Calculus of Variations: Circle Minimizing Arc Length

We rephrase the problem as finding a function \(f(x)\) that minimizes the arc length while satisfying a certain area constraint. Specifically, we seek the function \(f(x) \in C^1([a,b])\) that:

  1. Satisfies the boundary conditions:

    \[ f(a) = f(b) = 0 \]

  2. The area under the curve:

    \[ \int_a^b f(x)\, dx \]

    is fixed

The target is to minimize the arc length:

\[ L[f] = \int_a^b \sqrt{1 + (f'(x))^2}\, dx \]

The Proof

Proof:

To incorporate the integral constraint, introduce a Lagrange multiplier \(\lambda\), and define the augmented functional:

\[ \mathcal{J}[f] = \int_a^b \left[ \sqrt{1 + (f'(x))^2} - \lambda f(x) \right] dx \]

We now treat this as a standard variational problem with Lagrangian:

\[ \mathcal{L}(x, f, f') = \sqrt{1 + (f')^2} - \lambda f \]

Since:

\[\begin{split} \begin{aligned} \frac{\partial \mathcal{L}}{\partial f'} &= \frac{f'}{\sqrt{1 + (f')^2}} \\ \frac{\partial \mathcal{L}}{\partial f} &= -\lambda \end{aligned} \end{split}\]

By Euler–Lagrange equation

\[ \frac{d}{dx} \left( \frac{\partial \mathcal{L}}{\partial f'} \right) = \frac{\partial \mathcal{L}}{\partial f} \]

We have:

\[ \frac{d}{dx} \left( \frac{f'}{\sqrt{1 + (f')^2}} \right) = - \lambda \]

Define:

\[ u = \frac{f'}{\sqrt{1 + (f')^2}} \Rightarrow f' = \frac{u}{\sqrt{1 - u^2}} \]

we have:

\[ f' = \frac{u}{\sqrt{1 - u^2}} \tag{1} \]

and

\[ u' = -\lambda \tag{2} \]

Integrating (1) gives:

\[ f(x) = \int_a^x \frac{u(t)}{\sqrt{1 - u^2(t)}}\, dt \]

By (2), we have:

\[ u(t) = -\lambda t + C_1 \]
\[ f(x) = \int_{u(a)}^{u(x)} \frac{u}{\sqrt{1 - u^2}} \cdot (-\frac{1}{\lambda})\, du \]
\[ \therefore f(x) = \frac{1}{\lambda} \int_{u(x)}^{u(a)} \frac{u}{\sqrt{1 - u^2}}\, du \]
\[ \because \int \frac{u}{\sqrt{1 - u^2}}\, du = -\sqrt{1 - u^2} + C \]

Therefore:

\[\begin{split} \begin{aligned} f(x) &= \frac{1}{\lambda} \left[ -\sqrt{1 - u^2} \right]_{u(x)}^{u(a)} \\ &= \frac{1}{\lambda} \left[ \sqrt{1 - u^2} \right]_{u(a)}^{u(x)} \\ &= \frac{1}{\lambda} \left[ \sqrt{1 - (-\lambda x + C_1)^2} - \sqrt{1 - (-\lambda a + C_1)^2} \right] \end{aligned} \end{split}\]
\[ \because f(b) = 0 \]
\[ \therefore (-\lambda b + C_1)^2 = (-\lambda a + C_1)^2 \Rightarrow C_1 = \frac{\lambda(a + b)}{2} \]

Let \(c = \frac{a + b}{2}\) and \(R = \frac{1}{\lambda}\), then:

\[ f(x) = \sqrt{R^2 - (x - c)^2} - \sqrt{R^2 - (a - c)^2} \]

This is indeed a segment of a circle with radius \(R\).

\[\tag*{$\blacksquare$}\]

Interpretation

In fact, the circle is centered at \((c, -\sqrt{R^2 - (a - c)^2})\), which is below the \(x\)-axis, and the segment \(ab\) is a chord of the circle.

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