91Ó°ÊÓ

A functional is a map that takes a function as an input and returns a real number. It is a key concept in the calculus of variations, where we aim to find the function that optimizes the value of the functional. In this exercise, the functional is given by:\[ J[y] = \int_{0}^{4} \left( x y' - (y')^2 \right) dx \]Here, the function \( y(x) \) is the unknown we are trying to determine, \( y' \) is the derivative of \( y \) with respect to \( x \), and the integrand \( F(x, y, y') = x y' - (y')^2 \) depends on \( x \) and \( y' \). The goal is to find a function \( y(x) \) that makes this functional stationary, meaning the value of \( J[y] \) does not change for small variations of \( y(x) \), given specific boundary conditions.

The Euler-Lagrange Equation is a foundational tool in finding the stationary function that makes a functional reach its optimal value. For a functional of the form:\[ \int F(x, y, y') \; dx \]The Euler-Lagrange equation is expressed as:\[ \frac{d}{dx} \left( \frac{\partial F}{\partial y'} \right) - \frac{\partial F}{\partial y} = 0 \]In our specific problem, because the functional integrates \( F(x, y, y') = x y' - (y')^2 \), we apply the Euler-Lagrange formula to calculate:

• \( \frac{\partial F}{\partial y'} = x - 2y' \)
• \( \frac{\partial F}{\partial y} = 0 \) (since \( F \) does not explicitly depend on \( y \))

Substituting into the Euler-Lagrange Equation yields:\[ \frac{d}{dx}(x - 2y') = 0 \]Solving this leads to the differential equation \( x - 2y' = C \), where \( C \) is a constant to be determined.

Boundary conditions are additional constraints that a solution must satisfy. These conditions are crucial as they help determine the constants of integration when solving differential equations. In this exercise, the boundary conditions are:

• \( y(0) = 0 \)
• \( y(4) = 3 \)

After finding the general solution of the differential equation derived from the Euler-Lagrange equation, we use these boundary conditions to determine the specific values of the constants. Here, the process is as follows:

• From the boundary condition \( y(0) = 0 \), we deduced \( D = 0 \).
• Applying \( y(4) = 3 \), we solved for \( C \) and found it to be \( \frac{1}{2} \).

Thus, the function \( y(x) = \frac{x^2}{4} - \frac{x}{4} \) satisfies both the Euler-Lagrange equation and the boundary conditions, confirming it as the correct stationary function.