Why does $\vec \nabla \times \vec F=0 $ mean that $\exists \Phi: \nabla \Phi=\vec F$?

Question

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.

Answer 1

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.

Answer 2

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.