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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 $\nabla\cdot\mathbf{A}=\partial_i\mathbf{A}_i$.

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

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  • $\begingroup$ 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

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