\(\)
Problem 7-2
If \(F[x]=x(\tau)\), show that
\begin{equation}
\frac{\delta F}{\delta x(s)}=\delta(\tau-s)
\tag{7-26}
\end{equation}
\frac{\delta F}{\delta x(s)}=\delta(\tau-s)
\tag{7-26}
\end{equation}
( 解答 ) \(F[x]=x(\tau)\) のとき， 1次までの式 (7-20) は次となる：
\begin{equation}
F[x+\eta]-F[x]=\bigl\{x(\tau)+\eta(\tau)\bigr\}-x(\tau)=\eta(\tau)
=\int \frac{\delta x(\tau)}{\delta x(s)}\eta(s)\,ds
\tag{1}
\end{equation}
F[x+\eta]-F[x]=\bigl\{x(\tau)+\eta(\tau)\bigr\}-x(\tau)=\eta(\tau)
=\int \frac{\delta x(\tau)}{\delta x(s)}\eta(s)\,ds
\tag{1}
\end{equation}
これが成り立つには, 式 (7-26) であれば十分であることは明らかである (\(\delta\)-関数の性質を利用して)：
\begin{equation}
\int \frac{\delta x(\tau)}{\delta x(s)}\eta(s)\,ds=\int \eta(s)\,\delta(s-\tau)\,ds
=\eta(\tau)
\tag{2}
\end{equation}
\int \frac{\delta x(\tau)}{\delta x(s)}\eta(s)\,ds=\int \eta(s)\,\delta(s-\tau)\,ds
=\eta(\tau)
\tag{2}
\end{equation}
または， M.S.Swanson の汎関数微分の定義式 (sw-1) を用いるならば, 式 (7-26) は容易に示すことが出来る：
\begin{align}
&\frac{\delta \mathcal{F}}{\delta g(y)}=\lim_{\epsilon\to 0}\frac{\mathcal{F}[g(x)+\epsilon\,\delta(x-y)]
-\mathcal{F}[g(x)]}{\epsilon},\ \mathcal{F}[g(x)]\equiv g(x)=x(\tau),\ g(y)=x(s)\\
&\rightarrow\ \frac{\delta x(\tau)}{\delta x(s)}=\lim_{\epsilon\to 0}
\frac{x(\tau)+\epsilon\delta(\tau-s)-x(\tau)}{\epsilon}=\delta(\tau-s)
\tag{3}
\end{align}
&\frac{\delta \mathcal{F}}{\delta g(y)}=\lim_{\epsilon\to 0}\frac{\mathcal{F}[g(x)+\epsilon\,\delta(x-y)]
-\mathcal{F}[g(x)]}{\epsilon},\ \mathcal{F}[g(x)]\equiv g(x)=x(\tau),\ g(y)=x(s)\\
&\rightarrow\ \frac{\delta x(\tau)}{\delta x(s)}=\lim_{\epsilon\to 0}
\frac{x(\tau)+\epsilon\delta(\tau-s)-x(\tau)}{\epsilon}=\delta(\tau-s)
\tag{3}
\end{align}
