Finite Difference Discretization of a Fourth Order Partial Differential Term

[Hypatio]

What is a finite-difference discretization for the partial differential term:

$$\frac{\partial^4\phi}{\partial x^2\partial y^2}$$

Thanks in advance.

 

[lurflurf]

There are as usual an infinite number, the choice of which depends upon the situation. A common choice would be
u(x-1,y+1) -2u(x+0,y+1) +u(x+1,y+1)
-2u(x-1,y+0) +4u(x+0,y+0) -2u(x+1,y+0)
+u(x-1,y-1) -2u(x+0,y-1) +u(x+1,y-1)

 

[Hypatio]

There are as usual an infinite number, the choice of which depends upon the situation. A common choice would be
u(x-1,y+1) -2u(x+0,y+1) u(x+1,y+1)
-2u(x-1,y+0) +4u(x+0,y+0) -2u(x+1,y+0)
u(x-1,y-1) -2u(x+0,y-1) u(x+1,y-1)

Thanks, but is there a formulation which does not use bivariate (?) terms? eg. 2u(x+0,y+1)*u(x+1,y+1)

I do not understand how it is possible to create a solvable matrix with a mix of bivariate terms in it, which is what I am trying to to do (stress function solution using gaussian elimination).

 

[lurflurf]

^Sorry that was implied addition not implied multiplication. Writen out in full we have
16u_xxyy~u(x-1,y+1)-2u(x+0,y+1)+u(x+1,y+1)-2u(x-1,y+0)+4u(x+0,y+0)-2u(x+1,y+0)+u(x-1,y-1) -2u(x+0,y-1) +u(x+1,y-1)

or unscaled

(4st)² u_xxyy~u(x-s,y+t)-2u(x+0,y+t)+u(x+s,y+t)-2u(x-s,y+0)+4u(x+0,y+t)-2u(x+s,y+0)+u(x-s,y-t) -2u(x+0,y-t) +u(x+s,y-t)

A higher order or biased expansion could be used if needed.

 

[Hypatio]

Ah, thank you for alleviating my fears, and for the help.