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}
\begin{split}
D(\mathcal{A})=\{h\in Z | h\in \mathcal{H}^4 \text{with} \\ M_{x}(0,.)={Q}_{x}(0,.)=0,
M_{x}(1,.)={Q}_{x}(1,.)=0,\\ M_{y}(.,0)={Q}_{y}(.,0)=0, M_{y}(.,1)={Q}_{y}(.,1)=0 \}.
\end{split}
\end{equation}
where
\begin{equation}
Q_{x}=-\frac{\partial M_{x}}{\partial x}-2\frac{\partial M_{xy}}{\partial y},
\end{equation}
\begin{equation}
Q_{y}=-\frac{\partial M_{y}}{\partial y}-2\frac{\partial M_{xy}}{\partial x},
\end{equation}
\begin{equation}
M_{x}=\frac{\partial^2 h }{\partial x^2}+\nu\frac{\partial^2 h}{\partial y^2},
\end{equation}
\begin{equation}
M_{y}=\nu\frac{\partial^2 h }{\partial x^2}+\frac{\partial^2 h}{\partial y^2},
\end{equation}
Boundary components are the be
nding moment $M_{x}$, $M_{y}$ and shear forces ${Q}_{x}$ and ${Q}_{y}$ and all equal to zero.
Note that $\mathcal{H}^4(\Omega)$ 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 prove the self-adjointness of the operator, I have to show that it's symmetric and (surjective)onto. I've shown the symmetry by using 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)?
Thanks in advance Fatemeh
