I am constructing a smooth function $f(x)\equiv f(u(x),v(x))$, such that $u(x)$ and $v(x)$ are some trial parameters. I have the following integral $$G=\int_{x_i}^{x_f} f(u(x),v(x)) \mathrm{d}x.$$ My aim: To find the trial functions $u(x)$ and $v(x)$ using the stationary principle. Therefore, essentially I want $u(x)$ and $v(x)$ to be such that $G$ is minimum.

My question: Am I allowed to use: $$\frac{\partial f}{\partial u}=\frac{\partial f}{\partial v}=0$$ to solve for $u(x)$ and $v(x)$?

  • 1
    $\begingroup$ You don't have $\frac{\partial f}{\partial u}$, you essentially have $\frac{\partial G}{\partial u}$. But that derivative is different from familiar derivatives, because it is a derivative with respect to a function. You also should say something about constraints on $u$ and $v$. (You also needn't think of $u,v$ as separate variables, they can be joined together into a single vector-valued function of $x$.) $\endgroup$ – Ian Jun 7 '16 at 13:53
  • $\begingroup$ Well, the integral is supposed to be definite (sorry for the confusion, I have corrected it), and therefore, G is merely some constant. Thus, it does not make sense to have $\partial G/\partial u$. $\endgroup$ – titanium Jun 7 '16 at 13:57
  • $\begingroup$ Sure it does; $G$ is a function of the functions $u,v$, which for minimization purposes should be considered to be variables themselves. $\endgroup$ – Ian Jun 7 '16 at 13:58
  • $\begingroup$ $u$ is a function of $x$, and once you integrate over $x$, there is no information of $u$ that is left anymore. So, I am not sure how you could say $G$ is a function of $u, v$. $\endgroup$ – titanium Jun 7 '16 at 13:59
  • 1
    $\begingroup$ As for the derivation of the Euler-Lagrange-like equation, you take $G(g)=\int_a^b f(g(x)) dx$. Consider $h(x)=g(x)+\epsilon a(x)$ where we assume $a$ is bounded and $h$ is admissible. Then $G(h) = G(g)+\epsilon \int_a^b f'(g(x)) a(x) dx + o(\epsilon)$. This means that the functional derivative of $G$ is the linear map $a \mapsto \int_a^b f'(g(x)) a(x) dx$. For this to vanish means that $f'(g(x))$ must be $L^2$-perpendicular to all admissible perturbations $a$. With no constraints this means that you must have $f'(g(x))=0$ for every $x$; with constraints you can have a nontrivial situation. $\endgroup$ – Ian Jun 7 '16 at 14:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.