Timeline for "Integration by parts" formula for functionals
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 20, 2013 at 18:51 | comment | added | Deane Yang | Why can't you use essentially the same definitions and proof used for the Hilbert space result? | |
| Jan 20, 2013 at 18:30 | comment | added | Chris | @Deane Thanks for response. I did not consider weak derivatives of $f(t)$. But I don't know why $f'(t)$ would exist or why the formula you state is true (or false for that matter). Is it a standard result? | |
| Jan 20, 2013 at 18:06 | comment | added | Deane Yang | Yes, I know. So why isn't $$\frac{d}{dt}\langle f(t),u(t)\rangle = \langle f'(t),u(t)\rangle + \langle f(t),u'(t)\rangle$$ | |
| Jan 20, 2013 at 17:33 | comment | added | Chris | @Deane by $f(t)(u(t))$ I mean $\langle f(t), u(t) \rangle_{V^*, V}$, the pairing. | |
| Jan 20, 2013 at 17:30 | comment | added | Deane Yang | Why doesn't the usual product formula work? | |
| Jan 20, 2013 at 17:28 | comment | added | Chris | @Liviu Apologies, $v(t) \in V$. I edited. | |
| Jan 20, 2013 at 17:28 | history | edited | Chris | CC BY-SA 3.0 |
edited body
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| Jan 20, 2013 at 16:47 | comment | added | Liviu Nicolaescu | $u'(t)\in V^*$ and $v(t)\in H$. How do you define $v'(t)(\; v(t)\;)$? | |
| Jan 20, 2013 at 16:06 | history | asked | Chris | CC BY-SA 3.0 |