I've just been introduced to the Poisson's equation. I've never had the need to dealt with PDE, so I'm a bit lost.

Apparently we can compute an approximate solution of the Poisson's equation

$$\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} u(x, y) = f(x, y)$$

by discretizing the 2D Poisson's equation using finite differences. Here's a picture of that discretization taken from the website I've just linked you to:

enter image description here

According to that website, we have:

The above linear equation relating $U(i,j)$ and the value at its neighbors (indicated by the blue stencil) must hold for $1 <= i,j <= n$, giving us $N=n^2$ equations in $N$ unknowns.

where "by above linear equation" I guess they are referring to

$$−4u_{i,j}+u_{i+1,j}+u_{i−1,j}+u_{i,j+1}+u_{i,j−1} = b_{i,j}$$ $$1 \leq i, j \leq n$$

When $(i,j)$ is adjacent to a boundary ($i=1$ or $j=1$ or $i=n$ or $j=n$), one or more of the $U(i+-1, j+-1)$ values is on the boundary and therefore $0$. $$b(i,j) = -f(i*h,j*h)*h^2$$ the scaled value of the right-hand-side function $f(x,y)$ at the corresponding grid point $(i,j)$.


  1. Where does the $-4$ in front of $u_{i,j}$ in the equation above comes from?

  2. What's $b(i, j)$? Why is it equal to $−4u_{i,j}+u_{i+1,j}+u_{i−1,j}+u_{i,j+1}+u_{i,j−1}$? I mean, I don't understand where does it come from. Usually $b$ refers to a right-hand side, but...

  3. Why do we have $b(i,j) = -f(i*h,j*h)*h^2$?


1 Answer 1


The answers are all there in the link


so this is only a short guide to what you should be looking for.

  1. The derivatives have been discretized as

$\frac{\partial^2 u}{\partial x^2} = \frac{u(i+1,j) - 2u(i,j) + u(i-1, j)}{h^2}$

$\frac{\partial^2 u}{\partial y^2} = \frac{u(i,j+1) - 2u(i,j) + u(i, j-1)}{h^2}$

So, add the two together and you get

$\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = \frac{u(i+1,j) - 2u(i,j) + u(i-1, j)}{h^2} + \frac{u(i,j+1) - 2u(i,j) + u(i, j-1)}{h^2}$


$\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = \frac{u(i+1,j) + u(i-1, j) + u(i,j+1) + u(i, j-1) - 4u(i,j)}{h^2} $

This gives you the -4


$\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = f$


$ \frac{u(i+1,j) + u(i-1, j) + u(i,j+1) + u(i, j-1) - 4u(i,j)}{h^2} = f$

Now just take the $h^2$ to the other side and you get b which is also your third question

  • 1
    $\begingroup$ I don't think you'll get the why there. You'll only get the how. To get the why you'll have to start reading a numerical analysis book. If you only want a basic introduction, even Kreyszig should be enough. $\endgroup$
    – Vikram
    Commented Nov 17, 2016 at 9:44

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