I'v seen the regularity results for biharmonic equation: $$ \left\{\begin{array}{l} \Delta^2 u=f, \quad \text { in } \Omega, \\ \left.u\right|_{\partial \Omega}=\left.\frac{\partial u}{\partial \boldsymbol{\nu}}\right|_{\partial \Omega}=0 . \end{array}\right. $$ If $\Omega$ is a convex polyhedroid in $\mathbb{R}^2$, then for any $f\in H^{-1}(\Omega)$, the solution of the equation is in $H^3(\Omega) $ and $\|u\|_{3,\Omega} \leqslant C\|f\|_{-1,\Omega}$.
I hope to consider the elliptic equation like $$ \left\{\begin{array}{l} \Delta( \beta(x,y) \Delta u)=f, \quad \text { in } \Omega, \\ \left.u\right|_{\partial \Omega}=\left.\frac{\partial u}{\partial \boldsymbol{\nu}}\right|_{\partial \Omega}=0 . \end{array}\right. $$ Still suppose that $\Omega$ is a convex polyhedroid in $\mathbb{R}^2$, and assume that $\beta\geq \beta_{1}\geq 0$, and $\beta \in C^{\infty}$, can I prove that $\|u\|_{3,\Omega} \leqslant C\|f\|_{-1,\Omega}$?