Timeline for Eigenfunction on surface with boundary
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Dec 7, 2014 at 16:31 | comment | added | Robert Bryant | Let us continue this discussion in chat. | |
| Dec 6, 2014 at 23:42 | comment | added | Robert Bryant | @Sim2004: I have no idea about this more general case. | |
| Dec 6, 2014 at 23:38 | comment | added | Slm2004 | Yes. you are absolutely right. what about a general manifold with $K\geq 0$ and $k\geq 0$? Show we also have a two dimensional solution space? | |
| Dec 6, 2014 at 22:03 | comment | added | Robert Bryant | @Slm2004: I've had a chance to think about this and realized that, under your conditions $K\ge0$ and $\kappa\ge0$, this only allows the cases of a disk in the Euclidean plane ($K=0$) and of a convex spherical cap ($K>0$). I don't see how you can get anything else connected that has $\kappa\ge0$ on all the boundary components. In these cases, there is always a $2$-dimensional space of solutions, namely the linear functions vanishing at the center of the disk or cap, where 'linear' refers to the standard embedding of the sphere or plane into $\mathbb{E}^3$. Possibly there are no other solutions. | |
| Dec 6, 2014 at 17:10 | comment | added | Robert Bryant | Correction: Above, when I wrote "there are exactly two solutions" I should have written "the space of solutions has dimension $2$". Sorry about the careless phrasing! | |
| Dec 6, 2014 at 16:26 | comment | added | Robert Bryant | In the upper unit hemisphere case, yes, there are exactly two solutions: If the surface is $x^2+y^2+z^2=1$ where $z\ge0$, then the functions $f = ax+by$ are solutions of your problem for any constants $a$ and $b$. Existence is trivial, of course, and uniqueness follows because, again $f$ will reflect to the lower hemisphere to define a global solution to $\Delta_gf = -2f$, and these are well-known to be the restrictions of the linear functions $ax+by+cz$ to the $2$-sphere, of which, only the ones with $c=0$ will satisfy the boundary condition. Other values with $K>0$ follow by scaling. | |
| Dec 6, 2014 at 16:10 | comment | added | Slm2004 | I mean $K\geq 0$ and $k\geq 0$. | |
| Dec 6, 2014 at 16:02 | comment | added | Slm2004 | Thank you very much. I am more interested in the $K>0$ and $k>0$ case. In the upper hemisphere case, $K\equiv 1$ and $k\equiv 0$ there also exists two linear independent solutions. Can you prove the existence? | |
| Dec 6, 2014 at 11:49 | history | edited | Robert Bryant | CC BY-SA 3.0 |
added some details about the reflection principle being invoked to construct a global $f$
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| Dec 5, 2014 at 19:39 | history | edited | Robert Bryant | CC BY-SA 3.0 |
added information about a simpler example
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| Dec 5, 2014 at 17:19 | history | answered | Robert Bryant | CC BY-SA 3.0 |