10
$\begingroup$

Let $\mathbf{v}=(a,b)$ be a smooth vector field on the unit circle $\mathbb{S}^{1}$ such that $a^{2}+b^{2}\neq0$ everywhere in $\mathbb{S}^{1}$ with degree $\deg\mathbf{v}=0$. Suppose also that $\int\limits_{\mathbb{S} ^{1}}a\mathtt{dx}+b\mathtt{dy}=0$. My question is whether the field $\mathbf{v}$ may be extended to a nonzero gradient vector field $\overline {\mathbf{v}}=(A,B)$ on the unit disk $\mathbb{D}$, i.e. whether there exist smooth functions $A=A(x,y)$, $B=B(x,y)$, $\ (x,y)\in\mathbb{D}$, such that $A|_{\mathbb{S}^{1}}=a$, $B|_{\mathbb{S}^{1}}=b$, $A^{2}+B^{2}\neq0$ everywhere in $\mathbb{D}$ and finally $\frac{\partial B}{\partial x} =\frac{\partial A}{\partial y}$ in $\mathbb{D}$.

Let me make some remarks.

  1. The condition $\deg\mathbf{v}=0$ is necessary, for the field $\mathbf{v}$ to have an everywhere nonzero extension in the unit disk. The degree is defined as usually as the degree of $\mathbf{v/}\left\Vert \mathbf{v} \right\Vert $ considered as a map $\mathbb{S}^{1}\rightarrow\mathbb{S}^{1}$.

  2. The condition $\int\limits_{\mathbb{S}^{1}}a\mathtt{dx}+b\mathtt{dy}=0$ is also necessary for $\mathbf{v}$ to have a gradient extension $\overline {\mathbf{v}}$, as following by Green's Theorem.

  3. I suppose that this proposition should have some elegant proof (if true :)) and may be probably something well-known, but I have only few examples, not a proof. So, any references are welcome as well. Note also that this is somehow a "global" proposition, not a "local" one. Thanks in advance.

  • 0
    @Leonid Kovalev: thanks for reopening this discussion... I had forgotten about it, and see that one of the points I didn't discuss in enough detail is actually false.2012-06-16

2 Answers 2