Einführung in die Finite Element Methode Projekt 2 · Einführung in die Finite Element Methode...

Einführung in die Finite Element MethodeProjekt 2

Juri Schmelzer und Fred Brockstedt

17.7.2014

Juri Schmelzer und Fred Brockstedt Einführung in die Finite Element Methode Projekt 2 17.7.2014 1 / 29

Project 2 -Tasks

1 Study Section 3.4 of the lecture notes2a) Write conforming mesh generator2b) Write non conforming mesh generator3 Rewrite the solver for hanging nodes4 Convergence analysis

Agenda

1 Tasks

2 Mesh generator

3 Adapting the finite element solver

4 Convergence analysis

5 Demo

6 Questions

Mesh generator

TaskWrite a conforming/non-conforming mesh generator.

Definition (Mesh)

A MeshM of a computational domain Ω ⊂ R2 is a collection of cellsKkMk=1 und M = |M|.

Algorithm1 Generate nodes2 Find the corresponding elements (triangles)3 Find hanging nodes

Conforming mesh generatorSeries 4In series 4 we already wrote a mesh generator for conforming meshes, withthe help of Mesh2d.

Adaptation of a conforming mesh generator

ProblemWith every uniform refinement, refine.m squares the amount of nodes.

GoalMesh with fewer nodes. Refine towards the origin using Mesh2d.

Implementation

M = size(t,1); % find the triangle at the origint = sortrows(t, 1);ti = false(M,1);ti(1) = true;[p t] = refine(p,t,ti); % refine towards the

% triangle at the origin

ProblemMesh2d will malfunction. So patch line 129 in Mesh2d/refine.m with

t e ( s p l i t 1 , : )= s o r t ( t e ( s p l i t 1 , : ) ) ;

Non conforming mesh generator

Definition (hanging node)A node of a meshM that is located in the interior of a geometric face ofone of its cells is known as hanging node.

Definition (non conforming mesh)A non conforming mesh is a mesh with hanging nodes.

Special case for unit square refined towards the originhandle special nodes (origin, first diagonal point) that are alwayspresent.find the corresponding triangles in the lower left and upper right.find the hanging nodes on the subdiagonal.

Non conforming mesh generatorImplementation

r = r +1; % amount o f r e f i n emen t snodes = 1 . / 2 . ^ 0 : r ; % nodes on the d i a g on a l w i thou t

% 0 and 1 i n r e v e r s e o r d e rp = [0 0 ] ; % s t a r t a t the o r i g i n%% c r e a t e the t h r e e s p e z i a l nodes ,% numbered coun t e r c l o c k w i s ep ( end+1, : ) = [ nodes ( end ) 0 ] ;p ( end+1, : ) = [ nodes ( end ) nodes ( end ) ] ;p ( end+1, : ) = [0 nodes ( end ) ] ;

%% c r e a t e the t r i a n g l e st = [ 1 2 4

2 3 4] ;

the first onion ringCreate the 5 new points and the 6 new triangles.

%% c r e a t e 5 new nodesp ( end+1, : ) = [ nodes ( k ) 0 ] ; % boundary x−a x i s node

% save the p o i s i t i o n o f the f i r s t new% node f o r our t r i a n g l e c r e a t i o nnode Index = l e n g t h ( p ) ;

p ( end+1, : ) = [ nodes ( k ) nodes ( k +1) ] ;p ( end+1, : ) = [ nodes ( k ) nodes ( k ) ] ;p ( end+1, : ) = [ nodes ( k+1) nodes ( k ) ] ;p ( end+1, : ) = [0 nodes ( k ) ] ;

%% wr i t e down the 6 new t r i a n g l e s% the 2 bottom r i g h t t r i a n g l e st ( end+1, : ) = [ nodeIndex−5 node Index nodeIndex −3] ;t ( end+1, : ) = [ node Index node Index+1 nodeIndex −3] ;

% the 2 top r i g h t t r i a n g l e st ( end+1, : ) = [ nodeIndex−3 node Index+1 node Index +3] ;t ( end+1, : ) = [ node Index+1 node Index+2 node Index +3] ;

% 2 t r i a n g l e s to the top l e f tt ( end+1, : ) = [ nodeIndex−1 nodeIndex−3 node Index +4] ;t ( end+1, : ) = [ nodeIndex−3 node Index+3 node Index +4] ;

%% save the hang ing nodeshn ( end+1, : ) = [ node Index+1 node Index node Index +2] ;hn ( end+1, : ) = [ node Index+3 node Index+2 node Index +4] ;

AlgorithmIf the above points and elements have been generated, apply the followingalgorithm to refine the mesh.

1 take point on the diagonal [ 12n ,

2 make boundary node and hanging node to the left and the bottom3 create the corresponding triangles4 save the hanging nodes and their neighbours

after the last iteration remove the two hanging nodes

Generating the non conforming meshRepeat for r refinements.

Memory usage

DefinitionP is a hanging Node if the following conditions are fullfilled:

P is part of a triangle in the 2D meshP lies on a side of an other triangle and is not on of its corner

In our case they are the middlepoint of two non-hanging Nodes.The assembling part of stiffness, mass, load and neumannload needs to berewritten for it.

In our case they are the middlepoint of two non-hanging Nodes.

The assembling part of stiffness, mass, load and neumannload needs to berewritten for it.

In our case they are the middlepoint of two non-hanging Nodes.The assembling part of stiffness, mass, load and neumannload needs to berewritten for it.

How to compute the T-matrix?

The T-Matrix can be computed with the following steps:

1 If Pj(K ) is not hanging Node, than Ti ,j = 1 if Pj(K ) = Pi (M)

2 If Pj(K ) is a hanging Node, than get the neighbours of Pj(K ) i.e.Q1,Q2 and set Ti ,j = 1

2 if P(Q1) = Pi (M) ∨ P(Q2) = Pi (M) withP(Qi ) describing the index of Qi in the mesh

3 Any other entry is set to 0.With this we get a sparse matrix with 5 entries at most for our mesh.

The T-Matrix can be computed with the following steps:1 If Pj(K ) is not hanging Node, than Ti ,j = 1 if Pj(K ) = Pi (M)

2 If Pj(K ) is a hanging Node, than get the neighbours of Pj(K ) i.e.Q1,Q2 and set Ti ,j = 1

2 if P(Q1) = Pi (M) ∨ P(Q2) = Pi (M) withP(Qi ) describing the index of Qi in the mesh

3 Any other entry is set to 0.With this we get a sparse matrix with 5 entries at most for our mesh.

defining the coordinates of the triangle, initialising the T-Matrix andsearching for hanging nodes in the triangle and which of them is a hangingNode

f o r k=1:Mc o o r d i n a t e s = p ( t ( k , : ) , : ) ;containHangNode=ismember ( t ( k , : ) , hang ( : , 1 ) ) ;T=z e r o s ( l e n g t h ( p ) , 3 ) ;m=f i n d ( containHangNode ) ;Q=[1 , 2 , 3 ] ;Q(m)= [ ] ;

If triangle without hanging node than creating the usual T-Matrix

i f ( abs ( containHangNode)==0)T( t ( k , 1 ) , 1 )=1 ;T( t ( k , 2 ) , 2 )=1 ;T( t ( k , 3 ) , 3 )=1 ;

setting 1 for nonhanging nodes in the triangle and 12 on the neighbours of

the hanging Nodes.

e l s ef o r l =1:3−max( s i z e (m) )T( t ( k ,Q( l ) ) ,Q( l ))=1;

endf o r l =1:max( s i z e (m) )

[~ , hang index ]= ismember ( t ( k ,m( l ) ) , hang ( : , 1 ) ) ;T( hang ( hang index , 2 ) ,m( l ))=1/2;T( hang ( hang index , 3 ) ,m( l ))=1/2;

endend

Calculating∑

K TKBKTTK = B and removing of hanging node rows and

columns

A=A+spa r s e (T)∗ e l emS t i f f n e s sP 1 ( c o o r d i n a t e s ) ∗ . . .s p a r s e (T ’ ) ;

endA( hang ( : , 1 ) , : ) = [ ] ; A ( : , hang ( : , 1 ) ) = [ ] ;

Analytic solution

Problem

−∆u + u = f in Ω = [0, 1]2

∇u · n = g on ∂Ω

TaskChoose f and g such that u := e(−γ(x+y))

Solutionf = e(−γ(x+y)) − 2γ2e−γ(x+y)

(−e−(x+y)γγ

−e−(x+y)γγ

Exact solution of u

Analytic solution

TaskChoose γ > 0 and compute the discretization error in the energy norm.

Remember Series 6

||en||e :=√a(u − un, u − un)

u is known and un is the solution of the solver.

a(u − un, u − un) =

∫Ωfudx︸︷︷︸

solve analyticaly

−∫

Ωfundx︸︷︷︸u·f

Solution∫ 10

∫ 10 f (x , y)u(x , y)dxdy = 1

((1− 2γ2)(e−2γ − 1)2)

Convergence Analysis

Convegence Analysis

nodes conform hanging uniform9 0.0188 0.0072 0.007215 0.0857 0.0042 ≈ 0.007018 0.0549 0.0040 -24 0.0224 0.0040 ≈ 0.006027 0.0181 0.0040 -36 0.0163 0.0040 0.005148 0.0162 0.0040 ≈ 0.004451 0.0162 0.0040 -

Questions

Questions, remarks and comments!

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