I'm working through Gelfand & Fomin's Calculus of Variations by myself, and could use the guidance of someone familiar with the subject. The problem I'm on now is the following:
"Given two points $A$ and $B$ in the $xy$-plane, let $\gamma$ be a fixed curve joining them. Among all curves of length $\ell$ joining $A$ and $B$, find the curve which together with $\gamma$ encloses the greatest area."
I've spent a lot of time working this out, and I have a number of questions about the path I took. Letting the curve we're solving for be $C(t)=(x(t),y(t))$, we have to somehow express the area between $C$ and $\gamma$. I elected to assume that our curves satisfy the hypotheses of Green's Theorem, in order to calculate the area by the line integral: $$A=\int_{\gamma \cup C} x \, dy - y \, dx = \int_{\gamma} x \, dy-y \, dx + \int_C x \, dy-y \, dx$$
We could express these integrals in terms of $t$ by letting $\gamma[0,1]\rightarrow \mathbb{R}^2$, with $\gamma(0)=A$, $\gamma(1) = B$, and $C(1)= B$, $C(2) = A$. Because we want our curves to have length $\ell$, we have to consider solutions to the the Euler-Lagrange equations associated with the functional: $$A(x,y,x',y') = \int_{\gamma} x \, dy-y \, dx + \int_1^2 \left( x(t) y'(t) -y(t)x'(t) + \lambda \sqrt{x'(t)^2+y'(t)^2} \right) \, dt$$
The integral over $\gamma$ will be a constant, since that curve is fixed.
Question 1: Is this a reasonable way to proceed?
Question 2: Once I calculated the Euler-Lagrange equations, the curve $\gamma$ no longer had a role to play. This seems odd, since I feel like $C$ must depend upon the initial choice of $\gamma$ somehow.
If I'm on the right track with the above, then I also have:
Question 3: After much wailing and gnashing of teeth, I derived the relationship $$2(x'^2+y'^2)^{3/2} + \lambda (x' y'' -y' x'') = 0$$ for the extremals of $A$. Is this correct, and if so, can I go any further?
Q1: This is a somewhat reasonable approach. However, consider that there are infinitely many parametrizations of the extremal curve (whatever it is), and each of these will satisfy the Euler-Lagrange equation. This means the equation will be difficult to deal with, as it has such a large solution space. More on this later.
Q2: The shape of $\gamma$ does not affect the solution (unless we insist on $C$ not crossing $\gamma$, which leads to a difficult obstacle problem; I'm not sure the authors intended this). For illustration: if you push a spherical air balloon against a wall or some other object, the free part of its surface will not reflect the shape of the obstacle.
Q3: You should have two equations, not one. And as I mentioned earlier, this approach is difficult due to the large solution space.
Typically, one replaces length $\int |C'(t)|\,dt$ by energy $\int |C'(t)|^2\,dt$. This is justified as follows:
Indeed, one direction follows from the Cauchy-Schwarz while the other follows by considering the constant-speed parametrization.
Since we can always increase the energy (say, by moving back and forth) without changing the area, the area can be maximized among the curves of energy $=\ell^2$.
This leads to the quadratic functional $$A(x,y) = \int_{\gamma} x \, dy-y \, dx + \int_1^2 \left( x(t) y'(t) -y(t)x'(t) + \lambda (x'(t)^2+y'(t)^2) \right) \, dt$$ for which the Euler-Lagrange system is linear: $$ x'+\lambda y''=0,\qquad y'-\lambda x''=0 $$ (To obtain these, write out the variation $A(x+\delta x,y+\delta y)$, extract linear terms, and observe that $\delta x$ and $\delta y$ can be chosen independently: each of them yields an equation.)
The system is easy to solve: $x$ and $y$ are trigonometric functions, plus constants. Consequently, the curve is an arc of a circle, parametrized at constant speed.
The passage from length to energy yields uniqueness of minimizer because the functional now penalizes not only deviations from optimal shape, but also deviations from constant-speed parametrization. See also: first paragraph here, or search "length, minimize, energy".