91Ó°ÊÓ

In calculus of variations, the Euler-Lagrange equation is a foundational tool used to find functions that provide stationary values to functionals. A functional is an expression that takes a function as input and returns a number. When a function makes a functional stationary, it could mean a minimum, maximum, or saddle point.

If you have a functional of the form \[ J[y] = \int_{a}^{b} F(x, y, y') \, dx \]
the Euler-Lagrange equation is given by:\[ \frac{\partial F}{\partial y} - \frac{d}{dx}\left( \frac{\partial F}{\partial y'} \right) = 0.\]
In the original problem, the functional is \( J[y] = \int_{0}^{1} (y')^3 \, dx \), so \( F(y, y') = (y')^3 \).

The partial derivative of \( F \) with respect to \( y \) is \( 0 \), and with respect to \( y' \) it is \( 3(y')^2 \). Substituting into the Euler-Lagrange equation, we solve \( \frac{d}{dx}((y')^3) = 0 \), leading to \((y')^3 = C \), where \( C \) is a constant.

A strong extremum in calculus of variations is a situation where not only the first variation of a functional vanishes, but the second variation is positive, indicating a minimum or negative for a maximum. The functional achieves a strong extremum when it doesn't just level off but actively bends upwards (or downwards), signaling a true minimum or maximum.

For the problem at hand, we focused on a strong minimum. After finding that \( y(x) = x \) satisfies the Euler-Lagrange equation, we need to ensure the second variation is positive.

If the second variation, \( \delta^2 J \), is calculated to always be positive (in this case \( \delta^2 J = \int_{0}^{1} 6 (\delta y')^2 \, dx > 0 \)), it confirms a strong minimum for the problem. A positive second variation indicates a strong extremum.

The second variation is crucial when confirming whether a functional's extremum is strong. It builds upon the first variation by examining the functional's curvature—not just where it flattens out.

For a functional \( J[y] \), if the first variation is zero at a certain function (meaning it's stationary), you look to the second variation to decide the nature (min or max) of this stationary point.

• If the second variation \( \delta^2 J \) is positive, the functional has a strong minimum
• If negative, it implies a strong maximum.

In the exercise, for \( y(x) = x \) and \( y' = 1 \), the second variation was computed:\[ \delta^2 J = \int_{0}^{1} 6 (\delta y')^2 \, dx \]which is greater than zero for all non-zero \( \delta y' \), confirming a strong minimum. This positive value reassures that the given functional doesn't just flatten but looks "concave up" around this solution.