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I want to discretize the following problem

\begin{cases} \mu \nabla^2u+(\lambda+\mu)\nabla \nabla\cdot u = \rho \frac{\partial^2u }{\partial t^2 } + \beta \frac{\partial u}{\partial t}\\ u(x,y,t) = c, (x,y) \in \partial( [0,L_{x}]\times [0,L_{y}]), t \in [0,T]\\ u(x,y,0) = u_{0}(x,y), (x,y) \in [0,L_{x}]\times [0,L_{y}]\\ \frac{\partial u(x,y,0)}{\partial t} = u_{1}(x,y), (x,y) \in [0,L_{x}]\times [0,L_{y}]\\ \end{cases}

where $T,L_{x},L_{y} >0$, $c\in \mathbb{R}$ and $$ u_{0},u_{1} \in C^2( [0,L_{x}]\times [0,L_{y}], \mathbb{R^2} ) $$ $$u \in C^2( [0,L_{x}]\times [0,L_{y}]\times [0,T], \mathbb{R^2} ) $$

My attempt. Let $u = ( u_{x} \ u_{y} )^T$

$$ \mu \nabla^2u = ( \mu \nabla^2 u_{x} \ \ \mu \nabla^2u_{y} )^T = $$ $$\mu \begin{pmatrix} \nabla^2 & 0 \\ 0 & \nabla^2 \end{pmatrix}\begin{pmatrix} u_{x}\\ u_{y} \end{pmatrix} $$

where the laplacian in the Kronecker product is: $$ \nabla^2 = L_{2}\otimes I_{n_{x}} + I_{n_{y}} \otimes L_{2} $$

\begin{equation} L_{2} = \frac{1}{h^2}\left[\begin{matrix} -2 & 1 & & 0\\ 1 & \ddots & \ddots & \\ & \ddots & \ddots & 1 \\ 0 & & 1 & -2 \end{matrix} \right] \end{equation}

$h$ is the step size or mesh size... Now

$$ (\lambda+\mu)\nabla \nabla \cdot u = $$ $$ \begin{pmatrix} (\lambda+\mu)\frac{\partial^2 u_{x}}{\partial x^2} + (\lambda+\mu)\frac{\partial^2 u_{y}}{\partial x \partial y}\\ (\lambda+\mu)\frac{\partial^2 u_{x}}{\partial y \partial x}+(\lambda+\mu)\frac{\partial^2 u_{y}}{\partial y^2} \end{pmatrix} = $$

$$ (\lambda+\mu) \begin{pmatrix} \frac{\partial^2 }{\partial x^2} & \frac{\partial^2 }{\partial x \partial y} \\ \frac{\partial^2 }{\partial y \partial x} & \frac{\partial^2 }{\partial y^2} \end{pmatrix} \begin{pmatrix} u_{x} \\ u_{y} \end{pmatrix} $$

$$ \frac{\partial^2 }{\partial x^2} = L_{2}\otimes I_{n_{x}} $$ $$ \frac{\partial^2 }{\partial y^2} = I_{n_{y}} \otimes L_{2}$$

My question is how do I represent the following operators in terms of $L_{2}, I, \otimes$ or is not possible ? if it is not possible, what is the matrix representation of...?

$$\frac{\partial^2 }{\partial x \partial y} = \textbf{ ? }$$ $$\frac{\partial^2 }{\partial y \partial x} = \textbf{ ? }$$

Thank you!

EDIT: PHYSICAL INTERPRETATION This is known as the wave equation in elastodynamic (a more general form)

where $\lambda$ and $\mu$ are the Lame’s coefficients given by

$$ \mu = \frac{E}{2(1+\upsilon)} $$ and $$ \lambda = \frac{E\upsilon}{(1+\upsilon)(1-2\upsilon)} $$

where $E$ is the modulus of elasticity (Young modulus) and $\upsilon$ is the Poisson’s ratio of the elastic material. $\rho$ is the linear density.

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  • $\begingroup$ Which physical problem does this equation represent ? $\endgroup$ – Mathnoob Nov 13 '18 at 10:32
  • $\begingroup$ The term $\upsilon\frac{\partial u}{\partial t}$ is unusual; this could represent some kind of viscous damping. In any event, the coefficient would not be Poisson's ratio. You probably want to recheck this. $\endgroup$ – Bill Greene Nov 13 '18 at 12:26
  • $\begingroup$ ok, I have changed the coefficient's name, maybe to simplify the problem we can consider $\beta = 0 $...what I need is the matrix representation of the mixed derivatives, because I have not worked with that operator. I'm stuck in that part of the discretization. $\endgroup$ – tnt235711 Nov 13 '18 at 13:49
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In 1D, if we define $u_{k} := u(x_{k}) $; $\ \ x_{k} = kh $ and $ \ \ k = 0,1,2,...,N$. $h$ is known as the mesh size or step size.
I want to approximate the first derivative using central difference:

$$ \frac{du_{k}}{dx} \approx \frac{ u_{k+1}-u_{k-1} }{ 2h } = \frac{ -u_{k-1} +u_{k+1} }{ 2h } $$

In this problem (1D) if $u(0) = u_{0} = 0$ and $ u(L) = u(x_{N}) = u_{N} = 0 $ we get the following matrix representation of the first derivative

\begin{equation} \frac{d}{dx} \approx L_{1} = \frac{1}{2h}\left(\begin{matrix} 0 & 1 & & 0\\ -1 & \ddots & \ddots & \\ & \ddots & \ddots & 1 \\ 0 & & -1 & 0 \end{matrix} \right) \end{equation}

Now using $L_{1}$:

$$\frac{\partial^2 }{\partial x \partial y} = \Big((L_{1})_{n_{x}} \otimes I_{n_{y}} \Big)\Big( I_{n_{x}} \otimes (L_{1})_{n_{y}} \Big) = { (L_{1})_{n_{x}} \otimes (L_{1})_{n_{y}} } $$

I have used the mixed-product property for kronecker product. $$ (\mathbf{A} \otimes \mathbf{B})(\mathbf{C} \otimes \mathbf{D}) = (\mathbf{AC}) \otimes (\mathbf{BD})$$.

In a similar way we get:

$$\frac{\partial^2 }{\partial y \partial x} = (L_{1})_{ n_{y} } \otimes (L_{1})_{n_{x}} $$

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