Mixing
The Coercivity of uniformly positive definite Matrix of Sobolev function
$begingroup$
For $u=(u^1,ldots, u^N)in W^{1,2}(Omega,R^N)$ where $Omega$ is bounded. We define
$$ E[u]=int_Omega g_{ij}(u)nabla u^inabla u^jdx$$
where $G=(g_{ij})_{1leq i,jleq N}$ is an given uniformly positive definite matrix, i.e. $xi^Tcdot G(u)cdot xigeq alpha|xi|^2>0$ no matter the value of $u$. Now my text book conclude that
$$ E[u]geq lambda|nabla u|_{L^2}^2 $$
without prove. May it is a simple fact but I can not work it out. Could somebody help me to write done the details? Thx!
matrices sobolev-spaces
asked Dec 10 '14 at 23:18
spatiallyspatially
2,43831123
$endgroup$
add a comment |
$begingroup$
For $u=(u^1,ldots, u^N)in W^{1,2}(Omega,R^N)$ where $Omega$ is bounded. We define
$$ E[u]=int_Omega g_{ij}(u)nabla u^inabla u^jdx$$
where $G=(g_{ij})_{1leq i,jleq N}$ is an given uniformly positive definite matrix, i.e. $xi^Tcdot G(u)cdot xigeq alpha|xi|^2>0$ no matter the value of $u$. Now my text book conclude that
$$ E[u]geq lambda|nabla u|_{L^2}^2 $$
without prove. May it is a simple fact but I can not work it out. Could somebody help me to write done the details? Thx!
matrices sobolev-spaces
asked Dec 10 '14 at 23:18
spatiallyspatially
2,43831123
$endgroup$
add a comment |
$begingroup$
For $u=(u^1,ldots, u^N)in W^{1,2}(Omega,R^N)$ where $Omega$ is bounded. We define
$$ E[u]=int_Omega g_{ij}(u)nabla u^inabla u^jdx$$
where $G=(g_{ij})_{1leq i,jleq N}$ is an given uniformly positive definite matrix, i.e. $xi^Tcdot G(u)cdot xigeq alpha|xi|^2>0$ no matter the value of $u$. Now my text book conclude that
$$ E[u]geq lambda|nabla u|_{L^2}^2 $$
without prove. May it is a simple fact but I can not work it out. Could somebody help me to write done the details? Thx!
matrices sobolev-spaces
asked Dec 10 '14 at 23:18
spatiallyspatially
2,43831123
$endgroup$
For $u=(u^1,ldots, u^N)in W^{1,2}(Omega,R^N)$ where $Omega$ is bounded. We define
$$ E[u]=int_Omega g_{ij}(u)nabla u^inabla u^jdx$$
where $G=(g_{ij})_{1leq i,jleq N}$ is an given uniformly positive definite matrix, i.e. $xi^Tcdot G(u)cdot xigeq alpha|xi|^2>0$ no matter the value of $u$. Now my text book conclude that
$$ E[u]geq lambda|nabla u|_{L^2}^2 $$
without prove. May it is a simple fact but I can not work it out. Could somebody help me to write done the details? Thx!
matrices sobolev-spaces
matrices sobolev-spaces
asked Dec 10 '14 at 23:18
spatiallyspatially
2,43831123
asked Dec 10 '14 at 23:18
spatiallyspatially
2,43831123
edited Dec 12 '14 at 1:48
spatially
asked Dec 10 '14 at 23:18
spatiallyspatially
2,43831123
asked Dec 10 '14 at 23:18
spatiallyspatially
2,43831123
asked Dec 10 '14 at 23:18
spatiallyspatially
2,43831123
2,43831123
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
I think that you forgot $|xi|^2$ in the definition the uniformly positive definite matrix. With this adjustment what you want to prove is straightforward.
answered Dec 11 '14 at 5:14
Neutral ElementNeutral Element
9351817
$endgroup$
$begingroup$
No I didn't... This is why I confused...
$endgroup$
– spatially
Dec 11 '14 at 19:51
$begingroup$
@wisher So is your definition correct? Take $xi=0$ see what happens.
$endgroup$
– Neutral Element
Dec 11 '14 at 20:57
$begingroup$
I see what your point. Sorry it is an typo. But what I want to ask is that, as $u^j$ $u^i$ are different functions, we can not just apply the definition of positive definite matrix.
$endgroup$
– spatially
Dec 12 '14 at 1:48
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
$begingroup$
I think that you forgot $|xi|^2$ in the definition the uniformly positive definite matrix. With this adjustment what you want to prove is straightforward.
answered Dec 11 '14 at 5:14
Neutral ElementNeutral Element
9351817
$endgroup$
$begingroup$
No I didn't... This is why I confused...
$endgroup$
– spatially
Dec 11 '14 at 19:51
$begingroup$
@wisher So is your definition correct? Take $xi=0$ see what happens.
$endgroup$
– Neutral Element
Dec 11 '14 at 20:57
$begingroup$
I see what your point. Sorry it is an typo. But what I want to ask is that, as $u^j$ $u^i$ are different functions, we can not just apply the definition of positive definite matrix.
$endgroup$
– spatially
Dec 12 '14 at 1:48
add a comment |
$begingroup$
I think that you forgot $|xi|^2$ in the definition the uniformly positive definite matrix. With this adjustment what you want to prove is straightforward.
answered Dec 11 '14 at 5:14
Neutral ElementNeutral Element
9351817
$endgroup$
$begingroup$
No I didn't... This is why I confused...
$endgroup$
– spatially
Dec 11 '14 at 19:51
$begingroup$
@wisher So is your definition correct? Take $xi=0$ see what happens.
$endgroup$
– Neutral Element
Dec 11 '14 at 20:57
$begingroup$
I see what your point. Sorry it is an typo. But what I want to ask is that, as $u^j$ $u^i$ are different functions, we can not just apply the definition of positive definite matrix.
$endgroup$
– spatially
Dec 12 '14 at 1:48
add a comment |
$begingroup$
I think that you forgot $|xi|^2$ in the definition the uniformly positive definite matrix. With this adjustment what you want to prove is straightforward.
answered Dec 11 '14 at 5:14
Neutral ElementNeutral Element
9351817
$endgroup$
I think that you forgot $|xi|^2$ in the definition the uniformly positive definite matrix. With this adjustment what you want to prove is straightforward.
answered Dec 11 '14 at 5:14
Neutral ElementNeutral Element
9351817
answered Dec 11 '14 at 5:14
Neutral ElementNeutral Element
9351817
answered Dec 11 '14 at 5:14
Neutral ElementNeutral Element
9351817
answered Dec 11 '14 at 5:14
Neutral ElementNeutral Element
9351817
9351817
$begingroup$
No I didn't... This is why I confused...
$endgroup$
– spatially
Dec 11 '14 at 19:51
$begingroup$
@wisher So is your definition correct? Take $xi=0$ see what happens.
$endgroup$
– Neutral Element
Dec 11 '14 at 20:57
$begingroup$
I see what your point. Sorry it is an typo. But what I want to ask is that, as $u^j$ $u^i$ are different functions, we can not just apply the definition of positive definite matrix.
$endgroup$
– spatially
Dec 12 '14 at 1:48
add a comment |
$begingroup$
No I didn't... This is why I confused...
$endgroup$
– spatially
Dec 11 '14 at 19:51
$begingroup$
@wisher So is your definition correct? Take $xi=0$ see what happens.
$endgroup$
– Neutral Element
Dec 11 '14 at 20:57
$begingroup$
I see what your point. Sorry it is an typo. But what I want to ask is that, as $u^j$ $u^i$ are different functions, we can not just apply the definition of positive definite matrix.
$endgroup$
– spatially
Dec 12 '14 at 1:48
$begingroup$
No I didn't... This is why I confused...
$endgroup$
– spatially
Dec 11 '14 at 19:51
$begingroup$
No I didn't... This is why I confused...
$endgroup$
– spatially
Dec 11 '14 at 19:51
$begingroup$
@wisher So is your definition correct? Take $xi=0$ see what happens.
$endgroup$
– Neutral Element
Dec 11 '14 at 20:57
$begingroup$
@wisher So is your definition correct? Take $xi=0$ see what happens.
$endgroup$
– Neutral Element
Dec 11 '14 at 20:57
$begingroup$
I see what your point. Sorry it is an typo. But what I want to ask is that, as $u^j$ $u^i$ are different functions, we can not just apply the definition of positive definite matrix.
$endgroup$
– spatially
Dec 12 '14 at 1:48
$begingroup$
I see what your point. Sorry it is an typo. But what I want to ask is that, as $u^j$ $u^i$ are different functions, we can not just apply the definition of positive definite matrix.
$endgroup$
– spatially
Dec 12 '14 at 1:48
add a comment |
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