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What are the necessary and sufficient conditions on a vector field $F$ for the divergence $\nabla\cdot F$ to exist at a given point.

EDIT

In Divergence in the second line under the heading "Application in Cartesian coordinates", why is it assumed that $\vec{F}$ to be a continuously differentiable vector field ?

EDIT 2

Ideally one would expect each component of $F$ to be differentiable at a given point $\vec{a}$ no matter through which continuous contour you traverse the point $\vec{a}$.

EDIT 3

Or is it that each component of $F$ to be differentiable and the derivative being continuous at a given point $\vec{a}$ no matter through which continuous contour you traverse the point $\vec{a}$.

Rajesh D
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    By definition the divergence of the vector field $\mathbf{F}$ with differentiable components

    $\mathbf{F}=F_{x}(x,y,z)\mathbf{i}+F_{y}(x,y,z)\mathbf{j}+F_{z}(x,y,z)% \mathbf{k}$

    is $\text{div}\mathbf{F=\nabla \cdot F}=\dfrac{\partial F_{x}}{\partial x}+\dfrac{\partial F_{y}}{\partial y}+\dfrac{\partial F_{z}}{\partial z}$

    where $F_{x},F_{y},F_{z}$ are the components of $\mathbf{F}$ in the $xyz-$coordinate system.

     – Américo Tavares Nov 29 '10 at 15:20
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    @Américo: the divergence also has a coordinate-free definition which is given on the Wikipedia article. It is not clear to me that the existence of the coordinate-free limit is equivalent to the existence of your three partial derivatives. (And note that the existence of these three partial derivatives does not guarantee the existence of all directional derivatives, so if you use your definition then the divergence will not always be computable in every coordinate system.) – Qiaochu Yuan Nov 29 '10 at 15:47
  • @Qiaochu: You have raised two strong points: 1) The divergence defined by a limiting process in which a region shrinks down to a fixed point requires the additional condition that $\mathbf{F}$ is continuosly differentiable, if we want the divergence theorem be applied. Without using this theorem I do not see how we could give a coordinate-free definition. 2) Concerning the directional derivatives you are quite right. 3) So, the definition I took from a Calculus book has all the limitations you pointed out. – Américo Tavares Nov 29 '10 at 16:21

1 Answers1

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Perhaps the point is that existence of partial derivatives is not enough for a function of several variables to be differentiable in the multivariable sense, or even necessarily continuous: see this MO question for a standard example of a function of two variables all of whose partial (and even all directional) derivatives exist but is not even continuous at the origin.

[Edit: the following paragraph was added after reading Qiaochu's comment.]

It is also possible for all the partial derivatives in a given coordinate system to exist -- e.g. $\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}$ -- but for directional derivatives in other directions not to exist. (For instance, let $f: \mathbb{R}^3 \rightarrow \mathbb{R}$ be the function which is $0$ at $(x,y,z)$ if at least two of the variables are zero or if all of $x$,$y$,$z$ have rational coordinates and which is $1$ at all other points.) Here the divergence will be defined with respect to the standard coordinate system but not with respect to a different coordinate system.

Anyway, in general a function which has a continuous derivative (or partial derivatives) is much better behaved than a merely differentiable function. Inserting the hypothesis that a function is $C^1$ -- unless you really need to be considering weaker hypotheses than that for a specific application -- is generally a prudent practice.

Community
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Pete L. Clark
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  • @Pete L. Clark : Ideally one would expect each component of $F$ to be differentiable at a given point $\vec{a}$ no matter through which continuous contour you traverse the point $\vec{a}$. any way what do you mean by continuously differentiable ? (just partial derivatives being continuously differentiable may not be good enough.) – Rajesh D Nov 29 '10 at 16:01
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    Continuously differentiable means that the derivative is a continuous function. For functions of several variables, the derivative (or "total derivative") at a point is a linear transformation, not just a number. It is well-known that the continuity of the total derivative is equivalent to each of the partial derivatives in any coordinate system existing and being continuous functions of a single variable. – Pete L. Clark Nov 29 '10 at 16:09
  • @Pete L. Clark : Sorry, I mistook continuously differentiable to be equivalent to being smooth.Thank you for the clarification. I guess you mean to say that total derivative of each of the components of $F$ should be continuous ? Could you please explain meaning of 'partial derivatives in any co-ordinate system'. how many co-ordinate systems are there ? – Rajesh D Nov 29 '10 at 17:03
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    @Rajesh D: No, the total derivative of a function $F: \mathbb{R}^n \rightarrow \mathbb{R}^m$ (which you are calling a "vector field with components $f_1,\ldots,f_n$") at a point $x \in \mathbb{R}^n$ is an $m \times n$ matrix. There are as many coordinate systems as there are choices of basis on $\mathbb{R}^m$ and on $\mathbb{R}^n$. – Pete L. Clark Nov 29 '10 at 18:22
  • Small typo in my last comment: for $F$ as above, there are $m$ components, not $n$. – Pete L. Clark Nov 29 '10 at 18:49