What is the divergence of a matrix valued function?

Question

The divergence of a continuously differentiable tensor field $\underline{\underline{\epsilon}}$ is:

$$\overrightarrow{\operatorname{div}}\,(\mathbf{\underline{\underline{\epsilon}}}) = \begin{bmatrix} \frac{\partial \epsilon_{xx}}{\partial x} +\frac{\partial \epsilon_{xy}}{\partial y} +\frac{\partial \epsilon_{xz}}{\partial z} \\ \frac{\partial \epsilon_{yx}}{\partial x} +\frac{\partial \epsilon_{yy}}{\partial y} +\frac{\partial \epsilon_{yz}}{\partial z} \\ \frac{\partial \epsilon_{zx}}{\partial x} +\frac{\partial \epsilon_{zy}}{\partial y} +\frac{\partial \epsilon_{zz}}{\partial z} \end{bmatrix} $$

How do you get this formula from the definition of divergence? Either formally, or with some abuse of notation?

Also you can work it out from the definition of $\nabla$ notation. — , Nov 04 '11 at 02:54
I don't have a high enough reputation to comment, but note that the equation in the original post is not correct: $\epsilon_{ij}$ should be replaced with $\epsilon_{ji}$ in all of the term numerators. — masonlr, Aug 22 '17 at 08:54

score 6 · Accepted Answer · edited Nov 10 '11 at 22:00

6

If S a matrix, with columns $S^{j}$, $j=1$, $n$ then $\mathrm{div}(S)_{j} = \mathrm{div}(S^{j})$.

edited Nov 10 '11 at 22:00

hmakholm left over Monica

286,031

answered Nov 10 '11 at 20:56

user16847

304

4

+1. More generally, for any linear operator that acts on vectors, the natural extension to matrices is to apply the operator column-wise. – user7530 Nov 10 '11 at 22:02
2

hmmm are you sure? I think this is not (always) right. consider a 3x2 matrix that depends on two variables. – johannes_lalala Sep 20 '16 at 00:11

What is the divergence of a matrix valued function?

1 Answers1

Linked