TITLE: A Lagrangian problem with a countable family of local extrema ?
QUESTION [7 upvotes]: Dear MO contributors,
let $r > 0, L > 0$. I am interested in maximizing the integral:
$$
\int_0^{2\pi} \frac{f(\alpha)^2 f'(\alpha)^2}{\sqrt{f(\alpha)^2 + f'(\alpha)^2}} \ \mathrm{d} \alpha
$$
over continuous and piecewise smooth functions $f : [0,2\pi] \to \mathbf{R}$, subject to three the constraints:
$$
f(0) = f(2\pi), \quad f \ge r,  \quad \int_0^{2\pi} \sqrt{f(\alpha)^2 + f'(\alpha)^2} \ \mathrm{d} \alpha = L.
$$
Although not directly relevant to the question, let me mention that the function $f$ represents a curve in the plane, where $f(\alpha)$ is the distance from the origin at angle $\alpha$ from one axis, and the aim is to maximize a certain torque for a given length of curve.
When I try to write down the Euler-Lagrange equation, I end up with a not-so-inspiring differential equation that is a polynomial of degree 7 in $f$, $f'$ and $f''$. 
Heuristically, I expect an extremal function to develop a number of peaks, and I conjecture that the family of extremal functions can be indexed by the number of peaks (by peaks, I mean local maxima, probably with a discontinuous derivative). More precisely, is it true that there exists at most one extremal function with a given number of peaks ? How does the set
$$
\{n \in \mathbf{N} : \text{there exists an extremal function with } n \text{ peaks} \}
$$
look like ? How many peaks does the global maximizer have ? What is its rough shape ? How much do all this depend on $r$ and $L$ ? (of course, if $L < 2\pi r$, then there are no solutions, and there is a unique solution for $L = 2 \pi r$). Are there good ways to investigate these questions numerically ? Any suggestion about the relevant litterature would also be very welcome.

REPLY [3 votes]: Your problem does not have a maximizing solution.  Here is why:  
Assuming that $f$ is piecewise smooth and not identically constant, we can solve the Euler-Lagrange equations in an interval where $f$ is smooth and $f'$ is nonvanishing as follows (taking advantage of the fact that $\alpha$ does not explicitly appear in the Lagrangian or the constraint):  Suppose that $I=[\alpha_0,\alpha_1]$ is an interval in which $f$ is smooth and such that $f'(\alpha)\not=0$ for $\alpha\in I$.  For simplicity, I'll assume that $f'>0$ on $I$, but the argument is essentially the same for $f'<0$.  Then we have $\mathrm{d}\alpha = \mathrm{d}f/f'$ so that the integral to be maximized over this interval becomes 
(after setting $f_i = f(\alpha_i)$)
$$
I = \int_{f_0}^{f_1} \frac{f^2(f')^2}{\sqrt{f^2+(f')^2}}\,\frac{\mathrm{d}f}{f'}
= \int_{f_0}^{f_1} \frac{f^2\,\mathrm{d}f}{g}\,,
$$
where I have set $g = \sqrt{(f/f')^2+1}$.  Using the same notation, the constraint integral becomes
$$
C = \int_{f_0}^{f_1} g\,\mathrm{d}f\,.
$$
Thus, we are trying to choose $g$ as a function of $f$ so as to maximize $I$ subject to the condition that $C$ is to be held constant.  Once we do this, we can recover $\alpha$ as a function of $f$ by using the fact that
$$
\mathrm{d}\alpha = \frac{\mathrm{d}f}{f'} = \frac{\sqrt{g^2{-}1}}{f}\,\mathrm{d}f\,.
$$
By the method of Lagrange multipliers, we need to choose $g = g(f)>1$ on the interval $[f_0,f_1]$ so as to render the integral $I_\lambda$ stationary, where $\lambda$ is a constant and
$$
I_\lambda = \int_{f_0}^{f_1} \left(\frac{f^2}{g}+\lambda g\right)\,\mathrm{d}f.
$$
Since $g'$ does not appear in the integrand, the Euler-Lagrange equation is particularly simple:  We must have $(f/g)^2 = \lambda$.  Setting $\lambda = c^2>0$, this gives $g = f/c$, so $c < f_0$ (since $g>1$) and we have 
$$
\mathrm{d}\alpha = \frac{\sqrt{f^2{-}c^2}}{cf}\,\mathrm{d}f\,
$$
Setting $f = c\cosh\tau$ where $\tau\in[\tau_0,\tau_1]$ with $0<\tau_0<\tau_1$,
we see that 
$$
\mathrm{d}\alpha = c \frac{\sinh^2\tau}{\cosh\tau}\,\mathrm{d}\tau
= c\,\mathrm{d}\left(\sinh\tau - 2\arctan(e^\tau)\right).
$$
Thus, the parametric curves (for $a$ and $c$ constants)
$$
\bigl(\alpha,f(\alpha)\bigr) = \bigl(a {+} c\sinh\tau {-} 2c\arctan(e^\tau),\ c\cosh\tau\bigr) 
$$
are the solutions to the Euler-Lagrange equations on the intervals where $f'>0$
(with a similar formula when $f'<0$).
Unfortunately, it turns out that these extrema are always local minima, not local maxima, as is easily verified:  Let $g = f/c + \epsilon h(f)$ where the integral of $h(f)$ over $[f_0,f_1]$ is zero and $h(f_0)=h(f_1)=0$.  This is the general perturbation that does not change the constraint integral $C$.  However, the integral $I$ then has the expansion 
$$
I = \tfrac12 c(f_1{-}f_0)^2 + \epsilon^2 c^3\int_{f_0}^{f_1} \frac{h(f)^2}{f}\,\mathrm{d}f + O(\epsilon^3).
$$
Thus, $\epsilon=0$ is always a strict local minimum.
Thus, one concludes that your original maximization problem has no solution.  (Meanwhile, the minimum, namely $I=0$, is clearly achieved by taking $f$ to be an appropriate constant.)