Mixing
Does $nabla u$ have null tangential component to $mathrm{supp} u$?
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Let $u: mathbb{R}^N to mathbb{R}$ be a smooth function. Does $nabla u$ have null tangential component to $mathrm{supp}, u $ ? That is, do we have
$$frac{partial u}{partial nu} = 0 Rightarrow |nabla u| = 0$$
on $mathrm{supp} u$, where $nu $ denotes the outer normal to $mathrm{supp} u$?
calculus multivariable-calculus derivatives differential-geometry
asked Jun 20 '18 at 12:53
DalDal
1,50322470
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add a comment |
$begingroup$
Let $u: mathbb{R}^N to mathbb{R}$ be a smooth function. Does $nabla u$ have null tangential component to $mathrm{supp}, u $ ? That is, do we have
$$frac{partial u}{partial nu} = 0 Rightarrow |nabla u| = 0$$
on $mathrm{supp} u$, where $nu $ denotes the outer normal to $mathrm{supp} u$?
calculus multivariable-calculus derivatives differential-geometry
asked Jun 20 '18 at 12:53
DalDal
1,50322470
$endgroup$
add a comment |
$begingroup$
Let $u: mathbb{R}^N to mathbb{R}$ be a smooth function. Does $nabla u$ have null tangential component to $mathrm{supp}, u $ ? That is, do we have
$$frac{partial u}{partial nu} = 0 Rightarrow |nabla u| = 0$$
on $mathrm{supp} u$, where $nu $ denotes the outer normal to $mathrm{supp} u$?
calculus multivariable-calculus derivatives differential-geometry
asked Jun 20 '18 at 12:53
DalDal
1,50322470
$endgroup$
Let $u: mathbb{R}^N to mathbb{R}$ be a smooth function. Does $nabla u$ have null tangential component to $mathrm{supp}, u $ ? That is, do we have
$$frac{partial u}{partial nu} = 0 Rightarrow |nabla u| = 0$$
on $mathrm{supp} u$, where $nu $ denotes the outer normal to $mathrm{supp} u$?
calculus multivariable-calculus derivatives differential-geometry
calculus multivariable-calculus derivatives differential-geometry
asked Jun 20 '18 at 12:53
DalDal
1,50322470
asked Jun 20 '18 at 12:53
DalDal
1,50322470
edited Jan 13 at 1:01
Dal
asked Jun 20 '18 at 12:53
DalDal
1,50322470
asked Jun 20 '18 at 12:53
DalDal
1,50322470
asked Jun 20 '18 at 12:53
DalDal
1,50322470
1,50322470
add a comment |
add a comment |
1 Answer
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$begingroup$
$u$ is smooth on all of $Bbb R^n$, including on the boundary of $text{supp }u$. Outside of $text{supp }u$, we obviously have $nabla u = 0$. Therefore by continuity, we have $nabla u = 0$ on the boundary of $text{supp }u$ as well.
answered Jun 20 '18 at 23:53
Paul SinclairPaul Sinclair
19.7k21442
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1 Answer
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1 Answer
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$begingroup$
$u$ is smooth on all of $Bbb R^n$, including on the boundary of $text{supp }u$. Outside of $text{supp }u$, we obviously have $nabla u = 0$. Therefore by continuity, we have $nabla u = 0$ on the boundary of $text{supp }u$ as well.
answered Jun 20 '18 at 23:53
Paul SinclairPaul Sinclair
19.7k21442
$endgroup$
add a comment |
$begingroup$
$u$ is smooth on all of $Bbb R^n$, including on the boundary of $text{supp }u$. Outside of $text{supp }u$, we obviously have $nabla u = 0$. Therefore by continuity, we have $nabla u = 0$ on the boundary of $text{supp }u$ as well.
answered Jun 20 '18 at 23:53
Paul SinclairPaul Sinclair
19.7k21442
$endgroup$
add a comment |
$begingroup$
$u$ is smooth on all of $Bbb R^n$, including on the boundary of $text{supp }u$. Outside of $text{supp }u$, we obviously have $nabla u = 0$. Therefore by continuity, we have $nabla u = 0$ on the boundary of $text{supp }u$ as well.
answered Jun 20 '18 at 23:53
Paul SinclairPaul Sinclair
19.7k21442
$endgroup$
$u$ is smooth on all of $Bbb R^n$, including on the boundary of $text{supp }u$. Outside of $text{supp }u$, we obviously have $nabla u = 0$. Therefore by continuity, we have $nabla u = 0$ on the boundary of $text{supp }u$ as well.
answered Jun 20 '18 at 23:53
Paul SinclairPaul Sinclair
19.7k21442
answered Jun 20 '18 at 23:53
Paul SinclairPaul Sinclair
19.7k21442
answered Jun 20 '18 at 23:53
Paul SinclairPaul Sinclair
19.7k21442
answered Jun 20 '18 at 23:53
Paul SinclairPaul Sinclair
19.7k21442
19.7k21442
add a comment |
add a comment |
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