Euler-Lagrange equation
Question:
假設路徑\[y=f(x)\]，從\[(x_1,f(x_1))\]到\[(x_2,f(x_2))\]，試求\[f(x)\]之微分方程，使\[H(y’,y,x)=\int_{x_1}^{x_2}F(y’,y,x)\ dx\]有最小值。
Answer:
p.s. 此非正規證明，僅是一個簡易推導 因為要求\[H\]有最小值，因此對於任意微小偏移量\[\eta(x)\]\[\text{（}\because\text{所有路徑皆必須有相同起始點和終點，}\therefore \eta(x_1)=0\text{且}\eta(x_2)=0 \text{）}\]，\[H{之變化量（稱為}H\text{的變分}\delta H\text{）應為0。}\]
$$ \delta H=\int_{x_1}^{x_2}F((y+\eta)’,y+\eta,x)-F(y’,y,x)\ dx\ $$
$$ =\int_{x_1}^{x_2}F(y’+\eta’,y+\eta,x)-F(y’,y,x)\ dx\ $$
$$ =\int_{x_1}^{x_2}\frac{\partial F}{\partial y}\eta+\frac{\partial F}{\partial y’}\eta’\ dx\ $$
$$ =\int_{x_1}^{x_2}\frac{\partial F}{\partial y}\eta \ dx+[\frac{\partial F}{\partial y’}\eta]_{x_1}^{x_2} $$
$$ -\int_{x_1}^{x_2}\frac{d}{dx}\frac{\partial F}{\partial y’}\eta\ dx $$
$$ =\int_{x_1}^{x_2}(\frac{\partial F}{\partial y} -\frac{d}{dx}\frac{\partial F}{\partial y’})\eta\ dx $$
\[\because\text{我們要求}y=f(x)\text{使得}H(y’,y,x)\text{有最小值}\]
$$ \therefore \forall \eta,\delta H=0\ $$
$$ \Rightarrow \frac{\partial F}{\partial y} -\frac{d}{dx}\frac{\partial F}{\partial y’}=0 $$ 此式稱為歐拉─拉格朗日方程式（Euler-Lagrange equation）。