R. Abgrall 18/9/2008

Transcription

1 order HJ on Institut de Mathématiques de Bordeaux et INRIA Bordeaux Sud Ouest 18/9/2008

2 order 1 order 2 3

3 order Collaboration with J.D. Benamou, former Otto project, INRIA Roquencourt, now at Rice D. Ribereau and S. Ballereau, SNPE St Médard en Jalles S. Augoula former PhD student Financial support from SNPE, CNES

4 order order

5 Motivations order Not done before ( 93 94). Work of C.W. Shu and S.J. Osher on Cartesian, J.D. Benamou (rapport d activité INRIA Projet Ondes, 93?) Multivalued solutions for Eikonal equation, Computation of multivalued solutions, Big Ray method (Benamou) Computation for Ariane V boosters : what is the flight time lenght of a booster (SNPE CNES ALENIA) General domain, accuracy.

6 Cauchy problem for HJ in R d order u + H(x,u, u) = 0 t u(x,t = 0) = u 0 (x,t) Hamiltonian H uniformly continuous u 0 uniformly continuous Seek for continuous solutions

7 order Link with hyperbolic problems (2D) Differenciate u t set (p,q) T = (u x,u y ) T t ( ) ( p Hp 0 + q 0 H p + H(x,u, u) = 0 )( ) ( p Hq 0 + q 0 H x q )( ) p = 0 q y Problem :Weakly hyperbolic system (one eigenvalue) Problem : Which entropy condition(s)?

8 Non uniqueness order Solutions u(x,t) = u 0 (x) etc u t + (u ) 2 2 { = 0 x R x if x > 0 u 0 (x) = x else { u0 (x) if x > t u(x,t) = t x2 2 else

9 Viscosity solutions, 1 order Let ε > 0, u ε solution of ( uε ) t + H(x,u ε, u ε ) = ε u ε,u ε (x,0) = u 0 (x). Let ε 0. Take φ C 2 (R R + ) such that u ε ϕ maximal at some point x 0,t 0. Then u ε (t 0,x 0 ) = φ(x 0,t 0 ) supplementary assumption, free u ε t (x 0,t 0 ) = φ t (x 0,t 0 ) u ( ε (x 0,t 0 ) = φ(x 0,t 0 ) u ε φ (x 0,t 0 ) 0

10 Viscosity solutions, 1 order Hence ( φ ) t + H(x,φ, φ) 0 at (x 0,t 0 ).

11 Viscosity solution, 3 order Similarly, take φ such that u ε φ is minimal at some (x 0,t 0 ) Then ( φ ) t + H(x,φ, φ) 0 at (x 0,t 0 ).

12 u Sub solution order For any φ C 2 (R d R + ). If u φ has a local maximum at (x 0,t 0 ) then ( φ ) t + H(x,φ, φ) 0 at (x 0,t 0 ).

13 u Super solution order For any φ C 2 (R d R + ). If u φ has a local minimum at (x 0,t 0 ) then ( φ ) t + H(x,φ, φ) 0 at (x 0,t 0 ).

14 u solution order if u is a sub- and super solution

15 order Standard results Exsitence and uniqueness results if Assumptions H(x, u, p) H(x, v, p) γ R (u v) γ R > 0, R v u R, p H(x, u, p) H(y, u, p) C x y (1 + p ) Assumption H convex there exists a strict sub solution H(x,u,p) + when p + uniformly in x,u + H1

16 Uniqueness principle order Under any of the previous assumptions sub solution super solution

17 Numerical approximation order u t + H( u) = 0 u(x,0) = u 0 (x) {u 0 i } vertices u n+1 i = u n i th(u n i, {u n j } j Vi )

18 Constraints order Consistancy If ui 0 = C + p 0 x i, then H(ui n, {un j } j V i ) = H(p 0 ) Monotonicity If i, ui n vi n then i, u n+1 i v n+1 i under possible CFL, Accuracy Convergence

19 Numerical approximation, Idea 1 order Take the hyperbolic system ( ) ( p Hp 0 + t q 0 H p )( ) ( p Hq 0 + q 0 H x q )( ) p = 0 q y discretise (finite difference/finite volume) Unknown (u x,u y ) T Integrate (u x,u y ) T u i Possible for Cartesian, difficult for

20 idea 1 order

21 Idea 2 : Hopf formulas order define Ψ convex Ψ (p) = sup x R d ( ) x p Ψ(x) define Ψ concave Ψ (p) = inf x R d ( ) x p + Ψ(x)

22 H uniformly Lipschitz order u 0 convex, u 0 concave u(x,t) = sup p R d u(x,t) = inf p R d (x p u 0 (p) th(p) ) ( ) x p + u0(p) th(p)

23 u 0 uniformly continuous order H concave u(x,t) = inf y R d ( ( )) x y u 0 (y) + th t H convex u(x,t) = sup y R d ( ( )) y x u 0 (y) + th t

24 First formula (H Lispchitz) order Osher Bardi, Siam J. Numer Anal., 1989 Decompose u 0 = u convex + u concave u 0 u convex x p + ( u concave ) (p) = v p (x) Solve the Cauchy problem u t + H( u) = 0, Comparison principle u(x,t) sol of above problem p := V p (x,t) u(x,t) inf p V p (x,t) IC v p (x)

25 Osher Bardi order If u 0 defined on vertices of a Cartesian mesh, then uniqueness of decomposition, V p (x,t) easily computable at vertices

26 Case of Unstructured order u 0 defined at vertices (piecewise linear interpolation) u 0 = u convex + u concave Uniqueness only for cartesian Leads to unconsistant Hamiltonians

27 Idea 2 : Huygens principle order Assumption H = H 1 + H 2, H 1 convex in p, H 2 concave in p Dp 2 H bounded,h 2 = α p 2,H 1 = H + α p 2. H 2 (p) p p + (H 2 ) (p ) p H H 1 + p p + (H 2 ) (p ) p Comparison principle?

28 Comparison principle order If H(p) = H 1 (p) + H 2 (p), H 1 convex in p, H 2 concave in p. H, H 1, H 2 uniformly Lipschitz continuous Cauchy problem u t + H( u) = 0 u(x,0) = u 0 (x) Solution u(x,t) Then Φ 2 (x,t) u(x,t) Φ 1 (x,t)

29 Comparison principle order with Φ 1 (x,t) = inf sup q R d y R d [ ( u 0 (y) + th1(q) + th2 q + y x )] t Φ 2 (x,t) = inf sup q R d y R d [ ( u 0 (y) + th2(q) + th1 q + x y )] t

30 Idea of proof order H(p) H 1 (p) p p + H 2 (p ) = G p (p) G p is convex in p u t + H( u) u t + G p ( u) u sub-solution of u t + G p ( u) = 0 u(x,t) V p (x,t) take the inf.

31 Application order {ui 0 } vertex values Construct piecewise linear interpolation u 0 use one of the formula to approximate u 1 i repeat

32 Application order i i Ω l u Ω 0 l (M) = ui 0 + U i M i M

33 Application order Φ(M i, t) = ui 0 t min max set H i = min max q R 2 0 l k i q R 2 0 l k i ) sup (U l (z q) H 1 (z) H 2 (q) z Ω l +q ( ) sup U l (z q) H1(z) H2(q) z Ω l +q Consistant, Monotone (L 1 + L 2 ) t h 1 2

34 Case of H convex smooth order ( ) H i = min inf U i z H (z) 0 l k l z Ω l H ( x(p) ) + H ( p(x) ) = x(p) p(x) p(x) = x H ( x(p) ) x(p) = p H ( p(x) ) ) 1 x(u l ) Ω l, sup x Ωl (U i z H (z) = H(U i ) 2 x(u l ) Ω l, sup reached on boundary of ( Ω l )

35 Eikonal order ( H i = max max 0, u k u i, u ) j u i 0 l k i ki ji k j i

36 Properties order Consistant, Monotone Intrinsic k j i

37 Result, Crandall Lions extended order Let H : R d R uniformly continuous, u 0 bounded uniformly continuous, lipschitz constant L. Consider T triangulation, h = max{radius}, uniform. Numerical Hamiltonian H (1) consistant, (2)monotone, (3) intrinsic Initial condition u 0, ui 0 = u 0 (M i ), u n+1 i = ui n th i. u the viscosity solution of u t + H( u) = 0 u(x,0) = u 0 (x) then for all T > 0 and n, n t T max i un i u(m i,n t) C t

38 Other examples : Lax Friedrichs order H i =H ε ( ) u(m) ui dm ρ C ρ =H ε u ndm ρ C ρ with ǫ L π ( udx ) D ρ H = H πρ 2 or D ρ H( u)dx πρ 2

39 Properties order Consistant, Monotone under CFL Intrinsic by construction k j i

40 Numerical examples order distance functions various geometry

41 dist1 order

42 mesh order

43 dist2 order

44 lens order

45 order

46 order Overview What is the problem for boundary? Viscosity solutions and boundary Numerical approximation

47 The problem order Take Solution u 1 = εu xx x [0,1], u(0) = 0 u(1) = 2 > 1 u ε (x) = x + exp ( ) ( ) x 1 ε exp 1 ε 1 exp ( 1 ) ε

48 order

49 More serious situation sol.di.desc order

50 condition order Define H(x,u(x), u(x)) = 0 x Ω F(x,u(x), u(x) = 0 x Ω { G H(x,t,p) if x Ω (x,t,p) = max(f(x,t,p),h(x,t,p)) if x Ω { H(x,t,p) if x Ω G (x,t,p) = min(f(x,t,p),h(x,t,p)) if x Ω

51 Viscosity sub-solution order u super semi continuous function (semi continue supérieurement) φ C 2 (Ω), if x 0 Ω is a local maximum of u φ, then G (x 0,u ( x 0 ), φ(x 0 )) 0.

52 Viscosity sub-solution order u lower semi continuous function φ C 2 (Ω), if x 0 Ω is a local minimum of u φ, then G (x 0,u ( x 0 ), φ(x 0 )) 0.

53 Dirichlet order G (x,t,p) = G (x,t,p) = { H(x,t,p) if x Ω max(h(x,t,p),u(x) g(x)) if x Ω { H(x,t,p) if x Ω min(h(x,t,p),u(x) g(x)) if x Ω

54 Numerical approximation order H(x,u(x), u(x)) = 0 x Ω F(x,u(x), u(x) = 0 x Ω F(x,t,p) = t g(x) (Dirichlet) or F(x,t,p) = p n g(x) Neuman

55 Numerics order S H (x i,u(x i ), {u j ;j V i }) = 0 x i Ω S(ρ,x i,u(x i ),u) = max(s Hb (x i,u(x i ), {u j ;j V i }, S F (x i,u(x i ), {u j ;j V i }) = 0 x i Ω Under which, does the scheme converge to the solution of the problem.

56 General convergence result order Barles and Souganidis result for S(ρ,x,t,u) Monotonicity: if u v, for all ρ 0, x Ω, t R and u,v L (Ω), then S(ρ,x,t,u) S(ρ,x,t,v). Stability: for all ρ > 0, there exists a solution u ρ L (Ω) to (1) with a bound independant of ρ, Consistency: for all x Ω and φ Cb (Ω), lim sup S(ρ,y,φ(y) + ξ,φ + ξ) G (x,φ(x),dφ(x)) ρ 0,y x,ξ 0 lim inf S(ρ,y,φ(y) + ξ,φ + ξ) G (x,φ(x),dφ(x)) ρ 0,y x,ξ 0

57 General convergence result order and Strong uniqueness principle: if u L (Ω) is a subsolution and v L (Ω) is a supersolution, then u v on Ω.

58 Barles and Souganidis order Theorem (Barles, Souganidis) Assuming the monotonicity, the consistency, the stability and the strong uniqueness properties, then, when ρ 0, the solution u ρ of converges locally uniformly to the unique continuous viscosity solution of H(x,u(x), u) = 0.

59 Application order Assume that 1 H b H, S H, S Hb and S F are monotone and stable, 2 for all φ Cb (Ω), we have, For any x Ω or in a neighborhodd of Ω lim S H(ρ, y, ϕ(y) + ξ, ϕ + ξ) = H(x, ϕ(x), Dϕ(x)) ρ 0,y x,ξ 0 lim S H b (ρ, y, ϕ(y) + ξ, ϕ + ξ) = H b (x, ϕ(x), Dϕ(x)) ρ 0,y x,ξ 0 lim S F(ρ, y, ϕ(y) + ξ, ϕ + ξ) = F(x, ϕ(x), Dϕ(x)) ρ 0,y x,ξ 0 3 Problem has a uniqueness principle,

60 Application order Then the family u ρ defined by { 0 = S(ρ,x,u ρ (x),u ρ SH (ρ,x,u ) = ρ (x),u ρ ) max(s Hb (ρ,x,u ρ (x),u ρ ),S F (ρ,x,u ρ (x) converges locally uniformly to the solution of H(x,u, u) = 0 in Ω F(x,u, u) = 0 in Ω

61 Visual proof order Convex Hamiltonians + Dirichlet + Dirichlet H(x,u,p) = sup { b(x,v).p + λu f (x,v)} v V 0 = u(x) inf v(.) [ min(t,τ) 0 f (y x (t), v(t))e λt dt + 1 {T<τ } u(y x (T))e λt + 1 {T τ } ϕ(y x (τ))e λτ ]

62 Examples order H b = impose strongly boundary condition, Case of convex hamiltonians ( H b = max max Ui y H (y) ) angular sectors y ang. sect i.e. do nothing, etc

63 Numerical examples order u = 1 in Ω u = 0 for x = 0.5 u = C for x = 1. The constant C takes the values displayed in Table 1 Case C Table: condition for the Dirichlet b.c.s

64 Examples order (a) (b) Figure: Dirichlet, case A, (a) weak, (b) comparaison weak ( ) strong ( )

65 order (a) (b) Figure: Dirichlet, case B, (a) weak, (b) comparaison weak ( ) strong ( )

66 order (a) (b) Figure: Dirichlet, case C, (a) weak, (b) comparaison weak ( ) strong ( )

67 order (a) (b) Figure: Dirichlet, case D, (a) weak, (b) comparaison weak ( ) strong ( )

68 Improved convergence result order For convex Hamiltonians, take the Godunov boundary Hamitonian. No boundary layers : control of u.

69 Soner Bcs sol.soner.desc order

70 Dirichlet Bcs order sol.di.desc

71 order Non convex Hamiltonians Situation much less clear...

72 order

73 Scheme : 2 points of view order solve unsteady problem : u t + H(x,u, u) = 0 u(x,0) = u 0 (x) du dt + H(x i,u(x i ), {u j } j Vi ) and Runge Kutta type integration Solve steady problem directly H(x,u, u) = 0 unsteady one : u n+1 u n t u n+1 u n + H(x,u n+1, u n+1 ) = 0 t ( ) H(x,u n+1, u n+1 ) + H(x,u n, u n ) =

74 Overview order 1 ENO type scheme (WENO) 2 Tuning of artificial dissipation 3 Schemes for steady problems

75 order ENO WENO type du dt + H(x i,u(x i ), {u j } j Vi ) Runge Kutta (Shu Osher for preservation of stability properties of H) Take first order scheme H i = H(x i,u i, {u j } j Vi ) H(x i,u i, u T1,, u Tk ) i i Ω l Provide high order accuracy approximations of u T1,, u Tk

76 order Algorithm need to define a family of stencils, compute the one that minimise oscillation

77 approximation of u T order start from node values (Lagrange) k-th order approximation of u T : polynomial of degree k + 1, need (k + 1) ( (k + 1) + 1 ) /2 points construction of stencils with (k + 1) ( (k + 1) + 1 ) /2 need M i inside (k + 1) ( (k + 1) + 1 ) /2 1 additional points

78 Construction of potential stencils order All directions needed, As few stencils as possible, What is oscillation

79 Hierarchical algorithm order Step 1 Consider all triangles that surround M i (about 6 for regular ): u T Choose T min that minimize u T Step 2 : T min known construct 3 more points

80 Hierarchical algorithm, 2 order Centered

81 Facts and justifications order Compute stencils : not always possible (think of 3 aligned points) In practice always a solution P P k interpolation for f smooth max f (r) P (r) h p+1 x K CM p+1 ρ r f not smooth in the stencil of P P = r a ij (x x 0 ) i (y y 0 ) j l=0 i+j=r i+j=r ] a ij C [f (r) h p r

82 Numerical results order H(p,x,y) = p n(x,y) { 1 n(x,y) = ( ) 1 cos 2πd if d 2 = ( ) x 2 ( x 2 0.2) 1 1 else

83 Problems order Non compact stencil, not obvious to increase order (even for WENO) Not obvious for 3D, computational not very efficient

84 order Construction of Compact, problems Discontinuous Galerkin like (see Shu et al., JSC, SIAM SISC, etc), How to have provable convergent (no Lax Wendroff like theorem )? i.e How to have automatic entropy Do this directly in the scheme, i.e. start from a viscosity solution framework. Different from Shu et al.

85 Solution 1 order S. Augoula s thesis. Available at abgrall/these.tar.gz J. Scientific Computing, 2000, vol 2

86 order Solution 2, steady problems, SIAM SISC, in revision Compute H i H(x,u(x), u) as H i = l ( H i ) L + (1 l) ( Hi ) H ( H i ) (L) first order, monotone, approximation of H(x, u(x), u) ( H i ) H high order approximation of H(x,u(x), u)

87 Properties of l, 1 order Steady problems H(x, u(x), u(x)) = 0 u smooth ( H i ) H = O(h p ), p 2 l(u) = O(h p ) then H i = O(h p )

88 Properties of l, 2 order Iterative scheme u n+1 i = u n i th i ( H i ) (1) monotone under t Ch ( H i ) (H) L p stable for t C h l [0,1] same stability range for blended scheme

89 Compute l order l ( ) ) L ( ) H (Hi ) L H i + (1 l) Hi = (l + (1 l)r so ask r = H(H) H (L) l + (1 l)r 0

90 order ell 1 range 1 r

91 accuracy order l ( ( ) ) L ( ) H l (Hi H i + (1 l) Hi = r + (1 l) ) (H) r = H(H) H (L) If l [0,1] and l r C for some C > 0, then l( ) L ( ) H H i + (1 l) Hi C H (H). and high order accuracy.

92 order Solution Hyperbola l + (1 l)r = 0 slope 1 at (0,0), C 1 or l = { 0 if r > 0 min(1,c r ) if r 0 l = min(1,c r ), C 1

93 Minmod like order ell 1 range 1 r

94 order Proof for convergence for H(x, u(x), u) = 0 Barles and Souganidis result for S(ρ,x,t,u) H ρ (x,u(x), {u j } j Vi ) = 0 (1) 1 Monotonicity: if u v, for all ρ 0, x Ω, t R and u, v L (Ω), then S(ρ, x, t, u) S(ρ, x, t, v). 2 Stability: for all ρ > 0, there exists a solution u ρ L (Ω) to (1) with a bound independant of ρ, 3 Consistency: for all x Ω and φ C b (Ω), lim S(ρ, y, φ(y) + ξ, φ + ξ) = H(x, φ(x), Dφ(x)) ρ 0,y x,ξ 0 4 Strong uniqueness principle: if u L (Ω) is a subsolution and v L (Ω) is a supersolution, then u v on Ω.

95 Barles and Souganidis order Theorem (Barles, Souganidis) Assuming the monotonicity, the consistency, the stability and the strong uniqueness properties, then, when ρ 0, the solution u ρ of converges locally uniformly to the unique continuous viscosity solution of H(x,u(x), u) = 0.

96 order Sketch, 1 Assume H (L) monotone l as before assume iterative convergence reached

97 Sketch, 2 order l [0,1] L : bound on the solution (CFL condition) Take l = min(1, r ), same CFL number as low order scheme.

98 Sketch, 3 order u ϕ maximum at x 0 : u(x 0 ) ϕ(x 0 ) ( ) 0 = lh (L) + (1 l)h (H) = l + (1 l)r (u)h (L) (u) ( ) l + (1 l)r (u) H (L) (ϕ) so H (L) (ϕ) 0 if ( l + (1 l)r ) (u) 0 i.e. H (H) (u) 0 then u = limu h super solution of H(x,u(x), u) = 0

99 order Example u 1 = 0 x Ω u = 0 x Γ Ω

100 Examples order first example : Godunov scheme H (H) i = max i u ang sec (x i ) K (k) i = angular sectors K(k) max(0, u p u i ) else pi ( ) if u ang sec (x i ) angular sector

101 order Godunov :Practical implementation H=-1.e+30! Initialisation of the numerical Hamiltonian DO jt=1,nt! loop over element is(:)=nu(:,jt)! index of the 3 vertices of the element jt compute gradient of u compute H T H(is(:)) = max(h(is(:)),h T ) ENDOO

102 Lax Friedrich order H LF ρ ( u Ω1,, u Ωki ) = D h H( u) πh 2 ǫ h C h [u(m) u(m i )]dl.

103 Implementation order H=0! Initialisation of the numerical Hamiltonian DO jt=1,nt! loop over element is(:)=nu(:,jt)! index of the 3 vertices of the element jt compute gradient of u compute the angles θ 1 jt,θ2 jt,θ3 jt compute H T s and D T s H(is(1)) = H(is(1)) + H 1 T + D1 T H(is(2)) = H(is(2)) + H 2 T + D2 T H(is(3)) = H(is(3)) + H 3 T + D3 T

104 Lax Friedrichs 2 order H LF ( u Ω1,, u Ωki ) = T H( u T ) + α (u i u j ) T M i M j T T T and α h T max p p H where h T is the largest edge of T.

105 order Implementation H=0! Initialisation of the numerical Hamiltonian air=0! Initialisation of the area DO jt=1,nt! loop over element is(:)=nu(:,jt)! index of the 3 vertices of the element jt compute gradient of u compute the angles θ 1 jt,θ2 jt,θ3 jt compute H T s and D T s ENDDO H(is(1)) = H(is(1)) + T (H T + DT 1 ) H(is(2)) = H(is(2)) + T (H T + DT 2 ) H(is(3)) = H(is(3)) + T (H T + DT 3 ) air(is(1)) = air(is(1)) + T air(is(2)) = air(is(2)) + T air(is(3)) = air(is(3)) + T

106 order any of the above with P 2 interpolation i p

107 or SUPG like order H σ (σ,u σ,u ξξ Vσ ) := + h Ω ( H(x, u h )N σ dx Ω ] [ p H(x, u h ) p H(x, u h ) N σ / N σ dx Ω H(x, u h )dx )

108 One dimensional examples order u n(x) = 0 x [0,1] u(0) = u(1) = 0 where n(x) = 3x 2 + a with a = 1 2.x3 0 2x 0 1 and x 0 = The solution is u(x) = { x 3 + ax if x [0,x 0 ] 1 + a ax x 3 if x [x 0,1]

109 numerics order First order Blended scheme Second order Exact Comparison of the first order, second order and blended, 50 mesh points, zoom in [0.4,0.8]

110 ... order First order Second order, blended Second order, unlimited Exact solution 0.75 First order Second order, blended Second order, unlimited Exact solution

111 Errors order First order Second order unlimited Second order limited Slope 1 Slope 2 5 First order Second order unlimited Second order limited Slope 2 slope L 1 norm L norm

112 Non homogeneous Hamiltonian order The problem reads H(x,p) = cos(p) 2 + p. H(x,u ) = 0 in x [0,1],with u(0) = u(1) =

113 order Numerics numericaly checked that the gradient of the solution may be larger that 1 in absolute value, so that non convex effect are real First order Second order limited Second order unlimited

114 Distance function order

115 numerics order First order solution (blue isolines), second order unlimited solution (gree isolines) and second order limited solution (red isolines).

116 Non homogeneous case order Ω = [0,1] [0,1] C where C = {(x,y) [0,1/2] [0,1/2] and (x 1/2) 2 + (y 1/2) 2 1}, ψ( u) 1 = 0 in Ω and u = 0 on Ω. p ψ(p) = 2 p if p 1 else.

117 numerics order

118 order numerics First order Second order

119 numerics order

120 Distance order

121 order First order solution

122 order Second order solution, no limitation

123 Second order, no limitation order

124 Second order, no limitation order

125 Second order solution, limitation order

126 3D case order

127 second order, no limitation order

128 second order, limitation order

129 Error order

130 Error order