# Divergence of the product of a tensor and a vector field

Let $$\mathbf u$$ and $$\mathbf S$$ be smooth fields with $$\mathbf u$$ vector valued and $$\mathbf S$$ tensor valued. I would like to prove the following identity:

$$\operatorname{div}\mathbf S\mathbf{u}=\mathbf S^T\cdot\operatorname{grad}\mathbf u+\mathbf u\cdot\operatorname{div}\mathbf S$$

My attempt:

\begin{align} \operatorname{div}\mathbf S\mathbf{u}=\frac{\partial}{\partial x_i}(S_{ij}u_j)\mathbf{e}_j&=\frac{\partial S_{ij}}{\partial x_i}u_j\mathbf{e}_j+S_{ij}\frac{\partial u_j}{\partial x_i}\mathbf{e}_j\\ &=\frac{\partial S_{ij}}{\partial x_i}u_j\mathbf{e}_j+S_{ji}^T\frac{\partial u_j}{\partial x_i}\mathbf{e}_j\\ &=\mathbf u\cdot\operatorname{div}\mathbf S+S_{ji}^T\frac{\partial u_j}{\partial x_i}\mathbf{e}_j \end{align}

Here, I have used the following definition: $$\operatorname{div}\mathbf S=\frac{\partial S_{ij}}{\partial x_i}\mathbf{e}_j$$.

But I am stuck on the last expression, $$S_{ji}^T\frac{\partial u_j}{\partial x_i}\mathbf{e}_j$$, since the definition of the gradient of a vector field is $$\operatorname{grad}\mathbf u=\frac{\partial u_j}{\partial x_i}\mathbf{e}_i$$.

I would appreciate any help or hint. Thank you.

• You should only have two of the same index in a term when using summation convention. Also, you should have the same free indices on both sides of an equation, which you don't in several places. Commented Apr 9, 2017 at 23:41
• And what sort of product is $\mathbf{Su}$? Is it an inner product $S_{ij}u_j$ or an outer product $S_{ij}u_k$? Commented Apr 9, 2017 at 23:43
• @Chappers It is a inner product. As for your first comment I follow the definitions of my lecture notes so I am not sure what I am doing wrong. Commented Apr 10, 2017 at 12:01
• For operators like div and grad use \operatorname, not \mathrm. Commented Apr 25, 2021 at 22:55

Assuming $\mathbf{S}$ is a matrix and $\mathbf{u}$ is a vector, then $\operatorname{div}\mathbf{Su}$ is a scalar. Hence, we have \begin{align} \operatorname{div}\mathbf{Su} = \frac{\partial}{\partial x_i}(S_{ij}u_j) = \frac{\partial S_{ij}}{\partial x_i}u_j+S_{ij}\frac{\partial u_j}{\partial x_i}. \end{align}

Note that \begin{align} \frac{\partial S_{ij}}{\partial x_i}u_j=\frac{\partial S_{ij}}{\partial x_i}\mathbf{e}_j\cdot u_j\mathbf{e}_j=\mathbf{u}\cdot \operatorname{div}\mathbf{S}. \end{align} Next, observe \begin{align} S_{ij}\frac{\partial u_j}{\partial x_i}= \sum_j \sum_i S_{ij}\frac{\partial u_j}{\partial x_i} = \sum_j \sum_i S_{ji}^Tu_{j, i} = \mathbf{S}^T\cdot \operatorname{grad}\mathbf{u}. \end{align} Here we used the fact that \begin{align} \text{d}\mathbf{u}= (\operatorname{grad}\mathbf{u})^T \end{align} otherwise, we would have the expression \begin{align} \text{d}\mathbf{u}\cdot \mathbf{S}. \end{align}

I think, the right expression is:

$$\operatorname{div}(\mathbf{A} \mathbf{v})=\mathbf{v} \cdot \operatorname{div} \mathbf{A}^{\top}+\operatorname{tr}(\mathbf{A} \mathrm{g r a d}\mathbf{v}).$$

II found it here and in a slightly different form here.

First of all, $$\operatorname{div}(\mathbf{A})$$ and $$\nabla \cdot \mathbf{A}$$ are equal only for symmetric tensors (see here), since

$$\operatorname{div}(\mathbf{A})=\frac{\partial A_{i k}}{\partial x_{k}} \mathbf{e}_{i}=A_{i k, k} \mathbf{e}_{i}=\left[\begin{array}{l} \frac{\partial A_{11}}{\partial x_{1}}+\frac{\partial A_{12}}{\partial x_{2}}+\frac{\partial A_{13}}{\partial x_{3}} \\ \frac{\partial A_{21}}{\partial x_{1}}+\frac{\partial A_{22}}{\partial x_{2}}+\frac{\partial A_{23}}{\partial x_{3}} \\ \frac{\partial A_{31}}{\partial x_{1}}+\frac{\partial A_{32}}{\partial x_{2}}+\frac{\partial A_{33}}{\partial x_{3}} \end{array}\right]$$ and $$\nabla \cdot \mathbf{A}=\frac{\partial A_{k i}}{\partial x_{k}} \mathbf{e}_{i}=A_{k i, k} \mathbf{e}_{i}=\left[\begin{array}{l} \frac{\partial A_{11}}{\partial x_{1}}+\frac{\partial A_{21}}{\partial x_{2}}+\frac{\partial A_{31}}{\partial x_{3}} \\ \frac{\partial A_{12}}{\partial x_{1}}+\frac{\partial A_{22}}{\partial x_{2}}+\frac{\partial A_{32}}{\partial x_{3}} \\ \frac{\partial A_{13}}{\partial x_{1}}+\frac{\partial A_{23}}{\partial x_{2}}+\frac{\partial A_{33}}{\partial x_{3}} \end{array}\right].$$ We have a relation $$\operatorname{div}(\mathbf{A}^{\top}) = \nabla \cdot \mathbf{A}$$

Below is the proof of $$\operatorname{div}(\mathbf{A} \mathbf{v})=\mathbf{v} \cdot \operatorname{div} \mathbf{A}^{\top}+\operatorname{tr}(\mathbf{A} \mathrm{g r a d}\mathbf{v})$$ in Cartesian coordinate system: $$\operatorname{div}(\mathbf{A u})=\frac{\partial}{\partial x_{i}}\left(A_{i j} v_{j}\right)=\frac{\partial A_{i j}}{\partial x_{i}} v_{j}+A_{i j} \frac{\partial v_{j}}{\partial x_{i}}$$ For the first term we have: $$\begin{array}{c} \frac{\partial A_{i j}}{\partial x_{i}} v_{j}=\frac{\partial A_{1 j}}{\partial x_{1}} v_{j}+\frac{\partial A_{2 j}}{\partial x_{2}} v_{j}+\frac{\partial A_{3 j}}{\partial x_{3}} v_{j} \\ =\frac{\partial A_{11}}{\partial x_{1}} v_{1}+\frac{\partial A_{12}}{\partial x_{1}} v_{2}+\frac{\partial A_{13}}{\partial x_{1}} v_{3}+\frac{\partial A_{21}}{\partial x_{2}} v_{1}+\frac{\partial A_{22}}{\partial x_{2}} v_{2}+\frac{\partial A_{23}}{\partial x_{2}} v_{3}+\frac{\partial A_{31}}{\partial x_{3}} v_{1}+\frac{\partial A_{32}}{\partial x_{3}} v_{2} \\ +\frac{\partial A_{33}}{\partial x_{3}} v_{3}= \\ \frac{\partial A_{11}}{\partial x_{1}} v_{1}+\frac{\partial A_{21}}{\partial x_{2}} v_{1}+\frac{\partial A_{31}}{\partial x_{3}} v_{1}+\frac{\partial A_{12}}{\partial x_{1}} v_{2}+\frac{\partial A_{22}}{\partial x_{2}} v_{2}+\frac{\partial A_{32}}{\partial x_{3}} v_{2}+\frac{\partial A_{13}}{\partial x_{1}} v_{3}+\frac{\partial A_{23}}{\partial x_{2}} v_{3}+\frac{\partial A_{33}}{\partial x_{3}} v_{3}= \\ v_{1}\left(\frac{\partial A_{11}}{\partial x_{1}}+\frac{\partial A_{21}}{\partial x_{2}}+\frac{\partial A_{31}}{\partial x_{3}}\right)+v_{2}\left(\frac{\partial A_{12}}{\partial x_{1}}+\frac{\partial A_{22}}{\partial x_{2}}+\frac{\partial A_{32}}{\partial x_{3}}\right)+v_{3}\left(\frac{\partial A_{13}}{\partial x_{1}}+\frac{\partial A_{23}}{\partial x_{2}}+\frac{\partial A_{33}}{\partial x_{3}}\right)= \\ \left(\begin{array}{l} \frac{\partial A_{11}}{\partial x_{1}}+\frac{\partial A_{21}}{\partial x_{2}}+\frac{\partial A_{31}}{\partial x_{3}} \\ \frac{\partial A_{12}}{\partial x_{1}}+\frac{\partial A_{22}}{\partial x_{2}}+\frac{\partial A_{32}}{\partial x_{3}} \\ \frac{\partial A_{13}}{\partial x_{1}}+\frac{\partial A_{23}}{\partial x_{2}}+\frac{\partial A_{33}}{\partial x_{3}} \end{array}\right)\left(\begin{array}{l} v_{1} \\ v_{2} \\ v_{3} \end{array}\right)=\operatorname{div}\left(\mathbf{A}^{\mathrm{T}}\right) \cdot \mathbf{v} \end{array}$$ And for the second term: \begin{aligned} A_{i j} \frac{\partial v_{j}}{\partial x_{i}}=\sum_{j=1}^{j=3}(&\left.A_{1 j} \frac{\partial v_{j}}{\partial x_{1}}+A_{2 j} \frac{\partial v_{j}}{\partial x_{2}}+A_{3 j} \frac{\partial v_{j}}{\partial x_{3}}\right) \\ &=A_{11} \frac{\partial v_{1}}{\partial x_{1}}+A_{21} \frac{\partial v_{1}}{\partial x_{2}}+A_{31} \frac{\partial v_{1}}{\partial x_{3}}+A_{12} \frac{\partial v_{2}}{\partial x_{1}}+A_{22} \frac{\partial v_{2}}{\partial x_{2}}+A_{32} \frac{\partial v_{2}}{\partial x_{3}}+A_{13} \frac{\partial v_{3}}{\partial x_{1}}+A_{23} \frac{\partial v_{3}}{\partial x_{2}} \\ &+A_{33} \frac{\partial v_{3}}{\partial x_{3}} \end{aligned} Now, if we consider the product $$\mathbf{T} = \left(\begin{array}{lll} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33} \end{array}\right)\left(\begin{array}{lll} \frac{\partial v_{1}}{\partial x_{1}} & \frac{\partial v_{1}}{\partial x_{2}} & \frac{\partial v_{1}}{\partial x_{3}} \\ \frac{\partial v_{2}}{\partial x_{1}} & \frac{\partial v_{2}}{\partial x_{2}} & \frac{\partial v_{2}}{\partial x_{3}} \\ \frac{\partial v_{3}}{\partial x_{1}} & \frac{\partial v_{3}}{\partial x_{2}} & \frac{\partial v_{3}}{\partial x_{3}} \end{array}\right)$$ then $$A_{11} \frac{\partial v_{1}}{\partial x_{1}} + A_{12} \frac{\partial v_{2}}{\partial x_{1}} + A_{13} \frac{\partial v_{3}}{\partial x_{1}}$$ are correspondingly (1,1), $$A_{21} \frac{\partial v_{1}}{\partial x_{2}} + A_{22} \frac{\partial v_{2}}{\partial x_{2}} + A_{23} \frac{\partial v_{3}}{\partial x_{2}}$$ - (2,2) and $$A_{31} \frac{\partial v_{1}}{\partial x_{3}} + A_{32} \frac{\partial v_{2}}{\partial x_{3}} + A_{33} \frac{\partial v_{3}}{\partial x_{3}}$$ - (3,3) elements of $$\mathbf{T}$$. Thus, $$A_{i j} \frac{\partial v_{j}}{\partial x_{i}}=\operatorname{tr}(\mathbf{A} \mathrm{g r a d} \mathbf{v})$$.

But I don't know how to prove this for any coordinate system...

I agree with this answer, but I think it would be better to demonstrate the answer in a tensorial way.

Assuming that $$\textbf{S}$$ is a second order tensor and $$\textbf{u}$$ a first order tensor (vector), we have: $$\text{div} \left(\textbf{S} \cdot \textbf{u} \right) = \dfrac{\partial}{\partial x_i} \left( S_{ij} u_j \right) = \dfrac{\partial S_{ij}}{\partial x_i} u_j + S_{ij}\dfrac{\partial u_{j}}{\partial x_i}$$ $$\text{div} \left(\textbf{S} \cdot \textbf{u} \right) = \text{div} \left(\textbf{S}^T\right) \cdot \textbf{u} + \textbf{S} : \left(\text{grad} \ \textbf{u}\right)^T$$ or, equivalently, $$\text{div} \left(\textbf{S} \cdot \textbf{u} \right) = \textbf{u} \cdot \text{div} \left(\textbf{S}^T\right) + \text{tr}\left(\textbf{S} \cdot \text{grad} \ \textbf{u}\right)$$ where, considering two second order tensors $$\textbf{A}$$ and $$\textbf{B}$$, the double dot product is defined as follows: $$\textbf{A} : \textbf{B} = A_{ij}B_{ij}$$ ** Note that $$\textbf{A} : \textbf{B}$$ can be defined in two different ways (see this post).

It is known, but worth to remark, that dot product between first order tensors commute. From the first term on the right in the equations above, we have: $$\text{div} \left(\textbf{S}^T\right) \cdot \textbf{u} = \dfrac{\partial S_{ij}}{\partial x_i} \underline{e}_j \cdot u_k \underline{e}_k = \dfrac{\partial S_{ij}}{\partial x_i} u_j,$$ but also $$\textbf{u} \cdot \text{div} \left(\textbf{S}^T\right) = u_i\underline{e}_i \cdot \dfrac{\partial S_{lk}}{\partial x_l}\underline{e}_k = u_i \dfrac{\partial S_{ji}}{\partial x_j} = \dfrac{\partial S_{ij}}{\partial x_i} u_j$$ As a result, $$\text{div} \left(\textbf{S}^T\right) \cdot \textbf{u} = \textbf{u} \cdot \text{div} \left(\textbf{S}^T\right)$$.

In order to demonstrate the equality $$(\textbf{S} : \left(\text{grad} \ \textbf{u}\right)^T = \text{tr}\left(\textbf{S} \cdot \text{grad} \ \textbf{u}\right))$$, note that: $$\text{tr}\left(\textbf{S} \cdot \text{grad} \ \textbf{u}\right) = \left( \textbf{S} \cdot \text{grad} \ \textbf{u} \right) : \textbf{1} = \left(S_{ij} \underline{e}_i \otimes \underline{e}_{j} \cdot \dfrac{\partial u_k}{\partial x_l} \underline{e}_{k} \otimes \underline{e}_{l} \right) : \underline{e}_{p}\otimes \underline{e}_{p}$$ $$\text{tr}\left(\textbf{S} \cdot \text{grad} \ \textbf{u}\right) = \left(S_{ij} \dfrac{\partial u_j}{\partial x_l} \underline{e}_{i} \otimes \underline{e}_{l} \right) : \underline{e}_{p}\otimes \underline{e}_{p} = S_{ij}\dfrac{\partial u_{j}}{\partial x_i}$$ and $$\textbf{S} : \left(\text{grad} \ \textbf{u}\right)^T = S_{ij} \underline{e}_i \otimes \underline{e}_{j} : \dfrac{\partial u_l}{\partial x_k} \underline{e}_k \otimes \underline{e}_{l} = S_{ij}\dfrac{\partial u_{j}}{\partial x_i}$$ The references cited by this post are very good, by the way.