Mixing
What does Arg {Inf I(d)} means?
$begingroup$
I am currently studying the phase-field method for fracture modeling. In an article by Miehe -"Thermodynamically consistent phase-field models of fracture: Variational principles and multi-field FE implementation", I came across on an identity that I don't understand basically it says:
$$
d = mbox{Arg}left{inf_{d in W} I(d)right}.
$$
This expression came from the Euler type differential equation
$$
d(x) - l^2d''(x) = 0,
$$
which has the solution as an exponential function
$$
d(x) = exp(-|x|/l)
$$
Variational principle of this differential equation is ,
$$
d = mbox{Arg}left{inf_{d in W} I(d)right}.
$$
with $$ W = {d | d(0)=1, d(pm infty)=0}$$
where I(d) is a functional defined as an integral
$$
I(d) = frac{1}{2}int [d^2 + l^2d'^2] dV,
$$
and $l$ is a parameter.
Can anyone tell me what does that mean?
supremum-and-infimum
edited Jan 17 at 12:11
postmortes
2,04131119
asked Jan 17 at 9:37
Cro Simpson2.0Cro Simpson2.0
12
$endgroup$
add a comment |
$begingroup$
I am currently studying the phase-field method for fracture modeling. In an article by Miehe -"Thermodynamically consistent phase-field models of fracture: Variational principles and multi-field FE implementation", I came across on an identity that I don't understand basically it says:
$$
d = mbox{Arg}left{inf_{d in W} I(d)right}.
$$
This expression came from the Euler type differential equation
$$
d(x) - l^2d''(x) = 0,
$$
which has the solution as an exponential function
$$
d(x) = exp(-|x|/l)
$$
Variational principle of this differential equation is ,
$$
d = mbox{Arg}left{inf_{d in W} I(d)right}.
$$
with $$ W = {d | d(0)=1, d(pm infty)=0}$$
where I(d) is a functional defined as an integral
$$
I(d) = frac{1}{2}int [d^2 + l^2d'^2] dV,
$$
and $l$ is a parameter.
Can anyone tell me what does that mean?
supremum-and-infimum
edited Jan 17 at 12:11
postmortes
2,04131119
asked Jan 17 at 9:37
Cro Simpson2.0Cro Simpson2.0
12
$endgroup$
$begingroup$
Can you provide more context please? What is $I$? Maybe a function?
$endgroup$
– Dog_69
Jan 17 at 9:42
2
$begingroup$
Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
$endgroup$
– José Carlos Santos
Jan 17 at 9:43
$begingroup$
A pdf of the mentioned article is available here.
$endgroup$
– Mees de Vries
Jan 17 at 9:47
$begingroup$
It appears to simply be saying: find the infimum of the set $I(d)$ and return the argument, $d$, for the $I(d)$ that is the infimum.
$endgroup$
– Andy Walls
Jan 17 at 13:20
add a comment |
$begingroup$
I am currently studying the phase-field method for fracture modeling. In an article by Miehe -"Thermodynamically consistent phase-field models of fracture: Variational principles and multi-field FE implementation", I came across on an identity that I don't understand basically it says:
$$
d = mbox{Arg}left{inf_{d in W} I(d)right}.
$$
This expression came from the Euler type differential equation
$$
d(x) - l^2d''(x) = 0,
$$
which has the solution as an exponential function
$$
d(x) = exp(-|x|/l)
$$
Variational principle of this differential equation is ,
$$
d = mbox{Arg}left{inf_{d in W} I(d)right}.
$$
with $$ W = {d | d(0)=1, d(pm infty)=0}$$
where I(d) is a functional defined as an integral
$$
I(d) = frac{1}{2}int [d^2 + l^2d'^2] dV,
$$
and $l$ is a parameter.
Can anyone tell me what does that mean?
supremum-and-infimum
edited Jan 17 at 12:11
postmortes
2,04131119
asked Jan 17 at 9:37
Cro Simpson2.0Cro Simpson2.0
12
$endgroup$
I am currently studying the phase-field method for fracture modeling. In an article by Miehe -"Thermodynamically consistent phase-field models of fracture: Variational principles and multi-field FE implementation", I came across on an identity that I don't understand basically it says:
$$
d = mbox{Arg}left{inf_{d in W} I(d)right}.
$$
This expression came from the Euler type differential equation
$$
d(x) - l^2d''(x) = 0,
$$
which has the solution as an exponential function
$$
d(x) = exp(-|x|/l)
$$
Variational principle of this differential equation is ,
$$
d = mbox{Arg}left{inf_{d in W} I(d)right}.
$$
with $$ W = {d | d(0)=1, d(pm infty)=0}$$
where I(d) is a functional defined as an integral
$$
I(d) = frac{1}{2}int [d^2 + l^2d'^2] dV,
$$
and $l$ is a parameter.
Can anyone tell me what does that mean?
supremum-and-infimum
supremum-and-infimum
edited Jan 17 at 12:11
postmortes
2,04131119
asked Jan 17 at 9:37
Cro Simpson2.0Cro Simpson2.0
12
edited Jan 17 at 12:11
postmortes
2,04131119
asked Jan 17 at 9:37
Cro Simpson2.0Cro Simpson2.0
12
edited Jan 17 at 12:11
postmortes
2,04131119
edited Jan 17 at 12:11
postmortes
2,04131119
edited Jan 17 at 12:11
postmortes
2,04131119
2,04131119
asked Jan 17 at 9:37
Cro Simpson2.0Cro Simpson2.0
12
asked Jan 17 at 9:37
Cro Simpson2.0Cro Simpson2.0
12
asked Jan 17 at 9:37
Cro Simpson2.0Cro Simpson2.0
12
12
$begingroup$
Can you provide more context please? What is $I$? Maybe a function?
$endgroup$
– Dog_69
Jan 17 at 9:42
2
$begingroup$
Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
$endgroup$
– José Carlos Santos
Jan 17 at 9:43
$begingroup$
A pdf of the mentioned article is available here.
$endgroup$
– Mees de Vries
Jan 17 at 9:47
$begingroup$
It appears to simply be saying: find the infimum of the set $I(d)$ and return the argument, $d$, for the $I(d)$ that is the infimum.
$endgroup$
– Andy Walls
Jan 17 at 13:20
add a comment |
$begingroup$
Can you provide more context please? What is $I$? Maybe a function?
$endgroup$
– Dog_69
Jan 17 at 9:42
2
$begingroup$
Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
$endgroup$
– José Carlos Santos
Jan 17 at 9:43
$begingroup$
A pdf of the mentioned article is available here.
$endgroup$
– Mees de Vries
Jan 17 at 9:47
$begingroup$
It appears to simply be saying: find the infimum of the set $I(d)$ and return the argument, $d$, for the $I(d)$ that is the infimum.
$endgroup$
– Andy Walls
Jan 17 at 13:20
$begingroup$
Can you provide more context please? What is $I$? Maybe a function?
$endgroup$
– Dog_69
Jan 17 at 9:42
$begingroup$
Can you provide more context please? What is $I$? Maybe a function?
$endgroup$
– Dog_69
Jan 17 at 9:42
2
2
$begingroup$
Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
$endgroup$
– José Carlos Santos
Jan 17 at 9:43
$begingroup$
Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
$endgroup$
– José Carlos Santos
Jan 17 at 9:43
$begingroup$
A pdf of the mentioned article is available here.
$endgroup$
– Mees de Vries
Jan 17 at 9:47
$begingroup$
A pdf of the mentioned article is available here.
$endgroup$
– Mees de Vries
Jan 17 at 9:47
$begingroup$
It appears to simply be saying: find the infimum of the set $I(d)$ and return the argument, $d$, for the $I(d)$ that is the infimum.
$endgroup$
– Andy Walls
Jan 17 at 13:20
$begingroup$
It appears to simply be saying: find the infimum of the set $I(d)$ and return the argument, $d$, for the $I(d)$ that is the infimum.
$endgroup$
– Andy Walls
Jan 17 at 13:20
add a comment |
0
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
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%2f3076785%2fwhat-does-arg-inf-id-means%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
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%2f3076785%2fwhat-does-arg-inf-id-means%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
$begingroup$
Can you provide more context please? What is $I$? Maybe a function?
$endgroup$
– Dog_69
Jan 17 at 9:42
2
$begingroup$
Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
$endgroup$
– José Carlos Santos
Jan 17 at 9:43
$begingroup$
A pdf of the mentioned article is available here.
$endgroup$
– Mees de Vries
Jan 17 at 9:47
$begingroup$
It appears to simply be saying: find the infimum of the set $I(d)$ and return the argument, $d$, for the $I(d)$ that is the infimum.
$endgroup$
– Andy Walls
Jan 17 at 13:20