Superconvergence in the finite element method

Zbigniew Leyk

Mathematica Applicanda (1982)

  • Volume: 10, Issue: 20
  • ISSN: 1730-2668

Abstract

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For some variants of the finite element method there exist points having a remainder value or a derivation remainder remarkably less than those given by global norms. This phenomenon is called superconvergence and the points are called superconvergence points. The generalized problem corresponding to (1) is as follows: Let Hk(Ω) be Sobolev space and Hk0(Ω) the completion of the space C∞0(Ω) with norm ∥⋅∥k,Ω. Find u∈H10(Ω) such that for each v∈H10(Ω), (2) a(u,v)=(f,v)0 holds, where a(u,v)=∫Ω(∑n|α|=0aα(x)DαuDαv)dx, (f,v)0=∫Ωf(x)v(x)dx, Dα=Dα11⋯Dαnn,1.5pt Dαiiu=∂αiu/∂xαii, i=1,n¯¯¯¯¯¯¯¯. The approximate problem of the finite element variant considered is the following: Find uh∈Vh such that (3) for all v∈Vh, a(uh,v)=(f,v)0. The main result is the theorem: Let ai∈C(Ω¯), D1ai,D2ai∈L∞(Ω),i=1,2,∥σ∥∞L(Ω)≤σ,f∈L2(Ω). Suppose the eigenvalues of the operator L are different from zero, and u∈H4(Ω)∩H10(Ω). Then there exists h0 such that for h≤h0, h2∑P∈G|grad(u−uh)(P)|≤Ch3(|u|3+|u|4), where u and uh are the solutions of problems (2) and (3), respectively, and C is some constant independent of h. Further, |u|k={∫Ω(∑|α|=k(Dαu)2)dx}1/2, G=⋃N1N2i=1Fi(R), R={(±3√/3,±3√/3)} is a Gauss point set in the quadrant S={(ξ1,ξ2):|ξk|≤1,k=1,2}, and Fi(F(1)i,F(2)i):S→ei, ei an element; F(1)i(ξ1,ξ2)=x(i)0+h1ξ1/2, F(2)i(ξ1,ξ2)=y(i)0+h2ξ2/2, and (x(i)0,y(i)0) is the middle element.

How to cite

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Zbigniew Leyk. "Superconvergence in the finite element method." Mathematica Applicanda 10.20 (1982): null. <http://eudml.org/doc/292686>.

@article{ZbigniewLeyk1982,
abstract = {For some variants of the finite element method there exist points having a remainder value or a derivation remainder remarkably less than those given by global norms. This phenomenon is called superconvergence and the points are called superconvergence points. The generalized problem corresponding to (1) is as follows: Let Hk(Ω) be Sobolev space and Hk0(Ω) the completion of the space C∞0(Ω) with norm ∥⋅∥k,Ω. Find u∈H10(Ω) such that for each v∈H10(Ω), (2) a(u,v)=(f,v)0 holds, where a(u,v)=∫Ω(∑n|α|=0aα(x)DαuDαv)dx, (f,v)0=∫Ωf(x)v(x)dx, Dα=Dα11⋯Dαnn,1.5pt Dαiiu=∂αiu/∂xαii, i=1,n¯¯¯¯¯¯¯¯. The approximate problem of the finite element variant considered is the following: Find uh∈Vh such that (3) for all v∈Vh, a(uh,v)=(f,v)0. The main result is the theorem: Let ai∈C(Ω¯), D1ai,D2ai∈L∞(Ω),i=1,2,∥σ∥∞L(Ω)≤σ,f∈L2(Ω). Suppose the eigenvalues of the operator L are different from zero, and u∈H4(Ω)∩H10(Ω). Then there exists h0 such that for h≤h0, h2∑P∈G|grad(u−uh)(P)|≤Ch3(|u|3+|u|4), where u and uh are the solutions of problems (2) and (3), respectively, and C is some constant independent of h. Further, |u|k=\{∫Ω(∑|α|=k(Dαu)2)dx\}1/2, G=⋃N1N2i=1Fi(R), R=\{(±3√/3,±3√/3)\} is a Gauss point set in the quadrant S=\{(ξ1,ξ2):|ξk|≤1,k=1,2\}, and Fi(F(1)i,F(2)i):S→ei, ei an element; F(1)i(ξ1,ξ2)=x(i)0+h1ξ1/2, F(2)i(ξ1,ξ2)=y(i)0+h2ξ2/2, and (x(i)0,y(i)0) is the middle element.},
author = {Zbigniew Leyk},
journal = {Mathematica Applicanda},
keywords = {Finite elements, Rayleigh-Ritz and Galerkin methods, finite methods},
language = {eng},
number = {20},
pages = {null},
title = {Superconvergence in the finite element method},
url = {http://eudml.org/doc/292686},
volume = {10},
year = {1982},
}

TY - JOUR
AU - Zbigniew Leyk
TI - Superconvergence in the finite element method
JO - Mathematica Applicanda
PY - 1982
VL - 10
IS - 20
SP - null
AB - For some variants of the finite element method there exist points having a remainder value or a derivation remainder remarkably less than those given by global norms. This phenomenon is called superconvergence and the points are called superconvergence points. The generalized problem corresponding to (1) is as follows: Let Hk(Ω) be Sobolev space and Hk0(Ω) the completion of the space C∞0(Ω) with norm ∥⋅∥k,Ω. Find u∈H10(Ω) such that for each v∈H10(Ω), (2) a(u,v)=(f,v)0 holds, where a(u,v)=∫Ω(∑n|α|=0aα(x)DαuDαv)dx, (f,v)0=∫Ωf(x)v(x)dx, Dα=Dα11⋯Dαnn,1.5pt Dαiiu=∂αiu/∂xαii, i=1,n¯¯¯¯¯¯¯¯. The approximate problem of the finite element variant considered is the following: Find uh∈Vh such that (3) for all v∈Vh, a(uh,v)=(f,v)0. The main result is the theorem: Let ai∈C(Ω¯), D1ai,D2ai∈L∞(Ω),i=1,2,∥σ∥∞L(Ω)≤σ,f∈L2(Ω). Suppose the eigenvalues of the operator L are different from zero, and u∈H4(Ω)∩H10(Ω). Then there exists h0 such that for h≤h0, h2∑P∈G|grad(u−uh)(P)|≤Ch3(|u|3+|u|4), where u and uh are the solutions of problems (2) and (3), respectively, and C is some constant independent of h. Further, |u|k={∫Ω(∑|α|=k(Dαu)2)dx}1/2, G=⋃N1N2i=1Fi(R), R={(±3√/3,±3√/3)} is a Gauss point set in the quadrant S={(ξ1,ξ2):|ξk|≤1,k=1,2}, and Fi(F(1)i,F(2)i):S→ei, ei an element; F(1)i(ξ1,ξ2)=x(i)0+h1ξ1/2, F(2)i(ξ1,ξ2)=y(i)0+h2ξ2/2, and (x(i)0,y(i)0) is the middle element.
LA - eng
KW - Finite elements, Rayleigh-Ritz and Galerkin methods, finite methods
UR - http://eudml.org/doc/292686
ER -

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