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arunsmec August 17, 2013 04:46

Compact finite difference
 
I have some difficulty in understanding the implementation of compact difference schemes.
I understood that compact difference scheme gives a system of equations of the form Af' = Bf.
Let's say if i intend to solve an equation of the form df/dt+c df/dx = 0.
How do I implement the CDS in this equation?
Is it by simply substituting for f'=A^(-1)Bf in the discretized equation?

RameshK August 17, 2013 08:44

Compact Difference
 
Compact difference is one way of getting higher order of accuracy with a relatively small stencil. As it is aware that if we want a higher order of accuracy the stencil size increases as the order of accuracy increases. Compact schemes requires to solve a tri-diagonal system in the case of 1D and penta diagonal system in the case of 2D to get the coefficients of the variables getting into stencil. The key concept lies in the getting the coefficients. You should be careful once the boundary points are reached.

arunsmec August 17, 2013 11:02

Thank you Ramesh.
The coefficients can be obtained by matching terms of same order. Once I do that I get a system of equations of the form Af' = Bf.
How do I solve this equation since both f and f' are unknowns?

FMDenaro August 17, 2013 11:53

Quote:

Originally Posted by arunsmec (Post 446307)
Thank you Ramesh.
The coefficients can be obtained by matching terms of same order. Once I do that I get a system of equations of the form Af' = Bf.
How do I solve this equation since both f and f' are unknowns?


In the system, f is assumed to be the known term. Hence, start from assigned values f(x,0), compute f' and solve the equation df/dt + c f'=0. Repeat for each time step. Be careful in the BCs.

arunsmec August 18, 2013 04:53

Does that mean I have to simultaneously solve both the equations, and the scheme is implicit?

FMDenaro August 18, 2013 05:20

Quote:

Originally Posted by arunsmec (Post 446388)
Does that mean I have to simultaneously solve both the equations, and the scheme is implicit?

implicit or explicit depends on what you decide to do...for an explicit scheme you compute f' at time tn with f(tn) known term and then solve explicitly for f(tn+1).


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