# Extremising $\int_0^1 f(x) f(1-x) \ \mathrm{d}x$ subject to length of $f$ and endpoints

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!

• 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 $x=1/2$? Sep 25, 2021 at 13:44
• @achillehui Yeah, I specified $\ell \geq 1$ just so that the function would exist. Sep 25, 2021 at 13:48
• $$F= \int_0^1 y(x) y(1-x) dx$$ $$dF = \int_0^1 \left( y+ \epsilon \eta(x) \right) \left[ y(1-x) + \epsilon \eta(1-x) \right] - y(x)y(1-x) dx = \int_0^1 \epsilon \left[ \eta(x)y(1-x)+ y(x) \eta(1-x) \right] + O(\epsilon^2) dx$$ We set the first order variation to zero: $$\eta(x) y(1-x) + y(x) \eta(1-x) = 0$$ Sep 25, 2021 at 13:49
• We need to find $y$ such the last statement is true for any $\eta$ to my understanding Sep 25, 2021 at 13:49
• Do you mean $\ell$ is a fixed number, and you minimize along $f$ with a fixed $\ell$? Sep 25, 2021 at 15:00

$$J[f]~=~\int_0^1\!\mathrm{d}x\left( f(x)f(1-x)+\lambda \sqrt{1+f^{\prime}(x)^2} \right) -\lambda \ell,$$
where $$\lambda$$ is a Lagrange multiplier. The EL equation becomes non-local:
$$2f(1-x)~=~\lambda\frac{d}{dx}\frac{f^{\prime}(x)}{\sqrt{1+f^{\prime}(x)^2}}.$$