# O(dtdx)=O(dxdx)?

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 April 29, 2000, 23:08 O(dtdx)=O(dxdx)? #1 H.Y.Zhao Guest   Posts: n/a When the impicit scheme is employed for the computation of unsteady flow,the taylor expansion of the flux linearization usually leads to the terms like O(dtdx).Some consider the O(dtdx) is equivalent to the O(dtdt) or O(dxdt) based on the CFL condition,so the spatial or the temporal accuracy is second order,but others think the accuracy is independent to the stability condition so that the spatial or temporal accuracy are only one order.I wonder which is right?Please give me your comments on it,thank you!

 April 30, 2000, 00:48 Re: O(dtdx)=O(dxdx)? #2 John C. Chien Guest   Posts: n/a (1). In the Taylor's series expansion, the truncation error represents the terms dropped, without knowing the exact nature of the function. (2). So, it is simply another way to say the same thing. Beyond that, there is no special meaning to it. (3). Naturally, you can also create other names for it, just like what you have mentioned. (4). So, as long as you know exactly where the Taylor's series is chopped off, it really doesn't matter how you call it. This is because at the time of Taylor's series expansion, we know nothing about the function itself in terms of the independent variables, such as t,x,y,... or dt,dx,dy,... (5). Order of magnitude is just another way to say that. (6). So, before you use the equality sign, you need to define it first.

 May 2, 2000, 16:07 Re: O(dtdx)=O(dxdx)? #3 clifford bradford Guest   Posts: n/a many implicit schemes such as ADI use approximate factorization which leads to factorization error in addition to truncation error. usually these complicated looking error terms arise because of this it indicates that the even though the scheme may be unconditionally stable then time-step may not be chosen to be arbitrarily large. also with these schemes applied to nonlinear equation the temporal and spatial accuracy cannot be specified independently of each other. typically for an uncoupled scheme (where timeintegration and spatial differencing are not linked) a plot of log(error) vs log (grid size), with cfl fixed, will be a line with a gradient of approximately equal to the spatial order of accuracy. for your scheme it may not be the case because the order of accuracy contains these product terms

 May 8, 2000, 21:53 Re: O(dtdx)=O(dxdx)? #4 H.Y.Zhao Guest   Posts: n/a I want to clarify my intentions.My implicit scheme is LU approximate factorization.For one dimension,the implicit side is [I+(A(+)i-A(+)(i-1))/dx][I+(A(-)(i+1)-A(-)i)/dx]dU. Because the truncation error of dU is O(dt), and the truncation error of spatial discretization is O(dx),the gross truncation error is O(dtdx).I don't know whether the spatial accuracy is second order or not?

 May 9, 2000, 15:29 Re: O(dtdx)=O(dxdx)? #5 Adrin Gharakhani Guest   Posts: n/a I don't know the details of your analysis. But if you _limit_ your dt variation to O(dx), then clearly O(dtdx)=O(dx^2). Adrin Gharakhani

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