Timeline for Green function of an elliptic operator
Current License: CC BY-SA 4.0
18 events
| when toggle format | what | by | license | comment | |
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| Sep 10 at 12:54 | comment | added | Ryo Ken | I can't find it | |
| Aug 21 at 15:39 | comment | added | Ryo Ken | Thank you@ Daniele Tampieri. | |
| Aug 20 at 19:56 | comment | added | Igor Khavkine | My last remark is that the reformulation of the problem in terms of the "entire solution" or the interior Dirichlet Green function leaves the OP none the wiser, as both reformulations are equally hard and are at least as hard if not harder than finding the desired "Poisson kernel". | |
| Aug 20 at 14:38 | comment | added | Daniele Tampieri | Since @IgorKhavkine has perfectly described what I meant, the only thing I can say is that the procedure is completely described in the first chapter of Carlo Miranda's Partial differential equations of elliptic type, 2nd rev. ed. Translated from the Italian by Zane C. Motteler. Ergebnisse der Mathematik und ihrer Grenzgebiete, Band 2, Berlin-Heidelberg-New York: Springer-Verlag pp. XII+370 (1970) MR0284700, Zbl 0198.14101 (beware, the notation is old fashioned). | |
| Aug 18 at 23:00 | comment | added | Ryo Ken | @IgorKhavkine. Thank you for your help. | |
| Aug 18 at 10:41 | comment | added | Igor Khavkine | The notation and terminology here is a bit confusing. I think that the starting point to match the OP's question to Daniele's discussion is the fundamental solution $E(z-w) \text{ "=" } K_1(|z-w|^2) e^{i\operatorname{Im} z\cdot\bar{w}}$, which already doesn't work because the rhs is not translation invariant, so one would have to write the lhs as $E(z,w)$. Next, the correct interior Dirichlet Green function is $\mathscr{G}(z,w) = E(z,w) - G(z,w)$, where the crucially necessary $G(z,w)$ is the unknown "entire solution". The "Poisson kernel" is then the differentiated $\mathscr{G}$. | |
| Aug 18 at 9:34 | comment | added | Ryo Ken | Iin my case $G(z,w)=K_1(|z-w|^2)e^{i Imz.\overline{w}}$. So, $\mathscr G(x,y)= E(x-y) - G(x,y)=G(x,y/|y|)-G(x,y)$. Is it true. @DanieleTampieri | |
| Aug 17 at 21:47 | comment | added | Daniele Tampieri | Ryo, in the above case, the Green function is defined as $$\mathscr G(x,y)= E(x-y) - G(x,y)$$ and the Poisson kernel is given by $$\frac{\partial}{\partial \hat{n}} \mathscr G(x, y))$$ where $\hat{n}$ is the inner normal vector to the boundary of the associated domain, but at this point I should explain better, since I'm doing things remembering what happens for self-adjoint $L$'s and I am possibly doing some slips. I need a couple of days to arrange a nice answer. | |
| Aug 17 at 19:46 | history | edited | Michael Hardy | CC BY-SA 4.0 |
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| Aug 17 at 16:31 | comment | added | Ryo Ken | @DanieleTampieri. in the above case, how the Poisson Kernel is defined? | |
| Aug 17 at 15:43 | comment | added | Daniele Tampieri | Ryo, am I missing something? $G(z,w)$ does not vanish on $\partial B$ while a Green function should vanish on the boundary of the domain associated to it: $G(z,w)$ seems more a fundamental solution. | |
| Aug 17 at 14:34 | history | edited | Ryo Ken | CC BY-SA 4.0 |
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| Aug 17 at 14:32 | comment | added | Ryo Ken | Thank you for your response@Daniele Tampieri. In above we have $LK_1(|z-w|^2)=0$, but the Green function is not $G(z,w)=K_1(|z-w|^2)$, the Green function is the kernel of the operator $L^{-1}$. In our case $G(z,w)=K_1(|z-w|^2)e^{i Imz.\overline{w}}$ | |
| Aug 17 at 14:18 | comment | added | Daniele Tampieri | @GiorgioMetafune, assuming that we know a fundamental solution of the elliptic operator $L$, i.e. a solution of the non homogeneous equation $$ L E(x-y) = \delta (x-y),$$ if moreover we know an "entire solution" of the homogeneous one, i.e. a solution of the following equation $$ L G (x,y)=0$$ such that $$ E(x-y) =G(x,y)\quad x\in \partial B,\, y\in B^\circ,$$ where $B^\circ$ is the interior of $B$, then we know the Green function $\mathscr{G}(x,y)$. I apologise for the non-standard locution: when $L=\Delta$, $G$ is an harmonic function respect to the $x$ variable. | |
| Aug 17 at 13:30 | comment | added | Giorgio Metafune | @DanieleTampieri Can you explain in more detail the last part? | |
| Aug 17 at 11:54 | history | edited | Daniele Tampieri | CC BY-SA 4.0 |
Minor formatting and Math Jaxing
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| Aug 17 at 11:51 | comment | added | Daniele Tampieri | Related, but dealing with the construction of fundamental solution instead of the construction of the Green function. Remember that once you have a fundamental solution, you can construct a Green function for a given domain if you know an entire solution for the same operator | |
| Aug 17 at 9:57 | history | asked | Ryo Ken | CC BY-SA 4.0 |