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The textbook on calculus of variations by Liberson gives the following definition of "first variation":

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It also gives the definition of the "Gateaux derivative"

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I want to prove that if $G$ is the first variation of $J$, it is also the Gateaux derivative of $J$. This seems very simple, but I'm not sure about one particular step.

My derivation:

Define $A(\alpha, \eta)=\frac{J(y+\alpha \eta)-J(y)}{\alpha}+\frac {o(\alpha)} \alpha$

From the definition of first variation, it follows that $\forall \alpha \forall \eta:\delta J|_y(\eta)=A(\alpha, \eta)$.

$(1)$ Here is the step I'm not sure about: Since $\delta J|_y(\eta)=A(\alpha, \eta)$ holds for all $\alpha$, it must in particular also hold for $\alpha$ as it tends to $0$. Therefore:

$$\delta J|_y(\eta)=\lim_{\alpha \to 0}A(\alpha, \eta)$$

$(2)$ By the definition of little-oh, this is equal to $\lim_{\alpha \to 0}\frac{J(y+\alpha \eta)-J(y)}{\alpha}$, which is the definition of the Gateaux derivative.

Is step $(1)$ in particular, and my derivation in general, rigorous?


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EDIT: I have now tried to apply the same principle to the second variation, which the book defines as:

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(just a small point: shouldn't there be a $\frac 1 2 $ before the second variation there?)

However, this gives me the following nonsensical result:

Define $A(\alpha, \eta)=\frac {J(y+\alpha \eta)-J(y)-\alpha \cdot\delta J|y(\eta)} {\alpha^2}+\frac {o(\alpha^2)} {\alpha^2}$

From the definition of second variation, it follows that $\forall \alpha \forall \eta:\delta^2 J|_y(\eta)=A(\alpha, \eta)$.

$(1)$ Hence by the same step I used before:

$$\delta^2 J|_y(\eta)=\lim_{\alpha\to 0} A(\alpha, \eta)= \lim_{\alpha\to 0} \left( \frac{J(y+\alpha \eta)-J(y)-\alpha \lim_{\alpha\to 0} \left ( \frac{J(y+\alpha \eta)-J(y)}{\alpha}\right)}{\alpha^2}\right)= \lim_{\alpha\to 0} \left( \frac{J(y+\alpha \eta)-J(y)-J(y+\alpha \eta)+J(y)}{\alpha^2}\right)=0$$

What am I doing wrong there? Am I not allowed to remove the limit inside the limit?

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    $\begingroup$ Maybe there are different definitions of Gâteaux derivative but as far as I remember is the main difference between Gâteaux and first variation that the Gâteaux derivative is a bounded linear functional (in $\eta$). The first variation must neither be linear nor continuous. $\endgroup$ May 22, 2017 at 10:03
  • $\begingroup$ Now I realized that your source calls Gâteaux derivative what I am used to call first variation. Anyway there is something strange in your steps: you say $\delta J|_y(\eta) = A(\alpha,\eta)$ for all $\alpha$. Since the left-hand-side is independent of $\alpha$ the right-hand-side has also to be independent of $\alpha$ but this is not true! So your first claim doesn't hold. $\endgroup$ May 22, 2017 at 10:11
  • $\begingroup$ the right hand side can still be independent of $\alpha$, even though $alpha$ is in it. In the same way that $f(x)=5$ is independent of $x$. The thing is that there is an $o(\alpha)$ in $A$. $\endgroup$
    – user56834
    May 22, 2017 at 10:21
  • $\begingroup$ You are right that there are cases where it is independent but in general this doesn't hold. I think I don't have to post an example to convince you. $\endgroup$ May 22, 2017 at 11:07
  • $\begingroup$ I don't understand why giving an example would invalidate the derivation? Also, are you saying the definition in the book is meaningless, because I've just used $A$ as shorthand. other than that I've just copied the definition in the book. $\endgroup$
    – user56834
    May 22, 2017 at 11:17

1 Answer 1

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Okay I am giving an example of what my concern is about. Since this wouldn't fit into a comment I write it as an answer.

$\newcommand{\R}{\mathbb{R}}$ Lets regard $J:\R \to \R, x\mapsto x^2$ then for a fixed $\eta$ we know \begin{align*} J(x+\alpha \eta) = J(x) + J'(x)\cdot\eta\cdot\alpha + \underbrace{o(\alpha)}_{\alpha^2\eta^2} . \end{align*} This means $J|_x(\eta)=2x\eta$, clearly independent of $\alpha$. Now \begin{align} A(\alpha, \eta) =\frac{J(x+\alpha \eta)-J(x)}{\alpha}+\frac {o(\alpha)} \alpha = \frac{2x\alpha\eta + \alpha^2\eta^2}{\alpha} + %\frac{\alpha^2\eta^2}{\alpha} \frac{o(\alpha)}{\alpha} = 2x\eta + \alpha\eta^2 + \frac{o(\alpha)}{\alpha}. \end{align} Now you are saying $J|_x(\eta)= A(\alpha,\eta)$ which is a little bit confusing. You can say this is true if you choose the right function in $o(\alpha)$ but it isn't clear that you can find such a function in $o(\alpha)$. This would mean that $J|_x(\eta)-\frac{J(x+\alpha \eta)-J(x)}{\alpha} = \frac{o(\alpha)}{\alpha}$ which is true but you didn't say why.

edit the definition of $A$ is somehow a problem because $o(\alpha)$ isn't a function but a set. You can say you choose the same function for $o(\alpha)$ as in the definition of the first variation but in this case your claim doesn't hold as my example demonstrates.

In my first equation the right choice of $o(\alpha)$ is $\alpha^2\eta^2$ but in order to receive the equalities for the function $A$, you have to choose $o(\alpha)$ as $-\alpha^2\eta^2$.

Of course everything works fine somehow because the result is right but it seems you are doing circular arguments. I just wanted to warn you about a bad way of proving things. It isn't clear from the beginning that there is a function in $o(\alpha)$ such that $\delta J|_y(\eta) = A(\alpha,\eta)$ holds. Nevertheless it is easy to show but you can't start with that because you are already done if you start with that assumption.

Maybe this seems like bean-counting but this is what mathematics is all about ;)


In order to give a rigorous prove I would do the following \begin{align*} \lim_{\alpha\to 0} \frac{f(y+\alpha\eta)-f(y)}{\alpha} = \lim_{\alpha\to 0} \frac{f(y)+\delta J|_y(\eta)\alpha+o(\alpha)-f(y)}{\alpha} \end{align*} Where $\delta J|_y(\eta)$ is the first variation. I just used the definition of the first variation. Now I am doing some simplification \begin{align*} = \lim_{\alpha\to 0} \delta J|_y(\eta)+\frac{o(\alpha)}{\alpha} =\delta J|_y(\eta)+\lim_{\alpha\to 0}\frac{o(\alpha)}{\alpha} =\delta J|_y(\eta) \end{align*} So this justifies to use the same symbol for both definitions.

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  • $\begingroup$ You said "but in this case your claim doesn't hold as my example demonstrates". How does your example demonstrate that? if we take $\alpha^2 \eta ^2 $ as our $o(\alpha)$, as you've done, then the equation holds. and by the definition of $\delta J|y(\eta)$, there is always some function $f(\alpha)$ such that the mentioned equation holds. $\endgroup$
    – user56834
    May 22, 2017 at 14:11
  • $\begingroup$ In general, if I understand correctly, $f(x)=g(x)+o(h(x))$ means that there is some function $k(x)\in o(h(x))$, such that $f(x)=g(x)+k(x)$ (where $o(h(x))$ is a set of all functions $k$ such that $\lim_{x\to a}\left|\frac k h (x)\right|=0$, with $a=0$ in this case). $\endgroup$
    – user56834
    May 22, 2017 at 14:14
  • $\begingroup$ Now about your new proof. Thank you, this is helpful. by the way, doesn't this contradict your earlier statement that: "the main difference between Gâteaux and first variation that the Gâteaux derivative is a bounded linear functional (in $\eta$). The first variation must neither be linear nor continuous." $\endgroup$
    – user56834
    May 22, 2017 at 14:19
  • $\begingroup$ you have to choose $-\alpha^2\eta^2$ for $o(x)$ (the minus is important) in order that everything fit together. I would be careful by using $o(x)$ in proofs, this could lead to some wrong conclusion. When you define some function $f(x) := g(x) + o(x)$ then this isn't a proper definition because you didn't specify $o(x)$. As I already said the definition of Gâteaux derivative may differ in literature. I know the definition you stated as first variation. As you see from your question the definition of first variation you stated is equivalent to your definition of Gâteaux derivative. $\endgroup$ May 22, 2017 at 14:28
  • $\begingroup$ So there is actually no need to differ between them $\endgroup$ May 22, 2017 at 14:43

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