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 |
Questions about partial differential equations of elliptic type. Often used in combination with the top-level tag ap.analysis-of-pdes.
2
votes
1
answer
99
views
Estimate for elliptic problem on continuous functions
For an elliptic operator
$$ Lu = (a^{ij} D_iD_j + b^i D_i + c)u = f,$$
with suitable assumptions on the coefficients, one usually has Schauder estimates of the form
$$ \|u\|_{C^{2, \alpha}} \leq C(\|f …
1
vote
How to define the square root of $1-\Delta $?
You are done once you know that you have a functional calulus for the Laplace-Beltrami operator on $M$. For this, show that it is self-adjoint and has nonpositive spectrum (there are various ways to d …
7
votes
2
answers
1k
views
First order Elliptic operator
Assume that there exists a first order elliptic operator $D$ acting on functions from $\mathbb{R}^n$ to some vector space $V$. What can we conclude about $V$?
For example, is the dimension of $V$ alw …
4
votes
Generalized Hodge Decomposition on Manifolds with Boundary
The answer to this question as asked is no. However, you generally obtain something similar.
Consider $D = Q + Q^*$. By standard arguments, $$\ker(D) = \ker(Q)\cap \ker(Q^*).$$
(Of course $D\Phi = 0$ …
3
votes
2
answers
361
views
Exponential decay of resolvent kernel
For the integral kernel of the Laplacian $\Delta$ on $\mathbb{R}^n$, consider the resolvent $R(\lambda) := (\lambda - \Delta)^{-1}$ and let $R(\lambda; x, y)$ be its kernel, which is a smooth function …
2
votes
2
answers
433
views
Principal bundles and Subriemannian Geometry
In sub-Riemannian geometry, one considers manifolds $P$ equipped with a subbundle $\mathcal{H}$ of $TP$, the horizontal distribution. One then has a Riemannian metric only on this distribution $\mathc …