# Calculus of Variations: Nature of the Functional

1. Apr 27, 2014

### devd

Let $\normalsize S[y] = \int ^{a}_{b} f[y, \dot{y}, x] dx$ be the functional i want to minimize. Why does $\normalsize f$ (inside the integral) take this specific form?

Would i not be able to minimize the integral, $\normalsize S$ , if $f$ had any other form instead of $f = f[x, y, \dot{y}]$?

2. Apr 28, 2014

### HallsofIvy

Staff Emeritus
Do you understand what "f[x,y,y˙]" means? f is a function that can depend upon x, y, or the derivative of y but the "dependence" on any one can be 0- that is, this includes f(x), with f depending on x only, f(y) with f depending on y only, or f(y') with f depending on the derivative of y only. What more generality do you want?

3. Apr 28, 2014

### pasmith

Mathematically one can consider a functional of the form $$S[y] = \int_a^b f(x,y,y', \dots, y^{(n)})\,dx$$ for any $n \geq 1$, where the optimal solution satisfies $$\sum_{k=0}^n (-1)^k \frac{d^k}{dx^k}\left( \frac{\partial f}{\partial y^{(k)}}\right) = 0,$$ which is in principle a $2n$-order ODE subject to boundary conditions on $y$, $y'$, ..., $y^{(n-1)}$ at both $x = a$ and $x = b$. However in physical applications one generally has
$$\frac{\partial f}{\partial y^{(k)}} = 0$$ for $k \geq 2$ so there is no point in going beyond $n = 1$. Also the method of deriving the above ODE does not involve any ideas which are not required for the derivation of the Euler-Lagrange equation for the case $n = 1$; it just requires more integrations by parts.

4. Apr 28, 2014

### devd

Yes, i understand what f(x, y, y') means here. I was thinking about generalizations of the form that pasmith mentioned. Most of the texts are physically motivated, i guess. Probably that's why i didn't find the general form. Thanks, all! :)