Transcript 偏微分方程 PARTIAL DIFFIERENTIAL EQUATION (P.D.E)

Orthogonal Function Expansion
正交函數展開
•Introduction of the Eigenfunction Expansion
•Abstract Space
•Function Sapce
•Linear Operator and Orthogonal Function
Introduction -The Eigenfunction Expansion
Consider the equation :
y  d1 ( x) y  d2 ( x) y  0, a  x  b
With the b.c’s :
y(a)  y(b)  0
The g.s. is y(x) c1u1 (x) c2u 2 (x) where u1,u2 are linearly index. Fucs. And C1,C2
are arb consts.
For b.c’s :
c1u1 (a)  c2u 2 (a)  0
c1u1 (b)  c2u 2 (b)  0
The condition of nontrivial sol. of c1,c2 to be existed if  :
u1 ( a )u1 ( a )
u 2 (b)u 2 (b)
0
The Euler Column
d 2v
P
2
2


v

0
,
where
,


dx2
EI
The g.s.
v( x)  c1 sin x  c2 cosx
:v(0)  v(l )  0  c2  0
For the b.c.’s
And c1 sin x  0 for nontrivial solution c1  0
 sin l  0
i.e. l  n , n  1,2,3,4.....
So that we obtain the Eigen values n 
n
, n  1,2,3......
l
And the corresponding eigenfucs (nontrivial sols) are vn ( x)  sin n x  n ( x)
According the analysis, we will have v( x)  0 unless the end force P such
that: n 2  0

b
a
f ( x) g ( x)  ?  Function Space
Abstract Space
(N, )為一良序半群 (N,)為良序可交換單子
Topological Space
Metric Space
Normed Space
Inner Product Space
Rn
Q
Z
N
(Z ,)為一良序可交換群 (Z,)為良序可交換單子
(Z,  ,)為一良序可交換單子環
(Q,  ,)為一有序域
(R,  ,)為一完備的有序域
連續的有序域
有序 : m  n, m  n, m  n
ml  l n  m  n
ml  l  n  m  n
Hilbert Space
Banach Space
由R→Rn(有序性喪失)
代數的原型:   , ( - ,  )
完備性:每一Cauchy系列均收斂
Topological Space
Definition
A topological space is a non-empty set E together with a family X  (Ui i  I )
of subsets of E satisfying the following axioms:
(1) E  X ,0  X
(2) The union of any number of sets in X belongs to X i.e.
J finite, J  I   U i  X
i j
(3) The intersection of any finite number of sets in X belongs to X i.e.
J  I  Ui  X
i j
Metric Space
Definition
A metric space is a 2-tuple (X,d) where X is a set and d is a metric on X, that is,
a function d : X × X → R, such that
d(x, y) ≥ 0
(non-negativity)
d(x, y) = 0 if and only if x = y
d(x, y) = d(y, x)
d(x, z) ≤ d(x, y) + d(y, z)
(identity)
(symmetry)
(triangle inequality).
Cauchy Sequence
Definition: Complete space
A sequence (Xn):in a metric space X=(X,d) is said to be Cauchy if for every
there is an N=N(e) such that
for m,n>N 
d ( xm , xn )  e
x is called the limit of (Xn) and we write
lim xn  x ;
or, simply,
xn  x
n
Definition: Completeness
Any Cauchy Sequence in X is convergence
d < xn, x >0
Ball and Sphere
Definition:
Given a point x0  X and a real number r>0, we define three of sets:
(a) B( x0 ; r)  x  X d ( x, x0 )  r


S ( x ; r)  x  X d ( x, x )  r
(Open ball)
~
(b) B
( x0 ; r )  x  X d ( x, x0 )  r
(Closed ball)
(c)
(Sphere)
0
0
In all three case, x0 is called the center and r the radius.
Furthermore, the definition immediately implies that
~
S ( x0 ; r)  B( x0 ; r)  B( x0 ; r)
Definition (Open set and closed set):
A subset M of a metric space X is said to be open if is contains a ball about each of
its points. A subset K of X is said to be closed if its complement (in X ) is open, that
is, Kc=X-K is open.
A mapping from a normed space X into a normed space Y is called an
operator. A mapping from X into the scalar filed R or C is called a functional.
The set of all biunded linear operator from a given normed space X into a given
normed space Y can be made into a normed space, which is denoted by B(x,y).
Similarly, the set of all bounded linear functionals on X becomes a normed space,
which is called the dual space X’ of X.
Normed Space
Definition of Normed Space
Here a norm on a vector space X is a real-value function on X whose value
at an x  X is denoted by
x
x 0
x 0 x0
x   y
x y  x  y
Here x and y are arbitrary vector in X and  is any scalar
Definition of Banach Space
A Banach space is a complete normed space.
Lemma (Translation invariance)
A metric d induced by a norm on a normed space X satisfies
(a) d(x+a,y+a)=d(x,y)
(b) d (x,y)   d ( x, y)
For all x, y a  X and every scalar 
Proof.
d ( x  a, y  a)  x  a  ( y  a)  x  y  d ( x, y)
d (x,y)  x  y   x  y   d ( x, y)
Let X be the vector space of all ordered pairs x  (1 , 2 )
y  (1 ,2 ) of real
numbers. Show norms on X are defined by
x 1  1  2
2 12
x 2  (1   2 )
2
x
 (1   2 )
p
p
p
1
p
x   max1 , 2 
The sphere
S (0; r)  x  X , x  1
In a normed space X is called the unit sphere.
x  1
x 4 1
x 2 1
x 1 1
Unit Sphere in LP
If a normed space X contains a sequence (en) with the property that every x  X
there is a uniquie sequence of scalars (an) such that
x  (1e1  .......nen )  0
as
Then (en) is called basis for X. series

 e
k k
k 1
Which has the sum x is then called the expansion of x

x    k ek
k 1
n 
Inner Product Space
A inner product space is a vector space X with an inner product define on X.
Here, the inner product <x,y> is the mapping of   into the scale filed, such
that
 x  y, z  x, z    y, z 
  x, y    x, y 
 x, y   y, x 
 x, x  0
 x, y  0  x  0
Hilbert Space
A Hilbert space is a complete inner product space.
x 
 x, x 
d ( x, y)  x  y 
(Norm)
 x  y, x  y 
(Metric)
Hence inner product spaces are normed spaces, and Hilber spaces are Banach spaces.
Examples of finite-dimensional Hilbert spaces include
1. The real numbers  x with (v, u ) the vector dot product of v and u
2. The complex numbers C x with (v, u ) the vector dot product of
and the complex conjugate of u
.
v
Euclidean space Rn
The space Rn is a Hilbert space with inner product define by
 x, y  11  .... nn
Where
x  (1 )  (1 ,.....n )
y  (1 )  (1 ,.....n )
and
1
x  x, x   (1 ,..... n )
2
2
1
1
2
d ( x, y)  x  y  x  y, x  y   [(1  1)  ..... ( n  n ) ]
2
2
2
1
2
Space L2[a,b]
b
 x, y   x(t ) y(t )dt
a
b
x  (  x(t ) dt)
2
1
2
a
Hilbert sequence space l2
With the inner product

 x, y    j j
j 1
The norm

2
x  x, x   (  j )
1
2
1
2
j 1
Space lp
The space lp with p  0 is not inner product space, hence not a Hilbert space
Orthonormal Sets and Sequences
Orthogonality of elements plays a basis role in inner product and Hilbert spaces.
The vectors form a basis for R3, so that every x  R 3 has a unique representation.
x  1e1   2e2  3e3
 x, e1  1  1e1  2  2e2  3  3e3  1
Continuous functions
Let X be the inner product space of all real-valued continuous functions on [0,2π]
with inner product defined by
2
 x, y   x(t ) y(t )dt
a
An orthogonal sequence in X is (un), where
un (i)  cosnt
n  0,1,.......
Another orthogonal sequence in X is (vn), where
vn (i)  sin nt
0
2

 u m ,u n   cos m t cos ntdt  
0
2

n  1,2,.......
if
mn
if m  n  1,2,.......
if
mn0
Hence an orthonormal sequence sequence is (en)
e0 (t ) 
1
2
en (t ) 
un (t ) cos nt

un

From (vn) we obtain the orthonormal sequence ( en ) where
vn (t ) sin nt
en (t ) 

vn

~
Homework
1. Does d(x,y)=(x-y)2 define a metric on the set of all real numbers?
2. Show that d ( x, y ) 
x y
defines a metric on the set of all real
numbers.
3. Let
a, b  R
and
ab
Show that the open interval (a,b) is an
incomplete subspace of R, whereas the closed interval [a,b] is complete.
4. Prove that the eigenfunction and eigenvalue are orthogonalization and real for
the the Sturm-Lioville System.
5. Show the following when linear second-order difference equation is expressed
in self-adjoint form:
(a) The Wronskian is equal to constant divided by the initial coefficient p
C
p ( x)
W [ y1 , y2 ] 
(b) A second solution is given by
y2 ( x)  Cy1 ( x) 
6. For the very special case
equation becomes
x
dt
p[ y1 (t )]2
  0 and q( x)  0, the self-adjoint eigenvalue
d
du ( x)
[ p( x)
]0
dx
dx
Use this obtain a “second” solution of the following
(a ) Legendre’s equation
(b ) Laguerre’s equation
(c ) Hermite’s equation
Function Space
(A) L2 [a,b] space:
Space of real fucs. f(x) which is define on [a,b] and square integrable i.e .
f
2
b
 ( f , f )   f 2 ( x)dx  
a
In the language of vector space, we say that
“any n linearly indep vectors form a basis in E ”space”. Similarly, in function space
It is possible to choose a set of basis function such that any function, satisfying
Appropriate condition can be expressed as a linear combination to a basis in L2[a,b]
Certainly, any such set of fucs. Must have infinitely many numbers; that is, such a
L2[a,b] comprises infinitely many dimensions.
(B) Schwarz Inequality:
Given f(x), g(x) in L2[a,b], Define ( f , g ) 

b
a
f ( x) g ( x)dx, then, ( f , g ) 2  ( f , f )(g , g )
Proof: ( f , g )  ( f  ag, f  ag)  a 2 ( g, g )  2 ( f , g )  ( f , f )  0
 ( f , g ) 2  ( f , f )(g , g )  0
 ( f , g ) 2  ( f , f )(g , g )
(C) Linear Dependence, Independence:
Criterion: A set of fucs. 1 ( x),......n ( x) In L2 [a,b] is linear dep.(indep.) if its
If its Gramian (G) vanishes (does not vanish), where
(1 , 1 )(1 , 2 ).........(n , n )
G
(2 , 1 )(2 , 2 ).........(2 , n )
................
(n , 1 )....................(n , n )
The proof is the same as in linear vector space.
(D) The orthogonal System
A set of real fucs. 1 ( x),......n ( x) …….is called an orthogonal set of fucs.
In L2[a,b] if these fucs. are define in L2[a,b] if all the integral (m ( x),n ( x))
exist and are zero for all pairs of distinct
Properties of Complete System
Theorem: Let f(x), F(x) be defined on L2 [a,b] for which
n
f ( x)  lim  ckk .
n 
k 1
n
F ( x)  lim  Ckk
n 
Then we have

b
a
k 1

f ( x) F ( x)dx   ck Ck
k 1
Proof:
Since f+F, and f-F are square integer able, from the completeness relation


 [ f  F ] dx   (c
b
2
a
k 1
k
 [ f  F ] dx   (c
2
a
k 1
 Ck )
2

b
2
k
 Ck )

 4 f ( x)  F ( x)dx  4 ck Ck
b
a
k 1
Theorem:
Every square integer able fnc. f(x) is uniquely determined (except for its value
at a finite number of points) by its Fourier series.
Proof:
Suppose there are two fucs. f(x),g(x) having the identical Fourier series
representation
n
lim  [ f ( x)   ckk ( x)]2 dx  0
b
i.e.
n  a
k 1
n
lim  [ g ( x)   ckk ( x)]2 dx  0
b
n  a
k 1
Then using (   )  2(   ) we find
2
2
2
2
0   [ g ( x)  f ( x)] dx   [(g ( x)   c k k )  ( c k k )  f ( x)] dx
b
b
2
a
a
2
2
 2 [(g ( x)   c k k )]  2 ( c k k )  f ( x)] dx  0   [(g ( x)  f ( x)]  0
b
a
2
b
a
b
a
g(x)=f(x) at the pts of continuity of the integrand
g(x) and f(x) coincide everywhere, except possibly at a finite number of pts. of
discontinuity
Theorem:
An continuous fuc. f(x) which is orthogonal to all the fucs. of the complete system
must be identically zero.
Proof: Since…
n
lim  [ f ( x)   ckk ( x)]2 dx  0
b
n  a
k 1
n
And let
g ( x )   ck  k ( x )
We can prove f(x)=g(x) at every point.
k 1
n

f ( x )   ck  k ( x )
k 1
Theorem:
The fourier series of every square integer able fuc. f(x) can be integrated term by term.
In other words, if
f ( x) ~ c11 ( x)  ...... cnn ( x)  ......
Then

x2
x2
x2
x1
x1
f ( x)dx  c1  1 ( x)dx  .... cn  n ( x)dx  .......
x1
Where x1,x2 are any points on the inteval [a,b]
Proof: Since…
Assume x2>x1

x2
x1
n
fdx   ck  k dx  
k 1
x2
x2
x1
x1

Take
n 

limn
n
b
n
k 1
a
k 1
f   ckk dx   f   ckk dx
2
n
f   ckk dx 1dx
b
a
b
a
k 1

x2
x1
n
fdx   ck  k dx  0
k 1
x2
x1
The Sturm-Liouville Problem
Self-adjoint Operator
d2
d2
Lu( x)  p0 ( x) 2 u ( x)  P1 ( x) 2 u ( x)  p2 ( x)u ( x)
dx
dx
For a linear operator L the analog of a quadratic form for a matrix is the integral
 u L u  u Lu   u( x) Lu( x)dx  up0u  p1u  p2udx
b
b
a
a
Because of the analogy with the transposed matrix, it is convenient to define the
linear operator
2


d
d
b
 u L u  u Lu  [u ( x)( p1  p0 )u ( x)]x a    2 [ pou ]  [ p1u ]  p2u udx
a dx
dx


b
Comparing the integrands
u( p0  p1)u  2u( p0  p1 )u  0
p0 ( x)  P1 ( x)
d2
d
d 2u
du
Lu  2 [ p0u ]  [ p1u ]  p2u  p0 2  (2 p0  p1 )
 ( p0  p1  p2 )u
dx
dx
dx
dx
As the adjoint operator L. The necessary and sufficient condition that L  L
Lu  Lu 
d
du ( x)
[ p( x)
]  q( x)u ( x)
dx
dx
The operator L is said to be self-adjoint.
The Sturm-Liouville Boundary Value Problem
A differential equation defined on the interval a  x  b having the form of
d 
dy 
p
(
x
)
 [q( x)  w( x)] y  0


dx 
dx 
and the boundary conditions
a1 y ( a )  a2 y ' ( a )  0

b1 y (b)  b2 y ' (b)  0
is called as Sturm-Liouville boundary value problem or Sturm-Liouville system,
where ,
; the weighting function r(x)>0 are given functions; a1 , a2 ,
b1 , b2 are given constants; and the eigenvalue is an unspecified parameter.
The Regular Sturm-Liouville Equation
It is a special kind of boundary value problem which consists of a second-order
homogeneous linear differential equation and linear homogeneous boundary
conditions of the form
d 
dy 
p
(
x
)
 q( x)  w( x)y  0


dx 
dx 
d 
dy 
L( y ) 
p ( x)   q ( x) y ( x)

dx 
dx 
where the p, q and r are real and continuous functions such that p has a
continuous derivative, and p(x) > 0, r(x) > 0 for all x on a real interval a  x  b;
and  is a parameter independent of x. L is the linear homogeneous differential
operator defined by L(y) = [p(x)y´]´+q(x)y.And two supplementary
boundary conditions
A1y(a)+A2y´(a) = 0
B1y(b)+B2y´(b) = 0 .
where A1 , A2 , B1 and B2 are real constants such that A1 and A2 not both zero
and B1 and B2 are not both zero.
Definition 1.1 : Consider the Sturm-Liouville problem consisting of the differ
entail equation and supplementary conditions. The value of the parameter
 in for which there exists nontrivial solution of the problem is called
 the eigenvalue of the problem. The corresponding nontrivial solution
is called the eigenfunction of the problem. The Sturm-Liouville problem is also
called an eigenvalue problem.
The Nonhomogeneous Sturm-Liouville Problems
Consider boundary value problem consisting of the nonhomogeneous
differential equation
L[y] = - [p(x)y´]´+q(x)y = w(x)y+f(x),
where  is a given constant and f is a given function on a a  x  b
and the boundary conditions
A1y(a)+A2y´(a)=0
B1y(b)+B2y´(b)=0 .
And as in regular Sturm-Liouville problems we assume that p, p, q, and r are
continuous on a  x  b and p(x) > 0, r(x) > 0 there.We solve the problem
by making use of the eigenfunctions of the corresponding homogeneous problem
consisting of the differential equation
The Bessel's Differential Equation
In the Sturm-Liouville Boundary Value Problem, there is an important
special case called Bessel's Differential Equation which arises in numerous
problems, especially in polar and cylindrical coordinates. Bessel's Differential
Equation is defined as:
x 2 y  xy  ( x 2  n2 ) y  0
where
is a non-negative real number. The solutions of this equation are
called Bessel Functions of order n . Although the order n can be any real number,
.
the scope of this section is limited to non-negative integers, i.e.,
,
unless specified otherwise. Since Bessel's differential equation is a second
order ordinary differential equation, two sets of functions, the Bessel function
of the first kind Jn(x) and the Bessel function of the second kind
(also known as the Weber Function) Yn(x) , are needed to form the general
solution:
y( x)  c1J n( x)  c2Yn ( x)
Five Approaches
The Bessel functions are introduced here by means of a generating function.
Other approaches are possible. Listing the various possibilities, we have
1.
Gram-Schmidt Orthogonalization

b
a
i 2 wdx  Ni 2
We now demand that each solution i be multiplied by Ni

b
a
1
i 2 ( x)w( x)dx  1
b
  ( x) ( x)w( x)dx  
a
i
j
ij
The presence of the new un(x) will guarantee linear independence.
We star with n=0, letting
Then normalize
 0 ( x)  u0 ( x)
 0 ( x) 
 0 ( x)
[  0 wdx]
2
1
2
.
Fro n=1, let
 1 ( x)  u1 ( x)  a100 ( x)
This demand of orthogonality leads to
2


wdx

u

wdx

a

10  0 wdx  0
 10
 10
As 0 is normalized to unity, we have
a10    u1 0 wdx
Fixing the value of a10. Normalizing, we have
 1 ( x)
1 ( x) 
1
2
(  1 wdx) 2
We demand that  1 ( x) be orthogonal to 0 ( x)
1 ( x) 
Where
 1 ( x)
(  1 ( x) w( x)dx)
2
1
2
 1 ( x)  ui  ai 00  ai11  ........ aii1i 1
The coefficients aij are given by aij    ui j wdx
If some order normalization is selected

b
a
[ j ( x)]2 w( x)dx  N j
2
The equation can be replaced by
i ( x)  N i
And aij becomes
aij
 i ( x)
(  1 wdx)
2
1
2
u  wdx


i
i
Nj
2
p j ui ( x)  { ui (t ) j (t ) w(t )dt } j ( x)
i 1
 i ( x)  {1   Pj }ui ( x)
j 1
Orthogonal polynomial Generated by Gram-Schmidt Orthogonalization of
un ( x)  X n , n  0,1,2........
2. Series solution of Bessel’s differential equation
x 2 y  xy  ( x 2  n2 ) y  0
Using y’ for dy/dx and y for d2y/dx2 . Again, assuming a solution of the form

y( x)   a x k 
 0
Inserting these coefficients in our assumed series solution, we have
2
2
n
!
x
n
!
x
y( x)  a0 x n [1  2
 4
 .......]
2 1!(n  1) 2 2!(n  2)!
Inserting these coefficients in our assumed series solution, we have

y( x)  a0 2 n! (1) j
n
j 0
1
x
( ) n 2 j
j!(n  j )! 2
With the result that…..
J n ( x)  (1)n J n( x)
3. Generating function
g ( x, t )  e
x
1
( )( t  )
2
t
Expanding this function in a Laurent series, we obtain
e
x
1
( )( t  )
2
t


n
J
(
x
)
t
 n
n  
It is instructive to compare.
The coefficient of tn, Jn(x), is defined to be Bessel function of the first kind of
integral order n. Expanding the exponential, we have a product of Maclaurin
series in xt/2 and –x/2t, respectively.
xt
2
e e

x
2t
x t
x s t s
 ( )
(1) ( )

r! s 0
2 s!
r 0 2

r
r

s
For a given s we get tn(n>=0) from r=n+s;
s
x n s t n s
s x s t
( )
(1) ( )
2
(n  s)!
2 s!
The coefficient tn is then
(1) s x n 2 s
xn
x n2
J n ( x)  
( )
 n  n 2
 ......
2 n! 2 (n  1)!
s 0 s!(n  s )! 2

Bessel function J0(x), J1(x) and J2(x)
4. Contour integral: Some writers prefer to start with contour integral
definitions of the Hankel function, and develop the Bessel function Jv(x) from
the Hankel functions.
The integral representation
g ( x, t )  e
x
1
( )( t  )
2
t
may easily be established as a Cauchy integral for v=n, that is , an integer.
[Recognizing that the numerator is the generating function and integrating
around the origin]
1
( x )( t 1 ) dt
J v ( x) 
e 2 t v 1

2i
t
Cut line
(Schlsefli integral)
1 ei ( x 2 )( t 1t ) dt
( x)   e
i 0
t v 1
Hv
(1)
Hv
( 2)
1 0 ( x 2 )( t 1t ) dt
( x )   i e
i e
t v 1
1
(1)
( 2)
J v ( x)  [ H v ( x)  H v ( x)]
2
1
(1)
( 2)
N v ( x)  [ H v ( x)  H v ( x)]
2i
5. Direct solution of physical problems, Fraunhofer diffraction with a circular
aperture illusterates this. Incidentally, can be treated by series expansion if
desired. Feynman develop Bessel function from a consideration of cavity
resonators.
In the theory of diffraction through a circular aperture we encounter the integral
~
a
0

2
0
eibr cos drdr
The parameter B us given by
b
2

sin 
a
 ~ 2  J 0 (br)rdr
0
Feynman develop Bessel function
from a consideration of cavity
resonators. (Homework 1)
~
2ab
a
2a
J
(
ab
)
~
J
(
sin  )
1
1
2
b
sin 

The intensity of the light in the diffraction pattern is proportional to Ф2 and
2a ) sin  ]
J
[(
1

2 ~{
}2
sin 
2a

sin   3.8317 .....
Fro green light   5.5 105 cm Hence, if a=0.5 cm
  sin   6.7 10  5(radian)  14s.
of
arc.
Homework
(2) Using only the generating function
e
(
x
1
)( t  )
2
t


n
J
(
x
)
t
 n
n  
Explicit series form Jn(x), shoe that Jn(x) has odd or even parity according
to whether n is odd or even, this
J n ( x)  (1)n J n ( x)
(3) Show by direct differentiation that
(1) s
x  2s
J v ( x)  
( )
s
!
(
s


)!
2
s 0

Satisfies the two recurrence relations
2
J v ( x)
x
J v 1 ( x)  J v 1 ( x)  2 J v ( x)
J v 1 ( x)  J v 1 ( x) 
And bessel‘s differential equation
x 2 J v( x)  xJ v ( x)  ( x 2  2 ) J v ( x)  0
(4) Show that
e
(
x
1
)( t  )
2
t


n
I
(
x
)
t
 n
n  
Thus generating modified Bessel function In(x)
(5) The chebyshev polynomials (typeII) are generated,

1
n

U
(
x
)
t
 n
1  2 xt  t 2 n 0
Using the techniques for transforming series, develop a series representation of
Un(x)
PS:請參考補充講義