My question is regarding the Kirchhoff plate pde which is the so-called biharmonic operator.
Let $\Omega=[a,b]\times[a,b]\subset \mathbb{R}^2$, and $Z$ be the Hilbert space $Z=L^2[\Omega]$. The operator is given by the following:
\begin{equation}
\mathcal{A}: D(A) \rightarrow Z
\end{equation}
\begin{equation}
\mathcal{A}h= \frac{\partial^4}{\partial x^4}+2\frac{\partial^4}{\partial x^2 y^2}+\frac{\partial^4}{\partial y^4}
\end{equation}
with domain
\begin{equation}
D(\mathcal{A})=\{h\in Z | h\in \mathcal{H}^4(\Omega)\}
\end{equation}
where $\mathcal{H}^4(\Omega)$ is the Sobolev space and consists of all functions in $L^2(\Omega)$ which are four times differentiable and whose derivatives up to the 4th order also lies in $L^2(\Omega)$.
In order to proof the self-adjointness of the operator, I have to show that it's symmetric and (subjective) onto. I've shown the symmetry by using the inner product which is given by the following; \begin{equation} \langle \mathcal{A}x,y \rangle= \langle x, \mathcal{A}y \rangle \end{equation} My problem is about the surjectivity. How can I show that this operator is surjective (in other words the inverse exists)? Note that all the boundary conditions equal zero.
Thanks in advance Fatemeh