我想分享關於ode的variation of parameters中
不需要預先猜測把原本線性組合的常數換成函數
也能推導出相同結果的想法:
考慮一階ode形如: x'(t) = A(t)x(t)+b(t), 其中x, b為n維向量，A為nxn矩陣
唯一的變數為t
(Case i)
如果A(t)是常數，即A(t)≡A for some nxn matrix A.
那我們可以模仿解純量一階ode的想法:
x'(t)+a(t)x(t)=b(t)
∫a(t)dt
兩邊同乘積分因子: I(t)=e
原方程式變為: (I(t)x(t))'=I(t)b(t) 然後積分後除上I(t)就是答案
向量版本也一樣:
x'(t)-Ax(t) = b(t)
-At -At -At
同乘 I(t) = e => (e x(t))' = e b(t)
At -At
=> x(t) = e ∫e b(t) dt
這個長得跟variation of parameters的公式是一樣的
那麼... A(t)若不是常數的話呢?
我們試看看找出積分因子:
x'(t)-A(t)x(t) = b(t)
I(t)x'(t)-I(t)A(t)x(t) = I(t)b(t)
我們希望左式變成(I(t)x(t))'= I(t)x'(t)+I'(t)x(t)
可能的選擇是I'(t) = -I(t)A(t)
-∫A(t)dt
這讓我們也許會想說是不是I(t)=e 呢?
但這不總是可以這樣，只有:
(Case ii) 若A(s)A(t) = A(t)A(s)成立
-∫A(t)dt ∞ (-∫A(t)dt)^n
I(t) = e = Σ ---------------
n=0 n!
∞ n(-∫A(t)dt)^(n-1) ∞ (-∫A(t)dt)^n
=>I'(t) = -Σ ------------------- A(t) = -Σ -------------- A(t)=-I(t)A(t)
n=1 n! n=0 n!
(注意:可以這樣做的原因是因為A(s)A(t) = A(t)A(s)，在套n次product rule時，
這些含有積分的n-1項才會相等)
-∫A(t)dt
所以可以發現這個情況下，真的就是上面說的e
也就是我們可以把原式化成: (I(t)x(t))'=I(t)b(t)
-1 ∫A(t)dt -∫A(t)dt
=> x(t) = I (t)∫I(t)b(t)dt = e ∫e b(t)dt
這個也跟variation of parameters的公式長得很像
但若是A(s)A(t) = A(t)A(s)不成立，我們不能把它合成像上面那樣，例如n=2時，
t t
d(-∫A(s)ds)^2 d(∫A(s)ds)(∫A(t)dt) t t
--------------- = ---------------------- = A(t)∫A(s)ds+∫A(s)ds A(t)
dt dt
t t
這不能變成2A(t)∫A(s)ds或是2∫A(s)dsA(t) 因為A(t)A(s)≠A(s)A(t)
那這時候要怎麼辦呢? 答案是把積分因子換成fundamental matrix
A fundamental matrix of system X'(t) = A(t)X(t) is a solution to this system
with each columns being linearly independent.(X,A都是nxn矩陣)
也可以看出fundamenal matrix的column構成x'(t)=A(t)x(t)(A是上面那個矩陣，
x是n維向量)解空間的basis
對於x'(t)=A(t)x(t)+b(t)，x,b為n維向量，A為nxn矩陣.令Φ(t)是fundamental matrix
=> Φ'(t)=A(t)Φ(t)
-1
=> A(t) = Φ'(t)Φ (t) (因為根據定義Φ的column線性獨立，所以可逆)
照前面所說的，我們想要找出I'(t)=-I(t)A(t)的I，把上面的關係式帶進去，
-1
我們想要找出I，使得I'(t)=-I(t)Φ'(t)Φ (t)成立
=> I'(t)Φ(t)=-I(t)Φ'(t)
=> I'(t)Φ(t)+I(t)Φ'(t) = O
=> (I(t)Φ(t))' = O (by product rule)
=> I(t)Φ(t) = C for some nxn constant matrix C
-1
=> I(t) = CΦ (t)
-1
Now set C = I, then I(t) = Φ (t)
-1 -1
所以原方程式可變成(Φ (t)x(t))' = Φ (t)b(t)
-1
=> x(t) = Φ(t)∫Φ (t)b(t)dt
這個就是我們熟悉的variation of parameters的公式
注意到我沒有先假設特解要長成Φ(t)c(t)然後把c(t)解出來
我只假設了I'(t) = -I(t)A(t)成立(矩陣基本上不能直接消AC=BC不一定A=B，
所以這個是我假設可以這樣，然後做看看)
這也可以解釋為什麼書本要介紹fundamental matrix的概念
fundamental matrix不只是為了解X'(t)=A(t)X(t), X,A為矩陣而要引進
也不是為了能把x'(t)=A(t)x(t), A為矩陣, x為向量表示得比較好看而引進
-∫A(t)dt
我覺得最重要的原因是因為在找積分因子的過程，無法用e 當積分因子
但我們還是希望有如純量一階ode的結果，所以把那個積分因子換成fundamental matrix
來解決這個問題，所以才需要引進
而這個矩陣向量版的variation of parameters也可以幫助我們推出n階線性方程的
variation of parameters，並且不用突然設定奇怪的
(k) (k) (k)
c'(t)x (t)+c'(t)x (t)+...+c'(t)x (t) = 0, k < n-2
1 1 2 2 n n
這種條件，然後解出c1,...cn
過程如下:
(n) (n-1) (n-2)
考慮x (t) +a (t)x (t)+a (t)x (t) + ... a (t)x(t) = f(t)
n n-1 1
已知x (t),x (t), ... ,x (t)為n linearly independent solutions of the
1 2 n
corresponding homogeneos equation:
(n) (n-1) (n-2)
x (t) + a (t)x (t) + a x (t) + ... + a (t)x(t) = 0
n n-1 1
Then, set
(n-1)
y(t) = [y (t)], where y (t) = x(t), y (t) = x'(t),..., y (t) = x (t)
1 1 2 n
[y (t)]
2
...
[y (t)]
n
A(t) = [ 0 1 0 ... 0 ] [ 0 ]
[ 0 0 1 ... 0 ] [ 0 ]
[ ... ], b(t) = ...
[ 0 0 ...0 1 ] [ 0 ]
[-a (t) -a (t) ... -a (t)] [f(t)]
1 2 n
則觀察可得:
(n)
y'(t) = x'(t) = y (t), y'(t) = x''(t) = y (t), ... y'(t) = x (t)
1 2 2 3 n
(n) (n-1)
x (t) = -a (t)x(t)-a (t)x'(t)-...-a (t)x (t) + f(t)
1 2 n
=-a (t)y (t)-a (t)y (t)-...-a (t)y (t) + f(t)
1 1 2 2 n n
y'(t) = A(t)y(t)+b(t)
而[x (t) ] [x (t) ] [x (t) ]
1 2 n
[x'(t) ] [x'(t) ] [x'(t) ]
1 2 n 為 n linearly independent solution of
... ... ... ...
(n-1) (n-1) (n-1)
[x (t)] [x (t)] [x (t)]
1 2 n
the system y'(t) = A(t)y(t)
所以Φ(t) = [x (t) x (t) ... x (t) ]
1 2 n
[x'(t) x'(t) ... x'(t) ] 為fundamental matrix of the system
1 2 n
...
(n-1) (n-1) (n-1)
[x (t) x (t) ... x (t)]
1 2 n
Y'(t) = A(t)Y(t), Y(t)為nxn矩陣
利用variation of parameters:
-1
y(t) = Φ(t)∫Φ (t)b(t)dt
-1 -1
Set c(t) = ∫Φ (t)b(t)dt => c'(t) = Φ (t)b(t) => Φ(t)c'(t) = b(t)
用矩陣乘法乘開，第一個分量就是:
y(t) = c (t)x (t) + c (t)x (t) + ... + c (t)x (t)
1 1 2 2 n n
而Φ(t)c'(t) = b(t)把左式用矩陣乘法乘開和右邊比較係數可得:
c '(t)x (t) + c'(t)x (t) + ... + c'(t)x (t) = 0
1 1 2 2 n n
c'(t)x' (t) + c'(t)x'(t) + ... + c'(t)x'(t) = 0
1 1 2 2 n n
...
(n-2) (n-2) (n-2)
c'(t)x (t) + c'(t)x (t) + ... + c'(t)x (t) = 0
1 1 2 2 n n
(n-1) (n-1) (n-1)
c'(t)x (t) + c'(t)x (t) + ... + c'(t)x (t) = f(t)
1 1 2 2 n n
這個就是我們在variation of parameters所設定的n個條件
我認為這種講法會讓做variation of parameters變得更加自然
不需要做過多的猜測
希望這篇文章對各位有幫助~~
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