Timeline for The solvability of a Hölder ODE [closed]
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Jul 12, 2013 at 5:54 | history | closed |
Michael Renardy Andrey Rekalo Daniel Moskovich Willie Wong Andrés E. Caicedo |
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| Jul 7, 2013 at 9:48 | history | edited | Thomas | CC BY-SA 3.0 |
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| Jul 7, 2013 at 9:46 | comment | added | Thomas | You are right. But what I need is the existence of such a function. | |
| Jul 6, 2013 at 15:33 | answer | added | Michael Renardy | timeline score: 1 | |
| Jul 6, 2013 at 8:46 | comment | added | Andrew | For any such function the right derivative would be equal $+\infty$ for $x\ne0$: $$ \lim\limits_{h\rightarrow 0^+}\frac{u(x+h)-u(x)}{h}= \lim\limits_{h\rightarrow 0^+}\sqrt{|x|}{h^{-1/2}}=+\infty. $$ So it is not of $C^1$. | |
| Jul 6, 2013 at 5:08 | comment | added | Thomas | I think I made a mistake. There should be a exponent $\alpha$. | |
| Jul 6, 2013 at 5:06 | history | edited | Thomas | CC BY-SA 3.0 |
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| Jul 6, 2013 at 5:04 | comment | added | Thomas | You are right.I just want to construct such a function. And the function $u$ satisfies the equation I described is one. | |
| Jul 6, 2013 at 0:45 | history | edited | user9072 | CC BY-SA 3.0 |
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| Jul 5, 2013 at 16:25 | review | Close votes | |||
| Jul 12, 2013 at 5:57 | |||||
| Jul 5, 2013 at 16:21 | comment | added | Andrew | Your second statement is rather strange. If $u$ is an arbitrary function from $\text{C}^{0,1/2}(\Omega)$ then $v\in \text{C}^{0,1/2}(\Omega_h)$, $\Omega_h=\{x\in\Omega\,|\,\text{dist}(x,\partial \Omega)>h\}$. But, generally speaking, $v$ wouldn't belong to $\text{C}^{1}(\Omega_h)$. | |
| Jul 5, 2013 at 15:55 | review | First posts | |||
| Jul 5, 2013 at 16:23 | |||||
| Jul 5, 2013 at 15:38 | history | asked | Thomas | CC BY-SA 3.0 |