Timeline for Fourier Transform; half space baby problem (new)
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 21, 2021 at 15:42 | comment | added | Christian Remling | Yes, actually an issue with the extension is that we must preserve $\Delta u=0$ (again, secretly we know that odd extension will work). | |
| Jan 21, 2021 at 15:33 | comment | added | Math604 | Regarding your prior comment. I agree that making sense of (star) is a problem. Suppose we had limited $ \gamma$ to smooth compactly supported (instead of the slighty larger class) can we not view (star) as some distributional pairing (not claiming I can do anything with it) | |
| Jan 21, 2021 at 15:31 | comment | added | Math604 | Is that the proper extension (or do we want the odd extension in $y$). In any case for the equation I have in mind it has some coefficients in $y$ and so it might be quite complicated to do a Fourier transform in $y$ also. So, if possible, I really preferred to keep it to a transform just in $x$ | |
| Jan 21, 2021 at 15:26 | comment | added | Christian Remling | Why don't you just take the FT in both variables (declaring $u=0$ for $y<0$) to conclude that $\widehat{u}$ is supported by $\{0\}$, so $u$ is a linear combination of derivatives of $\delta$, but then only $\widehat{u} = C \partial_y \delta$ is consistent with the imposed bound + boundary condition. | |
| Jan 21, 2021 at 15:21 | comment | added | Christian Remling | On second thoughts, I think we are up against similar issues: since we only know that $x\mapsto u(x,y)$ is bounded, $\widehat{u}(\xi,y)$ is not guaranteed to be a function (and in fact we secretly know that it isn't, $\widehat{u}=Cy\delta$), and then it's not clear what (*) means. | |
| Jan 21, 2021 at 15:01 | comment | added | Math604 | Thanks for the comment. Recall that we imposed $ u(x,y)$ is bounded for each $y$ (sorry, i know its kinda buried in the post). | |
| Jan 21, 2021 at 9:45 | history | asked | Math604 | CC BY-SA 4.0 |