Discontinuous Galerkin Methods: An Overview and Some Applications. Daya Reddy UNIVERSITY OF CAPE TOWN

Transcription

1 Discontinuous Galerkin Methods: An Overview and Some Applications Daya Reddy UNIVRSITY OF CAP TOWN Joint work with Beverley Grieshaber and Andrew McBride SANUM, Stellenbosch, March 2016

2 Structure of talk A model elliptic problem: weak or variational formulations The Galerkin finite element method: analysis and approximations Discontinuous Galerkin (DG) formulations Near-incompressibility in elasticity

3 Model problem: deformation of a membrane Minimization of an energy min J(v) J(v) = 1 v 2 rv 2 dx fvdx f The solution u satisfies the weak problem ru rvdx= fvdx for all functions v which satisfy v = 0 on the boundary Sufficiently smooth u satisfies the Poisson equation and boundary condition u = f on u = 0 on u 2 2

4 The membrane problem, continued min v2v J(v) := 1 2 ru rv dx= rv 2 dx fv dx fv dx Define the bilinear form a(, ) and linear functional `( ) a : V V! R, a(u, v) = ru rv dx ` : V! R, `(v) = fv dx Thus the above problem is min v2v 1 2a(v, v) `(v) or equivalently a(u, v) =`(v) 8v 2 V

5 Interlude: the Sobolev spaces H m () Built from the Lebesgue space of square-integrable functions: L 2 () = v : v 2 dx := kvk 2 0 < 1 Define, for integer m 0, H m () = Seminorm v : D v 2 L 2 (), apple m v 2 m = X D v 2 dx =m D v = for problem in R 2 Hilbert space with induced norm kvk 2 m = X e.g. kvk 2 1 = applem v 2 m h v i dx We will also need H 1 0 () ={v 2 H 1 () : v = 0 on }

6 Well-posedness of the variational problem min w2w 1 2a(w, w) `(w) This problem has a unique solution if: W is a closed subspace of a Hilbert space H a is coercive or W-elliptic: a(w, w) kwk 2 H a is continuous: a(w, z) apple Mkwk H kzk H ` is continuous: `(z) apple Ckzk H The model problem has a unique solution in H 1 0 ()

7 Finite element approximations Aim: to pose the variational problem on a finite-dimensional subspace V h V 1. Partition the domain into subdomains or finite elements 2. Construct a basis {' i } N i=1 for V h comprising continuous functions that are polynomials on each element portion of hip replacement: physical object and finite element model

8 The Galerkin finite element method 3. The piecewise-polynomial approximations can be written u h = X i ' i (x)u i 'u ' i v h = X i ' i (x)v i 'v 4. Substitute in the weak formulation a(u h,v h )=`(v h ) to obtain X X v i [a(' i, ' j )u i `(' j )] = 0 a(' i, ' j ) u i = `(' j ) {z } {z } i,j i K ji F j Ku = F

9 Convergence of finite element approximations Construct V h V and seek u h 2 V h such that for all v h 2 V h a(u h,v h )=`(v h ) for all v h 2 V h Ku = F h T = diameter of T mesh size h = max T 2T h T h T Define the error by sense that u u h : under what conditions do we have convergence in the lim u h = u? h!0 Orthogonality of the error: a(u u h,v h ) = a(u, v h ) a(u h,v h ) = `(v h ) `(v h ) = 0

10 An a priori estimate ku u h k 2 V apple a(u u h,u u h ) = a(u u h,u v h )+a(u u h,v h u h ) (V- ellipticity) apple a(u u h,u v h ) apple Mku u h k V ku v h k V (continuity) (orthogonality of error) ku u h k 1 apple C inf v h 2V h ku v h k 1 Céa s lemma Strategy for obtaining error bound: a) choose v h to be the interpolate of uin V h b) use interpolation error estimate to bound actual error

11 Finite element interpolation theory Ciarlet and Raviart Arch. Rat. Mech. Anal h T = diameter of T h T T =sup{diameter of B; B a ball contained in T } T T = h T / T Let T be a triangulation of a bounded domain with polygonal boundary: = [ T 2T T Define the mesh size h = max T 2T h T A family of triangulations is regular as h! 0 if there exists > 0 such that T apple for all T 2 T h

12 Finite element interpolation theory (Local estimate) For a regular triangulation with v 2 H k+1 (T ), k +1 m and the interpolation operator which maps functions to polynomials of degree apple k, v v m,t apple Ch k+1 m v k+1,t (Global estimate) Let T domain. Define be a uniformly regular triangulation of a polygonal V h = {v h 2 C( ) :v h T 2 P k } \ V h : H 2 ()! V h h = max T h T (global interpolator) h v v h T kv h vk m, apple Ch k+1 m v k+1, m =0, 1

13 Convergence of finite element approximations ku u h k V apple C inf vh 2V h ku v h k V apple Cku h uk V apple Ch min(k,r 1) u r So for the simplest approximation, by piecewise-linear simplices, ku u h k 1 apple Ch u 2 for the second-order elliptic equations Could use piecewise-quadratic simplices (k = 2) in which case ku u h k 1 = O(h 2 ) -- provided that the solution is smooth enough to belong to H 3 ()!

14 Discontinuous Galerkin (DG) methods

15 Discontinuous Galerkin (DG) methods Drop the condition of continuity So V h 62 V A member v h 2 V h is now a polynomial on each element, but not continuous across element boundaries An immediate consequence: much greater number of degrees of freedom u { u} N 1 K 1 K 2 N 2 e 12

16 What do solutions look like? 4.3 Implementation of the predictor corrector solution strategy (a) Conforming approximation (b) Discontinuous Galerkin approximation (c) Discontinuous Galerkin approximation with a 101

17 Why bother with DG? Can accommodate hanging nodes Useful in adaptive mesh refinement log(energy error) log(interface jump)

18 fficiency and accuracy = = = 10 Conforming DG Displacement rror (m) 1 ten yck and Lew 2010 (nonlinear elas7city) Total DOF Figure 6. Plot of the L 2 -norm over B 0 of the error in the displacement field as a function of the total number of degrees of freedom for the test in Section 6.1. Curves are shown for the discontinuous Galerkin method as well as for the conforming one. Remarkably, the two curves overlap, indicating that both methods provide the same accuracy for the same computational cost, i.e. for this example they are equally efficient.

19 Our objective For the problem of deformations of elastic bodies DG methods show good behaviour for near-incompressibility with the use of low-order triangles in two dimensions, and tetrahedra in three DG with quadrilaterals and hexahedra less straightforward: poor behaviour, including locking, for low-order approximations We show why this is so, and propose some remedies

20 First, review derivation of heat equation Heat equation: Balance of energy + div q = s q = s = heat flux vector heat source Fourier s law q = kr# give the (steady) heat equation k # = s

21 Governing equations for elasticity displacement vector u x u(x) stress tensor or matrix σ lasticity: quilibrium + div = f give Navier s equation Hooke s law = div u +(ru +[ru] T ) u = [ u +(1+ )rdiv u] =f

22 quivalent to minimization problem min u J(u) = 1 2 r s u 2 +(1+ )(div u) 2 dv f u dv Define a(u, v) = [r s u : r s v +(1+ )div u div v] dx `(v) = f v dx r s u = 1 2 [ru +(ru)t ] (r s u) ij = j i Then the minimization problem is or equivalently min v2v 1 2a(v, v) `(v) a(u, v) =`(v) which has a unique solution in V := [H 1 0 ()] d Furthermore (Brenner and Sung 1992) the solution satisfies u 2 [H 2 ()] 2 kuk H 2 + k div uk H 1 apple Ckfk L 2

23 Locking The use of low-order elements leads to a non-physical solution in the incompressible limit a(u, v) = [r s u : r s v +(1+ )div u div v] dx!1 div u! 0 Q 1 0,8 0,7 0,6 0,5 0,4 Case III 0,3 Case V (AS) 0,2 MHW Case 1 0,1 HW Case I or Q1 0-0,6-0,4-0,2 0-0,1 0,2 0,4 0,6-0,2 We explore the use of DG methods as a remedy for locking Q 1

24 DG formulation: some definitions Jumps and averages: on an interior edge of element T, n + T + T n [v] = v + n + + v n, [[v]] = v + n + + v n, {v} = 1 2 (v+ + v ) [[ ]] = + n + + n, { } = 1 2 ( + + ) For example [[ ]] =( + )n + On an edge that forms part of the boundary, [[v]] = v n { } = n

25 Setting it up take the inner product with a test function integrate, and integrate by parts div v dv = f v dv u = [ u +(1+ {z )rdiv u] } div V = div u +(ru +[ru] T ) = f Consider the left-hand side: div v dv = X div T T v dv div v dv = X n v ds X T T : r s v dv

26 Thus weak equation is now X n v ds + X T T : r s v dv = f v Next, we need the magic formula T X v n ds = X [[v]] : { } ds + X int int {v} [[ ]] ds

27 This gives X [[v]] : { } ds X int int {v} [[ ]] ds + : r s v dv = f v dv Since the exact solution is smooth ( u 2 [H 2 ()] d ) and the stress is continuous we can assume that [[ ]] = 0 which leaves X [[v]] : { } ds + : r s v dv = f v dv

28 xploiting again the smoothness of the exact solution we can add a term to symmetrize the problem: X [[v]] : { (u)} ds X [[u]] : { (v)} ds + X [[u]] = 0 : r s v dv = f v dv Finally, we may need to stabilize the problem: this gives us the symmetric interior penalty (SIP) formulation (Douglas and Dupont 1976) X [[v]] : { (u)} ds X : { (v)} ds + k [[u]] X + X : r s v dv = 1 h [[u]][[v]] f v dv

29 The DG formulation V h = v h : v h 2 [L 2 ()] d, v h T 2 P 1 (T ) or Q 1 (T ) P 1 (T ): v h = a 0 + a 1 x + a 2 y Q 1 (T ): v h = a 0 + a 1 x + a 2 y + a 3 xy

30 V h = v h : v h 2 [L 2 ()] d, v h T 2 P 1 (T ) or Q 1 (T ) a h (u h, v h )=`(v h ) X T T (u) :r s v dv + X a h (u, v) [[u]] : { (v)} ds X +k X [[v]] : { (u)} ds 1 h [[u]][[v]] = f v dv = 8 >< >: Nonsymmetric Interior Penalty Galerkin (NIPG) RIVIÈR & WHLR 1999, 2000; ODN, BABUSKA & BAUMANN 1998 Symmetric Interior Penalty Galerkin (SIPG) DOUGLAS & DUPONT 1976, ARNOLD 1982, HANSBO & LARSON 2002 Incomplete Interior Penalty Galerkin (IIPG) DAWSON, SUN AND WHLR 2004

31 DG with triangles is uniformly convergent WIHLR 2002 y 62 Chapter 3. Locking-Freeh DGFM for lasticity Problems DGFM on graded mesh λ=1 λ=100 λ=500 λ=1000 λ= ements O x absolute L 2 error Figure 3.2: Graded mesh with refinement towards the origin (γ = (1/2, 0, 0, 0, 0)) number of degrees of freedom Figure 3.3: Uniform mesh (i.e. graded mesh with γ = (0, 0, 0, 0, 0)). Figure 3.5: Performance of the DGFM on the L-sh mesh) and with γ = 0 (uniform mesh). in order to obtain the optimal convergence rate, a graded mesh with refinement towards the origin must be used for the numerical simulations. The first picture of Figure 3.4 shows the errors of the DGFM for in the energy norm u 2 h = K T λ {1, 100, 500, 1000, 5000} (µ = 1) ϵ(u) 2 L 2 (K ) + 1 m elast e int,d e 1 e [u] 2 ds on a graded mesh with grading vector γ = (1/2, 0, 0, 0, 0) (cf. Figure 3.2). Ob-

32 Some computational results with quads appear to show locking-free behaviour OBB SIPG NIPG IIPG Fig. 8. Tensile stress contours ðr y > 0:15 MPaÞ of DG solutions of beam problem with Poisson s ratio (a) OBB; (b) SIPG; d p ¼ 30; (c) NIPG; d p ¼ 10; and (d) IIPG; d p ¼ 50. LIU, WHLR AND DAWSON 2009 These authors report good results using all methods

33 but not so in other cases Grieshaber, McBride and R. (2015) ALL ALL

34 We know that DG works for simplicial elements Some authors report good results for quads We find locking for a simple example What s going on?

35 Let s see why DG with triangles works a h (u, v) := X T 2T h +k µ µ X 2 T (u) :r s v + X 2 1 h [[u]] : [[v]] + k [[u]] : { (v)} X 1 h 2 X 2 [[u]] : [[v]] { (u)} :[[v]] := 2µ + " X " T 2T h + X T 2T h T r s u : r s v dx + X 2 T (div u)(div v)+ X 2 [[u]] : {r s v} (note the use of two stabilization terms) X 2 (I) (II) (III) 2 [[u]] : {(div v)i} X {ru} :[[v]] + k µ X 2 {(div u)i} :[[v]] ds +k X 2 1 h # 1 [[u]] : [[v]] 2h [[u]] : [[v]] (IV) #

36 rror analysis for linear triangles: NIPG Approximation error e = u u h u = interpolant of u = u u {z } interpolation error + u u h {z } Use Crouzeix-Raviart interpolant: linear on triangles, continuous at midpoints of edges Properties: (u u) n ds =0 T div(u u) dv =0

37 e = + = u u = u u h 2 V h ) div, ( ) const. λ-terms in error bound can be handled as follows: for example, (I) = X T to give eventually T (div )(div ) ds = X T div div ds =0 kek 2 DG apple Ch 2 kuk 2 H 2 () + 2 k div uk 2 H 1 () T Use estimate kuk H 2 + k div uk H 1 apple Ckfk L 2 to get uniform convergence

38 Convergence analysis for quadrilaterals First need to construct a suitable interpolant u 2 V h As before e = u u h = u u {z } + u u h {z } and u must satisfy (u u) n ds [[u u]] n ds =0 (u u) n ds =0

39 The interpolant Inspired by DOUGLAS, SANTOS, SHN & Y 1999; CIA, DOUGLAS, Y 1999: Construct interpolant with span 1,x,y, (x)+apple(y) 1,x,y,apple(x)+ (y) (x) =3x 2 10x 4 +7x 6 apple(x) = 3x 2 +5x 4 Define u as orthogonal projection on element

40 rror bound ventually get kũ u h k 2 DG apple C X T h 2 T kuk 2 H 2 (T ) + 2 kuk 2 H 2 (T ) + 2 k div uk 2 H 1 (T ) so not possible to bound λ-dependent term

41 xample: square plate BRNNR SINUM 1993 IIPG SIPG NIPG ALL =(0, 1) 2 µ =1 u 1 (x, y) = sin2 y 1 + cos 2 x u 2 (x, y) = sin2 x 1 cos 2 y sin x sin y sin x sin y

42 A remedy: selective reduced integration (SRI) Recall classical SRI for elasticity problem: to overcome volumetric locking one replaces rs u h : r s v h + (div u)(div v) dv {z } volumetric term by or rs u h : r s v h dv + rs u h : r s v h dv + X w T div u(x T )divv(x T ) T {z } underintegration of volumetric term X T T ( 0 div u)( 0 div v) dv 0 div v = 1 T T div v dv

43 Underintegration applied to (II) gives [[ ]]{div } ds ' 0 [[ ]] 0 {div } ds = 0 {div } 0 [[ ]] ds = 0 {div } [[ ]] ds =0 Replace (IV) with k X 0 [[ ]] 0 [[ ]] = k X = k X [[ 0 ]] [[ ]] 0 [[ ]] [[ ]] = 0 But we now have to check coercivity and consistency of the modified bilinear form!

44 Underintegration applied to (II) gives [[ ]]{div } ds ' = 0 {div } = 0 {div } ventually get 0 [[ ]] 0 {div } ds 0 [[ ]] ds [[ ]] ds =0 a h (, ) apple Ck k DG P T h2 T kuk 2 H kdiv uk 2 H 1 {z } appleckfk 2 1/2 But we now have to check coercivity and consistency of the modified bilinear form!

45 Coercivity NIPG: OK for SRI on terms (II) and (III) SIPG: OK for SRI on terms (II), (III), (IV) IIPG: OK for SRI on terms (III) and (IV) Term (III) when underintegrated leads to a consistency error RI (u, ) = X 0 {div u} tr 0 [[ ]] {div u} tr[[ ]] ds which can be controlled

46 Locking-free (uniformly convergent) behaviour P 1 Q 1 Q 1 +SRI SIP NIP IIP

47 Square plate: DG with triangles quads with and without SRI Standard finite elements / DG quads without SRI DG triangles / quads with SRI

48 NIPG IIPG DG DG DG+SRI DG+SRI SIPG DG DG+SRI

49 Cube with prescribed body force u 1 = (cos 2 x 1)(sin 2 y sin z sin y sin 2 z) u 2 = (cos 2 y 1)(sin 2 z sin x sin z sin 2 x) u 3 = (cos 2 z 1)(sin 2 x sin y sin x sin 2 y) sin x sin y sin z, sin x sin y sin z, sin x sin y sin z.

50 10 2 IIPG 10 1 Q 1 full SG, P 1 full 10 2 NIPG log(h1 error) 10 0 P 1 SRI Q 1 SRI log(h1 error) SG P 1 full Q 1 full Q 1 SRI log(h) Q1 full Q1 UI P1 full SG log(h1 error) log(h) SIPG SG, P 1 full, Q 1 full P 1 SRI Q 1 SRI Q1 full Q1 UI P1 full P1 UI SG Q1 full Q1 UI P1 full P1 UI SG

51 Back to the T-bar example LIU. WHLR, DAWSON 2009 appeared to show good behaviour for IIPG 3D problem but effectively plane stress (Neumann or flux-free boundary condition) in out-of-plane direction, along ABCD Here treated as plane strain (Dirichlet or constrained displacement) along ABCD A C B C Without SRI With SRI

52 Concluding remarks For near-incompressibility DG with quadrilateral elements is not straightforward A remedy, viz. selective under-integration λ-dependent edge terms, has been proposed, analysed, and shown to converge uniformly at the optimal rate Arbitrary quadrilaterals: numerical experiments indicate behaviour similar to that for rectangles. Analysis would be quite complex Current work: nonlinear problems; other parameter-dependent problems References Grieshaber BJ, McBride AT and Reddy BD, Uniformly convergent interior penalty approximations using multilinear approximations for problems in elasticity. SIAM J. Numer. Anal. 53 (2015) Grieshaber BJ, McBride AT and Reddy BD, Computational aspects of Discontinuous Galerkin approximations using quadrilateral and hexahedral elements. In review.

53 Alternative approach: P 1 approximation on quads Much of the manipulations carry over, all the bounds are as before, but this time = u u h 2 V h P 1 u h 2 P 1 Problematic terms: (II) = } ds = [[ ]]{div X ]{div } [[ ]] ds = 0 given properties of interpolant But (IV) = k k X 1 [[ [[ ]] [[w]] [[ ]] ds 6= 6= 0 h so remains a problem for SIPG and IIPG, but is absent for NIPG

54 Square plate, DG with quads and P 1 NIPG IIPG SIPG