Divergence and tensor

Question

I am studying pde's and in an example appear the following expression: $$ div(f\otimes g), $$ where $f,\, g: \mathbb{R}^3 \to \mathbb{R}$.

Does anyone know this notation? What is the explicit form of that expression?

Thanks for your comments.

Answer 1

One possible motivation for the divergence acting on tensors comes from thinking of the divergence as the adjoint of the negative gradient, so in your case if we have a $C^1$ and compactly supported vector field $X: \mathbb{R}^3 \to \mathbb{R}^3$, then $$ \int_{\mathbb{R}^3} \nabla X : f \otimes g = \int_{\mathbb{R}^3} \sum_{i,j=1}^3 (\nabla X)_{ij} (f\otimes g)_{ij} = \int_{\mathbb{R}^3} \sum_{i,j=1}^3 \partial_j X_i f_i g_j \\ = \int_{\mathbb{R}^3} \sum_{i,j=1}^3 -X_i \partial_j(f_i g_j) := - \int_{\mathbb{R}^3} X \cdot \text{div}(f\otimes g). $$ Hence we take $\text{div}(f\otimes g) : \mathbb{R}^3 \to \mathbb{R}^3$ to be the vector field with $i^{th}$ component $$ (\text{div}(f\otimes g))_i = \sum_{j=1}^3 \partial_j(f_i g_j). $$ This generalizes nicely with the same motivation: for a map $T: \mathbb{R}^n \to \mathbb{R}^{m \times n}$ we define $\text{div}T: \mathbb{R}^n \to \mathbb{R}^m$ via $$ (\text{div}T)_i = \sum_{j=1}^n \partial_j T_{ij} \text{ for } 1 \le i \le m. $$ We can think of this as applying the usual divergence operator along each row of $T$ and putting the resulting $m$ scalars into an $m-$vector.