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Fluids andParticles:

Doodads andKinetics

PeterConstantin

Introduction

ComplexMicroscopicSystems

Examples

OnsagerEquation

Relaxationkinetics

Fluids and Particles: Doodads and Kinetics

Peter Constantin

Department of MathematicsThe University of Chicago

CIME, Cetraro September 2010

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

Outline:

1 Equilibrium: Onsager Equation on Metric Spaces

2 Kinetics: Nonlinear Fokker-Planck Equation

3 Gradient System in Metric Spaces

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

Outline:

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

Outline:

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

Major Problems

1 Modeling of interactions in the appropriate phase space

2 Relaxation mechanisms in the appropriate phase space

3 Derivation of Micro-Macro coupling

4 PDE existence theory for coupled system

5 Dimension reduction

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

Major Problems

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

Major Problems

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

Major Problems

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

Major Problems

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

• Configuration space: M = (locally) compact, separable,metric space, d distance. m ∈ M = corpus = doodad.

• Reference measure: dµ – Borel Probability on M.• Corpora state f (m)dµ(m) – Probabililty, AC w.r. dµ.• Interaction kernel k : M ×M → R+,• symmetric: k(m, p) = k(p,m)• uniformly bi-Lipschitz:

|k(m, n)− k(p, n)| ≤ Ld(m, p)

• Potential U[f ](m) =∫M k(m, p)f (p)dµ(p)

• Potential U = micro-micro interaction• Free Energy

E [f ] =

f log fdµ +1

U[f ]fdµ

• Minima of Free Energy: Onsager Equation

f = Z−1e−U[f ].

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

• Reference measure: dµ – Borel Probability on M.

• Corpora state f (m)dµ(m) – Probabililty, AC w.r. dµ.• Interaction kernel k : M ×M → R+,• symmetric: k(m, p) = k(p,m)• uniformly bi-Lipschitz:

|k(m, n)− k(p, n)| ≤ Ld(m, p)

E [f ] =

f log fdµ +1

U[f ]fdµ

f = Z−1e−U[f ].

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

• Reference measure: dµ – Borel Probability on M.• Corpora state f (m)dµ(m) – Probabililty, AC w.r. dµ.

• Interaction kernel k : M ×M → R+,• symmetric: k(m, p) = k(p,m)• uniformly bi-Lipschitz:

|k(m, n)− k(p, n)| ≤ Ld(m, p)

E [f ] =

f log fdµ +1

U[f ]fdµ

f = Z−1e−U[f ].

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

• Reference measure: dµ – Borel Probability on M.• Corpora state f (m)dµ(m) – Probabililty, AC w.r. dµ.• Interaction kernel k : M ×M → R+,

• symmetric: k(m, p) = k(p,m)• uniformly bi-Lipschitz:

|k(m, n)− k(p, n)| ≤ Ld(m, p)

E [f ] =

f log fdµ +1

U[f ]fdµ

f = Z−1e−U[f ].

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

• Reference measure: dµ – Borel Probability on M.• Corpora state f (m)dµ(m) – Probabililty, AC w.r. dµ.• Interaction kernel k : M ×M → R+,• symmetric: k(m, p) = k(p,m)

• uniformly bi-Lipschitz:

|k(m, n)− k(p, n)| ≤ Ld(m, p)

E [f ] =

f log fdµ +1

U[f ]fdµ

f = Z−1e−U[f ].

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

|k(m, n)− k(p, n)| ≤ Ld(m, p)

E [f ] =

f log fdµ +1

U[f ]fdµ

f = Z−1e−U[f ].

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

|k(m, n)− k(p, n)| ≤ Ld(m, p)

E [f ] =

f log fdµ +1

U[f ]fdµ

f = Z−1e−U[f ].

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

|k(m, n)− k(p, n)| ≤ Ld(m, p)

• Potential U = micro-micro interaction

• Free Energy

E [f ] =

f log fdµ +1

U[f ]fdµ

f = Z−1e−U[f ].

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

|k(m, n)− k(p, n)| ≤ Ld(m, p)

E [f ] =

f log fdµ +1

U[f ]fdµ

f = Z−1e−U[f ].

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

|k(m, n)− k(p, n)| ≤ Ld(m, p)

E [f ] =

f log fdµ +1

U[f ]fdµ

f = Z−1e−U[f ].

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

Examples: packing of soft spheres, low energy many particlequantum systems, ensembles of stick-and-ball models ofmolecules.

m is the system not the particle.

Goals:

1 Existence theory for solutions of Onsager’s equation

2 Classification of high intensity limits

3 Selection mechanisms for high intensity limits

4 Gradient system relaxation

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

Examples: packing of soft spheres, low energy many particlequantum systems, ensembles of stick-and-ball models ofmolecules. m is the system not the particle.

Goals:

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

Goals:

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

Goals:

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

Example: Rods, Maier-Saupe potential

M = Sn−1, dµ = area.

U[f ](p) = −b

∫Sn−1

((p · q)2 − 1

)f (q)dµ

b = intensity, inverse temperature.

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

Dimension Reduction, Maier-Saupe

n × n symmetric, traceless matrix S :

S 7→ Z (S)

Z (S) =

∫Sn−1

eb(S ijmimj )dµ.

fS(m) = (Z (S))−1eb(S ijmimj )

σ(S)ij =

∫Sn−1

(mimj −

)fS(m)dµ.

TheoremOnsager’s equation with Maier-Saupe potential is equivalent to

σ(S) = S .

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

S 7→ Z (S)

Z (S) =

∫Sn−1

eb(S ijmimj )dµ.

σ(S)ij =

∫Sn−1

(mimj −

)fS(m)dµ.

σ(S) = S .

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

S 7→ Z (S)

Z (S) =

∫Sn−1

eb(S ijmimj )dµ.

σ(S)ij =

∫Sn−1

(mimj −

)fS(m)dµ.

σ(S) = S .

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

S 7→ Z (S)

Z (S) =

∫Sn−1

eb(S ijmimj )dµ.

σ(S)ij =

∫Sn−1

(mimj −

)fS(m)dµ.

σ(S) = S .

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

Onsager Equation, Maier-Saupe n = 3.

S ij = λiδij

λi ∈ [−13 , 2

3 ],λ1 + λ2 + λ3 = 0.

2(λ1 + λ2), v2 =

2(λ1 − λ2).

y1(p) = 1− 3p2

y2(p, t) = (1− p2) cos t

for (p, t) ∈ K = [−1, 1]× [0, 2π].

y = y(p, t) = (y1(p), y2(p, t)), v = (v1, v2).

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

S ij = λiδij

λi ∈ [−13 , 2

3 ],λ1 + λ2 + λ3 = 0.

2(λ1 + λ2), v2 =

2(λ1 − λ2).

y1(p) = 1− 3p2

y2(p, t) = (1− p2) cos t

for (p, t) ∈ K = [−1, 1]× [0, 2π].

y = y(p, t) = (y1(p), y2(p, t)), v = (v1, v2).

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

S ij = λiδij

λi ∈ [−13 , 2

3 ],λ1 + λ2 + λ3 = 0.

2(λ1 + λ2), v2 =

2(λ1 − λ2).

y1(p) = 1− 3p2

y2(p, t) = (1− p2) cos t

for (p, t) ∈ K = [−1, 1]× [0, 2π].

y = y(p, t) = (y1(p), y2(p, t)), v = (v1, v2).

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

TheoremLet

Z2(v) =

ebv ·y(p,t)dpdt

F(v) = log(Z2(v))− b(3v2

1 + v22

Onsager’s equation: critical points of F , v ∈ [−13 , 2

3 ]× [0, 12 ],

i.e.: 6v1 = [y1](v)2v2 = [y2](v)

where, for any φ : K → R,

[φ](v) = (Z2(v))−1

φ(p, t)ebv ·y(p,t)dpdt

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

Limit b →∞

[φ] =

φ(m)f (m)dµ.

Isotropic:

limb→∞

[φ] =1

φ(p)dµ

Oblate:

limb→∞

[φ] =1

∫ 2π

0φ(cos ϕ, sin ϕ, 0)dϕ

Prolate:lim

b→∞[φ] = φ(m), m ∈ S2.

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

Limit b →∞

[φ] =

φ(m)f (m)dµ.

Isotropic:

limb→∞

[φ] =1

φ(p)dµ

Oblate:

limb→∞

[φ] =1

∫ 2π

Prolate:lim

b→∞[φ] = φ(m), m ∈ S2.

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

Limit b →∞

[φ] =

φ(m)f (m)dµ.

Isotropic:

limb→∞

[φ] =1

φ(p)dµ

Oblate:

limb→∞

[φ] =1

∫ 2π

Prolate:lim

b→∞[φ] = φ(m), m ∈ S2.

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

Limit b →∞

[φ] =

φ(m)f (m)dµ.

Isotropic:

limb→∞

[φ] =1

φ(p)dµ

Oblate:

limb→∞

[φ] =1

∫ 2π

Prolate:lim

b→∞[φ] = φ(m), m ∈ S2.

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

Freely Articulated N-corpora

M = M1 × · · · ×MN , dµ = Πdµj

u(p1, q1, p2, q2, . . . ) =∑

ui (pi , qi )

U[f ] =N∑

Ui [f ], with

Ui [f ](pi ) =

∫eM ui (pi , qi )f (q1, . . . qN)dµ(q)

Onsager Equation f = Z−1e−eUef

Z = ΠNj=1Zj , with Zj =

e−Uj [fj ]dµj , fj = (Zj)−1e−Uj [fj ]

f (p1, . . . pN) = f1(p1)f (p2) . . . fN(pN) product measure

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

M = M1 × · · · ×MN , dµ = Πdµj

u(p1, q1, p2, q2, . . . ) =∑

ui (pi , qi )

U[f ] =N∑

Ui [f ], with

Ui [f ](pi ) =

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

M = M1 × · · · ×MN , dµ = Πdµj

u(p1, q1, p2, q2, . . . ) =∑

ui (pi , qi )

U[f ] =N∑

Ui [f ],

Ui [f ](pi ) =

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

M = M1 × · · · ×MN , dµ = Πdµj

u(p1, q1, p2, q2, . . . ) =∑

ui (pi , qi )

U[f ] =N∑

Ui [f ], with

Ui [f ](pi ) =

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

M = M1 × · · · ×MN , dµ = Πdµj

u(p1, q1, p2, q2, . . . ) =∑

ui (pi , qi )

U[f ] =N∑

Ui [f ], with

Ui [f ](pi ) =

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

M = M1 × · · · ×MN , dµ = Πdµj

u(p1, q1, p2, q2, . . . ) =∑

ui (pi , qi )

U[f ] =N∑

Ui [f ], with

Ui [f ](pi ) =

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

M = M1 × · · · ×MN , dµ = Πdµj

u(p1, q1, p2, q2, . . . ) =∑

ui (pi , qi )

U[f ] =N∑

Ui [f ], with

Ui [f ](pi ) =

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

Example of Interacting Corpora

M = S1, M = S1 × S1.

U[f ](p1, p2) =b

∫T2 ‖e(p1) ∧ e(p2)− e(q1) ∧ e(q2)‖2f (q1, q2)dq1dq2

with e(p) = (cos p, sin p) if p ∈ [0, 2π].

‖e(p1)∧e(p2)−e(q1)∧e(q2)‖2 = (sin(p1 − p2)− sin(q1 − q2))2

Dimension reduction: Onsager’s equation f = Z−1e−U[f ]

reduces toa = [sin θ](a)

with [φ](a) =

∫ 2π0 φ(θ)g(θ)dθ

g(θ) = Z−1e−b(sin(θ)−a)2

Z =∫ 2π0 e−b(sin(θ)−a)2dθ

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

M = S1, M = S1 × S1.

U[f ](p1, p2) =b

with [φ](a) =

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

M = S1, M = S1 × S1.

U[f ](p1, p2) =b

with [φ](a) =

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

M = S1, M = S1 × S1.

U[f ](p1, p2) =b

with [φ](a) =

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

M = S1, M = S1 × S1.

U[f ](p1, p2) =b

with [φ](a) =

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

M = S1, M = S1 × S1.

U[f ](p1, p2) =b

with [φ](a) =

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

The solution is f (θ1, θ2) = g(θ1 − θ2).

φ(θ, a) = sin θ − a,

and let

[φ](b, a) =

∫ 2π0 φ(θ, a)e−bφ2(θ,a)dθ∫ 2π

0 e−bφ2(θ,a)dθ.

The Onsager equation is equivalent to

[φ](b, a) = 0.

This determines a, which in turn determines g , f .a = 0 always a solution. It yields

f0(p1, p2) = Z−1e−b sin2(p1−p2).

As b →∞ this tends to δ((p1 − p2)modπ).

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

The solution is f (θ1, θ2) = g(θ1 − θ2). Let

and let

[φ](b, a) =

[φ](b, a) = 0.

f0(p1, p2) = Z−1e−b sin2(p1−p2).

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

and let

[φ](b, a) =

[φ](b, a) = 0.

f0(p1, p2) = Z−1e−b sin2(p1−p2).

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

and let

[φ](b, a) =

[φ](b, a) = 0.

f0(p1, p2) = Z−1e−b sin2(p1−p2).

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

and let

[φ](b, a) =

[φ](b, a) = 0.

This determines a, which in turn determines g , f .

a = 0 always a solution. It yields

f0(p1, p2) = Z−1e−b sin2(p1−p2).

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

and let

[φ](b, a) =

[φ](b, a) = 0.

f0(p1, p2) = Z−1e−b sin2(p1−p2).

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

and let

[φ](b, a) =

[φ](b, a) = 0.

f0(p1, p2) = Z−1e−b sin2(p1−p2).

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

Consider

λ(a, τ) = b12

∫ 2π

0e−b(sin θ−a)2dθ

with τ = b−1.

[φ] =1

∂aλ

∂τλ =1

limτ→0

λ(a, τ) = 2√

π1√

1− a2, 0 < a < 1.

Increasing. But things are subtle, ∂λ∂a (1, τ) < 0.

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

Consider

λ(a, τ) = b12

∫ 2π

with τ = b−1.Note

[φ] =1

∂aλ

∂τλ =1

limτ→0

λ(a, τ) = 2√

π1√

1− a2, 0 < a < 1.

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

Consider

λ(a, τ) = b12

∫ 2π

[φ] =1

∂aλ

∂τλ =1

limτ→0

λ(a, τ) = 2√

π1√

1− a2, 0 < a < 1.

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

Consider

λ(a, τ) = b12

∫ 2π

[φ] =1

∂aλ

∂τλ =1

limτ→0

λ(a, τ) = 2√

π1√

1− a2, 0 < a < 1.

Increasing.

But things are subtle, ∂λ∂a (1, τ) < 0.

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

Consider

λ(a, τ) = b12

∫ 2π

[φ] =1

∂aλ

∂τλ =1

limτ→0

λ(a, τ) = 2√

π1√

1− a2, 0 < a < 1.

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

In fact, phase transition at positive τ

∂aλ((a(τ), τ) = 0

and limit limτ→0 a(τ) = 1.

Consider

hb(a) =

∫ 2π

0ue−bu2

with u(θ, a) = sin θ − a. We seek zeros of hb(a). Clearly,hb(1) < 0. ∃ 0 < a < 1 such that hb(a) > 0: Changingvariables:

hb(a) = 2

∫ 1−a

−1−ae−bu2 udu√

1− (u + a)2

hb(a) = 4∫ 1−a0 e−bu2

1−(u+a)2− 1√

1−(a−u)2

−2∫ 1+a1−a e−bu2 udu√

1−(a−u)2

= 16a∫ 1−a0

e−bu2u2du√

(1−(u+a)2)(1−(u−a)2)“√

1−(u−a)2+√

1−(u+a)2”

−2∫ 1+a1−a e−bu2 udu√

1−(a−u)2.

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

∂aλ((a(τ), τ) = 0

and limit limτ→0 a(τ) = 1. Consider

hb(a) =

∫ 2π

0ue−bu2

with u(θ, a) = sin θ − a.

We seek zeros of hb(a). Clearly,hb(1) < 0. ∃ 0 < a < 1 such that hb(a) > 0: Changingvariables:

hb(a) = 2

∫ 1−a

1− (u + a)2

hb(a) = 4∫ 1−a0 e−bu2

1−(u+a)2− 1√

1−(a−u)2

−2∫ 1+a1−a e−bu2 udu√

1−(a−u)2

= 16a∫ 1−a0

e−bu2u2du√

(1−(u+a)2)(1−(u−a)2)“√

1−(u−a)2+√

1−(u+a)2”

−2∫ 1+a1−a e−bu2 udu√

1−(a−u)2.

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

∂aλ((a(τ), τ) = 0

hb(a) =

∫ 2π

0ue−bu2

with u(θ, a) = sin θ − a. We seek zeros of hb(a).

Clearly,hb(1) < 0. ∃ 0 < a < 1 such that hb(a) > 0: Changingvariables:

hb(a) = 2

∫ 1−a

1− (u + a)2

hb(a) = 4∫ 1−a0 e−bu2

1−(u+a)2− 1√

1−(a−u)2

−2∫ 1+a1−a e−bu2 udu√

1−(a−u)2

= 16a∫ 1−a0

e−bu2u2du√

(1−(u+a)2)(1−(u−a)2)“√

1−(u−a)2+√

1−(u+a)2”

−2∫ 1+a1−a e−bu2 udu√

1−(a−u)2.

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

∂aλ((a(τ), τ) = 0

hb(a) =

∫ 2π

0ue−bu2

with u(θ, a) = sin θ − a. We seek zeros of hb(a). Clearly,hb(1) < 0.

∃ 0 < a < 1 such that hb(a) > 0: Changingvariables:

hb(a) = 2

∫ 1−a

1− (u + a)2

hb(a) = 4∫ 1−a0 e−bu2

1−(u+a)2− 1√

1−(a−u)2

−2∫ 1+a1−a e−bu2 udu√

1−(a−u)2

= 16a∫ 1−a0

e−bu2u2du√

(1−(u+a)2)(1−(u−a)2)“√

1−(u−a)2+√

1−(u+a)2”

−2∫ 1+a1−a e−bu2 udu√

1−(a−u)2.

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

∂aλ((a(τ), τ) = 0

hb(a) =

∫ 2π

0ue−bu2

with u(θ, a) = sin θ − a. We seek zeros of hb(a). Clearly,hb(1) < 0. ∃ 0 < a < 1 such that hb(a) > 0:

Changingvariables:

hb(a) = 2

∫ 1−a

1− (u + a)2

hb(a) = 4∫ 1−a0 e−bu2

1−(u+a)2− 1√

1−(a−u)2

−2∫ 1+a1−a e−bu2 udu√

1−(a−u)2

= 16a∫ 1−a0

e−bu2u2du√

(1−(u+a)2)(1−(u−a)2)“√

1−(u−a)2+√

1−(u+a)2”

−2∫ 1+a1−a e−bu2 udu√

1−(a−u)2.

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

∂aλ((a(τ), τ) = 0

hb(a) =

∫ 2π

0ue−bu2

hb(a) = 2

∫ 1−a

1− (u + a)2

hb(a) = 4∫ 1−a0 e−bu2

1−(u+a)2− 1√

1−(a−u)2

−2∫ 1+a1−a e−bu2 udu√

1−(a−u)2

= 16a∫ 1−a0

e−bu2u2du√

(1−(u+a)2)(1−(u−a)2)“√

1−(u−a)2+√

1−(u+a)2”

−2∫ 1+a1−a e−bu2 udu√

1−(a−u)2.

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

∂aλ((a(τ), τ) = 0

hb(a) =

∫ 2π

0ue−bu2

hb(a) = 2

∫ 1−a

1− (u + a)2

hb(a) = 4∫ 1−a0 e−bu2

1−(u+a)2− 1√

1−(a−u)2

−2∫ 1+a1−a e−bu2 udu√

1−(a−u)2

= 16a∫ 1−a0

e−bu2u2du√

(1−(u+a)2)(1−(u−a)2)“√

1−(u−a)2+√

1−(u+a)2”

−2∫ 1+a1−a e−bu2 udu√

1−(a−u)2.

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

The last integral is negative but exponentially small as b →∞.

The positive integral is larger than a fixed multiple of∫ δ0 u2e−bu2

du for small fixed δ (depending on a). This integral

asymptotically equals Cb−3/2 with positive C , as b →∞.Thus, for any fixed 0 < a < 1 and b large enough, we havehb(a) > 0. This proves for all large enough b the existence ofa(b) > 0 such that hb(a(b)) = 0. Moreover, because 0 < a < 1was arbitrary, this also proves limb→∞ a(b) = 1 for any sucha(b) > 0.

limb→∞

f (p1 − p2) = δ((

p1 − p2 −π

)modπ

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

The last integral is negative but exponentially small as b →∞.The positive integral is larger than a fixed multiple of∫ δ0 u2e−bu2

du for small fixed δ (depending on a).

This integral

limb→∞

f (p1 − p2) = δ((

p1 − p2 −π

)modπ

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

asymptotically equals Cb−3/2 with positive C , as b →∞.

Thus, for any fixed 0 < a < 1 and b large enough, we havehb(a) > 0. This proves for all large enough b the existence ofa(b) > 0 such that hb(a(b)) = 0. Moreover, because 0 < a < 1was arbitrary, this also proves limb→∞ a(b) = 1 for any sucha(b) > 0.

limb→∞

f (p1 − p2) = δ((

p1 − p2 −π

)modπ

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

asymptotically equals Cb−3/2 with positive C , as b →∞.Thus, for any fixed 0 < a < 1 and b large enough, we havehb(a) > 0.

This proves for all large enough b the existence ofa(b) > 0 such that hb(a(b)) = 0. Moreover, because 0 < a < 1was arbitrary, this also proves limb→∞ a(b) = 1 for any sucha(b) > 0.

limb→∞

f (p1 − p2) = δ((

p1 − p2 −π

)modπ

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

limb→∞

f (p1 − p2) = δ((

p1 − p2 −π

)modπ

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

limb→∞

f (p1 − p2) = δ((

p1 − p2 −π

)modπ

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

More degrees of freedom

M = [0, L]× [0, L]× [0, π], dµ = 1πL2 dx1dx2dθ.

U[f ](x1, x2, θ) =

(x1x2 sin(θ)− y1y2 sin(φ))2f (y1, y2, φ)dµ

The solutions of Onsager’s equation are of the form

g(x1, x2, θ) = Z−1e−b(x1x2 sin θ−a)2

with Z determined by the requirement of normalization∫M gdµ = 1, a determined by

(x1x2 sin θ)g(x1, x2, θ)dµ

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

M = [0, L]× [0, L]× [0, π], dµ = 1πL2 dx1dx2dθ.

U[f ](x1, x2, θ) =

g(x1, x2, θ) = Z−1e−b(x1x2 sin θ−a)2

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

M = [0, L]× [0, L]× [0, π], dµ = 1πL2 dx1dx2dθ.

U[f ](x1, x2, θ) =

g(x1, x2, θ) = Z−1e−b(x1x2 sin θ−a)2

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

M = [0, L]× [0, L]× [0, π], dµ = 1πL2 dx1dx2dθ.

U[f ](x1, x2, θ) =

g(x1, x2, θ) = Z−1e−b(x1x2 sin θ−a)2

with Z determined by the requirement of normalization∫M gdµ = 1,

a determined by

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

M = [0, L]× [0, L]× [0, π], dµ = 1πL2 dx1dx2dθ.

U[f ](x1, x2, θ) =

g(x1, x2, θ) = Z−1e−b(x1x2 sin θ−a)2

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

Letφ(x1, x2, θ, a) = x1x2 sin θ − a

[φ] =

φgdµ

a is determined by [φ] = 0.

λ(a, τ) = τ−1/2

e−φ2/τdµ

obeys the heat equation

∂τλ =1

with τ = b−1.

[φ] =1

2b∂a log λ.

a → 0, as b →∞.

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

[φ] =

φgdµ

λ(a, τ) = τ−1/2

e−φ2/τdµ

∂τλ =1

with τ = b−1.

[φ] =1

2b∂a log λ.

a → 0, as b →∞.

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

[φ] =

φgdµ

λ(a, τ) = τ−1/2

e−φ2/τdµ

∂τλ =1

with τ = b−1.

[φ] =1

2b∂a log λ.

a → 0, as b →∞.

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

[φ] =

φgdµ

λ(a, τ) = τ−1/2

e−φ2/τdµ

∂τλ =1

with τ = b−1.

[φ] =1

2b∂a log λ.

a → 0, as b →∞.

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

[φ] =

φgdµ

λ(a, τ) = τ−1/2

e−φ2/τdµ

∂τλ =1

with τ = b−1.

[φ] =1

2b∂a log λ.

a → 0, as b →∞.

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

[φ] =

φgdµ

λ(a, τ) = τ−1/2

e−φ2/τdµ

∂τλ =1

with τ = b−1.

[φ] =1

2b∂a log λ.

a → 0, as b →∞.

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

[φ] =

φgdµ

λ(a, τ) = τ−1/2

e−φ2/τdµ

∂τλ =1

with τ = b−1.

[φ] =1

2b∂a log λ.

a → 0, as b →∞.

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

Even More Degrees of Freedom...

V (r) nonnegative, nonincreasing, compactly supported.p = (x1, . . . xN), xi ∈ Ω ⊂ Rn.

Packing energy:

F (p) =∑i<j

V (|xi − xj |).

M = Ω× · · · × Ω ∩ F ≤ F0.

U[f ](p) =

[|F (p)− F (q)|2 + d2(p, q)

]f (q)dq

Connection to the example of freely articulated 2n doodads,jamming, perhaps...

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

V (r) nonnegative, nonincreasing, compactly supported.p = (x1, . . . xN), xi ∈ Ω ⊂ Rn. Packing energy:

F (p) =∑i<j

V (|xi − xj |).

M = Ω× · · · × Ω ∩ F ≤ F0.

U[f ](p) =

[|F (p)− F (q)|2 + d2(p, q)

]f (q)dq

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

F (p) =∑i<j

V (|xi − xj |).

M = Ω× · · · × Ω ∩ F ≤ F0.

U[f ](p) =

[|F (p)− F (q)|2 + d2(p, q)

]f (q)dq

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

F (p) =∑i<j

V (|xi − xj |).

M = Ω× · · · × Ω ∩ F ≤ F0.

U[f ](p) =

[|F (p)− F (q)|2 + d2(p, q)

]f (q)dq

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

F (p) =∑i<j

V (|xi − xj |).

M = Ω× · · · × Ω ∩ F ≤ F0.

U[f ](p) =

[|F (p)− F (q)|2 + d2(p, q)

]f (q)dq

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

TheoremFor any b > 0 there exists a solution g that minimizes theenergy:

E [g ] = minf≥0,

RM fdµ=1

E [f ]

The function g solves the Onsager equation

g(x) = (Z (b))−1e−bU(x)

Z (b) =

e−bU(x)dµ(x)

U(x) =

k(x , y)g(y)dµ(y).

The function g is normalized∫

gdµ = 1, strictly positive andLipschitz continuous.

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

E [g ] = minf≥0,

RM fdµ=1

E [f ]

g(x) = (Z (b))−1e−bU(x)

Z (b) =

e−bU(x)dµ(x)

U(x) =

k(x , y)g(y)dµ(y).

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

E [g ] = minf≥0,

RM fdµ=1

E [f ]

g(x) = (Z (b))−1e−bU(x)

Z (b) =

e−bU(x)dµ(x)

U(x) =

k(x , y)g(y)dµ(y).

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

E [g ] = minf≥0,

RM fdµ=1

E [f ]

g(x) = (Z (b))−1e−bU(x)

Z (b) =

e−bU(x)dµ(x)

U(x) =

k(x , y)g(y)dµ(y).

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

E [g ] = minf≥0,

RM fdµ=1

E [f ]

g(x) = (Z (b))−1e−bU(x)

Z (b) =

e−bU(x)dµ(x)

U(x) =

k(x , y)g(y)dµ(y).

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

E [g ] = minf≥0,

RM fdµ=1

E [f ]

g(x) = (Z (b))−1e−bU(x)

Z (b) =

e−bU(x)dµ(x)

U(x) =

k(x , y)g(y)dµ(y).

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

Idea of Proof0 ≤ U[f ](p) ≤ ‖u‖∞

|U[f ](p)− U[f ](q)| ≤ Ld(p, q)

Jensen, µ(M) = 1 ⇒ ∫M

f log fdµ ≥ 0.

Consequently:Eb[f ] ≥ 0.

Minimizing sequence fj ,

a = inff≥0,

RM fdµ=1

Eb[f ] = limj→∞

Eb[fj ].

WLOG fjdµ converge weakly to a measure dν. Uj = U[fj ]converge uniformly to a non-negative Lipschitz continuousfunction U. Then it follows that

U(p) =

k(p, q)dν(q)

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

|U[f ](p)− U[f ](q)| ≤ Ld(p, q)

f log fdµ ≥ 0.

a = inff≥0,

RM fdµ=1

Eb[f ] = limj→∞

Eb[fj ].

U(p) =

k(p, q)dν(q)

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

|U[f ](p)− U[f ](q)| ≤ Ld(p, q)

f log fdµ ≥ 0.

a = inff≥0,

RM fdµ=1

Eb[f ] = limj→∞

Eb[fj ].

U(p) =

k(p, q)dν(q)

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

|U[f ](p)− U[f ](q)| ≤ Ld(p, q)

f log fdµ ≥ 0.

a = inff≥0,

RM fdµ=1

Eb[f ] = limj→∞

Eb[fj ].

U(p) =

k(p, q)dν(q)

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

|U[f ](p)− U[f ](q)| ≤ Ld(p, q)

f log fdµ ≥ 0.

a = inff≥0,

RM fdµ=1

Eb[f ] = limj→∞

Eb[fj ].

WLOG fjdµ converge weakly to a measure dν.

Uj = U[fj ]converge uniformly to a non-negative Lipschitz continuousfunction U. Then it follows that

U(p) =

k(p, q)dν(q)

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

|U[f ](p)− U[f ](q)| ≤ Ld(p, q)

f log fdµ ≥ 0.

a = inff≥0,

RM fdµ=1

Eb[f ] = limj→∞

Eb[fj ].

U(p) =

k(p, q)dν(q)

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

We claim dν AC wr dµ, i.e. dν = gdµ with g ≥ 0 andg ∈ L1(dµ).

Indeed, this is because the sequence fjdµ isuniformly absolutely continuous. The latter is proved usingthe convexity of the function y log y and Jensen’s inequality

µ(A)−1

fj log fjdµ ≥ m log m,

where m = m(A, j) = µ(A)−1∫A fjdµ. Thus for all A, j ,

m log m ≤ C

with a fixed C ≥ 1. Let us choose R = R(A) ≥ 1 so thatR log R = C/µ(A). Then m ≤ R and so∫A fjdµ ≤ µ(A)R = C/ log R. This means∫

Afjdµ ≤ δ(µ(A))

with limx→0 δ(x) = 0 and δ(·) independent of A and j . Itfollows that ν is absolutely continuous with respect to dµ.

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

We claim dν AC wr dµ, i.e. dν = gdµ with g ≥ 0 andg ∈ L1(dµ). Indeed, this is because the sequence fjdµ isuniformly absolutely continuous.

The latter is proved usingthe convexity of the function y log y and Jensen’s inequality

µ(A)−1

m log m ≤ C

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

We claim dν AC wr dµ, i.e. dν = gdµ with g ≥ 0 andg ∈ L1(dµ). Indeed, this is because the sequence fjdµ isuniformly absolutely continuous. The latter is proved usingthe convexity of the function y log y and Jensen’s inequality

µ(A)−1

where m = m(A, j) = µ(A)−1∫A fjdµ.

Thus for all A, j ,

m log m ≤ C

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

µ(A)−1

m log m ≤ C

with a fixed C ≥ 1.

Let us choose R = R(A) ≥ 1 so thatR log R = C/µ(A). Then m ≤ R and so∫A fjdµ ≤ µ(A)R = C/ log R. This means∫

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

µ(A)−1

m log m ≤ C

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

In general, weak convergence of measures is not enough toshow lower semicontinuity of nonlinear integrals or almosteverywhere convergence.

We claim however that, in fact, theconvergence fn → g takes place strongly in L1(dµ):

limn→∞

∫M|fn(p)− g(p)|dµ(p) = 0.

In order to prove this, we prove that fn is a Cauchy sequencein L1(dµ). We take ε > 0 and choose N large enough so that

supp∈M

∣∣Un(p)− U(p)∣∣ ≤ ε2

Eb[fn] ≤ a +ε2

holds for n ≥ N.

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

In general, weak convergence of measures is not enough toshow lower semicontinuity of nonlinear integrals or almosteverywhere convergence. We claim however that, in fact, theconvergence fn → g takes place strongly in L1(dµ):

limn→∞

∫M|fn(p)− g(p)|dµ(p) = 0.

supp∈M

∣∣Un(p)− U(p)∣∣ ≤ ε2

Eb[fn] ≤ a +ε2

holds for n ≥ N.

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

limn→∞

∫M|fn(p)− g(p)|dµ(p) = 0.

In order to prove this, we prove that fn is a Cauchy sequencein L1(dµ).

We take ε > 0 and choose N large enough so that

supp∈M

∣∣Un(p)− U(p)∣∣ ≤ ε2

Eb[fn] ≤ a +ε2

holds for n ≥ N.

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

limn→∞

∫M|fn(p)− g(p)|dµ(p) = 0.

supp∈M

∣∣Un(p)− U(p)∣∣ ≤ ε2

Eb[fn] ≤ a +ε2

holds for n ≥ N.

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

limn→∞

∫M|fn(p)− g(p)|dµ(p) = 0.

supp∈M

∣∣Un(p)− U(p)∣∣ ≤ ε2

Eb[fn] ≤ a +ε2

holds for n ≥ N.

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

Let s(p) = 12(fn(p) + fm(p)) with n,m ≥ N. Then

∫M sdµ = 1,

s ≥ 0, soa ≤ Eb[s].

Therefore1

2Eb[fn] + Eb[fm] − Eb[s] ≤

On the other hand,∫M

12 (fn log fn + fm log fm)− s log s

≤ 12 Eb[fn] + Eb[fm] − Eb[s] + ε2

using U[s] = 12(U[fn] + U[fm]) so,∫

2(fn log fn + fm log fm)− s log s

dµ ≤ ε2

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

∫M sdµ = 1,

s ≥ 0, soa ≤ Eb[s].

Therefore1

≤ 12 Eb[fn] + Eb[fm] − Eb[s] + ε2

dµ ≤ ε2

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

∫M sdµ = 1,

s ≥ 0, soa ≤ Eb[s].

Therefore1

≤ 12 Eb[fn] + Eb[fm] − Eb[s] + ε2

using U[s] = 12(U[fn] + U[fm])

so,∫M

dµ ≤ ε2

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

∫M sdµ = 1,

s ≥ 0, soa ≤ Eb[s].

Therefore1

≤ 12 Eb[fn] + Eb[fm] − Eb[s] + ε2

dµ ≤ ε2

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

Denote χ = fn−fmfn+fm

and note that −1 ≤ χ ≤ 1 holds µ - a.e.Also, elementary calculations show that

2G (χ)

holds with

G (χ) = log(1− χ2) + χ log

(1 + χ

1− χ

G is even on (−1, 1), G ′(χ) = log(

1+χ1−χ

), G (0) = G ′(0) = 0

and G ′′(χ) = 21−χ2 ≥ 2 on (−1, 1). Consequently,

0 ≤ χ2 ≤ G (χ)

holds for −1 ≤ χ ≤ 1.

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

It follows that we have∫M

(fn − fm)2

fn + fmdµ ≤ ε2

Writing |fn − fm| =√

fn + fm|fn−fm|√

fn+fmand using the Schwartz

inequality we deduce ∫M|fn − fm|dµ ≤ ε.

Therefore the sequence fn is Cauchy in L1(dµ). This provesthat the weak limit fndµ → gdµ is actually strong fn → g inL1(dµ). By passing to a subsequence if necessary, we mayassume that fn → g holds also µ- a.e. Then from Fatou’sLemma, ∫

Mg log gdµ ≤ lim

j→∞

fj log fjdµ.

and thus g is a minimizer of Eb with Eb[g ] = a. It also followsthat g ≥ δ where δ > 0 is such that (x log x)′ < −3‖u‖∞ forall x ≤ δ.

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

The dad-doodad

Let M be a compact metrizable space and let k(x , y) besymmetric, bi-Lipschitz and bounded below.

In addition,assume:

k(x , x) = 0.

Theorem(C-Zlatos) Let ν be any weak limit of a sequence fndµ ofminima of the free energy E corresponding to bn →∞. Thenthere exists m ∈ M such that ν is concentrated on the level setΣ(m) = p | k(m, p) = 0.Explains the selection of prolate states.Pattern recognition example: Rhombi centered at the origin,with k the area of the symmetric difference. The dad-rhombusis the square.

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

The dad-doodad

Let M be a compact metrizable space and let k(x , y) besymmetric, bi-Lipschitz and bounded below. In addition,assume:

k(x , x) = 0.

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

The dad-doodad

k(x , x) = 0.

Theorem(C-Zlatos) Let ν be any weak limit of a sequence fndµ ofminima of the free energy E corresponding to bn →∞. Thenthere exists m ∈ M such that ν is concentrated on the level setΣ(m) = p | k(m, p) = 0.

Explains the selection of prolate states.Pattern recognition example: Rhombi centered at the origin,with k the area of the symmetric difference. The dad-rhombusis the square.

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

The dad-doodad

k(x , x) = 0.

Theorem(C-Zlatos) Let ν be any weak limit of a sequence fndµ ofminima of the free energy E corresponding to bn →∞. Thenthere exists m ∈ M such that ν is concentrated on the level setΣ(m) = p | k(m, p) = 0.Explains the selection of prolate states.

Pattern recognition example: Rhombi centered at the origin,with k the area of the symmetric difference. The dad-rhombusis the square.

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

The dad-doodad

k(x , x) = 0.

Theorem(C-Zlatos) Let ν be any weak limit of a sequence fndµ ofminima of the free energy E corresponding to bn →∞. Thenthere exists m ∈ M such that ν is concentrated on the level setΣ(m) = p | k(m, p) = 0.Explains the selection of prolate states.Pattern recognition example: Rhombi centered at the origin,with k the area of the symmetric difference.

The dad-rhombusis the square.

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

The dad-doodad

k(x , x) = 0.

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

Idea of proof: First establish

limb→∞

f >0,RM fdµ=1

E [f ]

∫ ∫k(p,q)≥ε

f (p)dµ(p)f (q)dµ(q) ≤ 2

bE [f ].

if ε2n = 2bnE [fn], 0 < εn → 0, and

Q(p, ε) = q|k(p, q) ≤ ε,

then ∫M

[∫Q(p,εn)

fn(q)dµ(q)

]dµ(p) ≥ 1− εn

∃ pn,∫Q(pn,εn)

fn(q)dµ(q) ≥ 1− 2εn. Pass to subsequencepn → p.

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

limb→∞

f >0,RM fdµ=1

E [f ]

∫ ∫k(p,q)≥ε

bE [f ].

Q(p, ε) = q|k(p, q) ≤ ε,

then ∫M

[∫Q(p,εn)

fn(q)dµ(q)

]dµ(p) ≥ 1− εn

∃ pn,∫Q(pn,εn)

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

limb→∞

f >0,RM fdµ=1

E [f ]

∫ ∫k(p,q)≥ε

bE [f ].

if ε2n = 2bnE [fn], 0 < εn → 0,

Q(p, ε) = q|k(p, q) ≤ ε,

then ∫M

[∫Q(p,εn)

fn(q)dµ(q)

]dµ(p) ≥ 1− εn

∃ pn,∫Q(pn,εn)

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

limb→∞

f >0,RM fdµ=1

E [f ]

∫ ∫k(p,q)≥ε

bE [f ].

Q(p, ε) = q|k(p, q) ≤ ε,

then ∫M

[∫Q(p,εn)

fn(q)dµ(q)

]dµ(p) ≥ 1− εn

∃ pn,∫Q(pn,εn)

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

limb→∞

f >0,RM fdµ=1

E [f ]

∫ ∫k(p,q)≥ε

bE [f ].

Q(p, ε) = q|k(p, q) ≤ ε,

then ∫M

[∫Q(p,εn)

fn(q)dµ(q)

]dµ(p) ≥ 1− εn

∃ pn,∫Q(pn,εn)

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

limb→∞

f >0,RM fdµ=1

E [f ]

∫ ∫k(p,q)≥ε

bE [f ].

Q(p, ε) = q|k(p, q) ≤ ε,

then ∫M

[∫Q(p,εn)

fn(q)dµ(q)

]dµ(p) ≥ 1− εn

∃ pn,∫Q(pn,εn)

fn(q)dµ(q) ≥ 1− 2εn.

Pass to subsequencepn → p.

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

limb→∞

f >0,RM fdµ=1

E [f ]

∫ ∫k(p,q)≥ε

bE [f ].

Q(p, ε) = q|k(p, q) ≤ ε,

then ∫M

[∫Q(p,εn)

fn(q)dµ(q)

]dµ(p) ≥ 1− εn

∃ pn,∫Q(pn,εn)

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

Principle: if a µ measure-preserving transformation Texists such that locally around p = p0,k(Tp, Tq) ≤ ck(p, q) with c < 1, then p0 cannot be adad-doodad.

If a local k-preserving transformation around p = p0 hasthe property that µ(T (B)) ≥ Cµ(B) for small balls aroundp0, with C > 1, then p0 cannot be a dad-doodad.

Explains a number of examples with multiple states.

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

Principle: if a µ measure-preserving transformation Texists such that locally around p = p0,k(Tp, Tq) ≤ ck(p, q) with c < 1, then p0 cannot be adad-doodad.If a local k-preserving transformation around p = p0 hasthe property that µ(T (B)) ≥ Cµ(B) for small balls aroundp0, with C > 1, then p0 cannot be a dad-doodad.

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

Principle: if a µ measure-preserving transformation Texists such that locally around p = p0,k(Tp, Tq) ≤ ck(p, q) with c < 1, then p0 cannot be adad-doodad.If a local k-preserving transformation around p = p0 hasthe property that µ(T (B)) ≥ Cµ(B) for small balls aroundp0, with C > 1, then p0 cannot be a dad-doodad.

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

Theorem(Zlatos-C) Let A0,A1 ⊆ M be compacts such that k(p, q) = 0for any p, q ∈ Aj ( j = 0, 1) and letBj(ε) = p ∈ M | d(p,Aj) < ε. Assume that for some εj > 0there is a 1-1 map T : B1(ε1) → B0(ε0)

such that T ,T−1 aremeasurable and there is c > 1 such that

∀p, q ∈ B1(ε1) : k(T (p),T (q)) ≤ k(p, q)

∀B ⊆ B1(ε1) measurable : µ(T (B)) ≥ cµ(B).

Assume also that for each p ∈ A1, q ∈ M \ B1(ε1) we havek(p, q) > 0.Then ν(A1) < 1 for each measure ν that is a zero temperatureweak limit of minimizers.

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

Theorem(Zlatos-C) Let A0,A1 ⊆ M be compacts such that k(p, q) = 0for any p, q ∈ Aj ( j = 0, 1) and letBj(ε) = p ∈ M | d(p,Aj) < ε. Assume that for some εj > 0there is a 1-1 map T : B1(ε1) → B0(ε0) such that T ,T−1 aremeasurable

and there is c > 1 such that

∀p, q ∈ B1(ε1) : k(T (p),T (q)) ≤ k(p, q)

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

Theorem(Zlatos-C) Let A0,A1 ⊆ M be compacts such that k(p, q) = 0for any p, q ∈ Aj ( j = 0, 1) and letBj(ε) = p ∈ M | d(p,Aj) < ε. Assume that for some εj > 0there is a 1-1 map T : B1(ε1) → B0(ε0) such that T ,T−1 aremeasurable and there is c > 1 such that

∀p, q ∈ B1(ε1) : k(T (p),T (q)) ≤ k(p, q)

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

∀p, q ∈ B1(ε1) : k(T (p),T (q)) ≤ k(p, q)

Assume also that for each p ∈ A1, q ∈ M \ B1(ε1) we havek(p, q) > 0.

Then ν(A1) < 1 for each measure ν that is a zero temperatureweak limit of minimizers.

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

∀p, q ∈ B1(ε1) : k(T (p),T (q)) ≤ k(p, q)

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

Kinetics in Riemannian Setting

M compact, connected Riemannian manifold of dimension d

with metric gαβ , g = det(gαβ), (gαβ) = (gαβ)−1, volumeelement dµ =

√gdφ in local coordinates φ. Generalized

Doi-Smoluchowski equation

∂t f = ∆g f + divg (f∇gU)

∆g Laplace-Beltrami,

divg (f∇gU) =1√

g∂α

(√ggαβf ∂βU

U[f ](p) =

k(p, q)f (q)dµ(q)

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

M compact, connected Riemannian manifold of dimension dwith metric gαβ ,

g = det(gαβ), (gαβ) = (gαβ)−1, volumeelement dµ =

divg (f∇gU) =1√

g∂α

(√ggαβf ∂βU

U[f ](p) =

k(p, q)f (q)dµ(q)

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

M compact, connected Riemannian manifold of dimension dwith metric gαβ , g = det(gαβ),

(gαβ) = (gαβ)−1, volumeelement dµ =

divg (f∇gU) =1√

g∂α

(√ggαβf ∂βU

U[f ](p) =

k(p, q)f (q)dµ(q)

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

M compact, connected Riemannian manifold of dimension dwith metric gαβ , g = det(gαβ), (gαβ) = (gαβ)−1,

volumeelement dµ =

divg (f∇gU) =1√

g∂α

(√ggαβf ∂βU

U[f ](p) =

k(p, q)f (q)dµ(q)

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

M compact, connected Riemannian manifold of dimension dwith metric gαβ , g = det(gαβ), (gαβ) = (gαβ)−1, volumeelement dµ =

√gdφ in local coordinates φ.

GeneralizedDoi-Smoluchowski equation

divg (f∇gU) =1√

g∂α

(√ggαβf ∂βU

U[f ](p) =

k(p, q)f (q)dµ(q)

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

divg (f∇gU) =1√

g∂α

(√ggαβf ∂βU

U[f ](p) =

k(p, q)f (q)dµ(q)

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

divg (f∇gU) =1√

g∂α

(√ggαβf ∂βU

U[f ](p) =

k(p, q)f (q)dµ(q)

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

divg (f∇gU) =1√

g∂α

(√ggαβf ∂βU

U[f ](p) =

k(p, q)f (q)dµ(q)

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

divg (f∇gU) =1√

g∂α

(√ggαβf ∂βU

U[f ](p) =

k(p, q)f (q)dµ(q)

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

divg (f∇gU) =1√

g∂α

(√ggαβf ∂βU

U[f ](p) =

k(p, q)f (q)dµ(q)

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

Dissipative Structure

Free Energy:

E [f ] =

log f +

2U[f ]

∂t f = divg ·(

(δE [f ]

))Lyapunov functional:

dtE [f ] = −

∣∣∣∣∇g

(δE [f ]

)∣∣∣∣2 dµ

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

Free Energy:

E [f ] =

log f +

2U[f ]

∂t f = divg ·(

(δE [f ]

dtE [f ] = −

∣∣∣∣∇g

(δE [f ]

)∣∣∣∣2 dµ

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

Free Energy:

E [f ] =

log f +

2U[f ]

∂t f = divg ·(

(δE [f ]

dtE [f ] = −

∣∣∣∣∇g

(δE [f ]

)∣∣∣∣2 dµ

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

Free Energy:

E [f ] =

log f +

2U[f ]

∂t f = divg ·(

(δE [f ]

Lyapunov functional:

dtE [f ] = −

∣∣∣∣∇g

(δE [f ]

)∣∣∣∣2 dµ

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

Free Energy:

E [f ] =

log f +

2U[f ]

∂t f = divg ·(

(δE [f ]

dtE [f ] = −

∣∣∣∣∇g

(δE [f ]

)∣∣∣∣2 dµ

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

Connection to Onsager’s equation

For Doi-Smoluchowski

δE [f ]

δf= log f + U[f ]

Time independent solutions: Onsager equation:

f = Z−1e−U[f ]

Dynamics: nontrivial. Multiple steady states, gradient system,finite dimensional attractor. Inertial Manifolds: Vukadinovic(2008-9).

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

δE [f ]

δf= log f + U[f ]

f = Z−1e−U[f ]

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

δE [f ]

δf= log f + U[f ]

Time independent solutions:

Onsager equation:

f = Z−1e−U[f ]

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

δE [f ]

δf= log f + U[f ]

f = Z−1e−U[f ]

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

δE [f ]

δf= log f + U[f ]

f = Z−1e−U[f ]

Dynamics: nontrivial. Multiple steady states, gradient system,finite dimensional attractor.

Inertial Manifolds: Vukadinovic(2008-9).

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

δE [f ]

δf= log f + U[f ]

f = Z−1e−U[f ]

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

Gradient System in Metric Space

M metric. Length space:∀m, n ∈ M ∃p ∈ M, d(p,m) = d(p, n) = 1

2d(m, n).P2(M), probability space with the Wasserstein 2 distance (it isa length space if M is).

Problem: define

∂t f = −grad(E [f ])

DeGiorgi, Ambrosio et al: Energy Dissipation Identity (EDI):

dtE [f ] ≤ −1

2|f ′(t)|2 − 1

2|∂E [f ]|2

|f ′(t)| = lims→0

d2(f (t + s), f (t))

|s|and descending slope

|∂E [f ]| = lim supg→f

(E [f ]− E [g ])+d2(f , g)

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

2d(m, n).P2(M), probability space with the Wasserstein 2 distance (it isa length space if M is). Problem: define

dtE [f ] ≤ −1

2|f ′(t)|2 − 1

2|∂E [f ]|2

|f ′(t)| = lims→0

d2(f (t + s), f (t))

(E [f ]− E [g ])+d2(f , g)

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

dtE [f ] ≤ −1

2|f ′(t)|2 − 1

2|∂E [f ]|2

|f ′(t)| = lims→0

d2(f (t + s), f (t))

(E [f ]− E [g ])+d2(f , g)

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

dtE [f ] ≤ −1

2|f ′(t)|2 − 1

2|∂E [f ]|2

|f ′(t)| = lims→0

d2(f (t + s), f (t))

(E [f ]− E [g ])+d2(f , g)

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

dtE [f ] ≤ −1

2|f ′(t)|2 − 1

2|∂E [f ]|2

|f ′(t)| = lims→0

d2(f (t + s), f (t))

(E [f ]− E [g ])+d2(f , g)

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

dtE [f ] ≤ −1

2|f ′(t)|2 − 1

2|∂E [f ]|2

|f ′(t)| = lims→0

d2(f (t + s), f (t))

and descending slope

(E [f ]− E [g ])+d2(f , g)

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

dtE [f ] ≤ −1

2|f ′(t)|2 − 1

2|∂E [f ]|2

|f ′(t)| = lims→0

d2(f (t + s), f (t))

(E [f ]− E [g ])+d2(f , g)

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

dtE [f ] ≤ −1

2|f ′(t)|2 − 1

2|∂E [f ]|2

|f ′(t)| = lims→0

d2(f (t + s), f (t))

(E [f ]− E [g ])+d2(f , g)

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

Implicit Euler

Piece-wise constant fτ (t) = fk

for t ∈ (kτ, (k + 1)τ ]

fk+1 := argmin

(E [g ] +

2τd22 (g , fk)

Limits as τ → 0 exist in general, and givegradient flows in the sense of the energy dis-sipation identity EDI.

Motivation for definition:

In metric space, kinetic equation = EDI ofE in P2(M).

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

Implicit Euler

Piece-wise constant fτ (t) = fk for t ∈ (kτ, (k + 1)τ ]

fk+1 := argmin

(E [g ] +

2τd22 (g , fk)

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

Implicit Euler

fk+1 := argmin

(E [g ] +

2τd22 (g , fk)

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

Implicit Euler

fk+1 := argmin

(E [g ] +

2τd22 (g , fk)

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

Implicit Euler

fk+1 := argmin

(E [g ] +

2τd22 (g , fk)

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

Implicit Euler

fk+1 := argmin

(E [g ] +

2τd22 (g , fk)

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

Evolutional Variational Inequalities

In the convex energy, Hilbert space case: well understood.

Assume E [f ] = E [f ] + Q[f ] with E convex andQ[f ] = 1

2(Lf , f ), with L bounded, selfadjoint in Hilbert spaceH.Definition: Evolutional Variational Inequality (EVI)

dt‖f (t)−g‖2 ≤ E [g ]−E [f (t)]−Q[f (t)−g ], ∀g

recover classical definition for Q = 0. Interesting for L notpositive. Uniqueness of solutions of initial value problem canbe obtained in the non-convex, Hilbert case.

• Open: Define Evolutional Variational Inequality EVI fornon-convex energies in length spaces.

• Open: EDI implies uniqueness of solutions of initial valueproblem in the nonconvex case in length spaces.

• Open: Time asymptotics to Onsager solution, in P2(M).

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

In the convex energy, Hilbert space case: well understood.Assume E [f ] = E [f ] + Q[f ]

with E convex andQ[f ] = 1

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

In the convex energy, Hilbert space case: well understood.Assume E [f ] = E [f ] + Q[f ] with E convex and

Q[f ] = 12(Lf , f ), with L bounded, selfadjoint in Hilbert space

H.Definition: Evolutional Variational Inequality (EVI)

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

In the convex energy, Hilbert space case: well understood.Assume E [f ] = E [f ] + Q[f ] with E convex andQ[f ] = 1

2(Lf , f ),

with L bounded, selfadjoint in Hilbert spaceH.Definition: Evolutional Variational Inequality (EVI)

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

2(Lf , f ), with L bounded, selfadjoint

in Hilbert spaceH.Definition: Evolutional Variational Inequality (EVI)

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

2(Lf , f ), with L bounded, selfadjoint in Hilbert spaceH.

Definition: Evolutional Variational Inequality (EVI)

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

recover classical definition for Q = 0.

Interesting for L notpositive. Uniqueness of solutions of initial value problem canbe obtained in the non-convex, Hilbert case.

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

recover classical definition for Q = 0. Interesting for L notpositive.

Uniqueness of solutions of initial value problem canbe obtained in the non-convex, Hilbert case.

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

Doodads andKinetics

PeterConstantin

Introduction

Examples

OnsagerEquation

Relaxationkinetics

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