Mean value theorem for vector laplacian
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It is well known that all solutions of the Laplace equation $nabla^2 u = 0$ satisfy the mean value theorem: the average value of $u$ over a sphere equals its value at the center of the sphere.
My question is: does the same theorem hold in case of vector laplacian? Let the vector field $vecv$ satisfy $nabla^2 vecv = 0$. Is it true that the mean value of $vecv$ over a sphere equals its value at the center of the sphere?
I would imagine the answer to be positive. For example, writing $vecv$ in cartesian basis reduces the vector laplacian $nabla^2 vecv = 0$ to three ordinary laplacians: $nabla^2 v_x = nabla^2 v_y = nabla^2 v_z = 0$ and makes the statement trivial. However, one can imagine using some other, curvilinear basis, in which the statement is far from trivial. Therefore, I would appreciate some insight or a coordinate-free proof of the theorem.
multivariable-calculus vector-analysis coordinate-systems harmonic-functions laplacian
asked Aug 29 at 10:09
Fizikus
122113
add a comment |Â
up vote
3
down vote
favorite
It is well known that all solutions of the Laplace equation $nabla^2 u = 0$ satisfy the mean value theorem: the average value of $u$ over a sphere equals its value at the center of the sphere.
My question is: does the same theorem hold in case of vector laplacian? Let the vector field $vecv$ satisfy $nabla^2 vecv = 0$. Is it true that the mean value of $vecv$ over a sphere equals its value at the center of the sphere?
I would imagine the answer to be positive. For example, writing $vecv$ in cartesian basis reduces the vector laplacian $nabla^2 vecv = 0$ to three ordinary laplacians: $nabla^2 v_x = nabla^2 v_y = nabla^2 v_z = 0$ and makes the statement trivial. However, one can imagine using some other, curvilinear basis, in which the statement is far from trivial. Therefore, I would appreciate some insight or a coordinate-free proof of the theorem.
multivariable-calculus vector-analysis coordinate-systems harmonic-functions laplacian
asked Aug 29 at 10:09
Fizikus
122113
First thing first, what is $nabla^2 vec v$?
– xbh
Aug 29 at 10:10
@xbh en.wikipedia.org/wiki/Vector_Laplacian
– Fizikus
Aug 29 at 10:12
Thanks, my bad.
– xbh
Aug 29 at 10:13
Sorry if this question is trivial, but is there a situation when we are forced not to use Cartesian basis? I mean, the PDE is defined in a basis-independent way, so I would imagine that we are free to choose any basis we want to prove basis-independent statements. Since change of basis does not alter the function itself, but only the representation. Or are you interested in a basis-free proof just to see if it is possible and how?
– Aleksejs Fomins
Sep 7 at 9:28
@aleksejsfomins: I am interested in the latter.
– Fizikus
Sep 7 at 12:08
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
It is well known that all solutions of the Laplace equation $nabla^2 u = 0$ satisfy the mean value theorem: the average value of $u$ over a sphere equals its value at the center of the sphere.
My question is: does the same theorem hold in case of vector laplacian? Let the vector field $vecv$ satisfy $nabla^2 vecv = 0$. Is it true that the mean value of $vecv$ over a sphere equals its value at the center of the sphere?
I would imagine the answer to be positive. For example, writing $vecv$ in cartesian basis reduces the vector laplacian $nabla^2 vecv = 0$ to three ordinary laplacians: $nabla^2 v_x = nabla^2 v_y = nabla^2 v_z = 0$ and makes the statement trivial. However, one can imagine using some other, curvilinear basis, in which the statement is far from trivial. Therefore, I would appreciate some insight or a coordinate-free proof of the theorem.
multivariable-calculus vector-analysis coordinate-systems harmonic-functions laplacian
asked Aug 29 at 10:09
Fizikus
122113
It is well known that all solutions of the Laplace equation $nabla^2 u = 0$ satisfy the mean value theorem: the average value of $u$ over a sphere equals its value at the center of the sphere.
My question is: does the same theorem hold in case of vector laplacian? Let the vector field $vecv$ satisfy $nabla^2 vecv = 0$. Is it true that the mean value of $vecv$ over a sphere equals its value at the center of the sphere?
I would imagine the answer to be positive. For example, writing $vecv$ in cartesian basis reduces the vector laplacian $nabla^2 vecv = 0$ to three ordinary laplacians: $nabla^2 v_x = nabla^2 v_y = nabla^2 v_z = 0$ and makes the statement trivial. However, one can imagine using some other, curvilinear basis, in which the statement is far from trivial. Therefore, I would appreciate some insight or a coordinate-free proof of the theorem.
multivariable-calculus vector-analysis coordinate-systems harmonic-functions laplacian
asked Aug 29 at 10:09
Fizikus
122113
edited Aug 30 at 10:19
asked Aug 29 at 10:09
Fizikus
122113
asked Aug 29 at 10:09
Fizikus
122113
asked Aug 29 at 10:09
Fizikus
122113
122113
First thing first, what is $nabla^2 vec v$?
– xbh
Aug 29 at 10:10
@xbh en.wikipedia.org/wiki/Vector_Laplacian
– Fizikus
Aug 29 at 10:12
Thanks, my bad.
– xbh
Aug 29 at 10:13
Sorry if this question is trivial, but is there a situation when we are forced not to use Cartesian basis? I mean, the PDE is defined in a basis-independent way, so I would imagine that we are free to choose any basis we want to prove basis-independent statements. Since change of basis does not alter the function itself, but only the representation. Or are you interested in a basis-free proof just to see if it is possible and how?
– Aleksejs Fomins
Sep 7 at 9:28
@aleksejsfomins: I am interested in the latter.
– Fizikus
Sep 7 at 12:08
add a comment |Â
First thing first, what is $nabla^2 vec v$?
– xbh
Aug 29 at 10:10
@xbh en.wikipedia.org/wiki/Vector_Laplacian
– Fizikus
Aug 29 at 10:12
Thanks, my bad.
– xbh
Aug 29 at 10:13
Sorry if this question is trivial, but is there a situation when we are forced not to use Cartesian basis? I mean, the PDE is defined in a basis-independent way, so I would imagine that we are free to choose any basis we want to prove basis-independent statements. Since change of basis does not alter the function itself, but only the representation. Or are you interested in a basis-free proof just to see if it is possible and how?
– Aleksejs Fomins
Sep 7 at 9:28
@aleksejsfomins: I am interested in the latter.
– Fizikus
Sep 7 at 12:08
First thing first, what is $nabla^2 vec v$?
– xbh
Aug 29 at 10:10
First thing first, what is $nabla^2 vec v$?
– xbh
Aug 29 at 10:10
@xbh en.wikipedia.org/wiki/Vector_Laplacian
– Fizikus
Aug 29 at 10:12
@xbh en.wikipedia.org/wiki/Vector_Laplacian
– Fizikus
Aug 29 at 10:12
Thanks, my bad.
– xbh
Aug 29 at 10:13
Thanks, my bad.
– xbh
Aug 29 at 10:13
Sorry if this question is trivial, but is there a situation when we are forced not to use Cartesian basis? I mean, the PDE is defined in a basis-independent way, so I would imagine that we are free to choose any basis we want to prove basis-independent statements. Since change of basis does not alter the function itself, but only the representation. Or are you interested in a basis-free proof just to see if it is possible and how?
– Aleksejs Fomins
Sep 7 at 9:28
Sorry if this question is trivial, but is there a situation when we are forced not to use Cartesian basis? I mean, the PDE is defined in a basis-independent way, so I would imagine that we are free to choose any basis we want to prove basis-independent statements. Since change of basis does not alter the function itself, but only the representation. Or are you interested in a basis-free proof just to see if it is possible and how?
– Aleksejs Fomins
Sep 7 at 9:28
@aleksejsfomins: I am interested in the latter.
– Fizikus
Sep 7 at 12:08
@aleksejsfomins: I am interested in the latter.
– Fizikus
Sep 7 at 12:08
add a comment |Â
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First thing first, what is $nabla^2 vec v$?
– xbh
Aug 29 at 10:10
@xbh en.wikipedia.org/wiki/Vector_Laplacian
– Fizikus
Aug 29 at 10:12
Thanks, my bad.
– xbh
Aug 29 at 10:13
Sorry if this question is trivial, but is there a situation when we are forced not to use Cartesian basis? I mean, the PDE is defined in a basis-independent way, so I would imagine that we are free to choose any basis we want to prove basis-independent statements. Since change of basis does not alter the function itself, but only the representation. Or are you interested in a basis-free proof just to see if it is possible and how?
– Aleksejs Fomins
Sep 7 at 9:28
@aleksejsfomins: I am interested in the latter.
– Fizikus
Sep 7 at 12:08