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

    5) About Clements interpolation

    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

  • About Clements interpolation

    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

  • About Clements interpolation

    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)

    See also Gmati (1992), Bernardi et al. (1999)

    May be all the material presented here is well known ?!

    ETH Zurich, 2-3 June 2006