Functional Foundations Slides Calculus of Variations
Functionals and Variation
Transitioning from the optimization of points to the optimization of paths, shapes, and functions.
"Nature always acts by the shortest paths." — Pierre de Fermat
Standard Calculus vs. Variational Calculus
Standard Calculus
Input: A scalar variable \( x \in \mathbb{R} \)
\( f: \mathbb{R} \to \mathbb{R} \)
Find the point \( x \) that minimizes or maximizes the value of the function \( f(x) \).
Variational Calculus
Input: A function \( y(x) \in \mathcal{C}^2 \)
\( J: \mathcal{F} \to \mathbb{R} \)
Find the function \( y(x) \) that minimizes or maximizes the value of the functional \( J[y] \).
What is a Functional?
A functional is a mapping from a set of functions to the real numbers. In this course, we focus on functionals defined by integrals:
\[ J[y] = \int_{x_1}^{x_2} f(x, y(x), y'(x)) \, dx \]
Arc Length
\( \int \sqrt{1 + (y')^2} \, dx \)
Action (Physics)
\( \int (T - V) \, dt \)
Surface Area
\( 2\pi \int y \sqrt{1 + (y')^2} \, dx \)
The Concept of Variation
To optimize \( J[y] \), we consider a small "variation" in the function \( y(x) \).
Let \( \hat{y}(x) = y(x) + \epsilon \eta(x) \)
\( y(x) \) is the extremal function we seek.
\( \eta(x) \) is a smooth "test" function where \( \eta(x_1) = \eta(x_2) = 0 \).
\( \epsilon \) is a small scalar.
Necessary Condition
\( \frac{d}{d\epsilon} J[y + \epsilon \eta] \Big|_{\epsilon=0} = 0 \)
The first variation must vanish for an extremum.
The Path of a Photon
Fermat's Principle states that light follows the path of least time . If light travels through a medium where speed \( v(y) \) depends on depth, what is the functional we are trying to minimize?
\( J[y] = \int_{x_1}^{x_2} \frac{\sqrt{1 + (y')^2}}{v(y)} \, dx \)
Discuss: How does this differ from minimizing a simple distance?
Functional Foundations Worksheet Functional Foundations
Calculus of Variations: Lesson 1
Scholar:
Date:
The Variational Mapping
In standard calculus, we find a point \( x^* \) that extremizes a function \( f(x) \). In the calculus of variations, we seek a function \( y^*(x) \) that extremizes a functional \( J[y] \). Most functionals in physics and geometry take the form: \[ J[y] = \int_{x_1}^{x_2} L(x, y, y') \, dx \] where \( L \) is the Lagrangian (or the integrand function).
1
Setting the Stage: Geometric Functionals
Write the functional \( J[y] \) that represents the following quantities. Express the Lagrangian \( L \) explicitly in terms of \( x, y, y' \).
A. The arc length of a curve \( y(x) \) between \( x=a \) and \( x=b \).
B. The surface area of the solid formed by rotating \( y(x) \) around the \( x \)-axis.
C. The time taken for a particle to travel from \( (x_1, y_1) \) to \( (x_2, y_2) \) along path \( y(x) \) with velocity \( v(y) = \sqrt{2gy} \).
2
The Perturbed Path
Consider the candidate extremal function \( y(x) \) and a perturbed path \( \hat{y}(x) = y(x) + \epsilon \eta(x) \), where \( \eta(x_1) = \eta(x_2) = 0 \).
Explain conceptually: Why must the test function \( \eta(x) \) vanish at the boundaries?
For the functional \( J[y] = \int_{0}^{1} (y'^2 + y^2) \, dx \), write the perturbed functional \( J[y + \epsilon \eta] \).
Apply the chain rule to compute \( \frac{d}{d\epsilon} J[y + \epsilon \eta] \Big|_{\epsilon=0} \). Do not simplify or integrate by parts yet.
Mathematical Foreshadowing
In standard calculus, we set the derivative to zero. In variational calculus, we will soon see that the condition \( \delta J = 0 \) leads to a second-order differential equation. Looking at your answer for Problem 2, can you guess how we might isolate \( \eta(x) \) to find a condition that depends only on \( y \)?
Functionals Facilitation Guide Facilitation Guide
Lesson 1: Functionals and Variation
Context
Graduate Mathematics / Physics
Instructional Purpose
The primary goal of this lesson is to shift student intuition from static optimization (finding points) to dynamic optimization (finding paths). Students often struggle with the abstraction of a "function of functions." This guide focuses on bridging that gap through geometric intuition and physical analogy.
Key Objectives
Define a functional \( J[y] \).
Formulate physical problems as functionals.
Apply the first variation principle.
Discussion: The Anatomy of a Functional
Prompt: "In multivariable calculus, we optimize \( f(x, y, z) \). How is this different from optimizing a functional \( J[y] \)?"
Student Misconception
Treating \( y \) as a single value or variable. They may try to take a partial derivative \( \partial J / \partial y \) without considering the integral.
Facilitator Guidance
Emphasize that \( y \) represents an infinite number of degrees of freedom. We aren't choosing a height; we are choosing an entire history/path.
Concept: Why Fixed Endpoints?
When introducing \( \hat{y}(x) = y(x) + \epsilon \eta(x) \), students often ask why \( \eta(x_1) = \eta(x_2) = 0 \) is strictly required.
"If we allow the endpoints to move, we are changing the problem from 'find the best path between A and B' to 'find the best path between two vertical lines'."
Visual Analogy: Imagine a guitar string pinned at both ends. The vibration (variation) can change the shape in the middle, but the ends are fixed by the bridge and nut.
[Illustration: Two curves starting and ending at same points, diverging in middle]
The "Light in a Medium" Challenge
This is the critical "hook" for physics students. Explain that Fermat's Principle is the root of Snell's Law.
1. Time \( T = \int \frac{ds}{v} \)
2. Element of arc length \( ds = \sqrt{dx^2 + dy^2} = \sqrt{1 + (y')^2} \, dx \)
3. If \( v \) varies with \( y \), we have \( T[y] = \int \frac{\sqrt{1 + (y')^2}}{v(y)} \, dx \)
Note: Use this to show that standard distance minimization is just the special case where \( v(y) = c \).
Worksheet Answer Key Notes
1A. Arc Length
\( J[y] = \int_{a}^{b} \sqrt{1 + (y')^2} \, dx \). Here, \( L = \sqrt{1 + (y')^2} \).
1C. Brachistochrone Intro
\( v = \sqrt{2gy} \implies J[y] = \int \frac{\sqrt{1+(y')^2}}{\sqrt{2gy}} \, dx \).
Euler-Lagrange Equation Slides \[ \frac{\partial L}{\partial y} - \frac{d}{dx} \left( \frac{\partial L}{\partial y'} \right) = 0 \]
\[ \int \eta \phi \, dx = 0 \]
Lesson 2
The Master Equation
Deriving the Euler-Lagrange equation: The necessary condition for a functional extremum.
"It is glorious to behold the power of the soul in the invention of these formulas." — Leonhard Euler
THE OBJECTIVE
From Lesson 1, we know that for an extremum, the first variation must vanish:
\[ \delta J = \int_{x_1}^{x_2} \left( \frac{\partial L}{\partial y} \eta + \frac{\partial L}{\partial y'} \eta' \right) dx = 0 \]
Problem: We need a condition that only involves \( y \), not \( \eta \).
Step 1: Integration by Parts
Focus on the second term in the integrand: \( \int_{x_1}^{x_2} \frac{\partial L}{\partial y'} \eta' \, dx \)
Let \( u = \frac{\partial L}{\partial y'} \) and \( dv = \eta' \, dx \).
\[ \left[ \frac{\partial L}{\partial y'} \eta \right]_{x_1}^{x_2} - \int_{x_1}^{x_2} \eta \frac{d}{dx} \left( \frac{\partial L}{\partial y'} \right) dx \]
Boundary Effect
Since \( \eta(x_1) = \eta(x_2) = 0 \), the boundary term vanishes!
Terms outside the integral = 0
The Fundamental Lemma (FLVC)
If \( \int_{x_1}^{x_2} \phi(x) \eta(x) \, dx = 0 \) for every smooth function \( \eta(x) \) that vanishes at the endpoints...
THEN \( \phi(x) = 0 \)
Intuition
If the "average" of \( \phi \) weighted by any choice of \( \eta \) is zero, then \( \phi \) must be zero everywhere.
This lemma is the "key" that unlocks the differential equation.
The Euler-Lagrange Equation
Necessary Condition
\[ \frac{\partial L}{\partial y} - \frac{d}{dx} \left( \frac{\partial L}{\partial y'} \right) = 0 \]
Converts a functional problem into an ordinary differential equation (ODE).
This is a second-order ODE for \( y(x) \).
Master Equation Workshop Guide Derivation Workshop
The Euler-Lagrange Equation
Researcher:
The Challenge
Derive the necessary condition for a function \( y(x) \) to be an extremum of the functional \( J[y] = \int_{x_1}^{x_2} L(x, y, y') \, dx \).
Step 1: The Vanishing Variation
Start by writing the expression for the first variation \( \delta J \) using the partial derivatives of the Lagrangian \( L \). Recall \( \delta J = \frac{d}{d\epsilon} J[y + \epsilon \eta] \Big|_{\epsilon=0} \).
[Write \(\delta J = \int ...\) here]
Step 2: Integration by Parts
Identify the term containing \( \eta'(x) \) and perform integration by parts on it. Show the boundary terms and the resulting integral.
Boundary Logic:
Why do the boundary terms vanish? State the condition on \( \eta(x) \):
Step 3: Recombination
Substitute your result from Step 2 back into the expression for \( \delta J \). Factor out the test function \( \eta(x) \).
[Resulting integral]
Step 4: The Fundamental Lemma
State the final differential equation that must be satisfied for the integral to vanish for any choice of \( \eta(x) \).
The Euler-Lagrange Equation
First Application: The Straight Line
Minimize the arc length functional \( J[y] = \int_{x_1}^{x_2} \sqrt{1 + (y')^2} \, dx \).
1. Identify Lagrangian \( L \):
2. Calculate \( \frac{\partial L}{\partial y} \):
3. Calculate \( \frac{\partial L}{\partial y'} \):
4. Plug into Euler-Lagrange:
5. Solve the resulting ODE to find \( y(x) \):
Euler-Lagrange Answer Key Answer Key & Solutions
The Master Equation Workshop
Lesson
02
The Core Derivation
Variation Formulation
\[ \delta J = \int_{x_1}^{x_2} \left( \frac{\partial L}{\partial y} \eta(x) + \frac{\partial L}{\partial y'} \eta'(x) \right) dx = 0 \]
Integration by Parts Step
For the term \( \int \frac{\partial L}{\partial y'} \eta' dx \):
\[ \left[ \frac{\partial L}{\partial y'} \eta \right]_{x_1}^{x_2} - \int_{x_1}^{x_2} \eta(x) \frac{d}{dx} \left( \frac{\partial L}{\partial y'} \right) dx \]
Boundary terms vanish because \( \eta(x_1) = \eta(x_2) = 0 \). The variation must preserve the fixed endpoints of the physical path.
Final Combination & Lemma
\[ \int_{x_1}^{x_2} \eta(x) \left( \frac{\partial L}{\partial y} - \frac{d}{dx} \left( \frac{\partial L}{\partial y'} \right) \right) dx = 0 \]
Applying the Fundamental Lemma , the term in parentheses must vanish identically for all \( x \):
\[ \frac{\partial L}{\partial y} - \frac{d}{dx} \left( \frac{\partial L}{\partial y'} \right) = 0 \]
Application: The Shortest Path
Objective: Prove the shortest distance between two points is a straight line.
1. Lagrangian:
\( L = \sqrt{1 + (y')^2} \)
2. Partial Derivatives:
\( \frac{\partial L}{\partial y} = 0 \)
\( \frac{\partial L}{\partial y'} = \frac{1}{2}(1 + (y')^2)^{-1/2} (2y') = \frac{y'}{\sqrt{1 + (y')^2}} \)
3. Substitute into Euler-Lagrange:
\( 0 - \frac{d}{dx} \left( \frac{y'}{\sqrt{1 + (y')^2}} \right) = 0 \)
This implies the interior term is constant with respect to \( x \).
4. Solution:
\( \frac{y'}{\sqrt{1 + (y')^2}} = C \)
Squaring and solving for \( y' \):
\( y'^2 = C^2(1 + y'^2) \implies y'^2 (1 - C^2) = C^2 \implies y'^2 = \frac{C^2}{1-C^2} = K^2 \)
\( y' = K \implies y(x) = Kx + b \)
The extremal path is a straight line.
Brachistochrone Case Study Slides Lesson 3
Fastest Descent
"I recognize the lion by his paw." — Johann Bernoulli (on Newton's anonymous solution)
THE PROBLEM
Find the path \( y(x) \) between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) such that a particle sliding under gravity travels in the shortest possible time .
Not a Straight Line
The Brachistochrone
Gravity acts downward
Physics of the Descent
Conservation
Energy conservation from rest at origin:
\( \frac{1}{2}mv^2 = mgy \)
\( v = \sqrt{2gy} \)
Time Integral
Total time \( T \) is the integral of \( dt \):
\( T = \int \frac{ds}{v} \)
where \( ds = \sqrt{1 + y'^2} \, dx \)
The Functional
The Brachistochrone Functional \( T[y] \):
\[ \int \frac{\sqrt{1 + y'^2}}{\sqrt{2gy}} \, dx \]
Applying Euler-Lagrange
Our Lagrangian is:
\( L = \sqrt{\frac{1 + y'^2}{y}} \)
Wait! \( L \) does not depend on \( x \).
This suggests there might be a shortcut to solving the ODE.
Key Insights
1 The particle needs to drop rapidly at first to gain speed.
2 Excessive length is traded for higher velocity.
3 The solution is not a parabola, but a cycloid .
The Solution: The Brachistochrone Curve
The path traced by a point on the rim of a rolling circle.
Parametric Equations:
\( x = R(\theta - \sin\theta) \)
Next: We will perform the modeling to reach this result.
Fastest Descent Modeling Lab Modeling Lab
The Brachistochrone Problem
Analyst:
Phase 1: Physical Foundations
1. Velocity Derivation
Assuming a particle starts from rest at the origin \( (0,0) \) and drops to a depth \( y \), use conservation of energy to find its velocity \( v \) in terms of \( y \) and gravity \( g \).
[Show derivation here]
2. Element of Time
The time \( dt \) taken to travel a distance \( ds \) is \( dt = ds/v \). Express \( dt \) entirely in terms of \( x, y, y', g \).
Phase 2: The Functional
The Time Functional
\[ T[y] = \int_{x_1}^{x_2} \text{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_} \, dx \]
Identify the Lagrangian \( L \):
Phase 3: Seeking the Differential Equation
Apply the Euler-Lagrange equation \( \frac{\partial L}{\partial y} - \frac{d}{dx} \left( \frac{\partial L}{\partial y'} \right) = 0 \) to your functional. Compute the partial derivatives:
A. \( \frac{\partial L}{\partial y} \)
B. \( \frac{\partial L}{\partial y'} \)
Analytical Roadblock
The resulting second-order ODE is highly non-linear and difficult to solve directly. However, notice that the Lagrangian \( L \) is independent of \( x \) . In Lesson 4, we will learn a "shortcut" for such systems called the Beltrami Identity.
Conceptual Inquiry
If gravity was turned off and the velocity was constant \( v = c \), what path would your functional yield? Verify this mathematically using the same Lagrangian framework.
Cycloid Reference Sheet Geometric Reference
The Cycloid
The Geometry of Fastest Descent
Definition
A cycloid is the curve traced by a point on the rim of a circular wheel as the wheel rolls along a straight line without slipping.
Parametric Equations
\[ x(\theta) = a(\theta - \sin\theta) \]
\[ y(\theta) = a(1 - \cos\theta) \]
where \( a \) is the radius of the rolling circle and \( \theta \) is the angle of rotation.
Physical Insight
For the Brachistochrone problem, we use an inverted cycloid starting at the origin. The "rolling" occurs along the underside of the x-axis.
Isochronous
A bead sliding on an inverted cycloid takes the same amount of time to reach the bottom regardless of where it starts on the curve.
Brachistochrone
It is the curve of fastest descent between two points under the influence of gravity alone (neglecting friction).
Tautochrone
From the Greek tauto (same) and chronos (time). It defines the path for a pendulum whose period is independent of amplitude.
Cycloid Facts
8a
Total length of one complete arch of the cycloid.
3\(\pi\)a\(^2\)
Total area under one arch of the cycloid (exactly 3x the rolling circle area).
Beltrami Identity Slides Lesson 4
The Beltrami Identity
Leveraging symmetry and conservation to simplify variational problems.
"Symmetry is a key which has opened many doors." — Herman Weyl
THE SPECIAL CASE
In many physical problems (like the Brachistochrone), the Lagrangian \( L \) does not depend explicitly on the independent variable \( x \):
\( L = L(y, y') \)
\( \frac{\partial L}{\partial x} = 0 \)
This implies a conservation law, similar to conservation of energy in time-independent systems.
Deriving the Shortcut
Consider the total derivative \( \frac{d L}{d x} \):
\[ \frac{d L}{d x} = \frac{\partial L}{\partial x} + \frac{\partial L}{\partial y} y' + \frac{\partial L}{\partial y'} y'' \]
Substitute \( \frac{\partial L}{\partial x} = 0 \) and use the Euler-Lagrange equation to replace \( \frac{\partial L}{\partial y} \):
\[ \frac{d L}{d x} = \frac{d}{dx} \left( y' \frac{\partial L}{\partial y'} \right) \]
The Beltrami Identity
First-Order Integral
\[ L - y' \frac{\partial L}{\partial y'} = C \]
Reduces a second-order ODE to a first-order ODE immediately.
Essential for solving geodesics and minimum surface problems.
Brachistochrone Revisited
With Beltrami, the problem becomes:
\[ \frac{\sqrt{1+y'^2}}{\sqrt{y}} - y' \frac{\partial}{\partial y'} \left( \frac{\sqrt{1+y'^2}}{\sqrt{y}} \right) = C \]
After some algebra:
\[ y(1 + y'^2) = k \]
This is much easier to separate and integrate than the full E-L expansion!
Geodesic Explorer Worksheet Geodesic Explorer
Applications of the Beltrami Identity
Geometer:
The Cylindrical Manifold
A geodesic is the shortest path between two points on a curved surface. For a cylinder of radius \( R \), the metric is \( ds^2 = R^2 d\theta^2 + dz^2 \).
1
Formulating the Functional
We want to find the path \( z(\theta) \) that minimizes the total arc length \( J \). Write the functional \( J[z] \) using the metric above.
\[ J[z] = \int_{\theta_1}^{\theta_2} \sqrt{\text{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}} \, d\theta \]
Identify the Lagrangian \( L(z, z') \):
2
The Beltrami Step
Verify that \( \frac{\partial L}{\partial \theta} = 0 \). Then, apply the Beltrami Identity: \( L - z' \frac{\partial L}{\partial z'} = C \).
Space for algebraic manipulation:
3
Interpretation of the Geodesic
Show that the solution to the resulting first-order ODE is \( z(\theta) = k\theta + z_0 \). What is the physical name of this curve on a cylinder?
Bonus Challenge: The Sphere
On a sphere, the metric is \( ds^2 = R^2 d\phi^2 + R^2 \sin^2\phi \, d\theta^2 \). If we look for paths \( \phi(\theta) \), we find \( L(\phi, \phi') \). What physical curves do you expect to find as the geodesics?
Note: The Beltrami identity essentially "integrates" the Euler-Lagrange equation once by exploiting the symmetry of the system.
Beltrami Shortcut Cheat Sheet Variational Shortcuts
The Beltrami Identity Reference
\( \delta \)
Standard Euler-Lagrange
Use when \( L = L(x, y, y') \)
\[ \frac{\partial L}{\partial y} - \frac{d}{dx} \left( \frac{\partial L}{\partial y'} \right) = 0 \]
Result: A second-order ODE.
Beltrami Identity
Use when \( \partial L / \partial x = 0 \)
\[ L - y' \frac{\partial L}{\partial y'} = C \]
Result: A first-order ODE (first integral).
Common Lagrangian Forms
Problem Type Lagrangian \( L \) Beltrami Result Shortest Path \( \sqrt{1+y'^2} \) \( y' = \text{const} \) (Lines) Fastest Descent \( \sqrt{\frac{1+y'^2}{y}} \) \( y(1+y'^2) = k \) (Cycloids) Surface Revolution \( y\sqrt{1+y'^2} \) \( \frac{y}{\sqrt{1+y'^2}} = C \) (Catenaries)
Physical Intuition
The quantity \( L - y' \frac{\partial L}{\partial y'} \) is the Hamiltonian of the system. If the Lagrangian is independent of "time" \( x \), the Hamiltonian is conserved. This is why we get a constant \( C \)!
Optimal Boundaries Slides \[ \int y \, dx \]
\[ \lambda \]
Lesson 5
Optimal Boundaries
Constrained optimization: Solving the Isoperimetric Problem with Lagrange Multipliers.
"Of all shapes with the same perimeter, the circle has the largest area." — Ancient Greek Maxim
The Integral Constraint
In many problems, we want to extremize a functional \( J[y] \) subject to a fixed value of another functional \( G[y] \).
The Math Setup
Extremize \( J[y] = \int_{x_1}^{x_2} f(x, y, y') \, dx \)
Subject to \( G[y] = \int_{x_1}^{x_2} g(x, y, y') \, dx = L \)
Classic Example: Dido's Problem
Princess Dido was offered "as much land as could be enclosed by a bull's hide." She cut the hide into a thin rope.
Question: What shape encloses the maximum area for a fixed length of rope?
The Method of Multipliers
Just like in multivariable calculus, we construct a new functional \( J^* \):
\[ J^*[y] = J[y] + \lambda G[y] \]
or more simply, optimize the new Lagrangian:
\[ L^* = f + \lambda g \]
The stationary path \( y(x) \) must satisfy the Euler-Lagrange equation for \( L^* \).
Solving Dido's Rope
1. Area: \( J = \int y \, dx \)
2. Perimeter: \( G = \int \sqrt{1 + y'^2} \, dx \)
3. Combined: \( L^* = y + \lambda \sqrt{1 + y'^2} \)
Wait... \( L^* \) does not depend on \( x \). Beltrami Identity applies!
The Resulting Path:
\[ (x-h)^2 + (y-k)^2 = \lambda^2 \]
It's a Circle!
Path of Least Resistance
From the shortest line to the Brachistochrone, the Beltrami shortcut, and constrained isoperimetry—you have mastered the foundations of functional optimization.
01
Formulate
02
Optimize
03
Understand
Isoperimetric Project Guide The Dido Project
Isoperimetric Optimization
Background
"How can one enclose the maximum possible area with a boundary of fixed length?" This challenge, known as the Isoperimetric Problem, is one of the oldest in the history of mathematics. Using the tools of variational calculus and Lagrange multipliers, you will prove that the answer is a circular arc.
01 Defining the Objectives
A. Area Functional (Maximize)
\( A[y] = \int_{x_1}^{x_2} y \, dx \)
B. Constraint Functional (Fixed)
\( L[y] = \int_{x_1}^{x_2} \sqrt{1+y'^2} \, dx = \mathcal{L} \)
02 Construct the Augmented Lagrangian
Define \( L^* = f + \lambda g \), where \( \lambda \) is the Lagrange Multiplier.
\( L^* = \) ________________________________________
03 Applying the Beltrami Shortcut
Since \( L^* \) does not depend explicitly on \( x \), use the identity \( L^* - y' \frac{\partial L^*}{\partial y'} = C \).
Perform the partial derivative \( \frac{\partial L^*}{\partial y'} \):
Plug into the Beltrami identity and simplify:
04 Geometrical Conclusion
Show that the simplified equation takes the form \( (y - C)^2 = \lambda^2 (1 + y'^2)^{-1} \dots \) and ultimately describes a circle of radius \( \lambda \).
Final Reflection
In physics, the "Surface Tension" of a liquid droplet essentially acts as a constraint on surface area for a fixed volume. Based on your findings today, why are small rain droplets spherical?
Final Variational Quiz Path of Least Resistance
Final Variational Assessment
Scholar:
CALCULUS OF VARIATIONS UNIT
Section 1: Foundations
1. Define a functional \( J[y] \) in your own words. How does it differ from a standard multivariable function?
2. In the context of the first variation \( \delta J \), why is it mathematically necessary for the test function \( \eta(x) \) to satisfy \( \eta(x_1) = \eta(x_2) = 0 \)?
Section 2: The Master Equation
3. State the Euler-Lagrange equation for the functional \( J[y] = \int L(x, y, y') \, dx \):
4. Consider the Lagrangian \( L = \sqrt{x} (y'^2 + y^2) \). Write out the explicit second-order ODE that must be satisfied for an extremum. (Do not solve.)
Section 3: Symmetry and Shortcuts
5. State the Beltrami Identity and the specific condition under which it is valid.
Formula & Condition:
6. The Brachistochrone problem leads to a Lagrangian \( L = \sqrt{\frac{1+y'^2}{y}} \). Briefly explain why this curve drops vertically at first before curving toward the target.
Section 4: Constraints
7. You wish to find the function \( y(x) \) that minimizes the "Action" functional \( S[y] = \int \mathcal{L} \, dt \), subject to the constraint that the average value of the function over interval \( [0, T] \) is \( K \). Formulate the augmented Lagrangian \( L^* \).
\( L^* = \) ________________________________________
Section 5: Matching Applications
Match the physical problem with its corresponding extremal shape.
A. Shortest distance on a sphere
B. Fastest descent under gravity
C. Minimum surface area of revolution
D. Maximum area for fixed perimeter
Cycloid
Great Circle
Circle
Catenary
