Error In L2
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consists of computing, at each node: the difference the relative difference the percentage where is the l2 norm exact solution and is the calculated solution, and printing, for l2 error definition each degree of freedom: the error norm the relative error norm the relative error norm the
L2 Error Wiki
maximum error by indicating in addition the number and coordinates of the node where the maximum occurs. Preprocessor NORMXX compares the calculated solution with the exact
Root Mean Square Error
solution for those cases where the solution to a problem is known analytically. It calls module NORME: SUBROUTINE NORME (M,XM,DM,NFMAIL,NIMAIL,NFCOOR,NICOOR,NFB,NIB, + NFBS,NIBS,INDICB,NSM,FONINT,SOLEX,DSOLEX) C ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ C AIM : IPRINT THE EXACT SOLUTION, THE CALCULATED SOLUTION, C --- THE ABSOLUTE AND RELATIVE DIFFERENCES BETWEEN THEM, C THE L1,L2 ERRORS AND MAX C ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ where M, l2 error matlab XM and DM designates the super array, NFMAIL, NIMAIL are the file number and level of structure MAIL, NFCOOR, NICOOR are the file number and level of structure COOR, NFB, NIB are the file number and level of structure B, NFBS, NIBS are the file number and level of structure B on exit, INDICB is the save option: 1 : the error is stored in NFBS (used in this case), 0 if not, NSM is the number of the load case to consider (between 1 and NDSM), FONINT is a logical set to .TRUE. if functions SOLEX or DSOLEX are input as interpreted functions, and set to .FALSE. if they are input in the classical manner, SOLEX, DSOLEX are the functions used to input the exact solution (in single or double precision). Depending on the value of FONINT, functions SOLEX or DSOLEX must be written using the following format: FUNCTION SOLEX(I,X,Y,Z) DOUBLE PRECISION FUNCTION DSOLEX(I,X,Y,Z) where I
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L2 Cache Error
more about hiring developers or posting ads with us Computational Science beta Questions Tags Users l2 error code dryer Badges Unanswered Ask Question _ Computational Science Stack Exchange is a question and answer site for scientists using computers to solve scientific problems. critical error l2 interlude Join them; it only takes a minute: Sign up Here's how it works: Anybody can ask a question Anybody can answer The best answers are voted up and rise to the top Why a finite difference scheme would https://www.rocq.inria.fr/modulef/Doc/GB/Guide6-10/node21.html give second order of accuracy in norm L2 but 1.5 with L1 (while 1 with Linf)? up vote 2 down vote favorite My finite difference scheme for the 2D Euler equations is second order accurate in theory, since all the terms are second order accurate, with the advective terms being third order. So I expect a rate of convergence of about 2 when using successive refinements. However, I know that the discretization at the boundary reduces http://scicomp.stackexchange.com/questions/19746/why-a-finite-difference-scheme-would-give-second-order-of-accuracy-in-norm-l2-bu to first order (I am working on that). These are the values I compute (Dx=4 m, and coarser grid is 40X40 cells): |------|----------------------------------------------------------------- | grid | Norm L1 | Rate L1 | Norm L2 | Rate L2 | Norm Linf | Rate Linf |------|---------------------------------------------------------------- |Dx | 0.0264 | / |0.0017799 | / | 0.149 | / |Dx/2 | 0.0088 | 1.58 |0.0003712 | 2.26 | 0.075 | 1.00 |Dx/4 | 0.0032 | 1.40 |0.0000858 | 2.09 | 0.043 | 0.78 |Dx/8 | 0.0009 | 1.70 |0.0000159 | 2.38 | 0.024 | 0.90 |Dx/16 | 0.0003 | 1.45 |0.0000038 | 2.10 | 0.013 | 0.74 |Dx/32 | 0.0001 | 1.51 |0.0000007 | 2.19 | 0.009 | 0.82 Therefore, I do understand that my Linf norm is first order, since the max errors are at the boundary where the scheme is only first order accurate. Also, from my results it looks like the L2 norm averages the boundary error through the domain and blurs the information, therefore still giving a global second order of accuracy (or slightly above 2). What I do not understand is: why is the L1 norm converging with a rate of about 1.5? Or better why does the norm L2 blur the error at the boundary why the L1 does not? Note: I compute the norms with: \begin{align} L_1 &= \frac{1}{N}\sum_{j=1}^N|u^{numerical}_j-u^{exact}_j| \\ L_2 &= \frac{1}{N}
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