# Product rule for divergence of T grad phi

It is easy to verify that

$$$$\nabla \cdot \beta(\mathbf x) \nabla \phi = \beta(\mathbf x) \nabla^2 \phi + \nabla \beta(\mathbf x) \cdot \nabla \phi$$$$

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

$$$$\nabla \cdot T(\mathbf x) \nabla \phi = ... ??$$$$

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.

• 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. Nov 5, 2023 at 8:33
• 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. Nov 5, 2023 at 16:22

$$\nabla\cdot T\nabla\phi = \partial^i\left(T_{ij}\partial^j\phi\right) = \left(\partial^iT_{ij}\right)\partial^j\phi + T_{ij}\partial^i\partial^j\phi = \left(\nabla\cdot T\right)\cdot\nabla\phi + T:\nabla^2\phi,$$
Where I used that the divergence operates on the first index of the tensor $$T$$ and $$\nabla^2\phi$$ denotes the Hessian of $$\phi$$.
• 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. Nov 7, 2023 at 12:39
• 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. Nov 7, 2023 at 15:17