# Is there a free-boundary elliptic estimate in $L^p$?

Let $$\mathbb{D}^2$$ be the closed two-dimensional unit disk. Let $$V \in \Omega^1(\mathbb{D}^2)$$ be a smooth one-form.

Does there exist a constant $$C$$, such that

$$\vert\vert V\vert \vert_{W^{2,p}(\Omega)} \leq C(\vert\vert \Delta V \vert\vert_{L^p(\Omega)} + \vert\vert V\vert\vert_{L^p(\Omega)}),$$

holds for all one-forms $$V$$, where $$C$$ does not depend on the boundary values of $$V$$?

Here $$\Delta=\delta d+d\delta$$ is the usual Hodge Laplacian on differential forms*.

The point is that I don't assume any specific boundary conditions on $$V$$. (So, what is written here for instance does not seem to apply).

Comment:

*I am actually interested in a more general case where $$\mathbb{D}^2$$ is endowed with some smooth Riemannian metric $$g$$, and the Euclidean Laplacian $$\Delta$$ is replaced by the Riemannian one w.r.t $$g$$ (the adjoint $$\delta$$ depends on $$g$$).

No, there is no such $$C.$$ Define $$V_n(x,y)=\mathrm{Re}[f_n(x+iy)]\;dx + \mathrm{Im}[f_n(x+iy)]\;dy$$ where $$f_n(z)=z^n.$$ Then

$$\nabla V=0$$ and

\begin{align*} \|V_n\|_{L^p(\Omega)} &=\left(\int_\Omega |f_n(x+iy)|^p\right)^{1/p}\\ &=\left(\int_\Omega r^{np}\right)^{1/p}\\ &=\left(2\pi \int_0^1 r^{1+np}dr\right)^{1/p}\\ &=\left(\frac{2\pi}{2+np}\right)^{1/p}\\ &\sim c_p n^{-1/p}\text{ for some constant c_p} \end{align*}

But

\begin{align*} \|V_n\|_{W^{2,p}(\Omega)} &\geq c'\left(\int_\Omega \left|\frac{\partial^2 f_n}{\partial x^2}(x+iy)\right|^p\right)^{1/p}\\ &= c'_pn(n-1)\left(\int_\Omega r^{np-2}\right)^{1/p}\\ &\sim c'_pc_p n^{2-1/p} \end{align*} with some other constant $$c'_p$$ depending on how the Sobolev norm is defined. So $$\|V_n\|_{W^{2,p}(\Omega)}\gg \|\nabla V_n\|_{L^p(\Omega)} +\|V_n\|_{L^p(\Omega)}.$$