[微積] Wronskian

[yueayase (scrya)]

(1)
有關Wronskian本身和一組函數是否相依或獨立,
感覺上要先從獨立和相依的定義著手:

Definition(Linear Dependence of Functions):The n functions
f1(t), f2(t),...,fn(t) are said to be linearly dependent on
the interval I if there exists constants c1,c2,...,cn such
that c1*f1(t)+c2*f2(t)+...+cn*f(n) = 0 for all t in I. If
they are not linearly depenedent on I, it is said to be
linearly independent.

由此可知,n個函數要線性相依必須要存在一組不全為零的常數使得
c1*f1(t)+c2*f2(t)+...+cn*f(n) = 0 且要在整個區間都成立
所以如果對於整個區間,不存在不全為零的的一組常數,使得
c1*f1(t)+c2*f2(t)+...+cn*f(n) = 0,則不是線性相依(就是線性獨立,
和定義完全等價)

所以,如果說x,|x|在整條實數線是否相依,
則要去解
c1*x+c2*|x|=0

For x>=0 => (c1+c2)x=0
For x<0  => (c1-c2)x=0

=> c1+c2=0, c1-c2=0 =>c1=c2=0 => x,|x|在整條實數線上線性獨立

但是如果我改成x,|x|在x>0,是否相依呢?

c1*x+c2*|x| = (c1+c2)x = 0

=> c1=1, c2=-1,不全為零,則相依

同理,x,|x|在x<0,相依(c1*x+c2*|x|=(c1-c2)x=0, set c1=c2=1)

這會產生一個有趣的現象,似乎只要改變區間範圍,結果就會不一樣?

考慮x,|x|的Wronskian:
                             2                 2
W(x,|x|) = | x    |x|  |  = x / |x| - |x| = |x| / |x| - |x| = 0
           | 1   x/|x| |

for all x except for x = 0.

所以 Wronskian of x and |x| is identically zero for all x except for x=0.

看起來W(x) = 0,根本無法判斷相依或獨立


但"If there is a value b such that W(x) is not zero, f1,f2,...,fn are
linearly independent."這句話好像就是定義的等價敘述,所以

W(x) ≠ 0 for all x implies f1(x),f2(x),...,fn(x) are linearly independent.

好像是對的

----------------------------------------------------------------------------
(2) 微分方程有一個Wronskian和相依獨立的定理:

Theorem: Suppose y1(t),y2(t),...yn(t) are solutions of the homogenerous n-order
linear equation

         (n)            (n-1)               '
        y     + p  (t)y      + ... + p (t) y  + p (t) y = 0
                 n-1                  1          0

on an open interval I and p   i = 0,1,..,(n-1) are continuous on I.
                           i,

Let W = W(y1(t),y2(t),...,yn(t)).

(a) If y1,y2,...,yn are linearly dependent, W = 0 for all t on I
(b) If y1,y2,...,yn are linearly independent, W≠0 for all t on I.

Proof:

(a) Assume for t=a, W ≠ 0 and y1,y2,..yn are linearly depedent.

    The system
    c1*y1(a)+c2*y2(a)+ ... + cn*yn(a) = 0

         '        '               '
    c1*y1(a)+c2*y2(a)+ ... + cn*yn(a) = 0

    ....

         (n-1)        (n-1)             (n-1)
    c1*y1    (a)+c2*y2   (a)+ ... +cn*yn   (a) = 0


    has a unique solution c1=c2=...=cn=0.
    (det(A) ≠ 0 <=> Ax=0 has a unique solution x=0)

    By the definition of linear dependency,
    y1,y2,...,yn are linearly independent.(Contradiction!)
    So, if y1, y2,...,yn are linearly dependent ,  W = 0 for all t on I.

(b) Assume for t=a, W = 0 and y1,...,yn are linearly independent.
    Then, the system

    c1*y1(a)+c2*y2(a)+ ... + cn*yn(a) = 0

         '        '               '
    c1*y1(a)+c2*y2(a)+ ... + cn*yn(a) = 0

    ....

         (n-1)        (n-1)             (n-1)
    c1*y1    (a)+c2*y2   (a)+ ... +cn*yn   (a) = 0


    has a solution for c1,c2,..,cn and not all zeros.


    Use these constants c1,c2,...,cn and use the original differential equation

                     '            (n-1)
    and let y(a) = y (a) = ... = y    (a) = 0 to construct a initial value

    problem.

    By Supeposition Principle, Y(t) = c1*y1(t)+c2*y2(t)+...+cn*yn(t) are

    a general solution of the  original differential equation since

                                                     '           (n-1)
    y1,y2,...,yn are solutions. Furthermore, Y(a) = Y (a) = ...=Y     (a) = 0

    by our preceding setting of c1,c2,..,cn.Therefore, Y(t) is a solution

    to the I.V.P.

    However, by Exsitence and Uniqueness Theorem,  the I.V.P



         (n)            (n-1)               '
        y     + p  (t)y      + ... + p (t) y  + p (t) y = 0
                 n-1                  1          0

            '             (n-1)
       y(a) = y (a) = ... = y    (a) = 0

    has a unique solution.

    Then, Y(t) = c1*y1(t)+c2*y2(t)+...+cn*yn(t) = 0 for all t on I.

    Since c1,c2,...,cn are not all zeros, y1(t),y2(t),...yn(t) are

    linearly dependent and it leads to contradiction!

    So, statement (b) is true.

   看看(b)的等價敘述,不是指如果W(a) = 0 for some a on I,y1,y2,...,yn是
   線性相依嗎?(因為相依和獨立是二分的)

   ------------------------------------------------------------------------

   所以 Wronskian 等於0可不可以得到相依,是要看條件和你的定義域 I 而定的,
   不然不是沒有意義?

   我說的對不對??

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※ 編輯: yueayase        來自: 111.251.178.124      (11/15 07:52)
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