Bound gradient in $H^2_0(Omega)$ by Laplacian
$begingroup$
Let $Omegasubseteq mathbb{R}^n$ be an open set and show that
$$
lVert{nabla urVert}_{L^2}^2 leq epsilonlVert{Delta urVert}_{L_2}^2 + frac{1}{4epsilon}lVert{urVert}_{L^2}^2
$$
for any $epsilon>0$ and $uin H_0^2(Omega)$.
It would suffice to porve the inequality for smooth compactly supported functions since these are dense in $H_0^2(Omega)$. Other than the fact that I will probably need Green's identity, I am not sure how to proceed. Where does the $epsilon$ come into play?
Thanks!
real-analysis pde sobolev-spaces
asked Dec 6 '18 at 5:42
QuokaQuoka
1,240312
$endgroup$
add a comment |
$begingroup$
Let $Omegasubseteq mathbb{R}^n$ be an open set and show that
$$
lVert{nabla urVert}_{L^2}^2 leq epsilonlVert{Delta urVert}_{L_2}^2 + frac{1}{4epsilon}lVert{urVert}_{L^2}^2
$$
for any $epsilon>0$ and $uin H_0^2(Omega)$.
It would suffice to porve the inequality for smooth compactly supported functions since these are dense in $H_0^2(Omega)$. Other than the fact that I will probably need Green's identity, I am not sure how to proceed. Where does the $epsilon$ come into play?
Thanks!
real-analysis pde sobolev-spaces
asked Dec 6 '18 at 5:42
QuokaQuoka
1,240312
$endgroup$
add a comment |
$begingroup$
Let $Omegasubseteq mathbb{R}^n$ be an open set and show that
$$
lVert{nabla urVert}_{L^2}^2 leq epsilonlVert{Delta urVert}_{L_2}^2 + frac{1}{4epsilon}lVert{urVert}_{L^2}^2
$$
for any $epsilon>0$ and $uin H_0^2(Omega)$.
It would suffice to porve the inequality for smooth compactly supported functions since these are dense in $H_0^2(Omega)$. Other than the fact that I will probably need Green's identity, I am not sure how to proceed. Where does the $epsilon$ come into play?
Thanks!
real-analysis pde sobolev-spaces
asked Dec 6 '18 at 5:42
QuokaQuoka
1,240312
$endgroup$
Let $Omegasubseteq mathbb{R}^n$ be an open set and show that
$$
lVert{nabla urVert}_{L^2}^2 leq epsilonlVert{Delta urVert}_{L_2}^2 + frac{1}{4epsilon}lVert{urVert}_{L^2}^2
$$
for any $epsilon>0$ and $uin H_0^2(Omega)$.
It would suffice to porve the inequality for smooth compactly supported functions since these are dense in $H_0^2(Omega)$. Other than the fact that I will probably need Green's identity, I am not sure how to proceed. Where does the $epsilon$ come into play?
Thanks!
real-analysis pde sobolev-spaces
real-analysis pde sobolev-spaces
asked Dec 6 '18 at 5:42
QuokaQuoka
1,240312
asked Dec 6 '18 at 5:42
QuokaQuoka
1,240312
asked Dec 6 '18 at 5:42
QuokaQuoka
1,240312
asked Dec 6 '18 at 5:42
QuokaQuoka
1,240312
asked Dec 6 '18 at 5:42
QuokaQuoka
1,240312
1,240312
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
By Green's identity and the Cachy-Schwarz inequality
$$
int_Omega|nabla u|^2=-int_Omega u,Delta u le int_Omega |u,Delta u|leBigl(int_Omega|u|^2Bigr)^{1/2}Bigl(int_Omega|Delta u|^2Bigr)^{1/2},
$$
that is,
$$
|nabla u|_{L^2}^2le|u|_{L^2}|Delta u|_{L^2}.
$$
From the inequality
$$
a,blefrac12(a^2+b^2),
$$
we get for any $epsilon>0$
$$
a,b=(sqrt{2,epsilon},a)Bigl(frac{1}{sqrt{2,epsilon}},bBigr)leepsilon,a^2+frac{1}{4,epsilon},b^2.
$$
edited Dec 12 '18 at 14:37
ares
19710
answered Dec 6 '18 at 8:53
Julián AguirreJulián Aguirre
68.1k24094
$endgroup$
$begingroup$
Thank you very much! I also wanted to ask: If $Omega$ is just an open interval in $mathbb{R}$, then can we also obtain this inequality (or a very similar one) for all $uin H^2(Omega)$?
$endgroup$
– Quoka
Dec 7 '18 at 0:06
$begingroup$
In that case there are boundary terms cropping around.
$endgroup$
– Julián Aguirre
Dec 10 '18 at 12:28
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
By Green's identity and the Cachy-Schwarz inequality
$$
int_Omega|nabla u|^2=-int_Omega u,Delta u le int_Omega |u,Delta u|leBigl(int_Omega|u|^2Bigr)^{1/2}Bigl(int_Omega|Delta u|^2Bigr)^{1/2},
$$
that is,
$$
|nabla u|_{L^2}^2le|u|_{L^2}|Delta u|_{L^2}.
$$
From the inequality
$$
a,blefrac12(a^2+b^2),
$$
we get for any $epsilon>0$
$$
a,b=(sqrt{2,epsilon},a)Bigl(frac{1}{sqrt{2,epsilon}},bBigr)leepsilon,a^2+frac{1}{4,epsilon},b^2.
$$
edited Dec 12 '18 at 14:37
ares
19710
answered Dec 6 '18 at 8:53
Julián AguirreJulián Aguirre
68.1k24094
$endgroup$
$begingroup$
Thank you very much! I also wanted to ask: If $Omega$ is just an open interval in $mathbb{R}$, then can we also obtain this inequality (or a very similar one) for all $uin H^2(Omega)$?
$endgroup$
– Quoka
Dec 7 '18 at 0:06
$begingroup$
In that case there are boundary terms cropping around.
$endgroup$
– Julián Aguirre
Dec 10 '18 at 12:28
add a comment |
$begingroup$
By Green's identity and the Cachy-Schwarz inequality
$$
int_Omega|nabla u|^2=-int_Omega u,Delta u le int_Omega |u,Delta u|leBigl(int_Omega|u|^2Bigr)^{1/2}Bigl(int_Omega|Delta u|^2Bigr)^{1/2},
$$
that is,
$$
|nabla u|_{L^2}^2le|u|_{L^2}|Delta u|_{L^2}.
$$
From the inequality
$$
a,blefrac12(a^2+b^2),
$$
we get for any $epsilon>0$
$$
a,b=(sqrt{2,epsilon},a)Bigl(frac{1}{sqrt{2,epsilon}},bBigr)leepsilon,a^2+frac{1}{4,epsilon},b^2.
$$
edited Dec 12 '18 at 14:37
ares
19710
answered Dec 6 '18 at 8:53
Julián AguirreJulián Aguirre
68.1k24094
$endgroup$
$begingroup$
Thank you very much! I also wanted to ask: If $Omega$ is just an open interval in $mathbb{R}$, then can we also obtain this inequality (or a very similar one) for all $uin H^2(Omega)$?
$endgroup$
– Quoka
Dec 7 '18 at 0:06
$begingroup$
In that case there are boundary terms cropping around.
$endgroup$
– Julián Aguirre
Dec 10 '18 at 12:28
add a comment |
$begingroup$
By Green's identity and the Cachy-Schwarz inequality
$$
int_Omega|nabla u|^2=-int_Omega u,Delta u le int_Omega |u,Delta u|leBigl(int_Omega|u|^2Bigr)^{1/2}Bigl(int_Omega|Delta u|^2Bigr)^{1/2},
$$
that is,
$$
|nabla u|_{L^2}^2le|u|_{L^2}|Delta u|_{L^2}.
$$
From the inequality
$$
a,blefrac12(a^2+b^2),
$$
we get for any $epsilon>0$
$$
a,b=(sqrt{2,epsilon},a)Bigl(frac{1}{sqrt{2,epsilon}},bBigr)leepsilon,a^2+frac{1}{4,epsilon},b^2.
$$
edited Dec 12 '18 at 14:37
ares
19710
answered Dec 6 '18 at 8:53
Julián AguirreJulián Aguirre
68.1k24094
$endgroup$
By Green's identity and the Cachy-Schwarz inequality
$$
int_Omega|nabla u|^2=-int_Omega u,Delta u le int_Omega |u,Delta u|leBigl(int_Omega|u|^2Bigr)^{1/2}Bigl(int_Omega|Delta u|^2Bigr)^{1/2},
$$
that is,
$$
|nabla u|_{L^2}^2le|u|_{L^2}|Delta u|_{L^2}.
$$
From the inequality
$$
a,blefrac12(a^2+b^2),
$$
we get for any $epsilon>0$
$$
a,b=(sqrt{2,epsilon},a)Bigl(frac{1}{sqrt{2,epsilon}},bBigr)leepsilon,a^2+frac{1}{4,epsilon},b^2.
$$
edited Dec 12 '18 at 14:37
ares
19710
answered Dec 6 '18 at 8:53
Julián AguirreJulián Aguirre
68.1k24094
edited Dec 12 '18 at 14:37
ares
19710
edited Dec 12 '18 at 14:37
ares
19710
edited Dec 12 '18 at 14:37
ares
19710
19710
answered Dec 6 '18 at 8:53
Julián AguirreJulián Aguirre
68.1k24094
answered Dec 6 '18 at 8:53
Julián AguirreJulián Aguirre
68.1k24094
answered Dec 6 '18 at 8:53
Julián AguirreJulián Aguirre
68.1k24094
68.1k24094
$begingroup$
Thank you very much! I also wanted to ask: If $Omega$ is just an open interval in $mathbb{R}$, then can we also obtain this inequality (or a very similar one) for all $uin H^2(Omega)$?
$endgroup$
– Quoka
Dec 7 '18 at 0:06
$begingroup$
In that case there are boundary terms cropping around.
$endgroup$
– Julián Aguirre
Dec 10 '18 at 12:28
add a comment |
$begingroup$
Thank you very much! I also wanted to ask: If $Omega$ is just an open interval in $mathbb{R}$, then can we also obtain this inequality (or a very similar one) for all $uin H^2(Omega)$?
$endgroup$
– Quoka
Dec 7 '18 at 0:06
$begingroup$
In that case there are boundary terms cropping around.
$endgroup$
– Julián Aguirre
Dec 10 '18 at 12:28
$begingroup$
Thank you very much! I also wanted to ask: If $Omega$ is just an open interval in $mathbb{R}$, then can we also obtain this inequality (or a very similar one) for all $uin H^2(Omega)$?
$endgroup$
– Quoka
Dec 7 '18 at 0:06
$begingroup$
Thank you very much! I also wanted to ask: If $Omega$ is just an open interval in $mathbb{R}$, then can we also obtain this inequality (or a very similar one) for all $uin H^2(Omega)$?
$endgroup$
– Quoka
Dec 7 '18 at 0:06
$begingroup$
In that case there are boundary terms cropping around.
$endgroup$
– Julián Aguirre
Dec 10 '18 at 12:28
$begingroup$
In that case there are boundary terms cropping around.
$endgroup$
– Julián Aguirre
Dec 10 '18 at 12:28
add a comment |
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