Solving PDEs with freefem++

Transcription

1 Solving PDEs with freefem++ Tutorials at Basque Center BCA Olivier Pironneau 1 with Frederic Hecht, LJLL-University of Paris VI 1 March 13, 2011 Do not forget That everything about freefem++ is at Check for new release every month Now in 3D and in MPI OP ( www/ann.jussieu.fr/pironneau) Solving PDEs with freefem++ March 13, / 38

2 Outline I 1 Part I: Algorithms Introduction Blocks Rannacher s Projection Scheme for Navier-Stokes eqs Floating Object Fluids and Temperature A Large Example 2 Part II: Playing with Meshes Mesh Adaptivity Domain Decomposition and zoom Application to Hydrology Discontinuous Galerkin freefem++ and Option Pricing OP ( www/ann.jussieu.fr/pironneau) Solving PDEs with freefem++ March 13, / 38

3 Introduction freefem++ has two building blocks An elliptic PDE system solver problem, solve A convection module convect Most complex nonlinear problems can be solved by appropriate algorithmic combination of both Except hyperbolic problems with shocks The power of freefem++ is also due to the capacity to handle multiple meshes in one script freefem++ has an advanced mesh adaptivity module OP ( www/ann.jussieu.fr/pironneau) Solving PDEs with freefem++ March 13, / 38

4 A simple elliptic problem u = f in Ω, u Ω = u Γ Ω u û f û = 0, Ω û H 1 0 (Ω) Take Ω the unit circle, f a Dirac at x 0, y 0 and u Γ = 0. The problems is solved in variational form approximated by a finite element method : u H 1 0 (Ω) V h = {û C 0 (Ω h ) : u h Tk P 2 triangle in the mesh} Execute OP ( www/ann.jussieu.fr/pironneau) Solving PDEs with freefem++ March 13, / 38

5 A simple convection problem Let u be solution of D t u := t u + a u = 0, u(0) = u 0 in Ω, t < T One has D t u(x 0, t 0 ) := lim δt [u(χ(t 0), t 0 ) u(χ(t 0 δt), t 0 δt)] where χ(t) = a(χ(t), t), χ(t 0 ) = x 0 δt 0 1 Hence D t u(x 0 ) 1 δt (um+1 (x 0 ) u m (X m (x 0 ))) with X m (x 0 ) := χ(t δt) x 0 a(x 0 )δt (1) A possible scheme is u m+1 (x) = u m (X m (x)) Π h u m (x a(x)δt) The keyword convect([a1,a2],-dt,u) returns uoχ. Note that the result is not a piecewise polynomial function even when u is. The difference with u(x adt) is a that x adt is not a precise computation of χ(t dt), especially near boundaries. OP ( www/ann.jussieu.fr/pironneau) Solving PDEs with freefem++ March 13, / 38

6 A simple convection problem (II) Execute OP ( www/ann.jussieu.fr/pironneau) Solving PDEs with freefem++ March 13, / 38

7 A convection diffusion problem Let u be solution of the convection-diffusion-dissipation equation t u + a u (µ u) + bu = 0, u(0) = u 0 in Ω, t < T It can be rewritten as D t u (µ u) + bu = f, u(0) = u 0 where D t u is the convective derivative. Let us use an implicit Euler scheme in time combined with the previous methods [ u m+1 u m ox m ] û + µ u m+1 û + bu m+1 û = 0, û H0 1 (Ω) δt Ω OP ( www/ann.jussieu.fr/pironneau) Solving PDEs with freefem++ March 13, / 38

8 Convection Diffusion problem (II) Execute OP ( www/ann.jussieu.fr/pironneau) Solving PDEs with freefem++ March 13, / 38

9 Publish the results. e.g. Evolution of u and u 2 This exercise will show how to communicate with the outside world Execute Use gnuplot to create a one dimensional graph >cd /Users/pironneau/Desktop/bilbaoFreefem/edp >gnuplot plot "norms.txt"using 1:2 w l, "norms.txt"using 1:3 w l OP ( www/ann.jussieu.fr/pironneau) Solving PDEs with freefem++ March 13, / 38

10 Rannacher s Projection Scheme Her we used Neumann bdy conditions for the PDE instead of Dirichlet to illustrate the fact that u remains constant and u 2 decays exponentially OP ( www/ann.jussieu.fr/pironneau) Solving PDEs with freefem++ March 13, / 38

11 Rannacher s Projection Scheme Consider t u + u u + p ν u = 0, u = 0 with u(0) = u 0 and u Γ = u Γ. With X (x) = x u m (x)δt, the following is first order in time: 1 δt (um+1/2 u m ox ) + p m ν u m+1/2 = 0 q = u m+1/2 u m+1/2 u m+1 = u m+1/2 + qδt, p m+1 = p m q p m q Exercise Compare this scheme with a fully coupled scheme like 1 δt (um+1 u m ox ) + p m+1 ν u m+1 = 0, u m+1 = 0 on the driven cavity problem. OP ( www/ann.jussieu.fr/pironneau) Solving PDEs with freefem++ March 13, / 38

12 Freefem Script for Fully Implicit and Rannacher s OP ( www/ann.jussieu.fr/pironneau) Solving PDEs with freefem++ March 13, / 38

13 Floating Object Consider a cylinder C of radius R and mass µ and center q 0 := x 0 (t), y 0 (t) floating in the flow of a driven cavity with viscosity ν. To find its motion q 0 and its angular velocity ω we must apply Newtons law: µ d2 q 0 dt 2 = d ω dt = q C [ν( u + u T ) p)] n [ ] [ν( u + u T ) p] n q C The no-slip boundary condition on C requires to impose on C a velocity for the fluid equal to q 0 + ω (q q 0 ). The mesh must be rebuilt at each time step because C moves. we shall use the trick u=u to project u defined on the old triangulation onto a u defined on the new one. OP ( www/ann.jussieu.fr/pironneau) Solving PDEs with freefem++ March 13, / 38

14 Execute OP ( www/ann.jussieu.fr/pironneau) Solving PDEs with freefem++ March 13, / 38

15 Fluids at Varying Temperature The Boussinesq approximation assumes that the density changes slightly with the temperature θ and the buyancy is a force in the fluid: t u + u u + p ν u = aθ e y, u = 0 t θ + u θ κ θ = 0 with u(0) = u 0, θ(0) = θ 0 and u Γ = u Γ, θ Γ = θ Γ. Let us study the case of a box with a hot (θ Γ = 1) and a cold wall (θ Γ = 0) t=0.107 IsoValue t=0.13 IsoValue Vec Value Vec Value OP ( www/ann.jussieu.fr/pironneau) Solving PDEs with freefem++ March 13, / 38

16 Execute OP ( www/ann.jussieu.fr/pironneau) Solving PDEs with freefem++ March 13, / 38

17 Fluids, Temperature and Turbulence t θ + u θ (κ m T θ) = 0 t u + u u (µ T u) + p + e(θ θ 0 ) e 2, u = 0 k 2 µ T = c µ ɛ, κ T = κµ T t k + u k + ɛ (µ T k) = µ T 2 u + ut 2 ɛ 2 t ɛ + u ɛ + c 2 k c ɛ (µ T ɛ) = c 1 c µ 2 k u + ut 2 = 0 1 δt (θm+1 θ m X m ) (κ m T θ m+1 ) = 0 1 δt (um+1 u m X m ) (µ m T u m+1 ) + p m+1 + e(θ m+1 θ 0 ) e 2, u m+1 1 δt (km+1 k m X m ) + k 1 δt (ɛm+1 ɛ m X m ) + c 2 ɛ µ m+1 T = c µ k m+12 ɛ m+1, m+1 ɛm k m (µm T k m+1 ) = µm T 2 um + u mt 2 m+1 ɛm κm+1 T = κµ m+1 T k m c ɛ (µ m T ɛ m+1 ) = c 1 c µ 2 km u m + u mt 2 OP ( www/ann.jussieu.fr/pironneau) Solving PDEs with freefem++ March 13, / 38

18 Fluids, Temperature and Turbulence (II) IsoValue Execute OP ( www/ann.jussieu.fr/pironneau) Solving PDEs with freefem++ March 13, / 38

19 Conclusion of Part I 1 Learn Variational Formulations 2 Learn how to devise Algorithms 3 Choose your finite element spaces Thank you for your attention OP ( www/ann.jussieu.fr/pironneau) Solving PDEs with freefem++ March 13, / 38

20 Outline I 1 Part I: Algorithms Introduction Blocks Rannacher s Projection Scheme for Navier-Stokes eqs Floating Object Fluids and Temperature A Large Example 2 Part II: Playing with Meshes Mesh Adaptivity Domain Decomposition and zoom Application to Hydrology Discontinuous Galerkin freefem++ and Option Pricing OP ( www/ann.jussieu.fr/pironneau) Solving PDEs with freefem++ March 13, / 38

21 Local Error Minimized: Estimated Results OP ( www/ann.jussieu.fr/pironneau) Solving PDEs with freefem++ March 13, / 38

22 Mesh Refinement OP ( www/ann.jussieu.fr/pironneau) Solving PDEs with freefem++ March 13, / 38

23 Local Error and Adaptivity OP ( www/ann.jussieu.fr/pironneau) Solving PDEs with freefem++ March 13, / 38

24 Mesh Refinement OP ( www/ann.jussieu.fr/pironneau) Solving PDEs with freefem++ March 13, / 38

25 Domain Decomposition with Overlapping Schwarz Algoritm To Solve u = f in Ω, u Γ = 0 in a domain Ω = Ω 1 Ω 2, on computes u as the limit of u k = f in Ω 1, u k Γ1 = u k 1 u k = f in Ω 2, u k Γ2 = u k 1 OP ( www/ann.jussieu.fr/pironneau) Solving PDEs with freefem++ March 13, / 38

26 file ddm.edp Execute OP ( www/ann.jussieu.fr/pironneau) Solving PDEs with freefem++ March 13, / 38

27 The Schwarz-Zoom Method γ H (resp γ h ) is the interpolation operator on V H (resp V h ) Find u m+1 H V H, u m+1 h V h, such that w H V 0H, w h V 0h a H (u m+1 H, w H ) = (f, w H ), u m+1 H SH = γ H uh m, um+1 H ΓH = g H, a h (u m+1 h, w h ) = (f, w h ), u m+1 h Sh = γ h uh m, um+1 Γh = g h h OP ( www/ann.jussieu.fr/pironneau) Solving PDEs with freefem++ March 13, / 38

28 Hilbert Space Decomposition Method (JL. Lions) Model problem (for instance K = I, f = 1 + δ( x x 0 )): (K v) = f in Ω, u Γ = 0 v = U + u, U H 1 0 (Ω), u Λ H 1 0 (Λ) β(u n+1 U n ) (U n+1 + u n ) = f in Ω β(u n+1 u n ) (U n + u n+1 ) = f in Λ "usmall.gnu" using 1:3 "ularge.gnu" using 1:3 "u.gnu" using 1:3 "log.gnu" using 1: OP ( www/ann.jussieu.fr/pironneau) Solving PDEs with freefem++ March 13, / 38

29 The COUPLEX I Test Case Figure: A 2D multilayered geometry 20km long, 500m high with permeability variations K + K = O(10 9 ). Hydrostatic pressure by a FEM. (K H) = 0, H or H given on Γ n OP ( www/ann.jussieu.fr/pironneau) Solving PDEs with freefem++ March 13, / 38

30 COUPLEX I : Concentration of Radio-Nucleides Figure: Concentration at 4 times with Discontinuous Galerkin FEM (Apoung-Despré). r t c + λc + u c (K c) = q(t)δ(x x R ) OP ( www/ann.jussieu.fr/pironneau) Solving PDEs with freefem++ March 13, / 38

31 Solution by Zoom Execute OP ( www/ann.jussieu.fr/pironneau) Solving PDEs with freefem++ March 13, / 38

32 Execute OP ( www/ann.jussieu.fr/pironneau) Solving PDEs with freefem++ March 13, / 38

33 Discontinuous Galerkin (II) Execute OP ( www/ann.jussieu.fr/pironneau) Solving PDEs with freefem++ March 13, / 38

34 Pricing a Basket Option Find Given U h (T ) = (K S 1 S 2 ) + find U h H h such that V h H h ( t U h )V h [ [ 1 2 S i (σi 2 Si 2 V h ) Si U h µ i S i ( Si U h )V h ] Ω Ω i=1,2 S1 (ρσ 1 σ 2 S 1 S 2 V ) h S2 U h ] Execute OP ( www/ann.jussieu.fr/pironneau) Solving PDEs with freefem++ March 13, / 38

35 Heston Model t P (yx)2 2 xxp (yµ)2 yy P ρy 2 xµ xy P r x P + rp = 0 2 P(x, y, T ) = (K x y) + P(0, y, t) = Ke rt P(+, y, t) = 0 y P(x, 0, t) = y P(x, +, t) = 0 Execute OP ( www/ann.jussieu.fr/pironneau) Solving PDEs with freefem++ March 13, / 38

36 American Options u t σ2 x 2 2 u 2 x 2 rx u x + ru 0 u u, one is = with u(t, 0) = Ke rt, u(0, x) = u (x) := (K x) + { V = v L 2 (R + ), x v } x L2 (R + ), K = {v L 2 (0, T; V), v u }. Find u K C 0 ([0, T ]; L 2 (R + u )), t L2 (0, T ; V ), s.t. ( u t, v u) + a(u, v u) 0, v K, u(t = 0) = u, with ( σ 2 v w a(v, w) = x2 2 x x + (σ2 + x σ ) x r)x v x w + rvw dx. R + Assume σ [σ, σ] and x x σ 2 < M, then the solution exists and is unique. OP ( www/ann.jussieu.fr/pironneau) Solving PDEs with freefem++ March 13, / 38

37 Semi-Smooth Newton Method Kunisch et al suggested to reformulate the problem as a(u, v) (λ, v) = (f, v) v H 1 (R + ), λ min{0, λ + c(u φ)} = 0, i.e.au λ = f The last eq. is equivalent to λ 0, λ λ + c(u φ) i.e. u φ, λ 0, with equality on one of them for each S. This problem is equivalent for any real constant c > 0 because λ is the Lagrange multiplier of the constraint. Newton s algorithm applied to (3) gives Choose c > 0,, u 0, λ 0, set k = 0. Determine A k := {S : λ k (S) + c(u k (S) φ(s)) < 0} Set u k+1 = arg min u H 1 (R + ){ 1 2 a(u, u) (f, u) : u = φ on A k} Set λ k+1 = f Au k+1 OP ( www/ann.jussieu.fr/pironneau) Solving PDEs with freefem++ March 13, / 38

38 Results "exercise_250" Best of put basket option, σ 1 = 0.2, σ 2 = 0.1, ρ = 0.8 Execute OP ( www/ann.jussieu.fr/pironneau) Solving PDEs with freefem++ March 13, / 38