# how to understand the differential operator acting on functions that are not scalar

Quite often these days I find myself in a situation where I'd like to understand differential operators. One bit that is particularly subtle to me at the moment is how a differential operator is to be understood when it is supposed to act on vector - valued, or matrix - valued functions.

For example, suppose we are given a general linear partial differential operator $$D = \sum_{|\alpha| \leq m} a_\alpha(x)\partial^\alpha$$

where $\alpha = (\alpha_1, \dots \alpha_n)$ denotes a multi-index, $m$ is some positive integer, $x \in \mathbb{R}^n$, $\partial^\alpha := \partial^{|\alpha|}/(\partial^{\alpha_1} x_1 \dots \partial^{\alpha_n} x_n)$ denotes a mixed partial derivative, and the functions $a_\alpha$ are smooth. In various contexts they might be vector- or matrix valued. This is already where I am having difficulties, because usually it is assumed the reader knows how to apply these operators, and from this I guess one could deduce what kind of functions these $a_\alpha$ are ..

How is such an operator supposed to act on vector - valued or matrix valued functions $f : \mathbb{R}^m \to \mathbb{R}^k$ or $F: GL(n,\mathbb{R}) \to GL(k,\mathbb{R})$ ?

Unfortunately my Calculus classes didn't cover much beyond the one - variable setting so I am shaky on these grounds. I am aware there are differential operators for non - scalar functions, such as div, curl, grad. All of these act in a specific way. But the operator above is none of these so I am a bit lost ..

Sorry for being so confused about this - in case the question is unclear I am happy to try my best and improve the post, many thanks !

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Componentwise. If $f=(f_1 \ldots f_n)$ then $Df$ usually means $(Df_1 \ldots Df_n)$. If coefficients $a_\alpha$ are matrices, then $a_\alpha \partial^{\alpha}(f_1 \ldots f_n)=a_\alpha (\partial ^{\alpha}f_1 \ldots \partial^{\alpha}f_n)$. In $\mathbb{R}^n$ it is that simple. Things get worse when you move into more complicated manifolds. –  Giuseppe Negro Apr 10 '12 at 17:47

If $V$ and $W$ are vector spaces and $f:V \rightarrow W$ is a $C^\infty$ funtion, then the derivative $Df: V \rightarrow \mathrm{Hom}(V,W)$ can be regarded as a function that takes an element $v \in V$ as input and returns a linear map from $V$ (thought of its own tangent space at $v$) to $W$ (thought of as the tangent space to $W$ at $f(v)$.

The map $Df$ is now $C^\infty$ and the process can be repeated (thinking of $\mathrm{Hom}(V,W)$ as the new $W$). The whole procedure globalizes to manifolds with the obvious changes: the derivative becomes a map between tangent bundles.

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In the case of $$f : \mathbb{R}^m \to \mathbb{R}^k,$$ the answer is that the value of $f'(\mathbb{x})$ is the "Jacobian matrix". Suppose $$\mathbb{y}=\begin{bmatrix} y_1 \\ \vdots \\ y_k \end{bmatrix} = f(\mathbb{x}) = f\left(\begin{bmatrix} x_1 \\ \vdots \\ x_m \end{bmatrix}\right).$$ The Jacobian matrix is $$J= \begin{bmatrix} \frac{\partial y_1}{\partial x_1}, & \ldots, & \frac{\partial y_1}{\partial x_m} \\ \vdots & & \vdots \\ \frac{\partial y_k}{\partial x_1}, & \ldots, & \frac{\partial y_k}{\partial x_m} \end{bmatrix} \in \mathbb{R}^{k\times m}.$$ This is a $k\times m$ matrix. The idea is that if $$d\mathbb{x} = \begin{bmatrix} dx_1 \\ \vdots \\ dx_m \end{bmatrix} \in \mathbb{R}^{m\times 1}$$ is an infinitesimal change in $x$, then $$d\mathbb{y} = J\,d\mathbb{x} \in \mathbb{R}^k$$ is the corresponding infinitely small change in $\mathbb{y}$.
hm .. ok, but then thing I still have trouble with is - what kind of object then is a higher - order derivative of f ? That is, where $|\alpha| > n$ ? –  harlekin Apr 10 '12 at 17:56
If $f:\mathbb{R}^{m} \rightarrow \mathbb{R}$ the second order derivative of $f$ is the Hessian matrix (en.wikipedia.org/wiki/Hessian_matrix). When $f:\mathbb{R}^{m} \rightarrow \mathbb{R}^{k}$ this generalises to a third order tensor. –  in_wolfram_we_trust Jun 11 '12 at 6:32