Search Results
| Search type | Search syntax |
|---|---|
| Tags | [tag] |
| Exact | "words here" |
| Author |
user:1234 user:me (yours) |
| Score |
score:3 (3+) score:0 (none) |
| Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
| Views | views:250 |
| Code | code:"if (foo != bar)" |
| Sections |
title:apples body:"apples oranges" |
| URL | url:"*.example.com" |
| Saves | in:saves |
| Status |
closed:yes duplicate:no migrated:no wiki:no |
| Types |
is:question is:answer |
| Exclude |
-[tag] -apples |
| For more details on advanced search visit our help page | |
Results tagged with differential-operators
Search options not deleted
user 121820
Elliptic, parabolic and hyperbolic operators. Laplace, Laplace-Beltrami, Schrödinger, Dirac. Exterior derivative and Lie derivative operators.
6
votes
Simplification of integral on the sphere
You can also derive the lemma from the Bochner formula, which in general dimensions can be written
$$ \frac{1}{2}\Delta\lvert\nabla u\rvert^2 = \lvert\nabla^2u\rvert^2 - (\Delta u)^2 + \delta\left((\ …
- 1,713
5
votes
Accepted
Hodge decomposition for non-elliptic complexes
This is false. Equip $T^2 = S^1 \times S^1$ with the Lorentzian metric $g = -ds^2 + dt^2$. For concreteness, regard $S^1 = \mathbb{R} / \mathbb{Z}$. Consider the de Rham complex. The Hodge Laplaci …
- 1,713