Suppose that $u $ is a smooth function real valued function defined on an open neighborhood of the unit disc $ \mathbb{D} $, which satisfies a second order elliptic partial differential equation; \begin{equation*} \mathcal{L}[u] = A \frac{\partial^2 u}{\partial x^2} +2B \frac{\partial^2u}{\partial x \partial y} + C \frac{\partial^2 u}{\partial y^2} = 0. \end{equation*}
Then it is quite straightforward to verify that the vector field \begin{equation*} F = \begin{pmatrix} \frac{\partial u }{\partial x} \frac{\partial u }{\partial y} \end{pmatrix} \begin{pmatrix} B & -A \\ C & -B \end{pmatrix} \end{equation*} is conservative, therefore there exists some $v$ such that $\nabla v = F $. The potential $v$ is defined up to a constant, so if we normalize so that, lets say, $ v(0,0) = 0 $ we obtain a linear transformation $H_\mathcal{L}u := v$. When $\mathcal{L}$ is the Laplace operator $H_\Delta$ is just the ``Hilbert transform''.
Question: Is it true that for $1<p<\infty$ there exists $C=C(p,\mathcal{L})>0$ such that \begin{equation*} \int_{0}^{2\pi}| H_\mathcal{L}u(e^{i\theta})|^p d\theta \leq C \int_0^{2\pi} |u(e^{i\theta})|^p d\theta, \end{equation*} for all such $u$ ? More precicely, is it the case that the transformation $ u|_\mathbb{\partial\mathbb{D}} \mapsto H_\mathcal{L}u|_\mathbb{\partial \mathbb{D}} $ is a Fourier multiplier ?
