Mixing
Divergence theorem for inner product of Tensor
$begingroup$
Let $Omegasubset mathbb{R}^3$ be bounded domain and $f$ be smooth vector field on $Omega$ and $f|_{partialOmega=0}$. I not able to compute $int _{Omega} grad~f:(grad~f)^{T}$, where $:$ is inner product define by $A:B=Tr(AB^{T})$. I would like to use Divergence theorem, but I do not know hoe to do it.
real-analysis vector-analysis trace divergence
asked Jan 2 at 18:26
lojdmojlojdmoj
877
$endgroup$
add a comment |
$begingroup$
Let $Omegasubset mathbb{R}^3$ be bounded domain and $f$ be smooth vector field on $Omega$ and $f|_{partialOmega=0}$. I not able to compute $int _{Omega} grad~f:(grad~f)^{T}$, where $:$ is inner product define by $A:B=Tr(AB^{T})$. I would like to use Divergence theorem, but I do not know hoe to do it.
real-analysis vector-analysis trace divergence
asked Jan 2 at 18:26
lojdmojlojdmoj
877
$endgroup$
$begingroup$
Without having a specific $f$ you cannot get a value. For example, if $f$ satisfies the conditions then so does $lambda f$ and the integral for this will be $|lambda|^2$ times the one for $f$.
$endgroup$
– md2perpe
Jan 3 at 19:51
add a comment |
$begingroup$
Let $Omegasubset mathbb{R}^3$ be bounded domain and $f$ be smooth vector field on $Omega$ and $f|_{partialOmega=0}$. I not able to compute $int _{Omega} grad~f:(grad~f)^{T}$, where $:$ is inner product define by $A:B=Tr(AB^{T})$. I would like to use Divergence theorem, but I do not know hoe to do it.
real-analysis vector-analysis trace divergence
asked Jan 2 at 18:26
lojdmojlojdmoj
877
$endgroup$
Let $Omegasubset mathbb{R}^3$ be bounded domain and $f$ be smooth vector field on $Omega$ and $f|_{partialOmega=0}$. I not able to compute $int _{Omega} grad~f:(grad~f)^{T}$, where $:$ is inner product define by $A:B=Tr(AB^{T})$. I would like to use Divergence theorem, but I do not know hoe to do it.
real-analysis vector-analysis trace divergence
real-analysis vector-analysis trace divergence
asked Jan 2 at 18:26
lojdmojlojdmoj
877
asked Jan 2 at 18:26
lojdmojlojdmoj
877
asked Jan 2 at 18:26
lojdmojlojdmoj
877
asked Jan 2 at 18:26
lojdmojlojdmoj
877
asked Jan 2 at 18:26
lojdmojlojdmoj
877
877
$begingroup$
Without having a specific $f$ you cannot get a value. For example, if $f$ satisfies the conditions then so does $lambda f$ and the integral for this will be $|lambda|^2$ times the one for $f$.
$endgroup$
– md2perpe
Jan 3 at 19:51
add a comment |
$begingroup$
Without having a specific $f$ you cannot get a value. For example, if $f$ satisfies the conditions then so does $lambda f$ and the integral for this will be $|lambda|^2$ times the one for $f$.
$endgroup$
– md2perpe
Jan 3 at 19:51
$begingroup$
Without having a specific $f$ you cannot get a value. For example, if $f$ satisfies the conditions then so does $lambda f$ and the integral for this will be $|lambda|^2$ times the one for $f$.
$endgroup$
– md2perpe
Jan 3 at 19:51
$begingroup$
Without having a specific $f$ you cannot get a value. For example, if $f$ satisfies the conditions then so does $lambda f$ and the integral for this will be $|lambda|^2$ times the one for $f$.
$endgroup$
– md2perpe
Jan 3 at 19:51
add a comment |
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$begingroup$
Without having a specific $f$ you cannot get a value. For example, if $f$ satisfies the conditions then so does $lambda f$ and the integral for this will be $|lambda|^2$ times the one for $f$.
$endgroup$
– md2perpe
Jan 3 at 19:51