Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Given a smooth tensor valued function $\sigma:R^2\rightarrow R^{2\times2}$, I'm trying to show that

$\int_\Omega \nabla\cdot\sigma=\int_{\partial\Omega}\sigma n$,

where $\Omega$ is a connected simple region in $R^2$, $\partial\Omega$ is a closed simple curve along boundary of $\Omega$, and $n$ is a unit vector.

I'm struggling with several points in this problem:

  1. How can I represent $\sigma$ explicitly?
  2. How can I explicitly represent $\sigma n$?
  3. Given that $\nabla\cdot\sigma$ is a vector, would the differential of the left hand side of the equation still be $dxdy$?
  4. I am told that I can use divergence theorem explicitly, but the only version I'm familiar with applies to integrals whose integrands are scalar quantities? What is the tensor-divergence theorem equivalent?

Any help would be greatly appreciated! :) Thanks.

share|cite|improve this question
up vote 2 down vote accepted

The divergence theorem for a tensor could be written as

$$ \int_\Omega\sum_{i=1}^2\frac{\partial}{\partial x_i}\sigma_{ij}dxdy=\int_{\partial\Omega}\sum_{i=1}^2 n_i\sigma_{ij}ds\qquad j=1,2 $$

in practice you apply the usual divergence theorem to each vector $\boldsymbol{\sigma}_j=(\sigma_{1j},\sigma_{2j})$.

share|cite|improve this answer
So, you mean "the usual divergence theorem" for each column vector of $\sigma$? – Paul Sep 14 '12 at 20:56
Yes, essentially it is right. – enzotib Sep 14 '12 at 20:59
In the definition of $\sigma$, it maps a vector to a matrix. But in your notation of the right hand side, wouldn't $\sigma n$ be a vector? Isn't this a contradiction? – Paul Sep 14 '12 at 21:29
No, the notation $\sigma:R^2\rightarrow R^{2\times2}$ means that in each point of $\Omega$ (two coordinates) is defined a matrix (four numbers in a square). Like each $2\times2$ matrix, $\sigma$ can be considered, in each point, as a linear map from $R^2$ to $R^2$, i.e. it maps vectors to vectors (by a matrix multiplication). – enzotib Sep 14 '12 at 21:35
Oh... ok... So, at each point, a matrix is assigned. Could I also consider it as if $\sigma$ were a matrix whose elements are functions of variables in $R^2$? – Paul Sep 14 '12 at 21:40

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.