Mixing
In Evans (section on parabolic equations), why can he assume that $u(x_0, t_0)$ is positive?
$begingroup$
In Evan's proof for the Weak maximum principle
He simply assumes that $u(x_0, t_0) > 0$.
Why can he make this assumption? I have spent some time on this but unfortunately I don't see it. Does it follow from some result for parabolic equations or am I just missing something?
real-analysis pde proof-explanation parabolic-pde
asked Jan 26 at 5:17
QuokaQuoka
1,356316
$endgroup$
add a comment |
$begingroup$
In Evan's proof for the Weak maximum principle
He simply assumes that $u(x_0, t_0) > 0$.
Why can he make this assumption? I have spent some time on this but unfortunately I don't see it. Does it follow from some result for parabolic equations or am I just missing something?
real-analysis pde proof-explanation parabolic-pde
asked Jan 26 at 5:17
QuokaQuoka
1,356316
$endgroup$
add a comment |
$begingroup$
In Evan's proof for the Weak maximum principle
He simply assumes that $u(x_0, t_0) > 0$.
Why can he make this assumption? I have spent some time on this but unfortunately I don't see it. Does it follow from some result for parabolic equations or am I just missing something?
real-analysis pde proof-explanation parabolic-pde
asked Jan 26 at 5:17
QuokaQuoka
1,356316
$endgroup$
In Evan's proof for the Weak maximum principle
He simply assumes that $u(x_0, t_0) > 0$.
Why can he make this assumption? I have spent some time on this but unfortunately I don't see it. Does it follow from some result for parabolic equations or am I just missing something?
real-analysis pde proof-explanation parabolic-pde
real-analysis pde proof-explanation parabolic-pde
asked Jan 26 at 5:17
QuokaQuoka
1,356316
asked Jan 26 at 5:17
QuokaQuoka
1,356316
asked Jan 26 at 5:17
QuokaQuoka
1,356316
asked Jan 26 at 5:17
QuokaQuoka
1,356316
asked Jan 26 at 5:17
QuokaQuoka
1,356316
1,356316
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
By continuity, $u$ assumes its max on $bar U_T$. If it is assumed on the parabolic boundary, there is nothing to prove. If it is assumed inside and it is $le 0$ there is nothing to prove either because the max on the right hand side is $ge 0$. The only case to be considered is that of a positive max inside, and that actually never happens, as shown.
answered Jan 26 at 5:59
GReyesGReyes
2,22315
$endgroup$
$begingroup$
I didn't notice the right hand side had a plus sign, i.e. $u_+$ not simply $u$. Thanks!
$endgroup$
– Quoka
Jan 26 at 6:22
add a comment |
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%2f3087927%2fin-evans-section-on-parabolic-equations-why-can-he-assume-that-ux-0-t-0%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
By continuity, $u$ assumes its max on $bar U_T$. If it is assumed on the parabolic boundary, there is nothing to prove. If it is assumed inside and it is $le 0$ there is nothing to prove either because the max on the right hand side is $ge 0$. The only case to be considered is that of a positive max inside, and that actually never happens, as shown.
answered Jan 26 at 5:59
GReyesGReyes
2,22315
$endgroup$
$begingroup$
I didn't notice the right hand side had a plus sign, i.e. $u_+$ not simply $u$. Thanks!
$endgroup$
– Quoka
Jan 26 at 6:22
add a comment |
$begingroup$
By continuity, $u$ assumes its max on $bar U_T$. If it is assumed on the parabolic boundary, there is nothing to prove. If it is assumed inside and it is $le 0$ there is nothing to prove either because the max on the right hand side is $ge 0$. The only case to be considered is that of a positive max inside, and that actually never happens, as shown.
answered Jan 26 at 5:59
GReyesGReyes
2,22315
$endgroup$
$begingroup$
I didn't notice the right hand side had a plus sign, i.e. $u_+$ not simply $u$. Thanks!
$endgroup$
– Quoka
Jan 26 at 6:22
add a comment |
$begingroup$
By continuity, $u$ assumes its max on $bar U_T$. If it is assumed on the parabolic boundary, there is nothing to prove. If it is assumed inside and it is $le 0$ there is nothing to prove either because the max on the right hand side is $ge 0$. The only case to be considered is that of a positive max inside, and that actually never happens, as shown.
answered Jan 26 at 5:59
GReyesGReyes
2,22315
$endgroup$
By continuity, $u$ assumes its max on $bar U_T$. If it is assumed on the parabolic boundary, there is nothing to prove. If it is assumed inside and it is $le 0$ there is nothing to prove either because the max on the right hand side is $ge 0$. The only case to be considered is that of a positive max inside, and that actually never happens, as shown.
answered Jan 26 at 5:59
GReyesGReyes
2,22315
answered Jan 26 at 5:59
GReyesGReyes
2,22315
answered Jan 26 at 5:59
GReyesGReyes
2,22315
answered Jan 26 at 5:59
GReyesGReyes
2,22315
2,22315
$begingroup$
I didn't notice the right hand side had a plus sign, i.e. $u_+$ not simply $u$. Thanks!
$endgroup$
– Quoka
Jan 26 at 6:22
add a comment |
$begingroup$
I didn't notice the right hand side had a plus sign, i.e. $u_+$ not simply $u$. Thanks!
$endgroup$
– Quoka
Jan 26 at 6:22
$begingroup$
I didn't notice the right hand side had a plus sign, i.e. $u_+$ not simply $u$. Thanks!
$endgroup$
– Quoka
Jan 26 at 6:22
$begingroup$
I didn't notice the right hand side had a plus sign, i.e. $u_+$ not simply $u$. Thanks!
$endgroup$
– Quoka
Jan 26 at 6:22
add a comment |
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%2f3087927%2fin-evans-section-on-parabolic-equations-why-can-he-assume-that-ux-0-t-0%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
