Timeline for Decay of eigenfunctions for Laplacian
Current License: CC BY-SA 4.0
4 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 30, 2019 at 18:29 | comment | added | Mateusz Kwaśnicki | For the interval $(0, a)$, these barriers would be $f(x) = c_1 x$ (which has Laplacian equal to zero) and $g(x) = c_2 x - c_3 x^2$ (which has Laplacian equal to $-2c_3$). One shows that the first eigenfunction must be between $f$ and $g$ for appropriately chosen $c_1, c_2, c_3$. The above is a simple discretisation of this argument. (2/2) | |
| Jul 30, 2019 at 18:28 | comment | added | Mateusz Kwaśnicki | Well, the proof of (1) uses concavity of $v_1$. This is no longer true if you add a scalar potential (unless it has correct sign). Neither it does extend to general graphs with boundaries. The idea is borrowed from the continuous setting, where one typically searches for barriers, which allows one to bound the solution by some kind of comparison principle. (1/2) | |
| Jul 30, 2019 at 18:14 | comment | added | Yannis Pimalis | thank you, I am a bit puzzled what you actually use to get the lower bound. It seems your lower bound only uses positivity? Do you agree that you get this lower bound on any graph? Where do you use for example that you have zero potential? Wouldn't this bound also generalize to a model with order on the diagonal?-In which case eigenvectors decay exponentially? | |
| Jul 30, 2019 at 8:32 | history | answered | Mateusz Kwaśnicki | CC BY-SA 4.0 |