[變分法] 離散泛函極值的必要條件

此文主要討論離散泛函的極值與其必要條件，也就是所謂的離散版本的 Euler-Lagrange Equation，推薦讀者可先複習先前介紹過的 連續泛函 的極值與必要條件的相關知識，整體推導而言可謂非常類似。

考慮離散泛函
\[\left\{ \begin{align*}
  &J\left( x \right): = \sum\limits_{k = 0}^{N - 1} {F\left( {x\left( k \right),x\left( {k + 1} \right),k} \right)} ; \hfill \\
  &x\left( {{0}} \right) = {x_0};x\left( {{N}} \right) = {x_1} \hfill \\
\end{align*}  \right.
\]其中 $F(x,y,t), \frac{{\partial F}}{{\partial x}}, \frac{{\partial F}}{{\partial y}}$ 在其定義域上連續函數。我們的目標是求序列 $x(0), x(1),...,x(N)$ 使得上述泛函 $J(x)$ 達到極值。

Comments:
前述設定中的離散狀態 $x(k) := x(t_k)$ 其中 $t_k = kT$ 且 $T$ 為取樣週期 (sampling period)

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Theorem: 離散版本的 Euler-Lagrange Equation
若 $x(1),...,x(N - 1) $ 使得上述泛函 $J(x)$ 達到極值，則對任意 $k=1,2,...,N-1$
\[\frac{{\partial F\left( {x(k),x(k + 1),k} \right)}}{{\partial x\left( k \right)}} + \frac{{\partial F\left( {x(k - 1),x(k),k - 1} \right)}}{{\partial x(k)}} =0\]=====================================

Proof: 令 $\delta x(k)$ 為 $x(k)$ 的變分，由於 $x(0) = x_0$ 與 $x(N)=x_1$，故 $\delta x(0) = \delta x(N) =0$ ，現在我們觀察 $J(x)$ ，由假設可知  $x(1),...,x(N - 1) $ 使得泛函 $J$ 達到極值，故
\[
J(x + \alpha \delta x) \geq J(x)
\]且此表明 $ J\left( {x + \alpha \delta x} \right)$ 在 $\alpha =0$ 處達到極值，由變分與泛函極值關係可知
\[
\delta J\left( {x\left( k \right)} \right) = {\left. {\frac{\partial }{{\partial \alpha }}J\left( {x\left( k \right) + \alpha \delta x\left( k \right)} \right)} \right|_{\alpha  = 0}} = 0
\]其中
\[J\left( {x(k) + \alpha \delta x(k)} \right) = \sum\limits_{k = 0}^{N - 1} {F\left( {x(k) + \alpha \delta x(k),x(k + 1) + \alpha \delta x(k + 1),k} \right)}
\]因此
\[
\delta J\left( {x\left( k \right)} \right) = {\left. {\frac{\partial }{{\partial \alpha }}\sum\limits_{k = 0}^{N - 1} {F\left( {x(k) + \alpha \delta x(k),x(k + 1) + \alpha \delta x(k + 1),k} \right)} } \right|_{\alpha  = 0}} = 0
\]故我們可推得
\[\begin{align*}
  &{\left. {\frac{\partial }{{\partial \alpha }}\sum\limits_{k = 0}^{N - 1} {F\left( {x(k) + \alpha \delta x(k),x(k + 1) + \alpha \delta x(k + 1),k} \right)} } \right|_{\alpha  = 0}} = 0 \hfill \\
 &  \Rightarrow {\left. {\sum\limits_{k = 0}^{N - 1} {\frac{\partial }{{\partial \alpha }}F\left( {x(k) + \alpha \delta x(k),x(k + 1) + \alpha \delta x(k + 1),k} \right)} } \right|_{\alpha  = 0}} \hfill \\
 &  \Rightarrow {\left. {\sum\limits_{k = 0}^{N - 1} {\frac{{\partial F}}{{\partial x\left( k \right)}}\delta x(k) + \frac{{\partial F}}{{\partial x(k + 1)}}\delta x(k + 1)} } \right|_{\alpha  = 0}} = 0  \;\;\;\; (*)
\end{align*}
\] 現在觀察上式的第二項，利用變數變換 定義 $k:=m-1$ 則我們可改寫為
\[\begin{gathered}
  \sum\limits_{k = 0}^{N - 1} {\frac{{\partial F\left( {x(k),x(k + 1),k} \right)}}{{\partial x(k + 1)}}\delta x(k + 1)}  = \sum\limits_{m = 1}^N {\frac{{\partial F\left( {x(m - 1),x(m),m - 1} \right)}}{{\partial x(m)}}\delta x(m)}  \hfill \\
   = \frac{{\partial F\left( {x(N - 1),x(N),N - 1} \right)}}{{\partial x(N)}}\delta x(N) + \sum\limits_{m = 1}^{N - 1} {\frac{{\partial F\left( {x(m - 1),x(m),m - 1} \right)}}{{\partial x(m)}}\delta x(m)}  \hfill \\
   = \frac{{\partial F\left( {x(N - 1),x(N),N - 1} \right)}}{{\partial x(N)}}\delta x(N) + \sum\limits_{k = 1}^{N - 1} {\frac{{\partial F\left( {x(k - 1),x(k),k - 1} \right)}}{{\partial x(k)}}\delta x(k)}  \hfill \\
\end{gathered} \]現在將上述結果代回 $(*)$，故可得
\[\small \begin{align*}
  \delta J\left( {x\left( k \right)} \right) &= \sum\limits_{k = 0}^{N - 1} {\frac{{\partial F}}{{\partial x\left( k \right)}}\delta x(k) + \frac{{\partial F}}{{\partial x(k + 1)}}\delta x(k + 1)}  \hfill \\
  & = \sum\limits_{k = 0}^{N - 1} {\frac{{\partial F}}{{\partial x\left( k \right)}}\delta x(k) + \frac{{\partial F\left( {x(N - 1),x(N),N - 1} \right)}}{{\partial x(N)}}\delta x(N) + \frac{{\partial F\left( {x(k - 1),x(k),k - 1} \right)}}{{\partial x(k)}}\delta x(k)}  \hfill \\
 &  = \frac{{\partial F\left( {x(N - 1),x(N),N - 1} \right)}}{{\partial x(N)}}\delta x(N) + \sum\limits_{k = 0}^{N - 1} {\left( {\frac{{\partial F}}{{\partial x\left( k \right)}} + \frac{{\partial F\left( {x(k - 1),x(k),k - 1} \right)}}{{\partial x(k)}}} \right)\delta x(k)}  \hfill \\
\end{align*}
\]注意到上式中 $\delta x(N) =0$ 且由於 $\delta x(k)$ 為任意變分，故由 $\delta J = 0$ 我們可知
\[\begin{align*}
 & \frac{{\partial F\left( {x(N - 1),x(N),N - 1} \right)}}{{\partial x(N)}}\delta x(N) +  \hfill \\
  \begin{array}{*{20}{c}}
  {}&{}
\end{array}&\;\;\;\; \sum\limits_{k = 0}^{N - 1} {\left( {\frac{{\partial F\left( {x(k),x(k + 1),k} \right)}}{{\partial x\left( k \right)}} + \frac{{\partial F\left( {x(k - 1),x(k),k - 1} \right)}}{{\partial x(k)}}} \right)\delta x(k)}  = 0
\end{align*} \]亦即對任意 $k=0,1,...,N-1$，
\[\frac{{\partial F\left( {x(k),x(k + 1),k} \right)}}{{\partial x\left( k \right)}} + \frac{{\partial F\left( {x(k - 1),x(k),k - 1} \right)}}{{\partial x(k)}} = 0\;\;\;\; \square
\]