Is every 2D vector field locally integrable up to a rescaling?

Question

Let $\Omega$ be a region of the plane (simply connected, say, if it matters), and $v(x,y)$ a smooth vector field on $\Omega$.

Does there always exist a scalar field $s:\Omega\to\mathbb{R}$ so that $sv$ is locally integrable: so that for every point $(x_0,y_0)\in\Omega$, in some neighborhood of that point, $$s(x,y)v(x,y) = \nabla f(x,y)$$ for some $f:\Omega\to\mathbb{R}$?

Equivalently, does there always exist an $s$ with $\nabla \times (sv) =0$?

EDIT: As pointed out below, we want $s$ not identically zero, as this is clearly a trivial solution. The below answer also gives a counterexample, though it seems to rely fundamentally on $v$ having a singularity of nontrivial index; is the claim true if we require $v$ to be nowhere-vanishing?

Answer 1

We may w.l.o.g. use the musical isomorphism to reformulate OP's question as follows:

Question: Given a nowhere vanishing smooth one-form $$\omega~=~f\mathrm{d}x+ g\mathrm{d}y~\in~ \Omega^1(\mathbb{R}^2)\tag{1}$$ in the plane$^1$ $\mathbb{R}^2$, does there exist a globally defined smooth integrating factor$^2$ $\lambda$ such that $\lambda \omega $ is an exact one-form $\mathrm{d}h$?

Answer: No, the existence of an integrating factor $\lambda$ is only guaranteed in local neighborhoods.

Counterexample: Ref. 1 shows on p. 53 that the one-form $$\omega~=~f\mathrm{d}x+ g\mathrm{d}y ~=~y^3(1-y)^2 \mathrm{d}x+ (y^3-2(1-y)^2)\mathrm{d}y\tag{2}$$ has no globally defined smooth integrating factor $\lambda$.

Indirect proof: Assume that there exist $\lambda,h\in C^{\infty}(\mathbb{R}^2)$ such that $$\lambda\omega~=~\mathrm{d}h. \tag{3}$$ Restrict to the open strip $\Omega :=\mathbb{R}\times ]0,1[$. Then $$ \frac{1}{f}\frac{\partial h}{\partial x}~=~\lambda~=~\frac{1}{g}\frac{\partial h}{\partial y}, \qquad (x,y)~\in~\Omega. \tag{4}$$ Perform bi-smooth coordinate transformation: $\Omega \to \Omega$ $$ (\widetilde{x},\widetilde{y})~=~\left(x+\frac{1}{y^2}+\frac{1}{1-y},y\right)\qquad\Leftrightarrow \qquad (x,y)~=~\left(\widetilde{x}-\frac{1}{\widetilde{y}^2}-\frac{1}{1-\widetilde{y}},\widetilde{y}\right) .\tag{5}$$ Conclude that $h$ does not depend on $\widetilde{y}$: $$\frac{\partial h}{\partial \widetilde{y}} ~=~\frac{\partial y}{\partial \widetilde{y}}\frac{\partial h}{\partial y}+ \frac{\partial x}{\partial \widetilde{y}} \frac{\partial h}{\partial x} ~\stackrel{(5)}{=}~\frac{\partial h}{\partial y}+ \left( \frac{2}{y^3}- \frac{1}{(1-y)^2} \right) \frac{\partial h}{\partial x} ~\stackrel{(2)}{=}~\frac{\partial h}{\partial y}- \frac{g}{f} \frac{\partial h}{\partial x} ~\stackrel{(4)}{=}~0,\tag{6} $$ i.e. $h$ is only a function of $\widetilde{x}$. Hence $$\lambda ~\stackrel{(4)}{=}~\frac{h^{\prime}(\widetilde{x})}{f}.\tag{7}$$ This leads to the contradiction: $$ 0~\neq~\lambda(x,y=1)~\stackrel{(7)}{=}~\lim_{y\to 0^+} \frac{h^{\prime}(\widetilde{x})}{f(x,y)} ~=~\lim_{\widetilde{x}\to \infty}\widetilde{x}^{3/2} h^{\prime}(\widetilde{x}),\tag{8} $$ vs. $$ 0~\neq~\lambda(x,y=0)~\stackrel{(7)}{=}~\lim_{y\to 1^-} \frac{h^{\prime}(\widetilde{x})}{f(x,y)} ~=~\lim_{\widetilde{x}\to \infty}\widetilde{x}^{2} h^{\prime}(\widetilde{x}), \tag{9}$$ $\Box$

References:

1. J.B. Boyling, Carathéodory's principle and the existence of global integrating factors, Commun. Math. Phys. 10 (1968) 52. Project Euclid.

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$^1$ We chose the plane $\mathbb{R}^2$ as an example of a contractible space. It is easy to come up with examples of closed one-forms that are not exact in non-contractible spaces. (OP indirectly asks if it matters if the region is not simply connected? Yes, it does!)

$^2$ An integrating factor $\lambda$ is by definition nowhere vanishing.

Answer 2

If you take $s=0$ this is always so. Thats cheating, as you may agree. We could ask for $s$ to never be zero, or to not be zero unless $v$ is also zero.

With those restrictions it is false. Let $v= r\,e_\varphi$ on $\Bbb R^2$ in polar coordinates. If you integrate $sv$ along a circle of Radius $R$ centered around zero you will get $$\int_0^{2\pi} R\,s(R\cos\varphi,R\sin\varphi)\,d\varphi.$$ Note that because we are not allowing $s$ to become zero on this circle, that the sign of the integrand must remain constant. For that reason this integral is never zero. However, if $sv=\nabla f$ then the integral of $sv$ around a closed curve must be zero. Since our domain $\Bbb R^2$ is contractible, the global question of whether or not there exists an $f$ with $\nabla f=sv$ is the same as the local question.

Answer 3

As s.harp has pointed out this is not always possible, as illustrated by his example where the integral curves of $v$ swirl around some point. However it is always possible to find such a factor provided $v$ is non-zero. As you pointed out since the domain is simple connected it suffices to find $s$ such that $\nabla \times (sv) =0$, which is equivalent to $$ \partial_y(sv^1)-\partial_x(sv^2)=v^1\partial_ys-v^2\partial_xs+(\partial_yv^1-\partial_xv^2)s=0$$ which we write as $$v^\perp\cdot\nabla s + (\partial_yv^1-\partial_xv^2)s=0$$

This is a linear transport equation for $s$, and can be solved by the method of characteristics. The characteristics are the integral curves of $v^\perp=(-v^2,v^1)$, and its easy to see we can solve it provided these curves don't loop around. So long as $v(x_0,y_0)\neq 0$ then certainly you can solve it in a neighbourhood of $(x_0,y_0)$, and I think probably you can even argue somehow that it can be solved globally provided $v$ is nowhere vanishing (or making some other statement ensuring the existence of a Cauchy surface for the integral curves of $v^\perp$) .

Added in note: For global existence it turns out that a stronger condition than nowhere-vanishing $v$ is needed. See the comments below.