1

As I began to teach myself in differential geometry, I finally used to use the Nabla-Operator.

I know and understand its usage as in

$$ \nabla f := \left( \begin{matrix} \frac{∂f}{∂x_1} & \frac{∂f}{∂x_2} & \cdots & \frac{∂f}{∂x_n} \end{matrix} \right)^\intercal $$

but in many books I read a pure definition of $\nabla$: $$ \nabla := \left( \begin{matrix} \frac{∂}{∂x_1} & \frac{∂}{∂x_2} & \cdots & \frac{∂}{∂x_n} \end{matrix} \right) ^ \intercal $$

which seems to be just a visualisation of the content, because it's mathematically false $-$ an equation needs to have two evaluatable terms on both sides, but an operator is not a value.

For example, the derivation operator can conformly be defined as $$ \frac{∂}{∂x_i}: ℝ → ℝ, \quad f ↦ \frac{∂f}{∂x_i} := \lim_{x_i→0}{\frac{f(x_1,\cdots,x_i+h,\cdots,x_n)-f(_1,\cdots,x_i,\cdots,x_n)}{h}} $$

But the Nabla-Operator is applied in multiple ways; therefore, one cannot define it as a function.

Do I suppose rightly that there does not exists an explicit definition, or does there exist some kind of ‘trick’?

Milo Brandt
  • 60,888
Lukas Juhrich
  • 3,365
  • What kind of function is the function $f$ that you want to apply the nabla operator to? – BIS HD Nov 15 '13 at 14:03
  • I don't really want to apply the nabla operator to anything, I just want to define it explicitly (if possible)… but in most cases, $ f: ℝ^3 → ℝ^3 $. – Lukas Juhrich Nov 15 '13 at 14:05
  • 1
    The differentiation operator maps the some space of functions into some other space of functions. It does not make any sense to write: $\frac{\partial}{\partial x_i}:\mathbb{R}\rightarrow\mathbb{R}$. – Sergio Parreiras Nov 15 '13 at 14:27
  • OK, I'm sorry for that, it was just a try to define the partial derivative operator. But the basic question is: How can one define an operator? – Lukas Juhrich Nov 15 '13 at 14:36
  • I don't agree that the "pure" definition is "mathematically false" - the distinction between an operator and a value is a completely artificial one. At best it's possibly a slight abuse of notation; but if you define the action of a vector of operators to be the componentwise action then it's fine. Anyway, what's wrong with taking the definition you gave for the partial derivative and inserting it $n$ times into the first definition you gave for $\nabla$? – Anthony Carapetis Nov 15 '13 at 14:39
  • Your last idea sounds great, but how would it look like? – Lukas Juhrich Nov 15 '13 at 14:57
  • @SergioParreiras How would a ‘space of functions’ be written? Sorry for the question, these are just the small things I have (and want) to improve. – Lukas Juhrich Nov 15 '13 at 15:02
  • 2
    We can define $D: C^k(\mathbb{R})\rightarrow C^{k-1}(\mathbb{R})$ by $D(f)=f^\prime$ where $C^k$ is the set of functions that have a continuous k-th derivative. – Sergio Parreiras Nov 15 '13 at 15:23
  • 2
    Why can't you define $\nabla: C^k(\mathbb{R}^n)\rightarrow \left( C^k(\mathbb{R}^n)\right)^n$ by $\nabla \left(f\right)=(\frac{\partial}{\partial_1}f,\ldots,\frac{\partial}{\partial_n}f)$ ? – Sergio Parreiras Nov 15 '13 at 21:12
  • @SergioParreiras: do you really think that your definition is even well-defined? – Mister Benjamin Dover Dec 27 '14 at 00:27
  • This might be helpful. – JohnD Dec 27 '14 at 01:02
  • @Laters there is a typo should be $C^{k-1}$ in the image but I really think with this correction the definition is OK. What am I missing? – Sergio Parreiras Jan 06 '15 at 23:16

2 Answers2

2

The keyword is Operator Calculus or alternatively Operational Calculus. Here is an introductory (PDF) document . Other references are easily found on the internet, such as Fractional Calculus (Wikipedia), What is operator calculus? (MSE), How to make sense of this calculus notation, Advanced College Level (MSE).

Han de Bruijn
  • 17,070
1

I find this an interesting question, because it appeared to me also when I studied multivariable calculus. First of all, "an equation needs to have two evaluatable terms on both sides, but an operator is not a value" is meaningless (and also false, no matter what meaning you give to "evaluatable terms"). What does this even mean? Also $\nabla:=(\partial_1,\ldots,\partial_n)^T$ is not a mere equation, it is a definition. Your only objection can be to the meaning of the right hand sie.

"But the Nabla-Operator is applied in multiple ways; therefore, one cannot define it as a function. Do I suppose rightly that there does not exists an explicit definition, or does there exist some kind of ‘trick’?" Doesn't make any sense either, in particular "therefore, one cannot define it as a function" is a non sequitur. Counterexample: let $\Omega\subset\mathbf{R}^n$ be open, and $X$ be the set of all functions $\Omega\rightarrow\mathbf{R}$ such that $\partial_i f$ exists for all $i$. Also let $Y$ be the set of all maps $\Omega\rightarrow\mathbf{R}^n$. Then we define $\nabla:X\rightarrow Y$ by $\nabla f:=(\partial_1 f,\ldots,\partial_n f)^T$. It is a perfectly ok definition. You see, in the definition it is UNDERSTOOD (IMPLICIT) how the function $\nabla$ is to be defined (one does not need "operator calculus" or whatever), writing $\nabla:=(\partial_1,\ldots,\partial_n)^T$ already suggests this definition. The symbol $(\partial_1,\ldots,\partial_n)^T$ should not be understood as an element of $\mathbf{R}^n$ (because it isn't one), but in the sense I made precise.

Mister Benjamin Dover
  • 1,120