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I've been wondering why can we always find a scalar field $\Phi$ for any vector field: $\vec F$ being continuous on $\Bbb R^3$ and satisfying the condition: $\vec\nabla\times\vec F=(\frac{\partial F_z}{\partial y}-\frac{\partial F_y}{\partial z})\vec i+(\frac{\partial F_x}{\partial z}-\frac{\partial F_z}{\partial x})\vec j+(\frac{\partial F_y}{\partial x}-\frac{\partial F_x}{\partial y})\vec k=0 $ such that $\vec F$ can be expressed as $\nabla \Phi=\vec F$ ? I am looking for an algebraic proof. Any help is appreciated.

Edit:I am looking for a calculus-like proof which will emerge from calculations related to the three relations that we get by the fact that the field has curl equal to zero everywhere.

Dan Asimov
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Kani Pen
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  • this is true, for example, on any star shaped open region; this includes the entire space. In Spivak's little book it is called the Poincare Lemma; page 94 – Will Jagy Jun 01 '23 at 21:09
  • https://math.stackexchange.com/questions/2798424/why-does-vec-nabla-times-vec-a-0-imply-vec-a-nabla-b?noredirect=1&lq=1 – A. Goodier Jun 01 '23 at 21:10
  • @A.Goodier Its really the same question but I am looking for a proof that will consider the three relations that emerge from the fact that the field is curless and will give the result requested. – Kani Pen Jun 01 '23 at 21:14
  • https://math.stackexchange.com/questions/638099/why-curl-free-field-implies-existence-of-potential-function The key of the proof is Stokes theorem, which is a form of partial integration. – Gribouillis Jun 01 '23 at 21:25
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    Note that whether this is true depends on the domain of $\mathbf{F}$. For example, $\mathbf{F}(x,y,z) = \frac{-y}{x^2+y^2} \mathbf{i} + \frac{x}{x^2+y^2} \mathbf{j}$ has zero curl, but there is no function $\Phi$ defined on all of $\mathbb{R}^3$ minus the $z$-axis with $\nabla\Phi = \mathbf{F}$. – Daniel Schepler Jun 01 '23 at 22:02
  • @Gribouillis I disagree. Stokes’s Theorem is hardly involved at all. It may give a theoretical justification of path-independence (which has some nontrivial topology buried in it), but one just writes down explicit integrals in the star-shaped region. – Ted Shifrin Jun 01 '23 at 22:32
  • @TedShifrin All right. Do you have a link to an explicit computation? You still need to differentiate the integral to show that $\nabla \Phi = F$. Doesn't this involve Stokes' Theorem, perhaps in an infinitesimal form? – Gribouillis Jun 02 '23 at 05:26

2 Answers2

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The solution appears in many textbooks (including my own). I will do the $2$-dimensional case here, but the $n$-dimensional case is only notationally a bit more complicated. I work on a region that is star-shaped with respect to the origin. (The result is false without some topological restrictions.)

We are given a $C^1$ vector field $\vec F = (P,Q)$ with $\dfrac{\partial Q}{\partial x} = \dfrac{\partial P}{\partial y}$. We consider the path $\vec g(t) = t(x,y)$, $0\le t\le 1$, from $(0,0)$ to $(x,y)$. Then $\vec g{}'(t)=(x,y)$. Define $$\Phi(x,y) = \int_0^1 \vec F(\vec g(t))\cdot\vec g{}'(t)\,dt = \int_0^1 \big(P(tx,ty)x + Q(tx,ty)y\big)dt.$$ Now let's calculate $\dfrac{\partial\Phi}{\partial x}$. We differentiate under the integral sign (1), we use the chain rule and then the curl hypothesis (2), and at the end we use the chain rule (3), integration by parts, and the Fundamental Theorem of Calculus (4): \begin{align*} \frac{\partial\Phi}{\partial x} &\overset{(1)}= \int_0^1 \frac{\partial}{\partial x}\big(P(tx,ty)x + Q(tx,ty)y\big)\,dt \\ &= \int_0^1 \big(P(tx,ty) + tx\frac{\partial P}{\partial x}(tx,ty) + ty\frac{\partial Q}{\partial x}(tx,ty)\big)dt \\ &\overset{(2)}= \int_0^1 \big(P(tx,ty) + tx\frac{\partial P}{\partial x}(tx,ty) + ty\frac{\partial P}{\partial y}(tx,ty)\big)dt \\ &\overset{(3)}= \int_0^1 \big(P(tx,ty) + t\frac d{dt}P(tx,ty)\big)dt \\ &\overset{(4)}= \int_0^1 P(tx,ty)\,dt + tP(tx,ty)\Big]_{t=0}^1 - \int_0^1 P(tx,ty)\,dt \\ &= P(x,y), \end{align*} as required. The other partial derivative follows analogously.

Ted Shifrin
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  • Ok, this is not Stokes' theorem, but this is still integration by parts. If you apply Stokes theorem to the closed path $(0,0), (x+\Delta x, y), (x, y), (0,0) $ and you let $\Delta x$ tend to $0$, it also gives the result! – Gribouillis Jun 02 '23 at 06:17
  • @Gribouillis I totally disagree with you on philosophical grounds. Integration by parts is the Fundamental Theorem of Calculus together with the product rule. Green's Theorem in the plane — to which you are appealing — is of course just the Fundamental Theorem of Calculus if your region is nice enough. Moreover, the OP wanted a straightforward "calculus-like proof," so throwing fancy theorems around defeats the purpose. – Ted Shifrin Jun 02 '23 at 06:22
  • If I find the time to revise Stokes theorem, I may be able to discuss this philosophical matter with you. When I was a student, there was a large inscription on a wall of my math laboratory that just read $\int_\Omega d\omega = \int_{\partial \Omega} \omega$. It was a striking sight. – Gribouillis Jun 02 '23 at 06:36
  • @Gribouillis Oh, in that framework it is indeed one of my favorite theorems, and it leads to the Gauss-Bonnet(-Chern) Theorem, which is my favorite! – Ted Shifrin Jun 02 '23 at 16:10
  • @TedShifrin "$\vec g (t)=t(x,y)$" I don't understand that notation and how a path is expressed in that way. I am aware of the notation where $\vec g (t)=(x(t),y(t))$ but the one you use is completely new to me. Can you explain? – Kani Pen Jun 02 '23 at 16:38
  • We’re fixing the point $(x,y)$ and taking the line segment from the origin to that point. – Ted Shifrin Jun 02 '23 at 16:42
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On 3-space minus the z-axis, let a vector field be defined by

V(x,y,z) = (-y,x,0) / (x^2 + y^2).

Then curl(V) = (0, 0, 0) at all (x,y,z) such that (x,y) ≠ (0,0).

But the line integral of V around the unit circle (cos(t), sin(t), 0) for 0 ≤ t ≤ 2π is nonzero, which proves that V cannot be the gradient of any function.

Dan Asimov
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