# Matrix Representation of a Discretization for a Partial Differential Equation

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.

• Which physical problem does this equation represent ? – Mathnoob Nov 13 '18 at 10:32
• 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. – Bill Greene Nov 13 '18 at 12:26
• 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. – tnt235711 Nov 13 '18 at 13:49

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:

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