In mathematics, the discrete Poisson equation is the finite difference analog of the Poisson equation. In it, the discrete Laplace operator takes the place of the Laplace operator. The discrete Poisson equation is frequently used in numerical analysis as a stand-in for the continuous Poisson equation, although it is also studied in its own right as a topic in discrete mathematics.

Using the finite difference numerical method to discretize the 2-dimensional Poisson equation (assuming a uniform spatial discretization, ${\displaystyle \Delta x=\Delta y}$) on an m × n grid gives the following formula: $${\displaystyle ({\nabla }^{2}u)_{ij}={\frac {1}{\Delta x^{2}}}(u_{i+1,j}+u_{i-1,j}+u_{i,j+1}+u_{i,j-1}-4u_{ij})=g_{ij}}$$ where ${\displaystyle 2\leq i\leq m-1}$ and ${\displaystyle 2\leq j\leq n-1}$. The preferred arrangement of the solution vector is to use natural ordering which, prior to removing boundary elements, would look like: $${\displaystyle \mathbf {u} ={\begin{bmatrix}u_{11},u_{21},\ldots ,u_{m1},u_{12},u_{22},\ldots ,u_{m2},\ldots ,u_{mn}\end{bmatrix}}^{\mathsf {T}}}$$

This will result in an mn × mn linear system: $${\displaystyle A\mathbf {u} =\mathbf {b} }$$ where $${\displaystyle A={\begin{bmatrix}~D&-I&~0&~0&~0&\cdots &~0\\-I&~D&-I&~0&~0&\cdots &~0\\~0&-I&~D&-I&~0&\cdots &~0\\\vdots &\ddots &\ddots &\ddots &\ddots &\ddots &\vdots \\~0&\cdots &~0&-I&~D&-I&~0\\~0&\cdots &\cdots &~0&-I&~D&-I\\~0&\cdots &\cdots &\cdots &~0&-I&~D\end{bmatrix}},}$$

${\displaystyle I}$ is the m × m identity matrix, and ${\displaystyle D}$, also m × m, is given by: $${\displaystyle D={\begin{bmatrix}~4&-1&~0&~0&~0&\cdots &~0\\-1&~4&-1&~0&~0&\cdots &~0\\~0&-1&~4&-1&~0&\cdots &~0\\\vdots &\ddots &\ddots &\ddots &\ddots &\ddots &\vdots \\~0&\cdots &~0&-1&~4&-1&~0\\~0&\cdots &\cdots &~0&-1&~4&-1\\~0&\cdots &\cdots &\cdots &~0&-1&~4\end{bmatrix}},}$$ and ${\displaystyle \mathbf {b} }$ is defined by $${\displaystyle \mathbf {b} =-\Delta x^{2}{\begin{bmatrix}g_{11},g_{21},\ldots ,g_{m1},g_{12},g_{22},\ldots ,g_{m2},\ldots ,g_{mn}\end{bmatrix}}^{\mathsf {T}}.}$$

For each ${\displaystyle u_{ij}}$ equation, the columns of ${\displaystyle D}$ correspond to a block of ${\displaystyle m}$ components in ${\displaystyle u}$: $${\displaystyle {\begin{bmatrix}u_{1j},&u_{2j},&\ldots ,&u_{i-1,j},&u_{ij},&u_{i+1,j},&\ldots ,&u_{mj}\end{bmatrix}}^{\mathsf {T}}}$$ while the columns of ${\displaystyle I}$ to the left and right of ${\displaystyle D}$ each correspond to other blocks of ${\displaystyle m}$ components within ${\displaystyle u}$: $${\displaystyle {\begin{bmatrix}u_{1,j-1},&u_{2,j-1},&\ldots ,&u_{i-1,j-1},&u_{i,j-1},&u_{i+1,j-1},&\ldots ,&u_{m,j-1}\end{bmatrix}}^{\mathsf {T}}}$$ and $${\displaystyle {\begin{bmatrix}u_{1,j+1},&u_{2,j+1},&\ldots ,&u_{i-1,j+1},&u_{i,j+1},&u_{i+1,j+1},&\ldots ,&u_{m,j+1}\end{bmatrix}}^{\mathsf {T}}}$$

respectively.

From the above, it can be inferred that there are ${\displaystyle n}$ block columns of ${\displaystyle m}$ in ${\displaystyle A}$. It is important to note that prescribed values of ${\displaystyle u}$ (usually lying on the boundary) would have their corresponding elements removed from ${\displaystyle I}$ and ${\displaystyle D}$. For the common case that all the nodes on the boundary are set, we have ${\displaystyle 2\leq i\leq m-1}$ and ${\displaystyle 2\leq j\leq n-1}$, and the system would have the dimensions (m − 2)(n − 2) × (m− 2)(n − 2), where ${\displaystyle D}$ and ${\displaystyle I}$ would have dimensions (m − 2) × (m − 2).

For a 3×3 ( ${\displaystyle m=3}$ and ${\displaystyle n=3}$ ) grid with all the boundary nodes prescribed, the system would look like: $${\displaystyle {\begin{bmatrix}U\end{bmatrix}}={\begin{bmatrix}u_{22},u_{32},u_{42},u_{23},u_{33},u_{43},u_{24},u_{34},u_{44}\end{bmatrix}}^{\mathsf {T}}}$$ with $${\displaystyle A=\left[{\begin{array}{ccc|ccc|ccc}~4&-1&~0&-1&~0&~0&~0&~0&~0\\-1&~4&-1&~0&-1&~0&~0&~0&~0\\~0&-1&~4&~0&~0&-1&~0&~0&~0\\\hline -1&~0&~0&~4&-1&~0&-1&~0&~0\\~0&-1&~0&-1&~4&-1&~0&-1&~0\\~0&~0&-1&~0&-1&~4&~0&~0&-1\\\hline ~0&~0&~0&-1&~0&~0&~4&-1&~0\\~0&~0&~0&~0&-1&~0&-1&~4&-1\\~0&~0&~0&~0&~0&-1&~0&-1&~4\end{array}}\right]}$$ and