Finite volume method for linear and non linear elliptic problems - PowerPoint PPT Presentation

Finite volume method for linear and non linear elliptic problems with discontinuities Franck BOYER et Florence HUBERT L.A.T.P . - Marseille, FRANCE DD17 - 2006 1/ 37

Plan P LAN 1 I NTRODUCTION 2 T HE DDFV SCHEME 3 T HE 1D PROBLEM 4 T HE 2D PROBLEM 5 A SADDLE - POINT ALGORITHM 6 N UMERICAL RESULTS 2/ 37

Introduction P LAN 1 I NTRODUCTION The DDFV scheme 2 The 1D problem 3 The 2D problem 4 A saddle-point algorithm 5 Numerical results 6 3/ 37

Introduction I NTRODUCTION . ◮ DDFV scheme (D ISCRETE D UALITY F INITE V OLUME ) for � − div ( ϕ ( z , ∇ u e ( z ))) = f ( z ) , in Ω , (1) u e = 0 , on ∂ Ω , Ω polygonal open set in R 2 . u �→ − div ( ϕ ( · , ∇ u )) is coercitive, monotonous (of Leray-Lions type). ϕ presents discontinuities with respect to the space variable z (Transmission problem). Example : Ω 1 Ω 2 “ ” − div |∇ u | p − 2 ∇ u = f 1 k ( x ) |∇ u | p − 2 ∇ u − div = f 2 4/ 37

Introduction I NTRODUCTION With discontinuities the DDFV scheme converges but slowly : In the linear case : 1D : − ( λ ± u e ′ ) ′ = f Arythmetic mean-value : order 1 2 Harmonic mean-value : order 1 16 mailles, avec lambda=1si x<0.5 et 10 sinon 0.045 exact moyenne arithmétique 0.04 moyenne harmonique 0.035 0.03 0.025 0.02 0.015 0.01 0.005 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 5/ 37

Introduction I NTRODUCTION With discontinuities, the DDFV scheme converges but slowly : In the linear case : 2D : − div ( A ( z ) ∇ u e ) = f DDFV scheme : order 1 2 Improved DDFV scheme (Hermeline, BH) : order 1 −1 −2 −1 10 10 10 ordre 0.52 ordre 0.96 ordre 1.04 ordre 1 ordre 1.99 ordre 2.02 −2 10 −3 10 −3 −2 10 10 −4 10 −4 10 −5 −5 −3 10 10 10 −3 −2 −1 −3 −2 −1 −3 −2 −1 10 10 10 10 10 10 10 10 10 L ∞ error L 2 error H 1 error � 10 � 2 A = . 2 1 6/ 37

Introduction B IBLIOGRAPHIE Anisotropy problems with discontinuities EGH (00), Hermeline (03) Gradient reconstruction problems “O-scheme”, “U-scheme” Aavatsmark, Lepotier,...... Gradient FV schemes. Eymard, Gallou¨ et, Herbin,... Mixed FV schemes. Droniou, Eymard (06) DDFV schemes. Coudi` ere (99), Hermeline (00), Domelevo & Omn` es (05), ABH (06), Pierre (06), Delcourte & al (06)....... 7/ 37

Introduction H YPOTH ` ESES SUR ϕ Let p ∈ ] 1 , ∞ [ , p ′ = p − 1 and f ∈ L p ′ (Ω) . ◮ p ≥ 2 in this talk. p ϕ : Ω × R 2 → R 2 is a Caratheodory function such that : | ξ | p − 1 � � ( ϕ ( z , ξ ) , ξ ) ≥ C ϕ , ( H 1 ) � | ξ | p − 1 + 1 � | ϕ ( z , ξ ) | ≤ C ϕ . ( H 2 ) ( ϕ ( z , ξ ) − ϕ ( z , η ) , ξ − η ) ≥ 1 | ξ − η | p . ( H 3 ) C ϕ 1 + | ξ | p − 2 + | η | p − 2 � � | ϕ ( z , ξ ) − ϕ ( z , η ) | ≤ C ϕ | ξ − η | . ( H 4 ) ϕ is piecewisely lipschitz in z ⇒ Assumption ( H 5 ) . 8/ 37

The DDFV scheme P LAN Introduction 1 2 T HE DDFV SCHEME The 1D problem 3 The 2D problem 4 A saddle-point algorithm 5 Numerical results 6 9/ 37

The DDFV scheme T HE DDFV SCHEME The discrete unknowns : where u M = ( u K ) K ∈ M , u M ∗ = ( u K∗ ) K∗ ∈ M ∗ u T = u M , u M ∗ � � The discrete gradient : ∇ T u T is constant on diamond cells � u L − u K ν + u L∗ − u K∗ � 1 D u T = ν ∗ ∇ T , ∀D . | σ ∗ | sin α D | σ | x K∗ u K + u K∗ 2 u L + u K∗ 2 x K x L u K + u L∗ 2 u L + u L∗ 2 x L∗ 10/ 37

The DDFV scheme T HE DDFV MESHES primal, dual and “diammond”. x L∗ L∗ x L x K L K∗ K x K∗ maillage M ∗ maillage D maillage M 11/ 37

The DDFV scheme T HE DDFV SCHEME � D u T ) , ν K ) = � | σ | ( ϕ D ( ∇ T − f ( z ) dz , ∀ K ∈ M , K D σ,σ ∗ ∩ K � = ∅ � D u T ) , ν K∗ ) = K∗ f ( z ) dz , ∀ K∗ ∈ M ∗ , � | σ ∗ | ( ϕ D ( ∇ T − D σ,σ ∗ ∩ K∗ � = ∅ with ϕ D ( ξ ) = 1 � ϕ ( z , ξ ) dz . | D | D Discrete Duality formulation : � � D u T ) , ∇ T D v T ) = fv M ∗ dz , ∀ v T ∈ R T . � fv M dz + | D | ( ϕ D ( ∇ T 2 Ω Ω D ∈ D 12/ 37

The DDFV scheme K NOWN RESULTS Case p = 2 : Domelevo & Omn` es ⇒ Estimate in O ( h ) for a large class of meshes. General case : Andreianov, Boyer & Hubert ⇒ The scheme converges. ⇒ If u e ∈ W 2 , p (Ω) and � � ∂ϕ � 1 + | ξ | p − 1 � � � , ∀ ξ ∈ R 2 . ϕ Lip. on Ω , with ∂ z ( z , ξ ) � ≤ C ϕ ( H 5 ) � � � Then 1 � u e − u M � L p + � u e − u M ∗ � L p + �∇ u e −∇ T u T � L p ≤ C size ( T ) p − 1 . 13/ 37

The DDFV scheme Z OOM ON THE DIAMOND CELLS Diamond cells are supposed to be convex. Each diamond cell is cut into four triangles Q . x K∗ x K∗ | σ K∗ | Q K , K∗ ν x K x K Q L , K∗ τ ν ∗ τ ∗ x D Q K , L∗ Q L , L∗ α D | σ L∗ | x L x L | σ K | x L∗ x L∗ | σ L | 14/ 37

The DDFV scheme A IMS If ϕ presents some discontinuities on a curve Γ u e �∈ W 2 , p (Ω) . The numerical fluxes are no more consistent along Γ . We assume that on each Q , ϕ is Lip. and saitifies ( H 5 ) . ◮ We want to get the consistency of the fluxes on Γ . ◮ We contruct a new approximation ϕ N D of the non linearity on the diamond cell D . � D u T ) , ν K ) = � | σ | ( ϕ N D ( ∇ T − f ( z ) dz , ∀ K ∈ M K D σ,σ ∗ ∩ K � = ∅ � D u T ) , ν K∗ ) = K∗ f ( z ) dz , ∀ K∗ ∈ M ∗ � | σ ∗ | ( ϕ N D ( ∇ T − D σ,σ ∗ ∩ K∗ � = ∅ 15/ 37

The 1D problem P LAN Introduction 1 The DDFV scheme 2 3 T HE 1D PROBLEM The 2D problem 4 A saddle-point algorithm 5 Numerical results 6 16/ 37

The 1D problem T HE 1D PROBLEM � ϕ − ( · ) , if x < 0 , Ω =] − 1 , 1 [ , ϕ ( x , · ) = . ϕ + ( · ) , if x > 0 Let x 0 = − 1 < . . . < x N = 0 < . . . < x N + M = 1 a discretization of [ − 1 , 1 ] . The 1D FV scheme reads for i ∈ { 0 , N + M − 1 } : � x i + 1 − F i + 1 + F i = f ( x ) dx . (2) x i with u i + 1 2 − u i − 1 F i = ϕ ( x i , ∇ i u T ) , ∇ i u T = 2 , ∀ i � = N , (3) x i + 1 2 − x i − 1 2 QUESTION : how to define F N ? 17/ 37

The 1D problem T HE NEW GRADIENT We look for ˜ u such that 2 − ˜ ˜ u N + 1 u u − u N − 1 N u T = N u T = ∇ + ∇ − 2 , , h − h + N N we have N u T ) = ϕ + ( ∇ − N u T ) . ϕ − ( ∇ + u ˜ u N + 1 δ 2 u ¯ u N − 1 2 x N = 0 h − h + N N 18/ 37

The 1D problem T HE NEW GRADIENT We look for ˜ u in the form h − 2 + h + N u N + 1 N u N − 1 u = ¯ ˜ u + δ, with ¯ u = 2 . h − N + h + N that is N u T = ∇ N u T − δ N u T = ∇ N u T + δ ∇ + , and ∇ − . h − h + N N u ˜ u N + 1 δ 2 u ¯ u N − 1 2 x N = 0 h − h + N N 18/ 37

The 1D problem T HE NEW GRADIENT T HEOREM For all u T ∈ R N , there exists a unique δ N ( ∇ N u T ) such that ∇ N u T + δ N ( ∇ N u T ) ∇ N u T − δ N ( ∇ N u T ) � � � � def F N = ϕ − = ϕ + , h − h + N N The new scheme admits a unique solution. 1 p − 1 . The flux F N is consistent with an error in h 19/ 37

The 1D problem E XAMPLE For two p − laplacian fluxes ϕ − ( ξ ) = k − | ξ + G − | p − 2 ( ξ + G − ) , and ϕ + ( ξ ) = k + | ξ + G + | p − 2 ( ξ + G + ) , where k − , k + ∈ R + and G − , G + ∈ R 2 . We obtain p − 1 �  1 1  p − 1 p − 1 ( h − N + h + N ) p − 2 �  k k � ∇ N u T + G � ∇ N u T + G � − + F N = , � �  1 1 � p − 1 + h − p − 1 h + N k N k − + where G is some arythmetic mean value of G − and G + defined by G = h − N G − + h + N G + . h − N + h + N Warning : the fluxes are not explicit in general ! 20/ 37

The 2D problem P LAN Introduction 1 The DDFV scheme 2 The 1D problem 3 4 T HE 2D PROBLEM A saddle-point algorithm 5 Numerical results 6 21/ 37

The 2D problem T HE NEW GRADIENT D u T is constant on each quarter of diamond ◮ ∇ N D u T = Q u T , � ∇ N 1 Q ∇ N Q ∈ Q D x K 1 2 ( u K + u K∗ ) � u σ K∗ − 1 2 ( u K + u K∗ ) 2 u σ K T = ∇ QK , K∗ u N ν x σ K sin α D | σ K | Q K , K∗ + u σ K − 1 2 ( u K + u K∗ ) ν ∗ � x K∗ x σ K∗ x D | σ K∗ | u σ K∗ That is Q u T = ∇ T D u T + B Q δ D , δ D ∈ R 4 , ∇ N where δ D is to be determined 1 | QK , K∗ | ( | σ K | ν ∗ , 0 , | σ K∗ | ν , 0 ) B QK , K∗ = 22/ 37

The 2D problem C ONSISTENCY OF THE FLUX We note � ϕ Q ( ξ ) = ϕ ( z , ξ ) d µ ¯ Q ( z ) . Q and choose δ D ∈ R 4 such that � D u T + B QK , K∗ δ D ) , ν ∗ � � D u T + B QK , L∗ δ D ) , ν ∗ � ϕ QK , K∗ ( ∇ T ϕ QK , L∗ ( ∇ T = � D u T + B QL , K∗ δ D ) , ν ∗ � � D u T + B QL , L∗ δ D ) , ν ∗ � ϕ QL , K∗ ( ∇ T ϕ QL , L∗ ( ∇ T = � D u T + B QK , K∗ δ D ) , ν � � D u T + B QL , K∗ δ D ) , ν � ϕ QK , K∗ ( ∇ T ϕ QL , K∗ ( ∇ T = � D u T + B QK , L∗ δ D ) , ν � � D u T + B QL , L∗ δ D ) , ν � ϕ QK , L∗ ( ∇ T ϕ QL , L∗ ( ∇ T = ⇒ For all u T ∈ R T , and all diamond cell D , there exists a unique D u T ) ∈ R 4 that ensures such equalities. δ D ( ∇ T 23/ 37