Numerical explorations of a forward-backward diffusion equation

Transcription

1 Numerical explorations of a forward-backward diffusion equation Newton Institute KIT Programme Pauline Lafitte 1 C. Mascia 2 1 SIMPAF - INRIA & U. Lille 1, France 2 Univ. La Sapienza, Roma, Italy September 14th, 2010 September 14th, / 39

2 Plan 1 Introduction 2 Explicit solution : Riemann problem 3 Semi-discrete scheme September 14th, / 39

3 Plan 1 Introduction 2 Explicit solution : Riemann problem 3 Semi-discrete scheme September 14th, / 39

4 Phase transition Motivation : modeling of mass transport in a relaxing medium, such as polymers near and below glass transition Problem : understand the modeling of the transition and numerically approach it Figure: Two-phase polymer September 14th, / 39

5 Model Modeling by Jäckle and Frisch (1985) : u t = 2 v x 2 B where v potential defined by c b a d u t v = φ(u)+ θ(t ) ( ) ψ(u) φ(u) s A with θ a memory function and φ cubic-shaped, loc. Lipschitz Figure: diffusion function φ September 14th, / 39

6 Simplifications : θ(t) = exp( t/τ) and ψ : u εu/τ τ 2 u t 2 + u ( t = 2 x 2 φ(u) + ε u ) t Case τ 0 + (no memory), ε > 0 : Novick-Cohen and Pego (1991) ( u t = x u 0 L (Ω) φ(u) + ε u t ), x Ω R d, t > 0, global existence and uniqueness of the C 1 solution. September 14th, / 39

7 1d nonlinear diffusion equation Case with no memory (τ = 0) and ε 0 + u t = 2 φ(u) x 2 September 14th, / 39

8 1d nonlinear diffusion equation Case with no memory (τ = 0) and ε 0 + φ u t = 2 φ(u) x 2 S- c b B U a d u Cauchy problem : φ non monotone ill-posedness A S+ September 14th, / 39

9 1d nonlinear diffusion equation Case with no memory (τ = 0) and ε 0 + φ u t = 2 φ(u) x 2 B Cauchy problem : φ non monotone ill-posedness S- c b U a d S+ u For ε > 0, linearized problem around u U : exponential exit... A September 14th, / 39

10 Analytical results u t = 2 φ(u) x 2 September 14th, / 39

11 Analytical results u t = 2 φ(u) x 2 Plotnikov ( ) : description by Young measures u = λ (φ ) 1 (v) + λ u (φ u ) 1 (v) + λ + (φ + ) 1 (v) with λ (x, t)+λ u (x, t)+λ + (x, t) = 1, { λ (x, t) = 1 if v(x, t) < A, λ + (x, t) = 1 if v(x, t) > B. September 14th, / 39

12 Analytical results u t = 2 φ(u) x 2 Plotnikov ( ) : description by Young measures u = λ (φ ) 1 (v) + λ u (φ u ) 1 (v) + λ + (φ + ) 1 (v) with λ (x, t)+λ u (x, t)+λ + (x, t) = 1, { λ (x, t) = 1 if v(x, t) < A, λ + (x, t) = 1 if v(x, t) > B. For all g C 1, increasing, G := s g φ and G (v) := λ G (φ ) 1 + λ u G (φ u ) 1 + λ + G (φ + ) 1, we have G (v) t ( x g(v) v x ) + g (v) ( ) v 2 0 x September 14th, / 39

13 A moving interface? t R : u(x, t) < b γ = {(ξ(t), t)} R + : u(ξ(t), t) > a x September 14th, / 39

14 Concept of solution Evans and Portilheiro (2004) : two-phase (weak) solutions piecewise smooth solutions + «entropy inequality» September 14th, / 39

15 Concept of solution Evans and Portilheiro (2004) : two-phase (weak) solutions piecewise smooth solutions + «entropy inequality» At the interface γ : Transmission conditions : [ ] φ(u) [φ(u)] γ = 0, ξ (t) [u] γ + x γ = 0 September 14th, / 39

16 Concept of solution Evans and Portilheiro (2004) : two-phase (weak) solutions piecewise smooth solutions + «entropy inequality» At the interface γ : Transmission conditions : [ ] φ(u) [φ(u)] γ = 0, ξ (t) [u] γ + x γ = 0 Entropy conditions : φ(u(ξ(t), t)) [A, B] and ξ (t) 0 if φ(u(ξ(t), t)) = A, ξ (t) 0 if φ(u(ξ(t), t)) = B, ξ (t) = 0 if φ(u(ξ(t), t)) (A, B). September 14th, / 39

17 Concept of solution Evans and Portilheiro (2004) : two-phase (weak) solutions piecewise smooth solutions + «entropy inequality» At the interface γ : Transmission conditions : [ ] φ(u) [φ(u)] γ = 0, ξ (t) [u] γ + x γ = 0 Entropy conditions : φ(u(ξ(t), t)) [A, B] and ξ (t) 0 if φ(u(ξ(t), t)) = A, ξ (t) 0 if φ(u(ξ(t), t)) = B, ξ (t) = 0 if φ(u(ξ(t), t)) (A, B). Mascia, Terracina and Tesei (2009) : local existence and uniqueness. September 14th, / 39

18 Steady interface T = φ(u(ξ(t), t)) (A, B) t γ R R + T (A, B) x September 14th, / 39

19 Steady case : In the phase space Á(u) u September 14th, / 39

20 Left-to-right interface T = φ(u(ξ(t), t)) = A t γ T = A R R + T (A, B) x September 14th, / 39

21 Left-to-right case : In the phase space Á(u) u September 14th, / 39

22 Right-to-left interface T = φ(u(ξ(t), t)) = B t γ T = B T = A R R + T (A, B) x September 14th, / 39

23 Right-to-left case : In the phase space Á(u) u September 14th, / 39

24 t T = A T = B γ T (A, B) T = A T = B R R + T = A T (A, B) x September 14th, / 39

25 Plan 1 Introduction 2 Explicit solution : Riemann problem 3 Semi-discrete scheme September 14th, / 39

26 Self similar solutions Validation of the numerical schemes : computation of the solution of the Riemann problem u t = 2 φ(u) x 2 u(, 0) = u 0 : x { u b, x < 0 u + a, x > 0 September 14th, / 39

27 Self similar solutions Validation of the numerical schemes : computation of the solution of the Riemann problem u t = 2 φ(u) x 2 u(, 0) = u 0 : x { u b, x < 0 u + a, x > 0 Self-similar solution : (x, t) (λx, λ 2 t) so that u(x, t) = f (ζ), ζ = x t Discontinuity curve : ξ(t) = ζ t φ(f ) ζf = 0. September 14th, / 39

28 Simplified case Symmetric piecewise linear case m, q > 0 : { q, if u b, φ(u) = mu + q, if u a. September 14th, / 39

29 Simplified case Symmetric piecewise linear case m, q > 0 : { q, if u b, φ(u) = mu + q, if u a. Define E m (ζ) := 1 4 π m ζ e y 2 /4m dy, E m + (ζ) := 1 E m (ζ). September 14th, / 39

30 Simplified case Symmetric piecewise linear case m, q > 0 : { q, if u b, φ(u) = mu + q, if u a. Define E m (ζ) := 1 4 π m ζ e y 2 /4m dy, E m + (ζ) := 1 E m (ζ). General form : u(x, t) = f (x/ t), with f (ζ) := { (φ ) 1 (g(ζ)) ζ < 0, (φ + ) 1 (g(ζ)) ζ > 0, { φ(u )E + m (ζ) + φ(u + )E g(ζ) := m (ζ) ζ < ξ(t), φ(u ) E + m (ζ) + φ(u + ) E + m (ζ) ζ > ξ(t), + September 14th, / 39

31 Characteristic condition : T := 1 2 (φ(u ) + φ(u + )) [A, B], September 14th, / 39

32 Characteristic condition : T := 1 2 (φ(u ) + φ(u + )) [A, B], iff true, the interface is steady, i.e. ζ = 0 Ex. : m = 2, q ± = 3, u = 1.5. u + = T= u phi(u) x x September 14th, / 39

33 September 14th, / 39

34 iff false, the interface is moving and ζ is given by the implicit relation g( ζ) = B if T > B or g( ζ) = A if T < A. Ex. : m = 2, q ± = 3, u = 2.1. u + = T= u 1 phi(u) x x September 14th, / 39

35 September 14th, / 39

36 Plan 1 Introduction 2 Explicit solution : Riemann problem 3 Semi-discrete scheme September 14th, / 39

37 Space discretization Classic 2nd order centered space discretization, uniform mesh x U 0 := u.. u u +.. u + R J, A := Semi-discrete scheme : du ( 1A dt = x 2 I + ε A) φ(u) =: A x 2,εφ(U) Regular limit : ε = 0. September 14th, / 39

38 General case Problem : φ is mostly linear, but not always... Reinterpretation in nonautonomous terms : du 2 (τ) = M(τ)U(τ) + W (τ), τ = t/ x dτ M(τ) R J J and W (τ) R J piecewise constant. Jumps occur when u j changes phases. September 14th, / 39

39 General case Problem : φ is mostly linear, but not always... Reinterpretation in nonautonomous terms : du 2 (τ) = M(τ)U(τ) + W (τ), τ = t/ x dτ M(τ) R J J and W (τ) R J piecewise constant. Jumps occur when u j changes phases. Special symmetric case : m = 2, q = 3 Easy generalization through a matrix product. September 14th, / 39

40 Steady interface : u + + u [ 1, 1] Then V = φ(u) satisfies dv dτ = 2A V, since exp( 2τA) (R + ) J J, exp( 2τA) 1. Spectral analysis : lim V j(t/h 2 ) = φ(u+ ) + φ(u ) = v, t + 2 and U j = { (φ ) 1 (v ) if j K, (φ + ) 1 (v ) if j K + 1. September 14th, / 39

41 Steady interface : u + + u [ 1, 1] Then V = φ(u) satisfies dv dτ = 2A V, since exp( 2τA) (R + ) J J, exp( 2τA) 1. Spectral analysis : lim V j(t/h 2 ) = φ(u+ ) + φ(u ) = v, t + 2 and U j = { (φ ) 1 (v ) if j K, (φ + ) 1 (v ) if j K + 1. Fully implicit scheme preferred over semi-implicit or fully explicit : same order of convergence (0.5 in t / 1 in x) but no CFL. September 14th, / 39

42 Numerical results implicit (var t) implicit (var x) Explicit ( t = x 2 /4) K J error cpu K J error cpu J error cpu Table: Comparison of L 2 errors of φ(u) and execution times (cpu) for the implicit and Explicit schemes wrt t = 2 K, x = 2 J. September 14th, / 39

43 Moving interface : u + + u > 1 5 T= u 1 phi(u) x x September 14th, / 39

44 What are M(τ) and W (τ)? September 14th, / 39

45 What are M(τ) and W (τ)? Model initial condition, δ > 0 : 1 if j L Uj 0 = 2 if j = L 2 + δ if j L + 1 Taylor expansion : locally in time, we have M(τ) = AD =: B and W (τ) = (0,..., 0, 3, 0, 3, 0,..., 0) T =: B, with D = diag( 2,..., 1,..., 2). September 14th, / 39

46 Solution : Spectral analysis : U j L 1 non increasing in S, U L increasing through the unstable phase U, U j L+2 non increasing in S +. ( t ) U(τ) = exp( τb)u(0) + exp( sb)ds B. 0 Sp(B) = {µ 1 < 0} {0} {µ 1,..., µ J 2 > 0}. Exit time (P µ spectral projectors) : ( log 1 [P 0 U(0)](L) ) [P µ B](L) µ Sp(B)\{0} µ τ ( log ([P µ 1 U(0) B )] ). (L) µ 1 September 14th, / 39

47 u u No implicit scheme! The explicit scheme captures the transition, but allows U L to go through U. 4 explicit schem e - u at tim e dx= L 1/255 S- S+ 2.0 explicit schem e - u at tim e dx= 1/255 S- S x x Figure: Small time evolution in (x, u) of the Riemann data ( 2, 4). September 14th, / 39

48 phi 6 5 phi num 0.6 phi phi num phi(u) 2 phi(u) u u Figure: Small time evolution in (u, φ(u)) of the Riemann data ( 2, 4). September 14th, / 39

49 t Movement of the interface 0.18 Movement of the boundary x September 14th, / 39

50 Two-phase scheme Idea : use the entropy condition of Evans and Portilheiro Define the projection matrix 0 j Π j := 0 I J j 1 Let U R J, j {1,..., J} and C := C(U, j ) the transition value C = C(U, j ) := 1 2 (φ (U j 1) + φ + (U j +2)). If C > 1, truncation : C = 1 & U j = 1 & U j +1 = 2 If C < 1, truncation : C = 1 & U j = 2 & U j +1 = 1 If C 1, C = C transmission conditions : φ(u j ) = φ(u j +1) = 1 2 (φ(u j 1) + φ(u j +2)) = C(U, j ). September 14th, / 39

51 Let F(U, j ) := (0,..., 0, (φ ) 1 (C) }{{}, (φ +) 1 (C), 0,..., 0) j -th element The algorithm reads, for a given U n and Ũ n := F(U n, j n ) + (I Π jn ) U n, Π j n U n+1 = F(U n, j n ), (I Π j n )U n+1 = (I Π j n ) CFL condition t = O( x 2 ) No point in the unstable phase Generalization to the cubic case ( U n t ) x 2 A φ(ũn ) September 14th, / 39

52 4.0 two-phase 3.5 explicit 3.0 relative error (%) t Figure: Evolution of the relative error of the position of the interface for (u, u + ) = ( 2, 4) September 14th, / 39

53 Perspectives Perform the analysis for the cubic diffusion equation. Go to higher space dimensions. Find adapted numerical methods. September 14th, / 39