# Can we approximate a vector field on the plane with non-vanishing vector fields in $L^2$?

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 $$\mathbb{D}^2$$?

• Dear Asaf, I knew this came from you, way before seeing your signature. :-) – Giuseppe Negro Nov 13 '19 at 12:20
• 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. – Asaf Shachar Nov 13 '19 at 12:23
• It is a very interesting question, but I am afraid it will not earn enough popularity. – Slup Nov 13 '19 at 13:50
• @AsafShachar: No, that was a reboot. I deleted the community wiki, with the rough idea, and submitted what I think is a proper answer. – Giuseppe Negro Nov 16 '19 at 16:55

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.

• 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. – Giuseppe Negro Nov 17 '19 at 12:35
• @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. – Dap Nov 18 '19 at 10:17

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 $$R(t, x)=(4\pi t)^{-\frac d 2}\int_{\mathbb R^d} R(y)e^{-\frac{|x-y|^2}{4t}}\, dy.$$

• 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. – Giuseppe Negro Nov 16 '19 at 17:27
• 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. – Giuseppe Negro Nov 16 '19 at 17:34
• 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. – Giuseppe Negro Nov 16 '19 at 19:03