆ1๏ฟฝ +โฏ๏ฟฝ + 2 ๏ฟฝ๐ฆ0 + ๐๐ฆ1 + ๐2๐ฆ2 +โฏ๏ฟฝ = 1 (3)
Collecting terms with ๐๏ฟฝ๐โ1๏ฟฝ gives
๐ฆโฒ0 โผ1๐2๐ฆ20
This is separable
๏ฟฝ๐๐ฆ0๐ฆ20
โผ๏ฟฝ1๐2๐๐
โ๐ฆโ10 โผ โ๐โ1 + ๐ถ1๐ฆ0
โผ1๐โ ๐ถ
1๐ฆ0
โผ1 โ ๐ถ๐๐
๐ฆ๐๐0 โผ๐
1 โ ๐๐ถNow we use matching with ๐ฆ๐๐ข๐ก to find ๐ถ. We have found before that ๐ฆ๐๐ข๐ก0 (๐ฅ) โผ 1 therefore
lim๐โโ
๐ฆ๐๐0 (๐) + ๐ (๐) = lim๐ฅโ0
๐ฆ๐๐ข๐ก0 (๐ฅ) + ๐ (๐)
lim๐โโ
๐1 โ ๐๐ถ
= 1 + ๐ (๐)
lim๐โโ
(โ๐ถ) + ๐ ๏ฟฝ๐โ1๏ฟฝ + ๐ (๐) = 1 + ๐ (๐)
โ๐ถ = 1
Therefore
๐ฆ๐๐0 (๐) โผ๐
1 + ๐(4)
Therefore,
๐ฆ๐ข๐๐๐๐๐๐ = ๐ฆ๐๐0 + ๐ฆ๐๐ข๐ก0 โ ๐ฆ๐๐๐ก๐โ
=
๐ฆ๐๐๏ฟฝ๐1+ ๐
+๐ฆ๐๐ข๐กโ1 โ 1
Since ๐ฆ๐๐๐ก๐โ = 1 (this is what lim๐โโ ๐ฆ๐๐0 is). Writing everything in ๐ฅ, using ๐ = ๐ฅ๐ the above
becomes
๐ฆuniform =๐ฅ๐
1 + ๐ฅ๐
=๐ฅ
๐ + ๐ฅ
139
3.3. HW3 CHAPTER 3. HWS
The following is a plot of the above, using ๐ = 1100 to compare with the exact solution.,
In[189]:= y[x_, eps_] :=x
x + eps
p1 = Plot[{exactSol, y[x, 1 / 100]}, {x, 0.02, 1}, PlotRange โ All,
Frame โ True, GridLines โ Automatic, GridLinesStyle โ LightGray,
FrameLabel โ
{{"y(x)", None},
{"x", "Exact solution vs. uniform solution found. 9.6 part(c)"}},
BaseStyle โ 14, PlotStyle โ {Red, Blue}, ImageSize โ 400,
PlotLegends โ {"exact", "uniform approximation"}]
Plot[y[x, 1 / 100], {x, 0, 1}, PlotRange โ {{.05, 1}, Automatic}]
Out[190]=
0.0 0.2 0.4 0.6 0.8 1.00.65
0.70
0.75
0.80
0.85
0.90
0.95
1.00
x
y(x)
Exact solution vs. uniform solution found. 9.6 part(c)
exact
uniform approximation
3.3.3.4 Part (d)
Now we will obtain ๐ฆ๐๐1 solution inside the boundary layer. Using (3) we found in part (c),reproduced here
๏ฟฝ๐ฆโฒ0 + ๐๐ฆโฒ1 + ๐2๐ฆโฒ2 +โฏ๏ฟฝ ๐โ1 โ ๏ฟฝ1 +๐โ1
๐2 ๏ฟฝ๏ฟฝ๐ฆ20 + ๐ ๏ฟฝ2๐ฆ0๐ฆ1๏ฟฝ +โฏ๏ฟฝ + 2 ๏ฟฝ๐ฆ0 + ๐๐ฆ1 + ๐2๐ฆ2 +โฏ๏ฟฝ = 1 (3)
But now collecting all terms of order of ๐ (1), this results in
๐ฆโฒ1 โ ๐ฆ20 โ2๐2๐ฆ0๐ฆ1 + 2๐ฆ0 โผ 1
Using ๐ฆ๐๐0 found in part (c) into the above gives
๐ฆโฒ1 โ2๐ ๏ฟฝ
11 + ๐๏ฟฝ
๐ฆ1 โผ 1 โ 2 ๏ฟฝ๐
1 + ๐๏ฟฝ+ ๏ฟฝ
๐1 + ๐๏ฟฝ
2
๐ฆโฒ1 โ ๏ฟฝ2
๐ (1 + ๐)๏ฟฝ๐ฆ1 โผ
1(๐ + 1)2
140
3.3. HW3 CHAPTER 3. HWS
This can be solved using integrating factor ๐ = ๐โซ โ2
๐+๐2๐๐
using partial fractions gives ๐ =exp (โ2 ln ๐ + 2 ln (1 + ๐)) or ๐ = 1
๐2(1 + ๐)2. Hence we obtain
๐๐๐ฅ๏ฟฝ๐๐ฆ1๏ฟฝ โผ ๐
1(๐ + 1)2
๐๐๐ฅ ๏ฟฝ
1๐2(1 + ๐)2 ๐ฆ1๏ฟฝ โผ
1๐2
Integrating1๐2(1 + ๐)2 ๐ฆ1 โผ๏ฟฝ
1๐2๐๐
1๐2(1 + ๐)2 ๐ฆ1 โผ
โ1๐+ ๐ถ2
(1 + ๐)2 ๐ฆ1 โผ โ๐ + ๐2๐ถ2
๐ฆ1 โผโ๐ + ๐2๐ถ2
(1 + ๐)2
Therefore, the inner solution becomes
๐ฆ๐๐ (๐) = ๐ฆ0 + ๐๐ฆ1
=๐
1 + ๐๐ถ1+ ๐
๐2๐ถ2 โ ๐(1 + ๐)2
To find ๐ถ1, ๐ถ2 we do matching with with ๐ฆ๐๐ข๐ก that we found in part (a) which is ๐ฆ๐๐ข๐ก (๐ฅ) โผ1 + ๐ ๏ฟฝ1 โ 1
๐ฅ๏ฟฝ
lim๐โโ
๏ฟฝ๐
1 + ๐๐ถ1+ ๐
๐2๐ถ2 โ ๐(1 + ๐)2
๏ฟฝ โผ lim๐ฅโ0
1 + ๐ ๏ฟฝ1 โ1๐ฅ๏ฟฝ
Doing long division ๐1+๐๐ถ1
= 1๐ถ1โ 1
๐๐ถ21+ 1
๐2๐ถ31+โฏ and ๐2๐ถ2โ๐
(1+๐)2= ๐ถ2 โ
2๐ถ2+1๐ โโฏ, hence the above
becomes
lim๐โโ
๏ฟฝ๏ฟฝ1๐ถ1
โ1๐๐ถ2
1+
1๐2๐ถ3
1+โฏ๏ฟฝ + ๏ฟฝ๐๐ถ2 โ ๐
2๐ถ2 + 1๐
+โฏ๏ฟฝ๏ฟฝ โผ lim๐ฅโ0
1 + ๐ ๏ฟฝ1 โ1๐ฅ๏ฟฝ
1๐ถ1
+ ๐๐ถ2 โผ lim๐ฅโ0
1 + ๐ ๏ฟฝ1 โ1๐ฅ๏ฟฝ
Using ๐ฅ = ๐๐ on the RHS, the above simplifies to
1๐ถ1
+ ๐๐ถ2 โผ lim๐โโ
1 + ๐ ๏ฟฝ1 โ1๐๐๏ฟฝ
โผ lim๐โโ
1 + ๏ฟฝ๐ โ1๐๏ฟฝ
โผ 1 + ๐
141
3.3. HW3 CHAPTER 3. HWS
Therefore, ๐ถ1 = 1 and ๐ถ2 = 1. Hence the inner solution is
๐ฆ๐๐ (๐) = ๐ฆ0 + ๐๐ฆ1
=๐
1 + ๐+ ๐
๐2 โ ๐(1 + ๐)2
Therefore
๐ฆuniform = ๐ฆ๐๐ + ๐ฆ๐๐ข๐ก โ ๐ฆ๐๐๐ก๐โ
=
๐ฆ๐๐๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๐1 + ๐
+ ๐๐2 โ ๐(1 + ๐)2
+
๐ฆ๐๐ข๐ก๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ1 + ๐ ๏ฟฝ1 โ
1๐ฅ๏ฟฝโ (1 + ๐)
Writing everything in ๐ฅ, using ๐ = ๐ฅ๐ the above becomes
๐ฆuniform =๐ฅ๐
1 + ๐ฅ๐
+ ๐๐ฅ2
๐2 โ๐ฅ๐
๏ฟฝ1 + ๐ฅ๐๏ฟฝ2 + 1 + ๐ ๏ฟฝ1 โ
1๐ฅ๏ฟฝโ (1 + ๐)
=๐ฅ
๐ + ๐ฅ+
๐ฅ2 โ ๐ฅ๐
๐ ๏ฟฝ1 + ๐ฅ๐๏ฟฝ2 + 1 + ๐ โ
๐๐ฅโ 1 โ ๐
=๐ฅ
๐ + ๐ฅ+
๐ฅ2 โ ๐ฅ๐
๐ ๏ฟฝ1 + ๐ฅ๐๏ฟฝ2 โ
๐๐ฅ+ ๐ ๏ฟฝ๐2๏ฟฝ
The following is a plot of the above, using ๐ = 1100 to compare with the exact solution.
In[192]:= y[x_, eps_] :=x
x + eps+
x2 - x eps
eps 1 +x
eps2-eps
x
p1 = Plot[{exactSol, y[x, 1 / 100]}, {x, 0.02, 1}, PlotRange โ All, Frame โ True,
GridLines โ Automatic, GridLinesStyle โ LightGray,
FrameLabel โ {{"y(x)", None}, {"x", "Exact solution vs. uniform solution found. 9.6 part(d)"}},
BaseStyle โ 14, PlotStyle โ {Red, Blue}, ImageSize โ 400,
PlotLegends โ {"exact", "uniform approximation"}]
Plot[y[x, 1 / 100], {x, 0, 1}, PlotRange โ {{.05, 1}, Automatic}]
Out[193]=
0.0 0.2 0.4 0.6 0.8 1.0
0.2
0.4
0.6
0.8
1.0
x
y(x)
Exact solution vs. uniform solution found. 9.6 part(d)
exact
uniform approximation
142
3.3. HW3 CHAPTER 3. HWS
Let us check if ๐ฆuniform (๐ฅ) satisfies ๐ฆ (1) = 1 or not.
๐ฆuniform (1) =1
๐ + 1+
1 โ ๐
๐ ๏ฟฝ1 + 1๐๏ฟฝ2 โ ๐ + ๐ ๏ฟฝ๐
2๏ฟฝ
=1 โ ๐3 โ 3๐2 + ๐
(๐ + 1)2
Taking the limit ๐ โ 0 gives 1. Therefore ๐ฆuniform (๐ฅ) satisfies ๐ฆ (1) = 1.
3.3.4 problem 9.9
problem Use boundary layer methods to find an approximate solution to initial value problem
๐๐ฆโฒโฒ + ๐ (๐ฅ) ๐ฆโฒ + ๐ (๐ฅ) ๐ฆ = 0 (1)
๐ฆ (0) = 1๐ฆโฒ (0) = 1
And ๐ > 0. Show that leading order uniform approximation satisfies ๐ฆ (0) = 1 but not ๐ฆโฒ (0) = 1for arbitrary ๐. Compare leading order uniform approximation with the exact solution tothe problem when ๐ (๐ฅ) , ๐ (๐ฅ) are constants.
Solution
Since ๐ (๐ฅ) > 0 then we expect the boundary layer to be at ๐ฅ = 0. We start by finding ๐ฆ๐๐ข๐ก (๐ฅ).
๐ฆ๐๐ข๐ก (๐ฅ) =โ๏ฟฝ๐=0
๐๐๐ฆ๐ = ๐ฆ0 + ๐๐ฆ1 + ๐2๐ฆ2 +โฏ
Substituting this into (1) gives
๐ ๏ฟฝ๐ฆโฒโฒ0 + ๐๐ฆโฒโฒ1 + ๐2๐ฆโฒโฒ2 +โฏ๏ฟฝ + ๐ ๏ฟฝ๐ฆโฒ0 + ๐๐ฆโฒ1 + ๐2๐ฆโฒ2 +โฏ๏ฟฝ + ๐ ๏ฟฝ๐ฆ0 + ๐๐ฆ1 + ๐2๐ฆ2 +โฏ๏ฟฝ = 0
Collecting terms with ๐ (1) results in
๐๐ฆโฒ0 โผ โ๐๐ฆ0๐๐ฆ0๐๐ฅ
โผ โ๐๐๐ฆ0
This is separable
๏ฟฝ๐๐ฆ0๐ฆ0
โผ โ๏ฟฝ๐ (๐ฅ)๐ (๐ฅ)
๐๐ฅ
ln ๏ฟฝ๐ฆ0๏ฟฝ โผ โ๏ฟฝ๐ (๐ฅ)๐ (๐ฅ)
๐๐ฅ + ๐ถ
๐ฆ0 โผ ๐ถ๐โโซ๐ฅ1
๐(๐ )๐(๐ )๐๐
Now we find ๐ฆ๐๐. First we introduce interval variable
๐ =๐ฅ๐๐
143
3.3. HW3 CHAPTER 3. HWS
And transform the ODE. Since ๐๐ฆ๐๐ฅ =
๐๐ฆ๐๐
๐๐๐๐ฅ then ๐๐ฆ
๐๐ฅ =๐๐ฆ๐๐๐
โ๐. Hence ๐๐๐ฅ โก ๐
โ๐ ๐๐๐
๐2
๐๐ฅ2=๐๐๐ฅ
๐๐๐ฅ
= ๏ฟฝ๐โ๐๐๐๐๏ฟฝ ๏ฟฝ
๐โ๐๐๐๐๏ฟฝ
= ๐โ2๐๐2
๐๐2
Therefore ๐2๐ฆ๐๐ฅ2 = ๐
โ2๐ ๐2๐ฆ๐๐2 and the ODE becomes
๐๐โ2๐๐2๐ฆ๐๐2
+ ๐ (๐)๐๐ฆ๐๐๐โ๐ + ๐ (๐) ๐ฆ = 0
๐1โ2๐๐ฆโฒโฒ + ๐๐โ๐๐ฆโฒ + ๐๐ฆ = 0
Balancing 1 โ 2๐ with โ๐ shows that
๐ = 1
Hence
๐โ1๐ฆโฒโฒ + ๐๐โ1๐ฆโฒ + ๐๐ฆ = 0
Substituting ๐ฆ๐๐ = โโ๐=0 ๐
๐๐ฆ๐ = ๐ฆ0 + ๐๐ฆ1 + ๐2๐ฆ2 +โฏ in the above gives
๐โ1 ๏ฟฝ๐ฆโฒโฒ0 + ๐๐ฆโฒโฒ1 + ๐2๐ฆโฒโฒ2 +โฏ๏ฟฝ + ๐๐โ1 ๏ฟฝ๐ฆโฒ0 + ๐๐ฆโฒ1 + ๐2๐ฆโฒ2 +โฏ๏ฟฝ + ๐ ๏ฟฝ๐ฆ0 + ๐๐ฆ1 + ๐2๐ฆ2 +โฏ๏ฟฝ = 0
Collecting terms with order ๐๏ฟฝ๐โ1๏ฟฝ gives
๐ฆโฒโฒ0 โผ โ๐๐ฆโฒ0Assuming ๐ง = ๐ฆโฒ0 then the above becomes ๐งโฒ โผ โ๐๐ง or ๐๐ง
๐๐ โผ โ๐๐ง. This is separable. The
solution is ๐๐ง๐ง โผ โ๐๐๐ or
ln |๐ง| โผ โ๏ฟฝ๐
0๐ (๐ ) ๐๐ + ๐ถ1
๐ง โผ ๐ถ1๐โโซ
๐0
๐(๐ )๐๐
Hence๐๐ฆ0๐๐
โผ ๐ถ1๐โโซ
๐0
๐(๐ )๐๐
๐๐ฆ0 โผ ๏ฟฝ๐ถ1๐โโซ
๐0
๐(๐ )๐๐ ๏ฟฝ ๐๐
Integrating again
๐ฆ๐๐0 โผ๏ฟฝ๐
0๏ฟฝ๐ถ1๐
โโซ๐0
๐(๐ )๐๐ ๏ฟฝ ๐๐ + ๐ถ2
Applying initial conditions at ๐ฆ (0) since this is where the ๐ฆ๐๐ exist. Using ๐ฆ๐๐ (0) = 1 then theabove becomes
1 = ๐ถ2
144
3.3. HW3 CHAPTER 3. HWS
Hence the solution becomes
๐ฆ๐๐0 โผ๏ฟฝ๐
0๏ฟฝ๐ถ1๐
โโซ๐0
๐(๐ )๐๐ ๏ฟฝ ๐๐ + 1
To apply the second initial condition, which is ๐ฆโฒ (0) = 1, we first take derivative of the abovew.r.t. ๐
๐ฆโฒ0 โผ ๐ถ1๐โโซ
๐0
๐(๐ )๐๐
Applying ๐ฆโฒ0 (0) = 1 gives
1 = ๐ถ1
Hence
๐ฆ๐๐0 โผ 1 +๏ฟฝ๐
0๐โโซ
๐0
๐(๐ )๐๐ ๐๐
Now to find constant of integration for ๐ฆ๐๐ข๐ก from earlier, we need to do matching.
lim๐โโ
๐ฆ๐๐0 โผ lim๐ฅโ0
๐ฆ๐๐ข๐ก0
lim๐โโ
1 +๏ฟฝ๐
0๐โโซ
๐0
๐(๐ )๐๐ ๐๐ โผ lim๐ฅโ0
๐ถ๐โโซ๐ฅ1
๐(๐ )๐(๐ )๐๐
On the LHS the integral โซ๐
0๐โโซ
๐0
๐(๐ )๐๐ ๐๐ since ๐ > 0 and negative power on the exponential.So as ๐ โ โ the integral value is zero. So we have now
1 โผ lim๐ฅโ0
๐ถ๐โโซ๐ฅ1
๐(๐ )๐(๐ )๐๐
Let lim๐ฅโ0 ๐โโซ
๐ฅ1
๐(๐ )๐(๐ )๐๐ โ ๐ธ, where ๐ธ is the value of the definite integral ๐ถ๐โโซ
01
๐(๐ )๐(๐ )๐๐ . Another
constant, which if we know ๐ (๐ฅ) , ๐ (๐ฅ) we can evaluate. Hence the above gives the value of ๐ถas
๐ถ =1๐ธ
The uniform solution can now be written as
๐ฆuniform = ๐ฆ๐๐ + ๐ฆ๐๐ข๐ก โ ๐ฆmatch
= 1 +๏ฟฝ๐
0๐โโซ
๐0
๐(๐ )๐๐ ๐๐ +1๐ธ๐โโซ
๐ฅ1
๐(๐ )๐(๐ )๐๐ โ 1
= ๏ฟฝ๐
0๐โโซ
๐0
๐(๐ )๐๐ ๐๐ +1๐ธ๐โโซ
๐ฅ1
๐(๐ )๐(๐ )๐๐ (2)
Finally, we need to show that ๐ฆuniform (0) = 1 but not ๐ฆโฒuniform (0) = 1. From (2), at ๐ฅ = 0 whichalso means ๐ = 0, since boundary layer at left side, equation (2) becomes
๐ฆuniform (0) = 0 +1๐ธ
lim๐ฅโ0
๐โโซ๐ฅ1
๐(๐ )๐(๐ )๐๐
But we said that lim๐ฅโ0 ๐โโซ
๐ฅ1
๐(๐ )๐(๐ )๐๐ = ๐ธ, therefore
๐ฆuniform (0) = 1
145
3.3. HW3 CHAPTER 3. HWS
Now we take derivative of (2) w.r.t. ๐ฅ and obtain
๐ฆโฒuniform (๐ฅ) =๐๐๐ฅ
โโโโโ๏ฟฝ
๐ฅ๐
0๐โโซ
๐0
๐(๐ )๐๐ ๐๐โโโโโ +
1๐ธ๐๐๐ฅ ๏ฟฝ
๐โโซ๐ฅ1
๐(๐ )๐(๐ )๐๐ ๏ฟฝ
= ๐โโซ๐ฅ๐
0๐(๐ )๐๐ โ
1๐ธ๐ (๐ฅ)๐ (๐ฅ)
๐โโซ๐ฅ1
๐(๐ )๐(๐ )๐๐
And at ๐ฅ = 0 the above becomes
๐ฆโฒuniform (0) = 1 โ1๐ธ๐ (0)๐ (0)
The above is zero only if ๐ (0) = 0 (since we know ๐ (0) > 0). Therefore, we see that๐ฆโฒuniform (0) โ 1 for any arbitrary ๐ (๐ฅ). Which is what we are asked to show.
Will now solve the whole problem again, when ๐, ๐ are constants.
๐๐ฆโฒโฒ + ๐๐ฆโฒ + ๐๐ฆ = 0 (1A)
๐ฆ (0) = 1๐ฆโฒ (0) = 1
And ๐ > 0. And compare leading order uniform approximation with the exact solution tothe problem when ๐ (๐ฅ) , ๐ (๐ฅ) are constants. Since ๐ > 0 then boundary layer will occur at๐ฅ = 0. We start by finding ๐ฆ๐๐ข๐ก (๐ฅ).
๐ฆ๐๐ข๐ก (๐ฅ) =โ๏ฟฝ๐=0
๐๐๐ฆ๐ = ๐ฆ0 + ๐๐ฆ1 + ๐2๐ฆ2 +โฏ
Substituting this into (1) gives
๐ ๏ฟฝ๐ฆโฒโฒ0 + ๐๐ฆโฒโฒ1 + ๐2๐ฆโฒโฒ2 +โฏ๏ฟฝ + ๐ ๏ฟฝ๐ฆโฒ0 + ๐๐ฆโฒ1 + ๐2๐ฆโฒ2 +โฏ๏ฟฝ + ๐ ๏ฟฝ๐ฆ0 + ๐๐ฆ1 + ๐2๐ฆ2 +โฏ๏ฟฝ = 0
Collecting terms with ๐ (1) results in
๐๐ฆโฒ0 โผ โ๐๐ฆ0๐๐ฆ0๐๐ฅ
โผ โ๐๐๐ฆ0
This is separable
๏ฟฝ๐๐ฆ0๐ฆ0
โผ โ๐๐๐๐ฅ
ln ๏ฟฝ๐ฆ0๏ฟฝ โผ โ๐๐๐ฅ + ๐ถ
๐ฆ๐๐ข๐ก0 โผ ๐ถ1๐โ ๐๐๐ฅ
Now we find ๐ฆ๐๐. First we introduce internal variable ๐ =๐ฅ๐๐ and transform the ODE as we
did above. This results in
๐โ1 ๏ฟฝ๐ฆโฒโฒ0 + ๐๐ฆโฒโฒ1 + ๐2๐ฆโฒโฒ2 +โฏ๏ฟฝ + ๐๐โ1 ๏ฟฝ๐ฆโฒ0 + ๐๐ฆโฒ1 + ๐2๐ฆโฒ2 +โฏ๏ฟฝ + ๐ ๏ฟฝ๐ฆ0 + ๐๐ฆ1 + ๐2๐ฆ2 +โฏ๏ฟฝ = 0
Collecting terms with order ๐๏ฟฝ๐โ1๏ฟฝ gives
๐ฆโฒโฒ0 โผ โ๐๐ฆโฒ0
146
3.3. HW3 CHAPTER 3. HWS
Assuming ๐ง = ๐ฆโฒ0 then the above becomes ๐งโฒ โผ โ๐๐ง or ๐๐ง๐๐ โผ โ๐๐ง. This is separable. The
solution is ๐๐ง๐ง โผ โ๐๐๐ or
ln |๐ง| โผ โ๐๐ + ๐ธ1๐ง โผ ๐ธ1๐โ๐๐
Hence๐๐ฆ0๐๐
โผ ๐ธ1๐โ๐๐
๐๐ฆ0 โผ ๐ธ1๐โ๐๐๐๐
Integrating again
๐ฆ๐๐0 โผ ๐ธ1 ๏ฟฝโ1๐ ๏ฟฝ
๐โ๐๐ + ๐ธ2
Applying initial conditions at ๐ฆ (0) since this is where the ๐ฆ๐๐ exist. Using ๐ฆ๐๐ (0) = 1 then theabove becomes
1 = ๐ธ1 ๏ฟฝโ1๐ ๏ฟฝ
+ ๐ธ2
๐ (๐ธ2 โ 1) = ๐ธ1Hence the solution becomes
๐ฆ๐๐0 โผ (1 โ ๐ธ2) ๐โ๐๐ + ๐ธ2 (1B)
To apply the second initial condition, which is ๐ฆโฒ (0) = 1, we first take derivative of the abovew.r.t. ๐
๐ฆโฒ0 โผ โ๐ (1 โ ๐ธ2) ๐โ๐๐
Hence ๐ฆโฒ (0) = 1 gives
1 = โ๐ (1 โ ๐ธ2)1 = โ๐ + ๐๐ธ2
๐ธ2 =1 + ๐๐
And the solution ๐ฆ๐๐ in (1B) becomes
๐ฆ๐๐0 โผ ๏ฟฝ1 โ1 + ๐๐ ๏ฟฝ ๐โ๐๐ +
1 + ๐๐
โผ ๏ฟฝโ1๐ ๏ฟฝ
๐โ๐๐ +1 + ๐๐
โผ(1 + ๐) โ ๐โ๐๐
๐
147
3.3. HW3 CHAPTER 3. HWS
Now to find constant of integration for ๐ฆ๐๐ข๐ก (๐ฅ) from earlier, we need to do matching.
lim๐โโ
๐ฆ๐๐0 โผ lim๐ฅโ0
๐ฆ๐๐ข๐ก0
lim๐โโ
(1 + ๐) โ ๐โ๐๐
๐โผ lim
๐ฅโ0๐ถ1๐
โ ๐๐๐ฅ
1 + ๐๐
โผ ๐ถ1
Hence now the uniform solution can be written as
๐ฆuniform (๐ฅ) โผ ๐ฆ๐๐ + ๐ฆ๐๐ข๐ก โ ๐ฆmatch
โผ
๐ฆ๐๐๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ(1 + ๐) โ ๐โ๐
๐ฅ๐
๐+
๐ฆ๐๐ข๐ก๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ1 + ๐๐
๐โ๐๐๐ฅ โ
1 + ๐๐
โผ(1 + ๐)๐
โ๐โ๐
๐ฅ๐
๐+1 + ๐๐
๐โ๐๐๐ฅ โ
1 + ๐๐
โผ โ๐โ๐
๐ฅ๐
๐+1 + ๐๐
๐โ๐๐๐ฅ
โผ1๐๏ฟฝ(1 + ๐) ๐โ
๐๐๐ฅ โ ๐โ๐
๐ฅ๐ ๏ฟฝ (2A)
Now we compare the above, which is the leading order uniform approximation, to the exactsolution. Since now ๐, ๐ are constants, then the exact solution is
๐ฆ๐๐ฅ๐๐๐ก (๐ฅ) = ๐ด๐๐1๐ฅ + ๐ต๐๐2๐ฅ (3)
Where ๐1,2 are roots of characteristic equation of ๐๐ฆโฒโฒ + ๐๐ฆโฒ + ๐๐ฆ = 0. These are ๐ = โ๐2๐ ยฑ
12๐โ๐
2 โ 4๐๐. Hence
๐1 =โ๐2+12โ๐2 โ 4๐๐
๐2 =โ๐2โ12โ๐2 โ 4๐๐
Applying initial conditions to (3). ๐ฆ (0) = 1 gives
1 = ๐ด + ๐ต๐ต = 1 โ ๐ด
And solution becomes ๐ฆ๐๐ฅ๐๐๐ก (๐ฅ) = ๐ด๐๐1๐ฅ + (1 โ ๐ด) ๐๐2๐ฅ. Taking derivatives gives
๐ฆโฒ๐๐ฅ๐๐๐ก (๐ฅ) = ๐ด๐1๐๐1๐ฅ + (1 โ ๐ด) ๐2๐๐2๐ฅ
Using ๐ฆโฒ (0) = 1 gives
1 = ๐ด๐1 + (1 โ ๐ด) ๐21 = ๐ด (๐1 โ ๐2) + ๐2
๐ด =1 โ ๐2๐1 โ ๐2
148
3.3. HW3 CHAPTER 3. HWS
Therefore, ๐ต = 1 โ 1โ๐2๐1โ๐2
and the exact solution becomes
๐ฆ๐๐ฅ๐๐๐ก (๐ฅ) =1 โ ๐2๐1 โ ๐2
๐๐1๐ฅ + ๏ฟฝ1 โ1 โ ๐2๐1 โ ๐2
๏ฟฝ ๐๐2๐ฅ
=1 โ ๐2๐1 โ ๐2
๐๐1๐ฅ + ๏ฟฝ(๐1 โ ๐2) โ (1 โ ๐2)
๐1 โ ๐2๏ฟฝ ๐๐2๐ฅ
=1 โ ๐2๐1 โ ๐2
๐๐1๐ฅ + ๏ฟฝ๐1 โ 1๐1 โ ๐2
๏ฟฝ ๐๐2๐ฅ (4)
While the uniform solution above was found to be 1๐๏ฟฝ(1 + ๐) ๐โ
๐๐๐ฅ โ ๐โ๐
๐ฅ๐ ๏ฟฝ. Here is a plot
of the exact solution above, for ๐ = {1/10, 1/50, 1/100} , and for some values for ๐, ๐ such as๐ = 1, ๐ = 10 in order to compare with the uniform solution. Note that the uniform solutionis ๐ (๐). As ๐ becomes smaller, the leading order uniform solution will better approximatethe exact solution. At ๐ = 0.01 the uniform approximation gives very good approximation.This is using only leading term approximation.
In[134]:= ClearAll[x, y]
eps = 1/ 10; a = 1; b = 10;
mySol = 1/ a ((1 + a)* Exp[-b/ a x] - Exp[-a x/ eps]);
sol = y[x] /. First@DSolve[{eps y''[x] + a y'[x] + b y[x] โฉต 0, y[0] โฉต 1, y'[0] โฉต 1}, y[x], x]
Out[137]=15โ -5 x
5 Cos5 3 x + 2 3 Sin5 3 x
In[125]:= Plot[{sol, mySol}, {x, 0, 1}, PlotRange โ All, PlotStyle โ {Blue, Red}, PlotLegends โ {"Exact", "approximation"},
Frame โ True, FrameLabel โ {{"y(x)", None}, {"x", Row[{"Exact vs. approximation for ฯต =", eps}]}},
BaseStyle โ 14, GridLines โ Automatic, GridLinesStyle โ LightGray]
Out[125]=
0.0 0.2 0.4 0.6 0.8 1.0-0.2
0.0
0.2
0.4
0.6
0.8
1.0
x
y(x)
Exact vs. approximation for ฯต = 1
10
Exact
approximation
149
3.3. HW3 CHAPTER 3. HWS
In[126]:= ClearAll[x, y]
eps = 1/ 50; a = 1; b = 10;
sol = y[x] /. First@DSolve[{eps y''[x] + a y'[x] + b y[x] โฉต 0, y[0] โฉต 1, y'[0] โฉต 1}, y[x], x]
Plot[{sol, mySol}, {x, 0, 1}, PlotRange โ All, PlotStyle โ {Blue, Red}, PlotLegends โ {"Exact", "approximation"},
Frame โ True, FrameLabel โ {{"y(x)", None}, {"x", Row[{"Exact vs. approximation for ฯต =", eps}]}},
BaseStyle โ 14, GridLines โ Automatic, GridLinesStyle โ LightGray]
Out[128]=150
25 โ -25-5 5 x- 26 5 โ
-25-5 5 x+ 25 โ -25+5 5 x
+ 26 5 โ -25+5 5 x
Out[129]=
0.0 0.2 0.4 0.6 0.8 1.0
0.0
0.2
0.4
0.6
0.8
1.0
x
y(x)
Exact vs. approximation for ฯต = 1
50
Exact
approximation
In[130]:= ClearAll[x, y]
eps = 1/ 100; a = 1; b = 10;
sol = y[x] /. First@DSolve[{eps y''[x] + a y'[x] + b y[x] โฉต 0, y[0] โฉต 1, y'[0] โฉต 1}, y[x], x]
Plot[{sol, mySol}, {x, 0, 1}, PlotRange โ All, PlotStyle โ {Blue, Red}, PlotLegends โ {"Exact", "approximation"},
Frame โ True, FrameLabel โ {{"y(x)", None}, {"x", Row[{"Exact vs. approximation for ฯต =", eps}]}},
BaseStyle โ 14, GridLines โ Automatic, GridLinesStyle โ LightGray]
Out[132]=1100
50 โ -50-10 15 x- 17 15 โ
-50-10 15 x+ 50 โ -50+10 15 x
+ 17 15 โ -50+10 15 x
Out[133]=
0.0 0.2 0.4 0.6 0.8 1.0
0.0
0.2
0.4
0.6
0.8
1.0
x
y(x)
Exact vs. approximation for ฯต = 1
100
Exact
approximation
3.3.5 problem 9.15(b)
Problem Find first order uniform approximation valid as ๐ โ 0+ for 0 โค ๐ฅ โค 1
๐๐ฆโฒโฒ + ๏ฟฝ๐ฅ2 + 1๏ฟฝ ๐ฆโฒ โ ๐ฅ3๐ฆ = 0 (1)
๐ฆ (0) = 1๐ฆ (1) = 1
Solution
Since ๐ (๐ฅ) = ๏ฟฝ๐ฅ2 + 1๏ฟฝ is positive for 0 โค ๐ฅ โค 1, therefore we expect the boundary layer to be
150
3.3. HW3 CHAPTER 3. HWS
on the left side at ๐ฅ = 0. Assuming this is the case for now (if it is not, then we expect notto be able to do the matching). We start by finding ๐ฆ๐๐ข๐ก (๐ฅ).
๐ฆ๐๐ข๐ก (๐ฅ) =โ๏ฟฝ๐=0
๐๐๐ฆ๐ = ๐ฆ0 + ๐๐ฆ1 + ๐2๐ฆ2 +โฏ
Substituting this into (1) gives
๐ ๏ฟฝ๐ฆโฒโฒ0 + ๐๐ฆโฒโฒ1 + ๐2๐ฆโฒโฒ2 +โฏ๏ฟฝ + ๏ฟฝ๐ฅ2 + 1๏ฟฝ ๏ฟฝ๐ฆโฒ0 + ๐๐ฆโฒ1 + ๐2๐ฆโฒ2 +โฏ๏ฟฝ โ ๐ฅ3 ๏ฟฝ๐ฆ0 + ๐๐ฆ1 + ๐2๐ฆ2 +โฏ๏ฟฝ = 0 (2)
Collecting terms with ๐ (1) results in๏ฟฝ๐ฅ2 + 1๏ฟฝ ๐ฆโฒ0 โผ ๐ฅ3๐ฆ0
๐๐ฆ0๐๐ฅ
โผ๐ฅ3
๏ฟฝ๐ฅ2 + 1๏ฟฝ๐ฆ0
This is separable.
๏ฟฝ๐๐ฆ0๐ฆ0
โผ๏ฟฝ๐ฅ3
๏ฟฝ๐ฅ2 + 1๏ฟฝ๐๐ฅ
ln ๏ฟฝ๐ฆ0๏ฟฝ โผ๏ฟฝ๐ฅ โ๐ฅ
1 + ๐ฅ2๐๐ฅ
โผ๐ฅ2
2โ12
ln ๏ฟฝ1 + ๐ฅ2๏ฟฝ + ๐ถ
Hence
๐ฆ0 โผ ๐๐ฅ22 โ 1
2 ln๏ฟฝ1+๐ฅ2๏ฟฝ+๐ถ
โผ๐ถ๐
๐ฅ22
โ1 + ๐ฅ2
Applying ๐ฆ๐๐ข๐ก0 (1) = 1 to the above (since this is where the outer solution is), we solve for ๐ถ
1 โผ๐ถ๐
12
โ2
๐ถ โผ โ2๐โ12
Therefore
๐ฆ๐๐ข๐ก0 โผ โ2๐โ12 ๐
๐ฅ22
โ1 + ๐ฅ2
โผ๏ฟฝ2๐
๐๐ฅ22
โ1 + ๐ฅ2
Now we need to find ๐ฆ๐๐ข๐ก1 . From (2), but now collecting terms in ๐ (๐) gives
๐ฆโฒโฒ0 + ๏ฟฝ๐ฅ2 + 1๏ฟฝ ๐ฆโฒ1 โฝ ๐ฅ3๐ฆ1 (3)
151
3.3. HW3 CHAPTER 3. HWS
In the above ๐ฆโฒโฒ0 is known.
๐ฆโฒ0 (๐ฅ) = ๏ฟฝ2๐๐๐๐ฅ
โโโโโโโโโ
๐๐ฅ22
โ1 + ๐ฅ2
โโโโโโโโโ
=๏ฟฝ2๐
๐ฅ3๐๐ฅ22
๏ฟฝ1 + ๐ฅ2๏ฟฝ32
And
๐ฆโฒโฒ0 (๐ฅ) = ๏ฟฝ2๐๐ฅ2๐
๐ฅ22 ๏ฟฝ๐ฅ4 + ๐ฅ2 + 3๏ฟฝ
๏ฟฝ1 + ๐ฅ2๏ฟฝ52
Hence (3) becomes
๏ฟฝ๐ฅ2 + 1๏ฟฝ ๐ฆโฒ1 โฝ ๐ฅ3๐ฆ1 โ ๐ฆโฒโฒ0
๏ฟฝ๐ฅ2 + 1๏ฟฝ ๐ฆโฒ1 โฝ ๐ฅ3๐ฆ1 โ๏ฟฝ2๐๐ฅ2๐
๐ฅ22 ๏ฟฝ๐ฅ4 + ๐ฅ2 + 3๏ฟฝ
๏ฟฝ1 + ๐ฅ2๏ฟฝ52
๐ฆโฒ1 โ๐ฅ3
๏ฟฝ๐ฅ2 + 1๏ฟฝ๐ฆ1 โฝ โ๏ฟฝ
2๐๐ฅ2๐
๐ฅ22 ๏ฟฝ๐ฅ4 + ๐ฅ2 + 3๏ฟฝ
๏ฟฝ1 + ๐ฅ2๏ฟฝ72
Integrating factor is ๐ = ๐โโซ ๐ฅ3
๏ฟฝ๐ฅ2+1๏ฟฝ๐๐ฅ= ๐โ
๐ฅ22 + 1
2 ln๏ฟฝ1+๐ฅ2๏ฟฝ = ๏ฟฝ1 + ๐ฅ2๏ฟฝ12 ๐
โ๐ฅ22 , hence the above becomes
๐๐๐ฅ ๏ฟฝ
๏ฟฝ1 + ๐ฅ2๏ฟฝ12 ๐
โ๐ฅ22 ๐ฆ1๏ฟฝ โฝ โ๏ฟฝ
2๐๏ฟฝ1 + ๐ฅ2๏ฟฝ
12 ๐
โ๐ฅ22๐ฅ2๐
๐ฅ22 ๏ฟฝ๐ฅ4 + ๐ฅ2 + 3๏ฟฝ
๏ฟฝ1 + ๐ฅ2๏ฟฝ72
โฝ โ๏ฟฝ2๐๐ฅ2 ๏ฟฝ๐ฅ4 + ๐ฅ2 + 3๏ฟฝ
๏ฟฝ1 + ๐ฅ2๏ฟฝ3
Integrating gives (with help from CAS)
๏ฟฝ1 + ๐ฅ2๏ฟฝ12 ๐
โ๐ฅ22 ๐ฆ1 (๐ฅ) โฝ โ๏ฟฝ
2๐ ๏ฟฝ
๐ฅ2 ๏ฟฝ๐ฅ4 + ๐ฅ2 + 3๏ฟฝ
๏ฟฝ1 + ๐ฅ2๏ฟฝ3 ๐๐ฅ
โฝ โ๏ฟฝ2๐ ๏ฟฝ
1 โ3
๏ฟฝ1 + ๐ฅ2๏ฟฝ3 +
4
๏ฟฝ1 + ๐ฅ2๏ฟฝ2 โ
21 + ๐ฅ2
๐๐ฅ
โฝ โ๏ฟฝ2๐
โโโโโโโโ๐ฅ โ
3๐ฅ
4 ๏ฟฝ1 + ๐ฅ2๏ฟฝ2 +
7๐ฅ8 ๏ฟฝ1 + ๐ฅ2๏ฟฝ
โ 9arctan (๐ฅ)
8
โโโโโโโโ + ๐ถ1
152
3.3. HW3 CHAPTER 3. HWS
Hence
๐ฆ๐๐ข๐ก1 (๐ฅ) โฝ โ๏ฟฝ2๐
๐๐ฅ22
๏ฟฝ1 + ๐ฅ2๏ฟฝ12
โโโโโโโโ๐ฅ โ
3๐ฅ
4 ๏ฟฝ1 + ๐ฅ2๏ฟฝ2 +
7๐ฅ8 ๏ฟฝ1 + ๐ฅ2๏ฟฝ
โ 9arctan (๐ฅ)
8
โโโโโโโโ + ๐ถ1
๐๐ฅ22
๏ฟฝ1 + ๐ฅ2๏ฟฝ12
Now we find ๐ถ1 from boundary conditions ๐ฆ1 (1) = 0. (notice the BC now is ๐ฆ1 (1) = 0 andnot ๐ฆ1 (1) = 1, since we used ๐ฆ1 (1) = 1 already).
๏ฟฝ2๐
๐12
(1 + 1)12๏ฟฝ1 โ
34 (1 + 1)2
+7
8 (1 + 1)โ 9
arctan (1)8 ๏ฟฝ = ๐ถ1
๐12
(1 + 1)12
๏ฟฝ2๐๐12
โ2๏ฟฝ1 โ
316+716โ98
arctan (1)๏ฟฝ = ๐ถ1๐12
โ2Simplifying
1 โ316+716โ98
arctan (1) = ๐ถ1๐12
โ2
๐ถ1 = ๏ฟฝ2๐ ๏ฟฝ54โ98
arctan (1)๏ฟฝ
๐ถ1 = ๏ฟฝ2๐ ๏ฟฝ54โ932๐๏ฟฝ
= 0.31431
Hence
๐ฆ๐๐ข๐ก1 โผ โ๏ฟฝ2๐
๐๐ฅ22
๏ฟฝ๏ฟฝ1 + ๐ฅ2๏ฟฝ
โโโโโโโโ๐ฅ โ
3๐ฅ
4 ๏ฟฝ1 + ๐ฅ2๏ฟฝ2 +
7๐ฅ8 ๏ฟฝ1 + ๐ฅ2๏ฟฝ
โ98
arctan (๐ฅ)
โโโโโโโโ +๏ฟฝ
2๐ ๏ฟฝ54โ932๐๏ฟฝ
๐๐ฅ22
๏ฟฝ๏ฟฝ1 + ๐ฅ2๏ฟฝ
โผ๏ฟฝ2๐
๐๐ฅ22
๏ฟฝ๏ฟฝ1 + ๐ฅ2๏ฟฝ
โโโโโโโโ๏ฟฝ54โ932๐๏ฟฝ โ
โโโโโโโโ๐ฅ โ
3๐ฅ
4 ๏ฟฝ1 + ๐ฅ2๏ฟฝ2 +
7๐ฅ8 ๏ฟฝ1 + ๐ฅ2๏ฟฝ
โ98
arctan (๐ฅ)
โโโโโโโโ
โโโโโโโโ
โผ๏ฟฝ2๐
๐๐ฅ22
๏ฟฝ๏ฟฝ1 + ๐ฅ2๏ฟฝ
โโโโโโโโ54โ932๐ โ ๐ฅ +
3๐ฅ
4 ๏ฟฝ1 + ๐ฅ2๏ฟฝ2 โ
7๐ฅ8 ๏ฟฝ1 + ๐ฅ2๏ฟฝ
+98
arctan (๐ฅ)
โโโโโโโโ
153
3.3. HW3 CHAPTER 3. HWS
Hence
๐ฆ๐๐ข๐ก (๐ฅ) โผ ๐ฆ๐๐ข๐ก0 + ๐๐ฆ๐๐ข๐ก1
โผ๏ฟฝ2๐
๐๐ฅ22
โ1 + ๐ฅ2+ ๐
๏ฟฝ2๐
๐๐ฅ22
๏ฟฝ๏ฟฝ1 + ๐ฅ2๏ฟฝ
โโโโโโโโ54โ932๐ โ ๐ฅ +
3๐ฅ
4 ๏ฟฝ1 + ๐ฅ2๏ฟฝ2 โ
7๐ฅ8 ๏ฟฝ1 + ๐ฅ2๏ฟฝ
+98
arctan (๐ฅ)
โโโโโโโโ + ๐ ๏ฟฝ๐
2๏ฟฝ
โผ๏ฟฝ2๐
๐๐ฅ22
โ1 + ๐ฅ2
โโโโโโโโ1 + ๐
โโโโโโโโ54โ932๐ โ ๐ฅ +
3๐ฅ
4 ๏ฟฝ1 + ๐ฅ2๏ฟฝ2 โ
7๐ฅ8 ๏ฟฝ1 + ๐ฅ2๏ฟฝ
+98
arctan (๐ฅ)
โโโโโโโโ
โโโโโโโโ + ๐ ๏ฟฝ๐
2๏ฟฝ
(3A)
Now that we found ๐ฆ๐๐ข๐ก (๐ฅ), we need to find ๐ฆ๐๐ (๐ฅ) and then do the matching and the finduniform approximation. Since the boundary layer at ๐ฅ = 0, we introduce inner variable ๐ = ๐ฅ
๐๐and then express the original ODE using this new variable. We also need to determine ๐ inthe above expression. Since ๐๐ฆ
๐๐ฅ =๐๐ฆ๐๐
๐๐๐๐ฅ then ๐๐ฆ
๐๐ฅ =๐๐ฆ๐๐๐
โ๐. Hence ๐๐๐ฅ โก ๐
โ๐ ๐๐๐
๐2
๐๐ฅ2=๐๐๐ฅ
๐๐๐ฅ
= ๏ฟฝ๐โ๐๐๐๐๏ฟฝ ๏ฟฝ
๐โ๐๐๐๐๏ฟฝ
= ๐โ2๐๐2
๐๐2
Therefore ๐2๐ฆ๐๐ฅ2 = ๐
โ2๐ ๐2๐ฆ๐๐2 and the ODE ๐๐ฆโฒโฒ + ๏ฟฝ๐ฅ2 + 1๏ฟฝ ๐ฆโฒ โ ๐ฅ3๐ฆ = 0 now becomes
๐๐โ2๐๐2๐ฆ๐๐2
+ ๏ฟฝ(๐๐๐)2 + 1๏ฟฝ ๐โ๐๐๐ฆ๐๐
โ (๐๐๐)3 ๐ฆ = 0
๐1โ2๐๐2๐ฆ๐๐2
+ ๏ฟฝ๐2๐๐ + ๐โ๐๏ฟฝ๐๐ฆ๐๐
โ ๐3๐3๐๐ฆ = 0
The largest terms are ๏ฟฝ๐1โ2๐, ๐โ๐๏ฟฝ, therefore matching them gives ๐ = 1. The ODE nowbecomes
๐โ1๐2๐ฆ๐๐2
+ ๏ฟฝ๐2๐ + ๐โ1๏ฟฝ๐๐ฆ๐๐
โ ๐3๐3๐ฆ = 0 (4)
Assuming that
๐ฆ๐๐ (๐ฅ) =โ๏ฟฝ๐=0
๐๐๐ฆ๐ = ๐ฆ0 + ๐๐ฆ1 + ๐2๐ฆ2 +โฏ
And substituting the above into (4) gives
๐โ1 ๏ฟฝ๐ฆโฒโฒ0 + ๐๐ฆโฒโฒ1 +โฏ๏ฟฝ + ๏ฟฝ๐2๐ + ๐โ1๏ฟฝ ๏ฟฝ๐ฆโฒ0 + ๐๐ฆโฒ1 +โฏ๏ฟฝ โ ๐3๐3 ๏ฟฝ๐ฆ0 + ๐๐ฆ1 +โฏ๏ฟฝ = 0 (4A)
Collecting terms in ๐๏ฟฝ๐โ1๏ฟฝ gives
๐ฆโฒโฒ0 โฝ โ๐ฆโฒ0
154
3.3. HW3 CHAPTER 3. HWS
Letting ๐ง = ๐ฆโฒ0, the above becomes๐๐ง๐๐
โฝ โ๐ง
๐๐ง๐งโฝ โ๐๐
ln |๐ง| โฝ โ๐ + ๐ถ1
๐ง โฝ ๐ถ1๐โ๐
Hence๐๐ฆ0๐๐
โฝ ๐ถ1๐โ๐
๐ฆ0 โฝ ๐ถ1๏ฟฝ๐โ๐๐๐ + ๐ถ2
โฝ โ๐ถ1๐โ๐ + ๐ถ2 (5)
Applying boundary conditions ๐ฆ๐๐0 (0) = 1 gives
1 = โ๐ถ1 + ๐ถ2
๐ถ2 = 1 + ๐ถ1
And (5) becomes
๐ฆ๐๐0 (๐) โฝ โ๐ถ1๐โ๐ + (1 + ๐ถ1)
โฝ 1 + ๐ถ1 ๏ฟฝ1 โ ๐โ๐๏ฟฝ (6)
We now find ๐ฆ๐๐1 . Going back to (4) and collecting terms in ๐ (1) gives the ODE
๐ฆโฒโฒ1 โฝ ๐ฆโฒ1This is the same ODE we solved above. But it will have di๏ฟฝerent B.C. Hence
๐ฆ๐๐1 โฝ โ๐ถ3๐โ๐ + ๐ถ4
Applying boundary conditions ๐ฆ๐๐1 (0) = 0 gives
0 = โ๐ถ3 + ๐ถ4
๐ถ3 = ๐ถ4
Therefore
๐ฆ๐๐1 โฝ โ๐ถ3๐โ๐ + ๐ถ3
โฝ ๐ถ3 ๏ฟฝ1 โ ๐โ๐๏ฟฝ
Now we have the leading order ๐ฆ๐๐
๐ฆ๐๐ (๐) = ๐ฆ๐๐0 + ๐๐ฆ๐๐1= 1 + ๐ถ1 ๏ฟฝ1 โ ๐โ๐๏ฟฝ + ๐๐ถ3 ๏ฟฝ1 โ ๐โ๐๏ฟฝ + ๐ ๏ฟฝ๐2๏ฟฝ (7)
155
3.3. HW3 CHAPTER 3. HWS
Now we are ready to do the matching between (7) and (3A)
lim๐โโ
1 + ๐ถ1 ๏ฟฝ1 โ ๐โ๐๏ฟฝ + ๐๐ถ3 ๏ฟฝ1 โ ๐โ๐๏ฟฝ โผ
lim๐ฅโ0๏ฟฝ
2๐
๐๐ฅ22
โ1 + ๐ฅ2
โโโโโโโโ1 + ๐
โโโโโโโโ54โ932๐ โ ๐ฅ +
3๐ฅ
4 ๏ฟฝ1 + ๐ฅ2๏ฟฝ2 โ
7๐ฅ8 ๏ฟฝ1 + ๐ฅ2๏ฟฝ
+98
arctan (๐ฅ)
โโโโโโโโ
โโโโโโโโ
Or
1+๐ถ1+๐๐ถ3 โผ๏ฟฝ2๐
lim๐ฅโ0
๐๐ฅ22
โ1 + ๐ฅ2+๏ฟฝ2๐๐ lim๐ฅโ0
๐๐ฅ22
โ1 + ๐ฅ2
โโโโโโโโ54โ932๐ โ ๐ฅ +
3๐ฅ
4 ๏ฟฝ1 + ๐ฅ2๏ฟฝ2 โ
7๐ฅ8 ๏ฟฝ1 + ๐ฅ2๏ฟฝ
+98
arctan (๐ฅ)
โโโโโโโโ
But lim๐ฅโ0๐๐ฅ22
โ1+๐ฅ2โ 1, lim๐ฅโ0
3๐ฅ
4๏ฟฝ1+๐ฅ2๏ฟฝ2 โ 0 lim๐ฅโ0
7๐ฅ8๏ฟฝ1+๐ฅ2๏ฟฝ
โ 0 therefore the above becomes
1 + ๐ถ1 + ๐๐ถ3 โผ๏ฟฝ2๐+๏ฟฝ2๐๐ ๏ฟฝ54โ932๐๏ฟฝ
Hence
1 + ๐ถ1 = ๏ฟฝ2๐
๐ถ1 = ๏ฟฝ2๐โ 1
๐ถ3 = ๏ฟฝ2๐ ๏ฟฝ54โ932๐๏ฟฝ
This means that
๐ฆ๐๐ (๐) โผ 1 + ๐ถ1 ๏ฟฝ1 โ ๐โ๐๏ฟฝ + ๐๐ถ3 ๏ฟฝ1 โ ๐โ๐๏ฟฝ
โผ 1 +โโโโโโ๏ฟฝ
2๐โ 1โโโโโโ ๏ฟฝ1 โ ๐
โ๐๏ฟฝ + ๐๏ฟฝ2๐ ๏ฟฝ54โ932๐๏ฟฝ ๏ฟฝ1 โ ๐โ๐๏ฟฝ
Therefore
๐ฆuniform(๐ฅ) โผ ๐ฆ๐๐ (๐) + ๐ฆ๐๐ข๐ก (๐ฅ) โ ๐ฆ๐๐๐ก๐โ
โผ 1 +โโโโโโ๏ฟฝ
2๐โ 1โโโโโโ ๏ฟฝ1 โ ๐
โ๐๏ฟฝ + ๐๏ฟฝ2๐ ๏ฟฝ54โ932๐๏ฟฝ ๏ฟฝ1 โ ๐โ๐๏ฟฝ
+๏ฟฝ2๐
๐๐ฅ22
โ1 + ๐ฅ2
โโโโโโโโ1 + ๐
โโโโโโโโ54โ932๐ โ ๐ฅ +
3๐ฅ
4 ๏ฟฝ1 + ๐ฅ2๏ฟฝ2 โ
7๐ฅ8 ๏ฟฝ1 + ๐ฅ2๏ฟฝ
+98
arctan (๐ฅ)
โโโโโโโโ
โโโโโโโโ
โโโโโโโ๏ฟฝ
2๐+๏ฟฝ2๐๐ ๏ฟฝ54โ932๐๏ฟฝ
โโโโโโ
156
3.3. HW3 CHAPTER 3. HWS
Or (replacing ๐ by ๐ฅ๐ and simplifying)
๐ฆuniform(๐ฅ) โผ 1 +โโโโโโ๏ฟฝ
2๐โ 1โโโโโโ ๏ฟฝ1 โ ๐
โ๐๏ฟฝ โ ๐โ๐๐๏ฟฝ2๐ ๏ฟฝ54โ932๐๏ฟฝ
+๏ฟฝ2๐
๐๐ฅ22
โ1 + ๐ฅ2
โโโโโโโโ1 + ๐
โโโโโโโโ54โ932๐ โ ๐ฅ +
3๐ฅ
4 ๏ฟฝ1 + ๐ฅ2๏ฟฝ2 โ
7๐ฅ8 ๏ฟฝ1 + ๐ฅ2๏ฟฝ
+98
arctan (๐ฅ)
โโโโโโโโ
โโโโโโโโ
โ๏ฟฝ2๐
Or
๐ฆuniform(๐ฅ) โผ โ๏ฟฝ2๐๐โ๐ + ๐โ๐ โ ๐โ๐๐
๏ฟฝ2๐ ๏ฟฝ54โ932๐๏ฟฝ
+๏ฟฝ2๐
๐๐ฅ22
โ1 + ๐ฅ2
โโโโโโโโ1 + ๐
โโโโโโโโ54โ932๐ โ ๐ฅ +
3๐ฅ
4 ๏ฟฝ1 + ๐ฅ2๏ฟฝ2 โ
7๐ฅ8 ๏ฟฝ1 + ๐ฅ2๏ฟฝ
+98
arctan (๐ฅ)
โโโโโโโโ
โโโโโโโโ
To check validity of the above solution, the approximate solution is plotted against thenumerical solution for di๏ฟฝerent values of ๐ = {0.1, 0.05, 0.01}. This shows very good agreementwith the numerical solution. At ๐ = 0.01 the solutions are almost the same.
ClearAll[x, y, ฯต];
ฯต = 0.01;
r =2
Exp[1];
ode = ฯต y''[x] + x2 + 1 y'[x] - x3 y[x] โฉต 0;
sol = First@NDSolve[{ode, y[0] โฉต 1, y[1] โฉต 1}, y, {x, 0, 1}];
p1 = Plot[Evaluate[y[x] /. sol], {x, 0, 1}];
mysol[x_, ฯต_] := -r Exp-x
ฯต + Exp
-x
ฯต - ฯต r
5
4-
9
32ฯ + r
Exp x2
2
1 + x2
1 + ฯต5
4-
9
32ฯ - x +
3 x
4 1 + x2
2-
7 x
8 1 + x2
+9
8ArcTan[x] ;
p2 = Plot[{Evaluate[y[x] /. sol], mysol[x, ฯต]}, {x, 0, 1}, Frame โ True, PlotStyle โ {Red, Blue},
FrameLabel โ {{"y(x)", None}, {"x", Row[{"Comparing numerical and apprxoimation, using ฯต = ", N@ฯต}]}}, GridLines โ Automatic,
GridLinesStyle โ LightGray, PlotLegends โ {"Approximation", "Numerical"}, BaseStyle โ 14, ImageSize โ 400]
157
3.3. HW3 CHAPTER 3. HWS
Out[145]=
0.0 0.2 0.4 0.6 0.8 1.0
0.86
0.88
0.90
0.92
0.94
0.96
0.98
1.00
x
y(x)
Comparing numerical and apprxoimation, using ฯต = 0.1
Approximation
Numerical
Out[153]=
0.0 0.2 0.4 0.6 0.8 1.0
0.86
0.88
0.90
0.92
0.94
0.96
0.98
1.00
x
y(x)
Comparing numerical and apprxoimation, using ฯต = 0.05
Approximation
Numerical
158
3.3. HW3 CHAPTER 3. HWS
Out[161]=
0.0 0.2 0.4 0.6 0.8 1.0
0.86
0.88
0.90
0.92
0.94
0.96
0.98
1.00
x
y(x)
Comparing numerical and apprxoimation, using ฯต = 0.01
Approximation
Numerical
3.3.6 problem 9.19
Problem Find lowest order uniform approximation to boundary value problem
๐๐ฆโฒโฒ + (sin ๐ฅ) ๐ฆโฒ + ๐ฆ sin (2๐ฅ) = 0๐ฆ (0) = ๐๐ฆ (๐) = 0
Solution
We expect a boundary layer at left end at ๐ฅ = 0. Therefore, we need to find ๐ฆ๐๐ (๐) , ๐ฆ๐๐ข๐ก (๐ฅ) ,where ๐ is an inner variable defined by ๐ = ๐ฅ
๐๐ .
159
3.3. HW3 CHAPTER 3. HWS
x0 ฯ
ฮถ ฮท
leftboundarylayer
outer region
matching at these locations
rightboundarylayer
Finding ๐ฆ๐๐ (๐)
At ๐ฅ = 0, we introduce inner variable ๐ = ๐ฅ๐๐ and then express the original ODE using this
new variable. We also need to determine ๐ in the above expression. Since ๐๐ฆ๐๐ฅ =
๐๐ฆ๐๐
๐๐๐๐ฅ then
๐๐ฆ๐๐ฅ =
๐๐ฆ๐๐๐
โ๐. Hence ๐๐๐ฅ โก ๐
โ๐ ๐๐๐
๐2
๐๐ฅ2=๐๐๐ฅ
๐๐๐ฅ
= ๏ฟฝ๐โ๐๐๐๐๏ฟฝ ๏ฟฝ
๐โ๐๐๐๐๏ฟฝ
= ๐โ2๐๐2
๐๐2
Therefore ๐2๐ฆ๐๐ฅ2 = ๐
โ2๐ ๐2๐ฆ๐๐2 and the ODE ๐๐ฆโฒโฒ + (sin ๐ฅ) ๐ฆโฒ + ๐ฆ sin (2๐ฅ) = 0 now becomes
๐๐โ2๐๐2๐ฆ๐๐2
+ (sin (๐๐๐)) ๐โ๐ ๐๐ฆ๐๐
+ sin (2๐๐๐) ๐ฆ = 0
๐1โ2๐๐2๐ฆ๐๐2
+ (sin (๐๐๐)) ๐โ๐ ๐๐ฆ๐๐
+ sin (2๐๐๐) ๐ฆ = 0
160
3.3. HW3 CHAPTER 3. HWS
Expanding the sin terms in the above, in Taylor series around zero, sin (๐ฅ) = ๐ฅ โ ๐ฅ3
3! +โฏ gives
๐1โ2๐๐2๐ฆ๐๐2
+โโโโโ๐๐๐ โ
(๐๐๐)3
3!+โฏ
โโโโโ ๐โ๐
๐๐ฆ๐๐
+โโโโโ2๐๐๐ โ
(2๐๐๐)3
3!+โฏ
โโโโโ ๐ฆ = 0
๐1โ2๐๐2๐ฆ๐๐2
+ ๏ฟฝ๐ โ๐3๐2๐
3!+โฏ๏ฟฝ
๐๐ฆ๐๐
+โโโโโ2๐๐๐ โ
(2๐๐๐)3
3!+โฏ
โโโโโ ๐ฆ = 0
Then the largest terms are ๏ฟฝ๐1โ2๐, 1๏ฟฝ, therefore 1 โ 2๐ = 0 or
๐ = 12
The ODE now becomes
๐ฆโฒโฒ + ๏ฟฝ๐ โ๐3๐3!
+โฏ๏ฟฝ ๐ฆโฒ +
โโโโโโโโ2๐โ๐ โ
๏ฟฝ2๐โ๐๏ฟฝ3
3!+โฏ
โโโโโโโโ ๐ฆ = 0 (1)
Assuming that
๐ฆ๐๐๐๐ก (๐ฅ) =โ๏ฟฝ๐=0
๐๐๐ฆ๐ = ๐ฆ0 + ๐๐ฆ1 + ๐2๐ฆ2 +โฏ
Then (1) becomes
๏ฟฝ๐ฆโฒโฒ0 + ๐๐ฆโฒโฒ1 +โฏ๏ฟฝ + ๏ฟฝ๐ โ๐3๐3!
+โฏ๏ฟฝ ๏ฟฝ๐ฆโฒ0 + ๐๐ฆโฒ1 +โฏ๏ฟฝ +
โโโโโโโโ2๐โ๐ โ
๏ฟฝ2๐โ๐๏ฟฝ3
3!+โฏ
โโโโโโโโ ๏ฟฝ๐ฆ0 + ๐๐ฆ1 +โฏ๏ฟฝ = 0
Collecting terms in ๐ (1) gives the balance
๐ฆโฒโฒ0 (๐) โผ โ๐๐ฆโฒ0 (๐)๐ฆ0 (0) = ๐
Assuming ๐ง = ๐ฆโฒ0, then
๐งโฒ โผ โ๐๐ง๐๐ง๐ง
โผ โ๐
ln |๐ง| โผ โ๐2
2+ ๐ถ1
๐ง โผ ๐ถ1๐โ๐22
Therefore ๐ฆโฒ0 โผ ๐ถ1๐โ๐22 . Hence
๐ฆ0 (๐) โผ ๐ถ1๏ฟฝ๐
0๐โ๐ 22 ๐๐ + ๐ถ2
With boundary conditions ๐ฆ (0) = ๐. Hence
๐ = ๐ถ2
161
3.3. HW3 CHAPTER 3. HWS
And the solution becomes
๐ฆ๐๐0 (๐) โผ ๐ถ1๏ฟฝ๐
0๐โ๐ 22 ๐๐ + ๐ (2)
Now we need to find ๐ฆ๐๐ข๐ก (๐ฅ). Assuming that
๐ฆ๐๐ข๐ก (๐ฅ) =โ๏ฟฝ๐=0
๐๐๐ฆ๐ = ๐ฆ0 + ๐๐ฆ1 + ๐2๐ฆ2 +โฏ
Then ๐๐ฆโฒโฒ + (sin ๐ฅ) ๐ฆโฒ + ๐ฆ sin (2๐ฅ) = 0 becomes
๐ ๏ฟฝ๐ฆโฒโฒ0 + ๐๐ฆโฒโฒ1 +โฏ๏ฟฝ + sin (๐ฅ) ๏ฟฝ๐ฆโฒ0 + ๐๐ฆโฒ1 +โฏ๏ฟฝ + sin (2๐ฅ) ๏ฟฝ๐ฆ0 + ๐๐ฆ1 +โฏ๏ฟฝ = 0Collecting terms in ๐ (1) gives the balance
sin (๐ฅ) ๐ฆโฒ0 (๐ฅ) โผ โ sin (2๐ฅ) ๐ฆ0 (๐ฅ)๐๐ฆ0๐ฆ0
โผ โsin (2๐ฅ)sin (๐ฅ) ๐๐ฅ
ln ๏ฟฝ๐ฆ0๏ฟฝ โผ โ๏ฟฝsin (2๐ฅ)sin (๐ฅ) ๐๐ฅ
โผ โ๏ฟฝ2 sin ๐ฅ cos ๐ฅ
sin (๐ฅ) ๐๐ฅ
โผ โ๏ฟฝ2 cos ๐ฅ๐๐ฅ
โผ โ2 sin ๐ฅ + ๐ถ5
Hence
๐ฆ๐๐ข๐ก0 (๐ฅ) โผ ๐ด๐โ2 sin ๐ฅ
๐ฆ0 (๐) = 0
Therefore ๐ด = 0 and ๐ฆ๐๐ข๐ก0 (๐ฅ) = 0.Now that we found all solutions, we can do the matching.The matching on the left side gives
lim๐โโ
๐ฆ๐๐ (๐) = lim๐ฅโ0
๐ฆ๐๐ข๐ก (๐ฅ)
lim๐โโ
๐ถ1๏ฟฝ๐
0๐โ๐ 22 ๐๐ + ๐ = lim
๐ฅโ0๐ถ5๐โ2 sin ๐ฅ
lim๐โโ
๐ถ1๏ฟฝ๐
0๐โ๐ 22 ๐๐ + ๐ = 0 (3)
But
๏ฟฝ๐
0๐โ๐ 22 ๐๐ =
๏ฟฝ๐2
erf ๏ฟฝ๐
โ2๏ฟฝ
162
3.3. HW3 CHAPTER 3. HWS
And lim๐โโ erf ๏ฟฝ ๐
โ2๏ฟฝ = 1, hence (3) becomes
๐ถ1๏ฟฝ๐2+ ๐ = 0
๐ถ1 = โ๐๏ฟฝ2๐
= โโ2๐ (4)
Therefore from (2)
๐ฆ๐๐0 (๐) โผ โโ2๐๏ฟฝ๐
0๐โ๐ 22 ๐๐ + ๐ (5)
Near ๐ฅ = ๐, using ๐ = ๐โ๐ฅ๐๐ . Expansion ๐ฆ
๐๐ ๏ฟฝ๐๏ฟฝ โผ ๐ฆ0 ๏ฟฝ๐๏ฟฝ + ๐๐ฆ1 ๏ฟฝ๐๏ฟฝ + ๐ ๏ฟฝ๐2๏ฟฝ gives ๐ =12 . Hence
๐ (1) terms gives
๐ฆโฒโฒ0 ๏ฟฝ๐๏ฟฝ โผ ๐๐ฆโฒ0 ๏ฟฝ๐๏ฟฝ
๐ฆ๐๐0 (0) = 0
๐ฆ๐๐0 ๏ฟฝ๐๏ฟฝ โผ ๐ท๏ฟฝ๐
0๐โ๐ 22 ๐๐
And matching on the right side gives
lim๐โโ
๐ฆ๐๐ ๏ฟฝ๐๏ฟฝ = lim๐ฅโ๐
๐ฆ๐๐ข๐ก (๐ฅ)
lim๐โโ
๐ท๏ฟฝ๐
0๐โ๐ 22 ๐๐ = 0
๐ท = 0 (6)
Therefore the solution is
๐ฆ (๐ฅ) โผ ๐ฆ๐๐ (๐) + ๐ฆ๐๐ ๏ฟฝ๐๏ฟฝ + ๐ฆ๐๐ข๐ก (๐ฅ) โ ๐ฆ๐๐๐ก๐โ
โผ โโ2๐๏ฟฝ๐
0๐โ๐ 22 ๐๐ + ๐ + 0
โผ โโ2๐๏ฟฝ๐2
erf ๏ฟฝ๐
โ2๏ฟฝ + ๐
โผ โโ2๐๏ฟฝ๐2
erf ๏ฟฝ๐ฅ
โ2๐๏ฟฝ + ๐
โผ ๐ โ ๐ erf ๏ฟฝ๐ฅ
โ2๐๏ฟฝ (7)
The following plot compares exact solution with (7) for ๐ = 0.1, 0.05. We see from theseresults, that as ๐ decreased, the approximation solution improved.
163
3.3. HW3 CHAPTER 3. HWS
In[37]:= ClearAll[x, ฯต, y]
mySol[x_, ฯต_] := Pi - Pi Erfx
2 ฯต
;
ฯต = 0.1;
ode = ฯต y''[x] + Sin[x] y'[x] + y[x] Sin[2 x] โฉต 0;
sol = NDSolve[{ode, y[0] โฉต ฯ, y'[ฯ] โฉต 0}, y, {x, 0, ฯ}];
Plot[{mySol[x, ฯต], Evaluate[y[x] /. sol]}, {x, 0, Pi}, Frame โ True,
FrameLabel โ {{"y(x)", None}, {"x", Row[{"Numerical vs. approximation for ฯต=", ฯต}]}}, GridLines โ Automatic,
GridLinesStyle โ LightGray, BaseStyle โ 16, ImageSize โ 500, PlotLegends โ {"Approximation", "Numerical"},
PlotStyle โ {Red, Blue}, PlotRange โ All]
Out[42]=
0.0 0.5 1.0 1.5 2.0 2.5 3.0
0
1
2
3
4
x
y(x)
Numerical vs. approximation for ฯต=0.1
Approximation
Numerical
Out[54]=
0.0 0.5 1.0 1.5 2.0 2.5 3.0
0.0
0.5
1.0
1.5
2.0
2.5
3.0
x
y(x)
Numerical vs. approximation for ฯต=0.05
Approximation
Numerical
164
3.3. HW3 CHAPTER 3. HWS
3.3.7 key solution of selected problems
3.3.7.1 section 9 problem 9
Problem 9.9 asks us to use boundary layer theory to find the leading order solution to the initialvalue problem ฮตyโฒโฒ(x) + ayโฒ(x) + by(x) = 0 with y(0) = yโฒ(0) = 1 and a > 0. Then we are to compareto the exact solution. The problem is ambiguous as to whether a and b are functions or constants.For clarityโs sake, we assume a and b are constants, but all the following work can be generalized fornonconstant a and b as well.
Since a > 0, the boundary layer occurs at x = 0, where the initial conditions are specified. We setx = ฮตฮพ,1 and then in the inner region,
1
ฮตyโฒโฒin(ฮพ) +
a
ฮตyโฒin(ฮพ) + byin(ฮพ) = 0. (1)
Thus, to leading order, yโฒโฒin(ฮพ) โผ โayโฒin(ฮพ), which has solution yin(ฮพ) = C0 + C1eโaฮพ. Solving for C0 and
C1, we see that y(0) = 1 =โ C0 + C1 = 1, and
yโฒ(x)
โฃโฃโฃโฃx=0
= 1 =โ yโฒ(ฮพ)
โฃโฃโฃโฃฮพ=0
= ฮต =โ โaC1 = ฮต =โ C1 = โฮต/a = O(ฮต).
This means that C1eโaฮพ = O(ฮต) shouldnโt appear at this order in the expansion, and C1 = 0. We
should throw this information out because we have already thrown out information at O(ฮต) in solvingthe equation, and we have no guarantee that the O(ฮต) value for C1 is actually correct to O(ฮต).
So there is no boundary layer at leading order! The inner solution yin(x) = C0 = 1 does not changerapidly, and it will cancel when we match, just leaving the outer solution. This is okay and happensoccasionally when you get lucky.
In the outer region, we have yout โผ โ bayout, so yout(x) = Ceโbx/a + O(ฮต). Matching to the inner
solution, C = 1, so yuniform(x) = eโbx/a +O(ฮต). We note that yโฒuniform(0) = โ ba6= 1 in general.
Although the problem does not ask for it, we can also go to the next order in our asymptoticexpansion. And even though no boundary layer appeared at leading order, one will appear at O(ฮต).
Going back to the inner region, let yin(ฮพ) = Y0(ฮพ) + ฮตY1(ฮพ) + O(ฮต2). We already computed thatY0(ฮพ) = 1. Now looking at (1) at O(1), Y โฒโฒ1 (ฮพ) +aY โฒ1(ฮพ) + bY0(ฮพ) = 0 =โ Y โฒโฒ1 (ฮพ) +aY โฒ1(ฮพ) = โb. Solving,Y1(ฮพ) = C2 + C3e
โaฮพ โ baฮพ. We have initial conditions Y1(0) = 0, but since yโฒ(0) = 1 was not satisfied,
we note that Y โฒ1(0) = 1. Thus, C2 + C3 = 0 and โaC3 โ ba
= 1, so
C3 = โ1
aโ b
a2and C2 = โC3 =
1
a+
b
a2.
Therefore,
yin(ฮพ) = 1 + ฮต
((1
a+
b
a2
)(1โ eโaฮพ
)โ b
aฮพ
)+O(ฮต2),
so
yin(x) = 1 + ฮต
((1
a+
b
a2
)(1โ eโax/ฮต
)โ b
a
x
ฮต
)+O(ฮต2).
Turning to the outer solution, let yout(x) = y0(x) + ฮตy1(x) + O(ฮต2). We already found that y0(x) =eโbx/a. At order ฮต, the outer equation reads:
yโฒโฒ0(x) + ayโฒ1(x) + by1(x) = 0 =โ yโฒ1(x) +b
ay1(x) = โ1
aeโbx/a.
1We know that we can use ฮต instead of ฮตp for some unknown constant p because a is positive near zero, and p 6= 1 onlyoccurs when a(x)โ 0 at the boundary layer.
1
165
3.3. HW3 CHAPTER 3. HWS
Therefore,
y1(x) = C4eโbx/a โ x
aeโbx/a,
andyout(x) = eโbx/a + ฮต
((C4 โ
x
a
)eโbx/a
)+O(ฮต2).
Now we match. As xโ 0+, the outer solution goes to
yout(x) โผ 1 + C4ฮต+O(ฮต2),
while as ฮพ โโ, the inner solution approaches
yin(x) โผ 1 +
(1
a+
b
a2
)ฮต+O(ฮต2),
where I have cheated slightly.2 Matching,
C4 =1
a+
b
a2,
and thus,
yuniform(x) = eโbx/a + ฮต
(((1
a+
b
a2
)โ x
a
)eโbx/a โ
(1
a+
b
a2
)eโax/ฮต
)+O(ฮต2).
Let us now graphically compare the asymptotic solutions accurate to O(1) and O(ฮต) with the exactsolution:
yexact(x) =1
2โa2 โ 4bฮต
(โaeโ(a/ฮต+
โa2โ4bฮต/ฮต)x/2 โ 2ฮตeโ(a/ฮต+
โa2โ4bฮต/ฮต)x/2 +
โa2 โ 4bฮตeโ(a/ฮต+
โa2โ4bฮต/ฮต)x/2
+aeโ(a/ฮตโโa2โ4bฮต/ฮต)x/2 + 2ฮตeโ(a/ฮตโ
โa2โ4bฮต/ฮต)x/2 +
โa2 โ 4bฮตeโ(a/ฮตโ
โa2โ4bฮต/ฮต)x/2
).
For simplicity, take a = b = 1.
2I ignored the term which was linear in ฮพ, which blows up as ฮพ โโ. This term came from ignoring the outer expansion,which changes on the same order, and is called a secular term. This occurs because we ignored the outer expansion, as isconventional when working with the boundary layer, but the outer solution, when Taylor expanded about zero, matchesthis term exactly at order x, and so we do not have any problems. The outer solution changes at a slow rate compared toฮพ, and so the appearance of two time scales in the boundary layer causes problems with the match (which I swept underthe rug by cheating). This problem is therefore far better suited for multiscale methods, which form the subject matterof Chapter 11.
2
166
3.3. HW3 CHAPTER 3. HWS
For the first graph, ฮต = 0.1:
The blue curve represents the exact solution, the orange curve the uniform solution to leading order, andthe green curve the uniform solution accurate to O(ฮต). We see that the differences between the curvesremains small always, and that the higher order approximation is much closer to the exact solution. Theleading order solution never differs by more than about 0.1 โ ฮต, while the next order solution differsfrom the exact solution by a much smaller amount (approximately O(ฮต2)).
Now let ฮต = 0.05:
The color scheme is the same as before. We notice that the same qualitative observations from beforehold for this graph as well. The difference between the orange and blue curves is even half as much, ingood agreement with our O(ฮต) error estimation. This also validates our conclusion that the boundarylayer only appears at O(ฮต). It is somewhat more difficult to verify pictorially that the error for the greencurve is O(ฮต2), but it is also clear that the error in this plot is less than in the first plot, and by a greaterfactor than two.
3
167
3.3. HW3 CHAPTER 3. HWS
3.3.7.2 section 9 problem 15
168
3.3. HW3 CHAPTER 3. HWS
169
3.3. HW3 CHAPTER 3. HWS
170
3.3. HW3 CHAPTER 3. HWS
171
3.3. HW3 CHAPTER 3. HWS
3.3.7.3 section 9 problem 19
Here is an outline for the solution to BO 9.19.
The boundary value problem is
วซyโฒโฒ + sin(x)yโฒ + 2 sin(x) cos(x)y = 0, y(0) = ฯ, y(ฯ) = 0
using sin(2x) = 2 sin(x) cos(x). Our heuristic argument based on the 1D advection-diffusion equation suggests a boundary layer on the left at x = 0. An outer expansionyout(x) โผ yo(x) + วซy1(x) +O(วซ2) gives
O(1) : yโฒo โผ โ2 cos(x)yo
with solution
yo(x) โผ A exp[โ2 sin(x)]
which cannot satisfy y(ฯ) = 0 unless A = 0 (see below).
For the inner solution near x = 0 let ฮพ = x/วซp. The expansion yin(ฮพ) โผ yo(ฮพ)+วซy1(ฮพ)+O(วซ2)leads to p = 1/2 and
O(1) : yโฒโฒo (ฮพ) โผ โฮพyโฒo(ฮพ), y(ฮพ = 0) = ฯ
with solution
yo(ฮพ) โผ ฯ + C
โซ ฮพ
0
exp(โt2/2)dt
yo(ฮพ) โผ ฯ +โ2C
โซ ฮพ/โ2
0
exp(โs2)ds = ฯ +โ2C
โฯ
2erf(ฮพ/
โ2).
On the other side near x = ฯ, let ฮท = (ฯโ x)/วซp. The expansion yin(ฮท) โผ yo(ฮท)+ วซy1(ฮท)+O(วซ2) leads to p = 1/2 and
O(1) : yโฒโฒo (ฮท) โผ ฮทyโฒo(ฮท), y(ฮท = 0) = 0
using sin(ฯ โ วซ1/2ฮท) = sin(วซ1/2ฮท) โผ วซ1/2ฮท for วซ1/2ฮท โ 0. Then
yo(ฮท) โผ D
โซ ฮท
0
exp(t2/2)dt
Now matching
limฮพโโ
yin(ฮพ) = limxโ0
yout(x)
limฮทโโ
yin(ฮท) = limxโฯ
yout(x)
gives A = D = 0 and C = โโ2ฯ. Thus we obtain
y(x) โผ ฯ โ ฯ erf(x/โ2วซ).
1
172
3.3. HW3 CHAPTER 3. HWS
3.3.7.4 section 9 problem 4b
NEEP 548: Engineering Analysis II 2/27/2011
Problem 9.4-b: Bender & Orszag
Instructor: Leslie Smith
1 Problem Statement
Find the boundary layer solution for
วซyโฒโฒ + (x2 + 1)yโฒ โ x3y = 0 (1)
2 Solution
We can expect a boundary layer at x = 0.
2.1 Outer Solution
(x2 + 1)yโฒo โ x3yo = 0
yโฒo โ
(
x3
x2 + 1
)
yo = 0
This is in a form that we can find an integrating factor for,
ยต = exp
(โซ
x3
x2 + 1dx
)
= exp
[
โ1
2(ln[x2 + 1]โ x2)
]
= (x2 + 1)โ1/2eโx2/2
=โ [yo(x2 + 1)โ1/2eโx2/2]โฒ = 0
yo = C1(x2 + 1)โ1/2ex
2/2
B.C.: y(1) = 1 =โ 1 = C1(2)โ1/2e1/2 =โ C1 =
โ2eโ1/2. Therefore, the outer solution is:
yo =โ2eโ1/2(x2 + 1)โ1/2ex
2/2 (2)
1
173
3.3. HW3 CHAPTER 3. HWS
2.2 Inner Solution
Define ฮพ = x/วซ such that when x โผ วซ, ฮพ โผ 1
โy
โx=
dy
dฮพ
dฮพ
dx
โ2y
โx2=
1
วซ2d2y
dฮพ2
Plug this into the D. E. to get,
1
วซ
d2y
dฮพ2+ วซฮพ2
dy
dฮพ+
1
วซ
dy
dฮพโ ฮพ2วซ2y = 0 (3)
The let the inner solution take the form,
yin = yo(ฮพ) + วซy1(ฮพ) + วซ2y2(ฮพ) + ...
Balance terms, order by order,
O
(
1
วซ
)
: yโฒโฒo + yโฒo = 0 y(0) = 1
O(1) : yโฒโฒ1 + yโฒ1 = 0 y(0) = 0
To the lowest order,
yโฒโฒin + yโฒin = 0 let z = yโฒin
=โ ln z = โฮพ + C2
ln yโฒin = โฮพ + C2
yโฒin = C3eโฮพ
yin = C3eฮพ + C4
We can solve for one of the constants by imposing the boundary conditions, yin(0) = 1 =โ 1 =โC3 + C4 =โ C4 = 1 + C3
yin = 1 + C(
1โ eโฮพ)
(4)
The other constant we get by matching the โouterโ and โinnerโ solutions:
2
174
3.3. HW3 CHAPTER 3. HWS
limxโ0
yo = limฮพโโ
yin
limxโ0
โ2eโ1/2(x2 + 1)โ1/2ex
2/2 = limฮพโโ
[1 +C(1โ eโฮพ)]
โ2eโ1/2 = 1 +C
=โ C =โ2eโ1/2 โ 1
Note: ymatch =โ2eโ1/2
ytot = yin + yout โ ymatch (5)
ytot =โ2eโ1/2(x2 + 1)โ1/2ex
2/2 + eโx/วซ(1โโ2eโ1/2) (6)
3
175
3.3. HW3 CHAPTER 3. HWS
3.3.7.5 section 9 problem 6
Here is an outline for the solution to BO 9.6.
Write the equation as
yโฒ = (1 + วซxโ2)y2 โ 2y + 1, y(1) = 1
and we want to solve this on 0 โค x โค 1. For the outer solution, the expansion yout(x) โผyo(x) + วซy1(x) +O(วซ2) gives
O(1) : yโฒo โผ (yo โ 1)2, yo(1) = 1
O(วซ) : yโฒ1โผ 2yoy1 + y2ox
โ2โ 2y1, y1(1) = 0
with solutions
yo(x) โผ 1, y1(x) โผ 1โ xโ1
valid away from zero. So yout(x) โผ 1 + วซ(1โ xโ1) +O(วซ2).
For the inner solution let ฮพ = x/วซp. The expansion yout(ฮพ) โผ yo(ฮพ) + วซy1(ฮพ) +O(วซ2) leadsto ฮพ โ วซ (so we take ฮพ = วซ for convenience). The biggest terms are:
O(วซโ1) : yโฒo โผ y2oฮพโ2
with solution
yo(ฮพ) โผ ฮพ(1โ Aฮพ)โ1.
Now matching
limฮพโโ
yo(ฮพ) +O(วซ) = limxโ0
yo(x) +O(วซ)
limฮพโโ
ฮพ(1โ Aฮพ)โ1 +O(วซ) = limxโ0
1 +O(วซ)
limฮพโโ
โAโ1 +O(ฮพโ1) +O(วซ) = limxโ0
1 +O(วซ)
gives A = โ1 with matching region วซ โช x โช 1.
The next-order problem is
O(1) : yโฒ1โผ y2o + 2yoy1ฮพ
โ2โ 2yo + 1.
Use yo(ฮพ) from above and the integrating factor method to find
y1(ฮพ) โผ โฮพ(1 + ฮพ)โ2 + Cฮพ2(1 + ฮพ)โ2.
Now matching
1
176
3.3. HW3 CHAPTER 3. HWS
limฮพโโ
yo(ฮพ) + วซy1(ฮพ) +O(วซ2) = limxโ0
yo(x) + วซy1(x) +O(วซ2)
limฮพโโ
ฮพ(1 + ฮพ)โ1 + วซ
(
Cฮพ2(1 + ฮพ)โ2โ ฮพ(1 + ฮพ)โ2
)
+O(วซ2) = limxโ0
1 + วซ
(
1โ xโ1
)
+O(วซ2)
limฮพโโ
1
1 + ฮพโ1+ วซ
(
Cฮพ2(1 + ฮพ)โ2โ ฮพ(1 + ฮพ)โ2
)
+O(วซ2) = limxโ0
1 + วซโวซ
x+O(วซ2)
limฮพโโ
1โ ฮพโ1 +O(ฮพโ2) + วซC +O(วซฮพโ1) +O(วซ2) = limxโ0
1 + วซโวซ
x+O(วซ2)
and so we choose C = 1.
Question for you: what is the matching region? (I think วซ1/2 โช x โช 1). Is this correct?
Now we form the uniform approximation y โผ yin + yout โ ymatch +O(วซ2):
y โผ 1 + วซ(1โ xโ1) + ฮพ(1 + ฮพ)โ1 + วซ
(
ฮพ2(1 + ฮพ)โ2โ ฮพ(1 + ฮพ)โ2
)
โ
(
1 + วซโวซ
x
)
+O(วซ2)
with ฮพ = x/วซ and simplify. Notice that there is no singularity at x = 0 the solution isuniformly valid on 0 โค x โค 1.
More questions for you: Does the solution above satisfy the initial condition y(1) = 1?If not, can you add an O(วซ2) correction so that y(1) = 1? Will you still have a solutionuniformly valid to O(วซ2)?
I encourage you to make plots so that you can visualize the solution!
2
177
3.4. HW4 CHAPTER 3. HWS
3.4 HW4
3.4.1 problem 10.5 (page 540)
problem Use WKB to obtain solution to
๐๐ฆโฒโฒ + ๐ (๐ฅ) ๐ฆโฒ + ๐ (๐ฅ) ๐ฆ = 0 (1)
with ๐ (๐ฅ) > 0, ๐ฆ (0) = ๐ด, ๐ฆ (1) = ๐ต correct to order ๐.
solution
Assuming
๐ฆ (๐ฅ) โผ exp ๏ฟฝ1๐ฟ
โ๏ฟฝ๐=0
๐ฟ๐๐๐ (๐ฅ)๏ฟฝ ๐ฟ โ 0
Therefore
๐ฆโฒ (๐ฅ) โผ1๐ฟ
โ๏ฟฝ๐=0
๐ฟ๐๐โฒ๐ (๐ฅ) exp ๏ฟฝ1๐ฟ
โ๏ฟฝ๐=0
๐ฟ๐๐๐ (๐ฅ)๏ฟฝ
๐ฆโฒโฒ (๐ฅ) โผ1๐ฟ
โ๏ฟฝ๐=0
๐ฟ๐๐โฒโฒ๐ (๐ฅ) exp ๏ฟฝ1๐ฟ
โ๏ฟฝ๐=0
๐ฟ๐๐๐ (๐ฅ)๏ฟฝ + ๏ฟฝ1๐ฟ
โ๏ฟฝ๐=0
๐ฟ๐๐โฒ๐ (๐ฅ)๏ฟฝ2
exp ๏ฟฝ1๐ฟ
โ๏ฟฝ๐=0
๐ฟ๐๐๐ (๐ฅ)๏ฟฝ
Substituting the above into (1) and simplifying gives (writing = instead of โผ for simplicityfor now)
๐โกโขโขโขโขโฃ1๐ฟ
โ๏ฟฝ๐=0
๐ฟ๐๐โฒโฒ๐ (๐ฅ) + ๏ฟฝ1๐ฟ
โ๏ฟฝ๐=0
๐ฟ๐๐โฒ๐ (๐ฅ)๏ฟฝ2โคโฅโฅโฅโฅโฆ +
๐๐ฟ
โ๏ฟฝ๐=0
๐ฟ๐๐โฒ๐ (๐ฅ) + ๐ = 0
๐๐ฟ
โ๏ฟฝ๐=0
๐ฟ๐๐โฒโฒ๐ (๐ฅ) +๐๐ฟ2 ๏ฟฝ
โ๏ฟฝ๐=0
๐ฟ๐๐โฒ๐ (๐ฅ)โ๏ฟฝ๐=0
๐ฟ๐๐โฒ๐ (๐ฅ)๏ฟฝ +๐๐ฟ
โ๏ฟฝ๐=0
๐ฟ๐๐โฒ๐ (๐ฅ) + ๐ = 0
Expanding gives
๐๐ฟ๏ฟฝ๐โฒโฒ0 + ๐ฟ๐โฒโฒ1 + ๐ฟ2๐โฒโฒ2 +โฏ๏ฟฝ
+๐๐ฟ2๏ฟฝ๏ฟฝ๐โฒ0 + ๐ฟ๐โฒ1 + ๐ฟ2๐โฒ2 +โฏ๏ฟฝ ๏ฟฝ๐โฒ0 + ๐ฟ๐โฒ1 + ๐ฟ2๐โฒ2 +โฏ๏ฟฝ๏ฟฝ
+๐๐ฟ๏ฟฝ๐โฒ0 + ๐ฟ๐โฒ1 + ๐ฟ2๐โฒ2 +โฏ๏ฟฝ + ๐ = 0
Simplifying
๏ฟฝ๐๐ฟ๐โฒโฒ0 + ๐๐โฒโฒ1 + ๐๐ฟ๐โฒโฒ2 +โฏ๏ฟฝ
+ ๏ฟฝ๐๐ฟ2๏ฟฝ๐โฒ0๏ฟฝ
2+2๐๐ฟ๏ฟฝ๐โฒ0๐โฒ1๏ฟฝ + ๐ ๏ฟฝ2๐โฒ0๐โฒ2 + ๏ฟฝ๐โฒ1๏ฟฝ
2๏ฟฝ +โฏ๏ฟฝ
+ ๏ฟฝ๐๐ฟ๐โฒ0 + ๐๐โฒ1 + ๐๐ฟ๐โฒ2 +โฏ๏ฟฝ + ๐ = 0 (1A)
The largest terms in the left are ๐๐ฟ2๏ฟฝ๐โฒ0๏ฟฝ
2and ๐
๐ฟ๐โฒ0. By dominant balance these must be equal in
magnitude. Hence ๐๐ฟ2 = ๐๏ฟฝ
1๐ฟ๏ฟฝ or ๐
๐ฟ = ๐ (1). Therefore ๐ฟ is proportional to ๐ and for simplicity
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3.4. HW4 CHAPTER 3. HWS
๐ is taken as equal to ๐ฟ, hence (1A) becomes
๏ฟฝ๐โฒโฒ0 + ๐๐โฒโฒ1 + ๐2๐โฒโฒ2 +โฏ๏ฟฝ
+ ๏ฟฝ๐โ1 ๏ฟฝ๐โฒ0๏ฟฝ2+ 2๐โฒ0๐โฒ1 + ๐ ๏ฟฝ2๐โฒ0๐โฒ2 + ๏ฟฝ๐โฒ1๏ฟฝ
2๏ฟฝ +โฏ๏ฟฝ
+ ๏ฟฝ๐๐โ1๐โฒ0 + ๐๐โฒ1 + ๐๐๐โฒ2 +โฏ๏ฟฝ + ๐ = 0
Terms of ๐๏ฟฝ๐โ1๏ฟฝ give
๏ฟฝ๐โฒ0๏ฟฝ2+ ๐๐โฒ0 = 0 (2)
And terms of ๐ (1) give
๐โฒโฒ0 + 2๐โฒ0๐โฒ1 + ๐๐โฒ1 + ๐ = 0 (3)
And terms of ๐ (๐) give
2๐โฒ0๐โฒ2 + ๐๐โฒ2 + ๏ฟฝ๐โฒ1๏ฟฝ2+ ๐โฒโฒ1 = 0
๐โฒ2 = โ๏ฟฝ๐โฒ1๏ฟฝ
2+ ๐โฒโฒ1
๏ฟฝ๐ + 2๐โฒ0๏ฟฝ(4)
Starting with (2)
๐โฒ0 ๏ฟฝ๐โฒ0 + ๐๏ฟฝ = 0
There are two cases to consider.
case 1 ๐โฒ0 = 0. This means that ๐0 (๐ฅ) = ๐0. A constant. Using this result in (3) gives an ODEto solve for ๐1 (๐ฅ)
๐๐โฒ1 + ๐ = 0
๐โฒ1 = โ๐ (๐ฅ)๐ (๐ฅ)
๐1 โผ โ๏ฟฝ๐ฅ
0
๐ (๐ก)๐ (๐ก)
๐๐ก + ๐1
Using this result in (4) gives an ODE to solve for ๐2 (๐ฅ)
๐โฒ2 = โ๏ฟฝโ ๐(๐ฅ)
๐(๐ฅ)๏ฟฝ2+ ๏ฟฝโ ๐(๐ฅ)
๐(๐ฅ)๏ฟฝโฒ
๐ (๐ฅ)
= โ
๐2(๐ฅ)๐2(๐ฅ) โ ๏ฟฝ
๐โฒ(๐ฅ)๐(๐ฅ) โ
๐(๐ฅ)๐โฒ(๐ฅ)๐2(๐ฅ)
๏ฟฝ
๐ (๐ฅ)
= โ๐2(๐ฅ)๐2(๐ฅ) โ
๐(๐ฅ)๐โฒ(๐ฅ)๐2(๐ฅ) + ๐โฒ(๐ฅ)๐(๐ฅ)
๐2(๐ฅ)
๐ (๐ฅ)
=๐ (๐ฅ) ๐โฒ (๐ฅ) โ ๐2 (๐ฅ) โ ๐โฒ (๐ฅ) ๐ (๐ฅ)
๐3 (๐ฅ)
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3.4. HW4 CHAPTER 3. HWS
Therefore
๐2 = ๏ฟฝ๐ฅ
0
๐ (๐ก) ๐โฒ (๐ก) โ ๐2 (๐ก) โ ๐โฒ (๐ก) ๐ (๐ก)๐3 (๐ก)
๐๐ก + ๐2
For case one, the solution becomes
๐ฆ1 (๐ฅ) โผ exp ๏ฟฝ1๐ฟ
โ๏ฟฝ๐=0
๐ฟ๐๐๐ (๐ฅ)๏ฟฝ ๐ฟ โ 0
โผ exp ๏ฟฝ1๐๏ฟฝ๐0 (๐ฅ) + ๐๐1 (๐ฅ) + ๐2๐2 (๐ฅ)๏ฟฝ๏ฟฝ ๐ โ 0+
โผ exp ๏ฟฝ1๐๐0 (๐ฅ) + ๐1 (๐ฅ) + ๐๐2 (๐ฅ)๏ฟฝ
โผ exp ๏ฟฝ1๐๐0 โ๏ฟฝ
๐ฅ
0
๐ (๐ก)๐ (๐ก)
๐๐ก + ๐1 + ๐๏ฟฝ๐ฅ
0
๐ (๐ก) ๐โฒ (๐ก) โ ๐2 (๐ก) โ ๐ (๐ก)๐3 (๐ก)
๐๐ก + ๐2๏ฟฝ
โผ ๐ถ1 exp ๏ฟฝโ๏ฟฝ๐ฅ
0
๐ (๐ก)๐ (๐ก)
๐๐ก + ๐๏ฟฝ๐ฅ
0
๐ (๐ก) ๐โฒ (๐ก) โ ๐2 (๐ก) โ ๐ (๐ก)๐3 (๐ก)
๐๐ก๏ฟฝ (5)
Where ๐ถ1 is a constant that combines all ๐1๐ ๐0+๐1+๐2 constants into one. Equation (5) gives
the first WKB solution of order ๐ (๐) for case one. Case 2 now is considered.
case 2 In this case ๐โฒ0 = โ๐, therefore
๐0 = โ๏ฟฝ๐ฅ
0๐ (๐ก) ๐๐ก + ๐0
Equation (3) now gives
๐โฒโฒ0 + 2๐โฒ0๐โฒ1 + ๐๐โฒ1 + ๐ = 0โ๐โฒ โ 2๐๐โฒ1 + ๐๐โฒ1 + ๐ = 0
โ๐๐โฒ1 = ๐โฒ โ ๐
๐โฒ1 =๐ โ ๐โฒ
๐
๐โฒ1 =๐๐โ๐โฒ
๐Integrating the above results in
๐1 = ๏ฟฝ๐ฅ
0
๐ (๐ก)๐ (๐ก)
๐๐ก โ ln (๐) + ๐1
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3.4. HW4 CHAPTER 3. HWS
๐2 (๐ฅ) is now found from (4)
๐โฒ2 = โ๏ฟฝ๐โฒ1๏ฟฝ
2+ ๐โฒโฒ1
๏ฟฝ๐ + 2๐โฒ0๏ฟฝ
= โ๏ฟฝ ๐โ๐
โฒ
๐๏ฟฝ2+ ๏ฟฝ ๐โ๐
โฒ
๐๏ฟฝโฒ
(๐ + 2 (โ๐))
= โ๐2+(๐โฒ)2โ2๐๐โฒ
๐2 + ๐โฒโ๐โฒโฒ
๐ โ ๐โฒ๐โ(๐โฒ)2
๐2
โ๐
=๐2 + (๐โฒ)2 โ 2๐๐โฒ + ๐๐โฒ โ ๐๐โฒโฒ โ ๐โฒ๐ โ (๐โฒ)2
๐3
=๐2 โ 2๐๐โฒ + ๐๐โฒ โ ๐๐โฒโฒ โ ๐โฒ๐
๐3Hence
๐2 = ๏ฟฝ๐ฅ
0
๐2 (๐ก) โ 2๐ (๐ก) ๐โฒ (๐ก) + ๐ (๐ก) ๐โฒ (๐ก) โ ๐ (๐ก) ๐โฒโฒ (๐ก) โ ๐โฒ (๐ก) ๐ (๐ก)๐3 (๐ก)
๐๐ก + ๐2
Therefore for this case the solution becomes
๐ฆ2 (๐ฅ) โผ exp ๏ฟฝ1๐ฟ
โ๏ฟฝ๐=0
๐ฟ๐๐๐ (๐ฅ)๏ฟฝ ๐ฟ โ 0
โผ exp ๏ฟฝ1๐๏ฟฝ๐0 (๐ฅ) + ๐๐1 (๐ฅ) + ๐2๐2 (๐ฅ)๏ฟฝ๏ฟฝ ๐ โ 0+
โผ exp ๏ฟฝ1๐๐0 (๐ฅ) + ๐1 (๐ฅ) + ๐๐2 (๐ฅ)๏ฟฝ
Or
๐ฆ2 (๐ฅ) โผ exp ๏ฟฝโ1๐ ๏ฟฝ
๐ฅ
0๐ (๐ก) ๐๐ก + ๐0๏ฟฝ exp ๏ฟฝ๏ฟฝ
๐ฅ
0
๐ (๐ก)๐ (๐ก)
๐๐ก โ ln (๐) + ๐1๏ฟฝ
exp ๏ฟฝ๐๏ฟฝ๐ฅ
0
๐2 (๐ก) โ 2๐ (๐ก) ๐โฒ (๐ก) + ๐ (๐ก) ๐โฒ (๐ก) โ ๐ (๐ก) ๐โฒโฒ (๐ก) โ ๐โฒ (๐ก) ๐ (๐ก)๐3 (๐ก)
๐๐ก + ๐2๏ฟฝ
Which simplifies to
๐ฆ2 (๐ฅ) โผ๐ถ2๐
exp ๏ฟฝโ1๐ ๏ฟฝ
๐ฅ
0๐ (๐ก) ๐๐ก +๏ฟฝ
๐ฅ
0
๐ (๐ก)๐ (๐ก)
๐๐ก + ๐๏ฟฝ๐ฅ
0
๐2 (๐ก) โ 2๐ (๐ก) ๐โฒ (๐ก) + ๐ (๐ก) ๐โฒ (๐ก) โ ๐ (๐ก) ๐โฒโฒ (๐ก) โ ๐โฒ (๐ก) ๐ (๐ก)๐3 (๐ก)
๐๐ก๏ฟฝ
(6)
Where ๐ถ2 is new constant that combines ๐0, ๐1, ๐2 constants. The general solution is linearcombinations of ๐ฆ1, ๐ฆ2
๐ฆ (๐ฅ) โผ ๐ด๐ฆ1 (๐ฅ) + ๐ต๐ฆ2 (๐ฅ)
181
3.4. HW4 CHAPTER 3. HWS
Or
๐ฆ (๐ฅ) โผ ๐ถ1 exp ๏ฟฝโ๏ฟฝ๐ฅ
0
๐ (๐ก)๐ (๐ก)
๐๐ก + ๐๏ฟฝ๐ฅ
0
๐ (๐ก) ๐โฒ (๐ก) โ ๐2 (๐ก) โ ๐โฒ (๐ก) ๐ (๐ก)๐3 (๐ก)
๐๐ก๏ฟฝ
+๐ถ2๐ (๐ฅ)
exp ๏ฟฝโ1๐ ๏ฟฝ
๐ฅ
0๐ (๐ก) ๐๐ก +๏ฟฝ
๐ฅ
0
๐ (๐ก)๐ (๐ก)
๐๐ก + ๐๏ฟฝ๐ฅ
0
๐2 (๐ก) โ 2๐ (๐ก) ๐โฒ (๐ก) + ๐ (๐ก) ๐โฒ (๐ก) โ ๐ (๐ก) ๐โฒโฒ (๐ก) โ ๐โฒ (๐ก) ๐ (๐ก)๐3 (๐ก)
๐๐ก๏ฟฝ
Now boundary conditions are applied to find ๐ถ1, ๐ถ2. Using ๐ฆ (0) = ๐ด in the above gives
๐ด = ๐ถ1 +๐ถ2๐ (0)
(7)
And using ๐ฆ (1) = ๐ต gives
๐ต = ๐ถ1 exp ๏ฟฝโ๏ฟฝ1
0
๐ (๐ก)๐ (๐ก)
๐๐ก + ๐๏ฟฝ1
0
๐ (๐ก) ๐โฒ (๐ก) โ ๐2 (๐ก) โ ๐โฒ (๐ก) ๐ (๐ก)๐3 (๐ก)
๐๐ก๏ฟฝ
+๐ถ2๐ (1)
exp ๏ฟฝโ1๐ ๏ฟฝ
1
0๐ (๐ก) ๐๐ก +๏ฟฝ
1
0
๐ (๐ก)๐ (๐ก)
๐๐ก + ๐๏ฟฝ1
0
๐2 (๐ก) โ 2๐ (๐ก) ๐โฒ (๐ก) + ๐ (๐ก) ๐โฒ (๐ก) โ ๐ (๐ก) ๐โฒโฒ (๐ก) โ ๐โฒ (๐ก) ๐ (๐ก)๐3 (๐ก)
๐๐ก๏ฟฝ
Neglecting exponentially small terms involving ๐โ1๐ the above becomes
๐ต = ๐ถ1 exp ๏ฟฝโ๏ฟฝ1
0
๐ (๐ก)๐ (๐ก)
๐๐ก๏ฟฝ exp ๏ฟฝ๐๏ฟฝ1
0
๐ (๐ก) ๐โฒ (๐ก) โ ๐2 (๐ก) โ ๐โฒ (๐ก) ๐ (๐ก)๐3 (๐ก)
๐๐ก๏ฟฝ
+๐ถ2๐ (1)
exp ๏ฟฝ๏ฟฝ1
0
๐ (๐ก)๐ (๐ก)
๐๐ก๏ฟฝ exp ๏ฟฝ๐๏ฟฝ1
0
๐2 (๐ก) โ 2๐ (๐ก) ๐โฒ (๐ก) + ๐ (๐ก) ๐โฒ (๐ก) โ ๐ (๐ก) ๐โฒโฒ (๐ก) โ ๐โฒ (๐ก) ๐ (๐ก)๐3 (๐ก)
๐๐ก๏ฟฝ (8)
To simplify the rest of the solution which finds ๐ถ1, ๐ถ2, let
๐ง1 = ๏ฟฝ1
0
๐ (๐ก)๐ (๐ก)
๐๐ก
๐ง2 = ๏ฟฝ1
0
๐ (๐ก) ๐โฒ (๐ก) โ ๐2 (๐ก) โ ๐โฒ (๐ก) ๐ (๐ก)๐3 (๐ก)
๐๐ก
๐ง3 = ๏ฟฝ1
0
๐2 (๐ก) โ 2๐ (๐ก) ๐โฒ (๐ก) + ๐ (๐ก) ๐โฒ (๐ก) โ ๐ (๐ก) ๐โฒโฒ (๐ก) โ ๐โฒ (๐ก) ๐ (๐ก)๐3 (๐ก)
๐๐ก
Hence (8) becomes
๐ต = ๐ถ1๐โ๐ง1๐๐๐ง2 +๐ถ2๐ (1)
๐๐ง1๐๐๐ง3 (8A)
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3.4. HW4 CHAPTER 3. HWS
From (7) ๐ถ2 = ๐ (0) (๐ด โ ๐ถ1). Substituting this in (8A) gives
๐ต = ๐ถ1๐โ๐ง1๐๐๐ง2 +๐ (0) (๐ด โ ๐ถ1)
๐ (1)๐๐ง1๐๐๐ง3
= ๐ถ1๐โ๐ง1๐๐๐ง2 +๐ (0)๐ (1)
๐ด๐๐ง1๐๐๐ง3 โ๐ (0)๐ (1)
๐ถ1๐๐ง1๐๐๐ง3
๐ต = ๐ถ1 ๏ฟฝ๐โ๐ง1๐๐๐ง2 โ๐ (0)๐ (1)
๐๐ง1๐๐๐ง3๏ฟฝ + ๐ด๐ (0)๐ (1)
๐๐ง1๐๐๐ง3
๐ถ1 =๐ต โ ๐ด ๐(0)
๐(1)๐๐ง1๐๐๐ง3
๐โ๐ง1๐๐๐ง2 โ ๐(0)๐(1)๐
๐ง1๐๐๐ง3
=๐ (1) ๐ต โ ๐ด๐ (0) ๐๐ง1๐๐๐ง3
๐ (1) ๐โ๐ง1๐๐๐ง2 โ ๐ (0) ๐๐ง1๐๐๐ง3(9)
Using (7), now ๐ถ2 is found
๐ด = ๐ถ1 +๐ถ2๐ (0)
๐ด =๐ (1) ๐ต โ ๐ด๐ (0) ๐๐ง1๐๐๐ง3
๐ (1) ๐โ๐ง1๐๐๐ง2 โ ๐ (0) ๐๐ง1๐๐๐ง3+
๐ถ2๐ (0)
๐ถ2 = ๐ (0) ๏ฟฝ๐ด โ๐ (1) ๐ต โ ๐ด๐ (0) ๐๐ง1๐๐๐ง3
๐ (1) ๐โ๐ง1๐๐๐ง2 โ ๐ (0) ๐๐ง1๐๐๐ง3 ๏ฟฝ(10)
The constants ๐ถ1, ๐ถ2, are now found, hence the solution is now complete.
Summary of solution
๐ฆ (๐ฅ) โผ ๐ถ1 exp ๏ฟฝโ๏ฟฝ๐ฅ
0
๐ (๐ก)๐ (๐ก)
๐๐ก + ๐๏ฟฝ๐ฅ
0
๐ (๐ก) ๐โฒ (๐ก) โ ๐2 (๐ก) โ ๐โฒ (๐ก) ๐ (๐ก)๐3 (๐ก)
๐๐ก๏ฟฝ
+๐ถ2๐ (๐ฅ)
exp ๏ฟฝโ1๐ ๏ฟฝ
๐ฅ
0๐ (๐ก) ๐๐ก +๏ฟฝ
๐ฅ
0
๐ (๐ก)๐ (๐ก)
๐๐ก + ๐๏ฟฝ๐ฅ
0
๐2 (๐ก) โ 2๐ (๐ก) ๐โฒ (๐ก) + ๐ (๐ก) ๐โฒ (๐ก) โ ๐ (๐ก) ๐โฒโฒ (๐ก) โ ๐โฒ (๐ก) ๐ (๐ก)๐3 (๐ก)
๐๐ก๏ฟฝ
Where
๐ถ1 =๐ (1) ๐ต โ ๐ด๐ (0) ๐๐ง1๐๐๐ง3
๐ (1) ๐โ๐ง1๐๐๐ง2 โ ๐ (0) ๐๐ง1๐๐๐ง3
๐ถ2 = ๐ (0) ๏ฟฝ๐ด โ๐ (1) ๐ต โ ๐ด๐ (0) ๐๐ง1๐๐๐ง3
๐ (1) ๐โ๐ง1๐๐๐ง2 โ ๐ (0) ๐๐ง1๐๐๐ง3 ๏ฟฝ
And
๐ง1 = ๏ฟฝ1
0
๐ (๐ก)๐ (๐ก)
๐๐ก
๐ง2 = ๏ฟฝ1
0
๐ (๐ก) ๐โฒ (๐ก) โ ๐2 (๐ก) โ ๐โฒ (๐ก) ๐ (๐ก)๐3 (๐ก)
๐๐ก
๐ง3 = ๏ฟฝ1
0
๐2 (๐ก) โ 2๐ (๐ก) ๐โฒ (๐ก) + ๐ (๐ก) ๐โฒ (๐ก) โ ๐ (๐ก) ๐โฒโฒ (๐ก) โ ๐โฒ (๐ก) ๐ (๐ก)๐3 (๐ก)
๐๐ก
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3.4. HW4 CHAPTER 3. HWS
3.4.2 problem 10.6
problem Use second order WKB to derive formula which is more accurate than (10.1.31)
for the ๐๐กโ eigenvalue of the Sturm-Liouville problem in 10.1.27. Let ๐ (๐ฅ) = (๐ฅ + ๐)4 andcompare your formula with value of ๐ธ๐ in table 10.1
solution
Problem 10.1.27 is
๐ฆโฒโฒ + ๐ธ๐ (๐ฅ) ๐ฆ = 0
With ๐ (๐ฅ) = (๐ฅ + ๐)4 and boundary conditions ๐ฆ (0) = 0, ๐ฆ (๐) = 0. Letting
๐ธ =1๐
Then the ODE becomes
๐๐ฆโฒโฒ (๐ฅ) + (๐ฅ + ๐)4 ๐ฆ (๐ฅ) = 0 (1)
Physical optics approximation is obtained when ๐ โ โ or ๐ โ 0+. Since the ODE is linear,and the highest derivative is now multiplied by a very small parameter ๐, WKB can be usedto solve it. Assuming the solution is
๐ฆ (๐ฅ) โผ exp ๏ฟฝ1๐ฟ
โ๏ฟฝ๐=0
๐ฟ๐๐๐ (๐ฅ)๏ฟฝ ๐ฟ โ 0
Then
๐ฆโฒ (๐ฅ) โผ exp ๏ฟฝ1๐ฟ
โ๏ฟฝ๐=0
๐ฟ๐๐๐ (๐ฅ)๏ฟฝ ๏ฟฝ1๐ฟ
โ๏ฟฝ๐=0
๐ฟ๐๐โฒ๐ (๐ฅ)๏ฟฝ
๐ฆโฒโฒ (๐ฅ) โผ exp ๏ฟฝ1๐ฟ
โ๏ฟฝ๐=0
๐ฟ๐๐๐ (๐ฅ)๏ฟฝ ๏ฟฝ1๐ฟ
โ๏ฟฝ๐=0
๐ฟ๐๐โฒ๐ (๐ฅ)๏ฟฝ2
+ exp ๏ฟฝ1๐ฟ
โ๏ฟฝ๐=0
๐ฟ๐๐๐ (๐ฅ)๏ฟฝ ๏ฟฝ1๐ฟ
โ๏ฟฝ๐=0
๐ฟ๐๐โฒโฒ๐ (๐ฅ)๏ฟฝ
Substituting these into (1) and canceling the exponential terms gives
๐โโโโโโ๏ฟฝ1๐ฟ
โ๏ฟฝ๐=0
๐ฟ๐๐โฒ๐ (๐ฅ)๏ฟฝ2
+1๐ฟ
โ๏ฟฝ๐=0
๐ฟ๐๐โฒโฒ๐ (๐ฅ)โโโโโโ โผ โ (๐ฅ + ๐)4
๐๐ฟ2๏ฟฝ๐โฒ0 + ๐ฟ๐โฒ1 + ๐ฟ2๐โฒ2 +โฏ๏ฟฝ ๏ฟฝ๐โฒ0 + ๐ฟ๐โฒ1 + ๐ฟ2๐โฒ2 +โฏ๏ฟฝ +
๐๐ฟ๏ฟฝ๐โฒโฒ0 + ๐ฟ๐โฒโฒ1 + ๐ฟ2๐โฒโฒ2 +โฏ๏ฟฝ โผ โ (๐ฅ + ๐)4
๐๐ฟ2๏ฟฝ๏ฟฝ๐โฒ0๏ฟฝ
2+ ๐ฟ ๏ฟฝ2๐โฒ1๐โฒ0๏ฟฝ + ๐ฟ2 ๏ฟฝ2๐โฒ0๐โฒ2 + ๏ฟฝ๐โฒ0๏ฟฝ
2๏ฟฝ +โฏ๏ฟฝ +
๐๐ฟ๏ฟฝ๐โฒโฒ0 + ๐ฟ๐โฒโฒ1 + ๐ฟ2๐โฒโฒ2 +โฏ๏ฟฝ โผ โ (๐ฅ + ๐)4
๏ฟฝ๐๐ฟ2๏ฟฝ๐โฒ0๏ฟฝ
2+๐๐ฟ๏ฟฝ2๐โฒ1๐โฒ0๏ฟฝ + ๐ ๏ฟฝ2๐โฒ0๐โฒ2 + ๏ฟฝ๐โฒ0๏ฟฝ
2๏ฟฝ +โฏ๏ฟฝ + ๏ฟฝ
๐๐ฟ๐โฒโฒ0 + ๐2๐โฒโฒ1 +โฏ๏ฟฝ โผ โ (๐ฅ + ๐)4 (2)
The largest term in the left side is ๐๐ฟ2๏ฟฝ๐โฒ0๏ฟฝ
2. By dominant balance, this term has the same
order of magnitude as right side โ (๐ฅ + ๐)4. Hence ๐ฟ2 is proportional to ๐ and for simplicity,๐ฟ can be taken equal to โ๐ or
๐ฟ = โ๐
184
3.4. HW4 CHAPTER 3. HWS
Equation (2) becomes
๏ฟฝ๏ฟฝ๐โฒ0๏ฟฝ2+ โ๐ ๏ฟฝ2๐โฒ1๐โฒ0๏ฟฝ + ๐ ๏ฟฝ2๐โฒ0๐โฒ2 + ๏ฟฝ๐โฒ0๏ฟฝ
2๏ฟฝ +โฏ๏ฟฝ + ๏ฟฝโ๐๐โฒโฒ0 + ๐๐โฒโฒ1 +โฏ๏ฟฝ โผ โ (๐ฅ + ๐)4
Balance of ๐ (1) gives
๏ฟฝ๐โฒ0๏ฟฝ2โผ โ (๐ฅ + ๐)4 (3)
And Balance of ๐๏ฟฝโ๐๏ฟฝ gives
2๐โฒ1๐โฒ0 โผ โ๐โฒโฒ0 (4)
Balance of ๐ (๐) gives
2๐โฒ0๐โฒ2 + ๏ฟฝ๐โฒ1๏ฟฝ2โผ โ๐โฒโฒ1 (5)
Equation (3) is solved first in order to find ๐0 (๐ฅ). Therefore
๐โฒ0 โผ ยฑ๐ (๐ฅ + ๐)2
Hence
๐0 (๐ฅ) โผ ยฑ๐๏ฟฝ๐ฅ
0(๐ก + ๐)2 ๐๐ก + ๐ถยฑ
โผ ยฑ๐ ๏ฟฝ๐ก3
3+ ๐๐ก2 + ๐2๐ก๏ฟฝ
๐ฅ
0+ ๐ถยฑ
โผ ยฑ๐ ๏ฟฝ๐ฅ3
3+ ๐๐ฅ2 + ๐2๐ฅ๏ฟฝ + ๐ถยฑ (6)
๐1 (๐ฅ) is now found from (4), and since ๐โฒโฒ0 = ยฑ2๐ (๐ฅ + ๐) therefore
๐โฒ1 โผ โ12๐โฒโฒ0๐โฒ0
โผ โ12ยฑ2๐ (๐ฅ + ๐)ยฑ๐ (๐ฅ + ๐)2
โผ โ1
๐ฅ + ๐Hence
๐1 (๐ฅ) โผ โ๏ฟฝ๐ฅ
0
1๐ก + ๐
๐๐ก
โผ โ ln ๏ฟฝ๐ + ๐ฅ๐
๏ฟฝ (7)
๐2 (๐ฅ) is now solved from from (5)
2๐โฒ0๐โฒ2 + ๏ฟฝ๐โฒ1๏ฟฝ2โผ โ๐โฒโฒ1
๐โฒ2 โผโ๐โฒโฒ1 โ ๏ฟฝ๐โฒ1๏ฟฝ
2
2๐โฒ0
185
3.4. HW4 CHAPTER 3. HWS
Since ๐โฒ1 โผ โ 1๐ฅ+๐ , then ๐
โฒโฒ1 โผ 1
(๐ฅ+๐)2and the above becomes
๐โฒ2 โผโ 1(๐ฅ+๐)2
โ ๏ฟฝโ 1๐ฅ+๐
๏ฟฝ2
2 ๏ฟฝยฑ๐ (๐ฅ + ๐)2๏ฟฝ
โผโ 1(๐ฅ+๐)2
โ 1(๐ฅ+๐)2
ยฑ2๐ (๐ฅ + ๐)2
โผ ยฑ๐1
(๐ฅ + ๐)4
Hence
๐2 โผ ยฑ๐ ๏ฟฝ๏ฟฝ๐ฅ
0
1(๐ก + ๐)4
๐๐ก๏ฟฝ + ๐ยฑ
โผ ยฑ๐3 ๏ฟฝ
1๐3 โ
1(๐ + ๐ฅ)3
๏ฟฝ + ๐ยฑ
Therefore the solution becomes
๐ฆ (๐ฅ) โผ exp ๏ฟฝ1
โ๐๏ฟฝ๐0 (๐ฅ) + โ๐๐1 (๐ฅ) + ๐๐2 (๐ฅ)๏ฟฝ๏ฟฝ
โผ exp ๏ฟฝ1
โ๐๐0 (๐ฅ) + ๐1 (๐ฅ) + โ๐๐2 (๐ฅ)๏ฟฝ
But ๐ธ = 1๐ , hence โ๐ =
1
โ๐ธ, and the above becomes
๐ฆ (๐ฅ) โผ exp ๏ฟฝโ๐ธ๐0 (๐ฅ) + ๐1 (๐ฅ) +1
โ๐ธ๐2 (๐ฅ)๏ฟฝ
Therefore
๐ฆ (๐ฅ) โผ exp ๏ฟฝยฑ๐โ๐ธ ๏ฟฝ๐ฅ3
3+ ๐๐ฅ2 + ๐2๐ฅ๏ฟฝ + ๐ถยฑ โ ln ๏ฟฝ๐ + ๐ฅ
๐๏ฟฝ ยฑ ๐
1
โ๐ธ13 ๏ฟฝ
1๐3 โ
1(๐ + ๐ฅ)3
๏ฟฝ + ๐ยฑ๏ฟฝ
Which can be written as
๐ฆ (๐ฅ) โผ ๏ฟฝ๐ + ๐ฅ๐
๏ฟฝโ1๐ถ exp ๏ฟฝ๐ ๏ฟฝโ๐ธ ๏ฟฝ
๐ฅ3
3+ ๐๐ฅ2 + ๐2๐ฅ๏ฟฝ +
1
โ๐ธ13 ๏ฟฝ
1๐3 โ
1(๐ + ๐ฅ)3
๏ฟฝ๏ฟฝ๏ฟฝ
โ ๐ถ ๏ฟฝ๐ + ๐ฅ๐
๏ฟฝโ1
exp ๏ฟฝโ๐ ๏ฟฝโ๐ธ ๏ฟฝ๐ฅ3
3+ ๐๐ฅ2 + ๐2๐ฅ๏ฟฝ +
1
โ๐ธ13 ๏ฟฝ
1๐3 โ
1(๐ + ๐ฅ)3
๏ฟฝ๏ฟฝ๏ฟฝ
Where all constants combined into ยฑ๐ถ. In terms of sin / cos the above becomes
๐ฆ (๐ฅ) โผ๐๐ด๐ + ๐ฅ
cos ๏ฟฝโ๐ธ ๏ฟฝ๐ฅ3
3+ ๐๐ฅ2 + ๐2๐ฅ๏ฟฝ +
1
โ๐ธ13 ๏ฟฝ
1๐3 โ
1(๐ + ๐ฅ)3
๏ฟฝ๏ฟฝ
+๐๐ต๐ + ๐ฅ
sin ๏ฟฝโ๐ธ ๏ฟฝ๐ฅ3
3+ ๐๐ฅ2 + ๐2๐ฅ๏ฟฝ +
1
โ๐ธ13 ๏ฟฝ
1๐3 โ
1(๐ + ๐ฅ)3
๏ฟฝ๏ฟฝ (8)
186
3.4. HW4 CHAPTER 3. HWS
Boundary conditions ๐ฆ (0) = 0 gives
0 โผ ๐ด cos ๏ฟฝ0 +1
โ๐ธ๏ฟฝ1๐3 โ
1๐3 ๏ฟฝ๏ฟฝ + ๐ต sin ๏ฟฝ0 +
1
โ๐ธ13 ๏ฟฝ
1๐3 โ
1๐3 ๏ฟฝ๏ฟฝ
โผ ๐ด
Hence solution in (8) reduces to
๐ฆ (๐ฅ) โผ๐๐ต๐ + ๐ฅ
sin ๏ฟฝโ๐ธ ๏ฟฝ๐ฅ3
3+ ๐๐ฅ2 + ๐2๐ฅ๏ฟฝ +
1
โ๐ธ13 ๏ฟฝ
1๐3 โ
1(๐ + ๐ฅ)3
๏ฟฝ๏ฟฝ
Applying B.C. ๐ฆ (๐) = 0 gives
0 โผ๐๐ต๐ + ๐
sin ๏ฟฝโ๐ธ ๏ฟฝ๐3
3+ ๐๐2 + ๐2๐๏ฟฝ +
1
โ๐ธ13 ๏ฟฝ
1๐3 โ
1(๐ + ๐)3
๏ฟฝ๏ฟฝ
โผ๐ต2
sin ๏ฟฝโ๐ธ ๏ฟฝ73๐3๏ฟฝ +
1
โ๐ธ๏ฟฝ7
24๐3 ๏ฟฝ๏ฟฝ
For non trivial solution, therefore
๏ฟฝ๐ธ๐ ๏ฟฝ73๐3๏ฟฝ +
1
โ๐ธ๐๏ฟฝ7
24๐3 ๏ฟฝ = ๐๐ ๐ = 1, 2, 3,โฏ
Solving for โ๐ธ๐. Let โ๐ธ๐ = ๐ฅ, then the above becomes
๐ฅ2 ๏ฟฝ73๐3๏ฟฝ + ๏ฟฝ
724๐6 ๏ฟฝ = ๐ฅ๐๐
๐ฅ2 โ37๐2๐ฅ๐ +
18๐6 = 0
Solving using quadratic formula and taking the positive root, since ๐ธ๐ > 0 gives
๐ฅ =1
28๐3 ๏ฟฝโ2โ18๐2๐2 โ 49 + 6๐๐๏ฟฝ ๐ = 1, 2, 3,โฏ
๏ฟฝ๐ธ๐ =1
28๐3 ๏ฟฝโ2โ18๐2๐2 โ 49 + 6๐๐๏ฟฝ
๐ธ๐ = ๏ฟฝโ2โ18๐2๐2 โ 49 + 6๐๐๏ฟฝ2
Table 10.1 is now reproduced to compare the above more accurate ๐ธ๐. The following tableshows the actual ๐ธ๐ values obtained this more accurate method. Values computed fromabove formula are in column 3.
187
3.4. HW4 CHAPTER 3. HWS
In[207]:= sol = x /. Last@Solvex2 -3
7 Pi2x n +
1
8 Pi^6โฉต 0, x
Out[207]=3 n ฯ +
-49+18 n2 ฯ2
2
14 ฯ3
In[244]:= lam[n_] := (Evaluate@sol)^2
nPoints = {1, 2, 3, 4, 5, 10, 20, 40};
book = {0.00188559, 0.00754235, 0.0169703, 0.0301694, 0.0471397, 0.188559, 0.754235, 3.01694};
exact = {0.00174401, 0.00734865, 0.0167524, 0.0299383, 0.0469006, 0.188395, 0.753977, 3.01668};
hw = Table[N@(lam[n]), {n, nPoints}];
data = Table[{nPoints[[i]], book[[i]], hw[[i]], exact[[i]]}, {i, 1, Length[nPoints]}];
data = Join[{{"n", "En using S0+S1 (book)", "En using S0+S1+S2 (HW)", "Exact"}}, data];
Style[Grid[data, Frame โ All], 18]
Out[251]=
n En using S0+S1 (book) En using S0+S1+S2 (HW) Exact
1 0.0018856 0.0016151 0.001744
2 0.0075424 0.00728 0.0073487
3 0.01697 0.016709 0.016752
4 0.030169 0.029909 0.029938
5 0.04714 0.046879 0.046901
10 0.18856 0.1883 0.1884
20 0.75424 0.75398 0.75398
40 3.0169 3.0167 3.0167
The following table shows the relative error in place of the actual values of ๐ธ๐ to bettercompare how more accurate the result obtained in this solution is compared to the bookresult
In[252]:= data = Table[{nPoints[[i]],
100* Abs@(exact[[i]] - book[[i]])/ exact[[i]],
100* Abs@(exact[[i]] - hw[[i]])/ exact[[i]]}, {i, 1, Length[nPoints]}];
data = Join[{{"n", "En using S0+S1 (book) Rel error", "En using S0+S1+S2 (HW) Rel error"}}, data];
Style[Grid[data, Frame โ All], 18]
Out[254]=
n En using S0+S1 (book) Rel error En using S0+S1+S2 (HW) Rel error
1 8.1181 7.3927
2 2.6359 0.9343
3 1.3007 0.2576
4 0.77192 0.098497
5 0.5098 0.045387
10 0.087051 0.051101
20 0.034219 0.00021643
40 0.0086187 0.000055619
The above shows clearly that adding one more term in the WKB series resulted in moreaccurate eigenvalue estimate.
188
3.5. HW5 CHAPTER 3. HWS
3.5 HW5
NE548 Problem: Similarity solution for the 1D homogeneous heat equation
Due Thursday April 13, 2017
1. (a) Non-dimensionalize the 1D homogeneous heat equation:
โu(x, t)
โt= ฮฝ
โ2u(x, t)
โx2(1)
with โโ < x < โ, and u(x, t) bounded as x โ ยฑโ.
(b) Show that the non-dimensional equations motivate a similarity variable ฮพ = x/t1/2.
(c) Find a similarity solution u(x, t) = H(ฮพ) by solving the appropriate ODE for H(ฮพ).
(d) Show that the similarity solution is related to the solution we found in class on 4/6/17for initial condition
u(x, 0) = 0, x < 0 u(x, 0) = C, x > 0.
1
solution
3.5.1 Part a
Let ๏ฟฝ๏ฟฝ be the non-dimensional space coordinate and ๐ก the non-dimensional time coordinate.Therefore we need
๏ฟฝ๏ฟฝ =๐ฅ๐0
๐ก =๐ก๐ก0
๏ฟฝ๏ฟฝ =๐ข๐ข0
Where ๐0 is the physical characteristic length scale (even if this infinitely long domain, ๐0is given) whose dimensions is [๐ฟ] and ๐ก0 of dimensions [๐] is the characteristic time scaleand ๏ฟฝ๏ฟฝ (๏ฟฝ๏ฟฝ, ๐ก) is the new dependent variable, and ๐ข0 characteristic value of ๐ข to scale against(typically this is the initial conditions) but this will cancel out. We now rewrite the PDE๐๐ข๐๐ก = ๐
๐2๐ข๐๐ฅ2 in terms of the new dimensionless variables.
๐๐ข๐๐ก
=๐๐ข๐๏ฟฝ๏ฟฝ
๐๏ฟฝ๏ฟฝ๐ ๐ก๐ ๐ก๐๐ก
= ๐ข0๐๏ฟฝ๏ฟฝ๐ ๐ก
1๐ก0
(1)
189
3.5. HW5 CHAPTER 3. HWS
And๐๐ข๐๐ฅ
=๐๐ข๐๏ฟฝ๏ฟฝ
๐๏ฟฝ๏ฟฝ๐๏ฟฝ๏ฟฝ๐๏ฟฝ๏ฟฝ๐๐ฅ
= ๐ข0๐๏ฟฝ๏ฟฝ๐๏ฟฝ๏ฟฝ
1๐0
And๐2๐ข๐๐ฅ2
= ๐ข0๐2๏ฟฝ๏ฟฝ๐๏ฟฝ๏ฟฝ2
1๐20
(2)
Substituting (1) and (2) into ๐๐ข๐๐ก = ๐
๐2๐ข๐๐ฅ2 gives
๐ข0๐๏ฟฝ๏ฟฝ๐ ๐ก
1๐ก0= ๐๐ข0
๐2๏ฟฝ๏ฟฝ๐๏ฟฝ๏ฟฝ2
1๐20
๐๏ฟฝ๏ฟฝ๐ ๐ก
= ๏ฟฝ๐๐ก0๐20๏ฟฝ๐2๏ฟฝ๏ฟฝ๐๏ฟฝ๏ฟฝ2
The above is now non-dimensional. Since ๐ has units ๏ฟฝ๐ฟ2
๐ ๏ฟฝ and๐ก0๐20also has units ๏ฟฝ ๐๐ฟ2 ๏ฟฝ, therefore
the product ๐ ๐ก0๐20is non-dimensional quantity.
If we choose ๐ก0 to have same magnitude (not units) as ๐20, i.e. ๐ก0 = ๐20, then๐ก0๐20= 1 (with units
๏ฟฝ ๐๐ฟ2 ๏ฟฝ) and now we obtain the same PDE as the original, but it is non-dimensional. Where
now ๏ฟฝ๏ฟฝ โก ๏ฟฝ๏ฟฝ ( ๐ก, ๏ฟฝ๏ฟฝ).
3.5.2 Part (b)
I Will use the Buckingham ๐ theorem for finding expression for the solution in the form
๐ข (๐ฅ, ๐ก) = ๐ (๐) where ๐ is the similarity variable. Starting with ๐๐ข๐๐ก = ๐๐2๐ข
๐๐ฅ2 , in this PDE, thedi๏ฟฝusion substance is heat with units of Joule ๐ฝ. Hence the concentration of heat, whichis what ๐ข represents, will have units of [๐ข] = ๐ฝ
๐ฟ3 . (heat per unit volume). From physics, weexpect the solution ๐ข (๐ฅ, ๐ก) to depend on ๐ฅ, ๐ก, ๐ and initial conditions ๐ข0 as these are the onlyrelevant quantities involved that can a๏ฟฝect the di๏ฟฝusion. Therefore, by Buckingham theoremwe say
๐ข โก ๐ (๐ฅ, ๐ก, ๐, ๐ข0) (1)
190
3.5. HW5 CHAPTER 3. HWS
We have one dependent quantity ๐ข and 4 independent quantities.The units of each of theabove quantities is
[๐ข] =๐ฝ๐ฟ3
[๐ฅ] = ๐ฟ[๐ก] = ๐
[๐] =๐ฟ2
๐
[๐ข0] =๐ฝ๐ฟ3
Hence using Buckingham theorem, we write
[๐ข] = ๏ฟฝ๐ฅ๐๐ก๐๐๐๐ข๐0๏ฟฝ (2)
We now determine ๐, ๐, ๐, ๐, by dimensional analysis. The above is
๐ฝ๐ฟ3
= ๐ฟ๐๐๐ ๏ฟฝ๐ฟ2
๐ ๏ฟฝ๐
๏ฟฝ๐ฝ๐ฟ3 ๏ฟฝ
๐
(๐ฝ) ๏ฟฝ๐ฟโ3๏ฟฝ = ๏ฟฝ๐ฟ๐+2๐โ3๐๏ฟฝ ๏ฟฝ๐๐โ๐๏ฟฝ ๏ฟฝ๐ฝ๐๏ฟฝ
Comparing powers of same units on both sides, we see that
๐ = 1๐ โ ๐ = 0
๐ + 2๐ โ 3๐ = โ3
From second equation above, ๐ = ๐, hence third equation becomes
๐ + 2๐ โ 3๐ = โ3๐ + 2๐ = 0
Since ๐ = 1. Hence
๐ = โ๐2
๐ = โ๐2
Therefore, now that we found all the powers, (we have one free power ๐ which we can setto any value), then from equation (1)
[๐ข] = ๏ฟฝ๐ฅ๐๐ก๐๐๐๐ข๐0๏ฟฝ๐ข๐ข0
= ๏ฟฝ๏ฟฝ = ๐ฅ๐๐ก๐๐๐
Therefore ๏ฟฝ๏ฟฝ is function of all the variables in the RHS. Let this function be ๐ (This is thesame as ๐ป in problem statement). Hence the above becomes
๏ฟฝ๏ฟฝ = ๐ ๏ฟฝ๐ฅ๐๐กโ๐2๐โ
๐2 ๏ฟฝ
= ๐ ๏ฟฝ๐ฅ๐
๐๐2 ๐ก
๐2๏ฟฝ
191
3.5. HW5 CHAPTER 3. HWS
Since ๐ is free variable, we can choose
๐ = 1 (2)
And obtain
๏ฟฝ๏ฟฝ โก ๐ ๏ฟฝ๐ฅ
โ๐๐ก๏ฟฝ (3)
In the above ๐ฅ
โ๐๐กis now non-dimensional quantity, which we call, the similarity variable
๐ =๐ฅ
โ๐๐ก(4)
Notice that another choice of ๐ in (2), for example ๐ = 2 would lead to ๐ = ๐ฅ2
๐๐ก instead of๐ = ๐ฅ
โ๐๐กbut we will use the latter for the rest of the problem.
3.5.3 Part (c)
Using ๐ข โก ๐ (๐) where ๐ = ๐ฅ
โ๐๐กthen
๐๐ข๐๐ก
=๐๐๐๐๐๐๐๐ก
=๐๐๐๐
๐๐๐ก ๏ฟฝ
๐ฅ
โ๐๐ก๏ฟฝ
= โ12๐๐๐๐
โโโโโโโ
๐ฅ
โ๐๐ก32
โโโโโโโ
And๐๐ข๐๐ฅ
=๐๐๐๐๐๐๐๐ฅ
=๐๐๐๐
๐๐๐ฅ ๏ฟฝ
๐ฅ
โ๐๐ก๏ฟฝ
=๐๐๐๐
1
โ๐๐ก
192
3.5. HW5 CHAPTER 3. HWS
And๐2๐ข๐๐ฅ2
=๐๐๐ฅ ๏ฟฝ
๐๐๐๐
1
โ๐๐ก๏ฟฝ
=1
โ๐๐ก๐๐๐ฅ ๏ฟฝ
๐๐๐๐๏ฟฝ
=1
โ๐๐ก๏ฟฝ๐2๐๐๐2
๐๐๐๐ฅ๏ฟฝ
=1
โ๐๐ก๏ฟฝ๐2๐๐๐2
1
โ๐๐ก๏ฟฝ
=1๐๐ก๐2๐๐๐2
Hence the PDE ๐๐ข๐๐ก = ๐
๐2๐ข๐๐ฅ2 becomes
โ12๐๐๐๐
โโโโโโโ
๐ฅ
โ๐๐ก32
โโโโโโโ = ๐
1๐๐ก๐2๐๐๐2
1๐ก๐2๐๐๐2
+12
๐ฅ
โ๐๐ก32
๐๐๐๐
= 0
๐2๐๐๐2
+12๐ฅ
โ๐๐ก๐๐๐๐
= 0
But ๐ฅ
โ๐๐ก= ๐, hence we obtain the required ODE as
๐2๐ (๐)๐๐2
+12๐๐๐ (๐)๐๐
= 0
๐โฒโฒ +๐2๐โฒ = 0
We now solve the above ODE for ๐ (๐). Let ๐โฒ = ๐ง, then the ODE becomes
๐งโฒ +๐2๐ง = 0
Integrating factor is ๐ = ๐โซ๐2 ๐๐ = ๐
๐24 , hence
๐๐๐๏ฟฝ๐ง๐๏ฟฝ = 0
๐ง๐ = ๐1
๐ง = ๐1๐โ๐24
Therefore, since ๐โฒ = ๐ง, then
๐โฒ = ๐1๐โ๐24
193
3.5. HW5 CHAPTER 3. HWS
Integrating gives
๐ (๐) = ๐2 + ๐1๏ฟฝ๐
0๐โ๐ 24 ๐๐
3.5.4 Part (d)
For initial conditions of step function
๐ข (๐ฅ, 0) =
โงโชโชโจโชโชโฉ0 ๐ฅ < 0๐ถ ๐ฅ > 0
The solution found in class was
๐ข (๐ฅ, ๐ก) =๐ถ2+๐ถ2
erf ๏ฟฝ๐ฅ
โ4๐๐ก๏ฟฝ (1)
Where erf ๏ฟฝ ๐ฅ
โ4๐๐ก๏ฟฝ = 2
โ๐โซ
๐ฅโ4๐๐ก
0๐โ๐ง2๐๐ง. The solution found in part (c) earlier is
๐ (๐) = ๐1๏ฟฝ๐
0๐โ๐ 24 ๐๐ + ๐2
Let ๐ = โ4๐ง, then๐๐ ๐๐ง = โ4, when ๐ = 0, ๐ง = 0 and when ๐ = ๐, ๐ง = ๐
โ4, therefore the integral
becomes
๐ (๐) = ๐1โ4๏ฟฝ๐โ4
0๐โ๐ง2๐๐ง + ๐2
But 2
โ๐โซ
๐โ4
0๐โ๐ง2๐๐ง = erf ๏ฟฝ ๐
โ4๏ฟฝ, hence โซ
๐โ4
0๐โ๐ง2๐๐ง = โ๐
2 erf ๏ฟฝ ๐
โ4๏ฟฝ and the above becomes
๐ (๐) = ๐1โ๐ erf ๏ฟฝ๐
โ4๏ฟฝ + ๐2
= ๐3 erf ๏ฟฝ๐
โ4๏ฟฝ + ๐2
Since ๐ = ๐ฅ
โ๐๐ก, then above becomes, when converting back to ๐ข (๐ฅ, ๐ก)
๐ข (๐ฅ, ๐ก) = ๐3 erf ๏ฟฝ๐ฅ
โ4๐๐ก๏ฟฝ + ๐2 (2)
Comparing (1) and (2), we see they are the same. Constants of integration are arbitraryand can be found from initial conditions.
194
3.6. HW6 CHAPTER 3. HWS
3.6 HW6
3.6.1 Problem 1
NE548 Problems
Due Tuesday April 25, 2017
1. Use the Method of Images to solve
โu
โt= ฮฝ
โ2u
โx2+Q(x, t), x > 0
(a)u(0, t) = 0, u(x, 0) = 0.
(b)u(0, t) = 1, u(x, 0) = 0.
(c)ux(0, t) = A(t), u(x, 0) = f(x).
2. (a) Solve by the Method of Characteristics:
โ2u
โt2= c2
โ2u
โx2, x โฅ 0, t โฅ 0
u(x, 0) = f(x),โu(x, 0)
โt= g(x),
โu(0, t)
โx= h(t)
(b) For the special case h(t) = 0, explain how you could use a symmetry argument to helpconstruct the solution.
(c) Sketch the solution if g(x) = 0, h(t) = 0 and f(x) = 1 for 4 โค x โค 5 and f(x) = 0otherwise.
1
Note, I will use ๐ in place of ๐ since easier to type.
3.6.1.1 Part (a)
๐๐ข๐๐ก
= ๐๐2๐ข๐๐ฅ2
+ ๐ (๐ฅ, ๐ก)
๐ฅ > 0๐ข (0, ๐ก) = 0๐ข (๐ฅ, 0) = 0
Multiplying both sides by ๐บ (๐ฅ, ๐ก; ๐ฅ0, ๐ก0) and integrating over the domain gives (where in thefollowing ๐บ is used instead of ๐บ (๐ฅ, ๐ก; ๐ฅ0, ๐ก0) for simplicity).
๏ฟฝโ
๐ฅ=0๏ฟฝ
โ
๐ก=0๐บ๐ข๐ก ๐๐ก๐๐ฅ = ๏ฟฝ
โ
๐ฅ=0๏ฟฝ
โ
๐ก=0๐๐ข๐ฅ๐ฅ๐บ ๐๐ก๐๐ฅ +๏ฟฝ
โ
๐ฅ=0๏ฟฝ
โ
๐ก=0๐๐บ ๐๐ก๐๐ฅ (1)
For the integral on the LHS, we apply integration by parts once to move the time derivativefrom ๐ข to ๐บ
๏ฟฝโ
๐ฅ=0๏ฟฝ
โ
๐ก=0๐บ๐ข๐ก ๐๐ก๐๐ฅ = ๏ฟฝ
โ
๐ฅ=0[๐ข๐บ]โ๐ก=0 ๐๐ฅ โ๏ฟฝ
โ
๐ฅ=0๏ฟฝ
โ
๐ก=0๐บ๐ก๐ข ๐๐ก๐๐ฅ
195
3.6. HW6 CHAPTER 3. HWS
And the first integral in the RHS of (1) gives, after doing integration by parts two times
๏ฟฝโ
๐ฅ=0๏ฟฝ
โ
๐ก=0๐๐ข๐ฅ๐ฅ๐บ ๐๐ก๐๐ฅ = ๏ฟฝ
โ
๐ก=0[๐ข๐ฅ๐บ]
โ๐ฅ=0 ๐๐ก โ๏ฟฝ
โ
๐ฅ=0๏ฟฝ
โ
๐ก=0๐๐ข๐ฅ๐บ๐ฅ ๐๐ก๐๐ฅ
= ๏ฟฝโ
๐ก=0[๐ข๐ฅ๐บ]
โ๐ฅ=0 ๐๐ก โ ๏ฟฝ๏ฟฝ
โ
๐ก=0[๐ข๐บ๐ฅ]
โ๐ฅ=0 ๐๐ก โ๏ฟฝ
โ
๐ฅ=0๏ฟฝ
โ
๐ก=0๐๐ข๐บ๐ฅ๐ฅ ๐๐ก๐๐ฅ๏ฟฝ
= ๏ฟฝโ
๐ก=0๏ฟฝ[๐ข๐ฅ๐บ]
โ๐ฅ=0 โ [๐ข๐บ๐ฅ]
โ๐ฅ=0๏ฟฝ ๐๐ก +๏ฟฝ
โ
๐ฅ=0๏ฟฝ
โ
๐ก=0๐๐ข๐บ๐ฅ๐ฅ ๐๐ก๐๐ฅ
= ๏ฟฝโ
๐ก=0[๐ข๐ฅ๐บ โ ๐ข๐บ๐ฅ]
โ๐ฅ=0 ๐๐ก +๏ฟฝ
โ
๐ฅ=0๏ฟฝ
โ
๐ก=0๐๐ข๐บ๐ฅ๐ฅ ๐๐ก๐๐ฅ
= โ๏ฟฝโ
๐ก=0[๐ข๐บ๐ฅ โ ๐ข๐ฅ๐บ]
โ๐ฅ=0 ๐๐ก +๏ฟฝ
โ
๐ฅ=0๏ฟฝ
โ
๐ก=0๐๐ข๐บ๐ฅ๐ฅ ๐๐ก๐๐ฅ
Hence (1) becomes
๏ฟฝโ
๐ฅ=0[๐ข๐บ]โ๐ก=0 ๐๐ฅโ๏ฟฝ
โ
๐ฅ=0๏ฟฝ
โ
๐ก=0๐บ๐ก๐ข ๐๐ก๐๐ฅ = ๏ฟฝ
โ
๐ก=0[๐ข๐ฅ๐บ โ ๐ข๐บ๐ฅ]
โ๐ฅ=0 ๐๐ก+๏ฟฝ
โ
๐ฅ=0๏ฟฝ
โ
๐ก=0๐๐ข๐บ๐ฅ๐ฅ ๐๐ก๐๐ฅ+๏ฟฝ
โ
๐ฅ=0๏ฟฝ
โ
๐ก=0๐บ๐ ๐๐ก๐๐ฅ
Or
๏ฟฝโ
๐ฅ=0๏ฟฝ
โ
๐ก=0โ๐บ๐ก๐ข โ ๐๐ข๐บ๐ฅ๐ฅ ๐๐ก๐๐ฅ = โ๏ฟฝ
โ
๐ฅ=0[๐ข๐บ]โ๐ก=0 ๐๐ฅ โ๏ฟฝ
โ
๐ก=0[๐ข๐บ๐ฅ โ ๐ข๐ฅ๐บ]
โ๐ฅ=0 ๐๐ก +๏ฟฝ
โ
๐ฅ=0๏ฟฝ
โ
๐ก=0๐บ๐ ๐๐ก๐๐ฅ
(2)
We now want to choose ๐บ (๐ฅ, ๐ก; ๐ฅ0, ๐ก0) such that
โ๐บ๐ก๐ข โ ๐๐ข๐บ๐ฅ๐ฅ = ๐ฟ (๐ฅ โ ๐ฅ0) ๐ฟ (๐ก โ ๐ก0)โ๐บ๐ก๐ข = ๐๐ข๐บ๐ฅ๐ฅ + ๐ฟ (๐ฅ โ ๐ฅ0) ๐ฟ (๐ก โ ๐ก0) (3)
This way, the LHS of (2) becomes ๐ข (๐ฅ0, ๐ก0). Hence (2) now becomes
๐ข (๐ฅ0, ๐ก0) = โ๏ฟฝโ
๐ฅ=0[๐ข๐บ]โ๐ก=0 ๐๐ฅ โ๏ฟฝ
๐ก0
๐ก=0[๐ข๐บ๐ฅ โ ๐ข๐ฅ๐บ]
โ๐ฅ=0 ๐๐ก +๏ฟฝ
โ
๐ฅ=0๏ฟฝ
๐ก0
๐ก=0๐บ๐ ๐๐ก๐๐ฅ (4)
We now need to find the Green function which satisfies (3). But (3) is equivalent to solutionof problem of
โ๐บ๐ก๐ข = ๐๐ข๐บ๐ฅ๐ฅ
๐บ (๐ฅ, 0) = ๐ฟ (๐ฅ โ ๐ฅ0) ๐ฟ (๐ก โ ๐ก0)โโ < ๐ฅ < โ
๐บ (๐ฅ, ๐ก; ๐ฅ0, ๐ก0) = 0 ๐ก > ๐ก0๐บ (ยฑโ, ๐ก; ๐ฅ0, ๐ก0) = 0๐บ (๐ฅ, ๐ก0; ๐ฅ0, ๐ก0) = ๐ฟ (๐ฅ โ ๐ฅ0)
But the above problem has a known fundamental solution which we found, but for theforward heat PDE instead of the reverse heat PDE. The fundamental solution to the forwardheat PDE is
๐บ (๐ฅ, ๐ก) =1
โ4๐๐ (๐ก โ ๐ก0)exp
โโโโโโ (๐ฅ โ ๐ฅ0)
2
4๐ (๐ก โ ๐ก0)
โโโโโ 0 โค ๐ก0 โค ๐ก
196
3.6. HW6 CHAPTER 3. HWS
Hence for the reverse heat PDE the above becomes
๐บ (๐ฅ, ๐ก) =1
โ4๐๐ (๐ก0 โ ๐ก)exp
โโโโโโ (๐ฅ โ ๐ฅ0)
2
4๐ (๐ก0 โ ๐ก)
โโโโโ 0 โค ๐ก โค ๐ก0 (5)
The above is the infinite space Green function and what we will use in (4). Now we goback to (4) and simplify the boundary conditions term. Starting with the term โซโ
๐ฅ=0[๐ข๐บ]โ๐ก=0 ๐๐ฅ.
Since ๐บ (๐ฅ,โ; ๐ฅ0, ๐ก0) = 0 then upper limit is zero. At ๐ก = 0 we are given that ๐ข (๐ฅ, 0) = 0, hencethis whole term is zero. So now (4) simplifies to
๐ข (๐ฅ0, ๐ก0) = โ๏ฟฝโ
๐ก=0[๐ข๐บ๐ฅ โ ๐ข๐ฅ๐บ]
โ๐ฅ=0 ๐๐ก +๏ฟฝ
โ
๐ฅ=0๏ฟฝ
โ
๐ก=0๐บ๐ ๐๐ก๐๐ฅ (6)
We are told that ๐ข (0, ๐ก) = 0. Hence
[๐ข๐บ๐ฅ โ ๐ข๐ฅ๐บ]โ๐ฅ=0 = (๐ข (โ, ๐ก) ๐บ๐ฅ (โ, ๐ก) โ ๐ข๐ฅ (โ, ๐ก) ๐บ (โ, ๐ก)) โ (๐ข (0, ๐ก) ๐บ๐ฅ (0, ๐ก) โ ๐ข๐ฅ (0, ๐ก) ๐บ (0, ๐ก))
= (๐ข (โ, ๐ก) ๐บ๐ฅ (โ, ๐ก) โ ๐ข๐ฅ (โ, ๐ก) ๐บ (โ, ๐ก)) + ๐ข๐ฅ (0, ๐ก) ๐บ (0, ๐ก)
We also know that ๐บ (ยฑโ, ๐ก; ๐ฅ0, ๐ก0) = 0, Hence ๐บ (โ) = 0 and also ๐บ๐ฅ (โ) = 0, hence the abovesimplifies
[๐ข๐บ๐ฅ โ ๐ข๐ฅ๐บ]โ๐ฅ=0 = ๐ข๐ฅ (0, ๐ก) ๐บ (0, ๐ก)
To make ๐บ (0, ๐ก) = 0 we place an image impulse at โ๐ฅ0 with negative value to the impulse at๐ฅ0. This will make ๐บ at ๐ฅ = 0 zero.
xx = 0x0
โx0
+
โ
Therefore the Green function to use is, from (5) becomes
๐บ (๐ฅ, ๐ก) =1
โ4๐๐ (๐ก0 โ ๐ก)
โโโโโexp
โโโโโโ (๐ฅ โ ๐ฅ0)
2
4๐ (๐ก0 โ ๐ก)
โโโโโ โ exp
โโโโโโ (๐ฅ + ๐ฅ0)
2
4๐ (๐ก0 โ ๐ก)
โโโโโ
โโโโโ 0 โค ๐ก โค ๐ก0
Therefore the solution, from (4) becomes
๐ข (๐ฅ0, ๐ก0) = ๏ฟฝโ
๐ฅ=0๏ฟฝ
๐ก0
๐ก=0
1
โ4๐๐ (๐ก0 โ ๐ก)
โโโโโexp
โโโโโโ (๐ฅ โ ๐ฅ0)
2
4๐ (๐ก0 โ ๐ก)
โโโโโ โ exp
โโโโโโ (๐ฅ + ๐ฅ0)
2
4๐ (๐ก0 โ ๐ก)
โโโโโ
โโโโโ ๐ (๐ฅ, ๐ก) ๐๐ก๐๐ฅ (7)
Switching the order of ๐ฅ0, ๐ก0 with ๐ฅ, ๐ก gives
๐ข (๐ฅ, ๐ก) = ๏ฟฝโ
๐ฅ0=0๏ฟฝ
๐ก
๐ก0=0
1
โ4๐๐ (๐ก โ ๐ก0)
โโโโโexp
โโโโโโ (๐ฅ0 โ ๐ฅ)
2
4๐ (๐ก โ ๐ก0)
โโโโโ โ exp
โโโโโโ (๐ฅ0 + ๐ฅ)
2
4๐ (๐ก โ ๐ก0)
โโโโโ
โโโโโ ๐ (๐ฅ0, ๐ก0) ๐๐ก0๐๐ฅ0 (8)
Notice, for the terms (๐ฅ0 โ ๐ฅ)2 , (๐ฅ0 + ๐ฅ)
2, since they are squared, the order does not matter,so we might as well write the above as
๐ข (๐ฅ, ๐ก) = ๏ฟฝโ
๐ฅ0=0๏ฟฝ
๐ก
๐ก0=0
1
โ4๐๐ (๐ก โ ๐ก0)
โโโโโexp
โโโโโโ (๐ฅ โ ๐ฅ0)
2
4๐ (๐ก โ ๐ก0)
โโโโโ โ exp
โโโโโโ (๐ฅ + ๐ฅ0)
2
4๐ (๐ก โ ๐ก0)
โโโโโ
โโโโโ ๐ (๐ฅ0, ๐ก0) ๐๐ก0๐๐ฅ0 (8)
197
3.6. HW6 CHAPTER 3. HWS
3.6.1.2 Part (b)
๐๐ข๐๐ก
= ๐๐2๐ข๐๐ฅ2
+ ๐ (๐ฅ, ๐ก)
๐ฅ > 0๐ข (0, ๐ก) = 1๐ข (๐ฅ, 0) = 0
Everything follows the same as in part (a) up to the point where boundary condition termsneed to be evaluated.
๐ข (๐ฅ0, ๐ก0) = โ๏ฟฝโ
๐ฅ=0[๐ข๐บ]โ๐ก=0 ๐๐ฅ โ๏ฟฝ
๐ก0
๐ก=0[๐ข๐บ๐ฅ โ ๐ข๐ฅ๐บ]
โ๐ฅ=0 ๐๐ก +๏ฟฝ
โ
๐ฅ=0๏ฟฝ
๐ก0
๐ก=0๐บ๐ ๐๐ก๐๐ฅ (4)
Where
๐บ (๐ฅ, ๐ก) =1
โ4๐๐ (๐ก0 โ ๐ก)exp
โโโโโโ (๐ฅ โ ๐ฅ0)
2
4๐ (๐ก0 โ ๐ก)
โโโโโ 0 โค ๐ก โค ๐ก0 (5)
Starting with the term โซโ
๐ฅ=0[๐ข๐บ]โ๐ก=0 ๐๐ฅ. Since ๐บ (๐ฅ,โ; ๐ฅ0, ๐ก0) = 0 then upper limit is zero. At
๐ก = 0 we are given that ๐ข (๐ฅ, 0) = 0, hence this whole term is zero. So now (4) simplifies to
๐ข (๐ฅ0, ๐ก0) = โ๏ฟฝโ
๐ก=0[๐ข๐บ๐ฅ โ ๐ข๐ฅ๐บ]
โ๐ฅ=0 ๐๐ก +๏ฟฝ
โ
๐ฅ=0๏ฟฝ
โ
๐ก=0๐บ๐ ๐๐ก๐๐ฅ (6)
Now
[๐ข๐บ๐ฅ โ ๐ข๐ฅ๐บ]โ๐ฅ=0 = (๐ข (โ, ๐ก) ๐บ๐ฅ (โ, ๐ก) โ ๐ข๐ฅ (โ, ๐ก) ๐บ (โ, ๐ก)) โ (๐ข (0, ๐ก) ๐บ๐ฅ (0, ๐ก) โ ๐ข๐ฅ (0, ๐ก) ๐บ (0, ๐ก))
We are told that ๐ข (0, ๐ก) = 1, we also know that ๐บ (ยฑโ, ๐ก; ๐ฅ0, ๐ก0) = 0, Hence ๐บ (โ, ๐ก) = 0 andalso ๐บ๐ฅ (โ, ๐ก) = 0, hence the above simplifies
[๐ข๐บ๐ฅ โ ๐ข๐ฅ๐บ]โ๐ฅ=0 = โ๐บ๐ฅ (0, ๐ก) + ๐ข๐ฅ (0, ๐ก) ๐บ (0, ๐ก)
To make ๐บ (0) = 0 we place an image impulse at โ๐ฅ0 with negative value to the impulse at๐ฅ0. This is the same as part(a)
xx = 0x0
โx0
+
โ
๐บ (๐ฅ, ๐ก) =1
โ4๐๐ (๐ก0 โ ๐ก)
โโโโโexp
โโโโโโ (๐ฅ โ ๐ฅ0)
2
4๐ (๐ก0 โ ๐ก)
โโโโโ โ exp
โโโโโโ (๐ฅ + ๐ฅ0)
2
4๐ (๐ก0 โ ๐ก)
โโโโโ
โโโโโ 0 โค ๐ก โค ๐ก0
Now the boundary terms reduces to just
[๐ข๐บ๐ฅ โ ๐ข๐ฅ๐บ]โ๐ฅ=0 = โ๐บ๐ฅ (0, ๐ก)
We need now to evaluate ๐บ๐ฅ (0, ๐ก), the only remaining term. Since we know what ๐บ (๐ฅ, ๐ก) from198
3.6. HW6 CHAPTER 3. HWS
the above, then
๐๐๐ฅ๐บ (๐ฅ, ๐ก) =
1
โ4๐๐ (๐ก0 โ ๐ก)
โโโโโโ (๐ฅ โ ๐ฅ0)2๐ (๐ก0 โ ๐ก)
expโโโโโโ (๐ฅ โ ๐ฅ0)
2
4๐ (๐ก0 โ ๐ก)
โโโโโ +
2 (๐ฅ + ๐ฅ0)4๐ (๐ก0 โ ๐ก)
expโโโโโโ (๐ฅ + ๐ฅ0)
2
4๐ (๐ก0 โ ๐ก)
โโโโโ
โโโโโ
And at ๐ฅ = 0 the above simplifies to
๐บ๐ฅ (0, ๐ก) =1
โ4๐๐ (๐ก0 โ ๐ก)๏ฟฝ
๐ฅ02๐ (๐ก0 โ ๐ก)
exp ๏ฟฝโ๐ฅ20
4๐ (๐ก0 โ ๐ก)๏ฟฝ +
๐ฅ02๐ (๐ก0 โ ๐ก)
exp ๏ฟฝโ๐ฅ20
4๐ (๐ก0 โ ๐ก)๏ฟฝ๏ฟฝ
=1
โ4๐๐ (๐ก0 โ ๐ก)๏ฟฝ
๐ฅ0๐ (๐ก0 โ ๐ก)
exp ๏ฟฝโ๐ฅ20
4๐ (๐ก0 โ ๐ก)๏ฟฝ๏ฟฝ
Hence the complete solution becomes from (4)
๐ข (๐ฅ0, ๐ก0) = โ๏ฟฝโ
๐ฅ=0[๐ข๐บ]โ๐ก=0 ๐๐ฅ โ๏ฟฝ
๐ก0
๐ก=0[๐ข๐บ๐ฅ โ ๐ข๐ฅ๐บ]
โ๐ฅ=0 ๐๐ก +๏ฟฝ
โ
๐ฅ=0๏ฟฝ
๐ก0
๐ก=0๐บ๐ ๐๐ก๐๐ฅ
= โ๏ฟฝ๐ก0
๐ก=0โ๐บ๐ฅ (0, ๐ก) ๐๐ก +๏ฟฝ
โ
๐ฅ=0๏ฟฝ
๐ก0
๐ก=0๐บ๐ ๐๐ก๐๐ฅ
Substituting ๐บ and ๐บ๐ฅ in the above gives
๐ข (๐ฅ0, ๐ก0) = ๏ฟฝ๐ก0
๐ก=0
1
โ4๐๐ (๐ก0 โ ๐ก)๏ฟฝ
๐ฅ0๐ (๐ก0 โ ๐ก)
exp ๏ฟฝโ๐ฅ20
4๐ (๐ก0 โ ๐ก)๏ฟฝ๏ฟฝ ๐๐ก
+๏ฟฝโ
๐ฅ=0๏ฟฝ
๐ก0
๐ก=0
1
โ4๐๐ (๐ก0 โ ๐ก)
โโโโโexp
โโโโโโ (๐ฅ โ ๐ฅ0)
2
4๐ (๐ก0 โ ๐ก)
โโโโโ โ exp
โโโโโโ (๐ฅ + ๐ฅ0)
2
4๐ (๐ก0 โ ๐ก)
โโโโโ
โโโโโ ๐ (๐ฅ, ๐ก) ๐๐ก๐๐ฅ
Switching the order of ๐ฅ0, ๐ก0 with ๐ฅ, ๐ก
๐ข (๐ฅ, ๐ก) = ๏ฟฝ๐ก
๐ก0=0
1
โ4๐๐ (๐ก โ ๐ก0)๏ฟฝ
๐ฅ๐ (๐ก โ ๐ก0)
exp ๏ฟฝโ๐ฅ2
4๐ (๐ก โ ๐ก0)๏ฟฝ๏ฟฝ ๐๐ก0
+๏ฟฝโ
๐ฅ0=0๏ฟฝ
๐ก
๐ก0=0
1
โ4๐๐ (๐ก โ ๐ก0)
โโโโโexp
โโโโโโ (๐ฅ0 โ ๐ฅ)
2
4๐ (๐ก โ ๐ก0)
โโโโโ โ exp
โโโโโโ (๐ฅ0 + ๐ฅ)
2
4๐ (๐ก โ ๐ก0)
โโโโโ
โโโโโ ๐ (๐ฅ0, ๐ก0) ๐๐ก0๐๐ฅ0 (7)
But
๏ฟฝ๐ก
๐ก0=0
1
โ4๐๐ (๐ก โ ๐ก0)๏ฟฝ
๐ฅ๐ (๐ก โ ๐ก0)
exp ๏ฟฝโ๐ฅ2
4๐ (๐ก โ ๐ก0)๏ฟฝ๏ฟฝ ๐๐ก0 = erfc ๏ฟฝ
๐ฅ2โ๐ก
๏ฟฝ
Hence (7) becomes
๐ข (๐ฅ, ๐ก) = erfc ๏ฟฝ๐ฅ2โ๐ก
๏ฟฝ +๏ฟฝโ
๐ฅ0=0๏ฟฝ
๐ก
๐ก0=0
1
โ4๐๐ (๐ก โ ๐ก0)
โโโโโexp
โโโโโโ (๐ฅ0 โ ๐ฅ)
2
4๐ (๐ก โ ๐ก0)
โโโโโ โ exp
โโโโโโ (๐ฅ0 + ๐ฅ)
2
4๐ (๐ก โ ๐ก0)
โโโโโ
โโโโโ ๐ (๐ฅ0, ๐ก0) ๐๐ก0๐๐ฅ0
The only di๏ฟฝerence between this solution and part(a) solution is the extra term erfc ๏ฟฝ ๐ฅ2โ๐ก๏ฟฝ
due to having non-zero boundary conditions in this case.
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3.6. HW6 CHAPTER 3. HWS
3.6.1.3 part(c)
๐๐ข๐๐ก
= ๐๐2๐ข๐๐ฅ2
+ ๐ (๐ฅ, ๐ก)
๐ฅ > 0๐๐ข (0, ๐ก)๐๐ฅ
= ๐ด (๐ก)
๐ข (๐ฅ, 0) = ๐ (๐ฅ)
Everything follows the same as in part (a) up to the point where boundary condition termsneed to be evaluated.
๐ข (๐ฅ0, ๐ก0) = โ๏ฟฝโ
๐ฅ=0[๐ข๐บ]โ๐ก=0 ๐๐ฅ โ๏ฟฝ
๐ก0
๐ก=0[๐ข๐บ๐ฅ โ ๐ข๐ฅ๐บ]
โ๐ฅ=0 ๐๐ก +๏ฟฝ
โ
๐ฅ=0๏ฟฝ
๐ก0
๐ก=0๐บ๐ ๐๐ก๐๐ฅ (4)
Starting with the term โซโ
๐ฅ=0[๐ข๐บ]โ๐ก=0 ๐๐ฅ. Since ๐บ (๐ฅ,โ; ๐ฅ0, ๐ก0) = 0 then upper limit is zero. At
๐ก = 0 we are given that ๐ข (๐ฅ, 0) = ๐ (๐ฅ), hence
๏ฟฝโ
๐ฅ=0[๐ข๐บ]โ๐ก=0 ๐๐ฅ = ๏ฟฝ
โ
๐ฅ=0โ๐ข (๐ฅ, 0) ๐บ (๐ฅ, 0) ๐๐ฅ
= ๏ฟฝโ
๐ฅ=0โ๐ (๐ฅ)๐บ (๐ฅ, 0) ๐๐ฅ
Looking at the second term in RHS of (4)
[๐ข๐บ๐ฅ โ ๐ข๐ฅ๐บ]โ๐ฅ=0 = (๐ข (โ, ๐ก) ๐บ๐ฅ (โ, ๐ก) โ ๐ข๐ฅ (โ, ๐ก) ๐บ (โ, ๐ก)) โ (๐ข (0, ๐ก) ๐บ๐ฅ (0, ๐ก) โ ๐ข๐ฅ (0, ๐ก) ๐บ (0, ๐ก))
We are told that ๐ข๐ฅ (0, ๐ก) = ๐ด (๐ก), we also know that ๐บ (ยฑโ, ๐ก; ๐ฅ0, ๐ก0) = 0, Hence ๐บ (โ, ๐ก) = 0 andalso ๐บ๐ฅ (โ, ๐ก) = 0. The above simplifies
[๐ข๐บ๐ฅ โ ๐ข๐ฅ๐บ]โ๐ฅ=0 = โ (๐ข (0, ๐ก) ๐บ๐ฅ (0, ๐ก) โ ๐ด (๐ก) ๐บ (0, ๐ก)) (5)
We see now from the above, that to get rid of the ๐ข (0, ๐ก) ๐บ๐ฅ (0, ๐ก) term (since we do not knowwhat ๐ข (0, ๐ก) is), then we now need
๐บ๐ฅ (0, ๐ก) = 0
This means we need an image at โ๐ฅ0 which is of same sign as at +๐ฅ0 as shown in this diagram
xx = 0x0โx0
++
Which means the Green function we need to use is the sum of the Green function solutionsfor the infinite domain problem
๐บ (๐ฅ, ๐ก) =1
โ4๐๐ (๐ก0 โ ๐ก)
โโโโโexp
โโโโโโ (๐ฅ โ ๐ฅ0)
2
4๐ (๐ก0 โ ๐ก)
โโโโโ + exp
โโโโโโ (๐ฅ + ๐ฅ0)
2
4๐ (๐ก0 โ ๐ก)
โโโโโ
โโโโโ 0 โค ๐ก โค ๐ก0 (6)
200
3.6. HW6 CHAPTER 3. HWS
The above makes ๐บ๐ฅ (0, ๐ก) = 0 and now equation (5) reduces to
[๐ข๐บ๐ฅ โ ๐ข๐ฅ๐บ]โ๐ฅ=0 = ๐ด (๐ก) ๐บ (0, ๐ก)
= ๐ด (๐ก)โโโโโ
1
โ4๐๐ (๐ก0 โ ๐ก)
โโโโโexp
โโโโโโ (โ๐ฅ0)
2
4๐ (๐ก0 โ ๐ก)
โโโโโ + exp
โโโโโโ (๐ฅ0)
2
4๐ (๐ก0 โ ๐ก)
โโโโโ
โโโโโ
โโโโโ
=๐ด (๐ก)
โ4๐๐ (๐ก0 โ ๐ก)๏ฟฝexp ๏ฟฝ
โ๐ฅ204๐ (๐ก0 โ ๐ก)
๏ฟฝ + exp ๏ฟฝโ๐ฅ20
4๐ (๐ก0 โ ๐ก)๏ฟฝ๏ฟฝ
=๐ด (๐ก)
โ๐๐ (๐ก0 โ ๐ก)exp ๏ฟฝ
โ๐ฅ204๐ (๐ก0 โ ๐ก)
๏ฟฝ
We now know all the terms needed to evaluate the solution. From (4)
๐ข (๐ฅ0, ๐ก0) = โ๏ฟฝโ
๐ฅ=0โ๐ (๐ฅ)๐บ (๐ฅ, 0) ๐๐ฅ โ๏ฟฝ
๐ก0
๐ก=0๐ด (๐ก) ๐บ (0, ๐ก) ๐๐ก +๏ฟฝ
โ
๐ฅ=0๏ฟฝ
๐ก0
๐ก=0๐บ๐ ๐๐ก๐๐ฅ (7)
Using the Green function we found in (6), then (7) becomes
๐ข (๐ฅ0, ๐ก0) = ๏ฟฝโ
๐ฅ=0๐ (๐ฅ)
1
โ4๐๐๐ก0
โโโโโexp
โโโโโโ (๐ฅ โ ๐ฅ0)
2
4๐๐ก0
โโโโโ + exp
โโโโโโ (๐ฅ + ๐ฅ0)
2
4๐๐ก0
โโโโโ
โโโโโ ๐๐ฅ
โ๏ฟฝ๐ก0
๐ก=0
๐ด (๐ก)
โ๐๐ (๐ก0 โ ๐ก)exp ๏ฟฝ
โ๐ฅ204๐ (๐ก0 โ ๐ก)
๏ฟฝ ๐๐ก
+๏ฟฝโ
๐ฅ=0๏ฟฝ
๐ก0
๐ก=0
1
โ4๐๐ (๐ก0 โ ๐ก)
โโโโโexp
โโโโโโ (๐ฅ โ ๐ฅ0)
2
4๐ (๐ก0 โ ๐ก)
โโโโโ + exp
โโโโโโ (๐ฅ + ๐ฅ0)
2
4๐ (๐ก0 โ ๐ก)
โโโโโ
โโโโโ ๐ ๐๐ก๐๐ฅ
Changing the roles of ๐ฅ, ๐ก and ๐ฅ0, ๐ก0 the above becomes
๐ข (๐ฅ, ๐ก) =1
โ4๐๐๐ก๏ฟฝ
โ
๐ฅ0=0๐ (๐ฅ0)
โโโโโexp
โโโโโโ (๐ฅ0 โ ๐ฅ)
2
4๐๐ก
โโโโโ + exp
โโโโโโ (๐ฅ0 + ๐ฅ)
2
4๐๐ก
โโโโโ
โโโโโ ๐๐ฅ0
โ๏ฟฝ๐ก
๐ก0=0
๐ด (๐ก0)
โ๐๐ (๐ก โ ๐ก0)exp ๏ฟฝ
โ๐ฅ2
4๐ (๐ก โ ๐ก0)๏ฟฝ ๐๐ก0
+๏ฟฝโ
๐ฅ0=0๏ฟฝ
๐ก
๐ก0=0
1
โ4๐๐ (๐ก โ ๐ก0)
โโโโโexp
โโโโโโ (๐ฅ0 โ ๐ฅ)
2
4๐ (๐ก โ ๐ก0)
โโโโโ + exp
โโโโโโ (๐ฅ0 + ๐ฅ)
2
4๐ (๐ก โ ๐ก0)
โโโโโ
โโโโโ ๐ (๐ฅ0, ๐ก0) ๐๐ก0๐๐ฅ0
Summary
๐ข (๐ฅ, ๐ก) =1
โ4๐๐๐กฮ1 โ ฮ2 + ฮ3
Where
ฮ1 = ๏ฟฝโ
๐ฅ0=0๐ (๐ฅ0)
โโโโโexp
โโโโโโ (๐ฅ0 โ ๐ฅ)
2
4๐๐ก
โโโโโ + exp
โโโโโโ (๐ฅ0 + ๐ฅ)
2
4๐๐ก
โโโโโ
โโโโโ ๐๐ฅ0
ฮ2 = ๏ฟฝ๐ก
๐ก0=0
๐ด (๐ก0)
โ๐๐ (๐ก โ ๐ก0)exp ๏ฟฝ
โ๐ฅ2
4๐ (๐ก โ ๐ก0)๏ฟฝ ๐๐ก0
ฮ3 = ๏ฟฝโ
๐ฅ0=0๏ฟฝ
๐ก
๐ก0=0
1
โ4๐๐ (๐ก โ ๐ก0)
โโโโโexp
โโโโโโ (๐ฅ0 โ ๐ฅ)
2
4๐ (๐ก โ ๐ก0)
โโโโโ + exp
โโโโโโ (๐ฅ0 + ๐ฅ)
2
4๐ (๐ก โ ๐ก0)
โโโโโ
โโโโโ ๐ (๐ฅ0, ๐ก0) ๐๐ก0๐๐ฅ0
201
3.6. HW6 CHAPTER 3. HWS
Where ฮ1 comes from the initial conditions and ฮ2 comes from the boundary conditionsand ฮ3 comes from for forcing function. It is also important to note that ฮ1 is valid for only๐ก > 0 and not for ๐ก = 0.
3.6.2 Problem 2
NE548 Problems
Due Tuesday April 25, 2017
1. Use the Method of Images to solve
โu
โt= ฮฝ
โ2u
โx2+Q(x, t), x > 0
(a)u(0, t) = 0, u(x, 0) = 0.
(b)u(0, t) = 1, u(x, 0) = 0.
(c)ux(0, t) = A(t), u(x, 0) = f(x).
2. (a) Solve by the Method of Characteristics:
โ2u
โt2= c2
โ2u
โx2, x โฅ 0, t โฅ 0
u(x, 0) = f(x),โu(x, 0)
โt= g(x),
โu(0, t)
โx= h(t)
(b) For the special case h(t) = 0, explain how you could use a symmetry argument to helpconstruct the solution.
(c) Sketch the solution if g(x) = 0, h(t) = 0 and f(x) = 1 for 4 โค x โค 5 and f(x) = 0otherwise.
1
3.6.2.1 Part (a)
The general solution we will use as starting point is
๐ข (๐ฅ, ๐ก) = ๐น (๐ฅ โ ๐๐ก) + ๐บ (๐ฅ + ๐๐ก)
Where ๐น (๐ฅ โ ๐๐ก) is the right moving wave and ๐บ (๐ฅ + ๐๐ก) is the left moving wave. Applying๐ข (๐ฅ, 0) = ๐ (๐ฅ) gives
๐ (๐ฅ) = ๐น (๐ฅ) + ๐บ (๐ฅ) (1)
And๐๐ข (๐ฅ, ๐ก)๐๐ก
=๐๐น
๐ (๐ฅ โ ๐๐ก)๐ (๐ฅ โ ๐๐ก)
๐๐ก+
๐๐บ๐ (๐ฅ + ๐๐ก)
๐ (๐ฅ + ๐๐ก)๐๐ก
= โ๐๐นโฒ + ๐๐บโฒ
Hence from second initial conditions we obtain
๐ (๐ฅ) = โ๐๐นโฒ + ๐๐บโฒ (2)
Equation (1) and (2) are for valid only for positive argument, which means for ๐ฅ โฅ ๐๐ก.๐บ (๐ฅ + ๐๐ก) has positive argument always since ๐ฅ โฅ 0 and ๐ก โฅ 0, but ๐น (๐ฅ โ ๐๐ก) can have negativeargument when ๐ฅ < ๐๐ก. For ๐ฅ < ๐๐ก, we will use the boundary conditions to define ๐น (๐ฅ โ ๐๐ก).Therefore for ๐ฅ โฅ ๐๐ก we solve (1,2) for ๐บ, ๐น and find
๐น (๐ฅ โ ๐๐ก) =12๐ (๐ฅ โ ๐๐ก) โ
12๐ ๏ฟฝ
๐ฅโ๐๐ก
0๐ (๐ ) ๐๐ (2A)
๐บ (๐ฅ + ๐๐ก) =12๐ (๐ฅ + ๐๐ก) +
12๐ ๏ฟฝ
๐ฅ+๐๐ก
0๐ (๐ ) ๐๐ (2B)
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3.6. HW6 CHAPTER 3. HWS
This results in
๐ข (๐ฅ, ๐ก) =๐ (๐ฅ + ๐๐ก) + ๐ (๐ฅ โ ๐๐ก)
2+12๐ ๏ฟฝ
๐ฅ+๐๐ก
๐ฅโ๐๐ก๐ (๐ ) ๐๐ ๐ฅ โฅ ๐๐ก (3)
The solution (3) is only valid for arguments that are positive. This is not a problem for๐บ (๐ฅ + ๐๐ก) since its argument is always positive. But for ๐น (๐ฅ โ ๐๐ก) its argument can becomenegative when 0 < ๐ฅ < ๐๐ก. So we need to obtain a new solution for ๐น (๐ฅ โ ๐๐ก) for the case when๐ฅ < ๐๐ก. First we find ๐๐ข(๐ฅ,๐ก)
๐๐ฅ
๐๐ข (๐ฅ, ๐ก)๐๐ฅ
=๐๐น
๐ (๐ฅ โ ๐๐ก)๐ (๐ฅ โ ๐๐ก)๐๐ฅ
+๐๐บ
๐ (๐ฅ + ๐๐ก)๐ (๐ฅ + ๐๐ก)๐๐ฅ
=๐๐น (๐ฅ โ ๐๐ก)๐ (๐ฅ โ ๐๐ก)
+๐๐บ (๐ฅ + ๐๐ก)๐ (๐ฅ + ๐๐ก)
Hence at ๐ฅ = 0
โ (๐ก) =๐๐น (โ๐๐ก)๐ (โ๐๐ก)
+๐๐บ (๐๐ก)๐ (๐๐ก)
๐๐น (โ๐๐ก)๐ (โ๐๐ก)
= โ (๐ก) โ๐๐บ (๐๐ก)๐ (๐๐ก)
(4)
Let ๐ง = โ๐๐ก, therefore (4) becomes, where ๐ก = โ๐ง๐ also
๐๐น (๐ง)๐๐ง
= โ ๏ฟฝโ๐ง๐๏ฟฝ โ
๐๐บ (โ๐ง)๐ (โ๐ง)
๐ง < 0
๐๐น (๐ง)๐๐ง
= โ ๏ฟฝโ๐ง๐๏ฟฝ +
๐๐บ (โ๐ง)๐๐ง
๐ง < 0
To find ๐น (๐ง), we integrate the above which gives
๏ฟฝ๐ง
0
๐๐น (๐ )๐๐
๐๐ = ๏ฟฝ๐ง
0โ ๏ฟฝโ
๐ ๐๏ฟฝ ๐๐ +๏ฟฝ
๐ง
0
๐๐บ (โ๐ )๐๐
๐๐
๐น (๐ง) โ ๐น (0) = ๏ฟฝ๐ง
0โ ๏ฟฝโ
๐ ๐๏ฟฝ ๐๐ + ๐บ (โ๐ง) โ ๐บ (0)
Ignoring the constants of integration ๐น (0) , ๐บ (0) gives
๐น (๐ง) = ๏ฟฝ๐ง
0โ ๏ฟฝโ
๐ ๐๏ฟฝ ๐๐ + ๐บ (โ๐ง) (4A)
Replacing ๐ง = ๐ฅ โ ๐๐ก in the above gives
๐น (๐ฅ โ ๐๐ก) = ๏ฟฝ๐ฅโ๐๐ก
0โ ๏ฟฝโ
๐ ๐๏ฟฝ ๐๐ + ๐บ (๐๐ก โ ๐ฅ)
Let ๐ = โ ๐ ๐ then when ๐ = 0, ๐ = 0 and when ๐ = ๐ฅ โ ๐๐ก then ๐ = โ๐ฅโ๐๐ก
๐ = ๐๐กโ๐ฅ๐ . And ๐๐
๐๐ = โ1๐ .
Using these the integral in the above becomes
๐น (๐ฅ โ ๐๐ก) = ๏ฟฝ๐๐กโ๐ฅ๐
0โ (๐) (โ๐๐๐) + ๐บ (๐๐ก โ ๐ฅ)
= โ๐๏ฟฝ๐กโ ๐ฅ
๐
0โ (๐) ๐๐ + ๐บ (๐๐ก โ ๐ฅ) (5)
203
3.6. HW6 CHAPTER 3. HWS
The above is ๐น (โ ) when its argument are negative. But in the above ๐บ (๐๐ก โ ๐ฅ) is the same aswe found above in (2b), which just replace it argument in 2B which was ๐ฅ+ ๐๐ก with ๐๐ก โ ๐ฅ andobtain
๐บ (๐๐ก โ ๐ฅ) =12๐ (๐๐ก โ ๐ฅ) +
12๐ ๏ฟฝ
๐๐กโ๐ฅ
0๐ (๐ ) ๐๐ ๐ฅ < ๐๐ก
Therefore (5) becomes
๐น (๐ฅ โ ๐๐ก) = โ๐๏ฟฝ๐กโ ๐ฅ
๐
0โ (๐) ๐๐ +
12๐ (๐๐ก โ ๐ฅ) +
12๐ ๏ฟฝ
๐๐กโ๐ฅ
0๐ (๐ ) ๐๐ ๐ฅ < ๐๐ก (7)
Hence for ๐ฅ < ๐๐ก
๐ข (๐ฅ, ๐ก) = ๐น (๐ฅ โ ๐๐ก) + ๐บ (๐ฅ + ๐๐ก) (8)
But in the above ๐บ (๐ฅ + ๐๐ก) do not change, and we use the same solution for ๐บ for ๐ฅ โฅ ๐๐กwhich is in (2B), given again below
๐บ (๐ฅ + ๐๐ก) =12๐ (๐ฅ + ๐๐ก) +
12๐ ๏ฟฝ
๐ฅ+๐๐ก
0๐ (๐ ) ๐๐ (9)
Hence, plugging (7,9) into (8) gives
๐ข (๐ฅ, ๐ก) = โ๐๏ฟฝ๐๐กโ๐ฅ๐
0โ (๐ ) ๐๐ +
12๐ (๐๐ก โ ๐ฅ) +
12๐ ๏ฟฝ
๐๐กโ๐ฅ
0๐ (๐ ) ๐๐ +
12๐ (๐ฅ + ๐๐ก) +
12๐ ๏ฟฝ
๐ฅ+๐๐ก
0๐ (๐ ) ๐๐ ๐ฅ < ๐๐ก
= โ๐๏ฟฝ๐๐กโ๐ฅ๐
0โ (๐ ) ๐๐ +
๐ (๐๐ก โ ๐ฅ) + ๐ (๐ฅ + ๐๐ก)2
+12๐ ๏ฟฝ๏ฟฝ
๐๐กโ๐ฅ
0๐ (๐ ) ๐๐ +๏ฟฝ
๐ฅ+๐๐ก
0๐ (๐ ) ๐๐ ๏ฟฝ
The above for ๐ฅ < ๐๐ก. Therefore the full solution is
๐ข (๐ฅ, ๐ก) =
โงโชโชโชโจโชโชโชโฉ
๐(๐ฅ+๐๐ก)+๐(๐ฅโ๐๐ก)2 + 1
2๐โซ๐ฅ+๐๐ก
๐ฅโ๐๐ก๐ (๐ ) ๐๐ ๐ฅ โฅ ๐๐ก
โ๐โซ๐กโ ๐ฅ
๐0
โ (๐ ) ๐๐ + ๐(๐๐กโ๐ฅ)+๐(๐ฅ+๐๐ก)2 + 1
2๐๏ฟฝโซ๐๐กโ๐ฅ
0๐ (๐ ) ๐๐ + โซ
๐ฅ+๐๐ก
0๐ (๐ ) ๐๐ ๏ฟฝ ๐ฅ < ๐๐ก
(10)
3.6.2.2 Part(b)
From (10) above, for โ (๐ก) = 0 the solution becomes
๐ข (๐ฅ, ๐ก) =
โงโชโชโชโจโชโชโชโฉ
๐(๐ฅ+๐๐ก)+๐(๐ฅโ๐๐ก)2 + 1
2๐โซ๐ฅ+๐๐ก
๐ฅโ๐๐ก๐ (๐ ) ๐๐ ๐ฅ โฅ ๐๐ก
๐(๐๐กโ๐ฅ)+๐(๐ฅ+๐๐ก)2 + 1
2๐๏ฟฝโซ๐๐กโ๐ฅ
0๐ (๐ ) ๐๐ + โซ
๐ฅ+๐๐ก
0๐ (๐ ) ๐๐ ๏ฟฝ ๐ฅ < ๐๐ก
(1)
The idea of symmetry is to obtain the same solution (1) above but by starting from dโAlmbertsolution (which is valid only for positive arguments)
๐ข (๐ฅ, ๐ก) =๐ (๐ฅ + ๐๐ก) + ๐ (๐ฅ โ ๐๐ก)
2+12๐ ๏ฟฝ
๐ฅ+๐๐ก
๐ฅโ๐๐ก๐ (๐ ) ๐๐ ๐ฅ > ๐๐ก (1A)
But by using ๐๐๐ฅ๐ก, ๐๐๐ฅ๐ก in the above instead of ๐, ๐, where the dโAlmbert solution becomesvalid for ๐ฅ < ๐๐ก when using ๐๐๐ฅ๐ก, ๐๐๐ฅ๐ก
๐ข (๐ฅ, ๐ก) =๐๐๐ฅ๐ก (๐ฅ + ๐๐ก) + ๐๐๐ฅ๐ก (๐ฅ โ ๐๐ก)
2+12๐ ๏ฟฝ
๐ฅ+๐๐ก
๐ฅโ๐๐ก๐๐๐ฅ๐ก (๐ ) ๐๐ ๐ฅ < ๐๐ก (2)
204
3.6. HW6 CHAPTER 3. HWS
Then using (1A) and (2) we show it is the same as (1). We really need to show that (2) leadsto the second part of (1), since (1A) is the same as first part of (1).
The main issue is how to determine ๐๐๐ฅ๐ก, ๐๐๐ฅ๐ก and determine if they should be even or oddextension of ๐, ๐ . From boundary conditions, in part (a) equation (4A) we found that
๐น (๐ง) = ๏ฟฝ๐ง
0โ ๏ฟฝโ
๐ ๐๏ฟฝ ๐๐ + ๐บ (โ๐ง)
But now โ (๐ก) = 0, hence
๐น (๐ง) = ๐บ (โ๐ง) (3)
Now, looking at the first part of the solution in (1). we see that for positive argument thesolution has ๐ (๐ฅ โ ๐๐ก) for ๐ฅ > ๐๐ก and it has ๐ (๐๐ก โ ๐ฅ) when ๐ฅ < ๐๐ก. So this leads us to pick ๐๐๐ฅ๐กbeing even as follows. Let
๐ง = ๐ฅ โ ๐๐ก
Then we see that
๐ (๐ฅ โ ๐๐ก) = ๐ (โ (๐ฅ โ ๐๐ก))๐ (๐ง) = ๐ (โ๐ง)
Therefore we need ๐๐๐ฅ๐ก to be an even function.
๐๐๐ฅ๐ก (๐ง) =
โงโชโชโจโชโชโฉ๐ (๐ง) ๐ง > 0๐ (โ๐ง) ๐ง < 0
But since ๐น (๐ง) = ๐บ (โ๐ง) from (3), then ๐๐๐ฅ๐ก is also even function, which means
๐๐๐ฅ๐ก (๐ง) =
โงโชโชโจโชโชโฉ๐ (๐ง) ๐ง > 0๐ (โ๐ง) ๐ง < 0
Now that we found ๐๐๐ฅ๐ก, ๐๐๐ฅ๐ก extensions, we go back to (2). For negative argument
๐ข (๐ฅ, ๐ก) =๐๐๐ฅ๐ก (๐ฅ + ๐๐ก) + ๐๐๐ฅ๐ก (๐ฅ โ ๐๐ก)
2+12๐ ๏ฟฝ
๐ฅ+๐๐ก
๐ฅโ๐๐ก๐๐๐ฅ๐ก (๐ ) ๐๐ ๐ฅ < ๐๐ก
Since ๐๐๐ฅ๐ก, ๐๐๐ฅ๐ก are even, then using ๐๐๐ฅ๐ก (๐ง) = ๐ (โ๐ง) since now ๐ง < 0 and using ๐๐๐ฅ๐ก (๐ง) = ๐ (โ๐ง)since now ๐ง < 0 the above becomes
๐ข (๐ฅ, ๐ก) =๐ (โ (๐ฅ + ๐๐ก)) + ๐ (โ (๐ฅ โ ๐๐ก))
2+12๐ ๏ฟฝ
๐ฅ+๐๐ก
๐ฅโ๐๐ก๐๐๐ฅ๐ก (๐ ) ๐๐ ๐ฅ < ๐๐ก
But ๐ (โ (๐ฅ + ๐๐ก)) = ๐ (๐ฅ + ๐๐ก) since even and ๐ (โ (๐ฅ โ ๐๐ก)) = ๐ (๐๐ก โ ๐ฅ), hence the above becomes
๐ข (๐ฅ, ๐ก) ==๐ (๐ฅ + ๐๐ก) + ๐ (๐๐ก โ ๐ฅ)
2+12๐ ๏ฟฝ๏ฟฝ
0
๐ฅโ๐๐ก๐ (โ๐ ) ๐๐ +๏ฟฝ
๐ฅ+๐๐ก
0๐ (๐ ) ๐๐ ๏ฟฝ ๐ฅ < ๐๐ก
Let ๐ = โ๐ , then ๐๐๐๐ = โ1. When ๐ = ๐ฅ โ ๐๐ก, ๐ = ๐๐ก โ ๐ฅ and when ๐ = 0, ๐ = 0. Then the first
205
3.6. HW6 CHAPTER 3. HWS
integral on the RHS above becomes
๐ข (๐ฅ, ๐ก) =๐ (๐ฅ + ๐๐ก) + ๐ (๐๐ก โ ๐ฅ)
2+12๐ ๏ฟฝ๏ฟฝ
0
๐๐กโ๐ฅ๐ (๐) (โ๐๐) +๏ฟฝ
๐ฅ+๐๐ก
0๐ (๐ ) ๐๐ ๏ฟฝ
=๐ (๐ฅ + ๐๐ก) + ๐ (๐๐ก โ ๐ฅ)
2+12๐ ๏ฟฝ๏ฟฝ
๐๐กโ๐ฅ
0๐ (๐) ๐๐ +๏ฟฝ
๐ฅ+๐๐ก
0๐ (๐ ) ๐๐ ๏ฟฝ
Relabel ๐ back to ๐ , then
๐ข (๐ฅ, ๐ก) =๐ (๐ฅ + ๐๐ก) + ๐ (๐๐ก โ ๐ฅ)
2+12๐ ๏ฟฝ๏ฟฝ
๐๐กโ๐ฅ
0๐ (๐ ) ๐๐ +๏ฟฝ
๐ฅ+๐๐ก
0๐ (๐ ) ๐๐ ๏ฟฝ ๐ฅ < ๐๐ก (5)
Looking at (5) we see that this is the same solution in (1) for the case of ๐ฅ < ๐๐ก. Hence (1A)and (5) put together give
๐ข (๐ฅ, ๐ก) =
โงโชโชโชโจโชโชโชโฉ
๐(๐ฅ+๐๐ก)+๐(๐ฅโ๐๐ก)2 + 1
2๐โซ๐ฅ+๐๐ก
๐ฅโ๐๐ก๐ (๐ ) ๐๐ ๐ฅ โฅ ๐๐ก
๐(๐๐กโ๐ฅ)+๐(๐ฅ+๐๐ก)2 + 1
2๐๏ฟฝโซ๐๐กโ๐ฅ
0๐ (๐ ) ๐๐ + โซ
๐ฅ+๐๐ก
0๐ (๐ ) ๐๐ ๏ฟฝ ๐ฅ < ๐๐ก
(1)
Which is same solution obtain in part(a).
3.6.2.3 Part(c)
For ๐ (๐ฅ) = 0, โ (๐ก) = 0 and ๐ (๐ฅ) as given, the solution in equation (10) in part(a) becomes
๐ข (๐ฅ, ๐ก) =
โงโชโชโจโชโชโฉ
๐(๐ฅ+๐๐ก)+๐(๐ฅโ๐๐ก)2 ๐ฅ โฅ ๐๐ก
๐(๐๐กโ๐ฅ)+๐(๐ฅ+๐๐ก)2 ๐ฅ < ๐๐ก
(10)
A small program we written to make few sketches important time instantces. The left movingwave ๐บ (๐ฅ + ๐๐ก) hits the boundary at ๐ฅ = 0 but do not reflect now as the case with Dirichletboundary conditions, but instead it remains upright and turns around as shown.
206
3.6. HW6 CHAPTER 3. HWS
time = 0.00
2 4 6 8 10 120.0
0.2
0.4
0.6
0.8
1.0
time = 0.25
2 4 6 8 10 120.0
0.2
0.4
0.6
0.8
1.0
time = 0.40
2 4 6 8 10 120.0
0.2
0.4
0.6
0.8
1.0
time = 0.55
2 4 6 8 10 120.0
0.2
0.4
0.6
0.8
1.0
time = 0.59
2 4 6 8 10 120.0
0.2
0.4
0.6
0.8
1.0
time = 0.80
2 4 6 8 10 120.0
0.2
0.4
0.6
0.8
1.0
time = 1.20
2 4 6 8 10 120.0
0.2
0.4
0.6
0.8
1.0
time = 2.50
2 4 6 8 10 120.0
0.2
0.4
0.6
0.8
1.0
time = 3.90
2 4 6 8 10 120.0
0.2
0.4
0.6
0.8
1.0
time = 4.20
2 4 6 8 10 120.0
0.2
0.4
0.6
0.8
1.0
time = 5.40
2 4 6 8 10 120.0
0.2
0.4
0.6
0.8
1.0
time = 7.00
2 4 6 8 10 120.0
0.2
0.4
0.6
0.8
1.0
207
3.6. HW6 CHAPTER 3. HWS
208
Chapter 4
Study notes
Local contents4.1 question asked 2/7/2017 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2104.2 note on Airy added 1/31/2017 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2114.3 note p1 added 2/3/2017 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2134.4 note added 1/31/2017 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2204.5 note p3 figure 3.4 in text (page 92) reproduced in color . . . . . . . . . . . . . . . 2254.6 note p4 added 2/15/17 animations of the solutions to ODE from lecture 2/9/17
for small parameter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2274.7 note p5 added 2/15/17 animation figure 9.5 in text page 434 . . . . . . . . . . . . 2314.8 note p7 added 2/15/17, boundary layer problem solved in details . . . . . . . . . 2374.9 note p9. added 2/24/17, asking question about problem in book . . . . . . . . . . 2434.10 note p11. added 3/7/17, Showing that scaling for normalization is same for any ๐ 2454.11 note p12. added 3/8/17, comparing exact solution to WKB solution using Math-
ematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2474.12 Convert ODE to Liouville form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2524.13 note p13. added 3/15/17, solving ๐2๐ฆโณ(๐ฅ) = (๐ + ๐ฅ + ๐ฅ2)๐ฆ(๐ฅ) in Maple . . . . . . . . 254
209
4.1. question asked 2/7/2017 CHAPTER 4. STUDY NOTES
4.1 question asked 2/7/2017
Book at page 80 says "it is usually true that ๐โฒโฒ โ (๐โฒ)2" as ๐ฅ โ ๐ฅ0. What are the exceptionsto this? Since it is easy to find counter examples.
For ๐ฆโฒโฒ = โ๐ฅ๐ฆ, as ๐ฅ โ 0+. Using ๐ฆ (๐ฅ) = ๐๐0(๐ฅ) and substituting, gives the ODE
๐โฒโฒ0 + ๏ฟฝ๐โฒ0๏ฟฝ2= โ๐ฅ (1)
If we follow the book, and drop ๐โฒโฒ0 relative to (๐โฒ)2 then
๏ฟฝ๐โฒ0๏ฟฝ2= ๐ฅ
12
๐โฒ0 = ๐๐ฅ14
So, lets check the assumption. Since ๐โฒโฒ0 = ๐14๐ฅ
โ 34 . Therefore (ignoring all multiplicative
constants)
๐โฒโฒ0?โ ๏ฟฝ๐โฒ0๏ฟฝ
2
1
๐ฅ34
?โ ๐ฅ
12
No. Did not check out. Since when ๐ฅ โ 0+ the LHS blow up while the RHS goes to zero.So in this example, this "rule of thumb" did not work out, and it is the other way around.Assuming I did not make a mistake, when the book says "it is usually true", it will good toknow under what conditions is this true or why did the book say this?
This will help in solving these problem.
210
4.2. note on Airy added 1/31/2017 CHAPTER 4. STUDY NOTES
4.2 note on Airy added 1/31/2017
Note on using CAS for working with book AiryAI/AiryBI
termsCAS such as Mathematica or Maple can actually handle very large numbers. It is possible to calculate
individual terms of the Airy series given by the book (but no sum it). Calculating individual terms, up to
10 million terms for x=1,100 and x=10000 is shown below as illustration. It takes few minutes to do each
term
define small function to calculate one term
bookAi[x_Integer, n_Integer] :=
3-2
3x3 n
9n Factorial[n] Gamman + 2 3- 3
-4
3x3 n+1
9n Factorial[n] Gamman + 4 3;
This is for x=1, N=10,000,000
In[3]:= bookAi[1, 10 000 000] // N
Out[3]= 5.7642115 ร 10-140 856 542
This is for x=100, N=10,000,000
In[4]:= bookAi[100, 10 000 000] // N
Out[4]= 5.7583009 ร 10-80 856 542
This is for x=10,000 and N=10,000,000
In[5]:= Timing[bookAi[10000, 10 000000] // N]
Out[5]= 582.3829332, 5.167240 ร 10-20 856 542
The above took 582 seconds to complete! The largest number it can handle on my PC is
In[6]:= $MaxNumber
Out[6]= 1.605216761933662 ร 101355718576 299 609
Let compare the above to Gamma[10,000,000]
In[8]:= Gamma[10 000000.0]
Out[8]= 1.2024234 ร 1065657052
In[9]:= Gamma[10 000000.0] < $MaxNumber
Out[9]= True
We see that Gamma[10,000,000] is much less than the largest number it can handle. Here is Gamma
Printed by Wolfram Mathematica Student Edition
211
4.2. note on Airy added 1/31/2017 CHAPTER 4. STUDY NOTES
for 10 billion
In[10]:= Gamma[10 000000 000.0]
Out[10]= 2.3258 ร 1095657055176
And 10 billion Factorial
In[11]:= Factorial[10 000000 000.0]
Out[11]= 2.3258 ร 1095657055186
So CAS can handle these terms. But not the complete series summation as given in
the book
2 airy.nb
Printed by Wolfram Mathematica Student Edition
212
4.3. note p1 added 2/3/2017 CHAPTER 4. STUDY NOTES
4.3 note p1 added 2/3/2017
Current rules I am using in simplifications are
1. ๐โฒ0 โ ๐โฒ1
2. ๐โฒ0๐โฒ1 โ ๏ฟฝ๐โฒ1๏ฟฝ2
3. ๐0 โ ๐1
4. (๐0)๐ โ ๐(๐)0 (the first is power, the second is derivative order).
Verify the above are valid for ๐ฅ โ 0+ and well for ๐ฅ โ โ. What can we say about (๐โฒ)compared to (๐โฒ)2?
4.3.1 problem (a), page 88
๐ฆโฒโฒ =1๐ฅ5๐ฆ
Irregular singular point at ๐ฅ โ 0+. Let ๐ฆ = ๐๐0(๐ฅ) and the above becomes
๐ฆ (๐ฅ) = ๐๐0(๐ฅ)
๐ฆโฒ (๐ฅ) = ๐โฒ0๐๐
๐ฆโฒโฒ = ๐โฒโฒ0 ๐๐0 + ๏ฟฝ๐โฒ0๏ฟฝ2๐๐0
= ๏ฟฝ๐โฒโฒ0 + ๏ฟฝ๐โฒ0๏ฟฝ2๏ฟฝ ๐๐0
Substituting back into ๐2
๐๐ฅ2๐ฆ = ๐ฅโ5๐ฆ gives
๏ฟฝ๐โฒโฒ0 + ๏ฟฝ๐โฒ0๏ฟฝ2๏ฟฝ ๐๐ = ๐ฅโ5๐๐0(๐ฅ)
๐โฒโฒ0 + ๏ฟฝ๐โฒ0๏ฟฝ2= ๐ฅโ5
Before solving for ๐0, we can do one more simplification. Using the approximation that
๏ฟฝ๐โฒ0๏ฟฝ2โ ๐โฒโฒ0 for ๐ฅ โ ๐ฅ0, the above becomes
๏ฟฝ๐โฒ0๏ฟฝ2โผ ๐ฅโ5
213
4.3. note p1 added 2/3/2017 CHAPTER 4. STUDY NOTES
Now we are ready to solve for ๐0
๐โฒ0 โผ ๐๐ฅโ52
๐0 โผ ๐๏ฟฝ๐ฅโ52๐๐ฅ
โผ ๐๐ฅโ
32
โ32
โผ โ23๐๐ฅโ
32
To find leading behavior, let
๐ (๐ฅ) = ๐0 (๐ฅ) + ๐1 (๐ฅ)
Then ๐ฆ (๐ฅ) = ๐๐0(๐ฅ)+๐1(๐ฅ) and hence now
๐ฆโฒ (๐ฅ) = (๐0 (๐ฅ) + ๐1 (๐ฅ))โฒ ๐๐0+๐1
๐ฆโฒโฒ (๐ฅ) = ๏ฟฝ(๐0 + ๐1)โฒ๏ฟฝ2๐๐0+๐1 + (๐0 + ๐1)
โฒโฒ ๐๐0+๐1
Using the above, the ODE ๐2
๐๐ฅ2๐ฆ = ๐ฅโ5๐ฆ now becomes
๏ฟฝ(๐0 + ๐1)โฒ๏ฟฝ2๐๐0+๐1 + (๐0 + ๐1)
โฒโฒ ๐๐0+๐1 โผ ๐ฅโ5๐๐0+๐1
๏ฟฝ(๐0 + ๐1)โฒ๏ฟฝ2+ (๐0 + ๐1)
โฒโฒ โผ ๐ฅโ5
๏ฟฝ๐โฒ0 + ๐โฒ1๏ฟฝ2+ ๐โฒโฒ0 + ๐โฒโฒ1 โผ ๐ฅโ5
๏ฟฝ๐โฒ0๏ฟฝ2+ ๏ฟฝ๐โฒ1๏ฟฝ
2+ 2๐โฒ0๐โฒ1 + ๐โฒโฒ0 + ๐โฒโฒ1 โผ ๐ฅโ5
But ๐โฒ0 = ๐๐ฅโ 52 , found before, hence ๏ฟฝ๐โฒ0๏ฟฝ
2= ๐ฅโ5 and the above simplifies to
๏ฟฝ๐โฒ1๏ฟฝ2+ 2๐โฒ0๐โฒ1 + ๐โฒโฒ0 + ๐โฒโฒ1 = 0
Using approximation ๐โฒ0๐โฒ1 โ ๏ฟฝ๐โฒ1๏ฟฝ2the above simplifies to
2๐โฒ0๐โฒ1 + ๐โฒโฒ0 + ๐โฒโฒ1 = 0
Finally, using approximation ๐โฒโฒ0 โ ๐โฒโฒ1 , the above becomes
2๐โฒ0๐โฒ1 + ๐โฒโฒ0 = 0
๐โฒ1 โผ โ๐โฒโฒ02๐โฒ0
๐1 โผ โ12
ln ๐โฒ0 + ๐
๐โฒ1 โผ โ12
ln ๐ฅโ52 + ๐
๐โฒ1 โผ54
ln ๐ฅ + ๐
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4.3. note p1 added 2/3/2017 CHAPTER 4. STUDY NOTES
Hence, the leading behavior is
๐ฆ (๐ฅ) = ๐๐0(๐ฅ)+๐1(๐ฅ)
= exp ๏ฟฝโ23๐๐ฅโ
32 +
54
ln ๐ฅ + ๐๏ฟฝ
= ๐๐ฅ54 exp ๏ฟฝโ๐
23๐ฅโ
32 ๏ฟฝ (1)
To verify, using the formula 3.4.28, which is
๐ฆ (๐ฅ) โผ ๐๐1โ๐2๐ exp ๏ฟฝ๐๏ฟฝ
๐ฅ๐ (๐ก)
1๐ ๐๐ก๏ฟฝ
In this case, ๐ = 2, since the ODE ๐ฆโฒโฒ = ๐ฅโ5๐ฆ is second order. Here we have ๐ (๐ฅ) = ๐ฅโ5,therefore, plug-in into the above gives
๐ฆ (๐ฅ) โผ ๐ ๏ฟฝ๐ฅโ5๏ฟฝ1โ24 exp ๏ฟฝ๐๏ฟฝ
๐ฅ๏ฟฝ๐กโ5๏ฟฝ
12 ๐๐ก๏ฟฝ
โผ ๐ ๏ฟฝ๐ฅโ5๏ฟฝโ14 exp ๏ฟฝ๐๏ฟฝ
๐ฅ๐กโ
52๐๐ก๏ฟฝ
โผ ๐๐ฅ54 exp
โโโโโโโ๐
โโโโโโโ๐ฅโ
32
โ32
โโโโโโโ
โโโโโโโ
โผ ๐๐ฅ54 exp ๏ฟฝโ๐
23๐ฅโ
32 ๏ฟฝ (2)
Comparing (1) and (2), we see they are the same.
4.3.2 problem (b), page 88
๐ฆโฒโฒโฒ = ๐ฅ๐ฆ
Irregular singular point at ๐ฅ โ +โ. Let ๐ฆ = ๐๐0(๐ฅ) and the above becomes
๐ฆ (๐ฅ) = ๐๐0(๐ฅ)
๐ฆโฒ (๐ฅ) = ๐โฒ0๐๐0
๐ฆโฒโฒ = ๐โฒโฒ0 ๐๐0 + ๏ฟฝ๐โฒ0๏ฟฝ2๐๐0
= ๏ฟฝ๐โฒโฒ0 + ๏ฟฝ๐โฒ0๏ฟฝ2๏ฟฝ ๐๐0
๐ฆโฒโฒโฒ = ๏ฟฝ๐โฒโฒ0 + ๏ฟฝ๐โฒ0๏ฟฝ2๏ฟฝโฒ๐๐0 + ๏ฟฝ๐โฒโฒ0 + ๏ฟฝ๐โฒ0๏ฟฝ
2๏ฟฝ ๐โฒ0๐๐0
= ๏ฟฝ๐โฒโฒโฒ0 + 2๐โฒ0๐โฒโฒ0 ๏ฟฝ ๐๐0 + ๏ฟฝ๐โฒ0๐โฒโฒ0 + ๏ฟฝ๐โฒ0๏ฟฝ3๏ฟฝ ๐๐0
= ๏ฟฝ๐โฒโฒโฒ0 + 3๐โฒ0๐โฒโฒ0 + ๏ฟฝ๐โฒ0๏ฟฝ3๏ฟฝ ๐๐0
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4.3. note p1 added 2/3/2017 CHAPTER 4. STUDY NOTES
Substituting back into ๐ฆโฒโฒโฒ = ๐ฅ๐ฆ gives
๏ฟฝ๐โฒโฒโฒ0 + 3๐โฒ0๐โฒโฒ0 + ๏ฟฝ๐โฒ0๏ฟฝ3๏ฟฝ ๐๐0 = ๐ฅ๐๐0(๐ฅ)
๐โฒโฒโฒ0 + 3๐โฒ0๐โฒโฒ0 + ๏ฟฝ๐โฒ0๏ฟฝ3= ๐ฅ
Before solving for ๐0, we can do one more simplification. Using the approximation that
๏ฟฝ๐โฒ0๏ฟฝ3โ ๐โฒโฒโฒ0 for ๐ฅ โ ๐ฅ0, the above becomes
3๐โฒ0๐โฒโฒ0 + ๏ฟฝ๐โฒ0๏ฟฝ3โผ ๐ฅ
In addition, since ๐โฒ0 โ ๐โฒโฒ0 then we can use the approximation ๏ฟฝ๐โฒ0๏ฟฝ3โ ๐โฒ0๐โฒโฒ0 and the above
becomes
๏ฟฝ๐โฒ0๏ฟฝ3โผ ๐ฅ
๐โฒ0 โผ ๐๐ฅ13
๐0 โผ ๐๏ฟฝ๐ฅ13๐๐ฅ
๐0 โผ ๐34๐ฅ43
To find leading behavior, let
๐ (๐ฅ) = ๐0 (๐ฅ) + ๐1 (๐ฅ)
Then ๐ฆ (๐ฅ) = ๐๐0(๐ฅ)+๐1(๐ฅ) and hence now
๐ฆโฒ (๐ฅ) = (๐0 (๐ฅ) + ๐1 (๐ฅ))โฒ ๐๐0+๐1
๐ฆโฒโฒ (๐ฅ) = ๏ฟฝ(๐0 + ๐1)โฒ๏ฟฝ2๐๐0+๐1 + (๐0 + ๐1)
โฒโฒ ๐๐0+๐1
= ๏ฟฝ๐โฒ0 + ๐โฒ1๏ฟฝ2๐๐0+๐1 + ๏ฟฝ๐โฒโฒ0 + ๐โฒโฒ1 ๏ฟฝ ๐๐0+๐1
= ๏ฟฝ๏ฟฝ๐โฒ0๏ฟฝ2+ ๏ฟฝ๐โฒ1๏ฟฝ
2+ 2๐โฒ0๐โฒ1๏ฟฝ ๐๐0+๐1 + ๏ฟฝ๐โฒโฒ0 + ๐โฒโฒ1 ๏ฟฝ ๐๐0+๐1
= ๏ฟฝ๏ฟฝ๐โฒ0๏ฟฝ2+ ๏ฟฝ๐โฒ1๏ฟฝ
2+ 2๐โฒ0๐โฒ1 + ๐โฒโฒ0 + ๐โฒโฒ1 ๏ฟฝ ๐๐0+๐1
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4.3. note p1 added 2/3/2017 CHAPTER 4. STUDY NOTES
We can take the third derivative
๐ฆโฒโฒโฒ (๐ฅ) โผ ๏ฟฝ๏ฟฝ๐โฒ0๏ฟฝ2+ ๏ฟฝ๐โฒ1๏ฟฝ
2+ 2๐โฒ0๐โฒ1 + ๐โฒโฒ0 + ๐โฒโฒ1 ๏ฟฝ
โฒ๐๐0+๐1
+ ๏ฟฝ๏ฟฝ๐โฒ0๏ฟฝ2+ ๏ฟฝ๐โฒ1๏ฟฝ
2+ 2๐โฒ0๐โฒ1 + ๐โฒโฒ0 + ๐โฒโฒ1 ๏ฟฝ (๐0 + ๐1)
โฒ ๐๐0+๐1
โผ ๏ฟฝ2๐โฒ0๐โฒโฒ0 + 2๐โฒ1๐โฒโฒ1 + 2๐โฒโฒ0 ๐โฒ1 + 2๐โฒ0๐โฒโฒ1 + ๐โฒโฒโฒ0 + ๐โฒโฒโฒ1 ๏ฟฝ ๐๐0+๐1
+ ๏ฟฝ๏ฟฝ๐โฒ0๏ฟฝ2+ ๏ฟฝ๐โฒ1๏ฟฝ
2+ 2๐โฒ0๐โฒ1 + ๐โฒโฒ0 + ๐โฒโฒ1 ๏ฟฝ ๏ฟฝ๐โฒ0 + ๐โฒ1๏ฟฝ ๐๐0+๐
โผ ๏ฟฝ2๐โฒ0๐โฒโฒ0 + 2๐โฒ1๐โฒโฒ1 + 2๐โฒโฒ0 ๐โฒ1 + 2๐โฒ0๐โฒโฒ1 + ๐โฒโฒโฒ0 + ๐โฒโฒโฒ1 ๏ฟฝ ๐๐0+๐1
+ ๏ฟฝ๏ฟฝ๐โฒ0๏ฟฝ3+ ๐โฒ0 ๏ฟฝ๐โฒ1๏ฟฝ
2+ 2 ๏ฟฝ๐โฒ0๏ฟฝ
2๐โฒ1 + ๐โฒ0๐โฒโฒ0 + ๐โฒ0๐โฒโฒ1 ๏ฟฝ ๐๐0+๐1
+ ๏ฟฝ๐โฒ1 ๏ฟฝ๐โฒ0๏ฟฝ2+ ๏ฟฝ๐โฒ1๏ฟฝ
3+ 2๐โฒ0 ๏ฟฝ๐โฒ1๏ฟฝ
2+ ๐โฒ1๐โฒโฒ0 + ๐โฒ1๐โฒโฒ1 ๏ฟฝ ๐๐0+๐1
โผ ๏ฟฝ2๐โฒ0๐โฒโฒ0 + 2๐โฒ1๐โฒโฒ1 + 2๐โฒโฒ0 ๐โฒ1 + 2๐โฒ0๐โฒโฒ1 + ๐โฒโฒโฒ0 + ๐โฒโฒโฒ1 ๏ฟฝ ๐๐0+๐1
+ ๏ฟฝ๏ฟฝ๐โฒ0๏ฟฝ3+ 3๐โฒ0 ๏ฟฝ๐โฒ1๏ฟฝ
2+ 3 ๏ฟฝ๐โฒ0๏ฟฝ
2๐โฒ1 + ๐โฒ0๐โฒโฒ0 + ๐โฒ0๐โฒโฒ1 + ๏ฟฝ๐โฒ1๏ฟฝ
3+ ๐โฒ1๐โฒโฒ0 + ๐โฒ1๐โฒโฒ1 ๏ฟฝ ๐๐0+๐1
โผ ๏ฟฝ3๐โฒ0๐โฒโฒ0 + 3๐โฒ1๐โฒโฒ1 + 3๐โฒโฒ0 ๐โฒ1 + 3๐โฒ0๐โฒโฒ1 + ๐โฒโฒโฒ0 + ๐โฒโฒโฒ1 + ๏ฟฝ๐โฒ0๏ฟฝ3+ 3๐โฒ0 ๏ฟฝ๐โฒ1๏ฟฝ
2+ 3 ๏ฟฝ๐โฒ0๏ฟฝ
2๐โฒ1 + ๏ฟฝ๐โฒ1๏ฟฝ
3๏ฟฝ ๐๐0+๐1
โผ ๏ฟฝ๏ฟฝ๐โฒ0๏ฟฝ3+ ๏ฟฝ๐โฒ1๏ฟฝ
3+ 3๐โฒ0 ๏ฟฝ๐โฒ1๏ฟฝ
2+ 3 ๏ฟฝ๐โฒ0๏ฟฝ
2๐โฒ1 + 3๐โฒ0๐โฒโฒ0 + 3๐โฒ1๐โฒโฒ1 + 3๐โฒโฒ0 ๐โฒ1 + 3๐โฒ0๐โฒโฒ1 + ๐โฒโฒโฒ0 + ๐โฒโฒโฒ1 ๏ฟฝ ๐๐0+๐1
Lets go ahead and plug-in this into the ODE
๏ฟฝ๐โฒ0๏ฟฝ3+ ๏ฟฝ๐โฒ1๏ฟฝ
3+ 3๐โฒ0 ๏ฟฝ๐โฒ1๏ฟฝ
2+ 3 ๏ฟฝ๐โฒ0๏ฟฝ
2๐โฒ1 + 3๐โฒ0๐โฒโฒ0 + 3๐โฒ1๐โฒโฒ1 + 3๐โฒโฒ0 ๐โฒ1 + 3๐โฒ0๐โฒโฒ1 + ๐โฒโฒโฒ0 + ๐โฒโฒโฒ1 โผ ๐ฅ
Now we do some simplification. ๏ฟฝ๐โฒ0๏ฟฝ3โ ๐โฒโฒโฒ0 and ๏ฟฝ๐โฒ1๏ฟฝ
3โ ๐โฒโฒโฒ1 , hence above becomes
๏ฟฝ๐โฒ0๏ฟฝ3+ ๏ฟฝ๐โฒ1๏ฟฝ
3+ 3๐โฒ0 ๏ฟฝ๐โฒ1๏ฟฝ
2+ 3 ๏ฟฝ๐โฒ0๏ฟฝ
2๐โฒ1 + 3๐โฒ0๐โฒโฒ0 + 3๐โฒ1๐โฒโฒ1 + 3๐โฒโฒ0 ๐โฒ1 + 3๐โฒ0๐โฒโฒ1 โผ ๐ฅ
Also, since ๐โฒโฒ0 โ ๐โฒโฒ1 then 3๐โฒ0๐โฒโฒ0 โ 3๐โฒ0๐โฒโฒ1
๏ฟฝ๐โฒ0๏ฟฝ3+ ๏ฟฝ๐โฒ1๏ฟฝ
3+ 3๐โฒ0 ๏ฟฝ๐โฒ1๏ฟฝ
2+ 3 ๏ฟฝ๐โฒ0๏ฟฝ
2๐โฒ1 + 3๐โฒ0๐โฒโฒ0 + 3๐โฒ1๐โฒโฒ1 + 3๐โฒโฒ0 ๐โฒ1 โผ ๐ฅ
Also, since ๏ฟฝ๐โฒ0๏ฟฝ2โ ๐โฒโฒ0 then 3 ๏ฟฝ๐โฒ0๏ฟฝ
2๐โฒ1 โ 3๐โฒโฒ0 ๐โฒ1
๏ฟฝ๐โฒ0๏ฟฝ3+ ๏ฟฝ๐โฒ1๏ฟฝ
3+ 3๐โฒ0 ๏ฟฝ๐โฒ1๏ฟฝ
2+ 3 ๏ฟฝ๐โฒ0๏ฟฝ
2๐โฒ1 + 3๐โฒ0๐โฒโฒ0 + 3๐โฒ1๐โฒโฒ1 โผ ๐ฅ
Also since ๐โฒ1 โ ๐โฒโฒ1 then 3 ๏ฟฝ๐โฒ0๏ฟฝ2๐โฒ1 โ 3๐โฒ1๐โฒโฒ1
๏ฟฝ๐โฒ0๏ฟฝ3+ ๏ฟฝ๐โฒ1๏ฟฝ
3+ 3๐โฒ0 ๏ฟฝ๐โฒ1๏ฟฝ
2+ 3 ๏ฟฝ๐โฒ0๏ฟฝ
2๐โฒ1 + 3๐โฒ0๐โฒโฒ0 โผ ๐ฅ
But ๐โฒ0 โผ ๐ฅ13 hence๏ฟฝ๐โฒ0๏ฟฝ
3โผ ๐ฅ and the above simplies to
๏ฟฝ๐โฒ1๏ฟฝ3+ 3๐โฒ0 ๏ฟฝ๐โฒ1๏ฟฝ
2+ 3 ๏ฟฝ๐โฒ0๏ฟฝ
2๐โฒ1 + 3๐โฒ0๐โฒโฒ0 = 0
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4.3. note p1 added 2/3/2017 CHAPTER 4. STUDY NOTES
Using 3๐โฒ0 ๏ฟฝ๐โฒ1๏ฟฝ2โ ๏ฟฝ๐โฒ1๏ฟฝ
3since ๐โฒ0 โ ๐โฒ1 then
3๐โฒ0 ๏ฟฝ๐โฒ1๏ฟฝ2+ 3 ๏ฟฝ๐โฒ0๏ฟฝ
2๐โฒ1 + 3๐โฒ0๐โฒโฒ0 = 0
Using 3 ๏ฟฝ๐โฒ0๏ฟฝ2๐โฒ1 โ ๐โฒ0 ๏ฟฝ๐โฒ1๏ฟฝ
2since ๏ฟฝ๐โฒ0๏ฟฝ
2โ ๐โฒ0 then
3 ๏ฟฝ๐โฒ0๏ฟฝ2๐โฒ1 + 3๐โฒ0๐โฒโฒ0 = 0
No more simplification. We are ready to solve for ๐1.
๐โฒ1 โผโ๐โฒ0๐โฒโฒ0๏ฟฝ๐โฒ0๏ฟฝ
2
โผโ๐โฒโฒ0๐โฒ0
Hence
๐1 โผ โ๏ฟฝ๐โฒโฒ0๐โฒ0๐๐ฅ
โผ โ ln ๐โฒ0 + ๐
Since ๐โฒ0 โผ ๐ฅ13 then the above becomes
๐1 โผ โ ln ๐ฅ13 + ๐
๐1 โผ โ13
ln ๐ฅ + ๐
Hence, the leading behavior is
๐ฆ (๐ฅ) = ๐๐0(๐ฅ)+๐1(๐ฅ)
= exp ๏ฟฝ๐34๐ฅ43 โ
13
ln ๐ฅ + ๐๏ฟฝ
= ๐๐ฅโ13 exp ๏ฟฝ๐
34๐ฅ43 ๏ฟฝ (1)
To verify, using the formula 3.4.28, which is
๐ฆ (๐ฅ) โผ ๐๐1โ๐2๐ exp ๏ฟฝ๐๏ฟฝ
๐ฅ๐ (๐ก)
1๐ ๐๐ก๏ฟฝ
In this case, ๐ = 3, since the ODE ๐ฆโฒโฒโฒ = ๐ฅ๐ฆ is third order. Here we have ๐ (๐ฅ) = ๐ฅ, therefore,
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4.3. note p1 added 2/3/2017 CHAPTER 4. STUDY NOTES
plug-in into the above gives
๐ฆ (๐ฅ) โผ ๐ (๐ฅ)1โ36 exp ๏ฟฝ๐๏ฟฝ
๐ฅ(๐ก)
13 ๐๐ก๏ฟฝ
โผ ๐๐ฅโ13 exp ๏ฟฝ๐
43๐ฅ43 ๏ฟฝ
Comparing (1) and (2), we see they are the same.
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4.4. note added 1/31/2017 CHAPTER 4. STUDY NOTES
4.4 note added 1/31/2017
This note solves in details the ODE
๐ฅ3๐ฆโฒโฒ (๐ฅ) = ๐ฆ (๐ฅ)
Using asymptoes method using what is called the dominant balance submethod where it isassumed that ๐ฆ (๐ฅ) = ๐๐(๐ฅ).
4.4.1 Solution
๐ฅ = 0 is an irregular singular point. The solution is assumeed to be ๐ฆ (๐ฅ) = ๐๐(๐ฅ). Therefore๐ฆโฒ = ๐โฒ๐๐(๐ฅ) and ๐ฆโฒโฒ = ๐โฒโฒ๐๐(๐ฅ) + (๐โฒ)2 ๐๐(๐ฅ) and the given ODE becomes
๐ฅ3 ๏ฟฝ๐โฒโฒ + (๐โฒ)2๏ฟฝ = 1 (1)
Assuming that
๐โฒ (๐ฅ) โผ ๐๐ฅ๐ผ
Hence ๐โฒโฒ โผ ๐๐ผ๐ฅ๐ผโ1. and (1) becomes
๐ฅ3 ๏ฟฝ๐๐ผ๐ฅ๐ผโ1 + (๐๐ฅ๐ผ)2๏ฟฝ โผ 1๐๐ผ๐ฅ๐ผ+2 + ๐2๐ฅ2๐ผ+3 โผ 1
Term ๐๐ผ๐ฅ๐ผ+2 โ ๐2๐ฅ2๐ผ+3, hence we set ๐ผ = โ32 to remove the subdominant term. Therefore
the above becomes, after substituting for the found ๐ผ๐ฅโ0
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ32๐๐ฅ
12 + ๐2 โผ 1
๐2 = 1
Therefore ๐ = ยฑ1. The result so far is ๐โฒ (๐ฅ) โผ ๐๐ฅโ32 . Now another term is added. Let
๐โฒ (๐ฅ) โผ ๐๐ฅโ32 + ๐ด (๐ฅ)
Now we will try to find ๐ด (๐ฅ). Hence ๐โฒโฒ (๐ฅ) โผ โ32 ๐๐ฅ
โ52 +๐ดโฒ and ๐ฅ3 ๏ฟฝ๐โฒโฒ + (๐โฒ)2๏ฟฝ = 1 now becomes
๐ฅ3 ๏ฟฝโ32๐๐ฅ
โ52 + ๐ดโฒ + ๏ฟฝ๐๐ฅ
โ32 + ๐ด๏ฟฝ
2
๏ฟฝ โผ 1
๐ฅ3 ๏ฟฝโ32๐๐ฅ
โ52 + ๐ดโฒ + ๐2๐ฅโ3 + ๐ด2 + 2๐ด๐๐ฅ
โ32 ๏ฟฝ โผ 1
๏ฟฝโ32๐๐ฅ
12 + ๐ฅ3๐ดโฒ + ๐2 + ๐ฅ3๐ด2 + 2๐ด๐๐ฅ
32 ๏ฟฝ โผ 1
Since ๐2 = 1 from the above, thenโ32๐๐ฅ
12 + ๐ฅ3๐ดโฒ + ๐ฅ3๐ด2 + 2๐ด๐๐ฅ
32 โผ 0
220
4.4. note added 1/31/2017 CHAPTER 4. STUDY NOTES
Dominant balance says to keep dominant term (but now looking at those terms in ๐ด only).From the above, since ๐ดโ ๐ด2 and ๐ด โ ๐ดโฒ then from the above, we can cross out ๐ด2 and๐ดโฒ resulting in
โ32๐๐ฅ
12 + 2๐ด๐๐ฅ
32 โผ 0
Hence we just need to find ๐ด to balance the aboveโ32๐๐ฅ
12 + 2๐ด๐๐ฅ
32 โผ 0
2๐ด๐๐ฅ32 โผ
32๐๐ฅ
12
๐ด โผ34๐ฅ
We found ๐ด (๐ฅ) for the second term. Therefore, so far we have
๐โฒ (๐ฅ) = ๐๐ฅโ32 +
34๐ฅ
Or
๐ (๐ฅ) = โ2๐๐ฅโ12 +
34
ln ๐ฅ + ๐ถ0
But ๐ถ0 can be dropped (subdominant to ln ๐ฅ when ๐ฅ โ 0) and so far then we can write thesolution as
๐ฆ (๐ฅ) = ๐๐(๐ฅ)๐(๐ฅ)
= ๐๐(๐ฅ)โ๏ฟฝ๐=0
๐๐๐ฅ๐๐
= exp ๏ฟฝโ2๐๐ฅโ12 +
34
ln ๐ฅ๏ฟฝโ๏ฟฝ๐=0
๐๐๐ฅ๐๐
= ๐โ2๐๐ฅโ12 ๐ฅ
34
โ๏ฟฝ๐=0
๐๐๐ฅ๐๐
= ๐โ 2๐โ๐ฅ
โ๏ฟฝ๐=0
๐๐๐ฅ๐๐+ 3
4
= ๐ยฑ 2โ๐ฅ
โ๏ฟฝ๐=0
๐๐๐ฅ๐๐+ 3
4
Since ๐ = ยฑ1. We can now try adding one more term to ๐ (๐ฅ). Let
๐โฒ (๐ฅ) = ๐๐ฅโ32 +
34๐ฅ+ ๐ต (๐ฅ)
Hence
๐โฒโฒ =โ32๐๐ฅ
โ52 โ
34๐ฅ2
+ ๐ตโฒ (๐ฅ)
221
4.4. note added 1/31/2017 CHAPTER 4. STUDY NOTES
And ๐ฅ3 ๏ฟฝ๐โฒโฒ + (๐โฒ)2๏ฟฝ โผ 1 now becomes
๐ฅ3โโโโโโ๏ฟฝโ32๐๐ฅ
โ52 โ
34๐ฅ2
+ ๐ตโฒ (๐ฅ)๏ฟฝ + ๏ฟฝ๐๐ฅโ32 +
34๐ฅ+ ๐ต (๐ฅ)๏ฟฝ
2โโโโโโ โผ 1
๐ฅ3 ๏ฟฝ๐2
๐ฅ3โ316๐ฅโ2 + 2๐๐ต๐ฅโ
32 +
32๐ต๐ฅโ1 + ๐ต2 + ๐ตโฒ๏ฟฝ โผ 1
โโโโโโ=1โ๐2 โ
316๐ฅ + 2๐๐ต๐ฅ
32 +
32๐ต๐ฅ2 + ๐ฅ3๐ต2 + ๐ฅ3๐ตโฒ
โโโโโโ โผ 1
โ316๐ฅ + 2๐๐ต๐ฅ
32 +
32๐ต๐ฅ2 + ๐ฅ3๐ต2 + ๐ฅ3๐ตโฒ โผ 0
From the above, since ๐ต (๐ฅ) โ ๐ต2 (๐ฅ) and ๐ต (๐ฅ) โ ๐ตโฒ (๐ฅ) and for small ๐ฅ, then we can crossout terms with ๐ต2 and ๐ตโฒ from above, and we are left with
โ316๐ฅ + 2๐๐ต๐ฅ
32 +
32๐ต๐ฅ2 โฝ 0
Between 2๐๐ต๐ฅ32 and 3
2๐ต๐ฅ2, for small ๐ฅ, then 2๐๐ต๐ฅ
32 โ 3
2๐ต๐ฅ2, so we can cross out 3
2๐ต๐ฅ2 from
above
โ316๐ฅ + 2๐๐ต๐ฅ
32 โฝ 0
2๐๐ต๐ฅ32 โฝ
316๐ฅ
๐ต โฝ332๐
๐ฅโ12
We found ๐ต (๐ฅ), Hence now we have
๐โฒ (๐ฅ) = ๐๐ฅโ32 +
34๐ฅ+
332๐
๐ฅโ12
Or
๐ (๐ฅ) = โ2๐๐ฅโ12 +
34
ln ๐ฅ + 316๐
๐ฅ12 + ๐ถ1
But ๐ถ1 can be dropped (subdominant to ln ๐ฅ when ๐ฅ โ 0) and so far then we can write thesolution as
๐ฆ (๐ฅ) = ๐๐(๐ฅ)๐(๐ฅ)
= ๐๐(๐ฅ)โ๏ฟฝ๐=0
๐๐๐ฅ๐๐
= exp ๏ฟฝโ2๐๐ฅโ12 +
34
ln ๐ฅ + 316๐
๐ฅ12 ๏ฟฝ
โ๏ฟฝ๐=0
๐๐๐ฅ๐๐
= ๐โ2๐๐ฅโ12 + 3
16๐๐ฅ12 ๐ฅ
34
โ๏ฟฝ๐=0
๐๐๐ฅ๐๐
= ๐โ2๐๐ฅโ12 + 3
16๐๐ฅ12
โ๏ฟฝ๐=0
๐๐๐ฅ๐๐+ 3
4
222
4.4. note added 1/31/2017 CHAPTER 4. STUDY NOTES
For ๐ = 1
๐ฆ1 (๐ฅ) = ๐โ2๐ฅ
โ12 + 3
16๐ฅ12
โ๏ฟฝ๐=0
๐๐๐ฅ๐๐+ 3
4
For ๐ = โ1
๐ฆ2 (๐ฅ) = ๐2๐ฅ
โ12 โ 3
16๐ฅ12
โ๏ฟฝ๐=0
๐๐๐ฅ๐๐+ 3
4
Hence
๐ฆ (๐ฅ) โผ ๐ด๐ฆ1 (๐ฅ) + ๐ต๐ฆ2 (๐ฅ)
Reference
1. Page 80-82 Bender and Orszag textbook.
2. Lecture notes, Lecture 5, Tuesday Janurary 31, 2017. EP 548, University of Wisconsin,Madison by Professor Smith.
3. Lecture notes from http://www.damtp.cam.ac.uk/
223
4.4. note added 1/31/2017 CHAPTER 4. STUDY NOTES
224
4.5. note p3 figure 3.4 in text (page 92) . . . CHAPTER 4. STUDY NOTES
4.5 note p3 ๏ฟฝgure 3.4 in text (page 92) reproduced incolor
figure 3.4 in text (page 92) was not too clear (it is black and white in my text book that I am using). So I
reproduced it in color. This type of plot is needed to check what happens with the improvement in
approximation for problem 3.42 (a) as more terms are added to leading behavior.
It is done in Wolfram Mathematica. Used 300 terms to find y(x). We see the red line, which is leading
behavior/y(x) ratio going to 1 for large x as would be expected.
In[1]:= leading[x] := 1 2 Pi^-1 2 x^-1 4 Exp2 x^1 2;
y[x_, max_] := Sumx^n Factorial[n]^2, {n, 0, max};
LogLinearPlotEvaluateleading[x] y[x, 300], y[x, 10] y[x, 300], {x, 0.001, 70 000},
PlotRange โ All, Frame โ True, GridLines โ Automatic, GridLinesStyle โ LightGray,
PlotLegends โ {"leading behavior/y(x)", "truncated Taylor/y(x)"}, FrameLabel โ
{{None, None}, {"x", "Reproducing figure 3.4 in text book"}}, PlotStyle โ {Red, Blue}
Out[3]=
0.01 1 100 104
0.0
0.5
1.0
1.5
x
Reproducing figure 3.4 in text book
leading behavior/y(x)
truncated Taylor/y(x)
Printed by Wolfram Mathematica Student Edition
225
4.5. note p3 figure 3.4 in text (page 92) . . . CHAPTER 4. STUDY NOTES
226
4.6. note p4 added 2/15/17 animations of . . . CHAPTER 4. STUDY NOTES
4.6 note p4 added 2/15/17 animations of the solutions toODE from lecture 2/9/17 for small parameter
These are small animations of the solutions to the 2 ODEโs from lecture 2/9/17 for the small parameter.
Nasser M. Abbasi
2/20/2017, EP 548, Univ. Wisconsin Madison.
Manipulate
Labeled
Plot-1 + โ x/z
-1 + โ 1
z
, {x, 0, 2}, PlotRange โ {{0, 1.1}, {0, 1}},
GridLines โ Automatic, GridLinesStyle โ LightGray, PlotStyle โ Red,
Row[{"solution to ", TraditionalForm[ฯต y''[x] - y'[x] โฉต 0]}], Top,
{{z, 0.1, "ฯต"}, .1, 0.001, -0.001, Appearance โ "Labeled"}
ฯต 0.1
solution to ฯต yโฒโฒ(x) - yโฒ(x) 0
0.0 0.2 0.4 0.6 0.8 1.00.0
0.2
0.4
0.6
0.8
1.0
For eps*yโโ+y=0 ode, we notice global variations showing up in frequency as well as in amplitude as
eps become very small
Printed by Wolfram Mathematica Student Edition
227
4.6. note p4 added 2/15/17 animations of . . . CHAPTER 4. STUDY NOTES
Manipulate
Labeled
Plot
Sin x
z
Sin 1
z
, {x, 0, 2}, PlotRange โ {{0, 2}, {-10, 10}},
GridLines โ Automatic, GridLinesStyle โ LightGray, PlotStyle โ Red,
Style[Row[{"solution to ", TraditionalForm[z y''[x] + y[x] โฉต 0]}], 16], Top,
{{z, 0.01, "ฯต"}, 0.01, 0.0001, -0.0001}
ฯต
solution to 0.01 yโฒโฒ(x) + y(x) 0
0.5 1.0 1.5 2.0
-10
-5
0
5
10
2 p4.nb
Printed by Wolfram Mathematica Student Edition
228
4.6. note p4 added 2/15/17 animations of . . . CHAPTER 4. STUDY NOTES
First animation
second animation
229
4.6. note p4 added 2/15/17 animations of . . . CHAPTER 4. STUDY NOTES
230
4.7. note p5 added 2/15/17 animation . . . CHAPTER 4. STUDY NOTES
4.7 note p5 added 2/15/17 animation ๏ฟฝgure 9.5 in textpage 434
Animation of figure 9.5Nasser M. Abbasi, Feb 2017, EP 548 course
Animation of figure 9.5 in text page 434, compare higher order boundary layer solutions
This is an animation of the percentage error between boundary layer solution and the
exact solution for ฯตyโฒโฒ+(1+x)yโฒ+y=0 for different orders. Orders are given in text up y0,y1,y2,y3. We
see the error is smaller when order increases as what one would expect.
The percentage error is largest around x=0.1 than other regions. Why? Location of matching between
yin and yout. Due
to approximation made when doing matching? But all boundary layer solutions underestimate the exact
solution since error is negative.
The red color is the most accurate. 3rd order.
First animation
In[45]:= Clear["Global`*"];
yout[x_] := 2 1 + x;
yin[x_, eps_] :=
Plot[2 - Exp[-x / eps], {x, 0, 1}, AxesOrigin โ {0, 1}, PlotStyle โ Black];
p1 = Plot[yout[x], {x, 0, 1}, PlotStyle โ Blue];
combine[x_, eps_] := Plot[yout[x] - Exp[-x / eps], {x, 0, 1}, PlotStyle โ {Thick, Red},
AxesOrigin โ {0, 1}, GridLines โ Automatic, GridLinesStyle โ LightGray];
Animate
GridTraditionalForm
NumberForm[eps, {Infinity, 4}] HoldFormy''[x] + 1 + x y'[x] + y[x] โฉต 0,
{Show[
Legended[combine[x, eps], Style["combinbed solution", Red]],
Legended[p1, Style["Outer solution", Blue]],
Legended[yin[x, eps], Style["Inner solution", Black]]
, PlotRange โ {{0, 1}, {1, 2}}, ImageSize โ 400]
}, {eps, 0.001, .1, .00001}
Printed by Wolfram Mathematica Student Edition
231
4.7. note p5 added 2/15/17 animation . . . CHAPTER 4. STUDY NOTES
second animation, y0 vs. exactIn[51]:= Clear["Global`*"];
yout[x_] := 2 1 + x;
sol =
y[x] /. First@DSolveฯต y''[x] + 1 + x y'[x] + y[x] โฉต 0, y[0] โฉต 1, y[1] โฉต 1, y[x], x;
exact[x_, ฯต_] := Evaluate@sol;
yin[x_, eps_] :=
Plot[2 - Exp[-x / eps], {x, 0, 1}, AxesOrigin โ {0, 1}, PlotStyle โ Black];
p1 = Plot[yout[x], {x, 0, 1}, PlotStyle โ Blue];
combine[x_, eps_] := Plot[yout[x] - Exp[-x / eps], {x, 0, 1}, PlotStyle โ {Thick, Red},
AxesOrigin โ {0, 1}, GridLines โ Automatic, GridLinesStyle โ LightGray];
Animate
Grid
TraditionalForm
NumberForm[ฯต, {Infinity, 4}] HoldFormy''[x] + 1 + x y'[x] + y[x] โฉต 0,
{Row[{"exact solution ", TraditionalForm[sol]}]},
{Show[
Legended[combine[x, ฯต], Style["Boundary layer solution, zero order", Red]],
Legended[Plot[Evaluate@exact[x, ฯต], {x, 0, 1}, AxesOrigin โ {0, 0}, PlotStyle โ Blue],
Style["Exact analytical solution", Blue]]
, PlotRange โ {{0, 1}, {1, 2}}, ImageSize โ 400]
}, {ฯต, 0.0001, .1, .00001}
2 p5.nb
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232
4.7. note p5 added 2/15/17 animation . . . CHAPTER 4. STUDY NOTES
add more terms
In[60]:= Clear["Global`*"];
sol =
y[x] /. First@DSolveฯต y''[x] + 1 + x y'[x] + y[x] โฉต 0, y[0] โฉต 1, y[1] โฉต 1, y[x], x;
exact[x_, ฯต_] := Evaluate@sol;
boundary0[x_, eps_] :=2
1 + x- Exp[-x / eps] ;
boundary1[x_, eps_] :=
2
1 + x- Exp[-x / eps] + eps
2
1 + x3-
1
2 1 + x+
1
2(x / eps)2 -
3
2Exp[-x / eps] ;
boundary2[x_, eps_] :=2
1 + x- Exp[-x / eps] +
eps2
1 + x3-
1
2 1 + x+
1
2(x / eps)2 -
3
2Exp[-x / eps] +
eps26
1 + x5-
1
2 1 + x3-
1
4 1 + x-
1
8(x / eps)4 -
3
4(x / eps)2 +
21
4Exp[-x / eps] ;
boundary3[x_, eps_] :=2
1 + x- Exp[-x / eps] +
eps2
1 + x3-
1
2 1 + x+
1
2(x / eps)2 -
3
2Exp[-x / eps] +
eps26
1 + x5-
1
2 1 + x3-
1
4 1 + x-
1
8(x / eps)4 -
3
4(x / eps)2 +
21
4Exp[-x / eps] +
eps330
1 + x7-
3
2 1 + x5-
1
4 1 + x3-
5
16 1 + x+
1
48(x / eps)6 -
3
16(x / eps)4 +
21
8(x / eps)2 -
1949
72Exp[-x / eps] ;
p5.nb 3
Printed by Wolfram Mathematica Student Edition
233
4.7. note p5 added 2/15/17 animation . . . CHAPTER 4. STUDY NOTES
In[67]:= Animate
ex = exact[x, ฯต];
Grid
{Style[
"percentage relative error, exact vs. boundary layer for different order", 14]},
Show
LegendedPlot100 *boundary3[x, ฯต] - ex
ex, {x, 0, 1},
PlotStyle โ {Thick, Red}, AxesOrigin โ {0, 1}, GridLines โ Automatic,
GridLinesStyle โ LightGray, PlotRange โ {Automatic, {-4, .5}}, Frame โ True,
Epilog โ Text[Style[Row[{"ฯต = ", NumberForm[ฯต, {Infinity, 4}]}], 16], {.8, -3}],
Style"Boundary layer 3rd order error", Red,
LegendedPlot100 *boundary2[x, ฯต] - ex
ex, {x, 0, 1}, PlotStyle โ {Thick, Black},
AxesOrigin โ {0, 1}, GridLines โ Automatic, GridLinesStyle โ LightGray,
Style"Boundary layer 2nd order error", Black,
LegendedPlot100 *boundary1[x, ฯต] - ex
ex, {x, 0, 1}, PlotStyle โ {Thick, Blue},
AxesOrigin โ {0, 1}, GridLines โ Automatic, GridLinesStyle โ LightGray,
Style"Boundary layer 1st order error", Blue,
LegendedPlot100 *boundary0[x, ฯต] - ex
ex, {x, 0, 1}, PlotStyle โ {Thick, Magenta},
AxesOrigin โ {0, 1}, GridLines โ Automatic, GridLinesStyle โ LightGray,
Style["Boundary layer 0 order error", Magenta]
, PlotRange โ {{0, 1}, {-4, .5}}, ImageSize โ 400
, {ฯต, 0.001, .1, .0001}
4 p5.nb
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234
4.7. note p5 added 2/15/17 animation . . . CHAPTER 4. STUDY NOTES
Out[67]=
ฯต
percentage relative error, exact vs. boundary layer for different order
0.0 0.2 0.4 0.6 0.8 1.0-4
-3
-2
-1
0
ฯต = 0.0656
Boundary layer 3rd order error
Boundary layer 2nd order error
Boundary layer 1st order errorBoundary layer 0 order error
p5.nb 5
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235
4.7. note p5 added 2/15/17 animation . . . CHAPTER 4. STUDY NOTES
Animation of figure 9.5 in text page 434, compare higher order boundary layer solutions
This is an animation of the percentage error between boundary layer solution and the exactsolution for ๐๐ฆโณ+(1+๐ฅ)๐ฆโฒ+๐ฆ = 0 for di๏ฟฝerent orders. Orders are given in text up ๐ฆ0, ๐ฆ1, ๐ฆ2, ๐ฆ3.We see the error is smaller when the order increases as what one would expect.
The percentage error is largest around ๐ฅ = 0.1 than other regions. Why? Location of matchingis between ๐ฆ๐๐ and ๐ฆ๐๐ข๐ก. Due to approximation made when doing matching? But all boundarylayer solutions underestimate the exact solution since error is negative.
The red color is the most accurate. 3rd order.
First animation
second animation
third animation
236
4.8. note p7 added 2/15/17, boundary layer . . . CHAPTER 4. STUDY NOTES
4.8 note p7 added 2/15/17, boundary layer problemsolved in details
This note solves
๐๐ฆโฒโฒ (๐ฅ) + (1 + ๐ฅ) ๐ฆโฒ (๐ฅ) + ๐ฆ (๐ฅ) = 0๐ฆ (0) = 1๐ฆ (1) = 1
Where ๐ is small parameter, using boundary layer theory.
4.8.0.1 Solution
Since (1 + ๐ฅ) > 0 in the domain, we expect boundary layer to be on the left. Let ๐ฆ๐๐ข๐ก (๐ฅ) bethe solution in the outer region. Starting with ๐ฆ (๐ฅ) = โโ
๐=0 ๐๐๐ฆ๐ and substituting back into
the ODE gives
๐ ๏ฟฝ๐ฆโฒโฒ0 + ๐๐ฆโฒโฒ1 +โฏ๏ฟฝ + (1 + ๐ฅ) ๏ฟฝ๐ฆโฒ0 + ๐๐ฆโฒ1 +โฏ๏ฟฝ + ๏ฟฝ๐ฆ0 + ๐๐ฆ1 +โฏ๏ฟฝ = 0
๐ (1) terms
Collecting all terms with zero powers of ๐(1 + ๐ฅ) ๐ฆโฒ0 + ๐ฆ0 = 0
The above is solved using the right side conditions, since this is where the outer region islocated. Solving the above using ๐ฆ0 (1) = 1 gives
๐ฆ๐๐ข๐ก0 (๐ฅ) =2
1 + ๐ฅNow we need to find ๐ฆ๐๐ (๐ฅ). To do this, we convert the ODE using transformation ๐ = ๐ฅ
๐ .
Hence ๐๐ฆ๐๐ฅ = ๐๐ฆ
๐๐๐๐๐๐ฅ = ๐๐ฆ
๐๐1๐ . Hence the operator ๐
๐๐ฅ โก 1๐
๐๐๐ . This means the operator ๐2
๐๐ฅ2 โก
๏ฟฝ1๐
๐๐๐๏ฟฝ ๏ฟฝ1
๐๐๐๐๏ฟฝ = 1
๐2๐2
๐๐2 . The ODE becomes
๐1๐2๐2๐ฆ (๐)๐๐2
+ (1 + ๐๐)1๐๐๐ฆ (๐)๐๐
+ ๐ฆ (๐) = 0
1๐๐ฆโฒโฒ + ๏ฟฝ
1๐+ ๐๏ฟฝ ๐ฆโฒ + ๐ฆ = 0
Plugging ๐ฆ (๐) = โโ๐=0 ๐
๐๐ฆ (๐)๐ into the above gives
1๐๏ฟฝ๐ฆโฒโฒ0 + ๐๐ฆโฒโฒ1 +โฏ๏ฟฝ + ๏ฟฝ
1๐+ ๐๏ฟฝ ๏ฟฝ๐ฆโฒ0 + ๐๐ฆโฒ1 +โฏ๏ฟฝ + ๏ฟฝ๐ฆ0 + ๐๐ฆ1 +โฏ๏ฟฝ = 0
1๐๏ฟฝ๐ฆโฒโฒ0 + ๐๐ฆโฒโฒ1 +โฏ๏ฟฝ +
1๐๏ฟฝ๐ฆโฒ0 + ๐๐ฆโฒ1 +โฏ๏ฟฝ + ๐ ๏ฟฝ๐ฆโฒ0 + ๐๐ฆโฒ1 +โฏ๏ฟฝ + ๏ฟฝ๐ฆ0 + ๐๐ฆ1 +โฏ๏ฟฝ = 0
237
4.8. note p7 added 2/15/17, boundary layer . . . CHAPTER 4. STUDY NOTES
Collecting all terms with smallest power of ๐ , which is ๐โ1 in this case, gives1๐๐ฆโฒโฒ0 +
1๐๐ฆโฒ0 = 0
๐ฆโฒโฒ0 + ๐ฆโฒ0 = 0
Let ๐ง = ๐ฆโฒ0, the above becomes
๐งโฒ + ๐ง = 0
๐ ๏ฟฝ๐๐๐ง๏ฟฝ = 0๐๐๐ง = ๐๐ง = ๐๐โ๐
Hence ๐ฆโฒ0 (๐) = ๐๐โ๐. Integrating
๐ฆ๐๐0 (๐) = โ๐๐โ๐ + ๐1 (1A)
Since ๐ is arbitrary constant, the negative sign can be removed, giving
๐ฆ๐๐0 (๐) = ๐๐โ๐ + ๐1 (1A)
This is the lowest order solution for the inner ๐ฆ๐๐ (๐). We have two boundary conditions, butwe can only use the left side one, where ๐ฆ๐๐ (๐) lives. Hence using ๐ฆ0 (๐ = 0) = 1, the abovebecomes
1 = ๐ + ๐1๐1 = 1 โ ๐
The solution (1A) becomes
๐ฆ0 (๐) = ๐๐โ๐ + (1 โ ๐)
= 1 + ๐ ๏ฟฝ๐โ๐ โ 1๏ฟฝ
Let ๐ = ๐ด0 to match the book notation.
๐ฆ0 (๐) = 1 + ๐ด0 ๏ฟฝ๐โ๐ โ 1๏ฟฝ
To find ๐ด0, we match ๐ฆ๐๐0 (๐) with ๐ฆ๐๐ข๐ก0 (๐ฅ)
lim๐โโ
1 + ๐ด0 ๏ฟฝ๐โ๐ โ 1๏ฟฝ = lim๐ฅโ0+
21 + ๐ฅ
1 โ ๐ด0 = 2๐ด0 = โ1
Hence
๐ฆ๐๐0 (๐) = 2 โ ๐โ๐
๐ (๐) terms
We now repeat the process to find ๐ฆ๐๐1 (๐) and ๐ฆ๐๐ข๐ก1 (๐ฅ) . Starting with ๐ฆ๐๐ข๐ก (๐ฅ)
๐ ๏ฟฝ๐ฆโฒโฒ0 + ๐๐ฆโฒโฒ1 +โฏ๏ฟฝ + (1 + ๐ฅ) ๏ฟฝ๐ฆโฒ0 + ๐๐ฆโฒ1 +โฏ๏ฟฝ + ๏ฟฝ๐ฆ0 + ๐๐ฆ1 +โฏ๏ฟฝ = 0
238
4.8. note p7 added 2/15/17, boundary layer . . . CHAPTER 4. STUDY NOTES
Collecting all terms with ๐1 now
๐๐ฆโฒโฒ0 + (1 + ๐ฅ) ๐๐ฆโฒ1 + ๐๐ฆ1 = 0๐ฆโฒโฒ0 + (1 + ๐ฅ) ๐ฆโฒ1 + ๐ฆ1 = 0
But we know ๐ฆ0 =2
1+๐ฅ , from above. Hence ๐ฆโฒโฒ0 =4
(1+๐ฅ)3and the above becomes
(1 + ๐ฅ) ๐ฆโฒ1 + ๐ฆ1 = โ4
(1 + ๐ฅ)3
๐ฆโฒ1 +๐ฆ11 + ๐ฅ
= โ4
(1 + ๐ฅ)4
Integrating factor ๐ = ๐โซ1
1+๐ฅ๐๐ฅ = ๐ln(1+๐ฅ) = 1 + ๐ฅ and the above becomes
๐๐๐ฅ๏ฟฝ๐๐ฆ1๏ฟฝ = โ๐
4(1 + ๐ฅ)4
๐๐๐ฅ๏ฟฝ(1 + ๐ฅ) ๐ฆ1๏ฟฝ = โ
4(1 + ๐ฅ)3
Integrating
(1 + ๐ฅ) ๐ฆ1 = โ๏ฟฝ4
(1 + ๐ฅ)3๐๐ฅ + ๐
=2
(1 + ๐ฅ)2+ ๐
Hence
๐ฆ1 (๐ฅ) =2
(1 + ๐ฅ)3+
๐1 + ๐ฅ
Applying ๐ฆ (1) = 0 (notice the boundary condition now becomes ๐ฆ (1) = 0 and not ๐ฆ (1) = 1,since we have already used ๐ฆ (1) = 1 to find leading order). From now on, all boundaryconditions will be ๐ฆ (1) = 0.
0 =2
(1 + 1)3+
๐1 + 1
๐ = โ12
Hence
๐ฆ1 (๐ฅ) =2
(1 + ๐ฅ)3โ12
1(1 + ๐ฅ)
Now we need to find ๐ฆ๐๐1 (๐). To do this, starting from
1๐๏ฟฝ๐ฆโฒโฒ0 + ๐๐ฆโฒโฒ1 +โฏ๏ฟฝ + ๏ฟฝ
1๐+ ๐๏ฟฝ ๏ฟฝ๐ฆโฒ0 + ๐๐ฆโฒ1 +โฏ๏ฟฝ + ๏ฟฝ๐ฆ0 + ๐๐ฆ1 +โฏ๏ฟฝ = 0
1๐๏ฟฝ๐ฆโฒโฒ0 + ๐๐ฆโฒโฒ1 +โฏ๏ฟฝ +
1๐๏ฟฝ๐ฆโฒ0 + ๐๐ฆโฒ1 +โฏ๏ฟฝ + ๐ ๏ฟฝ๐ฆโฒ0 + ๐๐ฆโฒ1 +โฏ๏ฟฝ + ๏ฟฝ๐ฆ0 + ๐๐ฆ1 +โฏ๏ฟฝ = 0
239
4.8. note p7 added 2/15/17, boundary layer . . . CHAPTER 4. STUDY NOTES
But now collecting all terms with ๐ (1) order, (last time, we collected terms with ๐๏ฟฝ๐โ1๏ฟฝ ).
๐ฆโฒโฒ1 + ๐ฆโฒ1 + ๐๐ฆโฒ0 + ๐ฆ0 = 0๐ฆโฒโฒ1 + ๐ฆโฒ1 = โ๐๐ฆโฒ0 โ ๐ฆ0 (1)
But we found ๐ฆ๐๐0 earlier which was
๐ฆ๐๐0 (๐) = 1 + ๐ด0 ๏ฟฝ๐โ๐ โ 1๏ฟฝ
Hence ๐ฆโฒ0 = โ๐ด0๐โ๐ and the ODE (1) becomes
๐ฆโฒโฒ1 + ๐ฆโฒ1 = ๐๐ด0๐โ๐ โ ๏ฟฝ1 + ๐ด0 ๏ฟฝ๐โ๐ โ 1๏ฟฝ๏ฟฝ
We need to solve this with boundary conditions ๐ฆ1 (0) = 0. (again, notice change in B.C. aswas mentioned above). The solution is
๐ฆ1 (๐) = โ๐ + ๐ด0 ๏ฟฝ๐ โ12๐2๐โ๐๏ฟฝ + ๐ด1 ๏ฟฝ1 โ ๐โ๐๏ฟฝ
= โ๐ + ๐ด0 ๏ฟฝ๐ โ12๐2๐โ๐๏ฟฝ โ ๐ด1 ๏ฟฝ๐โ๐ โ 1๏ฟฝ
Since ๐ด1 is arbitrary constant, and to match the book, we can call ๐ด2 = โ๐ด1 and thenrename ๐ด2 back to ๐ด1 and obtain
๐ฆ1 (๐) = โ๐ + ๐ด0 ๏ฟฝ๐ โ12๐2๐โ๐๏ฟฝ + ๐ด1 ๏ฟฝ๐โ๐ โ 1๏ฟฝ
This is to be able to follow the book. Therefore, this is what we have so far
๐ฆ๐๐ข๐ก = ๐ฆ๐๐ข๐ก0 + ๐๐ฆ๐๐ข๐ก1
=2
1 + ๐ฅ+ ๐ ๏ฟฝ
2(1 + ๐ฅ)3
โ1
2 (1 + ๐ฅ)๏ฟฝ
And
๐ฆ๐๐ (๐) = ๐ฆ๐๐0 + ๐๐ฆ๐๐1
= ๏ฟฝ1 + ๐ด0 ๏ฟฝ๐โ๐ โ 1๏ฟฝ๏ฟฝ + ๐ ๏ฟฝโ๐ + ๐ด0 ๏ฟฝ๐ โ12๐2๐โ๐๏ฟฝ + ๐ด1 ๏ฟฝ๐โ๐ โ 1๏ฟฝ๏ฟฝ
= ๐ด0๐โ๐ โ ๐๐ โ ๐๐ด1 โ ๐ด0 + ๐๐ด1๐โ๐ + ๐๐๐ด0 โ12๐2๐๐ด0๐โ๐ + 1
To find ๐ด0, ๐ด1, we match ๐ฆ๐๐ with ๐ฆ๐๐ข๐ก, therefore
lim๐โโ
๐ฆ๐๐ = lim๐ฅโ0
๐ฆ๐๐ข๐ก
Or
lim๐โโ
๏ฟฝ๐ด0๐โ๐ โ ๐๐ โ ๐๐ด1 โ ๐ด0 + ๐๐ด1๐โ๐ + ๐๐๐ด0 โ12๐2๐๐ด0๐โ๐ + 1๏ฟฝ =
lim๐ฅโ0
21 + ๐ฅ
+ ๐ ๏ฟฝ2
(1 + ๐ฅ)3โ
12 (1 + ๐ฅ)๏ฟฝ
Which simplifies to
โ๐๐ โ ๐๐ด1 โ ๐ด0 + ๐๐๐ด0 + 1 = lim๐ฅโ0
21 + ๐ฅ
+ ๐ ๏ฟฝ2
(1 + ๐ฅ)3โ
12 (1 + ๐ฅ)๏ฟฝ
240
4.8. note p7 added 2/15/17, boundary layer . . . CHAPTER 4. STUDY NOTES
It is easier now to convert the LHS to use ๐ฅ instead of ๐ so we can compare. Since ๐ = ๐ฅ๐ ,
then the above becomes
โ๐ฅ โ ๐๐ด1 โ ๐ด0 + ๐ฅ๐ด0 + 1 = lim๐ฅโ0
21 + ๐ฅ
+ ๐ ๏ฟฝ2
(1 + ๐ฅ)3โ
12 (1 + ๐ฅ)๏ฟฝ
Using Taylor series on the RHS
1 โ ๐ด0 โ ๐ฅ + ๐ด0๐ฅ + ๐ด1๐ = lim๐ฅโ0
2 ๏ฟฝ1 โ ๐ฅ + ๐ฅ2 +โฏ๏ฟฝ
+ 2๐ ๏ฟฝ1 โ ๐ฅ + ๐ฅ2 +โฏ๏ฟฝ ๏ฟฝ1 โ ๐ฅ + ๐ฅ2 +โฏ๏ฟฝ ๏ฟฝ1 โ ๐ฅ + ๐ฅ2 +โฏ๏ฟฝ โ๐2๏ฟฝ1 โ ๐ฅ + ๐ฅ2 +โฏ๏ฟฝ
Since we have terms on the the LHS of only ๐ (1) ,๐ (๐ฅ) , ๐ (๐), then we need to keep at leastterms with ๐ (1) ,๐ (๐ฅ) , ๐ (๐) on the RHS and drop terms with ๐๏ฟฝ๐ฅ2๏ฟฝ , ๐ (๐๐ฅ) , ๐ ๏ฟฝ๐2๏ฟฝ to be ableto do the matching. So in the above, RHS simplifies to
โ๐ฅ โ ๐๐ด1 โ ๐ด0 + ๐ฅ๐ด0 + 1 = 2 (1 โ ๐ฅ) + 2๐ โ๐2
โ๐ฅ โ ๐๐ด1 โ ๐ด0 + ๐ฅ๐ด0 + 1 = 2 โ 2๐ฅ + 2๐ โ๐2
โ๐๐ด1 โ ๐ด0 + ๐ฅ (๐ด0 โ 1) + 1 = 2 โ 2๐ฅ +32๐
Comparing, we see that
๐ด0 โ 1 = โ2๐ด0 = โ1
We notice this is the same ๐ด0 we found for the lowest order. This is how it should always comeout. If we get di๏ฟฝerent value, it means we made mistake. We could also match โ๐ด0 + 1 = 2which gives ๐ด0 = โ1 as well. Finally
โ๐๐ด1 =32๐
๐ด1 = โ32
So we have used matching to find all the constants for ๐ฆ๐๐. Here is the final solution so far
๐ฆ๐๐ข๐ก (๐ฅ) =
๐ฆ0๏ฟฝ21+ ๐ฅ
+ ๐
๐ฆ1๏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
2(1 + ๐ฅ)3
โ1
2 (1 + ๐ฅ)๏ฟฝ
๐ฆ๐๐ (๐) =
๐ฆ0๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ1 + ๐ด0 ๏ฟฝ๐โ๐ โ 1๏ฟฝ + ๐
๐ฆ1๏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๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ๐ + ๐ด0 ๏ฟฝ๐ โ
12๐2๐โ๐๏ฟฝ + ๐ด1 ๏ฟฝ๐โ๐ โ 1๏ฟฝ๏ฟฝ
= 1 โ ๏ฟฝ๐โ๐ โ 1๏ฟฝ + ๐ ๏ฟฝโ๐ โ ๏ฟฝ๐ โ12๐2๐โ๐๏ฟฝ โ
32๏ฟฝ๐โ๐ โ 1๏ฟฝ๏ฟฝ
=32๐ โ ๐โ๐ โ 2๐๐ โ
32๐๐โ๐ +
12๐2๐๐โ๐ + 2
241
4.8. note p7 added 2/15/17, boundary layer . . . CHAPTER 4. STUDY NOTES
In terms of ๐ฅ, since Since ๐ = ๐ฅ๐ the above becomes
๐ฆ๐๐ (๐ฅ) =32๐ โ ๐โ
๐ฅ๐ โ 2๐ฅ โ
32๐๐โ
๐ฅ๐ +
12๐ฅ2
๐๐โ
๐ฅ๐ + 2
= 2 โ 2๐ฅ +32๐ + ๐โ
๐ฅ๐ ๏ฟฝ12๐ฅ2
๐โ32๐ โ 1๏ฟฝ
Hence
๐ฆ๐ข๐๐๐๐๐๐ = ๐ฆ๐๐ + ๐ฆ๐๐ข๐ก โ ๐ฆ๐๐๐ก๐โWhere
๐ฆ๐๐๐ก๐โ = lim๐โโ
๐ฆ๐๐
= 2 โ 2๐ฅ +32๐
Hence
๐ฆ๐ข๐๐๐๐๐๐ = 2 โ 2๐ฅ +32๐ + ๐โ
๐ฅ๐ ๏ฟฝ12๐ฅ2
๐โ32๐ โ 1๏ฟฝ +
21 + ๐ฅ
+ ๐ ๏ฟฝ2
(1 + ๐ฅ)3โ
12 (1 + ๐ฅ)๏ฟฝ
โ ๏ฟฝ2 โ 2๐ฅ +32๐๏ฟฝ
= ๐โ๐ฅ๐ ๏ฟฝ12๐ฅ2
๐โ32๐ โ 1๏ฟฝ +
21 + ๐ฅ
+ ๐ ๏ฟฝ2
(1 + ๐ฅ)3โ
12 (1 + ๐ฅ)๏ฟฝ
= ๏ฟฝ2
1 + ๐ฅโ ๐โ
๐ฅ๐ +
12๐ฅ2
๐๐โ
๐ฅ๐ ๏ฟฝ + ๐ ๏ฟฝ
2(1 + ๐ฅ)3
โ1
2 (1 + ๐ฅ)โ32๐โ
๐ฅ๐ ๏ฟฝ
Which is the same as
๐ฆ๐ข๐๐๐๐๐๐ = ๏ฟฝ2
1 + ๐ฅโ ๐โ๐ +
12๐ฅ2
๐๐โ๐๏ฟฝ + ๐ ๏ฟฝ
2(1 + ๐ฅ)3
โ1
2 (1 + ๐ฅ)โ32๐โ๐๏ฟฝ
= ๏ฟฝ2
1 + ๐ฅโ ๐โ๐๏ฟฝ + ๐ ๏ฟฝ
2(1 + ๐ฅ)3
โ1
2 (1 + ๐ฅ)โ32๐โ๐ +
12๐2๐โ๐๏ฟฝ
= ๏ฟฝ2
1 + ๐ฅโ ๐โ๐๏ฟฝ + ๐ ๏ฟฝ
2(1 + ๐ฅ)3
โ1
2 (1 + ๐ฅ)+ ๏ฟฝ
12๐2 โ
32๏ฟฝ๐โ๐๏ฟฝ (1)
Comparing (1) above, with book result in first line of 9.3.16, page 433, we see the sameresult.
4.8.0.2 References
1. Advanced Mathematica methods, Bender and Orszag. Chapter 9.
2. Lecture notes. Feb 16, 2017. By Professor Smith. University of Wisconsin. NE 548
242
4.9. note p9. added 2/24/17, asking . . . CHAPTER 4. STUDY NOTES
4.9 note p9. added 2/24/17, asking question aboutproblem in book
How did the book, on page 429, near the end, arrive at ๐ด1 = โ๐ ? I am not able to see it.This is what I tried. The book does things little di๏ฟฝerent than what we did in class. Thebook does not do
lim๐โโ
๐ฆ๐๐ โผ lim๐ฅโ0
๐ฆ๐๐ข๐ก
But instead, book replaces ๐ฅ in the ๐ฆ๐๐ข๐ก (๐ฅ) solution already obtained, with ๐๐, and rewrites๐ฆ๐๐ข๐ก (๐ฅ), which is what equation (9.2.14) is. So following this, I am trying to verify the bookresult for ๐ด1 = โ๐, but do not see how. Using the book notation, of using ๐ in place of ๐,we have
๐1 (๐) = (๐ด1 + ๐ด0) ๏ฟฝ1 โ ๐โ๐๏ฟฝ โ ๐๐
Which is the equation in the book just below 9.2.14. The goal now is to find ๐ด1. Book alreadyfound ๐ด0 = ๐ earlier. So we write
lim๐โโ
๐1(๐)
๏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(๐ด1 + ๐ด0) ๏ฟฝ1 โ ๐โ๐๏ฟฝ โ ๐๐ โผ ๐ฆ๐๐ข๐ก (๐ฅ)
lim๐โโ
(๐ด1 + ๐ด0) ๏ฟฝ1 โ ๐โ๐๏ฟฝ โ ๐๐ โผ ๐ ๏ฟฝ1 โ ๐๐ +๐2๐2
2!โโฏ๏ฟฝ
So far so good. But now the book says "comparing ๐1 (๐ฅ) when ๐ โ โ with the second termin 9.2.14 gives ๐ด1 = โ๐". But how? If we take ๐ โ โ on the LHS above, we get
lim๐โโ
(๐ด1 + ๐ด0) โ ๐๐ โผ ๐ ๏ฟฝ1 โ ๐๐ +๐๐2
2!โโฏ๏ฟฝ
But ๐ด0 = ๐, so
lim๐โโ
๐ด1 + ๐ โ ๐๐ โผ ๐ โ ๐๐๐ + ๐๐๐2
2!โโฏ
lim๐โโ
๐ด1 โ ๐๐ โผ โ๐๐๐ + ๐๐๐2
2!โโฏ
How does the above says that ๐ด1 = โ๐ ? If we move โ๐๐ to the right sides, it becomes
๐ด1 โผ ๐๐ โ ๐๐๐ + ๐๐๐2
2!โโฏ
๐ด1 โผ ๐ (๐ โ ๐๐) + ๐๐๐2
2!โโฏ
I do not see how ๐ด1 = โ๐. Does any one see how to get ๐ด1 = โ๐?
Let redo this using the class method
243
4.9. note p9. added 2/24/17, asking . . . CHAPTER 4. STUDY NOTES
lim๐โโ
๐1(๐)
๏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(๐ด1 + ๐ด0) ๏ฟฝ1 โ ๐โ๐๏ฟฝ โ ๐๐ โผ lim๐ฅโ0
๐ฆ๐๐ข๐ก (๐ฅ)
lim๐โโ
(๐ด1 + ๐ด0) ๏ฟฝ1 โ ๐โ๐๏ฟฝ โ ๐๐ โผ lim๐ฅโ0
๐ ๏ฟฝ1 โ ๐ฅ +๐ฅ2
2!โโฏ๏ฟฝ
lim๐โโ
๐ด1 + ๐ โ ๐๐ โผ ๐
lim๐โโ
๐ด1 โผ ๐๐
How does the above says that ๐ด1 = โ๐? and what happend to the lim๐โโ of ๐ which remainsthere?
244
4.10. note p11. added 3/7/17, Showing . . . CHAPTER 4. STUDY NOTES
4.10 note p11. added 3/7/17, Showing that scaling fornormalization is same for any ๐
This small computation verifies that normalization constant for the S-L from lecture 3/2/2017for making all eigenfunction orthonormal is the same for each ๐. Its numerical value is0.16627. Here is the table generated for ๐ = 1, 2, โฆ , 6.
245
4.10. note p11. added 3/7/17, Showing . . . CHAPTER 4. STUDY NOTES
246
4.11. note p12. added 3/8/17, comparing . . . CHAPTER 4. STUDY NOTES
4.11 note p12. added 3/8/17, comparing exact solutionto WKB solution using Mathematica
Comparing Exact to WKB solution
for ODE in lecture 3/2/2017by Nasser M. Abbasi EP 548, Spring 2017.
This note shows how to obtain exact solution for the ODE given in lecture 3/2/2017, EP 548, and to
compare it to the WKB solution for different modes. This shows that the WKB becomes very close to
the exact solution for higher modes.
Obtain the exact solution, in terms of BesselJ functionsIn[16]:= ClearAll[n, c, y, m, lam];
lam[n_] :=9 n^2
49 Pi^4; (*eigenvalues from WKB solution*)
c =6
7 Pi^3; (*normalization value found for WKB*)
y[n_, x_] := c1
Pi + xSinn x^3 + 3 x^2 Pi + 3 Pi^2 x 7 Pi^2;
(*WKB solution found*)
Find exact solution
In[18]:= ode = y''[x] + lam x + Pi^4 y[x] โฉต 0;
solExact = y[x] /. First@DSolve[{ode, y[0] โฉต 0}, y[x], x] // TraditionalForm
Out[19]//TraditionalForm=
1
66
lam24
J 16
lam ฯ3
3
c1 ฮ5
6 lam (x + ฯ)4
8J 1
6
lam ฯ3
3J-
16
lam x4 + 4 lam ฯ x3 + 6 lam ฯ2x
2 + 4 lam ฯ3x + lam ฯ4
34
3 lam4
-
J-
16
lam ฯ3
3J 1
6
lam x4 + 4 lam ฯ x3 + 6 lam ฯ2x
2 + 4 lam ฯ3x + lam ฯ4
34
3 lam4
Make function which normalizes the exact solution eigenfunctions and plot each
mode eigenfunction with the WKB on the same plot
Printed by Wolfram Mathematica Student Edition
247
4.11. note p12. added 3/8/17, comparing . . . CHAPTER 4. STUDY NOTES
In[65]:= compare[modeNumber_] :=
Module{solExact1, int, cFromExact, eigenvalueFromHandSolution, flip},
eigenvalueFromHandSolution = lam[modeNumber];
solExact1 = solExact /. lam โ eigenvalueFromHandSolution;
int = IntegratesolExact1^2 * x + Pi^4, {x, 0, Pi};
cFromExact = First@NSolve[int โฉต 1, C[1]];
solExact1 = solExact1 /. cFromExact;
If[modeNumber > 5, flip = -1, flip = 1];
Plot {y[modeNumber, x], flip * solExact1}, {x, 0, Pi},
PlotStyle โ {Red, Blue}, Frame -> True, FrameLabel โ {{"y(x)", None},
{"x", Row[{"Comparing exact solution with WKB for mode ", modeNumber}]}},
GridLines โ Automatic, GridLinesStyle โ LightGray, BaseStyle โ 12, ImageSize โ 310,
FrameTicks โ {Automatic, None}, 0, Pi 4, Pi 2, 3 4 Pi, Pi, None
;
Generate 4 plots, for mode 1, up to 6
These plots show that after mode 5 or 6, the two eigenfunctions are almost exact
2 p12.nb
Printed by Wolfram Mathematica Student Edition
248
4.11. note p12. added 3/8/17, comparing . . . CHAPTER 4. STUDY NOTES
In[66]:= plots = Table[compare[n], {n, 6}];
Grid[Partition[plots, 2]]
Out[67]=
0 ฯ
4
ฯ
2
3ฯ
4ฯ
0.00
0.01
0.02
0.03
x
y(x)
Comparing exact solution with WKB for mode 1
0 ฯ
4
ฯ
2
3ฯ
4ฯ
-0.03
-0.02
-0.01
0.00
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y(x)
Comparing exact solution with WKB for mode 2
0 ฯ
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ฯ
2
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-0.02
0.00
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y(x)
Comparing exact solution with WKB for mode 3
0 ฯ
4
ฯ
2
3ฯ
4ฯ
-0.02
0.00
0.02
0.04
x
y(x)
Comparing exact solution with WKB for mode 4
0 ฯ
4
ฯ
2
3ฯ
4ฯ
-0.04
-0.02
0.00
0.02
0.04
x
y(x)
Comparing exact solution with WKB for mode 5
0 ฯ
4
ฯ
2
3ฯ
4ฯ
-0.04
-0.02
0.00
0.02
0.04
x
y(x)
Comparing exact solution with WKB for mode 6
p12.nb 3
Printed by Wolfram Mathematica Student Edition
249
4.11. note p12. added 3/8/17, comparing . . . CHAPTER 4. STUDY NOTES
Generate the above again, but now using relative error between the exact and
WKB for each mode, to make it more clear
In[68]:= compareError[modeNumber_] :=
Module{solExact1, eigenvalueFromHandSolution, int, cFromExact, flip},
eigenvalueFromHandSolution = lam[modeNumber];
solExact1 = solExact /. lam โ eigenvalueFromHandSolution;
int = IntegratesolExact1^2 * x + Pi^4, {x, 0, Pi};
cFromExact = First@NSolve[int โฉต 1, C[1]];
solExact1 = solExact1 /. cFromExact;
If[modeNumber > 5, flip = -1, flip = 1];
Plot 100 * Absflip * solExact1 - y[modeNumber, x], {x, 0, Pi}, PlotStyle โ
{Red, Blue}, Frame -> True, FrameLabel โ {{"relative error percentage", None},
{"x", Row[{"Absolute error. Exact solution vs WKB for mode ", modeNumber}]}},
GridLines โ Automatic, GridLinesStyle โ LightGray, BaseStyle โ 12, ImageSize โ 310,
FrameTicks โ {Automatic, None}, 0, Pi 4, Pi 2, 3 4 Pi, Pi, None,
PlotRange โ {Automatic, {0, 0.3}}
;
In[69]:= plots = Table[compareError[n], {n, 10}]; (*let do 10 modes*)
Grid[Partition[plots, 2]]
0 ฯ
4
ฯ
2
3ฯ
4ฯ
0.00
0.05
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0.15
0.20
0.25
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x
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Absolute error. Exact solution vs WKB for mode 1
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Absolute error. Exact solution vs WKB for mode 2
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4 p12.nb
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250
4.11. note p12. added 3/8/17, comparing . . . CHAPTER 4. STUDY NOTES
Out[69]=
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Absolute error. Exact solution vs WKB for mode 5
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Absolute error. Exact solution vs WKB for mode 7
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Absolute error. Exact solution vs WKB for mode 8
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p12.nb 5
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251
4.12. Convert ODE to Liouville form CHAPTER 4. STUDY NOTES
4.12 Convert ODE to Liouville form
Handy image to remember. Thanks to http://people.uncw.edu/hermanr/mat463/ODEBook/Book/SL.pdf
252
4.12. Convert ODE to Liouville form CHAPTER 4. STUDY NOTES
253
4.13. note p13. added 3/15/17, solving . . . CHAPTER 4. STUDY NOTES
4.13 note p13. added 3/15/17, solving๐2๐ฆโณ(๐ฅ) = (๐ + ๐ฅ + ๐ฅ2)๐ฆ(๐ฅ) in Maple
> >
> >
> >
> >
> >
Define the ODEassume(a>0 and 'real');ode:=epsilon^2*diff(y(x),x$2)=(x^2+a*x)*y(x);
Solve without giving any B.C.sol:=dsolve(ode,y(x));
Solve with one B.C. at infinity givensol:=dsolve({ode,y(infinity)=0},y(x));
Now solve giving B.C. at -infinitysol:=dsolve({ode,y(-infinity)=0},y(x));
Now solve by giving both B.C.at both endssol:=dsolve({ode,y(infinity)=0,y(-infinity)=0},y(x));
254
Chapter 5
Exams
Local contents5.1 Exam 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2565.2 Exam 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284
255
5.1. Exam 1 CHAPTER 5. EXAMS
5.1 Exam 1
5.1.1 problem 3.26 (page 139)
Problem Perform local analysis solution to (๐ฅ โ 1) ๐ฆโฒโฒ โ ๐ฅ๐ฆโฒ + ๐ฆ = 0 at ๐ฅ = 1. Use the resultof this analysis to prove that a Taylor series expansion of any solution about ๐ฅ = 0 has aninfinite radius of convergence. Find the exact solution by summing the series.
solution
Writing the ODE in standard form
๐ฆโฒโฒ (๐ฅ) + ๐ (๐ฅ) ๐ฆโฒ (๐ฅ) + ๐ (๐ฅ) ๐ฆ (๐ฅ) = 0 (1)
๐ฆโฒโฒ โ๐ฅ
(๐ฅ โ 1)๐ฆโฒ +
1(๐ฅ โ 1)
๐ฆ = 0 (2)
Where ๐ (๐ฅ) = โ๐ฅ(๐ฅโ1) , ๐ (๐ฅ) =
1(๐ฅโ1) . The above shows that ๐ฅ = 1 is singular point for both ๐ (๐ฅ)
and ๐ (๐ฅ). The next step is to classify the type of the singular point. Is it regular singularpoint or irregular singular point?
lim๐ฅโ1
(๐ฅ โ 1) ๐ (๐ฅ) = lim๐ฅโ1
(๐ฅ โ 1)โ๐ฅ
(๐ฅ โ 1)= โ1
And
lim๐ฅโ1
(๐ฅ โ 1)2 ๐ (๐ฅ) = lim๐ฅโ1
(๐ฅ โ 1)21
(๐ฅ โ 1)= 0
Because the limit exist, then ๐ฅ = 1 is a regular singular point. Therefore solution is assumedto be a Frobenius power series given by
๐ฆ (๐ฅ) =โ๏ฟฝ๐=0
๐๐ (๐ฅ โ 1)๐+๐
Substituting this in the original ODE (๐ฅ โ 1) ๐ฆโฒโฒ โ ๐ฅ๐ฆโฒ + ๐ฆ = 0 gives
๐ฆโฒ (๐ฅ) =โ๏ฟฝ๐=0
(๐ + ๐) ๐๐ (๐ฅ โ 1)๐+๐โ1
๐ฆโฒโฒ (๐ฅ) =โ๏ฟฝ๐=0
(๐ + ๐) (๐ + ๐ โ 1) ๐๐ (๐ฅ โ 1)๐+๐โ2
In order to move the (๐ฅ โ 1) inside the summation, the original ODE (๐ฅ โ 1) ๐ฆโฒโฒ โ ๐ฅ๐ฆโฒ + ๐ฆ = 0is first rewritten as
(๐ฅ โ 1) ๐ฆโฒโฒ โ (๐ฅ โ 1) ๐ฆโฒ โ ๐ฆโฒ + ๐ฆ = 0 (3)
256
5.1. Exam 1 CHAPTER 5. EXAMS
Substituting the Frobenius series into the above gives
(๐ฅ โ 1)โ๏ฟฝ๐=0
(๐ + ๐) (๐ + ๐ โ 1) ๐๐ (๐ฅ โ 1)๐+๐โ2
โ (๐ฅ โ 1)โ๏ฟฝ๐=0
(๐ + ๐) ๐๐ (๐ฅ โ 1)๐+๐โ1
โโ๏ฟฝ๐=0
(๐ + ๐) ๐๐ (๐ฅ โ 1)๐+๐โ1
+โ๏ฟฝ๐=0
๐๐ (๐ฅ โ 1)๐+๐ = 0
Orโ๏ฟฝ๐=0
(๐ + ๐) (๐ + ๐ โ 1) ๐๐ (๐ฅ โ 1)๐+๐โ1
โโ๏ฟฝ๐=0
(๐ + ๐) ๐๐ (๐ฅ โ 1)๐+๐
โโ๏ฟฝ๐=0
(๐ + ๐) ๐๐ (๐ฅ โ 1)๐+๐โ1
+โ๏ฟฝ๐=0
๐๐ (๐ฅ โ 1)๐+๐ = 0
Adjusting all powers of (๐ฅ โ 1) to be the same by rewriting exponents and summation indicesgives
โ๏ฟฝ๐=0
(๐ + ๐) (๐ + ๐ โ 1) ๐๐ (๐ฅ โ 1)๐+๐โ1
โโ๏ฟฝ๐=1
(๐ + ๐ โ 1) ๐๐โ1 (๐ฅ โ 1)๐+๐โ1
โโ๏ฟฝ๐=0
(๐ + ๐) ๐๐ (๐ฅ โ 1)๐+๐โ1
+โ๏ฟฝ๐=1
๐๐โ1 (๐ฅ โ 1)๐+๐โ1 = 0
Collecting terms with same powers in (๐ฅ โ 1) simplifies the above toโ๏ฟฝ๐=0
((๐ + ๐) (๐ + ๐ โ 1) โ (๐ + ๐)) ๐๐ (๐ฅ โ 1)๐+๐โ1 โ
โ๏ฟฝ๐=1
(๐ + ๐ โ 2) ๐๐โ1 (๐ฅ โ 1)๐+๐โ1 = 0 (4)
Setting ๐ = 0 gives the indicial equation((๐ + ๐) (๐ + ๐ โ 1) โ (๐ + ๐)) ๐0 = 0
((๐) (๐ โ 1) โ ๐) ๐0 = 0
Since ๐0 โ 0 then the indicial equation is
(๐) (๐ โ 1) โ ๐ = 0๐2 โ 2๐ = 0๐ (๐ โ 2) = 0
257
5.1. Exam 1 CHAPTER 5. EXAMS
The roots of the indicial equation are therefore
๐1 = 2๐2 = 0
Each one of these roots generates a solution to the ODE. The next step is to find the solution๐ฆ1 (๐ฅ) associated with ๐ = 2. (The largest root is used first). Using ๐ = 2 in equation (4) gives
โ๏ฟฝ๐=0
((๐ + 2) (๐ + 1) โ (๐ + 2)) ๐๐ (๐ฅ โ 1)๐+1 โ
โ๏ฟฝ๐=1
๐๐๐โ1 (๐ฅ โ 1)๐+1 = 0
โ๏ฟฝ๐=0
๐ (๐ + 2) ๐๐ (๐ฅ โ 1)๐+1 โ
โ๏ฟฝ๐=1
๐๐๐โ1 (๐ฅ โ 1)๐+1 = 0 (5)
At ๐ โฅ 1, the recursive relation is found and used to generate the coe๏ฟฝcients of the Frobeniuspower series
๐ (๐ + 2) ๐๐ โ ๐๐๐โ1 = 0
๐๐ =๐
๐ (๐ + 2)๐๐โ1
Few terms are now generated to see the pattern of the series and to determine the closedform. For ๐ = 1
๐1 =13๐0
For ๐ = 2
๐2 =2
2 (2 + 2)๐1 =
2813๐0 =
112๐0
For ๐ = 3
๐3 =3
3 (3 + 2)๐2 =
315
112๐0 =
160๐0
For ๐ = 4
๐4 =4
4 (4 + 2)๐3 =
16160๐0 =
1360
๐0
And so on. From the above, the first solution becomes
๐ฆ1 (๐ฅ) =โ๏ฟฝ๐=0
๐๐ (๐ฅ โ 1)๐+2
= ๐0 (๐ฅ โ 1)2 + ๐1 (๐ฅ โ 1)
3 + ๐2 (๐ฅ โ 1)4 + ๐3 (๐ฅ โ 1)
4 + ๐4 (๐ฅ โ 1)5 +โฏ
= (๐ฅ โ 1)2 ๏ฟฝ๐0 + ๐1 (๐ฅ โ 1) + ๐2 (๐ฅ โ 1)2 + ๐3 (๐ฅ โ 1)
3 + ๐4 (๐ฅ โ 1)4 +โฏ๏ฟฝ
= (๐ฅ โ 1)2 ๏ฟฝ๐0 +13๐0 (๐ฅ โ 1) +
112๐0 (๐ฅ โ 1)
2 +160๐0 (๐ฅ โ 1)
3 +1360
๐0 (๐ฅ โ 1)4 +โฏ๏ฟฝ
= ๐0 (๐ฅ โ 1)2 ๏ฟฝ1 +
13(๐ฅ โ 1) +
112(๐ฅ โ 1)2 +
160(๐ฅ โ 1)3 +
1360
(๐ฅ โ 1)4 +โฏ๏ฟฝ (6)
To find closed form solution to ๐ฆ1 (๐ฅ), Taylor series expansion of ๐๐ฅ around ๐ฅ = 1 is found
258
5.1. Exam 1 CHAPTER 5. EXAMS
first
๐๐ฅ โ ๐ + ๐ (๐ฅ โ 1) +๐2(๐ฅ โ 1)2 +
๐3!(๐ฅ โ 1)3 +
๐4!(๐ฅ โ 1)4 +
๐5!(๐ฅ โ 1)5 +โฏ
โ ๐ + ๐ (๐ฅ โ 1) +๐2(๐ฅ โ 1)2 +
๐6(๐ฅ โ 1)3 +
๐24(๐ฅ โ 1)4 +
๐120
(๐ฅ โ 1)5 +โฏ
โ ๐ ๏ฟฝ1 + (๐ฅ โ 1) +12(๐ฅ โ 1)2 +
16(๐ฅ โ 1)3 +
124(๐ฅ โ 1)4 +
1120
(๐ฅ โ 1)5 +โฏ๏ฟฝ
Multiplying the above by 2 gives
2๐๐ฅ โ ๐ ๏ฟฝ2 + 2 (๐ฅ โ 1) + (๐ฅ โ 1)2 +13(๐ฅ โ 1)3 +
112(๐ฅ โ 1)4 +
160(๐ฅ โ 1)5 +โฏ๏ฟฝ
Factoring (๐ฅ โ 1)2 from the RHS results in
2๐๐ฅ โ ๐ ๏ฟฝ2 + 2 (๐ฅ โ 1) + (๐ฅ โ 1)2 ๏ฟฝ1 +13(๐ฅ โ 1) +
112(๐ฅ โ 1)2 +
160(๐ฅ โ 1)3 +โฏ๏ฟฝ๏ฟฝ (6A)
Comparing the above result with the solution ๐ฆ1 (๐ฅ) in (6), shows that the (6A) can be writtenin terms of ๐ฆ1 (๐ฅ) as
2๐๐ฅ = ๐ ๏ฟฝ2 + 2 (๐ฅ โ 1) + (๐ฅ โ 1)2 ๏ฟฝ๐ฆ1 (๐ฅ)
๐0 (๐ฅ โ 1)2 ๏ฟฝ๏ฟฝ
Therefore
2๐๐ฅ = ๐ ๏ฟฝ2 + 2 (๐ฅ โ 1) +๐ฆ1 (๐ฅ)๐0
๏ฟฝ
2๐๐ฅโ1 = 2 + 2 (๐ฅ โ 1) +๐ฆ1 (๐ฅ)๐0
2๐๐ฅโ1 โ 2 โ 2 (๐ฅ โ 1) =๐ฆ1 (๐ฅ)๐0
Solving for ๐ฆ1 (๐ฅ)
๐ฆ1 (๐ฅ) = ๐0 ๏ฟฝ2๐๐ฅโ1 โ 2 โ 2 (๐ฅ โ 1)๏ฟฝ
= ๐0 ๏ฟฝ2๐๐ฅโ1 โ 2 โ 2๐ฅ + 2๏ฟฝ
= ๐0 ๏ฟฝ2๐๐ฅโ1 โ 2๐ฅ๏ฟฝ
=2๐0๐๐๐ฅ โ 2๐0๐ฅ
Let 2๐0๐ = ๐ถ1 and โ2๐0 = ๐ถ2, then the above solution can be written as
๐ฆ1 (๐ฅ) = ๐ถ1๐๐ฅ + ๐ถ2๐ฅ
Now that ๐ฆ1 (๐ฅ) is found, which is the solution associated with ๐ = 2, the next step is to findthe second solution ๐ฆ2 (๐ฅ) associated with ๐ = 0. Since ๐2 โ ๐1 = 2 is an integer, the solutioncan be either case ๐ผ๐ผ(๐) (๐) or case ๐ผ๐ผ (๐) (๐๐) as given in the text book at page 72.
From equation (3.3.9) at page 72 of the text, using ๐ = 2 since ๐ = ๐2โ๐1 and where ๐ (๐ฅ) = โ๐ฅand ๐ (๐ฅ) = 1 in this problem by comparing our ODE with the standard ODE in (3.3.2) at
259
5.1. Exam 1 CHAPTER 5. EXAMS
page 70 given by
๐ฆโฒโฒ +๐ (๐ฅ)(๐ฅ โ ๐ฅ0)
๐ฆโฒ +๐ (๐ฅ)
(๐ฅ โ ๐ฅ0)2๐ฆ = 0
Expanding ๐ (๐ฅ) , ๐ (๐ฅ) in Taylor series
๐ (๐ฅ) =โ๏ฟฝ๐=0
๐๐ (๐ฅ โ 1)๐
๐ (๐ฅ) =โ๏ฟฝ๐=0
๐๐ (๐ฅ โ 1)๐
Since ๐ (๐ฅ) = โ๐ฅ in our ODE, then ๐0 = โ1 and ๐1 = โ1 and all other terms are zero. For ๐ (๐ฅ),which is just 1 in our ODE, then ๐0 = 1 and all other terms are zero. Hence
๐0 = โ1๐1 = โ1๐0 = 1๐ = 2๐ = 0
The above values are now used to evaluate RHS of 3.3.9 in order to find which case it is.(book uses ๐ผ for ๐)
0๐๐ = โ๐โ1๏ฟฝ๐=0
๏ฟฝ(๐ + ๐) ๐๐โ๐ + ๐๐โ๐๏ฟฝ ๐๐ (3.3.9)
Since ๐ = 2 the above becomes
0๐2 = โ1๏ฟฝ๐=0
๏ฟฝ(๐ + ๐) ๐2โ๐ + ๐2โ๐๏ฟฝ ๐๐
Using ๐ = 0, since this is the second root, gives
0๐2 = โ1๏ฟฝ๐=0
๏ฟฝ๐๐2โ๐ + ๐2โ๐๏ฟฝ ๐๐
= โ ๏ฟฝ๏ฟฝ0๐2โ0 + ๐2โ0๏ฟฝ ๐0 + ๏ฟฝ๐2โ1 + ๐2โ1๏ฟฝ ๐1๏ฟฝ
= โ ๏ฟฝ๏ฟฝ0๐2 + ๐2๏ฟฝ ๐0 + ๏ฟฝ๐1 + ๐1๏ฟฝ ๐1๏ฟฝ
= โ ๏ฟฝ0 + ๐2๏ฟฝ ๐0 โ ๏ฟฝ๐1 + ๐1๏ฟฝ ๐1Since ๐2 = 0, ๐1 = โ1, ๐1 = 1, therefore
0๐2 = โ (0 + 0) ๐0 โ (โ1 + 1) ๐1= 0
The above shows that this is case ๐ผ๐ผ (๐) (๐๐), because the right side of 3.3.9 is zero. This meansthe second solution ๐ฆ2 (๐ฅ) is also a Fronbenius series. If the above was not zero, the methodof reduction of order would be used to find second solution.
260
5.1. Exam 1 CHAPTER 5. EXAMS
Assuming ๐ฆ2 (๐ฅ) = โ๐๐ (๐ฅ โ 1)๐+๐, and since ๐ = 0, therefore
๐ฆ2 (๐ฅ) =โ๏ฟฝ๐=0
๐๐ (๐ฅ โ 1)๐
Following the same method used to find the first solution, this series is now used in the ODEto determine ๐๐.
๐ฆโฒ2 (๐ฅ) =โ๏ฟฝ๐=0
๐๐๐ (๐ฅ โ 1)๐โ1 =
โ๏ฟฝ๐=1
๐๐๐ (๐ฅ โ 1)๐โ1 =
โ๏ฟฝ๐=0
(๐ + 1) ๐๐+1 (๐ฅ โ 1)๐
๐ฆโฒโฒ2 (๐ฅ) =โ๏ฟฝ๐=0
๐ (๐ + 1) ๐๐+1 (๐ฅ โ 1)๐โ1 =
โ๏ฟฝ๐=1
๐ (๐ + 1) ๐๐+1 (๐ฅ โ 1)๐โ1 =
โ๏ฟฝ๐=0
(๐ + 1) (๐ + 2) ๐๐+2 (๐ฅ โ 1)๐
The ODE (๐ฅ โ 1) ๐ฆโฒโฒ โ (๐ฅ โ 1) ๐ฆโฒ โ ๐ฆโฒ + ๐ฆ = 0 now becomes
(๐ฅ โ 1)โ๏ฟฝ๐=0
(๐ + 1) (๐ + 2) ๐๐+2 (๐ฅ โ 1)๐
โ (๐ฅ โ 1)โ๏ฟฝ๐=0
(๐ + 1) ๐๐+1 (๐ฅ โ 1)๐
โโ๏ฟฝ๐=0
(๐ + 1) ๐๐+1 (๐ฅ โ 1)๐
+โ๏ฟฝ๐=0
๐๐ (๐ฅ โ 1)๐ = 0
Orโ๏ฟฝ๐=0
(๐ + 1) (๐ + 2) ๐๐+2 (๐ฅ โ 1)๐+1
โโ๏ฟฝ๐=0
(๐ + 1) ๐๐+1 (๐ฅ โ 1)๐+1
โโ๏ฟฝ๐=0
(๐ + 1) ๐๐+1 (๐ฅ โ 1)๐
+โ๏ฟฝ๐=0
๐๐ (๐ฅ โ 1)๐ = 0
Henceโ๏ฟฝ๐=1
(๐) (๐ + 1) ๐๐+1 (๐ฅ โ 1)๐ โ
โ๏ฟฝ๐=1
๐๐๐ (๐ฅ โ 1)๐ โ
โ๏ฟฝ๐=0
(๐ + 1) ๐๐+1 (๐ฅ โ 1)๐ +
โ๏ฟฝ๐=0
๐๐ (๐ฅ โ 1)๐ = 0
๐ = 0 gives
โ (๐ + 1) ๐๐+1 + ๐๐ = 0โ๐1 + ๐0 = 0
๐1 = ๐0
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5.1. Exam 1 CHAPTER 5. EXAMS
๐ โฅ 1 generates the recursive relation to find all remaining ๐๐ coe๏ฟฝcients
(๐) (๐ + 1) ๐๐+1 โ ๐๐๐ โ (๐ + 1) ๐๐+1 + ๐๐ = 0(๐) (๐ + 1) ๐๐+1 โ (๐ + 1) ๐๐+1 = ๐๐๐ โ ๐๐๐๐+1 ((๐) (๐ + 1) โ (๐ + 1)) = ๐๐ (๐ โ 1)
๐๐+1 = ๐๐(๐ โ 1)
(๐) (๐ + 1) โ (๐ + 1)Therefore the recursive relation is
๐๐+1 =๐๐๐ + 1
Few terms are generated to see the pattern and to find the closed form solution for ๐ฆ2 (๐ฅ).For ๐ = 1
๐2 = ๐112=12๐0
For ๐ = 2
๐3 =๐23=1312๐0 =
16๐0
For ๐ = 3
๐4 =๐33 + 1
=1416๐0 =
124๐0
For ๐ = 4
๐5 =๐44 + 1
=15124๐0 =
1120
๐0
And so on. Therefore, the second solution is
๐ฆ2 (๐ฅ) =โ๏ฟฝ๐=0
๐๐ (๐ฅ โ 1)๐
= ๐0 + ๐1 (๐ฅ โ 1) + ๐2 (๐ฅ โ 1)2 +โฏ
= ๐0 + ๐0 (๐ฅ โ 1) +12๐0 (๐ฅ โ 1)
2 +16๐0 (๐ฅ โ 1)
3 +124๐0 (๐ฅ โ 1)
4 +1120
๐0 (๐ฅ โ 1)5 +โฏ
= ๐0 ๏ฟฝ1 + (๐ฅ โ 1) +12(๐ฅ โ 1)2 +
16(๐ฅ โ 1)3 +
124(๐ฅ โ 1)4 +
1120
(๐ฅ โ 1)5 +โฏ๏ฟฝ (7A)
The Taylor series for ๐๐ฅ around ๐ฅ = 1 is
๐๐ฅ โ ๐ + ๐ (๐ฅ โ 1) +๐2(๐ฅ โ 1)2 +
๐6(๐ฅ โ 1)3 +
๐24(๐ฅ โ 1)4 +
๐120
(๐ฅ โ 1)5 +โฏ
โ ๐ ๏ฟฝ1 + (๐ฅ โ 1) +12(๐ฅ โ 1)2 +
16(๐ฅ โ 1)3 +
124(๐ฅ โ 1)4 +
1120
(๐ฅ โ 1)5 +โฏ๏ฟฝ (7B)
Comparing (7A) with (7B) shows that the second solution closed form is
๐ฆ2 (๐ฅ) = ๐0๐๐ฅ
๐Let ๐0
๐ be some constant, say ๐ถ3, the second solution above becomes
๐ฆ2 (๐ฅ) = ๐ถ3๐๐ฅ
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5.1. Exam 1 CHAPTER 5. EXAMS
Both solutions ๐ฆ1 (๐ฅ) , ๐ฆ2 (๐ฅ) have now been found. The final solution is
๐ฆ (๐ฅ) = ๐ฆ1 (๐ฅ) + ๐ฆ2 (๐ฅ)
=๐ฆ1(๐ฅ)
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๐ถ1๐๐ฅ + ๐ถ2๐ฅ +๐ฆ2(๐ฅ)๏ฟฝ๐ถ3๐๐ฅ
= ๐ถ4๐๐ฅ + ๐ถ2๐ฅ
Hence, the exact solution is
๐ฆ (๐ฅ) = ๐ด๐๐ฅ + ๐ต๐ฅ (7)
Where ๐ด,๐ต are constants to be found from initial conditions if given. Above solution is nowverified by substituting it back to original ODE
๐ฆโฒ = ๐ด๐๐ฅ + ๐ต๐ฆโฒโฒ = ๐ด๐๐ฅ
Substituting these into (๐ฅ โ 1) ๐ฆโฒโฒ โ ๐ฅ๐ฆโฒ + ๐ฆ = 0 gives(๐ฅ โ 1)๐ด๐๐ฅ โ ๐ฅ (๐ด๐๐ฅ + ๐ต) + ๐ด๐๐ฅ + ๐ต๐ฅ = 0๐ฅ๐ด๐๐ฅ โ ๐ด๐๐ฅ โ ๐ฅ๐ด๐๐ฅ โ ๐ฅ๐ต + ๐ด๐๐ฅ + ๐ต๐ฅ = 0
โ๐ด๐๐ฅ โ ๐ฅ๐ต + ๐ด๐๐ฅ + ๐ต๐ฅ = 00 = 0
To answer the final part of the question, the above solution (7) is analytic around ๐ฅ = 0 withinfinite radius of convergence since exp (โ ) is analytic everywhere. Writing the solution as
๐ฆ (๐ฅ) = ๏ฟฝ๐ดโ๏ฟฝ๐=0
๐ฅ๐
๐! ๏ฟฝ+ ๐ต๐ฅ
The function ๐ฅ have infinite radius of convergence, since it is its own series. And the ex-ponential function has infinite radius of convergence as known, verified by using standardratio test
๐ด lim๐โโ
๏ฟฝ๐๐+1๐๐
๏ฟฝ = ๐ด lim๐โโ
๏ฟฝ๐ฅ๐+1๐!
(๐ + 1)!๐ฅ๐ ๏ฟฝ= ๐ด lim
๐โโ๏ฟฝ๐ฅ๐!
(๐ + 1)!๏ฟฝ = ๐ด lim
๐โโ๏ฟฝ๐ฅ
๐ + 1๏ฟฝ = 0
For any ๐ฅ. Since the ratio is less than 1, then the solution ๐ฆ (๐ฅ) expanded around ๐ฅ = 0 hasan infinite radius of convergence.
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5.1. Exam 1 CHAPTER 5. EXAMS
5.1.2 problem 9.8 (page 480)
Problem Use boundary layer to find uniform approximation with error of order ๐๏ฟฝ๐2๏ฟฝ forthe problem ๐๐ฆโฒโฒ + ๐ฆโฒ + ๐ฆ = 0 with ๐ฆ (0) = ๐, ๐ฆ (1) = 1. Compare your solution to exact solution.Plot the solution for some values of ๐.
solution
๐๐ฆโฒโฒ + ๐ฆโฒ + ๐ฆ = 0 (1)
Since ๐ (๐ฅ) = 1 > 0, then a boundary layer is expected at the left side, near ๐ฅ = 0. Matchingwill fail if this was not the case. Starting with the outer solution near ๐ฅ = 1. Let
๐ฆ๐๐ข๐ก (๐ฅ) =โ๏ฟฝ๐=0
๐๐๐ฆ๐ (๐ฅ)
Substituting this into (1) gives
๐ ๏ฟฝ๐ฆโฒโฒ0 + ๐๐ฆโฒโฒ1 + ๐2๐ฆโฒโฒ2 +โฏ๏ฟฝ + ๏ฟฝ๐ฆโฒ0 + ๐๐ฆโฒ1 + ๐2๐ฆโฒ2 +โฏ๏ฟฝ + ๏ฟฝ๐ฆ0 + ๐๐ฆ1 + ๐2๐ฆ2 +โฏ๏ฟฝ = 0 (2)
Collecting powers of ๐๏ฟฝ๐0๏ฟฝ results in the ODE
๐ฆโฒ0 โผ โ๐ฆ0๐๐ฆ0๐ฆ0
โผ โ๐๐ฅ
ln ๏ฟฝ๐ฆ0๏ฟฝ โผ โ๐ฅ + ๐ถ1
๐ฆ๐๐ข๐ก0 (๐ฅ) โผ ๐ถ1๐โ๐ฅ + ๐ (๐) (3)
๐ถ1 is found from boundary conditions ๐ฆ (1) = 1. Equation (3) gives
1 = ๐ถ1๐โ1
๐ถ1 = ๐
Hence solution (3) becomes
๐ฆ๐๐ข๐ก0 (๐ฅ) โผ ๐1โ๐ฅ
๐ฆ๐๐ข๐ก1 (๐ฅ) is now found. Using (2) and collecting terms of ๐๏ฟฝ๐1๏ฟฝ gives the ODE
๐ฆโฒ1 + ๐ฆ1 โผ โ๐ฆโฒโฒ0 (4)
But
๐ฆโฒ0 (๐ฅ) = โ๐1โ๐ฅ
๐ฆโฒโฒ0 (๐ฅ) = ๐1โ๐ฅ
Using the above in the RHS of (4) gives
๐ฆโฒ1 + ๐ฆ1 โผ โ๐1โ๐ฅ
264
5.1. Exam 1 CHAPTER 5. EXAMS
The integrating factor is ๐๐ฅ, hence the above becomes๐๐๐ฅ๏ฟฝ๐ฆ1๐๐ฅ๏ฟฝ โผ โ๐๐ฅ๐1โ๐ฅ
๐๐๐ฅ๏ฟฝ๐ฆ1๐๐ฅ๏ฟฝ โผ โ๐
Integrating both sides gives
๐ฆ1๐๐ฅ โผ โ๐๐ฅ + ๐ถ2
๐ฆ๐๐ข๐ก1 (๐ฅ) โผ โ๐ฅ๐1โ๐ฅ + ๐ถ2๐โ๐ฅ (5)
Applying boundary conditions ๐ฆ (1) = 0 to the above gives
0 = โ1 + ๐ถ2๐โ1
๐ถ2 = ๐
Hence the solution in (5) becomes
๐ฆ๐๐ข๐ก1 (๐ฅ) โผ โ๐ฅ๐1โ๐ฅ + ๐1โ๐ฅ
โผ (1 โ ๐ฅ) ๐1โ๐ฅ
Therefore the outer solution is
๐ฆ๐๐ข๐ก (๐ฅ) = ๐ฆ0 + ๐๐ฆ1= ๐1โ๐ฅ + ๐ (1 โ ๐ฅ) ๐1โ๐ฅ (6)
Now the boundary layer (inner) solution ๐ฆ๐๐ (๐ฅ) near ๐ฅ = 0 is found. Let ๐ = ๐ฅ๐๐ be the inner
variable. The original ODE is expressed using this new variable, and ๐ is found. Since๐๐ฆ๐๐ฅ =
๐๐ฆ๐๐
๐๐๐๐ฅ then ๐๐ฆ
๐๐ฅ =๐๐ฆ๐๐๐
โ๐. The di๏ฟฝerential operator is ๐๐๐ฅ โก ๐
โ๐ ๐๐๐ therefore
๐2
๐๐ฅ2=๐๐๐ฅ
๐๐๐ฅ
= ๏ฟฝ๐โ๐๐๐๐๏ฟฝ ๏ฟฝ
๐โ๐๐๐๐๏ฟฝ
= ๐โ2๐๐2
๐๐2
Hence ๐2๐ฆ๐๐ฅ2 = ๐
โ2๐ ๐2๐ฆ๐๐2 and ๐๐ฆ
โฒโฒ + ๐ฆโฒ + ๐ฆ = 0 becomes
๐ ๏ฟฝ๐โ2๐๐2๐ฆ๐๐2 ๏ฟฝ
+ ๐โ๐๐๐ฆ๐๐
+ ๐ฆ = 0
๐1โ2๐๐ฆโฒโฒ + ๐โ๐๐ฆโฒ + ๐ฆ = 0 (7A)
The largest terms are ๏ฟฝ๐1โ2๐, ๐โ๐๏ฟฝ, balance gives 1 โ 2๐ = โ๐ or
๐ = 1
The ODE (7A) becomes
๐โ1๐ฆโฒโฒ + ๐โ1๐ฆโฒ + ๐ฆ = 0 (7)
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5.1. Exam 1 CHAPTER 5. EXAMS
Assuming that solution is
๐ฆ๐๐ (๐ฅ) =โ๏ฟฝ๐=0
๐๐๐ฆ๐ = ๐ฆ0 + ๐๐ฆ1 + ๐2๐ฆ2 +โฏ
Substituting the above into (7) gives
๐โ1 ๏ฟฝ๐ฆโฒโฒ0 + ๐๐ฆโฒโฒ1 +โฏ๏ฟฝ + ๐โ1 ๏ฟฝ๐ฆโฒ0 + ๐๐ฆโฒ1 +โฏ๏ฟฝ + ๏ฟฝ๐ฆ0 + ๐๐ฆ1 +โฏ๏ฟฝ = 0 (8)
Collecting terms with ๐๏ฟฝ๐โ1๏ฟฝ gives the first order ODE to solve
๐ฆโฒโฒ0 โผ โ๐ฆโฒ0Let ๐ง = ๐ฆโฒ0, the above becomes
๐งโฒ โผ โ๐ง๐๐ง๐ง
โผ โ๐๐
ln |๐ง| โผ โ๐ + ๐ถ4
๐ง โผ ๐ถ4๐โ๐
Hence
๐ฆโฒ0 โผ ๐ถ4๐โ๐
Integrating
๐ฆ๐๐0 (๐) โผ ๐ถ4๏ฟฝ๐โ๐๐๐ + ๐ถ5
โผ โ๐ถ4๐โ๐ + ๐ถ5 (9)
Applying boundary conditions ๐ฆ (0) = ๐ gives
๐ = โ๐ถ4 + ๐ถ5
๐ถ5 = ๐ + ๐ถ4
Equation (9) becomes
๐ฆ๐๐0 (๐) โผ โ๐ถ4๐โ๐ + ๐ + ๐ถ4
โผ ๐ถ4 ๏ฟฝ1 โ ๐โ๐๏ฟฝ + ๐ (10)
The next leading order ๐ฆ๐๐1 (๐) is found from (8) by collecting terms in ๐๏ฟฝ๐0๏ฟฝ, which resultsin the ODE
๐ฆโฒโฒ1 + ๐ฆโฒ1 โผ โ๐ฆ0Since ๐ฆ๐๐0 (๐) โผ ๐ถ4 ๏ฟฝ1 โ ๐โ๐๏ฟฝ + ๐, therefore ๐ฆโฒ0 โผ ๐ถ4๐โ๐ and the above becomes
๐ฆโฒโฒ1 + ๐ฆโฒ1 โผ โ๐ถ4๐โ๐
The homogenous solution is found first, then method of undetermined coe๏ฟฝcients is usedto find particular solution. The homogenous ODE is
๐ฆโฒโฒ1,โ โผ ๐ฆโฒ1,โThis was solved above for ๐ฆ๐๐0 , and the solution is
๐ฆ1,โ โผ โ๐ถ5๐โ๐ + ๐ถ6
266
5.1. Exam 1 CHAPTER 5. EXAMS
To find the particular solution, let ๐ฆ1,๐ โผ ๐ด๐๐โ๐, where ๐ was added since ๐โ๐ shows up inthe homogenous solution. Hence
๐ฆโฒ1,๐ โผ ๐ด๐โ๐ โ ๐ด๐๐โ๐
๐ฆโฒโฒ1,๐ โผ โ๐ด๐โ๐ โ ๏ฟฝ๐ด๐โ๐ โ ๐ด๐๐โ๐๏ฟฝ
โผ โ2๐ด๐โ๐ + ๐ด๐๐โ๐
Substituting these in the ODE ๐ฆโฒโฒ1,๐ + ๐ฆโฒ1,๐ โผ โ๐ถ4๐โ๐ results in
โ2๐ด๐โ๐ + ๐ด๐๐โ๐ + ๐ด๐โ๐ โ ๐ด๐๐โ๐ โผ โ๐ถ4๐โ๐
โ๐ด = โ๐ถ4
๐ด = ๐ถ4
Therefore the particular solution is
๐ฆ1,๐ โผ ๐ถ4๐๐โ๐
And therefore the complete solution is
๐ฆ๐๐1 (๐) โผ ๐ฆ1,โ + ๐ฆ1,๐โผ โ๐ถ5๐โ๐ + ๐ถ6 + ๐ถ4๐๐โ๐
Applying boundary conditions ๐ฆ (0) = 0 to the above gives
0 = โ๐ถ5 + ๐ถ6
๐ถ6 = ๐ถ5
Hence the solution becomes
๐ฆ๐๐1 (๐) โผ โ๐ถ5๐โ๐ + ๐ถ5 + ๐ถ4๐๐โ๐
โผ ๐ถ5 ๏ฟฝ1 โ ๐โ๐๏ฟฝ + ๐ถ4๐๐โ๐ (11)
The complete inner solution now becomes
๐ฆ๐๐ (๐) โผ ๐ฆ๐๐0 + ๐๐ฆ๐๐1โผ ๐ถ4 ๏ฟฝ1 โ ๐โ๐๏ฟฝ + ๐ + ๐ ๏ฟฝ๐ถ5 ๏ฟฝ1 โ ๐โ๐๏ฟฝ + ๐ถ4๐๐โ๐๏ฟฝ (12)
There are two constants that need to be determined in the above from matching with theouter solution.
lim๐โโ
๐ฆ๐๐ (๐) โผ lim๐ฅโ0
๐ฆ๐๐ข๐ก (๐ฅ)
lim๐โโ
๐ถ4 ๏ฟฝ1 โ ๐โ๐๏ฟฝ + ๐ + ๐ ๏ฟฝ๐ถ5 ๏ฟฝ1 โ ๐โ๐๏ฟฝ + ๐ถ4๐๐โ๐๏ฟฝ โผ lim๐ฅโ0
๐1โ๐ฅ + ๐ (1 โ ๐ฅ) ๐1โ๐ฅ
๐ถ4 + ๐ + ๐๐ถ5 โผ ๐ + ๐๐
The above shows that
๐ถ5 = ๐๐ถ4 + ๐ = ๐
๐ถ4 = 0
267
5.1. Exam 1 CHAPTER 5. EXAMS
This gives the boundary layer solution ๐ฆ๐๐ (๐) as
๐ฆ๐๐ (๐) โผ ๐ + ๐๐ ๏ฟฝ1 โ ๐โ๐๏ฟฝ
โผ ๐ ๏ฟฝ1 + ๐ ๏ฟฝ1 โ ๐โ๐๏ฟฝ๏ฟฝ
In terms of ๐ฅ, since ๐ = ๐ฅ๐ , the above can be written as
๐ฆ๐๐ (๐ฅ) โผ ๐ ๏ฟฝ1 + ๐ ๏ฟฝ1 โ ๐โ๐ฅ๐ ๏ฟฝ๏ฟฝ
The uniform solution is therefore
๐ฆuniform (๐ฅ) โผ ๐ฆ๐๐ (๐ฅ) + ๐ฆ๐๐ข๐ก (๐ฅ) โ ๐ฆ๐๐๐ก๐โ
โผ
๐ฆ๐๐
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๐ ๏ฟฝ1 + ๐ ๏ฟฝ1 โ ๐โ
๐ฅ๐ ๏ฟฝ๏ฟฝ +
๐ฆ๐๐ข๐ก
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๐1โ๐ฅ + ๐ (1 โ ๐ฅ) ๐1โ๐ฅ โ (๐ + ๐๐)
โผ ๐ + ๐๐ ๏ฟฝ1 โ ๐โ๐ฅ๐ ๏ฟฝ + ๐1โ๐ฅ + (๐ โ ๐๐ฅ) ๐1โ๐ฅ โ (๐ + ๐๐)
โผ ๐ + ๐๐ โ ๐๐1โ๐ฅ๐ + ๐1โ๐ฅ + ๐๐1โ๐ฅ โ ๐๐ฅ๐1โ๐ฅ โ ๐ โ ๐๐
โผ โ๐๐1โ๐ฅ๐ + ๐1โ๐ฅ + ๐๐1โ๐ฅ โ ๐๐ฅ๐1โ๐ฅ
โผ ๐1โ๐ฅ ๏ฟฝโ๐๐โ1๐ + 1 + ๐ โ ๐๐ฅ๏ฟฝ
Or
๐ฆuniform (๐ฅ) โผ ๐1โ๐ฅ ๏ฟฝ1 + ๐ ๏ฟฝ1 โ ๐ฅ โ ๐โ1๐ ๏ฟฝ๏ฟฝ
With error ๐๏ฟฝ๐2๏ฟฝ.
The above solution is now compared to the exact solution of ๐๐ฆโฒโฒ + ๐ฆโฒ + ๐ฆ = 0 with ๐ฆ (0) =๐, ๐ฆ (1) = 1. Since this is a homogenous second order ODE with constant coe๏ฟฝcient, it iseasily solved using characteristic equation.
๐๐2 + ๐ + 1 = 0
The roots are
๐ =โ๐2๐
ยฑ โ๐2 โ 4๐๐2๐
=โ12๐
ยฑ โ1 โ 4๐2๐
Therefore the solution is
๐ฆ (๐ฅ) = ๐ด๐๐1๐ฅ + ๐ต๐๐2๐ฅ
= ๐ด๐๏ฟฝโ12๐+
โ1โ4๐2๐ ๏ฟฝ๐ฅ
+ ๐ต๐๏ฟฝโ12๐ โ
โ1โ4๐2๐ ๏ฟฝ๐ฅ
= ๐ด๐โ๐ฅ2๐ ๐
โ1โ4๐2๐ ๐ฅ + ๐ต๐
โ๐ฅ2๐ ๐
โโ1โ4๐2๐ ๐ฅ
= ๐โ๐ฅ2๐ ๏ฟฝ๐ด๐
โ1โ4๐2๐ ๐ฅ + ๐ต๐
โโ1โ4๐2๐ ๐ฅ๏ฟฝ
268
5.1. Exam 1 CHAPTER 5. EXAMS
Applying first boundary conditions ๐ฆ (0) = ๐ to the above gives
๐ = ๐ด + ๐ต๐ต = ๐ โ ๐ด
Hence the solution becomes
๐ฆ (๐ฅ) = ๐โ๐ฅ2๐ ๏ฟฝ๐ด๐
โ1โ4๐2๐ ๐ฅ + (๐ โ ๐ด) ๐
โโ1โ4๐2๐ ๐ฅ๏ฟฝ
= ๐โ๐ฅ2๐ ๏ฟฝ๐ด๐
โ1โ4๐2๐ ๐ฅ + ๐1โ
โ1โ4๐2๐ ๐ฅ โ ๐ด๐
โโ1โ4๐2๐ ๐ฅ๏ฟฝ
= ๐โ๐ฅ2๐ ๏ฟฝ๐ด ๏ฟฝ๐
โ1โ4๐2๐ ๐ฅ โ ๐
โโ1โ4๐2๐ ๐ฅ๏ฟฝ + ๐1โ
โ1โ4๐2๐ ๐ฅ๏ฟฝ (13)
Applying second boundary conditions ๐ฆ (1) = 1 gives
1 = ๐โ12๐ ๏ฟฝ๐ด ๏ฟฝ๐
โ1โ4๐2๐ โ ๐
โโ1โ4๐2๐ ๏ฟฝ + ๐1โ
โ1โ4๐2๐ ๏ฟฝ
๐12๐ = ๐ด๏ฟฝ๐
โ1โ4๐2๐ โ ๐
โโ1โ4๐2๐ ๏ฟฝ + ๐1โ
โ1โ4๐2๐
๐ด =๐
12๐ โ ๐1โ
โ1โ4๐2๐
๐โ1โ4๐2๐ โ ๐
โโ1โ4๐2๐
(14)
Substituting this into (13) results in
๐ฆexact (๐ฅ) = ๐โ๐ฅ2๐ ๏ฟฝ๐ด ๏ฟฝ๐
โ1โ4๐2๐ ๐ฅ โ ๐
โโ1โ4๐2๐ ๐ฅ๏ฟฝ + ๐1โ
โ1โ4๐2๐ ๐ฅ๏ฟฝ
Where ๐ด is given in (14). hence
๐ฆexact (๐ฅ) = ๐โ๐ฅ2๐
โโโโโโโโโ
โโโโโโโโโ๐
12๐ โ ๐1โ
โ1โ4๐2๐
๐โ1โ4๐2๐ โ ๐
โโ1โ4๐2๐
โโโโโโโโโ ๏ฟฝ๐โ1โ4๐2๐ ๐ฅ โ ๐
โโ1โ4๐2๐ ๐ฅ๏ฟฝ + ๐1โ
โ1โ4๐2๐ ๐ฅ
โโโโโโโโโ
In summary
exact solution asymptotic solution
๐โ๐ฅ2๐
โโโโโโโ
โโโโโโโ
๐12๐ โ๐1โ
โ1โ4๐2๐
๐โ1โ4๐2๐ โ๐
โโ1โ4๐2๐
โโโโโโโ ๏ฟฝ๐
โ1โ4๐2๐ ๐ฅ โ ๐
โโ1โ4๐2๐ ๐ฅ๏ฟฝ + ๐1โ
โ1โ4๐2๐ ๐ฅ
โโโโโโโ ๐1โ๐ฅ + ๐๐1โ๐ฅ ๏ฟฝ1 โ ๐ฅ โ ๐โ
1๐ ๏ฟฝ + ๐ ๏ฟฝ๐2๏ฟฝ
The following plot compares the exact solution with the asymptotic solution for ๐ = 0.1
269
5.1. Exam 1 CHAPTER 5. EXAMS
Exact solution
Asymptotic
0.0 0.2 0.4 0.6 0.8 1.0
1.0
1.5
2.0
2.5
3.0
x
y(x)
Exact solution vs. two terms asymptotic ฯต = 0.1
The following plot compares the exact solution with the asymptotic solution for ๐ = 0.01.The di๏ฟฝerence was too small to notice in this case, the plot below is zoomed to be near ๐ฅ = 0
Out[21]=
Exact solution
Asymptotic
0.00 0.05 0.10 0.15 0.20
2.3
2.4
2.5
2.6
2.7
x
y(x)
Exact solution vs. two terms asymptotic ฯต = 0.01
At ๐ = 0.001, the di๏ฟฝerence between the exact and the asymptotic solution was not noticeable.Therefore, to better compare the solutions, the following plot shows the relative percentage
270
5.1. Exam 1 CHAPTER 5. EXAMS
error given by
100 ๏ฟฝ๐ฆexact โ ๐ฆuniform
๐ฆexact ๏ฟฝ%
For di๏ฟฝerent ๐.
ฯต=0.1
ฯต=0.05
ฯต=0.01
0.0 0.2 0.4 0.6 0.8 1.0
0
2
4
6
8
10
x
%relativeerror
Percentage relative error between exact solution and asymptotic solution
Some observations: The above plot shows more clearly how the di๏ฟฝerence between theexact solution and the asymptotic solution became smaller as ๐ became smaller. The plotalso shows that the boundary layer near ๐ฅ = 0 is becoming more narrow are ๐ becomessmaller as expected. It also shows that the relative error is smaller in the outer region thanin the boundary layer region. For example, for ๐ = 0.05, the largest percentage error inthe outer region was less than 1%, while in the boundary layer, very near ๐ฅ = 0, the errorgrows to about 5%. Another observation is that at the matching location, the relative errorgoes down to zero. One also notices that the matching location drifts towards ๐ฅ = 0 as ๐becomes smaller because the boundary layer is becoming more narrow. The following tablesummarizes these observations.
๐ % error near ๐ฅ = 0 apparent width of boundary layer
0.1 10 0.20.05 5 0.120.01 1 0.02
271
5.1. Exam 1 CHAPTER 5. EXAMS
5.1.3 problem 3
Problem (a) Find physical optics approximation to the eigenvalue and eigenfunctions of theSturm-Liouville problem are ๐ โ โ
โ๐ฆโฒโฒ = ๐ (sin (๐ฅ) + 1)2 ๐ฆ๐ฆ (0) = 0๐ฆ (๐) = 0
(b) What is the integral relation necessary to make the eigenfunctions orthonormal? Forsome reasonable choice of scaling coe๏ฟฝcient (give the value), plot the eigenfunctions for๐ = 5, ๐ = 20.
(c) Estimate how large ๐ should be for the relative error of less than 0.1%
solution
5.1.3.1 Part a
Writing the ODE as
๐ฆโฒโฒ + ๐ (sin (๐ฅ) + 1)2 ๐ฆ = 0Let1
๐ = 1๐2
Then the given ODE becomes
๐2๐ฆโฒโฒ (๐ฅ) + (sin (๐ฅ) + 1)2 ๐ฆ (๐ฅ) = 0 (1)
Physical optics approximation is obtained when ๐ โ โ which implies ๐ โ 0+. Since theODE is linear and the highest derivative is now multiplied by a very small parameter ๐,WKB can therefore be used to solve it. WKB starts by assuming that the solution has theform
๐ฆ (๐ฅ) โผ exp ๏ฟฝ1๐ฟ
โ๏ฟฝ๐=0
๐ฟ๐๐๐ (๐ฅ)๏ฟฝ ๐ฟ โ 0
Therefore, taking derivatives and substituting back in the ODE results in
๐ฆโฒ (๐ฅ) โผ exp ๏ฟฝ1๐ฟ
โ๏ฟฝ๐=0
๐ฟ๐๐๐ (๐ฅ)๏ฟฝ ๏ฟฝ1๐ฟ
โ๏ฟฝ๐=0
๐ฟ๐๐โฒ๐ (๐ฅ)๏ฟฝ
๐ฆโฒโฒ (๐ฅ) โผ exp ๏ฟฝ1๐ฟ
โ๏ฟฝ๐=0
๐ฟ๐๐๐ (๐ฅ)๏ฟฝ ๏ฟฝ1๐ฟ
โ๏ฟฝ๐=0
๐ฟ๐๐โฒ๐ (๐ฅ)๏ฟฝ2
+ exp ๏ฟฝ1๐ฟ
โ๏ฟฝ๐=0
๐ฟ๐๐๐ (๐ฅ)๏ฟฝ ๏ฟฝ1๐ฟ
โ๏ฟฝ๐=0
๐ฟ๐๐โฒโฒ๐ (๐ฅ)๏ฟฝ
1๐ = 1๐could also be used. But the book uses ๐2.
272
5.1. Exam 1 CHAPTER 5. EXAMS
Substituting these into (1) and canceling the exponential terms gives
๐2โโโโโโ๏ฟฝ1๐ฟ
โ๏ฟฝ๐=0
๐ฟ๐๐โฒ๐ (๐ฅ)๏ฟฝ2
+1๐ฟ
โ๏ฟฝ๐=0
๐ฟ๐๐โฒโฒ๐ (๐ฅ)โโโโโโ โผ โ (sin (๐ฅ) + 1)2
๐2
๐ฟ2๏ฟฝ๐โฒ0 + ๐ฟ๐โฒ1 +โฏ๏ฟฝ ๏ฟฝ๐โฒ0 + ๐ฟ๐โฒ1 +โฏ๏ฟฝ +
๐2
๐ฟ๏ฟฝ๐โฒโฒ0 + ๐ฟ๐โฒโฒ1 +โฏ๏ฟฝ โผ โ (sin (๐ฅ) + 1)2
๐2
๐ฟ2๏ฟฝ๏ฟฝ๐โฒ0๏ฟฝ
2+ ๐ฟ ๏ฟฝ2๐โฒ1๐โฒ0๏ฟฝ +โฏ๏ฟฝ +
๐2
๐ฟ๏ฟฝ๐โฒโฒ0 + ๐ฟ๐โฒโฒ1 +โฏ๏ฟฝ โผ โ (sin (๐ฅ) + 1)2
๏ฟฝ๐2
๐ฟ2๏ฟฝ๐โฒ0๏ฟฝ
2+2๐2
๐ฟ๐โฒ1๐โฒ0 +โฏ๏ฟฝ + ๏ฟฝ
๐2
๐ฟ๐โฒโฒ0 + ๐2๐โฒโฒ1 +โฏ๏ฟฝ โผ โ (sin (๐ฅ) + 1)2 (2)
The largest term in the left side is ๐2
๐ฟ2๏ฟฝ๐โฒ0๏ฟฝ
2. By dominant balance, this term has the same
order of magnitude as the right side โ (sin (๐ฅ) + 1)2. This implies that ๐ฟ2 is proportional to๐2. For simplicity (following the book) ๐ฟ can be taken as equal to ๐
๐ฟ = ๐
Using the above in equation (2) results in
๏ฟฝ๏ฟฝ๐โฒ0๏ฟฝ2+ 2๐๐โฒ1๐โฒ0 +โฏ๏ฟฝ + ๏ฟฝ๐๐โฒโฒ0 + ๐2๐โฒโฒ1 +โฏ๏ฟฝ โผ โ (sin (๐ฅ) + 1)2
Balance of ๐ (1) gives
๏ฟฝ๐โฒ0๏ฟฝ2โผ โ (sin (๐ฅ) + 1)2 (3)
Balance of ๐ (๐) gives
2๐โฒ1๐โฒ0 โผ โ๐โฒโฒ0 (4)
Equation (3) is solved first in order to find ๐0 (๐ฅ).
๐โฒ0 โผ ยฑ๐ (sin (๐ฅ) + 1)Hence
๐0 (๐ฅ) โผ ยฑ๐๏ฟฝ๐ฅ
0(sin (๐ก) + 1) ๐๐ก + ๐ถยฑ
โผ ยฑ๐ (๐ก โ cos (๐ก))๐ฅ0 + ๐ถยฑ
โผ ยฑ๐ (1 + ๐ฅ โ cos (๐ฅ)) + ๐ถยฑ (5)
๐1 (๐ฅ) is now found from (4) and using ๐โฒโฒ0 = ยฑ๐ cos (๐ฅ) gives
๐โฒ1 โผ โ12๐โฒโฒ0๐โฒ0
โผ โ12
ยฑ๐ cos (๐ฅ)ยฑ๐ (sin (๐ฅ) + 1)
โผ โ12
cos (๐ฅ)(sin (๐ฅ) + 1)
Hence the solution is
๐1 (๐ฅ) โผ โ12
ln (1 + sin (๐ฅ)) (6)
273
5.1. Exam 1 CHAPTER 5. EXAMS
Having found ๐0 (๐ฅ) and ๐1 (๐ฅ), the leading behavior is now obtained from
๐ฆ (๐ฅ) โผ exp ๏ฟฝ1๐ฟ
โ๏ฟฝ๐=0
๐ฟ๐๐๐ (๐ฅ)๏ฟฝ
โผ exp ๏ฟฝ1๐(๐0 (๐ฅ) + ๐๐1 (๐ฅ)) +โฏ๏ฟฝ
โผ exp ๏ฟฝ1๐๐0 (๐ฅ) + ๐1 (๐ฅ) +โฏ๏ฟฝ
The leading behavior is only the first two terms (called physical optics approximation inWKB), therefore
๐ฆ (๐ฅ) โผ exp ๏ฟฝ1๐๐0 (๐ฅ) + ๐1 (๐ฅ)๏ฟฝ
โผ exp ๏ฟฝยฑ๐๐(1 + ๐ฅ โ cos (๐ฅ)) + ๐ถยฑ โ
12
ln (1 + sin (๐ฅ))๏ฟฝ
โผ1
โ1 + sin ๐ฅexp ๏ฟฝยฑ
๐๐(1 + ๐ฅ โ cos (๐ฅ)) + ๐ถยฑ๏ฟฝ
Which can be written as
๐ฆ (๐ฅ) โผ๐ถ
โ1 + sin ๐ฅexp ๏ฟฝ
๐๐(1 + ๐ฅ โ cos (๐ฅ))๏ฟฝ โ
๐ถ
โ1 + sin ๐ฅexp ๏ฟฝ
โ๐๐(1 + ๐ฅ โ cos (๐ฅ))๏ฟฝ
In terms of sin and cos the above becomes (using the standard Euler relation simplifications)
๐ฆ (๐ฅ) โผ๐ด
โ1 + sin ๐ฅcos ๏ฟฝ
1๐(1 + ๐ฅ โ cos (๐ฅ))๏ฟฝ +
๐ต
โ1 + sin ๐ฅsin ๏ฟฝ
1๐(1 + ๐ฅ โ cos (๐ฅ))๏ฟฝ
Where ๐ด,๐ต are the new constants. But ๐ = 1๐2 , and the above becomes
๐ฆ (๐ฅ) โผ๐ด
โ1 + sin ๐ฅcos ๏ฟฝโ๐ (1 + ๐ฅ โ cos (๐ฅ))๏ฟฝ + ๐ต
โ1 + sin ๐ฅsin ๏ฟฝโ๐ (1 + ๐ฅ โ cos (๐ฅ))๏ฟฝ (7)
Boundary conditions are now applied to determine ๐ด,๐ต.
๐ฆ (0) = 0๐ฆ (๐) = 0
First B.C. applied to (7) gives (where now โผ is replaced by = for notation simplicity)
0 = ๐ด cos ๏ฟฝโ๐ (1 โ cos (0))๏ฟฝ + ๐ต sin ๏ฟฝโ๐ (1 โ cos (0))๏ฟฝ
0 = ๐ด cos (0) + ๐ต sin (0)0 = ๐ด
Hence solution (7) becomes
๐ฆ (๐ฅ) โผ๐ต
โ1 + sin ๐ฅsin ๏ฟฝโ๐ (1 + ๐ฅ โ cos (๐ฅ))๏ฟฝ
274
5.1. Exam 1 CHAPTER 5. EXAMS
Applying the second B.C. ๐ฆ (๐) = 0 to the above results in
0 =๐ต
โ1 + sin๐sin ๏ฟฝโ๐ (1 + ๐ โ cos (๐))๏ฟฝ
0 = ๐ต sin ๏ฟฝโ๐ (1 + ๐ + 1)๏ฟฝ
= ๐ต sin ๏ฟฝ(2 + ๐)โ๐๏ฟฝ
Hence, non-trivial solution implies that
(2 + ๐)๏ฟฝ๐๐ = ๐๐ ๐ = 1, 2, 3,โฏ
๏ฟฝ๐๐ =๐๐2 + ๐
The eigenvalues are
๐๐ =๐2๐2
(2 + ๐)2๐ = 1, 2, 3,โฏ
Hence ๐๐ โ ๐2 for large ๐. The eigenfunctions are
๐ฆ๐ (๐ฅ) โผ๐ต๐
โ1 + sin ๐ฅsin ๏ฟฝ๏ฟฝ๐๐ (1 + ๐ฅ โ cos (๐ฅ))๏ฟฝ ๐ = 1, 2, 3,โฏ
The solution is therefore a linear combination of the eigenfunctions
๐ฆ (๐ฅ) โผโ๏ฟฝ๐=1
๐ฆ๐ (๐ฅ)
โผโ๏ฟฝ๐=1
๐ต๐โ1 + sin ๐ฅ
sin ๏ฟฝ๏ฟฝ๐๐ (1 + ๐ฅ โ cos (๐ฅ))๏ฟฝ (7A)
This solution becomes more accurate for large ๐ or large ๐.
5.1.3.2 Part b
For normalization, the requirement is that
๏ฟฝ๐
0๐ฆ2๐ (๐ฅ)
weight
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ(sin (๐ฅ) + 1)2๐๐ฅ = 1
Substituting the eigenfunction ๐ฆ๐ (๐ฅ) solution obtained in first part in the above results in
๏ฟฝ๐
0๏ฟฝ
๐ต๐โ1 + sin ๐ฅ
sin ๏ฟฝ๏ฟฝ๐๐ (1 + ๐ฅ โ cos (๐ฅ))๏ฟฝ๏ฟฝ2
(sin (๐ฅ) + 1)2 ๐๐ฅ โผ 1
The above is now solved for constant ๐ต๐. The constant ๐ต๐ will the same for each ๐ fornormalization. Therefore any ๐ can be used for the purpose of finding the scaling constant.
275
5.1. Exam 1 CHAPTER 5. EXAMS
Selecting ๐ = 1 in the above gives
๏ฟฝ๐
0๏ฟฝ
๐ต
โ1 + sin ๐ฅsin ๏ฟฝ ๐
2 + ๐(1 + ๐ฅ โ cos (๐ฅ))๏ฟฝ๏ฟฝ
2
(sin (๐ฅ) + 1)2 ๐๐ฅ โผ 1
๐ต2๏ฟฝ๐
0
11 + sin ๐ฅ sin2 ๏ฟฝ
๐2 + ๐
(1 + ๐ฅ โ cos (๐ฅ))๏ฟฝ (sin (๐ฅ) + 1)2 ๐๐ฅ โผ 1
๐ต2๏ฟฝ๐
0sin2 ๏ฟฝ
๐2 + ๐
(1 + ๐ฅ โ cos (๐ฅ))๏ฟฝ (sin (๐ฅ) + 1) ๐๐ฅ โผ 1 (8)
Letting ๐ข = ๐2+๐
(1 + ๐ฅ โ cos (๐ฅ)), then๐๐ข๐๐ฅ
=๐
2 + ๐(1 + sin (๐ฅ))
When ๐ฅ = 0, then ๐ข = ๐2+๐
(1 + 0 โ cos (0)) = 0 and when ๐ฅ = ๐ then ๐ข = ๐2+๐
(1 + ๐ โ cos (๐)) =๐
2+๐(2 + ๐) = ๐, hence (8) becomes
๐ต2๏ฟฝ๐
0sin2 (๐ข)
2 + ๐๐
๐๐ข๐๐ฅ๐๐ฅ = 1
2 + ๐๐
๐ต2๏ฟฝ๐
0sin2 (๐ข) ๐๐ข = 1
But sin2 (๐ข) = 12 โ
12 cos 2๐ข, therefore the above becomes
2 + ๐๐
๐ต2๏ฟฝ๐
0๏ฟฝ12โ12
cos 2๐ข๏ฟฝ ๐๐ข = 1
122 + ๐๐
๐ต2 ๏ฟฝ๐ข โsin 2๐ข2 ๏ฟฝ
๐
0= 1
2 + ๐2๐
๐ต2 ๏ฟฝ๏ฟฝ๐ โsin 2๐2 ๏ฟฝ โ ๏ฟฝ0 โ
sin 02 ๏ฟฝ๏ฟฝ = 1
2 + ๐2๐
๐ต2๐ = 1
๐ต2 =2
2 + ๐Therefore
๐ต =๏ฟฝ
2๐ + 2
= 0.62369
Using the above for each ๐ต๐ in the solution obtained for the eigenfunctions in (7A), andpulling this scaling constant out of the sum results in
๐ฆnormalized โผ๏ฟฝ
2๐ + 2
โ๏ฟฝ๐=1
1
โ1 + sin ๐ฅsin ๏ฟฝ๏ฟฝ๐๐ (1 + ๐ฅ โ cos (๐ฅ))๏ฟฝ (9)
Where
๏ฟฝ๐๐ =๐๐2 + ๐
๐ = 1, 2, 3,โฏ
The following are plots for the normalized ๐ฆ๐ (๐ฅ) for ๐ values it asks to show.
276
5.1. Exam 1 CHAPTER 5. EXAMS
0 ฯ
4ฯ
23ฯ4
ฯ
-0.4
-0.2
0.0
0.2
0.4
x
y(x)
yn(x) for n =5
0 ฯ
4ฯ
23ฯ4
ฯ
-0.6
-0.4
-0.2
0.0
0.2
0.4
0.6
x
y(x)
yn(x) for n =20
The following shows the ๐ฆ (๐ฅ) as more eigenfunctions are added up to 55.
277
5.1. Exam 1 CHAPTER 5. EXAMS
0 ฯ
4ฯ
23ฯ4
ฯ
0.0
0.5
1.0
1.5
2.0
x
y(x)
sum of first 5 eigenfunction
0 ฯ
4ฯ
23ฯ4
ฯ
0
1
2
3
4
5
6
xy(x)
sum of first 15 eigenfunction
0 ฯ
4ฯ
23ฯ4
ฯ
0
2
4
6
8
10
x
y(x)
sum of first 25 eigenfunction
0 ฯ
4ฯ
23ฯ4
ฯ
0
5
10
15
x
y(x)
sum of first 35 eigenfunction
0 ฯ
4ฯ
23ฯ4
ฯ
0
5
10
15
20
x
y(x)
sum of first 45 eigenfunction
0 ฯ
4ฯ
23ฯ4
ฯ
0
5
10
15
20
25
x
y(x)
sum of first 55 eigenfunction
278
5.1. Exam 1 CHAPTER 5. EXAMS
5.1.3.3 Part c
Since approximate solution is
๐ฆ (๐ฅ) โผ exp ๏ฟฝ1๐ฟ
โ๏ฟฝ๐=0
๐ฟ๐๐๐ (๐ฅ)๏ฟฝ ๐ฟ โ 0
โผ exp ๏ฟฝ1๐ฟ๐0 (๐ฅ) + ๐1 (๐ฅ) + ๐ฟ๐2 (๐ฅ) +โฏ๏ฟฝ (1)
And the physical optics approximation includes the first two terms in the series above, thenthe relative error between physical optics and exact solution is given by ๐ฟ๐2 (๐ฅ). But ๐ฟ = ๐.Hence (1) becomes
๐ฆ (๐ฅ) โผ exp ๏ฟฝ1๐๐0 (๐ฅ) + ๐1 (๐ฅ) + ๐๐2 (๐ฅ) +โฏ๏ฟฝ
Hence the relative error must be such that
|๐๐2 (๐ฅ)|max โค 0.001 (1A)
Now ๐2 (๐ฅ) is found. From (2) in part(a)
๐2
๐ฟ2๏ฟฝ๐โฒ0 + ๐ฟ๐โฒ1 + ๐ฟ2๐โฒ2 +โฏ๏ฟฝ ๏ฟฝ๐โฒ0 + ๐ฟ๐โฒ1 + ๐ฟ2๐โฒ2 +โฏ๏ฟฝ +
๐2
๐ฟ๏ฟฝ๐โฒโฒ0 + ๐ฟ๐โฒโฒ1 + ๐ฟ2๐โฒโฒ2 +โฏ๏ฟฝ โผ โ (sin (๐ฅ) + 1)2
๐2
๐ฟ2๏ฟฝ๏ฟฝ๐โฒ0๏ฟฝ
2+ ๐ฟ ๏ฟฝ2๐โฒ1๐โฒ0๏ฟฝ + ๐ฟ2 ๏ฟฝ2๐โฒ0๐โฒ2 + ๏ฟฝ๐โฒ1๏ฟฝ
2๏ฟฝ +โฏ๏ฟฝ +
๐2
๐ฟ๏ฟฝ๐โฒโฒ0 + ๐ฟ๐โฒโฒ1 + ๐ฟ2๐โฒโฒ2 +โฏ๏ฟฝ โผ โ (sin (๐ฅ) + 1)2
๏ฟฝ๏ฟฝ๐โฒ0๏ฟฝ2+ ๐ ๏ฟฝ2๐โฒ1๐โฒ0๏ฟฝ + ๐2 ๏ฟฝ2๐โฒ0๐โฒ2 + ๏ฟฝ๐โฒ1๏ฟฝ
2๏ฟฝ +โฏ๏ฟฝ + ๏ฟฝ๐๐โฒโฒ0 + ๐2๐โฒโฒ1 + ๐3๐โฒโฒ2 +โฏ๏ฟฝ โผ โ (sin (๐ฅ) + 1)2
A balance on ๐๏ฟฝ๐2๏ฟฝ gives the ODE to solve to find ๐2
2๐โฒ0๐โฒ2 โผ โ ๏ฟฝ๐โฒ1๏ฟฝ2โ ๐โฒโฒ1 (2)
But
๐โฒ0 โผ ยฑ๐ (1 + sin (๐ฅ))
๏ฟฝ๐โฒ1๏ฟฝ2โผ ๏ฟฝโ
12
cos (๐ฅ)(sin (๐ฅ) + 1)๏ฟฝ
2
โผ14
cos2 (๐ฅ)(1 + sin (๐ฅ))2
๐โฒโฒ1 โผ โ12๐๐๐ฅ ๏ฟฝ
cos (๐ฅ)(1 + sin (๐ฅ))๏ฟฝ
โผ12 ๏ฟฝ
11 + sin (๐ฅ)๏ฟฝ
279
5.1. Exam 1 CHAPTER 5. EXAMS
Hence (2) becomes
2๐โฒ0๐โฒ2 โผ โ ๏ฟฝ๐โฒ1๏ฟฝ2โ ๐โฒโฒ1
๐โฒ2 โผ โ๏ฟฝ๏ฟฝ๐โฒ1๏ฟฝ
2+ ๐โฒโฒ1 ๏ฟฝ
2๐โฒ0
โผ โ๏ฟฝ14
cos2(๐ฅ)(1+sin(๐ฅ))2
+ 12๏ฟฝ 11+sin(๐ฅ)๏ฟฝ๏ฟฝ
ยฑ2๐ (1 + sin (๐ฅ))
โผ ยฑ๐ ๏ฟฝ14
cos2(๐ฅ)(sin(๐ฅ)+1)2
+ 12๏ฟฝ 11+sin(๐ฅ)๏ฟฝ๏ฟฝ
2 (sin (๐ฅ) + 1)
โผ ยฑ๐14 ๏ฟฝ
cos2(๐ฅ)+2(1+sin(๐ฅ))(sin(๐ฅ)+1)2
๏ฟฝ
2 (sin (๐ฅ) + 1)
โผ ยฑ๐8
cos2 (๐ฅ) + 2 (1 + sin (๐ฅ))(1 + sin (๐ฅ))3
Therefore
๐2 (๐ฅ) โผ ยฑ๐8 ๏ฟฝ
๐ฅ
0
cos2 (๐ก) + 2 (1 + sin (๐ก))(1 + sin (๐ก))3
๐๐ก
โผ ยฑ๐8 ๏ฟฝ๏ฟฝ
๐ฅ
0
cos2 (๐ก)(1 + sin (๐ก))3
๐๐ก + 2๏ฟฝ๐ฅ
0
1(1 + sin (๐ก))2
๐๐ก๏ฟฝ (3)
To do โซ๐ฅ
0cos2(๐ก)
(1+sin(๐ก))3๐๐ก, I used a lookup integration rule from tables which says โซ cos๐ (๐ก) (๐ + sin ๐ก)๐ ๐๐ก =
1(๐)(๐) cos๐+1 (๐ก) (๐ + sin ๐ก)๐, therefore using this rule the integral becomes, where now ๐ =โ3, ๐ = 2, ๐ = 1,
๏ฟฝ๐ฅ
0
cos2 ๐ก(1 + sin ๐ก)3
๐๐ก =1โ3 ๏ฟฝ
cos3 ๐ก(1 + sin ๐ก)3
๏ฟฝ๐ฅ
0
=1โ3 ๏ฟฝ
cos3 ๐ฅ(1 + sin ๐ฅ)3
โ 1๏ฟฝ
=13 ๏ฟฝ1 โ
cos3 ๐ฅ(1 + sin ๐ฅ)3
๏ฟฝ
And for โซ 1(1+sin(๐ฅ))2
๐๐ฅ, half angle substitution can be used. I do not know what other substi-
tution to use. Using CAS for little help on this, I get
๏ฟฝ๐ฅ
0
1(1 + sin ๐ก)2
๐๐ก = ๏ฟฝโcos ๐ก
3 (1 + sin ๐ก)2โ13
cos ๐ก1 + sin ๐ก๏ฟฝ
๐ฅ
0
= ๏ฟฝโcos ๐ฅ
3 (1 + sin ๐ฅ)2โ13
cos ๐ฅ1 + sin ๐ฅ๏ฟฝ โ ๏ฟฝโ
13โ13๏ฟฝ
=23โ
cos ๐ฅ3 (1 + sin ๐ฅ)2
โ13
cos ๐ฅ1 + sin ๐ฅ
280
5.1. Exam 1 CHAPTER 5. EXAMS
Hence from (3)
๐2 (๐ฅ) โผ ยฑ๐8 ๏ฟฝ
13 ๏ฟฝ1 โ
cos3 (๐ฅ)(1 + sin (๐ฅ))3
๏ฟฝ + 2 ๏ฟฝ23โ
cos ๐ฅ3 (1 + sin ๐ฅ)2
โ13
cos ๐ฅ1 + sin ๐ฅ๏ฟฝ๏ฟฝ
โผ ยฑ๐8 ๏ฟฝ
13โ13
cos3 (๐ฅ)(1 + sin (๐ฅ))3
+43โ
2 cos ๐ฅ3 (1 + sin ๐ฅ)2
โ23
cos ๐ฅ1 + sin ๐ฅ๏ฟฝ
โผ ยฑ๐24 ๏ฟฝ
1 โcos3 (๐ฅ)
(1 + sin (๐ฅ))3+ 4 โ
2 cos ๐ฅ(1 + sin ๐ฅ)2
โ2 cos ๐ฅ1 + sin ๐ฅ๏ฟฝ
Therefore, from (1A)
|๐๐2 (๐ฅ)|max โค 0.001
๏ฟฝ๐๐24 ๏ฟฝ
1 โcos3 (๐ฅ)
(1 + sin (๐ฅ))3+ 4 โ
2 cos ๐ฅ(1 + sin ๐ฅ)2
โ2 cos ๐ฅ1 + sin ๐ฅ๏ฟฝ๏ฟฝ
maxโค 0.001
124 ๏ฟฝ
๐ ๏ฟฝ1 โcos3 (๐ฅ)
(1 + sin (๐ฅ))3+ 4 โ
2 cos ๐ฅ(1 + sin ๐ฅ)2
โ2 cos ๐ฅ1 + sin ๐ฅ๏ฟฝ๏ฟฝ
maxโค 0.001
๏ฟฝ๐ ๏ฟฝ1 โcos3 (๐ฅ)
(1 + sin (๐ฅ))3+ 4 โ
2 cos ๐ฅ(1 + sin ๐ฅ)2
โ2 cos ๐ฅ1 + sin ๐ฅ๏ฟฝ๏ฟฝ
maxโค 0.024 (2)
The maximum value of ๏ฟฝ1 โ cos3(๐ฅ)(1+sin(๐ฅ))3
+ 4 โ 2 cos ๐ฅ(1+sin ๐ฅ)2
โ 2 cos ๐ฅ1+sin ๐ฅ
๏ฟฝ between ๐ฅ = 0 and ๐ฅ = ๐ is now
found and used to find ๐. A plot of the above shows the maximum is maximum at the end,at ๐ฅ = ๐ (Taking the derivative and setting it to zero to determine where the maximum iscan also be used).
In[324]:= myResult = 1 -Cos[x]3
(1 + Sin[x])3+ 4 -
2 Cos[x]
(1 + Sin[x])2-
2 Cos[x]
1 + Sin[x];
Plot[myResult, {x, 0, Pi}, PlotRange โ All, Frame โ True, GridLines โ Automatic, GridLinesStyle โ LightGray,
FrameLabel โ {{"s2(x)", None}, {"x", "Finding where maximum S2(x) is, part(c)"}}, PlotStyle โ Red, BaseStyle โ 14,
FrameTicks โ {Automatic, {{0, Pi / 4, Pi / 2, 3 / 4 Pi, Pi}, None}}, ImageSize โ 400]
Out[325]=
0 ฯ
4
ฯ
2
3ฯ
4ฯ
0
2
4
6
8
10
x
s2(x)
Finding where maximum S2(x) is, part(c)
Therefore, at ๐ฅ = ๐
๏ฟฝ1 โcos3 ๐ฅ
(1 + sin ๐ฅ)3+ 4 โ
2 cos ๐ฅ(1 + sin ๐ฅ)2
โ2 cos ๐ฅ1 + sin ๐ฅ๏ฟฝ
๐ฅ=๐
= ๏ฟฝ1 โcos3 (๐)
(1 + sin๐)3+ 4 โ
2 cos๐(1 + sin๐)2
โ2 cos๐1 + sin๐๏ฟฝ
= 10
281
5.1. Exam 1 CHAPTER 5. EXAMS
Hence (2) becomes
10๐ โค 0.024๐ โค 0.0024
But since ๐ = 1๐2 the above becomes
1
โ๐โค 0.0024
โ๐ โฅ1
0.0024โ๐ โฅ 416.67
Hence
๐ โฅ 17351.1
To find which mode this corresponds to, since ๐๐ =๐2๐2
(2+๐)2, then need to solve for ๐
17351.1 =๐2๐2
(2 + ๐)2
๐2๐2 = (17351.1) (2 + ๐)2
๐ =๏ฟฝ(17351.1) (2 + ๐)2
๐2
= 215.58
Hence the next largest integer is used
๐ = 216
To have relative error less than 0.1% compared to exact solution. Therefore using the resultobtained in (9) in part (b) the normalized solution needed is
๐ฆnormalized โผ๏ฟฝ
2๐ + 2
216๏ฟฝ๐=1
1
โ1 + sin ๐ฅsin ๏ฟฝ ๐๐
2 + ๐(1 + ๐ฅ โ cos (๐ฅ))๏ฟฝ
The following is a plot of the above solution adding all the first 216 modes for illustration.
282
5.1. Exam 1 CHAPTER 5. EXAMS
In[42]:= ClearAll[x, n, lam]
mySol[x_, max_] := Sqrt2
Pi + 2 Sum
1
Sqrt[1 + Sin[x]]Sin
n Pi
2 + Pi(1 + x - Cos[x]), {n, 1, max};
In[46]:= p[n_] := Plot[mySol[x, n], {x, 0, Pi}, PlotRange โ All, Frame โ True,
FrameLabel โ {{"y(x)", None}, {"x", Row[{"yn(x) for n =", n}]}}, BaseStyle โ 14, GridLines โ Automatic,
GridLinesStyle โ LightGray, ImageSize โ 600, PlotStyle โ Red,
FrameTicks โ { {Automatic, None}, {{0, Pi / 4, Pi / 2, 3 / 4 Pi, Pi}, None}}, PlotRange โ All]
In[47]:= p[216]
Out[47]=
0 ฯ
4
ฯ
2
3ฯ
4ฯ
0
20
40
60
80
100
x
y(x)
yn(x) for n =216
283
5.2. Exam 2 CHAPTER 5. EXAMS
5.2 Exam 2
5.2.1 problem 3
2. Consider the 1D heat equation in a semi-infinite domain:
โu
โt= ฮฝ
โ2u
โx2, x โฅ 0
with boundary conditions: u(0, t) = exp(โiฯt) and u(x, t) bounded as x โ โ. In orderto construct a real forcing, we need both positive and negative real values of ฯ. Considerthat this forcing has been and will be applied for all time. This โpure boundary valueproblemโ could be an idealization of heating the surface of the earth by the sun (periodicforcing). One could then ask, how far beneath the surface of the earth do the periodicfluctuations of the heat propagate?
(a) Consider solutions of the form u(x, t;ฯ) = exp(ikx) exp(โiฯt). Find a single expressionfor k as a function of (given) ฯ real, sgn(ฯ) and ฮฝ real.
Write u(x, t;ฯ) as a function of (given) ฯ real, sgn(ฯ) and ฮฝ real. To obtain the mostgeneral solution by superposition, one would next integrate over all values of ฯ, โโ <ฯ <โ (do not do this).
(b) The basic solution can be written as u(x, t;ฯ) = exp(โiฯt) exp(โฯx) exp(iฯ sgn(ฯ) x).Find ฯ in terms of |ฯ| and ฮฝ.
(c) Make an estimate for the propagation depth of daily temperature fluctuations.
3. Here we study the competing effects of nonlinearity and diffusion in the context ofBurgerโs equation
โu
โt+ u
โu
โx= ฮฝ
โ2u
โx2(3a)
which is the simplest model equation for diffusive waves in fluid dynamics. It can be solvedexactly using the Cole-Hopf transformation
u = โ2ฮฝฯxฯ
(3b)
as follows (with 2 steps to achieve the transformation (3b)).
(a) Let u = ฯx (where the subscript denotes partial differentiation) and integrate oncewith respect to x.
(b) Let ฯ = โ2ฮฝ ln(ฯ) to get the diffusion equation for ฯ.
(c) Solve for ฯ with ฯ(x, 0) = ฮฆ(x), โโ < x < โ. In your integral expression for ฯ, usedummy variable ฮท to facilitate the remaining parts below.
(d) Show that
ฮฆ(x) = exp
[
โ1
2ฮฝ
โซ x
xo
F (ฮฑ)dฮฑ
]
where u(x, 0) = F (x), with xo arbitrary which we will take to be positive for conveniencebelow (xo > 0).
2
(e) Write your expression for ฯ(x, t) in terms of
f(ฮท, x, t) =
โซ ฮท
xo
F (ฮฑ)dฮฑ+(xโ ฮท)2
2t.
(f) Find ฯx(x, t) and then use equation (3b) to find u(x, t).
4. (a) Use the Method of Images to solve
โu
โt= ฮฝ
โ2u
โx2+Q(x, t), 0 โค x โค L, t โฅ 0
u(x, 0) = f(x),โu
โx(0, t) = 0, u(L, t) = A
(b) For A = 0, compare your expression for the solution in (a) to the eigenfunction solution.
5. This problem is a simple model for diffraction of light passing through infinitesimallysmall slits separated by a distance 2a.
Solve the diffraction equation
โu
โt=iฮป
4ฯ
โ2u
โx2(1)
with initial source u(x, 0) = f(x) = ฮด(xโ a) + ฮด(x+ a), a > 0.
Show that the solution u(x, t) oscillates wildly, but that the intensity |u(x, t)|2 is well-behaved. The intensity |u(x, t)|2 shows that the diffraction pattern at a distance t consistsof a series of alternating bright and dark fringes with period ฮปt/(2a).
3
5.2.1.1 Part (a)
๐๐ข๐๐ก
+ ๐ข๐๐ข๐๐ฅ
= ๐๐2๐ข๐๐ฅ2
(1)
284
5.2. Exam 2 CHAPTER 5. EXAMS
Let
๐ข = โ2๐๐๐ฅ๐ฅ
=๐๐๐ฅ
๏ฟฝโ2๐ ln๐๏ฟฝ (1A)
5.2.1.2 Part(b)
Let
๐ = โ2๐ ln๐ (2)
Hence (1A) becomes
๐ข =๐๐๐ฅ๐
We now substitute the above back into (1) noting first that
๐๐ข๐๐ก
=๐๐๐ก๐๐๐๐ฅ
Interchanging the order gives
๐๐ข๐๐ก
=๐๐๐ฅ๐๐๐๐ก
=๐๐๐ฅ๐๐ก
And๐๐ข๐๐ฅ
=๐๐๐ฅ๐๐๐๐ฅ
= ๐๐ฅ๐ฅ
And๐2๐ข๐๐ฅ2
= ๐๐ฅ๐ฅ๐ฅ
Hence the original PDE (1) now can be written in term of ๐ as the new dependent variableas
๐๐๐ฅ๐๐ก + ๐๐ฅ ๏ฟฝ๐๐ฅ๐ฅ๏ฟฝ = ๐๐๐ฅ๐ฅ๐ฅ (3)
But
๐๐ฅ ๏ฟฝ๐๐ฅ๐ฅ๏ฟฝ =12๐๐๐ฅ
๏ฟฝ๐2๐ฅ๏ฟฝ
285
5.2. Exam 2 CHAPTER 5. EXAMS
Using the above in (3), then (3) becomes
๐๐๐ฅ๐๐ก +
12๐๐๐ฅ
๏ฟฝ๐2๐ฅ๏ฟฝ = ๐๐๐ฅ๐ฅ๐ฅ
๐๐๐ฅ๐๐ก +
12๐๐๐ฅ
๏ฟฝ๐2๐ฅ๏ฟฝ โ ๐
๐๐๐ฅ
๏ฟฝ๐๐ฅ๐ฅ๏ฟฝ = 0
๐๐๐ฅ ๏ฟฝ
๐๐ก +12๐2๐ฅ โ ๐๐๐ฅ๐ฅ๏ฟฝ = 0
Therefore
๐๐ก +12๐2๐ฅ โ ๐๐๐ฅ๐ฅ = 0 (4)
But from (2) ๐ = โ2๐ ln๐, then using this in (4), we now rewrite (4) in terms of ๐
๐๐๐ก๏ฟฝโ2๐ ln๐๏ฟฝ + 1
2 ๏ฟฝ๐๐๐ฅ
๏ฟฝโ2๐ ln๐๏ฟฝ๏ฟฝ2
โ ๐๐2
๐๐ฅ2๏ฟฝโ2๐ ln๐๏ฟฝ = 0
๏ฟฝโ2๐๐๐ก๐ ๏ฟฝ
+12 ๏ฟฝโ2๐
๐๐ฅ๐ ๏ฟฝ
2
โ ๐๐๐๐ฅ ๏ฟฝ
โ2๐๐๐ฅ๐ ๏ฟฝ
= 0
โ2๐๐๐ก๐+ 2๐2 ๏ฟฝ
๐๐ฅ๐ ๏ฟฝ
2
+ 2๐2๐๐๐ฅ ๏ฟฝ
๐๐ฅ๐ ๏ฟฝ
= 0
But ๐๐๐ฅ๏ฟฝ๐๐ฅ๐๏ฟฝ = ๐๐ฅ๐ฅ
๐ โ ๐2๐ฅ๐2 , hence the above becomes
โ2๐๐๐ก๐+ 2๐2 ๏ฟฝ
๐๐ฅ๐ ๏ฟฝ
2
+ 2๐2 ๏ฟฝ๐๐ฅ๐ฅ๐
โ๐2๐ฅ๐2 ๏ฟฝ = 0
โ2๐๐๐ก๐+ 2๐2 ๏ฟฝ
๐๐ฅ๐ ๏ฟฝ
2
+ 2๐2๐๐ฅ๐ฅ๐
โ 2๐2๐2๐ฅ๐2 = 0
โ2๐๐๐ก๐+ 2๐2
๐๐ฅ๐ฅ๐
= 0
โ๐๐ก๐+ ๐
๐๐ฅ๐ฅ๐
= 0
Since ๐ โ 0 identically, then the above simplifies to the heat PDE
๐๐ก = ๐๐๐ฅ๐ฅ (5)
๐ (๐ฅ, 0) = ฮฆ (๐ฅ)โโ < ๐ฅ < โ
5.2.1.3 Part (c)
Now we solve (5) for ๐ (๐ฅ, ๐ก) and then convert the solution back to ๐ข (๐ฅ, ๐ก) using the Cole-Hopftransformation. This infinite domain heat PDE has known solution (as ๐ (ยฑโ, ๐ก) is boundedwhich is
๐ (๐ฅ, ๐ก) = ๏ฟฝโ
โโฮฆ๏ฟฝ๐๏ฟฝ
1
โ4๐๐๐กexp
โโโโโโโโโ ๏ฟฝ๐ฅ โ ๐๏ฟฝ
2
4๐๐ก
โโโโโโโโ ๐๐ (6)
286
5.2. Exam 2 CHAPTER 5. EXAMS
5.2.1.4 Part(d)
Now
ฮฆ (๐ฅ) = ๐ (๐ฅ, 0) (7)
But since ๐ข (๐ฅ, ๐ก) = ๐๐๐ฅ๏ฟฝโ2๐ ln๐๏ฟฝ, then integrating
๏ฟฝ๐ฅ
๐ฅ0๐ข (๐ผ, ๐ก) ๐๐ผ = โ2๐ ln๐
ln๐ = โ12๐ ๏ฟฝ
๐ฅ
๐ฅ0๐ข (๐ผ, ๐ก) ๐๐ผ
๐ (๐ฅ, ๐ก) = exp ๏ฟฝโ12๐ ๏ฟฝ
๐ฅ
๐ฅ0๐ข (๐ผ, ๐ก) ๐๐ผ๏ฟฝ
Hence at ๐ก = 0 the above becomes
๐ (๐ฅ, 0) = exp ๏ฟฝโ12๐ ๏ฟฝ
๐ฅ
๐ฅ0๐ข (๐ผ, 0) ๐๐ผ๏ฟฝ
= exp ๏ฟฝโ12๐ ๏ฟฝ
๐ฅ
๐ฅ0๐น (๐ผ) ๐๐ผ๏ฟฝ
Where ๐น (๐ฅ) = ๐ข (๐ฅ, 0). Hence from the above, comparing it to (6) we see that
ฮฆ (๐ฅ) = exp ๏ฟฝโ12๐ ๏ฟฝ
๐ฅ
๐ฅ0๐น (๐ผ) ๐๐ผ๏ฟฝ (8)
5.2.1.5 Part(e)
From (6), we found ๐ (๐ฅ, ๐ก) = โซโ
โโฮฆ๏ฟฝ๐๏ฟฝ 1
โ4๐๐๐กexp
โโโโโโ๏ฟฝ๐ฅโ๐๏ฟฝ
2
4๐๐ก
โโโโโ ๐๐. Plugging (8) into this expression
gives
๐ (๐ฅ, ๐ก) =1
โ4๐๐๐ก๏ฟฝ
โ
โโexp ๏ฟฝ
โ12๐ ๏ฟฝ
๐
๐ฅ0๐น (๐ผ) ๐๐ผ๏ฟฝ exp
โโโโโโโโโ ๏ฟฝ๐ฅ โ ๐๏ฟฝ
2
4๐๐ก
โโโโโโโโ ๐๐
=1
โ4๐๐๐ก๏ฟฝ
โ
โโexp
โโโโโโโโโ12๐ ๏ฟฝ
๐
๐ฅ0๐น (๐ผ) ๐๐ผ โ
๏ฟฝ๐ฅ โ ๐๏ฟฝ2
4๐๐ก
โโโโโโโโ ๐๐
=1
โ4๐๐๐ก๏ฟฝ
โ
โโexp
โโโโโโโโโ12๐
โกโขโขโขโขโขโขโฃ๏ฟฝ
๐
๐ฅ0๐น (๐ผ) ๐๐ผ +
๏ฟฝ๐ฅ โ ๐๏ฟฝ2
2๐ก
โคโฅโฅโฅโฅโฅโฅโฆ
โโโโโโโโ ๐๐ (9)
Let
๏ฟฝ๐
๐ฅ0๐น (๐ผ) ๐๐ผ +
๏ฟฝ๐ฅ โ ๐๏ฟฝ2
2๐ก= ๐ ๏ฟฝ๐, ๐ฅ, ๐ก๏ฟฝ
287
5.2. Exam 2 CHAPTER 5. EXAMS
Hence (9) becomes
๐ (๐ฅ, ๐ก) = ๏ฟฝโ
โโ
1
โ4๐๐๐ก๐โ๐๏ฟฝ๐,๐ฅ,๐ก๏ฟฝ
2๐ ๐๐ (10)
5.2.1.6 Part(f)
From (10)
๐๐๐๐ฅ
= ๏ฟฝโ
โโ
1
โ4๐๐๐ก๐๐๐ฅ
โโโโโ๐
โ๐๏ฟฝ๐,๐ฅ,๐ก๏ฟฝ2๐
โโโโโ ๐๐
= ๏ฟฝโ
โโ
1
โ4๐๐๐ก
โโโโโ๐๐๐ฅ๐ ๏ฟฝ๐, ๐ฅ, ๐ก๏ฟฝ ๐
โ๐๏ฟฝ๐,๐ฅ,๐ก๏ฟฝ2๐
โโโโโ ๐๐
= ๏ฟฝโ
โโ
1
โ4๐๐๐ก
โโโโโโโโ๐๐๐ฅ
โกโขโขโขโขโขโขโฃ๏ฟฝ
๐
๐ฅ0๐น (๐ผ) ๐๐ผ +
๏ฟฝ๐ฅ โ ๐๏ฟฝ2
2๐ก
โคโฅโฅโฅโฅโฅโฅโฆ ๐
โ๐๏ฟฝ๐,๐ฅ,๐ก๏ฟฝ2๐
โโโโโโโโ ๐๐
Using Leibniz integral rule the above simplifies to
๐๐๐๐ฅ
= ๏ฟฝโ
โโ
1
โ4๐๐๐ก
โโโโโโ๏ฟฝ๐ฅ โ ๐๏ฟฝ๐ก
๐โ๐๏ฟฝ๐,๐ฅ,๐ก๏ฟฝ
2๐
โโโโโโ ๐๐ (11)
But
๐ข = โ2๐๐๐ฅ๐
Hence, using (10) and (11) in the above gives
๐ข = โ2๐โซโ
โโ1
โ4๐๐๐ก
โโโโโ๏ฟฝ๐ฅโ๐๏ฟฝ
๐ก ๐โ๐๏ฟฝ๐,๐ฅ,๐ก๏ฟฝ
2๐
โโโโโ ๐๐
โซโ
โโ1
โ4๐๐๐ก๐โ๐๏ฟฝ๐,๐ฅ,๐ก๏ฟฝ
2๐ ๐๐
Hence the solution is
๐ข (๐ฅ, ๐ก) = โ2๐โซโ
โโ
๏ฟฝ๐ฅโ๐๏ฟฝ
๐ก ๐โ๐๏ฟฝ๐,๐ฅ,๐ก๏ฟฝ
2๐ ๐๐
โซโ
โโ๐โ๐๏ฟฝ๐,๐ฅ,๐ก๏ฟฝ
2๐ ๐๐
Where
๐ ๏ฟฝ๐, ๐ฅ, ๐ก๏ฟฝ = ๏ฟฝ๐
๐ฅ0๐น (๐ผ) ๐๐ผ +
๏ฟฝ๐ฅ โ ๐๏ฟฝ2
2๐ก
288
5.2. Exam 2 CHAPTER 5. EXAMS
5.2.2 problem 4
(e) Write your expression for ฯ(x, t) in terms of
f(ฮท, x, t) =
โซ ฮท
xo
F (ฮฑ)dฮฑ+(xโ ฮท)2
2t.
(f) Find ฯx(x, t) and then use equation (3b) to find u(x, t).
4. (a) Use the Method of Images to solve
โu
โt= ฮฝ
โ2u
โx2+Q(x, t), 0 โค x โค L, t โฅ 0
u(x, 0) = f(x),โu
โx(0, t) = 0, u(L, t) = A
(b) For A = 0, compare your expression for the solution in (a) to the eigenfunction solution.
5. This problem is a simple model for diffraction of light passing through infinitesimallysmall slits separated by a distance 2a.
Solve the diffraction equation
โu
โt=iฮป
4ฯ
โ2u
โx2(1)
with initial source u(x, 0) = f(x) = ฮด(xโ a) + ฮด(x+ a), a > 0.
Show that the solution u(x, t) oscillates wildly, but that the intensity |u(x, t)|2 is well-behaved. The intensity |u(x, t)|2 shows that the diffraction pattern at a distance t consistsof a series of alternating bright and dark fringes with period ฮปt/(2a).
3
5.2.2.1 Part(a)
๐๐ข๐๐ก
= ๐๐2๐ข๐๐ฅ2
+ ๐ (๐ฅ, ๐ก) (1)
0 โค ๐ฅ โค ๐ฟ๐ก โฅ 0
Initial conditions are
๐ข (๐ฅ, 0) = ๐ (๐ฅ)
Boundary conditions are
๐๐ข (0, ๐ก)๐๐ฅ
= 0
๐ข (๐ฟ, ๐ก) = ๐ด
Multiplying both sides of (1) by ๐บ (๐ฅ, ๐ก; ๐ฅ0, ๐ก0) and integrating over the domain gives (wherein the following ๐บ is used instead of ๐บ (๐ฅ, ๐ก; ๐ฅ0, ๐ก0) for simplicity).
๏ฟฝ๐ฟ
๐ฅ=0๏ฟฝ
โ
๐ก=0๐บ๐ข๐ก ๐๐ก๐๐ฅ = ๏ฟฝ
๐ฟ
๐ฅ=0๏ฟฝ
โ
๐ก=0๐๐ข๐ฅ๐ฅ๐บ ๐๐ก๐๐ฅ +๏ฟฝ
๐ฟ
๐ฅ=0๏ฟฝ
โ
๐ก=0๐๐บ ๐๐ก๐๐ฅ (1)
For the integral on the LHS, we apply integration by parts once to move the time derivativefrom ๐ข to ๐บ
๏ฟฝ๐ฟ
๐ฅ=0๏ฟฝ
โ
๐ก=0๐บ๐ข๐ก ๐๐ก๐๐ฅ = ๏ฟฝ
๐ฟ
๐ฅ=0[๐ข๐บ]โ๐ก=0 ๐๐ฅ โ๏ฟฝ
๐ฟ
๐ฅ=0๏ฟฝ
โ
๐ก=0๐บ๐ก๐ข ๐๐ก๐๐ฅ (1A)
289
5.2. Exam 2 CHAPTER 5. EXAMS
And the first integral in the RHS of (1) gives, after doing integration by parts two times onit
๏ฟฝ๐ฟ
๐ฅ=0๏ฟฝ
โ
๐ก=0๐๐ข๐ฅ๐ฅ๐บ ๐๐ก๐๐ฅ = ๏ฟฝ
โ
๐ก=0[๐ข๐ฅ๐บ]
๐ฟ๐ฅ=0 ๐๐ก โ๏ฟฝ
๐ฟ
๐ฅ=0๏ฟฝ
โ
๐ก=0๐๐ข๐ฅ๐บ๐ฅ ๐๐ก๐๐ฅ
= ๏ฟฝ๐ฟ
๐ก=0[๐ข๐ฅ๐บ]
๐ฟ๐ฅ=0 ๐๐ก โ ๏ฟฝ๏ฟฝ
โ
๐ก=0[๐ข๐บ๐ฅ]
๐ฟ๐ฅ=0 ๐๐ก โ๏ฟฝ
๐ฟ
๐ฅ=0๏ฟฝ
โ
๐ก=0๐๐ข๐บ๐ฅ๐ฅ ๐๐ก๐๐ฅ๏ฟฝ
= ๏ฟฝโ
๐ก=0๏ฟฝ[๐ข๐ฅ๐บ]
๐ฟ๐ฅ=0 โ [๐ข๐บ๐ฅ]
๐ฟ๐ฅ=0๏ฟฝ ๐๐ก +๏ฟฝ
๐ฟ
๐ฅ=0๏ฟฝ
โ
๐ก=0๐๐ข๐บ๐ฅ๐ฅ ๐๐ก๐๐ฅ
= ๏ฟฝโ
๐ก=0[๐ข๐ฅ๐บ โ ๐ข๐บ๐ฅ]
๐ฟ๐ฅ=0 ๐๐ก +๏ฟฝ
๐ฟ
๐ฅ=0๏ฟฝ
โ
๐ก=0๐๐ข๐บ๐ฅ๐ฅ ๐๐ก๐๐ฅ
= โ๏ฟฝโ
๐ก=0[๐ข๐บ๐ฅ โ ๐ข๐ฅ๐บ]
๐ฟ๐ฅ=0 ๐๐ก +๏ฟฝ
๐ฟ
๐ฅ=0๏ฟฝ
โ
๐ก=0๐๐ข๐บ๐ฅ๐ฅ ๐๐ก๐๐ฅ (1B)
Substituting (1A) and (1B) back into (1) results in
๏ฟฝ๐ฟ
๐ฅ=0[๐ข๐บ]โ๐ก=0 ๐๐ฅโ๏ฟฝ
๐ฟ
๐ฅ=0๏ฟฝ
โ
๐ก=0๐บ๐ก๐ข ๐๐ก๐๐ฅ = ๏ฟฝ
โ
๐ก=0[๐ข๐ฅ๐บ โ ๐ข๐บ๐ฅ]
๐ฟ๐ฅ=0 ๐๐ก+๏ฟฝ
๐ฟ
๐ฅ=0๏ฟฝ
โ
๐ก=0๐๐ข๐บ๐ฅ๐ฅ ๐๐ก๐๐ฅ+๏ฟฝ
๐ฟ
๐ฅ=0๏ฟฝ
โ
๐ก=0๐บ๐ ๐๐ก๐๐ฅ
Or
๏ฟฝ๐ฟ
๐ฅ=0๏ฟฝ
โ
๐ก=0โ๐บ๐ก๐ข โ ๐๐ข๐บ๐ฅ๐ฅ ๐๐ก๐๐ฅ = โ๏ฟฝ
๐ฟ
๐ฅ=0[๐ข๐บ]โ๐ก=0 ๐๐ฅ โ๏ฟฝ
โ
๐ก=0[๐ข๐บ๐ฅ โ ๐ข๐ฅ๐บ]
๐ฟ๐ฅ=0 ๐๐ก +๏ฟฝ
๐ฟ
๐ฅ=0๏ฟฝ
โ
๐ก=0๐บ๐ ๐๐ก๐๐ฅ
(2)
We now want to choose ๐บ (๐ฅ, ๐ก; ๐ฅ0, ๐ก0) such that
โ๐บ๐ก๐ข โ ๐๐ข๐บ๐ฅ๐ฅ = ๐ฟ (๐ฅ โ ๐ฅ0) ๐ฟ (๐ก โ ๐ก0)โ๐บ๐ก๐ข = ๐๐ข๐บ๐ฅ๐ฅ + ๐ฟ (๐ฅ โ ๐ฅ0) ๐ฟ (๐ก โ ๐ก0) (3)
This way, the LHS of (2) becomes just ๐ข (๐ฅ0, ๐ก0). Hence (2) now (after the above choice of๐บ) reduces to
๐ข (๐ฅ0, ๐ก0) = โ๏ฟฝ๐ฟ
๐ฅ=0[๐ข๐บ]โ๐ก=0 ๐๐ฅ โ๏ฟฝ
๐ก0
๐ก=0[๐ข๐บ๐ฅ โ ๐ข๐ฅ๐บ]
๐ฟ๐ฅ=0 ๐๐ก +๏ฟฝ
๐ฟ
๐ฅ=0๏ฟฝ
๐ก0
๐ก=0๐บ๐ ๐๐ก๐๐ฅ (4)
We now need to find the Green function which satisfies (3). But (3) is equivalent to solutionof problem
โ๐บ๐ก๐ข = ๐๐ข๐บ๐ฅ๐ฅ
๐บ (๐ฅ, 0) = ๐ฟ (๐ฅ โ ๐ฅ0) ๐ฟ (๐ก โ ๐ก0)โโ < ๐ฅ < โ
๐บ (๐ฅ, ๐ก; ๐ฅ0, ๐ก0) = 0 ๐ก > ๐ก0๐บ (ยฑโ, ๐ก; ๐ฅ0, ๐ก0) = 0๐บ (๐ฅ, ๐ก0; ๐ฅ0, ๐ก0) = ๐ฟ (๐ฅ โ ๐ฅ0)
The above problem has a known fundamental solution which we found before, but for theforward heat PDE instead of the reverse heat PDE as it is now. The fundamental solution
290
5.2. Exam 2 CHAPTER 5. EXAMS
to the forward heat PDE is
๐บ (๐ฅ, ๐ก) =1
โ4๐๐ (๐ก โ ๐ก0)exp
โโโโโโ (๐ฅ โ ๐ฅ0)
2
4๐ (๐ก โ ๐ก0)
โโโโโ 0 โค ๐ก0 โค ๐ก
Therefore, for the reverse heat PDE the above becomes
๐บ (๐ฅ, ๐ก) =1
โ4๐๐ (๐ก0 โ ๐ก)exp
โโโโโโ (๐ฅ โ ๐ฅ0)
2
4๐ (๐ก0 โ ๐ก)
โโโโโ 0 โค ๐ก โค ๐ก0 (5)
We now go back to (4) and try to evaluate all terms in the RHS. Starting with the first termโซ๐ฟ
๐ฅ=0[๐ข๐บ]โ๐ก=0 ๐๐ฅ. Since ๐บ (๐ฅ,โ; ๐ฅ0, ๐ก0) = 0 then the upper limit is zero. But at lower limit ๐ก = 0
we are given that ๐ข (๐ฅ, 0) = ๐ (๐ฅ), hence this term becomes
๏ฟฝ๐ฟ
๐ฅ=0[๐ข๐บ]โ๐ก=0 ๐๐ฅ = ๏ฟฝ
๐ฟ
๐ฅ=0โ๐ข (๐ฅ, 0) ๐บ (๐ฅ, 0) ๐๐ฅ
= ๏ฟฝ๐ฟ
๐ฅ=0โ๐ (๐ฅ)๐บ (๐ฅ, 0) ๐๐ฅ
Now looking at the second term in RHS of (4), we expand it and find
[๐ข๐บ๐ฅ โ ๐ข๐ฅ๐บ]๐ฟ๐ฅ=0 = (๐ข (๐ฟ, ๐ก) ๐บ๐ฅ (๐ฟ, ๐ก) โ ๐ข๐ฅ (๐ฟ, ๐ก) ๐บ (๐ฟ, ๐ก)) โ (๐ข (0, ๐ก) ๐บ๐ฅ (0, ๐ก) โ ๐ข๐ฅ (0, ๐ก) ๐บ (0, ๐ก))
We are told that ๐ข๐ฅ (0, ๐ก) = 0 and ๐ข (๐ฟ, ๐ก) = ๐ด, then the above becomes
[๐ข๐บ๐ฅ โ ๐ข๐ฅ๐บ]๐ฟ๐ฅ=0 = ๐ด๐บ๐ฅ (๐ฟ, ๐ก) โ ๐ข๐ฅ (๐ฟ, ๐ก) ๐บ (๐ฟ, ๐ก) โ ๐ข (0, ๐ก) ๐บ๐ฅ (0, ๐ก) (5A)
There are still two terms above we do not know. We do not know ๐ข๐ฅ (๐ฟ, ๐ก) and we also donot know ๐ข (0, ๐ก). If we can configure, using method of images, such that ๐บ (๐ฟ, ๐ก) = 0 and๐บ๐ฅ (0, ๐ก) = 0 then we can get rid of these two terms and end up only with [๐ข๐บ๐ฅ โ ๐ข๐ฅ๐บ]
โ๐ฅ=0 =
๐ด๐บ๐ฅ (๐ฟ, ๐ก) which we can evaluate once we know what ๐บ (๐ฅ, ๐ก) is.
This means we need to put images on both sides of the boundaries such that to force๐บ (๐ฟ, ๐ก) = 0 and also ๐บ๐ฅ (0, ๐ก) = 0.
We see that this result agrees what we always did, which is, If the prescribed boundary
conditions on ๐ข are such that ๐ข = ๐ด, then we want ๐บ = 0 there. And if it is ๐๐ข๐๐ฅ = ๐ด, then we
want ๐๐บ๐๐ฅ = 0 there. And this is what we conclude here also from the above. In other words,
the boundary conditions on Green functions are always the homogeneous version of theboundary conditions given on ๐ข.
โuโx = 0โGโx = 0
x = 0 x = L
u(L, t) = A
G(L, t) = 0
291
5.2. Exam 2 CHAPTER 5. EXAMS
To force ๐บ๐ฅ (0, ๐ก) = 0, we need to put same sign images on both sides of ๐ฅ = 0. So we endup with this
x0x = 0 x = L
โGโx = 0
G = 0โx0
The above makes ๐บ๐ฅ (0, ๐ก) = 0 at ๐ฅ = 0. Now we want to make ๐บ = 0 at ๐ฅ = ๐ฟ. Then we updatethe above and put a negative image at ๐ฅ = 2๐ฟ โ ๐ฅ0 to the right of ๐ฅ = ๐ฟ as follows
x0x = 0 x = L
โGโx = 0
G = 0โx0
2Lโ x0
But now we see that the image at ๐ฅ = โ๐ฅ0 has a๏ฟฝected condition of ๐บ = 0 at ๐ฅ = ๐ฟ and willmake it not zero as we wanted. So to counter e๏ฟฝect this, we have to add another negativeimage at distance ๐ฅ = 2๐ฟ + ๐ฅ0 to cancel the e๏ฟฝect of the image at ๐ฅ = โ๐ฅ0. We end up withthis setup
x0x = 0 x = L
โGโx = 0
G = 0โx0
2Lโ x0 2L+ x0
But now we see that the two negative images we added to the right will no longer make๐บ๐ฅ (0, ๐ก) = 0, so we need to counter e๏ฟฝect this by adding two negative images to the left sideto keep ๐บ๐ฅ (0, ๐ก) = 0. So we end up with
292
5.2. Exam 2 CHAPTER 5. EXAMS
x0x = 0 x = L
โGโx = 0
G = 0โx0
2Lโ x0 2L+ x0โ2Lโ x0 โ2L+ x0
But now we see that by putting these two images on the left, we no longer have ๐บ = 0 at๐ฅ = ๐ฟ. So to counter e๏ฟฝect this, we have to put copies of these 2 images on the right againbut with positive sign, as follows
x0x = 0 x = L
โGโx = 0
G = 0โx0
2Lโ x0 2L+ x0โ2Lโ x0 โ2L+ x0
4Lโ x0 4L+ x0
But now these two images on the right, no longer keep ๐บ๐ฅ (0, ๐ก) = 0, so we have to put samesign images to the left, as follows
x0x = 0 x = L
โGโx = 0
G = 0
โx0
2Lโ x0
2L+ x0โ2Lโ x0
โ2L+ x0
4Lโ x0
4L+ x0
โ4L+ x0
โ4Lโ x0
And so on. This continues for infinite number of images. Therefore we see from the above,for the positive images, we have the following sum
๏ฟฝ4๐๐ (๐ก0 โ ๐ก)๐บ (๐ฅ, ๐ก; ๐ฅ0, ๐ก0) = expโโโโโโ (๐ฅ โ ๐ฅ0)
2
4๐ (๐ก0 โ ๐ก)
โโโโโ + exp
โโโโโโ (๐ฅ + ๐ฅ0)
2
4๐ (๐ก0 โ ๐ก)
โโโโโ + exp
โโโโโโ (๐ฅ โ (4๐ฟ โ ๐ฅ0))
2
4๐ (๐ก0 โ ๐ก)
โโโโโ
+ expโโโโโโ (๐ฅ โ (โ4๐ฟ + ๐ฅ0))
2
4๐ (๐ก0 โ ๐ก)
โโโโโ + exp
โโโโโโ (๐ฅ โ (4๐ฟ + ๐ฅ0))
2
4๐ (๐ก0 โ ๐ก)
โโโโโ
+ expโโโโโโ (๐ฅ โ (โ4๐ฟ โ ๐ฅ0))
2
4๐ (๐ก0 โ ๐ก)
โโโโโ +โฏ
Or
๐บ (๐ฅ, ๐ก; ๐ฅ0, ๐ก0) =1
โ4๐๐ (๐ก0 โ ๐ก)
โโโโโ
โ๏ฟฝ๐=โโ
expโโโโโโ (๐ฅ โ (4๐๐ฟ โ ๐ฅ0))
2
4๐ (๐ก0 โ ๐ก)
โโโโโ + exp
โโโโโโ (๐ฅ โ (โ4๐๐ฟ + ๐ฅ0))
2
4๐ (๐ก0 โ ๐ก)
โโโโโ
โโโโโ (6)
293
5.2. Exam 2 CHAPTER 5. EXAMS
The above takes care of the positive images. For negative images, we have this sum of images
๏ฟฝ4๐๐ (๐ก0 โ ๐ก)๐บ (๐ฅ, ๐ก; ๐ฅ0, ๐ก0) = expโโโโโโ (๐ฅ โ (2๐ฟ โ ๐ฅ0))
2
4๐ (๐ก0 โ ๐ก)
โโโโโ + exp
โโโโโโ (๐ฅ โ (โ2๐ฟ + ๐ฅ0))
2
4๐ (๐ก0 โ ๐ก)
โโโโโ
+ expโโโโโโ (๐ฅ โ (2๐ฟ + ๐ฅ0))
2
4๐ (๐ก0 โ ๐ก)
โโโโโ + exp
โโโโโโ (๐ฅ โ (โ2๐ฟ โ ๐ฅ0))
2
4๐ (๐ก0 โ ๐ก)
โโโโโ
+ expโโโโโโ (๐ฅ โ (6๐ฟ โ ๐ฅ0))
2
4๐ (๐ก0 โ ๐ก)
โโโโโ + exp
โโโโโโ (๐ฅ โ (โ6๐ฟ + ๐ฅ0))
2
4๐ (๐ก0 โ ๐ก)
โโโโโ
+ expโโโโโโ (๐ฅ โ (6๐ฟ + ๐ฅ0))
2
4๐ (๐ก0 โ ๐ก)
โโโโโ + exp
โโโโโโ (๐ฅ โ (โ6๐ฟ โ ๐ฅ0))
2
4๐ (๐ก0 โ ๐ก)
โโโโโ โฏ
Or
๏ฟฝ4๐๐ (๐ก0 โ ๐ก)๐บ (๐ฅ, ๐ก; ๐ฅ0, ๐ก0) = โโโโโโ
โ๏ฟฝ๐=โโ
expโโโโโโ (๐ฅ โ ((4๐ โ 2) ๐ฟ + ๐ฅ0))
2
4๐ (๐ก0 โ ๐ก)
โโโโโ + exp
โโโโโโ (๐ฅ โ ((4๐ โ 2) ๐ฟ โ ๐ฅ0))
2
4๐ (๐ก0 โ ๐ก)
โโโโโ
โโโโโ
(7)
Hence the Green function to use is (6)+(7) which gives
๐บ (๐ฅ, ๐ก; ๐ฅ0, ๐ก0) =1
โ4๐๐ (๐ก0 โ ๐ก)
โโโโโ
โ๏ฟฝ๐=โโ
expโโโโโโ (๐ฅ โ (4๐๐ฟ โ ๐ฅ0))
2
4๐ (๐ก0 โ ๐ก)
โโโโโ + exp
โโโโโโ (๐ฅ โ (โ4๐๐ฟ + ๐ฅ0))
2
4๐ (๐ก0 โ ๐ก)
โโโโโ
โโโโโ
โ1
โ4๐๐ (๐ก0 โ ๐ก)
โโโโโ
โ๏ฟฝ๐=โโ
expโโโโโโ (๐ฅ โ ((4๐ โ 2) ๐ฟ + ๐ฅ0))
2
4๐ (๐ก0 โ ๐ก)
โโโโโ + exp
โโโโโโ (๐ฅ โ ((4๐ โ 2) ๐ฟ โ ๐ฅ0))
2
4๐ (๐ก0 โ ๐ก)
โโโโโ
โโโโโ
(7A)
Using the above Green function, we go back to 5A and finally are able to simplify it
[๐ข๐บ๐ฅ โ ๐ข๐ฅ๐บ]๐ฟ๐ฅ=0 = ๐ด๐บ๐ฅ (๐ฟ, ๐ก) โ ๐ข๐ฅ (๐ฟ, ๐ก) ๐บ (๐ฟ, ๐ก) โ ๐ข (0, ๐ก) ๐บ๐ฅ (0, ๐ก)
The above becomes now (with the images in place as above)
[๐ข๐บ๐ฅ โ ๐ข๐ฅ๐บ]๐ฟ๐ฅ=0 = ๐ด
๐๐บ (๐ฟ, ๐ก; ๐ฅ0, ๐ก0)๐๐ฅ
(8)
Since now we know what ๐บ (๐ฅ, ๐ก; ๐ฅ0, ๐ก0), from (7), we can evaluate its derivative w.r.t. ๐ฅ. (brokenup, so it fits on one page)
๐๐บ (๐ฅ, ๐ก; ๐ฅ0, ๐ก0)๐๐ฅ
=1
โ4๐๐ (๐ก0 โ ๐ก)
โ๏ฟฝ๐=โโ
โ (๐ฅ โ (4๐๐ฟ โ ๐ฅ0))2๐ (๐ก0 โ ๐ก)
expโโโโโโ (๐ฅ โ (4๐๐ฟ โ ๐ฅ0))
2
4๐ (๐ก0 โ ๐ก)
โโโโโ
+1
โ4๐๐ (๐ก0 โ ๐ก)
โ๏ฟฝ๐=โโ
โ (๐ฅ โ (โ4๐๐ฟ + ๐ฅ0))2๐ (๐ก0 โ ๐ก)
expโโโโโโ (๐ฅ โ (โ4๐๐ฟ + ๐ฅ0))
2
4๐ (๐ก0 โ ๐ก)
โโโโโ
โ1
โ4๐๐ (๐ก0 โ ๐ก)
โ๏ฟฝ๐=โโ
โ (๐ฅ โ ((4๐ โ 2) ๐ฟ + ๐ฅ0))2๐ (๐ก0 โ ๐ก)
expโโโโโโ (๐ฅ โ ((4๐ โ 2) ๐ฟ + ๐ฅ0))
2
4๐ (๐ก0 โ ๐ก)
โโโโโ
โ1
โ4๐๐ (๐ก0 โ ๐ก)
โ๏ฟฝ๐=โโ
โ (๐ฅ โ ((4๐ โ 2) ๐ฟ โ ๐ฅ0))2๐ (๐ก0 โ ๐ก)
expโโโโโโ (๐ฅ โ ((4๐ โ 2) ๐ฟ โ ๐ฅ0))
2
4๐ (๐ก0 โ ๐ก)
โโโโโ
294
5.2. Exam 2 CHAPTER 5. EXAMS
At ๐ฅ = ๐ฟ, the above derivative becomes
๐๐บ (๐ฟ, ๐ก; ๐ฅ0, ๐ก0)๐๐ฅ
=1
โ4๐๐ (๐ก0 โ ๐ก)
โ๏ฟฝ๐=โโ
โ (๐ฟ โ (4๐๐ฟ โ ๐ฅ0))2๐ (๐ก0 โ ๐ก)
expโโโโโโ (๐ฟ โ (4๐๐ฟ โ ๐ฅ0))
2
4๐ (๐ก0 โ ๐ก)
โโโโโ
+1
โ4๐๐ (๐ก0 โ ๐ก)
โ๏ฟฝ๐=โโ
โ (๐ฟ โ (โ4๐๐ฟ + ๐ฅ0))2๐ (๐ก0 โ ๐ก)
expโโโโโโ (๐ฟ โ (โ4๐๐ฟ + ๐ฅ0))
2
4๐ (๐ก0 โ ๐ก)
โโโโโ
โ1
โ4๐๐ (๐ก0 โ ๐ก)
โ๏ฟฝ๐=โโ
โ (๐ฟ โ ((4๐ โ 2) ๐ฟ + ๐ฅ0))2๐ (๐ก0 โ ๐ก)
expโโโโโโ (๐ฟ โ ((4๐ โ 2) ๐ฟ + ๐ฅ0))
2
4๐ (๐ก0 โ ๐ก)
โโโโโ
โ1
โ4๐๐ (๐ก0 โ ๐ก)
โ๏ฟฝ๐=โโ
โ (๐ฟ โ ((4๐ โ 2) ๐ฟ โ ๐ฅ0))2๐ (๐ก0 โ ๐ก)
expโโโโโโ (๐ฟ โ ((4๐ โ 2) ๐ฟ โ ๐ฅ0))
2
4๐ (๐ก0 โ ๐ก)
โโโโโ (9)
From (4), we now collect all terms into the solution
๐ข (๐ฅ0, ๐ก0) = โ๏ฟฝ๐ฟ
๐ฅ=0[๐ข๐บ]โ๐ก=0 ๐๐ฅ โ๏ฟฝ
๐ก0
๐ก=0[๐ข๐บ๐ฅ โ ๐ข๐ฅ๐บ]
๐ฟ๐ฅ=0 ๐๐ก +๏ฟฝ
๐ฟ
๐ฅ=0๏ฟฝ
๐ก0
๐ก=0๐บ๐ ๐๐ก๐๐ฅ (4)
We found โซ๐ฟ
๐ฅ=0[๐ข๐บ]โ๐ก=0 ๐๐ฅ = โซ๐ฟ
๐ฅ=0โ๐ (๐ฅ)๐บ (๐ฅ, 0) ๐๐ฅ and now we know what ๐บ is. Hence we can
find ๐บ (๐ฅ, 0; ๐ฅ0, ๐ก0). It is, from (7A)
๐บ (๐ฅ, 0; ๐ฅ0, ๐ก0) =1
โ4๐๐๐ก0
โโโโโ
โ๏ฟฝ๐=โโ
expโโโโโโ (๐ฅ โ (4๐๐ฟ โ ๐ฅ0))
2
4๐๐ก0
โโโโโ + exp
โโโโโโ (๐ฅ โ (โ4๐๐ฟ + ๐ฅ0))
2
4๐๐ก0
โโโโโ
โโโโโ
โ1
โ4๐๐๐ก0
โโโโโ
โ๏ฟฝ๐=โโ
expโโโโโโ (๐ฅ โ ((4๐ โ 2) ๐ฟ + ๐ฅ0))
2
4๐๐ก0
โโโโโ + exp
โโโโโโ (๐ฅ โ ((4๐ โ 2) ๐ฟ โ ๐ฅ0))
2
4๐๐ก0
โโโโโ
โโโโโ
(10)
And we now know what [๐ข๐บ๐ฅ โ ๐ข๐ฅ๐บ]๐ฟ๐ฅ=0 is. It is ๐ด
๐๐บ(๐ฟ,๐ก;๐ฅ0,๐ก0)๐๐ฅ . Hence (4) becomes
๐ข (๐ฅ0, ๐ก0) = ๏ฟฝ๐ฟ
๐ฅ=0๐ (๐ฅ)๐บ (๐ฅ, 0; ๐ฅ0, ๐ก0) ๐๐ฅโ๏ฟฝ
๐ก0
๐ก=0๐ด๐๐บ (๐ฟ, ๐ก; ๐ฅ0, ๐ก0)
๐๐ฅ๐๐ก+๏ฟฝ
๐ฟ
๐ฅ=0๏ฟฝ
๐ก0
๐ก=0๐บ (๐ฅ, ๐ก; ๐ฅ0, ๐ก0) ๐ (๐ฅ, ๐ก) ๐๐ก๐๐ฅ
Changing the roles of ๐ฅ0, ๐ก0
๐ข (๐ฅ, ๐ก) = ๏ฟฝ๐ฟ
๐ฅ0=0๐ (๐ฅ0) ๐บ (๐ฅ0, ๐ก0; ๐ฅ, 0) ๐๐ฅ0 โ๏ฟฝ
๐ก
๐ก=0๐ด๐๐บ (๐ฅ0, ๐ก0; ๐ฟ, ๐ก)
๐๐ฅ0๐๐ก0 +๏ฟฝ
๐ฟ
๐ฅ0=0๏ฟฝ
๐ก
๐ก0=0๐บ (๐ฅ0, ๐ก0; ๐ฅ, ๐ก) ๐ (๐ฅ0, ๐ก0) ๐๐ก0๐๐ฅ0
(11)
This completes the solution.
Summary
The solution is
๐ข (๐ฅ, ๐ก) = ๏ฟฝ๐ฟ
๐ฅ0=0๐ (๐ฅ0) ๐บ (๐ฅ0, ๐ก0; ๐ฅ, 0) ๐๐ฅ0โ๏ฟฝ
๐ก
๐ก=0๐ด๐๐บ (๐ฅ0, ๐ก0; ๐ฟ, ๐ก)
๐๐ฅ0๐๐ก0+๏ฟฝ
๐ฟ
๐ฅ0=0๏ฟฝ
๐ก
๐ก0=0๐บ (๐ฅ0, ๐ก0; ๐ฅ, ๐ก) ๐ (๐ฅ0, ๐ก0) ๐๐ก0๐๐ฅ0
Where ๐บ (๐ฅ0, ๐ก0; ๐ฅ, 0) is given in (10) (after changing roles of parameters):
295
5.2. Exam 2 CHAPTER 5. EXAMS
๐บ (๐ฅ0, ๐ก0; ๐ฅ, 0) =1
โ4๐๐๐ก
โโโโโ
โ๏ฟฝ๐=โโ
expโโโโโโ (๐ฅ0 โ (4๐๐ฟ โ ๐ฅ))
2
4๐๐ก
โโโโโ + exp
โโโโโโ (๐ฅ0 โ (โ4๐๐ฟ + ๐ฅ))
2
4๐๐ก
โโโโโ
โโโโโ
โ1
โ4๐๐๐ก
โโโโโ
โ๏ฟฝ๐=โโ
expโโโโโโ (๐ฅ0 โ ((4๐ โ 2) ๐ฟ + ๐ฅ))
2
4๐๐ก
โโโโโ + exp
โโโโโโ (๐ฅ0 โ ((4๐ โ 2) ๐ฟ โ ๐ฅ))
2
4๐๐ก
โโโโโ
โโโโโ
(10A)
and ๐๐บ(๐ฅ0,๐ก0;๐ฟ,๐ก)๐๐ฅ0
is given in (9) (after also changing roles of parameters):
๐๐บ (๐ฅ0, ๐ก0; ๐ฟ, ๐ก)๐๐ฅ0
=1
โ4๐๐ (๐ก โ ๐ก0)
โ๏ฟฝ๐=โโ
โ (๐ฟ โ (4๐๐ฟ โ ๐ฅ))2๐ (๐ก โ ๐ก0)
expโโโโโโ (๐ฟ โ (4๐๐ฟ โ ๐ฅ))2
4๐ (๐ก โ ๐ก0)
โโโโโ
+1
โ4๐๐ (๐ก โ ๐ก0)
โ๏ฟฝ๐=โโ
โ (๐ฟ โ (โ4๐๐ฟ + ๐ฅ))2๐ (๐ก โ ๐ก0)
expโโโโโโ (๐ฟ โ (โ4๐๐ฟ + ๐ฅ))2
4๐ (๐ก โ ๐ก0)
โโโโโ
โ1
โ4๐๐ (๐ก โ ๐ก0)
โ๏ฟฝ๐=โโ
โ (๐ฟ โ ((4๐ โ 2) ๐ฟ + ๐ฅ))2๐ (๐ก โ ๐ก0)
expโโโโโโ (๐ฟ โ ((4๐ โ 2) ๐ฟ + ๐ฅ))2
4๐ (๐ก โ ๐ก0)
โโโโโ
โ1
โ4๐๐ (๐ก โ ๐ก0)
โ๏ฟฝ๐=โโ
โ (๐ฟ โ ((4๐ โ 2) ๐ฟ โ ๐ฅ))2๐ (๐ก โ ๐ก0)
expโโโโโโ (๐ฟ โ ((4๐ โ 2) ๐ฟ โ ๐ฅ))2
4๐ (๐ก โ ๐ก0)
โโโโโ (9A)
and ๐บ (๐ฅ0, ๐ก0; ๐ฅ, ๐ก) is given in (7A), but with roles changed as well to become
๐บ (๐ฅ0, ๐ก0; ๐ฅ, ๐ก) =1
โ4๐๐ (๐ก โ ๐ก0)
โโโโโ
โ๏ฟฝ๐=โโ
expโโโโโโ (๐ฅ0 โ (4๐๐ฟ โ ๐ฅ))
2
4๐ (๐ก โ ๐ก0)
โโโโโ + exp
โโโโโโ (๐ฅ0 โ (โ4๐๐ฟ + ๐ฅ))
2
4๐ (๐ก โ ๐ก0)
โโโโโ
โโโโโ
(7AA)
โ1
โ4๐๐ (๐ก โ ๐ก0)
โโโโโ
โ๏ฟฝ๐=โโ
expโโโโโโ (๐ฅ0 โ ((4๐ โ 2) ๐ฟ + ๐ฅ))
2
4๐ (๐ก โ ๐ก0)
โโโโโ + exp
โโโโโโ (๐ฅ0 โ ((4๐ โ 2) ๐ฟ โ ๐ฅ))
2
4๐ (๐ก โ ๐ก0)
โโโโโ
โโโโโ
5.2.2.2 Part(b)
When ๐ด = 0, the solution in part (a) becomes
๐ข (๐ฅ, ๐ก) = ๏ฟฝ๐ฟ
๐ฅ0=0๐ (๐ฅ0) ๐บ (๐ฅ0, ๐ก0; ๐ฅ, 0) ๐๐ฅ0 +๏ฟฝ
๐ฟ
๐ฅ0=0๏ฟฝ
๐ก
๐ก0=0๐บ (๐ฅ0, ๐ก0; ๐ฅ, ๐ก) ๐ ๐๐ก0๐๐ฅ0
Where ๐บ (๐ฅ0, ๐ก0; ๐ฅ, 0) is given in (10A) in part (a), and ๐บ (๐ฅ0, ๐ก0; ๐ฅ, ๐ก) is given in (7AA) in part(a). Now we find the eigenfunction solution for this problem order to compare it with theabove green function images solution. Since ๐ด = 0 then the PDE now becomes
๐๐ข๐๐ก
= ๐๐2๐ข๐๐ฅ2
+ ๐ (๐ฅ, ๐ก) (1)
0 โค ๐ฅ โค ๐ฟ๐ก โฅ 0
Initial conditions
๐ข (๐ฅ, 0) = ๐ (๐ฅ)
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5.2. Exam 2 CHAPTER 5. EXAMS
Boundary conditions
๐๐ข (0, ๐ก)๐๐ฅ
= 0
๐ข (๐ฟ, ๐ก) = 0
Since boundary conditions has now become homogenous (thanks for ๐ด = 0), we can useseparation of variables to find the eigenfunctions, and then use eigenfunction expansion.Let the solution be
๐ข๐ (๐ฅ, ๐ก) = ๐๐ (๐ก) ๐๐ (๐ฅ)
๐ข (๐ฅ, ๐ก) =โ๏ฟฝ๐=1
๐๐ (๐ก) ๐๐ (๐ฅ) (1A)
Where ๐๐ (๐ก) are eigenfunctions for the associated homogeneous PDE ๐๐ข๐๐ก = ๐
๐2๐ข๐๐ฅ2 which can
be found from separation of variables. To find ๐๐ (๐ฅ), we start by separation of variables.Let ๐ข (๐ฅ, ๐ก) = ๐ (๐ฅ) ๐ (๐ก) and we plug this solution back to the PDE to obtain
๐๐โฒ = ๐๐โฒโฒ๐1๐ฃ๐โฒ
๐=๐โฒโฒ
๐= โ๐
Hence the spatial ODE is ๐โฒโฒ
๐ = โ๐ or ๐โฒโฒ + ๐๐ = 0 with boundary conditions
๐โฒ (0) = 0๐ (๐ฟ) = 0
case ๐ = 0 The solution is ๐ = ๐1 + ๐2๐ฅ. Hence ๐โฒ = ๐2. Therefore ๐2 = 0. Hence ๐ = ๐1 = 0.Trivial solution. So ๐ = 0 is not possible.
case ๐ > 0 The solution is ๐ = ๐1 cos ๏ฟฝโ๐๐ฅ๏ฟฝ+๐2 sin ๏ฟฝโ๐๐ฅ๏ฟฝ. and ๐โฒ = โโ๐๐1 sin๐๐ฅ+๐2โ๐ cos๐๐ฅ.From first B.C. at ๐ฅ = 0 we find 0 = ๐2โ๐, hence ๐2 = 0 and the solution becomes ๐ =๐1 cos ๏ฟฝโ๐๐ฅ๏ฟฝ. At ๐ฅ = ๐ฟ, we have 0 = ๐1 cos ๏ฟฝโ๐๐ฟ๏ฟฝ which leads to โ๐๐ฟ = ๐
๐2 for ๐ = 1, 3, 5,โฏ or
๏ฟฝ๐๐ = ๏ฟฝ2๐ โ 12 ๏ฟฝ
๐๐ฟ
๐ = 1, 2, 3,โฏ
๐๐ = ๏ฟฝ2๐ โ 12
๐๐ฟ ๏ฟฝ
2
๐ = 1, 1, 2, 3,โฏ
Hence the ๐๐ (๐ฅ) solution is
๐๐ (๐ฅ) = ๐๐ cos ๏ฟฝ๏ฟฝ๐๐๐ฅ๏ฟฝ ๐ = 1, 2, 3,โฏ
The time ODE is now solved using the above eigenvalues. (we really do not need to do thispart, since ๐ด๐ (๐ก) will be solved for later, and ๐ด๐ (๐ก) will contain all the time dependent parts,
297
5.2. Exam 2 CHAPTER 5. EXAMS
including those that come from ๐ (๐ฅ, ๐ก), but for completion, this is done)1๐ฃ๐โฒ
๐= โ๐๐
๐โฒ + ๐ฃ๐๐๐ = 0๐๐๐= โ๐ฃ๐๐๐๐ก
ln |๐| = โ๐ฃ๐๐๐ก + ๐ถ๐ = ๐ถ๐๐๐ฃ๐๐๐ก
Hence the solution to the homogenous PDE is
๐ข๐ (๐ฅ, ๐ก) = ๐๐๐๐= ๐๐ cos ๏ฟฝ๏ฟฝ๐๐๐ฅ๏ฟฝ ๐๐ฃ๐๐๐ก
Where constants of integration are merged into one. Therefore
๐ข (๐ฅ, ๐ก) =โ๏ฟฝ๐=1
๐๐๐๐
=โ๏ฟฝ๐=1
๐๐ cos ๏ฟฝ๏ฟฝ๐๐๐ฅ๏ฟฝ ๐๐ฃ๐๐๐ก (2)
From the above we see that
๐๐ (๐ฅ) = cos ๏ฟฝโ๐๐๐ฅ๏ฟฝ
Using this, we now write the solution to ๐๐ข๐๐ก = ๐
๐2๐ข๐๐ฅ2 + ๐ (๐ฅ, ๐ก) using eigenfunction expansion
๐ข (๐ฅ, ๐ก) =โ๏ฟฝ๐=1
๐ด๐ (๐ก) ๐๐ (๐ฅ) (3)
Where now ๐ด๐ (๐ก) will have all the time dependent terms from ๐ (๐ฅ, ๐ก) as well from the time
solution from the homogenous PDE ๐๐ฃ๏ฟฝ 2๐โ12
๐๐ฟ ๏ฟฝ
2๐กpart. We will solve for ๐ด๐ (๐ก) now.
In this below, we will expand ๐ (๐ฅ, ๐ก) using these eigenfunctions (we can do this, since theeigenfunctions are basis for the whole solution space which the forcing function is in as well).We plug-in (3) back into the PDE, and since boundary conditions are now homogenous,
then term by term di๏ฟฝerentiation is justified. The PDE ๐๐ข๐๐ก = ๐
๐2๐ข๐๐ฅ2 + ๐ (๐ฅ, ๐ก) now becomes
๐๐๐ก
โ๏ฟฝ๐=1
๐ด๐ (๐ก) ๐๐ (๐ฅ) = ๐๐2
๐๐ฅ2โ๏ฟฝ๐=1
๐ด๐ (๐ก) ๐๐ (๐ฅ) +โ๏ฟฝ๐=1
๐๐ (๐ก) ๐๐ (๐ฅ) (4)
Where โโ๐=1 ๐๐ (๐ก) ๐๐ (๐ฅ) is the eigenfunction expansion of ๐ (๐ฅ, ๐ก). To find ๐๐ (๐ก) we apply
298
5.2. Exam 2 CHAPTER 5. EXAMS
orthogonality as follows
๐ (๐ฅ, ๐ก) =โ๏ฟฝ๐=1
๐๐ (๐ก) ๐๐ (๐ฅ)
๏ฟฝ๐ฟ
0๐๐ (๐ฅ)๐ (๐ฅ, ๐ก) ๐๐ฅ = ๏ฟฝ
๐ฟ
0๏ฟฝโ๏ฟฝ๐=1
๐๐ (๐ก) ๐๐ (๐ฅ)๏ฟฝ ๐๐ (๐ฅ) ๐๐ฅ
๏ฟฝ๐ฟ
0๐๐ (๐ฅ)๐ (๐ฅ, ๐ก) ๐๐ฅ =
โ๏ฟฝ๐=1
๏ฟฝ๐ฟ
0๐๐ (๐ก) ๐๐ (๐ฅ) ๐๐ (๐ฅ) ๐๐ฅ
= ๏ฟฝ๐ฟ
0๐๐ (๐ก) ๐2
๐ (๐ฅ) ๐๐ฅ
= ๐๐ (๐ก)๏ฟฝ๐ฟ
0cos2 ๏ฟฝ๏ฟฝ๐๐๐ฅ๏ฟฝ ๐๐ฅ
But
๏ฟฝ๐ฟ
0cos2 ๏ฟฝ๏ฟฝ๐๐๐ฅ๏ฟฝ ๐๐ฅ =
๐ฟ2
Hence
๐๐ (๐ก) =2๐ฟ ๏ฟฝ
๐ฟ
0๐๐ (๐ฅ)๐ (๐ฅ, ๐ก) ๐๐ฅ (5)
Now that we found ๐๐ (๐ก), we go back to (4) and simplifies it moreโ๏ฟฝ๐=1
๐ดโฒ๐ (๐ก) ๐๐ (๐ฅ) = ๐
โ๏ฟฝ๐=1
๐ด๐ (๐ก) ๐โฒโฒ๐ (๐ฅ) +
โ๏ฟฝ๐=1
๐๐ (๐ก) ๐๐ (๐ฅ)
๐ดโฒ๐ (๐ก) ๐๐ (๐ฅ) = ๐๐ด๐ (๐ก) ๐โฒโฒ
๐ (๐ฅ) + ๐๐ (๐ก) ๐๐ (๐ฅ)
But since ๐๐ (๐ฅ) = cos ๏ฟฝโ๐๐๐ฅ๏ฟฝ then
๐โฒ๐ (๐ฅ) = โ ๏ฟฝ๏ฟฝ๐๐๏ฟฝ sin ๏ฟฝ๏ฟฝ๐๐๐ฅ๏ฟฝ
And
๐โฒโฒ๐ (๐ฅ) = โ๐๐ cos ๏ฟฝ๏ฟฝ๐๐๐ฅ๏ฟฝ
= โ๐๐๐๐ (๐ฅ)
Hence the above ODE becomes
๐ดโฒ๐๐๐ = โ๐๐ด๐๐๐๐๐ + ๐๐๐๐
Canceling the eigenfunction ๐๐ (๐ฅ) (since not zero) gives
๐ดโฒ๐ (๐ก) + ๐๐ด๐ (๐ก) ๐๐ = ๐๐ (๐ก) (6)
We now solve this for ๐ด๐ (๐ก). Integrating factor is
๐ = exp ๏ฟฝ๏ฟฝ๐๐๐๐๐ก๏ฟฝ
= ๐๐๐๐๐ก
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5.2. Exam 2 CHAPTER 5. EXAMS
Hence (6) becomes๐๐๐ก๏ฟฝ๐๐ด๐ (๐ก)๏ฟฝ = ๐๐๐ (๐ก)
๐๐๐๐๐ก๐ด๐ (๐ก) = ๏ฟฝ๐ก
0๐๐๐๐๐ ๐๐ (๐ ) ๐๐ + ๐ถ
๐ด๐ (๐ก) = ๐โ๐๐๐๐ก๏ฟฝ๐ก
0๐๐๐๐๐ ๐๐ (๐ ) ๐๐ + ๐ถ๐โ๐๐๐๐ก (7)
Now that we found ๐ด๐ (๐ก), then the solution (3) becomes
๐ข (๐ฅ, ๐ก) =โ๏ฟฝ๐=1
โโโโโโ๐
โ๏ฟฝ 2๐โ12๐๐ฟ ๏ฟฝ
2๐๐ก๏ฟฝ
๐ก
0๐๏ฟฝ 2๐โ12
๐๐ฟ ๏ฟฝ
2๐๐ ๐๐ (๐ ) ๐๐ + ๐ถ๐
โ๏ฟฝ 2๐โ12๐๐ฟ ๏ฟฝ
2๐๐กโโโโโโ ๐๐ (๐ฅ) (8)
At ๐ก = 0 , we are given that ๐ข (๐ฅ, 0) = ๐ (๐ฅ), hence the above becomes
๐ (๐ฅ) =โ๏ฟฝ๐=1
๐ถ๐๐ (๐ฅ)
To find ๐ถ, we apply orthogonality again, which gives
๏ฟฝ๐ฟ
0๐ (๐ฅ) ๐๐ (๐ฅ) ๐๐ฅ =
โ๏ฟฝ๐=1
๏ฟฝ๐ฟ
0๐ถ๐๐ (๐ฅ) ๐๐ (๐ฅ) ๐๐ฅ
= ๐ถ๏ฟฝ๐ฟ
0๐2๐ (๐ฅ) ๐๐ฅ
๏ฟฝ๐ฟ
0๐ (๐ฅ) ๐๐ (๐ฅ) ๐๐ฅ =
๐ฟ2๐ถ
๐ถ =2๐ฟ ๏ฟฝ
๐ฟ
0๐ (๐ฅ) ๐๐ (๐ฅ) ๐๐ฅ
Now that we found ๐ถ, then the solution in (8) is complete. It is
๐ข (๐ฅ, ๐ก) =โ๏ฟฝ๐=1
๐ด๐ (๐ก) ๐๐ (๐ฅ)
=โ๏ฟฝ๐=1
๏ฟฝ๐โ๐๐๐๐ก๏ฟฝ๐ก
0๐๐๐๐๐ ๐๐ (๐ ) ๐๐ +
2๐ฟ๐โ๐๐๐๐ก๏ฟฝ
๐ฟ
0๐ (๐ฅ) ๐๐ (๐ฅ) ๐๐ฅ๏ฟฝ๐๐ (๐ฅ)
Or
๐ข (๐ฅ, ๐ก) =โ๏ฟฝ๐=1
๏ฟฝ๐๐ (๐ฅ) ๐โ๐๐๐๐ก๏ฟฝ๐ก
0๐๐๐๐๐ ๐๐ (๐ ) ๐๐ ๏ฟฝ
+2๐ฟ
โ๏ฟฝ๐=1
๏ฟฝ๐๐ (๐ฅ) ๐โ๐๐๐๐ก๏ฟฝ๐ฟ
0๐ (๐ฅ) ๐๐ (๐ฅ) ๐๐ฅ๏ฟฝ
Summary
The eigenfunction solution is
๐ข (๐ฅ, ๐ก) =โ๏ฟฝ๐=1
๏ฟฝ๐๐ (๐ฅ) ๐โ๐๐๐๐ก๏ฟฝ๐ก
0๐๐๐๐๐ ๐๐ (๐ ) ๐๐ ๏ฟฝ
+2๐ฟ
โ๏ฟฝ๐=1
๏ฟฝ๐๐ (๐ฅ) ๐โ๐๐๐๐ก๏ฟฝ๐ฟ
0๐ (๐ฅ) ๐๐ (๐ฅ) ๐๐ฅ๏ฟฝ
300
5.2. Exam 2 CHAPTER 5. EXAMS
Where
๐๐ (๐ฅ) = cos ๏ฟฝ๏ฟฝ๐๐๐ฅ๏ฟฝ ๐ = 1, 2, 3,โฏ
๐๐ = ๏ฟฝ2๐ โ 12
๐๐ฟ ๏ฟฝ
2
๐ = 1, 1, 2, 3,โฏ
And
๐๐ (๐ก) =2๐ฟ ๏ฟฝ
๐ฟ
0๐๐ (๐ฅ)๐ (๐ฅ, ๐ก) ๐๐ฅ
To compare the eigenfunction expansion solution and the Green function solution, we seethe following mapping of the two solutions
2๐ฟโโ๐=1๏ฟฝ๐๐(๐ฅ)๐
โ๐๐๐๐กโซ๐ฟ0
๐(๐ฅ)๐๐(๐ฅ)๐๐ฅ๏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
๐ฟ
0๐ (๐ฅ0) ๐บ (๐ฅ0, ๐ก0; ๐ฅ, 0) ๐๐ฅ0 +
โโ๐=1๏ฟฝ๐๐(๐ฅ)๐
โ๐๐๐๐กโซ๐ก0๐๐๐๐๐ ๐๐(๐ )๐๐ ๏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๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ
๐ฟ
0๏ฟฝ
๐ก
0๐บ (๐ฅ0, ๐ก0; ๐ฅ, ๐ก) ๐ (๐ฅ0, ๐ก0) ๐๐ก0๐๐ฅ0
Where the top expression is the eigenfunction expansion and the bottom expression is theGreen function solution using method of images. Where ๐บ (๐ฅ0, ๐ก0; ๐ฅ, ๐ก) in above contains theinfinite sums of the images. So the Green function solution contains integrals and insidethese integrals are the infinite sums. While the eigenfunction expansions contains two infinitesums, but inside the sums we see the integrals. So summing over the images seems to beequivalent to the operation of summing over eigenfunctions. These two solutions in thelimit should of course give the same result (unless I made a mistake somewhere). In thisexample, I found method of eigenfunction expansion easier, since getting the images incorrect locations and sign was tricky to get right.
301
5.2. Exam 2 CHAPTER 5. EXAMS
5.2.3 problem 5
(e) Write your expression for ฯ(x, t) in terms of
f(ฮท, x, t) =
โซ ฮท
xo
F (ฮฑ)dฮฑ+(xโ ฮท)2
2t.
(f) Find ฯx(x, t) and then use equation (3b) to find u(x, t).
4. (a) Use the Method of Images to solve
โu
โt= ฮฝ
โ2u
โx2+Q(x, t), 0 โค x โค L, t โฅ 0
u(x, 0) = f(x),โu
โx(0, t) = 0, u(L, t) = A
(b) For A = 0, compare your expression for the solution in (a) to the eigenfunction solution.
5. This problem is a simple model for diffraction of light passing through infinitesimallysmall slits separated by a distance 2a.
Solve the diffraction equation
โu
โt=iฮป
4ฯ
โ2u
โx2(1)
with initial source u(x, 0) = f(x) = ฮด(xโ a) + ฮด(x+ a), a > 0.
Show that the solution u(x, t) oscillates wildly, but that the intensity |u(x, t)|2 is well-behaved. The intensity |u(x, t)|2 shows that the diffraction pattern at a distance t consistsof a series of alternating bright and dark fringes with period ฮปt/(2a).
3
๐๐ข (๐ฅ, ๐ก)๐๐ก
=๐๐4๐
๐2๐ข (๐ฅ, ๐ก)๐๐ฅ2
(1)
๐ข (๐ฅ, 0) = ๐ (๐ฅ) = ๐ฟ (๐ฅ โ ๐) + ๐ฟ (๐ฅ + ๐)
I will Use Fourier transform to solve this, since this is for โโ < ๐ฅ < โ and the solution ๐ข (๐ฅ, ๐ก)is assumed bounded at ยฑโ (or goes to zero there), hence ๐ข (๐ฅ, ๐ก) is square integrable andtherefore we can assume it has a Fourier transform.
Let ๐ (๐, ๐ก) be the spatial part only Fourier transform of ๐ข (๐ฅ, ๐ก). Using the Fourier transformpairs defined as
๐ (๐, ๐ก) = โฑ (๐ข (๐ฅ, ๐ก)) = ๏ฟฝโ
โโ๐ข (๐ฅ, ๐ก) ๐โ๐2๐๐ฅ๐๐๐ฅ
๐ข (๐ฅ, ๐ก) = โฑ โ1 (๐ (๐, ๐ก)) = ๏ฟฝโ
โโ๐ (๐, ๐ก) ๐๐2๐๐ฅ๐๐๐
Therefore, by Fourier transform properties of derivatives
โฑ๏ฟฝ๐๐ข (๐ฅ, ๐ก)๐๐ฅ ๏ฟฝ = (2๐๐๐)๐ (๐, ๐ก)
โฑ ๏ฟฝ๐2๐ข (๐ฅ, ๐ก)๐๐ฅ2 ๏ฟฝ = (2๐๐๐)2๐ (๐, ๐ก) (2)
And
โฑ๏ฟฝ๐๐ข (๐ฅ, ๐ก)๐๐ก ๏ฟฝ =
๐๐ (๐, ๐ก)๐๐ก
(3)
Where in (3), we just need to take time derivative of ๐ (๐, ๐ก) since the transform is appliedonly to the space part. Now we take the Fourier transform of the given PDE and using (2,3)
302
5.2. Exam 2 CHAPTER 5. EXAMS
relations we obtain the PDE but now in Fourier space.
๐๐ (๐, ๐ก)๐๐ก
= ๏ฟฝ๐๐4๐๏ฟฝ
(2๐๐๐)2๐ (๐, ๐ก)
= โ ๏ฟฝ๐๐4๐๏ฟฝ
4๐2๐2๐ (๐, ๐ก)
= ๏ฟฝโ๐๐๐๐2๏ฟฝ๐ (๐, ๐ก) (4)
Equation (4) can now be easily solved for ๐ (๐, ๐ก) since it is separable.๐๐ (๐, ๐ก)๐ (๐, ๐ก)
= ๏ฟฝโ๐๐๐๐2๏ฟฝ ๐๐ก
Integrating
ln |๐ (๐, ๐ก)| = ๏ฟฝโ๐๐๐๐2๏ฟฝ ๐ก + ๐ถ
๐ (๐, ๐ก) = ๐ (๐, 0) ๐๏ฟฝโ๐๐๐๐2๏ฟฝ๐ก
Where ๐ (๐, 0) is the Fourier transform of ๐ข (๐ฅ, 0), the initial conditions, which is ๐ (๐ฅ) and isgiven in the problem. To go back to spatial domain, we now need to do the inverse Fouriertransform. By applying the convolution theorem, we know that multiplication in Fourierdomain is convolution in spatial domain, therefore
โฑ โ1 (๐ (๐, ๐ก)) = โฑ โ1 (๐ (๐, 0)) โ โฑ โ1 ๏ฟฝ๐โ๐๐๐๐2๐ก๏ฟฝ (5)
But
โฑ โ1 (๐ (๐, ๐ก)) = ๐ข (๐ฅ, ๐ก)โฑ โ1 (๐ (๐, 0)) = ๐ (๐ฅ)
And
โฑ โ1 ๏ฟฝ๐โ๐๐๐๐2๐ก๏ฟฝ = ๏ฟฝโ
โโ๐โ๐๐๐๐2๐ก๐2๐๐๐ฅ๐๐๐ (5A)
Hence (5) becomes
๐ข (๐ฅ, ๐ก) = ๐ (๐ฅ) โ โฑ โ1 ๏ฟฝ๐โ๐๐๐๐2๐ก๏ฟฝ
Here, I used Mathematica to help me with the above integral (5A) as I could not find it intables so far2. Here is the result
Find inverse Fourier transform, for problem 5, NE 548
In[13]:= InverseFourierTransform[ Exp[-I lam Pi z^2 t], z, x, FourierParameters -> {1, -2 * Pi}]
Out[13]=โ
โ ฯ x2
lam t
2 ฯ โ lam t
2Trying to do this integral by hand also, but so far having some di๏ฟฝculty..
303
5.2. Exam 2 CHAPTER 5. EXAMS
Therefore, from Mathematica, we see that
โฑ โ1 ๏ฟฝ๐โ๐๐๐๐2๐ก๏ฟฝ =๐๐๐๐ฅ2๐๐ก
โ2๐โ๐โ๐๐ก(6)
But3
โ๐ =1
โ2+ ๐
1
โ2
= cos ๏ฟฝ๐4๏ฟฝ + ๐ sin ๏ฟฝ๐
4๏ฟฝ
= ๐๐๐4
Hence (6) becomes
โฑ โ1 ๏ฟฝ๐โ๐๐๐๐2๐ก๏ฟฝ =1
๐๐๐4โ2๐๐๐ก
๐๐๐๐ฅ2๐๐ก (7)
Now we are ready to do the convolution in (5A) since we know everything in the RHS,hence
๐ข (๐ฅ, ๐ก) = ๐ (๐ฅ) โ1
๐๐๐4โ2๐๐๐ก
๐๐๐๐ฅ2๐๐ก
= (๐ฟ (๐ฅ โ ๐) + ๐ฟ (๐ฅ + ๐)) โ1
๐๐๐4โ2๐๐๐ก
๐๐๐๐ฅ2๐๐ก (8)
Applying convolution integral on (8), which says that
๐ (๐ฅ) = ๐1 (๐ฅ) โ ๐2 (๐ฅ)
= ๏ฟฝโ
โโ๐1 (๐ง) ๐2 (๐ฅ โ ๐ง) ๐๐ง
Therefore (8) becomes
๐ข (๐ฅ, ๐ก) = ๏ฟฝโ
โโ(๐ฟ (๐ง โ ๐) + ๐ฟ (๐ง + ๐))
1
๐๐๐4โ2๐๐๐ก
๐๐๐(๐ฅโ๐ง)2
๐๐ก ๐๐ง
=1
๐๐๐4โ2๐๐๐ก
๏ฟฝโ
โโ(๐ฟ (๐ง โ ๐) + ๐ฟ (๐ง + ๐)) ๐๐
๐(๐ฅโ๐ง)2
๐๐ก ๐๐ง
=1
๐๐๐4โ2๐๐๐ก
โโโโโ๏ฟฝ
โ
โโ๐ฟ (๐ง โ ๐) ๐๐
๐(๐ฅโ๐ง)2
๐๐ก ๐๐ง +๏ฟฝโ
โโ๐ฟ (๐ง + ๐) ๐๐
๐(๐ฅโ๐ง)2
๐๐ก ๐๐งโโโโโ
But an integral with delta function inside it, is just the integrand evaluated where the deltaargument become zero which is at ๐ง = ๐ and ๐ง = โ๐ in the above. (This is called the siftingproperty). Hence the above integrals are now easily found and we obtain the solution
๐ข (๐ฅ, ๐ก) =1
๐๐๐4โ2๐๐๐ก
โโโโโexp
โโโโโ๐๐ (๐ฅ โ ๐)2
๐๐ก
โโโโโ + exp
โโโโโ๐๐ (๐ฅ + ๐)2
๐๐ก
โโโโโ
โโโโโ
3Taking the positive root only.
304
5.2. Exam 2 CHAPTER 5. EXAMS
The above is the solution we need. But we can simplify it more by using Euler relation.
๐ข (๐ฅ, ๐ก) =1
๐๐๐4โ2๐๐๐ก
โโโโโโexp
โโโโโโ๐๐ ๏ฟฝ๐ฅ2 + ๐2 โ 2๐ฅ๐๏ฟฝ
๐๐ก
โโโโโโ + exp
โโโโโโ๐๐ ๏ฟฝ๐ฅ2 + ๐2 + 2๐๐ฅ๏ฟฝ
๐๐ก
โโโโโโ
โโโโโโ
=1
๐๐๐4โ2๐๐๐ก
โโโโโโexp
โโโโโโ๐๐ ๏ฟฝ๐ฅ2 + ๐2๏ฟฝ
๐๐ก
โโโโโโ exp ๏ฟฝ
โ๐2๐๐ฅ๐๐๐ก ๏ฟฝ + exp
โโโโโโ๐๐ ๏ฟฝ๐ฅ2 + ๐2๏ฟฝ
๐๐ก
โโโโโโ exp ๏ฟฝ
๐2๐๐๐ฅ๐๐ก ๏ฟฝ
โโโโโโ
Taking exp ๏ฟฝ๐๐๏ฟฝ๐ฅ2+๐2๏ฟฝ
๐๐ก ๏ฟฝ as common factor outside results in
๐ข (๐ฅ, ๐ก) =exp ๏ฟฝ
๐๐๏ฟฝ๐ฅ2+๐2๏ฟฝ
๐๐ก ๏ฟฝ
๐๐๐4โ2๐๐๐ก
๏ฟฝexp ๏ฟฝ๐2๐๐๐ฅ๐๐ก ๏ฟฝ + exp ๏ฟฝโ๐
2๐๐ฅ๐๐๐ก ๏ฟฝ๏ฟฝ
=2 exp ๏ฟฝ
๐๐๏ฟฝ๐ฅ2+๐2๏ฟฝ
๐๐ก ๏ฟฝ
๐๐๐4โ2๐๐๐ก
โโโโโโโโโโ
exp ๏ฟฝ๐2๐๐๐ฅ๐๐ก๏ฟฝ + exp ๏ฟฝโ๐2๐๐ฅ๐๐๐ก
๏ฟฝ
2
โโโโโโโโโโ
=2 exp ๏ฟฝ
๐๐๏ฟฝ๐ฅ2+๐2๏ฟฝ
๐๐ก ๏ฟฝ
๐๐๐4โ2๐๐๐ก
cos ๏ฟฝ2๐๐๐ฅ๐๐ก ๏ฟฝ
=โ2 exp ๏ฟฝ
๐๐๏ฟฝ๐ฅ2+๐2๏ฟฝ
๐๐ก โ ๐๐4 ๏ฟฝ
โ๐๐๐กcos ๏ฟฝ
2๐๐๐ฅ๐๐ก ๏ฟฝ
=โ2 exp ๏ฟฝ๐ ๏ฟฝ
๐๏ฟฝ๐ฅ2+๐2๏ฟฝ
๐๐ก โ ๐4 ๏ฟฝ๏ฟฝ
โ๐๐๐กcos ๏ฟฝ
2๐๐๐ฅ๐๐ก ๏ฟฝ
Hence the final solution is
๐ข (๐ฅ, ๐ก) = ๏ฟฝ2
๐๐๐ก exp ๏ฟฝ๐๐๏ฟฝ๐ฅ2+๐2๏ฟฝ
๐๐ก โ ๐4 ๏ฟฝ cos ๏ฟฝ2๐๐๐
๐ฅ๐ก๏ฟฝ (9)
Hence the real part of the solution is
โ (๐ข (๐ฅ, ๐ก)) =๏ฟฝ
2๐๐๐ก
cosโโโโโโ๐ ๏ฟฝ๐ฅ2 + ๐2๏ฟฝ
๐๐กโ๐4
โโโโโโ cos ๏ฟฝ
2๐๐๐
๐ฅ๐ก ๏ฟฝ
And the imaginary part of the solution is
โ (๐ข (๐ฅ, ๐ก)) =๏ฟฝ
2๐๐๐ก
sinโโโโโโ๐ ๏ฟฝ๐ฅ2 + ๐2๏ฟฝ
๐๐กโ๐4
โโโโโโ cos ๏ฟฝ
2๐๐๐
๐ฅ๐ก ๏ฟฝ
The ๐4 is just a phase shift. Here is a plot of the Real and Imaginary parts of the solution,
using for ๐ = 600 ร 10โ9๐๐๐ก๐๐, ๐ = 1000 ร 10โ9๐๐๐ก๐๐ at ๐ก = 1 second
305
5.2. Exam 2 CHAPTER 5. EXAMS
In[15]:= a = 1000 * 10^(-9);
ฮป = 600 * 10^(-9);
u[x_, t_] :=2
ฯ ฮป t
ExpIฯ x
2+ a2
ฮป t
-ฯ
4 Cos
2 ฯ a
ฮป
x
t
In[59]:= Clear[t];
Plot[Re@u[x, 1], {x, -3000 a, 3000 a}, Frame -> True,
FrameLabel โ {{"u(x,t)", None}, {"x", "Real part of solution at t= 1 second"}}, BaseStyle โ 12,
PlotStyle โ Red, PlotTheme โ "Classic"]
Out[60]=
-0.003 -0.002 -0.001 0.000 0.001 0.002 0.003
-400
-200
0
200
400
x
u(x,t)
Real part of solution at t= 1 second
In[61]:= Plot[Im@u[x, 1], {x, -3000 a, 3000 a}, Frame -> True,
FrameLabel โ {{"u(x,t)", None}, {"x", "Imaginary part of solution at t= 1 second"}}, BaseStyle โ 12,
PlotStyle โ Red, PlotTheme โ "Classic"]
Out[61]=
-0.003 -0.002 -0.001 0.000 0.001 0.002 0.003
-400
-200
0
200
400
x
u(x,t)
Imaginary part of solution at t= 1 second
We see the rapid oscillations as distance goes away from the origin. This is due to the ๐ฅ2
term making the radial frequency value increase quickly with ๐ฅ. We now plot the |๐ข (๐ฅ, ๐ก)|2.Looking at solution in (9), and since complex exponential is ยฑ1, then the amplitude is
governed by ๏ฟฝ2
๐๐๐ก cos ๏ฟฝ2๐๐๐๐ฅ๐ก๏ฟฝ part of the solution. Hence
|๐ข (๐ฅ, ๐ก)|2 = 2๐๐๐ก cos2 ๏ฟฝ2๐๐๐
๐ฅ๐ก๏ฟฝ
These plots show the intensity at di๏ฟฝerent time values. We see from these plots, that theintensity is well behaved in that it does not have the same rapid oscillations seen in the๐ข (๐ฅ, ๐ก) solution plots.
306
5.2. Exam 2 CHAPTER 5. EXAMS
In[154]:= a = 1000 * 10^(-9);
ฮป = 600 * 10^(-9);
intensity[x_, t_] :=2
ฯ ฮป t
Cos2 ฯ a
ฮป
x
t
2
p =
Plotintensity[x, #], {x, -3000 a, 3000 a}, Frame -> True,
FrameLabel โ {"u(x,t)", None}, "x", Row" intensity |u 2 of solution at t =", #, " second",
BaseStyle โ 12, PlotStyle โ Red, PlotTheme โ "Classic", ImageSize โ 300 & /@ Range[0.002, 0.012, 0.002];
Out[153]=
-0.003-0.002-0.001 0.000 0.001 0.002 0.003
0
1ร 108
2ร 108
3ร 108
4ร 108
5ร 108
x
u(x,t)
intensity |u 2 of solution at t =0.002 second
-0.003-0.002-0.001 0.000 0.001 0.002 0.003
0
5.0ร 107
1.0ร 108
1.5ร 108
2.0ร 108
2.5ร 108
xu(x,t)
intensity |u 2 of solution at t =0.004 second
-0.003-0.002-0.001 0.000 0.001 0.002 0.003
0
5.0ร 107
1.0ร 108
1.5ร 108
x
u(x,t)
intensity |u 2 of solution at t =0.006 second
-0.003-0.002-0.001 0.000 0.001 0.002 0.003
0
2.0ร 1074.0ร 1076.0ร 1078.0ร 1071.0ร 1081.2ร 1081.4ร 108
x
u(x,t)
intensity |u 2 of solution at t =0.008 second
-0.003-0.002-0.001 0.000 0.001 0.002 0.003
0
2ร 107
4ร 107
6ร 107
8ร 107
1ร 108
x
u(x,t)
intensity |u 2 of solution at t =0.01 second
-0.003-0.002-0.001 0.000 0.001 0.002 0.003
0
2ร 107
4ร 107
6ร 107
8ร 107
x
u(x,t)
intensity |u 2 of solution at t =0.012 second
Now Comparing argument to cosine in above to standard form in order to find the period:2๐๐๐
๐ฅ๐ก= 2๐๐๐ก
307
5.2. Exam 2 CHAPTER 5. EXAMS
Where ๐ is now in hertz, then when ๐ฅ = ๐ก, we get by comparing terms that2๐๐๐
1๐ก= 2๐๐
๐๐1๐ก= ๐
But ๐ = 1๐ where ๐ is the period in seconds. Hence ๐
๐1๐ก =
1๐ or
๐ = ๐๐ก๐
So period on intensity is ๐๐ก๐ at ๐ฅ = ๐ก (why problem statement is saying period is ๐๐ก
2๐?).
308
5.2. Exam 2 CHAPTER 5. EXAMS
5.2.4 problem 2 (optional)
2. Consider the 1D heat equation in a semi-infinite domain:
โu
โt= ฮฝ
โ2u
โx2, x โฅ 0
with boundary conditions: u(0, t) = exp(โiฯt) and u(x, t) bounded as x โ โ. In orderto construct a real forcing, we need both positive and negative real values of ฯ. Considerthat this forcing has been and will be applied for all time. This โpure boundary valueproblemโ could be an idealization of heating the surface of the earth by the sun (periodicforcing). One could then ask, how far beneath the surface of the earth do the periodicfluctuations of the heat propagate?
(a) Consider solutions of the form u(x, t;ฯ) = exp(ikx) exp(โiฯt). Find a single expressionfor k as a function of (given) ฯ real, sgn(ฯ) and ฮฝ real.
Write u(x, t;ฯ) as a function of (given) ฯ real, sgn(ฯ) and ฮฝ real. To obtain the mostgeneral solution by superposition, one would next integrate over all values of ฯ, โโ <ฯ <โ (do not do this).
(b) The basic solution can be written as u(x, t;ฯ) = exp(โiฯt) exp(โฯx) exp(iฯ sgn(ฯ) x).Find ฯ in terms of |ฯ| and ฮฝ.
(c) Make an estimate for the propagation depth of daily temperature fluctuations.
3. Here we study the competing effects of nonlinearity and diffusion in the context ofBurgerโs equation
โu
โt+ u
โu
โx= ฮฝ
โ2u
โx2(3a)
which is the simplest model equation for diffusive waves in fluid dynamics. It can be solvedexactly using the Cole-Hopf transformation
u = โ2ฮฝฯxฯ
(3b)
as follows (with 2 steps to achieve the transformation (3b)).
(a) Let u = ฯx (where the subscript denotes partial differentiation) and integrate oncewith respect to x.
(b) Let ฯ = โ2ฮฝ ln(ฯ) to get the diffusion equation for ฯ.
(c) Solve for ฯ with ฯ(x, 0) = ฮฆ(x), โโ < x < โ. In your integral expression for ฯ, usedummy variable ฮท to facilitate the remaining parts below.
(d) Show that
ฮฆ(x) = exp
[
โ1
2ฮฝ
โซ x
xo
F (ฮฑ)dฮฑ
]
where u(x, 0) = F (x), with xo arbitrary which we will take to be positive for conveniencebelow (xo > 0).
2
5.2.4.1 Part (a)
๐๐ข๐๐ก
= ๐๐2๐ข๐๐ฅ2
(1)
๐ข (0, ๐ก) = ๐โ๐๐๐ก
And ๐ฅ โฅ 0, ๐ข (โ, ๐ก) bounded. Let
๐ข (๐ฅ, ๐ก) = ๐๐๐๐ฅ๐โ๐๐๐ก
Hence๐๐ข (๐ฅ, ๐ก)๐๐ก
= โ๐๐๐๐๐๐ฅ๐โ๐๐๐ก
= โ๐๐๐ข (๐ฅ, ๐ก) (2)
And๐๐ข (๐ฅ, ๐ก)๐๐ฅ
= ๐๐๐๐๐๐ฅ๐โ๐๐๐ก
๐2๐ข (๐ฅ, ๐ก)๐๐ฅ2
= โ๐2๐๐๐๐ฅ๐โ๐๐๐ก
= โ๐2๐ข (๐ฅ, ๐ก) (3)
Substituting (2,3) into (1) gives
โ๐๐๐ข (๐ฅ, ๐ก) = โ๐๐2๐ข (๐ฅ, ๐ก)309
5.2. Exam 2 CHAPTER 5. EXAMS
Since ๐ข๐ (๐ฅ, ๐ก) can not be identically zero (trivial solution), then the above simplifies to
โ๐๐ = โ๐๐2
Or
๐2 = ๐๐๐ (4)
Writing
๐ = sgn (๐) |๐|Where
sgn (๐) =
โงโชโชโชโชโจโชโชโชโชโฉ
+1 ๐ > 00 ๐ = 0โ1 ๐ < 0
Then (4) becomes
๐2 =๐ sgn (๐) |๐|
๐
๐ = ยฑโ๐ โsgn (๐)โ|๐|
โ๐Since
โ๐ = ๐12 = ๏ฟฝ๐๐
๐2 ๏ฟฝ
12= ๐๐
๐4
Hence ๐ can be written as
๐ = ยฑ๐๐
๐4 โsgn (๐)โ|๐|
โ๐case A. Let start with the positive root hence
๐ =๐๐
๐4 โsgn (๐)โ|๐|
โ๐Case (A1) ๐ < 0 then the above becomes
๐ =๐๐๐
๐4 โ|๐|
โ๐=๐๐
๐2 ๐๐
๐4 โ|๐|
โ๐=๐๐
3๐4 โ|๐|
โ๐
= ๏ฟฝcos 34๐ + ๐ sin 3
4๐๏ฟฝ
โ|๐|
โ๐
= ๏ฟฝโ1
โ2+ ๐
1
โ2๏ฟฝโ|๐|
โ๐
=โโโโโโ
1
โ2โ|๐|
โ๐+ ๐
1
โ2โ|๐|
โ๐
โโโโโ
310
5.2. Exam 2 CHAPTER 5. EXAMS
And the solution becomes
๐ข (๐ฅ, ๐ก) = exp (๐๐๐ฅ) exp (โ๐๐๐ก)
= expโโโโโ๐โโโโโโ
1
โ2โ|๐|
โ๐+ ๐
1
โ2โ|๐|
โ๐
โโโโโ ๐ฅโโโโโ exp (โ๐๐๐ก)
= expโโโโโ
โโโโโโ
1
โ2โ|๐|
โ๐๐ โ
1
โ2โ|๐|
โ๐
โโโโโ ๐ฅโโโโโ exp (โ๐๐๐ก)
= expโโโโโโ
1
โ2โ|๐|
โ๐๐ฅโโโโโ exp
โโโโโโ
1
โ2โ|๐|
โ๐๐๐ฅโโโโโ exp (โ๐๐๐ก)
We are told that ๐ข (โ, ๐ก) is bounded. So for large ๐ฅ we want the above to be bounded. Thecomplex exponential present in the above expression cause no issue for large ๐ฅ since theyare oscillatory trig functions. We then just need to worry about exp ๏ฟฝโ 1
โ2โ|๐|
โ๐๐ฅ๏ฟฝ for large ๐ฅ.
This term will decay for large ๐ฅ since โ 1
โ2โ|๐|
โ๐is negative (assuming ๐ > 0 always). Hence
positive root hence worked OK when ๐ < 0. Now we check if it works OK also when ๐ > 0
case A2 When ๐ > 0 then ๐ now becomes
๐ =๐๐
๐4 โsgn (๐)โ|๐|
โ๐
=๐๐
๐4 โ๐
โ๐
= ๏ฟฝcos ๐4+ ๐ sin ๐
4๏ฟฝ โ๐
โ๐
= ๏ฟฝ1
โ2+ ๐
1
โ2๏ฟฝ โ๐
โ๐
=1
โ2โ๐
โ๐+ ๐
1
โ2โ๐
โ๐And the solution becomes
๐ข (๐ฅ, ๐ก) = exp (๐๐๐ฅ) exp (โ๐๐๐ก)
= exp ๏ฟฝ๐ ๏ฟฝ1
โ2โ๐
โ๐+ ๐
1
โ2โ๐
โ๐๏ฟฝ ๐ฅ๏ฟฝ exp (โ๐๐๐ก)
= exp ๏ฟฝ๏ฟฝ๐1
โ2โ๐
โ๐โ
1
โ2โ๐
โ๐๏ฟฝ ๐ฅ๏ฟฝ exp (โ๐๐๐ก)
= exp ๏ฟฝโ1
โ2โ๐
โ๐๐ฅ๏ฟฝ exp ๏ฟฝ๐
1
โ2โ๐
โ๐๐ฅ๏ฟฝ exp (โ๐๐๐ก)
We are told that ๐ข (โ, ๐ก) is bounded. So for large ๐ฅ we want the above to be bounded. Thecomplex exponential present in the above expression cause no issue for large ๐ฅ since theyare oscillatory trig functions. We then just need to worry about exp ๏ฟฝโ 1
โ2โ๐
โ๐๐ฅ๏ฟฝ for large ๐ฅ.
311
5.2. Exam 2 CHAPTER 5. EXAMS
This term will decay for large ๐ฅ since โ 1
โ2โ๐
โ๐is negative (assuming ๐ > 0 always). Hence
positive root hence worked OK when ๐ > 0 as well.
Let check what happens if we use the negative root.
case B. negative root hence
๐ = โ๐๐
๐4 โsgn (๐)โ|๐|
โ๐Case (A1) ๐ < 0 then the above becomes
๐ = โ๐๐๐
๐4 โ|๐|
โ๐= โ
๐๐๐2 ๐๐
๐4 โ|๐|
โ๐= โ
๐๐3๐4 โ|๐|
โ๐
= โ ๏ฟฝcos 34๐ + ๐ sin 3
4๐๏ฟฝ
โ|๐|
โ๐
= โ ๏ฟฝโ1
โ2+ ๐
1
โ2๏ฟฝโ|๐|
โ๐
= โโโโโโโ
1
โ2โ|๐|
โ๐+ ๐
1
โ2โ|๐|
โ๐
โโโโโ
=โโโโโ1
โ2โ|๐|
โ๐โ ๐
1
โ2โ|๐|
โ๐
โโโโโ
And the solution becomes
๐ข (๐ฅ, ๐ก) = exp (๐๐๐ฅ) exp (โ๐๐๐ก)
= expโโโโโ๐โโโโโ1
โ2โ|๐|
โ๐โ ๐
1
โ2โ|๐|
โ๐
โโโโโ ๐ฅโโโโโ exp (โ๐๐๐ก)
= expโโโโโ
โโโโโ๐1
โ2โ|๐|
โ๐+
1
โ2โ|๐|
โ๐
โโโโโ ๐ฅโโโโโ exp (โ๐๐๐ก)
= expโโโโโ+
1
โ2โ|๐|
โ๐๐ฅโโโโโ exp
โโโโโ1
โ2โ|๐|
โ๐๐๐ฅโโโโโ exp (โ๐๐๐ก)
We are told that ๐ข (โ, ๐ก) is bounded. So for large ๐ฅ we want the above to be bounded. Thecomplex exponential present in the above expression cause no issue for large ๐ฅ since theyare oscillatory trig functions. We then just need to worry about exp ๏ฟฝ+ 1
โ2โ|๐|
โ๐๐ฅ๏ฟฝ for large ๐ฅ.
This term will blow up for large ๐ฅ since + 1
โ2โ|๐|
โ๐is positive (assuming ๐ > 0 always). Hence
we reject the case of negative sign on ๐. And pick
๐ =๐๐
๐4 โsgn (๐)โ|๐|
โ๐
= ๏ฟฝ1
โ2+ ๐
1
โ2๏ฟฝโsgn (๐)โ|๐|
โ๐
312
5.2. Exam 2 CHAPTER 5. EXAMS
Therefore the solution is
๐ข (๐ฅ, ๐ก; ๐) = exp (๐๐๐ฅ) exp (โ๐๐๐ก)
= exp
โโโโโโโโโโ๐
โโโโโโโโโโ
๏ฟฝ 1
โ2+ ๐ 1
โ2๏ฟฝ โsgn (๐)โ|๐|
โ๐
โโโโโโโโโโ ๐ฅ
โโโโโโโโโโ exp (โ๐๐๐ก)
= exp ๏ฟฝ๐ ๏ฟฝ1
โ2๐๏ฟฝsgn (๐)โ|๐| + ๐
1
โ2๐๏ฟฝsgn (๐)โ|๐| ๏ฟฝ ๐ฅ๏ฟฝ exp (โ๐๐๐ก)
= exp ๏ฟฝ๐1
โ2๐๏ฟฝsgn (๐)โ|๐|๐ฅ โ
1
โ2๐๏ฟฝsgn (๐)โ|๐| ๐ฅ๏ฟฝ exp (โ๐๐๐ก)
Hence
๐ข (๐ฅ, ๐ก; ๐) = exp ๏ฟฝโโsgn(๐)โ|๐|โ2๐
๐ฅ๏ฟฝ exp ๏ฟฝ๐โsgn(๐)โ|๐|โ2๐
๐ฅ๏ฟฝ exp (โ๐๐๐ก)
The general solution ๐ข (๐ฅ, ๐ก) is therefore the integral over all ๐, hence
๐ข (๐ฅ, ๐ก) = ๏ฟฝโ
๐=โโ๐ข (๐ฅ, ๐ก; ๐) ๐๐
= ๏ฟฝโ
โโexp
โโโโโโโsgn (๐)โ|๐|
โ2๐๐ฅโโโโโ exp
โโโโโ๐โsgn (๐)โ|๐|
โ2๐๐ฅโโโโโ exp (โ๐๐๐ก) ๐๐
= ๏ฟฝโ
โโexp
โโโโโโโsgn (๐)โ|๐|
โ2๐๐ฅโโโโโ exp
โโโโโ๐โกโขโขโขโฃโsgn (๐)โ|๐|
โ2๐๐ฅ โ ๐๐ก
โคโฅโฅโฅโฆ
โโโโโ ๐๐
= ๏ฟฝโ
โโexp
โโโโโโโsgn (๐)โ|๐|
โ2๐๐ฅโโโโโ exp
โโโโโ๐๐ฅ
โกโขโขโขโฃโsgn (๐)โ|๐|
โ2๐โ ๐
๐ก๐ฅ
โคโฅโฅโฅโฆ
โโโโโ ๐๐
5.2.4.2 Part (b)
From part (a), we found that
๐ข (๐ฅ, ๐ก; ๐) = exp ๏ฟฝโโsgn(๐)โ|๐|โ2๐
๐ฅ๏ฟฝ exp ๏ฟฝ๐โsgn(๐)โ|๐|โ2๐
๐ฅ๏ฟฝ exp (โ๐๐๐ก) (1)
comparing the above to expression given in problem which is
๐ข (๐ฅ, ๐ก) = exp (โ๐๐ฅ) exp (๐๐ sgn (๐) ๐ฅ) exp (โ๐๐๐ก) (2)
Therefore, by comparing exp (โ๐๐ฅ) to exp ๏ฟฝโโsgn(๐)โ|๐|โ2๐
๐ฅ๏ฟฝ we see that
๐ = โsgn(๐)โ|๐|โ2๐
(3)
5.2.4.3 part (c)
Using the solution found in part (a)
๐ข (๐ฅ, ๐ก; ๐) = exp (โ๐๐ฅ) exp (๐๐ sgn (๐) ๐ฅ) exp (โ๐๐๐ก)
313
5.2. Exam 2 CHAPTER 5. EXAMS
To find numerical estimate, assuming ๐ > 0 for now
๐ข (๐ฅ, ๐ก; ๐) = exp (โ๐๐ฅ) exp (๐๐๐ฅ) exp (โ๐๐๐ก)= ๐โ๐๐ฅ (cos ๐๐ฅ + ๐ sin ๐๐ฅ) (cos๐๐ก โ ๐ sin๐๐ก)= ๐โ๐๐ฅ (cos ๐๐ฅ cos๐๐ก โ ๐ sin๐๐ก cos ๐๐ฅ + ๐ cos๐๐ก sin ๐๐ฅ + sin ๐๐ฅ sin๐๐ก)= ๐โ๐๐ฅ (cos (๐๐ฅ) cos (๐๐ก) + sin (๐๐ฅ) sin (๐๐ก) + ๐ (cos (๐๐ก) sin (๐๐ฅ) โ sin (๐๐ก) cos (๐๐ฅ)))= ๐โ๐๐ฅ (cos (๐ก๐ โ ๐ฅ๐) โ ๐ sin (๐ก๐ โ ๐ฅ๐))
Hence will evaluate
Re (๐ข (๐ฅ, ๐ก; ๐)) = ๐โ๐๐ฅ Re (cos (๐ก๐ โ ๐ฅ๐) โ ๐ sin (๐ก๐ โ ๐ฅ๐))= ๐โ๐๐ฅ cos (๐ก๐ โ ๐ฅ๐)
I assume here it is asking for numerical estimate. We only need to determine numericalestimate for ๐. For ๐, using the period ๐ = 24 hrs or ๐ = 86 400 seconds, then ๐ = 2๐
๐ is nowfound. Then we need to determine ๐, which is thermal di๏ฟฝusivity for earth crust. There doesnot seem to be an agreed on value for this and this value also changed with depth inside theearth crust. The value I found that seem mentioned more is 1.2 ร 10โ6 meter2 per second.Hence
๐ = ๏ฟฝ2๐
86 400
๏ฟฝ2 ๏ฟฝ1.2 ร 10โ6๏ฟฝ
= 5.505 per meter
Therefore
Re (๐ข (๐ฅ, ๐ก; ๐)) = ๐โ5.505๐ฅ cos (๐ก๐ โ 5.505๐ฅ)
Using ๐ = 2๐86 400 = 7.29 ร 10
โ5 rad/sec. Now we can use the above to estimate fluctuation ofheat over 24 hrs period. But we need to fix ๐ฅ for each case. Here ๐ฅ = 0 means on the earthsurface and ๐ฅ say 10, means at depth 10 meters and so on as I understand that ๐ฅ is starts at0 at surface or earth and increases as we go lower into the earth crust. Plotting at the abovefor ๐ฅ = 0, 0.5, 1, 1.5, 2 I see that when ๐ฅ > 2 then maximum value of ๐โ5.505๐ฅ cos (๐ก๐ โ 5.505๐ฅ)is almost zero. This seems to indicate a range of heat reach is about little more than2 meters below the surface of earth.
This is a plot of the fluctuation in temperature at di๏ฟฝerent ๐ฅ each in separate plot, then latera plot is given that combines them all.
314
5.2. Exam 2 CHAPTER 5. EXAMS
0 20000 40000 60000 80000
-1.0
-0.5
0.0
0.5
1.0
The time in the day (in seconds)
T(x,t)
Flucuation in temperature T at depth x = 0
0 20000 40000 60000 80000
-0.06
-0.04
-0.02
0.00
0.02
0.04
0.06
The time in the day (in seconds)
T(x,t)
Flucuation in temperature T at depth x = 0.5
0 20000 40000 60000 80000
-0.004
-0.002
0.000
0.002
0.004
The time in the day (in seconds)
T(x,t)
Flucuation in temperature T at depth x = 1
315
5.2. Exam 2 CHAPTER 5. EXAMS
0 20000 40000 60000 80000
-0.0002
-0.0001
0.0000
0.0001
0.0002
The time in the day (in seconds)
T(x,t)
Flucuation in temperature T at depth x = 1.5
0 20000 40000 60000 80000
-0.000015
-0.000010
-5.ร10-6
0.000000
5.ร10-6
0.000010
0.000015
The time in the day (in seconds)
T(x,t)
Flucuation in temperature T at depth x = 2
0 20000 40000 60000 80000
-1.ร10-6
-5.ร10-7
0
5.ร10-7
1.ร10-6
The time in the day (in seconds)
T(x,t)
Flucuation in temperature T at depth x = 2.5
This plot better show the di๏ฟฝerence per depth, as it combines all the plots into one.
316
5.2. Exam 2 CHAPTER 5. EXAMS
0 20000 40000 60000 80000
-1.0
-0.5
0.0
0.5
1.0
The time in the day (in seconds)
T(x,t)
Flucuation in temperature T at different depth x
u(0, t)
u(0.5, t)
u(1, t)
u(1.5, t)
317