The Isoperimetric Inequality

Why the circle?

Aryan Malhotra

mentor: Nathan Jackson

Summer 2026

Soap Films

img src: https://www.daviddarling.info/encyclopedia/B/bubbles.html

The Setup

F = -\frac{dU}{dr}

Molecules at the interface have the highest Potential Energy

\text{Total Surface Energy of an Interface} = \gamma \cdot \text{Area}

\gamma = \text{Surface Tension} = \frac{\text{force}}{\text{length}}

*the soap film has 2 interfaces

Backup

Soap Interface Force Field

import numpy as np
import matplotlib.pyplot as plt

# y = -1 is the bottom air interface; y = 0 is the center of the bulk liquid; y = 1 is the top air interface
y_points = np.linspace(-1.1, 1.1, 23)
x_points = np.linspace(-1, 1, 9)
X, Y = np.meshgrid(x_points, y_points)

# Define the Force Field
# The force always points towards the bulk (y=0)
# Above the center (y > 0), force is negative (pointing down)
# Below the center (y < 0), force is positive (pointing up)
# At the center (y = 0), force is zero.
F_y = -Y  
F_x = np.zeros_like(X) # for visualization, I scale the arrows down

# Mask forces outside the soap film for visual clarity (air has no cohesive pull)
mask = (Y >= -1) & (Y <= 1)
F_y_film = np.where(mask, F_y, 0)
F_y_film *= 0.2

# Define Potential Energy (U): Integrating F_y = -y gives U(y) = 0.5 * y^2
y_curve = np.linspace(-1, 1, 200)
U = 0.5 * y_curve**2

# Create the visualization
fig, (ax1, ax2) = plt.subplots(1, 2, figsize=(12, 6), sharey=True)

# Left Plot: Force Field Vector Map
ax1.quiver(X, Y, F_x, F_y_film, color='crimson', pivot='middle', scale=5)
ax1.set_title('Net Force Field (F)')
ax1.set_xlabel('Horizontal Position')
ax1.set_ylabel('Depth (-1 = Bottom Interface, 0 = Bulk, 1 = Top Interface)')
ax1.set_xlim(-1.2, 1.2)
ax1.set_ylim(-1.2, 1.2)

# Add boundaries for the interfaces and the bulk center
ax1.axhline(0, color='blue', linestyle='--', alpha=0.5, label='Bulk Center (F=0)')
ax1.axhline(1, color='lightblue', linestyle='-', linewidth=4, alpha=0.6, label='Top Air Interface')
ax1.axhline(-1, color='lightblue', linestyle='-', linewidth=4, alpha=0.6, label='Bottom Air Interface')
ax1.legend(loc='upper left', fontsize='small')

# Right Plot: The Potential Well
# Plotted sideways so the Y-axis maps directly to the physical depth of the film
ax2.plot(U, y_curve, color='purple', linewidth=3)
ax2.fill_betweenx(y_curve, 0, U, color='purple', alpha=0.2)
ax2.set_title('Potential Energy Well (U)')
ax2.set_xlabel('Stored Energy (U > 0)')
ax2.set_xlim(0, 0.6)
ax2.grid(True, alpha=0.3)

plt.suptitle('Soap Film as a Potential Well: Two Interfaces and a Central Bulk', fontsize=14)
plt.tight_layout()
plt.show()

\text{Total Surface Energy of an Interface} = \gamma \cdot \text{Area}

\gamma = \text{Surface Tension} = \frac{\text{force}}{\text{length}}

*the soap film has 2 interfaces

\text{Minimize Energy} \implies \text{Minimize Area of Soap Film}

Thread has a fixed length

Larger the hole, the smaller the Area of the Soap film

Fixed Perimeter
Maximize the Area

Isoperimetric Inequality

Source: How to make inverted bubbles by Steve Mould

How we'll prove it

4\pi A \,\color{red}{=}\, \color{black}L^2

only true for a circle

4\pi A \leq L^2

area is bounded for all shapes

4\pi A \leq L^2

Let C be a simple closed plane curve with length L, and let A be the area of the region bounded by C. Then

1. Isoperimetric Inequality

plane curve

\text{a continuous map} \, \alpha: [a, b] \to \mathbb{R}^2

closed

\alpha(a) = \alpha(b)

simple

(\forall t \in \mathbb{R})\, t_{1}, t_{2} \in (a,b) \implies \alpha(t_1) \neq \alpha(t_2)

4\pi A \leq L^2

Let C be a simple closed plane curve with length L, and let A be the area of the region bounded by C. Then

1. Isoperimetric Inequality

arc length

\alpha(t) = (x(t), y(t))

L = \int ds = \int_{a}^{b} \sqrt{x'(\tau)^2 + y'(\tau)^2} \, d\tau

s(t) = \int_{a}^{t} \sqrt{x'(\tau)^2 + y'(\tau)^2} \, d\tau

For a curve parameterized by arc length:

\boxed{s \text{ is the distance from } \alpha(0) \text{ to } \alpha(t)}

a=0 \qquad |ds| =\sqrt{x'(\tau)^2 + y'(\tau)^2}=1 \implies s(t) = \int_0^t d\tau = t

s \in [0, L]

What even is the Length of a curve?

4\pi A \leq L^2

Let C be a simple closed plane curve with length L, and let A be the area of the region bounded by C. Then

1. Isoperimetric Inequality

\alpha(t) = (x(t), y(t))

A = \int_{x_0}^{x_1} f_1(x)\,dx - \int_{x_0}^{x_1} f_2(x)\,dx

\begin{aligned} \int_{a}^{b} y(t)x'(t) dt &= \int_{a}^{t_1} y(t)x'(t) dt + \int_{t_1}^{t_2} y(t)x'(t) dt + \int_{t_2}^{t_3} y(t)x'(t) dt + \int_{t_3}^{b} y(t)x'(t) dt \\ &= \int_{x_1}^{x_0} f_1(x) dx + \int_{t_1}^{t_2} y(t)(0) dt + \int_{x_0}^{x_1} f_2(x) dx + \int_{t_3}^{b} y(t)(0) dt \end{aligned}

\int y \, dx \overset{\text{u-sub}}{=} \int y(t) x'(t) \, dt

\boxed{\sqrt{x'(\tau)^2 + y'(\tau)^2}=1 \implies s(t) = \int_0^t d\tau = t}

What even is the Area of a curve?

A = - \int_a^b y(t) x'(t) \, dt

4\pi A \leq L^2

Let C be a simple closed plane curve with length L, and let A be the area of the region bounded by C. Then

1. Isoperimetric Inequality

\alpha(t) = (x(t), y(t))

\boxed{\sqrt{x'(\tau)^2 + y'(\tau)^2}=1 \implies s(t) = \int_0^t d\tau = t}

What even is the Area of a curve?

A = - \int_a^b y(t) x'(t) \, dt

(xy)' = x'y + y'x

\int_a^b (xy)'\,dt = \int_a^b x'y\,dt + \int_a^b xy'\,dt

xy\Big|_a^b = \int_a^b x'y\,dt + \int_a^b xy'\,dt = 0

\boxed{ A=-\int_a^b y(t)x'(t)\,dt =\int_a^b x(t)y'(t)\,dt }

A cleaner way is to use Green's Theorem*

4\pi A \,\color{red}{=}\, \color{black}L^2

only true for a circle

4\pi A \leq L^2

Let C be a simple closed plane curve with length L, and let A be the area of the region bounded by C. Then

1. Isoperimetric Inequality

\alpha(t) = (x(t), y(t))

\boxed{\sqrt{x'(\tau)^2 + y'(\tau)^2}=1 \implies s(t) = \int_0^t d\tau = t}

What even is the Area of a curve?

Green's Theorem

\oint_C (P\,dx + Q\,dy) = \iint_D (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) dA

We want something that connects closed loop curves to 2D integrals:

\text{Area of closed plane curve} = A = \int 1 \, dA

Q = x

P = -y

\oint_C -y\,dx = -\oint_C yx'\,ds

\oint_C x\,dy = \oint_C xy'\,ds

two simple cases that satisfy that

1. Isoperimetric Inequality

\alpha(t) = (x(t), y(t))

\boxed{\color{green}\sqrt{x'(\tau)^2 + y'(\tau)^2}=1 \implies s(t) = \int_0^t d\tau = t}

\boxed{ A=-\int_a^b y(t)x'(t)\,dt =\int_a^b x(t)y'(t)\,dt }

\tilde{\alpha}(t) = (x(t), \tilde{y}(t))

Let C be parameterized by the arc length s , defined

The auxilary circle shares the parameter and function x(t)

\tilde{x} = x

\text{same parameter } s

4\pi A \leq L^2

Let C be a simple closed plane curve with length L, and let A be the area of the region bounded by C. Then

1. Isoperimetric Inequality

\alpha(t) = (x(t), y(t))

\boxed{\color{green}\sqrt{x'(\tau)^2 + y'(\tau)^2}=1 \implies s(t) = \int_0^t d\tau = t}

\boxed{ A=-\int_a^b y(t)x'(t)\,dt =\int_a^b x(t)y'(t)\,dt }

\tilde{\alpha}(t) = (x(t), \tilde{y}(t))

\tilde{x} = x

\text{same parameter } s

A = \int_0^L xy'\, ds

\pi r^2 = \tilde{A} = - \int_0^L \tilde{y} x\, ds

A + \tilde{A} = A + \pi r^2 = \int_0^L (xy' - \tilde{y}x)\,ds

\text{Observation: for vectors } \vec{u} = (x,-\tilde{y}) \text{ and } \vec{v} = (y', x')

\vec{u} \cdot \vec{v} = xy' - \tilde{y}x'

\implies A + \pi r^2 = \int_0^L (\vec{u} \cdot \vec{v}) \, ds \leq \int_0^L |\vec{u} \cdot \vec{v}| \, ds \leq \int_0^L \sqrt{x^2 + \tilde{y}^2} \, ds = Lr

\boxed{\text{Cauchy-Schwarz} \quad |\vec{u} \cdot \vec{v}| \leq |\vec{u}||\vec{v}|}

|\vec{v}| = x'^2 + y'^2 \color{green}= 1

4\pi A \leq L^2

Let C be a simple closed plane curve with length L, and let A be the area of the region bounded by C. Then

\implies A + \pi r^2 \leq Lr

1. Isoperimetric Inequality

\alpha(t) = (x(t), y(t))

\boxed{\sqrt{x'(\tau)^2 + y'(\tau)^2}=1 \implies s(t) = \int_0^t d\tau = t}

\boxed{ A=-\int_a^b y(t)x'(t)\,dt =\int_a^b x(t)y'(t)\,dt }

\tilde{\alpha}(t) = (x(t), \tilde{y}(t))

Let C be parameterized by the arc length s

The auxilary circle shares the parameter and function x(t)

\tilde{x} = x

\text{same parameter } s

\boxed{\text{AM-GM} \quad \color{green} \frac{x+y}{2} \geq \sqrt{xy}}

A + \pi r^2 \leq Lr

\boxed{\text{Cauchy-Schwarz} \quad |\vec{u} \cdot \vec{v}| \leq |\vec{u}||\vec{v}|}

\red{4\pi A \leq L^2}

Let C be a simple closed plane curve with length L, and let A be the area of the region bounded by C. Then

\sqrt{A \cdot \pi r^2} \color{green}\leq \color{black}\frac{A + \pi r^2}{2} \leq \frac{Lr}{2}

4 A \pi r^2\leq L^2 r^2

\boxed{\red{4 \pi A \leq L^2}}

square both sides

4\pi A \,\color{red}{=}\, \color{black}L^2

only true for a circle

1. Isoperimetric Inequality

\alpha(t) = (x(t), y(t))

\boxed{\sqrt{x'(\tau)^2 + y'(\tau)^2}=1 \implies s(t) = \int_0^t d\tau = t}

\boxed{ A=-\int_a^b y(t)x'(t)\,dt =\int_a^b x(t)y'(t)\,dt }

\tilde{\alpha}(t) = (x(t), \tilde{y}(t))

\tilde{x} = x

\text{same parameter } s

\boxed{\text{AM-GM} \quad \frac{x+y}{2} \geq \sqrt{xy}}

\boxed{\text{Cauchy-Schwarz} \quad |\vec{u} \cdot \vec{v}| \leq |\vec{u}||\vec{v}|}

\boxed{4\pi A \leq L^2}

Let C be a simple closed plane curve with length L, and let A be the area of the region bounded by C. Then

\sqrt{A \cdot \pi r^2} \leq \frac{A + \pi r^2}{2} \leq \frac{Lr}{2}

\sqrt{A\cdot\pi r^2} = \frac{A + \pi r^2}{2}

square
solve quadratic

AM-GM property
equality iff x=y

A = \pi r^2

\frac{A + \pi r^2}{2} \leq \frac{Lr}{2}

\implies L = 2\pi r

4\pi A \,\color{red}{=}\, \color{black}L^2

only true for a circle

1. Isoperimetric Inequality

\alpha(t) = (x(t), y(t))

\boxed{\sqrt{x'(\tau)^2 + y'(\tau)^2}=1 \implies s(t) = \int_0^t d\tau = t}

\boxed{ A=-\int_a^b y(t)x'(t)\,dt =\int_a^b x(t)y'(t)\,dt }

\tilde{\alpha}(t) = (x(t), \tilde{y}(t))

\tilde{x} = x

\text{same parameter } s

\boxed{\text{AM-GM} \quad \frac{x+y}{2} \geq \sqrt{xy}}

\boxed{\text{Cauchy-Schwarz} \quad |\vec{u} \cdot \vec{v}| \leq |\vec{u}||\vec{v}|}

\boxed{4\pi A \leq L^2}

\vec{u} = (x,-\tilde{y}) \text{ and } \vec{v} = (y', x')

\text{Cauchy-Schwarz} \quad |\vec{u} \cdot \vec{v}| \red{=} |\vec{u}||\vec{v}| \iff \vec{u} = \lambda \vec{v}

x = \lambda y' \quad -\tilde{y} =\lambda x'

x^2 + \tilde{y}^2 = \lambda^2 (y'^2 + x'^2)

\implies \lambda = \frac{\sqrt{x^2 + \tilde{y}^2}}{\sqrt{x'^2 + y'^2}}

\implies \lambda = \sqrt{x^2 + \tilde{y}^2} = \pm r

\therefore x = \pm r y'

By coordinate symmetry

y = \pm rx'

\implies x^2 + y^2 = r^2 (x'^2 + y'^2)

\text{Hence } x^2 + y^2 = r^2 \text{ for all points}

this condition is forced for the curve for which the equality holds

References

Differential Geometry of Curves and Surfaces by Manfredo P. do Carmo

How to make inverted bubbles by Steve Mould

https://math.stackexchange.com/questions/19997/a-proof-of-the-isoperimetric-inequality-how-does-it-work

(for additional insights)

DRP Summer 26

By Aryan Malhotra