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negative outward normal derivative implies positiveness near the boundary.
Clash Royale CLAN TAG#URR8PPP
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So, this seens to be simple but I could not handle to prove it:
Suppose $Omega$ is a bounded and open domain of $mathbb R^N$ with smooth boundary (as smooth as you want), $Ngeq 2$ and that $uin C^1(overlineOmega,mathbb R)$ is a function such that $partial u/partialnu<0$ all over $partialOmega$, where $nu$ denotes the outward normal unit vector on $partial Omega$.
Is it true that $u>0$ near the boundary? That means: can a $delta>0$ be chosen such that $u(x)>0$ for all $xinyinOmega; dist(y,partialOmega)<delta$?
calculus regularity-theory-of-pdes
asked Jul 30 at 19:45
bhcribeiro
1
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up vote
0
down vote
favorite
1
So, this seens to be simple but I could not handle to prove it:
Suppose $Omega$ is a bounded and open domain of $mathbb R^N$ with smooth boundary (as smooth as you want), $Ngeq 2$ and that $uin C^1(overlineOmega,mathbb R)$ is a function such that $partial u/partialnu<0$ all over $partialOmega$, where $nu$ denotes the outward normal unit vector on $partial Omega$.
Is it true that $u>0$ near the boundary? That means: can a $delta>0$ be chosen such that $u(x)>0$ for all $xinyinOmega; dist(y,partialOmega)<delta$?
calculus regularity-theory-of-pdes
asked Jul 30 at 19:45
bhcribeiro
1
Yes, it is true (if $partialOmega$ is at least $C^1$), but thinking about its proof: do you know "some" differential topology? (I mean, tubular neighborhood?).
– user539887
Jul 30 at 20:03
You will find some help math.stackexchange.com/questions/2839762/…
– Gustave
Jul 31 at 16:41
Yes, differential Topology is not a monster I haven't seen! But anyways, I forgot to mention that $uequiv 0$ on the boundary but I think you figured out that.
– bhcribeiro
Jul 31 at 17:34
Gustave, thanks for the link, but it seems that my question is exactly the opposite: I need positiveness of $u$ near the boundary knowing the condition over its normal derivative. And this is of course true near each point of the boundary but I need a kind of uniformity on this behaviour, as stated in the question.
– bhcribeiro
Jul 31 at 17:40
@bhcribeiro In all that talk about tubular neighborhood the essential thing is that the mapping $partialOmegatimes(-delta,delta)ni (x,s)mapsto x+snu(x)inmathbbR^N$ is, for $delta>0$ sufficiently small, a $C^1$ diffeomorphism of $partialOmegatimes(-delta,delta)$ onto its image (this follows by the inverse function theorem). In particular, every $yinOmega$ sufficiently close to $partialOmega$ can be uniquely written as $x+snu(x)$ for some $xinpartialOmega$ and $sin(-delta,0]$. If you have this, the remaining part is simple calculus.
– user539887
Jul 31 at 18:40
 |Â
show 1 more comment
up vote
0
down vote
favorite
1
up vote
0
down vote
favorite
1
1
So, this seens to be simple but I could not handle to prove it:
Suppose $Omega$ is a bounded and open domain of $mathbb R^N$ with smooth boundary (as smooth as you want), $Ngeq 2$ and that $uin C^1(overlineOmega,mathbb R)$ is a function such that $partial u/partialnu<0$ all over $partialOmega$, where $nu$ denotes the outward normal unit vector on $partial Omega$.
Is it true that $u>0$ near the boundary? That means: can a $delta>0$ be chosen such that $u(x)>0$ for all $xinyinOmega; dist(y,partialOmega)<delta$?
calculus regularity-theory-of-pdes
asked Jul 30 at 19:45
bhcribeiro
1
So, this seens to be simple but I could not handle to prove it:
Suppose $Omega$ is a bounded and open domain of $mathbb R^N$ with smooth boundary (as smooth as you want), $Ngeq 2$ and that $uin C^1(overlineOmega,mathbb R)$ is a function such that $partial u/partialnu<0$ all over $partialOmega$, where $nu$ denotes the outward normal unit vector on $partial Omega$.
Is it true that $u>0$ near the boundary? That means: can a $delta>0$ be chosen such that $u(x)>0$ for all $xinyinOmega; dist(y,partialOmega)<delta$?
calculus regularity-theory-of-pdes
asked Jul 30 at 19:45
bhcribeiro
1
asked Jul 30 at 19:45
bhcribeiro
1
asked Jul 30 at 19:45
bhcribeiro
1
asked Jul 30 at 19:45
bhcribeiro
1
1
Yes, it is true (if $partialOmega$ is at least $C^1$), but thinking about its proof: do you know "some" differential topology? (I mean, tubular neighborhood?).
– user539887
Jul 30 at 20:03
You will find some help math.stackexchange.com/questions/2839762/…
– Gustave
Jul 31 at 16:41
Yes, differential Topology is not a monster I haven't seen! But anyways, I forgot to mention that $uequiv 0$ on the boundary but I think you figured out that.
– bhcribeiro
Jul 31 at 17:34
Gustave, thanks for the link, but it seems that my question is exactly the opposite: I need positiveness of $u$ near the boundary knowing the condition over its normal derivative. And this is of course true near each point of the boundary but I need a kind of uniformity on this behaviour, as stated in the question.
– bhcribeiro
Jul 31 at 17:40
@bhcribeiro In all that talk about tubular neighborhood the essential thing is that the mapping $partialOmegatimes(-delta,delta)ni (x,s)mapsto x+snu(x)inmathbbR^N$ is, for $delta>0$ sufficiently small, a $C^1$ diffeomorphism of $partialOmegatimes(-delta,delta)$ onto its image (this follows by the inverse function theorem). In particular, every $yinOmega$ sufficiently close to $partialOmega$ can be uniquely written as $x+snu(x)$ for some $xinpartialOmega$ and $sin(-delta,0]$. If you have this, the remaining part is simple calculus.
– user539887
Jul 31 at 18:40
 |Â
show 1 more comment
Yes, it is true (if $partialOmega$ is at least $C^1$), but thinking about its proof: do you know "some" differential topology? (I mean, tubular neighborhood?).
– user539887
Jul 30 at 20:03
You will find some help math.stackexchange.com/questions/2839762/…
– Gustave
Jul 31 at 16:41
Yes, differential Topology is not a monster I haven't seen! But anyways, I forgot to mention that $uequiv 0$ on the boundary but I think you figured out that.
– bhcribeiro
Jul 31 at 17:34
Gustave, thanks for the link, but it seems that my question is exactly the opposite: I need positiveness of $u$ near the boundary knowing the condition over its normal derivative. And this is of course true near each point of the boundary but I need a kind of uniformity on this behaviour, as stated in the question.
– bhcribeiro
Jul 31 at 17:40
@bhcribeiro In all that talk about tubular neighborhood the essential thing is that the mapping $partialOmegatimes(-delta,delta)ni (x,s)mapsto x+snu(x)inmathbbR^N$ is, for $delta>0$ sufficiently small, a $C^1$ diffeomorphism of $partialOmegatimes(-delta,delta)$ onto its image (this follows by the inverse function theorem). In particular, every $yinOmega$ sufficiently close to $partialOmega$ can be uniquely written as $x+snu(x)$ for some $xinpartialOmega$ and $sin(-delta,0]$. If you have this, the remaining part is simple calculus.
– user539887
Jul 31 at 18:40
Yes, it is true (if $partialOmega$ is at least $C^1$), but thinking about its proof: do you know "some" differential topology? (I mean, tubular neighborhood?).
– user539887
Jul 30 at 20:03
Yes, it is true (if $partialOmega$ is at least $C^1$), but thinking about its proof: do you know "some" differential topology? (I mean, tubular neighborhood?).
– user539887
Jul 30 at 20:03
You will find some help math.stackexchange.com/questions/2839762/…
– Gustave
Jul 31 at 16:41
You will find some help math.stackexchange.com/questions/2839762/…
– Gustave
Jul 31 at 16:41
Yes, differential Topology is not a monster I haven't seen! But anyways, I forgot to mention that $uequiv 0$ on the boundary but I think you figured out that.
– bhcribeiro
Jul 31 at 17:34
Yes, differential Topology is not a monster I haven't seen! But anyways, I forgot to mention that $uequiv 0$ on the boundary but I think you figured out that.
– bhcribeiro
Jul 31 at 17:34
Gustave, thanks for the link, but it seems that my question is exactly the opposite: I need positiveness of $u$ near the boundary knowing the condition over its normal derivative. And this is of course true near each point of the boundary but I need a kind of uniformity on this behaviour, as stated in the question.
– bhcribeiro
Jul 31 at 17:40
Gustave, thanks for the link, but it seems that my question is exactly the opposite: I need positiveness of $u$ near the boundary knowing the condition over its normal derivative. And this is of course true near each point of the boundary but I need a kind of uniformity on this behaviour, as stated in the question.
– bhcribeiro
Jul 31 at 17:40
@bhcribeiro In all that talk about tubular neighborhood the essential thing is that the mapping $partialOmegatimes(-delta,delta)ni (x,s)mapsto x+snu(x)inmathbbR^N$ is, for $delta>0$ sufficiently small, a $C^1$ diffeomorphism of $partialOmegatimes(-delta,delta)$ onto its image (this follows by the inverse function theorem). In particular, every $yinOmega$ sufficiently close to $partialOmega$ can be uniquely written as $x+snu(x)$ for some $xinpartialOmega$ and $sin(-delta,0]$. If you have this, the remaining part is simple calculus.
– user539887
Jul 31 at 18:40
@bhcribeiro In all that talk about tubular neighborhood the essential thing is that the mapping $partialOmegatimes(-delta,delta)ni (x,s)mapsto x+snu(x)inmathbbR^N$ is, for $delta>0$ sufficiently small, a $C^1$ diffeomorphism of $partialOmegatimes(-delta,delta)$ onto its image (this follows by the inverse function theorem). In particular, every $yinOmega$ sufficiently close to $partialOmega$ can be uniquely written as $x+snu(x)$ for some $xinpartialOmega$ and $sin(-delta,0]$. If you have this, the remaining part is simple calculus.
– user539887
Jul 31 at 18:40
 |Â
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Yes, it is true (if $partialOmega$ is at least $C^1$), but thinking about its proof: do you know "some" differential topology? (I mean, tubular neighborhood?).
– user539887
Jul 30 at 20:03
You will find some help math.stackexchange.com/questions/2839762/…
– Gustave
Jul 31 at 16:41
Yes, differential Topology is not a monster I haven't seen! But anyways, I forgot to mention that $uequiv 0$ on the boundary but I think you figured out that.
– bhcribeiro
Jul 31 at 17:34
Gustave, thanks for the link, but it seems that my question is exactly the opposite: I need positiveness of $u$ near the boundary knowing the condition over its normal derivative. And this is of course true near each point of the boundary but I need a kind of uniformity on this behaviour, as stated in the question.
– bhcribeiro
Jul 31 at 17:40
@bhcribeiro In all that talk about tubular neighborhood the essential thing is that the mapping $partialOmegatimes(-delta,delta)ni (x,s)mapsto x+snu(x)inmathbbR^N$ is, for $delta>0$ sufficiently small, a $C^1$ diffeomorphism of $partialOmegatimes(-delta,delta)$ onto its image (this follows by the inverse function theorem). In particular, every $yinOmega$ sufficiently close to $partialOmega$ can be uniquely written as $x+snu(x)$ for some $xinpartialOmega$ and $sin(-delta,0]$. If you have this, the remaining part is simple calculus.
– user539887
Jul 31 at 18:40