Mixing
Compute the value of the Sobolev norm in $H^{-1/2}$
$begingroup$
I am working with finite elements using domain decomposition in 2D and one of the solutions I need to obtain is the co-normal derivative of the solution along a segment that is the intersection of two subdomains.
From trace theorems and Sobolev spaces, it is known that this derivative is in $H^{-1/2}$, and therefore convergence is also proven in that space.
My question is: once I have obtained my discrete FEM solution for this derivative, how can compute its $H^{-1/2}$ norm?
Since it is defined as the dual space of $H^{1/2}$, I have not found any way of effectively computing the $H^{-1/2}$ norm of a function.
Someone mentioned to me that the use of wavelets could be a possible way but I haven't found any resource backing up that claim.
I appreciate the help, thanks in advance.
sobolev-spaces norm
asked Mar 24 '15 at 15:47
MBenedettoMBenedetto
113
$endgroup$
add a comment |
$begingroup$
I am working with finite elements using domain decomposition in 2D and one of the solutions I need to obtain is the co-normal derivative of the solution along a segment that is the intersection of two subdomains.
From trace theorems and Sobolev spaces, it is known that this derivative is in $H^{-1/2}$, and therefore convergence is also proven in that space.
My question is: once I have obtained my discrete FEM solution for this derivative, how can compute its $H^{-1/2}$ norm?
Since it is defined as the dual space of $H^{1/2}$, I have not found any way of effectively computing the $H^{-1/2}$ norm of a function.
Someone mentioned to me that the use of wavelets could be a possible way but I haven't found any resource backing up that claim.
I appreciate the help, thanks in advance.
sobolev-spaces norm
asked Mar 24 '15 at 15:47
MBenedettoMBenedetto
113
$endgroup$
$begingroup$
The negative norm is non-computable, it is taking supreme over an infinite dimensional space. You can only approximate it using a mesh-dependent norm.
$endgroup$
– Shuhao Cao
Mar 24 '15 at 15:50
$begingroup$
First: are you sure that the trace theorem you are using is optimal? I know that the "basic" trace theorem loses a full derivative while the "sharp" trace theorem for the $H^k$ spaces actually only loses half a derivative. Second, all you can really do is use the definition: try a sequence of test functions of unit norm and conclude that the dual norm is at least the largest value you get from the test functions.
$endgroup$
– Ian
Mar 24 '15 at 15:51
$begingroup$
ShuhaoCao: Thank you for the suggestion. I have not used mesh dependent norms before. Ian: For the case we are considering, we are sure about the trace theorem. We have thought about trying with a sequence of functions, but with this approach we are always understimating the norm and we would like at least to overestimate it.
$endgroup$
– MBenedetto
Mar 24 '15 at 16:01
add a comment |
$begingroup$
I am working with finite elements using domain decomposition in 2D and one of the solutions I need to obtain is the co-normal derivative of the solution along a segment that is the intersection of two subdomains.
From trace theorems and Sobolev spaces, it is known that this derivative is in $H^{-1/2}$, and therefore convergence is also proven in that space.
My question is: once I have obtained my discrete FEM solution for this derivative, how can compute its $H^{-1/2}$ norm?
Since it is defined as the dual space of $H^{1/2}$, I have not found any way of effectively computing the $H^{-1/2}$ norm of a function.
Someone mentioned to me that the use of wavelets could be a possible way but I haven't found any resource backing up that claim.
I appreciate the help, thanks in advance.
sobolev-spaces norm
asked Mar 24 '15 at 15:47
MBenedettoMBenedetto
113
$endgroup$
I am working with finite elements using domain decomposition in 2D and one of the solutions I need to obtain is the co-normal derivative of the solution along a segment that is the intersection of two subdomains.
From trace theorems and Sobolev spaces, it is known that this derivative is in $H^{-1/2}$, and therefore convergence is also proven in that space.
My question is: once I have obtained my discrete FEM solution for this derivative, how can compute its $H^{-1/2}$ norm?
Since it is defined as the dual space of $H^{1/2}$, I have not found any way of effectively computing the $H^{-1/2}$ norm of a function.
Someone mentioned to me that the use of wavelets could be a possible way but I haven't found any resource backing up that claim.
I appreciate the help, thanks in advance.
sobolev-spaces norm
sobolev-spaces norm
asked Mar 24 '15 at 15:47
MBenedettoMBenedetto
113
asked Mar 24 '15 at 15:47
MBenedettoMBenedetto
113
edited Mar 24 '15 at 15:50
MBenedetto
asked Mar 24 '15 at 15:47
MBenedettoMBenedetto
113
asked Mar 24 '15 at 15:47
MBenedettoMBenedetto
113
asked Mar 24 '15 at 15:47
MBenedettoMBenedetto
113
113
$begingroup$
The negative norm is non-computable, it is taking supreme over an infinite dimensional space. You can only approximate it using a mesh-dependent norm.
$endgroup$
– Shuhao Cao
Mar 24 '15 at 15:50
$begingroup$
First: are you sure that the trace theorem you are using is optimal? I know that the "basic" trace theorem loses a full derivative while the "sharp" trace theorem for the $H^k$ spaces actually only loses half a derivative. Second, all you can really do is use the definition: try a sequence of test functions of unit norm and conclude that the dual norm is at least the largest value you get from the test functions.
$endgroup$
– Ian
Mar 24 '15 at 15:51
$begingroup$
ShuhaoCao: Thank you for the suggestion. I have not used mesh dependent norms before. Ian: For the case we are considering, we are sure about the trace theorem. We have thought about trying with a sequence of functions, but with this approach we are always understimating the norm and we would like at least to overestimate it.
$endgroup$
– MBenedetto
Mar 24 '15 at 16:01
add a comment |
$begingroup$
The negative norm is non-computable, it is taking supreme over an infinite dimensional space. You can only approximate it using a mesh-dependent norm.
$endgroup$
– Shuhao Cao
Mar 24 '15 at 15:50
$begingroup$
First: are you sure that the trace theorem you are using is optimal? I know that the "basic" trace theorem loses a full derivative while the "sharp" trace theorem for the $H^k$ spaces actually only loses half a derivative. Second, all you can really do is use the definition: try a sequence of test functions of unit norm and conclude that the dual norm is at least the largest value you get from the test functions.
$endgroup$
– Ian
Mar 24 '15 at 15:51
$begingroup$
ShuhaoCao: Thank you for the suggestion. I have not used mesh dependent norms before. Ian: For the case we are considering, we are sure about the trace theorem. We have thought about trying with a sequence of functions, but with this approach we are always understimating the norm and we would like at least to overestimate it.
$endgroup$
– MBenedetto
Mar 24 '15 at 16:01
$begingroup$
The negative norm is non-computable, it is taking supreme over an infinite dimensional space. You can only approximate it using a mesh-dependent norm.
$endgroup$
– Shuhao Cao
Mar 24 '15 at 15:50
$begingroup$
The negative norm is non-computable, it is taking supreme over an infinite dimensional space. You can only approximate it using a mesh-dependent norm.
$endgroup$
– Shuhao Cao
Mar 24 '15 at 15:50
$begingroup$
First: are you sure that the trace theorem you are using is optimal? I know that the "basic" trace theorem loses a full derivative while the "sharp" trace theorem for the $H^k$ spaces actually only loses half a derivative. Second, all you can really do is use the definition: try a sequence of test functions of unit norm and conclude that the dual norm is at least the largest value you get from the test functions.
$endgroup$
– Ian
Mar 24 '15 at 15:51
$begingroup$
First: are you sure that the trace theorem you are using is optimal? I know that the "basic" trace theorem loses a full derivative while the "sharp" trace theorem for the $H^k$ spaces actually only loses half a derivative. Second, all you can really do is use the definition: try a sequence of test functions of unit norm and conclude that the dual norm is at least the largest value you get from the test functions.
$endgroup$
– Ian
Mar 24 '15 at 15:51
$begingroup$
ShuhaoCao: Thank you for the suggestion. I have not used mesh dependent norms before. Ian: For the case we are considering, we are sure about the trace theorem. We have thought about trying with a sequence of functions, but with this approach we are always understimating the norm and we would like at least to overestimate it.
$endgroup$
– MBenedetto
Mar 24 '15 at 16:01
$begingroup$
ShuhaoCao: Thank you for the suggestion. I have not used mesh dependent norms before. Ian: For the case we are considering, we are sure about the trace theorem. We have thought about trying with a sequence of functions, but with this approach we are always understimating the norm and we would like at least to overestimate it.
$endgroup$
– MBenedetto
Mar 24 '15 at 16:01
add a comment |
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$begingroup$
The negative norm is non-computable, it is taking supreme over an infinite dimensional space. You can only approximate it using a mesh-dependent norm.
$endgroup$
– Shuhao Cao
Mar 24 '15 at 15:50
$begingroup$
First: are you sure that the trace theorem you are using is optimal? I know that the "basic" trace theorem loses a full derivative while the "sharp" trace theorem for the $H^k$ spaces actually only loses half a derivative. Second, all you can really do is use the definition: try a sequence of test functions of unit norm and conclude that the dual norm is at least the largest value you get from the test functions.
$endgroup$
– Ian
Mar 24 '15 at 15:51
$begingroup$
ShuhaoCao: Thank you for the suggestion. I have not used mesh dependent norms before. Ian: For the case we are considering, we are sure about the trace theorem. We have thought about trying with a sequence of functions, but with this approach we are always understimating the norm and we would like at least to overestimate it.
$endgroup$
– MBenedetto
Mar 24 '15 at 16:01