Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
@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. –  Sam Lisi Jun 16 '12 at 8:56

2 Answers 2

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:

enter image description here

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 $C=0$:

enter image description here

Now, observe that:

  1. The absolute minimum of $f$ on $\mathbb{S}^1$ occurs when $\theta=0$, i.e. at the point $(1,0)$.

  2. 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.

share|improve this answer
    
Thanks for giving a nice counterexample to what I claimed (in step 1). I'll edit my answer to point out where the problem shows up. –  Sam Lisi Jun 16 '12 at 8:57
    
Hi Jim! :) Thanks for settling this question. –  user31373 Jun 16 '12 at 16:11
    
I'm happy to help. This was a neat question! Thanks also to Sam Lisi for his answer. –  Jim Belk Jun 16 '12 at 18:57

EDIT: My original answer is wrong -- I addressed a slightly different question.

I originally wrote, "If you are still interested in the question, I would suggest reformulating it slightly as two different questions." That's not actually correct. Here are the two questions I originally suggested:

(1) Suppose you have a gradient vector field of a function $f$ on the disk, $\nabla f$ non-vanishing on the boundary and of degree $0$. Then, you can deform the function $f$, keeping it constant near the boundary of the disk, so that it has no critical points.

(2) Suppose you have a 1-form $\alpha = adx + b dy$, defined on the boundary of the disk, satisfying $\int_{S^1} \alpha = 0$. Then, you can extend $\alpha$ to a closed 1-form on the disk. This is therefore exact since we are on the disk. Then, of course, by taking the dual vector field, we are in the setting to apply (1).

The problem is that point (1) does not actually address the question of keeping the gradient of $f$ fixed on the boundary. That's not actually possible, as Jim Belk's example shows -- his gradient defined along the boundary points inwards at the boundary maximum, showing that there must be an interior max that we cannot deform away.

The argument I sketch below allows us to keep the function values fixed along the boundary, but requires us to let the gradient change.

With these caveats now, the rest of what I wrote before is correct, so I will leave it at least until I have time to illustrate the difference between the two problems.

I would prove (1) by using a Morse critical point cancellation. First, perturb $f$ so that it has non-degenerate critical points.
Then, because the degree of $\nabla f$ is zero, the sum of the degrees calculated on a small circle around each zero of $\nabla f$ in $D$ is also zero. Now note that at a local minimum or a local maximum, the corresponding degree is $+1$ and at a saddle, the degree is $-1$. This means we can pair each of {min or max} with a saddle and then cancel them pairwise.
There may be an easier proof, but I can't think of one right now.

I would prove (2) by first extending $\alpha$ to any 1-form on the disk. To make our life easier later, extend $\alpha$ to a closed $1$-form in a neighbourhood of the circle. This can be done easily in polar coordinates near the circle. Now extend to the disk by an arbitrary 1-form. Call this extension $\beta$. (Note that we can do this without any condition on $\alpha$.)

Now, we want to show that $d\beta = d \nu$ for a $1$-form $\nu$ with compact support. We may write $d\beta = g dx \wedge dy$. The function $g$ then has compact support in the interior of the disk. There are now a couple of ways of showing that there exists a compactly supported 1-form $\nu$ with the desired property.

(One is by showing the pairing between de Rham cohomology with compact support on the disk and homology of $D^2$ rel its boundary is non-degenerate... thus $d\beta$ represents the $0$ class in compactly supported de Rham cohomology, and thus is exact. Another is by solving $\Delta u = g$ in the disk with the boundary condition $u=0$. Then (modulo a sign question) take $\nu = du \circ i$, where $i$ is the standard complex structure on the disk. I'm sure there's a simpler way, by hand. I unfortunately don't have time to think about finding it now.)

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.