- #1
Hypatio
- 151
- 1
What is a finite difference discretization for the fourth-order partial differential terms
[itex]\frac{\partial u}{\partial x}k\frac{\partial u}{\partial x}\frac{\partial u}{\partial x}k(x,y)\frac{\partial u}{\partial x}[/itex]
and
[itex]\frac{\partial u}{\partial x}k(x,y) \frac{\partial u}{\partial y} \frac{\partial u}{\partial x} k(x,y)\frac{\partial u}{\partial y}[/itex]
with the variable coefficient k. I'm not certain if I have written them correctly. These terms should appear in the equation
[itex]\nabla\cdot\nabla(k\nabla^2 u)[/itex]
I know that for constant k the following respective discretizations work:
k[u(x-2,y+0) - 4u(x-1,y+0) + 6u(x+0,y+0) - 4u(x+1,y+0) + u(x+2,y+0)]
and
k[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)]
but what are the equivalent forms with variable k?
[itex]\frac{\partial u}{\partial x}k\frac{\partial u}{\partial x}\frac{\partial u}{\partial x}k(x,y)\frac{\partial u}{\partial x}[/itex]
and
[itex]\frac{\partial u}{\partial x}k(x,y) \frac{\partial u}{\partial y} \frac{\partial u}{\partial x} k(x,y)\frac{\partial u}{\partial y}[/itex]
with the variable coefficient k. I'm not certain if I have written them correctly. These terms should appear in the equation
[itex]\nabla\cdot\nabla(k\nabla^2 u)[/itex]
I know that for constant k the following respective discretizations work:
k[u(x-2,y+0) - 4u(x-1,y+0) + 6u(x+0,y+0) - 4u(x+1,y+0) + u(x+2,y+0)]
and
k[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)]
but what are the equivalent forms with variable k?