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Let take a square constant matrix $A\in \mathbb{R}^{d\times d}$, and a vector field $v:\mathbb{R}^d \to \mathbb{R}^d$. Is there a way to write the divergence of their product in terms of the divergence of $v$?

That is can we write

$$ \text{div}(Av) $$ as something involving $A$ and $\text{div}(v)$?

EditPiAf
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jay
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    This question is a special case of https://math.stackexchange.com/questions/2625185/divergence-of-matrix-vector-product If you take $A$ to be constant. – J.V.Gaiter Dec 04 '20 at 16:48
  • In a sense, the answer is no. It is possible that $\nabla \cdot (Av)$ depends on derivatives that do not appear in the sum that defines the divergence – Ben Grossmann Dec 04 '20 at 16:52
  • One perspective is as follows: The divergence of $v$ is the trace of the Jacobian matrix $dv$ of $v$. We compute the Jacobian matrix of $Av$ to be $$ d(Av) = A dv, $$ and the trace of this product is the divergence that we're looking for – Ben Grossmann Dec 04 '20 at 16:53

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Using index notation, divergence is interpreted as the trace of the Jacobian matrix of a vector i.e.

$$ \operatorname{div} \mathbf{v} = \partial_i v^i$$

with Einstein summation convention. Therefore we can correctly interpret the divergence of the product as follows

$$\partial_i(A_j^i v^j) = (\partial_i A_j^i)v^j + A_j^i (\partial_i v^j)$$

The object on the left is the dot product of $\mathbf{v}$ with a vector whose elements are the divergences of the column vectors of $\mathbf{A}$. The object on the right is the "matrix dot product" (element wise multiplication then adding) of $\mathbf{A}$ with the Jacobian matrix of $\mathbf{v}$.

Ninad Munshi
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