SICM Exercise 1.07
Solution to exercise 1.7 of Structure and Interpretation of Classical Mechanics by Gerald Jay Sussman and Jack Wisdom.
Properties of \(\delta\)
🛈 Note
Show that \(\delta\) has the properties 1.23–1.27.
Prove: \(\delta_\eta(fg)[q] = \delta_\eta f[q]g[q] + f[q] \delta_\eta g[q]\)
\begin{equation*}
\delta_\eta (fg)[q] = \lim_{\epsilon \to 0} \frac{f[q + \epsilon \eta] g[q + \epsilon \eta] - f[q]g[q]}{\epsilon}
= \lim_{\epsilon \to 0} \frac{f[q + \epsilon \eta] g[q + \epsilon \eta] - f[q + \epsilon \eta]g[q] + f[q + \epsilon \eta]g[q] - f[q]g[q]}{\epsilon}
= \lim_{\epsilon \to 0} \frac{f[q + \epsilon \eta](g[q + \epsilon \eta] - g[q]) + g[q](f[q + \epsilon \eta] - f[q])}{\epsilon}
= \lim_{\epsilon \to 0} \left(f[q + \epsilon \eta] \frac{g[q + \epsilon \eta] - g[q]}{\epsilon} + g[q]\frac{f[q + \epsilon \eta] - f[q]}{\epsilon}\right)
= \delta_\eta f[q]g[q] + f[q] \delta_\eta g[q]
\end{equation*}
Prove: \(\delta_\eta (f + g)[q] = \delta_\eta f[q] + \delta_\eta g[q]\)
\begin{equation*} \delta_\eta (f + g)[q] = \lim_{\epsilon \to 0} \frac{(f+g)[q + \epsilon \eta] - (f+g)[q]}{\epsilon} = \lim_{\epsilon \to 0} \frac{f[q + \epsilon \eta] + g[q + \epsilon \eta] - f[q] - g[q]}{\epsilon} = \lim_{\epsilon \to 0} \left( \frac{f[q + \epsilon \eta] - f[q]}{\epsilon} + \frac{g[q + \epsilon \eta] - g[q]}{\epsilon} \right) = \lim_{\epsilon \to 0} \frac{f[q + \epsilon \eta] - f[q]}{\epsilon} + \lim_{\epsilon \to 0} \frac{g[q + \epsilon \eta] - g[q]}{\epsilon} = \delta_\eta f[q] + \delta_\eta g[q] \end{equation*}
Prove: \(\delta_\eta (cf)[q] = c \delta_\eta f[q]\)
\begin{equation*} \delta_\eta (cf)[q] = \lim_{\epsilon \to 0} \frac{(cf)[q + \epsilon \eta] - (cf)[q]}{\epsilon} = \lim_{\epsilon \to 0} \frac{c \cdot f[q + \epsilon \eta] - c \cdot f[q]}{\epsilon} = c \cdot \lim_{\epsilon \to 0} \frac{f[q + \epsilon \eta] - f[q]}{\epsilon} = c \delta_\eta f[q] \end{equation*}
Prove: \(\delta_\eta h[q] = \left( DF \circ g[q] \right) \delta_\eta g[q]\)
\[\delta_\eta h[q] = \lim_{\epsilon \to 0} \frac{h[q + \epsilon \eta] - h[q]}{\epsilon}\]Substitute \(h[q] = F \circ g[q]\)
\[= \lim_{\epsilon \to 0} \frac{F \circ g[q + \epsilon \eta] - F \circ g[q]}{\epsilon} \]Substitute \(j(\epsilon) = g[q + \epsilon \eta]\)
\[= \lim_{\epsilon \to 0} \frac{F \circ j(\epsilon) - F \circ j(0)}{\epsilon} \]Using the derivative represention
\[= D(F \circ j(0)) \]Now we can use the chain rule
\[= (DF \circ j(0)) Dj(0) \]Finally, we use the variational representation for \(Dj(0)\) and we substitute \(g[q]\) for \(j(0)\)
\[= (DF \circ g[q]) \delta_\eta g[q]\]Prove: \(D \delta_\eta f[q] = \delta_\eta g[q]\)
TODO: Why can we take the derivative into the limit?
\[D \delta_\eta f[q] = D \lim_{\epsilon \to 0} \frac{f[q + \epsilon \eta] - f[q]}{\epsilon} \]We can move the derivative into the limit
\[= \lim_{\epsilon \to 0} \frac{D(f[q + \epsilon \eta]) - D(f[q])}{\epsilon} \]Using the definiton of the variation…
\[= \delta_\eta (D (f[q])) = \delta_\eta g[q]\]