1
$\begingroup$

It is easy to verify that

\begin{equation} \nabla \cdot \beta(\mathbf x) \nabla \phi = \beta(\mathbf x) \nabla^2 \phi + \nabla \beta(\mathbf x) \cdot \nabla \phi \end{equation}

for scalar valued functions $\beta(\mathbf x)$ and $\phi(\mathbf x)$, for $\mathbf x \in R^d$. Here, $\nabla \cdot$ is a divergence, and $\nabla \phi$ is a gradient.

My question is, what is the equivalent "product rule" if $\beta(\mathbf x)$ is replaced with a spatially varying tensor $T(\mathbf x)$? That is, what is the expansion of

\begin{equation} \nabla \cdot T(\mathbf x) \nabla \phi = ... ?? \end{equation}

where $T(\mathbf x) \in R^{d \times d}$. I have assumed everything is in Cartesian coordinates, so I can expand component-wise to get a scalar expression. However, I would welcome any input on how to formulate the problem using proper tensor notation. I suspect that the way I formulated the problem for scalar $\beta(\mathbf x)$ is misleading, and that in writing the tensor version the way I have, I am relying too much on intuition from linear algebra, which probably isn't helpful.

This post gave me some ideas, but it seems to be a different problem.

$\endgroup$
2
  • 1
    $\begingroup$ Before deriving a new product rule, you need to define the divergence of tensor field; yet, it can be done in several ways (see en.wikipedia.org/wiki/Divergence#Tensor_field). You may work component-wise. $\endgroup$
    – Abezhiko
    Nov 5, 2023 at 8:33
  • $\begingroup$ I don't want to derive a new product rule, just hoping to get some insight into the obvious one for the case I have. I've edited the post a bit to suggest that by writing the tensor case the way I did, I may be relying too much on linear algebra intuition. $\endgroup$
    – Donna
    Nov 5, 2023 at 16:22

1 Answer 1

1
$\begingroup$

The most straightforward way to derive results like this is to work in index notation and then interpret the resulting terms. To that end, notice that (Einstein summation assumed)

$$ \nabla\cdot T\nabla\phi = \partial^i\left(T_{ij}\partial^j\phi\right) = \left(\partial^iT_{ij}\right)\partial^j\phi + T_{ij}\partial^i\partial^j\phi = \left(\nabla\cdot T\right)\cdot\nabla\phi + T:\nabla^2\phi, $$

Where I used that the divergence operates on the first index of the tensor $T$ and $\nabla^2\phi$ denotes the Hessian of $\phi$.

$\endgroup$
2
  • $\begingroup$ Interesting! This is exactly what I was looking for. Is the $\nabla^2 \phi$ conventional notation for a Hessian? I understand it more commonly as a Laplacian. $\endgroup$
    – Donna
    Nov 7, 2023 at 12:39
  • 1
    $\begingroup$ In differential geometry and geometric PDE, $\Delta\phi$ would typically denote the Laplacian, whereas one of $\nabla^2\phi, \nabla d\phi$, or $\mathrm{Hess}\,\phi$ would denote the Hessian. $\endgroup$
    – MathIsArt
    Nov 7, 2023 at 15:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .