Chapter 8 Theories of Systems
8-1 Laplace Transform Solutions of Linear Systems

Linear Systems X  AX  F (t ) : Consider y(n)+an-1y(n-1)+an-2y(n-2) +…+a1y’+a0y=f(t).
Let
x1(t)=y(t), x2(t)=y’(t), …, xn(t)=y(n-1)(t)
 x1 '   0
 x '  0
 2 
   0
  
    
 x n '  a 0
1
0
0
1
0
0


 a1
 a2
0 
0 
0 
1 
0 

 
 
   a n 1 
0

 x1   0 
x   0 
 2 

    
  

  0 
 x n   f (t )
 x '   0 1   x1   0 
Eg. y”-3xy’+2y=sin(x) can be transformed into  1   
   

 x2 '  2 3t   x2  sin( t )
by t=x, x1(t)=y(x) and x2(t)=y’(x).
 x  2 x1  3x2
Eg. Solve  1
, x1(0)=8, x2(0)=3 by Laplace transform.
 x2  2 x1  x2
sX ( s)  x1 (0)  2 X 1 ( s)  3 X 2 ( s)
(Sol.) L[f’(t)]=sF(s)-f(0)   1
sX 2 ( s)  x2 (0)  2 X 1 ( s)  X 2 ( s)
5
3

 X 1 ( s)  s  1  s  4
x (t )  5e t  3e 4t
.
 
 1
t
4t
5
2
x
(
t
)

5
e

2
e
 X (s) 
2

 2
s 1 s  4
 x   2 x   3 y   2 y  4
Eg. Solve 
, x(0)=x’(0)=y(0)=y’(0)=0 by Laplace transform.
2 y   x  3 y  0

4
( s 2  2s) X ( s )  (3s  2)Y ( s) 
2
(Sol.) L[f’’(t)]=s F(s)-sf(0)-f’(0) 
s
 sX ( s)  (2s  3)Y ( s )  0
4s  6
s ( 2  s )( s  1)


4
Y (s) 
2 s ( s  2)( s  1)
X (s) 
2
7
1
10
x(t )    3t  e  2t  e t
2
6
3
1
2
y (t )  1  e  2t  e t
3
3
～～74
Eg. According to optical waveguide theory, the E-fields of two identical
waveguides A and B fulfill the coupled mode equations
 dE A
 dz   jE A  jE B  E A (0)  1
, 
, where κ is the coupling coefficient. Find
 dE
 B   jE B  jE A E B (0)  0
 dz
EA(z) and EB(z).
(Sol.) L(EA)=ΣA(s) and L(dEA/dz)=sΣA(s)-EA(0)=sΣA(s)-1.
Similarly, L(EB)=ΣB(s) and L(dEB/dz)=sΣB(s)-EB(0)=sΣB(s)
s  j

 A (s) 

 ( s  j )  A ( s)  j  B ( s)  1

( s  j ) 2   2
,


 j
 j  A ( s)  ( s  j )  B ( s)  0
 B ( s ) 

( s  j ) 2   2
s
a
] and sin( at )  L1 [ 2
]
∵ L-1[F(s+a)] = f(t)e-at, cos( at )  L1 [ 2
2
s a
s  a2
 E A ( z )  L1[ A (s)]  cos(z )e  jz and E B ( z )  L1[ B (s)]   j sin( z )e  jz .
In this case, the coupling length is Lc=π/2κ. While the waveguiding mode traverses a
distance of odd multiple of the coupling length (Lc, 3Lc, 5Lc, …, etc), the optical
power is completely transferred into the other waveguide. But it is back after a
distance of even multiple of the coupling lengths (2Lc, 4Lc, 6Lc, …, etc). If the
waveguiding mode traverses a distance of odd multiple of the half coupling length
(Lc/2, 3Lc/2, 5Lc/2, …, etc), the optical power is equally distributed in the two guides.
 y (t )  6 y (t )  x (t )
Eg. Solve 
, x(0)=2, y(0)=3. 【交大電信】
3x(t )  x (t )  2 y (t )
 x(t )  4e 2t  2e 3t
(Ans.) 
 y (t )  e 2t  2e 3t
～～75
8-2 Matrix Solutions of Linear Systems
Eg. Solve
d  x1 (t )  3  2  x1 (t ) 
.

dt  x2 (t ) 2  2  x2 (t )
2 
3  
2
(Sol.) det 
      2  0 , λ=2, -1  diagonalizable
2

2




 2   1  0
    2
3  2
λ1=2, 
    x1   1     , exp[  2dt ]  e 2t



 2  2  2  0
 2
 2  1
 2   1  0
   1
3  (1)
λ2=-1, 
    x2   1     , exp[  (1)dt ]  e t



 2  (1)  2  0
 2
 2  2
 x (t ) 
 2
1
∴  1   c1   e 2t  c2   e t
1 
 2
 x 2 (t )
Eg. Solve
d  x1 (t )  1  1  x1 (t ) 
.

dt  x2 (t ) 1 3   x2 (t )
1    1 
2
(Sol.) det 
  (   2)  0 , λ=2, 2
1
3




dim(V)-Rank(A-2I)=1  2=Multiplicity of (λ-2)2  not diagonalizable
   1 
1  2  1   1  0
λ=2, 
    x1   1    



3  2  2  0
 1
 2   1
    0 
1  2  1   1   1 
    x2   1    


 1

3  2  2   1

 2   1
 1 
 x (t ) 
1
0 
∴  1   c1   e 2t  c2   te 2t    e 2t 
 1
 1 
 x2 (t )
 1
 x1 (t ) 
0 1 1  x1 (t ) 
 x3 (t ) 
1 1 0  x3 (t ) 
Eg. Solve d  x (t )  1 0 1  x (t ) .
2
 
 2 
dt 
(Sol.) f(λ)=-λ3+3λ+2=0, λ=2, -1, -1.
For λ=-1, dim(V)-Rank(A+I)=2=Multiplicity of (λ+1)2  diagonalizable
λ=-1,
λ=2,
1 
0  1 1
 1
0  1 1 

 1
1
0  1
1  0
    0
 2  
 3  0
1
1 
0  2
 1
02
1 

 1
1
0  2
  1  0 
   0
 2  
 3  0


1
x2   0 
  1
or
1
x1  1
1
 x1 (t ) 
1
1
0






2
t

t
∴ x2 (t )  c1 1 e  c2 0 e  c3  1  et



 
 
 x3 (t ) 
1
 1
 1
～～76
0
1
 
 1
Eg. Solve
d  x1 (t )  1  1  x1 (t ) 
.

dt  x2 (t ) 5  3  x2 (t )
(Sol.) f ( )  2  2  2  0 ,   1  i
 1    1  2  i
 1   1  0
1  
λ=-1+i, 





 3     2   5
 2  i   2  0
 5
   1 
[ 1i ] dt
, e
 x1   1   
 e t  [cos(t )  i sin( t )]

 2  2  i 
 1   1   0
λ=-1-i, 1  
 3    2  0
 5
    1 
[ 1i ] dt
, e
 x1   1   
 e t  [cos(t )  i sin( t )]



 2  2  i 
 x1 (t ) 
 1  t
 1  t
 x (t )  c1 2  i  e  [cos( t )  i sin( t )]  c2 2  i  e  [cos( t )  i sin( t )]




 2 
cos(t )
sin( t )

 t

 t
 c1 
e  c2 

e
2 cos(t )  sin( t )
 cos(t )  2 sin( t )
Eg. Solve
d  x1 (t )   4 2   x1 (t )   1 / t 
.


dt  x2 (t )  2  1  x2 (t ) 2 / t  4
(Sol.) f ( )  2  5  0 ,   0, 5 : diagonalizable
1
  2
1  0  x1    and 2  5  x 2   
 2
1 
 x (t ) 
1
  2
Homogeneous solutions: X h   1c   c~1    c~2   e 5t
 2
1
 x2c (t )
Particular solution: Xp = TYp, Xp’=AXp+g(t)  TYp’=ATYp+g(t)  Yp’=T-1ATYp+ T-1g(t)
1  2 
1  1 2
Choose T such that T 1 AT  D , T  
, T 1  

5  2 1
2 1 

d  y1 (t )  0 0   y1 (t )  1  1 2  1 / t 


dt  y 2 (t ) 0  5  y 2 (t ) 5  2 1 2 / t  4

5 ln( t )  8t  c1 
 y1 (t )  1 5 ln( t )  8t  c1 
1  2  1 






4
X

TY



4
5t
p
p

2 1  5
 c2 e 5t 


 y 2 (t ) 5  5  c 2 e



 5


 x1 p (t )  1
8 1
4  2 c1 1 c2  2 5t
 Xp  
   ln( t )   t 

 1   5 2  5  1 e
x
(
t
)
2
2
5
25
2
p




 
 
 


1
1
  2
8 1
4   2
  x1 (t )     ln( t )   t     c1    c2  e 5t
5 2 25  1 
 2
1
 x 2 (t ) 2
～～77
Eg. There are two species of animals: wolf and fox, interacting within the same forest
ecosystem. Let w(t) and f(t) denote the wolf and fox population, respectively, at time t.
Suppose further that wolves might eat foxes as food and foxes might also eat wolves as
food, but only wolves are hunted by human. If there are no foxes, then one might expect
that the wolves, lacking an adequate food supply, would decline in number at a rate of
-11w(t). When foxes are present, there is a supply of food, and so wolves are added to the
forest at a rate of 3f(t). Furthermore the change rate of the wolf population also
positively depends on a seasonal hunting factor, denoted as 100sin(t). On the other hand,
if there is no wolves, then the foxes, lacking an adequate food supply, would decline in
number at a rate of -3f(t). But when wolves are present, the fox population is increased
by a rate of 3w(t). Please answer the following questions: (a) Formulate the above system
by a set of differential equations. (b) Use variation of parameters to solve the system. (c)
What are the steady-state populations of the wolf and fox, respectively? [台大電研]
(Sol.) (a)
d  w(t )   11 3   w(t )  100 sin( t )


.
0
dt  f (t )  3  3  f (t ) 

 11 3 
1
3
(b) A= 
, f(λ)= λ2+14λ+24=0, λ1= -2  x1=   , λ2= -12  x2=  

 3  3
3
1
w (t )
1
3
Homogeneous solution Xh: Xh=  h   c~1  e 2t  c~2   e 12t
3
 1
 f h (t ) 
Particular solution: Xp = TYp, Xp‘=AXp+g(t)  TYp’=ATYp+g(t)  Yp’=T-1ATYp+ T-1g(t)
 1 / 10 3 / 10
1  3
Let T fulfill T 1 AT  D , T  
, T 1  


 3 / 10 1 / 10 
3 1 

d  y1 (t )   2 0   y1 (t )   1 / 10 3 / 10 100 sin( t )



0
dt  y 2 (t )  0  12  y 2 (t )  3 / 10 1 / 10  

2 t

 y1 (t )   4 sin( t )  2 cos(t )  c1e

    72 sin( t )  6 cos(t )  c e 12t  =Yp, Xp=TYp
y
(
t
)
2
 2  
29
29

2 t

 w(t )  ~ 1 2t ~  3  12t 1  3  4 sin( t )  2 cos(t )  c1e


  c1 3 e  c2  1 e + 3 1   72 sin( t )  6 cos(t )  c e 12t 
f
(
t
)

2


 
 

 
29
29

1
3
1
 c1   e  2t  c 2   e 12t +
29
3
 1
332
1
276 sin( t ) - 29


 w(t )  1 332
1

sin( t ) (c) As t→∞, 



29
 f (t ) 29 276
 76 
168 cos(t )
 
 76 
168 cos(t ) : steady-state solution
 
～～78
Eg. Solve
d  x1 (t )  5 / t  1 / t   x1 (t ) 
.

dt  x2 (t ) 3 / t 1 / t   x2 (t )
2 4
(Sol.) det 5 / t    1 / t   0    ,
1 / t  
 3 / t
t t
1 
2 
1 / t
4
, ( A  2 I ) x 2  
t
3 / t
2
e
 t dt
 1 / t   1 
  1
 0  x1   1    



 1 / t   2 
 2  3
3 / t
2
, ( A  1 I ) x1  
t
3 / t
 1 / t   1 
   1
 0  x2   1    



 3 / t   2 
 2  1
4
 e 2 ln|t|  t 2 , e
 t dt
 e 4 ln|t|  t 4
 t2
 x (t ) 
1
1
∴  1   c1  t 2  c2  t 4   2
3
1
 x2 (t )
3t
Eg. Solve
t 4   c1 
    (t )  C
t 4  c2 
d  x1 (t )  5 / t  1 / t   x1 (t )  1

 .
dt  x2 (t ) 3 / t 1 / t   x2 (t ) t 
5 / t
(Sol.) Homogeneous solutions: Φ(t)．C. Note: Φ’(t)  
3 / t
Particular solution: X p (t )   (t )  u (t )
5 / t
X p ' (t )  (t )  u (t )  (t )  u (t )  
3 / t
 1/ t 
Φ(t)
1 / t 
1/ t
  u (t )  (t )  u (t )
1 / t 
5 / t  1 / t 
1
1

X p (t )      (t )u ' (t )   

3 / t 1 / t 
t 
t 
1 1  1 / t 2
d  u1 (t ) 
1
 u ' (t )  
  (t )     
4
dt u 2 (t )
t  2  3 / t
1 / t  1

 1 / t 4  t 
2
1
 1
 2t 2  2t 

3
1 
 4  3
2t 
 2t
1 1

 u1 (t )   2t  2 ln | t |



u 2 (t )   1  1 
 2t 3 4t 2 
t2 
1 1
  1 2

ln
|
t
|
t
ln
|
t
|


  2
t   2t 2
4 





2
2
t 4    1  1  t  3t ln( t )  t 
 2t 3 4t 2  
2
4 
 x1 p (t )   t
 X p (t )  
 2
x
(
t
)
2
p

 3t
2
 x (t )   t
∴  1  2
 x2 (t ) 3t
2
4
 1 2
t2 
t
ln
|
t
|


t   c1   2
4 

2
2

4 
t  c2  t  3t ln( t )  t 

2
4 
4
～～79
8-3 Nonlinear Systems and Phase Planes
 dx
 dt  F ( x, y )
Autonomous system: 
, where F(x,y) and G(x,y) are both independent
 dy  G ( x, y )
 dt
of t.
Phase plane: The xy-plane for the analysis of the autonomous systems.
 dx
 dt  F ( x, y )  ax  by  P( x, y )
P ( x, y )
Q ( x, y )
 lim
 0.
Theorem 
, lim
( x , y ) ( 0 , 0 )
x 2  y 2 ( x , y ) ( 0 , 0 ) x 2  y 2
 dy  G ( x, y )  cx  dy  Q( x, y )
 dt
We have λ2-(a+d)λ+ad-bc=0  λ=λ1, λ2
1. λ1≠λ2, λ1, λ2: real, λ1λ2>0, then (0,0) is a node.
Stable node in case of λ1<0 and λ2<0. Unstable node if λ1>0, λ2>0.
2. λ1≠λ2, λ1, λ2: real, λ1λ2<0, then (0,0) is a saddle point.
3. λ1, λ2: complex with nonzero real part, then (0,0) is the point, which a spiral
approaches it.
Stable spiral in case of Re(λ)<0. Unstable spiral if Re(λ)>0.
4. λ1, λ2: pure imaginary, then (0,0) is a center of a closed curve.
～～80
 dx
 dt  4 x  2 y  4 xy
Eg. 
,
 dy  x  6 y  8 x 2 y
 dt
lim
( x , y )( 0, 0 )
 4 xy
x2  y2

lim
( x , y )( 0, 0 )
 8x 2 y
x2  y2
0
λ2-10λ+26=0  λ=5  i, Re(λ)=5>0, ∴ (0,0) is an unstable spiral point.
 dx
 dt  3x  y
Eg. 
 λ2+6λ+8=0, λ=-4, -2, and (-4)(-2)=8>0: negative and distinct,
dy
  x  3y
 dt
 x(t )  c1e 4t  c 2 e 2t  0
∴ (0,0) is a stable node. Check: 
 y (t )  c1e  4t  c 2 e  2t  0
 dx
 dt   x  3 y
Eg. 
 λ2+3λ-4=0, λ=1, -4, 1  -4, and 1(-4)<0, ∴ (0,0) is a saddle
dy
  2x  2 y
 dt
 x(t )  c1e t  c 2 e 4t

point. Check: 
. If c1=0, (x(t),y(t))→(0,0). But in case of
2 t
 4t
 y (t )  c1e  c 2 e
3

c1  0 , ( x(t ), y (t ))  (, )
 dx
 dt  3x  y
Eg. 
 λ2+4=0, λ=  2i, ∴ (0,0) is a center of a closed curve.
dy
  13 x  3 y
 dt
 x(t )  c1 cos(2t )  c2 sin( 2t )
Check: 
 y(t )  (2c2  3c1 ) cos(2t )  (2c1  3c2 ) sin( 2t )
Critical point: (xc,yc) fulfills both F(xc,yc)=0 and G(xc,yc)=0.
 x   F ( x, y )
Theorem C is the closed trajectory of the autonomous system 
,
 y   G( x, y )
where F and G  C1(x,y), then there exists at least one critical point of the system
enclosed by C.
Three types of limit cycles: Stable, unstable, and semi-stable limit cycles.
～～81
～～82
 x   x  y  x x 2  y 2
Eg. Find the trajectory of the following system: 
.
 y    x  y  y x 2  y 2
(Sol.) Let x=rcosθ, y=rsinθ, x’=-rsinθ．θ’, y’=rcosθ．θ’
∵ r2=x2+y2,
∴ rr’=xx’+yy’= x2+y2- ( x 2  y 2 ) x 2  y 2 =r2-r3  r  r (1  r ) (1)
∵ yx’-xy’= rsinθ(-rsinθ．θ’)- rcosθ(rcosθ．θ’)=-r2θ’,
∴ -r2θ’=yx’-xy’=x2+y2= r2   r 2   r 2     1 (2)
   0  t

1

(1) and (2)  r 
r  1 t

1 0
e

r0
If r0  1  r  1 ; else if r0  1  r  1 ; else, r0  1  r  1 .
In this example, the unit circle r=1 is a stable limit cycle.
8-4 Some Approximate Solutions of Nonlinear Ordinary Differential Equations
Eg. For a simple pendulum of mass m with a thread of length l, show that
d 2 g
 sin   0 . Solve it and obtain its period. [台大電研]
dt 2 
1
d 2
) =Constant
(Sol.) Total energy is conservative: mg (l  l cos )  m(l
2
dt

Let u 
d 2 g
 sin   0
dt 2 
d
du g
d
g
du d 2
 sin  
0
, u 
 2  u ' sin   0  u
l
dt
dt 
dt
dt
dt
 udu  
g
u2 g
sin d 
 (cos   cos  0 ) , where  0 is the initial angle

2

d  2 g

  (cos   cos  0 )
u
dt  

 dt 
Period: T  4 T00 dt  4 00
If θ is small, sin    
d
1/ 2
 2g

  (cos   cos  0 )
d
1/ 2
2g
(cos   cos  0 )

 g 
 g 
d 2 g
t   c 2 sin 
t 
   0    c1 cos 
2


dt





～～83
Eg. Solve Van der Pol’s equation y (t )   [ y 2 (t )  1] y (t )  y(t )  0 .
(Sol.) For ε=0, y (t )  A sin( t   ) , y ' (t )  A cos(t   )
If ε is small, y (t )  A(t ) sin[ t   (t )] …(1), y (t )  A(t ) cos[t   (t )] …(2)
y (t )  A(t ) sin[ t   (t )]  A(t ) cos[t   (t )]  [1   (t )] …(3)
(2), (3)  A(t ) sin[ t   (t )]  A(t ) (t ) cos[t   (t )]  0 …(4)
(2)  y (t )  A(t ) cos[t   (t )]  A(t )  sin[ t   (t )]  [1   (t )] …(5)
(2), (5)  A(t ) cos[t   (t )]  A(t ) (t )  sin[ t   (t )]
  [1  A2 (t ) sin 2 [t   (t )]  A(t ) cos[t   (t )] (we set   t   (t ) )


A 3 (t ) 
A 3 (t )
   A(t ) 
cos(

)

cos(3 ) …(6)

4 
4







2
0
cos( )  (6) d  A(t )  A(t ) 
2
0
sin(  )  (6) d   (t )  0
1/ 2

A 3 (t )
1


： Bernoulli ' s equation   A(t )  2 
 2t 
4

1  ce 
 (t )  
0

 1 
 y (t )  2 
 2t 
1  ce 
1/ 2
 sin( t   0 )
d2y
Eg. Obtain the particular solution of
  2 y  y 3   cos( x) .
2
dx
(Sol.)
Suppose y ( x,  )  y 0 ( x)  y1 ( x)   2 y 2 ( x)  
[ y 0 ( x)  y1( x)   2 y 2 ( x)  ]   2 [ y 0 ( x)  y1 ( x)   2 y 2 ( x)  ]
 [ y 0 ( x)  y1 ( x)   2 y 2 ( x)  ]3   cos( x)
 [ y 0 ( x)   2 y 0 ( x)]   [ y1( x)   2 y1 ( x)  y 03 ( x)]
  2 [ y2 ( x)   2 y2 ( x)  3 y0 ( x) y1 ( x)]  O( 3 )   cos( x)
2
 y 0 ( x)   2 y 0 ( x)   cos( x) , y1( x)   2 y1 ( x)   y 03 ( x)
y 2 ( x)   2 y 2 ( x)  3 y 02 ( x) y1 ( x)

 Solve y0 ( x) , y1 ( x) , …
Eg. Obtain the approximate particular solution of y  
1
y  0.1 y 3  cos( x) .
4
(Sol.) y ( x;  )  y 0 ( x)  y1 ( x)   2 y 2 ( x)   3 y 3 ( x)    y0 ( x)  y1 ( x)
 y 0 "
4
1
y 0 ( x )  cos( x )  y 0   cos( x)
4
3
3
64
64
1
 4

cos(3x)
y1( x)  y1 ( x)    cos( x)  y1   cos( x) 
27
945
4
 3

 y( x)  y0 ( x)  0.1y1 ( x)  1.57 cos( x)  0.006 cos(3x)
～～84
8-5 Sturm-Liouville Theory
Consider y”+R(x)y’+[Q(x)+λP(x)]y=F(x), multiply it by e  R ( x ) dx , and then let
r(x)= e  R ( x ) dx , q(x)=Q(x) e  R ( x ) dx , p(x)=P(x) e  R ( x ) dx , and f(x)=F(x) e  R ( x ) dx
 Sturm-Liouville form: [r(x)y’(x)]’+[q(x)+λp(x)]y(x)=f(x)
Eg. Find the eigenvalues and eigenfunctions of y”-2y’+2(1+λ)y=0 with boundary
conditions y(0)=y(1)=0, and transform it into the Sturm-Liouville form. [台大造
船所]
(Sol.) r2-2r+2(1+λ)=0, r  1  1  2(1   )  1   (2  1)
1
If    , r=1, 1  y=Aex+Bxex
2
y(0)=y(1)=0  A=B=0, ∴ trivial solutions
1
If    , -(2λ+1)>0, r=1±k  y=Ae(1+k)x+Be(1-k)x
2
y(0)=y(1)=0  A=B=0, ∴ trivial solutions
1
If    , -(2λ+1)<0, r=1±ki  y=ex(Acoskx+Bsinkx)
2
 y (0)  0  A  0

x

n 2 2  1 , ∴ the corresponding eigenfunction is e sin(nπx)
y
(
1
)

0

k

n





2

e  2dx =e-2x, y”e-2x-2e-2xy’+2(1+λ)e-2xy=0  [e-2xy’(x)]’+[2e-2x+λ2e-2x]y(x)=0
Eg. Find the eigenvalues and eigenfunctions of x2y”+xy’-λy=0 with boundary
conditions y(1)=y(a)=0, 1<x<a. 【台大機研】
(Sol.) r2+(1-1)r-λ=0, r   
If λ=0, y( x)  c1 x 0  c2 x 0 n( x)  c1  c2 n( x) ,
c1  c2 n(1)  0
 c1=c2=0.

c1  c2 n(a)  0
d1  1  d 2  1  0
 d1=d2=0
If λ>0, set λ=k2, y( x)  d1 x k  d 2 x  k ,  k
k
d1a  d 2 a  0
If λ<0, set λ=-k2,
y( x )  e1 x ki  e2 x  ki  e1  eik ln( x )  e2  e ik ln( x )  h1 cos[k ln( x )]  h2 sin[ k ln( x)]
y(1)=0  h1=0, y(a)=0  k=
n ln( x)
n
n 2
   [
]  y ( x)  sin[
]
ln( a)
ln( a )
ln( a)
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Sturm-Liouville theorems
1. For the two distinct λn and λm of the Sturm-Liouville problem, with
corresponding functions φn and φm, then  p(x) fulfills  ba p( x) n ( x) m ( x)dx  0
if n  m.
2. For the regular Sturm-Liouville problem, and two eigenfunctions
corresponding to a given eigenvalue are linearly dependent.
Eg. For y”+λy=0, y(0)=y(π/2)=0.
1. If λ=0, y=ax+b  a=b=0: trivial solution
2. If λ=k2>0, y=acos(kx)+bsin(kx)  a=0, k=2n  λ=k2=4n2
3. If λ=-k2<0, y=aekx+be-kx  no solutions
And p(x)=1,

 /2
0
p ( x) sin( 2nx) sin( 2mx)dx  0 if n≠m.
Eg. The eigenfunctions and their corresponding eigenvalues of the stationary
 2  2
Helmholtz equation
=-k2ψ are presented as follows.

x 2 y 2
The first eigenfunction, k2 = 106.6774
The second eigenfunction, k2 = 254.2339
The third eigenfunction, k2 = 286.0975
The tenth eigenfunction, k2 = 960.0726
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Eg. Given a refractive index distribution n(x,y), the eigenmodal function Φ(x,y)
 2  2 2
of an optical waveguide fulfills
 2 +k n(x,y)2Φ=β2Φ, where β2 is the
2
x
y
eigenvalue and β represents the phase constant of the lossy waveguide or the
propagation constant of the lossless waveguide. The eigenmodes of some optical
waveguides are presented as follows.
If the eigenmode is injected into an
infinitely-long straight waveguide, it can
propagate along the waveguide without any
deformation. However, in case the input light
is not an eigenmode, some optical power loss
occurs and then it becomes the eigenmode
gradually.
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Eg. One-dimensional wave function Ψ(x) in a quantum well
 0, 0  x  L
with infinitely hard walls, V= 
, fulfills
, elsewhere
d2Ψ/dx2+2mEΨ/  2=0 in 0≦x≦L with boundary conditions
Ψ(0)=Ψ(L)=0. It can be transformed into the eigenvalue
2 d 2
problem as 
 E . It is proved that the
2m dx 2
n 2 2  2
eigenvalue E is quantized as En=
and the corresponding eigenfunction is
2mL2
Ψn(x)= 2 / L sin(nπx/L).
Eg. One-dimensional wave function Ψ(x) in a
quantum well with two finite potential walls,
 0, 0  x  L
V= 
, fulfills
V , elsewhere
 d 2 I 2m
 2 ( E  V ) I  0

2
dx


d 2 II 2m

 2 II  0
with boundary

2

 2 dx
 d III  2m ( E  V )   0
III
 dx 2
2

conditions: ΨI(0)=ΨII(0), ΨII(L)=ΨIII(L), ΨI’(0)=ΨII’(0), ΨII’(L)=ΨIII’(L). And the

I ( x)  Cex

2mE x
2mE x

)  B cos(
).
eigenfunctions have the forms as II ( x)  A sin(



III ( x)  De x

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Eg. Tunnel Effect: One-dimensional
wave function Ψ(x) in a quantum
V ( E ), 0  x  L
barrier, V(x)= 
,
elsewhere
 0,
fulfills
 d 2 I 2m
 2 E I  0

2
dx

 2
 d II 2m
 2 ( E  V )II  0
with

2

 dx 2
 d III  2m E  0
III

dx 2
2

boundary
conditions:
ΨI(0)=ΨII(0),
ΨII(L)=ΨIII(L),
ΨI’(0)=ΨII’(0),
ΨII’(L)=ΨIII’(L). And hence the eigenfunctions have the forms as
 I ( x)  Ae ik1x  Be  ik1x
2m(V  E )
2mE
2mE

ik x
ik x
, k2=
, k3=
=k1.
II ( x)  Ce 2  De 2 , where k1=



ik3 x

III ( x)  Fe

The quantum mechanics can prove that the transmission probability is
16
2 k2 L
 e 2 k 2 L .
T=|ΨIII+|2/|ΨI+|2=|F|2/|A|2  [ 4  ( K / K ) 2 ]  e
2
1
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