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I am reading a bit about calculus of variations, and I've encountered the following.

Suppose the given function $F(\cdot,\cdot,\cdot)$ is twice continuously differentiable with respect to all of its arguments. Among all functions/paths $y=y(x)$, which are twice continuously differentiable on the internal $[a,b]$ with $y(a)$ and $y(b)$ specified, find the one which extremizes the functional defined by $$J(y) := \int_a^b F(x, y, y_x) \, {\rm d} x$$

I have a bit of trouble understanding what this means exactly, as a lot of the lingo is quite new. As I currently understand it, the function $F$ is a function, of which is second derivative is constant (like for example $y=x^2)$, and it looks as if $F$ is a function of 3 variables? I'm not too sure what $F(x,y, y_x)$ means, since $y_x$ was earlier specified to mean $y'(x)$.

Also, what is this 'functional' intuitively? I have some trouble understanding it, mainly because of the aforementioned confusion.

Rodrigo de Azevedo
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Phaptitude
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  • just a name, special type of a mapping. functional could be replace by function. – Alexander Grothendieck Nov 18 '13 at 20:09
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    @000 But my question is what it actually represents. Do you just want that integral definite integral to be as big/small as possible? Also, I have some problems with the notation I'd like to be clarified, that's the meat of the question I'd say. – Phaptitude Nov 18 '13 at 20:10
  • by the way "of which is second derivative is constant" doesn't make sense to me? is it just me? – Alexander Grothendieck Nov 18 '13 at 21:45

2 Answers2

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Consider the expression $$ J(y) := \int_a^b F(x,y,y_x) dx. $$ Here, $y$ is a function: $y:[a,b]\rightarrow \mathbb{R}$. Note: $y_x$ is a common notation for the partial derivative. Since $y$ depends only on one variable, $y_x \equiv y'$. The above expression abuses notation (in a very common way), and a better version is $$ J(y) := \int_a^b F(x,y(x),y_x(x)) dx. $$ Note that $F$ is not required to have vanishing derivatives of any order; it is just required to be sufficiently smooth. The function $F : \mathbb{R}^3\rightarrow \mathbb{R}$, and is to be evaluated at $x$, $y(x)$, and $y_x(x)$. The integrand is a well-defined function whose integral is guaranteed to exist.

Thus, $J$ takes a function $y : [a,b] \rightarrow \mathbb{R}$ and computes a number $J(y)$ by the right hand side. Physicists often call such a "function of a function" by the name "functional". When analyzed mathematically, $y$ is often described as an element in an abstract vector space of infinite dimension, call it $X$. Thus, think of $J$ as a map that takes a vector $y$ in $X$(visualize a finite dimension if you like) and produces a number.

The regularity assumtion of $F$ and $y$ lets you compute the change in $J(y)$ when $y$ changes by a "small amount", meaning that a small function is added to $y$. Typically, physicists think intuitively here, but analytst turn to the space $X$ which comes equipped with a rigorous way to measure smallness, for example a norm.

In physics notation, $y(x) \rightarrow y(x) + \epsilon \eta(x)$ for such a small change, and then compute the Taylor expansion: $$ J(y + \epsilon \eta) \approx \int F(x,y+\epsilon\eta,y_x + \epsilon \eta_x) dx = \int F(x,y,y_x) dx + \int F_2(x,y,y_x)\epsilon \eta(x)dx + \int F_3(x,y,y_x)\epsilon y_x dx, $$ where $J_i$ is the partial derivative with respect to argument number $i$ of $F$. Physicists work from this expression and derives the Euler--Lagrange equations.

Jas Ter
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  • Physicist you should not try to be more rigorous than you, it is not credible to give a Taylor expansion argument and talk about vectors having "length" (nobody said that anything is normed) after saying "The integrand is a well-defined function whose integral is guaranteed to exist". – Alexander Grothendieck Nov 18 '13 at 22:35
  • I have edited the answer somewhat, to make it clearer. – Jas Ter Nov 19 '13 at 07:55
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Please, have a look at this answer for all details.

In summary, a functional $J$ is a map that takes functions from an appropriate functional space and returns numbers. If the functional $J$ is represented through an integral like in the OP, then the Lagrangian $F=F(x,y(x),y'(x))$ is seen as a "function" of the variable $x$ and the functions $y$, $y'$. In any case, $F$ cannot be a function by itself (like your proposed, i.e. $F=F(x)$, with $F(x)=x^2$) because $J$ would be no functional anymore. $F$ can be dependent of $x$ and $y$, i.e. $F=F(x,y(x))$, or dependent of higher derivatives, like in the $F=F(x,y(x),y'(x),y^{''}(x))$ case.

The estremals of $J$, if they exist, are solutions of the equation

$$\delta J=0,$$ with $$\delta J=\frac{dJ}{d\epsilon}|_{\epsilon=0}:=\lim_{\epsilon\rightarrow 0} \frac{J[y+\epsilon\varphi]-J[y]}{\epsilon}, $$

for all functions $\varphi$ "near to" $y$ (called variations), such that

$$y(a)\stackrel{!}{=}(y+\epsilon\varphi)(a), $$ $$y(b)\stackrel{!}{=}(y+\epsilon\varphi)(b). $$

In other words, we are searching for those functions perturbing $y$ w.r.t. which the rate of change of $J$ is zero. Expressing $\delta J=0$ w.r.t. the Lagrangian $F$ leads to the well known Euler Lagrange equations.

To finish this short motivation, I would suggest to focus on an easy example and try to understand the formalism (I recocmmend the geodesic problem). I hope it helps.

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