Let $A$ be continuously differentiable second-order tensor field defined as follows: A=\begin{bmatrix}A_{11}&A_{12}&A_{13}\\A_{21}&A_{22}&A_{23}\\A_{31}&A_{32}&A_{33}\end{bmatrix} the divergence in cartesian coordinate system is a first-order tensor field and can be defined as:\begin{bmatrix}{\dfrac {\partial A_{11}}{\partial x_{1}}}+{\dfrac {\partial A_{12}}{\partial x_{2}}}+{\dfrac {\partial A_{13}}{\partial x_{3}}}\\{\dfrac {\partial A_{21}}{\partial x_{1}}}+{\dfrac {\partial A_{22}}{\partial x_{2}}}+{\dfrac {\partial A_{23}}{\partial x_{3}}}\\{\dfrac {\partial A_{31}}{\partial x_{1}}}+{\dfrac {\partial A_{32}}{\partial x_{2}}}+{\dfrac {\partial A_{33}}{\partial x_{3}}}\end{bmatrix}
A)What is meant by 1st and 2nd order tensor?
B)And a second question is that I am little confused in general about $divU$, where $U$ is a vector or matrix, what shall be the nabla operator? i.e is it a column vector or a row vector? And how $divU$ is expressed in the two cases: 1)U is a matrix 2)U is a vector.
Thank you in advance!
A) A first order tensor can be represented by a vector. A second order tensor can be represented as a matrix.
B) The divergence of a matrix $A$ is a vector, given by the formule you wrote which can be written
$$div(A)=\begin{bmatrix}{ \nabla \cdot A_{1} \\ \nabla \cdot A_{2} \\ \nabla \cdot A_{3} } \end{bmatrix} = \begin{bmatrix}{\dfrac {\partial A_{11}}{\partial x_{1}}}+{\dfrac {\partial A_{12}}{\partial x_{2}}}+{\dfrac {\partial A_{13}}{\partial x_{3}}}\\{\dfrac {\partial A_{21}}{\partial x_{1}}}+{\dfrac {\partial A_{22}}{\partial x_{2}}}+{\dfrac {\partial A_{23}}{\partial x_{3}}}\\{\dfrac {\partial A_{31}}{\partial x_{1}}}+{\dfrac {\partial A_{32}}{\partial x_{2}}}+{\dfrac {\partial A_{33}}{\partial x_{3}}}\end{bmatrix}$$ where $A_i$ stand for the line $i$ of $A$ and $\nabla =(\partial_1, \partial_2, \partial_3)^t$.
The divergence of a vector $U=( u_1,_2,u_3)^t$ is a real, given by
$$div(U) = \nabla \cdot U= \partial_1 U_1 + \partial_2 U_2 + \partial_3 U_3$$