- #1

- 285

- 13

Given:

\begin{equation}

\begin{split}

F &=\int_{x_a}^{x_b} g(f,f_x,x) dx

\end{split}

\end{equation}

Thus, where ##\delta f## denotes an infinitesimal variation in the function ##f##:

\begin{equation}

\begin{split}

\delta F &= F[f+\delta f]-F[f]

\end{split}

\end{equation}Where:

\begin{equation}

\begin{split}

f &\rightarrow f+ \delta f \Rightarrow f_x \rightarrow f_x + (\delta f)'

\end{split}

\end{equation}Thus:

\begin{equation}

\begin{split}

\delta F &= \int_{x_a}^{x_b} g(f+\delta f,f_x+ (\delta f)',x) dx - \int_{x_a}^{x_b} g(f,f_x,x) dx

\end{split}

\end{equation}Where:

\begin{equation}

\begin{split}

g(f+\delta f,f_x+ (\delta f)',x) \rightarrow g(f,f_x,x) + \frac{\partial g}{\partial f} \partial f + \frac{\partial g}{\partial f_x} (\partial f)'

\end{split}

\end{equation}Such that:

\begin{equation}

\begin{split}

\partial F &= \int_{x_a}^{x_b} \frac{\partial g}{\partial f} \partial f dx + \int_{x_a}^{x_b} \frac{\partial g}{\partial f_x} (\partial f)' dx

\end{split}

\end{equation}Integration by parts:

\begin{equation}

\begin{split}

\int u dv &= u v - \int v du

\end{split}

\end{equation}Setting ##u = \frac{\partial g}{\partial f_x}## and ##dv = (\delta f)' dx##, such that:

\begin{equation}

\begin{split}

\int_{x_a}^{x_b} \frac{\partial g}{\partial f_x} (\partial f)' dx &= \frac{\partial g}{\partial f_x} (\delta f) \bigg\rvert_{x_a}^{x_b} - \int_{x_a}^{x_b} (\partial f) \frac{d}{dx}\frac{\partial g}{\partial f_x} dx

\end{split}

\end{equation}Thus:

\begin{equation}

\begin{split}

\partial F &= \int_{x_a}^{x_b} \frac{\partial g}{\partial f} \partial f dx + \frac{\partial g}{\partial f_x} (\delta f) \bigg\rvert_{x_a}^{x_b} - \int_{x_a}^{x_b} (\partial f) \frac{d}{dx}\frac{\partial g}{\partial f_x} dx

\\

&=\int_{x_a}^{x_b} \left( \frac{\partial g}{\partial f} \partial f - (\partial f) \frac{d}{dx}\frac{\partial g}{\partial f_x}\right) dx + \frac{\partial g}{\partial f_x} (\delta f) \bigg\rvert_{x_a}^{x_b}

\\

&=\int_{x_a}^{x_b} \left( \frac{\partial g}{\partial f} - \frac{d}{dx}\frac{\partial g}{\partial f_x}\right) \partial f dx + \frac{\partial g}{\partial f_x} (\delta f) \bigg\rvert_{x_a}^{x_b}

\\

&=\int_{x_a}^{x_b} \left( \frac{\partial g}{\partial f} - \frac{d}{dx}\frac{\partial g}{\partial f_x}\right) \partial f dx + \frac{\partial g}{\partial f_x} (\delta f) \bigg\rvert_{x_b} -\frac{\partial g}{\partial f_x} (\delta f) \bigg\rvert_{x_a}

\end{split}

\end{equation}At an extremum, ##\partial F=0##, such that:

\begin{equation}

\begin{split}

\int_{x_a}^{x_b} \left( \frac{\partial g}{\partial f} - \frac{d}{dx}\frac{\partial g}{\partial f_x}\right) \partial f dx + \frac{\partial g}{\partial f_x} (\delta f) \bigg\rvert_{x_b} -\frac{\partial g}{\partial f_x} (\delta f) \bigg\rvert_{x_a} &= 0

\end{split}

\end{equation}THIS IS THE STEP I DON'T UNDERSTAND. WHY IS THIS TRUE?:

\begin{equation}

\begin{split}

\frac{\partial g}{\partial f_x} (\delta f) \bigg\rvert_{x_b} -\frac{\partial g}{\partial f_x} (\delta f) \bigg\rvert_{x_a} &= 0

\end{split}

\end{equation}Such that:

\begin{equation}

\begin{split}

\int_{x_a}^{x_b} \left( \frac{\partial g}{\partial f} - \frac{d}{dx}\frac{\partial g}{\partial f_x}\right) \partial f dx &= 0

\end{split}

\end{equation}Thus:

\begin{equation}

\begin{split}

\frac{\partial g}{\partial f} - \frac{d}{dx}\frac{\partial g}{\partial f_x} &= 0

\end{split}

\end{equation}