# help!

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 March 10, 2004, 00:45 help! #1 lsm Guest   Posts: n/a who can tell me why the "middle point rule" in FVM is 2th precision?

 March 10, 2004, 02:49 Re: help! #2 sunil Guest   Posts: n/a jgjkagdxjagj

 March 10, 2004, 06:15 Re: help! #3 P. Birken Guest   Posts: n/a Any textbook or lecture notes on numerical methods for ODEs. Or a Taylor-expansion.

 March 10, 2004, 07:16 Re: help! #4 lsm Guest   Posts: n/a sorry,middle value rule and middle point rule are popular,but i can't see any analysis about precision.

 March 10, 2004, 08:33 Re: help! #5 Tom Guest   Posts: n/a It follows from the mean-value theorem (=> Taylors theorem) in any school (A-level in the UK) maths text book. Alternatively consider two points x_l and x_r with midpoint x_m = ( x_l + x_r )/2 and spacing h = x_r - x_l then the trapezoidal rule gives, for a function f, integral f = h( f_r + f_l )/2 + O(h^3) and by Taylors theorem f_m = ( f_r + f_l )/2 + O(h^2) which imples the result (assuming you accept the Trapezoidal rule is second order!)

 March 10, 2004, 10:38 Re: help! #6 lsm Guest   Posts: n/a yes integral f = f(average value at one point between x_l and x_r)*h=?f_middle-pint*h+o(h^3)

 March 11, 2004, 00:42 Re: help! #7 lsm Guest   Posts: n/a my puzzle is that f = f(average value at one point between x_l and x_r) by middle value rule is strict, but f(average value at one point between x_l and x_r)=?f_middle-pint+o(h^1) is 1th precise. how to explain it?

 March 11, 2004, 05:40 Re: help! #8 Tom Guest   Posts: n/a Basically the O(h) error term vanishes at the midpoint - look at a book on finite difference methods.

 March 11, 2004, 12:47 Re: help! #9 Hrvoje Jasak Guest   Posts: n/a Do integral of phi over a ceontrol volume and approximate it with Taylor series expansion. int_V phi(x) dV phi(x) = phi_P + (x - x_P) . grad(phi)_P + hot. (1) When you integrate you get int_V phi(x) dV = phi_P V + (int_V (x-x_P) dV) . grad(phi)_P + hot. The gradient comes out of the second integral because it is a constant. So, if the integral int_V (x-x_P) dV = 0 (3) the method will be second order, which is true if the point of expansion you are expressing phi in eq(1) around in the centroid of the cell. Fortunately, eq(3) is indeed the definition of the centroid so all is well. "... and there you are!" (quote from My Big Fat Greek Wedding) Hrv

 March 12, 2004, 00:59 Re: help! #10 lsm Guest   Posts: n/a thanks,but excuse me, let's do 1d case for 1d case your (3) equation int_V (x-x_P) dV = 0 (3) become int_V (x-x_P) dx = 1/2(x-xp)^2 it is o(2), not o(3)

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