Fourier law and phase transitions Errico Presutti Anogia. June - - PowerPoint PPT Presentation

Fourier law and phase transitions Errico Presutti

• Anogia. June 21-26, 2009.

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Contents

• Stochastic models, hydrodynamic limits
• Mesoscopic limits, an integro-differential equation
• Stationary solutions with non zero current
• Sharp interface limits, convergence to stationary Stefan problem

(in preparation with A. De Masi and D. Tsagkarojannis) 2

Symmetric simple exclusion process Particles on Zd. At most one particle per site. Each particle has a clock which rings at exponential times (clocks ring independently) When clock rings, particle chooses with equal probability a neighbor site Particle jumps to chosen site if empty, otherwise jump is suppressed: exclusion rule 3

Hydrodynamic equations. Exclusion rule gives drift towards regions with smaller density as shown by the hydrodynamic behavior. In fact, in the large space-time limit (parabolic scaling) the process converges to: ∂ρ ∂t = − ∂I ∂x , I = −∂ρ ∂x ρ particles density; I particles current Fourier law: current proportional to minus density gradient 4

Segregation phenomena Interaction: surrounding particles affect jump site selection. To contrast diffusion and favour segregation make jumps prefer regions with higher density. Density gradients determined by averages on “mesoscopic regions” (with many lattice sites). 5

Integro-differential evolution equation In the mesoscopic limit (scaling space time and interaction as well) Giacomin-Lebowitz have proved: ∂ρ ∂t = ∂ ∂x ∂ρ ∂x − 4βρ(1 − ρ)J ∗ ∂ρ ∂x

• J = J(x − y) probability density.

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Comparison with Cahn-Hilliard Stationary solutions: write m = 2ρ − 1, χ = 4βρ(1 − ρ) = β(1 − m2) 0 = ∂ ∂x

• χ ∂

∂x

• β−1arctanh(m) − J ∗ m
• Special solutions: β−1arctanh(m) − J ∗ m = 0.

Similar to Cahn-Hilliard 0 = − ∂2 ∂x2 ∂2u ∂x2 − V ′(u)

• and Allen-Cahn:

0 = ∂2u ∂x2 − V ′(u) 7

∂ρ ∂t = ∂ ∂x ∂ρ ∂x − 4βρ(1 − ρ)J ∗ ∂ρ ∂x

• If β < 1, 4βρ(1 − ρ) < 1 and Lebowitz, Orlandi, E.P. proved

convergence under parabolic rescaling of space-time to: ∂ρ ∂t = ∂ ∂x

• [1 − 4βρ(1 − ρ)]∂ρ

∂x

• Same result if initial datum is such that 4βρ(1 − ρ) < 1.

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The Lebowitz-Penrose free energy functional Change of variables, particles → spins: m = 2ρ − 1, ρ = 1 + m 2 . m = magnetization density ∂m ∂t = ∂ ∂x ∂m ∂x − β(1 − m2)J ∗ ∂m ∂x

• ∂m

∂t = − ∂ ∂x

• − χ(m) ∂

∂x δF(m) δm(x)

• χ(m) mobility coefficient; χ(m) = β(1 − m2).

F(m) the Lebowitz-Penrose free energy functional. 9

F(m) =

• {−m2

2 − S(m) β } + 1 4 J(x − y)

• m(x) − m(y)

2 S(m) entropy (of Bernoulli scheme) Non local version of Ginzburg-Landau functional. 10

Thermodynamics of L-P functional Thermodynamic free energy density fβ(m), m ∈ [−1, 1]: fβ(m) = CE

• − m2

2 − S(m) β

• fβ(s) = lim

Λ→Rd inf

FΛ(m) |Λ|

• −

Λ

m = s

• Figure:

fβ(m) when β < 1

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Figure: fβ(m) when β > 1.

Loss of strict convexity in the interval [−mβ, mβ] where fβ is constant indicates a phase transition. mβ = tanh{βmβ} 12

Axiomatic macroscopic theory The evolution is the conservation law: ∂m ∂t = − ∂ ∂x I Constitutive law for the current I: (gradient dynamics assumption) I = −χ(m) ∂ ∂x dfβ(m) dm

• Motion ruled by the parabolic equation

∂m ∂t = ∂ ∂x

• D(m) ∂m

∂x

• where D(m) = 0 when m ∈ [−mβ, mβ].

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Stationary states with current, the Macroscopic Theory Spatial domain: x ∈ [−1, 1]. Stationary states: m(x) solutions of ∂ ∂x

• D(m) ∂m

∂x

• = 0

(1) Dirichlet problem: fix m+ > mβ and m− < −mβ (β > 1). Select stationary state such that limx→±1 m(x) = m±, Stationary Stefan problem. Find x0 ∈ (−1, 1), j < 0 and m(x) so that D(m) ∂m ∂x = −j (−1, x0) ∪ (x0, 1) (2) any solution m(x) of (2) is also a solution of (1) and viceversa. 14

Stationary states with current, the Mesoscopic Theory Spatial domain: x ∈ ε−1[−1, 1]. Stationary states: m(x) solutions

• f

d dx dm dx − β(1 − m2)J ∗ dm dx

• = 0

(3) Dirichlet problem: fix m+ > mβ and m− < −mβ (β > 1). Select stationary state such that limx→±ε−1 m(x) = m±. Due to the convolution term we need to specify m outside ε−1[−1, 1]. We choose Neumann conditions obtained by reflecting m through the endpoints and write Jneum. Change of variables: given a function m define h as m = tanh{βJneum ∗ m + βh} Then the equation (3) becomes d dx

• χ(m)dh

dx

• = 0,

χ(m) = β(1 − m2) 15

The (j, xε) problem. Given j < 0, xε ∈ ε−1[−1, 1], find hε(x) and mε(x) so that χ(mε)dhε dx = −εj, hε(xε) = 0 (4) mε = tanh{βJneum ∗ mε + βhε} (5) 16

Theorem (DPT) The (j, xε) problem has a solution (mε, hε, xε). uε(r) := mε(ε−1r) → m(r), r ∈ (−1, 1), εxε → x0 m, j, x0 solve the macroscopic Stefan problem. 17

Outline of proofs Fixed point problem: T(h) = h where T(h) is defined by a two steps procedure:

• find m such that m = tanh{βJneum ∗ m + βh}
• Define T(h) as

T(h)(x) = −εj x

ε−1x0

χ(m)−1 x0 ∈ (−1, 1) the position of the macroscopic interface. Discuss in the sequel: easier case, x0 = 0. We can then restrict to space of antisymmetric functions. 18

Iterative scheme. Chosen h0, define hn+1 = T(hn), prove that hn → h. The limit h is the desired fixed point: h = T(h). Define m0 as :

• m0(x) = ¯

m(x), |x| ≤ c log ε−1. ¯ m(x) the instanton, ¯ m = tanh{βJ ∗ ¯ m} on R; ¯ m(x) antisymmetric, converges exponentially to mβ as x → ∞

• m0(x) = u(εx), |x| > c log ε−1, u solution of the macroscopic

problem. h0 such that m0 = tanh{βJneum ∗ m0 + βh0}. h1 = T(h0) is well defined. 19

To define T(h1) we need to find m1 = tanh{βJneum ∗ m1 + βh1}. Auxiliary dynamics. h(t) := th1 + (1 − t)h0, t ∈ [0, 1]. Suppose exists m(t) = tanh{βJneum ∗ m(t) + βh(t)}. Then Lt dm dt = −pt dh dt , pt = β(1 − m(t)2), Lt = (ptJneum∗) − 1 Theorem (Perron-Frobenius) Lt invertible as operator on the space

• f antisymmetric functions. L−1

t

bounded in the L∞ topology. Solve equations of motion dm dt = L−1

t

• − pt

dh dt

• , m(0) = m0

m1 := m(1): m1 = tanh{βJneum ∗ m1 + βh1}. 20