Timeline for Airy's equation on $\mathbb R_-$
Current License: CC BY-SA 3.0
11 events
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| Feb 5, 2020 at 13:12 | history | edited | YCor |
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| Jan 17, 2016 at 21:31 | vote | accept | Delio Mugnolo | ||
| Jan 14, 2016 at 19:46 | answer | added | ifw | timeline score: 3 | |
| Jan 13, 2016 at 23:58 | history | edited | Delio Mugnolo | CC BY-SA 3.0 |
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| Jan 13, 2016 at 21:57 | comment | added | Christian Remling | Yes, on $\mathbb R$ the $L^2$ theory is straightforward: $A=iD^3$ is essentially self-adjoint on $C_0^{\infty}$, so again $e^{itA}$ is a unitary group. | |
| Jan 13, 2016 at 20:04 | comment | added | Delio Mugnolo | @ChristianRemling Indeed! There are several works suggesting that two b.c. have to be imposed on the right and one on the left endpoints; but in these works I can never find a "clean" well-posedness theorem in a "nice" function space (ideally, a Sobolev or Lebesgue space, but I'd be happy even with one space from Triebel's zoo; ideally, by means of a unitary $C_0$-group). Remarkably, the equation on $\mathbb R$ is governed by a propagator that is just the convolution of an integral kernel based on Airy's function and the Fourier transform of the initial data, so a group might exist after all. | |
| Jan 13, 2016 at 20:00 | history | edited | Delio Mugnolo | CC BY-SA 3.0 |
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| Jan 13, 2016 at 18:54 | comment | added | Christian Remling | On a bounded interval, the deficiency index is $3$, so you need this many boundary conditions (not two). | |
| Jan 13, 2016 at 18:46 | comment | added | Christian Remling | On a bounded interval, $A=iD^3$ has self-adjoint realizations and then $e^{itA}$ is a unitary group. On a half line, $D^3$ has unequal deficiency indices $(2,1)$, so this approach won't work. | |
| Jan 13, 2016 at 18:42 | answer | added | András Bátkai | timeline score: 3 | |
| Jan 13, 2016 at 16:09 | history | asked | Delio Mugnolo | CC BY-SA 3.0 |