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 ap.analysis-of-pdes
Search options not deleted
user 10583
Partial differential equations (PDEs): Existence and uniqueness, regularity, boundary conditions, linear and non-linear operators, stability, soliton theory, integrable PDEs, conservation laws, qualitative dynamics.
0
votes
analysis question related to $L^p$ type inequalities
If $p=2n$, a positive even integer, then your inequality is true -- use polar coordinates.
In view of this fact, I think your inequality should be true for all $p>1$.
- 744
1
vote
1
answer
227
views
Double series solution of wave equation
Let $u(x,y,t)$ be the solution of wave equation $u_{tt}=u_{xx}+u_{yy}, 0\lt x\lt 1, 0\lt y\lt 1, t\ge 0,$ $u(x,y,0)=(x-x^2)(y-y^2), u_t(x,y,0)=0$ and $ u(x,y,t)=0$ on the boundary of the square. Then …
CommunityBot
- 1