Timeline for Non-linear hyperbolic PDE
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Jan 14, 2021 at 10:44 | history | edited | Robert Bryant | CC BY-SA 4.0 |
Fixed a typo where I had written v-u instead of u-v, which is what I intended.
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| Jan 14, 2021 at 8:24 | vote | accept | Daniel Castro | ||
| Jan 7, 2021 at 17:56 | comment | added | Daniel Castro | Yes, that's exactly my notation. | |
| Jan 7, 2021 at 11:57 | comment | added | Robert Bryant | @DanielCastro: Before I go into an explanation, let me just ask this for clarification and to be sure that our notations align: Does your rotation matrix have the formula $$R[\theta(x,y)] = \begin{pmatrix}\cos\theta(x,y)&-\sin\theta(x,y)\\ \sin\theta(x,y)&\cos\theta(x,y)\end{pmatrix}?$$ | |
| Jan 7, 2021 at 2:36 | comment | added | Daniel Castro | That is, I'm looking for explicit expressions for $u$ and $v$ as functions of $x$ and $y$, in order to construct $\theta(x,y)=u(x,y)-v(x,y)$. | |
| Jan 7, 2021 at 1:20 | comment | added | Robert Bryant | @DanielCastro: I don't know what you mean by writing $\mathrm{d}x = f\,\mathrm{d}u$ and $\mathrm{d}y = f_v\,\mathrm{d}v$. That could never happen, since it would imply that $\sin(u-v)=0$ and so $\cos(u-v) = \pm 1$. | |
| Jan 6, 2021 at 23:48 | comment | added | Daniel Castro | Thank you so much ! the analysis in both answers is really deep. I still have some questions, though. Considering $x$ and $y$ as functions of $u$ and $v$, if $dx=fdu$ and $dy=f_{v}dv$ it follows that $\frac{\partial x}{\partial u}=f, \frac{\partial x}{\partial v}=0$, and $\frac{\partial y}{\partial v}=f_{v}$, $\frac{\partial y}{\partial u}=0$, so $x=x(u)$ and $y=y(v)$ which implies that $f$ has to be trivial. Is that correct ? | |
| Jan 6, 2021 at 14:54 | comment | added | Robert Bryant | @DanielCastro: Yes, there are various ways to allow for mutliply-connected regions $D'$, and it can even happen that $D'$ is simply-connected while the image $D=(x,y)(D')$ is multiply-connected. There's really no reason to restrict to $D'$ being a domain in $\mathbb{R}^2$. All you really need is that $D'$ is a surface on which there exist independent functions $u$ and $v$ and the space of nonvanishing solutions of the equation $f_{uv}=f$ is nontrivial. You also don't need to restrict to smooth solutions if you want to construct solutions of the original equation of lower regularity. | |
| Jan 6, 2021 at 13:31 | comment | added | Daniel Castro | I wonder if the assumption of simply connectedness on $D'$ can be removed, or softened. The metric is not exactly flat, but may have isolated points where the curvature is not defined, like conical defects. | |
| Jan 6, 2021 at 1:22 | history | edited | Robert Bryant | CC BY-SA 4.0 |
Clarified some language
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| Jan 5, 2021 at 14:04 | history | edited | Robert Bryant | CC BY-SA 4.0 |
Fixed an error in the 'nonclosure' condition
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| Jan 5, 2021 at 13:42 | history | answered | Robert Bryant | CC BY-SA 4.0 |