Talk:Energetic space

Confusion

I can't parse the following, from the article:

B:D(B)\subseteq X\to X

Whats B? What's D(B)? I'm guessing B is an operator, but what is D(B)? domain of B ?? So B maps from the domain of B to X? It would be cleaner to say

Let $Y\subset X$ and let $B:Y\to X$ ... etc. that makes it comprehensible. linas 04:06, 15 December 2006 (UTC)[reply]

You guessed right. That's a clearer description. —Ben FrantzDale

Meaning

As far as I can tell, $\langle u,u\rangle _{E}$ (or sometimes the square root thereof) can represent the energy of a configuration u (at least in the area of elasticity). The thing I can't figure out is this: what does $\langle u,v\rangle _{E}$ when u and v are different configurations? I have a sense of what an inner product means between functions, but not in the energy space. Any thoughts? —Ben FrantzDale 05:56, 15 December 2006 (UTC)[reply]

Rewrite, and a question

I rewrote this article, primarily for style and to clarify things. I cut out the paragraph

That is,

B_{E}

is that linear functional which acts like B but has a domain of

X_{E}

—that is, its domain includes all limit points, u, of the domain of B for which Bu_n is bounded as

u_{n}\to u

.

since it is not clear what norm one uses to talk about convergence and boundedness (the original norm, or the energy norm?).

It would be nice if somebody else could also take a look at how the article looks now to make sure I didn't make any typos. Oleg Alexandrov (talk) 01:27, 7 May 2007 (UTC)[reply]

Cool. I'll fix anything I see, but when I started this page it was at the edge of my understanding (hence the {{expert}} tag). Thanks. —Ben FrantzDale 02:33, 7 May 2007 (UTC)[reply]

Sounds good. I added a physical example. I must say physics is not my strength, I hope I got all right from Zeidler's book. Extra eyes to look over that would be very welcome of course. Oleg Alexandrov (talk) 03:56, 7 May 2007 (UTC)[reply]

Woah... Integration by parts

Oleg, I just had an ah-ha regarding this line of your example:

(Bu|v)=-\int _{a}^{b}\!u''(x)v(x)\,dx=\int _{a}^{b}u'(x)v'(x)=(u|Bv)

It looks like that is saying that integration by parts is conceptually the same as distributing the metric tensor across both arguments to the inner product. Hua. —Ben FrantzDale 04:24, 7 May 2007 (UTC)[reply]

I guess. I don't know what "distributing the metric tensor across both arguments to the inner product" means, but I am glad it helped you. Oleg Alexandrov (talk) 05:05, 7 May 2007 (UTC)[reply]

It just means that I never liked this:

(Bu|v)=(u|Bv)

. If it doesn't matter which side the B goes, it seems much more symmetrical to think of it as

({\sqrt {B}}u|{\sqrt {B}}v)

. —Ben FrantzDale 05:09, 7 May 2007 (UTC)[reply]