# a new axisymmetric finite element

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European Finite Element Fair 4 ETH Z¨ urich, 2-3 June 2006 A natural ﬁnite element for axisymmetric problem Stefan Duprey University Henri Poincar´ e Nancy and EADS Suresnes.

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• European Finite Element Fair 4

ETH Zurich, 2-3 June 2006

A natural finite element for axisymmetric problem

Francois Dubois

CNAM Paris and University Paris South, Orsay

conjoint work with

Stefan Duprey

University Henri Poincare Nancy and EADS Suresnes.

• A natural finite element for axisymmetric problem

1) Axi-symmetric model problem

2) Axi-Sobolev spaces

3) Discrete formulation

4) Numerical results for an analytic test case

6) Numerical analysis

7) Conclusion

ETH Zurich, 2-3 June 2006

• Axi-symmetric model problem

Motivation : solve the Laplace equation in a axisymmetric domainFind a solution of the form u(r, z) exp(i )Change the notation : x z , y r

Consider the meridian plane of the axisymmetric domain

= 0 D N , 0 D = , 0 N = , D N = ,0 is the intersection of with the axis y = 0

Then the function u is solution of

2u

x2 1

y

y

(yu

y

)+

u

y2= f in

Boundary conditions : u = 0 on D ,u

n= g on N

ETH Zurich, 2-3 June 2006

• Axi-Sobolev spaces

Test function v null on the portion D of the boundaryIntegrate by parts relatively to the measure y dxdy.

Bilinear form a(u , v) =

yu v dxdy +

u v

ydxdy

Linear form < b , v >=

f v y dxdy +

N

g v y d .

Two notations:

u (x , y) =1yu(x , y) , u

(x , y) =

y u(x , y) , (x , y) .

Sobolev spaces:L2a() = {v : IR , v

L2()}H1a() = {v L2a() , v L2() , (v)

(L2())2}

H2a() =

{v H1a() , v L2() , (v) (L2())2 ,

(d2v) (L2())4

}.

ETH Zurich, 2-3 June 2006

• Axi-Sobolev spaces

Norms and semi-norms:

v20, a =

y |v|2 dxdy

|v|21, a =

(1

y|v|2 + y |v|2

)dxdy , v21, a = v20, a + |v|21, a

|v|22, a =

(1

y3|v|2 + 1

y|v|2 + y |d2v|2

)dxdy ,

v22, a = v21, a + |v|22, aThe condition u = 0 on 0is incorporated inside the choice of the axi-space H1a().

Sobolev space that takes into account the Dirichlet boundary conditionV = {v H1a() , v = 0 on D} .

ETH Zurich, 2-3 June 2006

• Axi-Sobolev spaces

Variational formulation:

{u Va(u , v) =< b , v > , v V .

We observe that a(v , v) = |v|21, a , v H1a() ,

The existence and uniqueness of the solution of problem is (relatively !)easy according to the so-called Lax-Milgram-Vishiks lemma.

See the article of B. Mercier and G. Raugel !

ETH Zurich, 2-3 June 2006

• Discrete formulation

Very simple, but fundamental remark

Consider v(x , y) =y (a x + b y + c) , (x , y) K T 2 ,

Then we havey v(x , y) = (a y , 1

2(a x + 3b y + c)

).

P1 : the space of polynomials of total degree less or equal to 1

We have v P1 = (v) (P1)2 .

A two-dimensional conforming mesh TT 0 set of verticesT 1 set of edgesT 2 set of triangular elements.

ETH Zurich, 2-3 June 2006

• Discrete formulation

Linear space P1 = {v, v P1}.

Degrees of freedom < S , v > for v regular, S T 0 : < S , v >= v (S)

Proposition 1. Unisolvance property.K T 2 be a triangle of the mesh T , the set of linear forms < S , >, S T 0 KP1 defined above.

Then the triple (K , , P1 ) is unisolvant.

Proposition 2. Conformity of the axi-finite element

The finite element (K , , P1 ) is conforming in space C0().

Proposition 3. Conformity in the axi-space H1a().

The discrete space HT is included in the axi-space H

1a() :HT H1a() .

ETH Zurich, 2-3 June 2006

• Numerical results for an analytic test case

=]0, 1[2 , D =

Parameters > 0, > 0,

Right hand side: f (y, x) y[(2 1) x

y2+ ( 1)x2

]Neumann datum:g(x, y) = if y = 1, yx1 if x = 0, y if x = 1.Solution: u(x, y) yx .

Comparison betweenthe present method (DD)the use of classical P1 finite elements (MR)

ETH Zurich, 2-3 June 2006

• Numerical results for an analytic test case

ETH Zurich, 2-3 June 2006

• Numerical results for an analytic test case

ETH Zurich, 2-3 June 2006

• Numerical results for an analytic test case

Numerical study of the convergence properties

Test cases : = 1/4, = 1/3, = 2/3 = 0, = 1, = 2

Three norms: v0, a |v|1, a v

Order of convergence easy (?) to see.

Example : = 0 and = 2/3:our axi-finite element has a rate of convergence 3 for the 0, a norm.

Synthesis of these experiments:

same order of convergence than with the classical approach

errors much more smaller!

ETH Zurich, 2-3 June 2006

• Numerical results for an analytic test case

ETH Zurich, 2-3 June 2006

• Numerical results for an analytic test case

ETH Zurich, 2-3 June 2006

• Analysis ?

Discrete space for the approximation of the variational problem:

VT = HT V .

Discrete variational formulation:

{uT VTa(uT , v) =< b , v > , v VT .

Estimate the error u uT 1, aStudy the interpolation error uT u 1, aWhat is the interpolate T u ??

Proposition 4. Lack of regularity.Hypothesis: u H2a().Then u belongs to the space H1() and u 1, C u2, a

ETH Zurich, 2-3 June 2006

• Analysis ?

Introduce v u . Small calculus: v = 12yyuy+ 1

yu .

Then

|v|2 dxdy

1

y|u|2 dxdy C u22, a

|v|2 dxdy 2

( 14y3

|u|2 + 1y|u|2

)dxdy C u22, a .

Derive (formally !) two times:

d2v =3

4 y2yuy y 1

yyu y + 1

yd2u

Even if u is regular, v has no reason to be continuous.

ETH Zurich, 2-3 June 2006

S

Vicinity S of the vertex S T 0.

Degree of freedom < CS , v >=1

|S |S

v(x) dxdy , S T 0

Clements interpolation: Cv =ST 0

< CS , v > S .

ETH Zurich, 2-3 June 2006

K

Vicinity ZK for a given triangle K T 2.|v Cv|

0,K C hT |v|1, ZK , |v

Cv|1,K

C |v|1, ZK

,

|v Cv|1,K

C hT |v|2, ZK .

ETH Zurich, 2-3 June 2006

• Numerical analysis

Interpolate u by conjugation: u =(Cu

)id est u(x, y) =

y(Cv

)(x, y) , (x, y) K T 2

Theorem 1. An interpolation result.Relatively strong hypotheses concerning the mesh TLet u H2a() and u defined above.Then we have uu1, a C hT u2, a .

1

y|uu|2 dxdy =

1

y|uy Cv|2 dxdy

=

|v Cv|2 dxdy = v Cv20, C h2T |v|21, C h2T u22, a

ETH Zurich, 2-3 June 2006

• Numerical analysis

(

y(v Cv)) = 1

2y

(v Cv)y + y (v Cv) .

y |(uu)|2 dxdy

|v Cv|2 dxdy + 2

y2 |(v Cv) |2 dxdy+ = {K T 2 , dist (ZK , 0) > 0} = \ + .

ETH Zurich, 2-3 June 2006

• Numerical analysis

S

0T

K

Triangle element K that belongs to the sub-domain +.

ETH Zurich, 2-3 June 2006

• Numerical analysis

Theorem 2. First order approximationrelatively strong hypotheses concerning the mesh Tu solution of the continuous problem: u H2a() ,Then we have u uT 1, a C hT u2, a .

Proof: classical with Ceas lemma!

ETH Zurich, 2-3 June 2006

• Conclusion

Axi-finite element

Interpolation properties founded of the underlying axi-Sobolev space

First numerical tests: good convergence properties

Numerical analysis based on Mercier-Raugel contribution (1982)