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 Thomashoffmann January 22, 2013 06:03

first and second order schemes

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Hi all.
I need a little help to understand how to decide if a scheme is first or second order accurate.. My supervisor tells me that I can see it from the Taylor series formulation, where if I subtract eq 3.2 from 3.3 I get eq 3.4, which is first order accurate. and If I add them I get eq. 3.5, which is second order accurate.
This far I understand it, because the first and second order derivative skibs out when adding/subtracting...

But for my thesis I written that in a simple case of constant grid I can estimate T_e:
T_e=(T_P+T_E)/2
and I have written that it is a first order scheme. My supervisor tells me that it is a second order scheme. Why is that?
I just don't understand how to my simple case can be compared with a taylor series expansion?
Can one of you guys explain it better for me?
Thanks

Thomas

 FMDenaro January 22, 2013 06:15

you are using a linear behavior (first degree Lagrangian polynomial) for the function, therefore your local truncation error for it is second order. You can compute the first derivative at first order (apart that in the centred point that is second order)

 RodriguezFatz January 22, 2013 08:42

In eq. 3.2 and 3.3 you leave the fourth term, which is proportional to "dx^3". Now, if you subtract the two equations and divide them by "dx" to get 3.4, your error term (the one you left) has a leading dx^2. Thus, it is second order accurate.
If you add them and divide by "dx^2" to get 3.5 you cancel out the second and the fourth term (both alternating +/- signs) and your leading error has "dx^4 / dx^2" which is also dx^2. Thus, this is also second order accurate.

In your case T_e = T_E or T_e = T_P is first order accurate.

 RodriguezFatz January 22, 2013 08:50

Now to your second question: Just change T_P to T_1, T_e to T_2 and T_E to T_3 and you have exactly what your equations from the book say. You make a Taylor series expansion at T_e, dx=1 in your case.

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