I am in Section 2.12.1 of Calculus of Variations by Gelfand & Fomin. I am attempting to follow the proof of the Euler equation for Constrained Variation (Theorem 1, pg. 42). However, I'm confused at certain steps (please see Proof and My Questions below) and would appreciate some help.
Theorem
"Given the functional
$$J[y]=\int_{a}^{b}F(x,y,y')dx$$
and admissible curves, $y=y(x)$, satisfying boundary conditions
\begin{aligned}
y(a)=A \\
y(b)=B \\
\end{aligned}
and constraint
$$K[y]=\int_{a}^{b}G(x,y,y')dx=C$$
for some functional $K$, integrand $G$, and constant $C$, let $J$ have an extremum for $y=y(x)$
where $y$ is not an extremal of $K$.Then there exists some constant, $\lambda$, such that $y$ is an
extremal of the functional
$$\int_{a}^{b}\left(F+\lambda G\right)dx$$
satisfying the constrained Euler equation
$$\left(F_{y}-\frac{d}{dx}F_{y'}\right)+\lambda \left(G_{y}-\frac{d}{dx}G_{y'}\right)=0."$$
Proof
Choose two points $x_{1}$ and $x_{2}$ in $[a,b]$ such that $x_{1}$ is arbitrary and $x_{2}$ satisfies
$$\left.\frac{\delta G}{\delta y}\right|_{x=x_{2}}\ne0.$$
(i.e. the variational derivative of $G$ at $x_{2}$ is nonzero). We know the point $x_{2}$ exists because $y$ is
not an extremal of $K$.
Give $y$ an increment $\Delta y$
$$\Delta y=\delta_{1}y(x)+\delta_{2}y(x)$$
where
$\delta_{1}y(x)\ne 0$ only in a neighborhood of $x_{1}$,
$\delta_{2}y(x)\ne 0$ only in a neighborhood of $x_{2}$.
(NOTE: The increment $h(x)$ of $J$ is also called the variation of $y(x)$, such that $h$ is written as
$$h=h(x)=\delta y(x).)$$
Since an arbitrary functional, $H$, can be written as
$$\Delta H \equiv H[y+h]-H[y]=\left\{\left.\frac{\delta H}{\delta y}\right|_{x=x_{0}}+\epsilon\right\}\Delta \sigma$$
if
$h(x)\ne 0$ in a neighborhood of $x_{0}$, and
$\Delta \sigma$ is the area between $y=h(x)$ and the $x$-axis
we can write $\Delta J$ as
$$\Delta J=\left\{\left.\frac{\delta F}{\delta y}\right|_{x=x_{1}}+\epsilon_{1}\right\}\Delta \sigma_{1}+\left\{\left.\frac{\delta F}{\delta y}\right|_{x=x_{2}}+\epsilon_{2}\right\}\Delta \sigma_{2}\tag{1}\label{eq1}$$
where
$$\Delta \sigma_{1}=\int_{a}^{b}\delta_{1}y(x)dx$$
$$\Delta \sigma_{2}=\int_{a}^{b}\delta_{2}y(x)dx$$
and
$$\lim_{\Delta \sigma_{1} \to 0}\epsilon_{1}=0,$$
$$\lim_{\Delta \sigma_{2} \to 0}\epsilon_{2}=0.$$
Let
$$y^{*}(x) \equiv y(x)+\Delta y = y +\delta_{1} y +\delta_{2}y$$
and require
$$K[y^{*}]=K[y].$$
Then
$$
\begin{align}
\Delta K & = K[y^{*}]-K[y] \\
& =K[y+\Delta y]-K[y] \\
& =0
\end{align}
$$
and
$$
\begin{align}
\Delta K & = \left\{\left.\frac{\delta G}{\delta y}\right|_{x=x_{1}}+\epsilon'_{1}\right\}\Delta \sigma_{1}+\left\{\left.\frac{\delta G}{\delta y}\right|_{x=x_{2}}+\epsilon'_{2}\right\}\Delta \sigma_{2}=0.
\end{align}\tag{2}\label{eq2}
$$
Since $\left.\frac{\delta G}{\delta y}\right|_{x=x_{2}}\ne0$ by our definition, we can write $\Delta \sigma_{2}$ as
$$
\Delta \sigma_{2} = -\left\{ \frac{\left.\frac{\delta G}{\delta y}\right|_{x=x_{1}}}{\left.\frac{\delta G}{\delta y}\right|_{x=x_{2}}} + \epsilon' \right\} \Delta \sigma_{1}\tag{3}\label{eq3}
$$
where
$$\lim_{\Delta \sigma_{1} \to 0}\epsilon'=0.$$
Defining
$$\lambda \equiv -\left\{ \frac{\left.\frac{\delta F}{\delta y}\right|_{x=x_{2}}}{\left.\frac{\delta G}{\delta y}\right|_{x=x_{2}}}\right\}$$
we can substitute \eqref{eq3} and $\lambda$ into our expression for $\Delta J$, \eqref{eq1}, to get
$$\Delta J = \left\{ \left.\frac{\delta F}{\delta y}\right|_{x=x_{1}} + \lambda \left.\frac{\delta G}{\delta y}\right|_{x=x_{1}} \right\}\Delta \sigma_{1} + \epsilon \Delta \sigma_{1}.\tag{4}\label{eq4}$$
The principal linear part of this term is the first term on the RHS. Hence,
$$\delta J = \left\{ \left.\frac{\delta F}{\delta y}\right|_{x=x_{1}} + \lambda \left.\frac{\delta G}{\delta y}\right|_{x=x_{1}} \right\}\Delta \sigma_{1}$$
and since the necessary condition for a weak extremum is $\delta J[h]=0$ and $\Delta \sigma_{1}$ is nonzero while
$x_{1}$ is arbitrary, we have
$$\left(F_{y}-\frac{d}{dx}F_{y'}\right)+\lambda \left(G_{y}-\frac{d}{dx}G_{y'}\right)=0.$$
My Questions
-
How do we get from \eqref{eq2} to \eqref{eq3}? I know it's not simple division since $\epsilon' \ne \frac{\epsilon'_{1}}{\epsilon'_{2}} $.
-
How do you substitute $\Delta \sigma_{2}$ \eqref{eq3} and $\lambda$ into $\Delta J$ \eqref{eq1} to get \eqref{eq4}? I have extra terms involving $\epsilon_{1}$, $\epsilon_{2}$, and $\epsilon'$ that aren't cancelling.
Best Answer
Hint.
Having in mind that $\epsilon_1',\epsilon_2'\to 0$ as $\Delta\sigma_1,\Delta\sigma_2\to 0$, from (3) if $\Delta\sigma_1\to 0$ then $\Delta\sigma_2\to 0$ and
$$ \frac{a+\epsilon_1'}{b+\epsilon_2'} = \frac ab + \epsilon' $$
with $\epsilon'=\frac{b\epsilon_1'-a\epsilon_2'}{b(b+\epsilon_2')}$ where $\epsilon'\to 0$ when $\Delta\sigma_1\to 0$