Let $\overline{\Omega}=\left([0,2]\times [0,1]\right)\cup \left([1,2]\times [1,2]\right)$ and $\Gamma=\Gamma_1\cup\ldots\cup\Gamma_6$ be defined as shown in the following picture:
I want to discretize the following heat equation: \begin{equation} \begin{split} -D\left(\frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2}\right) &= 0, \text{ for all }(x,y)\in\Omega\\ D\left(\frac{\partial u}{\partial x}n_1+\frac{\partial u}{\partial y}n_2\right) &= 0, \text{ for all }(x,y)\in\Gamma_2\cup\Gamma_3\cup\Gamma_5\cup\Gamma_6\\ u &= u_0, \text{ in }\Gamma_1\\ u &= u_1, \text{ in }\Gamma_4 \end{split} \end{equation} where $h$ is the step width of the used lattice and $D,u_0,u_1$ are given constants. I want to find a matrix $K_h$ and a vector $b_h$ such that $K_hu_h=b_h$, where $u_h$ is the approximate solution vector.
If $(x_i,y_j)$ is a point on the lattice and $u_{ij}=u(x_i,y_j)$, I want to use the approximations $$\Delta u_{ij}=\nabla^2u_{ij}\approx\frac{1}{h^2}\left(u_{i+1,j}+u_{i-1,j}-4u_{ij}+u_{i,j+1}+u_{i,j-1}\right)$$ and $$\frac{\partial}{\partial x}u_{ij}\approx\frac{u_{i+1,j}-u_{i-1,j}}{2h}$$ as well as $$\frac{\partial}{\partial y}u_{ij}\approx\frac{u_{i,j+1}-u_{i,j-1}}{2h}$$
Suppose we first consider $\Omega_1=(0,1)\times (0,1)$. We construct our grid $G^{(h)}$ by the tensor product of two grids of $(0,1)$: $$G_x^{(h)}:=\left\{x_i=(i-1)h:1\le i\le N\right\}\text{ and }G_y^{(h)}:=\left\{y_j=(i-1)h:1\le j\le N\right\}$$ where N:=\frac{h+1}{h}. Using our approximations, we've got the following equations:
\begin{equation} \begin{split} D\frac{1}{h^2}\left(4u_{ij}-u_{i+1,j}-u_{i-1,j}-u_{i,j+1}-u_{i,j-1}\right)&=0\text{ , for all }1\le i,j\le N\\ -D\frac{u_{i,2}-u_{i,0}}{2h}&=0\text{ , for all } 0\le i\le N+1\\ D\frac{u_{i,N+1}-u_{i,N-1}}{2h}&=0\text{ , for all } 0\le i\le N+1\\ u_{1j}&=u_1\text{ , for all } 1\le j\le N \end{split} \end{equation}
Please note that I've split the second equations for the half of $\Gamma_2$ and $\Gamma_6$. There some problems left: First of all, $u_{i,0}$ and $u_{i,N+1}$ are not well-defined. We can use the first equation for $j=1$ and $j=N$ in combination with the second and third equation, respectively. Insertion and scaling by a factor of $1/2$ yields $$D\frac{1}{h^2}\left(2u_{i,1}-\frac12 u_{i+1,1}-\frac12 u_{i-1,1}-u_{i,2}\right)=0$$ and $$D\frac{1}{h^2}\left(2u_{iN}-\frac12 u_{i+1,N}-\frac12 u_{i-1,N}-u_{i,N-1}\right)=0$$ Now there are still problems with the corner nodes $(0,0)$, $(0,1)$ and (maybe?) $(1,1)$.
For $(0,0)$ we consider $(i,j)=(0,0)$ and there are 3 equations involved: From the first equation we've got $$4u_{1,1}-u_{2,1}-u_{0,1}-u_{1,2}-u_{1,0}=0\tag{1}$$ From the second equation we've got: $$u_{1,2}-u_{1,0}=0\tag{2}$$ and the third equations yields $$u_{1,1}=u_1\tag{3}$$ However, we've got now information about $u_{0,1}$.
So there are three questions left:
- What can we say about $u_{0,1}$?
- How do we describe all these equations on $\Omega_1$ in a linear equation system "$Ax=b$"
- Suppose we've got all linear three linear equation systems for $\Omega_1$ as well as $\Omega_2=(1,2)\times (0,1)$ and $\Omega_3=(1,2)\times (1,2)$. How do we combine these systems into one system for $\Omega$?
Make a rectangular lattice on $[0,1]^2$ and on $[1,2] \times [0,2]$ separately. Merge them (in the process making the lattice points on $\{ 1 \} \times [0,1]$ only count once, of course). The awkward thing in this circumstance is that the "ghost points" have to be defined in a rather asymmetric fashion.