Evolution Under Constraints: Fate of Methane in Subsurface

Transcription

1 Evolution Under Constraints: Fate of Methane in Subsurface M. Peszyńska 1 Department of Mathematics, Oregon State University SIAM PDE. Nov Research supported by DOE Modeling, Analysis, and Simulation of Multiscale Preferential Flow, Fulbright Research Fellowship in Poland, NSF DMS Model adaptivity in porous media, NSF DMS Hybrid Modeling in porous media

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3 Why: system of conservation laws in ECBM Recovery Transport with adsorption in Enhanced CoalBed Methane processes U t + f (U) = 0 Ut z 8 } 9 { f (U) equilibrium >< 8 9 >= z } { < kinetic = MULTICOMPONENT + TRANSPORT = 0 non equilibrium double porosity >: : hysteretic ; >; {z } advection/ diffusion Multicomponent adsorption: (competitive/selective/preferential). Many processes are kinetic (e.g, coal swelling) Macropore + mesopore + micropore = double and triple porosity Adsorption-desorption hysteresis

4 Extend u t + f (u) x = 0 to adsorption models Scalar conservation law with adsorption equilibrium or non-equilibrium (relaxation) u t + v t + u x = 0, v = g(u) u t + v t + u x = 0, v t = 1 (g(u) v) τ [BarrKnab 9*],[KleinPesz 10] [TveitoWinth 97,Pember 93,MajdaCollRoyt ] Scalar conservation law with rate-dependent memory (double-porosity diffusion at a lower scale) αu t + β u t + f (u) x = 0 [Literature on ECBM], [Hyporheic exchange: Haggerty 95-02],[PeszShow 07,PeszYiShow[cw]] Conservation law with rate-independent memory: hysteresis G maximal monotone graph u t + v t + u x = 0, v G(u) [PeszynskaShowalter 97[D&IE]], [Literature on ECBM]

5 Extend u t + f (u) x = 0 to adsorption models Scalar conservation law with adsorption equilibrium or non-equilibrium (relaxation) u t + v t + u x = 0, v = g(u) u t + v t + u x = 0, v t = 1 (g(u) v) τ [BarrKnab 9*],[KleinPesz 10] [TveitoWinth 97,Pember 93,MajdaCollRoyt ] Scalar conservation law with rate-dependent memory (double-porosity diffusion at a lower scale) αu t + β u t + f (u) x = 0 [Literature on ECBM], [Hyporheic exchange: Haggerty 95-02],[PeszShow 07,PeszYiShow[cw]] Conservation law with rate-independent memory: hysteresis G maximal monotone graph u t + v t + u x = 0, v G(u) [PeszynskaShowalter 97[D&IE]], [Literature on ECBM]

6 Conservation law for adsorption hysteresis Adsorption equilibrium u t + v t + u x = 0 v = g(u) Adsorption kinetics: u t + v t + u x = 0 v t + 1 (v g(u)) = 0 τ Adsorption hysteresis u t + v t + u x = 0 v t + c(v u) 0 Here c( ) is a maximal monotone graph representing hysteresis. Preisach model of hysteresis uses differential equations (DE) Analysis in [PeszynskaShowalter 97[D&IE]]: continuity of solutions for convex-concave hysteresis graphs Applications: significant hysteresis in ECBM [Kovcsek et al 08]

7 Understand DE with hysteresis, Preisach construction Some notation and vocabulary for v (t) + c(v u) 0. Identify c : D(c) R (set-valued) with its graph c R R c is monotone if (v 1 v 2 )(u 1 u 2 ) 0. u 1, u 2 D(c), v 1 c(u 1 ), v 2 c(u 2 ) c is maximal if for any (some) α > 0, (I + αc) is onto R Follow semigroup theory for v (t) + c(v u) 0 [Kato, Brézis, Dorroh, Crandall-Liggett] Example: Graph c Resolvent J α := (I + αc) 1 Yosida appr. c α := 1 α (I Jα)

8 Preisach hysteresis via an auxiliary DE Given input u(t), find v: Graph c with D(c) := [ 1, 0] v (t) + c(v u) 0 Understand the evolution problem v u D(c) means 1 v u 0. v > 0 iff 1 = v u v < 0 iff v u = 0 v.. = iff <.. v... u.. <.. 0 Solution v

9 Numerical scheme(s), adsorption hysteresis Model problem u t + v t + u x = 0 v t + c(v u) 0 Approximation scheme [Peszynska 11[ms]] (1) u n j u n 1 j + v n j v n 1 j + λ(u n 1 j u n 1 j 1 ) = 0 (2) v n j vj n 1 + kc(v n j u n j ) }{{} HOW? 0 Understand the in the numerical scheme Use Yosida approximation c α, explicit or semi-implicit (2, YA) v n j v n 1 j + kc α (v n j Use Resolvent J k, semi-implicit (2, J k ) v n j u n j = Resolvent {}}{ (I + kc) 1 (vj n 1 u n j )=0 u n j ) := J k (v n 1 j u n j )

10 Theorem Stability of the scheme(s)[pesz[cw]] The [YA] explicit scheme is stable with k = k(α, h). Error: O(k + h + α) Proof: Note c α is ( 1 α )-Lipschitz; extend [TveitoWinther 96[SJMA]]( scalar laws with relaxation) Theorem The [Resolvent] implicit scheme is stable if λ = k h 1. Error: O(k + h) i) U n 1 + V n 1 U n V n 1 1 ii) TV(U n ) + TV(V n ) TV(U n 1 ) + TV(V n 1 ) Proof (i): (monotone operator techniques in L 1 ) u n j u n 1 j c(v n j u n j ) + λ(un 1 j v n j + c(v n j u n j ) vn 1 j sign(v n j ) add them together, note (sign(v n j ) sign(un j ))c(vn j u n j ) 0 to get u n 1 j 1 ) 0 sign(un j ) u n j + vn j un 1 j (1 λ) + λ u n 1 j 1 + vn 1 j Proof (ii): similar techniques, but harder

11 Order of convergence of the scheme Use ODE analysis via method of lines... (after spatial discretization) Rewrite as u j(t) + v j(t) + 1 h (u j(t) u j 1 (t)) = 0 v j(t) + c(v j u j ) 0 U (t) + V (t) = f (t, U) V (t) + c(v U) 0 For every h, use ODE analysis [Rulla 96[SJNA], RullaWalkington 96[SJNA]] to show that the schemes converge with O( k) or O(k). Convergence of fully discrete scheme as (h, k) 0 [Pesz[cw]] Consistency analysis in [Rulla 96[SJNA], RullaWalkington 96[SJNA]] only in Hilbert spaces Our stability analysis only in L 1. Convergence to entropy solutions (Kruzhkov-style)???

12 Animation example Numerical solution has the correct behavior Both u, v are smooth [PS 97,P 98] The constraint u(x, t) v(x, t) 1 holds v does not change when u(x, t) v(x, t) < 1

13 Algorithmic aspects of the semi-implicit scheme Given v n 1 j, u n 1 j, need to find u n j, vn j : (1) u n j u n 1 j + v n j vj n 1 + λ(uj n 1 u n 1 j 1 ) = 0 (2) v n j v n 1 j + kc(v n j u n j ) }{{} HOW? How do we actually solve this implicit problem? Recall Resolvent J k in z } { (2J) v n j u n j = (I + kc) 1 (v n 1 j u n j ) := J k(v n 1 j u n j ) 0 Newton methods 2 require J k C 2 but J k is only defined a.e. Have to satisfy the constraint 1 v n j u n j 0 (u n j, vn j ) D(c) Idea: use semi-smooth Newton methods These converge superlinearly with J k defined only a.e. [ItoKunisch 08,Ulbrich 11] 2 when she was good, she was really, really good... but when she was bad, she was really, really horrid

14 Methane Hydrates: Ice That Burns methane ice/clathrate { greenhouse gas, source of energy, drilling hazard } image from MH at surface. Hands: R. Collier (OSU), picture: M. Torres Thermodynamics constraints: x lm x EQ lm (P, T, x ls)

15 Model for evolution of Methane Hydrates Variables: (P, T), {X pc, S p, N C, u p} p,c Constraints: x lm x EQ lm (P, T, xls) Mass conservation: storage + advection + diffusion = source storage z } { advection diffusion z } { z } { (φsgρg + φs l ρ l X lm + φs h ρ h X hm ) + (φsgρgug + φs l ρ l X lm u l + φs h ρ h X hm ur) (DmφS l ρ l X lm ) = f t (φs l ρ l X lw + φs h ρ h X hl ) + (φs l ρ l X lw u l + φs h ρ h X hw ur) = 0 t (φs l ρ l X ls ) + (φs l ρ l X ls u l ) (DsφS l ρ l X ls ) = 0 t Energy equation Phase equilibria and volume, capillary and thermodynamic constraints

16 Obstacle problem as a minimization Elliptic (P)DE with constraint J(v) := 1 2 u = 1, x (0, 1), u(0) = u(1) = 0 Z 1 0 (v ) 2 dx Z 1 a(u, v) := R 1 uvdx, a(, ) bil., symm.,cont., coercive on V 0 Unconstrained: u = min v V J(v)... a(u, v) = f, u, v V 0 vdx, V = H 1 0(0, 1) Obstacle problem: min v K J(v) K := {v V : v 0.1} Finding inf u K J(u) is equivalent to Variational Inequality (VI) u K K : a(u K, v u K) < f, v u K >, v V

17 How to interpret and solve VI numerically? Interpret Example (): u ψ (VI) u K K : a(u K, v u K) < f, v u K >, v V as (VI) = 2 u K f & u K ψ, a.e.x Ω (VI) = ( 2 u K f )(u K ψ) = 0 a.e.x Ω Discretize (VI)... to have, in V = R N Au f = λ u ψ 0, λ 0, (u ψ, λ) = 0

18 Solvability conditions (VI) Discretization (VI) leads to... problem in V = R N Ex. (3-): u ψ Au f = λ u ψ 0, λ 0, (u ψ, λ) = 0 Solve as a Linear Complementarity Problem or (non)linear programming problem [Cottle, Gilbert] Solvability conditions (any A) sufficient: A T = A 0 sufficient and necessary: A is of type (all principal minors are positive) sufficient and necessary: real part of all eigenvalues of A and its principal submatrices are positive Solvability of nonlinear problem?

19 Nonlinear problems with nonlinear constraints Assume for simplicity ψ = 0... Au + ( λ) = f 8 < : u 0 λ 0 λ T u = 0 Linear problem with A 0 8 < : H(u, λ) = 0 F(u, λ) 0 G(u, λ) 0 F(u, λ)g(u, λ) = 0 Formulation: [Knabner/Eck 09, Ben-Gharbia/Gilbert/Jaffre 10] What properties ofh, F, G guarantee solvability? [P 11] Nonlinear problem with a solution Nonlinear problem without a solution

20 Constrained problem using monotone operator methods Au + ( λ) = f u 0, λ 0, (u, λ) V = 0 rewrite as Au + ( λ) = f λ B(u) then we can analyze solvability Theorem Au + B(u) f From monotone operator theory A accretive+lip. & B m-accretive =!u V Nonlinear problems analysis [P[cw]]

21 Applications problem: evolution of hydrates CH 4 [PTorresTrehu 10] Find (U(x, t), S(x, t)), x (0, 1), t > 0 8 < : (US + R(1 S)) (D(S) U) = 0, t U(x, t) U max (x) S(x, t) 1 (U max (x) U(x, t))(1 S(x, t)) = 0 8 < : H(U, S) = 0 F(U, S) 0 G(U, S) 0 F(U, S)G(U, S) = 0 Discretize (FEM) (x), backward euler (t) U n S n + R(1 S n ) + A(S n )U n = F n, n 1

22 Applications problem: evolution of hydrates CH 4 [PTorresTrehu 10] Find (U(x, t), S(x, t)), x (0, 1), t > 0 8 < : (US + R(1 S)) (D(S) U) = 0, t U(x, t) U max (x) S(x, t) 1 (U max (x) U(x, t))(1 S(x, t)) = 0 8 < : H(U, S) = 0 F(U, S) 0 G(U, S) 0 F(U, S)G(U, S) = 0 Discretize (FEM) (x), backward euler (t) U n S n + R(1 S n ) + A(S n )U n = F n, n 1

23 Applications problem: evolution of hydrates CH 4 [PTorresTrehu 10] Find (U(x, t), S(x, t)), x (0, 1), t > 0 8 < : (US + R(1 S)) (D(S) U) = 0, t U(x, t) U max (x) S(x, t) 1 (U max (x) U(x, t))(1 S(x, t)) = 0 8 < : H(U, S) = 0 F(U, S) 0 G(U, S) 0 F(U, S)G(U, S) = 0 Discretize (FEM) (x), backward euler (t) U n S n + R(1 S n ) + A(S n )U n = F n, n 1

24 Solving PDEs with inequality constraints Simplified model structure for unknowns (X, S) (PDE) : 8F(X, S) = 0 < X 0 Constraint : S 0 : XS = 0 j F(X, S) = 0 min(x, S) = 0 Replace Constraint with min(x, S) = 0 = Optimization very effective for variational inequalities arising in obstacle problems, optimal control, financial mathematics Optimization with inequality constraints well developed Complementarity Condition (primal-dual active set) [ItoKunisch 08,Ulbrich 11] System can be solved implicitly with semi-smooth Newton methods [EckKnabner 98-0*,Jaffre et al 09, BenGharbia 10]

25 Summary Discussed - evolution problems with constraints adsorption hysteresis in ECBM phase transitions & thermodynamics in MH Analysis: some possible (problem dependent) monotone operator techniques Solvers: semi-smooth Newton methods