The Non-homogeneous Diffusion Equation

Similar documents
PARTIAL DIFFERENTIAL EQUATIONS SEPARATION OF VARIABLES

Phys 6303 Final Exam Solutions December 19, 2012

1 1 2 = show that: over variables x and y. [2 marks] Write down necessary conditions involving first and second-order partial derivatives for ( x0, y

ECE Spring Prof. David R. Jackson ECE Dept. Notes 20

Fourier Series and their Applications

1. Hydrogen Atom: 3p State

Bernoulli Polynomials Talks given at LSBU, October and November 2015 Tony Forbes

The z-transform. 7.1 Introduction. 7.2 The z-transform Derivation of the z-transform: x[n] = z n LTI system, h[n] z = re j

COMM 602: Digital Signal Processing

Perturbation Theory, Zeeman Effect, Stark Effect

x !1! + 1!2!

SECTION 2.6 THE SECOND ALTERNATIVE

PAPER : IIT-JAM 2010

Orthogonal Functions

j=1 dz Res(f, z j ) = 1 d k 1 dz k 1 (z c)k f(z) Res(f, c) = lim z c (k 1)! Res g, c = f(c) g (c)

Problem Cosider the curve give parametrically as x = si t ad y = + cos t for» t» ß: (a) Describe the path this traverses: Where does it start (whe t =

Math 5C Discussion Problems 2

SphericalHarmonicY. Notations. Primary definition. Specific values. Traditional name. Traditional notation. Mathematica StandardForm notation

n n 2 + 4i = lim 2 n lim 1 + 4x 2 dx = 1 2 tan ( 2i 2 x x dx = 1 2 tan 1 2 = 2 n, x i = a + i x = 2i

A second look at separation of variables

Acoustic Field inside a Rigid Cylinder with a Point Source

Solution: APPM 1360 Final Spring 2013

PLEASE MARK YOUR ANSWERS WITH AN X, not a circle! 1. (a) (b) (c) (d) (e) 3. (a) (b) (c) (d) (e) 5. (a) (b) (c) (d) (e) 7. (a) (b) (c) (d) (e)

School of Mechanical Engineering Purdue University. ME375 Transfer Functions - 1

Chapter 4. Fourier Series

n 3 ln n n ln n is convergent by p-series for p = 2 > 1. n2 Therefore we can apply Limit Comparison Test to determine lutely convergent.

Transfer Function Analysis

Chapter 4 Postulates & General Principles of Quantum Mechanics

Assignment 1 : Real Numbers, Sequences. for n 1. Show that (x n ) converges. Further, by observing that x n+2 + x n+1

Orthogonal Function Solution of Differential Equations

Today in Physics 217: separation of variables IV

SCORE. Exam 2. MA 114 Exam 2 Fall 2017

MATH 10550, EXAM 3 SOLUTIONS

Assignment Number 3 Solutions

SOLUTIONS TO EXAM 3. Solution: Note that this defines two convergent geometric series with respective radii r 1 = 2/5 < 1 and r 2 = 1/5 < 1.

6.003 Homework #3 Solutions

Physics 219 Summary of linear response theory

Chapter 4 : Laplace Transform

Informal Notes: Zeno Contours, Parametric Forms, & Integrals. John Gill March August S for a convex set S in the complex plane.

In algebra one spends much time finding common denominators and thus simplifying rational expressions. For example:

Matsubara-Green s Functions

Lecture 4 Conformal Mapping and Green s Theorem. 1. Let s try to solve the following problem by separation of variables

Math 5C Discussion Problems 2 Selected Solutions

Quiz. Use either the RATIO or ROOT TEST to determine whether the series is convergent or not.

Definition of z-transform.

: Transforms and Partial Differential Equations

Solutions to Homework 1

Complete Solutions to Supplementary Exercises on Infinite Series

Integrals of Functions of Several Variables

Chapter 7: The z-transform. Chih-Wei Liu

Question 1: The magnetic case

Math 12 Final Exam, May 11, 2011 ANSWER KEY. 2sinh(2x) = lim. 1 x. lim e. x ln. = e. (x+1)(1) x(1) (x+1) 2. (2secθ) 5 2sec2 θ dθ.

Math 142, Final Exam. 5/2/11.

Math 113, Calculus II Winter 2007 Final Exam Solutions

REVIEW 1, MATH n=1 is convergent. (b) Determine whether a n is convergent.

MH1101 AY1617 Sem 2. Question 1. NOT TESTED THIS TIME

Math 163 REVIEW EXAM 3: SOLUTIONS

8.3 Perturbation theory

MATH 6101 Fall 2008 Newton and Differential Equations

MTH Assignment 1 : Real Numbers, Sequences

Chapter 10: Power Series

A PROOF OF THE THUE-SIEGEL THEOREM ABOUT THE APPROXIMATION OF ALGEBRAIC NUMBERS FOR BINOMIAL EQUATIONS

Honors Calculus Homework 13 Solutions, due 12/8/5

A Lattice Green Function Introduction. Abstract

Z - Transform. It offers the techniques for digital filter design and frequency analysis of digital signals.

Engineering Mechanics Dynamics & Vibrations. Engineering Mechanics Dynamics & Vibrations Plane Motion of a Rigid Body: Equations of Motion

MATH2007* Partial Answers to Review Exercises Fall 2004

PROBLEMS AND SOLUTIONS 2

Solutions to quizzes Math Spring 2007

(c) Write, but do not evaluate, an integral expression for the volume of the solid generated when R is

MIDTERM 3 CALCULUS 2. Monday, December 3, :15 PM to 6:45 PM. Name PRACTICE EXAM SOLUTIONS

Steady symmetrical temperature field in a hollow spherical particle with temperature-dependent thermal conductivity

Calculus 2 Test File Spring Test #1

Lesson 03 Heat Equation with Different BCs

Now we are looking to find a volume of solid S that lies below a surface z = f(x,y) and R= ab, cd,,[a,b] is the interval over

Ray Optics Theory and Mode Theory. Dr. Mohammad Faisal Dept. of EEE, BUET

Topic 5 [434 marks] (i) Find the range of values of n for which. (ii) Write down the value of x dx in terms of n, when it does exist.

Appendix: The Laplace Transform

Ma 4121: Introduction to Lebesgue Integration Solutions to Homework Assignment 5

3. Z Transform. Recall that the Fourier transform (FT) of a DT signal xn [ ] is ( ) [ ] = In order for the FT to exist in the finite magnitude sense,

UNIVERSITY OF CALIFORNIA - SANTA CRUZ DEPARTMENT OF PHYSICS PHYS 116C. Problem Set 4. Benjamin Stahl. November 6, 2014

Most text will write ordinary derivatives using either Leibniz notation 2 3. y + 5y= e and y y. xx tt t

Complex Analysis Spring 2001 Homework I Solution

Orthogonal transformations

Solutions to Homework 7

Generalizing the DTFT. The z Transform. Complex Exponential Excitation. The Transfer Function. Systems Described by Difference Equations

Fluid Physics 8.292J/12.330J % (1)

Jacobi symbols. p 1. Note: The Jacobi symbol does not necessarily distinguish between quadratic residues and nonresidues. That is, we could have ( a

De Moivre s Theorem - ALL

2D DSP Basics: Systems Stability, 2D Sampling

Streamfunction-Vorticity Formulation

Lecture 10: Bounded Linear Operators and Orthogonality in Hilbert Spaces

METHOD OF FUNDAMENTAL SOLUTIONS FOR HELMHOLTZ EIGENVALUE PROBLEMS IN ELLIPTICAL DOMAINS

Chapter 1. Complex Numbers. Dr. Pulak Sahoo

Calculus II exam 1 6/18/07 All problems are worth 10 points unless otherwise noted. Show all analytic work.

Notes 18 Green s Functions

Notes 8 Singularities

Section A assesses the Units Numerical Analysis 1 and 2 Section B assesses the Unit Mathematics for Applied Mathematics

Time-Domain Representations of LTI Systems

1. (25 points) Use the limit definition of the definite integral and the sum formulas 1 to compute

Transcription:

The No-hoogeeous Diffusio Equatio The o-hoogeeous diffusio equatio, with sources, has the geeral for, 2 r,t a 2 r,t Fr,t t a 2 is real ad The hoogeeous diffusio equatio, 2 r,t a 2 t r,t ca be solved by separatio of variables usig a separatio costat ik 2,wherek is real: r,t ht r 2 r r a2 ht ht ik2 t 2 r k 2 r t ht k2 /a 2 ht k k k x 2 k x 2 k 2 2 k 2 /a 2 p A geeral solutio to the hoogeeous diffusio equatioi Cartesia coordiates is obtaied by suig over all values of the separatio costat, k: r,t akexp ik rexp pt k If oe assues that k ad therefore p ca be cotious, oe has (i Cartesia coordiates) the followig for: p k 2 /a 2 k,pexp ik rexp ptd 3 kdp r,t 1 2 4 The solutios to r, give by expik r for a coplete orthooral set of states spaig all of space:

1 2 expik rexp ik rd 3 r k k 1 2 expik rexp ik r d 3 k r r.the Gree s Fuctio for the No-Hoogeeous Diffusio Equatio The Gree s fuctio satisfies the followig equatio: 2 Gr,r,t,t a 2 t Gr,r,t,t r r t t. Clearly we ca write the right had side of the equatio (the source) as follows: r r t t 1 2 4 expik rexp ik r d 3 k exp ipt t dp The Gree s fuctio for all of space ca be expaded i ters of the solutios to r,t, give by expik rexp ipt where k 2 /a 2 p : Gr,r,t,t 1 2 4 gk,p,r,t exp ik rexp iptd 3 kdp. To fid gk,p,r,t we put Gr,r,t,t ito the o-hoogeeous diffusio equatio for the Gree s fuctio: 1 gk,p,r,t 2 a 2 2 4 t exp ik rexp iptd3 kdp 1 expik rexp ik r d 3 k exp ipt t dp 2 4 Siplifyig, we have, gk,p,r,t k 2 ipa 2 expik r ptd 3 kdp exp ik r pt expik r ptd 3 kdp

gk,p,r,t k 2 ipa 2 exp ik r pt exp ik r ptd 3 kdp gk,p,r,t exp ik r pt k 2 ipa 2. Substitutig gk,p,r,t ito the expasio for Gr,r,t,t we have G i ters of r r ad t t. Gr r,t t 1 expik r r pt t dpd 3 k 2 4 k 2 ipa 2 Gr r,t t 1 ia 2 2 4 expik r r pt t p ik 2 /a 2 dpd 3 k Separatig the p itegratio, we fid that it has oly oe pole at p ik 2 /a 2 (ot o the real axis). Gr r,t t i expik r r d 3 k a 2 2 4 exp ipt t p ik 2 /a 2 dp As i the case of the wave equatio, we assue the causality coditio t t ad ust close the cotour i the lower half plae so that the sei circular path gives zero cotributio: G fk d 3 k closed exp ipt t exp ipt t dp dp p ik 2 /a 2 sei circle p ik 2 /a 2 Usig the Cauchy itegral theore we ca evaluate the p itegratio:.

G fk d 3 k closed exp ipt t p ik 2 /a 2 fk d 3 k 2iΘt t exp i ik 2 /a 2 t t 2iΘt t Θt t i a 2 2 4 2 a 2 2 4 dp expik r r exp k 2 /a 2 t t d 3 k expik r r exp k 2 /a 2 t t d 3 k Nowwedothekitegratioisphericalkspace, lettig r r be alog the ẑ directio: G Θt t 2 expik r r cos a 2 2 4 k exp k 2 /a 2 t t k 2 dkd k d cos k Θt t 1 1 expik r r cos a 2 k exp k 2 /a 2 t t k 2 dkdu; u cos k 2 1 Θt t 1 expik r r exp ik r r exp k 2 /a 2 t t k 2 dk a 2 2 ik r r Θt t 2 si k r r exp k 2 /a 2 t t kdk a 2 2 r r Θt t 1 si kexp k 2 2 kdk a 2 r r Θt t 1 a 2 r r Θt t 1 a 2 r r Θt t 4 3 exp 2 /4 2 Dwight "Tables of Itegrals" p. 236 r r a 3 4t t 3/2 exp a2 r r 2 4t t a 4t t 3/2 exp a2 r r 2 4t t Gr r,t t Θt t for V all space a 4t t 3/2 exp a2 r r 2 4t t For a Fiite Volue For a fiite volue oe fids the coplete orthooral set of states which satisfy : 2 k 2 u r u r u r r r

The, oe ca write the o-hoogeeous diffusio equatio for the Gree s fuctio as follows: 2 Gr,r,t,t a 2 t Gr,r,t,t The expasio for Gr,r,t,t isassuedtobe:, Gr,r,t,t Usig the differetial equatio 2 a 2 t Siplifyig, we have, u r u rt t. u r u r A pexp ipt t dp u r u r A pexp ipt t dp u r u r 1 2 exp ipt t dp A p 1 2ia 2 p ik 2 /a 2 The cotour itegratio of the p itegratio is doe the sae way, givig: Gr,r,t,t Θt t 1 a 2 u r u rexp k 2 /a 2 t t choose boudary coditios o u r to atch those of r,t Note this has sae for as derived for all of space where u r 1 expik r 3/2 2 Gr,r,t,t Θt t 1 a 2 2 3 expik r r exp k 2 /a 2 t t d 3 k Derivig the Solutio to the Diffusio Equatio Give

2 a 2 t r,t Fr,t Lr,t Fr,t 2 a 2 t Gr,r ;t,t r r t t I this case ad LGr,r ;t,t r r t t Gr,r ;t,t Gr,r; t,t reciprocity theore 2 a 2 Gr,r ;t,t r r t t t L Gr,r ;t,t r r t t For the diffusio equatio oe chooses to begi with the followig expressio: r,t 2 a 2 Gr,r ;t,t d 3 x dct t Gr,r ;t,t 2 a 2 t r,t d 3 x dct r,t r r t t d 3 x dct Gr,r ;t,t Fr,t d 3 x dct C1 r,t Gr,r ;t,t Fr,t d 3 x dct We also ca obtai the surface itegral ters as follows: r,t 2 Gr,r ;t,t Gr,r ;t,t 2 r,t d 3 x dct dct r,t Gr,r ;t,t Gr,r ;t,t r,t d 3 x dct r,t Gr,r ;t,t Gr,r ;t,t r,t s ds r The surface itegrals vaish for all space if r,t 1 r Gr,r ;t,t 1 r ad for a fiite volue if r s,t Gr,r s ;t,t

The ters ivolvig the itegratio over t becoe r,t a 2 t Gr,r ;t,t a 2 Gr,r ;t,t d 3 x dct t a 2 t r,t Gr,r ;t,t d 3 x a 2 r,t Gr,r ;t,t t r,t d 3 x dct t d 3 x Gr,r ;t,t fort t (violates Causality as t ) r,t att (o sigal before source is tured o) With these boudary coditios both the spatial ad tie surface ters vaish ad we have: r,t Gr,r ;t,t Fr,t d 3 x dt

Exaple. Diffusio Equatio: Usig the Gree s fuctio ethod for a fiite volue fid d 3 x r,t i the regio r ad t ifsatisfies the followig: (a) [ 2 a 2 t ]r,t o siei e t o j 1 k r; wherej 1 k (b) r, (c) r,t. Expad the Gree s fuctio for the fiite volue i orthooral solutios to the hoogeeous Helholtz equatio ad ake sure that the Gree s fuctio satisfies the coditio G(r,r ;t,t. Solutio: 1. We will use the followig for for the solutio with a (Dirichlet) Gree s fuctio zero wheever r or r (o the sphere of radius ) r,t dt sphere, radius Gr,r ;t,t r,t d 3 x dt r,t Gr,r ;t,t Gr,r ;t,t r,t sds r a 2 r,gr,r ;t,d 3 x 2. By coditio (b) r,, the last itegral o the right is zero ad the surface itegrals vaish because both ad G are zero o the spherical surface. Thus, r,t dt Gr,r ;t,t o si e i e t o j 1 k r r 2 dr d dt sphere of radius sphere of radius Gr,r ;t,t o 8 3 Y 1 1, e t o j 1 k r r 2 dr d 3. The Gree s fuctio has the for, Gr,r ;t,t 1 exp k 2 a 2 t t /a 2 u r u rθt t where Eq.1b #

2 u r k 2 u r u r u rd 3 x u r u r r r eigefuctio equatio orthooality coditio copleteess coditio 4. I spherical coordiates the geeral (o-diverget at r ) solutio to the Helholtz equatio, 2 k 2 u r isasfollows: u r C l j l k ry l, ; l,1,2,...;,1,2,..l. l, These fuctios will be zero o the sphere whe j l k, so is depedet oly o l, ad for each l, there will be a series of k. So each u r has a su over but ot over l: u r C l j l k ry l, We will oralize the u r later ad deterie the C l. The Gree s fuctio is Gr,r ;t,t 1 exp k 2 a 2 t t /a 2 C l j l k r Y l,l, 1 a 2, C l j l k ry l, Θt t exp k 2 t t /a 2 C l 2 j l k r j l k ry l, Y l,θt t,l, 5. Note that the Gree s fuctio will be set equal to zero o the sphere by suig oly over those k which satisfy k x l j l x l That is, the x l are the zeros of the spherical Bessel fuctio, j l. 5. Before akig the solutio ore coplicated tha it eed be, substitute the Gr,r ;t,t ito Eq. 1b ad use the orthogoality of the Y l :

r,t o 1 a 2,l, exp C l 2 2j 1 k r 8 3 Y l, jl k r j l k rexp k 2 t/a 2 r 2 dr 2 k a 1 2 o t Θt t dt uit sphere Y l, Y 1 1, d o 1 C a 2 l 2 4 3 Y l, 1 k2 exp,l, 1 a 2 o j l k r l1 1 r 2 j l k r j 1 k r dr k 2 a 2 1 o t 1 exp k 2 t/a 2 o 1 C a 2 11 2 8 3 Y 1 exp t/ o exp k 2 1, t/a 2 k2 1 a 2 o j 1 k r r 2 j 1 k r j 1 k r dr 6. Now we use the followig property of the spherical Bessel fuctios: 1 x 2 j i x j j x dx ij 2 j 1 i 2 where j i j j r 2 j 1 x 1 r j 1 k r dr 3 1 r 2 j 1 x 1 xj 1 k xdx 3 2 j 2k 2, where x k k k r,t o 1 C a 2 1 2 8 3 Y 1 exp t/ o exp k 2 1, t/a 2 k2 j 1 k r 3 1 2 j 2k 2, a 2 o o 3 2a 2 siei C 11 2 exp t/ o exp k k2 2 t/a 2 a 2 1 o j 1 k r j 2 k 2

7. Now deterie the C 11. C 11 2 r 2 j 1 x 1 r j 1 x 1 r dr 1 C 11 2 3 2 j 2x 1 2 1 C 11 2 3 2 j 2x 1 2 1 r,t o 3 2a 2 siei 3 2 j 2x 1 2 1 exp t/ o exp k 2 t/a 2 k2 j 1 k r j 2 k 2 1 a 2 o o exp t/ o exp k 2 t/a 2 a 2 siei k2 j 1 k r 1 a 2 o