2order discretization
 

sir, im new to cfd i just want to know how to apply d2u/dx2 at the boundary. usually we apply du/dx=0 for a fully developed flow at the boundary. inorder to have 2 order accuracy i want to give d2u/dx2 =0 at the boundary please tell.. thanks

Re: 2order discretization
 

d^2u/dx^2=(u_n-2u_n-1+u_n-2)/dx^2 + O(dx)

u_n=2u_n-1 - u_n-2

if this is what you are looking for.

Re: 2order discretization
 

thank you very much

Re: 2order discretization
 

Maybe we should point out what it is that we are doing here. The centered finite difference of the second-order derivative at n-1 is

(u_n -2u_n-1 +u_n-2)/dx^2

and Harish's approach to use the same term for the boundary (n) amounts to a zeroth-order extrapolation of the second derivative (in other words, you assume a zero third-order derivative). Now that you know where it comes from, you can probably come up with alternative methods. You have to use some sort of extrapolation, unless you know the conditions beyond the boundary. However, if you do know these conditions, you ought to make use of them.

Re: 2order discretization
 

yes mr mani i clearly understand what harish has written. actually im implementing fully developed condition at the exit. so i knw the values at n-1 and n-2. thanks a lot