Math 316 Notes

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Contents

1 Lecture 1 - Introduction to Partial Differential Equations 5

1.1 Modeling and Derivation of PDE: . . . . . . . . . . . . . . . . 6

1.2 The Wave Equation: . . . . . . . . . . . . . . . . . . . . . . . 91.3 The Drunkards Walk - The Heat Equation: . . . . . . . . . . 10

2 Lecture 2 - Preliminaries 132.1 Sequences and Series of Numbers: . . . . . . . . . . . . . . . . 132.2 Absolute and Conditional Convergence: . . . . . . . . . . . . 162.3 Power Series: . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

3 Lecture 3 - Review of Methods to Solve ODE 193.1 First Order ODE: . . . . . . . . . . . . . . . . . . . . . . . . . 193.2 Another Method - Series Solution: . . . . . . . . . . . . . . . 20

3.3 Second Order Constant Coefficient Linear Equations: . . . . . 213.4 Euler/Equidimensional Equations: . . . . . . . . . . . . . . . 23

4 Lectures 4,5 Ordinary Points and Singular Points 274.1 An Ordinary Point: . . . . . . . . . . . . . . . . . . . . . . . . 274.2 A Singular Point: . . . . . . . . . . . . . . . . . . . . . . . . . 284.3 The Airy equation: . . . . . . . . . . . . . . . . . . . . . . . . 304.4 The Hermite Equation: . . . . . . . . . . . . . . . . . . . . . . 31

5 Lecture 6 - Singular points 335.1 Radius of Convergence and Nearest Singular Points . . . . . . 335.2 Singular Points: . . . . . . . . . . . . . . . . . . . . . . . . . . 35

5.3 Regular Singular Points: . . . . . . . . . . . . . . . . . . . . . 355.4 More General Definition of a Regular Singular Point: . . . . . 36

6 Lecture 7 - Frobenius Series about Regular Singular Points 396.1 Series Expansion Summary: . . . . . . . . . . . . . . . . . . . 41

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CONTENTS

7 Bessels Equation 43

7.1 Bessels Function of Order / {. . . , 2, 1, 0, 1, 2 . . .}: . . . . 437.2 Bessels Function of Order = 0 - repeated roots: . . . . . . . 44

7.3 Bessels Function of Order = 12 : . . . . . . . . . . . . . . . . 46

7.4 Example - the roots differ by an integer . . . . . . . . . . . . 48

8 Separation of Variables 49

8.1 Types of Boundary Value Problems: . . . . . . . . . . . . . . 49

8.2 Separation of Variables - Fourier sine Series: . . . . . . . . . . 51

8.3 Heat Eq on a Circular Ring - Full Fourier Series . . . . . . . 57

9 Lecture 13 - Fourier Series 61

9.1 It can be useful to shift the interval of integration from [L, L]to [c, c + 2L] . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

9.2 Complex Form of Fourier Series . . . . . . . . . . . . . . . . . 65

10 Lecture 14 - Even and Odd Functions 67

10.1 Integrals of Even and Odd Functions . . . . . . . . . . . . . . 67

10.2 Consequences of Even/Odd Property for Fourier Series . . . . 68

10.3 Half-Range Expansions . . . . . . . . . . . . . . . . . . . . . . 70

11 Lecture 15 - Convergence of Fourier Series 73

11.1 Convergence of Fourier Series . . . . . . . . . . . . . . . . . . 76

11.2 Illustration of the Gibbs Phenomenon . . . . . . . . . . . . . 7711.3 Now consider the sum of the first N terms . . . . . . . . . . . 78

12 Lecture 16 - Parsevals Identity 81

12.1 Geometric Interpretation of Parsevals Formula . . . . . . . . 82

13 Lecture 17 - Solving the heat equation using finite differencemethods 85

13.1 Approximating the Derivatives of a Function by Finite Dif-ferences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

13.2 Heat Equation solution by Finite Differences . . . . . . . . . 87

14 Lecture 18 - Solving Laplaces Equation using finite differ-ences 91

14.1 Finite Difference approximation . . . . . . . . . . . . . . . . . 91

14.2 Solving the System of Equations by Jacobi Iteration . . . . . 93

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CONTENTS

15 Lecture 19 Further Heat Conduction Problems: Inhomoge-

neous BC 95

16 Lecture 20 - Inhomogeneous Derivative BC 101

17 Lecture 21 Distributed, Time Dependent Heat Sources -eigenfunction expansions 105

18 Lecture 22 More Eigenfunction Expansions - Time Depen-dent Boundary Conditions 111

19 Lecture 23 - 1D Wave Equation 117

19.1 Guitar String . . . . . . . . . . . . . . . . . . . . . . . . . . . 117

20 Lecture 24 - Space-Time Interpretation of DAlemberts So-lution 121

20.1 Characteristics . . . . . . . . . . . . . . . . . . . . . . . . . . 121

20.2 Region of Influence . . . . . . . . . . . . . . . . . . . . . . . . 122

20.3 Domain of Dependence . . . . . . . . . . . . . . . . . . . . . . 122

21 Lecture 25 Solution by separation of variables 125

21.1 Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126

21.2 Now we can use the trigonometric identities . . . . . . . . . . 127

22 Lecture 26 - Laplaces Equation 129

22.1 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130

22.2 Laplaces Equation . . . . . . . . . . . . . . . . . . . . . . . . 130

22.3 Rectangular Domains . . . . . . . . . . . . . . . . . . . . . . 130

22.4 Solution to Problem (1A) by Separation of Variables . . . . . 131

23 Lecture 27 - More Rectangular Domains and semi-infinitestrip problems 135

23.1 Solution to Problem (1B) by Separation of Variables . . . . . 135

23.2 Rectangular domains with mixed BC . . . . . . . . . . . . . . 136

23.3 Semi-infinite strip problems . . . . . . . . . . . . . . . . . . . 138

24 Lecture 28 - Neumann Problem - only flux BC and Circulardomains 141

24.1 Neumann Problem on a rectangle . . . . . . . . . . . . . . . . 141

24.2 General Analysis of Laplaces Equation on Circular Domains: 144

24.3 R Equation: . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144

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CONTENTS

24.4 Equation: . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145

24.5 For Different Boundary Conditions: . . . . . . . . . . . . . . . 14524.5.1 Notes: . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 45

25 Lecture 29 Wedge Problems 147

26 Lecture 30 Wedges with cut-outs, circles, holes and annuli 15326.1 Special Case - Electrical Impedance Tomography . . . . . . . 15726.2 Poissons Integral Formula: . . . . . . . . . . . . . . . . . . . 159

27 Lecture 31 Sturm-Liouville Theory 16127.1 Boundary value problems and Sturm-Liouville theory: . . . . 16127.2 The regular Sturm-Liouville problem: . . . . . . . . . . . . . 162

27.3 Properties of SL Problems . . . . . . . . . . . . . . . . . . . . 164

28 Lecture 32 Solving the heat equation with Robin BC 16728.1 Expansion in Robin Eigenfunctions . . . . . . . . . . . . . . . 16728.2 Solving the Heat Equation with Robin BC . . . . . . . . . . . 168

29 Lecture 33 Variable coefficient BVP - eigenfunctions involv-ing solutions to the Euler Equation: 17129.1 Cases: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17129.2 Solving the heat equation by expanding in eigenfunctions in-

volving solutions to an Euler Equation: . . . . . . . . . . . . 173

30 Lecture 34 Sturm Liouville Theory 17530.1 Properties of SL Problems: . . . . . . . . . . . . . . . . . . . 17530.2 Lagranges Identity: . . . . . . . . . . . . . . . . . . . . . . . 17630.3 Proofs to selected properties: . . . . . . . . . . . . . . . . . . 177

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Chapter 1

Lecture 1 - Introduction toPartial Differential Equations

ODE - Equations which define functions of a single independent variableby prescribing a relationship between the values of the function and itsderivatives.

EG:y(x) + ey(x) = 0. (1.1)

PDE - Involve multivariable functions u(x, t), u(x, y) that are determined byprescribing a relationship between the function value and its partial deriva-tives.

EG 1:

a

xu(x, y) + b

yu(x, y) = c First Order PDE (1.2)

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Lecture 1 - Introduction to Partial Differential Equations

EG 2: Some Classic Second Order PDEs:

Quadric Classification Eq. Name

T = X2 Parabolic t

u(x, t) = 2

x2u(x, t) Heat Equation or Diffusion Eq

X2 + Y2 = k Elliptic 2

x2u(x, y) +

2

y2u(x, y) = f(x, y)

Poisson Eq f 0Laplace Eq f = 0

T2

c2X2 = k Hyperbolic

2

t2 u(x, t)

c2

2

x2 u(x, t) = 0 The Wave Eq

By analogy with quadric surfaces aX2 + 2bXY + c2Y2 + = k that can bereduced to a standard form by coordinate rotation, the most general linear2nd order PDE

auxx + 2buxy + cuyy + (1.3)can be reduced by a transformation of coordinates to one of the Heat,Laplace or the Wave Eq.

1.1 Modeling and Derivation of PDE:

1D Conservation Law: Traffic flow on a highway.Consider the traffic flow on a highway and let u(x, t) be the density of

cars at x at time t.

[u] = # of cars/unit length. (1.4)

Let q(x, t) be the flux of cars at x at time t.

[q] = # of cars/unit time. (1.5)

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1.1. MODELING AND DERIVATION OF PDE:

{u(x, t + t) u(x, t)}x {q(x, t) q(x + x, t)}t (1.6)Let x 0 and t 0:

u

t+

q

x= 0 (1.7)

conservation of cars conservation of heat conservation of chemicals.

How does q change with u?

Convection - and the first order Wave Equation: Assume that qvaries linearly with u, i.e., q = cu from which it follows that

u

t+ c

u

x= 0 (1.8)

But this is just a wave equation. To see this consider the following movingcoordinate system.

x = x ct transformation of coordinates (1.9)Guess:

u(x, t) = f(x ct) solves ut + cux = 0ut = cf ux = f (1.10)

Therefore ut + cux =

cf + cf = 0.

Thus ut + cux = 0 has solutions of the form u(x, t) = f(x ct) whichrepresents a right moving wave. What happens if q = cu in which case

u

t cu

x= 0 (1.11)

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Exercise: Show that (1.11) has a solution of the form u(x, t) = f(x + ct)

which represents a left moving wave.Note:

t+ c

x

t c

x

u(x, t) =

2u

t2 c2

2u

x2= 0 (1.12)

is the 2nd order wave equation that has both left and right moving wavesolutions.Fouriers Law: Heat flows from hotter regions to colder ones?

q = 2 ux

(1.13)

In this case the conservation law reduces to the form:

u

t= 2

2u

x2The Heat Equation (1.14)

2D Heat Equation:u

t= 2

2u

x2+

2u

y 2

(1.15)

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1.2. THE WAVE EQUATION:

1.2 The Wave Equation:

Consider an elastic rod having a density and cross-sectional area A, and let(x, t) be the pressure in the rod at x at time t and u(x, t) the displacementof the rod from equilibrium.

Balance of Linear Momentum F = Ma.

(x + x, t)A (x, t)A = Ax2u

t2

(x + x, t) (x, t)x

= 2u

t2(1.16)

x 0 x

= 2u

t2(BLM) Balance of Linear Momentum

Hookes Law

= Eu

x(1.17)

Plug into (BLM) to obtain the 2nd order wave equation.

2u

t2= E

2u

x2= c2

2u

x2, where c = E (1.18)

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1.3 The Drunkards Walk - The Heat Equation:

Let u(x, t) be the density of fruit-flies at point x at time t. Find an equationfor the density of flies at t + t.

u(x, t + t) = pu(x + x, t) + (1 2p)u(x, t) +pu(x x, t)

= u(x, t) +px

[u(x + x, t)

u(x, t)]

x [u(x, t)

u(x

x, t)]

x

u(x, t) +px2

ux (x, t) ux (x x, t)

x

(1.19)

u(x, t) +px2 2u

x2

u(x, t + t) u(x, t)t

px2

t

2u

x2 u

t= 2

2u

x2The Heat Eq.

What is the Mean Absolute Deviation of the Drunkard?

sj = xtj = jt

xN = s1 + s2 + + sN 0 Expected Value (1.20)x2N = (s1 + + sN)2 = s21 + + s2N + 2(s1s2 + + sN1sN)

Nx2

Therefore

x2N

tNt

x2 = k2tN

|xN| k

tN

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1.3. THE DRUNKARDS WALK - THE HEAT EQUATION:

0 2 4 6 8 10 12 14 16 18 4

3

2

1

0

1

2

3

4

5

t

x

Drunkard Walk

Figure 1.1: Simulation with N = 1000 trajectories for 200 steps along withthe mean absolute deviation envelopes shown in red

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Chapter 2

Lecture 2 - Preliminaries

2.1 Sequences and Series of Numbers:

A Sequence of Numbers:

1, 12 ,13 , . . . ,

1n , . . .

1, 12 ,14 , . . . ,

12

n1, . . .

Notation{an} an = 1n{bn} bn =

12

n1A series of numbers:

1 + 12

+ 13

+ + 1n

+ = n=0

1n

n=0

an (2.1)

Does this infinite sum yield a finite result?

1 +

1

2

+

1

2

2+ +

1

2

n1=

n=1

1

2

n1 n=1

bn (2.2)

Note: In order to sum to a finite number the terms of the sequence musttend to 0 as n .Divergence Test:

lim an = 0 0

an diverges. (2.3)

EG: an = 1 1 + 1 + + 1 + .

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Integral Test: Does

n=11n converge?

Consider

1

dx

x< 1 +

1

2+

1

3+ + 1

n+ =

n=1

1

n(2.4)

Now

1

dx

x= lim

T

T1

dx

x= lim

T(ln T ln1) =

But

1

dx

x 1 p 1

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2.1. SEQUENCES AND SERIES OF NUMBERS:

p > 1:

n=2

1

np 1

p 1:

1

dx

xp 1

diverges p 1

Geometric Series - the G-Series:n=0

rn = 1 + r + r2 + + rn + (2.6)

For what values of r does the G-Series converge?

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Partial Sum:

SN =N

n=0

rn = 1 + r + + rN

=(1 + r + + rN)

(1 r) (1 r)

=1 + r + + rN r r2 rN rN+1

1 r (2.7)

=1 rN+1

1 rIf

|r

|< 1 then

limN

Nn=0

rn = limN

1 rN+11 r =

1

1 r (2.8)

If |r| 1, series diverges.G-Series:

n=0

rn =

11r |r| < 1 |r| 1 (2.9)

EG:

n=01

2n=

1

1

1/2= 2, r =

1

2. (2.10)

2.2 Absolute and Conditional Convergence:

Alternating Series Test:(1)nan an > 0

(a) an+1 an(b) lim

n an = 0

(1)nan converges:

EG:

n=1

(

1)

n

n1

converges.

Consider a series

n=0

(1)nan.

If

|an| < then

(1)nan is said to be absolutely convergent.

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2.3. POWER SERIES:

If |an| = but (1)nan < is conditionally convergent.

Ratio Test:

Consider

n=0

bn and let limn

bn+1bn = L. Then bn converges abso-

lutely if L < 1, diverges if L > 1. Test is inconclusive if L = 1.

EG 1:

n=1

(1)n1 2n

n2.

bn+1bn

=2n+1/(n + 1)2

2n/n2= 2

1 +

1

n

2 2 series diverges. (2.11)

EG 2:

n=1

n4en2

.

bn+1bn = (n + 1)4e(n+1)2n4en2 =

1 +

1

n

4e2n1 0 < 1 converges absolutely.

(2.12)

2.3 Power Series:

f(x) = a0 + a1x + a2x2 + + anxn polynomial approximation.Idea: Extend the polynomial to include # of terms.

f(x) = a0 + a1x + a2x2 + + anxn + Power Series

=

n=0

anxn (2.13)

EG: ex = 1 + x1! +x2

2! +x3

3! + + xn

n! + =

n=0xn

n! .

More General Power Series:

f(x) =

n=0

an(x x0)n = a0 + a1(x x0) + a2(x x0)2 + (2.14)

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Taylor Series - matching all the derivatives at a point:

f(x) =

n=0

an(x x0)n = a0 + a1(x x0) + a2(x x0)2 +

f(x) = a1 + 2a2(x x0) + 3a3(x x0)2 + + nan(x x0)n + f(x0) = a1

f(x) = 2a2 + 3.2a3(x x0) + + n(n 1)an(x x0)n2 + f(x0) = 2a2

f(3)(x) = 3!a3 + 4.3.2(x x0) + + n(n 1)(n 2)an(x x0)n3 + f(3)(x0) = 3!a3

f(n)

(x0) = n!an an =f(n)(x0)

n!

Therefore f(x) =

n=0

f(n)(x0)

n!(x x0)n (2.15)

Alternative Form of Taylor Series:

f(x0 + h) =

n=0

f(n)(x0)

n!hn (2.16)

EG 1:

ex = n=0

xn

n!

sin x =

n=0

(1)n x2n+1

(2n + 1)!sinh x =

n=0

x2n+1

(2n + 1)!(2.17)

cos x =

n=0

(1)n x2n

(2n)!cosh x =

n=0

x2n

(2n)!

ei = 1 + i +(i)2

2!+

(i)3

3!+

= 1 22!

+ 4

4! + i 3

3!+ (2.18)

= cos + i sin

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Chapter 3

Lecture 3 - Review ofMethods to Solve ODE

3.1 First Order ODE:

Separable Equations:

dy

dx= P(x)Q(y) (3.1)

dy

Q(y)=

P(x) dx + C

EG:dy

dx=

4y

x(y 3)y 3

y

dy =

4

xdx

y 3 ln |y| = 4 ln |x| + C (3.2)y = ln(x4y3) + C

Ax4y3 = ey

Linear First Order Eq. - The Integrating Factor:

y(x) + P(x)y = Q(x) (3.3)Can we find a function F(x) to multiply (4.3) by in order to turn the lefthand side into a derivative of a product:

F y + F P y = F Q (3.4)

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Lecture 3 - Review of Methods to Solve ODE

(F y) = F y + Fy = F Q (3.5)

So let F = F P which is a separable Eq.

dF

F(x)= P(x) dx

dF

F=

P(x) dx + C

Therefore ln F =

P(x) dx + C (3.6)

or F = AeR

P(x) dx choose A = 1

F = eR

P(x) dx integrating factor

Therefore

eR

P(x) dxy + eR

P(x) dxP(x)y = eR

P(x) dxQ(x)(eR

P(x) dxy)

= eR

P(x) dxQ(x)

y(x) = eR

P(x) dx

eRx P(t) dtQ(x) dx + C

(3.7)EG: 1

y + 2y = 0 (3.8)

F(x) = e2x e2xy + e2x2y = (e2xy) = 0e2xy =?c

y(x) = Ce2x

3.2 Another Method - Series Solution:

Since the unknown solution y(x) is defined implicitly by (3.8) let us look for

a series solution: y(x) =

n=0anx

n.

y =

n=1

annxn1 (3.9)

Therefore y + 2y =

n=1 annxn1 +

n=0 2anxn = 0

In the first sum let

m = n 1 n = 1 m = 0n = m + 1

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3.3. SECOND ORDER CONSTANT COEFFICIENT LINEAREQUATIONS:

Therefore

m=0 am+1(m + 1)xm +

n=0 2anxn = 0

n m :

m=0{am+1(m + 1) + 2am} xm = 0

(3.10)

am+1 = 2(m+1) am

a1 = 2a0, a2 = + 22 21 a0, a3 = 23 22 21 a0 = (1)3 23

3! a0,

. . . , am = (1)m 2mm! a0

Therefore y(x) = a0

m=0(2x)m

m! = a0e2x

(3.11)

EG 2: Solvedy

dx+ cot(x)y = 5ecos x, y(/2) = 4.

P(x) = cot x Q(x) = 5ecos x

F(x) = eR

cot x dx = eln(sin x) = sin x(3.12)

Therefore sin(x)y + cos(x)y = (sin(x)y) = 5ecos x sin x

sin(x)y = 5ecos x + C

y(x) =

5ecosxCsin x

4 = y(/2) = 5C1 C = 1

Therefore y(x) = 15ecosx

sin x

(3.13)

3.3 Second Order Constant Coefficient Linear Equa-tions:

Ly = ay + by + cy = 0Guess y = erx y = rerx y = r2erx

Ly = (ar2 + br + c)erx = 0

Indicial Eq.:

ar2 + br + c = 0 r = b

b24ac2a

a(r r1)(r r2) = 0 (3.14)

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Case I: = b2 4ac > 0, r1 = r2, y(x) = c1er1x + c2er2x is the generalsolution.Case II: = 0, r1 = r2, repeated roots Ly = a(r r1)2erx = 0. Then wehave one solution y(x) = er1x what about the second solution. Let

y(r, x) = erx

Ly(r, x) = a(r r1)2erxL

yr (r, x)

r=r1

= [2a(r r1)erx + 2a(r r1)xerx]r=r1 = 0Therefore

yr (r, x)r=r1 = xer1x is also a solution.(3.15)

Thus y(x) = c1er1x + c2xe

r1x is the general solution.

Another Method:

Consider a small perturbation to the double root case:

r (r1 + )

r (r1 )

= 0 (3.16)

y(x) = c1e(r1+)x + c2e

(r1)x

= e(r1+)xe(r1)x

2 c1 =1

2 = c2= er1x

exex

2

0 xer1x = r erxr=r1

(3.17)

Case III: Complex Conjugate Roots: = b2 4ac < 0

r = b2a

i4ac b21/2 = iy(x) = c1e

(+i)x + c2e(i)x (3.18)

= ex [A cos x + B sin x] .

EG 1:

Ly = y + y 6y = 0y = erx(r2 + r 6) = (r + 3)(r 2) = 0 (3.19)

y(x) = c1e3x + c2e2x

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3.4. EULER/EQUIDIMENSIONAL EQUATIONS:

EG 2:

Ly = y + 6y + 9y = 0y = erx(r + 3)2 = 0 (3.20)

y(x) = c1e3x + c2xe3x

EG 3:

Ly = y 4y + 13y = 0y = erx : r2 4r + 13 = 0

r = 4

16522 = 2 3i

Therefore y(x) = e2n [A cos3x + B sin3x] .

(3.21)

3.4 Euler/Equidimensional Equations:

Ly = x2y + xy + y = 0. (3.22)

Aside: Note if we let t = ln x or x = et thend

dx=

d

dt

dt

dx d

dt= x

d

dx.

d2

dt2= x

d

dx

x

d

dx

= x2

d2

dx2+ x

d

dx x2 d

2

dx2=

d2

dt2 d

dt(3.23)

Therefore y y + y + y = 0y + ( 1)y + y = 0 (3.24)

y = ert r2 + ( 1)r + = 0 Characteristic Eq.Back to (3.22): Guess y = xr, y = rxr1, and y = r(r 1)xr2.

Therefore {r(r 1) + r + } xr = 0f(r) = r2 + ( 1)r + = 0 as above. (3.25)

r =1

( 1)2 42 (3.26)Case 1: = ( 1)2 4 > 0 Two Distinct Real Roots r1, r2.

y = c1xr1 + c2x

r2 if r1 or r2 < 0 |y| as x 0. (3.27)

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Case 2: = 0 Double Root (r r1)2 = 0.

y = c1xr1

r L[x

r] = L

r x

r

= L[xr log x]

r {f(r)xr} = f(r)xr + f(r)xr log x = 0 since f(r) = (r r1)2.

(3.28)

General Solution: y(x) = (c1 + c2 log x)xr1 .

Check:

L(xr1 log x) = x2(xr log x) + x(xr log x) + (xr log x) = x

2 r(r 1)xr log x + rxr2 + (r 1)xr2 (3.29)+ x

rxr1 log x + xr1

+ (xr log x)

=

r2 + ( 1)r + xr log x + {2r 1 + } xr = 0Case 3: = ( 1)2 4 < 0.

r =(1 )

2 i [4 ( 1)

2]1/2

2= i

y(x) = c1x(+i) + c2x

(i) xr = er ln x

= c1e(+i) ln x + c2e

(i) ln x (3.30)

= x c1ei ln x + c2ei ln x= A1x

cos( ln x) + A2x sin( ln x)

Notes:

(1) Ifx < 0 replace by |x|.

(2)

w(y1, y2) =

y1 y2y1 y

2

= y1y2 y1y2 (see Section 3.2 p.143)

=

x cos( ln x)

log xx sin( ln x) + x1 cos( ln x)

x log x cos( ln x) x1 sin( ln x)

x sin( ln x)

= x21 independent for x = 0.

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3.4. EULER/EQUIDIMENSIONAL EQUATIONS:

EG 1:

x2y xy 2y = 0 y(1) = 0 y(1) = 1y = xr r(r 1) r 2 = 0 r2 2r 2 = 0

(r 1)2 = 3 r = 1 3(3.31)

y = c1x1+

3 + c2x13

y(1) = c1 + c2 = 0 c2 = c1y(x) = c1

x1+

3 x1

3

(3.32)

y(x) = c1

1 +

3

x

3

1

3

x

3

x=1= c12

3 = 1

Therefore y(x) = 12

3x1+3 x13 . (3.33)

EG 2:

x2y 3xy + 4y = 0 y(1) = 1 y(1) = 0y = xr = r(r 1) 3r + 4 = r2 4r + 4 = 0 (r 2)2 = 0 (3.34)

y(x) = c1x2 + c2x

2 log x

y(1) = c1 = 1 y(x) = 2x + c2 [2x log x + x]=1 (3.35)

= 2 + c2 = 0

Therefore y(x) = x2 2x2 log x.

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Chapter 4

Lectures 4,5 Ordinary Pointsand Singular Points

Lecture 4

Consider

P(x)y + Q(x)y + R(x)y = 0 Homogeneous Eq. (4.1)

Divide through by P(x):

Ly = y +p(x)y + q(x)y = 0 p(x) = Q/P, R/P (4.2)

4.1 An Ordinary Point:

x0 is said to be an ordinary point of (5.2) if p(x) = Q/P and q(x) = R/Pare analytic at x0.

i.e. p(x) = p0 +p1(x x0) + =

k=0pk(x x0)k

q(x) = q0 + q1(x x0) + =

k=0qk(x x0)k

Note:

(1) If P, Q and R are polynomials then a point x0 such that P(x0) = 0 isan ordinary point.

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Lectures 4,5 Ordinary Points and Singular Points

(2) Ifx0 = 0 is an ordinary point then we assume

y =

n=0

cnxn, yn =

n=1

cnnxn1, yn =

n=2

cnn(n 1)xn2

0 = Ly =

n=2

cnn(n 1)xn2 +

n=0

pnxn

n=1

ncnxn1 (4.3)

+

n=0

qnxn

n=0

cnxn

m=0(m + 2)(m + 1)cm+2 + p0(m + 1)cm+1 + +pmc1

+ (q0cm + + qmc0)} xm = 0 (4.4)yields a non-degenerate recursion for the cm.

At an ordinary point x0 we can obtain two linearly independent solu-tions by power series expansion.

About x0:

y(x) =

n=0

cn(x x0)n. (4.5)

(3) The radius of convergence of (4.5) is at least as large as the radius of

convergence of each of the series p(x) = Q/P q(x) = R/P.i.e. up to the closest singularity to x0.

4.2 A Singular Point:

If p(x) or q(x) are not analytic at x0, then x0 is said to be a singular pointof (4.2). For example if P, Q and R are polynomials and P(x0) = 0 andQ(x0) = 0 or R(x0) = 0 then x0 is a singular point.EG:

(x 1)y + y = 0 (4.6)x = 0 is an ordinary point.

x = 1 is a singular point.Expand around the ordinary point

y(x) =

n=0

cnxn, y =

n=1

ncnxn1, y =

n=2

cnn(n 1)xn2 (4.7)

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4.2. A SINGULAR POINT:

(x

1)

n=2

cnn(n

1)xn2 +

n=1

ncnxn1 = 0

n=2

cnn(n 1)xn2 +

n=2

cn {n(n 1) + n} xn1 + c1 = 0 (4.8)

m 1 = n 2 m = n 1 n = 2 m = 1 n = m + 1c2 2 1 + c1 +

m=2

cm+1(m + 1)m + cmm2xm1 = 0c0 Arbitrary:

cm+1 =m

m + 1cm m 2 c2 = c1

2

c3 =23

c2 =c13

c4 =34

c3 =c14

. . . cn =c1n

(4.9)

Therefore y(x) = c0 + c1

n=1

xn

n.

Recall

1

1 x = 1 + x + x2 +

1

1 x dx = ln |1 x| = x +x2

2+

x3

3+

y(x) = A + B ln |x 1| (4.10)

But (4.6) is also an Euler Equation:

y = (x 1)r r(r 1) + r = r2 = 0 r = 0, 0.y(x) = A + B ln(x 1) (4.11)

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Lecture 5

4.3 The Airy equation:

Consider the Airy equation y = xy.x = 0 is an ordinary point.

y =

n=0cnx

n, y =

n=1cnnx

n1, y =

n=2cnn(n 1)xn2

n=2

cnn(n 1)xn2 =

n=0cnx

n+1

m + 1 = n 2 n = m + 3 n = 2 m = 1c22x0 +

m=0

cm+3(m + 3)(m + 2) cmxm+1 = 0c2 = 0 cm+3 =

cm(m+3)(m+2) m = 0, 1, . . .

(4.12)

(1) c0 c3 c6.c3 =

c02.2

, c6 =c3

6.5=

c06.5.3.2

, c9 =c0

9.8.6.5.3.2

c3n =c0

(3n)(3n 1)(3n 3)(3n 4) . . . 9.8.6.5.3.2 (4.13)

y0(x) = 1 +x3

3.2+

x6

6.5+ + x

3n

(3n)(3n 1) . . . 3.2 + . . .

(2) c1

c4

c7

.

c4 =c1

4.3c7 =

c17.64.3

c10 =c1

(10.9)(7.6)(4.3)(4.14)

c3n+1 =c1

(3n + 1)(3n)(3n 2)(3n 3) . . . (7.6)(4.3)y1(x) = x +

x4

4.3+

x7

7.6.4.3+ + x

3n+1

(3n + 1)(3n) . . . 4.3(4.15)

y(x) = c0y0(x) + c1y1(x)

Radius of Convergence:

limmcm+3

cm |x|3

= lnm |x

|3

(m + 3)(m + 2) = 0 < 1 = . (4.16)See B&D for expansion of Airy Solution about x0 = 1 y(x) =

an(x 1)n.

It is useful to write x = (x 1) + 1.y = (x 1)y + y (4.17)

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4.4. THE HERMITE EQUATION:

4.4 The Hermite Equation:

Ly = y 2xy + y = 0.Since x = 0 is an ordinary point let y(x) =

n=0

anxn then

Ly =

n=2

ann(n 1)xn2 2

n=1

annxn +

n=0

anxn = 0. (4.18)

m = n 2 n = m + 2 m n m nn = 2 m = 0

Thereforem=1

am+2(m + 2)(m + 1) 2amm + am

xm + [a22 + a0] x

0 = 0. (4.19)

x0 :a2 = a0/2 (4.20)

xm :

am+2 =(2m )am

(m + 1)(m + 2)m 1 (4.21)

a0:

a2 =

2 a0, a4 =(4

)

4.3 a2 =(4

)(

)

4.3.2 a0, a6 =(8

)(4

)(

)

6.5.4.3.2 a0

a2k =[4(k 1) ][4(k 2) ] . . . (1))?a0

(2k)!(4.22)

y0 = a0

1

2x2 +

( 4)4!

x4 +(8 )(4 )()

6!x6 +

a1:

a3 =(2 )

3.2a1; a5 =

(6 )5!

(2 )a1; a7 = (10 )(6 )(2 )7!

a1, . . .(4.23)

y1 = a1 x +(2 )

3!x3 +

(6 )(2 )5!

x5 +(10 )(6 )(2 )x7

7!+

The general solution is of the form

y(x) = Ay0(x) + By1(x) (4.24)

Note:

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(a) If = 2n then the recursion yields am+2 = 0 = am+4 = for m = n.Thus if n is an even integer then the series solution y0 will terminateand become a polynomial of degree n.

In this case:

y0(x) = a0

1 nx2 + n(n 2)22 x

4

4! n(n 2)(n 4) 2

3x6

6!+

+ (1)n/2n(n 2) . . . 2.2n/2xnn!

. (4.25)

On the other hand ifn is an odd integer then the series solution y1(x)will terminate and become a polynomial of degree n. In this case

y1(x) = a1

x 2(n 1) x

3

3!+ 22(n 1)(n 3) x

5

5!

(n 1)(n 3)(n 5)23 x7

7!+ (4.26)

+ (n 1)(n 3) . . . 3.1(2) (n1)2 xn

n!

(b) For example in the special case = 4 = 2n then n = 2.

y0(x) = a0[1 2x2]. (4.27)

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Chapter 5

Lecture 6 - Singular points

5.1 Radius of Convergence and Nearest SingularPoints

EG. 1: (1 + x2)y + 2xy + 4x2y = 0.

(1) If we were given y(0) = 0 and y(0) = 1 then we would want a powerseries expansion of the form

y =

n=0

cnxn about x0 = 0. (5.1)

Roots of 1+ x2 = 0 are x = i, so we expect the radius of convergenceof the TS for 1

1+x2to be 1 since

11+x2 = 1 x2 + x4 1

liman+2an = 1 = 1.

(5.2)

(2) If we were given y(1) = 1, y(1) = 0 then a power series expansion ofthe form

cn(x 1)n is required. In this case =

2.

EG. 2: (x 1)(2x 1)y + 2xy 2y = 0.x = 0 is an ordinary point. x = 1 and x = 12 are singular points. Onesolution of this equation is

y(x) =1

x 1 = (1 + x + x2 + ) = 1. (5.3)

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This TS solution about the ordinary point x = 0 converges beyond the

singular point x =12 .

EG: (x2 2x)y + 5(x 1)y + 3y = 0 y(1) = 7 y(1) = 3.x = 1 is an ordinary point. x = 0 is a singular point

(x 1)2 1y +

5(x 1)y + 3y = 0.Let t = x 1 so that ddt = ddx and the equation is transformed to

(t2 1)y + 5ty + 3y = 0y =

n=0

cntn, y =

n=1

cnntn1, y =

n=2

cnn(n 1)tn2

n=2n(n 1)cntn

n=2

n(n 1)cntn2 + 5

n=1ncnt

n + 3

n=0cnt

n = 0

m = n

2 n = m + 2 n = 2 = m = 0

m=2

[cm+2(m + 2)(m + 1) + {m(m 1) + 5m + 3} cm] tm

2c2 + 3c0 + [c33.2 + 5c1 + 3c1] t = 0

t0 > c2 =32 c0

t1 > c3 =86 c1 =

43 c1

tm > cm+2 =cm(m+1)(m+3)

(m+1)(m+2) m 2.

(5.4)

c0:c4 =

5c24 =

54

32

c0, c6 =

76 c4 =

76

54

32 c0

y0(x) =

n=0

357...(2n+1)246...(2n) (x 1)2n

(5.5)

c1:

c5 =65 c3 =

65

43 c1 c2n+1 =

46...2n+235...2n+1 c1

y1(x) =

n=0

46...2n+235...2n+1 (x 1)2n+1

limncm+2

cm =

m + 3

m + 1 = 1 = 1 (5.6)

y(x) = c0y0(x) + c1y1(x)y(1) = c0 = 7 y

(1) = c1 = 3.(5.7)

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5.2. SINGULAR POINTS:

5.2 Singular Points:

Consider

P(x)y + Q(x)y + R(x)y = 0. (5.8)

IfP, Q and R are polynomials without common factors then singular pointsare points x0 at which P(x0) = 0.

Note: At singular points the solution is not necessarily analytic.

Examples:

1.x2y + xy = 0y = xr

r(r

1) + r = 0

y = c1 + c2 ln x

The x2y admits wild behaviour.

2.x2y 2y = 0y = xr r(r 1) 2 = 0 r = 2, 1 y = c1x2 + c2x1

Again the x2y admits wild behaviour.

3.x2y 2xy + 2y = 0y = xr r(r 1) 2r + 2 = 0 r = 1, 2 y = c1x + c2x2

In this case both solutions are analytic.

5.3 Regular Singular Points:

Notice that all these cases are equidimensional equations for which we canidentify solutions of the form xr or xr log x. There is a special class ofsingular points called regular singular points in which the singularities areno worse than those in the equidimensional equations.

y +

xy +

x2y = 0. (5.9)

If P, Q and R are polynomials and suppose P(x0) = 0 then 0 is a regularsingular point if

limxx0

(x x0) Q(x)P(x)

and limxx0

(x x0)2 R(x)P(x)

are finite. (5.10)

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I.E.Q(x)

P(x)

=p0

(x x0)+p1 +p2(x

x0) +

singularity no worse than 1xx0 (5.11)R(x)

P(x)=

q0(x x0)2 +

q1(x x0) + q2 +

singularity no worse than 1(xx0)2

Examples:

1.

(1 x2)y 2xy + 4y = 0P = 1

x2 P(

1) = 0 Q =

2x R = 4

limx1

(x 1) (2x)(1x)(1+x) = 1 limx1(x 1)2 4

(1+x)(1x) = 0 (5.12)

x = 1 is a R.S.P. (similarly for x = 1).2.

x3y y = 0P(x) = x3 Q = 0 R = 1

limx0 x

21

x3

=

(5.13)

Thus x = 0 is an irregular singular point. Actually y

x3/4e

2/x1/2 asx 0+ which is much wilder than the simple power law xr or xr log x.Note: Any singular point that is not a regular singular point is calledan irregular singular point.

3. 2(x 2)2xy + 3xy + (x 2)y = 0. Singular points at x = 0, 2. x = 0is a regular singular point. x = 2 is an irregular singular point.

5.4 More General Definition of a Regular SingularPoint:

If P, Q, and R are not limited to polynomials then consider

P(x)y + Q(x)y + R(x)y = 0or

x2y + x

xQP

y +

x2RP

y = 0

(5.14)

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5.4. MORE GENERAL DEFINITION OF A REGULAR SINGULARPOINT:

x = 0 is a regular singular point ifxQP and x2R

P are analytic atx = 0. I.E.xQ

P= p(x) = p0 +p1x + and x

2R

P= q(x) = q0 + q1x + . (5.15)

In this case

Ly = x2y + xp0y + q0y +

small as x0 xp1xy

+ q1y +

= 0. (5.16)

Then as x 0 x2y + xp0y + q0y 0 which is an Euler Equation whichhas solutions of the form y = xr.

Thus about a regular singular point we look for solutions of the form

y = xr

n=0

anxn =

n=0

anxn+r.

Our task is to determine;

(i) r

(ii) the coefficients an

(iii) the radius of convergence.

EG. 1: x2y + 2(ex 1)y + ex cos xy = 0 P = x2 Q = 2(ex 1) R =e

x

cos x.x = 0 is a singular point.

limx0

xQ

P= lim

k0x

2(ex 1)x2

= limx0

2(ex 1)x

00= lim

x02ex

1= 2 LHopital

limx0

x2R

P= lim

x0x2

ex cos xx2

= 1 < . (5.17)

Since the quotient functions p = xQ/P and q = x2R/P have Taylor Expan-sions about x = 0, x = 0 is a regular singular point.

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Chapter 6

Lecture 7 - Frobenius Seriesabout Regular SingularPoints

Example 1:

Ly = 2x2y xy + (1 + x)y = 0 x = 0 is a RSP.

y =

n=0

anxn+r (6.1)

Ly = 2x2 n=0

an(n + r)(n + r 1)xn+r2 x n=0

an(n + r)xn+r1

+ (1 + x)

n=0

anxn+r = 0

n=0

an {2(n + r)(n + r 1) (n + r) + 1} xn+r

+

n=0

anxn+r+1 = 0 (6.2)

m = n + 1 n = 0

m = 1

n = m 1Therefore a0 {2r(r 1) r + 1} xr +

n=1

[an {2(n + r)(n + r 1)

(n + r) + 1} + an1] xn+r = 0.

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Lecture 7 - Frobenius Series about Regular Singular Points

xr > Indicial Equation: 2r2 3r + 1 = (2r 1)(r 1) = 0 r = 12 , r = 1.a0 arbitraryRecursion

an =an1

(2n + 2r 3)(n + r) + 1 (6.3)

Let r = 1/2:

an =an1

(2n 2)(n + 1/2) + 1 =an1

(n 1)(2n + 1) + 1 =an1

n(2n 1)n = 1 : a1 =

a01

; n = 2 : a2 =a12.3

=+a02.3

a3 =a23.5

=a0

1.(2.3)(3.5); a4 =

a34.7

=+a0

1(2.3)(3.5)(4.7)(6.4)

an =(1)na0

n!1.3.5.(2n 1) =(1)n2(n1)a0

n(2n 1)!

y1(x) = x1/2

n=0

(1)n2(n1)n(2n 1)! x

n

r = 1:

an =an1

(2n 1)(n + 1) + 1 =an1

(2n + 1)n

a1 =a03.1

, a2 =a15.2

=+a0

(1.3)(2.5); a3 =

a23.7

=a0

(1.3)(2.5)(3.7)

an = (1)na0n!3.5.7.(2n + 1)

= (1)n2na0(2n + 1)!

(6.5)

y2(x) = x

n=0

(1)n2n(2n + 1)!

xn

General Solution: y(x) = c1y1(x) + c2y2(x)Radius of Convergence .

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6.1. SERIES EXPANSION SUMMARY:

6.1 Series Expansion Summary:

ConsiderP(x)y + Q(x)y + R(x)y = 0 (6.6)

Divide by P(x):

y +p(x)y + q(x)y = 0, p(x) =Q(x)

P(x), q(x) =

R(x)

P(x)(6.7)

Ordinary Points:x0 is an ordinary point if p(x) and q(x) are analytic at x0. I.E.

p(x) = p0 +p1(x x0) + q(x) = q0 + q1(x

x0) +

. (6.8)

About an ordinary point x0 we can obtain 2 linearly independent solutionsof the form

y(x) =

n=0

an(x x0)n (6.9)

whose radius of convergence is at least as large as those of p and q in (7.8)- up to the singularity closest to x0.Singular Points: If P(x0) = 0 then p(x) and q(x) may fail to be analyticin which case x0 is a singular point.Regular Singular Points:

A point x0 is a regular singular point if

(x x0) Q(x)P(x)

= p0 +p1(x x0) +

(x x0)2 R(x)P(x)

= q0 + q1(x x0) + (6.10)

are analytic at x0. In this case

(x x0)2y + (x x0)

(x x0) Q(x)P(x)

y + (x x0)2 R(x)

P(x)y = 0 (6.11)

has singularities no worse than the Euler Equation:

(x

x0)

2y + (x

x0)p0y

+ q0y = 0. (6.12)

In this case we look for solutions of the form

y(x) = (x x0)r

n=0

an(x x0)n. (6.13)

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Lecture 7 - Frobenius Series about Regular Singular Points

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Chapter 7

Bessels Equation

Lecture 8

7.1 Bessels Function of Order / {. . . , 2, 1, 0, 1, 2 . . .}:Ly = x2y + xy + (x2 2) y = 0 (7.1)

x = 0 is a regular Singular Point: therefore let y =

n=0

anxn+r.

0 =

n=0

an

(n + r)(n + r 1) + (n + r) 2xn+r + n=0

anxn+r+2 (7.2)

m = n + 2 n = m 2n = 0 m = 2

0 =

m=2

am

(m + r)2 2 + am2 xm+r + a0 r2 2xr (7.3)+ a1

(1 + r)2 2

xr+1

xr > a0 = 0 r = Indicial Eq. Rootsxr+1 > a1

(1 )2 2 = a1(1 2) = 0 provided = 12 .

xm+r > am = am2(m + r)2 2 m 2

(7.4)

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Bessels Equation

r = :

am = am2(m + )2 2 =

am2m2 + 2m

= am2m(m + 2)

a2 = a02(2 + 2)

= a022(1 + )

a4 = a24(4 + 2)

=(1)2a0

2.24(2 + )(1 + )

. . . a2m =(1)ma0

m!22m(1 + ) . . . (m + )(7.5)

y1(x) = x

m=0

(1)m(x/2)2mm!(1 + )(2 + ) . . . (m + )

x0 0

r = :

am = am2m(m 2)

a2 = a02(2 2) =

a022(1 ) , a4 =

a24(4 2) =

(1)2a0224(1 )(2 )

. . . a2m =(1)ma0

m!22m(1 ) . . . (m ) (7.6)

y2(x) = x

m=0

(1)m(x/2)2mm!(1 ) . . . (m )

x0

7.2 Bessels Function of Order = 0 - repeatedroots:

Ly = x2y + xy + x2y = 0.

y =

n=0

anxn+r

Ly =

n=0

an

(n + r)(n + r 1) + (n + r)xn+r + anxn+r+2 = 0m = n + 2 n = m 2 (7.7)

0 = n=2

an(n + r)2 + an2xn+r + a0r(r 1) + rxr + a1(r + 1)r + r + 1xr+1 = 0The indicial equation is: a0r

2 = 0 r1,2 = 0, 0 a double root.

r1 = 0 a1.1 = 0 a1 = 0.

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7.2. BESSELS FUNCTION OF ORDER = 0 - REPEATED ROOTS:

Recursion: an = an2(n + r)2

n 2.

a2 = a022

; a4 = a242

=a0

2242; a6 = a4

62= a0

224262; a8 =

a022426282

(7.8)

a2m =(1)m

22m(m!)2a0 (7.9)

y1(x) =

1 +

m=1

(1)mx2m22m(m!)2

= J0(x)

0 5 10 15 204

3

2

1

0

1

x

J

0(x)and

Y0

(x)

J0

Y

0

Figure 7.1: Zeroth order bessel functions j0(x) and Y0(x)

To get a second solution

y(x, r) = a0xr

1 x2

(2 + r)2+

x4

(2 + r)2(4 + r)2+ + (1)

mx2m

(2 + r)2(4 + r)2 . . . (2m + r)2

+

(7.10

y

r(x, r)

r=r1

= a0 log xy1(x) + a0xr

m=1

(1)mx2m r

1

(2 + r)2 . . . (2m + r)2

.

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Bessels Equation

Let

a2m(r) = { } ln a2m(r) = 2 ln(2 + r) . . . 2ln(2m + r) (7.11)a2m(0) =

2

2 + r 2

4 + r 2

(2m + r)

r=0

a2m(0)

=

1 1

2 . . . 1

m

a2m(0) = Hma2m(0).

Let Hm = 1 +1

2+ + 1

m. Therefore

y2(x) = J0(x) ln x +

m=1(1)m+1Hm

22m(m!)2x2m x > 0. (7.12)

It is conventional to define

Y0(x) =2

y2(x) + ( log 2)J0(x)

(7.13)

where

= limn(Hn log n) = 0.5772 Eulers Constant

y(x) = c1J0(x) + c2Y0(x). (7.14)

Lecture 9

7.3 Bessels Function of Order = 12:

Consider the case = 1/2 Ly = x2y + xy +

x2 14

y = 0. Let

y =

n=0

anxn+r (7.15)

Ly = n=0

an

(n + r)2 14

xn+r + n=0

anxn+r+2 = 0m = n + 2n = m 2n = 0 m = 2

(7.16)

Ly = a0

r2 1

4

+ a1

(r + 1)2 1

4

+

n=2

an

(n + r)2 1

4

+ an2

xn+r = 0.

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7.3. BESSELS FUNCTION OF ORDER = 12 :

Indicial Equation: r2

1

4

= 0, r =

1

2

Roots differ by an integer.

Recurrence: an = an2(n + r)2 14

n 2.

r1 = +1/2:

an = an2(n + 12 )

2 14= an2

(n + 1)nn 2;

9

4 1

4

a1 = 0 a1 = 0

a2 = a03.2

a4 =(1)2a05.4.3.2

. . . a2n =(1)na0(2n + 1)!

y1(x) = x12

n=0(1)nx2n(2n + 1)!

= x12

n=0(1)nx2n+1

(2n + 1)!= x

12 sin x

(7.17)

r2 = 12

:

an = an2(n 12 )2 14

= an2n(n 1) , n 2,

n = 1 a1

12

+ 1

2 1

4

= a1.0 = 0 a1 and a0 arbitrary.

(7.18)

a0:

a2 = a02.1

a4 =(1)2a04.3.2.1

. . . a2n =(1)na0

(2n)!(7.19)

a1:

a3 = a13.2

a5 =(1)2a15.4.3.2

a2n+1 =(1)na1(2n + 1)!

(7.20)

y2(x) = a0x 12 n=0

(1)n

x2n

(2n)!+ a1x 12

n=0

(1)n

x2n+1

(2n + 1)!

= a0x 1

2 cos x + a1x 1

2 sin x (7.21)

included in y1(x).

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Bessels Equation

7.4 Example - the roots differ by an integer

Let Ly = xy y = 0, x = 0 is a regular singular point.

y =

n=0

cnxn+

n=0

cn(n + )(n + 1)xn+1 cnxn+ = 0

(7.22)p 1 = n

n=1 {cn(x + )(n + 1) cn1} xn+1 + c0( 1)x1 = 0Indicial Equation: ( 1) = 0 = 0, 1 differ by integer.Recurrence Rel: cn =

cn1(n + )(n + 1) n 1.

Note: When = 0, c1 blows up!

Let = 1 c1 = c02

, c2 =c012

, . . ..

y1(x) = c0x

1 +

x

2+

x2

12+

= c0u1(x). (7.23)

Second Solution:

y(x, ) = y(x, ) = c0x

+

x

1 + +

x2

(1 + )(2 + )(1 + )+

y

= c0x

ln x

+

x

1 + +

(7.24)

+ c0x

1 x

(1 + )2 x

2

(1 + )2(2 + )

2

(1 + )+

1

(2 + )

+

y

=0

= c0

x +

x2

2+

x3

12+

ln x + c0

1 x 5

4x2

= c0u2.

Therefore y(x) = (A + B ln x)x +x2

2

+x2

12

+

+ B 1 x 5

4

x3

.

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Chapter 8

Separation of Variables

Lecture 10

8.1 Types of Boundary Value Problems:

Dirichlet Boundary Conditions

1. Heat Equation: 2 = Thermal Conductivity.

Heat Flow in a Bar Heat Flow on a Disk

2. Wave Equation: c = Wave Speed.

Vibration of a String

3. Laplaces Equation:

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Neuman Boundary Conditions: What do you expect the solution to

look like as t ?

Mixed Boundary Conditions:

Ice

Heat Bath u(0, t) = A u(L, t) = B Heat Bath 2.

Ice

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8.2. SEPARATION OF VARIABLES - FOURIER SINE SERIES:

8.2 Separation of Variables - Fourier sine Series:

Consider the heat conduction in an insulated rod whose endpoints are heldat zero degree for all time and within which the initial temperature is givenby f(x).

Fouriers Guess:

u(x, t) = X(x)T(t) (8.1)

ut = X(x)T(t) = 2uxx =

2X(x)T(t)

2XT:X(x)X(x)

=T(t)

2T(t)= Constant = 2. (8.2)

>

T(t) = 22T(t) dTT

= 22 dtln |T| = 22t + c

T(t) = De22t.

(8.3)

x >

X(x) + 2X(x) = 0Guess X(x) = erx (r2 + 2)erx = 0 r = i (8.4)

X = c1eix + c2?e

ix

= A sin x + B cos x.(8.5)

Impose the boundary conditions:

0 = u(0, t) = X(0)T(t) = BT(t) B = 00 = u(L, t) = X(L)T(t) = (A sin L)T(t).

(8.6)

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Now we do not want the trivial solution so A = 0. Thus we look for valuesof such that

sin L = 0 =n

L

n = 1, 2, . . . . (8.7)

Thus un(x, t) = e2(nL )

2t sin

nxL

n = 1, 2, . . .

are all solutions of ut = 2uxx. (8.8)

Since (8.8) (above eq. number) is linear, a linear combination of solutionsis again a solution. Thus the most general solution is

u(x, t) =

n=1

bn sinnx

L

e

2(nL )2

t. (8.9)

What about the initial condition u(x, 0) = f(x).

u(x, 0) = f(x) =

n=1

bn sinnx

L

. (8.10)

Given f(x) we need to find the bn such that the infinite series of functionsbn sin

nxL

agrees with f on [0, L].

Question: f(x) may not be periodic f(x + 2L) = f(x) but the series isperiodic since sin

nL

(x + 2L) = sin

nxL

.

Answer: In fact they do agree on [0, L] and are different elsewhere.

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Lecture 11

How do we find the bn?

Observe that we have a new type of eigenvalue problem subject toX(0) = 0 X(L) = 0. Just as in the case with matrices we obtain sequenceof eigenvalues which in this case is infinite:

n =n

L

n = 1, 2, . . . (8.11)

and corresponding eigenfunctions

xn(x) = sin nx = sinnx

L

sinx

L

, sin

2x

L

, sin

3x

L

, . . .

.

(8.12)

Recall that for symmetric matrices the eigenvectors form a basis.

Aside: How do we expand a vector?

Express f in terms of the basis vectors

v1, v2, v3

f = 1v1 + 2v2 + 3v3f vk = 1v1 vk + 2v2 vk + 3v3 vk v1 v1 v1 v2 v1 v3v1 v2 v2 v2 v2 v3

v1 v3 v2 v3 v3 v3

12

3

=

f v1f v2

f v3

(8.13)

If vk v, k = i.e. the vk are orthogonal

k =f vk

vk vk (8.14)

But functions are just infinite dimensional vectors:

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f [f1, f2, . . . , f N]g [g1, g2, . . . , gN]

f g = f1g1 + f2g2 + + fNgN x = LN

(8.15)

=N

k=1

f(xk)g(xk).

NowN

k=1

f(xk)g(xk)x L

0

f(x)g(x) dx = f, g. (8.16)

Back to finding bn:

f(x) =

n=1

bn sinnx

L

(8.17)

L0

f(x)sin

kx

L

dx =

n=1

bn

L0

sinnx

L

sin

kx

L

dx.

Recall sin(A)sin B =1

2{cos(A B) cos(A + B)}. Therefore

Ink =

L

0

sinnxL sinkx

L dx=

1

2

L0

cos(n k) xL

cos(n + k) xL

dx n = k

=1

2

sin(n k)x/L

(n k)/L sin(n + k)x/L

(n + k)/L

L0

= 0 (8.18)

Inn =

L0

sin2

nx

L dx =

1

2

L0

1 cos

2nx

L

dx

= L/2

Therefore bk =2

L

L0

f(x)sin

kx

L

dx.

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Example 8.1

f(x) = 2x 0 < x < 12 L = 1

2(1 x) 12 < x < 1

bn = 2

12

0

2x sin(nx) dx +

112

2(1 x)sin(nx) dx

= 8sin(n/2)

n22n = 1 2 3 4 5

sin

nL

1 0 1 0 1

Therefore u(x, t) =8

2

k=0

(1)k(2k + 1)2

sin

(2k + 1)x

e(2k+1)

22t. (8.19)

Observe as t u(x, t) 0 (all the heat leaks out).

u(x, 0) = 82

k=0

(1)k(2k + 1)2

sin

(2k + 1)x

.

2

8=

k=0

1

(2k + 1)2by letting x =

1

2 f(x) = 1.

2 1 0 1 21

0.5

0

0.5

1

1 terms of the Fourier Series

x

f(x)

2 1 0 1 21

0.5

0

0.5

1


x

f(x)

Example 8.2

f(x) = x 0 < x < 1 L = 1

bn = 2

10

x sin(nx) dx = 2 cos(n)n

= 2 (1)n+1n

Therefore u(x, t) =2

n=1

(1)n+1n

sin(nx)e(n)2t. (8.20)

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As t u(x, t) 0.

u(x, 0) = 2

n=1

(1)n+1n

sin(nx).

u

1

2, 0

=

1

2= 2

n=1

(1)n+1n sin(n/2)

=2

k=0

(1)k(2k + 1)

4= 1 1

3+

1

5 . . . .

k n sin

n2

0 1 1

2 01 3 1

4 02 5 1

(8.21)

2 1 0 1 21

0.5

0

0.5

1


x

f(x)

2 1 0 1 21

0.5

0

0.5

1


x

f(x)

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8.3. HEAT EQ ON A CIRCULAR RING - FULL FOURIER SERIES

Lecture 12

8.3 Heat Eq on a Circular Ring - Full Fourier Se-ries

Physical Interpretation: Consider a thin circular wire in which there is

no radial temperature dependence.u

r= 0.

ut = 2uxx (8.22)

BC:u(L, t) = u(L, t)u

x (L, t) =u

x (L, t) Periodic BC

IC: u(x, 0) = f(x)

Assume u(x, t) = X(x)T(t).

As before:X(x)X(x)

=T(t)

2T(t)= 2.

IVP:T(t)

2T(t)= 2 T(t) = ce2t.

r =2L

2=

L

= Constant.

The Laplacian becomes

u =2u

r2+

1

r

u

r+

1

r22u

2

=2u

(r)2(8.23)

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if we let x = r we obtain (1.1).

BVP:X +

2

X = 0X(L) = X(L)X(L) = X(L)

Eigenvalue Problemlook for such thatnontrivial x can be found.

X(x) = A cos(x) + B sin(x)

X(L) = A cos(L) B sin(L) = A cos(L) + B sin(L) = X(L)therefore 2B sin(L) = 0

X(x) = A sin x + B cos(x) (8.24)X(L) = +A sin(L) + B cos(L) = A sin(L) + B cos(L) = X(L)

therefore 2A sin(L) = 0

Therefore nL = (n) n = 0, 1, . . . .Solutions to (1.1) that satisfy the BC are thus of the form

un(x, t) = e(nL )

22t

An cos

nxL

+ Bn sin

nxL

. (8.25)

Superposition of all these solutions yields the general solution

u(x, t) = A0 +

n=1

An cos

nxL

+ Bn sin

nxL

e(

nL )

22t. (8.26)

In order to match the IC we have

f(x) = u(x, 0) = A0 +

n=1

An cosnx

L

+ Bn sin

nxL

. (8.27)

As before we obtain expressions for the An and Bn by projecting f(x) onto

the basis functions sinnx

L

and cos

nxL

.

LL

f(x)

sin

mx

L

cos

mx

L

dx = A0L

L

sin

mx

L

cos

mx

L

dx (8.28)

+

n=1 A

n

L

L

cosnxL sin mxL cos mxL dx

+

n=1

Bn

LL

sinnx

L

sin mxL cos

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L

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8.3. HEAT EQ ON A CIRCULAR RING - FULL FOURIER SERIES

As in the previous example we use the orthogonality relations:

LL

sinmx

L

sin

nxL

dx = Lmn

LL

cosmx

L

cos

nxL

dx = Lmn m and n = 0 (8.29)

= 2L m = n = 0L

L

sinmx

L

cos

nxL

dx = 0 m,n.

Plugging these orthogonality conditions into (1.6) we obtain

A0 =1

2L

LL

f(x) dx = average value off(x) on [L, L]

An =1

L

LL

f(x)cosnx

L

dx and Bn =

1

L

LL

f(x)sinnx

L

dx.

(8.30)

Note:

1. (1.6) and (1.9) [typists note: check re-numbering when equations arere-labeled, these could be renamed to (9.6) and (9.9) respectively]represent the full Fourier Series Expansion for f(x) on the interval[L, L].

2. By defining an =1

L

LL

f(x)cosnx

L

dx =

2A0

An

and bn = Bn

the Fourier Series (1.6) is often written in the form

f(x) =a02

+

n=1 an cosnx

L + bn sinnx

L . (8.31)

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Chapter 9

Lecture 13 - Fourier Series

We consider the expansion of the function f(x) of the form

f(x) a02

+

n=1

an cosnx

L

+ bn sin

nxL

= S(x) (9.1)

where

an =1

L

LL

f(x)cosnx

L

dx

a02

=1

2L

LL

f(x) dx = average value off.

bn = 1L

LL

f(x)sinnx

L

dx (9.2)

Note:

1. Note that cosn

L(x + )

= cos

nxL

provided

n

L= 2, =

2L

nand similarly sin

nL

(x + 2L)

= sinnx

L

. Thus each of the terms

of the Fourier Series S(x) on the RHS of (10.1) is a periodic functionhaving a period 2L. As a result the function S(x) is also periodic.

How does this relate to f(x) which may not be periodic?

The function S(x) represented by the series is known as the periodicextension of f on [L, L].

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2. Iff (or its periodic extension) is discontinuous at a point x0 then S(x)

converges to the average value of f across the discontinuity.

S(x0) =1

2

f(x+0 ) + f(x

0 )

(9.3)

Example 9.1

f(x) =

0 < x < 0 L = x 0 x (9.4)

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a0 =

1

f(x) dx =

1

0

x dx =

2 (9.5)

an =1

f(x)cos(nx) dx

=1

0

x cos(nx) dx

=1

xsin(nx)

n

0

1n

0

1. sin(nx) dx

=1

sinn

(n) + 1n2

cos(nx)

0

=

1

n2

(1)n 1 n 1 2 3 4(1)n 1 2 0 2 0 (9.6)

a2m+1 = 2(2m + 1)2

m = 0, 1, 2, . . . (9.7)

bn =1

f(x)sin(nx) dx

=

1

0

x sin(nx) dx

=1

x cos(nx)n

0

+1

n

0

1. cos(nx) dx

=1

cos(n)

n+

0. cos0

n+

1

n2sin(nx)

0

= (1)n+1/n (9.8)

f(x) =a02

+

n=1an cos(nx) + bn sin(nx)

=

4 2

m=0

cos (2m + 1)x(2m + 1)2

+

n=1

(1)n+1 sin(nx)n

(9.9)

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9.1 It can be useful to shift the interval of inte-

gration from [L, L] to [c, c + 2L]Since the periodic extension fe(x) is periodic with period 2L (as are the

basis functions cosnx

L

and sin

nxL

).

an = 1L

LL

f(x)cosnxL

dx = 1L

c+2Lc

fe(x)cosnxL

dx (9.10)bn =

1

L

LL

f(x)sinnx

L

dx =

1

L

c+2Lc

fe(x)sinnx

L

dx. (9.11)

Example 9.2 Previous Example:

f(x) =

0 < x < 0x 0 x (9.12)

On [, 3]

fe(x) =

0 < x < 2x 2 2 x 3 (9.13)

an =1

3

fe(x)cos(nx) dx= 1

32

(x 2)cos(nx) dx

= 1

0

t cos(nt) dt.

t = x 2 dx = dtx = t + 2 x = t = x = 3 t =

since cos n(t + 2) = cos t

(9.14)

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9.2. COMPLEX FORM OF FOURIER SERIES

9.2 Complex Form of Fourier Series

f(x) =a02

+

n=1

an cosnx

L

+ bn sin

nxL

cosnx

L

=

ei(nxL ) + ei(

nxL )

2; sin

nxL

=

ei(nxL ) ei(nxL )

2i

Therefore f(x) =a02

+

n=1

an2

ei(

nxL ) + ei(

nxL )

+

bn2i

ei(

nxL ) ei(nxL )

= a02 +

n=1

an ibn2 ei(nxL ) +an + ibn2 ei(nxL ) (9.15) c0 cn cn

=

n=

cnei(nxL )

cn = an ibn2

= 12L

LL

f(x)cosnxL

i sinnxL

dx (9.16)

=1

2L

LL

f(x)ei(nxL ) dx bn = bn (9.17)

Therefore

f(x) =

n=

cn

ei(nxL ) (9.18)

cn =1

2L

LL

f(x)ei(nxL ) dx. (9.19)

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Example 9.3

f(x) =

1 x < 01 0 < x <

L = (9.20)

cn =1

2

0L

einx dx +

0

einx dx

(9.21)

=1

2

e

inx0

(in) +einx

0

(in)

(9.22)

=i

2n 2 + e+in + ein

=

0 n even

(2/in) n odd(9.23)

Therefore

f(x) =

n=

2

i(2n + 1)ei

(2n+1)x

. (9.24)

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Chapter 10

Lecture 14 - Even and OddFunctions

Even: f(x) = f(x)Odd: f(x) = f(x)

10.1 Integrals of Even and Odd Functions

LL

f(x) dx =

0L

f(x) dx +

L0

f(x) dx (10.1)

=

L0

f(x) + f(x) dx (10.2)

=

2L0

f(x) dx f even

0 f odd.

(10.3)

Notes: Let E(x) represent an even function and O(x) an odd function.1. If f(x) = E(x) O(x) then f(x) = E(x)O(x) = E(x)O(x) =

f(x) f is odd.2. E1(x) E2(x) even.

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Lecture 14 - Even and Odd Functions

3. O1(x) O2(x) even.

4. Any function can be expressed as a sum of an even part and an oddpart:

f(x) =1

2

f(x) + f(x)

even part

+1

2

f(x) f(x)

odd part

. (10.4)

Check: Let E(x) =1

2f(x) + f(x)

. Then E(x) = 1

2f(x) + f(x)

=

E(x) even. Similarly let

O(x) =1

2

f(x) f(x) (10.5)

O(x) = 12

f(x) f(x) = O(x) odd. (10.6)

10.2 Consequences of Even/Odd Property for FourierSeries

(I) Let f(x) be Even-Cosine Series:

an =1

L

LL

f(x)cos even

nxL

dx =

2

L

L0

f(x)cosnx

L

dx(10.7)

bn =1

L

LL

f(x)sinnx

L

odd

dx = 0. (10.8)

Therefore

f(x) =a02

+

n=1

an cosnx

L

; an =

2

L

L0

f(x)cosnx

L

dx.(10.9)

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10.2. CONSEQUENCES OF EVEN/ODD PROPERTY FOR FOURIERSERIES

(II) Let f(x) be Odd-Sine Series:

an =1

L

LL

f(x)cosnx

L

odd

dx = 0 (10.10)

bn =1

L

LL

f(x)sinnx

L

even

dx =2

L

L0

f(x)sinnx

L

dx

Therefore

f(x) =

n=1

bn sinnxL ; bn = 2LL

0f(x)sinnxL dx.

(III) Since any function can be written as the sum of an even and odd part,we can interpret the cos and sin series as even/odd:

f(x) =even odd

1

2

f(x) + f(x) + 1

2

f(x) f(x) (10.11)

=

a02

+

n=1an cos

nx

L

+

n=1

bn sin

nx

L

where

an =2

L

L0

1

2

f(x) + f(x) cosnx

L

dx =

1

L

LL

f(x)cosnx

L

dx

bn =2

L

L0

1

2

f(x) f(x) sinnx

L

dx =

1

L

LL

f(x)sinnx

L

dx.

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10.3 Half-Range Expansions

If we are given a function f(x) on an interval [0, L] and we want to representf by a Fourier Series we have two choices - a Cosine Series or a Sine Series.

Cosine Series:

f(x) =a02

+

n=1

an cosnx

L

(10.12)

an =2

L

L0

f(x)cosnx

L

dx. (10.13)

Sine Series:

f(x) =

n=1

bn sinnx

L

(10.14)

bn =2

L

L0

f(x)sinnx

L

dx. (10.15)

Example 10.1 Expand f(x) = x 0 < x < 2 in a half-range (a) Sine Series,(b) Cosine Series.

(a)

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10.3. HALF-RANGE EXPANSIONS

bn =

2

0

f(t)sin

n

t dt (10.16)

=

20

t sinn

2t dt (10.17)

= t cosn2 t

n2

2

0

+2

n

20

cosn

2t dt (10.18)

= 4n

cos(n) +

2

n

2sin

n2

t

2

0(10.19)

= 4n

(1)n f(1) = 1 = 4 n=1 (1)n+1n sin n2 4 = 1 13 + 15 17 +

(10.20)

Therefore

f(t) =4

n=1

(1)n+1n

sinn

2t

. (10.21)

(b)

a0 =1

2

20

t dt =1

2

t2

2

20

= 1 (10.22)

an =

20

t cos n2

t dt = 2n t sin n

2t2

0 2

n 2

0

sin n2

t dt

= +

2

n

2cos

n

2t

2

0

=4

n22{cos n 1} (10.23)

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Therefore

f(t) = 1 +4

2

n=1

(1)n 1

n2cos

n

2t (10.24)

= 1 82

n=0

cos(2n + 1)

2t/(2n + 1)2. (10.25)

The cosine series converges faster than Sine Series.

f(2) = 2 = 1 +8

2

n=0

1

(2n + 1)22

8= 1 +

1

32+

1

52+

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Chapter 11

Lecture 15 - Convergence ofFourier Series

Example 11.1 (Completion of problem illustrating Half-range Expansions)Periodic Extension: Assume that f(x) = x, 0 < x < 2 represents one fullperiod of the function so that f(x + 2) = f(x). 2L = 2

L = 1.

a0 =1

L

LL

f(x) dx =

11

f(x) dx =

20

x dx =x2

2

20

= 2 (11.1)

since f(x + 2) = f(2). (11.2)

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Lecture 15 - Convergence of Fourier Series

n 1:

an =1

L

LL

f(x)cosnx

L

dx =

11

f(x) cos(nx) dx L = 1

=

20

x cos(nx) dx

=

x sin(nx)

n

20

1

n

20

sin(nx) dx

=

1

(n)2 cos(nx)2

0=

1

(n)2 cos(2n) 1 = 0 (11.3)bn =

1

L

LL

f(x)sinnx

L

dx =

11

f(x)sin(nx) dx

=

20

x sin(nx) dx =

xcos(nx)

n

20

+1

(n)

20

cos(nx) dx

=2n

+sin(nx)

(n)2

2

0=

2n

(11.4)

Therefore

f(x) =2

2 2

n=1

sin(nx)

n(11.5)

= 1 2

n=1

sin(nx)

n

(11.6)

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4 2 0 2 4

1

0

1

2

3

x

S(x)

Figure 11.1: Full Range Expansion SN(x) = 1 2N=20n=1

sin(nx)n

4 2 0 2 42

1

0

1

2

x

S(x)1

Figure 11.2: Full Range Expansion SN(x) 1 = 2N=20n=1

sin(nx)n

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11.1 Convergence of Fourier Series

What conditions do we need to impose on f to ensure that the FourierSeries converges to f.

We consider piecewise continuous functions:

Theorem 11.2 Let f and f be piecewise continuous functions on [L, L]and let f be periodic with period 2L, then f has a Fourier Series

f(x) = a02 +

n=1an cos

nxL

+ bn sin

nxL

= S(x)

where

an = 1L

LL

f(x)cos nxL dx and bn = 1L LL f(x)sin nxL dx.(11.7)

The Fourier Series converges to f(x) at all points at which f is continuous

and to1

2

f(x+) + f(x) at all points at which f is discontinuous.

Thus a Fourier Series converges to the average value of the left andright limits at a point of discontinuity of the function f(x).

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11.2. ILLUSTRATION OF THE GIBBS PHENOMENON

11.2 Illustration of the Gibbs Phenomenon

Near points of discontinuity truncated Fourier Series exhibit oscilla-tions - overshoot.

2 1 0 1 21.5

1

0.5

0

0.5

1

1.5

x/

SN

(x)forN=5

Figure 11.3: Fourier Series for a step function

Example 11.3 Consider the half-range sine series expansion of

f(x) = 1 on [0, ]. (11.8)

f(x) = 1 =

n=1bn sin(nx)

where bn =2

0

sin(nx) dx = 2 cos nxn 0 = 2n 1 (1)n

=

4/n n odd0 n even

Therefore f(x) = 4

n=1

n odd

sin(nx)n =

4

m=0

sin(2m+1)x(2m+1) .

(11.9)

Note:

1. f(/2) = 1 =4

m=0

sin

(2m + 1)/2

(2m + 1)=

4

1 1

3+

1

5

. There-

fore

4= 1 1

3+

1

5 .

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2. Recall the complex Fourier Series example for the function

f(x) =

1 x < 01 0 < x <

(11.10)

which turns out to be equivalent to the odd extension of the abovefunction represented by the half-range sine expansion, which we cansee from the following calculation

f(x) =

n=

n odd

2in e

inx = 4

n=1

n odd

einxeinx2in

= 4n=1

n odd

sin(nx)n .

(11.11)

11.3 Now consider the sum of the first N terms

SN(x) =4

Nm=0

sin(2m + 1)x

(2m + 1)=

4

Im

N

m=0

ei(2m+1)x

(2m + 1)

(11.12)

SN(x) =4

Im

N

m=0 iei(2m+1)x

(11.13)=

4

Im

ieix

Nm=0

ei2x

m(11.14)

=4

Im

ieix

1 + ei2x + + ei2xN

1 ei2x

(1 ei2x)

(11.15)

=4

Im

ieix

1 ei2(N+1)x

1 ei2x

(11.16)

=4

Im

i

1 ei2(N+1)x

eix

eix

(11.17)

=2

Im

ei2(N+1)x 1

sin x

(11.18)

=2

sin2(N + 1)x

sin x. (11.19)

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11.3. NOW CONSIDER THE SUM OF THE FIRST N TERMS

Therefore

t = 2(N + 1)u du = dt2(N+1)

SN(x) =

2

x0

sin 2(N+1)usin u du 2

2(N+1)x0

sint t dt

(11.20)

0

0.5

1

1.

5

2

1

0

5 0 5 1

0

x/

(2/) sin(2(N+1)x)/sin(x)

Figure 11.4: (2/)sin(2(N + 1)x)/sin(x) for N = 5

Observe SN(x) =2

sin2(N + 1)x

sin x= 0 when 2(N + 1)xN = thus the

maximum value of SN(x) occurs at

xN =

2(N + 1)(11.21)

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0

0.

5

1

1

.5

2

1.

51

0.

5 0

0.

5 1

1.

5

x/

(2/) 0

xsin 2(N+1)u /sin u du for N= 5

Figure 11.5: Integral of (2/)sin(2(N + 1)x)/sin(x)

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Chapter 12

Lecture 16 - ParsevalsIdentity

Lemma 12.1 (A version of Parsevals Identity)

Letf(x) =

n=1

bn sinnx

L

0 < x < L. Then

2

L

L0

f(x)

2dx =

n=1

b2n.

Proof:

L

0

f(x)2 dx =

m=1

n=1

bmbn

L

0

sinmxL sinnxL dx (12.1)=

m=1

n=1

bmbn mn L2

=L

2

n=1

b2n. (12.2)

For a full Fourier Series on [L, L] Parsevals Theorem assumes the form:

f(x) =a02

+

n=1

an cosnx

L

+ bn sin

nxL

(12.3)

1

L

L

L

f(x)2 dx = a20

2 +

n=1 a

2

n + b2

n. (12.4)

Example 12.2 Recall for x [0, 2] f(x) = x = 4

n=1

(1)n+1n

sinnx

2

.

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Lecture 16 - Parsevals Identity

Therefore

2L

L0

f(x)

2dx = 22

20

x2 dx = 4

2 n=1

1n2

x332

0=

4

2 n=1

1n2

2

6 =

n=1

1n2

(12.5)

Note:

n=1

1

(2n)2=

1

22

n=1

1

n2=

1

4

2

6

=

2

24.

Also note that

evens odds

2

6 = n=1

1n2 = m=1

1(2m)2 + m=0

1(2m+1)2

= 2

24 +

m=0

1(2m+1)2

Therefore

m=0

1

(2m + 1)2=

2

6

2

24=

2

8. (12.6)

12.1 Geometric Interpretation of Parsevals For-mula

f = b1e1 + b2e2 (12.7)

|f|2 = f f = b21e1 e1 + b22e2 e2 (12.8)= b21 + b

22 Pythagoras Theorem (12.9)

For Fourier Sine Components:

2

L

L0

f(x)

2dx =

n=1

b2n. (12.10)

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12.1. GEOMETRIC INTERPRETATION OF PARSEVALS FORMULA

Example 12.3 Consider f(x) = x2 < x < . The Fourier SeriesExpansion is:

x2 =2

3+ 4

n=1

(1)nn2

cos(nx). (12.11)

n 1 2 3 4cos

n2

0 1 0 1

Let

x = 2 2

4 =2

3 + 4

n=1

(1)nn2 cos

n2

212 = 4

k=1

(1)k(2k)2

(12.12)

Therefore

2

12 =

k=1

(

1)k+1

k2 . (12.13)

By Parsevals Formula:

2

0

x4 dx = 2

2

3

2+ 16

n=1

1n4

2

x5

5

0

= 24

9 + 16

n=1

1n4

9545 =

445 =

890

190

(12.14)

Therefore

4

90=

n=1

1

n4= ?(4). (12.15)

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Lecture 16 - Parsevals Identity

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Chapter 13

Lecture 17 - Solving the heatequation using finitedifference methods

13.1 Approximating the Derivatives of a Functionby Finite Differences

Recall that the derivative of a function was defined by taking the limit of adifference quotient:

f(x) = limx0

f(x + x) f(x)x

. (13.1)

Now to use the computer to solve differential equations we go in the oppositedirection - we replace derivatives by appropriate difference quotients. If weassume that the function can be differentiated many times then TaylorsTheorem is a very useful device in determining the appropriate differencequotient to use. For example consider

f(x + x) = f(x) + xf(x) +x2

2!f(x) +

x3

3!f(3)(x) +

x4

4!f(4)(x) + . . .(13.2)

Re-arranging terms in (2) and dividing by x we obtain

f(x + x) f(x)x

= f(x) +x

2f(x) +

x2

3!f(3)(x) + . . . .

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Lecture 17 - Solving the heat equation using finite difference methods

If we take the limit x 0 then we recover (1). But for our purposes it ismore useful to retain the approximation

f(x + x) f(x)x

= f(x) +x

2f() (13.3)

= f(x) + O(x).

We retain the termx

2f() in (3) as a measure of the error involved when

we approximate f(x) by the difference quotient

f(x + x) f(x)/x.Notice that this error depends on how large f is in the interval [x, x + x](i.e. on the smoothness of f) and on the size of x. Since we like to focuson that part of the error we can control we say that the error term is of the

order x denoted by O(x). Technically a term or function E(x) isO(x) if

E(x)

x

x0 const.

Now the difference quotient (3) is not the only one that can be used toapproximate f(x). Indeed if we consider the expansion of f(x x):

f(x x) = f(x) xf(x) + x2

2!f(x) x

3

3!f(3)(x) +

x4

4!f(4)(x) + . . . .(13.4)

and we subtract (4) from (2) and divide by (2x) we obtain:

f(x + x) f(x x)2x

= f(x) +x2

3!f(3)(). (13.5)

We notice that the error term associated with this form of difference ap-proximation is O(x2), which converges more rapidly to zero as x 0.

In order to obtain an approximation to f(x) we add (2) to (4) whichupon re-arrangement and dividing by x2 leads to:

f(x + x) 2f(x) + f(x x)x2

= f(x) +1

12x2f(4)(). (13.6)

Due to the symmetry of the difference approximations (5) and (6) about theexpansion point x these are called central difference approximations. Thedifference approximation (3) is known as a forward difference approximation.We note that the central difference schemes (5) and (6) are second orderaccurate while the forward difference scheme (3) is only O(x).

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13.2. HEAT EQUATION SOLUTION BY FINITE DIFFERENCES

13.2 Heat Equation solution by Finite Differences

Consider the following initial-boundary value problem for the heat equation

u

t= 2

2u

x20 < x < 1, t > 0 (13.7)

BC: u(0, t) = 0 u(1, t) = 0 (13.8)

IC: u(x, 0) = f(x). (13.9)

The basic idea is to replace the derivatives in the heat equation by differ-ence quotients. We consider the relationships between u at (x, t) and itsneighbours a distance x apart and at a time t later.

Corresponding to the difference quotient approximations introduced in

Section 1, we consider the following partial difference approximations.Forward Difference in Time:

u(x, t + t) = u(x, t) + tu

t(x, t) +

t2

2!

2u

x2(x, t) + .

After re-arrangement and division by t:


=u

t(u, t) + O(t). (13.10)

Central Differences in Space:

u(x + x, t) = u(x, t) + xux

(x, t) + x22!

2ux2

(u, t) + x33!

3ux3

(x, t) + x44!

4ux2

(x, t) +

u(x x, t) = u(x, t) xux

(x, t) +x2

2!

2u

x2(x, t) x

3

3!

3u

x3(x, t) +

x4

4!

4u

x4(x, t) +

Adding and re-arranging:

u(x + x, t) 2u(x, t) + u(x x, t)x2

=2u

x2(x, t) + O(x2). (13.11)

Substituting (2) and (3) into (1a) we obtain


= 2 u(x + x, t) 2u(x, t) + u(x x, t)x2 + O(t, x2).Re-arranging:

u(x, t + t) = u(x, t) + 2

t

x2

{u(x + x, t) 2u(x, t) + u(x x, t)} .(13.12)

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We subdivide the spatial interval [0, 1] into N + 1 equally spaced sample

points xn = nx. The time interval [0, T] is subdivided into M + 1 equaltime levels tk = kt. At each of these space-time sample points we introduceapproximations:

u(xn, tk) ukn.

u

u u

u

u u

u

uu

TTTTTT

-?

6

uk0 ukn1 u

kn u

kn+1 u

kN1 u

kN

tk+1

tk

uk+1Nuk+1nu

k+10

t

x

uk+1n = ukn +

2

t

x2

ukn+1 2ukn + ukn1

This is implemented in the spread sheets Heat0 and Heat.

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13.2. HEAT EQUATION SOLUTION BY FINITE DIFFERENCES

Implementing Derivative Boundary Conditions:

Assume that the boundary conditions (1b) are changed to

BC: u(0, t) = 0,u

x(1, t) = 0.

Consider a central difference approximation tou

x(1, t), where xN = Nx =

1,

u(xN + x, t) u(xN x, t)x

= 0.

Re-arranging we obtain:

u(xN + x, t) = u(xN x, t) ()

Since xN = 1 we observe that xN+x is outside the domain we introducean extra column uN+1 into which we copy the values uN1. In the columnxN we implement the same difference approximation for the Heat Equation,namely:

uk+1N = ukN +

2(t

x2)(ukN+1 2ukN + ukN1) ()

This is implemented in the spread sheet Heat0f.while u

kN+1 = u

kN1 (see (*) ) since column u

kN1 is copied to column u

kN+1.

Note that this BC could be implemented another way without introducingthe additional column, by eliminating uN+1 from () and ():

uk+1N = ukN + 2

2

t

x2

ukN1 ukN

.

If this latter equation is implemented at xN there is no need to introducean extra column UN+1 or to implement the difference equation given in (**)as the the derivative boundary condition is taken care of automatically.

Some EXERCISES and Observations: Heat Equation

1. Change the t in cell D1 from 0.001 to 0.05 and you will observe whatis known as a numerical instability. Now change t to 0.00625, whichis known as the stability boundary and observe what happens. Nowlet t = 0.006 and observe the abrupt change in the solution - it ismuch closer to what we would expect.

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2. The instability noted in 1. above is not the only source of error in the

numerical approximation. Although numerical instability is evident fora parameter choice that is unstable, the other type of error is presentin almost every type of numerical approximation scheme. This class oferror results from discarding the O(x2) and O(t) terms in (2) and(3) when we replace derivatives in (1a) by difference quotients. Thiserror is known as the truncation error. To determine the truncationerror change the spread sheet to implement the initial condition

f(x) =

2x 0 < x < 1/22(1 x) 1/2 x < 1 .

Now code up the Fourier Series (in another spread sheet) that is de-

rived on page 21 of the notes and compare the numerical solution tothe exact Fourier Series solution with 50 terms. The difference be-tween the two is mainly due to the truncation error since the round-offerror is about 1012 and does not grow if stable parameters are used.

3. Implement derivative boundary conditions on both endpoints x = 0 andx = 1. Check the numerical solution against the problem solved inHW1 #3.

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Chapter 14

Lecture 18 - Solving

Laplaces Equation usingfinite differences

14.1 Finite Difference approximation

Consider the boundary value problem

2u

x2+

2u

y 2= 0 0 < x, y < 1 (14.1)

BC: u(0, y) = 0; u(1, y) = 0; u(x, 0) = f(x); u(x, 1) = 0.(14.2)

-

6 6

-

-

6?

1

u(1, y) = 0

1 x1 xn xN = 1y0

y1

1 = yM

x

y

y

x x0

uxx + uyy = 0

u(x, 0) = f(x)

u(0, y) = 0

u(x, 1) = 0

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Lecture 18 - Solving Laplaces Equation using finite differences

As before we replace the second derivatives in (1a) by central difference

quotients that are second order accurate:

u(x + x, y) 2u(x, y) + u(x x, y)x2

=2u

x2(x, y) + O(x2)(14.3)

u(x, y + y) 2u(x, y) + u(x, y y)y2

=2u

y2(x, y) + O(y2).(14.4)

We partition the interval 0 x 1 into (N + 1) equally spaced nodesxn = nx and the interval 0 y 1 into (M + 1) equally spaced nodesym = my. Replacing the derivatives in (1a) by the difference quotients in(3) and (4) and representing the mesh values at (xn, ym) by unm u(xn, ym)we obtain:

un+1m 2unm + un1mx2

+unm+1 2unm + unm1

y2= (uxx + uyy)(xn,xm) + O(x

2, y2).

If we choose x = y then we obtain

un+1m + un1m + unm+1 + unm1 4unm = 0 1 n, m (N 1), (M 1).(14.5)

u

u u

u

u jj

j

j

j

1 1

1

1

-4

un1m unm un+1m

unm1

unm+1

This is known as the finite difference Stencil that relates unm to its 4nearest neighbours.

This is a system of (N 1) (M 1) unknowns for the values of unminterior to the domain - recall the boundary values are already specified!

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14.2. SOLVING THE SYSTEM OF EQUATIONS BY JACOBIITERATION

14.2 Solving the System of Equations by Jacobi

Iteration

This is a procedure to solve the system of Equation (3) by looping througheach of the mesh points and updating unm according to (3) assuming thatthe nearest neighbours already have values close to the exact solution. Thisprocedure is repeated until the changes that are made in each iteration fallsbelow a certain tolerance.

To implement this iterative procedure we observe that the discrete LaplaceEquation (5) can be re-written in the form:

u

k+1

nm =

ukn+1m + ukn

1m + u

knm+1 + u

knm

1

4 (14.6)

t t t

t

t

- ?

6 average

Thus unm is the average value of its nearest neighbours. Note that a newsuperscript index k has been introduced to represent the nodal values at thekth iteration. Thus iteration can be viewed as taking successive neighbouraverages until there is no change, at which point the value of umn equalsthe average of the values at its mesh neighbours. This mean value propertyis a discrete form of a fundamental property of any solution to LaplacesEquation.

To implement the iterative procedure (6) on a spread sheet, go to theTools Menu at the top of the screen and click on the Options Tab. Thenselect the Calculation Tab. Check the Iteration box. If you set the numberof iterations to 5 say, then if you start with zero values throughout theinterior of the domain (as you should if you cut and paste as demonstratedin class), you will see the values percolate 5 cells into the domain from thenon zero boundary condition f(x) = sin(x). You can choose a surface plot

to visualize the solution. Now hold down the F9key and watch the solutionmove to equilibrium. This iterative process essentially uses diffusion on apseudo time scale to take the solution to equilibrium.

EXERCISES and Notes for Laplaces Equation:

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Lecture 18 - Solving Laplaces Equation using finite differences

1. Implement a 0 derivative BC along the lines x = 0 and x = 1. Plot a

cross section of the results along y = 1/2. To ensure that ux (0, y) =

0 =u

x(1, y).

2. Implement an inhomogeneous term for Poissons Equation:

2u

x2+

2u

y2= f(x, y) 0 < x, y < 1.

Introduce finite difference quotients, assume x = y to arrive at theiterative formula:

uk+1nm = ukn+1m + ukn1m + uknm+1 + uknm1 x2f(xn, ym)4 . ()It may be useful to calculate the values of fnm on a separate sheet inwhich the same cell values as those for unm are maintained. Then thevalues offnm can be referenced in the calculation ofunm according to().

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Chapter 15

Lecture 19 Further HeatConduction Problems:Inhomogeneous BC

Example 15.1 Specified

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