Is this vectorial identity between operators true?
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$$nabla^2(mathbf{t} u(mathbf{r}) ) =mathbf{t} div(nabla u(mathbf{r}))$$
Where $mathbf{t}$ is constant vector and $nabla^2$ is the vector laplacian (defined here: https://en.wikipedia.org/wiki/Vector_Laplacian).
Thanks for your help.
vector-analysis
asked Nov 20 at 17:30
Landau
447
add a comment |
up vote
0
down vote
favorite
$$nabla^2(mathbf{t} u(mathbf{r}) ) =mathbf{t} div(nabla u(mathbf{r}))$$
Where $mathbf{t}$ is constant vector and $nabla^2$ is the vector laplacian (defined here: https://en.wikipedia.org/wiki/Vector_Laplacian).
Thanks for your help.
vector-analysis
asked Nov 20 at 17:30
Landau
447
1
Yes. It looks true.
– Jacky Chong
Nov 20 at 17:36
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
$$nabla^2(mathbf{t} u(mathbf{r}) ) =mathbf{t} div(nabla u(mathbf{r}))$$
Where $mathbf{t}$ is constant vector and $nabla^2$ is the vector laplacian (defined here: https://en.wikipedia.org/wiki/Vector_Laplacian).
Thanks for your help.
vector-analysis
asked Nov 20 at 17:30
Landau
447
$$nabla^2(mathbf{t} u(mathbf{r}) ) =mathbf{t} div(nabla u(mathbf{r}))$$
Where $mathbf{t}$ is constant vector and $nabla^2$ is the vector laplacian (defined here: https://en.wikipedia.org/wiki/Vector_Laplacian).
Thanks for your help.
vector-analysis
vector-analysis
asked Nov 20 at 17:30
Landau
447
asked Nov 20 at 17:30
Landau
447
asked Nov 20 at 17:30
Landau
447
asked Nov 20 at 17:30
Landau
447
asked Nov 20 at 17:30
Landau
447
447
1
Yes. It looks true.
– Jacky Chong
Nov 20 at 17:36
add a comment |
1
Yes. It looks true.
– Jacky Chong
Nov 20 at 17:36
1
1
Yes. It looks true.
– Jacky Chong
Nov 20 at 17:36
Yes. It looks true.
– Jacky Chong
Nov 20 at 17:36
add a comment |
1 Answer
1
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oldest
votes
up vote
1
down vote
accepted
Yes: $nabla^2 (tu) = partial_i^2 (t_i u_i) = t_i (partial_i^2 u_i) = t . operatorname{div} (nabla u)$.
answered Nov 20 at 17:37
anomaly
17.1k42662
Your demonstration holds only in cartesiancoordinates I'd say, is it correct? Otherwise you can't define the laplacian as you have done in your first passage.
– Landau
Nov 20 at 21:58
Yes, I was assuming you're in flat $mathbb{R}^n$. Otherwise, throw a bunch of metric coefficients in, choose a convenient local coordinate chart, etc.
– anomaly
Nov 20 at 22:03
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Yes: $nabla^2 (tu) = partial_i^2 (t_i u_i) = t_i (partial_i^2 u_i) = t . operatorname{div} (nabla u)$.
answered Nov 20 at 17:37
anomaly
17.1k42662
Your demonstration holds only in cartesiancoordinates I'd say, is it correct? Otherwise you can't define the laplacian as you have done in your first passage.
– Landau
Nov 20 at 21:58
Yes, I was assuming you're in flat $mathbb{R}^n$. Otherwise, throw a bunch of metric coefficients in, choose a convenient local coordinate chart, etc.
– anomaly
Nov 20 at 22:03
add a comment |
up vote
1
down vote
accepted
Yes: $nabla^2 (tu) = partial_i^2 (t_i u_i) = t_i (partial_i^2 u_i) = t . operatorname{div} (nabla u)$.
answered Nov 20 at 17:37
anomaly
17.1k42662
Your demonstration holds only in cartesiancoordinates I'd say, is it correct? Otherwise you can't define the laplacian as you have done in your first passage.
– Landau
Nov 20 at 21:58
Yes, I was assuming you're in flat $mathbb{R}^n$. Otherwise, throw a bunch of metric coefficients in, choose a convenient local coordinate chart, etc.
– anomaly
Nov 20 at 22:03
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Yes: $nabla^2 (tu) = partial_i^2 (t_i u_i) = t_i (partial_i^2 u_i) = t . operatorname{div} (nabla u)$.
answered Nov 20 at 17:37
anomaly
17.1k42662
Yes: $nabla^2 (tu) = partial_i^2 (t_i u_i) = t_i (partial_i^2 u_i) = t . operatorname{div} (nabla u)$.
answered Nov 20 at 17:37
anomaly
17.1k42662
answered Nov 20 at 17:37
anomaly
17.1k42662
answered Nov 20 at 17:37
anomaly
17.1k42662
answered Nov 20 at 17:37
anomaly
17.1k42662
17.1k42662
Your demonstration holds only in cartesiancoordinates I'd say, is it correct? Otherwise you can't define the laplacian as you have done in your first passage.
– Landau
Nov 20 at 21:58
Yes, I was assuming you're in flat $mathbb{R}^n$. Otherwise, throw a bunch of metric coefficients in, choose a convenient local coordinate chart, etc.
– anomaly
Nov 20 at 22:03
add a comment |
Your demonstration holds only in cartesiancoordinates I'd say, is it correct? Otherwise you can't define the laplacian as you have done in your first passage.
– Landau
Nov 20 at 21:58
Yes, I was assuming you're in flat $mathbb{R}^n$. Otherwise, throw a bunch of metric coefficients in, choose a convenient local coordinate chart, etc.
– anomaly
Nov 20 at 22:03
Your demonstration holds only in cartesiancoordinates I'd say, is it correct? Otherwise you can't define the laplacian as you have done in your first passage.
– Landau
Nov 20 at 21:58
Your demonstration holds only in cartesiancoordinates I'd say, is it correct? Otherwise you can't define the laplacian as you have done in your first passage.
– Landau
Nov 20 at 21:58
Yes, I was assuming you're in flat $mathbb{R}^n$. Otherwise, throw a bunch of metric coefficients in, choose a convenient local coordinate chart, etc.
– anomaly
Nov 20 at 22:03
Yes, I was assuming you're in flat $mathbb{R}^n$. Otherwise, throw a bunch of metric coefficients in, choose a convenient local coordinate chart, etc.
– anomaly
Nov 20 at 22:03
add a comment |
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1
Yes. It looks true.
– Jacky Chong
Nov 20 at 17:36