# Question on the Laplacian operator of a vector field

Let $\;u:\mathbb R^n \rightarrow \mathbb R^m\;$. I want to know what the Laplacian operator of $\;u\;$ will be like. Searching on the web, I found the following formula:

$\;{\nabla}^2 u=\nabla ( \nabla \cdot u)-\nabla \times (\nabla \times u)\;$

which I have trouble applying on $\;u\;$ for several reasons. For example, I don't know how to compute the gradient of the matrix $\; \nabla \cdot u \;$.

I know $\; \nabla \cdot u=\begin{pmatrix} \frac{\partial u_1}{\partial x_1} \dots \frac{\partial u_1}{\partial x_n} \\ \dots \\ \frac{\partial u_m}{\partial x_1} \dots \frac{\partial u_m}{\partial x_n}\\ \end{pmatrix}\;$

but I have no clue how should I compute the gradient of the above.

I've only seen the Laplacian operator for these kind of functions: $\;f:\mathbb R^n \rightarrow \mathbb R\;$, so I 'm a bit lost now.

Any help would be valuable. Any suggestions at books that I could study from, are also welcome.

• Is $m$ equal to $n$? The symbol $\cdot$ usually denotes inner product, in this case divergence. What are your sources ("the web")? – Michał Miśkiewicz Aug 19 '17 at 16:45
• @MichałMiśkiewicz No, $m$ isn't necessary equal to $n$. I found this formula here: en.wikipedia.org/wiki/Vector_Laplacian – kaithkolesidou Aug 19 '17 at 16:59
• The cross product $v \times w$ of two vectors is possible only in $\mathbb{R}^3$, so it seems like $n=m=3$ here. The formula you have looks similar to Laplace-Hodge operator on $1$-forms (or $2$-forms) on $3$-dimensional manifolds - you can find it e.g. in Jurgen Jost's Riemannian geometry and geometric analysis. – Michał Miśkiewicz Aug 19 '17 at 17:05