2

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!

megamence
  • 297

2 Answers2

2

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$$\nabla\cdot(T\cdot v)=\partial_i(T_{ij}v_j)=T_{ij}\partial_iv_j+v_j\partial_iT_{ij}=T:\nabla v+v\cdot(\nabla\cdot T).$$

J.G.
  • 115,835
  • Thanks for your reply @J.G.! I have a quick question regarding the first term in the RHS... Shouldn't $T:\nabla v = T_{ij}e_ie_j:(\frac{\partial v_m}{\partial x_n})e_n e_m$, and not $T_{ij}e_ie_j:(\frac{\partial v_m}{\partial x_n})e_m e_n$? – megamence Feb 05 '21 at 02:13
  • If the above is the case, I get $T_{ij}\frac{\partial v_i}{\partial x_j}$ for $T:\nabla v$, and that results in a mismatch of indices... – megamence Feb 05 '21 at 02:25
  • @megamence I think the issue is just as $e_i:e_j=\delta_{ij}=e_i^Te_j$ (sort of),$$e_ie_j:e_ne_m=(e_ie_j)^Te_ne_m=e_j^Te_i^Te_ne_m=e_j^T\delta_{in}e_m=\delta_{in}\delta_{jm}.$$ – J.G. Feb 05 '21 at 07:40
0

Okay, I figured it out...

I was making a mistake when taking the divergence of the tensor.

$$\nabla \cdot T = e_j \cdot \frac{\partial (T_{kl} e_k e_l)}{\partial x_j} \neq \frac{\partial (T_{kl} e_k e_l)}{\partial x_j} \cdot e_j$$

This resolves the issue

J.G.
  • 115,835
megamence
  • 297