people.maths.bris.ac.ukmb13434/balazs_warwick13.pdf · models hydrod. 2nd class before question...

Models Hydrod. 2nd class Before Question BCRW

Second class particles can perform simplerandom walks (in some cases)

Joint withGyorgy Farkas, Peter Kovacs and Attila Rakos

Marton Balazs

University of Bristol

Warwick Probability SeminarOctober 23, 2013.

The modelsAsymmetric simple exclusion processZero range processGeneralized ZRPBricklayers processStationary distributions

Hydrodynamics

The second class particle

Earlier results

The question

Branching coalescing random walk

Models Hydrod. 2nd class Before Question BCRW ASEP TAZRP TAGZRP TABLP Stati distr.

Asymmetric simple exclusion

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• • • •

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• • • •

Particles try to jump

to the right with rate p,to the left with rate q = 1 − p < p.

The jump is suppressed if the destination site is occupied byanother particle.

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• • • •

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• • • •

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• • • •

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• • • •

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• • • •

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• • • •

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• • • •

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• • • •

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• • • •

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• • • •

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• • • •

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• • • •

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• • • •

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• • • •

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• • • •

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• • • •

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• • • •

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• • • •

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• • • •

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• • • •

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• • • •

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• • • •

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• • • •

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• • • • •

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• • • • •

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• • • • •

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• • • • •

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• • • • •

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• • • • •

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• • • • •

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• • • • •

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• • • • •

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• • • • •

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• • • • •

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• • • • •

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• • • • •

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• • • •

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• • • •

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• • • •

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• • • •

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• • • •

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• • • •

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• • • •

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• • • •

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• • • •

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• • • •

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• • • •

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• • • •

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• • • •

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• • • •

The totally asymmetric zero range process

-3 -2 -1 0 1 2 3 4 i

•• •••

-3 -2 -1 0 1 2 3 4 i

•• •••

ω−3=2

-3 -2 -1 0 1 2 3 4 i

•• •••

ω−3=2

Particles jump to the right from site i with rate r(ωi)

(r non-decreasing).

-3 -2 -1 0 1 2 3 4 i

•• •••

ω−3=2

(r non-decreasing).

-3 -2 -1 0 1 2 3 4 i

•• •••

ω−3=2

(r non-decreasing).

-3 -2 -1 0 1 2 3 4 i

•• ••• •

ω−3=2

(r non-decreasing).

-3 -2 -1 0 1 2 3 4 i

•• •• • •

ω−3=2

(r non-decreasing).

-3 -2 -1 0 1 2 3 4 i

•• •• • •

ω−3=2

(r non-decreasing).

-3 -2 -1 0 1 2 3 4 i

•• •• ••

ω−3=2

(r non-decreasing).

-3 -2 -1 0 1 2 3 4 i

•• •• ••

ω−3=2

(r non-decreasing).

-3 -2 -1 0 1 2 3 4 i

•• •• ••

ω−3=2

(r non-decreasing).

-3 -2 -1 0 1 2 3 4 i

•• •• ••

ω−3=2

(r non-decreasing).

-3 -2 -1 0 1 2 3 4 i

•• •• ••

ω−3=2

(r non-decreasing).

-3 -2 -1 0 1 2 3 4 i

•• •• ••

ω−3=2

(r non-decreasing).

-3 -2 -1 0 1 2 3 4 i

•• •• ••

ω−3=2

(r non-decreasing).

-3 -2 -1 0 1 2 3 4 i

•• •• • •

ω−3=2

(r non-decreasing).

-3 -2 -1 0 1 2 3 4 i

•• •• • •

ω−3=2

(r non-decreasing).

-3 -2 -1 0 1 2 3 4 i

•• •• • •

ω−3=2

(r non-decreasing).

-3 -2 -1 0 1 2 3 4 i

•• •• • •

ω−3=2

(r non-decreasing).

-3 -2 -1 0 1 2 3 4 i

•• •• • •

ω−3=2

(r non-decreasing).

-3 -2 -1 0 1 2 3 4 i

•• •• • •

ω−3=2

(r non-decreasing).

-3 -2 -1 0 1 2 3 4 i

•• •• • •

ω−3=2

(r non-decreasing).

A generalized totally asymmetric zero range process:ωi ∈ Z

i i + 1

ωi+1=−1

i i + 1

ωi+1=−1

a brick is added with rate r(ωi).

(r non-decreasing).

i i + 1

ωi+1=−1

(r non-decreasing).

i i + 1

ωi+1=−1

(r non-decreasing).

i i + 1

ωi+1=−1

(r non-decreasing).

i i + 1

ωi+1=−1

(r non-decreasing).

i i + 1

ωi+1=−1

(r non-decreasing).

i i + 1

ωi+1=−1

(r non-decreasing).

i i + 1

ωi+1=−1

(r non-decreasing).

i i + 1

ωi − 1ωi+1 + 1

|| ||(

(r non-decreasing).

i i + 1

ωi − 1ωi+1 + 1

|| ||(

(r non-decreasing).

i i + 1

ωi − 1ωi+1 + 1

|| ||(

(r non-decreasing).

i i + 1

ωi − 1ωi+1 + 1

|| ||(

(r non-decreasing).

i i + 1

ωi − 1ωi+1 + 1

|| ||(

(r non-decreasing).

i i + 1

ωi − 1ωi+1 + 1

|| ||(

(r non-decreasing).

i i + 1

ωi − 1ωi+1 + 1

|| ||(

(r non-decreasing).

i i + 1

ωi − 1ωi+1 + 1

|| ||(

(r non-decreasing).

i i + 1

ωi − 1ωi+1 + 1

|| ||(

(r non-decreasing).

i i + 1

ωi − 1ωi+1 + 1

|| ||(

(r non-decreasing).

i i + 1

ωi − 1ωi+1 + 1

|| ||(

(r non-decreasing).

i i + 1

ωi − 1ωi+1 + 1

|| ||(

(r non-decreasing).

i i + 1

ωi − 1ωi+1 + 1

|| ||(

(r non-decreasing).

i i + 1

ωi − 1ωi+1 + 1

|| ||(

(r non-decreasing).

i i + 1

ωi − 1ωi+1 + 1

|| ||(

(r non-decreasing).

i i + 1

ωi − 1ωi+1 + 1

|| ||(

(r non-decreasing).

i i + 1

ωi − 1ωi+1 + 1

|| ||(

(r non-decreasing).

i i + 1

ωi − 1ωi+1 + 1

|| ||(

(r non-decreasing).

i i + 1

ωi − 1ωi+1 + 1

|| ||(

(r non-decreasing).

i i + 1

ωi − 1ωi+1 + 1

|| ||(

(r non-decreasing).

i i + 1

ωi − 1ωi+1 + 1

|| ||(

(r non-decreasing).

i i + 1

ωi − 1ωi+1 + 1

|| ||(

(r non-decreasing).

i i + 1

ωi − 1ωi+1 + 1

|| ||(

(r non-decreasing).

i i + 1

ωi − 1ωi+1 + 1

|| ||(

(r non-decreasing).

i i + 1

ωi − 1ωi+1 + 1

|| ||(

(r non-decreasing).

i i + 1

ωi − 1ωi+1 + 1

|| ||(

(r non-decreasing).

i i + 1

ωi − 1ωi+1 + 1

|| ||(

(r non-decreasing).

i i + 1

ωi − 1ωi+1 + 1

|| ||(

(r non-decreasing).

i i + 1

ωi − 1ωi+1 + 1

|| ||(

(r non-decreasing).

i i + 1

ωi − 1ωi+1 + 1

|| ||(

(r non-decreasing).

The totally asymmetric bricklayers process

i i + 1

ωi+1=−1

i i + 1

ωi+1=−1

a brick is added with rate r(ωi) + r(−ωi+1).

i i + 1

ωi+1=−1

i i + 1

ωi+1=−1

i i + 1

ωi+1=−1

i i + 1

ωi+1=−1

i i + 1

ωi+1=−1

i i + 1

ωi+1=−1

i i + 1

ωi+1=−1

i i + 1

ωi+1=−1

i i + 1

ωi+1=−1

i i + 1

ωi+1=−1

i i + 1

ωi+1=−1

i i + 1

ωi+1=−1

i i + 1

ωi+1=−1

i i + 1

ωi+1=−1

i i + 1

ωi − 1ωi+1 + 1

|| ||(

i i + 1

ωi − 1ωi+1 + 1

|| ||(

i i + 1

ωi − 1ωi+1 + 1

|| ||(

i i + 1

ωi − 1ωi+1 + 1

|| ||(

i i + 1

ωi − 1ωi+1 + 1

|| ||(

i i + 1

ωi − 1ωi+1 + 1

|| ||(

i i + 1

ωi − 1ωi+1 + 1

|| ||(

i i + 1

ωi − 1ωi+1 + 1

|| ||(

i i + 1

ωi − 1ωi+1 + 1

|| ||(

i i + 1

ωi − 1ωi+1 + 1

|| ||(

i i + 1

ωi − 1ωi+1 + 1

|| ||(

i i + 1

ωi − 1ωi+1 + 1

|| ||(

i i + 1

ωi − 1ωi+1 + 1

|| ||(

i i + 1

ωi − 1ωi+1 + 1

|| ||(

i i + 1

ωi − 1ωi+1 + 1

|| ||(

i i + 1

ωi − 1ωi+1 + 1

|| ||(

i i + 1

ωi − 1ωi+1 + 1

|| ||(

i i + 1

ωi − 1ωi+1 + 1

|| ||(

i i + 1

ωi − 1ωi+1 + 1

|| ||(

i i + 1

ωi − 1ωi+1 + 1

|| ||(

i i + 1

ωi − 1ωi+1 + 1

|| ||(

i i + 1

ωi − 1ωi+1 + 1

|| ||(

i i + 1

ωi − 1ωi+1 + 1

|| ||(

i i + 1

ωi − 1ωi+1 + 1

|| ||(

i i + 1

ωi − 1ωi+1 + 1

|| ||(

i i + 1

ωi − 1ωi+1 + 1

|| ||(

i i + 1

ωi − 1ωi+1 + 1

|| ||(

i i + 1

ωi − 1ωi+1 + 1

|| ||(

i i + 1

ωi − 1ωi+1 + 1

|| ||(

i i + 1

ωi − 1ωi+1 + 1

|| ||(

A mirror-symmetrized version of the extended zero range. Leftand right jumps of the dynamics cooperate, if(r(ω) · r(1 − ω) = 1; r non-decreasing).

Stationary product distributions

For the ASEP: the Bernoulli() distribution is time-stationary forany (0 ≤ ≤ 1).

For zero range, bricklayers: the product of marginals

µθ(ωi) =eθωi

r(ωi)!·

1Z (θ)

is stationary for any θ ∈ R that makes Z (θ) finite.

For zero range, bricklayers: the product of marginals

µθ(ωi) =eθωi

r(ωi)!·

1Z (θ)

is stationary for any θ ∈ R that makes Z (θ) finite.

Here r(0)! : = 1, and r(z + 1)! = r(z)! · r(z + 1) for all z ∈ Z.

Hydrodynamics (very briefly)

The density = (θ) : = Eθ(ω) and the hydrodynamic fluxH = Hθ : = Eθ[growth rate] both depend on a parameter or θof the stationary distribution.

◮ H() = Hθ() is the hydrodynamic flux function.

◮ H() = Hθ() is the hydrodynamic flux function.◮ If the process is locally in equilibrium, but changes over

some large scale (variables X = εi and T = εt), then

∂T(T , X ) + ∂X H((T , X )) = 0 (conservation law).

◮ H() = Hθ() is the hydrodynamic flux function.◮ If the process is locally in equilibrium, but changes over

some large scale (variables X = εi and T = εt), then

∂T(T , X ) + ∂X H((T , X )) = 0 (conservation law).

∂T+ ∂X H() = 0 = E(ω), H() = Eθ()[growth rate]

For the ASEP, H() = (p − q) · (1 − ), concave.

For the zero range and bricklayers, H() is convex/concave ifthe rate function r is convex/concave.

Special case: r(ω) = const. · eβω; H() is convex.

will be interested in TAGEZRP, TAEBLP.

Special case: r(ω) = const. · eβω; H() is convex.

will be interested in TAGEZRP, TAEBLP.

Either convex or concave, discontinuous shock solutionsexist.

Rarefaction wave

(T , X)

Traffic jam

Rarefaction wave

(T , X)

Traffic jam

Rarefaction wave

(T , X)

Traffic jam

Rarefaction wave

(T , X)

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Rarefaction wave

(T , X)

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Rarefaction wave

(T , X)

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Rarefaction wave

(T , X)

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(T , X)

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Rarefaction wave

(T , X)

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Rarefaction wave

(T , X)

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Rarefaction wave

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Rarefaction wave

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Rarefaction wave

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Rarefaction wave

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Rarefaction wave

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Rarefaction wave

(T , X)

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Rarefaction wave

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Rarefaction wave

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Rarefaction wave

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Rarefaction wave

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Rarefaction wave

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Rarefaction wave

(T , X)

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Rarefaction wave

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(T , X)

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Rarefaction wave

(T , X)

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(T , X)

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Shock wave

(T , X)

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(T , X)

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(T , X)

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(T , X)

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(T , X)

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(T , X)

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(T , X)

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Shock wave

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(T , X)

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Shock wave

(T , X)

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Discontinuous shock appears. Its velocity is given by theRankine-Hugoniot speed for densities left and right

R =H(left)− H(right)

left − right.

Shock wave

(T , X)

Traffic jam

Discontinuous shock appears. Its velocity is given by theRankine-Hugoniot speed for densities left and right

R =H(left)− H(right)

left − right.

Let’s look for the corresponding microscopic structure.

States ω and ω only differ at one site.

•• • •••• •• ••

Growth on the right:rate≤rate

•• • •••• •• ••

Growth on the right:rate≤ratewith rate:

•••• •• ••

•••

•••• •• ••

•••

•••• •• ••

•••

•••• • • ••

•••

•••• • • ••

•••

•••• • • •

•••

•••• • ••

•••

•••• • ••

•••

•••• • ••

•••

•••• • ••

•••

•• • •••• •• ••

Growth on the right:rate≤ratewith rate-rate:

•• • •••• •• ••

•• • •••• • • ••

•• • •••• • • •

•• • •••• • ••

Growth on the left:rate≥rate

•• • •••• •• ••

Growth on the left:rate≥ratewith rate:

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Growth on the left:rate≥ratewith rate-rate:

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A single discrepancy , the second class particle, is conserved.

Earlier results: as seen by the second class particle

From now on: ASEP, TAGEZRP, TAEBLP only; “E”=exponential.

Theorem (Derrida, Lebowitz, Speer ’97)For the ASEP, the Bernoulli product distribution with densities

-3 -2 -1 0 1 2 3

left 0

is stationary for the process, as seen from the second class particle, if

right · (1 − left)

left · (1 − right)=

From now on: ASEP, TAGEZRP, TAEBLP only; “E”=exponential.

Theorem (Derrida, Lebowitz, Speer ’97)For the ASEP, the Bernoulli product distribution with densities

-3 -2 -1 0 1 2 3

left 0

1right

is stationary for the process, as seen from the second class particle, if

right · (1 − left)

Earlier results: random walking shocks

Theorem (Belitsky and Schutz ’02)For the ASEP with the very same parameters, the Bernoulliproduct distribution µ0 with densities

-3 -2 -1 0 1 2 3

evolves according to

ddtµ0 = p ·

right· [µ

−1 − µ0] + q ·right

left· [µ1 − µ0].

Theorem (Belitsky and Schutz ’02)For the ASEP with the very same parameters, the Bernoulliproduct distribution µ0 with densities

-3 -2 -1 0 1 2 3

ddtµ0 = p ·

right· [µ

−1 − µ0] + q ·right

left· [µ1 − µ0].

Multiple shocks and their interactions are also handled.

Interpretation: random walking shock µ(t) = µX(t):

with rate p · leftright

-3 -2 -1 0 1 2 3

X (t) = 0µ0

ddtµ0 = p ·

right· [µ

−1 − µ0] + q ·right

left· [µ1 − µ0]

-3 -2 -1 0 1 2 3

ddtµ0 = p ·

right· [µ

−1 − µ0] + q ·right

-3 -2 -1 0 1 2 3

ddtµ0 = p ·

right· [µ

−1 − µ0] + q ·right

-3 -2 -1 0 1 2 3

ddtµ0 = p ·

right· [µ

−1 − µ0] + q ·right

-3 -2 -1 0 1 2 3

ddtµ0 = p ·

right· [µ

−1 − µ0] + q ·right

-3 -2 -1 0 1 2 3

X (t) = −1µ−1

ddtµ0 = p ·

right· [µ

−1 − µ0] + q ·right

with rate q ·rightleft

-3 -2 -1 0 1 2 3

X (t) = 0µ0

ddtµ0 = p ·

right· [µ

−1 − µ0] + q ·right

-3 -2 -1 0 1 2 3

ddtµ0 = p ·

right· [µ

−1 − µ0] + q ·right

-3 -2 -1 0 1 2 3

ddtµ0 = p ·

right· [µ

−1 − µ0] + q ·right

-3 -2 -1 0 1 2 3

ddtµ0 = p ·

right· [µ

−1 − µ0] + q ·right

-3 -2 -1 0 1 2 3

ddtµ0 = p ·

right· [µ

−1 − µ0] + q ·right

-3 -2 -1 0 1 2 3

X (t) = 1µ1

ddtµ0 = p ·

right· [µ

−1 − µ0] + q ·right

Theorem (B. ’01)For the TAEBLP, the product distribution of marginals µi withdensities

-3 -2 -1 0 1 2 3

is stationary for the process, as seen from the second classparticle, if

left − right = 1.

Theorem (B. ’01)For the TAEBLP, the product distribution of marginals µi withdensities

-3 -2 -1 0 1 2 3

is stationary for the process, as seen from the second classparticle, if

left − right = 1.

Theorem (B. ’04)For the very same parameters, the product distribution µ0 withdensities

-3 -2 -1 0 1 2 3

ddtµ0 = Cleft · [µ−1 − µ0] + Cright · [µ1 − µ0].

Theorem (B. ’04)For the very same parameters, the product distribution µ0 withdensities

-3 -2 -1 0 1 2 3

Multiple shocks and their interactions are also handled.

with rate Cleft:

-3 -2 -1 0 1 2 3

X (t) = 0µ0

with rate Cleft:

-3 -2 -1 0 1 2 3

with rate Cleft:

-3 -2 -1 0 1 2 3

with rate Cleft:

-3 -2 -1 0 1 2 3

with rate Cleft:

-3 -2 -1 0 1 2 3

with rate Cleft:

-3 -2 -1 0 1 2 3

X (t) = −1µ−1

with rate Cright:

-3 -2 -1 0 1 2 3

X (t) = 0µ0

with rate Cright:

-3 -2 -1 0 1 2 3

with rate Cright:

-3 -2 -1 0 1 2 3

with rate Cright:

-3 -2 -1 0 1 2 3

with rate Cright:

-3 -2 -1 0 1 2 3

with rate Cright:

-3 -2 -1 0 1 2 3

X (t) = 1µ1

The question

Of course, the drift of the walk X (t) is the same as theexpected drift of the second class particle in its stationaryshock distribution,

The question

and also agrees with the Rankine Hugoniot formula for thespeed of shocks.

The question

Is it the second class particle that performs the simple randomwalk in the middle of a shock?

The question

Is it the second class particle that performs the simple randomwalk in the middle of a shock?

In what sense? Annealed w.r.t. the initial shock distribution...But what does this mean?

Here is the question:

For the ASEP, let ν0 be the Bernoulli product distribution

ν0 =(

µleft

µright

µ(ω = ω) =

, if ω = 1,

1 − , if ω = 0;δ(0, 1) = 1.

-3 -2 -1 0 1 2 3

left 0

1right

For the ASEP, let ν0 be the Bernoulli product distribution

-3 -2 -1 0 1 2 3

left 0

1right

Does it satisfy

ddtν0 = p ·

right· [ν

−1 − ν0] + q ·right

left· [ν1 − ν0]

whenright · (1 − left)

For the TAEBLP, let ν0 be the product distribution

ν0 =(

µleft

δright

µright

µ(ω = ω) =eθ()·ω

r(ω)!·

1Z (θ())

δ(ω, ω + 1) =eθ()·ω

r(ω)!·

1Z (θ())

-3 -2 -1 0 1 2 3

Here is the question:For the TAEBLP, let ν0 be the product distribution

ν0 =(

µleft

δright

µright

-3 -2 -1 0 1 2 3

Does it satisfy

ddtν0 = Cleft · [ν−1 − ν0] + Cright · [ν1 − ν0].

whenleft − right = 1 ?

Here is the answer:

Theorem (Gy. Farkas, P. Kovacs, A. Rakos, B. ’10)Yes, and yes. Even more, the thing also works for theTAGEZRP.

Here is the answer:

The second class particle, annealed w.r.t. the initial shockproduct distribution, does perform a drifted simple random walkin these cases.

Here is the answer:

This explains both types of the previous results.

Here is the answer:

This explains both types of the previous results.

The presence of a second class particle in the measuresignificantly simplifies the computations. This is how wediscovered the TAGEZRP.

Nice, since

This might open up the path for applying methods physicistslike.

Nice, since

This might open up the path for applying methods physicistslike.

It also gives a rough tail bound for the second class particle in aflat initial distribution; essential in the t2/3 proofs for theexponential models.

Interactions:

We also see that shocks+second class particles◮ locally interact by exclusion in ASEP, and don’t locally

interact in TAGEZRP, TAEBLP, but

Interactions:

◮ their jump rates depend on their position, making themattract each other such that

Interactions:

◮ their center of mass has the Rankine-Hugoniot velocity forthe large shock they jointly represent.

Interactions:

◮ their center of mass has the Rankine-Hugoniot velocity forthe large shock they jointly represent.

Macroscopically it’s one shock after all.

A similar result: branching coalescing random walk

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• ••

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• ••

With rate p: jump to the right

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• ••

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• ••

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• ••

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• ••

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• ••

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• ••

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• ••

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• ••

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• ••

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• ••

With rate q: jump to the left

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• ••

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• ••

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• ••

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• ••

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• ••

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• ••

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• ••

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• ••

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• ••

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• ••

With rate cr : coalescence to the right

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• ••

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• ••

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• ••

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• ••

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• ••

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• ••

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• ••

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• ••

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• ••

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• ••

With rate bl : branching to the left

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• ••

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• ••

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• ••

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• ••

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• ••

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• ••

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• ••

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• ••

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• ••

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• ••

With rate cl : coalescence to the left

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• ••

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• ••

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• ••

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• ••

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• ••

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• ••

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• ••

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• ••

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• ••

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• ••

With rate br : branching to the right

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• ••

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• ••

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• ••

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• ••

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• ••

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• ••

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• ••

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• ••

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• ••

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• ••

-3 -2 -1 0 1 2 3 4 i

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦• ••

The Bernoulli(∗) distribution is stationary for

∗ =bl + br

bl + br + cl + cr.

Earlier results: as seen by the rightmost particle

TheoremFor the BCRW, the Bernoulli product distribution with densities

-3 -2 -1 0 1 2 3

is stationary for the process, as seen from the rightmostparticle.

Theorem (Krebs, Jafarpour and Schutz ’03)For the BCRW with the very same parameters, the Bernoulliproduct distribution µ0 with densities

-3 -2 -1 0 1 2 3

ddtµ0 =

cl · (bl + br ) + q · (cl + cr )

bl + br + cl + cr· [µ

−1 − µ0]

+ p ·bl + br + cl + cr

cl + cr· [µ1 − µ0].

with rate cl ·(bl+br )+q·(cl+cr )bl+br+cl+cr

-3 -2 -1 0 1 2 3

X (t) = 0µ0

ddtµ0 =

cl · (bl + br ) + q · (cl + cr )

−1 − µ0]

cl + cr· [µ1 − µ0].

-3 -2 -1 0 1 2 3

ddtµ0 =

cl · (bl + br ) + q · (cl + cr )

−1 − µ0]

cl + cr· [µ1 − µ0].

-3 -2 -1 0 1 2 3

ddtµ0 =

cl · (bl + br ) + q · (cl + cr )

−1 − µ0]

cl + cr· [µ1 − µ0].

-3 -2 -1 0 1 2 3

ddtµ0 =

cl · (bl + br ) + q · (cl + cr )

−1 − µ0]

cl + cr· [µ1 − µ0].

-3 -2 -1 0 1 2 3

ddtµ0 =

cl · (bl + br ) + q · (cl + cr )

−1 − µ0]

cl + cr· [µ1 − µ0].

-3 -2 -1 0 1 2 3

X (t) = −1µ−1

ddtµ0 =

cl · (bl + br ) + q · (cl + cr )

−1 − µ0]

cl + cr· [µ1 − µ0].

with rate p · bl+br+cl+crcl+cr

-3 -2 -1 0 1 2 3

X (t) = 0µ0

ddtµ0 =

cl · (bl + br ) + q · (cl + cr )

−1 − µ0]

cl + cr· [µ1 − µ0].

-3 -2 -1 0 1 2 3

ddtµ0 =

cl · (bl + br ) + q · (cl + cr )

−1 − µ0]

cl + cr· [µ1 − µ0].

-3 -2 -1 0 1 2 3

ddtµ0 =

cl · (bl + br ) + q · (cl + cr )

−1 − µ0]

cl + cr· [µ1 − µ0].

-3 -2 -1 0 1 2 3

ddtµ0 =

cl · (bl + br ) + q · (cl + cr )

−1 − µ0]

cl + cr· [µ1 − µ0].

-3 -2 -1 0 1 2 3

ddtµ0 =

cl · (bl + br ) + q · (cl + cr )

−1 − µ0]

cl + cr· [µ1 − µ0].

-3 -2 -1 0 1 2 3

X (t) = 1µ1

ddtµ0 =

cl · (bl + br ) + q · (cl + cr )

−1 − µ0]

cl + cr· [µ1 − µ0].

The question:

Is it the rightmost particle that performs the random walk?

Here is the question:For the BCRW, let ν0 be the Bernoulli product distribution

ν0 =(

µ∗)

where δ(0) = 1.

-3 -2 -1 0 1 2 3

Does it satisfy

ddtν0 =

cl · (bl + br ) + q · (cl + cr )

bl + br + cl + cr· [ν

−1 − ν0]

cl + cr· [ν1 − ν0]?

The answer

◮ ... is, of course, yes again. [Gy. Farkas, P. Kovacs, A.Rakos, B. ’10]

The answer

◮ Fronts of the other direction: 0 − 1 − ∗ can also behandled.

The answer

◮ Fronts of the other direction: 0 − 1 − ∗ can also behandled.

Thank you.

people.maths.bris.ac.ukmb13434/balazs_warwick13.pdf · models hydrod. 2nd class before question...

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