# Calculus of Variations (Gelfand & Fomin): Proof of Euler's Equation for Constrained Variation

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

1. How do we get from \eqref{eq2} to \eqref{eq3}? I know it's not simple division since $$\epsilon' \ne \frac{\epsilon'_{1}}{\epsilon'_{2}}$$.

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.

• Other texts (e.g. Calculus of Variations by R. Weinstock and Calculus of Variations by Bruce van Brunt) express $y^{*}$ as $y^{*}(x)=y(x)+\epsilon_{1} \eta_{1}(x) + \epsilon_{2} \eta_{2}(x)$ where $\eta_{1}(x_{1})=\eta_{1}(x_{2})=\eta_{2}(x_{1})=\eta_{2}(x_{2})=0$. They then plug $y^{*}$ and $y'^{*}$ into $J$ and $K$ and express them as functions of $\epsilon_{1}$ and $\epsilon_{2}$ for fixed choice of $\eta_{1}(x)$ and $\eta_{2}(x)$. Lastly, they state that $\epsilon_{1}$ and $\epsilon_{2}$ are functionally dependent by the implicit function theorem. – A. Hendry Mar 2 '19 at 23:42
• In the above, either $\epsilon_{2}=\epsilon_{2}(\epsilon_{1})$, or vice-versa, provided $\nabla K(x, y^{*}, y'^{*}) \ne 0$, where $\nabla$ denotes the operator $\left(\frac{\partial}{\partial \epsilon_{1}},\frac{\partial}{\partial \epsilon_{2}}\right).$ – A. Hendry Mar 2 '19 at 23:46
• Their functional dependence is made visually apparent by the fact that $K=K(\epsilon_{1},\epsilon_{2})=$ constant. – A. Hendry Mar 2 '19 at 23:51
• Do any of the 3 above comments help answer my questions? – A. Hendry Mar 2 '19 at 23:52

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 $$\Delta\sigma_1\to 0$$

• This is great! Would you be able to do the same for my second question? – A. Hendry Mar 4 '19 at 13:30
• Nevermind on that, I got it. The result of the second question follows naturally as it involves all terms that multiply an $\epsilon$, which can be redefined as a single $\epsilon$ and vanish with $\sigma_{1} \to 0$. Bravo! You are certainly a mathemagician. It was so simple it was hard!! A++! – A. Hendry Mar 4 '19 at 19:59

I believe I have an answer, and my comments did help.

Writing $$\Delta \sigma_{2}$$ as a function of $$\Delta \sigma_{1}$$ (by the Implicit Function theorem) and that $$\epsilon_{1}$$, $$\epsilon_{2}$$, $$\epsilon_{1}'$$, $$\epsilon_{2}'$$, $$\Delta \sigma_{1}$$, and $$\Delta \sigma_{2}$$ are all controlled by closeness to $$x_{1}$$ and $$x_{2}$$ (they are all non-zero ONLY in a neighborhood of $$x_{1}$$ and $$x_{2}$$, respectively), we can let $$x$$ approach $$x_{1}$$ so that $$\epsilon_{2}$$, $$\epsilon_{2}'$$, and $$\Delta \sigma_{2}$$ vanish. Then$$\Delta J$$ and $$\Delta \sigma_{2}$$ are functions only of variables near $$x_{1}$$ (i.e. $$\epsilon_{1}, \epsilon_{1}'$$, and $$\Delta \sigma_{1}$$). (See details).

• This might not actually be correct since $\Delta \sigma_{2}$ vanishes outside the neighborhood fo $x_{2}$ by definition... – A. Hendry Mar 3 '19 at 23:44