Numerical methods for conservation laws with a ... · PDF fileNumerical methods for conservation laws with a stochastically driven ux ... is di cult to work with: ... ku(t;) v(t;)k

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

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Overview

1 Problem description

2 Deterministic conservation lawsCharacteristics and shocksWell-posedness

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

4 Conclusion

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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.

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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.

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Motivation

For the mean-field SDE

dX i = σ

X i ,1

L− 1

∑j 6=i

δX j

◦ dW , for i = 1, 2, . . . , L,

with σ : R× P(R)→ R, one has that

1

L

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.

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Our contribution

Develop numerical methods for solving the SSCL

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

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x

0

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1

u(t

,x)

t=0.0000

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x

0

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1

u(t

,x)

t=0.2500

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x

0

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1

u(t

,x)

t=0.5000

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x

0

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1

u(t

,x)

t=0.7500

0 0.2 0.4 0.6 0.8 1

t

-1

-0.8

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0

0.2

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1

z(t

)

z(t)

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Overview




4 Conclusion

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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

d

dt

∫R

udx =

∫R

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.

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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

d

dt

∫R

udx =

∫R

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.

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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

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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) =⇒

∫R

∣∣∣∣∫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.




∫R

∣∣∣∣∫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.




∫R

∣∣∣∣∫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

)︸︷︷︸

=0

+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

d

dt

∫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

d

dt

∫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

∆τ

)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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z24

z25

z210

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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.

Then

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.

Then


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

unt +

∆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)

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Numerical schemes


unt +

∆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

Nk

f (u`m+1)− f (u`m−1)

2∆x, over `,m, k. (2)

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Numerical schemes


unt +

∆zk∆τ

∂x f (un) = 0, in (τk , τk+1]× R,




u`+1m =

u`m+1 + u`m−1

2− ∆zk

Nk

f (u`m+1)− f (u`m−1)

2∆x, over `,m, k. (2)

With initial data

u0m =

1

∆x

∫ xm+1/2

xm−1/2

u0(x)dx .

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Numerical schemes


unt +

∆zk∆τ

∂x f (un) = 0, in (τk , τk+1]× R,




u`+1m =

u`m+1 + u`m−1

2− ∆zk

Nk

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.

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Engquist–Osher solution

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x

0.0

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1.0

Lax–Friedrichs solution

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t

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−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0.0

0.2Driving rough path

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

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−1.0

−0.5

0.0

0.5

1.0

Engquist–Osher solution

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x

−1.0

−0.5

0.0

0.5

1.0

Lax–Friedrichs solution

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t

−1.4

−1.2

−1.0

−0.8

−0.6

−0.4

−0.2

0.0

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.

and


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.

and


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

− 1

αt≤ u(x + h, t)− u(x , t)

h∀h > 0 and t > 0.

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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

− 1

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)

1

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

h∀h > 0 and t ∈ (b, c).

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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).

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Running min and max

t0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

-1.4

-1.2

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

zn

M+

M-

24 / 33

An equivalent integration path

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

For

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 .)

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An equivalent integration path

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

For

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 .)

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The ORM function

t0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

-1.4

-1.2

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

zn

M+

M-

ORM

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

α)

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

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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

α)

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

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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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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

ut +1

2

(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

t

-0.5

0

0.5

1

1.5

2

z(t)

zorm

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Example

ut +1

2

(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

t

-0.5

0

0.5

1

1.5

2

z(t)

zorm

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

x

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

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

-1.4

-1.2

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

zn

M+

M-

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

-1.4

-1.2

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

zn

M+

M-



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t0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

-1.4

-1.2

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

zn

M+

M-



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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∑

i=1

∂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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