Approximate a one-form on the disk with nowhere vanishing one-forms satisfying an asymptotic vanishing of some derivatives

Question

Let $\mathbb{D}^2$ be the closed two-dimensional unit disk, and let $g:\mathbb{D}^2 \to \mathbb{R}$ be a non-constant harmonic function (smooth up to the boundary).

Does there exist a sequence of smooth one-forms $\sigma_n$ on $\mathbb{D}^2$ such that

1. $\sigma_n \to dg$ in $L^2$.
2. $\sigma_n$ do not vanish on $\mathbb{D}^2$.
3. $\|\delta d \sigma_n \|_{L^1} \to 0, \| d \delta \sigma_n \|_{L^1} \to 0$

That is, I want to approximate the one-form $dg$, which may have zeroes in the domain, with nowhere vanishing forms, such that certain second derivatives become negligible in the limit.

I know how to do that without achieving the third condition.

Let me explain a bit further on that condition:

Write $\sigma =dg \in \Omega^1(\mathbb{D}^2)$. Since $g$ is harmonic, $\delta \sigma=0$. Since $\sigma$ is exact, we also have $d\sigma=0$. In fact, for a one-form $\sigma \in \Omega^1(\mathbb{D}^2)$, $d\sigma=\delta \sigma=0$ is equivalent to the existence of a harmonic function $g$ on $\mathbb{D}^2$ such that $\sigma=dg$.

Thus, our desired limit form $\sigma =dg$ satisfies $\|\delta d \sigma \|_{L^1} = \| d \delta \sigma \|_{L^1} =0$.

Hence, if we could approximate $\sigma$ by nowhere-vanishing $\sigma_n$ in $W^{2,1}$, we were done. However, we cannot in general approximate even in $W^{1,1}$ while staying nowhere-vanishing. (See also this comment).

Two more comments:

1. Trying to approximate (in $L^2$) $g$ by harmonic functions $g_n$ in with non-vanishing differential also doesn't work: This creates a too fast approximation, due to rigidity properties of harmonic functions, again hitting the topological obstruction.

2. Even trying to approximate (again in $L^2$) $\sigma=dg$ with $\sigma_n$ satisfying $d\delta \sigma_n=0, \delta d \sigma_n=0$ cannot work: If $\sigma_n=f^1_ndx+f^2_ndy$, then these conditions imply that $f^1_n$ and $f^2_n$ are harmonic, which again implies a too fast convergence.

Answer 1

Your third condition implies

$$ \| (\delta d + d\delta) (\sigma_n - \sigma) \|_{L^1} \to 0 $$

Notice that up to constants, $\delta d + d\delta$ is identical (on the flat disk) to the standard Laplacian acting componentwise.

Since the disk is compact, you also have by your first condition and Holder

$$ \| \sigma_n - \sigma \|_{L^1} \lesssim \|\sigma_n - \sigma\|_{L^2} \to 0 $$

Now, it is well-known that elliptic regularity in $L^1$ generally fails, and so we cannot expect $\| \sigma_n - \sigma\|_{W^{2,1}}$ to go to zero. On the other hand, if we are willing to give up a little bit, it turns out some regularity can be recovered. (Note that since in your case you are just working essentially with the flat Laplacian, you can get the same result just by using the explicit formula of the Green's function on the disk without appealing to the high powered machinery.) In particular, one gets that in your case $ \sigma_n \overset{W^{1,1}}{\longrightarrow} \sigma $. (It maybe necessary to truncate and localize to compacts strictly contained in the unit ball, but this shouldn't matter for your problem.)

Together with your previous question this tells you what you want is unlikely to be possible.