Cut-off functions in Caccioppoli's inequality
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Caccioppoli's inequality states that the solution $u$ of the equation $-nablacdot(Anabla u)=0$ in some bounded domain $Omega$ satisfies
$$int_{B(0,rho)}|nabla u|^2dyleq frac{C}{(R-rho)^2}int_{B(0,R)}|u|^2~dy,$$ for $0<rho<R$, $2R$ should be smaller than diameter of $Omega$, etc. In the proof of Caccioppoli's inequality, a cut-off function $phi$ is constructed on $B(0,R)$ satisfying $|nabla phi|<frac{C}{R-rho}$.
Is it possible to construct a cut-off function satisfying $|nabla phi|<frac{C}{(R-rho)^2}$ or with higher powers of $R-rho$? What is the limit on this?
regularity-theory-of-pdes
asked Nov 11 '18 at 0:01
Tanuj DipshikhaTanuj Dipshikha
197210
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Caccioppoli's inequality states that the solution $u$ of the equation $-nablacdot(Anabla u)=0$ in some bounded domain $Omega$ satisfies
$$int_{B(0,rho)}|nabla u|^2dyleq frac{C}{(R-rho)^2}int_{B(0,R)}|u|^2~dy,$$ for $0<rho<R$, $2R$ should be smaller than diameter of $Omega$, etc. In the proof of Caccioppoli's inequality, a cut-off function $phi$ is constructed on $B(0,R)$ satisfying $|nabla phi|<frac{C}{R-rho}$.
Is it possible to construct a cut-off function satisfying $|nabla phi|<frac{C}{(R-rho)^2}$ or with higher powers of $R-rho$? What is the limit on this?
regularity-theory-of-pdes
asked Nov 11 '18 at 0:01
Tanuj DipshikhaTanuj Dipshikha
197210
$endgroup$
add a comment |
$begingroup$
Caccioppoli's inequality states that the solution $u$ of the equation $-nablacdot(Anabla u)=0$ in some bounded domain $Omega$ satisfies
$$int_{B(0,rho)}|nabla u|^2dyleq frac{C}{(R-rho)^2}int_{B(0,R)}|u|^2~dy,$$ for $0<rho<R$, $2R$ should be smaller than diameter of $Omega$, etc. In the proof of Caccioppoli's inequality, a cut-off function $phi$ is constructed on $B(0,R)$ satisfying $|nabla phi|<frac{C}{R-rho}$.
Is it possible to construct a cut-off function satisfying $|nabla phi|<frac{C}{(R-rho)^2}$ or with higher powers of $R-rho$? What is the limit on this?
regularity-theory-of-pdes
asked Nov 11 '18 at 0:01
Tanuj DipshikhaTanuj Dipshikha
197210
$endgroup$
Caccioppoli's inequality states that the solution $u$ of the equation $-nablacdot(Anabla u)=0$ in some bounded domain $Omega$ satisfies
$$int_{B(0,rho)}|nabla u|^2dyleq frac{C}{(R-rho)^2}int_{B(0,R)}|u|^2~dy,$$ for $0<rho<R$, $2R$ should be smaller than diameter of $Omega$, etc. In the proof of Caccioppoli's inequality, a cut-off function $phi$ is constructed on $B(0,R)$ satisfying $|nabla phi|<frac{C}{R-rho}$.
Is it possible to construct a cut-off function satisfying $|nabla phi|<frac{C}{(R-rho)^2}$ or with higher powers of $R-rho$? What is the limit on this?
regularity-theory-of-pdes
regularity-theory-of-pdes
asked Nov 11 '18 at 0:01
Tanuj DipshikhaTanuj Dipshikha
197210
asked Nov 11 '18 at 0:01
Tanuj DipshikhaTanuj Dipshikha
197210
asked Nov 11 '18 at 0:01
Tanuj DipshikhaTanuj Dipshikha
197210
asked Nov 11 '18 at 0:01
Tanuj DipshikhaTanuj Dipshikha
197210
asked Nov 11 '18 at 0:01
Tanuj DipshikhaTanuj Dipshikha
197210
197210
add a comment |
add a comment |
1 Answer
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In general no, you can't expect to improve the exponent. Given such a $phi,$ we have $phi(rho e_1) = 1$ and $phi(R e_1) = 0,$ where $e_1 = (1,0,dots,0)$ is the first of the standard basis vectors. Thus applying the mean value theorem to $t mapsto phi(te_1)$ we obtain $xi in (rho, R)$ such that,
$$ frac{partial phi}{partial x_1}(xi e_1) = frac1{R-rho}. $$
In particular, we see that,
$$ sup_{B_R} |nabla phi| geq frac1{R-rho}. $$
answered Dec 22 '18 at 11:54
ktoiktoi
2,4061618
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1 Answer
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1 Answer
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$begingroup$
In general no, you can't expect to improve the exponent. Given such a $phi,$ we have $phi(rho e_1) = 1$ and $phi(R e_1) = 0,$ where $e_1 = (1,0,dots,0)$ is the first of the standard basis vectors. Thus applying the mean value theorem to $t mapsto phi(te_1)$ we obtain $xi in (rho, R)$ such that,
$$ frac{partial phi}{partial x_1}(xi e_1) = frac1{R-rho}. $$
In particular, we see that,
$$ sup_{B_R} |nabla phi| geq frac1{R-rho}. $$
answered Dec 22 '18 at 11:54
ktoiktoi
2,4061618
$endgroup$
add a comment |
$begingroup$
In general no, you can't expect to improve the exponent. Given such a $phi,$ we have $phi(rho e_1) = 1$ and $phi(R e_1) = 0,$ where $e_1 = (1,0,dots,0)$ is the first of the standard basis vectors. Thus applying the mean value theorem to $t mapsto phi(te_1)$ we obtain $xi in (rho, R)$ such that,
$$ frac{partial phi}{partial x_1}(xi e_1) = frac1{R-rho}. $$
In particular, we see that,
$$ sup_{B_R} |nabla phi| geq frac1{R-rho}. $$
answered Dec 22 '18 at 11:54
ktoiktoi
2,4061618
$endgroup$
add a comment |
$begingroup$
In general no, you can't expect to improve the exponent. Given such a $phi,$ we have $phi(rho e_1) = 1$ and $phi(R e_1) = 0,$ where $e_1 = (1,0,dots,0)$ is the first of the standard basis vectors. Thus applying the mean value theorem to $t mapsto phi(te_1)$ we obtain $xi in (rho, R)$ such that,
$$ frac{partial phi}{partial x_1}(xi e_1) = frac1{R-rho}. $$
In particular, we see that,
$$ sup_{B_R} |nabla phi| geq frac1{R-rho}. $$
answered Dec 22 '18 at 11:54
ktoiktoi
2,4061618
$endgroup$
In general no, you can't expect to improve the exponent. Given such a $phi,$ we have $phi(rho e_1) = 1$ and $phi(R e_1) = 0,$ where $e_1 = (1,0,dots,0)$ is the first of the standard basis vectors. Thus applying the mean value theorem to $t mapsto phi(te_1)$ we obtain $xi in (rho, R)$ such that,
$$ frac{partial phi}{partial x_1}(xi e_1) = frac1{R-rho}. $$
In particular, we see that,
$$ sup_{B_R} |nabla phi| geq frac1{R-rho}. $$
answered Dec 22 '18 at 11:54
ktoiktoi
2,4061618
answered Dec 22 '18 at 11:54
ktoiktoi
2,4061618
answered Dec 22 '18 at 11:54
ktoiktoi
2,4061618
answered Dec 22 '18 at 11:54
ktoiktoi
2,4061618
2,4061618
add a comment |
add a comment |
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