Numerical methods for conservation laws with a stochastically driven flux

Transcription

1 Numerical methods for conservation laws with a stochastically driven flux Håkon Hoel, Kenneth Karlsen, Nils Henrik Risebro, Erlend Briseid Storrøsten Department of Mathematics, University of Oslo, Norway December 8, / 33

2 Overview 1 Problem description 2 Deterministic conservation laws Characteristics and shocks Well-posedness 3 Stochastic scalar conservation laws Definition and well-posedness Numerical methods Flow map cancellations 4 Conclusion 2 / 33

3 The model problem Consider the stochastic conservation law Regularity assumptions f C 2 (R; R) du + x f (u) dz = 0, in (0, T ] R, u(0, ) = u 0 (L 1 L )(R). z C 0,α ([0, T ]; R) for some α > 0. That is, z(t) z(s) sup s t [0,T ] t s α <. Examples z(t) = t, Wiener processes, fractional Brownian motions. 3 / 33

4 The model problem Consider the stochastic conservation law Regularity assumptions f C 2 (R; R) du + x f (u) dz = 0, in (0, T ] R, u(0, ) = u 0 (L 1 L )(R). z C 0,α ([0, T ]; R) for some α > 0. That is, z(t) z(s) sup s t [0,T ] t s α <. Examples z(t) = t, Wiener processes, fractional Brownian motions. 3 / 33

5 Motivation For the mean-field SDE dx i = σ X i 1, L 1 j i with σ : R P(R) R, one has that 1 L δ X j dw, for i = 1, 2,..., L, L δ X i (t) π(t) P(P(R)), as L. i=1 The measure s density satisfies the dynamics dρ π + x σ(x, ρ π ) dw = 0. 4 / 33

6 Our contribution Develop numerical methods for solving the SSCL Show that oscillations in z may lead to cancellations in the flow map. t= t= t= t= u(t,x) 0.6 u(t,x) 0.6 u(t,x) 0.6 u(t,x) x x x x z(t) z(t) t 5 / 33

7 Overview 1 Problem description 2 Deterministic conservation laws Characteristics and shocks Well-posedness 3 Stochastic scalar conservation laws Definition and well-posedness Numerical methods Flow map cancellations 4 Conclusion 6 / 33

8 The deterministic conservation law The equation u t + x f (u) = 0 takes its name from the property d udx = dt Classical notion of weak solutions φ t u + f (u)φ x dxdt + 0 R R in (0, ) R u(0, ) = u 0 (L 1 L )(R) R u t dx = f (u) x dx = 0. R R φ(0, x)u 0 (x)dx = 0, leads to existence, but not uniqueness, due to formation of shocks. φ D(R R), 7 / 33

9 The deterministic conservation law The equation u t + x f (u) = 0 takes its name from the property d udx = dt Classical notion of weak solutions φ t u + f (u)φ x dxdt + 0 R R in (0, ) R u(0, ) = u 0 (L 1 L )(R) R u t dx = f (u) x dx = 0. R R φ(0, x)u 0 (x)dx = 0, leads to existence, but not uniqueness, due to formation of shocks. φ D(R R), 7 / 33

10 Well-posedness Definition 1 (Kruzkov s entropy condition) t η(u) + x q(u) 0, φ D +(R R), holds for all smooth and convex η : R R, and q (u) := f (u)η (u). Theorem 2 Consider u t + x f (u) = 0, u(0, x) = u 0. in R + R Assume that u 0 (L 1 L )(R) and f C 2 (R; R). Then there exists a unique solution u C(R + ; L 1 (R)) L (R + R) which satisfies the Kruzkov entropy condition. Moreover, for any t 0, u(t) v(t) 1 u 0 v / 33

11 Well-posedness Definition 3 (Kruzkov s entropy condition for z C 1 ) t η(u) + ż x q(u) 0, φ D +(R R), holds for all smooth and convex η : R R, and q (u) := f (u)η (u). Theorem 4 Consider u t + żf (u) x = 0, u(0, x) = u 0. in R + R Assume that u 0 (L 1 L )(R), z piecewise C 1 and f C 2 (R; R). Then there exists a unique solution u C(R + ; L 1 (R)) L (R + R) which satisfies the Kruzkov entropy condition. Moreover, for any t 0, u(t) v(t) 1 u 0 v / 33

12 Overview 1 Problem description 2 Deterministic conservation laws Characteristics and shocks Well-posedness 3 Stochastic scalar conservation laws Definition and well-posedness Numerical methods Flow map cancellations 4 Conclusion 10 / 33

13 Definition The problem formulation du + x f (u) dz = 0, in (0, T ] R, u(0, ) = u 0 (L 1 L )(R). Kruzkov s entropy condition dη(u) + x q(u) dz 0, in D +(R R) is difficult to work with: If w is a standard Wiener process, then x q(u) dw =... + η (u)(f (u)) 2 (u x ) 2 dt. 11 / 33

14 Kinetic formulation Consider the kinetic formulation instead dχ + f (ξ)χ x dz = ξ m dt in D (R R R ξ ) for some non-negative, bounded measure m(t, x, ξ) and the constraint 1 if 0 ξ u(t, x) χ(t, x, ξ) = χ(ξ; u(t, x)) := 1 if u(t, x) ξ < 0 0 otherwise. Formal motivation for equivalence: χ t (ξ; u(t, x)) + f (ξ)χ x (ξ; u(t, x)) dz = δ(u = ξ)(u t + f (u)u x dz). And L 1 isometry: χ(ξ; u(t, x))dξ = u(t, x) = R χ(ξ; u) χ(ξ; v)dξ dx = u v dx. R R R 12 / 33

15 Kinetic formulation Consider the kinetic formulation instead dχ + f (ξ)χ x dz = ξ m dt in D (R R R ξ ) for some non-negative, bounded measure m(t, x, ξ) and the constraint 1 if 0 ξ u(t, x) χ(t, x, ξ) = χ(ξ; u(t, x)) := 1 if u(t, x) ξ < 0 0 otherwise. Formal motivation for equivalence: χ t (ξ; u(t, x)) + f (ξ)χ x (ξ; u(t, x)) dz = δ(u = ξ)(u t + f (u)u x dz). And L 1 isometry: χ(ξ; u(t, x))dξ = u(t, x) = R χ(ξ; u) χ(ξ; v)dξ dx = u v dx. R R R 12 / 33

16 Kinetic formulation Consider the kinetic formulation instead dχ + f (ξ)χ x dz = ξ m dt in D (R R R ξ ) for some non-negative, bounded measure m(t, x, ξ) and the constraint 1 if 0 ξ u(t, x) χ(t, x, ξ) = χ(ξ; u(t, x)) := 1 if u(t, x) ξ < 0 0 otherwise. Formal motivation for equivalence: χ t (ξ; u(t, x)) + f (ξ)χ x (ξ; u(t, x)) dz = δ(u = ξ)(u t + f (u)u x dz). And L 1 isometry: χ(ξ; u(t, x))dξ = u(t, x) = R χ(ξ; u) χ(ξ; v)dξ dx = u v dx. R R R 12 / 33

17 Notion of solution The term f (ξ)χ x dz is difficult to treat, even as distribution. Workaround: introduce ρ 0 D(R) and ρ(t, x, ξ; y) := ρ 0 (y x + f (ξ)z(t)), Then, if z C 1 ([0, T ]), dρ + f (ξ)ρ x dz = 0, in (0, T ] R x R ξ. 13 / 33

18 Notion of solution The term f (ξ)χ x dz is difficult to treat, even as distribution. Workaround: introduce ρ 0 D(R) and Then, if z C 1 ([0, T ]), ρ(t, x, ξ; y) := ρ 0 (y x + f (ξ)z(t)), dρ + f (ξ)ρ x dz = 0, in (0, T ] R x R ξ. Consequently, ( ) d(ρχ) + f (ξ)(ρχ) x dz = χ dρ + f (ξ)ρ x dz } {{ } =0 ( +ρ ) dχ + f (ξ)χ x dz = ρm ξ dt. } {{ } =m ξ dt 13 / 33

19 Notion of solution The term f (ξ)χ x dz is difficult to treat, even as distribution. Workaround: introduce ρ 0 D(R) and ρ(t, x, ξ; y) := ρ 0 (y x + f (ξ)z(t)), Then, if z C 1 ([0, T ]), dρ + f (ξ)ρ x dz = 0, in (0, T ] R x R ξ. Consequently, d(ρχ) + f (ξ)(ρχ) x dzdx = ρm ξ dtdx. R R 13 / 33

20 Notion of solution The term f (ξ)χ x dz is difficult to treat, even as distribution. Workaround: introduce ρ 0 D(R) and ρ(t, x, ξ; y) := ρ 0 (y x + f (ξ)z(t)), Then, if z C 1 ([0, T ]), dρ + f (ξ)ρ x dz = 0, in (0, T ] R x R ξ. Leads to condition d ρχdx = ρm ξ dx, in D (R t R ξ ). dt R R 13 / 33

21 Notion of solution The term f (ξ)χ x dz is difficult to treat, even as distribution. Workaround: introduce ρ 0 D(R) and Then, if z C 1 ([0, T ]), Leads to condition ρ(t, x, ξ; y) := ρ 0 (y x + f (ξ)z(t)), dρ + f (ξ)ρ x dz = 0, in (0, T ] R x R ξ. d dt ρχdx = ρm ξ dx in D ([0, T ] R ξ ). (1) R R Definition 5 (Pathwise entropy solution (PES)) u L 1 L ([0, T ] R) is a PES if there exists a non-negative, bounded measure m such that equation (1) holds for all ρ, as defined above. 13 / 33

22 Well-posedness Theorem 6 (Lions, Perthame, Souganidis, 2013) Assume f C 2 (R; R), z C([0, T ]; R) and u 0 (L 1 L )(R). Then, for all T > 0, there exists a unique PES u C([0, T ]; L 1 (R)) L ([0, T ] R). Furthermore, for two solutions u, v generated from the respective driving paths z, z and u 0, v 0 BV (R), u(t, ) v(t, ) 1 u 0 v C for a uniform constant C(u 0, v 0, f, f, f ) > 0. sup s (0,t) z(s) z(s), Note that if z n is a piecewise linear interpolation of z C 0,α using interpolation points with z n (t k ) = z(t k ) and u n := u(, ; z n ), then u(t, ) u n (t, ) 1 = O(n α/2 ). 14 / 33

23 Well-posedness Theorem 6 (Lions, Perthame, Souganidis, 2013) Assume f C 2 (R; R), z C([0, T ]; R) and u 0 (L 1 L )(R). Then, for all T > 0, there exists a unique PES u C([0, T ]; L 1 (R)) L ([0, T ] R). Furthermore, for two solutions u, v generated from the respective driving paths z, z and u 0, v 0 BV (R), u(t, ) v(t, ) 1 u 0 v C for a uniform constant C(u 0, v 0, f, f, f ) > 0. sup s (0,t) z(s) z(s), Note that if z n is a piecewise linear interpolation of z C 0,α using interpolation points with z n (t k ) = z(t k ) and u n := u(, ; z n ), then u(t, ) u n (t, ) 1 = O(n α/2 ). 14 / 33

24 Numerical solution approach (i) Approximate the rough path z by a piecewise linear interpolation ( z n (t) = 1 t τ ) k z(τ k ) + t τ k τ τ z(τ k+1), t [τ k, τ k+1 ], where τ k = k τ and τ = T /n (ii) Solve the conservation law with driving noise z n using a standard numerical method in classical Kruzkov entropy sense z 24 z 25 z t 15 / 33

25 Solution with approximated driving noise z n The problem to solve: Let S ( τ, z) ũ denote the solution of u n t + ż n x f (u n ) = 0, in (0, T ] R, u n (0, ) = u 0. u t + z τ xf (u) = 0, in (0, τ] R, u(0, ) = ũ. Then k 1 u n (τ k ) = S ( τ, z j ) u 0, for k = 0, 1,..., n, j=0 where z j := z(τ j+1 ) z(τ j ). 16 / 33

26 Solution with approximated driving noise z n The problem to solve: Let S ( τ, z) ũ denote the solution of u n t + ż n x f (u n ) = 0, in (0, T ] R, u n (0, ) = u 0. u t + z τ xf (u) = 0, in (0, τ] R, u(0, ) = ũ. Then k 1 u n (τ k ) = S ( τ, z j ) u 0, for k = 0, 1,..., n, j=0 where z j := z(τ j+1 ) z(τ j ). 16 / 33

27 Numerical schemes Solve iteratively k = 0, 1,..., n u n t + z k τ xf (u n ) = 0, in (τ k, τ k+1 ] R, with x = O(N 1 ) and time-steps t k = τ/n k. Numerical solution ū l m := ū(t l, x m ; z n ). Solve, for instance, by Lax Friedrichs (assuming t l (τ k, τ k+1 )), ū l+1 m = ūl m+1 + ūl m 1 2 t k z k τ f (ūm+1 l ) f (ūl m 1 ), over l, m, k. (2) 2 x 17 / 33

28 Numerical schemes Solve iteratively k = 0, 1,..., n u n t + z k τ xf (u n ) = 0, in (τ k, τ k+1 ] R, with x = O(N 1 ) and time-steps t k = τ/n k. Numerical solution ū l m := ū(t l, x m ; z n ). Solve, for instance, by Lax Friedrichs ū l+1 m = ūl m+1 + ūl m 1 2 z k f (ūm+1 l ) f (ūl m 1 ), over l, m, k. (2) N k 2 x 17 / 33

29 Numerical schemes Solve iteratively k = 0, 1,..., n u n t + z k τ xf (u n ) = 0, in (τ k, τ k+1 ] R, with x = O(N 1 ) and time-steps t k = τ/n k. Numerical solution ū l m := ū(t l, x m ; z n ). Solve, for instance, by Lax Friedrichs ū l+1 m = ūl m+1 + ūl m 1 2 z k f (ūm+1 l ) f (ūl m 1 ), over l, m, k. (2) N k 2 x With initial data ūm 0 = 1 xm+1/2 u 0 (x)dx. x x m 1/2 17 / 33

30 Numerical schemes Solve iteratively k = 0, 1,..., n u n t + z k τ xf (u n ) = 0, in (τ k, τ k+1 ] R, with x = O(N 1 ) and time-steps t k = τ/n k. Numerical solution ū l m := ū(t l, x m ; z n ). Solve, for instance, by Lax Friedrichs ū l+1 m = ūl m+1 + ūl m 1 2 CFL: żk n f t k L ( u 0, u 0 ) x 1 = N k = O So t k = τ/n k = O(n α 1 N 1 ) for all k. z k f (ūm+1 l ) f (ūl m 1 ), over l, m, k. (2) N k 2 x ( ) zk = O(n α N 1 ) x 17 / 33

31 Convergence rates If u 0 (L 1 BV )(R) and f C 2 (R; R), ū(t, ) u n (T, ) 1 ū 0 u n C x Recall further u(t, ) u n (T, ) 1 C Hence, Balance error contributions: n z k k=0 = O(N 1 ) + O(N 1/2 n 1 α ). sup s [0,T ] z(s) z n (s) = O(n α/2 ). u(t, ) ū(t, ) 1 = O(N 1/2 n 1 α + n α/2 ). (3) N(n) = O(n 2 α ). If u 0 has compact support, the cost of achieving O(ɛ) error in (3) O(ɛ (2/α)(5 3α) )! Which is O(ɛ 14 ) for z C 0,1/2 ([0, T ]). 18 / 33

32 Convergence rates If u 0 (L 1 BV )(R) and f C 2 (R; R), ū(t, ) u n (T, ) 1 ū 0 u n C x Recall further u(t, ) u n (T, ) 1 C Hence, Balance error contributions: n z k k=0 = O(N 1 ) + O(N 1/2 n 1 α ). sup s [0,T ] z(s) z n (s) = O(n α/2 ). u(t, ) ū(t, ) 1 = O(N 1/2 n 1 α + n α/2 ). (3) N(n) = O(n 2 α ). If u 0 has compact support, the cost of achieving O(ɛ) error in (3) O(ɛ (2/α)(5 3α) )! Which is O(ɛ 14 ) for z C 0,1/2 ([0, T ]). 18 / 33

33 Convergence rates If u 0 (L 1 BV )(R) and f C 2 (R; R), ū(t, ) u n (T, ) 1 ū 0 u n C x Recall further u(t, ) u n (T, ) 1 C Hence, Balance error contributions: n z k k=0 = O(N 1 ) + O(N 1/2 n 1 α ). sup s [0,T ] z(s) z n (s) = O(n α/2 ). u(t, ) ū(t, ) 1 = O(N 1/2 n 1 α + n α/2 ). (3) N(n) = O(n 2 α ). If u 0 has compact support, the cost of achieving O(ɛ) error in (3) O(ɛ (2/α)(5 3α) )! Which is O(ɛ 14 ) for z C 0,1/2 ([0, T ]). 18 / 33

34 Convergence rates If u 0 (L 1 BV )(R) and f C 2 (R; R), ū(t, ) u n (T, ) 1 ū 0 u n C x Recall further u(t, ) u n (T, ) 1 C Hence, Balance error contributions: n z k k=0 = O(N 1 ) + O(N 1/2 n 1 α ). sup s [0,T ] z(s) z n (s) = O(n α/2 ). u(t, ) ū(t, ) 1 = O(N 1/2 n 1 α + n α/2 ). (3) N(n) = O(n 2 α ). If u 0 has compact support, the cost of achieving O(ɛ) error in (3) O(ɛ (2/α)(5 3α) )! Which is O(ɛ 14 ) for z C 0,1/2 ([0, T ]). 18 / 33

35 Numerical example with u 0 = 1 x <0.5 and f (u) = u 2 / Engquist Osher solution Lax Friedrichs solution x Driving rough path t 19 / 33

36 Numerical example with u 0 = sign(x)1 x <0.5 and f (u) = u 2 / Engquist Osher solution Lax Friedrichs solution x Driving rough path t 20 / 33

37 Flow map cancellations Recall that S ( τ, z) ũ denotes the solution of and u t + z τ xf (u) = 0, in (0, τ] R, u(0, ) = ũ. k 1 u n (τ k ) = S ( τ, z j ) u 0, for k = 0, 1,..., n, j=0 where z j := z(τ j+1 ) z(τ j ). Provided u n (s, ) C(R) for all s (τ l, τ k ), then k 1 u n (τ k ) = S ( τ, z j ) u n (τ l ) = S ((k l) τ, k 1 j=l z j) u n (τ l ) j=l Benefit z(τ k ) z(τ l ) replaces k 1 j=l z j in the numerical error bound, CFL / 33

38 Flow map cancellations Recall that S ( τ, z) ũ denotes the solution of and u t + z τ xf (u) = 0, in (0, τ] R, u(0, ) = ũ. k 1 u n (τ k ) = S ( τ, z j ) u 0, for k = 0, 1,..., n, j=0 where z j := z(τ j+1 ) z(τ j ). Provided u n (s, ) C(R) for all s (τ l, τ k ), then k 1 u n (τ k ) = S ( τ, z j ) u n (τ l ) = S ((k l) τ, k 1 j=l z j) u n (τ l ) j=l Benefit z(τ k ) z(τ l ) replaces k 1 j=l z j in the numerical error bound, CFL / 33

39 Local continuity of solutions One-sided estimates deterministic setting (z(t) = t): If f α > 0 then u(x + h, t) u(x, t) h 1 αt h > 0, and t > 0 and if f α < 0 1 αt u(x + h, t) u(x, t) h h > 0 and t > / 33

40 Local continuity of solutions One-sided estimates deterministic setting (z(t) = t): If f α > 0 then u(x + h, t) u(x, t) h 1 αt h > 0, and t > 0 and if f α < 0 1 αt u(x + h, t) u(x, t) h h > 0 and t > 0. One-sided estimates: If f α > 0 and ż n > 0 for all t (a, b), then u n (x + h, t) u n (x, t) h 1 α(z n (t) z n (a)) h > 0 and t (a, b), and if ż n < 0 for t (b, c) 1 α(z n (t) z n (b)) un (x + h, t) u n (x, t) h h > 0 and t (b, c). 22 / 33

41 Continuity result Theorem 7 (Flow map product sum property) Consider Burgers equation, f (u) = u 2 /2. Let M + (t) := max s [0,t] zn (s), M (t) := min s [0,t] zn (s). For all t and all intervals s.t.: t (a, b) [0, T ] for which M (a) < z n (t) < M + (a), we have that u n (s, ) C(R), s (a, b), and u n (b) = S b a,zn (b) z n (a) u n (a). Secondly, whenever z k z k+1 > 0, then S ( τ, z2) S ( τ, z1) = S (2 τ, z1+ z2). 23 / 33

42 Running min and max z n M + M t 24 / 33

43 An equivalent integration path Theorem 8 (Oscillating running max/min (ORM) function) For y n (t) := { M + (t)1 s + (t) s (t) + M (t)1 s (t) s + (t) if t (0, T ) z(t ) if t = T with s + (t) = max{s t z n (s) = M + (s)} and s (t) = max{s t z n (s) = M (s)}. Then, for Burgers equation, (Note that S ( τ,0) = I.) n 1 k 1 S ( τ, z j ) = S ( τ, y n j ). j=0 j=0 25 / 33

44 An equivalent integration path Theorem 8 (Oscillating running max/min (ORM) function) For y n (t) := { M + (t)1 s + (t) s (t) + M (t)1 s (t) s + (t) if t (0, T ) z(t ) if t = T with s + (t) = max{s t z n (s) = M + (s)} and s (t) = max{s t z n (s) = M (s)}. Then, for Burgers equation, (Note that S ( τ,0) = I.) n 1 k 1 S ( τ, z j ) = S ( τ, y n j ). j=0 j=0 25 / 33

45 The ORM function z n M + M - ORM t 26 / 33

46 Numerical errors Numerical integration along the ORM yields ū(t, ) u n (T, ) 1 ū 0 u n C x = O(N 1/2 y n BV (0,T ) ), n y n k where x = O(N 1 ). Recall that integrating along z n yields O(N 1/2 z n BV (0,t) ) num error bound. Efficiency to be gained provided since, respectively y n BV (0,T ) z n BV (0,T ) = o(1), k=0 N(n) = O( z n 2 BV (0,T ) nα ), O( y n 2 BV (0,T ) nα ) and Cost(ū(T )) = O(N(n)n). 27 / 33

47 Numerical errors Numerical integration along the ORM yields ū(t, ) u n (T, ) 1 ū 0 u n C x = O(N 1/2 y n BV (0,T ) ), n y n k where x = O(N 1 ). Recall that integrating along z n yields O(N 1/2 z n BV (0,t) ) num error bound. Efficiency to be gained provided since, respectively y n BV (0,T ) z n BV (0,T ) = o(1), k=0 N(n) = O( z n 2 BV (0,T ) nα ), O( y n 2 BV (0,T ) nα ) and Cost(ū(T )) = O(N(n)n). 27 / 33

48 Bounded variation of ORM Theorem 9 (Bounded variation of Wiener path ORM) For standard Wiener paths w : [0, T ] R, the ORM function y n ( ) : [0, T ] R associated to w n fulfils and y n BV ([0, T ]) n > 0 almost surely, E [ y n BV [0,1] ] <, n > 0. The above also holds for the ORM y of w. Implication: Cost of achieving O(ɛ) approximation error is improved by this sharper bound from O(ɛ 14 ) to O(ɛ 4 ) for Burgers equation. 28 / 33

49 Bounded variation of ORM Theorem 9 (Bounded variation of Wiener path ORM) For standard Wiener paths w : [0, T ] R, the ORM function y n ( ) : [0, T ] R associated to w n fulfils and y n BV ([0, T ]) n > 0 almost surely, E [ y n BV [0,1] ] <, n > 0. The above also holds for the ORM y of w. Implication: Cost of achieving O(ɛ) approximation error is improved by this sharper bound from O(ɛ 14 ) to O(ɛ 4 ) for Burgers equation. 28 / 33

50 Example u t ( u 2 ) x dz = 0, u 0(x) = 1 x <1/2, t [0, 2] z orm 1 z(t) t 29 / 33

51 Example u t ( u 2 ) x dz = 0, u 0(x) = 1 x <1/2, t [0, 2] z orm u(2,x) z(t) initial data front tracking EO, using ORM EO, not using orm t x 29 / 33

52 Regularity of solutions Does the driving noise z have a regularizing effect on the solution? For Burgers, u n (t) can only be discontinuous at times when z n (t) = M + (t) and/or z n (t) = M (t): z n M + M t For Wiener processes {s [0, T ] w(s) = M + (s) and/or w(s) = M (s)} has Lebesgue measue 0. But, not (presently) clear if regularity behavior of u n extends to the limit solution. 30 / 33

53 Regularity of solutions Does the driving noise z have a regularizing effect on the solution? For Burgers, u n (t) can only be discontinuous at times when z n (t) = M + (t) and/or z n (t) = M (t): z n M + M t For Wiener processes {s [0, T ] w(s) = M + (s) and/or w(s) = M (s)} has Lebesgue measue 0. But, not (presently) clear if regularity behavior of u n extends to the limit solution. 30 / 33

54 Regularity of solutions Does the driving noise z have a regularizing effect on the solution? For Burgers, u n (t) can only be discontinuous at times when z n (t) = M + (t) and/or z n (t) = M (t): z n M + M t For Wiener processes {s [0, T ] w(s) = M + (s) and/or w(s) = M (s)} has Lebesgue measue 0. But, not (presently) clear if regularity behavior of u n extends to the limit solution. 30 / 33

55 Overview 1 Problem description 2 Deterministic conservation laws Characteristics and shocks Well-posedness 3 Stochastic scalar conservation laws Definition and well-posedness Numerical methods Flow map cancellations 4 Conclusion 31 / 33

56 Summary Developed a numerical method for solving stochastic scalar conservation laws. Identified cancellations of oscillations that in some settings lead to sharper error bounds and more efficient numerical algorithms. Future challenge: Develop numerics for higher dimensional version du + d xi f (x, u) dz i = 0, in (0, T ] R d, i=1 u(0, ) = u 0 (L 1 L )(R d ). 32 / 33

57 References 1 P-L Lions, B Perthame, P E Souganidis Scalar conservation laws with rough (stochastic) fluxes. Stoch. Partial Differ. Equ. Anal. Comput. 1 (2013), no. 4, B Gess, P E Souganidis Long-Time Behavior, invariant measures and regularizing effects for stochastic scalar conservation laws. Communications on Pure and Applied Mathematics (2016). 3 P-L Lions, B Perthame, P Souganidis Scalar conservation laws with rough (stochastic) fluxes: the spatially dependent case. Stochastic Partial Differential Equations: Analysis and Computations 2.4 (2014): Thank you for your attention! 33 / 33