Sturm-Liouville Theory

ECE 6382

Notes are from D. R. Wilton, Dept. of ECE

1

David R. Jackson

Fall 2019

Notes 18

Second-Order Linear Differential Equations (SOLDE)

2

0 1 22( ) ( ) ( ) ( )

( ) 0

d y dyp x p x p x y f xdx

f x

dx

=

+ + =

A SOLDE has the form

If , the equation is said to be "homogeneous."

The inhomogeneous equation can be solved once we know the solution to the homogeneo

( ) ( ) 0 ( )( ) ( ) 0 (

y a y by a y b

= =′ ′= =

Dirichlet

Neumann)

us equation using the method of Green's functions (discussed later).

Boundary conditions (BC) are usually of the form

2

Sturm-Liouville Form

1

0

1 1

0 0

0 1 2

( )( )

0( ) ( )( ) ( )1

20

( ) ( ) ( ) ( )

( )( )

( ) ( ) (( )

x

x x

p tdt

p t

p t p tdt dt

p t p t

p x y p x y p x y f x

ew xp x

p xe y e y p x wp x

′′ ′+ + =

∫= −

∫ ∫′′ ′− − +

If we multiply the general differential equation

by the integrating factor we have :

1

0

1

0

( )( )

2

( )( )

2

) ( ) ( )

( ) ( ) ( ) ( )

( )

1 ( )( ) ( ) ( ) ( )( )

( ) , ( ) ( )

x

x

p tdt

p t

p tdt

p t

x y w x f x

d e y p x w x y w x f xdx

w x

d dy xP x Q x y x f xw x dx dx

P x e Q x p x

=

∫ ′⇒ − + =

− + =

∫≡ ≡

Dividing this result by yields

where 3

Sturm-Liouville Operator

4

1 ( ) ( )( )

d dP x Q xw x dx dx

≡ − +

u f=

1 ( )( ) ( ) ( ) ( )( )

d dy xP x Q x y x f xw x dx dx

− + =

This is called the Sturm-Liouville or self-adjoint form of the differential equation:

or (using u instead of y):

Where is the (self-adjoint) “Sturm-Liouville” operator:

Note: The operator is assumed to be real here (w, P, Q are real). The solution u

does not have to be real (because f is allowed to be complex).

Inner Product Definition

*

(

, ( ) ( ) ( )

( )

)

b

a

u v

w

u x

x

v x w x d

w x

x< > ≡ ∫

We define an as

where is called a function.

Although the weight function is arbitrary, we will choose it to be the same as the function in the Stur

inner product

weight

m-Liouville equation. This will give us the nice "self - adjoint" properties as we will see.

5

An inner product between two functions is defined:

6

., ,u v u v =< > = < > †For a Sturm -Liouville operator so

, ,u v u v< > = < >†

1 ( ) ( )( )

d dP x Q xw x dx dx

= = − + †

Note: Self-adjoint operators have nice properties for eigenvalue problems,

which is discussed a little later.

The adjoint operator † is defined from

The Adjoint Problem

(proof given next)

Hence, the Sturm-Liouville operator is said to be self-adjoint:

Proof of Self-Adjoint Property

( ) ( )

( )

*

*

*

:

, ( ) ( ) ( )

1( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

b

ab

a

b

a

u v

u v v x w x u x dx

d dv x w x P x Q x u x dxw x dx dx

d dv x P x Q x w x u x dxdx dx

< > =

= − +

= − +

∫

∫

∫

Consider the inner product between the two functions and

The first term i

** *

**

( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )

( ) ( ) (( ) ( ) ( )

bb b

a aab

a

d du x du x du x dv xv x P x dx v x P x P x dxdx dx dx dx dx

du x dv x duv x P x P xdx dx

− = − +

= − +

∫ ∫

nside the square brackets is first integrated by parts, twice :

* **

)

( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( )

b

a

b b

aa

x dxdx

du x dv x d dv xv x P x P x u x u x P x dxdx dx dx dx

= − + −

∫

∫

7

( )

* **

*

( ) ( ) ( ), ( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( )

b b

aab

a

du x dv x d dv xu v v x P x P x u x u x P x dxdx dx dx dx

v x Q x u x w x dx

< > = − + −

+

∫

∫

8

Proof of Self-Adjoint Property (cont.)

Hence we have:

**

*

( ) ( ), ( ) ( ) ( ) ( )

1( ) ( ) ( ) ( ) ( )( )

b

ab

a

du x dv xu v v x P x P x u xdx dx

d du x w x P x Q x v x dxw x dx dx

< > = − +

+ − + ∫

or

Multiply and divide by w(x)

**

*

( ) ( ), ( ) ( ) ( ) ( )

1( ) ( ) ( ) ( ) ( )( )

b

ab

a

du x dv xu v v x P x P x u xdx dx

d du x P x Q x v x w x dxw x dx dx

< > = − +

+ − + ∫

9 ( )

( , ) 0, ,

b

aJ u vu v u v

=

⇒ < > = < > proof complete

Proof of Self-Adjoint Property (cont.)

( )( )

**v v=

=

Note :

real operator*

* ( ) ( )( , ) ( ) ( ) ( )du x dv xJ u v P x v x u xdx dx

≡ − +

This can be written as:

, ( , ) ,b

au v J u v u v< > = + < >

where

From boundary conditions we have:

Eigenvalue Problems

, 1,2,n nλ λ= =

10

u uλ=

We often encounter an eigenvalue problem of the form

(The operator can be the Sturm-Liouville operator, or any other operator here.)

The eigenvalue problem (with boundary conditions) is usually only satisfied for specific eigenvalues:

For each distinct eigenvalue, there corresponds an eigenfunction u = un satisfying the eigenvalue equation.

Property of Eigenvalues

11

The eigenvalues corresponding to a self-adjoint operator are real.

( ) * *

, ,

, ,

, ,

b b

a a

u u

u u wdx u u wdx

u u u u

u u u u

u u u u

λ

λ

λ

λ

λ

=

⇒ =

⇒ =

⇒ =

⇒ =

∫ ∫

†

Proof:

( )

( )

* * *

* * *

*, ,

b b

a a

u u

u u

u u wdx u u wdx

u u u u

λ

λ

λ

λ

=

⇒ =

⇒ =

⇒ =

∫ ∫

*λ λ=Hence:

Orthogonality of Eigenfunctions

:

1

2

1

,

,

m m m

n m

n n n

u u

n m

u u

λ

λ λ

λ

=

≠ ≠

=

Consider one solution of the eigenvalue problem : ( )

Next, consider another solution of the eigenvalue problem with a different index and distinct eigenvalue

( )

We multiply (

( ) ( )( ) ( )

*

** * *

)2

3

( ) ( , :n m

b b

m n m n m n m na a

u uw x x a b

u u u u wdx u u wdxλ λ

∈

− = −∫ ∫

) by , the conjugate of ( ) by , subtract the second from the first,weight the result by , and integrate on

( )

Next, we consider the LHS of the above equation. 12

The eigenfunctions corresponding to a self-adjoint operator equation are orthogonal if the eigenvalues are distinct.

Orthogonality of Eigenfunctions (cont.)

( ) ( )( )** , ,

, , ( )

, , ( )

0

b

m n m n m n m na

m n m n

m n m n

u u u u wdx u u u u

u u u u

u u u u

− = < > − < >

= < > − < >

= < > − < >

=

∫† from the definition of adjoint

from the self - adjoint property

13

The LHS is:

( ) ( )* * *0 0b b

m n m n m na a

u u wdx u u wdxλ λ =⇒− =∫ ∫ orthogonality

Hence, for the RHS we have

* ,n n m nλ λ λ λ= ≠since

Summary of Properties

14

Then we have:

,u uλ= = †

Assume an eigenvalue problem with a self-adjoint operator:

The eigenvalues are real.

The eigenfunctions corresponding to distinct eigenvalues are orthogonal.

*, 0,b

n m n m m na

u u u u wdx λ λ= = ≠∫

*m mλ λ=

That is,

Example

15

( ) ( ) ( )

( ) ( ) ( ) ( )

* 0 (

0 ( )

b

m na

m a m a

u x u x w x dx

u a u a u b u b

=

= = = =

∫ for a Sturm - Liouville problem)

Dirchlet boundary conditions

Recall :

if

( ) ( ) ( ), thm n nm nm nu x J p x p m J x= = root ofConsider :

0, 1a b= =Choose :

( ) ( )0 0, 1 0n nm n nmJ p J p= =

0 0n n≠ =(The first equation is true for . The case can be considered as a limiting case.)

Orthogonality of Bessel functions

What is w(x)? We need to identify the appropriate DE that u(x) satisfies in Sturm-Liouville form.

Example (cont.)

16

( ) ( )2 2 2 0, nt y ty t n y y J t′′ ′+ + − = =Bessel equation :

,1

nm nm

nm

t p x dt p dx

t p x

= =∂ ∂

⇒ =∂ ∂

( )2 2 2 2 0nmx y xy p x n y′′ ′+ + − =

22

2

1 0nmny y p y

x x ′′ ′+ + − =

( )n nmy J p x=

Use:

17

22

2

1 0nmny y p y

x x ′′ ′+ + − =

22

2

1nm

ny y y p yx x

′′ ′− − + =

Rearrange to put into Sturm-Liouville eigenvalue form:

( )2

22

1 , ,n nm nmny y y y y J p x p

x xλ λ

′′ ′− − + = = =

( )2

22

1 , ,n nm nmd d nx y y y J p x p

x dx dx xλ λ

− + = = =

Example (cont.)

Orthogonality of Eigenfunctions (cont.)

18

1 ( ) ( )( )

d dP x Q x y yw x dx dx

λ − + =

( ) ( ) ( )2

2, , nP x x w x x Q xx

= = =

Compare with our standard Sturm-Liouville form:

We can now see that the u(x) functions come from a Sturm-Liouville problem, and we can identify:

( )2

22

1 , ,n nm nmd d nx y y y J p x p

x dx dx xλ λ

− + = = =

Hence, we have:

Orthogonality of Eigenfunctions (cont.)

19

Hence we have:

( ) ( )1

0

0,n nm n nmJ p x J p x x dx m m′ ′= ≠∫

( ) ( )0 thn nm nm np m JJ p = = root of Bessel functiow re n he

( ) ( ) ( )* 0b

m na

u x u x w x dx =∫

Adjoint in Linear Algebra

( )

*

*

* *

(

, ,

,

,

t

t

i ii

ij ji j

A A

Au v u A v

a b a b a b

Au v A u

=

=

= ⋅ =

=

∑

∑

Theorem:† i.e., the conjugate of the trans

For a complex matrix, the adjoint

e

Proof :

is given by :

To show this, w need to show :

where

To show this :

pose)

( ) ( )

* *

* * * *,

i j ij ii j

t ti ij j i ji j j ij i

i j i j i ji j

v u A v

u A v u A v u A v u A v

=

= = =

∑ ∑∑

∑∑ ∑∑ ∑∑ relabeling and

20

Adjoint in Linear Algebra (cont.)

*tA A = †

21

For a complex matrix we have established that

Therefore, if a complex matrix is self-adjoint, this means that the matrix is Hermetian:

[ ] ( )* *: tij jiA A A A = = Definition of a Hermetian matrix

Note: For a real matrix, self-adjoint means that the matrix is symmetric.

[ ] *tA A A = = †

Orthogonality in Linear Algebra

22

The eigenvalues of a Hermetian matrix are real.

The eigenvectors of a Hermetian matrix corresponding to distinct eigenvalues are orthogonal.

The eigenvectors of a Hermetian matrix corresponding to the same eigenvalue can be chosen to be orthogonal (proof omitted).

Because a Hermetian matrix is self-adjoint, we have the following properties:

Diagonalizing a Matrix

23

[ ] [ ][ ] [ ] 1A e D e −=

[ ]1

2

0 0 00 0 00 0 00 0 0 N

D

λλ

λ

=

[ ] [ ] [ ] [ ] [ ]11 12 1

21 22 21 2

31 32 3

41 42 4

N

NN

N

N

e e ee e e

e e e ee e ee e e

= =

[ ] th n ne n λ= eigenvector (a column vector) corresponding to eigenvalue

(proof on next slide)

If the eigenvectors of an N×N matrix [A] are linearly independent, then it can be diagonalized as follows:

24

[ ][ ] [ ] [ ] [ ] ( )1 1 2 2 N Ne D e e eλ λ λ = from properties of a diagonal matrix

Diagonalizing a Matrix (cont.)

[ ][ ] [ ] [ ] [ ] ( )1 1 2 2 N NA e e e eλ λ λ = from properties of eigenvectors

Proof

Hence, we have

[ ][ ] [ ][ ]A e e D=

[ ] [ ][ ][ ] 1A e D e −=

so that

Note: The inverse will exist since the columns of

the matrix [e] are linearly independent

by assumption.

25

Diagonalizing a Matrix (cont.)

[ ] ( )1 *te e− = the eigenvalue matrix is unitary

If the matrix [A] is Hermetian, then:

Therefore, for a Hermetian (self-adjoint) matrix we have:

[ ] [ ][ ] *tA e D e =

This follows from the orthogonality property of the eigenvectors.

[ ] [ ] ( )*te e I = identitymatrix

Note: For the diagonal elements, the eigenvectors can always be scaled so that * 1n ne e⋅ =

A Hermetian matrix is always diagonalizable!

Hence for a Hermetian matrix we then have: