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
Existence and uniqueness [1]
Sturm-Liouville equation: 
d 
dy 
 p  x    q  x  y  Ew  x  y on interval  0, a 
dx 
dx 
Assumptions:
1
p  x  is continuous differentiable on closed interval 0, a , or say p  C1 0, a , and p  0 on  0, a 
lim p  x   p  0 
x 0
lim p  x   p  a 
lim p  x   p  0 
x 0
, and
lim p  x   p  a 
xa
, and
min  p  x  : 0  x  a  pmin  0
xa
2
q  x  is continuous on closed interval 0, a , or say q  C 0, a 
3
w  x   C 0, a  , and w  0 on  0, a 
d 
dy 
 p  x    q  x  y  Ew  x  y
dx 
dx 
dy
y  0   y0 ,
 0   y01
dx

Proposition 1.1: Sturm-Liouville initial value problem
under three assumptions.
is unique on
0, a
proof: as before, we re-formulate it as integral equation and apply “contraction mapping principle”
x
dy
1 
p  0  y01    q  s   Ew  s   y  s  ds 
 x 
0

dx
p  x  
y  x   y0  p  0  y01 
x
0
Let
x
1
1  t
ds  
 q  s   Ew  s   y  s  ds  dt


0 p t 
0
p s

C  0, a    f :  0, a   R is continuous be continuous space equipped with norm
f



 max f  x  : 0  x  a
Existence and uniqueness [2]

1
C  0, a  is complete under norm
2
define a mapping T : C 0, a  C 0, a 
f

 max f  x  : 0  x  a

Ty  x   y0  p  0 y01 0
x
by
1  t
q  s   Ew  s     s    s   ds  dt




0
p t  
T  x   T  x   0
x
extract supnorm
T  x   T  x   0
x

1  t
q
pmin  0
q
T  x   T  x  
T   T

q
E w
2 pmin
 q E w
 
2 pmin

T is a contraction mapping if
x
1
1  t
ds  
q  s   Ew  s   y  s  ds  dt



0
0


p s
p t 




E w
2 pmin


x 2  

a 2   

E w
 

  

ds  dt



a2  1
Existence and uniqueness
Fundamental matrix
Sturm-Liouville equation: 
d 
dy 
 p  x    q  x  y  Ew  x  y on interval  0, a 
dx 
dx 
Transform to ODE system by setting

0
d  y 
 
dx  z  
 q  x   Ew  x 
dy
dx
1 
 y1  0    1   y2  0    0 
 y1   y2 
 y
, 
p  x     has two fundamental solutions   ,   satisfying 


 z
z1  0    0   z2  0    1 
z
z


1


2

0   
d 
dy 
 p  x    q  x  y  Ew  x  y
dx 
dx 
is
dy
y  0   y0 ,
 0   y01
dx

Solution of initial value problem
Definition: fundamental matrix of 
satisfying
z  x  p  x
d 
dy 
y
 p  x    q  x  y  Ew  x  y is   x    1
dx 
dx 
 z1
 y  0
  0   1
 z1  0 
y2  0    1 

z2  0    1 
Question: Can we expect that we have two linear independent fundamental solutions, say
 y1  x    y2  x   0 on 0, a


y  x   y0 y1  x   p  0  y01 y2  x 
  0
y2   y1

z2   p  x  y1


p  x  y2 
y2
Abel’s formula
Consider general ODE system of dimension two:
 y
d
det   x   det  1
dx
 z1
[1]
d  y   a11  x  a12  x    y 
 
 
dx  z   a21  x  a22  x    z 
y2 
 y1

det

 
z2 
 z1
y2 

z2 
 y1
 z1
with   x   
y2 
 ,   0  I2
z2 
( from product rule)
 y
det  1
 z1
y2 
 a11 y1  a12 z1 a11 y2  a12 z2 
 a11 y1 a11 y2 
 y1
  det 
  det 
  a11 det 
z2 
z1
z2
z2 


 z1
 z1
y
det  1
 z1
y2 
y1
y2
y2 


 y
 det 
 det  1


  a22 det 
z2 
a
y

a
z
a
y

a
z
a
z
a
z
 21 1 22 1 21 2 22 2 
 22 1 22 2 
y2 
  a det 
z2  11
d
det   x    a11  x   a22  x   det   x   trA  det   x 
dx
First order ODE
det   x   det   0  exp
  trA  s  ds 
x
0
dy
   x  dx
y
dy
   x y
dx
x
dy

y0 y 0   s  ds
y
log
det   0  0  det   x   0
implies y1  x  , y2  x  are linearly independent
x
y  x
    s  ds
y  0 0
Abel’s formula

 y
d  y   a11  x  a12  x    y 
    a x a x     A x 
dx  z   21   22     z 
z
 y1
 z1
with fundamental matrix   x   
d 
dy 
 p  x    q  x  y  Ew  x  y
dx 
dx 
z  x  p  x
y2 
 , det  0   0
z2 
Abel’s formula: det   x   det   0  exp
[2]
  trA  s  ds 
x
0

0
d  y 
 
dx  z  
 q  x   Ew  x 
dy
dx
1 
 y
p  x  
 z
0   
det   x   det   0 x  0, a
Definition: Wronskian W 1 , 2 
y
  x   1
 z1
since trA  0
  
det  1 2  , then fundamental matrix of Sturm-Liouville equation can be expressed as
 1 2 
y2   y1

z2   p  x  y1

  p  x W  y1 , y2 
p  x  y2 
y2
In our time-independent Schrodinger equation, we focus on eigenvalue problem
d 
dy 
  p  x    q  x  y  Ew  x  y
dx 
dx 
Definition: boundary condition
1
p  x  , q  x  V , w x  1
2
 1 d2


V
x



  x   E  x  ,   0       0
2
 2 dx

  0      0 is called Dirichlet boundary condition
Question: What is “solvability condition” of Sturm-Liouville Dirichlet eigenvalue problem?
Dirichlet eigenvalue problem
Dirichlet eigenvalue problem: 
[1]
d 
dy 
 p  x    q  x  y  Ew  x  y , y  0   y  a   0
dx 
dx 
Observation:
1
If
 , E 
is eigen-pair of Dirichlet eigenvalue problem, then
  C 0, a ,in fact,   C 2 0, a
dp d
dx is continuous on closed interval 0, a
q  Ew 
d 
dx

2
dx 2
Under assumption
 |
 : analyitc
p, q, w : analyitc
Exercise:
2
p
W

a
0
w  x   C 0, a  , and w  0 on  0, a  , we can define inner-product
   x   x  w  x  dx
where
  x
is complex conjugate of
1
If w is not positive, then such definition is not an “inner-product”
2
 |
  x
is Dirac Notation, different from conventional form used by Mathematician
b

Thesis of Veerle Ledoux: yi , y j  a yi y j wdx

Matrix computation: u H v  u1
u

2
 v1 
 
u3   v2 
v 
 3
u 
v 
Dirac Notation:
 |
Dirichlet eigenvalue problem
3
4
1  d 
d 

  p  x    q  x   , then it is linear and
w  x   dx 
dx 

Define differential operator L 

d 
dy 
 p  x    q  x  y  Ew  x  y , y  0   y  a   0
dx 
dx 
 | L
Green’s identity:
 | L
 | L
 | L
W
W
W
a

  p 
  p
Ly  Ey, y  0  y  a   0
 pW  ,    |xx0a
 d 
d
   L wdx      p  x 
0
0
dx
 dx 
a
W
  | L
W
[2]



 d a
|0 
  q  x   dx   p
dx


 | L
W
0
 d  d
 
 p dx dx  q  x     dx



d a a  d  d
|0    p
 q  x    dx
0
dx
 dx dx

d  a
|0
dx


a
0
 d  d
 
 p dx dx  q  x     dx


The same
a
hence

a
  | L
W
 d 
d 
 p 

 pW  ,    |xx 0a

dx  0
 dx
Dirichlet boundary condition:
 | L
W
  | L
W
0
  0    a   0
  0    a   0
Dirichlet eigenvalue problem
2
5
Definition: operator L is called self-adjoint on inner-product space L
If Green’s identity is zero, say  | L
W
  | L
W
Matrix computation (finite dimension):
[3]
0, a , w   f : 0, a   C : 
a
0

f wdx  
2
0
Functional analysis (infinite dimension):
ei Ae j  Aij
ei  
e j Aei  A ji
e j 

Matrix A is Hermitian (self-adjoint): Aij  Aji
 | L
 | L
W
 Lij
W
 Lji
operator L is self-adjoint:  | L
W
  | L
W
0
Matrix A is Hermitian, then
1
A is diagonalizable and has real eigenvalue
2
eigenvectors are orthogonal
AV  V 
  diag  1 ,
A  V V
, n 
V H V  I : orthonormal
operator L is Hermitian, then
H
?
1
L is diagonalizable and has real eigenvalue
2
eigenvectors are orthogonal
Dirichlet eigenvalue problem
[4]
d 
dy 
Suppose  , E1  and  , E2  are two eigen-pair of   p  x    q  x  y  Ew  x  y , y  0   y  a   0
dx 
dx 
 , E1 
 | L
W
L  E1
and  , E2  are two eigen-pair
L  E2
L  E1
 | L
W
 E1  | 
W
L  E2
 | L
W
 E2  |
W
  | L
W
  E2  E1   |
 | L
W
  | L
W
0
W
0   E2  E1   | 
W
Theorem 1: all eigenvalue of Sturm-Liouville Dirichlet problem are real
<proof>
L
1  d 
d 

  p  x    q  x  
w  x   dx 
dx 

 L 
L  E1

p, q, w : real
E
 
1
L  L
L  L
L   E1 


Hence  , E1  is eigen-pair if and only if  , E1  is eigen-pair
0   E2  E

1
  |
E2  E1 ,    
W
0   E1  E1   | 
 0
W
E1  E1  E1  R
Dirichlet eigenvalue problem
[5]
d 
dy 
Theorem 2: if  , E1  and  , E2  are two eigen-pair of   p  x    q  x  y  Ew  x  y , y  0   y  a   0
dx 
dx 
If eigenvalue E1  E2 , then  |
W
 0 :  , are orthogonal in L2  0, a  , w 
<proof> from Theorem 1, we know E1 and E2 are real
0   E2  E1   | 
E1 , E2 : real
W
0   E2  E1   |
E1  E2
W
0   |
W
Theorem 3 (unique eigen-function): eigenfunction of Sturm-Liouville Dirichlet problem is unique,
in other words, eignevalue is simple.
<proof>
Abel’s formula

d 
dy 
 p  x    q  x  y  Ew  x  y
dx 
dx 
z  x  p  x

0
d  y 

 
dx  z  
 q  x   Ew  x 

Let   x   
 p 
dy
dx
1 
 y
p  x  
 z
0   
det   x   det   0 x  0, a
L  E
suppose
L  E
since trA  0
det   0  0
and
  0    a   0
  0    a   0
 
, then
p 
det   x   0 x  0, 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   u  1v is eigen-function with eigen-value E , satisfying
L  E
  u  1v
and
  0    a   0
Hence
 u, E  ,  v, E 
  u  1v
Lu  Eu
and
Lv  Ev
L  L
u  0  u  a   0
v  0  v  a   0
are both eigen-pair, from uniqueness of eigen-function, we have v  cu
v  cu


  1  c 1 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 E0  E1  E2 
2
2
Eigen-functions are orthogonal in L  0, a  , w  with inner-product  |
3
Eigen-functions are real and twice differentiable
Exercise (failure of uniqueness): consider

d2y
 Ey
dx 2
y     y   ,
W

a
0
   x   x  w  x  dx
, find eigen-pair and
d
d
y    
y  
dx
dx
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 , E1  , 2 , E2  be eigen-pair of Sturm-Liouville Dirichlet problem.
suppose E2  E1 , then 2 is more oscillatory than 1 . Precisely speaking
1
2
Between any consecutive two zeros of 1 , there is at least one zero of 2
<proof>
x1
1
2
Let x1 , x2 are consecutive zeros of 1 and 1  0 on  x1 , x2  as left figure
suppose 2  0 on  x1 , x2 
x2
1, E1  , 2 , E2 

d1 
d 
 p  x
  q  x  1  Ew  x  1 , 1  0   1  a   0
dx 
dx 

d2 
d 
 p  x
  q  x  2  Ew  x  2 , 2  0   2  a   0
dx 
dx 
are eigen-pair, then

define det   x   det  2
 p2
1 
  pW 2 , 1 
p1 
1  x1   0
det   x1   2  x1  p1 x1   1  x1  p2  x1   2  x1  p1 x1   0
1  x2   0
det   x2   2  x2  p1 x2   1  x2  p2  x2   2  x2  p1 x2   0
IVT
det   c   0, c   x1 , x2 
Dirichlet eigenvalue problem
 
d
det   x   det  2
dx
 p2
 2
1 


det

  p  
p1 
2

2

d
det   x   det 
dx
  q  E2 w 2
det   x1   0 and
1 
 2
  det 
  p  
 p1 
2

[8]
1
1 



p

 1  

d1 
d 
 p  x
  q  x  1  Ew  x  1
dx 
dx 

d2 
d 
 p  x
  q  x  2  Ew  x  2
dx 
dx 
x1
2
x2
1

  E2  E1  w12  0 on  x1 , x2 
 q  E1w 1 
d
det   x   0 on  x1 , x2  implies det   x   0 on  x1 , x2 
dx
det   c   0, c   x1 , x2 
Question: why 2 is more oscillatory than 1 , any physical interpretation?
Sturm-Liouville equation

Time-independent Schrodinger equation


      V  x    x   E  x 
 2 me

d 
dy 
 p  x   q  x y  E w x y
dx 
dx 
Definition: average quantity of operator Ô over interval I   x1 , x2  by
Oˆ
I
ˆ
  yOywdx
/  y 2 wdx
I
I
Dirichlet eigenvalue problem
1  d 
d 

  p  x    q  x    T  V
w  x   dx 
dx 

Express operator L as L 
 | L
L  E
where
 | L
d x2
  p
|x 
dx 1
I
I

x2
x1
 E  |
[9]
where T : kinetic operator, V : potential operator
I
 d 
p
 dx 
 dx 
2

x2
x1
q 2 dx
neglect boundary term
L I E
T
I
 V
I
E
where
T
I


 d 
p
 dx
 dx 
2
x2
x1

x2
x1
and
 wdx
2
V
I



x2
x1
x2
x1
q 2 dx
 2 wdx
Heuristic argument for Theorem 5
1, E1  , 2 , E2 
T
T
I
1   E1 
2   E2 
I
V
V
Average value of
I
E2  E1
T
I
2  
T
I
d 2
d1
 Average value of
dx
dx
2 is more oscillatory than 1
I
2 

x2
x1
x2
 d 
p  2  dx  
x1
 dx 
2
 d 
p  1  dx
 dx 
2
Prufer method [1]
Sturm-Liouville Dirichlet eigenvalue problem: 
d 
dy 
 p  x    q  x  y  Ew  x  y , y  0   y  a   0
dx 
dx 
Idea: introduce polar coordinate   ,  in the phase plane
y   sin 
z  py   cos 
 dy 
 dx   sin 
 
 dz   cos 
 
 dx 
 d 
 cos    dx 


  sin    d 


 dx 

0
d  y 

 
dx  z  
 q  x   Ew  x 
1 
 y
p  x  
 z
0   
Objective: find eigenvalue E such that
 y, z  py
 2  y2  z2
tan  
y
z
 d 
 dx   sin 

   cos 
 d  

  
 dx 
dy
cos    
 dx
sin    

dz
   
 dx 
d 1
 cos 2    Ew  q  sin 2 
dx p

1 d  1
    Ew  q   sin  cos 
 dx  p

Exercise: check it
Objective: find eigenvalue E such that
y  0  y  a   0
tan   0  tan   a   0
d 1
 cos 2    Ew  q  sin 2 
dx p
when solution   x; E  is found with condition tan   0; E   tan   a; E   0
Advantage of Prufer method: we only need to solve
 x


then   x     0  exp   1   Ew  t   q  t    sin   t ; E  cos   t; E  dt 


 0  p  t 


Prufer method [2]
Observation:
1
 2  y2  z2  0
Suppose   x0   0, 0  x0  a
y  x0   0, z  x0   0

0
d  y 

 
dx  z  
 q  x   Ew  x 
1 
 y
p  x  
 z
0   
has unique solution y  x   0, z  x   0
with initial condition y  x0   0, z  x0   0
2
fixed x  0,   x; E  is a strictly increasing function of variable E
d 
dy 
  p  x    q  x  y  Ew  x  y
dx 
dx 
d2y
 2  Ey
dx
y  0  0
p  1, q  0, w  1
d2y
 2  Ey is Hook’s law
dx
has general solution y  sin kx, k  E , E  0
d 1
 cos 2    Ew  q  sin 2 
dx p
d  x; k1 
 cos 2   k12 sin 2 
dx
d  x; k2 
 cos 2   k22 sin 2 
dx
p  1, q  0, w  1
k1  k2
d
 cos 2   E sin 2 
dx
and
tan   x; E  
slope : cos 2   k12 sin 2   cos 2   k22 sin 2 
implies   x; k1     x; k2   0
1
tan  kx 
k
Prufer method [3]
Question: Given energy E, how can we solve
d
 cos 2   E sin 2  ,   0   0 numerically
dx
Forward Euler method: consider first order ODE
dy
 f  x , y  , y  0   y0
dx
1
Uniformly partition domain 0, a as 0  x0  x1  x2 
2
k 
Let y  y  xk  and approximate
continuous equation:
 x j  jh 
2
dy
 xk  by one-side finite difference dy  x   y  x  h   y  x   h d y2  c 
dx
dx
h
2 dx
dy
 f  x , y  , y  0   y0
dx
y  x  h  y  x
dy
 x 
dx
h
 k 1
 yk 
discrete equation: y
 f xk , y  k  , y  0  y0
h

 xn  a

Exercise : use forward Euler method to solve angle equation
d  x; E 
 cos 2   x; E    E sin 2   x; E   ,   0; E   0
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]
y   sin 
z  py   cos 
3
d 
dy 
  p  x    q  x  y  Ew  x  y
dx 
dx 
y  0  y  a   0
y  x  0
Although
y   sin 
 0
d 1
 cos 2    Ew  q  sin 2 
dx p
  0  0
 2  y2  z2
tan  
y
z
d 1
 cos 2    Ew  q  sin 2 
dx p
tan   0    tan   a    0
sin   x; E    0
  x; E   n
has a unique solution   x; E  . But we want to find energy E
d 1
2
2
such that dx  p cos    Ew  q  sin  , that is   a; E   n is constraint.
  0   0, tan   a    0
Example: for model problem
    1.817  n
d
 cos 2    E sin 2   ,   0   0, E  3
dx
Prufer method [5]
4
d 1
 cos 2    Ew  q  sin 2 
dx p
y  xi   0    xi   n
y   sin 
y  x  0
d
1
0
 xi  
dx
p  xi 
  x; E   n
 0
 never decrease in a point where   xi   n
number of zeros of y in
 0, a  
number of multiples of  in   0 ,  a  
 n  1 
 n  1 
n
n
  x, E0 
tan   x  

 /2
x
y  x; E0 
 /2

y  x
py  x 
y  x  0
tan   x   0
y  x   0
tan   x   
  x   n


Ground state y  x; E0  has no zeros except end points.
x
1
  x   n  
2

Prufer method [6]
  x, E1 
2
First excited state y  x; E1  has one zero in  0,  
3 / 2
d 1
 cos 2    Ew  q  sin 2 
dx p

 /2
x
y  x; E1 




z  py   cos   0
d
 x j    Ew  q 
dx
x
model problem:
  x, E2 
1
  x j    n  
2
d
 xj   E  0
dx
second excited state y  x; E1  has two zeros in  0,  
3
5 / 2
2
3 / 2

conjecture: k-th excited state y  x; Ek  has k zeros in
 /2
x
y  x; E2 
x
 0,  
Prufer method [7]
Theorem 6 : consider Prufer equations of regular Sturm-Liouville Dirichlet problem

d 
dy 
 p  x    q  x  y  Ew  x  y , y  0   y  a   0
dx 
dx 
Let boundary values satisfy following normalization:
p  C1 0, a , q, w  C 0, a  and
p  0, w  0
d 1
 cos 2    Ew  q  sin 2 
dx p
  0   0,   a   
Then the kth eigenvalue Ek satisfies   0, Ek   0,   a, Ek     k
Moreover the kth eigen-function y  x, Ek   0 has k zeros in
 0, a 
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
y
1
 sin 
S
z  py  S  cos 
d
dx
1
2
y
where S  x; E   0 : scaling function, continuous differentiable
1
 sin 
S
d  y  S 
 
dx  z  2 S
z  py  S  cos 
 sin 

S
A 
 S cos 


 S sin 
1
A 
cos 

 S 

 cos 


S

 S  sin  

d 
S 
1  d  y 
   A   
dx   
 dx  z  2 S

 1
  sin 

sin

 S

S

 
 cos    S cos 
 1

  S sin   

 
 cos   

cos 
S
sin 

 S
 cos 

 d 
S
 dx   
 
 S  sin  






Theorem 6 holds for scaled Prufer equations
dy
z
dz

,
  q  x   Ew  x   z
dx p  x  dx
d S
Ew  q 2
S
 cos 2  
sin   sin  cos 
dx p
S
S
2 d   S Ew  q 
S
 
 sin 2  cos 2
 dx  p
S 
S
Exercise: use Symbolic toolbox in MATLAB to check
scaled Prufer transformation.
Prufer method [9]
Scaled Prufer transformation
Simple Prufer transformation
d S
Ew  q 2
S
 cos 2  
sin   sin  cos 
dx p
S
S
d 1
 cos 2    Ew  q  sin 2 
dx p
S 1

1 d  1
    Ew  q   sin  cos 
 dx  p

2 d   S Ew  q 
S
 
 sin 2  cos 2
 dx  p
S 
S
Recall time-independent Schrodinger’s equation
 1.0542 1034 J  s
2
d2

  x   V  x   E  x 
2m dx 2
me  mass of electron   9.11031 kg
x  ay, or say  x   a
dimensionless form
E   E, or say  E   
V    E 
 1

ma 2
ma 2
   y  2 V  ay    ay   2 E  ay 
 2


2
ma 2
reduce to 1D Dirichlet problem
 1 d2



V
x



  x   E  x 
2
 2 dx

  0       0
p
1
,q V,w 1
2

d 
dy 
 p  x    q  x  y  Ew  x  y , y  0   y  a   0
dx 
dx 
Exercise 1
[1]
Consider 1D Schrodinger equation with Dirichlet boundary condition
 1 d2

 V  x    x   E  x 

2
 2 dx

  0       0
1
V  x 
1
2  x  0.1
V : repulsive potential
2
use standard 3-poiint centered difference method to find eigenvalue

 2 1


 1  1 2
 2
 2h 


1
1




  diag V    E




2 

0
number of grids = 50, (n = 50)
1
x1
x2
x3
xn
x
xn 1  
0  E0  0.7597  E1 
all eigenvalues ae positive, can you explain this?
2
Ground state   x; E0    0 has no zeros in  0,  
1st excited state   x; E1   1 has 1 zero in  0,  
2nd excited state   x; E2    2 has 2 zero in  0,  
Check if k-th eigen-function has k zeros in  0,  
Exercise 1
2
[2]
solve simple Prufer equation for first 10 eigenvalues
 1 d2


V
x



  x   E  x  ,   0       0
2
 2 dx

d 1
 cos 2    Ew  q  sin 2 
dx p
1
p  ,q V,w 1
2
  0; E   0
d
 2 cos 2    E  V  sin 2 
dx
  0; E   0
E  0 : 9 
Forward Euler method: N = 200
dy
 f  x , y  , y  0   y0
dx
y  k 1  y  k 
 f xk , y  k  , y  0  y0
h


1
  , Ek   k   is consistent with that in theorem 6
2
  , Ek  increases gradually “staircase” shape with
“plateaus” at   , Ek   j and steep slope around


1
  , Ek    j   
2

Question: what is disadvantage of this “staircase” shape?
Exercise 1
Forward Euler method: N = 100
[3]
Forward Euler method: N = 200
  , E9   12   under Forward Euler method with number of grids, N = 100, this is wrong!
Question: Can you explain why   x, E9  is not good?
hint: see page 21 in reference
Veerle Ledoux, Study of Special Algorithms for solving Sturm-Liouville and Schrodinger Equations.
Exercise 2
Scaled Prufer transformation
[1]

d 1  S Ew  q   S Ew  q 
S
  
 
 cos 2  sin 2 
dx 2  p
S  p
S 
S

 A  x   B  x  cos 2  C  x  sin 2
d S
Ew  q 2
S
 cos 2  
sin   sin  cos 
dx p
S
S
 1 d2


V
x


1D Schrodinger:  
  x   E  x  ,   0       0
2
 2 dx

d
 A  x   B  x  cos 2  C  x  sin 2
dx
  0; E   0
p
1
,q V,w 1
2
d 1 
E V  
E V 
S

  2 S 
   2S 
 cos 2  sin 2 
dx 2 
S  
S 
S

  0; E   0
if E  V  x   1

1
S
x

 f  E V  x 
Suppose we choose   
E

V
x
if
E

V
x

1






1
if x  1
 x
if x  1
where f  x   
dS
Question: function f is continuous but not differentiable at x = 1. How can we obtain
dx
Although we have known
V  x 
dS
1 dV
 x 
dx
2S dx
1
2  x  0.1
dS
 x   0 on open set I1  x : E  V  x   1 and
dx
2
on open set I 2  x : E  V  x   1
E 1
x
x0
I1
I2
V  x0  
1
2  x0  0.1
2
 E 1
x0 
1
2  E  1
 0.1
is continuous
f  x
x
Exercise 2
[2]
solve simple Prufer equation for first 10 eigenvalues
 1 d2


V
x



  x   E  x  ,   0       0
2
 2 dx

d S
Ew  q 2
S
 cos 2  
sin   sin  cos 
dx p
S
S
p
1
,q V,w 1
2
  0; E   0
 0.5

S
x

Choose scaling function   

 0.5  E  V  x  
E  0 : 9 
d
E V
S
 2 S cos 2  
sin 2   sin  cos 
dx
S
S
  0; E   0
if E  V  x   1
if E  V  x   1
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