Energy and Momentum Conserving Algorithms in Continuum Mechanics. O. Gonzalez

Transcription

1 Energy and Momentum Conserving Algorithms in Continuum Mechanics O. Gonzalez Département de Mathématiques École Polytechnique Fédérale de Lausanne Lausanne, Switzerland

2 INTRODUCTION MAIN CONCERN Time integration of the general system ż = J(z) H(z) D(z) H(z) z IR m, H : IR m IR, J = J T IR m m, D = D T 0 IR m m includes general Hamiltonian systems analogous infinite-dimensional case

3 Introduction KEY FEATURES Energy decay/conservation Ḣ = H ż = H [J H D H] Ḣ = H D H 0 and Ḣ = 0 when D O Momentum conservation (linear/angular momentum) Suppose H = Ĥ ζ ζ : IR m IR k invariants under a symmetry group, e.g. ζ(z) = ( q 2, q p, p 2 ), z = (q, p) IR 6 rotational invariants F(z) such that J F ker[ ζ] Then F = 0 when D O

4 OBJECTIVE & MOTIVATION OBJECTIVE To develop time integration schemes with : mild or no restrictions on time step (stable implicit schemes) energy/momentum conservation when no damping present physical energy decay when damping present MOTIVATION classic schemes generally fail on all three points

5 Energy Energy Motivation EXAMPLE: Ḣ = 0 Nonlinear Shells (Simo & Tarnow 992) 9 C B 4 A 4 D 0 z y 3 x Energy vs Time: Mid-Point Rule & Trapezoidal Rule Midpoint Rule Trapezoidal Rule t = 0:002 t = 0: t = 0:004 t = 0:00 68 t = 0: t = 0: Time Time

6 OUTLINE I. Introduction and motivation II. Why does the Mid-Point Rule fail? Kepler model problem Elastodynamics model problem Illustration of difficulties Remedies III. Abstract framework for conserving schemes Discrete gradients General scheme IV. Closing remarks

7 Part II KEPLER MODEL PROBLEM m p R 3 q q = m p ṗ = V ( q ) q q V(q) z = (q, p), H(z) = 2 m p 2 + V ( q ), J = ( O ) I I O FEATURES H(z) invariant under rotation group H(z), j(z) = q p are integrals q, p IR 3 evolve in plane normal to µ = j(z 0 ) the radial variables λ = q and π = p q/ q satisfy λ = m π π = V µ (λ) reduced equations where V µ (λ) = V (λ) + 2 µ 2 /mλ 2 Smale s amended potential

8 Part II RELATIVE EQUILIBRIA steady, circular orbit µ = q x p p q = m p ṗ = V ( q ) q q R 3 λ * m q fixed point π λ = m π π = V µ(λ) V µ (λ ) = 0 λ ( λ *, 0 ) STABILITY stable relative equilibria = stable fixed point = local min of V µ (λ)

9 Part II Analysis of the Mid-Point Rule MID-POINT RULE q n+ q n t p n+ p n t = m p n+ 2 = V ( q n+ ) 2 q n+ q n+ 2 2 () where q n+ 2 = 2 (q n+ + q n ). preserves j(z n ) = q n p n does not preserve H(z n ) = 2 m p n 2 + V ( q n ). REDUCED EQUATIONS Introduce reduced variables as before λ n = q n, π n = p n q n / q n. Then () implies the implicit reduced equation G MP (λ n+, π n+ ; λ n, π n ) = 0. (2) look for fixed points ( λ, 0) relative equilibria study stability

10 Part II RESULTS Existence of Relative Equilibria: Fixed points λ are stationary points of a perturbed potential: V µ (λ, t) V µ (λ) Stability of Relative Equilibria: Fixed points λ are only conditionally stable in general: linearly stable when t < t critical unstable when t > t critical Two questions: What s causing the instability? Is there a cure?

11 Part II SOURCE OF INSTABILITY coupling between internal forces and rotations Exact Problem: V ( q ) q rotationally invariant when q = Λ(t)q 0 Mid-point discretization: V ( q n+ 2 ) q n+ 2 not rotationally invariant when q n+ = Λq n n+ n+/2 n A REMEDY replace V ( q n+ 2 ) q n+ 2 by V ( q n+ ) V ( q n ) q n+ q n 2 ( q n+ + q n )

12 Part II A Conserving Scheme CONSERVING SCHEME q n+ q n t p n+ p n t = m p n+ 2 = V ( q n+ ) V ( q n ) q n+ q n q n+ 2 2 ( q n+ + q n ) (3) preserves j(z n ) = q n p n preserves H(z n ) = 2 m p n 2 + V ( q n ). REDUCED EQUATIONS Introduce reduced variables as before λ n = q n, π n = p n q n / q n. Then (3) implies the implicit reduced equation G EM (λ n+, π n+ ; λ n, π n ) = 0 (4) look for fixed points ( λ, 0) relative equilibria study stability

13 Part II RESULTS Existence of Relative Equilibria: Fixed points λ are exact, i.e. stationary points of V µ (λ). Stability of Relative Equilibria: i. The reduced equations preserve the integral H µ (λ, π) = 2 m π 2 + V µ (λ) reduced Hamiltonian ii. V µ local min at λ H µ local min at ( λ, 0) H µ is a Lyapunov function nonlinear stability. Three remarks on conserving scheme: 2nd order accurate, like Mid-Point rule conserving modification cures coupling/stability problem similar problem/cure in more complex systems

14 Part II ELASTODYNAMICS MODEL PROBLEM ϕ (,t) * π (X,t) X ϕ (X,t) R 3 Ω NOTATION Deformation ϕ : Ω [0, T] IR 3 (Linear) Momentum Density π : Ω [0, T] IR 3 Deformation Gradient Cauchy Strain Strain Energy Function Second Piola-Kirchhoff Stress F(ϕ) = ϕ C(ϕ) = F(ϕ) T F(ϕ) W(C) Σ(ϕ) Σ(ϕ) = 2 DW(C(ϕ))

15 Part II Typical IBVP ϕ (,t) π (X,t) Γ ϕ X Ω Γ Σ ϕ (X,t) h R 3 N R 3 b Find ϕ, π : Ω [0, T] IR 3 such that ϕ = ρ π π = [F(ϕ)Σ(ϕ)] + b in Ω (0, T] in Ω (0, T] ϕ = g F(ϕ)Σ(ϕ)N = h in Γ ϕ (0, T] in Γ Σ (0, T] plus initial conditions

16 Part II Key Features Infinite-dimensional version of general system with H(z) = Typical integrals z = (ϕ, π), J = Ω [ ( O ) I I O ] 2 ρ π 2 + W(C(ϕ)) ϕ b dω ϕ h dγ Γ Σ Energy Linear Momentum H(z) l(z) = Ω π dω Angular Momentum j(z) = Ω ϕ π dω Note: For the Neumann problem with no external loads i.e. Γ Σ = Ω, h = 0 and b = 0 all of the above quantities are integrals.

17 Part II Mid-Point Rule Time Discretization Given ϕ n, π n find ϕ n+, π n+ such that ϕ n+ ϕ n t π n+ π n t = ρ π n+ 2 = [F(ϕ n+ 2 )Σ(ϕ n+ 2 )] + b in in Ω Ω ϕ n+ 2 = g F(ϕ n+ 2 )Σ(ϕ n+ 2 )N = h on on Γ ϕ Γ Σ where ( ) n+ 2 = 2 [( ) n + ( ) n+ ] and t > 0 is the time step. time discretization reduces the IBVP to a sequence of BVPs for ϕ n+, π n+ (n = 0,, 2,...) each BVP may be solved using a finite-element method scheme preserves l and j, but in general not H

18 Part II Numerical Example Pinched Rubber Quarter-Cylinder Initial Pinched Stored Energy Function: three-term Ogden model W(C) = 3 3 A= m= µ m α m [λ α m A (C) ] µ m ln[λ A (C)] where λ 2 A (C) > 0 (A =, 2, 3) are the eigenvalues of C S3 PD. Spatial Discretization: FEM with 52 trilinear bricks (2295 dof).

19 Implicit Mid-Point Rule with Time Step t = s S T R E S S 3 Min = -2.03E+06 Max =.24E+06 t = s -.56E E E E E E+05 S T R E S S 3 Min = -.92E+06 Max = 9.99E+05 t = s -.50E E E E+05.67E E+05 S T R E S S 3 Min = -.28E+06 Max = 9.57E+05 t = s -9.60E E E E E E+05 S T R E S S 3 Min = -.20E+06 Max = 8.0E+05 t = s -9.7E E E E E E+05

20 Implicit Mid-Point Rule with Time Step t = s Angular Momentum and Energy Histories 0 -e-06-2e-06-3e-06-4e-06 J -5e-06-6e-06-7e-06-8e-06-9e-06 -e t J t 0 -e-05-2e-05-3e-05 J3-4e-05-5e-05-6e-05-7e-05-8e t H t

21 Implicit Mid-Point Rule with Time Step t = s Angular Momentum and Energy Histories 0 0 -e-06-2e e-06-4e J -5e-06 J2-6e e-06-8e e-06 -e t t 0.7 -e-05.6 J3-2e-05-3e-05-4e-05-5e-05-6e-05 H e e t t

22 Character of Solution During Energy Growth S T R E S S 3 Min = -2.6E+06 Max =.37E+06 t = s -2.04E E E E E E+05 S T R E S S 3 Min = -.34E+06 Max =.34E+06 t = s -9.54E E E+05.92E E E+05 S T R E S S 3 Min = -6.8E+05 Max =.22E+06 t = 0.050s -3.56E E+04.69E E E E+05 S T R E S S 3 Min = -2.65E+06 Max = 9.58E+05 t = 0.055s -2.3E E E E E E+05

23 Part II A Familiar Difficulty Exact Problem: Σ(ϕ) = 2 DW(C(ϕ)) invariant when ϕ = Λ(t)ϕ 0 Mid-point discretization: Σ(ϕ n+ 2 ) = 2 DW(C(ϕ n+ 2 )) not invariant when ϕ n+ = Λϕ n ϕ n+ ϕ n+/2 ϕ n A REMEDY replace DW(C(ϕ n+ 2 )) by dw(c n, C n+ ) DW(C(ϕ n+ 2 )) where C n = C(ϕ n )

24 Part II A Conserving Scheme ϕ n+ ϕ n t π n+ π n t = ρ π n+ 2 = [F(ϕ n+ 2 ) Σ] + b in in Ω Ω ϕ n+ 2 = g F(ϕ n+ 2 ) ΣN = h on on Γ ϕ Γ Σ where Σ = 2 dw(c n, C n+ ) dw(c n, C n+ ) = DW(C n+ 2 ) + [ W(Cn+ ) W(C n ) DW(C n+ ) : M ] 2 M 2 M and M = C n+ C n. dw(c n, C n+ ) is a discrete gradient conserves l, j and H

25 Conserving Scheme with Time Step t = s Part II Angular Momentum and Energy Histories 0 -e-06-2e-06-3e-06-4e-06 J -5e-06-6e-06-7e-06-8e-06-9e-06 -e t J t 0 -e-05-2e-05-3e-05 J3-4e-05-5e-05-6e-05-7e-05-8e t H t

26 Comparison of Mid-Point and Conserving Schemes Mid-Point S T R E S S 3 Min = -2.6E+06 Max =.37E+06 Conserving S T R E S S 3 Min = -.58E+06 Max =.08E E E E E E E E E E E E E+05 t=0.0500s t=0.0495s S T R E S S 3 Min = -.34E+06 Max =.34E+06 S T R E S S 3 Min = -3.E+05 Max = 6.57E E E E+05.92E E E E E+04.04E E E E+05 t=0.0505s t=0.0500s S T R E S S 3 Min = -6.8E+05 Max =.22E+06 S T R E S S 3 Min = -8.8E+05 Max = 7.0E+05 t=0.050s -3.56E E+04.69E E E E+05 t=0.0505s -6.55E E E E E E+05 S T R E S S 3 Min = -2.65E+06 Max = 9.58E+05 S T R E S S 3 Min = -2.07E+06 Max = 9.43E+05 t=0.055s -2.3E E E E E E+05 t=0.050s -.64E E E E E E+05

27 Part II Two remarks on conserving scheme: same formal order of accuracy as Mid-Point rule conserving modification cures coupling/stability problem

28 GENERAL CONSERVING SCHEMES ż = J(z) H(z) D(z) H(z) z IR m, H : IR m IR, J = J T IR m m, D = D T 0 IR m m Abstract framework based on discrete gradients can treat arbitrary integrals carries over to infinite-dimensions

29 Part III Discrete Gradients Definition dh : IR m IR m IR m is a discrete gradient of H if ) dh(x, y) (y x) = H(y) H(x), x, y IR m 2) dh(x, x) = H(x), x IR m Example dh(x, y) = H( x + y 2 ) + [ H(y) H(x) H( x+y 2 y x 2 ] ) (y x) (y x)

30 Part III General Conserving Scheme To approximate ż = J(z) H(z) D(z) H(z) consider z n+ z n t = J(z n+ 2 )dh(z n, z n+ ) D(z n+ 2 )dh(z n, z n+ ) where z n+ 2 = 2 (z n+ + z n ). KEY FEATURES Energy decay/conservation H(z n+ ) H(z n ) t = dh(z n, z n+ ) (z n+ z n ) t = dh [JdH DdH] = dh DdH 0 and H(z n+ ) H(z n ) t = 0 when D O.

31 Part III Momentum conservation (linear/angular momentum) Result Suppose H = Ĥ ζ ζ : IR m IR k invariants under a symmetry group, e.g. ζ(z) = ( q 2, q p, p 2 ), z = (q, p) IR 6 rotational invariants F(z) an integral associated with ζ, i.e. J F ker[ ζ] ζ(z), F(z) at most quadratic composed discrete gradient d c H = dĥ ζ conservation of F, H when D O Two remarks: internal force dĥ(ζ n, ζ n+ ) depends only on invariants order of accuracy same as Mid-Point rule

32 Closing Remarks Why does the Mid-Point Rule fail? artificial coupling between internal forces/rotations Is there a cure? decoupled discretization discrete gradients Why preserve decay inequalities and integrals? enhanced stability useful for constrained systems: Lagrangian config constraints Hamiltonian integrals Other application for discrete gradients: discrete gradient flow z n+ z n t = df(z n+, z n ) generates strictly decreasing sequence F n+ F n = t df 2 < 0 df = 0 F = 0