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

d dyp x q x y Ew x y

dx dx

Sturm-Liouville equation:

Assumptions:

1 p x

on interval 0,a

is continuous differentiable on closed interval 0,a 1 0,p C a , or say , and 0 on 0,p a

0lim 0

limx

x a

p x p

p x p a

0lim 0

limx

x a

p x p

p x p a

minmin : 0 0p x x a p , and , and

2 q x is continuous on closed interval 0,a 0,q C a , or say

3 0,w x C a , and 0 on 0,w a

Proposition 1.1: Sturm-Liouville initial value problem

10 00 , 0

d dyp x q x y Ew x y

dx dx

dyy y y

dx

is unique on 0,a

under three assumptions.

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

10 0

10

xdyx p y q s Ew s y s ds

dx p x

10 0 0 0 0

1 10

x x ty x y p y ds q s Ew s y s ds dt

p s p t

Existence and uniqueness [1]

0, : 0, is continuousC a f a R Let be continuous space equipped with norm max : 0f f x x a

Existence and uniqueness [2]

0,C a is complete under norm max : 0f f x x a

define a mapping : 0, 0,T C a C a by

0 0 0

100

1 10

x x tTy x p ds q s Ew s y s ds d

p s p ty ty

1

2

0 0

1x tT x T x q s Ew s ds dts

p ts

extract supnorm

0 0

min

1x tT x T x q E w ds dt

p

2

min2

q E wT x T x x

p

2

min2

q E wT T a

p

T is a contraction mapping if 2

min

12

q E wa

p

Existence and uniqueness

Fundamental matrix

Sturm-Liouville equation: on interval 0,a

Transform to ODE system by setting dyz x p xdx

10

0

y yd p xz zdx

q x Ew x

10 00 , 0

d dyp x q x y Ew x y

dx dx

dyy y y

dx

has two fundamental solutions

d dyp x q x y Ew x y

dx dx

1 2

1 2

, y y

z z

satisfying

1 2

1 2

0 01 0,

0 00 1

y y

z z

Solution of initial value problem is 1 21

00 0y p yy x xyx y

Definition: fundamental matrix of d dyp x q x y Ew x y

dx dx

is 1 21 2

1 21 2

y yy yx

p x y p x yz z

satisfying

1 2

1 2

0 0 10

0 0 1

y y

z z

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

1 2 0 on 0,y x y x a 0

Abel’s formula [1]

11 12

21 22

a x a xy yda x a xz zdx

Consider general ODE system of dimension two: 1 2

21 2

, 0y y

x Iz z

with

1 2 1 2

1 2 1 2

det det dety y y yd

xz z z zdx

( from product rule)

121 2 111 11 11 1111

11

2 2 1 2 1 2

1 2 1 2 1 2

1

21

1

2det det det det dety y y z y z y y y y

z z z z z z z z

a aa aa

a aa

21 21

1 2 1 2 1 2

1 2 1 1 222

22 22 22 222 1 2

det det det dety y y y y y

z z y z y z za

a a aa za a

11 22det det detad

x x trA xd

x a xx

0expdet det 0

xx trA s ds

dyx y

dx

dyx dx

y

0 0

y x

y

dys ds

y

First order ODE

0log

0

xy xs ds

y

det 0 0 det 0x implies 1 2,y x y x are linearly independent

Abel’s formula [2]

11 12

21 22

a x a xy y ydA x

a x a xz z zdx

0expdet det 0

xx trA s ds

10

0

y yd p xz zdx

q x Ew x

d dyp x q x y Ew x y

dx dx

dyz x p xdx

det det 0 0,x x a since 0trA

1 2

1 2

, det 0 0y y

xz z

with fundamental matrix

Abel’s formula:

Definition: Wronskian 1 21 2

1 2

, detW

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

1 21 21 2

1 21 2

,y yy y

x p x W y yp x y p x yz z

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

d dyp x q x y Ew x y

dx dx

2

2

1, 0 0

2

dV x x E x

dx

1, , 1

2p x q x V w x

Definition: boundary condition 0 0 is called Dirichlet boundary condition

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

Dirichlet eigenvalue problem [1]

, 0 0d dy

p x q x y Ew x y y y adx dx

Dirichlet eigenvalue problem:

Observation:

1 If ,E is eigen-pair of Dirichlet eigenvalue problem, then 0,C a ,in fact, 2 0,C a

2

2

dp dq Ewd dx dx

dx p

is continuous on closed interval 0,a

Exercise: , , :p q w analyitc : analyitc

2 Under assumption 0,w x C a , and 0 on 0,w a , we can define inner-product

0

|a

Wx dxx w x where x is complex conjugate of x

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

, b

j ji iay yy y wdxThesis of Veerle Ledoux:

Matrix computation:

1

1 2 3 2

3

H

v

u v u u u v

v

Dirac Notation: | u

v

Dirichlet eigenvalue problem [2]

3 Define differential operator 1 d d

L p x q xw x dx dx

, then it is linear and

, 0 0d dy

p x q x y Ew x y y y adx dx

, 0 0Ly Ey y y a

4 Green’s identity: 0| | , |x axW W

L L pW

00 0 0| |

a a aa

W

d d d d dL L dx p x q x dx p p q x dx

dx dx dx dx dxw

0 0| |

aa

W

d d dL p p q x dx

dx dx dx

0 0| |

aa

W

d d dL p p q x dx

dx dx dx

0

0

| | , |a

x axW W

d dL L p pW

dx dx

The same

hence

Dirichlet boundary condition: 0 0a

0 0a

| | 0W W

L L

Dirichlet eigenvalue problem [3]

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

| | 0W W

L L

22

00, , : 0, :

aL a w f a C f wdx

If Green’s identity is zero, say

i j ije Ae A

j i jie Ae A

Matrix computation (finite dimension): Functional analysis (infinite dimension):

ij jiA AMatrix A is Hermitian (self-adjoint):

ie

je

| ijWL L

| jiWL L

operator L is self-adjoint: | | 0W W

L L

Matrix A is Hermitian, then

1 A is diagonalizable and has real eigenvalue

2 eigenvectors are orthogonal

HA V V

1, , ndiag

AV V

HV V I : orthonormal

operator L is Hermitian, then

1 L is diagonalizable and has real eigenvalue

2 eigenvectors are orthogonal

?

Dirichlet eigenvalue problem [4]

Suppose , 0 0d dy

p x q x y Ew x y y y adx dx

and 1,E 2,E are two eigen-pair of

<proof>

and 1,E 2,E are two eigen-pair 1

2

L E

L E

1| |W W

L E 1L E

2| |W W

L E 2L E

2 1| | |W W W

L L E E | | 0

W WL L

2 10 |W

E E

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

1L E 1L E L L

1L E

Hence 1,E is eigen-pair if and only if 1,E is eigen-pair

2 10 |W

E E 2 1 , E E 1 10 |

WE E

0 1 1 1E E E R

1 d dL p x q x

w x dx dx

L L, , :p q w real

Dirichlet eigenvalue problem [5]

If eigenvalue 1 2E E , then | 0W

: , are orthogonal in

2 0, ,L a w

Theorem 2: if , 0 0d dy

p x q x y Ew x y y y adx dx

and 1,E 2,E are two eigen-pair of

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

2 10 |W

E E 1 2, :E E real 2 10 |

WE E 1 2E E

0 |W

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

<proof>

10

0

y yd p xz zdx

q x Ew x

d dyp x q x y Ew x y

dx dx

dyz x p xdx

det det 0 0,x x a since 0trA

Abel’s formulaL E

L E

suppose and 0 0

0 0

a

a

Let xp p

, then

det 0 0 det 0 0,x x a

c for some constant c

Dirichlet eigenvalue problem [6]

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

real function. <proof> suppose 1u v is eigen-function with eigen-value E

L E and

0 0a

, satisfying

Lu Eu

Lv Ev

and 0 0

0 0

u u a

v v a

1u v

L L

Hence , , ,u E v E are both eigen-pair, from uniqueness of eigen-function, we have v cu

1u v v cu 1 1c u 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 0 1 2E E E

2 Eigen-functions are orthogonal in

2 0, ,L a w with inner-product 0

|a

Wx dxx w x

3 Eigen-functions are real and twice differentiable

Exercise (failure of uniqueness): consider

2

2

,

d yEy

dxd d

y y y ydx dx

, find eigen-pair and

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

Dirichlet eigenvalue problem [7]

Theorem 5 (Sturm’s Comparison theorem): let 1 1 2 2, , ,E E be eigen-pair of Sturm-Liouville Dirichlet problem.

suppose 2 1E E , then 2 is more oscillatory than 1 . Precisely speaking

Between any consecutive two zeros of 1 , there is at least one zero of 2

1 2

<proof>1

2

1x 2x

Let 1 2,x x are consecutive zeros of 1 and 1 0 on 1 2,x x

suppose 2 0 on 1 2,x x

as left figure

1 1 2 2, , ,E E are eigen-pair, then

11 1 1 1, 0 0

ddp x q x Ew x a

dx dx

22 2 2 2, 0 0

ddp x q x Ew x a

dx dx

define 2 12 1

2 1

det det ,x pWp p

11 2 1 1 1 2 11 2 1 1 1det 0xx x p x p x x p x

1 1 0x

12 2 2 1 2 2 22 2 2 1 2det 0xx x p x p x x p x

1 2 0x

IVT 1 2det 0, ,c c x x

Dirichlet eigenvalue problem [8]

2 1 2 12 1

2 1 2 1 2 1

det det det detd

xp pdx p p p p

11 1

ddp x q x Ew x

dx dx

22 2

ddp x q x Ew x

dx dx

2 12 1 1 2

2 2 1 1

det det 0d

x E E wq E w q E wdx

on 1 2,x x

12

1x 2x

1det 0x and det 0d

xdx

on 1 2,x x implies det 0x on 1 2,x x 1 2det 0, ,c c x x

Question: why 2 is more oscillatory than 1 , any physical interpretation?

dy

p xdx

dy w x y

dxEq x

2 e

xx EVm

x

Sturm-Liouville equation Time-independent Schrodinger equation

Definition: average quantity of operator O over interval 1 2,I x x by

2ˆ ˆ /I II

O yOywdx y wdx

Dirichlet eigenvalue problem [9]

2

1

2

1

2

1

22| |

x

x

x

x

xxI

dL p

dx

dp dx

dq x

xd

1 d dp q x Vx T

dx dxL

w x

Express operator L as where : kinetic operator, : potential operatorT V

L E

IL E

IIT EV

| |I I

L E

where

neglect boundary term

where

2

1

2

1

2

2

x

x

xI

x

dp dx

dxT

wdx

2

1

2

1

2

2

x

x

xI

x

q dxV

wdx

and

1 1 2 2, , ,E E 11I I

ET V

22I IET V

Heuristic argument for Theorem 5

2 1E E 2 2I I

T T 2 2

1 1

2 2

12x x

x x

d dp dx p dx

dx dx

Average value of 2d

dx

Average value of 1d

dx

2 is more oscillatory than 1

Prufer method [1]

, 0 0d dy

p x q x y Ew x y y y adx dx

Sturm-Liouville Dirichlet eigenvalue problem:Idea: introduce polar coordinate , in the phase plane ,y z py

sin

cos

y

z py

2 2 2

tan

y z

y

z

sin cos

cos sin

dy d

dx dxdz d

dx dx

sin cos

cos sin

d dy

dx dxd dz

dx dx

10

0

y yd p xz zdx

q x Ew x

2 2

1 1si

1cos si

n c

n

os

dEw q

dx

dq

p

Ewdx p

Objective: find eigenvalue E such that

0 0y y a

Objective: find eigenvalue E such that

tan 0 tan 0a

Advantage of Prufer method: we only need to solve

2 21cos sin

dEw q

dx p

when solution ;x E is found with condition tan 0; tan ; 0E a E

then 0

10 exp sin ; cos ;

xx Ew t q t t E t E dt

p t

Exercise: check it

Prufer method [2]Observation:

1 2 2 2 0y z

Suppose 0 00, 0x x a 0 00, 0y x z x

10

0

y yd p xz zdx

q x Ew x

0 00, 0y x z x with initial condition

has unique solution 0, 0y x z x

2 fixed 0, ;x x E is a strictly increasing function of variable E

d dyp x q x y Ew x y

dx dx

1, 0, 1p q w 2

2

d yEy

dx is Hook’s law

2

2

0 0

d yEy

dxy

has general solution sin , , 0y kx k E E

2 21cos sin

dEw q

dx p

1, 0, 1p q w 2 2cos sin

dE

dx

1 2 2 21

;cos sin

d x kk

dx

and 1tan ; tanx E kx

k

2 2 2 22

;cos sin

d x kk

dx

1 2k k2 2 2 2 2 2

1 2: cos sin cos sinslope k k

implies 1 2; ; 0x k x k

Prufer method [3]

2 2cos sin , 0 0d

Edx

Question: Given energy E, how can we solve

numerically

Forward Euler method: consider first order ODE

0, , 0dy

f x y y ydx

1 Uniformly partition domain 0,a as 0 1 20 j nx x x x jh x a

2 kky y xLet and approximate k

dyx

dx by one-side finite difference

2

22

y x h y xdy h d yx c

dx h dx

0, , 0dy

f x y y ydx

10

0, , k k

kk

y yf x y y y

h

y x h y xdyx

dx h

continuous equation:

discrete equation:

Exercise : use forward Euler method to solve angle equation

2 2;cos ; sin ; , 0; 0

d x Ex E E x E E

dx

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

It is clear that ;x E is a strictly increasing function of variable E

Prufer method [4]

3 sin

cos

y

z py

2 2 2

tan

y z

y

z

2 21cos sin

tan 0 tan 0

dEw q

dx p

a

0 0

d dyp x q x y Ew x y

dx dx

y y a

0y x siny

0 sin ; 0x E ;x E n

Although

2 21cos sin

0 0

dEw q

dx p

has a unique solution ;x E . But we want to find energy E

such that

2 21cos sin

0 0 , tan 0

dEw q

dx p

a

, that is ;a E n is constraint.

Example: for model problem 2 2cos sin , 0 0, 3d

E Edx

1.817 n

Prufer method [5]

4 2 21cos sin

dEw q

dx p

0y x siny

0 ;x E n

0i iy x x n 1

0ii

dx

dx p x

never decrease in a point where ix n

number of zeros of y in 0,a number of multiples of in 0 , a

n

1n

n

1n

/ 2 x

x

0,x E

/ 2

0;y x E

tany x

xpy x

0y x x n

0y x tan x 1

2x n

tan 0x

Ground state 0;y x E has no zeros except end points.

Prufer method [6]

x

x

1,x E

/ 2

1;y x E

3 / 2

2First excited state 1;y x E has one zero in 0,

2 21cos sin

dEw q

dx p

1

2jx n

cos 0z py

j

dx Ew q

dx

model problem: 0j

dx E

dx

x

x

2,x E

/ 2

2;y x E

3 / 2

2

5 / 2

3 second excited state 1;y x E has two zeros in 0,

conjecture: k-th excited state ; ky x E has k zeros in 0,

Prufer method [7]

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

0, 0p w

Let boundary values satisfy following normalization:

2 21cos sin

, 0 0

dw q

dx p

a

E

Then the kth eigenvalue kE satisfies 0, 0, ,k kE ka E

Moreover the kth eigen-function , 0ky x E has k zeros in 0,a

, 0 0d dy

p x q x y Ew x y y y adx dx

1 0, , , 0,p C a q w C a and

Remark: for detailed description, see Theorem 2.1 in

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

Prufer method [8]

Scaled Prufer transformation: a generalization of simple Prufer method

1sin

cos

yS

z py S

, dy z dz

q x Ew x zdx p x dx

; 0 :S x E scaling function, continuous differentiable where

sin cos1sin

2 cos cos sin

yd S dS SS

zdx dxSS S

1sin

cos

yS

z py S

d

dx

1sin cos

cos sin

S SA

S S

1

cossin

cos sin

SS

AS

S

2 1

1sin

2 cos

yd d SA S

zdx dx S

2 2cos sin sin cos

2sin 2 cos 2

d S Ew q S

dx p S S

d S Ew q S

dx p S S

Theorem 6 holds for scaled Prufer equations

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

Prufer method [9]

Scaled Prufer transformation

2 2cos sin sin cos

2sin 2 cos 2

d S Ew q S

dx p S S

d S Ew q S

dx p S S

2 21cos sin

1 1sin cos

dEw q

dx p

dEw q

dx p

1S

Simple Prufer transformation

Recall time-independent Schrodinger’s equation

2 2

22

dx V x E x

m dx

341.0542 10 J s

31mass of electron 9.1 10em kg

dimensionless form , or say x ay x a

, or say E E E V E

2 2

2 2

1

2 y

ma maV ay ay E ay

2

2ma

2

2

1

2

0 0

dV x x E x

dx

reduce to 1D Dirichlet problem

, 0 0d dy

p x q x y Ew x y y y adx dx

1, , 1

2p q V w

Exercise 1 [1]

Consider 1D Schrodinger equation with Dirichlet boundary condition

2

2

1

2

0 0

dV x x E x

dx

2

1

2 0.1V x

x

2

2 1

1 2 11

2

1 2

diag V Eh

x

0 1x 2x 3x 1nx

nx

use standard 3-poiint centered difference method to find eigenvalue1

: repulsive potentialV

1 0 10 0.7597E E

all eigenvalues ae positive, can you explain this?

2

number of grids = 50, (n = 50)

Ground state 0 0;x E has no zeros in 0,

1st excited state 1 1;x E has 1 zero in 0,

2nd excited state 2 2;x E has 2 zero in 0,

Check if k-th eigen-function has k zeros in 0,

Exercise 1 [2] solve simple Prufer equation for first 10 eigenvalues2

0 : 9E

2

2

1, 0 0

2

dV x x E x

dx

2 21cos sin

0; 0

dEw q

dx p

E

1, , 1

2p q V w

2 22cos sin

0; 0

dE V

dxE

Forward Euler method: N = 200

0, , 0dy

f x y y ydx

10

0, , k k

kk

y yf x y y y

h

1 , k kE is consistent with that in theorem 6

2 , kE increases gradually “staircase” shape with

“plateaus” at , kE j and steep slope around

1,

2kE j

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

Exercise 1 [3]Forward Euler method: N = 100 Forward Euler method: N = 200

9 1, 2E under Forward Euler method with number of grids, N = 100, this is wrong!

Question: Can you explain why

9,x E 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

2 2cos sin sin cosd S Ew q S

dx p S S

1cos 2 sin 2

2

cos 2 sin 2

d S Ew q S Ew q S

dx p S p S S

A x B x C x

2

2

1, 0 0

2

dV x x E x

dx

1, , 1

2p q V w

1D Schrodinger:

12 2 cos 2 sin 2

2

0; 0

d E V E V SS S

dx S S S

E

cos 2 sin 2

0; 0

dA x B x C x

dxE

1 if 1

if 1

E V xS x f E V x

E V x E V x

Suppose we choose where 1 if 1

if 1

xf x

x x

is continuous

Exercise 2 [1]

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

dx

Although we have known 0dS

xdx

on open set 1 : 1I x E V x and

1

2

dS dVx

dx S dx

on open set 2 : 1I x E V x

x

f x

x

1E

0x

1I2I

2

1

2 0.1V x

x

0 2

0

11

2 0.1V x E

x

0

10.1

2 1x

E

solve simple Prufer equation for first 10 eigenvalues

2

2

1, 0 0

2

dV x x E x

dx

Exercise 2 [2]

0 : 9E

2 2cos sin sin cos

0; 0

d S Ew q S

dx p S S

E

1, , 1

2p q V w

2 22 cos sin sin cos

0; 0

d E V SS

dx S SE

Choose scaling function

0.5 if 1

0.5 if 1

E V xS x

E V x E V x

Forward Euler method: N = 100

scaling function S

Forward Euler method: N = 100

NO scaling function S

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