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Let $V$ be a compactly-supported smooth vector field on $\mathbb{R}^2$, whose zeros inside some open neighbourhood of the closed unit disk $\mathbb{D}^2$ are isolated.

Does there exist a sequence of vector fields $V_n \in C^\infty \cap L^{2}$ on $\mathbb{R}^2$, such that $V_n \to V$ in $L^2$ and $V_n$ do not vanish on $\mathbb{D}^2$?

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  • $\begingroup$ Dear Asaf, I knew this came from you, way before seeing your signature. :-) $\endgroup$ – Giuseppe Negro Nov 13 '19 at 12:20
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    $\begingroup$ Lol:) Yeah, this is indeed similar to other questions I have asked before, about approximation of singular objects with non-singular objects. Most of these questions have been remained unanswered...However, this one seems simpler, so I hope there is a chance. $\endgroup$ – Asaf Shachar Nov 13 '19 at 12:23
  • $\begingroup$ It is a very interesting question, but I am afraid it will not earn enough popularity. $\endgroup$ – Slup Nov 13 '19 at 13:50
  • $\begingroup$ @AsafShachar: No, that was a reboot. I deleted the community wiki, with the rough idea, and submitted what I think is a proper answer. $\endgroup$ – Giuseppe Negro Nov 16 '19 at 16:55
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I'll assume the zeroes of $V$ are isolated on an open neighborhood of the unit disk.

Here is a procedure for approximating $V$ in $L^2$ by a vector field in the same class - $C^\infty_c$ with isolated zeros in a neighborhood of the unit disk - but with one fewer zero in $\mathbb D^2.$ The idea is to push the zero out. (This is basically what I meant in the linked answer by "composing with a suitable diffeo".)

Pick points $(x_0,y_0)$ and $(x_1,y_1)$ such that:

  • $(x_0,y_0)\in\mathbb D^2.$
  • $(x_1,y_1)\not\in\mathbb D^2.$
  • $V(x_0,y_0)=(0,0).$
  • The straight line segment from $(x_0,y_0)$ to $(x_1,y_1)$ contains no zeroes of $V$ except $(x_0,y_0).$

Let $C$ be the straight line segment from $(x_0,y_0)$ to $(x_1,y_1).$ Let $C_n$ be a sequence of open neighborhoods of $C$ such that $$\mu(C_n)\to 0$$ where $\mu$ is Lebesgue measure, and such that $C_n$ contains no zeros of $V$ except $(x_0,y_0).$ Using Whitney's extension theorem pick a function $s:\mathbb R^2\to\mathbb R$ such that:

  • $s(x,y)=1$ for $(x,y)\in C$
  • $s(x,y)=0$ for $(x,y)\not\in C_n$

Since $s$ is compactly supported, $(x,y)\mapsto s(x,y)(x_1-x_0,y_1-y_0)$ is a complete vector field and defines a flow globally: there are diffeomorphisms $\psi_t$ for $t\in\mathbb R$ where $\psi_0$ is the identity and $\frac{d}{dt}\psi_t(x,y)=s(x,y)(x_1-x_0,y_1-y_0).$ Define $$V_n=V \circ \psi_{-1}.$$ Since $$\|V-V_n\|_2^2\leq \mu(C_n)(2\max|V|)^2$$ we have $V_n\to V$ in $L^2.$

For $(x,y)\in C_n$ we have $V_n(x,y)=(0,0)$ if and only if $\psi_{-1}(x,y)=(x_0,y_0),$ which is equivalent to $(x,y)=(x_1,y_1).$ And outside $C_n,$ the vector fields $V_n$ and $V$ are the same. So the number of zeros inside the unit disk has decreased by one.

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  • $\begingroup$ Great answer. Thank you for writing it. I have a question, if you don't mind. Using this procedure, could you "push the zero off to infinity"? That could be an idea to approach this follow-up question. Thank you again. $\endgroup$ – Giuseppe Negro Nov 17 '19 at 12:35
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    $\begingroup$ @GiuseppeNegro: precomposing with a bijection can't work of course, but trying to precompose with an suitable injection $\mathbb R^2\to \{(x,y)\mid V(x,y)\}$ might be a good approach. I think the smoothness is a bit of a red herring - smooth functions should be dense in continuous functions in pretty much any sense you want. $\endgroup$ – Dap Nov 18 '19 at 10:17
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EDIT. Step 2 is wrong. I asked a separate question on this matter.

I submit that the answer is affirmative, with the $L^2$ convergence. The idea comes from a property of the heat equation, according to which, if a function has a local minimum, then letting it evolve according to the heat equation will fill in such minimum. We are going to apply this idea to the modulus $\lvert \vec{V}\rvert$ of the given vector field, whose minima are precisely the zeros of $\vec V$.


Step 1. This solves the approximation problem, but produces a non-smooth vector field at the zeros of $\vec V$. I attempted to solve the smoothness issue in the forthcoming Step 2, which however contains an error.

Assume that $\vec{V}$ is continuous and $\vec{V}\in L^2(\mathbb R^d; \mathbb R^d)$, and that $$Z:=\{x\in \mathbb R^d\ :\ \vec V(x)=0\}$$ is a set of measure zero; this is actually a weaker assumption than requested. Write $$ \vec{V}(x)=R(x)\omega(x), \quad \text{where }R(x):=\lvert \vec V(x)\rvert,\text{and } \omega(x)\in \mathbb S^{d-1}.$$ We remark that the definition of $\omega(x)$ is ambiguous on $Z$.

Now let $R(t, x)$ be the unique solution to $$ \begin{cases}  \partial_t R =\Delta R, & t>0, x\in\mathbb R^d,\\ R(0, x)=R(x). \end{cases}$$ Since $R(x)\ge 0$, by the minimum principle $R(t, x)>0$ for all $t>0$.$^{[1]}$ Moreover, $R(t, x)\to R(x)$ both pointwise and in $L^2$ sense. By all these considerations, the time-dependent vector field $$ \vec V(t, x):=R(t,x)\omega(x)$$ vanishes nowhere and converges to $\vec V$ as $t\downarrow 0$, pointwise and in $L^2$ sense.


Step 2. (wrong) The function $\vec{V}(t, x)$ of the previous step needs not be continuous for $t>0$ and $x\in Z$, where it will point in one of the "ambiguous" directions of $\omega(x)$.

To circumvent this difficulty, we introduce $\omega(t, x)$, defined as follows. Consider $\omega(x)\in\mathbb S^{d-1}$ as a vector in $\mathbb R^d$, and let $$\eta\colon [0, \infty)\times \mathbb R^d\to \mathbb R^d$$ be the unique solution to the vector-valued heat equation $$ \begin{cases} \partial_t \eta = \Delta \eta, &t>0, \\ \eta(0, x)=\omega(x). \end{cases}$$ This makes sense, because $\omega\in L^\infty(\mathbb R^d; \mathbb R^d)$.

Now, since $\omega(x)\ne 0$ at all $x\in\mathbb R^d$, and since $\eta$ is continuous in $t$, there is a $\delta >0$ such that $\eta(t,x)\ne 0$ for all $t\in [0, \delta]$ and $x\in \mathbb R^d$. (Warning: this needs not be true; for example, consider $\omega(x)=x/\lvert x \rvert$. See the follow-up question.)

It makes thus sense to define $$ \omega(t, x):=\frac{\eta(t, x)}{\lvert \eta(t, x)\rvert}, \qquad t>0.$$ Now, by a standard property of the heat equation known as instantaneous smoothing effect, the function $\eta(t, \cdot)$ is smooth for all $t>0$. Therefore, $\omega(t, \cdot)$ is also smooth.

Conclusion. By the same argument as in Step 1, the function $$\vec V(t, x):= R(t, x) \omega(t, x),\qquad t\in [0, \delta], $$ is such that $\vec V(t, x)\ne 0$ for all $t>0$ and $x\in \mathbb R^d$, and it converges to $\vec V(x)$, both pointwise and in the $L^2$ sense, as $t\downarrow 0$. (It is possible that this convergence can be upgraded with further regularity assumptions on $\vec V$). Also, $\vec V(t, \cdot)$ is smooth for all $t\in (0, \delta)$.


[1] Actually, this can also be seen as an immediate consequence of the explicit formula $$R(t, x)=(4\pi t)^{-\frac d 2}\int_{\mathbb R^d} R(y)e^{-\frac{|x-y|^2}{4t}}\, dy.$$

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  • $\begingroup$ Thank you for your interest! If you want smoothness, I suspect that this construction is incomplete, because it relies on arbitrary choices of the directions $\omega(x)$ at $x$ such that $\vec V(x)=0$. The complete construction must prescribe an evolution $\omega(t, x)$. The difficulty is that now this is a $\mathbb R\times \mathbb R^d\to \mathbb S^{d-1}$ map. I know there is a manifold-valued version of the heat equation, and maybe that's going to work here. This enters into serious mathematics. $\endgroup$ – Giuseppe Negro Nov 16 '19 at 17:27
  • $\begingroup$ Concerning the minimum/maximum principles; yes, you can prove these things in A LOT of different ways. The book of Evans uses the parabolic mean value formula. For classical/smooth solutions you can use the equation. And finally, for weak solutions, you can use measure theoretic tricks. I am not an expert on this but if you need further information let me know. Anyway, I think that the standard reference is the book "Partial differential equations of parabolic type" of Avner Friedman. $\endgroup$ – Giuseppe Negro Nov 16 '19 at 17:34
  • $\begingroup$ I have included a complete solution, but I have a feeling that I am on shaky ground. Step 1 is correct but of course the true difficulty comes from Step 2, and I am not sure that my answer is entirely right, I have added a WARNING sign. I have asked a follow-up question. $\endgroup$ – Giuseppe Negro Nov 16 '19 at 19:03

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