Timeline for The sum of linear partial differential operators of equal strength
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Jun 18, 2018 at 14:04 | comment | added | Alex M. | I am using the notations of Hörmander: If $P$ is a polynomial, then $P(D)$ is a differential operator, but $P(v)$ is its symbol. Therefore, $\partial^\alpha P(v)$ means $\partial _{v_1} ^{\alpha_1} \dots \partial _{v_m} ^{\alpha_m} (P(v_1, \dots, v_m))$. | |
| Jun 18, 2018 at 13:57 | comment | added | Ali Taghavi | Yes I know but what does it mean $\partial^{\alpha}P$ while P is itself a differential operator. For example what is ${|P^{\alpha}|}^2$ where $P$ is Laplacian and $\alpha=(2,0)$. I really do not understand your question. | |
| Jun 18, 2018 at 13:49 | comment | added | Alex M. | @AliTaghavi: $\alpha$ is a multi-index, i.e. $\alpha = (\alpha_1, \dots, \alpha_m) \in \mathbb N^m$. Then $\partial ^{\alpha} = \partial _{x_1} ^{\alpha_1} \dots \partial _{x_m} ^{\alpha_m}$. | |
| Jun 18, 2018 at 13:43 | comment | added | Ali Taghavi | Could you please more explain on $P^{\alpha}$ and its norm?What does $\partial^{\alpha}P$ mean?Is it $P$ a differential operator itself? Could you explain your notations by a precise operator, say Laplacian on the plane and $\alpha=(2,0)$?How it act on a given $v\in \mathbb{r}^2$ ? | |
| Apr 9, 2017 at 9:55 | history | asked | Alex M. | CC BY-SA 3.0 |