I have recently learnt some Calculus of Variations and was trying to apply this to a question I made:

Over all functions $f: [0, 1] \to \mathbb{R}$ satisfying $f(0) = f(1) = 0$ with fixed curve length $\ell \geq 1$ (i.e. $\int_0^1 \sqrt{1 + (f'(x))^2} \ \mathrm{d}x = \ell$), find $f$ which maximise and minimise \begin{align*} \int_0^1 f(x) f(1 - x) \ \mathrm{d}x. \end{align*}

Ordinarily, I would proceed by Lagrange Multipliers and use Euler-Lagrange equations to solve for $f$, but I'm not sure how this would work with $f$ being shifted above. I considered rederiving the Euler-Lagrange equation for this as well, but the fact that it is a shifted argument makes me think this would likely not be nice to work with.

Any help would be appreciated, thanks!

  • $\begingroup$ I know very little about calculus of variations, but couldn't you make the integral arbitrarily large and small? Large by choosing a large 'bump' function, symmetric about $x=1/2$, small with a double bump function, antisymmetric about $x=1/2$? $\endgroup$
    – Steven
    Sep 25, 2021 at 13:44
  • $\begingroup$ @achillehui Yeah, I specified $\ell \geq 1$ just so that the function would exist. $\endgroup$ Sep 25, 2021 at 13:48
  • 1
    $\begingroup$ $$ F= \int_0^1 y(x) y(1-x) dx$$ $$ dF = \int_0^1 \left( y+ \epsilon \eta(x) \right) \left[ y(1-x) + \epsilon \eta(1-x) \right] - y(x)y(1-x) dx = \int_0^1 \epsilon \left[ \eta(x)y(1-x)+ y(x) \eta(1-x) \right] + O(\epsilon^2) dx$$ We set the first order variation to zero: $$ \eta(x) y(1-x) + y(x) \eta(1-x) = 0$$ $\endgroup$ Sep 25, 2021 at 13:49
  • $\begingroup$ We need to find $y$ such the last statement is true for any $\eta$ to my understanding $\endgroup$ Sep 25, 2021 at 13:49
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    $\begingroup$ Do you mean $\ell$ is a fixed number, and you minimize along $f$ with a fixed $\ell$? $\endgroup$ Sep 25, 2021 at 15:00

1 Answer 1


Hint: The extended functional reads

$$ J[f]~=~\int_0^1\!\mathrm{d}x\left( f(x)f(1-x)+\lambda \sqrt{1+f^{\prime}(x)^2} \right) -\lambda \ell, $$

where $\lambda$ is a Lagrange multiplier. The EL equation becomes non-local:

$$ 2f(1-x)~=~\lambda\frac{d}{dx}\frac{f^{\prime}(x)}{\sqrt{1+f^{\prime}(x)^2}}.$$


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