Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 \begin{equation} D = \sum_{|\alpha| \leq m} a_\alpha(x)\partial^\alpha \end{equation}

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 !

share|improve this question
    
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

2 Answers 2

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.

share|improve this answer

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 $\mathbb{y}$.

share|improve this answer
    
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_wolframAlpha_we_trust Jun 11 '12 at 6:32

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.