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# Chapter 24 Sturm-Liouville problem - PowerPoint PPT Presentation

Chapter 24 Sturm-Liouville problem. Speaker: Lung-Sheng Chien. Reference: [1] Veerle Ledoux, Study of Special Algorithms for solving Sturm-Liouville and Schrodinger Equations.

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### Chapter 24 Sturm-Liouville problem

Speaker: Lung-Sheng Chien

Reference: [1] Veerle Ledoux, Study of Special Algorithms for solving Sturm-Liouville and Schrodinger Equations.

[2] 王信華教授, chapter 8, lecture note of Ordinary Differential equation

on interval

Sturm-Liouville equation:

Assumptions:

1

, and

is continuous differentiable on closed interval

, or say

, and

, and

2

is continuous on closed interval

, or say

3

, and

Proposition 1.1: Sturm-Liouville initial value problem

is unique on

under three assumptions.

proof: as before, we re-formulate it as integral equation and apply “contraction mapping principle”

Let

be continuous space equipped with norm

is complete under norm

1

by

define a mapping

2

extract supnorm

Existence and uniqueness

is a contraction mapping if

on interval

Sturm-Liouville equation:

Transform to ODE system by setting

has two fundamental solutions

satisfying

is

Solution of initial value problem

Definition: fundamental matrix of

is

satisfying

Question: Can we expect that we have two linear independent fundamental solutions, say

Consider general ODE system of dimension two:

with

( from product rule)

First order ODE

implies

are linearly independent

with fundamental matrix

Abel’s formula:

since

Definition: Wronskian

, then fundamental matrix of Sturm-Liouville equation can be expressed as

In our time-independent Schrodinger equation, we focus on eigenvalue problem

Definition: boundary condition

is called Dirichlet boundary condition

Question: What is “solvability condition” of Sturm-Liouville Dirichlet eigenvalue problem?

Dirichlet eigenvalue problem:

Observation:

1

,in fact,

If

is eigen-pair of Dirichlet eigenvalue problem, then

is continuous on closed interval

Exercise:

2

Under assumption

, and

, we can define inner-product

where

is complex conjugate of

1

If w is not positive, then such definition is not an “inner-product”

2

is Dirac Notation, different from conventional form used by Mathematician

Thesis of Veerle Ledoux:

Matrix computation:

Dirac Notation:

3

Define differential operator

, then it is linear and

4

Green’s identity:

The same

hence

Dirichlet boundary condition:

5

Definition: operator L is called self-adjoint on inner-product space

If Green’s identity is zero, say

Matrix computation (finite dimension):

Functional analysis (infinite dimension):

Matrix A is Hermitian, then

1

A is diagonalizable and has real eigenvalue

operator L is Hermitian, then

eigenvectors are orthogonal

2

1

L is diagonalizable and has real eigenvalue

eigenvectors are orthogonal

2

: orthonormal

and

are two eigen-pair of

Suppose

and

are two eigen-pair

Theorem 1: all eigenvalue of Sturm-Liouville Dirichlet problem are real

<proof>

Hence

is eigen-pair if and only if

is eigen-pair

and

are two eigen-pair of

Theorem 2: if

If eigenvalue

, then

are orthogonal in

<proof> from Theorem 1, we know E1 and E2 are real

Theorem 3 (unique eigen-function): eigenfunction of Sturm-Liouville Dirichlet problem is unique, in other words, eignevalue is simple.

<proof>

Abel’s formula

suppose

and

Let

, then

for some constant c

since

Theorem 4 (eigen-function is real): eigenfunction of Sturm-Liouville Dirichlet problem can be chosen as real function.

<proof>

suppose

is eigen-function with eigen-value

, satisfying

and

and

are both eigen-pair, from uniqueness of eigen-function, we have

Hence

we can choose real function u as eigenfunction

So far we have shown that Sturm-Liouville Dirichlet problem has following properties

1

Eigenvalues are real and simple, ordered as

Eigen-functions are orthogonal in

with inner-product

2

Eigen-functions are real and twice differentiable

3

, find eigen-pair and

Exercise (failure of uniqueness): consider

show “eigenvalue is not simple”, can you explain this? (compare it with Theorem 3)

Theorem 5 (Sturm’s Comparison theorem): let

be eigen-pair of Sturm-Liouville Dirichlet problem.

. Precisely speaking

suppose

, then

is more oscillatory than

Between any consecutive two zeros of

, there is at least one zero of

<proof>

Let

are consecutive zeros of

and

on

as left figure

suppose

on

are eigen-pair, then

define

IVT

on

on

and

on

implies

Question: why

, any physical interpretation?

is more oscillatory than

Sturm-Liouville equation

Time-independent Schrodinger equation

Definition: average quantity of operator

over interval

by

Express operator L as

where

where

neglect boundary term

where

and

Heuristic argument for Theorem 5

Average value of

Average value of

is more oscillatory than

Sturm-Liouville Dirichlet eigenvalue problem:

Idea: introduce polar coordinate

in the phase plane

Exercise: check it

Objective: find eigenvalue E such that

Objective: find eigenvalue E such that

Advantage of Prufer method: we only need to solve

when solution

is found with condition

then

Observation:

1

Suppose

has unique solution

with initial condition

fixed

is a strictly increasing function of variable

2

is Hook’s law

has general solution

and

implies

numerically

Question: Given energy E, how can we solve

Forward Euler method: consider first order ODE

Uniformly partition domain

as

1

2

Let

and approximate

by one-side finite difference

continuous equation:

discrete equation:

Exercise : use forward Euler method to solve angle equation

for different E = 1, 4, 9, 16 as right figure

It is clear that

is a strictly increasing function of variable

3

Although

has a unique solution

. But we want to find energy E

such that

, that is

is constraint.

Example: for model problem

4

never decrease in a point where

number of zeros of

in

number of multiples of

in

Ground state

has no zeros except end points.

First excited state

has one zero in

model problem:

second excited state

has two zeros in

conjecture: k-th excited state

has k zeros in

Theorem 6 : consider Prufer equations of regular Sturm-Liouville Dirichlet problem

and

Let boundary values satisfy following normalization:

Then the kth eigenvalue

satisfies

Moreover the kth eigen-function

has k zeros in

Remark: for detailed description, see Theorem 2.1 in

Veerle Ledoux, Study of Special Algorithms for solving Sturm-Liouville and Schrodinger Equations.

Scaled Prufer transformation: a generalization of simple Prufer method

scaling function, continuous differentiable

where

1

2

Theorem 6 holds for scaled Prufer equations

Exercise: use Symbolic toolbox in MATLAB to check scaled Prufer transformation.

Scaled Prufer transformation

Simple Prufer transformation

Recall time-independent Schrodinger’s equation

dimensionless form

reduce to 1D Dirichlet problem

Consider 1D Schrodinger equation with Dirichlet boundary condition

1

use standard 3-poiint centered difference method to find eigenvalue

number of grids = 50, (n = 50)

1

all eigenvalues ae positive, can you explain this?

2

Ground state

has no zeros in

1st excited state

has 1 zero in

2nd excited state

has 2 zero in

Check if k-th eigen-function has k zeros in

2

solve simple Prufer equation for first 10 eigenvalues

Forward Euler method: N = 200

is consistent with that in theorem 6

1

2

“plateaus” at

and steep slope around

Question: what is disadvantage of this “staircase” shape?

Forward Euler method: N = 100

Forward Euler method: N = 200

under Forward Euler method with number of grids, N = 100, this is wrong!

Question: Can you explain why

is not good?

hint: see page 21 in reference

Veerle Ledoux, Study of Special Algorithms for solving Sturm-Liouville and Schrodinger Equations.

Scaled Prufer transformation

1D Schrodinger:

where

is continuous

Suppose we choose

Question: function f is continuous but not differentiable at x = 1. How can we obtain

and

Although we have known

on open set

on open set

solve simple Prufer equation for first 10 eigenvalues

Choose scaling function

Forward Euler method: N = 100

Forward Euler method: N = 100

scaling function S

NO scaling function S

Compare both figures and interpret why “staircase” disappear when using scaling function