Show that the function $y$ which minimizes $\int_C F(y)\,ds$ satisfies $\frac{F(y)}{\sqrt{1+\left(y'\right)^2}}=C$

Question

Say we want to minimize $$\int_C F(y)\, ds$$ for some function $F(y)>0$. Using some physics, you can derive that if $y$ is a function for which the above functional is stationary, then $$\frac{F(y)}{\sqrt{1+\left(y'\right)^2}}=C$$ The derivation of this can be found here. I wanted to prove this in a more rigorous manner using Euler-Lagrange equations. First, I rewrote the integral: $$\int_CF(y)ds=\int_a^bF(y)\sqrt{1+\left(y'\right)^2}\,dx$$ Assuming that $F(y)$ is solely a function of $y$, we can write the Euler-Lagrange equation for the above functional. $$\frac{dF(y)}{dy}\sqrt{1+\left(y'\right)^2}-\frac{d}{dx}\left[\frac{F(y)y'}{\sqrt{1+\left(y'\right)^2}}\right]=0$$ I can rewrite this equation as follows: \begin{align*} \frac{d}{dx}\left[\frac{F(y)y'}{\sqrt{1+\left(y'\right)^2}}\right]&=\frac{dF(y)}{dy}\sqrt{1+\left(y'\right)^2}\\ \frac{d}{dx}\left[\frac{F(y)}{\sqrt{1+\left(y'\right)^2}}\right]y'+\frac{F(y)y''}{\sqrt{1+\left(y'\right)^2}}&=\frac{dF(y)}{dy}\sqrt{1+\left(y'\right)^2}\\ \frac{d}{dx}\left[\frac{F(y)}{\sqrt{1+\left(y'\right)^2}}\right]&=\frac{\frac{dF(y)}{dy}\left[1+\left(y'\right)^2\right]-F(y)y''}{y'\sqrt{1+\left(y'\right)^2}} \end{align*} From here it is sufficient to prove that $$\frac{dF(y)}{dy}\left[1+\left(y'\right)^2\right]-F(y)y''=0$$ Unfortunately, I don't know how to proceed.

Answer 1

This is easiest to prove using the Beltrami identity. Suppose we want to extremize the quantity $$ S = \int_a^b f(y, y') \, dx, $$ for which $f$ does not explicitly depend on $x$, i.e., $\partial f/\partial x = 0$. This implies that $$ \frac{df}{dx} = \frac{\partial f}{\partial y} \frac{dy}{dx} + \frac{\partial f}{\partial y'} \frac{dy'}{dx}. $$ According to the Euler-Lagrange equation, we will have $$ \frac{d}{dx} \left( \frac{\partial f}{\partial y'} \right) = \frac{\partial f}{\partial y}, $$ and so $$ \frac{df}{dx} = \frac{d}{dx} \left( \frac{\partial f}{\partial y'} \right) y' + \frac{\partial f}{\partial y'} \frac{dy'}{dx} = \frac{d}{dx} \left( \frac{\partial f}{\partial y'} y' \right) $$ This then implies that $$ \frac{d}{dx} \left( f - \frac{\partial f}{\partial y'} y' \right) = 0 \quad \Rightarrow \quad f - \frac{\partial f}{\partial y'} y' = C, $$ where $C$ is a constant.

In your case, you have $f(y,y') = F(y) \sqrt{ 1 + (y')^2}$, and so $$ f - \frac{\partial f}{\partial y'} y' = F(y) \sqrt{ 1 + (y')^2} - \frac{F(y) y '}{\sqrt{ 1 + (y')^2}} y' = \frac{F(y)}{ \sqrt{ 1 + (y')^2}} \left( 1 + (y')^2 - (y')^2 \right) \\= \frac{F(y)}{ \sqrt{ 1 + (y')^2}}, $$ and the fact that this is a constant follows from the Beltrami identity.

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Alternately: if you wish to proceed from the last step of your derivation, note that $$ \frac{d}{dx} \left[ \frac{F(y)}{\sqrt{1 + (y')^2}} \right] = \left( 1 + (y')^2 \right)^{-3/2} \left[ \frac{\partial F}{\partial y} \left( 1 + (y')^2 \right) - F(y) y'' \right] y'. $$ The quantity in square brackets is what you want to vanish. Combining this equation with the last equation in your derivation, you can obtain $$ \frac{d}{dx} \left[ \frac{F(y)}{\sqrt{1 + (y')^2}} \right] = \frac{\left( 1 + (y')^2 \right)}{(y')^2} \frac{d}{dx} \left[ \frac{F(y)}{\sqrt{1 + (y')^2}} \right], $$ which (since the pre-factor is never zero) implies the result sought. (But the Beltrami identity is so useful in calculus of variations and in physics that I much prefer the derivation that uses it.)