ACM95b Notes

Giordon Stark ACM95b Notes March 13, 2010

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TABLE OF CONTENTS

1st Order ODEs ...........................................................................................................................................................................................4 Linear 1st Order ODEs ................................................................................................................................................................................4

General Strategy (Leibniz) .........................................................................................................................................4 Existence and Uniqueness .........................................................................................................................................5 Complex Case ............................................................................................................................................................5

2nd Order Linear IVPs ..................................................................................................................................................................................6 Homogeneous 2nd Order Linear IVP ...........................................................................................................................................................7

Solution Properties ....................................................................................................................................................7 Homogeneous 2nd Order Linear ODEs with constant coefficients .............................................................................................................8 Nonhomogeneous 2nd Order Linear ODEs ................................................................................................................................................10 Dirac Delta Function .................................................................................................................................................................................11 Heaviside Step Function ...........................................................................................................................................................................11 Green’s Function for IVP ..........................................................................................................................................................................11 IVPs with Constant Coefficients ...............................................................................................................................................................13

Solution Strategy .....................................................................................................................................................14 Uniqueness ..............................................................................................................................................................14

Convolution Integrals ...............................................................................................................................................................................15 Shifting Theorems ....................................................................................................................................................................................16

s-shifting ..................................................................................................................................................................16 t-shifting ..................................................................................................................................................................17

IVPs with Discontinuous Forcing ..............................................................................................................................................................18 IVPs with Impulsive Forcing ......................................................................................................................................................................19 Inverting Laplace Transforms ...................................................................................................................................................................20 Evaluating Bromwich Contour Integrals ...................................................................................................................................................21 Nonlinear 1st Order ODEs .........................................................................................................................................................................23

Existence and Uniqueness .......................................................................................................................................23 Picard’s Theorem ......................................................................................................................................................................................23 Numerical Methods for IVP ......................................................................................................................................................................23

Euler’s Method ........................................................................................................................................................24 Alternative derivations ............................................................................................................................................24 Other Possibilities ....................................................................................................................................................24

Truncation Error .......................................................................................................................................................................................25 Explicit Euler ............................................................................................................................................................26


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Implicit Euler ...........................................................................................................................................................26 Trapezoid Rule .........................................................................................................................................................26

Removing Implicitness..............................................................................................................................................................................28 Explicit Runge-Kutta Methods ..................................................................................................................................................................29

1-Stage Scheme .......................................................................................................................................................29 2-Stage Scheme .......................................................................................................................................................29 4-Stage Scheme .......................................................................................................................................................29

nth Order ODEs ........................................................................................................................................................................................30 Linear Stability Analysis ............................................................................................................................................................................30

1-Stage Runge-Kutta (explicit Euler) ........................................................................................................................30 4-Stage Runge-Kutta ................................................................................................................................................31 Backward Euler ........................................................................................................................................................31

Linear Equations with Analytic Coefficients .............................................................................................................................................31 Existence of Series Solution......................................................................................................................................................................32 Series Solution Near an Ordinary Point ....................................................................................................................................................32

Example: (Airy’s Equation) ..................................................................................................................................32 Behavior near Singular Points ..................................................................................................................................................................34 Regular Singular Points of Second Order Equation ..................................................................................................................................35

Euler Equations .......................................................................................................................................................36 The Method of Frobenius .........................................................................................................................................................................37 2nd Order Linear Boundary Value Problem (BVP) .....................................................................................................................................38 Simple Eigenvalue Problems ....................................................................................................................................................................40 Fourier Series ...........................................................................................................................................................................................42

Periodicity ...........................................................................................................................................................42 Orthogonality Properties ....................................................................................................................................42

Convergence of Fourier Series .................................................................................................................................................................44 Fourier Cosine and Sine Series .................................................................................................................................................................45 Solving PDes .............................................................................................................................................................................................46 The Heat Equation ....................................................................................................................................................................................48 Revisiting Solution to the Heat Equation .................................................................................................................................................49

Justification of Termwise Differentiation ................................................................................................................49 Complex Form of Fourier Series ...............................................................................................................................................................50

Useful Terminology .............................................................................................................................................51 Representing Non-Periodic Functions ......................................................................................................................................................52 Fourier Transform ....................................................................................................................................................................................52 Spectral Analysis Using the Fourier Transform ........................................................................................................................................52


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Evaluating Fourier Transforms .................................................................................................................................................................53 Properties of the Fourier Transform .......................................................................................................................55

Convolution Theorem .........................................................................................................................................55 Derivatives ..........................................................................................................................................................56 Linearity ..............................................................................................................................................................56

Solving Differential Equations on an Infinite Domain using the Fourier Transform ................................................................................56 More General Eigenvalue Problems .........................................................................................................................................................57 The Adjoint Problem ................................................................................................................................................................................58

Self-Adjointness .......................................................................................................................................................58 Regular Sturm-Liouville Eigenvalue Problems ..........................................................................................................................................59 Orthogonality of Eigenfunctions ..............................................................................................................................................................59

The Eigenvalues Are Real ........................................................................................................................................60 Simplicity of Eigenvalues .........................................................................................................................................60 Sequence of Eigenvalues .........................................................................................................................................61 Eigenfunction Expansions ........................................................................................................................................61 Orthonormal Eigenfunctions ...................................................................................................................................61 Convergence of Eigenfunction Expansions .............................................................................................................62

Nonhomogeneous Boundary Value Problems .........................................................................................................................................62 The Fredholm Alternative ........................................................................................................................................................................64 Green’s Functions for BVP ........................................................................................................................................................................64

Reciprocity ...............................................................................................................................................................65 Singular Sturm-Liouville Eigenvalue Problems .........................................................................................................................................69

Properties of Singular Sturm-Liouville Eigenvalue Problems ..................................................................................70 Heat Equation in a Disk with Radial Symmetry ........................................................................................................................................70 ACM Final Review .....................................................................................................................................................................................71

I N T RO D UC T I O N

These notes were taken during the 2010 Winter term of ACM95. At the time, Niles Pierce was the instructor and these notes come from the lectures he held. This version of the notes is meant to be the finalized version, with (hopefully) no typos, and less spaces so that it can be ‘read like a novel’. Similarly, Sean Mach’s textbook (found in the Underground) will do just as well. Originally, I had no purpose other than to electronic-ize all my work (problem sets, for example). However, I realized that after a while, a lot of people started getting interested in having electronic notes. Getting some inspiration from other students who had done similar things, I decided to consistently post all my notes on my ITS student account. Without further ado, enjoy!

- Giordon


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L EC T URE 1

1ST ORDER ODES

𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

= 𝑓𝑓(𝑑𝑑, 𝑑𝑑)

Visualize Solution Field

Pick an integral curve (1,2,3, for example) by providing initial condition 𝑑𝑑(0) = 𝑑𝑑0.

LINEAR 1ST ORDER ODES

General Form


+ 𝑝𝑝(𝑑𝑑)𝑑𝑑 = ℎ(𝑑𝑑), 𝑑𝑑(𝑑𝑑0) = 𝑑𝑑0

With 𝑝𝑝(𝑑𝑑), ℎ(𝑑𝑑) continuous

EXAMPLE


= 𝑑𝑑, 𝑑𝑑(0) = 1

So, 𝑑𝑑𝑑𝑑𝑑𝑑

= 𝑑𝑑𝑑𝑑 → ln|𝑑𝑑| = 𝑑𝑑 + 𝑐𝑐0 → |𝑑𝑑| = 𝑒𝑒𝑐𝑐0𝑒𝑒𝑑𝑑

𝑑𝑑 = 𝑐𝑐𝑒𝑒𝑑𝑑 , 𝑐𝑐 = 𝑒𝑒𝑐𝑐0

𝑑𝑑(0) = 𝑐𝑐𝑒𝑒0 = 1 ⇒ 𝑐𝑐 = 1

Our specific solution to the ODE is then 𝑑𝑑 = 𝑒𝑒𝑑𝑑 .

GENERAL STRATEGY (LEIBNIZ)

Find an integrating factor 𝜇𝜇(𝑑𝑑) that makes

𝜇𝜇(𝑑𝑑) �𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

+ 𝑝𝑝(𝑑𝑑)𝑑𝑑 = ℎ(𝑑𝑑)� (1)


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Equivalent to

𝑑𝑑𝑑𝑑𝑑𝑑

(𝜇𝜇𝑑𝑑) = 𝜇𝜇ℎ (2)

And then integrate as in example. How do we find 𝜇𝜇(𝑑𝑑)? Compare (1) and (2).

𝜇𝜇𝑑𝑑′ + 𝜇𝜇𝑝𝑝𝑑𝑑 = 𝜇𝜇ℎ (1)

𝜇𝜇𝑑𝑑′ + 𝜇𝜇′𝑑𝑑 = 𝜇𝜇ℎ (2)

Thus we need some 𝜇𝜇 with 𝜇𝜇′ = 𝑝𝑝𝜇𝜇. So 𝜇𝜇 = 𝑒𝑒∫ 𝑝𝑝(𝑠𝑠)𝑑𝑑𝑠𝑠 𝑑𝑑𝑎𝑎 𝑤𝑤ith a constant, choose 𝑎𝑎 = 𝑑𝑑0 for convenience.

𝜇𝜇(𝑑𝑑) = 𝑒𝑒∫ 𝑝𝑝(𝑠𝑠)𝑑𝑑𝑠𝑠 𝑑𝑑𝑑𝑑0 , 𝜇𝜇(𝑑𝑑0) = 1

Then solve 𝑑𝑑𝑑𝑑𝑑𝑑

(𝜇𝜇𝑑𝑑) = 𝜇𝜇ℎ. So 𝜇𝜇𝑑𝑑 = ∫ 𝜇𝜇ℎ𝑑𝑑𝑠𝑠𝑑𝑑𝑑𝑑0

+ 𝐶𝐶. Choose 𝐶𝐶 to satisfy initial condition

𝑑𝑑(𝑑𝑑) =1

𝜇𝜇(𝑑𝑑)�𝜇𝜇(𝑠𝑠)ℎ(𝑠𝑠)𝑑𝑑𝑠𝑠𝑑𝑑

𝑑𝑑0

+𝑑𝑑0

𝜇𝜇(𝑑𝑑)

EXISTENCE AND UNIQUENESS

𝑑𝑑′ + 𝑝𝑝(𝑥𝑥)𝑑𝑑 = ℎ(𝑥𝑥), 𝑑𝑑(𝑥𝑥0) = 𝑑𝑑0

Unique solution exists throughout an open interval containing 𝑥𝑥0 in which 𝑝𝑝(𝑥𝑥) and ℎ(𝑥𝑥) are continuous (proof is constructive).

Moral: just inspect ODE to identify interval of guaranteed existence and uniqueness. You may lose existence and uniqueness at a singular point where 𝑝𝑝(𝑥𝑥) or ℎ(𝑥𝑥) is not continous.

EXAMPLE

𝑑𝑑′ =𝑑𝑑𝑥𝑥

, 𝑑𝑑(𝑥𝑥0) = 𝑑𝑑0

So ℎ(𝑥𝑥) = 0, 𝑝𝑝(𝑥𝑥) = −1𝑥𝑥

Then 𝑑𝑑(𝑥𝑥) = 𝑐𝑐𝑥𝑥

If 𝑥𝑥0 ≠ 0: 𝑑𝑑(𝑥𝑥) =𝑑𝑑0

𝑥𝑥0𝑥𝑥, "all is well"

If 𝑥𝑥0 = 0: �𝑑𝑑(𝑥𝑥0) ≠ 0 → Solution DNE

𝑑𝑑(𝑥𝑥0) = 0 → infinite solutions → not unique → all integral curves pass through origin�

COMPLEX CASE

𝑑𝑑𝑤𝑤𝑑𝑑𝑑𝑑

+ 𝑝𝑝(𝑑𝑑)𝑤𝑤 = ℎ(𝑑𝑑)

𝑝𝑝(𝑑𝑑), ℎ(𝑑𝑑) analytic in a simply connected domain 𝐷𝐷


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𝜇𝜇(𝑑𝑑) = 𝑒𝑒∫ 𝑝𝑝(𝑠𝑠)𝑑𝑑(𝑠𝑠)𝑑𝑑𝑑𝑑0

→analytic independent of contour 𝑑𝑑0 → 𝑑𝑑 since 𝑝𝑝(𝑑𝑑) is analytic (Cauchy-Goursat and Equivalence Theorem)

𝑤𝑤(𝑑𝑑) =1

𝜇𝜇(𝑑𝑑) �𝜇𝜇(𝑠𝑠)ℎ(𝑠𝑠)𝑑𝑑𝑠𝑠𝑑𝑑

𝑑𝑑0

+𝑤𝑤(𝑑𝑑0)𝜇𝜇(𝑑𝑑)

Analytic in 𝐷𝐷 since 𝜇𝜇(𝑑𝑑) ≠ 0.

EXAMPLE


+𝛼𝛼𝑑𝑑𝑤𝑤 = 0

Note that 𝑝𝑝(𝑑𝑑) = 𝛼𝛼𝑑𝑑 has a simple pole at 𝑑𝑑 = 0. Solution has form

𝑤𝑤(𝑑𝑑) =𝑐𝑐𝑑𝑑𝛼𝛼

Behavior of 𝑤𝑤 at singular point 𝑑𝑑 = 0?

- 𝛼𝛼 = integer ≤ 0 → 𝑤𝑤(𝑑𝑑) is entire

- 𝛼𝛼 = integer > 0 → 𝑤𝑤(𝑑𝑑) has a pole of order 𝛼𝛼 at 𝑑𝑑 = 0

- 𝛼𝛼 = not integer → 𝑤𝑤(𝑑𝑑) has branch point at 𝑑𝑑 = 0

Complex plane gives insight into behavior near singularities

L EC T URE 2

2N D ORDER LINEAR IVPS

𝑑𝑑′′ + 𝑝𝑝(𝑑𝑑)𝑑𝑑′ + 𝑞𝑞(𝑑𝑑)𝑑𝑑 = ℎ(𝑑𝑑)

Homogeneous if ℎ(𝑑𝑑) = 0, nonhomogeneous otherwise. IVP requires two initial conditions (roughly, to fix two constants of integration)

𝑑𝑑(𝑑𝑑0) = 𝑑𝑑0, 𝑑𝑑′(𝑑𝑑0) = 𝑑𝑑0′

If 𝑝𝑝, 𝑞𝑞, ℎ are continuous in an open interval, then ∃ a unique solution to the IVP through the interval (proof Coddington)

EXAMPLE

(1 − 𝑑𝑑2)𝑑𝑑′′ + (1 + 𝑑𝑑)𝑑𝑑 = 1 − 𝑑𝑑, 𝑑𝑑(0) = 1, 𝑑𝑑′(0) = 1

Where can we guarantee existence and uniqueness?


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𝑝𝑝 = 0, 𝑞𝑞 =1

1 − 𝑑𝑑, ℎ =

11 + 𝑑𝑑

So −1 < 𝑑𝑑 < 1 ⇒ existence and uniqueness guaranteed

HOMOGENEOUS 2ND ORDER LINEAR IVP

Given the ODE 𝐿𝐿𝑑𝑑 ≡ 𝑑𝑑′′ + 𝑝𝑝(𝑑𝑑)𝑑𝑑′ + 𝑞𝑞(𝑑𝑑)𝑑𝑑 = 0 with initial conditions: 𝑑𝑑(𝑑𝑑0) = 𝑑𝑑0,𝑑𝑑′(𝑑𝑑0) = 𝑑𝑑0′ .

SOLUTION PROPERTIES

Suppose 𝑝𝑝 and 𝑞𝑞 are continous on interval 𝐼𝐼 and 𝑑𝑑1 and 𝑑𝑑2 are solutions to 1.

SUPERPOSITION

𝑐𝑐1𝑑𝑑1 + 𝑐𝑐2𝑑𝑑2 is also a solution to 1 by linearity

SATISFIABILITY

Can 𝑐𝑐1𝑑𝑑1 + 𝑐𝑐2𝑑𝑑2 solve IVP? Need to find 𝑐𝑐1, 𝑐𝑐2 by finding solution of:

𝑐𝑐1𝑑𝑑1(𝑑𝑑0) + 𝑐𝑐2𝑑𝑑2(𝑑𝑑0) = 𝑑𝑑0

𝑐𝑐1𝑑𝑑1′ (𝑑𝑑0) + 𝑐𝑐2𝑑𝑑2

′ (𝑑𝑑0) = 𝑑𝑑0′

Solution exists if determinant of coefficients is nonzero.

wronskian ≡ 𝑊𝑊(𝑑𝑑0) = �𝑑𝑑1(𝑑𝑑0) 𝑑𝑑2(𝑑𝑑0)𝑑𝑑1′ (𝑑𝑑0) 𝑑𝑑2

′ (𝑑𝑑0)� = 𝑑𝑑1𝑑𝑑2′ − 𝑑𝑑1

′ 𝑑𝑑2

If 𝑊𝑊(𝑑𝑑0) ≠ 0, then ∃ 𝑐𝑐1 and 𝑐𝑐2 such that 𝑐𝑐1𝑑𝑑1 + 𝑐𝑐2𝑑𝑑2 satisfies the IVP.

GENERALITY

Does 𝑐𝑐1𝑑𝑑1 + 𝑐𝑐2𝑑𝑑2 describe all solutions to 1?

Consider another solution 𝜙𝜙 and set initial conditions

𝑑𝑑0 = 𝜙𝜙(𝑑𝑑0), 𝑑𝑑0′ = 𝜙𝜙′(𝑑𝑑0)

So 𝜙𝜙 solves this new IVP. But if 𝑊𝑊(𝑑𝑑0) ≠ 0 then

𝑑𝑑 = 𝑐𝑐1𝑑𝑑1 + 𝑐𝑐2𝑑𝑑2

Also solves this IVP so by uniqueness theorem

𝜙𝜙(𝑑𝑑) = 𝑐𝑐1𝑑𝑑1(𝑑𝑑) + 𝑐𝑐2𝑑𝑑2(𝑑𝑑)

Hence if 𝑊𝑊(𝑑𝑑0) ≠ 0 for some 𝑑𝑑0, then general solution to 1 has form

𝑑𝑑 = 𝑐𝑐1𝑑𝑑1(𝑑𝑑) + 𝑐𝑐2𝑑𝑑2(𝑑𝑑)

With 𝑐𝑐1 and 𝑐𝑐2 arbitrary. We say 𝑑𝑑1 and 𝑑𝑑2 form a fundamental set of solutions to 1.


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WRONSKIAN BEHAVIOR (ABEL’S THEOREM)

𝑑𝑑1 and 𝑑𝑑2 satisfy 1 so

−𝑑𝑑2(𝑑𝑑1′′ + 𝑝𝑝𝑑𝑑1

′ + 𝑞𝑞𝑑𝑑1) = 0

𝑑𝑑1(𝑑𝑑2′′ + 𝑝𝑝𝑑𝑑2

′ + 𝑞𝑞𝑑𝑑2) = 0

(𝑑𝑑1𝑑𝑑2′′ − 𝑑𝑑1

′′ 𝑑𝑑2) + 𝑝𝑝(𝑑𝑑1𝑑𝑑2′ − 𝑑𝑑1

′ 𝑑𝑑2) = 𝑊𝑊′(𝑑𝑑) + 𝑝𝑝𝑊𝑊(𝑑𝑑) = 0

So 𝑊𝑊′ + 𝑝𝑝𝑊𝑊 = 0 or 𝑊𝑊(𝑑𝑑) = 𝑐𝑐𝑒𝑒−∫𝑝𝑝(𝑑𝑑)𝑑𝑑𝑑𝑑 . Constant depends on choice of 𝑑𝑑1 and 𝑑𝑑2 but not on 𝑑𝑑. Hence 𝑊𝑊(𝑑𝑑) = 0 ∀ 𝑑𝑑 ∈ 𝐼𝐼 → 𝑐𝑐 = 0 or is never zero in 𝐼𝐼 → 𝑐𝑐 ≠ 0.

EXAMPLE

Consider 𝑥𝑥2𝑑𝑑′′ − 3𝑥𝑥𝑑𝑑′ + 3𝑑𝑑 = 0, 𝑥𝑥 > 0 and solutions 𝑑𝑑1 = 𝑥𝑥,𝑑𝑑2 = 𝑥𝑥3.

𝑊𝑊 = �𝑥𝑥 𝑥𝑥3

1 3𝑥𝑥2� = 2𝑥𝑥3

𝑊𝑊 ≠ 0 for 𝑥𝑥 > 0. So 𝑑𝑑 = 𝑐𝑐1𝑑𝑑1 + 𝑐𝑐2𝑑𝑑2 is general solution for 𝑥𝑥 > 0.

LINEAR INDEPENDENCE

𝑑𝑑1 and 𝑑𝑑2 are linearly independent on 𝐼𝐼 if ∃ 𝑐𝑐1, 𝑐𝑐2 not both zero such that

𝑐𝑐1𝑑𝑑1(𝑑𝑑) + 𝑐𝑐2𝑑𝑑2(𝑑𝑑) = 0 (3)

∀ 𝑑𝑑 ∈ 𝐼𝐼. If (3) requires 𝑐𝑐1 = 𝑐𝑐2 = 0, then 𝑑𝑑1 and 𝑑𝑑2 are linearly independent on 𝐼𝐼. Can show: 𝑑𝑑1 and 𝑑𝑑2 are linearly independent on 𝐼𝐼 iff 𝑊𝑊(𝑑𝑑) ≠ 0 ∀ 𝑑𝑑 ∈ 𝐼𝐼. (proof Coddington)

SUMMARY

To find general solution to homogeneous 2nd order linear ODE, find two linearly independent solutions with 𝑊𝑊(𝑑𝑑) ≠ 0 in 𝐼𝐼.

L EC T URE 3

HOMOGENEOUS 2ND ORDER LINEAR ODES WITH CONSTANT COEFFICIENTS

𝐿𝐿𝑑𝑑 ≡ 𝑎𝑎𝑑𝑑′′ + 𝑏𝑏𝑑𝑑′ + 𝑐𝑐𝑑𝑑 = 0 (1)

Attempt solutions of form 𝑑𝑑 = 𝑒𝑒𝑟𝑟𝑥𝑥 . 𝐿𝐿𝑒𝑒𝑟𝑟𝑥𝑥 = (𝑎𝑎𝑟𝑟2 + 𝑏𝑏𝑟𝑟 + 𝑐𝑐)𝑒𝑒𝑟𝑟𝑥𝑥 = 0. We see that (1) is satisfied for 𝑟𝑟 = −𝑏𝑏±�𝑏𝑏2−4𝑎𝑎𝑐𝑐2𝑎𝑎

.

CASE 1 – REAL DISTINCT ROOTS (𝑏𝑏2 − 4𝑎𝑎𝑐𝑐 > 0)

𝑑𝑑1 = 𝑒𝑒𝑟𝑟1𝑥𝑥 , 𝑑𝑑2 = 𝑒𝑒𝑟𝑟2𝑥𝑥 , 𝑟𝑟1 ≠ 𝑟𝑟2


9

𝑊𝑊 = � 𝑒𝑒𝑟𝑟1𝑥𝑥 𝑒𝑒𝑟𝑟2𝑥𝑥

𝑟𝑟1𝑒𝑒𝑟𝑟1𝑥𝑥 𝑟𝑟2𝑒𝑒𝑟𝑟2𝑥𝑥� = (𝑟𝑟2 − 𝑟𝑟1)𝑒𝑒(𝑟𝑟1+𝑟𝑟2)𝑥𝑥 ≠ 0

General Solution: 𝑑𝑑 = 𝑐𝑐1𝑒𝑒𝑟𝑟1𝑥𝑥 + 𝑐𝑐2𝑒𝑒𝑟𝑟2𝑥𝑥

CASE 2 – COMPLEX CONJUGATE ROOTS (𝑏𝑏2 − 4𝑎𝑎𝑐𝑐 < 0)

𝑟𝑟1, 𝑟𝑟2 = 𝛼𝛼 ± 𝑖𝑖β, β ≠ 0, 𝑑𝑑1,𝑑𝑑2 = 𝑒𝑒(𝛼𝛼±𝑖𝑖𝑖𝑖 )𝑥𝑥

Use superposition to obtain fundamental solution set

𝑢𝑢 ≡𝑑𝑑1 + 𝑑𝑑2

2= 𝑒𝑒𝛼𝛼𝑥𝑥 cos𝑖𝑖𝑥𝑥 , 𝑣𝑣 =

𝑑𝑑1 − 𝑑𝑑2

2= 𝑒𝑒𝛼𝛼𝑥𝑥 sin𝑖𝑖𝑥𝑥

𝑊𝑊 = 𝑖𝑖𝑒𝑒2𝛼𝛼𝑥𝑥 ≠ 0, 𝑖𝑖 ≠ 0

General Solution: 𝑑𝑑 = 𝑐𝑐1𝑒𝑒𝛼𝛼𝑥𝑥 cos𝑖𝑖𝑥𝑥 + 𝑐𝑐2𝑒𝑒𝛼𝛼𝑥𝑥 sin𝑖𝑖𝑥𝑥

CASE 3 – REAL REPEATED ROOT (𝑏𝑏2 − 4𝑎𝑎𝑐𝑐 = 0)

𝑟𝑟1 = 𝑟𝑟2, 𝐿𝐿𝑒𝑒𝑟𝑟𝑥𝑥 = (𝑟𝑟 − 𝑟𝑟1)2𝑒𝑒𝑟𝑟𝑥𝑥

So 𝐿𝐿𝑒𝑒𝑟𝑟𝑥𝑥 �|𝑟𝑟=𝑟𝑟1 = 0 and also

�𝜕𝜕𝜕𝜕𝑟𝑟

(𝐿𝐿𝑒𝑒𝑟𝑟𝑥𝑥 )�𝑟𝑟=𝑟𝑟1

= 0

Also, note

𝜕𝜕𝜕𝜕𝑟𝑟�𝜕𝜕𝑘𝑘𝑒𝑒𝑟𝑟𝑥𝑥

𝜕𝜕𝑥𝑥𝑘𝑘� =

𝜕𝜕𝑘𝑘

𝜕𝜕𝑥𝑥𝑘𝑘�𝜕𝜕𝑒𝑒𝑟𝑟𝑥𝑥

𝜕𝜕𝑟𝑟�

So

𝜕𝜕𝜕𝜕𝑟𝑟

(𝐿𝐿𝑒𝑒𝑟𝑟𝑥𝑥 ) �|𝑟𝑟=𝑟𝑟1 = 𝐿𝐿 �𝜕𝜕𝜕𝜕𝑟𝑟𝑒𝑒𝑟𝑟𝑥𝑥 � �|𝑟𝑟=𝑟𝑟1 = 𝐿𝐿(𝑥𝑥𝑒𝑒𝑟𝑟𝑥𝑥 )�|𝑟𝑟=𝑟𝑟1 = 0

And so 𝑑𝑑1 = 𝑒𝑒𝑟𝑟1𝑥𝑥 , 𝑑𝑑2 = 𝑥𝑥𝑒𝑒𝑟𝑟1𝑥𝑥

𝑊𝑊 = � 𝑒𝑒𝑟𝑟1𝑥𝑥 𝑥𝑥𝑒𝑒𝑟𝑟1𝑥𝑥

𝑟𝑟1𝑒𝑒𝑟𝑟1𝑥𝑥 𝑟𝑟1𝑥𝑥𝑒𝑒𝑟𝑟1𝑥𝑥 + 𝑒𝑒𝑟𝑟1𝑥𝑥� = 𝑒𝑒2𝑟𝑟1𝑥𝑥 ≠ 0

General Solution: 𝑑𝑑 = 𝑐𝑐1𝑒𝑒𝑟𝑟1𝑥𝑥 + 𝑐𝑐2𝑥𝑥𝑒𝑒𝑟𝑟1𝑥𝑥

EXAMPLE

𝑑𝑑′′ − 8𝑑𝑑′ + 16𝑑𝑑 = 0, 𝑑𝑑(0) = 1, 𝑑𝑑′(0) = 5

(𝑟𝑟2 − 8𝑟𝑟 + 16)𝑒𝑒𝑟𝑟𝑥𝑥 = 0 → 𝑟𝑟1 = 𝑟𝑟2 = 4 → 𝑑𝑑1 = 𝑒𝑒4𝑥𝑥

Try 𝑑𝑑2 = 𝑣𝑣(𝑥𝑥)𝑒𝑒4𝑥𝑥 :

𝑣𝑣′′ + �𝑝𝑝 + 2𝑑𝑑1′

𝑑𝑑1� 𝑣𝑣′ = 0 → 𝑣𝑣′′ + (−8 + 8)𝑣𝑣′ = 0 → 𝑣𝑣′′ = 0 or 𝑣𝑣(𝑥𝑥) = 𝛼𝛼𝑥𝑥 + β

So 𝑑𝑑 = 𝑐𝑐1𝑒𝑒4𝑥𝑥 + 𝑐𝑐2𝑥𝑥𝑒𝑒4𝑥𝑥 . We plug in our initial conditions to get


10

𝑐𝑐1 + 𝑐𝑐2 ∗ 0 = 1

4𝑐𝑐1 + 𝑐𝑐2 = 5 ⇒ 𝑐𝑐1 = 1, 𝑐𝑐2 = 1

Thus our specific solution is 𝑑𝑑 = 𝑒𝑒4𝑥𝑥 + 𝑥𝑥𝑒𝑒4𝑥𝑥 .

NONHOMOGENEOUS 2N D ORDER LINEAR ODES

𝐿𝐿𝑑𝑑 ≡ 𝑑𝑑′′ + 𝑝𝑝(𝑥𝑥)𝑑𝑑′ + 𝑞𝑞(𝑥𝑥)𝑑𝑑 = ℎ(𝑥𝑥) (𝒊𝒊)

p,q,h continuous on interval 𝐼𝐼. Suppose 𝑑𝑑𝑝𝑝 is a solution to (𝒊𝒊) and 𝑑𝑑 is any other solution to (𝒊𝒊). Then 𝑙𝑙�𝑑𝑑 − 𝑑𝑑𝑝𝑝� = 0. So 𝑑𝑑 − 𝑑𝑑𝑝𝑝 = 𝑐𝑐1𝑑𝑑1(𝑥𝑥) + 𝑐𝑐2𝑑𝑑2(𝑥𝑥) with 𝑑𝑑1 and 𝑑𝑑2 linearly independent solutions to homogeneous problem

𝐿𝐿𝑑𝑑 ≡ 0 (𝒊𝒊𝒊𝒊)

Hence any solution to (𝒊𝒊) may be written (i.e., general solution to (𝒊𝒊𝒊𝒊)) 𝑑𝑑 = 𝑐𝑐1𝑑𝑑2 + 𝑐𝑐2𝑑𝑑2 + 𝑑𝑑𝑝𝑝 where 𝑑𝑑𝑐𝑐 = 𝑐𝑐1𝑑𝑑1 +𝑐𝑐2𝑑𝑑2 is the complementary solution and 𝑑𝑑𝑝𝑝 is any particular solution to (𝒊𝒊). How to find 𝑑𝑑𝑝𝑝?

EXAMPLE

𝑑𝑑′′ − 𝑑𝑑′ − 2𝑑𝑑 = 𝑒𝑒−𝑥𝑥 , 𝑑𝑑(0) = 2, 𝑑𝑑′(0) =23

(𝑟𝑟2 − 𝑑𝑑 − 2) = (𝑟𝑟 + 1)(𝑟𝑟 − 2)

So our two linearly independent solutions are 𝑑𝑑1 = 𝑒𝑒−𝑥𝑥 ,𝑑𝑑2 = 𝑒𝑒2𝑥𝑥 . Try a particular solution 𝑑𝑑𝑝𝑝 = 𝑢𝑢1𝑒𝑒−𝑥𝑥 + 𝑢𝑢2𝑒𝑒2𝑥𝑥 or

𝑢𝑢1′ 𝑒𝑒−𝑥𝑥 + 𝑢𝑢2

′ 𝑒𝑒2𝑥𝑥 = 0

−𝑢𝑢1′ 𝑒𝑒−𝑥𝑥 + 2𝑢𝑢2

′ 𝑒𝑒2𝑥𝑥 = 𝑒𝑒−𝑥𝑥

This implies 𝑢𝑢1′ = − 1

3 and 𝑢𝑢2

′ = 13𝑒𝑒−3𝑥𝑥 . So 𝑢𝑢1(𝑥𝑥) = − 𝑥𝑥

3,𝑢𝑢2(𝑥𝑥) = − 1

9𝑒𝑒−3𝑥𝑥 or 𝑑𝑑𝑝𝑝 = − 𝑥𝑥

3𝑒𝑒−𝑥𝑥 − 1

9𝑒𝑒−𝑥𝑥 . Note that

− 19𝑒𝑒−𝑥𝑥 is simply a multiple of 𝑑𝑑1 so our general solution is 𝑑𝑑 = 𝑐𝑐1𝑒𝑒−𝑥𝑥 + 𝑐𝑐2𝑒𝑒2𝑥𝑥 − 𝑥𝑥

3𝑒𝑒−𝑥𝑥 . Then set constants to

satisfy initial conditions 𝑐𝑐1 = 1, 𝑐𝑐2 = 1. We get 𝑑𝑑 = 𝑒𝑒−𝑥𝑥 + 𝑒𝑒2𝑥𝑥 − 𝑥𝑥3𝑒𝑒−𝑥𝑥 . The particular solution obtained by

variation of parameters can be written

𝑑𝑑𝑝𝑝(𝑥𝑥) = �𝑑𝑑1(𝑠𝑠)𝑑𝑑2(𝑥𝑥) − 𝑑𝑑1(𝑥𝑥)𝑑𝑑2(𝑠𝑠)

𝑊𝑊(𝑑𝑑1,𝑑𝑑2)(𝑠𝑠) ℎ(𝑠𝑠)𝑑𝑑𝑠𝑠𝑥𝑥

𝑥𝑥0

= �𝐺𝐺(𝑥𝑥, 𝑠𝑠)ℎ(𝑠𝑠)𝑑𝑑𝑠𝑠𝑥𝑥

𝑥𝑥0

Where 𝐺𝐺(𝑥𝑥, 𝑠𝑠) ≔ Green’s function, ℎ(𝑠𝑠) ≔ forcing term, with

𝐺𝐺(𝑥𝑥, 𝑠𝑠) =𝑑𝑑1(𝑠𝑠)𝑑𝑑2(𝑥𝑥) − 𝑑𝑑1(𝑥𝑥)𝑑𝑑2(𝑠𝑠)

𝑊𝑊(𝑑𝑑1,𝑑𝑑2)(𝑠𝑠)

What is the interpretation of the Green’s function?

L EC T URE 4


11

DIRAC DELTA FUNCTION

Consider the impulse

Define intuitively by

𝛿𝛿(𝑥𝑥 − 𝑥𝑥0) = �0, 𝑥𝑥 ≠ 𝑥𝑥0∞, 𝑥𝑥 = 𝑥𝑥0

� , � 𝛿𝛿(𝑥𝑥 − 𝑥𝑥0)𝑑𝑑𝑥𝑥∞

−∞

= 1

This generalized function has the sifting property

� 𝛿𝛿(𝑥𝑥 − 𝑥𝑥0)𝑓𝑓(𝑥𝑥)𝑑𝑑𝑥𝑥∞

−∞

= 𝑓𝑓(𝑥𝑥0)

Which sifts out the value of 𝑓𝑓(𝑥𝑥) at 𝑥𝑥 = 𝑥𝑥0.

HEAVISIDE STEP FUNCTION

Integrating the delta function yields the step functions

𝐻𝐻(𝑥𝑥 − 𝑥𝑥0) = �𝛿𝛿(𝜁𝜁 − 𝑥𝑥0)𝑑𝑑𝑥𝑥𝑥𝑥

−∞

= �1, 𝑥𝑥 > 𝑥𝑥00, 𝑥𝑥 < 𝑥𝑥0

�

So we have

𝑑𝑑𝐻𝐻(𝑥𝑥 − 𝑥𝑥0)𝑑𝑑𝑥𝑥

= 𝛿𝛿(𝑥𝑥 − 𝑥𝑥0)

GREEN’S FUNCTION FOR IVP

Consider the linear 2nd order nonhomogeneous IVP 𝐿𝐿𝑑𝑑 ≡ 𝑑𝑑′′ + 𝑝𝑝(𝑥𝑥)𝑑𝑑′ + 𝑞𝑞(𝑥𝑥)𝑑𝑑 = ℎ(𝑥𝑥) with 𝑝𝑝,𝑞𝑞, ℎ continuous and homogeneous initial conditions 𝑑𝑑(𝑥𝑥0) = 0, 𝑑𝑑′(𝑥𝑥0) = 0. We wish to find 𝐺𝐺(𝑥𝑥, 𝜁𝜁) such that

𝑑𝑑𝑝𝑝(𝑥𝑥) = �𝐺𝐺(𝑥𝑥, 𝜁𝜁)ℎ(𝜁𝜁)𝑑𝑑𝜁𝜁𝑥𝑥

𝑥𝑥0

(1)

Consider letting ℎ(𝜁𝜁) = 𝛿𝛿(𝜁𝜁 − 𝑥𝑥∗) so by (1)


12

𝑑𝑑𝑝𝑝(𝑥𝑥) = �𝐺𝐺(𝑥𝑥, 𝜁𝜁)𝛿𝛿(𝜁𝜁 − 𝑥𝑥∗)𝑑𝑑𝜁𝜁𝑥𝑥

𝑥𝑥0

And by the sifting property 𝑑𝑑𝑝𝑝(𝑥𝑥) = 𝐺𝐺(𝑥𝑥, 𝑥𝑥∗). So 𝐺𝐺(𝑥𝑥, 𝑥𝑥∗) is the response of the solution of 𝑥𝑥 due to an impulse at 𝑥𝑥∗. Hence Green’s function satisfies

𝐿𝐿[𝐺𝐺(𝑥𝑥, 𝜁𝜁)] = 𝛿𝛿(𝑥𝑥 − 𝜁𝜁) (2)

With homogeneous initial conditions

𝐺𝐺(𝑥𝑥, 𝜁𝜁) �|𝑥𝑥=𝑥𝑥0 = 0,𝜕𝜕𝐺𝐺𝜕𝜕𝑥𝑥

(𝑥𝑥, 𝜁𝜁) �|𝑥𝑥=𝑥𝑥0 = 0

So 𝐺𝐺(𝑥𝑥, 𝜁𝜁) = 0 for 𝑥𝑥 < 𝜁𝜁. What is the behavior of 𝐺𝐺(𝑥𝑥, 𝜁𝜁) at 𝑥𝑥 = 𝜁𝜁?

𝜕𝜕2𝐺𝐺𝜕𝜕𝑥𝑥2 + 𝑝𝑝(𝑥𝑥)

𝜕𝜕𝐺𝐺𝜕𝜕𝑥𝑥

+ 𝑞𝑞(𝑥𝑥)𝐺𝐺 = 𝛿𝛿(𝑥𝑥 − 𝜁𝜁)

Highest derivative balances 𝛿𝛿(𝑥𝑥 − 𝜁𝜁) so expect 𝜕𝜕𝐺𝐺𝜕𝜕𝑥𝑥

to jump because it represents an integral of an impulse, expect 𝐺𝐺 to be continuous because it represents an integral of a step. Integrate ODE from 𝑠𝑠 − 𝜖𝜖 to 𝑠𝑠 + 𝜖𝜖

�𝜕𝜕𝐺𝐺𝜕𝜕𝑥𝑥�𝑠𝑠−𝜖𝜖

𝑠𝑠+𝜖𝜖

+ � 𝑝𝑝(𝑥𝑥)𝜕𝜕𝐺𝐺𝜕𝜕𝑥𝑥

𝑑𝑑𝑥𝑥𝑠𝑠+𝜖𝜖

𝑠𝑠−𝜖𝜖

+ � 𝑞𝑞(𝑥𝑥)𝐺𝐺𝑑𝑑𝑥𝑥𝑠𝑠+𝜖𝜖

𝑠𝑠−𝜖𝜖

= 1

Both integrals approach zero as 𝜖𝜖 → 0. So

�𝜕𝜕𝐺𝐺𝜕𝜕𝑥𝑥�𝑠𝑠−

𝑠𝑠+

= 1, �𝐺𝐺|𝑠𝑠−𝑠𝑠+ = 0

Form of 𝐺𝐺(𝑥𝑥, 𝜁𝜁)? For 𝑥𝑥 < 𝜁𝜁,𝐺𝐺(𝑥𝑥, 𝜁𝜁) = 0. Now solve homogeneous IVP at 𝑥𝑥 = 𝑠𝑠+

𝑐𝑐1(𝜁𝜁)𝑑𝑑1(𝜁𝜁) + 𝑐𝑐2(𝜁𝜁)𝑑𝑑2(𝜁𝜁) = 0 ← function continuous

𝑐𝑐1(𝜁𝜁)𝑑𝑑1′ (𝜁𝜁) + 𝑐𝑐2(𝜁𝜁)𝑑𝑑2

′ (𝜁𝜁) = 1 ← derivative jumps

So

𝑐𝑐1(𝜁𝜁) = −𝑑𝑑2(𝜁𝜁)

𝑊𝑊(𝑑𝑑1,𝑑𝑑2)(𝜁𝜁) , 𝑐𝑐2(𝜁𝜁) =𝑑𝑑1(𝜁𝜁)

𝑊𝑊(𝑑𝑑1,𝑑𝑑2)(𝜁𝜁)

𝐺𝐺(𝑥𝑥, 𝜁𝜁) = �0, 𝑥𝑥 < 𝜁𝜁𝑑𝑑1(𝜁𝜁)𝑑𝑑2(𝑥𝑥) − 𝑑𝑑1(𝑥𝑥)𝑑𝑑2(𝜁𝜁)

𝑊𝑊(𝑑𝑑1,𝑑𝑑2)(𝜁𝜁) , 𝑥𝑥 > 𝜁𝜁�


𝑥𝑥0

+ � 𝐺𝐺(𝑥𝑥, 𝜁𝜁)ℎ(𝜁𝜁)𝑑𝑑𝜁𝜁∞

𝑥𝑥

The first integral is non-zero as 𝑥𝑥 > 𝜁𝜁 (𝑥𝑥 is upper limit) while the second integral is zero as 𝑥𝑥 < 𝜁𝜁 (𝑥𝑥 is lower limit).

𝑑𝑑𝑝𝑝(𝑥𝑥) = �𝑑𝑑1(𝜁𝜁)𝑑𝑑2(𝑥𝑥) − 𝑑𝑑1(𝑥𝑥)𝑑𝑑2(𝜁𝜁)

𝑊𝑊(𝑑𝑑1,𝑑𝑑2)(𝜁𝜁) ℎ(𝜁𝜁)𝑑𝑑𝜁𝜁𝑥𝑥

𝑥𝑥0


13

(We recover variation of parameters form)

EXAMPLE (REVISITED)

𝑑𝑑′′ − 𝑑𝑑′ − 2𝑑𝑑 = 𝑒𝑒−𝑥𝑥 , 𝑑𝑑1 = 𝑒𝑒−𝑥𝑥 , 𝑑𝑑2 = 𝑒𝑒2𝑥𝑥 , 𝑑𝑑0 = 0

So

𝜕𝜕2𝐺𝐺𝜕𝜕𝑥𝑥2 −

𝜕𝜕𝐺𝐺𝜕𝜕𝑥𝑥

− 2𝐺𝐺 = 𝛿𝛿(𝑥𝑥 − 𝜁𝜁)

𝐺𝐺(0, 𝜁𝜁) = 0, �𝜕𝜕𝐺𝐺𝜕𝜕𝑥𝑥�𝑥𝑥=0

= 0 ← definition of Green's function

At 𝑥𝑥 = 𝑠𝑠

�𝐺𝐺|𝑠𝑠−𝑠𝑠+ = 0, �𝜕𝜕𝐺𝐺

𝜕𝜕𝑥𝑥�𝑠𝑠−

𝑠𝑠+

= 1

So 𝐺𝐺 = 𝑐𝑐1(𝜁𝜁)𝑒𝑒−𝑥𝑥 + 𝑐𝑐2(𝜁𝜁)𝑒𝑒2𝑥𝑥 with 𝑐𝑐1(𝜁𝜁)𝑒𝑒−𝜁𝜁 + 𝑐𝑐2(𝜁𝜁)𝑒𝑒2𝜁𝜁 = 0 and −𝑐𝑐1(𝜁𝜁)𝑒𝑒−𝜁𝜁 + 2𝑐𝑐2(𝜁𝜁)𝑒𝑒2𝜁𝜁 = 1. So 𝑐𝑐2(𝜁𝜁) = 𝑒𝑒−2𝜁𝜁

3 and

𝑐𝑐1(𝜁𝜁) = − 𝑒𝑒𝜁𝜁

3.

𝐺𝐺(𝑥𝑥, 𝜁𝜁) = �0, 𝑥𝑥 < 𝜁𝜁

−13𝑒𝑒𝜁𝜁−𝑥𝑥 +

13𝑒𝑒2𝑥𝑥−2𝜁𝜁 , 𝑥𝑥 > 𝜁𝜁

�


𝑥𝑥0

+ �� 𝐺𝐺(𝑥𝑥, 𝜁𝜁)ℎ(𝜁𝜁)𝑑𝑑𝜁𝜁𝑑𝑑𝑥𝑥∞

𝑥𝑥

= 0�

= −𝑥𝑥3𝑒𝑒−𝑥𝑥 + (𝑐𝑐1𝑑𝑑1 + 𝑐𝑐2𝑑𝑑2 = 0 ← for convenience)

L EC T URE 5

IVPS WITH CONSTANT COEFFICIENTS

EXAMPLE

𝑑𝑑′′ − 𝑑𝑑′ − 6𝑑𝑑 = 0, 𝑑𝑑(0) = 2, 𝑑𝑑′(0) = 1

Suppose 𝑑𝑑(𝑑𝑑) exists satisfying (1c) and (2c) from handout ℒ{𝑑𝑑′′ } − ℒ{𝑑𝑑′} − 6ℒ{𝑑𝑑} = 0. So

𝑠𝑠2ℒ{𝑑𝑑} − 𝑠𝑠𝑑𝑑(0) − 𝑑𝑑′(0) − [𝑠𝑠ℒ{𝑑𝑑} − 𝑑𝑑(0)] − 6ℒ{𝑑𝑑} = 0 → (𝑠𝑠2 − 𝑠𝑠 − 6)ℒ{𝑑𝑑} + (1 − 𝑠𝑠)𝑑𝑑(0) − 𝑑𝑑′(0) = 0

Then

ℒ{𝑑𝑑} =2𝑠𝑠 − 1

𝑠𝑠2 − 𝑠𝑠 − 6≡ 𝑑𝑑(𝑠𝑠) (i)


14

It turns differential equation into algebraic expression. To solve IVP, need to find 𝑑𝑑 whose transform is (i). By partial fractions

𝑌𝑌(𝑠𝑠) = ℒ{𝑑𝑑′(𝑑𝑑)} =1

𝑠𝑠 − 3+

1𝑠𝑠 + 2

And recall from handout

ℒ{𝑒𝑒𝑎𝑎𝑑𝑑 } =1

𝑠𝑠 − 𝑎𝑎, 𝑠𝑠 > 𝑎𝑎

So by linearity 𝑑𝑑 = 𝑒𝑒3𝑑𝑑 + 𝑒𝑒−2𝑑𝑑 . Note 𝑑𝑑(𝑑𝑑) satisfies (1c) and (2c)

SOLUTION STRATEGY

t-space

𝑑𝑑′′ − 𝑑𝑑 = 𝑑𝑑, 𝑑𝑑(0) = 1, 𝑑𝑑′(0) = 1

t-spaceℒ→s-space (apply laplace transform)

Subsidiary equation: (𝑠𝑠2 − 1)𝑌𝑌 − 𝑠𝑠 − 1 = 1𝑠𝑠2

Some algebra: 𝑌𝑌 = 1𝑠𝑠−1

+ 1𝑠𝑠2−1

− 1𝑠𝑠2

s-spaceℒ−1

�⎯� 𝑑𝑑-space (apply inverse laplace transform)

Solution: 𝑑𝑑(𝑑𝑑) = 𝑒𝑒𝑑𝑑 + sinh 𝑑𝑑 − 𝑑𝑑

NOTE

a.) Initial conditions incorporated automatically during solution procedure

b.) Complementary and particular solutions found simultaneously

To solve IVP, need to invert the Laplace transform: find 𝑑𝑑(𝑑𝑑) that satisfies the integral equation

� 𝑒𝑒−𝑠𝑠𝑑𝑑𝑑𝑑(𝑑𝑑)𝑑𝑑𝑑𝑑∞

0

= 𝑌𝑌(𝑠𝑠)

Denoted

𝑑𝑑(𝑑𝑑) = ℒ−1{𝑌𝑌(𝑠𝑠)}

UNIQUENESS

Two functions with the same transform cannot differ over any interval of positive length (Lerch’s Theorem). Hence, if 𝑑𝑑(𝑑𝑑) is a continuous function with transform 𝑌𝑌(𝑠𝑠), then it is the only possible one. We can build lookup tables of 𝑓𝑓(𝑑𝑑) ⟷ 𝐹𝐹(𝑠𝑠) invert simple transforms.


15

CONVOLUTION INTEGRALS

Suppose 𝑓𝑓(𝑑𝑑),𝑔𝑔(𝑑𝑑), ℎ(𝑑𝑑) have transforms 𝐹𝐹(𝑠𝑠),𝐺𝐺(𝑠𝑠),𝐻𝐻(𝑠𝑠) for 𝑠𝑠 > 𝑎𝑎 ≥ 0. If 𝐻𝐻(𝑠𝑠) = 𝐹𝐹(𝑠𝑠)𝐺𝐺(𝑠𝑠), then ℎ(𝑑𝑑) is the convolution of 𝑓𝑓(𝑑𝑑) and 𝑔𝑔(𝑑𝑑).

ℎ(𝑑𝑑) = �𝑓𝑓(𝑑𝑑 − 𝜏𝜏)𝑔𝑔(𝜏𝜏)𝑑𝑑𝜏𝜏𝑑𝑑

0

= �𝑓𝑓(𝜏𝜏)𝑔𝑔(𝑑𝑑 − 𝜏𝜏)𝑑𝑑𝜏𝜏𝑑𝑑

0

Or ℎ = 𝑓𝑓 ∗ 𝑔𝑔 = 𝑔𝑔 ∗ 𝑓𝑓. So if 𝐻𝐻(𝑠𝑠) can be expressed as a product, then we can write ℎ(𝑑𝑑) as a convolution integral.

PROOF SKETCH

𝐻𝐻(𝑠𝑠) = 𝐹𝐹(𝑠𝑠)𝐺𝐺(𝑠𝑠) = � 𝑒𝑒−𝑠𝑠𝜁𝜁𝑓𝑓(𝜁𝜁)𝑑𝑑𝜁𝜁∞

0

� 𝑒𝑒−𝑠𝑠𝜏𝜏𝑔𝑔(𝜏𝜏)𝑑𝑑𝜏𝜏∞

0

= � 𝑔𝑔(𝜏𝜏) �� 𝑒𝑒−𝑠𝑠(𝜁𝜁+𝜏𝜏)𝑓𝑓(𝜁𝜁)𝑑𝑑𝜁𝜁∞

0

� 𝑑𝑑𝜏𝜏∞

0

Let 𝜁𝜁 = 𝑑𝑑 − 𝜏𝜏 so 𝑑𝑑𝜁𝜁 = 𝑑𝑑𝑑𝑑

= � 𝑔𝑔(𝜏𝜏) �� 𝑒𝑒−𝑠𝑠𝑑𝑑𝑓𝑓(𝑑𝑑 − 𝜏𝜏)𝑑𝑑𝜏𝜏∞

𝜏𝜏

� 𝑑𝑑𝜏𝜏∞

0

Integrate over triangle, so by reversing the order

= � 𝑒𝑒−𝑠𝑠𝑑𝑑∞

0

��𝑓𝑓(𝑑𝑑 − 𝜏𝜏)𝑔𝑔(𝜏𝜏)𝑑𝑑𝜏𝜏𝑑𝑑

0

� 𝑑𝑑𝑑𝑑

The portion in the brackets = ℎ(𝑑𝑑) by definition of Laplace Transform

= � 𝑒𝑒−𝑠𝑠𝑑𝑑ℎ(𝑑𝑑)𝑑𝑑𝑑𝑑∞

0


16

So

ℎ(𝑑𝑑) = �𝑓𝑓(𝑑𝑑 − 𝜏𝜏)𝑔𝑔(𝜏𝜏)𝑑𝑑𝜏𝜏𝑑𝑑

0

EXAMPLE

𝑌𝑌(𝑠𝑠) =𝑎𝑎𝑠𝑠

1𝑠𝑠 − 𝑎𝑎

, 𝐹𝐹(𝑠𝑠) =𝑎𝑎𝑠𝑠

, 𝐺𝐺(𝑠𝑠) =1

𝑠𝑠 − 𝑎𝑎

Then 𝑓𝑓(𝑑𝑑) = 𝑎𝑎,𝑔𝑔(𝑑𝑑) = 𝑒𝑒𝑎𝑎𝑑𝑑

ℎ(𝑑𝑑) = 𝑓𝑓 ∗ 𝑔𝑔 = �𝑎𝑎𝑒𝑒𝑎𝑎𝜏𝜏𝑑𝑑𝜏𝜏𝑑𝑑

0

= 𝑒𝑒𝑎𝑎𝑑𝑑 − 1

L EC T URE 6

SHIFTING THEOREMS

S-SHIFTING

If ℒ{𝑓𝑓′(𝑑𝑑)} = 𝐹𝐹(𝑠𝑠), 𝑠𝑠 > 𝛾𝛾, i.e. |𝑓𝑓(𝑑𝑑)| ≤ 𝐾𝐾𝑒𝑒𝛾𝛾𝑑𝑑 , 𝑑𝑑 ≥ 𝑚𝑚, then ℒ{𝑒𝑒𝑎𝑎𝑑𝑑 𝑓𝑓(𝑑𝑑)} = 𝐹𝐹(𝑠𝑠 − 𝑎𝑎), 𝑠𝑠 − 𝑎𝑎 > 𝛾𝛾 or

ℒ−1{𝐹𝐹(𝑠𝑠 − 𝑎𝑎)} = 𝑒𝑒𝑎𝑎𝑑𝑑 𝑓𝑓(𝑑𝑑)

To see this, note

𝐹𝐹(𝑠𝑠 − 𝑎𝑎) = � 𝑒𝑒−(𝑠𝑠−𝑎𝑎)𝑑𝑑𝑓𝑓(𝑑𝑑)𝑑𝑑𝑑𝑑∞

0

= � 𝑒𝑒−𝑠𝑠𝑑𝑑 [𝑒𝑒𝑎𝑎𝑑𝑑 𝑓𝑓(𝑑𝑑)]𝑑𝑑𝑑𝑑∞

0

= ℒ{𝑒𝑒𝑎𝑎𝑑𝑑 𝑓𝑓(𝑑𝑑)}

EXAMPLE

Note

ℒ−1 �𝑠𝑠

𝑠𝑠2 + 𝑏𝑏2� = cos 𝑏𝑏𝑑𝑑 , 𝑠𝑠 > 0


17

So if

𝐹𝐹(𝑠𝑠) =𝑠𝑠 + 1

(𝑠𝑠 + 1)2 + 22 , 𝑠𝑠 + 1 > 0

Then 𝑓𝑓(𝑑𝑑) = 𝑒𝑒−𝑑𝑑 cos 2𝑑𝑑.

T-SHIFTING

If ℒ{𝑓𝑓(𝑑𝑑)} = 𝐹𝐹(𝑠𝑠), 𝑠𝑠 > 𝛾𝛾, then for 𝑎𝑎 > 0: ℒ{𝑓𝑓(𝑑𝑑 − 𝑎𝑎)𝐻𝐻(𝑑𝑑 − 𝑎𝑎)} = 𝑒𝑒−𝑎𝑎𝑠𝑠𝐹𝐹(𝑠𝑠), 𝑠𝑠 > 𝛾𝛾 or

ℒ−1{𝑒𝑒−𝑎𝑎𝑠𝑠𝐹𝐹(𝑠𝑠)} = 𝑓𝑓(𝑑𝑑 − 𝑎𝑎)𝐻𝐻(𝑑𝑑 − 𝑎𝑎)

To see this, take the Laplace transform:

𝑒𝑒−𝑎𝑎𝑠𝑠𝐹𝐹(𝑠𝑠) = 𝑒𝑒−𝑎𝑎𝑠𝑠 � 𝑒𝑒−𝑠𝑠𝜏𝜏𝑓𝑓(𝜏𝜏)𝑑𝑑𝜏𝜏∞

0

= � 𝑒𝑒−𝑠𝑠(𝜏𝜏+𝑎𝑎)𝑓𝑓(𝜏𝜏)𝑑𝑑𝜏𝜏∞

0

Let 𝑑𝑑 = 𝜏𝜏 + 𝑎𝑎, so

𝑒𝑒−𝑎𝑎𝑠𝑠𝐹𝐹(𝑠𝑠) = � 𝑒𝑒−𝑠𝑠𝑑𝑑𝑓𝑓(𝑑𝑑 − 𝑎𝑎)𝑑𝑑𝑑𝑑 ∞

0

= � 𝑒𝑒−𝑠𝑠𝑑𝑑𝑓𝑓(𝑑𝑑 − 𝑎𝑎)𝐻𝐻(𝑑𝑑 − 𝑎𝑎)𝑑𝑑𝑑𝑑 ∞

0

= ℒ{𝑓𝑓(𝑑𝑑 − 𝑎𝑎)𝐻𝐻(𝑑𝑑 − 𝑎𝑎)}

Note:

ℒ{𝐻𝐻(𝑑𝑑 − 𝑎𝑎)} =𝑒𝑒−𝑎𝑎𝑠𝑠

𝑠𝑠, 𝑠𝑠 > 0

Because

ℒ{1} =1𝑠𝑠

, 𝑠𝑠 > 0, ℒ{𝐻𝐻(𝑑𝑑)} =1𝑠𝑠

, 𝑠𝑠 > 0

EXAMPLE

𝑓𝑓(𝑑𝑑) = � 0, 0 < 𝑑𝑑 < 2𝜋𝜋cos 𝑑𝑑 , 𝑑𝑑 > 2𝜋𝜋

� , 𝐹𝐹(𝑠𝑠) =?


18

𝑓𝑓(𝑑𝑑) = 𝐻𝐻(𝑑𝑑 − 2𝜋𝜋) cos 𝑑𝑑 = 𝐻𝐻(𝑑𝑑 − 2𝜋𝜋) cos(𝑑𝑑 − 2𝜋𝜋)

So

ℒ{𝑓𝑓(𝑑𝑑)} = 𝑒𝑒−2𝜋𝜋𝑠𝑠 𝑠𝑠𝑠𝑠2 + 1

IVPS WITH DISCONTINUOUS FORCING

EXAMPLE

𝑑𝑑′′ + 3𝑑𝑑′ + 2𝑑𝑑 = 𝑟𝑟(𝑑𝑑), 𝑑𝑑(0) = 0, 𝑑𝑑′(0) = 0

So 𝑟𝑟(𝑑𝑑) = 𝐻𝐻(𝑑𝑑) − 𝐻𝐻(𝑑𝑑 − 1). Transforming the equation gives 𝑠𝑠2𝑌𝑌 + 3𝑠𝑠𝑌𝑌 + 2𝑌𝑌 = 1𝑠𝑠

(1 − 𝑒𝑒−𝑠𝑠). So

𝑌𝑌(𝑠𝑠) = 𝐹𝐹(𝑠𝑠)(1 − 𝑒𝑒−𝑠𝑠)

With 𝐹𝐹(𝑠𝑠) = 1𝑠𝑠(𝑠𝑠+1)(𝑠𝑠+2)

. Then

𝐹𝐹(𝑠𝑠) =𝑌𝑌2

𝑠𝑠−

1𝑠𝑠 + 1

+𝑌𝑌2

𝑠𝑠 + 2

So 𝑓𝑓(𝑑𝑑) = 12− 𝑒𝑒−𝑑𝑑 + 1

2𝑒𝑒−2𝑑𝑑 . Also ℒ−1{𝑒𝑒−𝑠𝑠𝐹𝐹(𝑠𝑠)} = 𝑓𝑓(𝑑𝑑 − 1)𝐻𝐻(𝑑𝑑 − 1). So 𝑑𝑑(𝑑𝑑) = 𝑓𝑓(𝑑𝑑) = 𝑓𝑓(𝑑𝑑 − 1)𝐻𝐻(𝑑𝑑 − 1).

Note: 𝑑𝑑 is continuous at 𝑑𝑑 = 1, 𝑑𝑑′ is continuous at 𝑑𝑑 = 1,

𝑑𝑑′′ = �−𝑒𝑒−𝑑𝑑 + 2𝑒𝑒−2𝑑𝑑 , 𝑑𝑑 ≤ 1

(𝑒𝑒 − 1)𝑒𝑒−𝑑𝑑 − 2(𝑒𝑒2 − 1)𝑒𝑒−2𝑑𝑑 , 𝑑𝑑 > 1�

So

lim𝑑𝑑→1−

𝑑𝑑′′ (𝑑𝑑) =2𝑒𝑒2 −

1𝑒𝑒

, lim𝑑𝑑→1+

2𝑒𝑒2 −

1𝑒𝑒− 1


19

Jump in highest derivative. [𝑑𝑑′′ ] = −1 at 𝑑𝑑 = 1 (brackets denote “jump notation”) to match jump in forcing term [𝑟𝑟] = −1 at 𝑑𝑑 = 1.

IVPS WITH IMPULSIVE FORCING

Let ℎ(𝑑𝑑) = 𝛿𝛿(𝑑𝑑 − 𝑎𝑎). Delta function fails to satisfy previous sufficient conditions for existence of the Laplace Transform (proceed formally).

ℒ{𝛿𝛿(𝑑𝑑 − 𝑎𝑎)} = � 𝑒𝑒−𝑠𝑠𝑑𝑑𝛿𝛿(𝑑𝑑 − 𝑎𝑎)𝑑𝑑𝑑𝑑∞

0

= 𝑒𝑒−𝑎𝑎𝑠𝑠

EXAMPLE

𝐿𝐿𝑑𝑑 = 𝑑𝑑′′ + 3𝑑𝑑′ + 2𝑑𝑑 = 𝛿𝛿(𝑑𝑑 − 𝑎𝑎), 𝑑𝑑(0) = 0, 𝑑𝑑′(0) = 0

𝑠𝑠2𝑌𝑌 + 3𝑠𝑠𝑌𝑌 + 2𝑌𝑌 = 𝑒𝑒−𝑎𝑎𝑠𝑠

So 𝑌𝑌(𝑠𝑠) = 𝐹𝐹(𝑠𝑠)𝑒𝑒−𝑎𝑎𝑠𝑠 with

𝐹𝐹(𝑠𝑠) =1

𝑠𝑠 + 1−

1𝑠𝑠 + 2

𝑓𝑓(𝑑𝑑) = ℒ−1{𝐹𝐹(𝑠𝑠)} = 𝑒𝑒−𝑑𝑑 − 𝑒𝑒−2𝑑𝑑

So

𝑑𝑑(𝑑𝑑) = 𝑓𝑓(𝑑𝑑 − 𝑎𝑎)𝐻𝐻(𝑑𝑑 − 𝑎𝑎) = �0, 0 ≤ 𝑑𝑑 < 𝑎𝑎𝑒𝑒−(𝑑𝑑−𝑎𝑎) − 𝑒𝑒−2(𝑑𝑑−𝑎𝑎), 𝑑𝑑 > 𝑎𝑎

�

We note that: 𝑑𝑑 continuous at 𝑑𝑑 = 𝑎𝑎, [𝑑𝑑′ ] = 1 at 𝑑𝑑 = 𝑎𝑎, and 𝑑𝑑′′ behaves like 𝛿𝛿(𝑑𝑑 − 𝑎𝑎) at 𝑑𝑑 = 𝑎𝑎. We recognize that 𝑑𝑑(𝑑𝑑) for ℎ(𝑑𝑑) = 𝛿𝛿(𝑑𝑑 − 𝑎𝑎) is the Green’s function 𝐺𝐺(𝑑𝑑, 𝑎𝑎) = 𝑓𝑓(𝑑𝑑 − 𝑎𝑎)𝐻𝐻(𝑑𝑑 − 𝑎𝑎). So for a different ℎ(𝑑𝑑), 𝐿𝐿𝑑𝑑 = ℎ(𝑑𝑑), has particular solution

𝑑𝑑𝑝𝑝 = �𝐺𝐺(𝑑𝑑, 𝜁𝜁)ℎ(𝜁𝜁)𝑑𝑑𝜁𝜁 𝑑𝑑

0

L EC T URE 7


20

INVERTING LAPLACE TRANSFORMS

Start from Fourier Transform

𝑓𝑓(𝑑𝑑) = � 𝑓𝑓(𝑑𝑑)𝑒𝑒−𝑖𝑖𝑥𝑥𝑑𝑑 𝑑𝑑𝑥𝑥∞

−∞

And inverse Fourier Transform

𝑓𝑓(𝑥𝑥) =1

2𝜋𝜋� 𝑓𝑓(𝑑𝑑)𝑒𝑒𝑖𝑖𝑥𝑥𝑑𝑑 𝑑𝑑𝑑𝑑∞

−∞

Exists if 𝑓𝑓(𝑥𝑥) is absolutely integrable, i.e.

� |𝑓𝑓(𝑥𝑥)|𝑑𝑑𝑥𝑥∞

−∞

< ∞

And piecewise continuous (we’ll derive these later in the course). Consider a piecewise continuous 𝑓𝑓(𝑥𝑥) of exponential order (recall |𝑔𝑔(𝑑𝑑)| ≤ 𝐾𝐾𝑒𝑒𝑎𝑎𝑑𝑑 , 𝑑𝑑 → ∞): apply Fourier formulae to a modified function.

𝑓𝑓(𝑥𝑥) ∗ 𝑒𝑒−𝑐𝑐𝑥𝑥 ∗ 𝐻𝐻(𝑥𝑥)

- 𝑓𝑓(𝑥𝑥) could blow up as 𝑥𝑥 → +∞

- 𝑒𝑒−𝑐𝑐𝑥𝑥 prevents blow up as 𝑥𝑥 → +∞, causes blow up as 𝑥𝑥 → −∞; has a real constant 𝑐𝑐 > 𝑎𝑎

- 𝐻𝐻(𝑥𝑥) removes blow up as 𝑥𝑥 → −∞

Then

𝑓𝑓(𝑑𝑑) = � 𝑓𝑓(𝑥𝑥)𝑒𝑒−𝑐𝑐𝑥𝑥𝐻𝐻(𝑥𝑥)𝑒𝑒−𝑖𝑖𝑥𝑥𝑑𝑑 𝑑𝑑𝑥𝑥∞

−∞

= � 𝑓𝑓(𝑥𝑥)𝑒𝑒−(𝑐𝑐+𝑖𝑖𝑑𝑑)𝑥𝑥𝑑𝑑𝑥𝑥∞

0

With inverse transform

𝑓𝑓(𝑥𝑥)𝑒𝑒−𝑐𝑐𝑥𝑥𝐻𝐻(𝑥𝑥) =1

2𝜋𝜋� 𝑓𝑓(𝑑𝑑)𝑒𝑒(𝑐𝑐+𝑖𝑖𝑑𝑑)𝑥𝑥𝑑𝑑𝑑𝑑∞

−∞

Now let 𝑑𝑑 ≡ 𝑥𝑥, 𝑠𝑠 ≡ 𝑐𝑐 + 𝑖𝑖𝑑𝑑, and 𝐹𝐹(𝑠𝑠) ≡ 𝑓𝑓(𝑑𝑑)

𝐹𝐹(𝑠𝑠) = � 𝑓𝑓(𝑑𝑑)𝑒𝑒−𝑠𝑠𝑑𝑑𝑑𝑑𝑑𝑑∞

0

, ℜ(𝑠𝑠) = 𝑐𝑐 > 𝑎𝑎

We obtain the Laplace Transform. We can show 𝐹𝐹(𝑠𝑠) is analytic for ℜ(𝑠𝑠) > 𝑎𝑎 (Weinberger, Dover 1965). Analytically continue for ℜ(𝑠𝑠) ≤ 𝑎𝑎, changing variables

𝑑𝑑𝑑𝑑 =𝑑𝑑𝑠𝑠𝑖𝑖

, −∞ < 𝑑𝑑 < ∞ ⇒𝑐𝑐 − 𝑖𝑖∞𝑠𝑠↔𝑐𝑐 + 𝑖𝑖∞


21

𝑓𝑓(𝑑𝑑)𝐻𝐻(𝑑𝑑) =1

2𝜋𝜋𝑖𝑖� 𝐹𝐹(𝑠𝑠)𝑒𝑒𝑠𝑠𝑑𝑑𝑑𝑑𝑠𝑠𝑐𝑐+𝑖𝑖∞

𝑐𝑐−𝑖𝑖∞

Omit 𝐻𝐻(𝑑𝑑) in the future by remembering that 𝑓𝑓(𝑑𝑑) = 0 for 𝑑𝑑 < 0. This is the Mellin inversion formula

𝑓𝑓(𝑑𝑑) =1



The inverse Laplace transform.

If |𝑓𝑓(𝑑𝑑)| ≤ 𝐾𝐾𝑒𝑒𝑎𝑎𝑑𝑑 , 𝑑𝑑 → ∞, then 𝑐𝑐 > 𝑎𝑎 is to the right of all singularities in 𝐹𝐹(𝑠𝑠) since 𝐹𝐹(𝑠𝑠) is analytic for ℜ(𝑠𝑠) > 𝑎𝑎.

EVALUATING BROMWICH CONTOUR INTEGRALS

Recall a form of Jordan’s Lemma:

If 𝑓𝑓(𝑑𝑑) → 0 uniformly on a circular arc 𝐶𝐶𝑅𝑅 as 𝑅𝑅 → ∞. (|𝑓𝑓(𝑑𝑑)| ≤ 𝐾𝐾𝑅𝑅 ,𝐾𝐾𝑅𝑅 → 0 𝑎𝑎𝑠𝑠 𝑅𝑅 → ∞).

lim𝑅𝑅→∞

�𝑒𝑒𝑚𝑚𝑑𝑑 𝑓𝑓(𝑑𝑑)𝑑𝑑𝑑𝑑𝐶𝐶𝑅𝑅

= 0

Using

Suppose 𝐹𝐹(𝑠𝑠) → 0 uniformly as |𝑠𝑠| → ∞ and is analytic except for 𝑛𝑛 isolated singularities at 𝑆𝑆 = 𝑆𝑆𝑘𝑘 with ℜ(𝑆𝑆𝑘𝑘) < 𝑐𝑐, 𝑘𝑘 = 1,⋯𝑛𝑛


22

Let 𝐶𝐶 = 𝐵𝐵𝑅𝑅 + 𝐶𝐶1 + 𝐶𝐶𝑅𝑅 + 𝐶𝐶2. By the residue theorem for 𝑆𝑆𝑘𝑘 inside 𝐶𝐶, 𝑘𝑘 = 1,⋯𝑛𝑛

�𝑒𝑒𝑠𝑠𝑑𝑑𝐹𝐹(𝑠𝑠)𝑑𝑑𝑠𝑠𝐶𝐶

= 2𝜋𝜋𝑖𝑖�𝑅𝑅𝑒𝑒𝑠𝑠{𝑒𝑒𝑠𝑠𝑑𝑑𝐹𝐹(𝑠𝑠); 𝑆𝑆𝑘𝑘}𝑛𝑛

𝑘𝑘=1

If

� =𝐶𝐶𝑅𝑅

� =𝐶𝐶1

� →𝐶𝐶2

0, 𝑅𝑅 → ∞

∮ 𝑒𝑒𝑠𝑠𝑑𝑑𝐹𝐹(𝑠𝑠)𝑑𝑑𝑠𝑠𝐶𝐶𝑅𝑅→ 0 as 𝑅𝑅 → ∞ by Jordan’s Lemma

On 𝐶𝐶1

|𝑒𝑒𝑠𝑠𝑑𝑑 | = �𝑒𝑒𝑑𝑑(ℜ(𝑠𝑠)+𝑖𝑖𝑅𝑅)� = 𝑒𝑒𝑑𝑑ℜ(𝑠𝑠) ≤ 𝑒𝑒𝑑𝑑𝑐𝑐

And |𝐹𝐹(𝑠𝑠)| ≤ 𝐾𝐾𝑅𝑅 → 0 as 𝑅𝑅 → ∞ with 𝐿𝐿 = 𝑐𝑐 so by 𝑀𝑀𝐿𝐿-bound

�� 𝑒𝑒𝑠𝑠𝑑𝑑𝐹𝐹(𝑠𝑠)𝑑𝑑𝑠𝑠𝐶𝐶1

� ≤ 𝑐𝑐𝑒𝑒𝑑𝑑𝑐𝑐𝐾𝐾𝑅𝑅 → 0, 𝑅𝑅 → ∞

Likewise for 𝐶𝐶2.

This leaves

lim𝑅𝑅→∞

� 𝐹𝐹(𝑠𝑠)𝑒𝑒𝑠𝑠𝑑𝑑𝑑𝑑𝑠𝑠𝑐𝑐+𝑖𝑖𝑅𝑅

𝑐𝑐−𝑖𝑖𝑅𝑅

= 2𝜋𝜋𝑖𝑖�𝑅𝑅𝑒𝑒𝑠𝑠{𝑒𝑒𝑠𝑠𝑑𝑑𝐹𝐹(𝑠𝑠); 𝑆𝑆𝑘𝑘}𝑛𝑛

𝑘𝑘=1

𝑓𝑓(𝑑𝑑 > 0) = ℒ−1{𝐹𝐹(𝑠𝑠)} =1



= �𝑅𝑅𝑒𝑒𝑠𝑠{𝑒𝑒𝑠𝑠𝑑𝑑𝐹𝐹(𝑠𝑠);𝑆𝑆𝑘𝑘}𝑛𝑛

𝑘𝑘=1

EXAMPLE

𝐹𝐹(𝑠𝑠) =1

𝑠𝑠2 + 𝑏𝑏2

So |𝐹𝐹(𝑠𝑠)| ≤ 1|𝑠𝑠|2−𝑏𝑏2 ≡ 𝐾𝐾|𝑠𝑠| → 0 as |𝑠𝑠| → ∞. Simple poles at 𝑆𝑆1 = 𝑖𝑖𝑏𝑏, 𝑆𝑆2 = −𝑖𝑖𝑏𝑏.

𝑅𝑅𝑒𝑒𝑠𝑠 �𝑒𝑒𝑠𝑠𝑑𝑑

𝑠𝑠2 + 𝑏𝑏2 ; 𝑆𝑆𝑘𝑘� = �𝑒𝑒𝑠𝑠𝑑𝑑

2𝑠𝑠�𝑠𝑠=±𝑖𝑖𝑏𝑏

=𝑒𝑒±𝑖𝑖𝑏𝑏

±2𝑖𝑖𝑏𝑏

So

𝑓𝑓(𝑑𝑑) =𝑒𝑒𝑖𝑖𝑏𝑏𝑑𝑑

2𝑖𝑖𝑏𝑏−𝑒𝑒−𝑖𝑖𝑏𝑏𝑑𝑑

2𝑖𝑖𝑏𝑏=

sin 𝑏𝑏𝑑𝑑𝑏𝑏

, 𝑑𝑑 > 0


23

L EC T URE 8

NONLINEAR 1ST ORDER ODES

General Form


= 𝑓𝑓(𝑑𝑑,𝑑𝑑), 𝑑𝑑(𝑑𝑑0) = 𝑑𝑑0 (1)

Facts:

- No universal solution technique

- Exact solutions of some special cases

- Pursue approximate solutions using general numerical methods

- Analyze method performance

EXISTENCE AND UNIQUENESS

EXAMPLE

Consider 𝑑𝑑′ = 𝑑𝑑2, 𝑑𝑑(1) = −1. The solution is 𝑑𝑑(𝑑𝑑) = − 1𝑑𝑑. DNE at 𝑑𝑑 = 0 even though 𝑓𝑓(𝑑𝑑,𝑑𝑑) is well-behaved at

𝑑𝑑 = 0. Settle for local existence and uniqueness result near (𝑑𝑑0,𝑑𝑑0). Moral: simple inspection of ODE not sufficient to judge solution behavior.

PICARD’S THEOREM

Suppose that 𝑓𝑓(𝑑𝑑,𝑑𝑑) is a continuous function in the rectangle 𝑅𝑅 = {(𝑑𝑑,𝑑𝑑): |𝑑𝑑 − 𝑑𝑑0| ≤ 𝑎𝑎, |𝑑𝑑 − 𝑑𝑑0| ≤ 𝑏𝑏} (𝑎𝑎 > 0, 𝑏𝑏 > 0 constants) with |𝑓𝑓(𝑑𝑑,𝑑𝑑)| ≤ 𝑀𝑀 ∀ (𝑑𝑑,𝑑𝑑) ∈ 𝑅𝑅 and Lipschitz constant 𝐿𝐿 such that

|𝑓𝑓(𝑑𝑑,𝑑𝑑2) − 𝑓𝑓(𝑑𝑑,𝑑𝑑1)| ≤ 𝐿𝐿|𝑑𝑑2 − 𝑑𝑑1|

Whenever (𝑑𝑑,𝑑𝑑1) and (𝑑𝑑,𝑑𝑑2) lie in 𝑅𝑅. Then there exists a unique solution 𝑑𝑑(𝑑𝑑) satisfying the IVP (1) on the interval 𝐼𝐼 = {𝑑𝑑: |𝑑𝑑 − 𝑑𝑑0| ≤ min(𝑎𝑎, 𝑏𝑏

𝑀𝑀)}. (Proof Coddington). The proof relies on the Picard iteration

𝜙𝜙0(𝑑𝑑) = 𝑑𝑑0

𝜙𝜙𝑘𝑘+1(𝑑𝑑) = 𝑑𝑑0 + �𝑓𝑓(𝜁𝜁,𝜙𝜙𝑘𝑘(𝜁𝜁)𝑑𝑑𝜁𝜁𝑑𝑑

𝑑𝑑0

(2a)

(2b)

Which converges to the solution 𝑑𝑑(𝑑𝑑) in the interval 𝐼𝐼.

NUMERICAL METHODS FOR IVP


24

Consider


= 𝑓𝑓(𝑑𝑑,𝑑𝑑) (3)

𝑑𝑑(𝑑𝑑0) = 𝑑𝑑0 (4)

Solve on interval 𝑑𝑑0 ≤ 𝑑𝑑 ≤ 𝑑𝑑𝑓𝑓 . Discretize the interval into mesh points 𝑑𝑑 = 𝑑𝑑0 + 𝑛𝑛Δ𝑑𝑑, 𝑛𝑛 = 0,1,⋯ ,𝑁𝑁 with step size

Δ𝑑𝑑 = 𝑑𝑑𝑓𝑓−𝑑𝑑0

𝑁𝑁.

EULER’S METHOD

Discretize the derivative

𝑑𝑑𝑛𝑛+1 − 𝑑𝑑𝑛𝑛Δ𝑑𝑑

= 𝑓𝑓(𝑑𝑑𝑛𝑛 ,𝑑𝑑𝑛𝑛), 𝑛𝑛 = 0,1,⋯ ,𝑁𝑁 − 1

Or

𝑑𝑑𝑛𝑛+1 = 𝑑𝑑𝑛𝑛 + Δ𝑑𝑑 𝑓𝑓(𝑑𝑑𝑛𝑛𝑑𝑑𝑛𝑛) (5)

Method is explicit using only information that is already available: 𝑑𝑑𝑛𝑛 , 𝑓𝑓(𝑑𝑑𝑛𝑛 ,𝑑𝑑𝑛𝑛).

ALTERNATIVE DERIVATIONS

Integrate (3) between mesh points to obtain the integral equation:

𝑑𝑑(𝑑𝑑𝑛𝑛+1) = 𝑑𝑑(𝑑𝑑) + � 𝑓𝑓�𝑑𝑑,𝑑𝑑(𝑑𝑑)�𝑑𝑑𝑑𝑑

𝑑𝑑𝑛𝑛+1

𝑑𝑑𝑛𝑛

, 𝑛𝑛 = 0,1,⋯ ,𝑁𝑁 − 1 (6)

Use the numerical integration rule:

� 𝑓𝑓�𝑑𝑑,𝑑𝑑(𝑑𝑑)�𝑑𝑑𝑑𝑑

𝑑𝑑𝑛𝑛+1

𝑑𝑑𝑛𝑛

≅ Δ𝑑𝑑 𝑓𝑓�𝑑𝑑𝑛𝑛 ,𝑑𝑑(𝑑𝑑𝑛𝑛)�

To recover the explicit or forward Euler Method (5).

OTHER POSSIBILITIES

IMPLICIT OR BACKWARD EULER

𝑑𝑑(𝑑𝑑𝑛𝑛+1) = 𝑑𝑑(𝑑𝑑𝑛𝑛) + Δ𝑑𝑑 𝑓𝑓�𝑑𝑑𝑛𝑛+1,𝑑𝑑(𝑑𝑑𝑛𝑛+1)� (7)


25

TRAPEZOID RULE

𝑑𝑑(𝑑𝑑𝑛𝑛+1) = 𝑑𝑑(𝑑𝑑𝑛𝑛) +Δ𝑑𝑑2�𝑓𝑓�𝑑𝑑𝑛𝑛 ,𝑑𝑑(𝑑𝑑𝑛𝑛)� + 𝑓𝑓�𝑑𝑑𝑛𝑛+1,𝑑𝑑(𝑑𝑑𝑛𝑛+1)�� (8)

These methods are implicit: must solve implicit nonlinear equation to solve for 𝑑𝑑𝑛𝑛+1 given 𝑑𝑑𝑛𝑛 .

EXAMPLE


= −2 + 𝑑𝑑

𝑑𝑑2 + 3𝑑𝑑2 , 𝑑𝑑(0) = 1

Find solution 𝑑𝑑�𝑑𝑑𝑓𝑓� at 𝑑𝑑𝑓𝑓 = 10. Compare solution accuracy for:

a) Explicit Euler

b) Implicit Euler

c) Trapezoidal

On a sequence of meshes. The implicit methods were updated using a Picard iteration to solve the nonlinear equation within each time step.

L EC T URE 9

TRUNCATION ERROR

It is the degree to which the exact solution fails to satisfy the discretization. Consider the ODE 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

= 𝑓𝑓(𝑑𝑑,𝑑𝑑)

With discretization


= 𝑓𝑓(𝑑𝑑𝑛𝑛 ,𝑑𝑑𝑛𝑛) (Explicit Euler)

Substitute Taylor series for analytical solution into discrete operator to find the truncation error:


26

𝑇𝑇𝑛𝑛 =�𝑑𝑑(𝑑𝑑𝑛𝑛) + Δ𝑑𝑑𝑑𝑑′(𝑑𝑑𝑛𝑛) + Δ𝑑𝑑2

2 𝑑𝑑′′ (𝑑𝑑𝑛𝑛) + 𝑂𝑂(Δ𝑑𝑑3)� − 𝑑𝑑(𝑑𝑑𝑛𝑛)

Δ𝑑𝑑− 𝑓𝑓�𝑑𝑑𝑛𝑛 ,𝑑𝑑(𝑑𝑑𝑛𝑛)�

So

𝑇𝑇𝑛𝑛 = �𝑑𝑑′(𝑑𝑑𝑛𝑛) − 𝑓𝑓�𝑑𝑑𝑛𝑛 ,𝑑𝑑(𝑑𝑑𝑛𝑛)�� +Δ𝑑𝑑2𝑑𝑑′′ (𝑑𝑑𝑛𝑛) + 𝑂𝑂(Δ𝑑𝑑2)

Or by Taylor’s Theorem

𝑇𝑇𝑛𝑛 =12Δ𝑑𝑑𝑑𝑑′′ (𝜁𝜁𝑛𝑛), 𝜁𝜁𝑛𝑛 ∈ (𝑑𝑑𝑛𝑛 , 𝑑𝑑𝑛𝑛+1)

Assuming 𝑓𝑓 is sufficiently smooth to assume 𝑑𝑑′′ exists and is bounded on [𝑑𝑑𝑛𝑛 , 𝑑𝑑𝑛𝑛+1]

EXPLICIT EULER

𝑇𝑇𝑛𝑛 =12Δ𝑑𝑑𝑑𝑑′′ (𝜁𝜁) = 𝑂𝑂(Δ𝑑𝑑)

IMPLICIT EULER


= 𝑓𝑓(𝑑𝑑𝑛𝑛+1,𝑑𝑑𝑛𝑛+1)

𝑇𝑇𝑛𝑛 =𝑑𝑑(𝑑𝑑𝑛𝑛+1) − �𝑑𝑑(𝑑𝑑𝑛𝑛+1) − Δ𝑑𝑑𝑑𝑑′(𝑑𝑑𝑛𝑛+1) + Δ𝑑𝑑2

2 𝑑𝑑′′ (𝑑𝑑𝑛𝑛+1) − 𝑂𝑂(Δ𝑑𝑑3)�

Δ𝑑𝑑− 𝑓𝑓(𝑑𝑑𝑛𝑛+1,𝑑𝑑𝑛𝑛+1) = −

12Δ𝑑𝑑𝑑𝑑′′ (𝜁𝜁𝑛𝑛)

= 𝑂𝑂(Δ𝑑𝑑)

TRAPEZOID RULE


=12

[𝑓𝑓(𝑑𝑑𝑛𝑛 ,𝑑𝑑𝑛𝑛) + 𝑓𝑓(𝑑𝑑𝑛𝑛+1,𝑑𝑑𝑛𝑛+1)]some math�⎯⎯⎯⎯⎯� 𝑇𝑇𝑛𝑛 = 𝑂𝑂(Δ𝑑𝑑2)

A method has order of accuracy 𝑝𝑝 if |𝑇𝑇𝑛𝑛 | = 𝑂𝑂(Δ𝑑𝑑𝑝𝑝), 0 < Δ𝑑𝑑 ≤ Δ𝑑𝑑0 for 𝑑𝑑(𝑑𝑑) sufficiently smooth (i.e. |𝑇𝑇𝑛𝑛 | ≤ 𝐾𝐾Δ𝑑𝑑𝑝𝑝 for some 𝐾𝐾). In our numerical experiment, we examined the solution error

𝑒𝑒𝑛𝑛 ≡ 𝑑𝑑(𝑑𝑑𝑛𝑛) − 𝑑𝑑𝑛𝑛

Where 𝑑𝑑(𝑑𝑑𝑛𝑛) was the exact solution and 𝑑𝑑𝑛𝑛 was the numerical (or approximate) solution. If

|𝑒𝑒𝑛𝑛 | = 𝐶𝐶1Δ𝑑𝑑𝑝𝑝 = 𝐶𝐶2𝑁𝑁−𝑝𝑝 , Δ𝑑𝑑 = time step, 𝑁𝑁 = # of steps

Then log|𝑒𝑒𝑛𝑛 | = log𝐶𝐶1 + 𝑝𝑝 logΔ𝑑𝑑 = log𝐶𝐶2 − 𝑝𝑝 log𝑁𝑁. Plot log(error) versus log(steps) and find slope of the line to estimate 𝑝𝑝.We observed:

• explicit Euler, 𝑝𝑝 ≅ 1

• implicit Euler, 𝑝𝑝 ≅ 1

• Trapezoid rule, 𝑝𝑝 ≅ 2


27

The order of accuracy of the truncation error matches the empirical behavior of the solution error in our example. Can we predict this? Consider the explicit Euler method

𝑑𝑑𝑛𝑛+1 = 𝑑𝑑𝑛𝑛 + Δ𝑑𝑑𝑓𝑓(𝑑𝑑𝑛𝑛 ,𝑑𝑑𝑛𝑛) (1)

The solution error is defined by 𝑒𝑒𝑛𝑛 = 𝑑𝑑(𝑑𝑑𝑛𝑛) − 𝑑𝑑𝑛𝑛 and the truncation error is 𝑇𝑇𝑛𝑛 = 𝑑𝑑(𝑑𝑑𝑛𝑛+1)−𝑑𝑑(𝑑𝑑𝑛𝑛 )Δ𝑑𝑑

− 𝑓𝑓�𝑑𝑑𝑛𝑛 ,𝑑𝑑(𝑑𝑑𝑛𝑛)� or rearranging

𝑑𝑑(𝑑𝑑𝑛𝑛+1) = 𝑑𝑑(𝑑𝑑𝑛𝑛) + Δ𝑑𝑑𝑓𝑓�𝑑𝑑𝑛𝑛 ,𝑑𝑑(𝑑𝑑𝑛𝑛)� + Δ𝑑𝑑𝑇𝑇𝑛𝑛 (2)

Subtracting (1) from (2): 𝑒𝑒𝑛𝑛+1 = 𝑒𝑒𝑛𝑛 + Δ𝑑𝑑�𝑓𝑓�𝑑𝑑𝑛𝑛 ,𝑑𝑑(𝑑𝑑𝑛𝑛)� − 𝑓𝑓(𝑑𝑑𝑛𝑛 ,𝑑𝑑𝑛𝑛)� + Δ𝑑𝑑𝑇𝑇𝑛𝑛 . By the Lipschitz condition in rectangle R: �𝑓𝑓�𝑑𝑑𝑛𝑛 ,𝑑𝑑(𝑑𝑑𝑛𝑛)� − 𝑓𝑓(𝑑𝑑𝑛𝑛 ,𝑑𝑑𝑛𝑛)� ≤ 𝐿𝐿|𝑑𝑑(𝑑𝑑𝑛𝑛) − 𝑑𝑑𝑛𝑛 | = 𝐿𝐿|𝑒𝑒𝑛𝑛 |. So |𝑒𝑒𝑛𝑛+1| ≤ (1 + Δ𝑑𝑑𝐿𝐿)|𝑒𝑒𝑛𝑛 | + Δ𝑑𝑑|𝑇𝑇𝑛𝑛 |, 𝑛𝑛 = 0,1,⋯𝑁𝑁 − 1. Let 𝑇𝑇 = max|𝑇𝑇𝑛𝑛 | so 0 ≤ 𝑛𝑛 ≤ 𝑁𝑁 − 1.

|𝑒𝑒𝑛𝑛+1| ≤ (1 + Δ𝑑𝑑𝐿𝐿)|𝑒𝑒𝑛𝑛 | + Δ𝑑𝑑𝑇𝑇

Then by induction

|𝑒𝑒1| ≤ (1 + Δ𝑑𝑑𝐿𝐿)|𝑒𝑒0| + Δ𝑑𝑑𝑇𝑇

|𝑒𝑒2| ≤ (1 + Δ𝑑𝑑𝐿𝐿)2|𝑒𝑒0| + Δ𝑑𝑑(1 + Δ𝑑𝑑𝐿𝐿)𝑇𝑇 + Δ𝑑𝑑𝑇𝑇

⋮

|𝑒𝑒𝑛𝑛 | ≤ (1 + Δ𝑑𝑑𝐿𝐿)𝑛𝑛 |𝑒𝑒0| + Δ𝑑𝑑𝑇𝑇[1 + (1 + Δ𝑑𝑑𝐿𝐿) + (1 + Δ𝑑𝑑𝐿𝐿)2 + ⋯+ (1 + Δ𝑑𝑑𝐿𝐿)𝑛𝑛−1]

Note:

𝜌𝜌𝑛𝑛 − 1𝜌𝜌 − 1

= 1 + 𝜌𝜌 + 𝜌𝜌2 + ⋯+ 𝜌𝜌𝑛𝑛−1

[verify by multiplying by (𝜌𝜌 − 1) and telescoping]. So

(1 + Δ𝑑𝑑𝐿𝐿)𝑛𝑛 − 1(1 + Δ𝑑𝑑𝐿𝐿) − 1

= 1 + (1 + Δ𝑑𝑑𝐿𝐿) + ⋯+ (1 + Δ𝑑𝑑𝐿𝐿)𝑛𝑛−1

And hence |𝑒𝑒𝑛𝑛 | ≤ (1 + Δ𝑑𝑑𝐿𝐿)𝑛𝑛 |𝑒𝑒0| + 𝑇𝑇𝐿𝐿

[(1 + Δ𝑑𝑑𝐿𝐿)𝑛𝑛 − 1]. Note (1 + Δ𝑑𝑑𝐿𝐿) ≤ 𝑒𝑒Δ𝑑𝑑𝐿𝐿 . So (𝑛𝑛Δ𝑑𝑑 ≡ 𝑑𝑑𝑛𝑛 − 𝑑𝑑0)

|𝑒𝑒𝑛𝑛 | ≤𝑇𝑇𝐿𝐿�𝑒𝑒𝐿𝐿(𝑑𝑑𝑛𝑛−𝑑𝑑0) − 1� + 𝑒𝑒𝐿𝐿(𝑑𝑑𝑛𝑛−𝑑𝑑0)|𝑒𝑒0| (3)

Hence the solution error should decrease at the same rate as the truncation error as we refine the mesh (which we observed in our numerical example).

L EC T URE 1 0

EXAMPLE (ERROR CONTROL)

Solve 𝑑𝑑′ = arctan𝑑𝑑 , 𝑑𝑑(0) = 𝑑𝑑0 using explicit Euler with |𝑒𝑒𝑛𝑛 | ≤ 𝜖𝜖. By our error bound


|𝑒𝑒𝑛𝑛 | ≤𝑇𝑇𝐿𝐿

(𝑒𝑒𝐿𝐿𝑑𝑑𝑛𝑛 − 1) + 𝑒𝑒𝐿𝐿𝑑𝑑𝑛𝑛 |𝑒𝑒0| (4)

Set initial condition correctly so 𝑒𝑒0 = 0. We need to find valid values of 𝑇𝑇, 𝐿𝐿. The truncation error is

𝑇𝑇𝑛𝑛 =12Δ𝑑𝑑𝑑𝑑′′ (𝜁𝜁𝑛𝑛)

So 𝑇𝑇 ≤ Δ𝑑𝑑2𝐾𝐾, 𝐾𝐾 ≥ max|𝑑𝑑′′ (𝑑𝑑)| , t ∈ [0, tn] and

𝑑𝑑′′ =𝑑𝑑𝑑𝑑𝑑𝑑

(arctan𝑑𝑑) =𝑑𝑑′

1 + 𝑑𝑑2 =arctan𝑑𝑑1 + 𝑑𝑑2 , |𝑑𝑑′′ | ≤ |arctan𝑑𝑑| �

11 + 𝑑𝑑2� ≤

𝜋𝜋2∗ 1

Thus 𝑇𝑇 ≤ 𝜋𝜋Δ𝑑𝑑4

. To find 𝐿𝐿, note

𝑓𝑓(𝑑𝑑,𝑑𝑑1) − 𝑓𝑓(𝑑𝑑,𝑑𝑑2) = �𝜕𝜕𝑓𝑓𝜕𝜕𝑑𝑑

(𝑑𝑑,𝑑𝑑)𝑑𝑑𝑑𝑑

𝑑𝑑1

𝑑𝑑2

For 𝜕𝜕𝑓𝑓𝜕𝜕𝑑𝑑

(𝑑𝑑,𝑑𝑑) continuous in rectangle 𝑅𝑅. Hence |𝑓𝑓(𝑑𝑑,𝑑𝑑1) − 𝑓𝑓(𝑑𝑑,𝑑𝑑2)| ≤ 𝐿𝐿|𝑑𝑑1 − 𝑑𝑑2| holds with

𝐿𝐿 = max �𝜕𝜕𝑓𝑓𝜕𝜕𝑑𝑑� , (𝑑𝑑,𝑑𝑑) ∈ 𝑅𝑅

In our case

�𝜕𝜕𝑓𝑓𝜕𝜕𝑑𝑑� = �

11 + 𝑑𝑑2� ≤ 1

So 𝐿𝐿 = 1 is a valid choice (maybe not the best, but it works). By (4) we then have

|𝑒𝑒𝑛𝑛 | ≤𝜋𝜋4

(𝑒𝑒𝑑𝑑𝑛𝑛 − 1)Δ𝑑𝑑, 𝑛𝑛 = 1,⋯ ,𝑁𝑁

We now require

|𝑒𝑒𝑛𝑛 | ≤𝜋𝜋4

(𝑒𝑒𝑑𝑑𝑛𝑛 − 1)Δ𝑑𝑑 ≤ 𝜖𝜖

Or

Δ𝑑𝑑 ≤4𝜖𝜖

𝜋𝜋(𝑒𝑒𝑑𝑑𝑛𝑛 − 1)

REMOVING IMPLICITNESS

Trapezoidal method gives 2nd order accuracy but it is implicit – requires a nonlinear iteration at each step.

𝑑𝑑𝑛𝑛+1 = 𝑑𝑑𝑛𝑛 +Δ𝑑𝑑2

[𝑓𝑓(𝑑𝑑𝑛𝑛 ,𝑑𝑑𝑛𝑛) + 𝑓𝑓(𝑑𝑑𝑛𝑛+1,𝑑𝑑𝑛𝑛+1)]

We can try approximating 𝑑𝑑𝑛𝑛+1 by forward Euler step 𝑑𝑑𝑛𝑛+1 ≅ 𝑑𝑑𝑛𝑛 + Δ𝑑𝑑𝑓𝑓(𝑑𝑑𝑛𝑛 ,𝑑𝑑𝑛𝑛) to obtain the Improved Euler Method.


29

𝑑𝑑𝑛𝑛+1 = 𝑑𝑑𝑛𝑛 +Δ𝑑𝑑2�𝑓𝑓(𝑑𝑑𝑛𝑛 ,𝑑𝑑𝑛𝑛) + 𝑓𝑓�𝑑𝑑𝑛𝑛+1,𝑑𝑑𝑛𝑛 + Δ𝑑𝑑𝑓𝑓(𝑑𝑑𝑛𝑛 ,𝑑𝑑𝑛𝑛)��

This is a second order explicit method – a special case of Runge-Kutta method.

EXPLICIT RUNGE-KUTTA METHODS

We average the slope at various points in the interval [𝑑𝑑𝑛𝑛 , 𝑑𝑑𝑛𝑛+1] thus effectively gathering information about neighboring integral curves.

1-STAGE SCHEME

𝑘𝑘1 = 𝑓𝑓(𝑑𝑑𝑛𝑛 ,𝑑𝑑𝑛𝑛)

𝑑𝑑𝑛𝑛+1 = 𝑑𝑑𝑛𝑛 + Δ𝑑𝑑𝑘𝑘1

𝑇𝑇 = 𝑂𝑂(Δ𝑑𝑑)

1st Order Explicit Euler is a unique choice of the 1-stage scheme.

2-STAGE SCHEME


𝑘𝑘2 = 𝑓𝑓(𝑑𝑑𝑛𝑛 + Δ𝑑𝑑,𝑑𝑑𝑛𝑛 + Δ𝑑𝑑𝑘𝑘1)

𝑑𝑑𝑛𝑛+1 = 𝑑𝑑𝑛𝑛 + Δ𝑑𝑑 �𝑘𝑘1 + 𝑘𝑘2

2�

𝑇𝑇 = 𝑂𝑂(Δ𝑑𝑑2)

2nd Order Improved Euler is one example of the 2-stage scheme.

4-STAGE SCHEME


𝑘𝑘2 = 𝑓𝑓 �𝑑𝑑𝑛𝑛 +Δ𝑑𝑑2

,𝑑𝑑𝑛𝑛 +Δ𝑑𝑑2𝑘𝑘1�

𝑘𝑘3 = 𝑓𝑓 �𝑑𝑑𝑛𝑛 +Δ𝑑𝑑2

,𝑑𝑑𝑛𝑛 +Δ𝑑𝑑2𝑘𝑘2�

𝑘𝑘4 = 𝑓𝑓(𝑑𝑑𝑛𝑛 + Δ𝑑𝑑,𝑑𝑑𝑛𝑛 + Δ𝑑𝑑𝑘𝑘3)

𝑑𝑑𝑛𝑛+1 = 𝑑𝑑𝑛𝑛 + Δ𝑑𝑑 �𝑘𝑘1 + 2𝑘𝑘2 + 2𝑘𝑘3 + 𝑘𝑘4

6�

𝑇𝑇 = 𝑂𝑂(Δ𝑑𝑑4)

4th Order “classical” Runge-Kutta scheme is one example of the 4-stage scheme.


30

NTH ORDER ODES

An 𝑛𝑛𝑑𝑑ℎ order (possibly nonlinear) ODE of form 𝑑𝑑𝑛𝑛 = 𝐹𝐹�𝑑𝑑,𝑑𝑑,𝑑𝑑′ ,⋯ ,𝑑𝑑(𝑛𝑛−1)�

Can always be written as a system of 𝑛𝑛 1st order ODEs. Let 𝑥𝑥1 = 𝑑𝑑, 𝑥𝑥2 = 𝑑𝑑′ ,⋯ , 𝑥𝑥𝑛𝑛 = 𝑑𝑑(𝑛𝑛−1), then

𝑥𝑥1′ = 𝑥𝑥2

𝑥𝑥2′ = 𝑥𝑥3

⋮

𝑥𝑥𝑛𝑛−1′ = 𝑥𝑥𝑛𝑛

𝑥𝑥𝑛𝑛′ = 𝐹𝐹(𝑑𝑑, 𝑥𝑥1, 𝑥𝑥2,⋯ , 𝑥𝑥𝑛𝑛)

So 1st order systems are a unifying framework for ODE theory. It is not immediately obvious that Runge-Kutta methods should work for 1st order nonlinear systems – but it can be shown rigorously that they do (Lambert, 1991). What’s different? We treat 𝑑𝑑, 𝑘𝑘, 𝑓𝑓 as vector functions. The order of accuracy remains the same as the scalar case.

L EC T URE 1 1

LINEAR STABILITY ANALYSIS

Consider model problem


= 𝜆𝜆𝑑𝑑, 𝑑𝑑(0) = 𝑑𝑑0 ≠ 0, ℜ(𝜆𝜆) < 0

Analytical solution is 𝑑𝑑(𝑑𝑑) = 𝑑𝑑0𝑒𝑒𝜆𝜆𝑑𝑑 with lim𝑑𝑑→∞ 𝑑𝑑(𝑑𝑑) = 0. What is the limit on Δ𝑑𝑑 so integrating reproduces decaying behavior? If scheme has form 𝑑𝑑𝑛𝑛+1 = 𝜓𝜓(Δ𝑑𝑑𝜆𝜆)𝑑𝑑𝑛𝑛 where 𝜓𝜓(𝑥𝑥) is the amplification factor, then we’ll require lim𝑛𝑛→∞ 𝑑𝑑𝑛𝑛 = 0 ⇔ |𝜓𝜓(Δ𝑑𝑑𝜆𝜆)| < 1 which gives condition for absolute stability.

1-STAGE RUNGE-KUTTA (EXPLICIT EULER)

𝑑𝑑𝑛𝑛+1 = (1 + 𝜆𝜆Δ𝑑𝑑)𝑑𝑑𝑛𝑛 , 𝑛𝑛 = 0,1,⋯

So 𝜓𝜓(𝑑𝑑) = 1 + 𝑑𝑑 with 𝑑𝑑 ≡ 𝜆𝜆Δ𝑑𝑑. The stability region is |1 + 𝑑𝑑| < 1.

2.0 1.5 1.0 0.5 0.5 1.0Rez

1.5

1.0

0.5

0.5

1.0

1.5Imz


31

If 𝜆𝜆 ∈ ℝ, 𝜆𝜆 < 0 then absolute stability for Δ𝑑𝑑 ∈ �0, �2𝜆𝜆��.

4-STAGE RUNGE-KUTTA

𝑑𝑑𝑛𝑛+1 = 𝜓𝜓(𝜆𝜆Δ𝑑𝑑)𝑑𝑑𝑛𝑛 , |𝜓𝜓(𝑑𝑑)| < 1

We can choose a complex or imaginary 𝜆𝜆.

BACKWARD EULER

𝑑𝑑𝑛𝑛+1 = 𝑑𝑑𝑛𝑛 + 𝜆𝜆Δ𝑑𝑑𝑑𝑑𝑛𝑛+1 → 𝑑𝑑𝑛𝑛+1 =1

1 − 𝜆𝜆Δ𝑑𝑑𝑑𝑑𝑛𝑛

So 𝜓𝜓(𝑑𝑑) = 11−𝑑𝑑

with 𝑑𝑑 ≡ 𝜆𝜆Δ𝑑𝑑 and there is absolute stability for � 11−𝑑𝑑

� < 1 or |1 − 𝑑𝑑| > 1.

The stability region if 𝜆𝜆 ∈ ℝ, 𝜆𝜆 < 0 then Δ𝑑𝑑 ∈ (0,∞). Moral: In practice, we must limit the time step to ensure sufficient accuracy.

L EC T URE 1 2

LINEAR EQUATIONS WITH ANALYTIC COEFFICIENTS

Consider

𝐿𝐿𝑑𝑑 ≡ 𝑑𝑑′′ + 𝑝𝑝(𝑥𝑥)𝑑𝑑′ + 𝑞𝑞(𝑥𝑥)𝑑𝑑 = 0, 𝑑𝑑(𝑥𝑥0) = 0,𝑑𝑑′(𝑥𝑥0) = 0 (1)

If 𝑝𝑝(𝑑𝑑) and 𝑞𝑞(𝑑𝑑) are analytic at 𝑑𝑑 = 𝑥𝑥0 then 𝑥𝑥0 is an ordinary point. Otherwise, 𝑥𝑥0 is a singular point. Recall: unique solution exists within interval of continuity of 𝑝𝑝(𝑥𝑥) and 𝑞𝑞(𝑥𝑥) around 𝑥𝑥0. In general, 𝑑𝑑 cannot be expressed in terms of elementary functions. Strategy near an ordinary point 𝑥𝑥0:

1.0 0.5 0.5 1.0 1.5 2.0Rez

1.5

1.0

0.5

0.5

1.0

1.5Imz


32

a) Expand 𝑝𝑝(𝑥𝑥) and 𝑞𝑞(𝑥𝑥) as power series (analyticity ⇔ positive radius of convergence)

b) Seek power series solution for 𝑑𝑑(𝑥𝑥)

𝑑𝑑(𝑥𝑥) = �𝑎𝑎𝑘𝑘(𝑥𝑥 − 𝑥𝑥0)𝑘𝑘∞

𝑘𝑘=0

(2)

EXISTENCE OF SERIES SOLUTION

If 𝑝𝑝(𝑑𝑑) and 𝑞𝑞(𝑑𝑑) are analytic at 𝑑𝑑 = 𝑥𝑥0, with convergent power series for

|𝑥𝑥 − 𝑥𝑥0| < 𝑟𝑟0, 𝑟𝑟0 > 0 (3)

Then there exists a series solution to the IVP (1) of form (2) that converges at least throughout interval (3). (Proof Coddington). Note: find interval of convergence for 𝑝𝑝(𝑥𝑥) and 𝑞𝑞(𝑥𝑥) series by

a) Ratio test

b) Check distance to nearest singularity of 𝑝𝑝(𝑑𝑑) and 𝑞𝑞(𝑑𝑑) in complex plane

EXAMPLE

(1 + 𝑥𝑥2)𝑑𝑑′′ + 2𝑥𝑥𝑑𝑑′ + 4𝑥𝑥2𝑑𝑑 = 0, 𝑑𝑑(𝑥𝑥0) = 𝑑𝑑0,𝑑𝑑′(𝑥𝑥0) = 𝑑𝑑0′ , 𝑥𝑥0 = −

12

𝑝𝑝(𝑑𝑑) and 𝑞𝑞(𝑑𝑑) have simple poles at 𝑑𝑑 = ±𝑖𝑖.

∑ 𝑎𝑎𝑘𝑘 �𝑥𝑥 + 12�𝑘𝑘

∞𝑘𝑘=0 converges at least for �𝑥𝑥 + 1

2� < √5

2.

Note: 𝑝𝑝(𝑥𝑥) and 𝑞𝑞(𝑥𝑥) are continuous for −∞ < 𝑥𝑥 < ∞ so there exists a unique solution 𝑑𝑑(𝑥𝑥) for −∞ < 𝑥𝑥 < ∞ but may not have power series about 𝑥𝑥0 that converges for all 𝑥𝑥.

POWER SERIES OPERATIONS

Recall: termwise differentiation, addition, and multiplication are valid inside circle of convergence.

SERIES SOLUTION NEAR AN ORDINARY POINT

EXAMPLE: (AIRY’S EQUATION)


33

𝑑𝑑′′ − 𝑥𝑥𝑑𝑑 = 0,−∞ < 𝑥𝑥 < ∞ (4)

𝑝𝑝(𝑑𝑑) and 𝑞𝑞(𝑑𝑑) are analytic for all 𝑥𝑥0: every point is an ordinary point. Note:

𝑑𝑑′′ − 𝑑𝑑 = 0 → 𝑑𝑑1,2 = 𝑒𝑒±𝑥𝑥 , monotonic

𝑑𝑑′′ + 𝑑𝑑 = 0 → 𝑑𝑑1,2 = 𝑒𝑒±𝑖𝑖𝑥𝑥 , oscillatory

So expect: monotonic for 𝑥𝑥 > 0, oscillatory for 𝑥𝑥 < 0. Expand around 𝑥𝑥0 = 0:

𝑑𝑑 = 𝑎𝑎0 + 𝑎𝑎1𝑥𝑥 + 𝑎𝑎2𝑥𝑥2 + ⋯+ 𝑎𝑎𝑘𝑘𝑥𝑥𝑘𝑘 + ⋯ = �𝑎𝑎𝑘𝑘𝑥𝑥𝑘𝑘∞

𝑘𝑘=0

𝑑𝑑′ = 𝑎𝑎1 + 2𝑎𝑎2𝑥𝑥 + 3𝑎𝑎3𝑥𝑥2 + ⋯+ 𝑘𝑘𝑎𝑎𝑘𝑘𝑥𝑥𝑘𝑘−1 + ⋯ = �𝑘𝑘𝑎𝑎𝑘𝑘𝑥𝑥𝑘𝑘−1∞

𝑘𝑘=1

𝑑𝑑′′ = 2𝑎𝑎2 + ⋯+ 𝑘𝑘(𝑘𝑘 − 1)𝑎𝑎𝑘𝑘𝑥𝑥𝑘𝑘−2 + ⋯ = �𝑘𝑘(𝑘𝑘 − 1)𝑎𝑎𝑘𝑘𝑥𝑥𝑘𝑘−2∞

𝑘𝑘=2

For convenience, shift indices to get 𝑥𝑥𝑘𝑘 :

𝑑𝑑′′ = �(𝑘𝑘 + 2)(𝑘𝑘 + 1)𝑎𝑎𝑘𝑘+2𝑥𝑥𝑘𝑘∞

𝑘𝑘=0

, 𝑥𝑥𝑑𝑑 = �𝑎𝑎𝑘𝑘−1𝑥𝑥𝑘𝑘∞

𝑘𝑘=1

So to satisfy (4)

𝑑𝑑′′ − 𝑥𝑥𝑑𝑑 = 2𝑎𝑎2 + �[(𝑘𝑘 + 2)(𝑘𝑘 + 1)𝑎𝑎𝑘𝑘+2 − 𝑎𝑎𝑘𝑘−1]𝑥𝑥𝑘𝑘∞

𝑘𝑘=1

= 0

Coefficients of all powers must be zero

2𝑎𝑎2 = 0, (𝑘𝑘 + 2)(𝑘𝑘 + 1)𝑎𝑎𝑘𝑘+2 − 𝑎𝑎𝑘𝑘−1 = 0, 𝑘𝑘 = 1,2, …

𝑘𝑘 = 1: 𝑎𝑎3 =𝑎𝑎0

3 ∗ 2

𝑘𝑘 = 2: 𝑎𝑎4 =𝑎𝑎1

4 ∗ 3

𝑘𝑘 = 3: 𝑎𝑎5 =𝑎𝑎2

5 ∗ 4= 0

By induction

𝑎𝑎3𝑚𝑚 =𝑎𝑎0

[2 ∗ 3][5 ∗ 6] … [(3𝑚𝑚 − 1)(3𝑚𝑚)] , 𝑚𝑚 = 1,2, …

𝑎𝑎3𝑚𝑚+1 =𝑎𝑎1

[3 ∗ 4][6 ∗ 7] … [3𝑚𝑚(3𝑚𝑚 + 1)] , 𝑚𝑚 = 1,2, …

𝑎𝑎3𝑚𝑚+2 = 0, 𝑚𝑚 = 0,1,2 …

General Solution


34

𝑑𝑑 = 𝑎𝑎0 �1 + �𝑥𝑥3𝑘𝑘

[2 ∗ 3][5 ∗ 6] … [(3𝑘𝑘 − 1)3𝑘𝑘]

∞

𝑘𝑘=1

� + 𝑎𝑎1 �𝑥𝑥 + �𝑥𝑥3𝑘𝑘+1

[3 ∗ 4][6 ∗ 7] … [3𝑘𝑘(3𝑘𝑘 + 1)]

∞

𝑘𝑘=1

� (5)

= 𝑎𝑎0𝑑𝑑1(𝑥𝑥) + 𝑎𝑎1𝑑𝑑2(𝑥𝑥)

𝑎𝑎0, 𝑎𝑎1 two constants to satisfy 𝑑𝑑(0) = 𝑑𝑑0,𝑑𝑑′(0) = 𝑑𝑑0′ . To see that 𝑑𝑑1 and 𝑑𝑑2 are linearly independent, check the

Wronskian at a single point:

𝑑𝑑1(0) = 1, 𝑑𝑑2(0) = 0

𝑑𝑑1′ (0) = 0, 𝑑𝑑2

′ (0) = 1

So 𝑊𝑊(𝑑𝑑1,𝑑𝑑2)(0) = 1 ≠ 0. Series solution (5) converges for −∞ < 𝑥𝑥 < ∞ since 𝑝𝑝(𝑑𝑑) and 𝑞𝑞(𝑑𝑑) are entire.

L EC T URE 1 3

BEHAVIOR NEAR SINGULAR POINTS

Can sometimes hold greatest physical interest → look in complex plane


+ 𝑝𝑝(𝑤𝑤)𝑤𝑤 = 0 (1)

Suppose 𝑝𝑝(𝑑𝑑) is analytic with an isolated singularity as 𝑑𝑑 = 0. Solution has form:

𝑤𝑤(𝑑𝑑) = 𝑤𝑤(𝑑𝑑0)𝑒𝑒−∫ 𝑝𝑝(𝑑𝑑)𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑0 (2)

(for contours that avoid 𝑑𝑑 = 0). Start at 𝑑𝑑0, analytically continue 𝑤𝑤(𝑑𝑑) to 𝑑𝑑0𝑒𝑒2𝜋𝜋𝑖𝑖 .

In general, 𝑤𝑤(𝑑𝑑0𝑒𝑒2𝜋𝜋𝑖𝑖 ) ≠ 𝑤𝑤(𝑑𝑑0). From (2), 𝑤𝑤(𝑑𝑑0𝑒𝑒2𝜋𝜋𝑖𝑖) = 𝑤𝑤(𝑑𝑑0)𝑒𝑒−∮𝑝𝑝(𝑑𝑑)𝑑𝑑𝑑𝑑 = 𝑤𝑤(𝑑𝑑0)𝑒𝑒−2𝜋𝜋𝑖𝑖𝑅𝑅𝑒𝑒𝑠𝑠 (𝑝𝑝 ;0). Note: 𝜆𝜆 = 1 if 𝑝𝑝 analytic at 𝑑𝑑 = 0. Also, 𝜆𝜆 independent of 𝑑𝑑0 so 𝑤𝑤(𝑑𝑑𝑒𝑒2𝜋𝜋𝑖𝑖 ) = 𝜆𝜆𝑤𝑤(𝑑𝑑). Let 𝑝𝑝 = − log 𝜆𝜆

2𝜋𝜋𝑖𝑖. Then, 𝑤𝑤(𝑑𝑑)𝑑𝑑𝑝𝑝 is single-valued


35

and analytic for 0 < |𝑑𝑑| < 𝜖𝜖, since (𝑑𝑑𝑒𝑒2𝜋𝜋𝑖𝑖 )𝑝𝑝 = 𝑑𝑑𝑝𝑝𝑒𝑒2𝜋𝜋𝑖𝑖𝑝𝑝 = 𝑑𝑑𝑝𝑝𝑒𝑒− log 𝜆𝜆 = 𝑑𝑑𝑝𝑝

𝜆𝜆. Hence 𝑤𝑤(𝑑𝑑)𝑑𝑑𝑝𝑝 may be expanded as a

Laurent series at 𝑑𝑑 = 0:

𝑤𝑤(𝑑𝑑)𝑑𝑑𝑝𝑝 = � 𝑎𝑎𝑘𝑘𝑑𝑑𝑘𝑘∞

𝑘𝑘=−∞

, 0 < |𝑑𝑑| < 𝜖𝜖

If 𝑎𝑎𝑘𝑘 = 0 for 𝑘𝑘 < −𝑁𝑁 (𝑁𝑁 > 0) then 𝑤𝑤(𝑑𝑑) = 𝑑𝑑−𝑝𝑝−𝑁𝑁(𝑏𝑏0 + 𝑏𝑏1𝑑𝑑 + ⋯ ), 𝑏𝑏0 ≠ 0. Or

𝑤𝑤(𝑑𝑑) = 𝑑𝑑𝜃𝜃𝑔𝑔(𝑑𝑑), 𝜃𝜃 = −𝑝𝑝 − 𝑁𝑁 (3)

Where 𝑔𝑔(𝑑𝑑) is analytic in |𝑑𝑑| < 𝜖𝜖 with 𝑔𝑔(0) ≠ 0. If solution can be written in form (3), then 𝑑𝑑 = 0 is a regular singular point otherwise 𝑑𝑑 = 0 is an irregular singular point. What is the requirement on 𝑝𝑝(𝑑𝑑)? 𝑑𝑑 = 0 is a regular singular point ⇔ 𝑝𝑝(𝑑𝑑) has a simple pole at 𝑑𝑑 = 0.

PROOF (FORWARD)

Recall 𝑤𝑤′ + 𝑝𝑝(𝑑𝑑)𝑤𝑤 = 0 and suppose 𝑤𝑤(𝑑𝑑) = 𝑑𝑑𝜃𝜃𝑔𝑔(𝑑𝑑), 𝑔𝑔(0) ≠ 0. Then 𝑝𝑝(𝑑𝑑) = −𝑤𝑤′

𝑤𝑤= −�𝜃𝜃

𝑑𝑑+ 𝑔𝑔′

𝑔𝑔�. (See

Gee Ince, 1956 for reverse).

EXAMPLE

𝑝𝑝(𝑑𝑑) =1𝑑𝑑⇒ 𝑤𝑤(𝑑𝑑) = 𝑑𝑑−1𝑏𝑏0

𝑑𝑑 = 0 is regular singular point

𝑝𝑝(𝑑𝑑) =1𝑑𝑑2 ⇒ 𝑤𝑤(𝑑𝑑) = 𝑒𝑒

1𝑑𝑑

Essential singularity, 𝑑𝑑 = 0 is an irregular singular point

SOLUTION STRATEGY

Expand 𝑤𝑤(𝑑𝑑) at regular singular point 𝑑𝑑 = 0. Attempt solution form

𝑤𝑤(𝑑𝑑) = 𝑑𝑑𝜃𝜃 �𝑎𝑎𝑘𝑘𝑑𝑑𝑘𝑘∞

𝑘𝑘=0

REGULAR SINGULAR POINTS OF SECOND ORDER EQUATION

Consider

𝑑𝑑′′ +𝑝𝑝(𝑥𝑥)𝑥𝑥

𝑑𝑑′ +𝑞𝑞(𝑥𝑥)𝑥𝑥2 𝑑𝑑 = 0

(4)

At worst, 𝑝𝑝(𝑥𝑥)𝑥𝑥

is a simple pole, 𝑞𝑞(𝑥𝑥)𝑥𝑥2 is a 2nd order pole. 𝑥𝑥 = 0 is a regular singular point ⇔ it is not ordinary and 𝑝𝑝(𝑑𝑑)

and 𝑞𝑞(𝑑𝑑) are analytic at 𝑑𝑑 = 0. If 𝑥𝑥 = 0 is a regular singular point, (4) has at least one solution of form:


36

𝑑𝑑(𝑥𝑥) = 𝑥𝑥𝑟𝑟(𝑎𝑎0 + 𝑎𝑎1𝑥𝑥 + 𝑎𝑎2𝑥𝑥2 + ⋯ ) = 𝑥𝑥𝑟𝑟 �𝑎𝑎𝑘𝑘𝑥𝑥𝑘𝑘∞

𝑘𝑘=0

, 𝑥𝑥 ≠ 0 (5)

(by convention, define 𝑟𝑟 so 𝑎𝑎0 ≠ 0) What is form of 𝑑𝑑2? Consider simplest case of (4).

EULER EQUATIONS

𝐿𝐿𝑑𝑑 ≡ 𝑥𝑥2𝑑𝑑′′ + 𝑎𝑎𝑥𝑥𝑑𝑑′ + 𝑏𝑏𝑑𝑑 = 0, 𝑎𝑎,𝑏𝑏 constants

With 𝑥𝑥 = 0 a regular singular point. Attempt solution of form (5) with 𝑎𝑎𝑘𝑘 = 0 for 𝑘𝑘 ≥ 1, i.e.,

𝑑𝑑 = 𝑥𝑥𝑟𝑟 (6)

Then 𝑑𝑑′ = (𝑥𝑥𝑟𝑟)′ = 𝑟𝑟𝑥𝑥𝑟𝑟−1 and 𝑑𝑑′′ (𝑥𝑥𝑟𝑟)′′ = 𝑟𝑟(𝑟𝑟 − 1)𝑥𝑥𝑟𝑟−2. So 𝐿𝐿𝑥𝑥𝑟𝑟 = 𝑥𝑥𝑟𝑟[𝑟𝑟(𝑟𝑟 − 1) + 𝑎𝑎𝑟𝑟 + 𝑏𝑏] = 0. (6) is a solution if 𝑟𝑟 satisfies 𝐹𝐹(𝑟𝑟) ≡ 𝑟𝑟(𝑟𝑟 − 1) + 𝑎𝑎𝑟𝑟 + 𝑏𝑏 = 0 ≡ indicial equation.

2 CASES

DISTINCT ROOTS (𝑟𝑟1 ≠ 𝑟𝑟2)

𝑑𝑑1 = 𝑥𝑥𝑟𝑟1 , 𝑑𝑑2 = 𝑥𝑥𝑟𝑟2

𝑊𝑊(𝑥𝑥𝑟𝑟1 , 𝑥𝑥𝑟𝑟2) = (𝑟𝑟2 − 𝑟𝑟1)𝑥𝑥𝑟𝑟1+𝑟𝑟2−1 ≠ 0, 𝑥𝑥 ≠ 0

Linearly independent solution for 𝑥𝑥 < 0 or 𝑥𝑥 > 0 satisfy initial conditions in either interval.

REPEATED ROOTS (𝑟𝑟1 = 𝑟𝑟2)

�𝐿𝐿𝑥𝑥𝑟𝑟 |𝑟𝑟=𝑟𝑟1 = 0

And also

𝜕𝜕𝜕𝜕𝑟𝑟�[𝐿𝐿𝑥𝑥𝑟𝑟]|𝑟𝑟=𝑟𝑟1 = 𝐿𝐿

𝜕𝜕𝜕𝜕𝑟𝑟�[𝑥𝑥𝑟𝑟]|𝑟𝑟=𝑟𝑟1 = 0

But 𝜕𝜕𝜕𝜕𝑟𝑟𝑥𝑥𝑟𝑟 = 𝑥𝑥𝑟𝑟 log 𝑥𝑥 so �𝐿𝐿[𝑥𝑥𝑟𝑟 log 𝑥𝑥]|𝑟𝑟=𝑟𝑟1 = 0 ⇒ 𝑑𝑑2 = 𝑥𝑥𝑟𝑟1 log 𝑥𝑥 = 𝑑𝑑1 log 𝑥𝑥.

𝑊𝑊(𝑥𝑥𝑟𝑟1 , 𝑥𝑥𝑟𝑟1 log 𝑥𝑥) = 𝑥𝑥2𝑟𝑟1−1 ≠ 0, 𝑥𝑥 ≠ 0

EXAMPLE

2𝑥𝑥2𝑑𝑑′′ + 3𝑥𝑥𝑑𝑑′ − 𝑑𝑑 = 0

Try 𝑑𝑑 = 𝑥𝑥𝑟𝑟 ⇒ 𝑥𝑥𝑟𝑟(2𝑟𝑟2 + 𝑟𝑟 − 1) = 0. We have 𝑟𝑟1 = 12

, 𝑟𝑟2 = −1. Thus 𝑑𝑑 = 𝑐𝑐1𝑥𝑥12 + 𝑐𝑐2𝑥𝑥−1 is the general solution for

𝑥𝑥 < 0 or 𝑥𝑥 > 0. Real form ⇒ recall 𝑥𝑥𝛼𝛼 = 𝑒𝑒𝛼𝛼 log 𝑥𝑥 :

𝑥𝑥12 = 𝑒𝑒

12(Log 𝑥𝑥 + 𝑖𝑖 arg 𝑥𝑥) = 𝐾𝐾𝑒𝑒

12 Log 𝑥𝑥 = 𝐾𝐾|𝑥𝑥|

12

So real general solution is

𝑑𝑑 = 𝑘𝑘1|𝑥𝑥|12 + 𝑘𝑘2|𝑥𝑥|−1, 𝑥𝑥 < 0 or 𝑥𝑥 > 0


37

L EC T URE 1 4

THE METHOD OF FROBENIUS

Consider

𝑥𝑥2𝑑𝑑′′ + 𝑥𝑥𝑝𝑝(𝑥𝑥)𝑑𝑑′ + 𝑞𝑞(𝑥𝑥)𝑑𝑑 = 0 (1)

With 𝑝𝑝(𝑑𝑑) and 𝑞𝑞(𝑑𝑑) analytic at 𝑥𝑥0 = 0. Then

𝑝𝑝(𝑥𝑥) = �𝑝𝑝𝑘𝑘𝑥𝑥𝑘𝑘∞

𝑘𝑘=0

, 𝑞𝑞(𝑥𝑥) = �𝑞𝑞𝑘𝑘𝑥𝑥𝑘𝑘∞

𝑘𝑘=0

, |𝑥𝑥| < 𝜌𝜌,𝜌𝜌 > 0 (2)

And 𝑥𝑥 = 0 is a regular singular point. Note: if 𝑝𝑝𝑘𝑘 = 𝑞𝑞𝑘𝑘 = 0 for 𝑘𝑘 ≥ 1 this is just the Euler equation

𝑥𝑥2𝑑𝑑′′ + 𝑥𝑥𝑝𝑝0𝑑𝑑′ + 𝑞𝑞0𝑑𝑑 = 0 (3)

Seek solution to (1) of form

𝑑𝑑 = 𝑥𝑥𝑟𝑟 �𝑎𝑎𝑘𝑘𝑥𝑥𝑘𝑘∞

𝑘𝑘=0

= 𝑥𝑥𝑟𝑟(𝑎𝑎0 + 𝑎𝑎1𝑥𝑥 + ⋯ ) (4)

Where 𝑎𝑎0 ≠ 0. Then

𝑑𝑑′ = �(𝑟𝑟 + 𝑘𝑘)𝑎𝑎𝑘𝑘𝑥𝑥𝑟𝑟+𝑘𝑘−1∞

𝑘𝑘=0

= 𝑥𝑥𝑟𝑟−1[𝑟𝑟𝑎𝑎0 + (𝑟𝑟 + 1)𝑎𝑎1𝑥𝑥 + ⋯ ] (5)

And

𝑑𝑑′′ = �(𝑘𝑘 + 𝑟𝑟)(𝑘𝑘 + 𝑟𝑟 − 1)𝑎𝑎𝑘𝑘𝑥𝑥𝑘𝑘+𝑟𝑟−2∞

𝑘𝑘=0

(6)

Substituting (2),(4),(5) and (6) into (1)

𝑥𝑥𝑟𝑟[𝑟𝑟(𝑟𝑟 − 1)𝑎𝑎0 + ⋯ ] + (𝑝𝑝0 + 𝑝𝑝1𝑥𝑥 + ⋯ )𝑥𝑥𝑟𝑟[𝑟𝑟𝑎𝑎0 + (𝑟𝑟 + 1)𝑎𝑎1𝑥𝑥 + ⋯ ] + (𝑞𝑞0 + 𝑞𝑞1𝑥𝑥 + ⋯ ) = 0 (7)

Set coefficients of all powers of 𝑥𝑥 to zero. Smallest power is 𝑥𝑥𝑟𝑟 : 𝑎𝑎0[𝑟𝑟(𝑟𝑟 − 1) + 𝑝𝑝0𝑟𝑟 + 𝑎𝑎0] = 0. But 𝑎𝑎0 ≠ 0, so

𝐹𝐹(𝑟𝑟) = 𝑟𝑟(𝑟𝑟 − 1) + 𝑝𝑝0𝑟𝑟 + 𝑞𝑞0 = 0 (8)

(8) is known as the indicial equation. Note that it is identical to that for corresponding Euler equation (3). Roots of (8), 𝑟𝑟1 and 𝑟𝑟2, determine qualitative behavior of solution near singularity. Setting coefficients of 𝑥𝑥𝑟𝑟+𝑘𝑘 to zero for 𝑘𝑘 ≥ 1 gives

𝐹𝐹(𝑟𝑟 + 𝑘𝑘)𝑎𝑎𝑘𝑘 + � 𝑎𝑎𝑚𝑚 [(𝑟𝑟 + 𝑚𝑚)𝑝𝑝𝑘𝑘−𝑚𝑚 + 𝑞𝑞𝑘𝑘−𝑚𝑚 ]𝑘𝑘−1

𝑚𝑚=0

= 0 (9)


Recurrence relation: can computer 𝑎𝑎𝑘𝑘 using 𝑎𝑎0, 𝑎𝑎1, … , 𝑎𝑎𝑘𝑘−1 and 𝑝𝑝1, … , 𝑝𝑝𝑘𝑘 , 𝑞𝑞1, … 𝑞𝑞𝑘𝑘 as long as 𝐹𝐹(𝑟𝑟 + 𝑚𝑚) ≠ 0 for 𝑚𝑚 = 1, … ,𝑘𝑘. Note: 𝐹𝐹(𝑟𝑟) = 0 only for 𝑟𝑟 = 𝑟𝑟1, 𝑟𝑟2. Let 𝑟𝑟1 ≥ 𝑟𝑟2 if 𝑟𝑟1, 𝑟𝑟2 are real. Then 𝐹𝐹(𝑟𝑟1 + 𝑘𝑘) ≠ 0 for 𝑘𝑘 ≥ 1. Hence, can always determine one solution of form (4).

𝑑𝑑1(𝑥𝑥) = 𝑥𝑥𝑟𝑟1 �𝑎𝑎𝑘𝑘(𝑟𝑟1)𝑥𝑥𝑘𝑘∞

𝑘𝑘=0

Coefficients from (9) with 𝑟𝑟 = 𝑟𝑟1.

TROUBLE

a) If 𝑟𝑟1 = 𝑟𝑟2, (9) gives only one linearly independent solution

- Expect log 𝑥𝑥 term in 𝑑𝑑2 based on Euler equation experience

b) If 𝑟𝑟1 = 𝑟𝑟2 + 𝑁𝑁,𝑁𝑁 ∈ ℤ+ then 𝐹𝐹(𝑟𝑟2 + 𝑘𝑘) = 0 for 𝑘𝑘 = 𝑁𝑁 so (9) fails to define 𝑎𝑎𝑘𝑘(𝑟𝑟2). May require log 𝑥𝑥 term in 𝑑𝑑2.

Last ≅ 15 minutes of lecture was on the Method of Frobenius handout online: HTTP://WWW.ACM.CALTECH.EDU/~NILES/ACM95/FROBENIUS.PDF

L EC T URE 1 5

2N D ORDER LINEAR BOUNDARY VALUE PROBLEM (BVP)

Consider

𝑑𝑑′′ + 𝑝𝑝(𝑥𝑥)𝑑𝑑′ + 𝑞𝑞(𝑥𝑥)𝑑𝑑 = 0 (1)

With continuous 𝑝𝑝(𝑥𝑥) and 𝑞𝑞(𝑥𝑥) and general solution 𝑑𝑑 = 𝑐𝑐1𝑑𝑑1(𝑥𝑥) + 𝑐𝑐2𝑑𝑑2(𝑥𝑥). Does the BVP with boundary conditions 𝑑𝑑(𝑎𝑎) = 𝑐𝑐, 𝑑𝑑(𝑏𝑏) = 𝑑𝑑 have a unique solution?

EXAMPLE

𝑑𝑑′′ + 𝜆𝜆𝑑𝑑 = 0, 𝑎𝑎 = 0, 𝑏𝑏 = 1, (𝜆𝜆 > 0)

General solution is 𝑑𝑑 = 𝑐𝑐1 sin√𝜆𝜆𝑥𝑥 + 𝑐𝑐2 cos√𝜆𝜆𝑥𝑥. Now suppose 𝜆𝜆 = 1 so 𝑑𝑑 = 𝑐𝑐1 sin 𝑥𝑥 + 𝑐𝑐2 cos 𝑥𝑥.

WITH HOMOGENEOUS BOUNDARY CONDITIONS

𝑑𝑑(0) = 0, 𝑑𝑑(1) = 0

⇓ ⇓

𝑐𝑐2 = 0 𝑐𝑐1 = 0

So 𝑑𝑑 = 0, the trivial solution

WITH NONHOMOGENEOUS BOUNDARY CONDITIONS

http://www.acm.caltech.edu/~niles/acm95/frobenius.pdf�


39

𝑑𝑑(0) = 1, 𝑑𝑑(1) = 0

⇓ ⇓

𝑐𝑐2 = 1 𝑐𝑐1 sin 1 + cos 1 = 0

So 𝑑𝑑 = − sin 𝑥𝑥tan 1

+ cos 𝑥𝑥, a unique solution

Now suppose 𝜆𝜆 = 𝜋𝜋2 so 𝑑𝑑 = 𝑐𝑐1 sin𝜋𝜋𝑥𝑥 + 𝑐𝑐2 cos𝜋𝜋𝑥𝑥.

WITH HOMOGENEOUS BOUNDARY CONDITIONS

𝑑𝑑(0) = 0, 𝑑𝑑(1) = 0

⇓ ⇓

𝑐𝑐2 = 0 𝑐𝑐1 sin𝜋𝜋 = 0

So 𝑑𝑑 = 𝑐𝑐1 sin𝜋𝜋𝑥𝑥, infinitely many solutions

WITH NONHOMOGENEOUS BOUNDARY CONDITIONS

i) 𝑑𝑑(0) = 1, 𝑑𝑑(1) = 0

⇓ ⇓

𝑐𝑐2 = 1 𝑐𝑐1 sin𝜋𝜋 − 1 ≠ 0

So there is no solution.

ii) 𝑑𝑑(0) = 1, 𝑑𝑑(1) = −1

⇓ ⇓

𝑐𝑐2 = 1 𝑐𝑐1 sin𝜋𝜋 − 1 = −1

So 𝑑𝑑 = 𝑐𝑐1 sin𝜋𝜋𝑥𝑥 + cos𝜋𝜋𝑥𝑥, infinitely many solutions.

UNIQUE SOLUTION THEOREM

The BVP consisting of (1) plus homogeneous boundary conditions 𝑑𝑑(𝑎𝑎) = 𝑐𝑐,𝑑𝑑(𝑏𝑏) = 𝑑𝑑 has a unique solution if and only if the homogeneous boundary conditions are satisfied only by the trivial solution 𝑑𝑑 = 0.

PROOF

The nonhomogenous linear systems

𝑐𝑐1𝑑𝑑1(𝑎𝑎) + 𝑐𝑐2𝑑𝑑2(𝑎𝑎) = 𝑐𝑐

𝑐𝑐1𝑑𝑑1(𝑏𝑏) + 𝑐𝑐2𝑑𝑑2(𝑏𝑏) = 𝑑𝑑

Has a unique solution if and only if Det ≡ 𝑑𝑑1(𝑎𝑎)𝑑𝑑2(𝑏𝑏) − 𝑑𝑑1(𝑏𝑏)𝑑𝑑2(𝑎𝑎) ≠ 0. If det ≠ 0, the homogeneous system

𝑐𝑐1𝑑𝑑1(𝑎𝑎) + 𝑐𝑐2𝑑𝑑2(𝑎𝑎) = 0


40

𝑐𝑐1𝑑𝑑1(𝑏𝑏) + 𝑐𝑐2𝑑𝑑2(𝑏𝑏) = 0

Has only the trivial solution (𝑐𝑐1, 𝑐𝑐2) = (0,0). Alternatively, if det = 0, there is a nontrivial solution (𝑐𝑐1, 𝑐𝑐2) ≠ (0,0).

What’s going on? 𝑑𝑑′′ + 𝜆𝜆𝑑𝑑 = 0, 𝑑𝑑(0) = 0, 𝑑𝑑(1) = 0 is an eigenvalue problem. It has a nontrivial solution (called an eigenfunction) if and only if 𝜆𝜆 is an eigenvalue.

SIMPLE EIGENVALUE PROBLEMS

EXAMPLE 1

𝑑𝑑′′ + 𝜆𝜆𝑑𝑑 = 0, 𝑑𝑑(0) = 0, 𝑑𝑑(𝐿𝐿) = 0

Let’s find all the eigenvalues and eigenfunctions

CASE 1: 𝜆𝜆 > 0

𝑑𝑑𝑔𝑔𝑒𝑒𝑛𝑛𝑒𝑒𝑟𝑟𝑎𝑎𝑙𝑙 = 𝑐𝑐1 sin√𝜆𝜆𝑥𝑥 + 𝑐𝑐2 cos√𝜆𝜆𝑥𝑥

𝑑𝑑(0) = 0 ⇒ 𝑐𝑐2 = 0

𝑑𝑑(𝐿𝐿) = 0 ⇒ 𝑐𝑐1 sin√𝜆𝜆𝐿𝐿 = 0 ⇒ √𝜆𝜆 =𝑛𝑛𝜋𝜋𝐿𝐿

, 𝑛𝑛 = 1,2, …

So we have eigenvalues

𝜆𝜆𝑛𝑛 =𝑛𝑛2𝜋𝜋2

𝐿𝐿2 , 𝑛𝑛 = 1,2, …

And our corresponding eigenfunctions are

𝜙𝜙𝑛𝑛 = sin𝑛𝑛𝜋𝜋𝑥𝑥𝐿𝐿

, 𝑛𝑛 = 1,2, …

CASE 2: 𝜆𝜆 = 0

𝑑𝑑′′ = 0 → 𝑑𝑑𝑔𝑔𝑒𝑒𝑛𝑛𝑒𝑒𝑟𝑟𝑎𝑎𝑙𝑙 = 𝑐𝑐1 + 𝑐𝑐2𝑥𝑥

𝑑𝑑(0) = 0 ⇒ 𝑐𝑐1 = 0

𝑑𝑑(𝐿𝐿) = 0 ⇒ 𝑐𝑐2 = 0

Hence 𝜆𝜆 = 0 is not an eigenvalue.

CASE 3: 𝜆𝜆 < 0

Consider 𝑑𝑑′′ − 𝜃𝜃𝑑𝑑 = 0, 𝜃𝜃 = −𝜆𝜆 > 0


41

𝑑𝑑𝑔𝑔𝑒𝑒𝑛𝑛𝑒𝑒𝑟𝑟𝑎𝑎𝑙𝑙 = 𝑐𝑐1𝑒𝑒√𝜃𝜃𝑥𝑥 + 𝑐𝑐2𝑒𝑒−√𝜃𝜃𝑥𝑥 = 𝑐𝑐3 sinh√𝜃𝜃𝑥𝑥 + 𝑐𝑐4 cosh√𝜃𝜃𝑥𝑥

𝑑𝑑(0) = 0 ⇒ 𝑐𝑐4 = 0, 𝑑𝑑(𝐿𝐿) = 0 ⇒ 𝑐𝑐3 = 0

Hence no negative eigenvalues.

CASE 4: 𝜆𝜆 COMPLEX

We will later prove for a more general class of problems that there are no complex eigenvalues.

EXAMPLE 2

𝑑𝑑′′ + 𝜆𝜆𝑑𝑑 = 0, 𝑑𝑑′(0) = 0, 𝑑𝑑′(𝐿𝐿) = 0

Let’s find all the eigenvalues and eigenfunctions

CASE 1: 𝜆𝜆 > 0

𝑑𝑑𝑔𝑔𝑒𝑒𝑛𝑛𝑒𝑒𝑟𝑟𝑎𝑎𝑙𝑙 = 𝑐𝑐1 sin√𝜆𝜆𝑥𝑥 + 𝑐𝑐2 cos√𝜆𝜆𝑥𝑥

𝑑𝑑′(0) = 0 ⇒ 𝑐𝑐1 = 0

𝑑𝑑′(𝐿𝐿) = 0 ⇒ √𝜆𝜆𝑐𝑐2 sin√𝜆𝜆𝐿𝐿 = 0 ⇒ √𝜆𝜆 =𝑛𝑛𝜋𝜋𝐿𝐿

, 𝑛𝑛 = 1,2, …

So we have eigenvalues


𝐿𝐿2 , 𝑛𝑛 = 1,2, …

And our corresponding eigenfunctions are

𝜙𝜙𝑛𝑛 = cos𝑛𝑛𝜋𝜋𝑥𝑥𝐿𝐿

, 𝑛𝑛 = 1,2, …

CASE 2: 𝜆𝜆 = 0

𝑑𝑑𝑔𝑔𝑒𝑒𝑛𝑛𝑒𝑒𝑟𝑟𝑎𝑎𝑙𝑙 = 𝑐𝑐1 + 𝑐𝑐2𝑥𝑥

𝑑𝑑′(0) = 0 ⇒ 𝑐𝑐2 = 0

𝑑𝑑′(𝐿𝐿) = 0 ⇒ 𝑐𝑐2 = 0

So 𝜆𝜆0 = 0, 𝜙𝜙0 = 1. As before, there are no negative or complex eigenvalues, hence this eigenvalue problem gives us


𝐿𝐿2 , 𝑛𝑛 = 0,1, …


42

𝜙𝜙𝑛𝑛 = cos𝑛𝑛𝜋𝜋𝑥𝑥𝐿𝐿

L EC T URE 1 6

FOURIER SERIES

Represent a periodic function 𝑓𝑓(𝑥𝑥) with period 2𝐿𝐿 as an infinite series in terms of these eigenfunctions:

𝑓𝑓(𝑥𝑥) =𝑎𝑎0

2+ �𝑎𝑎𝑛𝑛 cos

𝑛𝑛𝜋𝜋𝑥𝑥𝐿𝐿

+ 𝑏𝑏𝑛𝑛 sin𝑛𝑛𝜋𝜋𝑥𝑥𝐿𝐿

∞

𝑛𝑛=1

(1)

Sum of eigenfunctions is an eigenfunction, so this is convenient.

PERIODICITY

𝑓𝑓(𝑥𝑥 + 𝑇𝑇) = 𝑓𝑓(𝑥𝑥); periodic function with period 𝑇𝑇. Smallest valid 𝑇𝑇 is the fundamental period.

sin𝑛𝑛𝜋𝜋𝑥𝑥𝐿𝐿

and cos𝑛𝑛𝜋𝜋𝑥𝑥𝐿𝐿

have fundamental period →2𝜋𝜋

�𝑛𝑛𝜋𝜋𝐿𝐿 �=

2𝐿𝐿𝑛𝑛

; all share period 2𝐿𝐿

Note: constant 𝑎𝑎0 has an arbitrary period, but no fundamental period.

ORTHOGONALITY PROPERTIES

Want to avoid redundancy in information: two real functions 𝜙𝜙(𝑥𝑥),𝜓𝜓(𝑥𝑥) are orthogonal on the interval 𝑥𝑥 ∈ [𝑎𝑎, 𝑏𝑏] if their inner product vanishes, ie:

(𝜙𝜙,𝜓𝜓) = �𝜙𝜙(𝑥𝑥)𝜓𝜓(𝑥𝑥)𝑑𝑑𝑥𝑥𝑏𝑏

𝑎𝑎

= 0

All sin and cos are orthogonal to each other for varying values of 𝑛𝑛 on 𝑥𝑥 = [−𝐿𝐿, 𝐿𝐿], 𝑛𝑛 = 1,2, …

� sin𝑛𝑛𝜋𝜋𝑥𝑥𝐿𝐿

sin𝑚𝑚𝜋𝜋𝑥𝑥𝐿𝐿

𝑑𝑑𝑥𝑥𝐿𝐿

−𝐿𝐿

= 𝐿𝐿𝛿𝛿𝑚𝑚 ,𝑛𝑛 , � cos𝑛𝑛𝜋𝜋𝑥𝑥𝐿𝐿



−𝐿𝐿

= 𝐿𝐿𝛿𝛿𝑚𝑚 ,𝑛𝑛

Note that 𝛿𝛿𝑚𝑚 ,𝑛𝑛 = �1, 𝑚𝑚 = 𝑛𝑛0, 𝑚𝑚 ≠ 𝑛𝑛

� (Kronecker-Delta).


cos𝑚𝑚𝜋𝜋𝑥𝑥𝐿𝐿


−𝐿𝐿

= 0

For example, to show the first case:


43




−𝐿𝐿

=12��cos

(𝑚𝑚 − 𝑛𝑛)𝜋𝜋𝑥𝑥𝐿𝐿

− cos(𝑚𝑚 + 𝑛𝑛)𝜋𝜋𝑥𝑥

𝐿𝐿� 𝑑𝑑𝑥𝑥

𝐿𝐿

−𝐿𝐿

=12𝐿𝐿𝜋𝜋�sin (𝑚𝑚 − 𝑛𝑛)𝜋𝜋𝑥𝑥

𝐿𝐿𝑚𝑚 − 𝑛𝑛

−sin (𝑚𝑚 + 𝑛𝑛)𝜋𝜋𝑥𝑥

𝐿𝐿𝑚𝑚 + 𝑛𝑛

�

−𝐿𝐿

𝐿𝐿

= 0

Similar methods can be used for the other two cases by using complex exponentials and rearranging. Now suppose the Fourier Series (1) converges to the function 𝑓𝑓(𝑥𝑥) on the interval 𝑥𝑥 ∈ [−𝐿𝐿, 𝐿𝐿]. How do you find the coefficients? Use orthgonality to find 𝑎𝑎𝑚𝑚 �𝑚𝑚 = 1,2, … ; the coefficient of cos𝑚𝑚𝜋𝜋𝑥𝑥

𝐿𝐿� then use (1).

�𝑓𝑓(𝑥𝑥) cos𝑚𝑚𝜋𝜋𝑥𝑥𝐿𝐿


−𝐿𝐿

=𝑎𝑎0

2� cos

𝑚𝑚𝜋𝜋𝑥𝑥𝐿𝐿


−𝐿𝐿

+ �𝑎𝑎𝑛𝑛 � cos𝑛𝑛𝜋𝜋𝑥𝑥𝐿𝐿



−𝐿𝐿

∞

𝑛𝑛=1

+ �𝑏𝑏𝑛𝑛 � sin𝑛𝑛𝜋𝜋𝑥𝑥𝐿𝐿



−𝐿𝐿

∞

𝑛𝑛=1

The first and last integrals go to zero by orthgonality conditions, the middle integral becomes 𝐿𝐿𝑎𝑎𝑛𝑛

�𝑓𝑓(𝑥𝑥)𝑑𝑑𝑥𝑥𝐿𝐿

−𝐿𝐿

=𝑎𝑎0

2�𝑑𝑑𝑥𝑥𝐿𝐿

−𝐿𝐿

+ �𝑎𝑎𝑛𝑛 � cos𝑛𝑛𝜋𝜋𝑥𝑥𝐿𝐿


−𝐿𝐿

∞

𝑛𝑛=1

+ �𝑏𝑏𝑛𝑛 � sin𝑛𝑛𝜋𝜋𝑥𝑥𝐿𝐿


−𝐿𝐿

∞

𝑛𝑛=1

The last two integrals go to zero by periodicity, the first integral becomes 𝐿𝐿𝑎𝑎0.

𝑎𝑎𝑛𝑛 =1𝐿𝐿�𝑓𝑓(𝑥𝑥) cos



−𝐿𝐿

, 𝑛𝑛 = 0,1,2 … , 𝑏𝑏𝑛𝑛 =1𝐿𝐿�𝑓𝑓(𝑥𝑥) sin



−𝐿𝐿

, 𝑛𝑛 = 1,2, …

If the series converges to 𝑓𝑓(𝑥𝑥) on 𝑥𝑥 ∈ [−𝐿𝐿, 𝐿𝐿] then it converges to the periodic extension of 𝑓𝑓(𝑥𝑥) outside the interval.

EXAMPLE

𝑓𝑓(𝑥𝑥) = 𝑥𝑥2, 𝑥𝑥 ∈ [−𝜋𝜋,𝜋𝜋]

𝑎𝑎0 =1𝜋𝜋� 𝑥𝑥2𝑑𝑑𝑥𝑥𝜋𝜋

−𝜋𝜋

=23𝜋𝜋2, 𝑎𝑎𝑛𝑛 =

1𝜋𝜋� 𝑥𝑥2 cos𝑛𝑛𝑥𝑥 𝑑𝑑𝑥𝑥𝜋𝜋

−𝜋𝜋

= (−1)𝑛𝑛4𝑛𝑛2 , 𝑛𝑛 = 1,2

𝑏𝑏𝑛𝑛 =1𝜋𝜋� 𝑥𝑥2 sin𝑛𝑛𝑥𝑥 𝑑𝑑𝑥𝑥𝜋𝜋

−𝜋𝜋

= 0


44

𝑥𝑥2 sin𝑛𝑛𝑥𝑥 is an even function times an odd function so it goes to zero by orthgonality. So

𝑔𝑔(𝑥𝑥) =𝜋𝜋2

3+ �(−1)𝑛𝑛

4𝑛𝑛2 cos𝑛𝑛𝑥𝑥

∞

𝑛𝑛=1

Is the periodic extension of 𝑓𝑓(𝑥𝑥). Note: it is also continuous. Also note coefficients decay by 1𝑛𝑛2. More smooth

would decay faster.

CONVERGENCE OF FOURIER SERIES

A function is piecewise continuous on 𝑥𝑥 ∈ [𝑎𝑎, 𝑏𝑏] if the interval can be split into a finite number of subintervals in which 𝑓𝑓(𝑥𝑥) is continuous with one-sided limits at the ends of each subinterval.

A function is piecewise smooth if it is piecewise continuous and 𝑓𝑓′(𝑥𝑥) is continuous in each subinterval.

THEOREM

If 𝑓𝑓(𝑥𝑥) is piecewise smooth on 𝑥𝑥 ∈ [−𝐿𝐿, 𝐿𝐿] then the Fourier series converges to the periodic extension of

𝑓𝑓(𝑥𝑥) where the extension is continous and equals �𝑓𝑓�𝑥𝑥+�−𝑓𝑓(𝑥𝑥−)�

2 at jump discontinuities in the periodic

extension (Proof Kaplan).

A power series fails with discontinuities. Note; Fourier series converges even for a function containing discontinuities (IE: it converges even if the function doesn’t converge). Compare this to power series.

L EC T URE 1 7

EXAMPLE

𝑓𝑓(𝑥𝑥) = �−1, 𝑥𝑥 < 01, 𝑥𝑥 > 0

�

On 𝑥𝑥 ∈ [−𝜋𝜋,𝜋𝜋].

𝑎𝑎0 =1𝜋𝜋�(−1)𝑑𝑑𝑥𝑥

0

−𝜋𝜋

+1𝜋𝜋�(+1)𝑑𝑑𝑥𝑥𝜋𝜋

0

= −1 + 1 = 0, 𝑎𝑎𝑛𝑛 =1𝜋𝜋�− cos𝑛𝑛𝑥𝑥 𝑑𝑑𝑥𝑥

0

−𝜋𝜋

+1𝜋𝜋� cos𝑛𝑛𝑥𝑥 𝑑𝑑𝑥𝑥𝜋𝜋

0

= 0


45

𝑏𝑏𝑛𝑛 =1𝜋𝜋�− sin 𝑛𝑛𝑥𝑥 𝑑𝑑𝑥𝑥

0

−𝜋𝜋

+1𝜋𝜋� sin𝑛𝑛𝑥𝑥 𝑑𝑑𝑥𝑥𝜋𝜋

0

=2𝜋𝜋� sin𝑛𝑛𝑥𝑥 𝑑𝑑𝑥𝑥𝜋𝜋

0

= �4𝑛𝑛𝜋𝜋

, 𝑛𝑛 odd

0, 𝑛𝑛 even�

𝑔𝑔(𝑥𝑥) =4𝜋𝜋�

12𝑛𝑛 + 1

sin[(2𝑛𝑛 + 1)𝑥𝑥]∞

𝑛𝑛=0

Periodic extension of 𝑓𝑓(𝑥𝑥)

Note: coefficients decay like 1𝑛𝑛

.

FOURIER COSINE AND SINE SERIES

Consider the Fourier coefficients formulae




−𝐿𝐿

, 𝑛𝑛 = 0,1,2 … , 𝑏𝑏𝑛𝑛 =1𝐿𝐿�𝑓𝑓(𝑥𝑥) sin



−𝐿𝐿

, 𝑛𝑛 = 1,2, …

If 𝑓𝑓(𝑥𝑥) is even for 𝑥𝑥 ∈ [−𝐿𝐿, 𝐿𝐿] then 𝑏𝑏𝑛𝑛 = 0 and




0

, 𝑛𝑛 = 0,1,2 …

Yields Fourier Cosine Series

𝑎𝑎0

2+ �𝑎𝑎𝑛𝑛 cos


∞

𝑛𝑛=1

For the periodic extension of 𝑓𝑓(𝑥𝑥). If 𝑓𝑓(𝑥𝑥) is odd for 𝑥𝑥 ∈ [−𝐿𝐿, 𝐿𝐿], then 𝑎𝑎𝑛𝑛 = 0 and

𝑏𝑏𝑛𝑛 =2𝐿𝐿�𝑓𝑓(𝑥𝑥) sin



0

, 𝑛𝑛 = 1,2, …

Which yields the Fourier Sine Series

�𝑏𝑏𝑛𝑛 sin𝑛𝑛𝜋𝜋𝑥𝑥𝐿𝐿

∞

𝑛𝑛=1

Now consider a function that is not necessarily even or odd on 𝑥𝑥 ∈ [−𝐿𝐿, 𝐿𝐿] and suppose we are only interested in the interval 𝑥𝑥 ∈ [0, 𝐿𝐿]. We have a choice of series representations.


46

EXAMPLE

Fourier Cosine is smoother so faster convergence. Choose based on convenience (e.g., rate of convergence or notational simplicity). How do eigenvalue problems and Fourier series arise in solving PDEs?

SOLVING PDES

EXAMPLE

Consider heat conduction along a perfectly insulated wire governed by the PDE

𝜕𝜕𝑢𝑢𝜕𝜕𝑑𝑑

= 𝛼𝛼2 𝜕𝜕2𝑢𝑢𝜕𝜕𝑥𝑥2 , 0 < 𝑥𝑥 < 𝐿𝐿, 𝑑𝑑 > 0

(1)

This is the “heat equation”, where 𝛼𝛼 is thermal diffusivity, with boundary conditions

𝑢𝑢(0, 𝑑𝑑) = 0, 𝑢𝑢(𝐿𝐿, 𝑑𝑑) = 0, 𝑑𝑑 > 0 (2)

And initial condition


47

𝑢𝑢(𝑥𝑥, 0) = 𝑓𝑓(𝑥𝑥), 0 ≤ 𝑥𝑥 ≤ 𝐿𝐿, 𝑑𝑑 = 0 (3)

This is an initial boundary value problem (IBVP).

SEPARATION OF VARIABLES

Attempt to find solution of the from 𝑢𝑢(𝑥𝑥, 𝑑𝑑) = 𝑋𝑋(𝑥𝑥)𝑇𝑇(𝑑𝑑). Substituting into (1) we find

𝑋𝑋′′

𝑋𝑋=

1𝛼𝛼2

𝑇𝑇′

𝑇𝑇

In order for this to hold (left side depends on 𝑥𝑥, right side depends on 𝑑𝑑) throughout the rectangle

𝑅𝑅 = {𝑥𝑥, 𝑑𝑑: 𝑥𝑥 ∈ (0, 𝐿𝐿), 𝑑𝑑 ∈ (0,∞)}

Both sides must be equal to a constant. Otherwise, equality would be violated by varying 𝑥𝑥 and holding 𝑑𝑑 constant (or vice-versa). Hence

𝑋𝑋′′

𝑋𝑋=

1𝛼𝛼2

𝑇𝑇′

𝑇𝑇= −𝜆𝜆

Note: −𝜆𝜆 for convenience. So we have

𝑋𝑋′′ + 𝜆𝜆𝑋𝑋 = 0, 0 < 𝑥𝑥 < 𝐿𝐿

𝑇𝑇′ + 𝛼𝛼2𝜆𝜆𝑇𝑇 = 0, 𝑑𝑑 > 0

Obtain two ODEs from original PDE.

To satisfy the boundary conditions, we require 𝑢𝑢(0, 𝑑𝑑) = 𝑋𝑋(0)𝑇𝑇(𝑑𝑑) = 0 → rejecting the trivial solution 𝑇𝑇(𝑑𝑑) = 0. We have 𝑋𝑋(0) = 0. Likewise 𝑢𝑢(𝐿𝐿, 𝑑𝑑) = 𝑋𝑋(𝐿𝐿)𝑇𝑇(𝑑𝑑) = 0 ⇒ 𝑋𝑋(𝐿𝐿) = 0. Hence, must solve the BVP

𝑋𝑋′′ + 𝜆𝜆𝑋𝑋 = 0

With boundary conditions

𝑋𝑋(0) = 𝑋𝑋(𝐿𝐿) = 0

This is an eigenvalue problem!

L EC T URE 1 8


48

THE HEAT EQUATION


= 𝛼𝛼2 𝜕𝜕2𝑢𝑢𝜕𝜕𝑥𝑥2

(1)

𝑢𝑢(0, 𝑑𝑑) = 𝑢𝑢(𝐿𝐿, 𝑑𝑑) = 0, 𝑑𝑑 > 0 (boundary conditions) (2)

𝑢𝑢(𝑥𝑥, 0) = 𝑓𝑓(𝑥𝑥), 0 ≤ 𝑥𝑥 ≤ 𝐿𝐿, 𝑑𝑑 = 0 (initial condition) (3)

Separation of variables

𝑢𝑢(𝑥𝑥, 𝑑𝑑) = 𝑋𝑋(𝑥𝑥)𝑇𝑇(𝑑𝑑) ⇒ � 𝑇𝑇′ + 𝛼𝛼2𝜆𝜆𝑇𝑇 = 0, 𝑑𝑑 > 0𝑋𝑋′′ + 𝜆𝜆𝑋𝑋 = 0, 0 ≤ 𝑥𝑥 ≤ 𝐿𝐿

�

With 𝑋𝑋(0) = 𝑋𝑋(𝐿𝐿) = 0. This defines our eigenvalue problem. The eigenvalues are 𝜆𝜆𝑛𝑛 = 𝑛𝑛2𝜋𝜋2

𝐿𝐿2 and corresponding

eigenfunctions 𝑋𝑋𝑛𝑛(𝑥𝑥) = sin 𝑛𝑛𝜋𝜋𝑥𝑥𝐿𝐿

for 𝑛𝑛 = 1,2, … So, solving for 𝑇𝑇(𝑑𝑑):

𝑇𝑇′ +𝑛𝑛2𝜋𝜋2𝛼𝛼2

𝐿𝐿2 𝑇𝑇 = 0 ⇒ 𝑇𝑇(𝑑𝑑) = 𝐶𝐶𝑒𝑒−𝑛𝑛2𝜋𝜋2𝛼𝛼2

𝐿𝐿2 𝑑𝑑

So

𝑢𝑢𝑛𝑛(𝑥𝑥, 𝑑𝑑) = 𝑒𝑒−𝑛𝑛2𝜋𝜋2𝛼𝛼2

𝐿𝐿2 𝑑𝑑 sin𝑛𝑛𝜋𝜋𝑥𝑥𝐿𝐿

, 𝑛𝑛 = 1,2,3, …

This satisfies PDE (1) and boundary conditions (2). To satisfy the initial condition (3), consider the infinite series obtained by superposition of 𝑢𝑢𝑛𝑛 (valid because PDE is linear)

𝑢𝑢(𝑥𝑥, 𝑑𝑑) = �𝑏𝑏𝑛𝑛𝑢𝑢𝑛𝑛(𝑥𝑥, 𝑑𝑑)∞

𝑛𝑛=1

(4)

Letting 𝑑𝑑 = 0 gives us a Fourier Sine Series for 𝑓𝑓(𝑥𝑥):

𝑢𝑢(𝑥𝑥, 0) = �𝑏𝑏𝑛𝑛 sin𝑛𝑛𝜋𝜋𝑥𝑥𝐿𝐿

∞

𝑛𝑛=1

= 𝑓𝑓(𝑥𝑥)

This implies the coefficients




0

(5)

Note: if working with a cosine series, would also have 𝑎𝑎0. Check that the infinite series PDE (1) solves, differentiate term-by-term:


= −�𝑛𝑛2𝜋𝜋2𝛼𝛼2

𝐿𝐿2 𝑏𝑏𝑛𝑛𝑒𝑒−�𝑛𝑛

2𝜋𝜋2𝛼𝛼2

𝐿𝐿2 �𝑑𝑑 sin𝑛𝑛𝜋𝜋𝑥𝑥𝐿𝐿

∞

𝑛𝑛=1

,𝜕𝜕2𝑢𝑢𝜕𝜕𝑥𝑥2 = −�

𝑛𝑛2𝜋𝜋2

𝐿𝐿2 𝑏𝑏𝑛𝑛𝑒𝑒−�𝑛𝑛

2𝜋𝜋2𝛼𝛼2


∞

𝑛𝑛=1


49

Hence the solution to the initial boundary value problem defined by (1), (2), (3) is given by (4) with coefficients (5); assuming termwise differentiation of Fourier series is justified.

EXAMPLE

𝑢𝑢𝑑𝑑 = 𝑢𝑢𝑥𝑥𝑥𝑥 , 𝑢𝑢(0, 𝑑𝑑) = 𝑢𝑢(𝜋𝜋, 𝑑𝑑) = 0, 𝑑𝑑 > 0

𝑓𝑓(𝑥𝑥) ≡ 𝑢𝑢(𝑥𝑥, 0) = 1, 0 ≤ 𝑥𝑥 ≤ 𝜋𝜋

𝑏𝑏𝑛𝑛 =2𝜋𝜋� sin𝑛𝑛𝑥𝑥 𝑑𝑑𝑥𝑥𝜋𝜋

0

= �−2𝑛𝑛𝜋𝜋

cos𝑛𝑛𝑥𝑥�0

𝜋𝜋

=2𝑛𝑛𝜋𝜋

(1 − cos𝑛𝑛𝜋𝜋) = �4𝑛𝑛𝜋𝜋

, 𝑛𝑛 odd

0, 𝑛𝑛 even�

𝑢𝑢(𝑥𝑥, 𝑑𝑑) = �4𝑛𝑛𝜋𝜋

𝑒𝑒−𝑛𝑛2𝑑𝑑 sin 𝑛𝑛𝑥𝑥∞

𝑛𝑛=1,3,5,…

As 𝑑𝑑 → large, it looks close to sin 𝑥𝑥.

REVISITING SOLUTION TO THE HEAT EQUATION

Note: for 𝑑𝑑 > 0, 𝑢𝑢 continuous and partials are continuous based on physical concept of diffusive interaction (at least in 𝑅𝑅 = {𝑥𝑥, 𝑑𝑑: 𝑥𝑥 ∈ (0, 𝐿𝐿), 𝑑𝑑 ∈ (0,∞)}).

Separation of variables ⇒

𝑢𝑢(𝑥𝑥, 𝑑𝑑) = �𝑏𝑏𝑛𝑛𝑒𝑒−�𝑛𝑛

2𝜋𝜋2𝜃𝜃2


∞

𝑛𝑛=1

(6)

With coefficients




0

JUSTIFICATION OF TERMWISE DIFFERENTIATION

Interpret (6) as the Fourier Sine Series

�𝐵𝐵𝑛𝑛(𝑑𝑑) sin𝑛𝑛𝜋𝜋𝑥𝑥𝐿𝐿

∞

𝑛𝑛=1

(7)

To differentiate it, need to have

𝜕𝜕𝑢𝑢(𝑥𝑥, 𝑑𝑑)𝜕𝜕𝑥𝑥

= �𝑛𝑛𝜋𝜋𝐿𝐿𝐵𝐵𝑛𝑛(𝑑𝑑) cos


∞

𝑛𝑛=1

By Theorem 3, we can differentiate this


50

𝑢𝑢𝑥𝑥𝑥𝑥 = −�𝑛𝑛2𝜋𝜋2

𝐿𝐿2 𝐵𝐵𝑛𝑛(𝑑𝑑) sin𝑛𝑛𝜋𝜋𝑥𝑥𝐿𝐿

∞

𝑛𝑛=1

We differentiated a Sine Series when the physical boundary conditions justified differentiation. We differentiated a Cosine Series when no boundary conditions were available or needed in order to justify differentiation.

By Theorem 4, termwise differentiation of (6) with respect to 𝑑𝑑 yields 𝜕𝜕𝑢𝑢𝜕𝜕𝑑𝑑

. Hence the series (6) solves the heat equation IBVP.

Need to check conditions to make sure we can take 𝑢𝑢𝜕𝜕2

𝜕𝜕𝑥𝑥2�� 𝑢𝑢𝑥𝑥𝑥𝑥 . Obtain Fourier Cosine Series for 𝑢𝑢𝑥𝑥 by termwise

differentiation of (7) to get

𝜕𝜕𝑢𝑢𝜕𝜕𝑥𝑥

= �𝑛𝑛𝜋𝜋𝐿𝐿𝐵𝐵𝑛𝑛(𝑑𝑑) cos


∞

𝑛𝑛=1

(8)

Because 𝑢𝑢(𝑥𝑥, 𝑑𝑑) satisfies homogeneous boundary conditions, 𝑢𝑢(0, 𝑑𝑑) = 𝑢𝑢(𝐿𝐿, 𝑑𝑑) = 0 for 𝑑𝑑 > 0, which are precisely the conditions required by Theorem 2 for differentiation of a Fourier Sine Series.

L EC T URE 1 9

COMPLEX FORM OF FOURIER SERIES

Consider Fourier Series


2+ ��𝑎𝑎𝑛𝑛 cos


+ 𝑏𝑏𝑛𝑛 sin𝑛𝑛𝜋𝜋𝑥𝑥𝐿𝐿�

∞

𝑛𝑛=1

With coefficients




−𝐿𝐿

(1a)




−𝐿𝐿

(1b)

Recall:

cos𝑛𝑛𝜋𝜋𝑥𝑥𝐿𝐿

=12�𝑒𝑒𝑖𝑖

𝑛𝑛𝜋𝜋𝑥𝑥𝐿𝐿 + 𝑒𝑒−𝑖𝑖

𝑛𝑛𝜋𝜋𝑥𝑥𝐿𝐿 � , sin


=12𝑖𝑖�𝑒𝑒𝑖𝑖

𝑛𝑛𝜋𝜋𝑥𝑥𝐿𝐿 − 𝑒𝑒−𝑖𝑖

𝑛𝑛𝜋𝜋𝑥𝑥𝐿𝐿 �


2+

12�(𝑎𝑎𝑛𝑛 − 𝑖𝑖 𝑏𝑏𝑛𝑛)𝑒𝑒𝑖𝑖


∞

𝑛𝑛=1

+12�(𝑎𝑎𝑛𝑛 + 𝑖𝑖 𝑏𝑏𝑛𝑛)𝑒𝑒−𝑖𝑖


∞

𝑛𝑛=1

Change the second series’ summation index from 𝑛𝑛 → −𝑛𝑛. It then becomes


51

12� (𝑎𝑎−𝑛𝑛 + 𝑖𝑖 𝑏𝑏−𝑛𝑛)𝑒𝑒𝑖𝑖


−∞

𝑛𝑛=−1

From (1a), (1b): 𝑎𝑎−𝑛𝑛 = 𝑎𝑎𝑛𝑛 since cos 𝑛𝑛𝜋𝜋𝑥𝑥𝐿𝐿

is even; 𝑏𝑏−𝑛𝑛 = −𝑏𝑏𝑛𝑛 since sin 𝑛𝑛𝜋𝜋𝑥𝑥𝐿𝐿

is odd. Hence defining

𝑐𝑐𝑛𝑛 =𝑎𝑎𝑛𝑛 − 𝑖𝑖𝑏𝑏𝑛𝑛

2

(1c)

We have

𝑓𝑓(𝑥𝑥) = � 𝑐𝑐𝑛𝑛𝑒𝑒𝑖𝑖𝑛𝑛𝜋𝜋𝑥𝑥𝐿𝐿

∞

𝑛𝑛=−∞

(2)

From (1)

𝑐𝑐𝑛𝑛 =1

2𝐿𝐿�𝑓𝑓(𝑥𝑥) �cos


− 𝑖𝑖 sin𝑛𝑛𝜋𝜋𝑥𝑥𝐿𝐿�𝑑𝑑𝑥𝑥

𝐿𝐿

−𝐿𝐿

Hence

𝑐𝑐𝑛𝑛 =1

2𝐿𝐿�𝑓𝑓(𝑥𝑥)𝑒𝑒−𝑖𝑖

𝑛𝑛𝜋𝜋𝑥𝑥𝐿𝐿 𝑑𝑑𝑥𝑥

𝐿𝐿

−𝐿𝐿

USEFUL TERMINOLOGY

𝑓𝑓(𝑑𝑑) – function of time with

− period 𝑇𝑇

− frequency 𝑓𝑓 = 1𝑇𝑇

− angular frequency 𝜔𝜔 = 2𝜋𝜋𝑓𝑓 = 2𝜋𝜋𝑇𝑇

𝑓𝑓(𝑥𝑥) – function of position with

− wavelength 𝜆𝜆

− wavenumber 𝑛𝑛 = 1𝜆𝜆

− angular wavenumber 𝑘𝑘 = 2𝜋𝜋𝑛𝑛 = 2𝜋𝜋𝜆𝜆

(note: usually called “wavenumber”)

Note: often abused – e.g. – “frequency” used to mean “wavenumber”

Question of the Day: A man is at the beach with his dog. His dog starts running around from him at 2𝑚𝑚𝑠𝑠

. The man has a whistle he blows to get the dog to come back. Each time the dog hears the whistle, he doubles his speed. Given that the speed of sound is ≅ 300𝑚𝑚

𝑠𝑠, how many whistles does the dog hear?

The answer is 15. The dog will hear whistles 4,8,16,32,64,128,256,512. At this point, he’s faster than the speed of sound, so he starts to catch up to the old whistles: 4,8,16,32,64,128,256.


52

REPRESENTING NON-PERIODIC FUNCTIONS

Consider the representation of a non-periodic function on 𝑥𝑥 ∈ (−∞,∞). Start with Fourier Series on 𝑥𝑥 ∈ (−𝐿𝐿, 𝐿𝐿) and let 𝐿𝐿 → ∞. By (2), the Fourier Series represents a function periodic on 𝑥𝑥 ∈ (−𝐿𝐿, 𝐿𝐿) as a sum of harmonics 𝑒𝑒𝑖𝑖𝑘𝑘𝑥𝑥 with angular wavenumbers 𝑘𝑘 = 𝑛𝑛𝜋𝜋

𝐿𝐿,𝑛𝑛 = −∞, … ,∞. The spectrum is discrete with gap Δ𝑘𝑘 = 𝜋𝜋

𝐿𝐿 between angular

wavenumbers. As 𝐿𝐿 → ∞,Δ𝑘𝑘 → 0. So spectrum becomes continuous and all 𝑘𝑘 are considered.

FOURIER TRANSFORM

Start with complex Fourier Series

𝑓𝑓(𝑥𝑥) = � 𝑐𝑐𝑛𝑛𝑒𝑒𝑖𝑖𝑛𝑛𝜋𝜋𝑥𝑥𝐿𝐿

∞

𝑛𝑛=−∞

, 𝑐𝑐𝑛𝑛 =1

2𝐿𝐿�𝑓𝑓(𝑥𝑥)𝑒𝑒−𝑖𝑖

𝑛𝑛𝜋𝜋𝑥𝑥𝐿𝐿 𝑑𝑑𝑥𝑥

𝐿𝐿

−𝐿𝐿

Let 𝑘𝑘 = 𝑛𝑛𝜋𝜋𝐿𝐿

and define 𝑐𝑐𝐿𝐿(𝑘𝑘) = 2𝐿𝐿𝑐𝑐𝑛𝑛 . So

𝑐𝑐𝐿𝐿(𝑘𝑘) = �𝑓𝑓(𝑥𝑥)𝑒𝑒−𝑖𝑖𝑘𝑘𝑥𝑥 𝑑𝑑𝑥𝑥𝐿𝐿

−𝐿𝐿

And

𝑓𝑓(𝑥𝑥) = � 𝑐𝑐𝑛𝑛𝑒𝑒𝑖𝑖𝑛𝑛𝜋𝜋𝑥𝑥𝐿𝐿 Δ𝑛𝑛

∞

𝑛𝑛=−∞

=1

2𝜋𝜋� 𝑐𝑐𝐿𝐿(𝑘𝑘)𝑒𝑒𝑖𝑖𝑘𝑘𝑥𝑥 Δ𝑘𝑘∞

𝐿𝐿𝑘𝑘𝜋𝜋 =−∞

Note: 𝑐𝑐𝑛𝑛 = 12𝐿𝐿𝑐𝑐𝐿𝐿(𝑘𝑘), Δ𝑛𝑛 = 𝐿𝐿Δ𝑘𝑘

𝜋𝜋. Let 𝐿𝐿 → ∞ and let 𝑐𝑐(𝑘𝑘) = lim𝐿𝐿→∞ 𝑐𝑐𝐿𝐿(𝑘𝑘)

𝑐𝑐(𝑘𝑘) = � 𝑓𝑓(𝑥𝑥)𝑒𝑒−𝑖𝑖𝑘𝑘𝑥𝑥 𝑑𝑑𝑥𝑥∞

−∞

Fourier Transform (3)


2𝜋𝜋� 𝑐𝑐(𝑘𝑘)𝑒𝑒𝑖𝑖𝑘𝑘𝑥𝑥 𝑑𝑑𝑘𝑘∞

−∞

Inverse Fourier Transform (4)

The Fourier Transform 𝑐𝑐(𝑘𝑘) exists if 𝑓𝑓(𝑥𝑥) is absolutely integrable �∫ |𝑓𝑓(𝑥𝑥)|𝑑𝑑𝑥𝑥∞−∞ < ∞� and piecewise continuous.

(e.g. – if 𝑓𝑓(𝑥𝑥) → 0 sufficiently fast as 𝑥𝑥 → ±∞).

SPECTRAL ANALYSIS USING THE FOURIER TRANSFORM

EXAMPLE


53

Note: if you can’t read the cut-off portion of the bottom left, it says “Fourier Transform”.

L EC T URE 2 0

Professor Note: Everything from this point on will appear in the final exam – crucial.

EVALUATING FOURIER TRANSFORMS

Denote the function 𝑓𝑓(𝑥𝑥) and the Fourier Transform 𝐹𝐹(𝑘𝑘) ≡ ℱ{𝑓𝑓(𝑥𝑥)} ≡ 𝑐𝑐(𝑘𝑘)

EXAMPLE 1: GAUSSIAN FUNCTION

𝑓𝑓(𝑥𝑥) = 𝑒𝑒−𝛼𝛼𝑥𝑥2 , 𝛼𝛼 > 0

𝐹𝐹(𝑘𝑘) = � 𝑒𝑒−𝛼𝛼𝑥𝑥2𝑒𝑒−𝑖𝑖𝑘𝑘𝑥𝑥 𝑑𝑑𝑥𝑥∞

−∞

= � 𝑒𝑒−𝛼𝛼𝑥𝑥2−𝑖𝑖𝑘𝑘𝑥𝑥 𝑑𝑑𝑥𝑥∞

−∞


54

Complete the square

−𝛼𝛼𝑥𝑥2 + 𝑖𝑖𝑘𝑘𝑥𝑥 = −�√𝛼𝛼𝑥𝑥 +𝑖𝑖𝑘𝑘

2√𝛼𝛼�

2

−𝑘𝑘2

4𝛼𝛼

𝐹𝐹(𝑘𝑘) = 𝑒𝑒−𝑘𝑘2

4𝛼𝛼 � 𝑒𝑒−�√𝛼𝛼𝑥𝑥+ 𝑖𝑖𝑘𝑘

2√𝛼𝛼�

2∞

−∞

𝑑𝑑𝑥𝑥

Let 𝑑𝑑 = √𝛼𝛼𝑥𝑥 + 𝑖𝑖 𝑘𝑘2√𝛼𝛼

so 𝑑𝑑𝑑𝑑 = √𝛼𝛼𝑑𝑑𝑥𝑥. So

𝐹𝐹(𝑘𝑘) =𝑒𝑒−

𝑘𝑘2

4𝛼𝛼

√𝛼𝛼� 𝑒𝑒−𝑑𝑑2𝑑𝑑𝑑𝑑

∞+𝑖𝑖 𝑘𝑘2√𝛼𝛼

−∞+𝑖𝑖 𝑘𝑘2√𝛼𝛼

� +𝐶𝐶1

� +𝐶𝐶2

� +𝐶𝐶3

� =𝐶𝐶4

0

Where

𝐹𝐹(𝑘𝑘) = �𝐶𝐶1

On 𝐶𝐶3,𝐶𝐶4, 𝐿𝐿 = 𝑘𝑘2√𝛼𝛼

: 𝑀𝑀 = �𝑒𝑒−𝑑𝑑2 � = �𝑒𝑒−�𝑥𝑥2−𝑑𝑑2+2𝑖𝑖𝑥𝑥𝑑𝑑 �� = �𝑒𝑒−�𝑥𝑥2−𝑑𝑑2�� → 0 as 𝑥𝑥 → ±∞. So

� =𝐶𝐶3

� =𝐶𝐶4

0

By M-L Bound. Hence

� +𝐶𝐶1

� =𝐶𝐶2

0 → 𝐹𝐹(𝑘𝑘) = −� =𝐶𝐶2

𝑒𝑒−𝑘𝑘2

4𝛼𝛼

√𝛼𝛼� 𝑒𝑒−𝑥𝑥2𝑑𝑑𝑥𝑥∞

−∞

= �𝜋𝜋𝛼𝛼𝑒𝑒−

𝑘𝑘2

4𝛼𝛼

For 𝑓𝑓(𝑥𝑥) = 𝑒𝑒−𝛼𝛼𝑥𝑥2. The Fourier Transform of a Gaussian is a Gaussian!


55

(Interpreting the picture. As 𝛼𝛼 decreases: 𝑓𝑓(𝑥𝑥) increases spread while 𝐹𝐹(𝑘𝑘) decreases spread, and vice-versa.) Small spread transforms to large spread and vice-versa. Also if 𝐹𝐹(𝑘𝑘) = 𝑒𝑒−𝛼𝛼𝑘𝑘2

then


√4𝜋𝜋𝑥𝑥𝑒𝑒−

𝑥𝑥2

4𝛼𝛼

EXAMPLE 2: DIRAC DELTA FUNCTION

𝑓𝑓(𝑥𝑥) = 𝛿𝛿(𝑥𝑥)

𝐹𝐹(𝑘𝑘) = � 𝛿𝛿(𝑥𝑥)𝑒𝑒−𝑖𝑖𝑘𝑘𝑥𝑥 𝑑𝑑𝑥𝑥∞

−∞

= 1

If we make a perfect spike, we get an equal amount of every component in the spectrum. All harmonics are equally strong in a delta function!

PROPERTIES OF THE FOURIER TRANSFORM

CONVOLUTION THEOREM

Suppose 𝐻𝐻(𝑘𝑘) = 𝐹𝐹(𝑘𝑘)𝐺𝐺(𝑘𝑘)

ℎ(𝑥𝑥) = ℱ−1{𝐹𝐹(𝑘𝑘)𝐺𝐺(𝑘𝑘)} =1

2𝜋𝜋� 𝐹𝐹(𝑘𝑘)𝐺𝐺(𝑘𝑘)𝑒𝑒𝑖𝑖𝑘𝑘𝑥𝑥 𝑑𝑑𝑘𝑘∞

−∞

=1

2𝜋𝜋� 𝐹𝐹(𝑘𝑘)𝑒𝑒𝑖𝑖𝑘𝑘𝑥𝑥 � � 𝑔𝑔(𝑥𝑥′)𝑒𝑒−𝑖𝑖𝑘𝑘𝑥𝑥′ 𝑑𝑑𝑥𝑥′

∞

−∞

� 𝑑𝑑𝑘𝑘∞

−∞

= � 𝑔𝑔(𝑥𝑥′)�1

2𝜋𝜋� 𝐹𝐹(𝑘𝑘)𝑒𝑒𝑖𝑖𝑘𝑘�𝑥𝑥−𝑥𝑥′ �𝑑𝑑𝑘𝑘∞

−∞

�𝑑𝑑𝑥𝑥′∞

−∞


56

= � 𝑔𝑔(𝑥𝑥′)𝑓𝑓(𝑥𝑥 − 𝑥𝑥′)𝑑𝑑𝑥𝑥′∞

−∞

Then

ℎ(𝑥𝑥) = 𝑓𝑓 ∗ 𝑔𝑔 = 𝑔𝑔 ∗ 𝑓𝑓 = � 𝑓𝑓(𝜁𝜁)𝑔𝑔(𝑥𝑥 − 𝜁𝜁)𝑑𝑑𝜁𝜁∞

−∞

DERIVATIVES

Suppose ℱ{𝑓𝑓′(𝑥𝑥)} exists with 𝑓𝑓(𝑥𝑥) → 0 as 𝑥𝑥 → ±∞. Then

ℱ{𝑓𝑓′(𝑥𝑥)} = � 𝑓𝑓′(𝑥𝑥)𝑒𝑒−𝑖𝑖𝑘𝑘𝑥𝑥 𝑑𝑑𝑥𝑥∞

−∞

= �𝑓𝑓(𝑥𝑥)𝑒𝑒−𝑖𝑖𝑘𝑘𝑥𝑥 �−∞∞ + 𝑖𝑖𝑘𝑘 � 𝑓𝑓(𝑥𝑥)𝑒𝑒−𝑖𝑖𝑘𝑘𝑥𝑥 𝑑𝑑𝑥𝑥

∞

−∞

Thus ℱ{𝑓𝑓′(𝑥𝑥)} = 𝑖𝑖𝑘𝑘ℱ{𝑓𝑓(𝑥𝑥)}. Similarly, if 𝑓𝑓(𝑥𝑥) has continuous derivatives through order 𝑛𝑛 that vanish as 𝑥𝑥 → ±∞:

ℱ�𝑓𝑓(𝑛𝑛)(𝑥𝑥)� = (𝑖𝑖𝑘𝑘)𝑛𝑛ℱ{𝑓𝑓(𝑥𝑥)}

LINEARITY

ℱ{𝑓𝑓(𝑥𝑥) + 𝑔𝑔(𝑥𝑥)} = ℱ{𝑓𝑓(𝑥𝑥)} + ℱ{𝑔𝑔(𝑥𝑥)}

SOLVING DIFFERENTIAL EQUATIONS ON AN INFINITE DOMAIN USING THE FOURIER TRANSFORM

EXAMPLE (HEAT EQUATION REVISITED)


= 𝛼𝛼2 𝜕𝜕2𝑢𝑢𝜕𝜕𝑥𝑥2 , −∞ < 𝑥𝑥 < ∞, 𝑑𝑑 > 0

𝑢𝑢(𝑥𝑥, 0) = 𝑓𝑓(𝑥𝑥) initial condition

𝑢𝑢(±∞, 𝑑𝑑) = 0 boundary conditions

Professors Note: Sometimes the boundary conditions are left unstated in books, etc… Fourier Transform the PDE in space

ℱ �𝜕𝜕𝑢𝑢𝜕𝜕𝑑𝑑� = �


𝑒𝑒−𝑖𝑖𝑘𝑘𝑥𝑥 𝑑𝑑𝑥𝑥∞

−∞

=𝜕𝜕𝜕𝜕𝑑𝑑

� 𝑢𝑢(𝑥𝑥, 𝑑𝑑)𝑒𝑒−𝑖𝑖𝑘𝑘𝑥𝑥 𝑑𝑑𝑥𝑥∞

−∞

=𝜕𝜕𝜕𝜕𝑑𝑑𝑢𝑢(𝑘𝑘, 𝑑𝑑)

ℱ �𝜕𝜕2𝑢𝑢𝜕𝜕𝑥𝑥2� = (𝑖𝑖𝑘𝑘)2𝑢𝑢(𝑘𝑘, 𝑑𝑑) = −𝑘𝑘2𝑢𝑢(𝑘𝑘, 𝑑𝑑)

𝜕𝜕𝜕𝜕𝑑𝑑𝑢𝑢(𝑘𝑘, 𝑑𝑑) = −𝛼𝛼2𝑘𝑘2𝑢𝑢(𝑘𝑘, 𝑑𝑑)


57

Solution is 𝑢𝑢(𝑘𝑘, 𝑑𝑑) = 𝑢𝑢(𝑘𝑘, 0)𝑒𝑒−𝛼𝛼2𝑘𝑘2𝑑𝑑 where

𝑢𝑢(𝑘𝑘, 0) = ℱ{𝑓𝑓(𝑥𝑥)} = � 𝑓𝑓(𝑥𝑥)𝑒𝑒−𝑖𝑖𝑘𝑘𝑥𝑥 𝑑𝑑𝑥𝑥∞

−∞

Hence 𝑢𝑢(𝑥𝑥, 𝑑𝑑) = ℱ−1�𝑢𝑢(𝑘𝑘, 0}𝑒𝑒−𝛼𝛼2𝑘𝑘2𝑑𝑑�. We have a Gaussian in 𝑘𝑘 with coefficient 𝛼𝛼2𝑑𝑑

ℱ−1�𝑒𝑒−𝛼𝛼2𝑘𝑘2𝑑𝑑� =1

√4𝜋𝜋𝛼𝛼2𝑑𝑑𝑒𝑒− 𝑥𝑥

2

4𝛼𝛼2𝑑𝑑

By the Convolution Theorem

𝑢𝑢(𝑥𝑥, 𝑑𝑑) = � 𝑓𝑓(𝜁𝜁)1

√4𝜋𝜋𝛼𝛼2𝑑𝑑𝑒𝑒− (𝑥𝑥−𝜁𝜁)2

4𝛼𝛼2𝑑𝑑 𝑑𝑑𝜁𝜁∞

−∞

Suppose we consider the diffusion of salt in a pipe from an initial point source 𝑓𝑓(𝑥𝑥) = 𝛿𝛿(𝑥𝑥 − 𝑥𝑥0) at 𝑥𝑥 = 𝑥𝑥0

𝑢𝑢(𝑥𝑥, 𝑑𝑑) =1

√4𝜋𝜋𝛼𝛼2𝑑𝑑𝑒𝑒− (𝑥𝑥−𝑥𝑥0)2

4𝛼𝛼2𝑑𝑑

𝑢𝑢(𝑥𝑥, 𝑑𝑑) > 0 for 𝑑𝑑 > 0; diffusion at infinite speed. (Picture interpretation: start with something that looks like an impulse at 𝑥𝑥 = 𝑥𝑥0. As 𝑑𝑑 → ∞, the spread of the Gaussian becomes infinitely large.)

L EC T URE 2 1

MORE GENERAL EIGENVALUE PROBLEMS

We have examined the eigenvalue problem 𝑢𝑢′′ + 𝜆𝜆𝑢𝑢 = 0, 𝑢𝑢(0) = 𝑢𝑢(𝐿𝐿) = 0 with 𝐿𝐿[𝑢𝑢] ≡ 𝑢𝑢′′ , i.e. 𝐿𝐿[𝑢𝑢] + 𝜆𝜆𝑢𝑢 = 0,

which arises in the solution of the heat equation using separation of variables. The eigenvalues are 𝜆𝜆𝑛𝑛 = 𝑛𝑛2𝜋𝜋2

𝐿𝐿2 with

eigenfunctions 𝜙𝜙𝑛𝑛 = sin 𝑛𝑛𝜋𝜋𝑥𝑥𝐿𝐿

for 𝑛𝑛 = 1,2, … We now examine the general case of 2nd order linear eigenvalue problem where 𝐿𝐿[𝑢𝑢] = 𝐴𝐴(𝑥𝑥)𝑢𝑢′′ + 𝐵𝐵(𝑥𝑥)𝑢𝑢′ + 𝐶𝐶(𝑥𝑥)𝑢𝑢 assuming 𝐴𝐴 ≠ 0 and 𝐴𝐴,𝐵𝐵,𝐶𝐶 are sufficiently smooth.


58

THE ADJOINT PROBLEM

Consider the “primal” operator

𝐿𝐿[𝑢𝑢] = 𝐴𝐴(𝑥𝑥)𝑢𝑢′′ + 𝐵𝐵(𝑥𝑥)𝑢𝑢′ + 𝐶𝐶(𝑥𝑥)𝑢𝑢, 𝑎𝑎 < 𝑥𝑥 < 𝑏𝑏 (1)

With boundary conditions

𝑢𝑢(𝑎𝑎) = 0, 𝑢𝑢(𝑏𝑏) = 0 (2)

Now consider inner product (𝑣𝑣, 𝐿𝐿[𝑢𝑢]) and integrate by parts to obtain

(𝑣𝑣, 𝐿𝐿[𝑢𝑢]) = (𝐿𝐿∗[𝑣𝑣],𝑢𝑢) + �𝐷𝐷(𝑥𝑥)|𝑎𝑎𝑏𝑏 (3)

Where 𝐿𝐿∗ defines “adjoint” or “dual” operator and 𝐷𝐷(𝑥𝑥) is boundary terms from integration by parts. Adjoint boundary conditions defined to make �𝐷𝐷(𝑥𝑥)|𝑎𝑎𝑏𝑏 = 0 for 𝑢𝑢 satisfying (2). Recall the inner product

(𝑢𝑢,𝑣𝑣) ≡ �𝑢𝑢(𝑥𝑥)𝑣𝑣(𝑥𝑥)𝑑𝑑𝑥𝑥𝑏𝑏

𝑎𝑎

For real functions. This yields the adjoint operator

𝐿𝐿∗[𝑣𝑣] = 𝐴𝐴𝑣𝑣′′ + (2𝐴𝐴′ − 𝐵𝐵)𝑣𝑣′ + (𝐴𝐴′′ − 𝐵𝐵′ + 𝐶𝐶)𝑣𝑣 (4)

And boundary terms

�𝐷𝐷(𝑥𝑥)|𝑎𝑎𝑏𝑏 = [𝐴𝐴(𝑢𝑢′𝑣𝑣 − 𝑣𝑣′𝑢𝑢) − (𝐴𝐴′ − 𝐵𝐵)𝑢𝑢𝑣𝑣|𝑎𝑎𝑏𝑏 (5)

Which vanish for any 𝑢𝑢 satisfying (2) if we choose the adjoint boundary conditions

𝑣𝑣(𝑎𝑎) = 0, 𝑣𝑣(𝑏𝑏) = 0 (6)

SELF-ADJOINTNESS

A problem is self-adjoint if 𝐿𝐿 = 𝐿𝐿∗ and the primal and adjoint boundary conditions are the same. This is equivalent to requiring that (3) becomes

(𝑣𝑣, 𝐿𝐿[𝑢𝑢]) − (𝐿𝐿[𝑣𝑣],𝑢𝑢) = 0 (7)

For any 𝑢𝑢(𝑥𝑥) and 𝑣𝑣(𝑥𝑥) satisfying the same primal/adjoint boundary conditions. Comparing (1) and (4), this requires 2𝐴𝐴′ − 𝐵𝐵 = 𝐵𝐵, 𝐴𝐴′′ − 𝐵𝐵′ + 𝐶𝐶 = 𝐶𝐶 or simply

𝐴𝐴′ = 𝐵𝐵 (8)

Note that 𝐴𝐴′ = 𝐵𝐵 ⇒ 𝐴𝐴′′ = 𝐵𝐵′ so 𝐿𝐿 = 𝐿𝐿∗ must have the form 𝐿𝐿[𝑢𝑢] = 𝐴𝐴𝑢𝑢′′ + 𝐴𝐴′𝑢𝑢′ + 𝐶𝐶𝑢𝑢 or equivalently

𝐿𝐿[𝑢𝑢] = (𝑝𝑝𝑢𝑢′)′ + 𝑞𝑞𝑢𝑢 (9)

With 𝑝𝑝(𝑥𝑥) ≡ 𝐴𝐴(𝑥𝑥) and 𝑞𝑞(𝑥𝑥) ≡ 𝑐𝑐(𝑥𝑥). This is the Sturm-Liouville Operator. In general, the second order linear operator (1) can be transformed to the self-adjoint Sturm-Liouville form (9) using an integrating factor.


59

REGULAR STURM-LIOUVILLE EIGENVALUE PROBLEMS

Consider the eigenvalue problem

(𝑝𝑝(𝑥𝑥)𝑢𝑢′)′ + 𝑞𝑞(𝑥𝑥)𝑢𝑢 + 𝜆𝜆𝑟𝑟(𝑥𝑥)𝑢𝑢 = 0, 𝑎𝑎 < 𝑥𝑥 < 𝑏𝑏 (10)

With the separated boundary conditions

𝛼𝛼1𝑢𝑢(𝑎𝑎) + 𝛼𝛼2𝑢𝑢′(𝑎𝑎) = 0, 𝑖𝑖1𝑢𝑢(𝑏𝑏) + 𝑖𝑖2𝑢𝑢′(𝑏𝑏) = 0 (11)

Where −∞ < 𝑎𝑎 < 𝑏𝑏 < ∞. 𝑝𝑝, 𝑝𝑝′ , 𝑞𝑞, 𝑟𝑟 are continuous with 𝑝𝑝(𝑥𝑥) > 0, 𝑟𝑟(𝑥𝑥) > 0 for 𝑎𝑎 ≤ 𝑥𝑥 ≤ 𝑏𝑏. 𝛼𝛼1,𝛼𝛼2,𝑖𝑖1,𝑖𝑖2 are real with

|𝛼𝛼1| + |𝛼𝛼2| > 0, |𝑖𝑖1| + |𝑖𝑖2| > 0

i.e. they can’t both be zero. For convenience, (10) may be written

𝐿𝐿[𝑢𝑢] + 𝜆𝜆𝑟𝑟(𝑥𝑥)𝑢𝑢 = 0 (12)

Where 𝐿𝐿 is given by (9). For this operator (9), we have already verified that 𝐿𝐿 = 𝐿𝐿∗. Also, (8) simplifies the boundary terms (5) to yield

�𝐷𝐷(𝑥𝑥)|𝑎𝑎𝑏𝑏 = [𝑝𝑝(𝑥𝑥)(𝑢𝑢′𝑣𝑣 − 𝑣𝑣′𝑢𝑢)|𝑎𝑎𝑏𝑏 (13)

Do these vanish for any 𝑢𝑢(𝑥𝑥) and 𝑣𝑣(𝑥𝑥) satisfying (11)? If 𝛼𝛼2 = 0 then 𝑢𝑢(𝑎𝑎) = 𝑣𝑣(𝑎𝑎) so 𝐷𝐷(𝑎𝑎) = 0. If 𝛼𝛼2 ≠ 0, then 𝑢𝑢′(𝑎𝑎) = − 𝛼𝛼1

𝛼𝛼2𝑢𝑢(𝑎𝑎), 𝑣𝑣′(𝑎𝑎) = −𝛼𝛼1

𝛼𝛼2𝑣𝑣(𝑎𝑎). So 𝑝𝑝(𝑥𝑥) �− 𝛼𝛼1

𝛼𝛼2𝑢𝑢(𝑎𝑎)𝑣𝑣(𝑎𝑎) + 𝛼𝛼1

𝛼𝛼2𝑣𝑣(𝑎𝑎)𝑢𝑢(𝑎𝑎)� = 0. The same result follows at

𝑥𝑥 = 𝑏𝑏. Hence (7) holds and the S-L problem (10),(11) is self-adjoint. Question to answer for next class: Why is this important?

L EC T URE 2 2

Note: This continues on from Lecture 21. From last time:

(𝑣𝑣, 𝐿𝐿[𝑢𝑢]) − (𝐿𝐿[𝑣𝑣],𝑢𝑢) = 0 (7)

𝐿𝐿[𝑢𝑢] + 𝜆𝜆𝑟𝑟(𝑥𝑥)𝑢𝑢 = 0 (12)

ORTHOGONALITY OF EIGENFUNCTIONS

THEOREM

Eigenfunctions of a regular S-L problem (10),(11) corresponding to different eigenvalues are orthogonal with respect to the weight function 𝑟𝑟(𝑥𝑥).

�𝑟𝑟(𝑥𝑥)𝜙𝜙𝑚𝑚 (𝑥𝑥)𝜙𝜙𝑛𝑛(𝑥𝑥)𝑑𝑑𝑥𝑥𝑏𝑏

𝑎𝑎

= 0 (14)


60

PROOF

Letting 𝑢𝑢 = 𝜙𝜙𝑚𝑚 , 𝑣𝑣 = 𝜙𝜙𝑛𝑛 , (7) gives: (𝜙𝜙𝑛𝑛 , 𝐿𝐿[𝜙𝜙𝑚𝑚 ]) − (𝐿𝐿[𝜙𝜙𝑛𝑛],𝜙𝜙𝑚𝑚) = 0. Using (12):

0 = (𝜙𝜙𝑛𝑛 ,−𝜆𝜆𝑚𝑚𝑟𝑟(𝑥𝑥)𝜙𝜙𝑚𝑚 ) − ( −𝜆𝜆𝑛𝑛𝑟𝑟(𝑥𝑥)𝜙𝜙𝑛𝑛 ,𝜙𝜙𝑚𝑚 ) = (𝜆𝜆𝑚𝑚 − 𝜆𝜆𝑛𝑛)�𝑟𝑟(𝑥𝑥)𝜙𝜙𝑚𝑚 (𝑥𝑥)𝜙𝜙𝑛𝑛(𝑥𝑥)𝑑𝑑𝑥𝑥𝑏𝑏

𝑎𝑎

𝜆𝜆𝑚𝑚 ≠ 𝜆𝜆𝑛𝑛 , so (14) follows.

THE EIGENVALUES ARE REAL

The inner product of two complex functions of a real variable is given by

(𝑢𝑢,𝑣𝑣) = �𝑢𝑢(𝑥𝑥)𝑣𝑣(𝑥𝑥)��𝑑𝑑𝑥𝑥𝑏𝑏

𝑎𝑎

(15)

Repeating the self-adjointness derivation using the definition shows that (7) remains valid.

THEOREM

The eigenvalues of a regular S-L problem (10),(11) are real.

PROOF

Consider possibly complex eigenvalue 𝜆𝜆 with eigenfunction 𝜙𝜙(𝑥𝑥). By (7) and (12), using (15)

�𝜆𝜆 − 𝜆𝜆̅� � 𝑟𝑟(𝑥𝑥)𝜙𝜙(𝑥𝑥)𝜙𝜙(𝑥𝑥)��𝑑𝑑𝑥𝑥𝑏𝑏

𝑎𝑎

= 0

So

�𝜆𝜆 − 𝜆𝜆̅� � 𝑟𝑟(𝑥𝑥)|𝜙𝜙(𝑥𝑥)|2𝑑𝑑𝑥𝑥𝑏𝑏

𝑎𝑎

= 0

This is positive because 𝑟𝑟(𝑥𝑥) > 0 hence 𝜆𝜆 − 𝜆𝜆̅ = 0 → 𝜆𝜆 is real. Note also, the eigenfunctions can be chosen to be real.

SIMPLICITY OF EIGENVALUES

THEOREM

Each eigenvalue of the regular S-L problem (10),(11) corresponds to one linearly independent eigenfunction.

PROOF

For a given 𝜆𝜆, consider two eigenfunctions 𝑢𝑢(𝑥𝑥) and 𝑣𝑣(𝑥𝑥) satisfying (10). The Wronskian is 𝑊𝑊(𝑢𝑢, 𝑣𝑣)(𝑥𝑥) =𝑢𝑢𝑣𝑣′ − 𝑢𝑢′𝑣𝑣 which, like 𝐷𝐷(𝑥𝑥) vanishes at 𝑥𝑥 = 𝑎𝑎. So 𝑢𝑢(𝑥𝑥), 𝑣𝑣(𝑥𝑥) are linearly dependent.


61

SEQUENCE OF EIGENVALUES

The regular S-L problem (10),(11) has an infinite sequence of eigenvalues 𝜆𝜆1 < 𝜆𝜆2 < ⋯ < 𝜆𝜆𝑛𝑛 < ⋯ with 𝜆𝜆𝑛𝑛 → ∞ as 𝑛𝑛 → ∞. The eigenfunction 𝜙𝜙𝑛𝑛 has 𝑛𝑛 − 1 zeros in 𝑎𝑎 < 𝑥𝑥 < 𝑏𝑏 (Proof: Birkhoff-Roots)

EXAMPLE

𝑑𝑑′′ + 𝜆𝜆𝑑𝑑 = 0, 𝑝𝑝(𝑥𝑥) = 1, 𝑞𝑞(𝑥𝑥) = 0, 𝑟𝑟(𝑥𝑥) = 1

𝛼𝛼1 = 𝑖𝑖1 = 1, 𝛼𝛼2 = 𝑖𝑖2 = 0, 𝑑𝑑(0) = 𝑑𝑑(𝐿𝐿) = 0

Real simple infinite sequence 𝜆𝜆𝑛𝑛 = 𝑛𝑛2𝜋𝜋2

𝐿𝐿2 , 𝜙𝜙𝑛𝑛 = sin 𝑛𝑛𝜋𝜋𝑥𝑥𝐿𝐿

is orthogonal, 𝑛𝑛 = 1,2, …

EIGENFUNCTION EXPANSIONS

These properties suggest the idea of generalizing Fourier Series to expansions of orthogonal eigenfunctions

𝑓𝑓(𝑥𝑥) = �𝐶𝐶𝑛𝑛𝜙𝜙𝑛𝑛(𝑥𝑥)∞

𝑛𝑛=1

, 𝑎𝑎 < 𝑥𝑥 < 𝑏𝑏 (16)

Find coefficients 𝐶𝐶𝑛𝑛 via orthogonality with respect to 𝑟𝑟(𝑥𝑥)

�𝑓𝑓(𝑥𝑥)𝜙𝜙𝑚𝑚 (𝑥𝑥)𝑟𝑟(𝑥𝑥)𝑑𝑑𝑥𝑥𝑏𝑏

𝑎𝑎

= �𝐶𝐶𝑛𝑛 �𝜙𝜙𝑚𝑚 (𝑥𝑥)𝜙𝜙𝑛𝑛(𝑥𝑥)𝑟𝑟(𝑥𝑥)𝑑𝑑𝑥𝑥𝑏𝑏

𝑎𝑎

∞

𝑛𝑛=1

= 𝐶𝐶𝑚𝑚 �𝜙𝜙𝑚𝑚 (𝑥𝑥)2𝑟𝑟(𝑥𝑥)𝑑𝑑𝑥𝑥𝑏𝑏

𝑎𝑎

Which is positive since 𝑟𝑟(𝑥𝑥) > 0.

𝐶𝐶𝑚𝑚 =∫ 𝑓𝑓(𝑥𝑥)𝜙𝜙𝑚𝑚 (𝑥𝑥)𝑟𝑟(𝑥𝑥)𝑑𝑑𝑥𝑥𝑏𝑏𝑎𝑎

∫ 𝜙𝜙𝑚𝑚 (𝑥𝑥)2𝑟𝑟(𝑥𝑥)𝑑𝑑𝑥𝑥𝑏𝑏𝑎𝑎

(17a)

ORTHONORMAL EIGENFUNCTIONS

For convenience, choose the arbitrary constant multiplying each 𝜙𝜙𝑚𝑚 so that

�𝜙𝜙𝑚𝑚(𝑥𝑥)2𝑟𝑟(𝑥𝑥)𝑑𝑑𝑥𝑥𝑏𝑏

𝑎𝑎

= 1

Then the Eigenfunctions form an orthonormal set with respect to 𝑟𝑟(𝑥𝑥) with

�𝜙𝜙𝑚𝑚(𝑥𝑥)𝜙𝜙𝑛𝑛(𝑥𝑥)𝑟𝑟(𝑥𝑥)𝑑𝑑𝑥𝑥𝑏𝑏

𝑎𝑎

= 𝛿𝛿𝑚𝑚𝑛𝑛


62

The coefficients of the eigenfunction expansion (16) becomes

𝐶𝐶𝑛𝑛 = �𝑓𝑓(𝑥𝑥)𝜙𝜙𝑛𝑛(𝑥𝑥)𝑟𝑟(𝑥𝑥)𝑑𝑑𝑥𝑥𝑏𝑏

𝑎𝑎

(17b)

CONVERGENCE OF EIGENFUNCTION EXPANSIONS

Let 𝜙𝜙1,𝜙𝜙2, … be the eigenfunctions of the regular S-L problem (10),(11) and let 𝑓𝑓(𝑥𝑥) be piecewise smooth on

𝑎𝑎 ≤ 𝑥𝑥 ≤ 𝑏𝑏. The series (16) with coefficients (17) converges to �𝑓𝑓�𝑥𝑥+�+𝑓𝑓(𝑥𝑥−)�

2 at each point on 𝑎𝑎 < 𝑥𝑥 < 𝑏𝑏.

L EC T URE 2 3

Note: this continues on from Lecture 22.

NONHOMOGENEOUS BOUNDARY VALUE PROBLEMS

Consider the nonhomogeneous differential equation

𝐿𝐿[𝑢𝑢] + 𝜇𝜇𝑟𝑟(𝑥𝑥)𝑢𝑢 = 𝑓𝑓(𝑥𝑥), 𝑎𝑎 < 𝑥𝑥 < 𝑏𝑏 (18)

Where 𝐿𝐿 is the Sturm-Liouville operator (9) and 𝜇𝜇 is a real constant. Furthermore, consider the homogeneous separated boundary conditions (11).

Attempt solution by expanding in the eigenfunctions of the corresponding S-L eigenvalue problem (10) given by

𝐿𝐿[𝑢𝑢] + 𝜆𝜆𝑟𝑟(𝑥𝑥) = 0 (19)

With boundary conditions (11). Denote the eigenvalues 𝜆𝜆1 < 𝜆𝜆2 < ⋯ (ordered) and orthonormal eigenfunctions 𝜙𝜙1,𝜙𝜙2, … Then let

𝑢𝑢(𝑥𝑥) = �𝑐𝑐𝑛𝑛𝜙𝜙𝑛𝑛(𝑥𝑥)∞

𝑛𝑛=1

(20)

Where 𝑐𝑐𝑛𝑛 is unknown. By (19)

𝐿𝐿[𝑢𝑢] = �𝑐𝑐𝑛𝑛𝐿𝐿[𝜙𝜙𝑛𝑛(𝑥𝑥)]∞

𝑛𝑛=1

= −�𝑐𝑐𝑛𝑛𝜆𝜆𝑛𝑛𝜙𝜙𝑛𝑛(𝑥𝑥)𝑟𝑟(𝑥𝑥)∞

𝑛𝑛=1

(21)

So substituting into (18)

−𝑟𝑟(𝑥𝑥)�𝜆𝜆𝑛𝑛𝑐𝑐𝑛𝑛𝜙𝜙𝑛𝑛(𝑥𝑥)∞

𝑛𝑛=1

+ 𝜇𝜇𝑟𝑟(𝑥𝑥)�𝑐𝑐𝑛𝑛𝜙𝜙𝑛𝑛(𝑥𝑥)∞

𝑛𝑛=1

= 𝑓𝑓(𝑥𝑥) (22)

To expand 𝑓𝑓(𝑥𝑥) so as to obtain a uniform factor of 𝑟𝑟(𝑥𝑥) in all terms we consider


63

𝑓𝑓(𝑥𝑥) =𝑟𝑟(𝑥𝑥)𝑓𝑓(𝑥𝑥)𝑟𝑟(𝑥𝑥) , (recall 𝑟𝑟(𝑥𝑥) > 0)

(23)

And expand

𝑓𝑓(𝑥𝑥)𝑟𝑟(𝑥𝑥) = �𝑑𝑑𝑛𝑛𝜙𝜙𝑛𝑛(𝑥𝑥)

∞

𝑛𝑛=1

With

𝑑𝑑𝑛𝑛 = �𝑓𝑓(𝑥𝑥)𝑟𝑟(𝑥𝑥) 𝜙𝜙𝑛𝑛

(𝑥𝑥)𝑟𝑟(𝑥𝑥)𝑑𝑑𝑥𝑥𝑏𝑏

𝑎𝑎

Or

𝑑𝑑𝑛𝑛 = �𝑓𝑓(𝑥𝑥)𝜙𝜙𝑛𝑛(𝑥𝑥)𝑑𝑑𝑥𝑥𝑏𝑏

𝑎𝑎

(25)

Substituting (23) and (24) into (22) we have

𝑟𝑟(𝑥𝑥) = �[(𝜇𝜇 − 𝜆𝜆𝑛𝑛)𝑐𝑐𝑛𝑛 − 𝑑𝑑𝑛𝑛 ]𝜙𝜙𝑛𝑛(𝑥𝑥)∞

𝑛𝑛=1

= 0 (26)

Taking the inner product with 𝜙𝜙𝑚𝑚 (𝑥𝑥). Orthogonality (with respect to 𝑟𝑟(𝑥𝑥)) gives

(𝜇𝜇 − 𝜆𝜆𝑛𝑛)𝑐𝑐𝑛𝑛 − 𝑑𝑑𝑛𝑛 = 0, 𝑛𝑛 = 1,2, … (27)

CASE 1

If 𝜇𝜇 is not an eigenvalue of the S-L problem (19),(11) (i.e. 𝜇𝜇 ≠ 𝜆𝜆𝑛𝑛 for 𝑛𝑛 = 1,2, …) then

𝑐𝑐𝑛𝑛 =𝑑𝑑𝑛𝑛

𝜇𝜇 − 𝜆𝜆𝑛𝑛

(28)

And the solution to the nonhomogeneous boundary value problem (18),(11) is

𝑢𝑢(𝑥𝑥) = �𝑑𝑑𝑛𝑛

𝜇𝜇 − 𝜆𝜆𝑛𝑛𝜙𝜙𝑛𝑛(𝑥𝑥)

∞

𝑛𝑛=1

(29)

CASE 2

If 𝜇𝜇 is an eigenvalue (say 𝜇𝜇 = 𝜆𝜆𝑚𝑚 for some 𝑚𝑚) then there are two possibilities

a) If 𝑑𝑑𝑚𝑚 ≠ 0, then (27) has no solution and the nonhomogeneous problem (18),(11) has no solution

b) If 𝑑𝑑𝑚𝑚 = 0, then (27) shows that 𝑐𝑐𝑚𝑚 is arbitrary. Hence the nonhomogeneous problem (18),(11) has infinitely many solutions

𝑢𝑢(𝑥𝑥) = 𝐾𝐾𝜙𝜙𝑚𝑚 (𝑥𝑥) + �𝑑𝑑𝑚𝑚


∞

𝑛𝑛=1𝑛𝑛≠𝑚𝑚

(30)


64

Containing an arbitrary multiple of the eigenfunction 𝜙𝜙𝑚𝑚(𝑥𝑥). From (25) we see that 𝑑𝑑𝑚𝑚 = 0 when

�𝑓𝑓(𝑥𝑥)𝜙𝜙𝑚𝑚 (𝑥𝑥)𝑑𝑑𝑥𝑥𝑏𝑏

𝑎𝑎

= 0 (31)

Hence if 𝜇𝜇 = 𝜆𝜆𝑚𝑚 , the nonhomogeneous boundary value problem (18),(11) can be solved (nonuniquely) only if 𝑓𝑓(𝑥𝑥) is orthogonal to the corresponding eigenfunction 𝜙𝜙𝑚𝑚(𝑥𝑥). To summarize, we have something called the Fredholm Alternative.

THE FREDHOLM ALTERNATIVE

For a given 𝜇𝜇 and continuous 𝑓𝑓(𝑥𝑥) either the nonhomogeneous problem (18),(11) has a unique solution or else the corresponding homogeneous problem has a non trivial solution. What it is saying: If 𝜇𝜇 is not an eigenvalue 𝜆𝜆𝑚𝑚 of the corresponding S-L eigenvalue problem (19),(11) (i.e. the eigenfunction 𝜙𝜙𝑚𝑚 (𝑥𝑥) corresponding to 𝜇𝜇 = 𝜆𝜆𝑚𝑚 , in which case the nonhomogeneous problem (18),(11) has no solution or an infinite number of solutions.

L EC T URE 2 4

Note: this continues from the past two or so lectures.. yes, I realize it’s a long thing on S-L problems and solving BVPs…

GREEN’S FUNCTIONS FOR BVP

Consider the nonhomogeneous problem (18),(11) when 𝜇𝜇 is not an eigenvalue of the corresponding eigenvalue problem (19),(11). Hence, the unique solution (29) is given by

𝑢𝑢(𝑥𝑥) = �𝑑𝑑𝑛𝑛


∞

𝑛𝑛=1

(32)

Where 𝜆𝜆𝑛𝑛 are the eigenvalues of (19),(11) and 𝜙𝜙𝑛𝑛 are the corresponding orthonormal eigenfunctions. By (25)

𝑑𝑑𝑛𝑛 = �𝑓𝑓(𝑥𝑥)𝜙𝜙𝑛𝑛(𝑥𝑥)𝑑𝑑𝑥𝑥𝑏𝑏

𝑎𝑎

So (32) may be written

𝑢𝑢(𝑥𝑥) = �𝐺𝐺(𝑥𝑥, 𝜁𝜁)𝑓𝑓(𝜁𝜁)𝑑𝑑𝜁𝜁𝑏𝑏

𝑎𝑎

(33)

With Green’s function

𝐺𝐺(𝑥𝑥, 𝜁𝜁) = �𝜙𝜙𝑛𝑛(𝑥𝑥)𝜙𝜙𝑛𝑛(𝜁𝜁)𝜇𝜇 − 𝜆𝜆𝑛𝑛

∞

𝑛𝑛=1

(34)


65

Note: the Green’s function does not exist if 𝜇𝜇 = 𝜆𝜆𝑛𝑛 for some 𝑛𝑛 = 1,2, … What are the properties of the Green’s function? Consider letting 𝑓𝑓(𝑥𝑥) = 𝛿𝛿(𝑥𝑥 − 𝑥𝑥∗) so by (33):

𝑢𝑢(𝑥𝑥) = �𝐺𝐺(𝑥𝑥, 𝜁𝜁)𝛿𝛿(𝜁𝜁 − 𝑥𝑥∗)𝑑𝑑𝜁𝜁𝑏𝑏

𝑎𝑎

= 𝐺𝐺(𝑥𝑥, 𝑥𝑥∗)

So 𝐺𝐺(𝑥𝑥, 𝑥𝑥∗) is the response of the solution at 𝑥𝑥 to an impulse at 𝑥𝑥∗. Hence, by (18) the Green’s function satisfies

𝐿𝐿[𝐺𝐺(𝑥𝑥, 𝜁𝜁)] + 𝜇𝜇𝑟𝑟(𝑥𝑥)𝐺𝐺(𝑥𝑥, 𝜁𝜁) = 𝛿𝛿(𝑥𝑥 − 𝜁𝜁), 𝑎𝑎 < 𝑥𝑥 < 𝑏𝑏 (35)

With homogeneous boundary conditions (11)

𝛼𝛼1𝐺𝐺(𝑎𝑎, 𝜁𝜁) + 𝛼𝛼2𝜕𝜕𝐺𝐺𝜕𝜕𝑥𝑥

(𝑎𝑎, 𝜁𝜁) = 0 (36a)

𝑖𝑖1𝐺𝐺(𝑏𝑏, 𝜁𝜁) + 𝑖𝑖2𝜕𝜕𝐺𝐺𝜕𝜕𝑥𝑥

(𝑏𝑏, 𝜁𝜁) = 0 (36b)

RECIPROCITY

From (34) we note the reciprocity relationship

𝐺𝐺(𝑥𝑥, 𝜁𝜁) = 𝐺𝐺(𝜁𝜁, 𝑥𝑥) (37)

The response at 𝑥𝑥 due to an impulse at 𝜁𝜁 is identical to the response at 𝜁𝜁 due to an impulse at 𝑥𝑥. Physically, this is a remarkable observation! Note: Reciprocity did not hold for the IVP Green’s function. What is the behavior of 𝐺𝐺(𝑥𝑥, 𝜁𝜁) at 𝑥𝑥 = 𝜁𝜁? (35) and (9) give

𝑑𝑑𝑑𝑑𝑥𝑥

�𝑝𝑝(𝑥𝑥)𝜕𝜕𝜕𝜕𝑥𝑥

𝐺𝐺(𝑥𝑥, 𝜁𝜁)� + [𝑞𝑞(𝑥𝑥) + 𝜇𝜇𝑟𝑟(𝑥𝑥)]𝐺𝐺(𝑥𝑥, 𝜁𝜁) = 𝛿𝛿(𝑥𝑥 − 𝜁𝜁)

As with IVP, expect �𝜕𝜕2𝐺𝐺𝜕𝜕𝑥𝑥2�𝑥𝑥=𝜁𝜁

to balance 𝛿𝛿(𝑥𝑥 − 𝜁𝜁). So �𝜕𝜕𝐺𝐺𝜕𝜕𝑥𝑥�𝑥𝑥=𝜁𝜁

has jump and �𝐺𝐺|𝑥𝑥=𝜁𝜁 is continuous. Integrate the ODE for

𝜁𝜁 − 𝜖𝜖 to 𝜁𝜁 + 𝜖𝜖.


𝐺𝐺(𝑥𝑥, 𝜁𝜁)�𝜁𝜁−𝜖𝜖

𝜁𝜁+𝜖𝜖

+ � [[𝑞𝑞(𝑥𝑥) + 𝜇𝜇𝑟𝑟(𝑥𝑥)]𝐺𝐺(𝑥𝑥, 𝜁𝜁)𝑑𝑑𝑥𝑥

𝜁𝜁+𝜖𝜖

𝜁𝜁−𝜖𝜖

= 1

Noting that the integral vanishes as 𝜖𝜖 → 0. So


𝐺𝐺(𝑥𝑥, 𝜁𝜁)�𝜁𝜁−

𝜁𝜁+

=1

𝑝𝑝(𝜁𝜁) (38)

And also

�𝐺𝐺(𝑥𝑥, 𝜁𝜁)|𝜁𝜁−𝜁𝜁+

= 0 (39)

How to find 𝐺𝐺(𝑥𝑥, 𝜁𝜁) assuming it exists (i.e. when 𝜇𝜇 is not an eigenvalue).

METHOD 1


66

Find 𝜆𝜆𝑛𝑛 and 𝜙𝜙𝑛𝑛 for S-L eigenvalue problem (19),(11) and use (34) or equivalently solve (35),(36) using eigenfunction expansion.

METHOD 2

Alternatively, solve (35),(36) by solving homogeneous problems in 𝑎𝑎 ≤ 𝑥𝑥 < 𝜁𝜁 and 𝜁𝜁 > 𝑥𝑥 ≥ 𝑏𝑏 and matching solutions at 𝑥𝑥 = 𝜁𝜁 using (38) and (39).

Consider homogeneous solutions 𝑢𝑢1(𝑥𝑥) satisfying boundary condition at 𝑥𝑥 = 𝑎𝑎 and 𝑢𝑢2(𝑥𝑥) satisfying boundary condition at 𝑥𝑥 = 𝑏𝑏. If 𝜇𝜇 is not an eigenvalue of (19),(11) then neither 𝑢𝑢1 or 𝑢𝑢2 satisfies both boundary conditions (11). So it can be shown that 𝑊𝑊(𝑢𝑢1,𝑢𝑢2)(𝑥𝑥) ≠ 0, 𝑎𝑎 ≤ 𝑥𝑥 ≤ 𝑏𝑏 and 𝑢𝑢1(𝑥𝑥) and 𝑢𝑢2(𝑥𝑥) are linearly independent. However, by (36a) 𝐺𝐺(𝑥𝑥, 𝜁𝜁) and 𝑢𝑢1(𝑥𝑥) do satisfy the same boundary condition at 𝑥𝑥 = 𝑎𝑎 so 𝑊𝑊(𝑢𝑢1,𝐺𝐺)(𝑎𝑎) = 0. They also satisfy the same homogeneous differential equation for 𝑎𝑎 < 𝑥𝑥 < 𝜁𝜁 so they are linearly dependent.

𝐺𝐺(𝑥𝑥, 𝜁𝜁) = 𝑐𝑐1(𝜁𝜁)𝑢𝑢1(𝑥𝑥), 𝑎𝑎 ≤ 𝑥𝑥 < 𝜁𝜁

Likewise,

𝐺𝐺(𝑥𝑥, 𝜁𝜁) = 𝑐𝑐2(𝜁𝜁)𝑢𝑢2(𝑥𝑥), 𝜁𝜁 < 𝑥𝑥 ≤ 𝑏𝑏

By the jump conditions (38) and (39)

𝑐𝑐2(𝜁𝜁)𝑢𝑢2(𝜁𝜁) − 𝑐𝑐1(𝜁𝜁)𝑢𝑢1(𝜁𝜁) = 0

𝑐𝑐2(𝜁𝜁)𝑢𝑢2′ (𝜁𝜁) − 𝑐𝑐1(𝜁𝜁)𝑢𝑢1

′ (𝜁𝜁) =1

𝑝𝑝(𝜁𝜁)

So

𝑐𝑐1(𝜁𝜁) =𝑢𝑢2(𝜁𝜁)

𝑝𝑝(𝜁𝜁)𝑊𝑊(𝜁𝜁) , 𝑐𝑐2(𝜁𝜁) =𝑢𝑢1(𝜁𝜁)

𝑝𝑝(𝜁𝜁)𝑊𝑊(𝜁𝜁)

Where 𝑊𝑊(𝜁𝜁) ≡ 𝑊𝑊(𝑢𝑢1,𝑢𝑢2)(𝜁𝜁) ≠ 0. By (41). Hence

𝐺𝐺(𝑥𝑥, 𝜁𝜁) =

⎩⎪⎨

⎪⎧𝑢𝑢1(𝑥𝑥)𝑢𝑢2(𝜁𝜁)𝑝𝑝(𝜁𝜁)𝑊𝑊(𝜁𝜁) , 𝑎𝑎 ≤ 𝑥𝑥 ≤ 𝜁𝜁

𝑢𝑢2(𝑥𝑥)𝑢𝑢1(𝜁𝜁)𝑝𝑝(𝜁𝜁)𝑊𝑊(𝜁𝜁) , 𝜁𝜁 ≤ 𝑥𝑥 ≤ 𝑏𝑏

�

NOTE

a) Equivalent to (34) despite different appearance

b) Evidently, 𝑝𝑝(𝜁𝜁)𝑊𝑊(𝜁𝜁) is independent of 𝜁𝜁 in order for reciprocity to hold

L EC T URE 2 5

EXAMPLE


67

Consider a string under tension 𝑇𝑇 carrying a vertical load 𝐹𝐹(𝑥𝑥)

𝑇𝑇𝑑𝑑2𝑢𝑢𝑑𝑑𝑥𝑥2 = 𝐹𝐹(𝑥𝑥), 𝑢𝑢(0) = 0, 𝑢𝑢(𝐿𝐿) = 0

Then

𝐿𝐿[𝑢𝑢] ≡𝑑𝑑2𝑢𝑢𝑑𝑑𝑥𝑥2 = 𝑓𝑓(𝑥𝑥) ≡

𝐹𝐹(𝑥𝑥)𝑇𝑇

Note: this is a nonhomogeneous S-L BVP with 𝑝𝑝(𝑥𝑥) = 1, 𝑞𝑞(𝑥𝑥) = 0, 𝛾𝛾(𝑥𝑥) = 1, 𝜇𝜇 = 0. We have two different methods of finding Green’s function 𝐺𝐺(𝑥𝑥, 𝜁𝜁) to solve this.

METHOD 1:

Solve

𝑑𝑑2

𝑑𝑑𝑥𝑥2 𝐺𝐺(𝑥𝑥, 𝜁𝜁) = 𝛿𝛿(𝑥𝑥 − 𝜁𝜁) (1a)

With boundary conditions 𝐺𝐺(0, 𝜁𝜁) = 𝐺𝐺(𝐿𝐿, 𝜁𝜁) = 0 by eigenfunction expansion. Consider corresponding S-L eigenvalue problem

𝐿𝐿[𝜙𝜙] + 𝜆𝜆𝜙𝜙 = 0, 𝜙𝜙(0) = 𝜙𝜙(𝐿𝐿) = 0 (1b)

Eigenvalues are 𝜆𝜆𝑛𝑛 = 𝑛𝑛2𝜋𝜋2

𝐿𝐿2 ,𝑛𝑛 = 1,2, … with orthonormal eigenfunctions 𝜙𝜙𝑛𝑛(𝑥𝑥) = �2𝐿𝐿

sin 𝑛𝑛𝜋𝜋𝑥𝑥𝐿𝐿

. Let

𝐺𝐺(𝑥𝑥, 𝜁𝜁) = �𝑐𝑐𝑛𝑛(𝜁𝜁)𝜙𝜙𝑛𝑛(𝑥𝑥)∞

𝑛𝑛=1

(2)

And

𝛿𝛿(𝑥𝑥 − 𝜁𝜁) = �𝑑𝑑𝑛𝑛(𝜁𝜁)𝜙𝜙𝑛𝑛(𝑥𝑥)∞

𝑛𝑛=1

(3)

So

�2𝐿𝐿� sin

𝑚𝑚𝜋𝜋𝑥𝑥𝐿𝐿

𝛿𝛿(𝑥𝑥 − 𝜁𝜁)𝑑𝑑𝑥𝑥𝐿𝐿

0

=2𝐿𝐿�𝑑𝑑𝑛𝑛(𝜁𝜁) sin



∞

𝑛𝑛=1

And by orthogonality 𝑑𝑑𝑚𝑚 (𝜁𝜁) = �2𝐿𝐿

sin 𝑚𝑚𝜋𝜋𝜁𝜁𝐿𝐿

. Now

𝐿𝐿[𝐺𝐺(𝑥𝑥, 𝜁𝜁)] = �𝑐𝑐𝑛𝑛(𝜁𝜁)𝐿𝐿[𝜙𝜙𝑛𝑛(𝑥𝑥)]∞

𝑛𝑛=1

= −�𝑐𝑐𝑛𝑛(𝜁𝜁)𝜆𝜆𝑛𝑛𝜙𝜙𝑛𝑛(𝑥𝑥)∞

𝑛𝑛=1


68

By (1b). So substituting into (1a)

�−𝑛𝑛2𝜋𝜋2

𝐿𝐿2 𝑐𝑐𝑛𝑛�2𝐿𝐿


∞

𝑛𝑛=1

= �2𝐿𝐿

sin𝑛𝑛𝜋𝜋𝜁𝜁𝐿𝐿


∞

𝑛𝑛=1

Note: this came up during class, and I thought it was a good question. We can’t just ignore the summations and equate coefficients, we need to justify ‘picking off individual terms’ which is why we talk about orthogonality and

inner products. Taking the inner product with respect to 𝜙𝜙𝑚𝑚(𝑥𝑥), by orthogonality we have −𝑛𝑛2𝜋𝜋2

𝐿𝐿2 �2𝐿𝐿𝑐𝑐𝑛𝑛 = 2

𝐿𝐿sin 𝑛𝑛𝜋𝜋𝜁𝜁

𝐿𝐿

sor 𝑐𝑐𝑛𝑛 = −�2𝐿𝐿

𝐿𝐿2

𝑛𝑛2𝜋𝜋2 sin 𝑛𝑛𝜋𝜋𝜁𝜁𝐿𝐿

, 𝑛𝑛 = 1,2, … and by (2)

𝐺𝐺(𝑥𝑥, 𝜁𝜁) = −2𝐿𝐿�

𝐿𝐿2

𝑛𝑛2𝜋𝜋2 sin𝑛𝑛𝜋𝜋𝜁𝜁𝐿𝐿


∞

𝑛𝑛=1

(4)

Professor Pierce made a point of saying that we can’t study the behavior of the solution from an infinite series of sine terms. Also, to get a particular solution, we take the inner product with 𝑓𝑓(𝑥𝑥).

METHOD 2:

Solve homogeneous problems

𝑑𝑑2𝐺𝐺𝑑𝑑𝑥𝑥2 = 0

(5)

In intervals 𝑎𝑎 ≤ 𝑥𝑥 < 𝜁𝜁 and 𝜁𝜁 ≤ 𝑥𝑥 < 𝑏𝑏 while satisfying boundary conditions

𝐺𝐺(0, 𝜁𝜁) = 0 (6a)

𝐺𝐺(𝐿𝐿, 𝜁𝜁) = 0 (6b)

And jump conditions

�𝑑𝑑𝐺𝐺𝑑𝑑𝑥𝑥�𝜁𝜁−

𝜁𝜁+

=1

𝑝𝑝(𝜁𝜁) = 1 (7)

�𝐺𝐺|𝜁𝜁−𝜁𝜁+

= 0 (8)

For 𝑎𝑎 ≤ 𝑥𝑥 < 𝜁𝜁: 𝐺𝐺(𝑥𝑥, 𝜁𝜁) = 𝑐𝑐1(𝜁𝜁)𝑥𝑥 satisfies (5) and (6a). For 𝜁𝜁 < 𝑥𝑥 ≤ 𝑏𝑏: 𝐺𝐺(𝑥𝑥, 𝜁𝜁) = 𝑐𝑐2(𝜁𝜁)(𝑥𝑥 − 𝐿𝐿) satisfies (5) and (6b). To satisfy (7) and (8), we must have both 𝑐𝑐2(𝜁𝜁)(𝜁𝜁 − 𝐿𝐿) − 𝑐𝑐1(𝜁𝜁)𝜁𝜁 = 0 and 𝑐𝑐2(𝜁𝜁) − 𝑐𝑐1(𝜁𝜁) = 1 This gives us 𝑐𝑐1(𝜁𝜁) = 𝜁𝜁−𝐿𝐿

𝐿𝐿 and 𝑐𝑐2(𝜁𝜁) = 𝜁𝜁

𝐿𝐿, so

𝐺𝐺(𝑥𝑥, 𝜁𝜁) = �

−𝑥𝑥(𝐿𝐿 − 𝜁𝜁)𝐿𝐿

, 0 ≤ 𝑥𝑥 ≤ 𝜁𝜁

𝜁𝜁(𝐿𝐿 − 𝑥𝑥)𝐿𝐿

, 𝜁𝜁 ≤ 𝑥𝑥 ≤ 𝐿𝐿

�

(9)

NOTES

a) We see that 𝐺𝐺(𝑥𝑥, 𝜁𝜁) has the following appearance:


69

b) (4) is the Fourier Sine series for (9) (i.e. the eigenfunction expansion).

L EC T URE 2 6

SINGULAR STURM-LIOUVILLE EIGENVALUE PROBLEMS

Again consider

𝐿𝐿[𝑢𝑢] + 𝜆𝜆𝑟𝑟(𝑥𝑥)𝑢𝑢 = 0, 𝑎𝑎 < 𝑥𝑥 < 𝑏𝑏 (1)

With S-L operator

𝐿𝐿[𝑢𝑢] = (𝑝𝑝(𝑥𝑥)𝑢𝑢′)′ + 𝑞𝑞(𝑥𝑥)𝑢𝑢 (2)

A singular S-L problem arises if the “regular” assumptions on 𝑝𝑝, 𝑞𝑞, 𝑟𝑟 are valid for 𝑎𝑎 < 𝑥𝑥 < 𝑏𝑏, but one or more of the following apply:

• 𝑎𝑎 → −∞, 𝑏𝑏 → ∞ (not a finite domain)

• 𝑝𝑝 → 0, 𝑟𝑟 → 0, 𝑞𝑞 unbounded (at 𝑎𝑎 or 𝑏𝑏)

Whatever the singularity, appropriate boundary conditions are chosen to ensure that the problem remains self-adjoint. Recall that integration by parts for the S-L operator yields

(𝑣𝑣, 𝐿𝐿[𝑢𝑢]) − (𝐿𝐿[𝑣𝑣],𝑢𝑢) = �𝐷𝐷(𝑥𝑥)|𝑎𝑎𝑏𝑏 (3)

�𝐷𝐷(𝑥𝑥)|𝑎𝑎𝑏𝑏 = [𝑝𝑝(𝑥𝑥)(𝑢𝑢′𝑣𝑣 − 𝑣𝑣′𝑢𝑢)|𝑎𝑎𝑏𝑏 (4)

Use separated boundary conditions such that any 𝑢𝑢(𝑥𝑥) and 𝑣𝑣(𝑥𝑥) satisfying the boundary conditions give 𝐷𝐷(𝑎𝑎) = 0 and 𝐷𝐷(𝑏𝑏) = 0. For example, if 𝑝𝑝(𝑥𝑥) → 0 as 𝑥𝑥 → 𝑎𝑎, then need 𝑢𝑢′𝑣𝑣 − 𝑣𝑣′𝑢𝑢 to be bounded. Then appropriate boundary conditions are 𝑢𝑢(𝑥𝑥), 𝑢𝑢′(𝑥𝑥) bounded as 𝑥𝑥 → 𝑎𝑎. 𝑖𝑖1𝑢𝑢(𝑏𝑏) + 𝑖𝑖2𝑢𝑢′(𝑏𝑏) = 0.


70

PROPERTIES OF SINGULAR STURM-LIOUVILLE EIGENVALUE PROBLEMS

a) Singular S-L problems remain self-adjoint with (𝑣𝑣, 𝐿𝐿[𝑢𝑢]) − (𝐿𝐿[𝑣𝑣],𝑢𝑢) = 0 for any 𝑢𝑢(𝑥𝑥) and 𝑣𝑣(𝑥𝑥) satisfying the same primal/adjoint boundary conditions.

b) Therefore, by similar arguments to the regular case, square integrable eigenfunctions corresponding to different eigenvalues, IE:

�𝜙𝜙2(𝑥𝑥)𝑟𝑟(𝑥𝑥)𝑑𝑑𝑥𝑥𝑏𝑏

𝑎𝑎

< ∞

Are orthogonal and the eigenvalues are real.

c) Suppose at least one of the boundary conditions is of the “regular S-L” type (say at 𝑥𝑥 = 0). Consider eigenvalue 𝜆𝜆 and eigenfunctions 𝑢𝑢 and 𝑣𝑣. Since 𝐷𝐷(𝑏𝑏) → 0 and 𝐷𝐷 = 𝑝𝑝(𝑥𝑥). 𝑊𝑊(𝑢𝑢, 𝑣𝑣) = 0, 𝑢𝑢 and 𝑣𝑣 are not linearly independent solutions (𝑢𝑢, 𝑣𝑣 linearly dependent).

d) Singular problems may have continuous, discrete, or mixed spectra.

HEAT EQUATION IN A DISK WITH RADIAL SYMMETRY


= 𝛼𝛼2 1𝑟𝑟𝜕𝜕𝜕𝜕𝑟𝑟�𝑟𝑟𝜕𝜕𝑢𝑢𝜕𝜕𝑟𝑟�

(3)

Boundary conditions

𝑢𝑢(𝑏𝑏, 𝑑𝑑) = 0 (4)

Initial conditions

𝑢𝑢(𝑟𝑟, 0) = 𝑓𝑓(𝑑𝑑) (5)

Using separation of variables: 𝑢𝑢(𝑟𝑟, 𝑑𝑑) = 𝑅𝑅(𝑟𝑟)𝑇𝑇(𝑑𝑑). Substituting into (3),

𝑅𝑅𝑇𝑇′ = 𝛼𝛼2 �𝑇𝑇𝑅𝑅′′ +𝑇𝑇𝑟𝑟𝑅𝑅′� ,

1𝛼𝛼2

𝑇𝑇′

𝑇𝑇=

1𝑅𝑅�𝑅𝑅′′ +

𝑅𝑅′

𝑟𝑟�

Both sides equal to a constant – 𝜆𝜆. We obtain

𝑇𝑇′ + 𝛼𝛼2𝜆𝜆𝑇𝑇 = 0 (6)

𝑑𝑑𝑑𝑑𝑟𝑟

𝑟𝑟 �𝑑𝑑𝑅𝑅𝑑𝑑𝑟𝑟� + 𝜆𝜆𝑟𝑟𝑅𝑅 = 0

(7)

This is a singular S-L problem. By (4):

𝑅𝑅(𝑏𝑏) = 0 (8)

Note 𝑝𝑝(𝑟𝑟) → 0, also weight→ 0 as 𝑟𝑟 → 0.

𝑅𝑅,𝑅𝑅′ bounded as 𝑟𝑟 → 0 (9)

To show 𝜆𝜆 > 0; multiply (7) by 𝑅𝑅 and integrate by parts


71

−�𝑟𝑟 �𝑑𝑑𝑅𝑅𝑑𝑑𝑟𝑟� 𝑑𝑑𝑟𝑟

𝑏𝑏

0

+ �𝑟𝑟𝑅𝑅𝑑𝑑𝑅𝑅𝑑𝑑𝑟𝑟�

0

𝑏𝑏

+ 𝜆𝜆�𝑟𝑟𝑅𝑅2𝑑𝑑𝑟𝑟𝑏𝑏

0

= 0

𝜆𝜆 =− �𝑟𝑟𝑅𝑅 𝑑𝑑𝑅𝑅𝑑𝑑𝑟𝑟�0

𝑏𝑏+ ∫ 𝑟𝑟 �𝑑𝑑𝑅𝑅𝑑𝑑𝑟𝑟�

2𝑑𝑑𝑟𝑟𝑏𝑏

0

∫ 𝑟𝑟𝑅𝑅2𝑑𝑑𝑟𝑟𝑏𝑏0

> 0

Let 𝑥𝑥 = √𝜆𝜆 𝑟𝑟 valid since 𝜆𝜆 > 0. (7) becomes 𝑥𝑥2 𝑑𝑑2𝑅𝑅𝑑𝑑𝑥𝑥2 + 𝑥𝑥 𝑑𝑑𝑅𝑅

𝑑𝑑𝑥𝑥+ (𝑥𝑥2 −𝑚𝑚2)𝑅𝑅 = 0 based on equation order 𝑚𝑚 = 0. By

the Method of Frobenius, the general solution is 𝑅𝑅(𝑥𝑥) = 𝑐𝑐1𝐽𝐽0(𝑥𝑥) + 𝑐𝑐2𝑌𝑌0(𝑥𝑥) where 𝐽𝐽0(𝑥𝑥),𝑌𝑌0(𝑥𝑥) are the Bessel functions of first and second kind, both of order 0. Other examples of “special functions” defined only as solutions to differential equations. We note that 𝐽𝐽0(𝑥𝑥), 𝐽𝐽0′ (𝑥𝑥) bounded as 𝑥𝑥 → 0 and |𝑌𝑌0(𝑥𝑥)| → ∞ as 𝑥𝑥 → 0. Also 𝐽𝐽0(𝑥𝑥) has infinitely many zeros for 𝑥𝑥 > 0 → 𝑌𝑌0 fails the conditions so set 𝑐𝑐2 = 0. The infinite sequence of eigenvalues is defined by 𝐽𝐽0��𝜆𝜆𝑛𝑛𝑏𝑏� = 0, 𝑛𝑛 = 1,2, … with eigenfunctions 𝜙𝜙𝑛𝑛(𝑥𝑥) = 𝐽𝐽0�√𝜆𝜆𝑟𝑟� that are orthogonal with respect to weight function 𝑟𝑟. By (6), 𝑇𝑇(𝑑𝑑) = 𝐶𝐶𝑒𝑒−𝛼𝛼2𝜆𝜆𝑑𝑑 . By superposition,

𝑢𝑢(𝑟𝑟, 𝑑𝑑) = �𝑎𝑎𝑛𝑛𝐽𝐽0��𝜆𝜆𝑛𝑛𝑟𝑟�𝑒𝑒−𝛼𝛼2𝜆𝜆𝑛𝑛 𝑑𝑑

∞

𝑛𝑛=1

Determine 𝑎𝑎𝑛𝑛 to satisfy the initial condition

𝑓𝑓(𝑟𝑟) = �𝑎𝑎𝑛𝑛𝐽𝐽0��𝜆𝜆𝑛𝑛𝑟𝑟�∞

𝑛𝑛=1

By orthogonality with respect to 𝑟𝑟

𝑎𝑎𝑛𝑛 =∫ 𝑓𝑓(𝑟𝑟)𝐽𝐽0��𝜆𝜆𝑛𝑛𝑟𝑟�𝑟𝑟𝑑𝑑𝑟𝑟𝑏𝑏

0

∫ 𝐽𝐽02��𝜆𝜆𝑛𝑛𝑟𝑟�𝑟𝑟𝑑𝑑𝑟𝑟𝑏𝑏

0

L EC T URE 2 7

ACM FINAL REVIEW

Topics:

- IVP: Euler equations, series solutions around regular and ordinary points, Frobenius method

- Representation of Functions: Fourier Series, Sine/Cosine series, Fourier Transform

- BVP: Linear 2nd order BVPs, eigenvalue problems, separation of variables for PDEs, adjoint operator and self-adjoint linear operators, S-L eigenvalue problems, Green’s function for BVPs

EXAMPLE: (E-‘STUFF’ FOR A BVP)


72

𝑑𝑑′′ +𝜆𝜆

(𝑥𝑥 + 1)2 𝑑𝑑 = 0, 0 < 𝑥𝑥 < 𝜋𝜋, 𝑑𝑑(0) = 𝑑𝑑(𝜋𝜋) = 0

Guess solution of form 𝑑𝑑 = (1 + 𝑥𝑥)𝜇𝜇 . 𝜇𝜇 may be complex: 𝜇𝜇 = 𝑎𝑎 + 𝑖𝑖𝑏𝑏.

𝑑𝑑′ = (𝑎𝑎 + 𝑖𝑖𝑏𝑏)(1 + 𝑥𝑥)𝑎𝑎+𝑖𝑖𝑏𝑏−1

𝑑𝑑′′ = (𝑎𝑎 + 𝑖𝑖𝑏𝑏)(𝑎𝑎 + 𝑖𝑖𝑏𝑏 − 1)(1 + 𝑥𝑥)𝑎𝑎+𝑖𝑖𝑏𝑏−2

Plugging into the ODE

(𝑎𝑎 + 𝑖𝑖𝑏𝑏)(𝑎𝑎 + 𝑖𝑖𝑏𝑏 − 1)(1 + 𝑥𝑥)𝑎𝑎+𝑖𝑖𝑏𝑏−2 +𝜆𝜆

(1 + 𝑥𝑥)2 (1 + 𝑥𝑥)𝑎𝑎+𝑖𝑖𝑏𝑏 = 0

(𝑎𝑎 + 𝑖𝑖𝑏𝑏)(𝑎𝑎 + 𝑖𝑖𝑏𝑏 − 1) + 𝜆𝜆 = 𝑏𝑏2 − 𝑏𝑏 − 𝑎𝑎2 + 2𝑎𝑎𝑏𝑏𝑖𝑖 + 𝑎𝑎𝑖𝑖 + 𝜆𝜆 = 0

In a S-L eigenvalue problem, 𝜆𝜆 real and positive

𝑖𝑖(2𝑎𝑎𝑏𝑏 − 𝑎𝑎) = 0 → 𝑏𝑏 = ±�𝜆𝜆 −14

, 𝑏𝑏2 − 𝑏𝑏 − 𝑎𝑎2 + 𝜆𝜆 = 0 → 𝑎𝑎 =12

→ 𝑑𝑑(𝑥𝑥) = (1 + 𝑥𝑥)𝑎𝑎+𝑖𝑖𝑏𝑏 = (1 + 𝑥𝑥)12(1 + 𝑥𝑥)±�𝜆𝜆−1

4

(1 + 𝑥𝑥)𝑖𝑖𝑎𝑎 = 𝑒𝑒𝑖𝑖𝑎𝑎 log (1+𝑥𝑥) = cos(𝑎𝑎 log(1 + 𝑥𝑥)) + 𝑖𝑖 sin(𝑎𝑎 log(1 + 𝑥𝑥))

𝑑𝑑1 = (1 + 𝑥𝑥)12 �cos��𝜆𝜆 −

14

log(1 + 𝑥𝑥)� + 𝑖𝑖 sin��𝜆𝜆 −14

log(1 + 𝑥𝑥)��

𝑑𝑑2 = (1 + 𝑥𝑥)12 �cos��𝜆𝜆 −

14

log(1 + 𝑥𝑥)� − 𝑖𝑖 sin��𝜆𝜆 −14

log(1 + 𝑥𝑥)��

𝑑𝑑 = 𝑐𝑐1𝑑𝑑1 + 𝑐𝑐2𝑑𝑑2

= (𝑐𝑐1 + 𝑐𝑐2)(1 + 𝑥𝑥)12 cos��𝜆𝜆 −

14

log(1 + 𝑥𝑥)� + 𝑑𝑑2(1 + 𝑥𝑥)12 sin��𝜆𝜆 −

14

log(1 + 𝑥𝑥)�

Let 𝑑𝑑1 = 𝑐𝑐1 + 𝑐𝑐2 so

𝑑𝑑(0) = 0 → 𝑑𝑑1 = 0, 𝑑𝑑(𝜋𝜋) = 0 → ��𝜆𝜆 −14

log(1 + 𝜋𝜋)� = 𝑛𝑛𝜋𝜋


(log(1 + 𝜋𝜋))2 +14

, 𝜙𝜙 = (1 + 𝑥𝑥)12 sin �

𝑛𝑛𝜋𝜋log(1 + 𝜋𝜋) log(1 + 𝑥𝑥)�

EXAMPLE (FOURIER TRANSFORM)

Use Fourier transform to find an integral representation for the solution of


73

𝑑𝑑′′ − 𝑥𝑥𝑑𝑑 = 0

lim𝑥𝑥→±∞

𝑑𝑑(𝑥𝑥) = 0, 𝑑𝑑(0) =1𝜋𝜋� cos�

𝑘𝑘3

3�𝑑𝑑𝑘𝑘

∞

0

𝐹𝐹[𝑥𝑥𝑑𝑑(𝑥𝑥)] = � 𝑥𝑥𝑑𝑑(𝑥𝑥)𝑒𝑒−𝑖𝑖𝑘𝑘𝑥𝑥 𝑑𝑑𝑥𝑥∞

−∞

= −1𝑖𝑖�

𝑑𝑑𝑑𝑑𝑘𝑘

𝑑𝑑(𝑘𝑘)𝑒𝑒−𝑖𝑖𝑘𝑘𝑥𝑥 𝑑𝑑𝑥𝑥∞

−∞

= 𝑖𝑖𝑑𝑑𝑑𝑑𝑘𝑘

� 𝑑𝑑(𝑥𝑥)𝑒𝑒−𝑖𝑖𝑘𝑘𝑥𝑥 𝑑𝑑𝑥𝑥∞

−∞

=𝑑𝑑𝑑𝑑𝑘𝑘

𝑌𝑌(𝑘𝑘)

Taking transform:

−𝑘𝑘2𝑌𝑌 − 𝑖𝑖𝑌𝑌′𝑘𝑘 = 0 →𝑌𝑌′

𝑌𝑌= −

𝑘𝑘2

𝑖𝑖= 𝑖𝑖𝑘𝑘2 → 𝑌𝑌(𝑘𝑘) = 𝑌𝑌(0)𝑒𝑒𝑖𝑖

𝑘𝑘3

3

Applying inversion:

𝑑𝑑(𝑥𝑥) =𝑐𝑐

2𝜋𝜋� 𝑒𝑒𝑖𝑖

𝑘𝑘3

3 𝑒𝑒𝑖𝑖𝑘𝑘𝑥𝑥 𝑑𝑑𝑘𝑘∞

−∞

=𝑐𝑐

2𝜋𝜋� 𝑒𝑒𝑖𝑖�

𝑘𝑘3

3 +𝑘𝑘𝑥𝑥�𝑑𝑑𝑘𝑘∞

−∞

=𝑐𝑐

2𝜋𝜋� �cos�

𝑘𝑘3

3+ 𝑘𝑘𝑥𝑥� + 𝑖𝑖 sin �

𝑘𝑘3

3+ 𝑘𝑘𝑥𝑥�� 𝑑𝑑𝑘𝑘

∞

−∞

By odd/even orthogonality

𝑐𝑐2𝜋𝜋

2� cos�𝑘𝑘3

3+ 𝑘𝑘𝑥𝑥� 𝑑𝑑𝑘𝑘

∞

0

From initial condition, 𝑐𝑐 = 1.

EXAMPLE (HEAT EQUATION)

𝑢𝑢𝑑𝑑 = 𝛼𝛼2𝑢𝑢𝑥𝑥𝑥𝑥 + 𝑐𝑐𝑢𝑢𝑥𝑥

With 𝑥𝑥 on real line −∞ < 𝑥𝑥 < ∞ and our initial condition 𝑢𝑢(𝑥𝑥, 0) = 𝑓𝑓(𝑥𝑥). Take fourier transform

𝑑𝑑𝑑𝑑𝑑𝑑𝑈𝑈(𝑘𝑘, 𝑑𝑑) = �−(𝑘𝑘𝛼𝛼)2 + 𝑖𝑖𝑘𝑘𝑐𝑐𝑈𝑈(𝑘𝑘, 𝑑𝑑)� → 𝑈𝑈(𝑘𝑘, 𝑑𝑑) = 𝐹𝐹(𝑘𝑘)𝑒𝑒−(𝑘𝑘𝛼𝛼 )2+𝑖𝑖𝑘𝑘𝑐𝑐𝑑𝑑

Invert using convolution and shifts theorem

𝑢𝑢(𝑥𝑥, 𝑑𝑑) = � 𝑓𝑓(𝜁𝜁)𝑒𝑒(−𝑐𝑐𝑑𝑑+𝑥𝑥−𝜁𝜁)2

(4𝛼𝛼2𝑑𝑑) 𝑑𝑑𝜁𝜁∞

−∞

EXAMPLE (S-L PROBLEM)

Consider

𝑥𝑥2𝑑𝑑′′ + 𝑥𝑥𝑑𝑑′ + 𝑑𝑑 = 𝜇𝜇𝑑𝑑, 𝑑𝑑(1) = 𝑑𝑑(2) = 0

a) Write as regular Sturm-Liouville Problem


74

ℎ(𝑥𝑥) = 𝑥𝑥 is integrating factor, so

[𝑥𝑥𝑑𝑑′ ] +1𝑥𝑥𝜆𝜆𝑑𝑑 = 0

b) Without solving, we can bound the eigenvalues using Rayleigh quotient

𝜆𝜆 = 0 → 𝜇𝜇 < 1

c) Find eigenvalues, eigenfunctions

Euler equation with initial equation

𝛼𝛼2 + 1 − 𝜇𝜇 = 0

→ 𝜆𝜆𝑛𝑛 = 1 − �𝑛𝑛𝜋𝜋

log 2�

2, 𝜙𝜙𝑛𝑛 = sin �

𝑛𝑛𝜋𝜋 log 𝑥𝑥log 2

� , 𝑛𝑛 = 1,2, …

d) Find Green’s Function for

𝑥𝑥𝑢𝑢′′ + 𝑢𝑢′ = 0, lim𝑥𝑥→0

𝑢𝑢(𝑥𝑥) < ∞

Two homogeneous solutions

𝑢𝑢1 = 1, 𝑢𝑢2 = log 𝑥𝑥

� 𝑑𝑑𝑑𝑑𝑥𝑥

𝐺𝐺�𝜁𝜁−

𝜁𝜁+

=1𝜁𝜁→ 𝐺𝐺(𝑥𝑥, 𝜁𝜁) = �log(𝜁𝜁) , 0 ≤ 𝑥𝑥 < 𝜁𝜁

log(𝑥𝑥) , 𝜁𝜁 < 𝑥𝑥 ≤ 1�

Documents

ACM95b Notes