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In the book about calculus of variation sand optimal control theory by Liberzon, he gives the following Taylor expansion in order to define the variations of a functional $J$:

enter image description here

However, for a normal $\mathbb R \to \mathbb R$ function, the Taylor expansion would have a $\frac 1 2 $ inserted there:

$$f(y+\alpha)=f(y)+f'(y)\alpha+\frac 1 2f''(y)\alpha ^2 + o(\alpha ^2)$$

Why is the $\frac 1 2$ not also next to the second-variation for the functional Taylor expansion?

I think that the result of this is that $$\delta ^2J|_y(\eta)=\frac 1 2\lim_{\alpha \to \infty}\left ( \frac{\delta J|_{y+\alpha \eta}(\eta)-\delta J|_y(\eta)}{\alpha}\right)$$

In other words, the $\frac 1 2 $ shows up here, where it shouldn't be according to my intuition.

Why is this?

EDIT: per request of @P.Siehr:

enter image description here

user56834
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  • Sorry maybe I am missing the point, but what difference could it make to incorporate the $\frac{1}{2}$ term in the expression for $\delta^2 J$, or not to do so and write it down in the expansion? – An aedonist Aug 31 '17 at 09:32
  • It is nice that you wrote everything in TeX, but here it might help if you could provide a picture of the page where he defines it. – P. Siehr Aug 31 '17 at 09:43
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    @Anaedonist, the difference is that according to this definition, all second variations are half the size of what I would intuitively say they should be. – user56834 Aug 31 '17 at 09:48
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    For the theory afterwards it does not matter (e.g. the term $δ^2J$ will still be positive definite). But I agree, that it seems to be odd. Have you read how it is done in "L. C. Young. Lectures on the Calculus of Variations and Optimal Control Theory." ? He, Liberzon, remarks [in "Notes and references for Chapter 1" , page 20] that there it is done in the same fashion. – P. Siehr Aug 31 '17 at 10:03

1 Answers1

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There is no particular reason that there is not a $1/2$ there; it's just a convention from the author (though perhaps an unconventional one).

I could do the same thing for a real-valued function $f$, and define the $n$th "variations" of $f$ to be the functions $f^{[n]}$ by $$f(x+\epsilon)=f(x)+f^{[1]}(x)\epsilon+f^{[2]}(x)\epsilon^2+\dots+f^{[n]}(x)\epsilon^n+o(\epsilon^n).$$ In this case Taylor's theorem tells us that $f^{[n]}(x)=f^{(n)}(x)/n!$, where $f^{(n)}$ indicates the $n$th derivative, so there is no need to introduce the new notation $f^{[n]}$.

That being said, I would guess the reason for omitting the $1/2$ in this case is that we don't care about the value of the second variation, only its sign. The usefulness of the second variation comes from looking at whether it is positive- or negative-definite, and an errant factor of $1/2$ doesn't affect that.

Carmeister
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