Mixing
Variational problem of Laplace equation
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1
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We know that:
$Delta u=-f, inOmega$
$u=0$ ,on $partialOmega$
Is equivalent to this variational problem:
find the minimum of $J(u)$ in $C^2(Omega)cap C^1(barOmega)$, where $J(u)=int_Omega[frac{1}{2}(u_x^2+u_y^2)-uf]dxdy$.
My question is:
why we need u belongs to $C^2(Omega)cap C^1(barOmega)$? specifically, why we need $u,Du$ continuos up to the boundary of $Omega$? Why can't we just use $uin C^2(Omega)cap C^1(Omega)$?
functional-analysis pde sobolev-spaces
edited yesterday
Mefitico
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chloe hj
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up vote
1
down vote
favorite
We know that:
$Delta u=-f, inOmega$
$u=0$ ,on $partialOmega$
Is equivalent to this variational problem:
find the minimum of $J(u)$ in $C^2(Omega)cap C^1(barOmega)$, where $J(u)=int_Omega[frac{1}{2}(u_x^2+u_y^2)-uf]dxdy$.
My question is:
why we need u belongs to $C^2(Omega)cap C^1(barOmega)$? specifically, why we need $u,Du$ continuos up to the boundary of $Omega$? Why can't we just use $uin C^2(Omega)cap C^1(Omega)$?
functional-analysis pde sobolev-spaces
edited yesterday
Mefitico
829117
asked yesterday
chloe hj
535
New contributor
chloe hj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
We know that:
$Delta u=-f, inOmega$
$u=0$ ,on $partialOmega$
Is equivalent to this variational problem:
find the minimum of $J(u)$ in $C^2(Omega)cap C^1(barOmega)$, where $J(u)=int_Omega[frac{1}{2}(u_x^2+u_y^2)-uf]dxdy$.
My question is:
why we need u belongs to $C^2(Omega)cap C^1(barOmega)$? specifically, why we need $u,Du$ continuos up to the boundary of $Omega$? Why can't we just use $uin C^2(Omega)cap C^1(Omega)$?
functional-analysis pde sobolev-spaces
edited yesterday
Mefitico
829117
asked yesterday
chloe hj
535
New contributor
chloe hj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
We know that:
$Delta u=-f, inOmega$
$u=0$ ,on $partialOmega$
Is equivalent to this variational problem:
find the minimum of $J(u)$ in $C^2(Omega)cap C^1(barOmega)$, where $J(u)=int_Omega[frac{1}{2}(u_x^2+u_y^2)-uf]dxdy$.
My question is:
why we need u belongs to $C^2(Omega)cap C^1(barOmega)$? specifically, why we need $u,Du$ continuos up to the boundary of $Omega$? Why can't we just use $uin C^2(Omega)cap C^1(Omega)$?
functional-analysis pde sobolev-spaces
functional-analysis pde sobolev-spaces
edited yesterday
Mefitico
829117
asked yesterday
chloe hj
535
New contributor
chloe hj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Mefitico
829117
asked yesterday
chloe hj
535
New contributor
chloe hj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Mefitico
829117
edited yesterday
Mefitico
829117
edited yesterday
Mefitico
829117
829117
asked yesterday
chloe hj
535
New contributor
chloe hj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
chloe hj
535
asked yesterday
chloe hj
535
535
New contributor
chloe hj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
chloe hj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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1 Answer
1
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2
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Typically the requirement is $uin C^2(Omega)cap C^0(bar{Omega})$ if you want a classical solution; you need this because you have boundary conditions and you want u to be continuous on the boundary (otherwise you could just find a solution without BC in the interior and define it as $0$ on the boundary). The (stronger) requirement $uin C^2(Omega)cap C^1(bar{Omega})$ is a bit weird actually, I've never heard about it, but it guarantees continuity at $partialOmega$ . However it may be useless.
answered 22 hours ago
Marco
1908
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Typically the requirement is $uin C^2(Omega)cap C^0(bar{Omega})$ if you want a classical solution; you need this because you have boundary conditions and you want u to be continuous on the boundary (otherwise you could just find a solution without BC in the interior and define it as $0$ on the boundary). The (stronger) requirement $uin C^2(Omega)cap C^1(bar{Omega})$ is a bit weird actually, I've never heard about it, but it guarantees continuity at $partialOmega$ . However it may be useless.
answered 22 hours ago
Marco
1908
add a comment |
up vote
2
down vote
Typically the requirement is $uin C^2(Omega)cap C^0(bar{Omega})$ if you want a classical solution; you need this because you have boundary conditions and you want u to be continuous on the boundary (otherwise you could just find a solution without BC in the interior and define it as $0$ on the boundary). The (stronger) requirement $uin C^2(Omega)cap C^1(bar{Omega})$ is a bit weird actually, I've never heard about it, but it guarantees continuity at $partialOmega$ . However it may be useless.
answered 22 hours ago
Marco
1908
add a comment |
up vote
2
down vote
up vote
2
down vote
Typically the requirement is $uin C^2(Omega)cap C^0(bar{Omega})$ if you want a classical solution; you need this because you have boundary conditions and you want u to be continuous on the boundary (otherwise you could just find a solution without BC in the interior and define it as $0$ on the boundary). The (stronger) requirement $uin C^2(Omega)cap C^1(bar{Omega})$ is a bit weird actually, I've never heard about it, but it guarantees continuity at $partialOmega$ . However it may be useless.
answered 22 hours ago
Marco
1908
Typically the requirement is $uin C^2(Omega)cap C^0(bar{Omega})$ if you want a classical solution; you need this because you have boundary conditions and you want u to be continuous on the boundary (otherwise you could just find a solution without BC in the interior and define it as $0$ on the boundary). The (stronger) requirement $uin C^2(Omega)cap C^1(bar{Omega})$ is a bit weird actually, I've never heard about it, but it guarantees continuity at $partialOmega$ . However it may be useless.
answered 22 hours ago
Marco
1908
answered 22 hours ago
Marco
1908
answered 22 hours ago
Marco
1908
answered 22 hours ago
Marco
1908
1908
add a comment |
add a comment |
chloe hj is a new contributor. Be nice, and check out our Code of Conduct.
chloe hj is a new contributor. Be nice, and check out our Code of Conduct.
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