Area between Smooth Curve and Line with Fixed Endpoints is Maximized by Arc of Circle

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Theorem

Let $y$ be a smooth curve, embedded in $2$-dimensional Euclidean space.

Let $y$ have a total length of $l$.

Let it be contained in the upper halfplane with an exception of endpoints, which are on the x-axis and are given.

Suppose, $y$, together with a line segment connecting $y$'s endpoints, maximizes the enclosed area.


Then $y$ is an arc of a circle.


Proof

Without loss of generality, we choose our point of reference such that $y$ intersect x-axis at points $\tuple {-a, 0}$ and $\tuple {a, 0}$ for some $a > 0$.

The area below the curve $y$ is a functional of the following form:

$\displaystyle A \sqbrk y = \int_{-a}^a y \rd x$

Furthermore, $y$ has to satisfy the following conditions:

$\map y {-a} = \map y a = 0$
$\displaystyle L \sqbrk y = \int_{-a}^a \sqrt {1 + y'^2} \rd x = l$

By Simplest Variational Problem with Subsidiary Conditions, there exists a constant $\lambda$ such that the functional:

$\displaystyle A \sqbrk y + \lambda L \sqbrk y = \int_{-a}^a \paren {y + \lambda \sqrt {1 + y'^2} } \rd x$

is extremized by the mapping $y$.

Corresponding Euler's Equation reads:

$1 + \lambda \dfrac \d {\d x} \dfrac {y'} {\sqrt {1 + y'^2} } = 0$

Integrating with respect to $x$ once yields:

$x + \lambda \dfrac {y'} {\sqrt {1 + y'^2} } = C_1$

Solve this for $y'$:

$\displaystyle y' = \pm \frac {c_1 - x} {\sqrt {\lambda^2 - c_1^2 + 2 c_1 x - x^2}}$.

Integration yields:

$\paren {x - C_1}^2 + \paren {y - C_2}^2 = \lambda^2$

This is an equation for a circle with radius $\lambda$ and center $\tuple {C_1, C_2}$.

To find $C_1, C_2, \lambda$, apply boundary conditions and the length constraint.

From the boundary conditions we have that:

\(\displaystyle \paren {-a - C_1}^2\) \(=\) \(\displaystyle \lambda^2\)
\(\displaystyle \paren {a - C_1}^2\) \(=\) \(\displaystyle \lambda^2\)

Take the difference of these two equations:

$4 a C_1 = 0 \implies C_1 = 0$

because $a > 0$.

Apply one of the boundary conditions again, that is, at $\tuple {a, 0}$:

$a^2 + C_2^2 = \lambda^2$

Then:

$C_2 = \pm \sqrt {\lambda^2 - a^2}$.

which can be used to get rid of $C_2$.

The last parameter to find is $\lambda$.

We have two cases:

the curve is an arc of the upper semicirle;
the curve is a union of upper semicirle with two arcs of lower semicirle.

In the first case the length constraint is:

$l = 2 \lambda \, \map \arctan {\dfrac a {\sqrt {\lambda^2 - a^2} } }$

For real $\lambda$, if $\lambda \ge a$, then $l \in \R$.

To find extremal values, consider the derivate $\dfrac {\d l} {\d \lambda}$:

\(\displaystyle \dfrac {\d l} {\d \lambda}\) \(=\) \(\displaystyle 2 \paren {\map {\arctan} {\frac 1 {\sqrt {\lambda^2 - 1} } } - \frac 1 {\sqrt {\lambda^2 - 1} } }\)
\(\displaystyle \) \(<\) \(\displaystyle 0\) Tangent Inequality

Hence the domain of $l$ is determined by boundary values.

At the boundary of $\lambda = a$ we have:

\(\displaystyle \lim_{\lambda \to a^+} l\) \(=\) \(\displaystyle \lim_{\lambda \to a^+} 2 \lambda \map {\arctan} {\frac a {\sqrt{\lambda^2 - a^2} } }\)
\(\displaystyle \) \(=\) \(\displaystyle 2 a \lim_{\lambda \to a^+} \map {\arctan} {\frac a {\sqrt{\lambda^2 - a^2} } }\) Product Rule for Limits

To calculate the limit of such composite function denote:

\(\displaystyle \map f y\) \(=\) \(\displaystyle \map {\arctan} y\)
\(\displaystyle \map g {\lambda}\) \(=\) \(\displaystyle \frac a {\sqrt {\lambda^2 - a^2} }\)

It follows that:

$\displaystyle \lim_{\lambda \to a^+} \map g {\lambda} = + \infty$
$\displaystyle \lim_{y \to \infty} \map f y = \frac \pi 2$

Arctangent is continuous for all $x \in \R$.

Then, by Limit of Composite Function:

$\displaystyle 2 a \lim_{\lambda \mathop \to a^+} \map \arctan {\frac a {\sqrt {\lambda^2 - a^2} } } = \pi a$.

At the boundary of $\lambda = + \infty$ we have:

\(\displaystyle \lim_{\lambda \mathop \to \infty} l\) \(=\) \(\displaystyle \lim_{\lambda \mathop \to \infty} 2 \lambda \map \arctan {\frac a {\sqrt {\lambda^2 - a^2} } }\)
\(\displaystyle \) \(=\) \(\displaystyle \lim_{\lambda \mathop \to \infty} 2 \frac {\map \arctan {\frac a {\sqrt {\lambda^2 - a^2} } } } {\frac 1 \lambda}\) Indeterminate limit $\infty \cdot 0$
\(\displaystyle \) \(=\) \(\displaystyle \lim_{\lambda \mathop \to \infty} 2 \frac {-\frac a {\lambda \sqrt {\lambda^2 - a^2} } } {-\frac 1 {\lambda^2} }\) L'Hôpital's Rule
\(\displaystyle \) \(=\) \(\displaystyle \lim_{\lambda \mathop \to \infty} 2 a \frac \lambda {\sqrt {\lambda^2 - a^2} }\)
\(\displaystyle \) \(=\) \(\displaystyle 2 a\)

In the second case the length constraint is:

$l = 2 \lambda \paren {\pi - \arctan \dfrac a {\sqrt {\lambda^2 - a^2} } }$

Similarly to the previous case:

\(\displaystyle \dfrac {\d l} {\d \lambda}\) \(=\) \(\displaystyle 2 \pi - 2 \map \arctan {\frac 1 {\sqrt {\lambda^2 - 1} } } + \frac 1 {\sqrt {\lambda^2 - 1} }\)
\(\displaystyle \) \(>\) \(\displaystyle 0\) Tangent Inequality

Hence the domain of $l$ is determined by boundary values.

At the boundary of $\lambda = a$ we have:

\(\displaystyle \lim_{\lambda \mathop \to a^+} l\) \(=\) \(\displaystyle 2 \lambda \paren {\pi - \arctan \dfrac a {\sqrt {\lambda^2 - a^2} } }\)
\(\displaystyle \) \(=\) \(\displaystyle 2 \pi a - \pi a\)
\(\displaystyle \) \(=\) \(\displaystyle \pi a\)

As $\lambda$ approaches the infinity we have:

\(\displaystyle \lim_{\lambda \mathop \to \infty} l\) \(=\) \(\displaystyle \lim_{\lambda \mathop \to \infty} 2 \lambda \paren {\pi - \arctan \dfrac a {\sqrt {\lambda^2 - a^2} } }\)
\(\displaystyle \) \(=\) \(\displaystyle \lim_{\lambda \mathop \to \infty} 2 \lambda \pi - 2 a\)
\(\displaystyle \) \(=\) \(\displaystyle \infty\)

Therefore:

$\forall l \ge 2 a: \exists \lambda \ge a$

Hence, within these constraints the real solution maximizing the area with fixed endpoints is an arc of a circle.

$\blacksquare$


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