Numerical methods for conservation laws with a ... · PDF fileNumerical methods for...

Numerical methods for conservation laws with astochastically driven flux

Hakon Hoel, Kenneth Karlsen, Nils Henrik Risebro, Erlend Briseid Storrøsten

Department of Mathematics, University of Oslo, Norway

December 8, 2016

1 / 33

Overview

1 Problem description

2 Deterministic conservation lawsCharacteristics and shocksWell-posedness

3 Stochastic scalar conservation lawsDefinition and well-posednessNumerical methodsFlow map cancellations

4 Conclusion

2 / 33

The model problem

Consider the stochastic conservation law

du + ∂x f (u) ◦ dz = 0, in (0,T ]× R,u(0, ·) = u0 ∈ (L1 ∩ L∞)(R).

Regularity assumptions

f ∈ C 2(R;R)

z ∈ C 0,α([0,T ];R) for some α > 0. That is,

sups 6=t∈[0,T ]

|z(t)− z(s)||t − s|α <∞.

Examples z(t) = t, Wiener processes, fractional Brownian motions.

3 / 33

The model problem

Consider the stochastic conservation law

Regularity assumptions

f ∈ C 2(R;R)

z ∈ C 0,α([0,T ];R) for some α > 0. That is,

sups 6=t∈[0,T ]

|z(t)− z(s)||t − s|α <∞.

Examples z(t) = t, Wiener processes, fractional Brownian motions.

3 / 33

Motivation

For the mean-field SDE

dX i = σ

X i ,1

L− 1

∑j 6=i

◦ dW , for i = 1, 2, . . . , L,

with σ : R× P(R)→ R, one has that

L∑i=1

δX i (t) → π(t) ∈ P(P(R)), as L→∞.

The measure’s density satisfies the dynamics

dρπ + ∂xσ(x , ρπ) ◦ dW = 0.

4 / 33

Our contribution

Develop numerical methods for solving the SSCL

Show that oscillations in z may lead to cancellations in the flow map.

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2

t=0.0000

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2

t=0.2500

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2

t=0.5000

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2

t=0.7500

0 0.2 0.4 0.6 0.8 1

5 / 33

Overview

4 Conclusion

6 / 33

The deterministic conservation law

The equation

ut + ∂x f (u) = 0 in (0,∞)× Ru(0, ·) = u0 ∈ (L1 ∩ L∞)(R)

takes its name from the property

utdx = −∫R

f (u)xdx = 0.

Classical notion of weak solutions∫ ∞0

∫Rφtu + f (u)φxdxdt +

∫Rφ(0, x)u0(x)dx = 0, ∀φ ∈ D(R× R),

leads to existence, but not uniqueness, due to formation of shocks.

7 / 33

The deterministic conservation law

The equation

ut + ∂x f (u) = 0 in (0,∞)× Ru(0, ·) = u0 ∈ (L1 ∩ L∞)(R)

takes its name from the property

utdx = −∫R

f (u)xdx = 0.

Classical notion of weak solutions∫ ∞0

∫Rφtu + f (u)φxdxdt +

∫Rφ(0, x)u0(x)dx = 0, ∀φ ∈ D(R× R),

leads to existence, but not uniqueness, due to formation of shocks.

7 / 33

Well-posedness

Definition 1 (Kruzkov’s entropy condition)

∂tη(u) + ∂xq(u) ≤ 0, φ ∈ D′+(R× R),

holds for all smooth and convex η : R→ R, and q′(u) := f ′(u)η′(u).

Theorem 2Consider

ut + ∂x f (u) = 0, in R+ × Ru(0, x) = u0.

Assume that u0 ∈ (L1 ∩ L∞)(R) and f ∈ C 2(R;R). Then there exists a uniquesolution u ∈ C (R+; L1(R)) ∩ L∞(R+ × R) which satisfies the Kruzkov entropycondition. Moreover, for any t ≥ 0,

‖u(t)− v(t)‖1 ≤ ‖u0 − v0‖1.

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Well-posedness

Definition 3 (Kruzkov’s entropy condition for z ∈ C 1)

∂tη(u) + z∂xq(u) ≤ 0, φ ∈ D′+(R× R),

holds for all smooth and convex η : R→ R, and q′(u) := f ′(u)η′(u).

Theorem 4Consider

ut + z f (u)x = 0, in R+ × Ru(0, x) = u0.

Assume that u0 ∈ (L1 ∩ L∞)(R), z piecewise C 1 and f ∈ C 2(R;R). Then thereexists a unique solution u ∈ C (R+; L1(R)) ∩ L∞(R+ × R) which satisfies theKruzkov entropy condition. Moreover, for any t ≥ 0,

‖u(t)− v(t)‖1 ≤ ‖u0 − v0‖1.

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Overview

4 Conclusion

10 / 33

Definition

The problem formulation

Kruzkov’s entropy condition

dη(u) + ∂xq(u) ◦ dz ≤ 0, in D′+(R× R)

is difficult to work with: If w is a standard Wiener process, then

∂xq(u) ◦ dw = . . .+ η′′(u)(f ′(u))2(ux)2dt.

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Kinetic formulation

Consider the kinetic formulation instead

dχ+ f ′(ξ)χx ◦ dz = ∂ξm dt in D′(R× R× Rξ)

for some non-negative, bounded measure m(t, x , ξ) and the constraint

χ(t, x , ξ) = χ(ξ; u(t, x)) :=

1 if 0 ≤ ξ ≤ u(t, x)

−1 if u(t, x) ≤ ξ < 0

0 otherwise.

Formal motivation for equivalence:

χt(ξ; u(t, x)) + f ′(ξ)χx(ξ; u(t, x)) ◦ dz = δ(u = ξ)(ut + f ′(u)ux ◦ dz).

And L1 isometry:∫Rχ(ξ; u(t, x))dξ = u(t, x) =⇒

∣∣∣∣∫Rχ(ξ; u)− χ(ξ; v)dξ

∣∣∣∣dx =

∫R|u − v |dx .

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Kinetic formulation

χ(t, x , ξ) = χ(ξ; u(t, x)) :=

1 if 0 ≤ ξ ≤ u(t, x)

−1 if u(t, x) ≤ ξ < 0

0 otherwise.

∣∣∣∣∫Rχ(ξ; u)− χ(ξ; v)dξ

∣∣∣∣dx =

∫R|u − v |dx .

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Kinetic formulation

χ(t, x , ξ) = χ(ξ; u(t, x)) :=

1 if 0 ≤ ξ ≤ u(t, x)

−1 if u(t, x) ≤ ξ < 0

0 otherwise.

∣∣∣∣∫Rχ(ξ; u)− χ(ξ; v)dξ

∣∣∣∣dx =

∫R|u − v |dx .

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Notion of solution

The term f ′(ξ)χx ◦ dz is difficult to treat, even as distribution.

Workaround: introduce ρ0 ∈ D(R) and

ρ(t, x , ξ; y) := ρ0(y − x + f ′(ξ)z(t)),

Then, if z ∈ C 1([0,T ]),

dρ+ f ′(ξ)ρx ◦ dz = 0, in (0,T ]× Rx × Rξ.

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Notion of solution

ρ(t, x , ξ; y) := ρ0(y − x + f ′(ξ)z(t)),

Then, if z ∈ C 1([0,T ]),

Consequently,

d(ρχ) + f ′(ξ)(ρχ)x ◦ dz = χ(dρ+ f ′(ξ)ρx ◦ dz

)︸︷︷︸

+ρ(dχ+ f ′(ξ)χx ◦ dz

)︸︷︷︸

=mξ dt

= ρmξdt.

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Notion of solution

ρ(t, x , ξ; y) := ρ0(y − x + f ′(ξ)z(t)),

Then, if z ∈ C 1([0,T ]),

Consequently, ∫R

d(ρχ) + f ′(ξ)(ρχ)x ◦ dzdx =

∫Rρmξdtdx .

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Notion of solution

ρ(t, x , ξ; y) := ρ0(y − x + f ′(ξ)z(t)),

Then, if z ∈ C 1([0,T ]),

Leads to condition

∫Rρχdx =

∫Rρmξdx , in D′(Rt × Rξ).

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Notion of solution

ρ(t, x , ξ; y) := ρ0(y − x + f ′(ξ)z(t)),

Then, if z ∈ C 1([0,T ]),

Leads to condition

∫Rρχdx =

∫Rρmξdx in D′([0,T ]× Rξ). (1)

Definition 5 (Pathwise entropy solution (PES))

u ∈ L1 ∩ L∞([0,T ]×R) is a PES if there exists a non-negative, bounded measurem such that equation (1) holds for all ρ, as defined above.

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Well-posedness

Theorem 6 (Lions, Perthame, Souganidis, 2013)

Assume f ∈ C 2(R;R), z ∈ C ([0,T ];R) and u0 ∈ (L1 ∩ L∞)(R). Then, for allT > 0, there exists a unique PES u ∈ C ([0,T ]; L1(R)) ∩ L∞([0,T ]× R).Furthermore, for two solutions u, v generated from the respective driving pathsz , z and u0, v0 ∈ BV (R),

‖u(t, ·)− v(t, ·)‖1 ≤ ‖u0 − v0‖1 + C√

sups∈(0,t)

|z(s)− z(s)|,

for a uniform constant C (u0, v0, f , f′, f ′′) > 0.

Note that if zn is a piecewise linear interpolation of z ∈ C 0,α using interpolationpoints with zn(tk) = z(tk) and un := u(·, ·; zn), then

‖u(t, ·)− un(t, ·)‖1 = O(n−α/2).

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Well-posedness

Theorem 6 (Lions, Perthame, Souganidis, 2013)

Assume f ∈ C 2(R;R), z ∈ C ([0,T ];R) and u0 ∈ (L1 ∩ L∞)(R). Then, for allT > 0, there exists a unique PES u ∈ C ([0,T ]; L1(R)) ∩ L∞([0,T ]× R).Furthermore, for two solutions u, v generated from the respective driving pathsz , z and u0, v0 ∈ BV (R),

‖u(t, ·)− v(t, ·)‖1 ≤ ‖u0 − v0‖1 + C√

sups∈(0,t)

|z(s)− z(s)|,

for a uniform constant C (u0, v0, f , f′, f ′′) > 0.

Note that if zn is a piecewise linear interpolation of z ∈ C 0,α using interpolationpoints with zn(tk) = z(tk) and un := u(·, ·; zn), then

‖u(t, ·)− un(t, ·)‖1 = O(n−α/2).

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Numerical solution approach

(i) Approximate the rough path z by a piecewise linear interpolation

zn(t) =

(1− t − τk

)z(τk) +

t − τk∆τ

z(τk+1), t ∈ [τk , τk+1],

where τk = k∆τ and ∆τ = T/n(ii) Solve the conservation law with driving noise zn using a standard numerical

method in classical Kruzkov entropy sense.

0.0 0.2 0.4 0.6 0.8 1.0t

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Solution with approximated driving noise zn

The problem to solve:

unt + zn∂x f (un) = 0, in (0,T ]× R,

un(0, ·) = u0.

Let S(∆τ,∆z)u denote the solution of

ut +∆z

∆τ∂x f (u) = 0, in (0,∆τ ]× R,

u(0, ·) = u.

un(τk) =k−1∏j=0

S(∆τ,∆zj )u0, for k = 0, 1, . . . , n,

where ∆zj := z(τj+1)− z(τj).

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Solution with approximated driving noise zn

The problem to solve:

unt + zn∂x f (un) = 0, in (0,T ]× R,

un(0, ·) = u0.

Let S(∆τ,∆z)u denote the solution of

ut +∆z

∆τ∂x f (u) = 0, in (0,∆τ ]× R,

u(0, ·) = u.

S(∆τ,∆zj )u0, for k = 0, 1, . . . , n,

where ∆zj := z(τj+1)− z(τj).

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Numerical schemes

Solve iteratively k = 0, 1, . . . , n

∆zk∆τ

∂x f (un) = 0, in (τk , τk+1]× R,

with ∆x = O(N−1) and time-steps ∆tk = ∆τ/Nk .

Numerical solution u`m := u(t`, xm; zn).

Solve, for instance, by Lax–Friedrichs (assuming t` ∈ (τk , τk+1)),

u`+1m =

u`m+1 + u`m−1

2−∆tk

∆zk∆τ

f (u`m+1)− f (u`m−1)

2∆x, over `,m, k. (2)

17 / 33

Numerical schemes

∆zk∆τ

∂x f (un) = 0, in (τk , τk+1]× R,

Solve, for instance, by Lax–Friedrichs

u`+1m =

u`m+1 + u`m−1

2− ∆zk

f (u`m+1)− f (u`m−1)

2∆x, over `,m, k. (2)

17 / 33

Numerical schemes

∆zk∆τ

∂x f (un) = 0, in (τk , τk+1]× R,

u`+1m =

u`m+1 + u`m−1

2− ∆zk

f (u`m+1)− f (u`m−1)

2∆x, over `,m, k. (2)

With initial data

∫ xm+1/2

xm−1/2

u0(x)dx .

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Numerical schemes

∆zk∆τ

∂x f (un) = 0, in (τk , τk+1]× R,

u`+1m =

u`m+1 + u`m−1

2− ∆zk

f (u`m+1)− f (u`m−1)

2∆x, over `,m, k. (2)

CFL: |znk |‖f ′‖L∞(−‖u0‖∞,‖u0‖∞)∆tk∆x≤ 1 =⇒ Nk = O

(|∆zk |∆x

)= O(n−αN−1)

So ∆tk = ∆τ/Nk = O(nα−1N−1) for all k .

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Convergence rates

If u0 ∈ (L1 ∩ BV )(R) and f ∈ C 2(R;R),

‖u(T , ·)− un(T , ·)‖1 ≤ ‖u0 − un0‖1 + C

√∆x

n∑k=0

|∆zk |

= O(N−1) + O(N−1/2n1−α).

Recall further

‖u(T , ·)− un(T , ·)‖1 ≤ C√

sups∈[0,T ]

|z(s)− zn(s)| = O(n−α/2).

Hence,‖u(T , ·)− u(T , ·)‖1 = O(N−1/2n1−α + n−α/2). (3)

Balance error contributions:

N(n) = O(n2−α).

If u0 has compact support, the cost of achieving O(ε) error in (3)O(ε−(2/α)(5−3α))!

Which is O(ε−14) for z ∈ C 0,1/2([0,T ]).18 / 33

Convergence rates

If u0 ∈ (L1 ∩ BV )(R) and f ∈ C 2(R;R),

‖u(T , ·)− un(T , ·)‖1 ≤ ‖u0 − un0‖1 + C

√∆x

n∑k=0

|∆zk |

= O(N−1) + O(N−1/2n1−α).

Recall further

‖u(T , ·)− un(T , ·)‖1 ≤ C√

sups∈[0,T ]

|z(s)− zn(s)| = O(n−α/2).

N(n) = O(n2−α).

Convergence rates

If u0 ∈ (L1 ∩ BV )(R) and f ∈ C 2(R;R),

‖u(T , ·)− un(T , ·)‖1 ≤ ‖u0 − un0‖1 + C

√∆x

n∑k=0

|∆zk |

= O(N−1) + O(N−1/2n1−α).

Recall further

‖u(T , ·)− un(T , ·)‖1 ≤ C√

sups∈[0,T ]

|z(s)− zn(s)| = O(n−α/2).

N(n) = O(n2−α).

Convergence rates

If u0 ∈ (L1 ∩ BV )(R) and f ∈ C 2(R;R),

‖u(T , ·)− un(T , ·)‖1 ≤ ‖u0 − un0‖1 + C

√∆x

n∑k=0

|∆zk |

= O(N−1) + O(N−1/2n1−α).

Recall further

‖u(T , ·)− un(T , ·)‖1 ≤ C√

sups∈[0,T ]

|z(s)− zn(s)| = O(n−α/2).

N(n) = O(n2−α).

Numerical example with u0 = 1|x|<0.5 and f (u) = u2/2.

−1.0 −0.5 0.0 0.5 1.00.0

Engquist–Osher solution

−1.0 −0.5 0.0 0.5 1.0

Lax–Friedrichs solution

0.0 0.2 0.4 0.6 0.8 1.0

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0.2Driving rough path

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Numerical example with u0 = sign(x)1|x|<0.5 and f (u) = u2/2.

−0.6 −0.4 −0.2 0.0 0.2 0.4 0.6

−1.0

−0.5

Engquist–Osher solution

−0.6 −0.4 −0.2 0.0 0.2 0.4 0.6

−1.0

−0.5

Lax–Friedrichs solution

0.0 0.2 0.4 0.6 0.8 1.0

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0.2Driving rough path

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Flow map cancellations

Recall that S(∆τ,∆z)u denotes the solution of

ut +∆z

∆τ∂x f (u) = 0, in (0,∆τ ]× R,

u(0, ·) = u.

S(∆τ,∆zj )u0, for k = 0, 1, . . . , n,

where ∆zj := z(τj+1)− z(τj).Provided un(s, ·) ∈ C (R) for all s ∈ (τ`, τk), then

un(τk) =k−1∏j=`

S(∆τ,∆zj )un(τ`) = S((k−`)∆τ,∑k−1

j=` ∆zj)un(τ`)

Benefit |z(τk)− z(τ`)| replaces∑k−1

j=` |∆zj | in the numerical error bound, CFL . . .

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Flow map cancellations

Recall that S(∆τ,∆z)u denotes the solution of

ut +∆z

∆τ∂x f (u) = 0, in (0,∆τ ]× R,

u(0, ·) = u.

S(∆τ,∆zj )u0, for k = 0, 1, . . . , n,

where ∆zj := z(τj+1)− z(τj).Provided un(s, ·) ∈ C (R) for all s ∈ (τ`, τk), then

un(τk) =k−1∏j=`

S(∆τ,∆zj )un(τ`) = S((k−`)∆τ,∑k−1

j=` ∆zj)un(τ`)

Benefit |z(τk)− z(τ`)| replaces∑k−1

j=` |∆zj | in the numerical error bound, CFL . . .

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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

αt≤ u(x + h, t)− u(x , t)

h∀h > 0 and t > 0.

22 / 33

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

αt≤ u(x + h, t)− u(x , t)

h∀h > 0 and t > 0.

One-sided estimates: If f ′′ ≥ α > 0 and zn > 0 for all t ∈ (a, b), then

un(x + h, t)− un(x , t)

h≤ 1

α(zn(t)− zn(a))∀h > 0 and t ∈ (a, b),

and if zn < 0 for t ∈ (b, c)

α(zn(t)− zn(b))≤ un(x + h, t)− un(x , t)

h∀h > 0 and t ∈ (b, c).

22 / 33

Continuity result

Theorem 7 (Flow map product sum property)

Consider Burgers’ equation, f (u) = u2/2. Let

M+(t) := maxs∈[0,t]

zn(s), M−(t) := mins∈[0,t]

zn(s).

For all t and all intervals s.t.: t ∈ (a, b) ⊆ [0,T ] for whichM−(a) < zn(t) < M+(a), we have that

un(s, ·) ∈ C (R), ∀s ∈ (a, b),

andun(b) = Sb−a,zn(b)−zn(a)un(a).

Secondly, whenever ∆zk∆zk+1 > 0, then

S(∆τ,∆z2)S(∆τ,∆z1) = S(2∆τ,∆z1+∆z2).

23 / 33

Running min and max

t0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

24 / 33

An equivalent integration path

Theorem 8 (Oscillating running max/min (ORM) function)

yn(t) :=

{M+(t)1s+(t)≥s−(t) + M−(t)1s−(t)≥s+(t) if t ∈ (0,T )

z(T ) if t = T

with s+(t) = max{s ≤ t|zn(s) = M+(s)} ands−(t) = max{s ≤ t|zn(s) = M−(s)}.Then, for Burgers’ equation,

n−1∏j=0

S(∆τ,∆zj ) =k−1∏j=0

S(∆τ,∆ynj ).

(Note that S(∆τ,0) = I .)

25 / 33

An equivalent integration path

Theorem 8 (Oscillating running max/min (ORM) function)

yn(t) :=

{M+(t)1s+(t)≥s−(t) + M−(t)1s−(t)≥s+(t) if t ∈ (0,T )

z(T ) if t = T

with s+(t) = max{s ≤ t|zn(s) = M+(s)} ands−(t) = max{s ≤ t|zn(s) = M−(s)}.Then, for Burgers’ equation,

n−1∏j=0

S(∆τ,∆zj ) =k−1∏j=0

S(∆τ,∆ynj ).

(Note that S(∆τ,0) = I .)

25 / 33

The ORM function

t0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

26 / 33

Numerical errors

Numerical integration “along” the ORM yields

‖u(T , ·)− un(T , ·)‖1 ≤ ‖u0 − un0‖1 + C

√∆x

n∑k=0

|∆ynk |

= O(N−1/2|yn|BV (0,T )),

where ∆x = O(N−1).Recall that integrating “along” zn yields O(N−1/2|zn|BV (0,t)) num error bound.Efficiency to be gained provided

|yn|BV (0,T )

|zn|BV (0,T )= o(1),

since, respectively

N(n) = O(|zn|2BV (0,T )nα),O(|yn|2BV (0,T )n

andCost(u(T )) = O(N(n)n).

27 / 33

Numerical errors

Numerical integration “along” the ORM yields

‖u(T , ·)− un(T , ·)‖1 ≤ ‖u0 − un0‖1 + C

√∆x

n∑k=0

|∆ynk |

= O(N−1/2|yn|BV (0,T )),

where ∆x = O(N−1).Recall that integrating “along” zn yields O(N−1/2|zn|BV (0,t)) num error bound.Efficiency to be gained provided

|yn|BV (0,T )

|zn|BV (0,T )= o(1),

since, respectively

N(n) = O(|zn|2BV (0,T )nα),O(|yn|2BV (0,T )n

andCost(u(T )) = O(N(n)n).

27 / 33

Bounded variation of ORM

Theorem 9 (Bounded variation of Wiener path ORM)

For standard Wiener paths w : [0,T ]→ R, the ORM function yn(·) : [0,T ]→ Rassociated to wn fulfils

yn ∈ BV ([0,T ]) ∀n > 0 almost surely,

andE[|yn|BV [0,1]

]<∞, ∀n > 0.

The above also holds for the ORM y of w.

Implication: Cost of achieving O(ε) approximation error is improved by thissharper bound from O(ε−14) to O(ε−4) for Burgers’ equation.

28 / 33

Bounded variation of ORM

Theorem 9 (Bounded variation of Wiener path ORM)

For standard Wiener paths w : [0,T ]→ R, the ORM function yn(·) : [0,T ]→ Rassociated to wn fulfils

yn ∈ BV ([0,T ]) ∀n > 0 almost surely,

andE[|yn|BV [0,1]

]<∞, ∀n > 0.

The above also holds for the ORM y of w.

Implication: Cost of achieving O(ε) approximation error is improved by thissharper bound from O(ε−14) to O(ε−4) for Burgers’ equation.

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Example

(u2)x◦ dz = 0, u0(x) = 1|x|<1/2, t ∈ [0, 2]

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

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Example

(u2)x◦ dz = 0, u0(x) = 1|x|<1/2, t ∈ [0, 2]

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1

u(2, x)

initial datafront trackingEO, using ORMEO, not using orm

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Regularity of solutions

Does the driving noise z have a regularizing effect on the solution?

For Burgers’, un(t) can only be discontinuous at times when zn(t) = M+(t)and/or zn(t) = M−(t):

t0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

For Wiener processes {s ∈ [0,T ]|w(s) = M+(s) and/or w(s) = M−(s)} hasLebesgue measue 0.

But, not (presently) clear if regularity behavior of un extends to the limitsolution.

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t0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

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t0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

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Overview

4 Conclusion

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Summary

Developed a numerical method for solving stochastic scalar conservation laws.

Identified cancellations of oscillations that in some settings lead to sharpererror 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 ]× Rd ,

u(0, ·) = u0 ∈ (L1 ∩ L∞)(Rd).

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References

1 P-L Lions, B Perthame, P E Souganidis Scalar conservation lawswith rough (stochastic) fluxes .Stoch. Partial Differ. Equ. Anal. Comput. 1 (2013), no. 4, 664-686.

2 B Gess, P E Souganidis Long-Time Behavior, invariant measures andregularizing effects for stochastic scalar conservation laws.Communications on Pure and Applied Mathematics (2016).

3 P-L Lions, B Perthame, P Souganidis Scalar conservation laws withrough (stochastic) fluxes: the spatially dependent case.Stochastic Partial Differential Equations: Analysis and Computations 2.4(2014): 517-538.

Thank you for your attention!

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