numerical methods for conservation laws with a ... · pdf filenumerical methods for...
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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
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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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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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 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
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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Definition
The problem formulation
du + ∂x f (u) ◦ dz = 0, in (0,T ]× R,u(0, ·) = u0 ∈ (L1 ∩ L∞)(R).
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
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
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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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
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.
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
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.
Consequently, ∫R
d(ρχ) + f ′(ξ)(ρχ)x ◦ dzdx =
∫Rρmξdtdx .
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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.
Leads to condition
d
dt
∫Rρχdx =
∫Rρmξdx , in D′(Rt × Rξ).
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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.
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
1.0
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0.8
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
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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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
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
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
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
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
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
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
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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
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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
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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
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Numerical example with u0 = 1|x|<0.5 and f (u) = u2/2.
−1.0 −0.5 0.0 0.5 1.00.0
0.2
0.4
0.6
0.8
1.0
Engquist–Osher solution
−1.0 −0.5 0.0 0.5 1.0
x
0.0
0.2
0.4
0.6
0.8
1.0
Lax–Friedrichs solution
0.0 0.2 0.4 0.6 0.8 1.0
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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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
0.0
0.5
1.0
Engquist–Osher solution
−0.6 −0.4 −0.2 0.0 0.2 0.4 0.6
x
−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
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
un(τk) =k−1∏j=0
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
un(τk) =k−1∏j=0
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-
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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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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
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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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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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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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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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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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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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