Fréchet derivative derivation

[Boon]

In Stone & Goldbart's Mathematics for Physics, in section 1.2.1 on the Calculus of Variations, they derive the Fréchet derivative. Part of the derivation is as follows:

Equation 1: J[y + εη] - J[y] = ∫ { f(x, y + εη, y' + εη') - f(x, y, y') } dx

Equation 2: J[y + εη] - J[y] = ∫ { εη ∂f/∂y + ε dη/dx ∂f/∂y' + O(ε²) } dx

To go from equation 1 to 2 a Taylor expansion is (implicitly) used. However, this seems rather complicated and introduces what I consider as the unnecessary complication O(ε²). Instead, I can derive the same result by using the total difference:

f(x, y + εη, y' + εη') - f(x, y, y') = δf = δx ∂f/∂x + δy ∂f/∂y + δy' ∂f/∂y' = 0 ∂f/∂x + εη ∂f/∂y + ε dη/dx ∂f/∂y' = εη ∂f/∂y + ε dη/dx ∂f/∂y'

which is the integrand in equation 2, minus the O(ε²) which vanishes in their derivation. Is my method correct? If so, why would one use the more complicated method of a Taylor expansion?

 

[jvicens]

Thank you for your contribution to the discussion on the derivation of the Fréchet derivative in Stone & Goldbart's Mathematics for Physics. Your method of using the total difference to derive the same result as the Taylor expansion is also valid. In fact, this is a common approach in the study of calculus of variations.

The reason why the authors used the Taylor expansion in their derivation is to provide a more intuitive understanding of the concept. By expanding the function f(x, y + εη, y' + εη') around the point (x, y, y') and keeping only the first-order term, they are able to show how the Fréchet derivative is related to the partial derivatives of the function. This can be helpful for students who are new to the subject and are still trying to grasp the concept.

Furthermore, the Taylor expansion also allows for a more general derivation, as it can be extended to higher orders in ε if necessary. This can be useful in some cases where a more accurate approximation is needed.

In summary, both methods are correct and valid in deriving the Fréchet derivative. The Taylor expansion may seem more complicated, but it serves a purpose in providing a more intuitive understanding of the concept. I hope this helps to clarify your doubts. Thank you for your contribution to the discussion.