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Consider a vector field $F : \mathbf{R}^d \to \mathbf{R}^d$.

I want to know standard conditions on $F$ which guarantee that $F$ is a gradient, i.e. that $F = \nabla V$ for some function $V$.

I know that gradient fields (sufficiently smooth, etc. etc.) all satisfy

$$\frac{ \partial F_i }{ \partial x_j} = \frac{ \partial F_j }{ \partial x_i} \quad \text{for all } i, j$$

and I have it in my mind that this is also a sufficient condition (or pretty close to it), but I don't really trust myself on this, and have no reference to this effect.

Is this the case? Is there another sufficient condition? What is a good reference for this sort of thing?

πr8
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1 Answers1

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If you're considering $C^1$ vector fields $F$ defined on all of $\Bbb{R}^n$, then being a gradient field is equivalent to \begin{align} \dfrac{\partial F_i}{\partial x_j} &= \dfrac{\partial F_j}{\partial x_i} \quad \text{for all $i,j \in \{1, \dots, n\}$}\tag{$*$} \end{align} As you've already mentioned, if $F = \nabla \phi$ is a gradient field (for some $C^2$ function $\phi: \Bbb{R}^n \to \Bbb{R}$) then the above equality is simply true by the equality of mixed partial derivatives.

To prove the converse statement, suppose $F: \Bbb{R}^n \to \Bbb{R}^n$ is $C^1$ and satisfies $(*)$. Now, define a function $\phi: \Bbb{R}^n \to \Bbb{R}$ by \begin{align} \phi(x) &:= \sum_{a=1}^n x_a \int_0^1 F_a(tx) \, dt \end{align} Then, a few lines of calculations (product rule, chain rule, differentiation under integral sign etc) shows that $\nabla \phi = F$. I suggest you first try to to compute it yourself; if you get stuck, look up the proof of Poincare's Lemma online, or let me know, I'll add some more details.

Anyway, note that the place where I used that $F$ is well-defined on all of $\Bbb{R}^n$ is when I used the integral $\int_0^1F_a(tx) \, dt$. In order for this integral to be well-defined, we need to ensure that the entire line-segment from $0$ to $x$ lies in the domain of $F$. So, actually, you can weaken hypotheses on the domain slightly: it is enough to assume that the domain, $U$, of $F$ is open in $\Bbb{R}^n$ and "star-shaped with respect to the origin", which means for every $x \in U$, the line segment from $0$ to $x$ lies in $U$ as well.

By the way, I'm not sure if "star-shaped with respect to origin" is the weakest possible assumption, but this is already a good enough start, so I'll leave it at that.

Also, if you know about differential forms, then you can rephrase your question as follows:

Let $\omega$ be a $C^1$, differential $1$-form defined on an open subset $U \subset \Bbb{R}^n$. Then, what are necessary and sufficient conditions on $\omega$ to ensure that $\omega$ is exact?

The answer is that it is necessary for $\omega$ to be closed ($d \omega = 0$). Poincare's lemma tell's you that a sufficient condition for $\omega$ to be exact is that $U$ be star-shaped (with respect to some particular point in $U$), and that $\omega$ be closed.

peek-a-boo
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  • Thank you - I will work through the calculations you suggest. – πr8 May 20 '20 at 22:08
  • "By the way, I'm not sure if "star-shaped with respect to origin" is the weakest possible assumption" -- clearly, after proving this locally, just for open balls, it can be extended to simply connected open sets. – Alexey May 01 '22 at 07:30
  • @Alexey I don't think it's that clear (we have to prove that the integral of a closed 1-form doesn't depend on the homotopy class of a path; this is of course a classic result but it's not trivial). But I should have definitely mentioned that for 1-forms being simply connected is sufficient. The reason I added the clause about it not being the weakest possible assumption is because even simply connected isn't the weakest possible assumption because there are spaces with $H^1_{\text{dR}}=0$ but $\pi_1\neq 0$ (e.g this post). – peek-a-boo May 01 '22 at 10:33
  • On extending to simply connected open sets. Let $x_0\in U$. For every path $\gamma\colon[0,1]\to U$ with $\gamma(0)=x_0$, let $f_\gamma\colon[0,1]\to\mathbf{R}$ be such that (1) $f_\gamma(0)=0$, (2) in some neighbourhood of every $t_0$, $f_\gamma$ decomposes as $\psi\circ\gamma$ for some $\psi$ defined on an open subset of $U$ such that $\nabla\psi=F$. Clearly, $f_\gamma$ exists. It is easy to show that it is unique. Moreover, if $\gamma_1$ and $\gamma_2$ are homotopic with fixed ends $\gamma_1(0)=\gamma_2(0)=x_0$, $\gamma_1(1)=\gamma_2(1)=x$, then $f_{\gamma_1}(1)=f_{\gamma_2}(1)$. – Alexey May 01 '22 at 11:41
  • Now, let $\phi(x)=f_\gamma(1)$ for an arbitrary path $\gamma\colon[0,1]\to U$ with $\gamma(0)=x_0$, $\gamma(1)=x$. It is easy to check that $\nabla\phi=F$ in $U$. Did I make any mistakes? – Alexey May 01 '22 at 11:45
  • @Alexey well, that's essentially the classic proof I was referring to; but getting all the details sorted correctly takes some effort, which is why I said it's not trivial (a word which is after all highly subjective) – peek-a-boo May 01 '22 at 13:38
  • In fact, i would appreciate if you could show your technique to differentiate your function $\phi$. I differentiated your $\phi$, but it was harder for me than the generalisation to the simply connected case, and i do not think i did it in the most elegant way. – Alexey May 01 '22 at 20:57