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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: $$T:\nabla v = T_{ij}e_i e_j : e_n\frac{\partial (v_m e_m)}{\partial x_n} = T_{ij}\frac{\partial v_m}{x_n}e_ie_j:e_ne_m = T_{ij}V_{mn}\delta_{jn}\delta_{im}=T_{qp}V_{qp}=T_{qp}\frac{\partial v_q}{\partial x_p}$$ The second term of the RHS: $$v\cdot(\nabla \cdot T)=(v_ie_i)\cdot\left(e_j \cdot \frac{\partial}{\partial x_j}(T_{kl} e_k e_l)\right)=(v_ie_i)\cdot\left(\frac{\partial T_{kl}}{\partial x_j} e_l\delta _{jk}\right) = (v_ie_i)\cdot\left(\frac{\partial T_{jl}}{\partial x_j} e_l\right) = v_i\frac{\partial T_{jl}}{\partial x_j} \delta _{il}=v_q\frac{\partial T_{pq}}{\partial x_p}$$ Where I replaced $m$ with $p$, and contracted $i$,$k$ with the index $p$ in the last statement.

As you can see, $$v_q\frac{\partial T_{pq}}{\partial x_{p}} + T_{qp}\frac{\partial v_{q}}{\partial x_{p}} \neq \frac{\partial T_{pq}v_q}{x_p}$$

What am I doing wrong here? Any advice you have would be appreciated!

Initially, I had made a mistake wrt to the divergence operator, but I realize now that the error is in the first term, not the second term.

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