# Compact finite difference

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 August 17, 2013, 04:46 Compact finite difference #1 New Member   Aru Join Date: Feb 2012 Location: Chennai, India Posts: 26 Rep Power: 6 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?

 August 17, 2013, 08:44 Compact Difference #2 Member   Ramesh K Join Date: Dec 2009 Location: Bangalore Posts: 65 Rep Power: 8 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.

 August 17, 2013, 11:02 #3 New Member   Aru Join Date: Feb 2012 Location: Chennai, India Posts: 26 Rep Power: 6 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?

August 17, 2013, 11:53
#4
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Filippo Maria Denaro
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Quote:
 Originally Posted by arunsmec 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.

 August 18, 2013, 04:53 #5 New Member   Aru Join Date: Feb 2012 Location: Chennai, India Posts: 26 Rep Power: 6 Does that mean I have to simultaneously solve both the equations, and the scheme is implicit?

August 18, 2013, 05:20
#6
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Filippo Maria Denaro
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Quote:
 Originally Posted by arunsmec 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).

 Tags compact difference, fdm, finite difference

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