Return to Question

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}=Q_{x}-\frac{\partial M_{xy}}{\partial y}=-\frac{\partial M_{x}}{\partial x}-2\frac{\partial M_{xy}}{\partial y}, \end{equation} Boundary components are the bending 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

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

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\} \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 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)? Note that all the boundary conditions equal to zero.

Thanks in advance Fatemeh

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}=Q_{x}-\frac{\partial M_{xy}}{\partial y}=-\frac{\partial M_{x}}{\partial x}-2\frac{\partial M_{xy}}{\partial y}, \end{equation} Boundary components are the bending 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

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\} \end{equation} where $\mathcal{H}^4(\Omega)$ is the Sobolev space and consists of all functions in $L^(\Omega)$ 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)? Note that all the boundary conditions equal to zero.

Thanks in advance Fatemeh

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\} \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 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)? Note that all the boundary conditions equal to zero.

Thanks in advance Fatemeh