Taylor Church

2022-09-06

$\mathrm{\Delta }\left(x\cdot \mathrm{\nabla }u\right)=\mathrm{\Delta }u+x\cdot \left(\mathrm{\nabla }\left(\mathrm{\Delta }u\right)\right)$
Does this hold for $x\in {\mathbb{R}}^{n}$ and $u:{\mathbb{R}}^{n}\to \mathbb{R}$ ? If it does, why does it?
I calculated the above formula by writing out all the variables from ${x}_{1}$ to ${x}_{2}$ and guessing that it must be correct. But I am not familiar with 'direct' calculations using $\mathrm{\nabla }$ and $\mathrm{\Delta }$, so I need some help.

barquegese2

Since there is no standard writing of the derivative of a vector field I don't know of a way to do this in a coordinate free manner. I'm coming to the following result which is different from your formula.
The j in the calculation is fixed, and ${\delta }_{ij}$ is the Kronecker delta. I write ${x}_{i}$ for the ${i}^{\mathrm{t}\mathrm{h}}$ component of x and ${u}_{.i}$ for the partial derivative of the scalar function u with respect to ${x}_{i}$. Then

Doing this a second time we obtain

Summing over j we find that in fact

Note that in the one-dimensional case we have

as obtained in (1)

Do you have a similar question?