Mixing
Uniform boundedness of weak solution
$begingroup$
Let $u_nin W_{0}^{1,p}(Omega)$ be a positive weak solution of the equation:
$$
-Delta_p u=frac{f_n(x)}{(u+frac{1}{n})^delta}text{ in }Omega.
$$
Let $p=N$ and $fin L^m(Omega)$ for some $m>1$. Then how $u_n$ is uniformly bounded in $W_{0}^{1,p}(Omega) $. This result is proved in Lemma 4.5 in the following article: https://link.springer.com/content/pdf/10.1007%2Fs00030-016-0361-6.pdf
Can you help me with it. It is written as a simple fact.
Here $Omega$ is a bounded smooth domain in $mathbb{R}^N$, $f$ is a nonnegative but not identically zero function in $Omega$.
pde sobolev-spaces regularity-theory-of-pdes parabolic-pde fractional-sobolev-spaces
asked Jan 13 at 20:00
MathloverMathlover
1558
$endgroup$
add a comment |
$begingroup$
Let $u_nin W_{0}^{1,p}(Omega)$ be a positive weak solution of the equation:
$$
-Delta_p u=frac{f_n(x)}{(u+frac{1}{n})^delta}text{ in }Omega.
$$
Let $p=N$ and $fin L^m(Omega)$ for some $m>1$. Then how $u_n$ is uniformly bounded in $W_{0}^{1,p}(Omega) $. This result is proved in Lemma 4.5 in the following article: https://link.springer.com/content/pdf/10.1007%2Fs00030-016-0361-6.pdf
Can you help me with it. It is written as a simple fact.
Here $Omega$ is a bounded smooth domain in $mathbb{R}^N$, $f$ is a nonnegative but not identically zero function in $Omega$.
pde sobolev-spaces regularity-theory-of-pdes parabolic-pde fractional-sobolev-spaces
asked Jan 13 at 20:00
MathloverMathlover
1558
$endgroup$
add a comment |
$begingroup$
Let $u_nin W_{0}^{1,p}(Omega)$ be a positive weak solution of the equation:
$$
-Delta_p u=frac{f_n(x)}{(u+frac{1}{n})^delta}text{ in }Omega.
$$
Let $p=N$ and $fin L^m(Omega)$ for some $m>1$. Then how $u_n$ is uniformly bounded in $W_{0}^{1,p}(Omega) $. This result is proved in Lemma 4.5 in the following article: https://link.springer.com/content/pdf/10.1007%2Fs00030-016-0361-6.pdf
Can you help me with it. It is written as a simple fact.
Here $Omega$ is a bounded smooth domain in $mathbb{R}^N$, $f$ is a nonnegative but not identically zero function in $Omega$.
pde sobolev-spaces regularity-theory-of-pdes parabolic-pde fractional-sobolev-spaces
asked Jan 13 at 20:00
MathloverMathlover
1558
$endgroup$
Let $u_nin W_{0}^{1,p}(Omega)$ be a positive weak solution of the equation:
$$
-Delta_p u=frac{f_n(x)}{(u+frac{1}{n})^delta}text{ in }Omega.
$$
Let $p=N$ and $fin L^m(Omega)$ for some $m>1$. Then how $u_n$ is uniformly bounded in $W_{0}^{1,p}(Omega) $. This result is proved in Lemma 4.5 in the following article: https://link.springer.com/content/pdf/10.1007%2Fs00030-016-0361-6.pdf
Can you help me with it. It is written as a simple fact.
Here $Omega$ is a bounded smooth domain in $mathbb{R}^N$, $f$ is a nonnegative but not identically zero function in $Omega$.
pde sobolev-spaces regularity-theory-of-pdes parabolic-pde fractional-sobolev-spaces
pde sobolev-spaces regularity-theory-of-pdes parabolic-pde fractional-sobolev-spaces
asked Jan 13 at 20:00
MathloverMathlover
1558
asked Jan 13 at 20:00
MathloverMathlover
1558
asked Jan 13 at 20:00
MathloverMathlover
1558
asked Jan 13 at 20:00
MathloverMathlover
1558
asked Jan 13 at 20:00
MathloverMathlover
1558
1558
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Okay, as you written the result is proved in the article for $1 < p < N$. They say that the case $p=N$ can be proved similarly. You are now asking about this case.
Remember that $Omega subset mathbb{R}^N$ and $Delta_N u = text{div}(|nabla u|^{N-2} nabla u).$ The Sobolev inequality tells us that $W^{1,N}_0$ is embedded in $L^q$ for all $qgeq 1$.
Then
$$begin{aligned}|nabla u_n|_{L^N}^N = int |nabla u_n|^{N-2} nabla u cdot nabla u= -int Delta_N u_n cdot u_n &=int frac{f_n u_n}{(u_n+frac{1}{n})^gamma}
\ &leq int f_n u_n^{1-gamma}
\ &leq |f|_{L^m} |u_n^{1-gamma}|_{L^{m'}}
\ &=|f|_{L^m} |u_n|_{L^{(1-gamma)m'}}^{1-gamma}
\ &leq C |f|_{L^m} |nabla u_n|_{L^N}^{1-gamma}
end{aligned}$$
Hence
$$|nabla u_n|_{L^N} leq (C |f|_{L^m})^{frac{1}{N+gamma-1}}.$$
answered Jan 14 at 21:45
MarvinMarvin
2,5363920
$endgroup$
$begingroup$
Ok. I agree but one confusion is that suppose we have given $m>1$ and $gammain(0,1)$. Then to us ethe above embedding theorem one need that $(1-gamma)m'geq 1$. Why this is true?
$endgroup$
– Mathlover
Jan 15 at 19:09
$begingroup$
Please help me with this argument. I have understood all the arguments except this. Thank you very much.
$endgroup$
– Mathlover
Jan 15 at 19:13
$begingroup$
I have got. If this is less than 1 then using Holder inequality one can get the uniform bound. Thanks.
$endgroup$
– Mathlover
Jan 18 at 7:05
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%2f3072456%2funiform-boundedness-of-weak-solution%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
$begingroup$
Okay, as you written the result is proved in the article for $1 < p < N$. They say that the case $p=N$ can be proved similarly. You are now asking about this case.
Remember that $Omega subset mathbb{R}^N$ and $Delta_N u = text{div}(|nabla u|^{N-2} nabla u).$ The Sobolev inequality tells us that $W^{1,N}_0$ is embedded in $L^q$ for all $qgeq 1$.
Then
$$begin{aligned}|nabla u_n|_{L^N}^N = int |nabla u_n|^{N-2} nabla u cdot nabla u= -int Delta_N u_n cdot u_n &=int frac{f_n u_n}{(u_n+frac{1}{n})^gamma}
\ &leq int f_n u_n^{1-gamma}
\ &leq |f|_{L^m} |u_n^{1-gamma}|_{L^{m'}}
\ &=|f|_{L^m} |u_n|_{L^{(1-gamma)m'}}^{1-gamma}
\ &leq C |f|_{L^m} |nabla u_n|_{L^N}^{1-gamma}
end{aligned}$$
Hence
$$|nabla u_n|_{L^N} leq (C |f|_{L^m})^{frac{1}{N+gamma-1}}.$$
answered Jan 14 at 21:45
MarvinMarvin
2,5363920
$endgroup$
$begingroup$
Ok. I agree but one confusion is that suppose we have given $m>1$ and $gammain(0,1)$. Then to us ethe above embedding theorem one need that $(1-gamma)m'geq 1$. Why this is true?
$endgroup$
– Mathlover
Jan 15 at 19:09
$begingroup$
Please help me with this argument. I have understood all the arguments except this. Thank you very much.
$endgroup$
– Mathlover
Jan 15 at 19:13
$begingroup$
I have got. If this is less than 1 then using Holder inequality one can get the uniform bound. Thanks.
$endgroup$
– Mathlover
Jan 18 at 7:05
add a comment |
$begingroup$
Okay, as you written the result is proved in the article for $1 < p < N$. They say that the case $p=N$ can be proved similarly. You are now asking about this case.
Remember that $Omega subset mathbb{R}^N$ and $Delta_N u = text{div}(|nabla u|^{N-2} nabla u).$ The Sobolev inequality tells us that $W^{1,N}_0$ is embedded in $L^q$ for all $qgeq 1$.
Then
$$begin{aligned}|nabla u_n|_{L^N}^N = int |nabla u_n|^{N-2} nabla u cdot nabla u= -int Delta_N u_n cdot u_n &=int frac{f_n u_n}{(u_n+frac{1}{n})^gamma}
\ &leq int f_n u_n^{1-gamma}
\ &leq |f|_{L^m} |u_n^{1-gamma}|_{L^{m'}}
\ &=|f|_{L^m} |u_n|_{L^{(1-gamma)m'}}^{1-gamma}
\ &leq C |f|_{L^m} |nabla u_n|_{L^N}^{1-gamma}
end{aligned}$$
Hence
$$|nabla u_n|_{L^N} leq (C |f|_{L^m})^{frac{1}{N+gamma-1}}.$$
answered Jan 14 at 21:45
MarvinMarvin
2,5363920
$endgroup$
$begingroup$
Ok. I agree but one confusion is that suppose we have given $m>1$ and $gammain(0,1)$. Then to us ethe above embedding theorem one need that $(1-gamma)m'geq 1$. Why this is true?
$endgroup$
– Mathlover
Jan 15 at 19:09
$begingroup$
Please help me with this argument. I have understood all the arguments except this. Thank you very much.
$endgroup$
– Mathlover
Jan 15 at 19:13
$begingroup$
I have got. If this is less than 1 then using Holder inequality one can get the uniform bound. Thanks.
$endgroup$
– Mathlover
Jan 18 at 7:05
add a comment |
$begingroup$
Okay, as you written the result is proved in the article for $1 < p < N$. They say that the case $p=N$ can be proved similarly. You are now asking about this case.
Remember that $Omega subset mathbb{R}^N$ and $Delta_N u = text{div}(|nabla u|^{N-2} nabla u).$ The Sobolev inequality tells us that $W^{1,N}_0$ is embedded in $L^q$ for all $qgeq 1$.
Then
$$begin{aligned}|nabla u_n|_{L^N}^N = int |nabla u_n|^{N-2} nabla u cdot nabla u= -int Delta_N u_n cdot u_n &=int frac{f_n u_n}{(u_n+frac{1}{n})^gamma}
\ &leq int f_n u_n^{1-gamma}
\ &leq |f|_{L^m} |u_n^{1-gamma}|_{L^{m'}}
\ &=|f|_{L^m} |u_n|_{L^{(1-gamma)m'}}^{1-gamma}
\ &leq C |f|_{L^m} |nabla u_n|_{L^N}^{1-gamma}
end{aligned}$$
Hence
$$|nabla u_n|_{L^N} leq (C |f|_{L^m})^{frac{1}{N+gamma-1}}.$$
answered Jan 14 at 21:45
MarvinMarvin
2,5363920
$endgroup$
Okay, as you written the result is proved in the article for $1 < p < N$. They say that the case $p=N$ can be proved similarly. You are now asking about this case.
Remember that $Omega subset mathbb{R}^N$ and $Delta_N u = text{div}(|nabla u|^{N-2} nabla u).$ The Sobolev inequality tells us that $W^{1,N}_0$ is embedded in $L^q$ for all $qgeq 1$.
Then
$$begin{aligned}|nabla u_n|_{L^N}^N = int |nabla u_n|^{N-2} nabla u cdot nabla u= -int Delta_N u_n cdot u_n &=int frac{f_n u_n}{(u_n+frac{1}{n})^gamma}
\ &leq int f_n u_n^{1-gamma}
\ &leq |f|_{L^m} |u_n^{1-gamma}|_{L^{m'}}
\ &=|f|_{L^m} |u_n|_{L^{(1-gamma)m'}}^{1-gamma}
\ &leq C |f|_{L^m} |nabla u_n|_{L^N}^{1-gamma}
end{aligned}$$
Hence
$$|nabla u_n|_{L^N} leq (C |f|_{L^m})^{frac{1}{N+gamma-1}}.$$
answered Jan 14 at 21:45
MarvinMarvin
2,5363920
answered Jan 14 at 21:45
MarvinMarvin
2,5363920
answered Jan 14 at 21:45
MarvinMarvin
2,5363920
answered Jan 14 at 21:45
MarvinMarvin
2,5363920
2,5363920
$begingroup$
Ok. I agree but one confusion is that suppose we have given $m>1$ and $gammain(0,1)$. Then to us ethe above embedding theorem one need that $(1-gamma)m'geq 1$. Why this is true?
$endgroup$
– Mathlover
Jan 15 at 19:09
$begingroup$
Please help me with this argument. I have understood all the arguments except this. Thank you very much.
$endgroup$
– Mathlover
Jan 15 at 19:13
$begingroup$
I have got. If this is less than 1 then using Holder inequality one can get the uniform bound. Thanks.
$endgroup$
– Mathlover
Jan 18 at 7:05
add a comment |
$begingroup$
Ok. I agree but one confusion is that suppose we have given $m>1$ and $gammain(0,1)$. Then to us ethe above embedding theorem one need that $(1-gamma)m'geq 1$. Why this is true?
$endgroup$
– Mathlover
Jan 15 at 19:09
$begingroup$
Please help me with this argument. I have understood all the arguments except this. Thank you very much.
$endgroup$
– Mathlover
Jan 15 at 19:13
$begingroup$
I have got. If this is less than 1 then using Holder inequality one can get the uniform bound. Thanks.
$endgroup$
– Mathlover
Jan 18 at 7:05
$begingroup$
Ok. I agree but one confusion is that suppose we have given $m>1$ and $gammain(0,1)$. Then to us ethe above embedding theorem one need that $(1-gamma)m'geq 1$. Why this is true?
$endgroup$
– Mathlover
Jan 15 at 19:09
$begingroup$
Ok. I agree but one confusion is that suppose we have given $m>1$ and $gammain(0,1)$. Then to us ethe above embedding theorem one need that $(1-gamma)m'geq 1$. Why this is true?
$endgroup$
– Mathlover
Jan 15 at 19:09
$begingroup$
Please help me with this argument. I have understood all the arguments except this. Thank you very much.
$endgroup$
– Mathlover
Jan 15 at 19:13
$begingroup$
Please help me with this argument. I have understood all the arguments except this. Thank you very much.
$endgroup$
– Mathlover
Jan 15 at 19:13
$begingroup$
I have got. If this is less than 1 then using Holder inequality one can get the uniform bound. Thanks.
$endgroup$
– Mathlover
Jan 18 at 7:05
$begingroup$
I have got. If this is less than 1 then using Holder inequality one can get the uniform bound. Thanks.
$endgroup$
– Mathlover
Jan 18 at 7:05
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%2f3072456%2funiform-boundedness-of-weak-solution%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
