Numerical methods for physics simulations.

Transcription

1 Numerical methods for physics simulations. Claude Lacoursière HPC2N/VRlab, Umeå Universiy, Umeå, Sweden, and CMLabs Simulations, Montréal, Canada May 8, 26 Typeset by FoilTEX May 5th 26 Outline Typeset by FoilTEX May 5th 26 1

2 Integrating Newton s second law Newton s law states dp dt = f, p = Mẋ for point masses. M is the mass matrix, x is the position vector, p is the momentum vector, and f is the force vector. Integrating these equations of motion means solving the following system of ordinary differential equations: ẋ = v v = M 1 f(x, v) (1) In general, forces depend on both position and velocities but keep it simple and assume f(x) for now. Typeset by FoilTEX May 5th 26 2 Simple Harmonic Oscillator Simple harmonic oscillator in 1D: f(x) = kx, M = m, (a scalar) f(x) m = k m x = ω2 x ω = k/m is the natural frequency, so: ẋ = v v = ω 2 x, (2) solution: x(t) = x() cos(ωt) + v() ω sin(ωt). Because any physical force is smooth, we can always expand: f(x) = a + b(x x ) +... Linear case is easy to study and gives good insight. Typeset by FoilTEX May 5th 26 3

3 Integration Balance: accuracy stability speed Accuracy can also impliy preservation of energy and other physical invariants (symmetry). We want to integrate indefinitely, without having to change the time step all the time so we need: fixed time step unconditional stability ease of implementation Investigate simple methods (first order) on spring systems without damping to get some good idea of what to expect. Typeset by FoilTEX May 5th 26 4 Integrating the Simple Harmonic Oscillator Ignore the literature and follow basic intuition. Discretize time : t = nh where h is small and n = 1, 2,..., x(t) = x(nh) = x n, and similarly for v(t) = v(nh) = v n. Simple formulae for derivatives: ẋ n+1 v n+1 x n+1 x n h v n+1 v n h so using the most naive integration formula, we have: and therefore: ẋ n+1 v n+1 x n+1 x n+1 x n h v n+1 v n h v n = x n + hv n ω 2 x n v n+1 = hω 2 x n + v n (if you stay tuned, we ll explain that this formula is bad for your health; it is called the Explicit Euler integration formula, applied here to the harmonic oscillator) Typeset by FoilTEX May 5th 26 5

4 Simple Harmonic Oscillator cont In matrix form, this is: x n+1 v n+1 = 1 h hω 2 1 which we can also write as z n+1 = Az n with: z n = x n v n ; A = x n v n 1 h hω 2 1 and so: z n = Az n 1 =... = A n z. This is a form of matrix iteration procedure (more on this later). We have stability if z n stays bounded for all n. So, our first interesting matrix problem of the day: what is lim n An? Note that for this case, the energy is E = 1 2 [mv2 + kx 2 ] α z 2 for some scalar α.. Typeset by FoilTEX May 5th 26 6 Intermission: the spectral radius Think of a matrix A and its eigenvalues and eigenvectors: λ i, v i. For any of these, we have: z n = A n v i = λ n i v i so that z n is bounded iff λ i < 1. The spectral radius states that for any consistent matrix norm, the following holds: ρ(a) = max( λ i ) = lim A n 1/n. i n Using that theorem, we can state the following: Theorem 1. Given a matrix A over the complex numbers, the iterations z n = A n z are bounded if and only if ρ(a) 1. Of course, an interesting limit case is when ρ(a) = 1 which leads to oscillations which are bounded forever. For the case where ρ(a) is strictly less than one, the iterates just decay to. To build good integrators, you need cases which have ρ(a) = 1. Typeset by FoilTEX May 5th 26 7

5 Back to Integration What are the eigenvalues of our iteration matrix? det(a λi) = λ h 1 hω 2 1 λ which means that λ ± = 1 ihω and so, = (1 λ) 2 + h 2 ω 2 = ρ(a) = 1 + h2 ω2 > 1. This is always unstable! It is easy to show that the energy increases at each step by the factor 1 + h 2 ω 2. A naïve analysis would suggest that keeping hω < 1 would keep things stable but that is wrong. Adding some damping force of the form γẋ can make the system stable but too much damping makes it unstable again (it is easy to do the analysis). Typeset by FoilTEX May 5th 26 8 Energy Energy as a function of time, hω =.15 γ = 2.5m/h γ =.1mω discretized time (units of hω) Figure 1: Energy of simple harmonic oscillator integrated explicitly. We either bet big oscillations when the oscillator is damped or a blowup. But the energy should strictly decrease at the rate γ. Typeset by FoilTEX May 5th 26 9

6 Intermission: standard stability analysis (simplified) Given ODE ż = f(z), integrator produces z n+1 = φ(h, f, z n, z n 1, ). For linear stability, look at the Dahlqvist (another famous Swede) test equation: ż = λz, z(t) = exp[λt]z(), where λ, z C. For this system, any integration method leads to a linear recurrence relation of the form: w n+1 = Aw n where A is a matrix function which depends on τ = λh and w is a generalized state vector of the form: w n = [z n, z n 1,...] T. Typeset by FoilTEX May 5th 26 1 Intermission: standard stability analysis (simplified) To use this analysis for our case, we need to redifine: and so: s = ωt, dt ds = 1 ω y(s) = x(t), w(s) = ω 1 v(t) y = dy ds = dx(t) dt dt ds = 1 ω v = w(s) dv w = dw ds = 1 ω dt dt ds = ω2 x(t) = x(t) = y(s) (3) If we then set z(s) = y(s) + iw(s) we find that this corresponds to: z = iz where i is the imaginary unit. The analysis is straight forward but it does not apply easily to all the integration cases we want to cover. A matrix format is more convenient. Typeset by FoilTEX May 5th 26 11

7 Discretizing Differently Actually, we had some choice in discretizing. Using Taylor expansions: x n+1 = x n + ẋ n h + O(h 2 ) (4) x n = x n+1 ẋ n+1 h + O(h 2 ) (5) (x n+1 x n )/h = ẋ n + O(h 2 ) = ẋ n+1 + O(h 2 ) (6) x n+1 xn h v n = αv n + (1 α)v n+1 v n+1 vn h ω 2 x n = ω 2 (βx n + (1 β)x n+1 ) and we can chose α 1, β 1 independently. Typeset by FoilTEX May 5th Discretizing Differently The general stepping equation is then: 1 h(1 α) hω 2 (1 β) 1 x n+1 v n+1 = 1 hα hω 2 β 1 The most interesting cases are the ones where α, β are either, 1 or 1/2. Three new stepping formulae: 1 h hω h 1 1 h 2 hω x n+1 v n+1 x n+1 v n+1 x n+1 v n+1 = = = 1 1 x n v n 1 hω 2 1 h 1 2 hω2 2 1, fully implicit x n v n x n v n x n v n, v implicit: the Verlet method, implicit midpoint rule (7) Typeset by FoilTEX May 5th 26 13

8 Discretizing Differently All of these are of the general form: Bz n+1 = Cz n and with a little algebra, we convert them to the form z n+1 = Az n. Computing these matrices is nothing complicated (but tedious). Once you get matrix A, you compute the eigenvalues, carefully check all cases to find out what values of h, ω produce ρ(a) 1. This is the stability regime. Some simple dimensional analysis shows that ρ(a) will always be a function of τ = hω. This is the natural time scale of the system. The following cases will occur: ρ(a) > 1 for all values of τ: unstable, no good (explicit Euler) ρ(a) = 1 for some values of < τ τ: conditionally satble ρ(a) < 1 for all values of τ: unconditionally stable but everything is damped: this is L-stability (implicit Euler) ρ(a) = 1 for all values of τ: unconditionally stable but nothing is damped: this is A-stability (implicit midpoint) Typeset by FoilTEX May 5th Discretizing Differently: Special Cases We cover the different cases and we use: τ = hω. Implicit stepper: Verlet stepper: 1 A ie = 1 + h 2 ω 2 A vv = 1 h hω h 2 ω 2 h hω 2 1, λ ± = 1, ρ(a) < τ 2, λ ± = (1 τ 2 ) ± τ τ In this case, as long as hω < 2, we have λ ± = 1 and ρ(a vv ) = 1. Otherwise, ρ(a v v) > 1. The implicit stepper is even more interesting: A mid = τ2 4 1 τ2 hω 2 4 h 1 τ2 4, λ ± = τ2 2 and in this case, λ ± = ρ(a mid ) = 1 for any value of τ = hω. 1 τ 2 2 iτ Typeset by FoilTEX May 5th 26 15

9 Some Results amplitude Damped Simple Harmonic Oscillator, 1D, c = 1E-1, d = time in units of natural period Variational Implicit midpoint time Vortex Implicit Euler Exact Figure 2: Simple harmonic oscillator in one dimension using different integration methods. At low frequency, they all perform reasonably well as expected but both the Vortex and the implicit Euler method, which produce identical output in this case, dissipate energy. Note that both the variational and the midpoint method produce identical output in this case. Typeset by FoilTEX May 5th Damped Simple Harmonic Oscillator, 1D, c = 1E-4, d = time in units of natural period amplitude Variational Implicit midpoint time Vortex Implicit Euler.6.7 Exact.8.9 Figure 3: Simple harmonic oscillator in one dimension using different integration methods. At higher frequencies, the Vortex and implicit methods dissipate energy very quickly. The midpoint and variational methods are producing the correct output qualitatively but of course, the phase of the signal is erratic. Typeset by FoilTEX May 5th 26 17

10 amplitude Damped Simple Harmonic Oscillator, 1D, c = 1E-6, d = time in units of natural period Variational Implicit midpoint.1.15 time.2 Vortex Implicit Euler.25.3 Exact Figure 4: Simple harmonic oscillator in one dimension using different integration methods. At even higher frequencies, the variational and midpoint methods show signs of instability. A simple computation shows that both integration methods produce an oscillatory signal which diverges linearly to infinity. Vortex and implicit Euler are intrinsically damped and just decay to Typeset by FoilTEX May 5th Summary of reasonably good methods Derivations can be found elsewhere but here is the summary of the one and only explicit method you should bother with at first order: Velocity Verlet: x n+1 = x n + hv n + h2 Verlet or Leap Frog (don t bother with v n ) 2 M 1 f(x n ) (8) v n+1 = v n + h 2 (f(x n) + f(x n+1 )) (9) x n+1 = x n + hv n+1/2 (1) v n+1/2 = v n 1/2 + hf(x n ) (11) Typeset by FoilTEX May 5th 26 19

11 Summary of reasonably good methods Original Verlet: don t even track velocity: x n+1 = 2x n x n 1 + M 1 h 2 f(x n ) Position Verlet, partially implicit: x n+1 = 2x n x n 1 + M 1 h 2 ((1 α)f(x n+1 ) + αf(x n )) Note that the half implicit Verlet method with α = 1/2 is of second order. Typeset by FoilTEX May 5th 26 2 Summary of reasonably good methods Two implicit methods that come in handy: Implicit Euler (dissipative): x n+1 = x n + hv n+1 v n+1 = v n + hf(x n+1 ) = v n + hf(x n + hv n+1 ) (12) Implicit midpoint: x n+1 = x n + h 2 [v n+1 + v n ] v n+1 = v n + hf( x n+1 + x n ) = v n + hf(x n + h 2 4 [v n+1 + v n ]) (13) Must do root finding on the velocity equation: more about this later. Typeset by FoilTEX May 5th 26 21

12 Conclusion of part I It is possible to find simple methods to integrate physical systems in a stable way but careful analysis is necessary. There is a multitude of integration method to choose from and there is no real clear cut method that will work in all cases, satisfying all requirements. It is possible to integrate high frequency problems without resolving them at the time scale of the simulation, and at least capturing some of the overall physics (the midpoint rule). However, there are problems and this is discussed later. Typeset by FoilTEX May 5th Constraints Introduction Simplify the machinery of Erleblen: only holonomic case for now: no velocity or time dependence, no friction constraints ignore rigid rotation details and think of point particles Typeset by FoilTEX May 5th 26 23

13 Constraints Introduction Physical constraints Φ(x) are limits of strong forces. Think of a spring f = kx. As k becomes large, we get the constraint Φ(x) = x = k m mg Figure 5: A stretchable pendulum Typeset by FoilTEX May 5th Constraints Introduction Geometric constraints produce forces. How much? Just enough to keep Φ(x) =. We are going to have to compute the constraint forces somehow. Since motion is tangential to Φ(x) =, a force in the direction normal to Φ does not work. This means that f c = J T λ. Φ/ x T = J T f e (x) External force Resulting force is tangent to the curve Φ(x) = and motion satisfies constraints Φ(x) = Constraint force f c = J T λ Cancel component of f that would violate the constraint Figure 6: A bead on a wire Typeset by FoilTEX May 5th 26 25

14 Constraints Introduction The equation of motion is now: Mẍ = f e + f c = f e + J T λ Φ(x) = (14) This is called a differential algebraic equation (DAE). The variable λ is implicitly defined now as we can rewrite these equations in the form F (ẍ, ẋ, x, λ) =, i.e., as a nonlinear equations in the varaibles ẍ, ẋ, x, λ. Since there is no differential equation for λ, we cannot use the stepping methods that work for ODEs. Typeset by FoilTEX May 5th How can we solve for λ? Let us try to find if there is a linear expression hiding here. Since Φ(x) =, then, Φ = const so that Φ = Φ =. Φ = d dt Φ Φ(x) = ẋ = Jv =. x Φ = d2 dt 2Φ(x) = d dt [Jẋ] = Jẍ + Jẋ = Jẍ + c =. Formally, the equation Φ(x) = is mathematically equivalent to Φ = Jẍ =. So: Mẍ J T λ = f e Jẍ = c (15) This is linear in ẍ and λ: M J T J ẍ λ = f e c Typeset by FoilTEX May 5th 26 27

15 How can we solve for λ? Solve first equation for ẍ and sub in second equation: ẍ = M 1 J T λ + M 1 f e Jẍ = JM 1 J T λ + JM 1 f e = c JM 1 J T λ = JM 1 f e c Put S = JM 1 J T, the Schur complement, and substitute back in: ẍ = M 1 J T S 1 Then use an ODE method, right? JM 1 fe + c + M 1 f e No. Numerically, enforcing Φ(x) = is not the same as Φ(x) =. Though we do follow the tangent of the constraint, we take finite steps and move off the surface. Hard problem but, fortunately, there are good simple methods (and many, many simple looking BAD methods: be careful!) Typeset by FoilTEX May 5th Solve for λ numerically the Rattle way. We can approximate ẍ numerically directly with: ẍ x n+1 2x n + x n 1 h 2 and put this in the equation in the equation of motion so we can write: Mx n+1 h 2 J T λ = M [2x n x n 1 ] + h 2 f e Φ(x n+1 ) = Φ(x n ) + J n (x n+1 x n ) Now, this is linear in x n+1. Remains to chose at which step to evaluate J in the first equation and we can at least try J n. Putting λ = h 2 λ, we have: M J n J T n x n+1 λ = M [2x n x n 1 ] + h 2 f e (x n ) Φ(x n ) + J n x n Typeset by FoilTEX May 5th 26 29

16 Solve for λ numerically the Rattle way. This will still drift though because we are not solving exactly for Φ(x n+1 = but it is still better than discretizing the ODE reduction directly. Curiously, Erleben insists on calling this a velocity based formulation as opposed to the previous format which he (and others) call an acceleration based formulation. To make this stable, we need to solve the nonlinear equations Φ(x n+1 ) =. To see how this is done, first remember that: x n+1 = z n + h 2 M 1 J T n λ where z n is the sum of all the terms that are evaluated at time n. Then: Φ(x n+1 ) = Φ(z n + h 2 M 1 J T n λ) = This is m nonlinear equations in m unknowns. Solve this with Newton iterations say: J ν M 1 J T n λ (ν+1) λ (ν) = 1 h 2Φ(z n + h 2 M 1 J T n λ(ν) ) Start with λ () =, say, and iterate over ν a few times. You could update J ν so that it is reevaluated at the new coordinate x n+1 = z n + h 2 M 1 J T λ (ν) but you can also keep it at J n. This last choice corresponds to a modified Newton method. (I m not going into details because you will not implement this). Typeset by FoilTEX May 5th 26 3 Is there another way? Constraints are the limit of strong forces. instead of constraints? Can we use strong forces numerically Potential: V (x) = k 2 ΦT Φ Force: f c (x) = V x = kj T Φ Maybe λ kφ? Remember that for implicit integration, we need to evaluate f (αx n + (1 α)x n+1 ). Approximate this linearly in x n+1 so that: f (αx n + (1 α)x n+1 ) = f(x n ) + (1 α) f x (x n x n+1 ) and then, using f = kj T Φ: K = 1 k f c x = J T Φ x + J T x Φ = J T J + 2 Φ x x T Typeset by FoilTEX May 5th 26 31

17 Is there another way? Substituting this in the implicit Verlet stepper: x n+1 = 2x n x n 1 + M 1 h 2 f (αx n + (1 α)x n+1 ) and accounting for all the terms, this becomes: I + h 2 k(1 α)m 1 Kn x n+1 = (16) 2x n x n 1 + M 1 h 2 f n + h 2 k(1 α)m 1 K n x n (17) This is essentially the Baraff [1] cloth method when α =. In general, the case with α = 1/2 is best as it corresponds to the midpont rule but one must explictly add some damping. The matrix multiplying x n+1 is symmetric but not necessarily positive definite. In addition, the factor of k is problematic as the problem is ill-posed when this gets too large. Matrix K contains second derivative terms and not well behaved. 2 Φ x x T which are expensive to compute Typeset by FoilTEX May 5th To do relax Rattle using λ = kφ show the difference between the two matrices show what happens in the limit k. Next time: methods. spring-damper systems, cable example, Gauss-Seidel iterations, direct References [1] David Baraff and Andrew Witkin. Large steps in cloth simulation. In Michael F. Cohen, editor, Computer graphics: proceedings: SIGGRAPH 98 Conference proceedings, July 19 24, 1998, Computer Graphics -proceedings- 1998, pages 43 54, New York, NY 136, USA and Reading, MA, USA, ACM Press and Addison-Wesley. ACM order number: Typeset by FoilTEX May 5th 26 33