I am asked to prove
$$\nabla \cdot (T \cdot v) = T : \nabla v + v\cdot (\nabla \cdot T)$$
Where $T$ is a order-2 tensor and $v$ is vector, in an orthogonal basis. Let $\delta _{ij}$ denote the Kronecker delta.
My proof:
$$T = T_{ij} e_i e_j, v = v_k e_k$$
where $e_i$ are unit vectors in an orthogonal basis.
So,
$$T \cdot v = T_{ij}v_k e_i e_j \cdot e_k = T_{ij}v_k e_i \delta _{jk} = T_{im}v_me_i \\
\implies \nabla \cdot (T \cdot v) = \frac{\partial (T_{im}v_m e_i)}{\partial x_n}\cdot e_n = \frac{\partial (T_{im}v_m)}{\partial x_n} \delta_{in}=\frac{\partial (T_{pm}v_m)}{\partial x_p} = \frac{\partial (T_{pq}v_q)}{\partial x_p} $$
Moving on to the RHS:
$$T:\nabla v = T_{ij}e_i e_j : \frac{\partial (v_m e_m)}{\partial x_n} e_n = T_{ij}\frac{\partial v_m}{x_n}e_ie_j:e_me_n = T_{ij}V_{mn}\delta_{jm}\delta_{in}=T_{pq}V_{qp}=T_{pq}\frac{\partial v_q}{\partial x_p}$$
The second term of the RHS:
$$v\cdot(\nabla \cdot T)=(v_ie_i)\cdot\left(\frac{\partial}{\partial x_j}(T_{kl} e_k e_l)\cdot e_j\right)=(v_ie_i)\cdot\left(\frac{\partial T_{kl}}{\partial x_j} e_k\delta _{lj}\right) = (v_ie_i)\cdot\left(\frac{\partial T_{km}}{\partial x_m} e_k\right)=v_i\frac{\partial T_{km}}{\partial x_m} \delta _{ik}=v_q\frac{\partial T_{qp}}{\partial x_p}$$
Where I replaced $m$ with $p$, and contracted $i$,$k$ with the index $p$ in the last statement.
As you can see, $$v_q\frac{\partial T_{qp}}{\partial x_{p}} + T_{pq}\frac{\partial v_{q}}{\partial x_{p}} \neq \frac{\partial T_{pq}v_q}{x_p}$$
What am I doing wrong here? Any advice you have would be appreciated!
Best Answer
Now you've found your mistake, let me suggest a slight change in proof technique that saves these difficulties. Rather than working with explicit $e_i$, write$$\nabla\cdot(T\cdot v)=\partial_i(T_{ij}v_j)=T_{ij}\partial_iv_j+v_j\partial_iT_{ij}=T:\nabla v+v\cdot(\nabla\cdot T).$$