Partititioned Methods for Multifield Problems

Transcription

1 Partititioned Methods for Multifield Problems Joachim Rang, Joachim Rang Partititioned Methods for Multifield Problems Seite 1

2 Non-matching matches usually the fluid and the structure mesh do not match transfer informations from fluid to structure and vice versa use an interpolation or projection step Examples: nearest neighbour interpolation, projection methods, scattered data interpolation Joachim Rang Partititioned Methods for Multifield Problems Seite 2

3 The setting boundary conditions at the interface: u f = u s and p s n s = p f n f u f, u s... displacements p f... fluid pressure p s... stress tensor n f, n s... outer normals Joachim Rang Partititioned Methods for Multifield Problems Seite 3

4 Discretisation discretisation of u and p n: N u u(x) = N j (x)u j, j=1 N p p(x) n(x) = D j (x)p j j=1 discrete form of interface conditions U f = H fs U s and P s = H sf P f where H fs R nf u,ns u and Hsf R ns p,nf p H... transformation matrices between fluid and structure mesh Joachim Rang Partititioned Methods for Multifield Problems Seite 4

5 Energy conservation Condition: u f p f n f ds = Γ f u s p s n s ds Γ s discrete form of lhs: N f N u p f ( ) u f p f n f ds Di f (x)nj f (x)ds Γ f Γ f j=1 Let M ij ff R nf p,n f u be given by Then M ij ff := i=1 Γ f D f i (x)n f j (x)ds Γ f u f p f n f ds [M ff U f ] P f U f j P f i, Joachim Rang Partititioned Methods for Multifield Problems Seite 5

6 Energy conservation Condition: discrete form of rhs: u f p f n f ds = u s p s n s ds Γ f Γ s Γ s u s p s n s ds [M ss U s ] P s where Mss ij R ns p,ns u be given by Mss ij := Di s (x)ns j (x)ds Γ s Joachim Rang Partititioned Methods for Multifield Problems Seite 6

7 Conservation of energy It follows Then Chose [M ff U f ] P f = [M ss U s ] P s P s = [ ] M ff H fs Mss 1 Pf H sf := [ ] M ff H fs Mss Joachim Rang Partititioned Methods for Multifield Problems Seite 7

8 Consistent interpolation Aim: constant displacement and constant pressure should be exactly interpolated over the interface row-sum of H fs and H sf = [ ] M ff H fs Mss 1 should be equal to one Note, this is not the case for general transformation matrices H sf Joachim Rang Partititioned Methods for Multifield Problems Seite 8

9 Coupling methods Nearest neighbour interpolation Weighted residual method scattered data interpolation Joachim Rang Partititioned Methods for Multifield Problems Seite 9

10 Nearest neighbour interpolation Idea: Find the nearest point x A in the mesh A and assign the variable x B with the value X A transformation matrix H BA... Boolean matrix condition for exact interpolation of pressure values n f u i=1 Hfs ik xi f xk s Example: (equidistant grid) = 1, k = 1,..., n s u n f u i=1 H ik fs = xs x f, condition not satisfied in general k = 1,..., ns u Joachim Rang Partititioned Methods for Multifield Problems Seite 1

11 Weighted residual method interface condition in weak formulation: (w B (x) w A (x))λ(x)dx =, where w {u, pn}. λ... Lagrangian multiplier discretisation w A (x) = w B (x) = n A j=1 n B i=1 Γ Ψ A j (x)w Aj, Ψ B i (x)w Bi Joachim Rang Partititioned Methods for Multifield Problems Seite 11

12 Weighted residual method Then Γ λ(x) n B i=1 Ψ B i (x)w Bi dx = Γ λ(x) n A j=1 Ψ A j (x)w Aj dx mortar approach: λ(x) = n α k=1 Ψ α k (x), α {A, B} optimal for non-conforming but matching grids Joachim Rang Partititioned Methods for Multifield Problems Seite 12

13 Weighted residual method Then n B i=1 ( ) Ψ A k (x)ψ B i (x)dx W Bi = Γ n A j=1 ( ) Ψ A j (x)ψ B k (x)dx Γ W Aj Let Then Cki αb := Ψ A k (x)ψ B i (x)dx, Γ C αb W B = C alphaa W A Ckj αa := Ψ A j (x)ψ B k (x)dx Γ Joachim Rang Partititioned Methods for Multifield Problems Seite 13

14 Transformation matrix Assume that W A is known and that W B is unknown. Then transformation matrix W B = C 1 αb C αaw A H BA = C 1 αb C αa Joachim Rang Partititioned Methods for Multifield Problems Seite 14

15 Consistent approach We require C BB β B = C BA β A, where β B R n B and β A R n A approach is consistent, if Γ A = Γ B Joachim Rang Partititioned Methods for Multifield Problems Seite 15

16 Conservative approach We start with H fs = C 1 ff C fs ] Pf. Then P s = [ M ff C 1 ff C fs Mss 1 Assumption: same discretisation for displacement and pressure in the fluid, i. e. N f = D f and nu f = np f. Then M ss P s = C sf P f Only if Γ f = Γ s the approach is consistent and energy conservative need the convergence to a matching mesh. A conforming mesh is not needed Joachim Rang Partititioned Methods for Multifield Problems Seite 16

17 Scattered Data Interpolation Let the points x i = (x (1) i,..., x (d) i ) R d, d > 1 and values y i R, i = 1,..., N be given. Aim: Find a function S : R d R, which yields a data value for each point of the observation space and which satisfies the interpolation condition S(x i ) = y i, i = 1,..., N. Idea: Try to reduce the numerical complexing in reducing the problem from R d to R by using so-called radial basis functions Joachim Rang Partititioned Methods for Multifield Problems Seite 17

18 Radial basis functions The function ˆϕ : R d R is called radial basis function if it satisfies the condition x y = ˆx y ˆϕ(x) = ˆϕ(ˆx). for all x, ˆx, y R d. Note that a function ϕ : R + R which depends on the scalar parameter r = x y is a radial basis function Joachim Rang Partititioned Methods for Multifield Problems Seite 18

19 Examples The Gauss function is given by ϕ : R [, 1], ϕ(r) = exp( αr 2 ), α R 1.9 α = 1/1 α = 1/1 α = φ r Joachim Rang Partititioned Methods for Multifield Problems Seite 19

20 Examples The multi-quadric from Hardy (1971) is given by ϕ : R [ δ, ), ϕ(r) = r 2 + δ 2, δ > 12 1 δ = 1/1 δ = 1 δ = 2 8 φ r Joachim Rang Partititioned Methods for Multifield Problems Seite 2

21 Examples The inverse multi-quadric from Hardy (1971) is given by ϕ : R [ δ, ), ϕ(r) = 1/ r 2 + δ 2, δ > 1 9 δ = 1/1 δ = 1 δ = φ r Joachim Rang Partititioned Methods for Multifield Problems Seite 21

22 Examples The volume spline function is given by ϕ : R R, ϕ(r) = r \varphi r Joachim Rang Partititioned Methods for Multifield Problems Seite 22

23 Examples Wendland uses the function ( r 3 ϕ : R R, ϕ(r) = π 12 α2 r + 4 ) 3 α3, where α > is given. 3 α = 1/1 α = 1/1 α = 1/2 25 α = \varphi r Joachim Rang Partititioned Methods for Multifield Problems Seite 23

24 Examples Wendland uses the function { ϕ : R R, ϕ(r) = (1 r) 2 (1 r) + = 2, if r 2, else \varphi r Joachim Rang Partititioned Methods for Multifield Problems Seite 24

25 Examples Wendland uses the function ϕ : R R, ϕ(r) = (1 r) 4 +(4r + 1) \varphi r Joachim Rang Partititioned Methods for Multifield Problems Seite 25

26 Ansatz For the spline function S(x) we make the ansatz S(x) := N a i ϕ( x x i ) + p(x), i=1 where x := (x 1,..., x d ) R d is a d-dimensional vector and p(x) : R d R is a low degree d-variate polynomial. d By x := j=1 x2 j we denote Euclidian norm. The function ϕ : R R is a radial basis functions Joachim Rang Partititioned Methods for Multifield Problems Seite 26

27 The coefficients The coefficients a i and the polynomial p are determined by the interpolation condition S(x i ) = w i, i = 1,..., N and the additional requirements N α j q(x j ) = j=1 for all polynomials q with deg q deg p Joachim Rang Partititioned Methods for Multifield Problems Seite 27

28 Polynomials For simplicity we consider only linear polynomials, i.e. q(x) = c + c 1 x 1 + c 2 x c d x d, x = (x 1,..., x d ). To determine the coefficients c i of q we need a basis of polynomials, i.e. the set of polynomials {1, x 1, x 2,..., x d } Joachim Rang Partititioned Methods for Multifield Problems Seite 28

29 Linear system Using theses monomials we obtain a linear system of equations N a i ϕ( x 1 x i ) + q(x 1 ) = w 1 i=1 N a i ϕ( x N x i ) + q(x N ) = w N i=1. a + + a N = N a i x (1) i = i=1 N. a i x (d) i =, Joachim Rang i=1 Partititioned Methods for Multifield Problems Seite 29

30 Linear system This system can be written as ( A B B with B = ) ( a c ) = ( y ϕ( x 1 x 1 )... ϕ( x 1 x N ) ϕ( x 2 x 1 )... ϕ( x 2 x N ) A =.. ϕ( x N x 1 )... ϕ( x N x N ) 1 x (1) 1 x (2) 1 x (3) x (1) N x (2) N x (3) N a = (a 1,..., a N ), R N,4 y = (y 1,..., y N ) ) RN,N Joachim Rang Partititioned Methods for Multifield Problems Seite 3

31 Strong positive definite matrices A matrix A R N,N is called strong positive definite on L R N, if A is symmetric and the inequality is valid. Remark: Ax, x > x L \ {} A is strong positive definite on R N, if all eigenvalues of A are positive. A is strong positive definite on R N, if all sub-determinants are positive Joachim Rang Partititioned Methods for Multifield Problems Seite 31

32 Strong negative definite matrices A matrix A R N,N is called strong negative definite on L R N, if A is symmetric and the inequality is valid. Ax, x < x L \ {} Joachim Rang Partititioned Methods for Multifield Problems Seite 32

33 Positive definite matrices A matrix A is called positive definite on L R N, if A is symmetric and the inequality Ax, x x L is valid. A is called negative definite on L R N, if A is symmetric and the inequality Ax, x x L is valid Joachim Rang Partititioned Methods for Multifield Problems Seite 33

34 Strong positive definite functions A function ϕ C[, ) is called (conditionally) strong positive definite of order m in R d, if for every m > 1 and all {x 1,..., x N } R d with x i x j for i j the symmetric matrix A = (ϕ( x j x i ) N i,j=1 is (strong) positive definite on the sub-space { N a R n, a = (a 1,..., a N ), a i Q n (x i ) =, i=1 n =,..., m } Joachim Rang Partititioned Methods for Multifield Problems Seite 34

35 Solvability The interpolation problem S(x i ) = w i, i = 1,..., N is uniquely solvable, if the matrix A = (ϕ( x j x i ) N i,j=1 is strong positive definite on the sub-space N L = {a R N, a i Q n (x i ) =, n =,..., M}. i= Joachim Rang Partititioned Methods for Multifield Problems Seite 35

36 Conditionally positive definite functions A function ϕ is conditionally positive definite in R d (d 1), if and only if it holds If further the condition ( 1) j ( d dt ) j ϕ( t), t >, j. ϕ( t) const is valid, then ϕ is conditionally strong positive definite in R N. see Micchelli (1986) Joachim Rang Partititioned Methods for Multifield Problems Seite 36

37 Example Let us consider the two-dimensional function f : (, 1) (, 1) (, 1/2) given by f(x, y) = xy 1 + y. 2 Here: N = Joachim Rang Partititioned Methods for Multifield Problems Seite 37

38 Example Let us consider the two-dimensional function f : (, 1) (, 1) (, 1/2) given by f(x, y) = xy 1 + y. 2 Here: N = Joachim Rang Partititioned Methods for Multifield Problems Seite 38

39 Example Let us consider the two-dimensional function f : (, 1) (, 1) (, 1/2) given by f(x, y) = xy 1 + y. 2 Here: N = Joachim Rang Partititioned Methods for Multifield Problems Seite 39

40 Example Let us consider the two-dimensional function f : (, 1) (, 1) (, 1/2) given by f(x, y) = xy 1 + y. 2 Here: N = Joachim Rang Partititioned Methods for Multifield Problems Seite 4

41 The choice of α Joachim Rang Partititioned Methods for Multifield Problems Seite

42 Application to non-matching meshes interpolation condition: ( WA ) = ( ΦAA Q A QA ) ( γ β ) Φ Ai,A j Then Finally := φ( x Ai x Aj ). ( γ W B = [Φ BA Q B ] β ( ΦAA Q W B = [Φ BA Q B ] A QA ) ) 1 ( WA ) Joachim Rang Partititioned Methods for Multifield Problems Seite 42

43 Transformation matrix Let H := [Φ BA Q B ] ( ΦAA Q A Q A ) 1 transformation matrix H BA... first n B rows and n A columns Remember, linear functions are interpolated exactly, if linear polynomials are included. Consistent approach: possible, if ansatz with linear polynomials is used Joachim Rang Partititioned Methods for Multifield Problems Seite 43

44 Conservative approach transformation of displacements: ( Φss Q U f = [Φ fs Q f ] s Qs transformation of pressure forces ( ) ( Fs Φss Q = s Qs condition: and finally F s = Φ 1 ss Φ fs F f ) 1 ( Us ) 1 ( Φ fs Q f M ss P s = Φ 1 ss Φ fs M ff P f conservative, if both meshes converge to a matching mesh ) ) F f Joachim Rang Partititioned Methods for Multifield Problems Seite 44