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I'm studying a course of calculus of variation, and I was wondering why we always want to minimise $$I(u)=\int_\Omega f(x,u(x),u'(x))dx\ \ ?$$

Indeed, as I read on the internet, we are interested on the minimisation of functional. So why in in calculus of variation we only consider function of this form $$I(u)=\int_\Omega f(x,u(x),u'(x))dx\ \ ?$$

(I mean, it can be also $\int f(x,u(x),\nabla u(x))dx$ or stuff of this form.) Does all functional can be express as such an integral ?

Edit

As I can see, my course is adapted for physics. But why do we use the term : Calculus of VARIATION, I don't see in what we compute variation.

user330587
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  • We don't. The calculus of variations is about optimizing functionals, not necessarily minimizing them. That said, a lot of the time we do want to minimize the functional. For instance, in a lot of problems that come from physics, we want to minimize energy. Nonetheless, here's an example where we want to maximize the functional: look at example 2.4 here. –  Jun 09 '17 at 15:42
  • $f(x,u(x),u'(x))dx$ is the one dimensional theory, meant for the times when we want to find optimal curves. Of course, there are similar variations for when what you're looking for is a surface, or a vector field, or something else, and in that case, $f$ is a function of the corresponding parameter family, point, and first degree derivatives. – Arthur Jun 09 '17 at 15:42
  • And there are certainly problems in which the function to be optimized involves $f(x,u(x), u'(x), u''(x))$ but in general, these would be too difficult for a first course (and even too difficult period). – Mark Fischler Jun 09 '17 at 15:46
  • @MarkFischler That's certainly true. However, don't forget that mechanics is usually a second order theory, which means the Lagrangian (which is to say $f$) will be a first order expression. So not only is it more difficult, it is also strictly more uncommon, at least in applied fields. – Arthur Jun 09 '17 at 15:48
  • @Bye_World calculus of variations is about finding the extrema of functions and functionals, regardless the type of extremum. For instance, if we're intrested in finding the maximum of a function f then we search the minimum of the function -f. We're never interested about the nature of an extremum. Same argument holds also for other function's properties as for example the monotonicity; for example if a function is monotone we don't care if it is monotolically increasing or decreasing. – Matheman Jun 09 '17 at 21:19
  • @twinprime Optimizing is a synonym of "finding the extrema". So in that sense you're not disagreeing with me. But to say that we never care about whether a solution is a maximum or minimum is patently false. Sometimes in physics we'll get multiple (weak or strong) local extrema and we know to throw some out based on whether we're looking for a max or a min. –  Jun 09 '17 at 21:49

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Basically, in variational methods, we are interested in devising functions $f$ (usually "energy", i.e., squared field) of a given function $u(x)$ (usually a field), in such a way that "small" variations $\epsilon = \delta u$ of $u(x)$ give even smaller variations $\delta f$, i.e., $\delta f \sim {\cal O}(|\delta u|^2)=\epsilon^2$. Effectively, this leads to selection of functions that are stationary around an optimum, "stable" function $u_0$, provided the optimum is a minimum or maximum (but not, for example, a saddle point obtained from second-order derivatives). So this $f[u(x)]$ reaches a maximum or minimum at the targeted $u_0$.

The solution is found by solving Euler's equations, ${\cal L} u_0 = g$, where ${\cal L}$ is a linear operator which depends on the choice of $f$. The easiest way to find such an optimum is by considering quadratic functions of $u$, i.e., energy functions as the square of the field (parabolic functionals), because they reach a single (and hence global) maximum or minimum of $f$ for the optimum $u$. So, just as with Taylor series expansions of functions, one can consider the expansion of functionals $f$ about the optimum function $u_0(x)$:

$$f(u) = f(u_0) + (u-u_0) \left ( \frac{d f(u)}{d u} \right )_{u=u_0} + \ldots $$

Limiting to the first two terms when we search for the optimum, the expansion shows that the optimum as the solution of $df/du = 0$ at $u=u_0$ will depend on $u(x)$ (and hence on $x$) and on its first derivative $u^\prime$. This is our search space $f$ for $u_0$.

In principle, you could consider more complicated functions of $u$, that is $f(x,u,u^\prime, u^{\prime\prime}, \ldots)$ depending on higher-order partial derivatives. But this will not guarantee that the optimum is global. It also makes the search space of higher dimensions, which complicates the search. But it may lead to other better solutions.

TeXCub
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