Prove: $\nabla\cdot(\mathbf{W^TA)}=\mathbf{A}\cdot(\nabla\cdot \mathbf{W})+\mathbf{W}\cdot(\nabla \mathbf{A})$

Where $\mathbf{A,W}$ are a vector and second order tensor field respectively.

I am having trouble choosing the indices for the L.H.S. in order to change the terms in the parenthesis into something I know how to write a divergence for such as $\nabla\cdot\mathbf{A}=\partial_i\mathbf{A}_i$.

  • $\begingroup$ Have you tried using $\mathbf{W^TA}=W_jA_{ji}$? $\endgroup$ – MasterYoda Sep 12 '17 at 6:03
  • $\begingroup$ Using $\mathbf{W^TA}=W_jA_{ji}$ leads to $\partial_k (W_jA_{ji})_k$ on the left hand side. The R.H.S. (I think) would then be $\mathbf{A_k(\partial_i \cdot W_{ij})_k + W_{ij} \cdot (\partial A_i)}_j$. Going from there I am unsure how to compound some of the terms on the RHS to get the equation right. $\endgroup$ – asuperbaname Sep 12 '17 at 6:19
  • $\begingroup$ Now I'm confused. I'm not sure if you are using your indices correctly. Which is your tensor, $\mathbf{W}$ or $\mathbf{A}$? Also, is there a typo in your problem statement? In particular, the first term on the RHS? $\endgroup$ – MasterYoda Sep 12 '17 at 7:33
  • $\begingroup$ That were many typos in the problem sorry, they should be fixed now. I'm not entirely sure I'm using indices correctly. W is the 2nd order tensor; A is the vector. $\endgroup$ – asuperbaname Sep 12 '17 at 8:17
  • $\begingroup$ Okay, then you should use $\mathbf{W}=W_{ij}$ and $\mathbf{A}=A_i$. See my answer below. $\endgroup$ – MasterYoda Sep 12 '17 at 16:37

Under traditional Einstein summation notation, the proof is as follows: $$\begin{align} \nabla\cdot\left(\mathbf{W^TA}\right)&=\nabla\cdot\left(\mathbf{A^TW}\right)=\partial_j\left(A_iW_{ij}\right)=A_i(\partial_jW_{ij})+W_{ij}\left(\partial_jA_{i}\right)\\&=\mathbf{A}\cdot\left(\nabla\cdot\mathbf{W}\right)+\mathbf{W}\cdot\left(\nabla\mathbf{A}\right) \end{align}$$

  • $\begingroup$ What is the property that let you change $\mathbf{W^T A}$ to $\mathbf{A^T W}$? Then in your next step, the expansion of the indicies is simply the product rule for partial derivatives correct? $\endgroup$ – asuperbaname Sep 13 '17 at 0:54
  • $\begingroup$ Notice that the quantity is a vector. Who cares if it is a row vector or a column vector? They contain the same components within: $\mathbf{W^TA} = \left(\mathbf{W^TA}\right)^T=\mathbf{A^T\left(W^T\right)^T}=\mathbf{A^TW}$. And yes, it is the product rule :) $\endgroup$ – MasterYoda Sep 13 '17 at 9:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.