Mixing
Estimation of relation between vertices of a triangle
up vote
0
down vote
favorite
Let $hat{t}$ be the reference triangle with the vertices $hat{A_1} = (0,0)$, $hat{A_2} = (1,0)$, $hat{A_3} = (0,1)$ and let $t$ be the triangle with the vertices $A_1 = (0,0)$, $A_2 = (h_1, 0)$ and $A_3 = (0, h_2)$. Let $hat{u} := u circ F_t$, where $F_t: hat{t} to t$ denotes the affine map with $F_t(hat{A_i}) = A_i$ for i = 1, 2, 3.
In my script, I read that the following estimate holds true:
$$ |u|_{H^k (t)} leq C^k h_t^{1-k} |hat{u}|_{H^k (hat{t})}$$
where $h_t$ is the diameter of the triangle $t$.
Why does this estimate hold true?
geometry numerical-methods estimation triangulation
asked 6 hours ago
StMan
1387
add a comment |
up vote
0
down vote
favorite
Let $hat{t}$ be the reference triangle with the vertices $hat{A_1} = (0,0)$, $hat{A_2} = (1,0)$, $hat{A_3} = (0,1)$ and let $t$ be the triangle with the vertices $A_1 = (0,0)$, $A_2 = (h_1, 0)$ and $A_3 = (0, h_2)$. Let $hat{u} := u circ F_t$, where $F_t: hat{t} to t$ denotes the affine map with $F_t(hat{A_i}) = A_i$ for i = 1, 2, 3.
In my script, I read that the following estimate holds true:
$$ |u|_{H^k (t)} leq C^k h_t^{1-k} |hat{u}|_{H^k (hat{t})}$$
where $h_t$ is the diameter of the triangle $t$.
Why does this estimate hold true?
geometry numerical-methods estimation triangulation
asked 6 hours ago
StMan
1387
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $hat{t}$ be the reference triangle with the vertices $hat{A_1} = (0,0)$, $hat{A_2} = (1,0)$, $hat{A_3} = (0,1)$ and let $t$ be the triangle with the vertices $A_1 = (0,0)$, $A_2 = (h_1, 0)$ and $A_3 = (0, h_2)$. Let $hat{u} := u circ F_t$, where $F_t: hat{t} to t$ denotes the affine map with $F_t(hat{A_i}) = A_i$ for i = 1, 2, 3.
In my script, I read that the following estimate holds true:
$$ |u|_{H^k (t)} leq C^k h_t^{1-k} |hat{u}|_{H^k (hat{t})}$$
where $h_t$ is the diameter of the triangle $t$.
Why does this estimate hold true?
geometry numerical-methods estimation triangulation
asked 6 hours ago
StMan
1387
Let $hat{t}$ be the reference triangle with the vertices $hat{A_1} = (0,0)$, $hat{A_2} = (1,0)$, $hat{A_3} = (0,1)$ and let $t$ be the triangle with the vertices $A_1 = (0,0)$, $A_2 = (h_1, 0)$ and $A_3 = (0, h_2)$. Let $hat{u} := u circ F_t$, where $F_t: hat{t} to t$ denotes the affine map with $F_t(hat{A_i}) = A_i$ for i = 1, 2, 3.
In my script, I read that the following estimate holds true:
$$ |u|_{H^k (t)} leq C^k h_t^{1-k} |hat{u}|_{H^k (hat{t})}$$
where $h_t$ is the diameter of the triangle $t$.
Why does this estimate hold true?
geometry numerical-methods estimation triangulation
geometry numerical-methods estimation triangulation
asked 6 hours ago
StMan
1387
asked 6 hours ago
StMan
1387
asked 6 hours ago
StMan
1387
asked 6 hours ago
StMan
1387
asked 6 hours ago
StMan
1387
1387
add a comment |
add a comment |
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3004629%2festimation-of-relation-between-vertices-of-a-triangle%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
