TITLE: Gradient nonzero extensions of a vector field on the circle
QUESTION [10 upvotes]: 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.

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 $\mathbb{S}^{1}\rightarrow\mathbb{S}^{1}$.
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.
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.

REPLY [3 votes]: No, not every vector field of this form can be extended in the way you desire.  In particular, it's sometimes possible to see from the vector field that there must be an interior critical point.
For example, consider the vector field $\mathbf{v}$ on $\mathbb{S}^1$ defined by
$$
\mathbf{v}(\theta) \;=\; \bigl(\sin \theta + 2\sin 2\theta \bigr)\,\mathbf{t}(\theta) \,+\, \bigl(1+2\cos\theta\bigr)\,\textbf{n}(\theta),
$$
where $\mathbf{t}(\theta) = (-\sin\theta,\cos\theta)$ is the counterclockwise unit tangent vector, and $\mathbf{n}(\theta) = (\cos\theta,\sin\theta)$ is the outward-pointing normal vector.  Here is a picture of this vector field:

It is easy to check that the degree of this vector field is zero, and it is clear from the vertical symmetry that $\displaystyle\oint_{\mathbb{S}^1} \!a\,dx+b\, dy = 0$.
Now suppose that $\mathbf{v}$ is the restriction to the unit circle of the gradient of some function $f\colon\mathbb{D}\to\mathbb{R}$.  Using the Gradient Theorem, we can compute the restriction of $f$ to the unit circle:
$$
f(\cos\theta,\sin\theta) \;=\; \int \bigl(\sin\theta + 2\sin 2\theta\bigr)\,d\theta \;=\; -\cos\theta - \cos 2\theta + C.
$$
Here is a plot of $f$ on the unit circle, assuming $C=0$:

Now, observe that:

The absolute minimum of $f$ on $\mathbb{S}^1$ occurs when $\theta=0$, i.e. at the point $(1,0)$.
At the point $(1,0)$, the gradient vector $\textbf{v}$ points directly to the right.

Thus, if $p = (1-\epsilon,0)$ is a point slightly to the left of $(1,0)$, then the value of $f$ at $p$ is less than the value of $f$ anywhere on the unit circle.  Therefore, $f$ obtains its minimum somewhere in the interior on the unit disk, so $f$ must have a critical point.