Consider the function u(x,y,z)
I am trying to approximate the partial derivative at point (i,j,k) by one sided finite difference method.
Now using one sided 2nd order finite difference approxmation for the first derivative, we have
$u_y(x,y,z) = \frac{3u(x,y,z) – 4u(x,y-\Delta{y},z) + u(x,y-2\Delta{y},z)}{\Delta{y}}$
Now applying this for second derivative, we have
$u_{xy} = \frac{3u_y(x,y,z) – 4u_y(x-\Delta{x},y,z) + u_y(x-2\Delta{x},y,z)}{\Delta{x}}$
$ = \frac{3[\frac{3u(x,y,z) – 4u(x,y-\Delta{y},z) + u(x,y-2\Delta{y},z)}{\Delta{y}}] – 4[\frac{3u(x-\Delta{x},y,z) – 4u(x-\Delta{x},y-\Delta{y},z) + u(x-\Delta{x},y-2\Delta{y},z)}{\Delta{y}} ] + [\frac{3u(x-2\Delta{x},y,z) – 4u(x-2\Delta{x},y-\Delta{y},z) + u(x-2\Delta{x},y-2\Delta{y},z)}{\Delta{y}}]}{\Delta{x}}$
$ = \frac{3[{3u(x,y,z) – 4u(x,y-\Delta{y},z) + u(x,y-2\Delta{y},z)}] – 4[{3u(x-\Delta{x},y,z) – 4u(x-\Delta{x},y-\Delta{y},z) + u(x-\Delta{x},y-2\Delta{y},z)} ] + [{3u(x-2\Delta{x},y,z) – 4u(x-2\Delta{x},y-\Delta{y},z) + u(x-2\Delta{x},y-2\Delta{y},z)}]}{\Delta{x}{\Delta{y}}} $
$ = $ resulting exp
My query is that for one sided approximation, is this the simplest form of mixed derivative that is obtainable or have I missed something that simplifies the resulting equation?
Best Answer
It's never too late for an answer.
The answer is YES. This is the simplest (=shortest) form of a second order finite difference for the mixed derivative in $x$ and $y$ direction, if you want both directions to be one-sided. (I assume this is what you want, although you are talking about a second derivitive.)
Since you have
$u(x,y,z), (x,y-\Delta{y},z), u(x,y-2\Delta{y},z), u(x-\Delta{x},y,z), u(x-\Delta{x},y-\Delta{y},z), u(x-\Delta{x},y-2\Delta{y},z), u(x-2\Delta{x},y,z), u(x-2\Delta{x},y-\Delta{y},z), u(x-2\Delta{x},y-2\Delta{y},z)$
only once in your representation, there cannot be a shorter (=simpler) one. The only thing you could do is to multiply out the brackets.