Simplest Variational Problem with Subsidiary Conditions

Theorem
Let $J\sqbrk y$ and $K\sqbrk y$ be functionals, such that


 * $\displaystyle J\sqbrk y=\int_{a}^{b}\map F{x,y,y'}\rd x,$


 * $K\sqbrk y=\int_{a}^{b}\map G {x,y,y'}\rd x=l$

where $l$ is a constant.

Let $y=\map y x$ be an extremum of $F\sqbrk y$, and satisfy boundary conditions


 * $\map y a=A,$


 * $\map y b=B$

Then, if $y=\map y x$ is not an extremal of $K\sqbrk y$, there exists a constant $\lambda$,

such that $y=\map y x$ is an extremal of the functional


 * $\displaystyle\int_{a}^{b}\paren{F+\lambda G}\rd x$

or, in other words, $y=\map y x$ satisfies


 * $\displaystyle F_y-\frac{\d}{\d x}F_{y'}+\lambda\paren{G_y-\frac{\d}{\d x}G_{y'} }=0$

Proof
Let $J\sqbrk y$ be a functional, for which $y=\map y x$ is an extremal with the boundary conditions $\map y a=A,~\map y b=B$.

Choose two points, $x_1$ and $x_2$ from the interval $\closedint a b$.

Let $\delta_1 \map y x$ and $\delta_2 \map y x$ be functions, different from zero only in the neighbourhood of $x_1$ and $x_2$ respectively.

Then we can exploit the definition of variational derivative in a following way:


 * $\displaystyle\Delta J\sqbrk {y;~\delta_1\map y x+\delta_2\map y x}=\paren{\frac{\delta F} {\delta y}\bigg\rvert_{x=x_1}+\epsilon_1}\Delta\sigma_1+\paren{\frac {\delta F} {\delta y}\bigg\rvert_{x=x_2}+\epsilon_2}\Delta\sigma_2$

where


 * $\displaystyle\Delta\sigma_1=\int_{a}^{b}\delta_1\map y x,~\Delta\sigma_2=\int_{a}^{b}\delta_2\map y x$

and $\epsilon_1,~\epsilon_2\to 0$ as $\Delta\sigma_1,~\Delta\sigma_2\to 0$.

We now require that the varied curve $y^*=\map y x+\delta_1\map y x+\delta_2\map y x$ satisfies the condition $K\sqbrk{y^*}=K\sqbrk y$.

This way we limit arbitrary varied curves to those who still satisfy the given functional constraint.

Similarly, write down $\Delta K\sqbrk y $ as


 * $\displaystyle\Delta K\sqbrk y= K\sqbrk{y^*}-K\sqbrk y=\paren{\frac{\delta G}{\delta y}\bigg\rvert_{x=x_1}+\epsilon_1'}\Delta\sigma_1+\paren{\frac{\delta G}{\delta y}\bigg\rvert_{x=x_2}+\epsilon_2'}\Delta\sigma_2$

where $\epsilon_1',~\epsilon_2'\to 0$ as $\Delta\sigma_1,~\Delta\sigma_2\to 0$.

Suppose $x_2$ is such that


 * $\displaystyle\frac{\delta G} {\delta y} \bigg\rvert_{x=x_2}\ne 0$

Such a point exists, because by assumption $\map y x$ is not an extremal of $K\sqbrk y$.

Since $\Delta K=0$, the previous equation can be rewritten as


 * $\displaystyle\Delta\sigma_2=-\paren{\frac{\frac{\delta G}{\delta y}\bigg\rvert_{x=x_1} }{\frac{\delta G}{\delta y}\bigg\rvert_{x=x_2} }+\epsilon'}\Delta\sigma_1$

where $\epsilon'\to 0$ as $\Delta\sigma_1\to 0$.

Set


 * $\displaystyle\lambda=-\frac{\frac{\delta F}{\delta y}\bigg\rvert_{x=x_2} }{\frac{\delta G}{\delta y}\bigg\rvert_{x=x_2} }$

Substitute this into formula for $\Delta J$:

where $\epsilon\to 0$ as $\Delta\sigma_1\to 0$.

Then the variation of the functional $J\sqbrk y$ at the point $x_1$ is:


 * $\delta J=\paren{\frac{\delta F}{\delta y}\bigg\vert_{x=x_1}+\lambda\frac{\delta G}{\delta y}\bigg\vert_{x=x_1} }\Delta\sigma$

A necessary condition for $\delta J$ vanish for any $\Delta\sigma$ and arbitrary $x_1$ is:


 * $\displaystyle\frac{\delta F}{\delta y}+\lambda\frac{\delta G}{\delta y}=F_y-\frac{\d}{\d x}F_{y'}+\lambda\left(G_y-\frac{\d}{\d x}G_{y'}\right)=0$