Functionals:
Variational Methods
Euler-Lagrange equation

Euler-Lagrange equation

From the functional S[y] defined by the integral:

S[y] = ∫ _x1^x2 f(y(x), y'(x), x) dx

and by using the variational approach, we will determine the Euler-Lagrange equation .

Let's suppose that we know the function yo(x), which makes the functional S extremal; that is S[yo] is a point extremum.

Since S[yo] is stationary, a small variation η(x) of the function y(x) around yo(x) involves a variation δS = 0, at the first order in η(x). That is:

y(x) = yo(x) + η(x)
Where η(x) << yo(x) for any x .

We will calculate the induced variation of the functional for a fixed value of x:

δS[y] = ∫ _x1^x2 δf(y(x), y'(x), x) dx =
∫ _x1^x2 [f(y(x), y'(x), x) - f(yo(x), y'o(x), x)] dx =
∫ _x1^x2 f(yo(x) + η(x), yo'(x) + η'(x), x) - f(yo(x), yo'(x), x)

In the first order in η(x) and η'(x), we have the following first terns of the Taylor series:

f(y(x), y'(x), x) =
f(yo(x), yo'(x), x) + (1/!1)(y - yo)∂f/∂y + (1/!1)(y' - yo')∂f/∂y' =
f(yo(x), yo'(x), x) + η(x)∂f/∂y + η'(x)∂f/∂y'

Therefore:

δS[y] = ∫ _x1^x2 [f(y(x), y'(x), x) - f(yo(x), y'o(x), x)] dx =
∫ _x1^x2 [f(yo(x), yo'(x), x) + η(x)∂f/∂y + η'(x)∂f/∂y' -
f(yo(x), y'o(x), x)] dx =
∫ _x1^x2 [ η(x)∂f/∂y + η'(x)∂f/∂y' ] dx .

δS[y] = ∫ _x1^x2 [η(x)∂f/∂y + η'(x)∂f/∂y' ] dx

The integration by part of the second term is:

∫ _x1^x2 [η'(x)∂f/∂y' ] dx = [η(x)∂f/∂y']_x1^x2 - ∫ _x1^x2 η(x) d[∂f/∂y']/dx

Therefore:

δS[y] = ∫ _x1^x2[ η(x)∂f/∂y - η(x) d[∂f/∂y']/dx] dx + [η(x)∂f/∂y']_x1^x2.

Since y(x1) and y(x2) are fixed, η(x1) = η(x2) = 0 and the last term of the equation vanishes. It remains :

δS[y] = ∫ _x1^x2 η(x)[∂f/∂y - d[∂f/∂y']/dx] dx

As δS = 0, regardless η(x), we obtain:

∫ _x1^x2 η(x)[∂f/∂y - d[∂f/∂y']/dx] dx = 0

∂f/∂y - d[∂f/∂y']/dx = 0

This is the Euler-Lagrange equation.