Why if $nabla(u-v)=0$ in $u<v$ imply $ugeq v$?

Clash Royale CLAN TAG#URR8PPP

up vote
-2
down vote

favorite

$A$ is a bounded domain in $mathbbR^N$ with $partial A$ of class $C^2$;

$u,v:overlineAtomathbbR$ - we write $uleq v$ if $u(x)leq v(x)$ for a.e. $xin A$ and $u(x)<v(x)$ if $u(x)leq v(x)$ and $u(x)<v(x)$ in a subset of $A$ having positive measure;

$|nabla u|_infty<1$, $|nabla v|_infty<1$;

$u,vin C^0,1(overlineA)$ and $u=0$ on $partial A$ and $vleq 0$ on $partial A$.

Why $nabla(u-v)=0$ in the set $u<v$ imply that $ugeq v$?

calculus real-analysis partial-derivative

share|cite|improve this question

edited Jul 16 at 21:21

asked Jul 16 at 19:10

BlackHawk

24

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Please clarify what is meant with $u<v$. Also, what are $u$ and $v$? Are those functions or variables?
  – Mefitico
  Jul 16 at 19:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Assuming that $u$ and $v$ are functions of $x,y,z$ in the $3$-dimensional real space and that $u<v$ is the set of $(x,y,z)$ such that $u(x,y,z)<v(x,y,z)$, what about $u=0$ and $v=1$ (i.e, they are constant functions)?
  – Batominovski
  Jul 16 at 19:19
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $A$ is a bounded domain in $mathbbR^N$ with $partial A$ of class $C^2$; $u,v:overlineAtomathbbR$ - we write $uleq v$ if $u(x)leq v(x)$ for a.e. $xin A$ and $u(x)<v(x)$ if $u(x)leq v(x)$ and $u(x)<v(x)$ in a subset of $A$ having positive measure.
  – BlackHawk
  Jul 16 at 19:26
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I think then u and v must vanish at the boundary then. @BlackHawk please give the precise space for u and v
  – Calvin Khor
  Jul 16 at 21:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  There must be more information missing. I can imagine either $u=v$ on the boundary, or $u$ is greater than $v$ somewhere in the domain...
  – Jeff
  Jul 16 at 21:05
  

  
  
  
  

 | 
show 1 more comment

up vote
-2
down vote

favorite

$A$ is a bounded domain in $mathbbR^N$ with $partial A$ of class $C^2$;

$u,v:overlineAtomathbbR$ - we write $uleq v$ if $u(x)leq v(x)$ for a.e. $xin A$ and $u(x)<v(x)$ if $u(x)leq v(x)$ and $u(x)<v(x)$ in a subset of $A$ having positive measure;

$|nabla u|_infty<1$, $|nabla v|_infty<1$;

$u,vin C^0,1(overlineA)$ and $u=0$ on $partial A$ and $vleq 0$ on $partial A$.

Why $nabla(u-v)=0$ in the set $u<v$ imply that $ugeq v$?

calculus real-analysis partial-derivative

share|cite|improve this question

edited Jul 16 at 21:21

asked Jul 16 at 19:10

BlackHawk

24

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Please clarify what is meant with $u<v$. Also, what are $u$ and $v$? Are those functions or variables?
  – Mefitico
  Jul 16 at 19:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Assuming that $u$ and $v$ are functions of $x,y,z$ in the $3$-dimensional real space and that $u<v$ is the set of $(x,y,z)$ such that $u(x,y,z)<v(x,y,z)$, what about $u=0$ and $v=1$ (i.e, they are constant functions)?
  – Batominovski
  Jul 16 at 19:19
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $A$ is a bounded domain in $mathbbR^N$ with $partial A$ of class $C^2$; $u,v:overlineAtomathbbR$ - we write $uleq v$ if $u(x)leq v(x)$ for a.e. $xin A$ and $u(x)<v(x)$ if $u(x)leq v(x)$ and $u(x)<v(x)$ in a subset of $A$ having positive measure.
  – BlackHawk
  Jul 16 at 19:26
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I think then u and v must vanish at the boundary then. @BlackHawk please give the precise space for u and v
  – Calvin Khor
  Jul 16 at 21:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  There must be more information missing. I can imagine either $u=v$ on the boundary, or $u$ is greater than $v$ somewhere in the domain...
  – Jeff
  Jul 16 at 21:05
  

  
  
  
  

 | 
show 1 more comment

up vote
-2
down vote

favorite

up vote
-2
down vote

favorite

$A$ is a bounded domain in $mathbbR^N$ with $partial A$ of class $C^2$;

$u,v:overlineAtomathbbR$ - we write $uleq v$ if $u(x)leq v(x)$ for a.e. $xin A$ and $u(x)<v(x)$ if $u(x)leq v(x)$ and $u(x)<v(x)$ in a subset of $A$ having positive measure;

$|nabla u|_infty<1$, $|nabla v|_infty<1$;

$u,vin C^0,1(overlineA)$ and $u=0$ on $partial A$ and $vleq 0$ on $partial A$.

Why $nabla(u-v)=0$ in the set $u<v$ imply that $ugeq v$?

calculus real-analysis partial-derivative

share|cite|improve this question

edited Jul 16 at 21:21

asked Jul 16 at 19:10

BlackHawk

24

$A$ is a bounded domain in $mathbbR^N$ with $partial A$ of class $C^2$;

$u,v:overlineAtomathbbR$ - we write $uleq v$ if $u(x)leq v(x)$ for a.e. $xin A$ and $u(x)<v(x)$ if $u(x)leq v(x)$ and $u(x)<v(x)$ in a subset of $A$ having positive measure;

$|nabla u|_infty<1$, $|nabla v|_infty<1$;

$u,vin C^0,1(overlineA)$ and $u=0$ on $partial A$ and $vleq 0$ on $partial A$.

Why $nabla(u-v)=0$ in the set $u<v$ imply that $ugeq v$?

calculus real-analysis partial-derivative

share|cite|improve this question

edited Jul 16 at 21:21

asked Jul 16 at 19:10

BlackHawk

24

share|cite|improve this question

share|cite|improve this question

edited Jul 16 at 21:21

edited Jul 16 at 21:21

edited Jul 16 at 21:21

asked Jul 16 at 19:10

BlackHawk

24

asked Jul 16 at 19:10

BlackHawk

24

asked Jul 16 at 19:10

BlackHawk

24

24

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Please clarify what is meant with $u<v$. Also, what are $u$ and $v$? Are those functions or variables?
  – Mefitico
  Jul 16 at 19:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Assuming that $u$ and $v$ are functions of $x,y,z$ in the $3$-dimensional real space and that $u<v$ is the set of $(x,y,z)$ such that $u(x,y,z)<v(x,y,z)$, what about $u=0$ and $v=1$ (i.e, they are constant functions)?
  – Batominovski
  Jul 16 at 19:19
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $A$ is a bounded domain in $mathbbR^N$ with $partial A$ of class $C^2$; $u,v:overlineAtomathbbR$ - we write $uleq v$ if $u(x)leq v(x)$ for a.e. $xin A$ and $u(x)<v(x)$ if $u(x)leq v(x)$ and $u(x)<v(x)$ in a subset of $A$ having positive measure.
  – BlackHawk
  Jul 16 at 19:26
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I think then u and v must vanish at the boundary then. @BlackHawk please give the precise space for u and v
  – Calvin Khor
  Jul 16 at 21:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  There must be more information missing. I can imagine either $u=v$ on the boundary, or $u$ is greater than $v$ somewhere in the domain...
  – Jeff
  Jul 16 at 21:05
  

  
  
  
  

 | 
show 1 more comment

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Please clarify what is meant with $u<v$. Also, what are $u$ and $v$? Are those functions or variables?
  – Mefitico
  Jul 16 at 19:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Assuming that $u$ and $v$ are functions of $x,y,z$ in the $3$-dimensional real space and that $u<v$ is the set of $(x,y,z)$ such that $u(x,y,z)<v(x,y,z)$, what about $u=0$ and $v=1$ (i.e, they are constant functions)?
  – Batominovski
  Jul 16 at 19:19
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $A$ is a bounded domain in $mathbbR^N$ with $partial A$ of class $C^2$; $u,v:overlineAtomathbbR$ - we write $uleq v$ if $u(x)leq v(x)$ for a.e. $xin A$ and $u(x)<v(x)$ if $u(x)leq v(x)$ and $u(x)<v(x)$ in a subset of $A$ having positive measure.
  – BlackHawk
  Jul 16 at 19:26
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I think then u and v must vanish at the boundary then. @BlackHawk please give the precise space for u and v
  – Calvin Khor
  Jul 16 at 21:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  There must be more information missing. I can imagine either $u=v$ on the boundary, or $u$ is greater than $v$ somewhere in the domain...
  – Jeff
  Jul 16 at 21:05
  

  
  
  
  

1

1

Please clarify what is meant with $u<v$. Also, what are $u$ and $v$? Are those functions or variables?
– Mefitico
Jul 16 at 19:16

Please clarify what is meant with $u<v$. Also, what are $u$ and $v$? Are those functions or variables?
– Mefitico
Jul 16 at 19:16

Assuming that $u$ and $v$ are functions of $x,y,z$ in the $3$-dimensional real space and that $u<v$ is the set of $(x,y,z)$ such that $u(x,y,z)<v(x,y,z)$, what about $u=0$ and $v=1$ (i.e, they are constant functions)?
– Batominovski
Jul 16 at 19:19

Assuming that $u$ and $v$ are functions of $x,y,z$ in the $3$-dimensional real space and that $u<v$ is the set of $(x,y,z)$ such that $u(x,y,z)<v(x,y,z)$, what about $u=0$ and $v=1$ (i.e, they are constant functions)?
– Batominovski
Jul 16 at 19:19

$A$ is a bounded domain in $mathbbR^N$ with $partial A$ of class $C^2$; $u,v:overlineAtomathbbR$ - we write $uleq v$ if $u(x)leq v(x)$ for a.e. $xin A$ and $u(x)<v(x)$ if $u(x)leq v(x)$ and $u(x)<v(x)$ in a subset of $A$ having positive measure.
– BlackHawk
Jul 16 at 19:26

$A$ is a bounded domain in $mathbbR^N$ with $partial A$ of class $C^2$; $u,v:overlineAtomathbbR$ - we write $uleq v$ if $u(x)leq v(x)$ for a.e. $xin A$ and $u(x)<v(x)$ if $u(x)leq v(x)$ and $u(x)<v(x)$ in a subset of $A$ having positive measure.
– BlackHawk
Jul 16 at 19:26

1

1

I think then u and v must vanish at the boundary then. @BlackHawk please give the precise space for u and v
– Calvin Khor
Jul 16 at 21:05

I think then u and v must vanish at the boundary then. @BlackHawk please give the precise space for u and v
– Calvin Khor
Jul 16 at 21:05

There must be more information missing. I can imagine either $u=v$ on the boundary, or $u$ is greater than $v$ somewhere in the domain...
– Jeff
Jul 16 at 21:05

There must be more information missing. I can imagine either $u=v$ on the boundary, or $u$ is greater than $v$ somewhere in the domain...
– Jeff
Jul 16 at 21:05

 | 
show 1 more comment

1 Answer
1

active

oldest

votes

up vote
0
down vote

Write $w=(v-u)_+$, where $t_+=maxt,0$. Then $win C^0,1(barA)$, $nabla w=0$ on $A$ and $wleq 0$ on $partial A$. It follows that $wleq 0$ on $A$.

share|cite|improve this answer

answered Jul 20 at 17:16

Jeff

3,094313

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',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

 

draft saved

draft discarded

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

Name Email

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2853744%2fwhy-if-nablau-v-0-in-uv-imply-u-geq-v%23new-answer', 'question_page');

);

Post as a guest

Name Email

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote

Write $w=(v-u)_+$, where $t_+=maxt,0$. Then $win C^0,1(barA)$, $nabla w=0$ on $A$ and $wleq 0$ on $partial A$. It follows that $wleq 0$ on $A$.

share|cite|improve this answer

answered Jul 20 at 17:16

Jeff

3,094313

add a comment | 

up vote
0
down vote

Write $w=(v-u)_+$, where $t_+=maxt,0$. Then $win C^0,1(barA)$, $nabla w=0$ on $A$ and $wleq 0$ on $partial A$. It follows that $wleq 0$ on $A$.

share|cite|improve this answer

answered Jul 20 at 17:16

Jeff

3,094313

add a comment | 

up vote
0
down vote

up vote
0
down vote

Write $w=(v-u)_+$, where $t_+=maxt,0$. Then $win C^0,1(barA)$, $nabla w=0$ on $A$ and $wleq 0$ on $partial A$. It follows that $wleq 0$ on $A$.

share|cite|improve this answer

answered Jul 20 at 17:16

Jeff

3,094313

Write $w=(v-u)_+$, where $t_+=maxt,0$. Then $win C^0,1(barA)$, $nabla w=0$ on $A$ and $wleq 0$ on $partial A$. It follows that $wleq 0$ on $A$.

share|cite|improve this answer

answered Jul 20 at 17:16

Jeff

3,094313

share|cite|improve this answer

share|cite|improve this answer

answered Jul 20 at 17:16

Jeff

3,094313

answered Jul 20 at 17:16

Jeff

3,094313

answered Jul 20 at 17:16

Jeff

3,094313

3,094313

add a comment | 

add a comment | 

 

draft saved

draft discarded

 

draft saved

draft discarded

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

Name Email

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2853744%2fwhy-if-nablau-v-0-in-uv-imply-u-geq-v%23new-answer', 'question_page');

);

Post as a guest

Name Email

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

Name Email

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

Name Email

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

Name Email

Name Email

Name

Email

Name Email

Name

Email