Understanding the Euler Equation in Calculus of Variations

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🧩 The Simplest Problem in Calculus of Variations: The Euler Equation

What’s the shortest path between two points? You might instinctively say: a straight line. But how do we prove that — mathematically, rigorously?

Enter the elegant world of the Calculus of Variations — where instead of finding a single number, we’re finding an entire function that minimizes (or maximizes) some quantity. And the star of the show? The Euler Equation.

🎯 Problem Statement

Find the curve y(x) that connects the two points:

  • A: (0, 0)
  • B: (1, 1)

such that the total length of the path is as short as possible.

⛳ Functional to Minimize

The total length of the path is given by:

L[y] = ∫₀¹ √(1 + (dy/dx)²) dx

We want to find the function y(x) that minimizes this.

🧠 Step-by-Step Using Euler’s Equation

1. Identify the Function F

This is in the standard form:

F(x, y, y') = √(1 + (y')²)

2. Apply Euler’s Equation

Euler’s equation is:

d/dx (∂F/∂y') = ∂F/∂y

Now calculate the derivatives:

  • ∂F/∂y’ = y’ / √(1 + (y’)²)
  • ∂F/∂y = 0 (since F doesn’t depend on y)

So the equation becomes:

d/dx [ y' / √(1 + (y')²) ] = 0

3. Solve the Equation

If the derivative is 0, then the expression is constant:

y' / √(1 + (y')²) = C

Solving this leads to:

y' = constant  →  y(x) = mx + b

4. Apply Boundary Conditions

We know:

  • y(0) = 0 → b = 0
  • y(1) = 1 → m = 1

So the final solution is:

✅ y(x) = x

📌 Final Thoughts

This example illustrates the beauty of the calculus of variations. We didn’t just guess that the shortest path is a straight line — we proved it using math.

And all it took was a little magic from the Euler Equation.

Key Takeaway: When optimizing a functional that depends on a function and its slope, the solution must satisfy the Euler-Lagrange equation — a powerful tool with endless applications in physics, engineering, and beyond.