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 $\vec{v}$ satisfy $\nabla^2 \vec{v} = 0$. Is it true that the mean value of $\vec{v}$ over a sphere equals its value at the center of the sphere?

I would imagine the answer to be positive. For example, writing $\vec{v}$ in cartesian basis reduces the vector laplacian $\nabla^2 \vec{v} = 0$ 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.

  • $\begingroup$ First thing first, what is $\nabla^2 \vec v$? $\endgroup$ – xbh Aug 29 '18 at 10:10
  • $\begingroup$ @xbh en.wikipedia.org/wiki/Vector_Laplacian $\endgroup$ – Fizikus Aug 29 '18 at 10:12
  • $\begingroup$ Thanks, my bad. $\endgroup$ – xbh Aug 29 '18 at 10:13
  • $\begingroup$ 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? $\endgroup$ – Aleksejs Fomins Sep 7 '18 at 9:28
  • $\begingroup$ @aleksejsfomins: I am interested in the latter. $\endgroup$ – Fizikus Sep 7 '18 at 12:08

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