\(\)
Problem 7-1
If \(\displaystyle{S[x(t)]=\int_{t_1}^{t_2} L(\dot{x},x,t)\,dt}\), show that, for any \(s\) inside the range \(t_1\) to \(t_2\),
\begin{equation}
\def\ppdiff#1#2{\frac{\partial #1}{\partial #2}}
\frac{\delta S}{\delta x(s)}=-\frac{d}{ds}\left(\ppdiff{L}{\dot{x}}\right)+\ppdiff{L}{x}
\tag{7-25}
\end{equation}
\def\ppdiff#1#2{\frac{\partial #1}{\partial #2}}
\frac{\delta S}{\delta x(s)}=-\frac{d}{ds}\left(\ppdiff{L}{\dot{x}}\right)+\ppdiff{L}{x}
\tag{7-25}
\end{equation}
where the partial derivatives are evaluated at \(t=s\).
( 解答 ) 関数 \(x(t)\) が任意の \(t=s\) において任意の小さな \(\eta(s)\) だけ変化したとき, すなわち \(\delta x(s)=\eta(s)\) であるとき, 作用 \(S\) の変分 \(\delta S\) は次である：
\begin{align}
\delta S &= S[x+\eta]-S[x]=\int_{t_1}^{t_2}L(\dot{x}+\dot{\eta},x+\eta,t)\,ds
-\int_{t_1}^{t_2}L(\dot{x},x,t)\,ds\\
&=\int_{t_1}^{t_2}\left\{L(\dot{x}+\dot{\eta},x+\eta,s)-L(\dot{x},x,s)\right\}\,ds\\
&=\int_{t_1}^{t_2}\left\{\ppdiff{L}{\dot{x}}\dot{\eta}+\ppdiff{L}{x}\eta\right\}\,ds
=\int_{t_1}^{t_2}\ppdiff{L}{\dot{x}}\dot{\eta}\,ds+\int_{t_1}^{t_2}\ppdiff{L}{x}\eta\,ds
\tag{1}
\end{align}
\delta S &= S[x+\eta]-S[x]=\int_{t_1}^{t_2}L(\dot{x}+\dot{\eta},x+\eta,t)\,ds
-\int_{t_1}^{t_2}L(\dot{x},x,t)\,ds\\
&=\int_{t_1}^{t_2}\left\{L(\dot{x}+\dot{\eta},x+\eta,s)-L(\dot{x},x,s)\right\}\,ds\\
&=\int_{t_1}^{t_2}\left\{\ppdiff{L}{\dot{x}}\dot{\eta}+\ppdiff{L}{x}\eta\right\}\,ds
=\int_{t_1}^{t_2}\ppdiff{L}{\dot{x}}\dot{\eta}\,ds+\int_{t_1}^{t_2}\ppdiff{L}{x}\eta\,ds
\tag{1}
\end{align}
この第 1 項は部分積分を行い, 端点で \(\eta(t_1)=\eta(t_2)=0\) であるとして次となる：
\begin{equation}
\int_{t_1}^{t_2}\ppdiff{L}{\dot{x}}\dot{\eta}\,ds=\left[\ppdiff{L}{\dot{x}}\eta
\right]_{t_1}^{t_2}-\int_{t_1}^{t_2}\frac{d}{ds}\left(\ppdiff{L}{\dot{x}}\right)\eta\,ds
=-\int_{t_1}^{t_2}\frac{d}{ds}\left(\ppdiff{L}{\dot{x}}\right)\eta\,ds
\tag{2}
\end{equation}
\int_{t_1}^{t_2}\ppdiff{L}{\dot{x}}\dot{\eta}\,ds=\left[\ppdiff{L}{\dot{x}}\eta
\right]_{t_1}^{t_2}-\int_{t_1}^{t_2}\frac{d}{ds}\left(\ppdiff{L}{\dot{x}}\right)\eta\,ds
=-\int_{t_1}^{t_2}\frac{d}{ds}\left(\ppdiff{L}{\dot{x}}\right)\eta\,ds
\tag{2}
\end{equation}
これを式 (1) に代入すると,
\begin{align}
\delta S&=-\int_{t_1}^{t_2}\frac{d}{dt}\left(\ppdiff{L}{\dot{x}}\right)\eta\,ds
+\int_{t_1}^{t_2}\ppdiff{L}{x}\eta\,ds
=\int_{t_1}^{t_2}\left\{-\frac{d}{ds}\left(\ppdiff{L}{\dot{x}}\right)+\ppdiff{L}{x}\right\}\eta\,ds
\tag{3}
\end{align}
\delta S&=-\int_{t_1}^{t_2}\frac{d}{dt}\left(\ppdiff{L}{\dot{x}}\right)\eta\,ds
+\int_{t_1}^{t_2}\ppdiff{L}{x}\eta\,ds
=\int_{t_1}^{t_2}\left\{-\frac{d}{ds}\left(\ppdiff{L}{\dot{x}}\right)+\ppdiff{L}{x}\right\}\eta\,ds
\tag{3}
\end{align}
任意の汎関数 \(F=S\) の1次の変分に対して成り立つ式 (7-24) と, この式を比較してみる：
\begin{equation}
\delta S=\int \frac{\delta S}{\delta x(s)}\delta x(s)\,ds\quad\leftrightarrow\quad
\delta S=\int_{t_1}^{t_2}\left\{-\frac{d}{ds}\left(\ppdiff{L}{\dot{x}}\right)+\ppdiff{L}{x}\right\}\eta\,ds
\tag{4}
\end{equation}
\delta S=\int \frac{\delta S}{\delta x(s)}\delta x(s)\,ds\quad\leftrightarrow\quad
\delta S=\int_{t_1}^{t_2}\left\{-\frac{d}{ds}\left(\ppdiff{L}{\dot{x}}\right)+\ppdiff{L}{x}\right\}\eta\,ds
\tag{4}
\end{equation}
この場合の \(\delta x(s)\) に相当するのは \(\eta\) であるから, 結局求めるべき式 (7-25) が言える：
\begin{equation}
\frac{\delta S}{\delta x(s)}=-\frac{d}{ds}\left(\ppdiff{L}{\dot{x}}\right)+\ppdiff{L}{x}
\tag{5}
\end{equation}
\frac{\delta S}{\delta x(s)}=-\frac{d}{ds}\left(\ppdiff{L}{\dot{x}}\right)+\ppdiff{L}{x}
\tag{5}
\end{equation}
この結果は, 前述したブログ記事中の M.S.Swanson の合成関数の微分則の式 (sw-12) を用いることでより簡便に求めることも出来る．\(\displaystyle{S=\int dt\mathcal{L}}\) において \(\mathcal{L}=\mathcal{L}[(x(t),\dot{x}(t))]\) であるから， 式(sw-12) より
\begin{align}
\delta S&=\int dt\,\delta \mathcal{L}
=\int dt\,\left\{\frac{\delta \mathcal{L}[x]}{\delta x(t)}\delta x(t)
+\frac{\delta\mathcal{L}[x]}{\delta \dot{x}(t)}\delta \dot{x}(t)\right\}\\
&=\int dt\,\frac{\delta \mathcal{L}}{\delta x(t)}\delta x(t)
+\int dt\,\frac{\delta \mathcal{L}}{\delta \dot{x}(t)}\frac{d}{dt}\delta x(t)
\tag{6}
\end{align}
\delta S&=\int dt\,\delta \mathcal{L}
=\int dt\,\left\{\frac{\delta \mathcal{L}[x]}{\delta x(t)}\delta x(t)
+\frac{\delta\mathcal{L}[x]}{\delta \dot{x}(t)}\delta \dot{x}(t)\right\}\\
&=\int dt\,\frac{\delta \mathcal{L}}{\delta x(t)}\delta x(t)
+\int dt\,\frac{\delta \mathcal{L}}{\delta \dot{x}(t)}\frac{d}{dt}\delta x(t)
\tag{6}
\end{align}
ここで第 2 項について部分積分を行うならば, 変分 \(\delta x\) は端点ではゼロであることから,
\begin{equation}
\int dt\,\frac{\delta \mathcal{L}}{\delta \dot{x}(t)}\frac{d}{dt}\delta x(t)
=\left.\frac{\delta \mathcal{L}}{\delta \dot{x}}\delta x\right|_{t_a}^{t_b}-\int dt\,\frac{d}{dt}
\frac{\delta \mathcal{L}}{\delta \dot{x}}\delta x
=-\int dt\,\frac{d}{dt}\frac{\delta \mathcal{L}}{\delta \dot{x}}\delta x
\tag{7}
\end{equation}
\int dt\,\frac{\delta \mathcal{L}}{\delta \dot{x}(t)}\frac{d}{dt}\delta x(t)
=\left.\frac{\delta \mathcal{L}}{\delta \dot{x}}\delta x\right|_{t_a}^{t_b}-\int dt\,\frac{d}{dt}
\frac{\delta \mathcal{L}}{\delta \dot{x}}\delta x
=-\int dt\,\frac{d}{dt}\frac{\delta \mathcal{L}}{\delta \dot{x}}\delta x
\tag{7}
\end{equation}
この結果を式 (6) に代入し, また式 (sw-6) を用いると,
\begin{align}
\delta S&=\int dt\,\frac{\delta \mathcal{L}}{\delta x(t)}\delta x(t)
-\int dt\,\frac{d}{dt}\frac{\delta \mathcal{L}}{\delta \dot{x}}\delta x
=\int dt\,\left\{\frac{\delta \mathcal{L}}{\delta x(t)}
-\frac{d}{dt}\frac{\delta \mathcal{L}}{\delta \dot{x}}\right\}\delta x(t)\\
&\equiv \int dt\,\frac{\delta S}{\delta x(t)}\delta x(t)
\tag{8}
\end{align}
\delta S&=\int dt\,\frac{\delta \mathcal{L}}{\delta x(t)}\delta x(t)
-\int dt\,\frac{d}{dt}\frac{\delta \mathcal{L}}{\delta \dot{x}}\delta x
=\int dt\,\left\{\frac{\delta \mathcal{L}}{\delta x(t)}
-\frac{d}{dt}\frac{\delta \mathcal{L}}{\delta \dot{x}}\right\}\delta x(t)\\
&\equiv \int dt\,\frac{\delta S}{\delta x(t)}\delta x(t)
\tag{8}
\end{align}
よって, 式 (7-25) がやはり成り立つと言える：
\begin{equation}
\frac{\delta S}{\delta x(t)}=\frac{\delta \mathcal{L}}{\delta x(t)}
-\frac{d}{dt}\frac{\delta \mathcal{L}}{\delta \dot{x}}
\tag{9}
\end{equation}
\frac{\delta S}{\delta x(t)}=\frac{\delta \mathcal{L}}{\delta x(t)}
-\frac{d}{dt}\frac{\delta \mathcal{L}}{\delta \dot{x}}
\tag{9}
\end{equation}
