Timeline for Linear elliptic equation
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Jun 13, 2023 at 19:13 | vote | accept | Samir | ||
| Feb 17, 2023 at 23:00 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Jan 18, 2023 at 21:27 | answer | added | Daniel Castro | timeline score: 1 | |
| Jan 13, 2023 at 22:53 | history | edited | YCor |
edited tags
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| Jan 13, 2023 at 22:19 | history | edited | Michael Hardy | CC BY-SA 4.0 |
added 2 characters in body
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| Jan 11, 2023 at 22:26 | comment | added | Mhamdi Med | Thank you Zachary for your answer, but I want to know, what about the tangential derivative $\partial_\theta u(re^{i\theta})$? Because the function $u$ depends on the variables $z$ and $\overline{z}$. knowing that: $$\partial_z=\frac{1}{2}e^{-i\theta}\bigg(\partial_r-\frac{i}{r}\partial_\theta\bigg)$$ and $$\partial_{\overline{z}}=\frac{1}{2}e^{i\theta}\bigg(\partial_r+\frac{i}{r}\partial_\theta\bigg).$$ So the Laplacian operator can be $$\Delta=\frac{\partial^2_r}{\partial r^2}+\frac{\partial_r}{r\partial r}+\frac{\partial^2_\theta}{\partial\theta^2}.$$ | |
| Jan 11, 2023 at 16:34 | comment | added | Dispersion | Wolfram gives the solution as a linear combination of the Meijer G-function $G^{2,0}_{2,2}$ and the hypergeometric function $_{2}F_1$ in the radial case. | |
| Jan 11, 2023 at 16:27 | comment | added | Dispersion | Letting $u=f(r)$ be radial, your PDE becomes the ODE $f''+\frac{1}{r}f'-\frac{a}{1-r^2}f=0$. | |
| Jan 11, 2023 at 15:22 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
a typo in the title
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| S Jan 11, 2023 at 15:14 | review | First questions | |||
| Jan 11, 2023 at 16:39 | |||||
| S Jan 11, 2023 at 15:14 | history | asked | Samir | CC BY-SA 4.0 |