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I'm beginner in Calculus of variation. I do not understand why a derivative is compulsory in the definition of the functional, as we see $y'(x)$ in $$ J(y) := \int_a^b F(x,y(x),y'(x)) dx. $$ How does $y'(x)$ make $J(y)$ a functional?

pluton
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Henry
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    It is not compulsory, it could be $F(x,y(x))$ or even just $F(y(x))$ like $J(y)=\int_a^by(x)^2,dx$ for $J$ to be a functional. The presence of $x$ and $y'(x)$ in $F$ is made to increase generality. In fact, in some problems it is even considered $F$ to include higher derivatives, for example, $F(x,y,y',y'')$. – A.Γ. Oct 05 '19 at 07:48

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A functional is simply a function that maps to $\Bbb{R}$. In this case the fact that this is a functional comes from the fact that you have an integral, returning a real number. The derivative comes from the nature of the Euler-Lagrange equation in your integrand and is not compulsory in the definition of a functional.

Dominic Petti
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