A natural finite element for axisymmetric problem Fran cois Dubois - PowerPoint PPT Presentation

European Finite Element Fair 4 ETH Z¨ urich, 2-3 June 2006 A natural finite element for axisymmetric problem Fran¸ cois Dubois CNAM Paris and University Paris South, Orsay conjoint work with Stefan Duprey University Henri Poincar´ e 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 Cl´ ement’s interpolation 6) Numerical analysis 7) Conclusion ETH Z¨ urich, 2-3 June 2006

 Axi-symmetric model problem Motivation : solve the Laplace equation in a axisymmetric domain Find a solution of the form u ( r, z ) exp( i θ ) Change the notation : x ≡ z , y ≡ r Consider the meridian plane Ω of the axisymmetric domain ∂ Ω = Γ 0 ∪ Γ D ∪ Γ N , Γ 0 ∩ Γ D = Ø , Γ 0 ∩ Γ N = Ø , Γ D ∩ Γ N = Ø , Γ 0 is the intersection of Ω with the “axis” y = 0 Then the function u is solution of � � − ∂ 2 u ∂x 2 − 1 ∂ y ∂u + u y 2 = f in Ω y ∂y ∂y ∂u Boundary conditions : u = 0 on Γ D , ∂n = g on Γ N ETH Z¨ urich, 2-3 June 2006

 Axi-Sobolev spaces Test function v null on the portion Γ D of the boundary Integrate by parts relatively to the measure y d x d y. � � u v Bilinear form a ( u , v ) = y ∇ u • ∇ v d x d y + d x d y y Ω Ω � � Linear form < b , v > = f v y d x d y + g v y d γ . Ω Γ N Two notations: √ ( x , y ) = √ y u ( x , y ) , 1 √ ( x , y ) = u √ y u ( x , y ) , u ( x , y ) ∈ Ω . Sobolev spaces: √ ∈ L 2 (Ω) } L 2 a (Ω) = { v : Ω − → I R , v √ ∈ (L 2 (Ω)) 2 } √ ∈ L 2 (Ω) , ( ∇ v ) H 1 a (Ω) = { v ∈ L 2 a (Ω) , v � v ∈ H 1 � √ ∈ (L 2 (Ω)) 2 , a (Ω) , v √ √ √ ∈ L 2 (Ω) , ( ∇ v ) H 2 a (Ω) = √ ∈ (L 2 (Ω)) 4 . (d 2 v ) ETH Z¨ urich, 2-3 June 2006

 Axi-Sobolev spaces Norms and semi-norms: � y | v | 2 d x d y � v � 2 0 , a = Ω � 1 � � y | v | 2 + y |∇ v | 2 | v | 2 � v � 2 1 , a = � v � 2 0 , a + | v | 2 1 , a = d x d y , 1 , a Ω � 1 � � y 3 | v | 2 + 1 y |∇ v | 2 + y | d 2 v | 2 | v | 2 2 , a = d x d y , Ω � v � 2 2 , a = � v � 2 1 , a + | v | 2 2 , a The condition u = 0 on Γ 0 is incorporated inside the choice of the axi-space H 1 a (Ω) . Sobolev space that takes into account the Dirichlet boundary condition V = { v ∈ H 1 a (Ω) , γv = 0 on Γ D } . ETH Z¨ urich, 2-3 June 2006

 Axi-Sobolev spaces � u ∈ V Variational formulation: a ( u , v ) = < b , v > , ∀ v ∈ V . a ( v , v ) = | v | 2 ∀ v ∈ H 1 We observe that 1 , a , a (Ω) , The existence and uniqueness of the solution of problem is (relatively !) easy according to the so-called Lax-Milgram-Vishik’s lemma. See the article of B. Mercier and G. Raugel ! ETH Z¨ urich, 2-3 June 2006

 Discrete formulation Very simple, but fundamental remark v ( x , y ) = √ y ( a x + b y + c ) , ( x , y ) ∈ K ∈ T 2 , Consider � � √ y ∇ v ( x , y ) = a y , 1 Then we have 2 ( a x + 3 b y + c ) . P 1 : the space of polynomials of total degree less or equal to 1 √ ∈ ( P 1 ) 2 . √ ∈ P 1 We have v = ⇒ ( ∇ v ) A two-dimensional conforming mesh T T 0 set of vertices T 1 set of edges T 2 set of triangular elements. ETH Z¨ urich, 2-3 June 2006

 Discrete formulation √ √ ∈ P 1 } . Linear space P = { v, v 1 δ S , v > for v regular, S ∈ T 0 : < � Degrees of freedom < � √ ( S ) δ S , v > = v Proposition 1 . Unisolvance property. K ∈ T 2 be a triangle of the mesh T , δ S , • >, S ∈ T 0 ∩ ∂K Σ the set of linear forms < � √ P defined above. 1 √ Then the triple ( K , Σ , P 1 ) is unisolvant. Proposition 2 . Conformity of the axi-finite element √ 1 ) is conforming in space C 0 (Ω) . The finite element ( K , Σ , P Conformity in the axi-space H 1 Proposition 3 . a (Ω) . √ is included in the axi-space H 1 The discrete space H a (Ω) : √ T ⊂ H 1 H a (Ω) . T ETH Z¨ urich, 2-3 June 2006

 Numerical results for an analytic test case Ω =]0 , 1[ 2 , Γ D = Ø Parameters α > 0 , β > 0, �� � � x β α 2 − 1 Right hand side: f ( y, x ) ≡ y α y 2 + β ( β − 1) x β − 2 Neumann datum: g ( x, y ) = α if y = 1, − βy α x β − 1 if x = 0, βy α if x = 1. Solution: u ( x, y ) ≡ y α x β . Comparison between the present method (DD) the use of classical P 1 finite elements (MR) ETH Z¨ urich, 2-3 June 2006

 Numerical results for an analytic test case ETH Z¨ urich, 2-3 June 2006

 Numerical results for an analytic test case ETH Z¨ urich, 2-3 June 2006

 Numerical results for an analytic test case Numerical study of the convergence properties Test cases : α = 1 / 4, α = 1 / 3, α = 2 / 3 β = 0, β = 1, β = 2 Three norms: � v � 0 , a | v | 1 , a � v � ℓ ∞ Order of convergence easy (?) to see. Example : β = 0 and α = 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 Z¨ urich, 2-3 June 2006

 Numerical results for an analytic test case ETH Z¨ urich, 2-3 June 2006

 Numerical results for an analytic test case ETH Z¨ urich, 2-3 June 2006

 Analysis ? Discrete space for the approximation of the variational problem: √ V T = H T ∩ V . � u T ∈ V T Discrete variational formulation: a ( u T , v ) = < b , v > , ∀ v ∈ V T . Estimate the error � u − u T � 1 , a Study the interpolation error � u − Π T u � 1 , a What is the interpolate Π T u ?? Proposition 4 . Lack of regularity. Hypothesis: u ∈ H 2 a (Ω) . √ belongs to the space H 1 (Ω) and √ � 1 , Ω ≤ C � u � 2 , a Then u � u ETH Z¨ urich, 2-3 June 2006

 Analysis ? 2 y √ y u ∇ y + 1 1 √ . Introduce v ≡ u Small calculus: ∇ v = − √ y ∇ u . � � 1 | v | 2 d x d y ≤ y | u | 2 d x d y ≤ C � u � 2 Then 2 , a Ω Ω � � � 1 y |∇ u | 2 � 4 y 3 | u | 2 + 1 |∇ v | 2 d x d y ≤ 2 d x d y ≤ C � u � 2 2 , a . Ω Ω Derive (formally !) two times: 3 1 1 d 2 v = √ y d 2 u 4 y 2 √ y u ∇ y • ∇ y − y √ y ∇ u • ∇ y + Even if u is regular, v has no reason to be continuous. ETH Z¨ urich, 2-3 June 2006

About Cl´  ement’s interpolation S Vicinity Ξ S of the vertex S ∈ T 0 . � 1 < δ C S ∈ T 0 Degree of freedom S , v > = v ( x ) d x d y , | Ξ S | Ξ S � Π C v = < δ C Cl´ ement’s interpolation: S , v > ϕ S . S ∈T 0 ETH Z¨ urich, 2-3 June 2006

About Cl´  ement’s interpolation K Vicinity Z K for a given triangle K ∈ T 2 . | v − Π C v | 0 , K ≤ C h T | v | 1 , Z K , | v − Π C v | 1 , K ≤ C | v | 1 , Z K , | v − Π C v | 1 , K ≤ C h T | v | 2 , Z K . ETH Z¨ urich, 2-3 June 2006

 Numerical analysis � √ � √ Π C u Interpolate Π u by conjugation: Π u = � � Π u ( x, y ) = √ y Π C v ( x, y ) ∈ K ∈ T 2 ( x, y ) , id est Theorem 1 . An interpolation result. Relatively strong hypotheses concerning the mesh T Let u ∈ H 2 a (Ω) and Π u defined above. Then we have � u − Π u � 1 , a ≤ C h T � u � 2 , a . � � 1 1 y | u − √ y Π C v | 2 d x d y y | u − Π u | 2 d x d y = Ω Ω � | v − Π C v | 2 d x d y = � v − Π C v � 2 = 0 , Ω Ω ≤ C h 2 T | v | 2 1 , Ω ≤ C h 2 T � u � 2 2 , a ETH Z¨ urich, 2-3 June 2006

 Numerical analysis � √ y �� � � � � � 1 ∇ y + √ y ∇ v − Π C v v − Π C v v − Π C v ∇ = . 2 √ y � � � | 2 d x d y ≤ y |∇ u − Π u � � Ω � � | v − Π C v | 2 d x d y + 2 y 2 |∇ | 2 d x d y v − Π C v ≤ Ω Ω Ω + = { K ∈ T 2 , dist ( Z K , Γ 0 ) > 0 } Ω − = Ω \ Ω + . ETH Z¨ urich, 2-3 June 2006

 Numerical analysis K S Γ T θ 0 Triangle element K that belongs to the sub-domain Ω + . ETH Z¨ urich, 2-3 June 2006

 Numerical analysis Theorem 2 . First order approximation relatively strong hypotheses concerning the mesh T u ∈ H 2 u solution of the continuous problem: a (Ω) , Then we have � u − u T � 1 , a ≤ C h T � u � 2 , a . Proof: classical with Cea’s lemma! ETH Z¨ urich, 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 Z¨ urich, 2-3 June 2006