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Proof of Irreversibility for Systems with Time

Reversal Symmetry

Ken-ichi Okubo1 and Ken Umeno2

Department of Applied Mathematics and Physics, Graduate School of Informatics,

Kyoto University, Yoshida-honmachi, Sakyo-ku, Kyoto 606-8501, Japan

E-mail: 1okubo.kenichi.65z@st.kyoto-u.ac.jp

E-mail: 2umeno.ken.8z@kyoto-u.ac.jp

March 2020

Abstract. The positivity of entropy change rate was rigorously proven in a certain

microscopic system associated with a Hamiltonian using only information about the

microscopic system. This microscopic system obeys an equation with time-reversal

symmetry. We proved that a symplectic map with time-reversal symmetry has the

same ergodic properties as an Anosov diffeomorphism. This result guarantees that any

initial density function defined except for a zero volume set converges to the unique

invariant density (uniform distribution) in the sense of mixing. Moreover, we proved

that the Kolmogorov–Sinai entropy, which is the entropy change rate, is positive using

the property of Anosov diffeomorphisms. In addition, we discovered a mathematical

structure connecting the convergence of initial density functions to the uniform

distribution and positivity of the entropy change rate (Anosov diffeomorphism).

1. Introduction

The problem of time’s arrow is considered controversial because it claims that

unidirectional time evolution occurs on a macroscopic scale although a microscopic

system obeys an equation with time-reversal symmetry. In discussions about time

irreversibility, it is important to specify time scales and which quantity evolves

irreversibly. In the 19th century, Boltzmann clarified the irreversible behavior of an

H-function by assuming the molecular chaos hypothesis [1, 2]. Based on Boltzmann’s

H-theorem, Loschmidt [3] claimed that in a system with time-reversal symmetry, the

time derivative of the entropy would be negative in a backward motion. In addition,

Zermelo [4] asserted that the value of the entropy would return to the initial value

by quoting Poincare’s recurrence theorem. Kac showed how this irreversible behavior

occurs in the Ehrenfest model when the initial condition is far from the equilibrium

state using the mean recurrence time [5, 6]. In the Ehrenfest model, the state of the

system evolves stochastically. In the 1970s, for nonequilibrium systems in stationary

and chaotic states, the idea of the SRB distribution was proposed by Sinai, Ruelle, and

Proof of Irreversibility for Systems with Time Reversal Symmetry 2

Bowen through mixing Anosov diffeomorphisms, in which it was proved that there is

a unique ergodic distribution [2, 7, 8]. If one proves that a dynamical system has an

SRB measure, then it is guaranteed that initial density functions converge to the SRB

density function (unique ergodic density function)[8, 9] in the sense of mixing. Using

the property of the Anosov diffeomorphisms and SRB measure, one obtains the positive

Kolmogorov–Sinai (KS) entropy hKS. It is known that for Anosov systems, by coarse-

graining, the Boltzmann entropy can be denoted as a linear function in terms of time

whose slope is represented by the Kolmogorov–Sinai entropy for certain time ranges by

assuming appropriate conditions [10, 11, 12].

As an example of a physical model having an SRB measure (a mixing invariant

density function), Sinai [13, 14] proved the hyperbolicity, mixing property, and ergodicity

of dynamical systems with scatterers (Sinai billiards)[15]. Bunimovich [14, 16] also

proved these properties in a stadium (Bunimovich stadium). Another example is the

Lorentz gas where a periodic continuation of Sinai billiards’ boundary is applied without

any walls [17].

The Sinai billiards, Bunimovich stadium and Lorentz gas are popular chaotic and

mixing dynamical models that have scatterers or walls. In these models, the properties of

the dynamical systems such as mixing are caused by the existence of obstacles, scatterers,

and their shapes and positions. Then, how about dynamical systems without obstacles,

scatterers, or walls? For symplectic maps, the sawtooth map [11, 18], Arnold’s cat map

[19] and multi-baker’s map [20] have been known as mixing symplectic maps without

scatterers. However, these three maps do not have time-reversal symmetry in that they

are not invariant under an operation R1 such as (p, q) 7→ (−p, q).

It is also important to distinguish time-reversal symmetry from reversibility. If a

map T has reversibility, it only means that the map T has an inverse T−1. However, in

the case of time-reversal symmetry, the form of R−11 ◦ T−1 ◦ R1 needs to be consistent

with that of T . Then, we can say that the standard map [11, 21, 22, 23] and tangent map

[24] do not have time-reversal symmetry although they are reversible. Table 1 compares

systems satisfying the conditions of paradoxes proposed by Loschmidt and Zermelo.

The condition of paradox proposed by Loschmidt is that the system has time-reversal

symmetry in the sense of an operator R1 : (p, q) 7→ (−p, q).

In this paper, our maps derived from a Hamiltonian have time-reversal symmetry

in the sense of an operator R1 : (p, q) 7→ (−p, q) without scatterers or walls, which is

different from the previous studies in which following maps or systems are used such as

maps without time-reversal symmetry or systems with scatterers or walls.

In this study, a “macroscopic unidirection” is defined by the convergence of initial

density functions to the uniform distribution and positivity of the entropy rate on the

time scale of relaxation. We propose a symplectic map with time-reversal symmetry

derived from a Hamiltonian without scatterers or walls. In our model, for certain

parameter ranges we proved that any initial density function defined except for a zero

volume set relaxes uniquely to the uniform distribution and that the system is mixed

using only information about the microscopic dynamical system [8] and the KS entropy

Proof of Irreversibility for Systems with Time Reversal Symmetry 3

Table 1: Comparison among systems satisfying the conditions of Loschmidt and Zermelo.

Tε,∆τ , Tε,∆τstandard map tangent map multi-baker’s map

(proposed in this paper)

Reversibility ◦ ◦ ◦ ◦Loschmidt (time-reversal

◦ (satisfied) × (not satisfied) × ×symmetry in the sense of

R1 : (p, q) 7→ (−p, q))

Zermelo ◦ ◦ ◦ ◦(measure preserving)

is positive. While proving these properties, we showed that the maps have Anosov

properties.

2. Mechanics

Let us introduce our Hamiltonian

H(p1, p2, q1, q2) ≡p212

+p222

+ V (q1, q2), (1)

V (q1, q2) ≡ − ε

πlog |cos {π(q1 − q2)}| . (2)

where ε ∈ R is a perturbation parameter, p1, p2 ∈ R and q1, q2 ∈ IδN,∆τ≡(

−12+

δN,∆τ

π, 12− δN,∆τ

π

]. On IδN,∆τ

, we identify −12+

δN,∆τ

πwith 1

2− δN,∆τ

π. We chose

δN,∆τ > 0 such that

|2(1− 2δN,∆τ

π

)− 2ε(∆τ)2

[tan(π2− δN,∆τ

)− tan

(−π

2+ δN,∆τ

)]|

(1− 2δN,∆τ

π

) = N ∈ N. (3)

Figures 1 and 2 show the shape of the potential V (q1, q2) when δN,∆τ is small enough at

ε = 2.0,−2.0, respectively.

Then we consider the following map Tε,∆τ obtained using the leap frog method

(second order time-reversal symmetric symplectic integrator) for the canonical equations

of motion associated with (1) .

Tε,∆τ : (R2 × I2δN,∆τ

)\Γ → (R2 × I2δN,∆τ)\Γ, (4)

p1(n+ 1)

p2(n+ 1)

q1(n+ 1)

q2(n+ 1)

=

p1(n)− ε∆τ tan[π{q1(n) +

∆τ2 p1(n)− q2(n)− ∆τ

2 p2(n)}]

p2(n) + ε∆τ tan[π{q1(n) +

∆τ2 p1(n)− q2(n)− ∆τ

2 p2(n)}]

q1(n) + p1(n)∆τ − ε(∆τ)2

2 tan[π{q1(n) +

∆τ2 p1(n)− q2(n)− ∆τ

2 p2(n)}]

mod IδN,∆τ

q2(n) + p2(n)∆τ + ε(∆τ)2

2 tan[π{q1(n) +

∆τ2 p1(n)− q2(n)− ∆τ

2 p2(n)}]

mod IδN,∆τ

,(5)

where ∆τ represents a step size. The definition of Γ is a set {(p1, p2, q1, q2)} in

which there exists an integer n such that T nε,∆τ(p1, p2, q1, q2) = (p′1, p′2, q

′1, q

′2), where

q′1 + (∆τ/2)p′1 − q′2 − (∆τ/2)p′2 = (2m + 1)/2, m ∈ Z. It then turns out that the

Lebesgue measure of Γ is zero. The definition of “ mod IδN,∆τ” is as follows.

Proof of Irreversibility for Systems with Time Reversal Symmetry 4

-0.50

0.5

-0.50

0.5

0

5

10

V(q1, q2)

q1q2

0246810

Figure 1: Shape of the periodic

potential V (q1, q2) for ε = 2.0

(repulsive force occurs).

-0.50

0.5

-0.50

0.5

-10

-5

0

V(q1, q2)

q1q2

-10-8-6-4-20

Figure 2: Shape of the periodic

potential V (q1, q2) for

ε = −2.0 (attractive force

occurs).

Definition 1 (mod IδN,∆τ). We define an operation mod IδN,∆τ

: R → IδN,∆τto be

x mod IδN,∆τ≡ x− n

(1− 2

πδN,∆τ

),

where − 12+

δN,∆τ

π+ n

(1− 2

πδN,∆τ

)< x ≤ 1

2− δN,∆τ

π+ n

(1− 2

πδN,∆τ

), n ∈ Z.

(6)

In this study, a map T was called to esnure time-reversal symmetry when the form

of R1 ◦T was the same as that of T−1 ◦R1 by changing the sign of the momentum. The

mathematical definition is provided in Section 3. That is, if one chooses one of the two

time directions as the “forward direction” and when entropy increases in that direction,

then the entropy decreases in the “inverse direction”. However, for systems with time-

reversal symmetry, this time evolution in the “inverse direction” occurs as time evolution

in the “forward direction” through the operator R1. In our study, we distinguished

between “reversibility” and “time-reversal symmetry”. The following standard map can

be illustrated as such an example as it does not have time-reversal symmetry despite

being reversible.

Tstandard

(pnxn

)=

(pn −K sin xnxn + pn+1

). (7)

The inverse map is such that

T−1standard

(pn+1

xn+1

)=

(pn+1 −K sin(xn+1 − pn+1)

xn+1 − pn+1

). (8)

R1 ◦ Tstandard is not consistent with T−1standard ◦ R1. Then, the standard map does not

satisfy the condition of Loschmidt (time-reversal symmetry in the sense of an operation

R1 : (p, q) 7→ (−p, q)).

In the case of tangent map [24], the time evolution is denoted as

Ttangent

(pnxn

)=

(pn +Kxnxn + c tan

{12(pn +Kxn)

}). (9)

Proof of Irreversibility for Systems with Time Reversal Symmetry 5

The inverse map is such that

T−1tangent

(pn+1

xn+1

)=

(−Kxn+1 + pn+1 +Kc tan

(pn+1

2

)

xn+1 − c tan(pn+1

2

)). (10)

Then, the tangent map does not satisfy the condition of Loschmidt as R1 ◦ Ttangent is

not consistent with T−1tangent ◦R1.

In the case of multi-baker’s map [20],

B(n, x, y) =

{ (n− 1, 2x, y

2

), 0 ≤ x < 1

2,(

n+ 1, 2x− 1, y+12

), 1

2< x < 1,

(11)

the inverse map is given by

B−1(n, x, y) =

{ (n+ 1, x

2, 2y), 0 ≤ y < 1

2,(

n− 1, x+12, 2y − 1

), 1

2< y < 1.

(12)

Note that there is an involution R′ instead of R1 such that

R′(n, x, y) ≡ (n, 1− y, 1− x), (13)

and it satisfies R′ ◦B = B−1 ◦R′. However, in multi-baker’s map, one cannot determine

which variable has a momentum dimension between x and y. That means we cannot

determine which variable represents momentum among x and y. Then, we can say that

multi-baker’s map does not have a pair of momentum and position (p, q) and this map

does not express the equation of motion directly. Thus, in this study, multi-baker’s map

is not considered to have time-reversal symmetry in the sense of Loschmidt (in the sense

of changing the sign of the momentum p → −p).

Similarly, Arnold’s cat map does not have time-reversal symmetry although they

are reversible. On the other hand, the map Tε,∆τ has time-reversal symmetry in terms

of p1 → −p1, p2 → −p2. The proof of this time-reversal symmetry is provided in Section

3.

The condition of paradox proposed by Zermelo is that the system satisfies the

condition of Poincare’s recurrence theorem. That is, the system needs to preserve the

invariant measure [25, 26]. For example, our map Tε,∆τ , Hamiltonian systems, Arnold’s

cat map [19], multi-baker’s map [19, 20], the standard map [11, 21, 22, 23] and tangent

map [24] preserve the Lebesgue measure. Therefore, these systems satisfy the condition

proposed by Zeromelo. Therefore, our map Tε,∆τ satisfies the conditions of paradoxes

proposed by both Loschmidt and Zermelo.

To prove the convergence of the initial density functions to the uniform distribution

and positivity of the entropy rate, we have to show that the dynamical system has

Anosov properties [8]. To this end, we constructed other forms of the symplectic map

denoted as Tε,∆τ and Tε,∆τ derived from Tε,∆τ to shorten the proof. For the Hamiltonian

H , maps Tε,∆τ , Tε,∆τ and Tε,∆τ , the following diagram holds:

H // Tε,∆τ // Tε,∆τ // Tε,∆τ

Proof of Irreversibility for Systems with Time Reversal Symmetry 6

Let us define the map Tε,∆τ . In the map Tε,∆τ , there are two relations such that

p1(n) + p2(n) = C ≡ p1(0) + p2(0),

q1(n) + q2(n) = q0 + nC,(14)

where q0 ≡ q1(0) + q2(0). Then, by eliminating p2 and q2 using the above equation, one

defines the map Tε,∆τ as follows.

Definition 2 (Tε,∆τ). A symplectic map Tε,∆τ :M′ ≡(R× IδN,∆τ

)\Γ′ →M ′ is defined

as(

p1(n+ 1)

q1(n+ 1)

)=

(p1(n)− ε(∆τ) tan

[π{2q1(n) + ∆τp1(n)− q0 − C∆τ

(n+ 1

2

)}]

q1(n) + p1(n)∆τ − ε(∆τ)2

2 tan[π{2q1(n) + ∆τp1(n)− q0 − C∆τ

(n+ 1

2

)}]mod IδN,∆τ

)(15)

where Γ′ represents a set {(p1(n), q1(n))} ⊂ (R×IδN,∆τ) such that there exists an integer

k that satisfies the condition

T kε,∆τ(p1(n), q1(n)) = (p1(n + k), q1(n + k)), (16)

where 2q1(n+ k) + ∆τp1(n + k)− q0 − C∆τ(n+ k + 1

2

)= 2m+1

2, m ∈ Z.

One can reconstruct the map Tε,∆τ using Tε,∆τ and the following relations:{p1(n) + p2(n) = C ≡ p1(0) + p2(0),

q1(n) + q2(n) = q0 + nC,(17)

where q0 ≡ q1(0)+ q2(0). The time-reversal symmetry of Tε,∆τ is explained in Section 3.Then, using the transformation H defined by(

sntn

)= H

(p1(n)

q1(n)

)=

(∆τp1(n) + 2q1(n)− q0 − C∆τ (n+ 1/2) mod IδN,∆τ

J(p1(n), q1(n))− q0 − C∆τ (n+ 3/2) mod IδN,∆τ

), (18)

where J(p1(n), q1(n)) ≡ ∆τp1(n + 1) + 2q1(n + 1), the map Tε,∆τ : M ≡ I2δN,∆τ→ M

was obtained from Tε,∆τ as follows.

Definition 3 (Tε,∆τ ). Let us define a symplectic map Tε,∆τ :M →M as(sn+1,

tn+1

)=

(g(tn)

h(sn, tn)

)=

(tn

2tn − sn + 2ε(∆τ)2 tan(πtn) mod IδN,∆τ

). (19)

where xn ≡ (sn, tn) ∈ I2δN,∆τ.

The symplectic map Tε,∆τ is topologically semi-conjugate with Tε,∆τ using a

transformation H . For the symplectic maps Tε,∆τ , Tε,∆τ and Tε,∆τ the following diagram

holds:(R

2 × I2δN,∆τ

)\Γ Tε,∆τ−−−→

(R

2 × I2δN,∆τ

)\Γ

(R× IδN,∆τ

)\Γ′ Tε,∆τ−−−→

(R× IδN,∆τ

)\Γ′

H

yyH

I2δN,∆τ

Tε,∆τ−−−→ I2δN,∆τ

Proof of Irreversibility for Systems with Time Reversal Symmetry 7

The Jacobian of the map Tε,∆τ is given by

J(xn) =

(0 1

−1 2− 2πε(∆τ)2

cos2(πtn)

), (20)

and it is not a constant matrix. The case of constant matrices was studied by Franks

[27]. The linear instability condition (for at least one eigenvalue of the Jacobian, its

absolute value is larger than unity) for the map Tε,∆τ is straightforwardly obtained as

ε < 0 or2

π(∆τ)2< ε, (21)

and this is related to the condition that the map is an Anosov diffeomorphism (Anosov

system).

As the map Tε,∆τ is symplectic, Tε,∆τ preserves the Lebesgue measure µ on I2δN,∆τ.

3. Time reversal symmetry

In this section, it is shown that the symplectic maps, Tε,∆τ and Tε,∆τ , used in this work

have time-reversal symmetry.

Definition 4. Let M be either(R

2 × I2δN,∆τ

)\Γ or M ′ and f : M → M be a map. It

is said that map f has time-reversal symmetry when there exists operator R1 such that

R1 ◦ f = f−1 ◦R1, (22)

where R1 satisfies R1(p, q) = (−p, q) [28].

When there is operator R1, the form of f−1◦R1 is equivalent to that of R1◦f . Therefore,any orbit obtained by {f−n}n∈Z is also obtained by {R1 ◦ fn ◦R−1

1 }n∈Z.For any point (p(n), q(n)), it holds that

R1 ◦ Tε,∆τ

p1(n)

p2(n)

q1(n)

q2(n)

=

−p1(n) + ε∆τ tan[π{q1(n) +

∆τ2 p1(n)− q2(n)− ∆τ

2 p2(n)}]

−p2(n)− ε∆τ tan[π{q1(n) +

∆τ2 p1(n)− q2(n)− ∆τ

2 p2(n)}]

q1(n) + p1(n)∆τ − ε(∆τ)2

2 tan[π{q1(n) +

∆τ2 p1(n)− q2(n)− ∆τ

2 p2(n)}]

mod IδN,∆τ

q2(n) + p2(n)∆τ + ε(∆τ)2

2 tan[π{q1(n) +

∆τ2 p1(n)− q2(n)− ∆τ

2 p2(n)}]

mod IδN,∆τ

.

(23)

One can also obtain

T−1ε,∆τ ◦R1

p1(n)

p2(n)

q1(n)

q2(n)

=

−p1(n) + ε∆τ tan[π{q1(n) +

∆τ2 p1(n)− q2(n)− ∆τ

2 p2(n)}]

−p2(n)− ε∆τ tan[π{q1(n) +

∆τ2 p1(n)− q2(n)− ∆τ

2 p2(n)}]

q1(n) + p1(n)∆τ − ε(∆τ)2

2 tan[π{q1(n) +

∆τ2 p1(n)− q2(n)− ∆τ

2 p2(n)}]

mod IδN,∆τ

q2(n) + p2(n)∆τ + ε(∆τ)2

2 tan[π{q1(n) +

∆τ2 p1(n)− q2(n)− ∆τ

2 p2(n)}]

mod IδN,∆τ

.

(24)

Therefore, from Definition 4, (23), and (24), it holds that map Tε,∆τ has time-

reversal symmetry.

Proof of Irreversibility for Systems with Time Reversal Symmetry 8

In the case of Tε,∆τ , the inverse, T−1ε,∆τ , is defined by

T−1ε,∆τ

(p1(n+ 1)

q1(n+ 1)

)

=

(p1(n+ 1) + ε∆τ tan

[π{2q1(n+ 1)−∆τp1(n+ 1)− q0 − C∆τ

(n+ 1

2

)}],

q1(n+ 1)− p1(n+ 1)∆τ − ε(∆τ)2

2 tan[π{2q1(n+ 1)−∆τp1(n+ 1)− q0 − C∆τ

(n+ 1

2

)}]).

(25)

Then, it holds that

R1 ◦ Tε,∆τ

(p1(n)

q1(n)

)

=

(−p1(n) + ε(∆τ) tan

[π{2q1(n) + ∆τp1(n)− q0 − C∆τ

(n+ 1

2

)}]

q1(n) + p1(n)∆τ − ε(∆τ)2

2 tan[π{2q1(n) + ∆τp1(n)− q0 − C∆τ

(n+ 1

2

)}]).

(26)

One can also obtain

T−1ε,∆τ ◦R1

(p1(n)

q1(n)

)

=

(−p1(n) + ε(∆τ) tan

[π{2q1(n) + ∆τp1(n)− q0 − C∆τ

(n+ 1

2

)}]

q1(n) + p1(n)∆τ − ε(∆τ)2

2 tan[π{2q1(n) + ∆τp1(n)− q0 − C∆τ

(n+ 1

2

)}]).

(27)

Therefore, from Definition 4, (26), and (27), it holds that map Tε,∆τ has time-

reversal symmetry.

In terms of convergence of the initial density function to the equilibrium denisty

function, when a dynamical system is exact, any initial density function converges to

the unique equilibrium function in the sense of L1 norm [29]. In the Appendix, we show

that the exactness is not compatible with time-reversal symmetry.

4. Compactness and diffeomorphism

In this section, it is shown that the domainM is compact and Tε,∆τ is a diffeomorphism

on M .

Domain IδN,∆τis homeomorphic to circle S because we identify −1

2+

δN,∆τ

πwith

12− δN,∆τ

π. Then, domainM is topologically conjugate with torus T2; thus,M is compact.

Next, we show that map Tε,∆τ is a diffeomorphism on M . To this end, one must show

that map Tε,∆τ is surjective, injective, and smooth and that its inverse is continuous.

(Proof of surjective): For any point (sn+1, tn+1) ∈ M , there is a point, (sn, tn) ∈ M ,

such that {sn = 2sn+1 + 2ε(∆τ)2 tan(πsn+1)− tn+1 mod IδN,∆τ

,

tn = sn+1.(28)

Thus, map Tε,∆τ is surjective.

(Proof of injective): We show that map Tε,∆τ is injective by contradiction. Let (sn, tn)

and (s′n, t′n) be two points that cannot be identified on M . If one assumes that image

(sn+1, tn+1) is identified with (s′n+1, t′n+1), then it holds that

sn+1 = s′n+1,

Thus, tn = t′n. (29)

Proof of Irreversibility for Systems with Time Reversal Symmetry 9

Based on the fact that 2t + 2ε(∆τ)2 tan(πt) is finite for t ∈ IδN,∆τand (29), one

obtains the relation sn = s′n. These results are in contradiction to the condition

(sn, tn) 6= (s′n, t′n). Therefore, map Tε,∆τ is injective.

(Proof of smoothness): It is evident that g, h, ∂g∂t, ∂h

∂s, and ∂h

∂tdefined in (19) are

continuous on M except for (s, t), where s = 12− δN,∆τ

πor t = 1

2− δN,∆τ

π. Let us

discuss the smoothness at 12− δN,∆τ

π. It is evident that g(t) = t and ∂g

∂tare continuous

at t = 12− δN,∆τ

πand that h(s, t) = 2t + 2ε(∆τ)2 tan(πt) − s and ∂h

∂sare continuous at

s = 12− δN,∆τ

πin a similar manner.

Let us discuss the continuity of h(s, t) and ∂h∂t

at t = 12− δN,∆τ

π. From the definition of

δN,∆τ , one obtains∣∣∣h(s,−1

2+

δN,∆τ

π

)− h

(s, 1

2− δN,∆τ

π

)∣∣∣ = N(1− 2δN,∆τ

π

),

Thus,∣∣∣h(s,−1

2+

δN,∆τ

π

)− h

(s, 1

2− δN,∆τ

π

)∣∣∣ mod IδN,∆τ= 0.

Then, h(s, t) is continuous at t = 12− δN,∆τ

πon torus M . As h(s, t) is an odd function

with t, one obtains ∂h∂t

(s,−1

2+

δN,∆τ

π

)= ∂h

∂t

(s, 1

2− δN,∆τ

π

). Thus, ∂h

∂tis continuous at

t = 12− δN,∆τ

π. Map T−1

ε,∆τ is also continuous on M .

Therefore, map Tε,∆τ is a diffeomorphism on M .

5. Mixing Anosov Property

The following theorems give the proof of convergence of initial density functions to the

uniform distribution and positivity of the entropy rate for certain parameter ranges.

Theorem A. Suppose that the condition ε < 0 or 2π(∆τ)2

< ε is satisfied. Map Tε,∆τ on

M is an Anosov diffeomorphism.

Theorem B. Suppose that the condition ε < 0 or 2π(∆τ)2

< ε is satisfied. The map Tε,∆τon M ′ has the same ergodic properties as an Anosov diffeomorphism.

The definition of an Anosov diffeomorphism [2, 8, 19, 26, 27] and the proofs of these

theorems are given as follows. Although Gallavotti assumed the chaotic hypothesis

[2, 8] in many-particle systems, we proved directly that the systems with few degrees of

freedom have Anosov properties when the condition (21) is satisfied.

5.1. Anosov diffeomorphism and mixing property

In this subsection, we prove that map Tε,∆τ defined by (19) is an Anosov diffeomorphism

for ε < 0 or 2π(∆τ)2

< ε. A point, (sn, tn), ofM is denoted as xn = (sn, tn). The Jacobian

of map Tε,∆τ is given by

J(xn) =

(0 1

−1 2− 2πε(∆τ)2

cos2(πtn)

), (30)

where the components of Jacobian J(xn) are on R. An Anosov diffeomorphism defined

according to [2, 8, 19, 26, 27] is given below.

Proof of Irreversibility for Systems with Time Reversal Symmetry 10

Definition 1 (Anosov diffeomorphism). Let M∗ be a compact manifold, ‖ · ‖ be a

norm, and f : M∗ → M∗ be a diffeomorphism. Consider the tangent space, TxM∗, at

each point x ∈ M∗. f is an Anosov diffeomorphism if it satisfies all of the following

conditions:

(I) There exist subspaces Eux and Es

x such that TxM∗ = Eux ⊕ Es

x,

(II) (Dxf)Eux = Eu

f(x), (Dxf)Esx = Es

f(x),

(III) There exist K > 0 and 0 < λ < 1 determined by x and not by ξ or n such that

ξ ∈ Eux =⇒ ‖(Dfn)ξ‖ ≥ K

(1

λ

)n‖ξ‖, (31)

ξ ∈ Esx =⇒ ‖(Df−n)ξ‖ ≥ K

(1

λ

)n‖ξ‖, (32)

The linear instability condition (for at least one eigenvalue of matrix (30), this

eigenvalue’s absolute value is larger than unity) for map Tε,∆τ is

ε < 0 or2

π(∆τ)2< ε. (33)

The linear instability condition is related to the condition that map Tε,∆τ is an Anosov

diffeomorphism for map Tε,∆τ .

We prove that map Tε,∆τ is an Anosov diffeomorphism for ε < 0, 2π(∆τ)2

< ε. This proof

is based on the idea used in [11]. The Jacobians for Arnold’s cat map and multi-baker’s

map are constant. However, the Jacobians for our maps are not constant for any point,

and the aforementioned maps are extremely ideal compared with our maps. Thus, it

is more difficult to prove that our maps are Anosov diffeomorphisms, as compared to

Arnold’s cat map and multi-baker’s map. To prove this, conditions (I), (II), and (III) in

Definition 1 should be satisfied. In order to prove that these conditions are satisfied, we

first introduce the definitions of cones L+ and L− and then show the following Lemmas:

Definition 2. For any two-dimensional nonzero vector on TxnM,xn ∈M,n ∈ Z,

an = (a1(n), a2(n)) 6= O,

we define cones L+ and L− as

L+(xn) = {(a1(n), a2(n)); ‖J(xn)an‖ > ‖an‖},L−(xn) = {(a1(n), a2(n)); ‖J(T−1

ε,∆τxn)−1an‖ > ‖an‖},

where ‖ · ‖ represents ‖a(n)‖ ≡√a21(n) + a22(n).

Lemma 3. When the condition ε < 0, 2π(∆τ)2

< ε is satisfied, it holds that

J(xn)an ∈ L+(Tε,∆τxn),∀an ∈ L+(xn), (34)

This Lemma corresponds to condition (II).

Proof of Irreversibility for Systems with Time Reversal Symmetry 11

Proof. It follows from simple calculation that an /∈ L+(xn) and an /∈ L−(xn) when

a2(n) = 0 and a1(n) = 0, respectively. Consider the case where a2(n) 6= 0 and a1(n) 6= 0.

The condition an ∈ L+(xn) in (34) is expressed as

an ∈ L+(xn) ⇔{

a1(n)a2(n)

> 1− πε(∆τ)2

cos2(πtn), ε > 2

π(∆τ)2,

a1(n)a2(n)

< 1− πε(∆τ)2

cos2(πtn), ε < 0.

(35)

Selecting an+1 as J(xn)an = (a1(n + 1), a2(n + 1)), one can rewrite the condition

an+1 ∈ L+(Tε,∆τxn) in (34) as

an+1 ∈ L+(Tε,∆τxn) = L+(xn+1) ⇔{

a1(n+1)a2(n+1)

> 1− πε(∆τ)2

cos2(πtn+1), ε > 2

π(∆τ)2,

a1(n+1)a2(n+1)

< 1− πε(∆τ)2

cos2(πtn+1), ε < 0.

(36)

To prove (34), we show below that (35) implies (36).

Substituting a1(n + 1) = a2(n) and a2(n + 1) = −a1(n) + 2(1− πε(∆τ)2

cos2(πtn)

)a2(n) into

(36), one obtains

a1(n+ 1)

a2(n+ 1)=

a2(n)

−a1(n) + 2(1− πε(∆τ)2

cos2(πtn)

)a2(n)

=1

−a1(n)a2(n)

+ 2(1− πε(∆τ)2

cos2(πtn)

) . (37)

Then, when condition (35) is satisfied, condition (36) holds because it inductively holds

that {1− πε(∆τ)2

cos2(πtk)< a1(k)

a2(k)< 0, ∀k ≥ n+ 1, ε > 2

π(∆τ)2,

0 < a1(k)a2(k)

< 1− πε(∆τ)2

cos2(πtk), ∀k ≥ n+ 1, ε < 0.

Therefore, condition (34) holds.

Lemma 4. When the condition ε < 0, 2π(∆τ)2

< ε is satisfied, it holds that

J(xn−1)−1an ∈ L−(T−1

ε,∆τxn),∀an ∈ L−(xn). (38)

Proof. The conditions an ∈ L−(xn) and J(xn−1)−1an ∈ L−(T−1

ε,∆τxn) in (38) are

expressed as

an ∈ L−(xn) ⇔{

a2(n)a1(n)

> 1− πε(∆τ)2

cos2(πtn−1), for ε > 2

π(∆τ)2,

a2(n)a1(n)

< 1− πε(∆τ)2

cos2(πtn−1), for ε < 0,

(39)

and

an−1 ≡ J(xn−1)−1an ∈ L−(T−1

ε,∆τxn) = L−(xn−1)

⇔{

a2(n−1)a1(n−1)

< 1− πε(∆τ)2

cos2(πtn−2), for ε > 2

π(∆τ)2,

a2(n−1)a1(n−1)

> 1− πε(∆τ)2

cos2(πtn−2), for ε < 0.

(40)

Substituting a1(n−1) = 2(1− πε(∆τ)2

cos2(πtn−1)

)a1(n)−a2(n) and a2(n−1) = a1(n) into

(40), one can obtain that condition (40) holds when condition (39) holds.

Proof of Irreversibility for Systems with Time Reversal Symmetry 12

Definition 5. Define a subset LL−1 (xn) of L

−(xn) to be

LL−1 (xn)

≡

{(a1(n), a2(n));−1 ≤ a2(n)

a1(n)< 0}

for ε > 2π(∆τ)2

,{(a1(n), a2(n)); 0 <

a2(n)a1(n)

≤ 1}

for ε < 0.

(41)

From a simple calculation, it holds that

J(xn−1)−1an ∈ LL−

1 (xn−1),∀an ∈ LL−

1 (xn). (42)

Lemma 6. For any point xn and for any integer k ∈ Z ≥ n, there exists a deviation

vector, an, such thatn∏

i=k

J(xi)an ∈ LL−1 (xk+1), an ∈ LL−

1 (xn). (43)

Proof. We prove this by contradiction. We assume that

∃yn,∃k′ ∈ Z ≥ n, ∀an ∈ LL−

1 (yn) s.t.n∏

i=k′

J(yi)an /∈ LL−1 (yk′+1). (44)

One can obtain an orbit, {ym}, including yn and yn′, where n′ > k′ + 1 > n. For all

points, there exists a cone, LL−1 . Hence, there exists a vector, An′ ∈ LL−

1 (yn′). Then,

one selects Ak′+1 and An as

Ak′+1 ≡n′−1∏

i=k′+1

J−1(yi)An′ ∈ LL−1 (yk′+1) ∵ (38),

An ≡n′−1∏

i=n

J−1(yi)An′ ∈ LL−1 (yn),

Thus,

n∏

i=k′

J(yi)An ∈ LL−1 (yk′+1). (45)

Equation (45) is contradictory to (44). Therefore, Lemma 6 holds.

From Lemmas 3, 4, and 6, we define LL−(xn) and LL+(xn) as follows to prove that

conditions (I), (II), and (III) are satisfied:

Definition 7. We define a subset, LL−(xn), of L−(xn) as

LL−(xn) ≡{(a1(n), a2(n));an ∈ LL−

1 (xn),

n∏

i=k

J(xi)an ∈ LL−1 (xk+1),

∀k ≥ n

}. (46)

From Lemma 6, one can obtain that

J(xn)an ∈ LL−(xn+1),∀an ∈ LL−(xn). (47)

Definition 8. We define a subset, LL+(xn), of L+(xn) as

LL+(xn) ≡

{(a1(n), a2(n));−1 ≤ a1(n)

a2(n)< 0}

for ε > 2π(∆τ)2

,{(a1(n), a2(n)); 0 <

a1(n)a2(n)

≤ 1}

for ε < 0.(48)

Proof of Irreversibility for Systems with Time Reversal Symmetry 13

Based on simple calculation, it holds that

J(xn)an ∈ LL+(xn+1),∀an ∈ LL+(xn). (49)

Lemma 9. When the condition ε < 0 or 2π(∆τ)2

< ε is satisfied, conditions (I) and (II)

are satisfied.

Proof. One can select any eigenvector spaces Eux and Es

x in LL+(xn) and LL−(xn),

respectively. Hence, combining (46) and (48) gives

Euxn

⊂ LL+(xn), Esxn

⊂ LL−(xn). (50)

Combining (47), (49), and (50), one obtains J(xn)Euxn

⊂ LL+(xn+1) and J(xn)Esxn

⊂LL−(xn+1). Then, one can inductively select Eu

xn+1and Es

xn+1by

Euxn+1

≡ J(xn)Euxn, Es

xn+1≡ J(xn)E

sxn. (51)

Subspaces Euxn

and Esxn

are linearly independent by definition, from which it holds that

TxnM = Euxn

⊕ Esxn.

Lemma 10. When the condition ε < 0, 2π(∆τ)2

< ε is satisfied, the condition (III) is

satisfied.

Proof. Let us define αn ≡(1− πε(∆τ)2

cos2(πtn)

)and the stretching rate σ(xn,an) ≡ ‖J(xn)an‖2

‖an‖2

which can be rewritten as

σ(xn,an) =a21 + a22 − 4αna1a2 + 4α2

na22

a21 + a22,

= 1− 4αna1a2a21 + a22

+ 4α2n

a22a21 + a22

,

= 1 + 4αn(αn sin

2 φn − sinφn cosφn),

where φn is defined such that sin2 φn =a22(n)

a21(n)+a22(n)

and sinφn cosφn = a1(n)a2(n)

a21(n)+a22(n)

, −π <

φ ≤ π. Then, if an ∈ LL+(xn), then it holds that

F (φn) ≡ αn sin2 φn − sinφn cos φn

{< 0, ε > 2

π(∆τ)2,

> 0, ε < 0.(52)

There are two cases for ε. The minimum value of σ for each case is calculated as

follows.

(i) Case of ε > 2π(∆τ)2

,

From the condition that an ∈ LL+(xn) ⇔ −1 ≤ a1(n)a2(n)

= cotφn < 0, φn is defined

in the range

−π2< φn ≤ −π

4, or

π

2< φn ≤ 3

4π. (53)

Proof of Irreversibility for Systems with Time Reversal Symmetry 14

Since F ′(φ) = sin(2φ) (αn − cot(2φ)) is positive in this range, it follows that

F (φn) ≤ F(−π

4

)= F

(34π)= 1

2(αn + 1) < 0,

∴ σ(xn,an) ≥ 1 + 2αn(αn + 1)

≥ 1 + 2 {1− πε(∆τ)2} {2− πε(∆τ)2} .(54)

Then by choosing K = 1, 1λ=√

1 + 2 {1− πε(∆τ)2} {2− πε(∆τ)2} > 1, one sees that

condition (31) is satisfied.

(ii) Case of ε < 0,

For 0 < cotφn ≤ 1, the domain of φn is

−3

4π ≤ φn < −π

2, or

π

4≤ φn <

π

2. (55)

Since F ′(φ) is positive in this range, one has that

F (φn) ≥ F(−3

4π)= F

(π4

)= 1

2(αn − 1) > 0,

∴ σ(xn,an) ≥ 1 + 2αn(αn − 1)

≥ 1− 2πε(∆τ)2 {1− πε(∆τ)2} .(56)

Then by choosing K = 1, 1λ

=√

1− 2πε(∆τ)2 {1− πε(∆τ)2} > 1, one sees that

condition (31) is satisfied. Besides, this relation about λ (λ → 1 − 0 as ε → −0) is

equivalent to the fact that the Lyapunov exponent converges to zero value as ε → −0.

Let us define the stretching rate σ′(xn,an) ≡‖J−1(T−1

ε,∆τxn)an‖2

‖an‖2 = ‖J−1(xn−1)an‖2‖an‖2 which

can be rewritten as

σ′(xn,an) =a21 + a22 − 4αn−1a1a2 + 4α2

n−1a21

a21 + a22,

= 1− 4αn−1a1a2a21 + a22

+ 4α2n−1

a21a21 + a22

,

= 1 + 4αn−1

(αn−1 sin

2 φ′n − sinφ′

n cosφ′n

),

where φ′n is defined such that sin2 φ′

n =a21(n)

a21(n)+a22(n)

and sinφ′n cosφ

′n = a1(n)a2(n)

a21(n)+a22(n)

, −π <

φ′ ≤ π. Then, if an ∈ LL−(xn), then it holds that

G(φ′n) ≡ αn−1 sin

2 φ′n − sin φ′

n cosφ′n

{< 0, ε > 2

π(∆τ)2,

> 0, ε < 0.(57)

There are two cases for ε. The minimum value of σ′ for each case is calculated as

follows.

(i) Case of ε > 2π(∆τ)2

,

From the condition that an ∈ LL−(xn) ⇔ −1 ≤ a2(n)a1(n)

= cotφ′n < 0, φ′

n is defined

in the range

−π2< φ′

n ≤ −π4, or

π

2< φ′

n ≤ 3

4π. (58)

Proof of Irreversibility for Systems with Time Reversal Symmetry 15

Then, in the same way as above discussion, it holds that

G(φ′n) ≤ 1

2(αn−1 + 1) < 0,

∴ σ′(xn,an) ≥ 1 + 2αn−1(αn−1 + 1)

≥ 1 + 2 {1− πε(∆τ)2} {2− πε(∆τ)2} .(59)

Then by choosing K = 1, 1λ=√

1 + 2 {1− πε(∆τ)2} {2− πε(∆τ)2} > 1, one sees that

condition (32) is satisfied.

(ii) Case of ε < 0,

For 0 < cotφ′n ≤ 1, the domain of φ′

n is

−3

4π ≤ φ′

n < −π2, or

π

4≤ φ′

n <π

2. (60)

Then, in the same way as above discussion, it holds that

G(φ′n) ≥ 1

2(αn−1 − 1) > 0,

∴ σ′(xn,an) ≥ 1 + 2αn−1(αn−1 − 1)

≥ 1− 2πε(∆τ)2 {1− πε(∆τ)2} .(61)

Then by choosing K = 1, 1λ

=√

1− 2πε(∆τ)2 {1− πε(∆τ)2} > 1, one sees that

condition (32) is satisfied.

Now, we state the first main theorem in this paper.

Theorem A. Suppose that the condition ε < 0 or 2π(∆τ)2

< ε is satisfied. Map Tε,∆τ on

M is an Anosov diffeomorphism.

Proof. From Lemma 9 and Lemma 10, map Tε,∆τ on M is an Anosov diffeomorphism.

According to [19], if maps are Anosov diffeomorphisms, then the corresponding

dynamical systems are K-systems. Therefore, the corollary provided below holds.

Corollary 18. Suppose that the condition ε < 0, 2π(∆τ)2

< ε is satisfied. Then, the

dynamical system, (M, Tε,∆τ , µ), is a K-system with µ as the Lebesgue measure. In

addition, (M, Tε,∆τ , µ) has a mixing property.

5.2. Anosov property and mixing property

In this subsection, we prove that map Tε,∆τ defined by (15) has the same ergodic

properties as an Anosov diffeomorphism for ε < 0 or 2π(∆τ)2

< ε. First, we show

that map Tε,∆τ satisfies the conditions (I), (II) and (III) in Definition 1 where M ′ is

not conpact. A point, (p1(n), q1(n)), on M ′, is denoted as Xn = (p1(n), q1(n)). The

Jacobian of map Tε,∆τ is given by

J(Xn) =

(1− πε(∆τ)2

cos2 θn−2πε(∆τ)

cos2 θn

∆τ − πε(∆τ)3

2 cos2 θn1− πε(∆τ)2

cos2 θn

), (62)

Proof of Irreversibility for Systems with Time Reversal Symmetry 16

where θn ≡ π{2q1(n) + ∆τp1(n)− q0 − C∆τ

(n+ 1

2

)}. The linear instability condition

for map Tε,∆τ is the same as that in (33).

Let us introduce a vector on TXnM ′, as follows:

Definition 19. For any two-dimensional non-zero vector on TXnM ′, Xn ∈M ′, n ∈ Z

a′n = (a′1(n), a

′2(n)) 6= O, (63)

we define cones l+ and l− as

l+(Xn) = {(a′1(n), a′2(n)); ‖J(Xn)a′n‖ > ‖a′

n‖},l−(Xn) = {(a′1(n), a′2(n)); ‖J(T−1

ε,∆τXn)−1a′

n‖ > ‖a′n‖},

where ‖ · ‖ represents ‖a′n‖ ≡

√a′21 (n) + a′22 (n).

To prove that map Tε,∆τ satisfies the conditions (I), (II) and (III) in Definition 1, we

must show the following Lemmas:

Lemma 20. When the condition ε < 0, 2π(∆τ)2

< ε is satisfied, there is a subset, ll+(Xn),

of l+(Xn), such that

J(Xn)a′n ∈ ll+(Tε,∆τXn),

∀a′n ∈ ll+(Xn) (64)

Proof. Let us consider a vector, a′n, such that a′2(n) 6= 0. The condition a′

n ∈ l+(Xn) is

expressed by

a′n ∈ l+(Xn) ⇔ An

(a′1(n)

a′2(n)

)2

+Bn

(a′1(n)

a′2(n)

)+ Cn > 0, (65)

where An, Bn, and Cn are defined as

An ≡(1− πε(∆τ)2

cos2 θn

)2+ (∆τ)2

(1− πε(∆τ)2

2 cos2 θn

)2− 1,

Bn ≡(1− πε(∆τ)2

cos2 θn

){−4πε(∆τ)

cos2 θn+ 2∆τ

(1− πε(∆τ)2

2 cos2 θn

)},

Cn ≡{

{2πε(∆τ)}2cos4 θn

+(1− πε(∆τ)2

cos2 θn

)2− 1

}.

(66)

For any n ∈ Z and Xn ∈M ′, if (33) is satisfied, then it holds that

An > 0, Bn > 0, Cn > 0,

B2n − 4AnCn =

[{(∆τ)2+4}πε(∆τ)2

cos2 θ−2(∆τ)2

]2

(∆τ)2≥ 0.

(67)

In this case, condition (65) is rewritten as

a′n(Xn) ∈ l+(Xn)

⇔ a′1(n)

a′2(n)<

−Bn−√B2

n−4AnCn

2Anor

−Bn+√B2

n−4AnCn

2An<

a′1(n)

a′2(n)

(68)

where−Bn−

√B2

n−4AnCn

2An<

−Bn+√B2

n−4AnCn

2An< 0.

Proof of Irreversibility for Systems with Time Reversal Symmetry 17

Selecting a′n+1 as J(Xn)a

′n+1 = (a1(n + 1), a2(n + 1)), one obtains that the condition

a′n+1 ∈ l+(Tε,∆τXn) is expressed as

a′n+1 ∈ l+(Tε,∆τXn) = l+(Xn+1)

⇔ a′1(n+1)

a′2(n+1)<

−Bn+1−√B2

n+1−4An+1Cn+1

2An+1or

−Bn+1+√B2

n+1−4An+1Cn+1

2An+1<

a′1(n+1)

a′2(n+1)

(69)

where−Bn+1−

√B2

n+1−4An+1Cn+1

2An+1<

−Bn+1+√B2

n+1−4An+1Cn+1

2An+1< 0.

Selecting a subset, ll+(Xn), of l+(Xn) as

ll+(Xn) =

{(a′1(n), a

′2(n));

a′1(n)

a′2(n)> 0

}, (70)

and assuming that a′2(n + 1) = 0 and a′n ∈ ll+(Xn), one obtains that

a′2(n + 1) = ∆τ

(1− πε(∆τ)2

2 cos2 θn

)a′1(n) +

(1− πε(∆τ)2

cos2 θn

)a′2(n + 1) = 0,

a′1(n)

a′2(n)= −

(1− πε(∆τ)2

cos2 θn

)

(1− πε(∆τ)2

2 cos2 θn

) < 0.

This is contradictory toa′1(n)

a′2(n)> 0. If a′

n ∈ ll+(Xn), then a′2(n+ 1) 6= 0. Substituting

a′1(n + 1) =

(1− πε(∆τ)2

cos2 θn

)a′1(n)−

2πε(∆τ)

cos2 θna′2(n), (71)

a′2(n + 1) = ∆τ

(1− πε(∆τ)2

2 cos2 θn

)a′1(n) +

(1− πε(∆τ)2

cos2 θn

)a′2(n) (72)

intoa′1(n+1)

a′2(n+1), one obtains

a′1(n + 1)

a′2(n + 1)=

(1− πε(∆τ)2

cos2 θn

)a′1(n)− 2πε(∆τ)

cos2 θna′2(n)

∆τ(1− πε(∆τ)2

2 cos2 θn

)a′1(n) +

(1− πε(∆τ)2

cos2 θn

)a′2(n)

,

=

(1− πε(∆τ)2

cos2 θn

)a′1(n)

a′2(n)− 2πε(∆τ)

cos2 θn

∆τ(1− πε(∆τ)2

2 cos2 θn

)a′1(n)

a′2(n)+(1− πε(∆τ)2

cos2 θn

) (73)

Here, if the conditiona′1(n)

a′2(n)> 0 ⇔ a′

n ∈ ll+(Xn) ⊂ l+(Xn) holds, then it holds that for

ε < 0, 2π(∆τ)2

< ε

a′1(n+1)

a′2(n+1)=

(1− πε(∆τ)2

cos2 θn

)a′1(n)

a′2(n)− 2πε(∆τ)

cos2 θn

∆τ(1− πε(∆τ)2

2 cos2 θn

)a′1(n)

a′2(n)+(1− πε(∆τ)2

cos2 θn

) > 0

⇒ a′n+1 ∈ ll+(Xn+1) ⊂ l+(Xn+1).

If a′n ∈ ll+(Xn) ⊂ l+(Xn), then a′

n+1 ∈ ll+(Xn+1) ⊂ l+(Xn+1).

Proof of Irreversibility for Systems with Time Reversal Symmetry 18

Lemma 21. When the condition ε < 0, 2π(∆τ)2

< ε is satisfied, there is a subset, ll−1 (Xn),

of l−(Xn), such that

J(Xn−1)−1a′

n ∈ ll−1 (T−1ε,∆τXn),

∀a′n ∈ ll−1 (Xn). (74)

Proof. Let us consider a vector, a′n, such that a′2(n) 6= 0. The condition a′

n ∈ l−(Xn) in

(74) is expressed as

a′n ∈ l−(Xn)

⇔ ‖a′n−1‖2 − ‖a′

n‖2 > 0,

⇔ An−1

(a′1(n)

a′2(n)

)2− Bn−1

(a′1(n)

a′2(n)

)+ Cn−1 > 0,

⇔ a′1(n)

a′2(n)<

Bn−1−√B2

n−1−4An−1Cn−1

2An−1or

Bn−1+√B2

n−1−4An−1Cn−1

2An−1<

a′1(n)

a′2(n)

(75)

where 0 <Bn−1 −

√B2n−1 − 4An−1Cn−1

2An−1

<Bn−1 +

√B2n−1 − 4An−1Cn−1

2An−1

.

The condition J(Xn−1)−1a′

n ∈ l−(T−1ε,∆τXn) in (74) is expressed as

a′n−1 ≡ J(Xn−1)

−1a′n ∈ l−(T−1

ε,∆τXn),

⇔ ‖a′n−2‖2 − ‖a′

n−1‖2 > 0,

⇔ a′1(n− 1)

a′2(n− 1)<Bn−2 −

√B2n−2 − 4An−2Cn−2

2An−2

or

Bn−2 +√B2n−2 − 4An−2Cn−2

2An−2<a′1(n− 1)

a′2(n− 1),

(76)

where 0 <Bn−2−

√B2

n−2−4An−2Cn−2

2An−2<

Bn−2+√B2

n−2−4An−2Cn−2

2An−2.

Choose a subset ll−1 (Xn) of l−(Xn) as

ll−1 (Xn) =

{(a′1(n), a

′2(n));

a′1(n)

a′2(n)< 0

}. (77)

It is evident that for a′n ∈ ll−1 (Xn), it holds that a′2(n − 1) 6= 0, as follows: If

a′n ∈ ll−1 (Xn), then substituting

a′1(n− 1) =(1− πε(∆τ)2

cos2 θn−1

)a′1(n) +

2πε(∆τ)cos2 θn−1

a′2(n),

a′2(n− 1) = −∆τ(1− πε(∆τ)2

2 cos2 θn−1

)a′1(n) +

(1− πε(∆τ)2

cos2 θn−1

)a′2(n)

(78)

into a′1(n− 1) and a′2(n− 1), one obtains

a′1(n−1)

a′2(n−1)=

(1− πε(∆τ)2

cos2 θn−1

)a′1(n)

a′2(n)+ 2πε(∆τ)

cos2 θn−1

−∆τ(1− πε(∆τ)2

2 cos2 θn−1

)a′1(n)

a′2(n)+(1− πε(∆τ)2

cos2 θn−1

) < 0

⇒ a′2(n− 1) 6= 0.

(79)

Therefore, it holds that a′n ∈ ll−1 (Xn) ⊂ l−(Xn) ⇒ J(Xn−1)

−1a′n = a′

n−1 ∈ ll−1 (Xn−1) ⊂l−(Xn−1).

Proof of Irreversibility for Systems with Time Reversal Symmetry 19

Lemma 22. For any point Xn and for any integer k ∈ Z ≥ n, there exists a deviation

vector, a′n, such that

n∏

i=k

J(Xi)a′n ∈ ll−1 (Xk+1), a

′n ∈ ll−1 (Xn). (80)

Proof. We prove this by contradiction. We assume that

∃Yn,∃k′ ∈ Z ≥ n, ∀a′

n ∈ ll−1 (Yn) s.t.n∏

i=k′

J(Yi)a′n /∈ ll−1 (Yk′+1). (81)

One can obtain an orbit, {Ym}, including Yn and Yn′, where n′ > k′ + 1 > n. For all

points, there exists a cone, ll−1 . Thus, there exists a vector, A′n′ ∈ ll−1 (Yn′). Then, one

selects A′k′+1 and A′

n as

A′k′+1 ≡

n′−1∏

i=k′+1

J−1(Yi)A′n′ ∈ ll−1 (Yk′+1) ∵ (74),

A′n ≡

n′−1∏

i=n

J−1(Yi)A′n′ ∈ ll−1 (Yn).

Thus,n∏

i=k′

J(Yi)A′n ∈ ll−1 (Yk′+1). (82)

Equation (82) is contradictory to (81). Therefore, Lemma 22 holds.

From Lemmas 20, 21, and 22, we define ll−(Xn) and ll+(Xn) as follows to prove that

conditions (I), (II), and (III) are satisfied:

Definition 23. We define a subset, ll−(Xn), of ll−1 (Xn) as

ll−(Xn) ≡{(a′1(n), a

′2(n));a

′n ∈ ll−1 (Xn),

k∏

i=n

J(Xi)a′n ∈ ll−1 (Xk),

∀k ≥ n

}. (83)

From Lemma 22, one can obtain that

J(Xn)a′n ∈ ll−(Xn+1),

∀an ∈ ll−(Xn). (84)

Definition 24. We define a subset, ll+(Xn), of l+(Xn) as

ll+(Xn) ≡{(a′1(n), a

′2(n));

a′1(n)

a′2(n)> 0

}. (85)

Based on simple calculation, one obtains

J(Xn)a′n ∈ ll+(Xn+1),

∀a′n ∈ ll+(Xn). (86)

Lemma 25. When the condition ε < 0, 2π(∆τ)2

< ε is satisfied, conditions (I) and (II)

are satisfied.

Proof of Irreversibility for Systems with Time Reversal Symmetry 20

Proof. One can select any eigenvector spaces EuX and Es

X in ll+(Xn) and ll−(Xn),

respectively. Hence, combining (83) and (85) gives

EuXn

⊂ ll+(Xn), EsXn

⊂ ll−(Xn). (87)

Combining (84), (86), and (87), one obtains J(Xn)EuXn

⊂ ll+(Xn+1) and J(Xn)EsXn

⊂ll−(Xn+1). Then, one can inductively select Eu

Xn+1and Es

Xn+1by

EuXn+1

≡ J(Xn)EuXn, Es

Xn+1≡ J(Xn)E

sXn. (88)

Subspaces EuXn

and EsXn

are linearly independent by definition, from which it holds that

TXnM = Eu

Xn⊕Es

Xn.

Lemma 26. When the condition ε < 0 or 2π(∆τ)2

< ε is satisfied, the condition (III) is

satisfied.

Proof. Let us define the stretching rate σ(Xn,a′n) ≡ ‖J(Xn)a′

n‖2‖a′

n‖2which can be rewritten

as

σ(Xn,a′n) =

(An + 1)a′21 +Bna′1a

′2 + (Cn + 1)a′22

a′21 + a′22

= 1 + Ana′21

a′21 + a′22+Bn

a′1a′2

a′21 + a′22+ Cn

a′22a′21 + a′22

= 1 + An cos2 φn +Bn sin φn cos φn + Cn sin

2 φn

where φn is defined such that cos2 φn =a′21 (n)

a′21 (n)+a′22 (n), sin2 φn =

a′22 (n)

a′21 (n)+a′22 (n)and

sin φn cos φn =a′1(n)a

′

2(n)

a′21 (n)+a′22 (n), − π < φ ≤ π. Then, if a′

n ∈ ll+(Xn), then

F (φn) ≡ An cos2 φn +Bn sin φn cos φn + Cn sin

2 φn > 0, ε < 0,2

π(∆τ)2< ε (89)

and sincea′1(n)

a′2(n)> 0, the variable φ is defined on (−π,−π

2) ∪ (0, π

2). There are two cases

for Xn. The minimum value of σ for each case is calculated as follows.

The derivative of F (φ) is rewritten in

F ′(φn)

= (Cn −An) sin 2φn +Bn cos 2φn

= [(Cn − An)2 +B2

n][

Cn−An

(Cn−An)2+B2nsin 2φn +

Bn

(Cn−An)2+B2ncos 2φn

]

= [(Cn − An)2 +B2

n][Pn sin 2φn +Qn cos 2φn

]

= [(Cn − An)2 +B2

n][sin(2φn + ψn

)]

(90)

Proof of Irreversibility for Systems with Time Reversal Symmetry 21

where Pn, Qn and ψ are defined by

Pn ≡ Cn −An(Cn − An)2 +B2

n

, (91)

Qn ≡ Bn

(Cn − An)2 +B2n

, (92)

tan ψn =Qn

Pn, ψn ∈

(−π2,π

2

). (93)

(i) Case of Pn ≥ 0 ⇔ Cn ≥ An,

It holds that

tan ψn =Qn

Pn> 0 ⇒ 0 < ψn <

π

2. (94)

The variable ψn depends on Xn not on a′n. In this case, F (φn) increases monotonously

for −π < φn < −π+ψn

2and for 0 < φn < π−ψn

2and decreases monotonously for

−π+ψn

2< φn < −π

2and for π−ψn

2< φn < π

2. Then the candidates for the minimal

value of F (φn) for −π < φn < −π2and 0 < φn <

π2are

F (−π) = F (0) = An, (95)

F(−π2

)= F

(π2

)= Cn. (96)

In this case, Cn ≥ An > 0. Then, the minimal value of σ(Xn,a′n) by varying the vector

a′n is obtained as 1 + An. With this minimal value and An > 0, one has

σ(Xn,a′n) ≥ 1 + An ≥

{1− πε(∆τ)2

}2+ (∆τ)2

{1− πε(∆τ)2

2

}2

> 1. (97)

(ii) Case of Pn < 0 ⇔ Cn < An,

It holds that

tan ψn =Qn

Pn< 0 ⇒ −π

2< ψn < 0. (98)

In this case, F (φn) decreases monotonously for −π < φn < −π− ψn

2and 0 < φn < − ψn

2

and increases monotonously for −π − ψn

2< φ < −π

2and − ψn

2< φn <

π2. Then the

candidates for the minimal value of F (φn) for −π < φn < −π2and 0 < φn <

π2are

F

(−π − ψn

2

)= F

(− ψn

2

)= An cos

2 ψn2

− Bn sinψn2

cosψn2

+ Cn sin2 ψn2. (99)

From An > Cn > 0, it holds that

F

(−π − ψn

2

)> Cn − Bn sin

ψn2

cosψn2> Cn, ∵ sin ψn < 0, Bn > 0. (100)

Then, the minimal value of σ(Xn,a′n) by varying the vector a′

n is obtained as 1 + Cn.

With this minimal value and Cn > 0, one has

σ(Xn,a′n) ≥ 1 + Cn ≥ {2πε(∆τ)}2 +

{1− πε(∆τ)2

}2> 1. (101)

Proof of Irreversibility for Systems with Time Reversal Symmetry 22

Thus from (97) and (101), by choosing

K = 1,

1λ

= min

√{1− πε(∆τ)2}2 + (∆τ)2

{1− πε(∆τ)2

2

}2

,√{2πε(∆τ)}2 + {1− πε(∆τ)2}2

, (102)

one sees that condition (31) is satisfied. Besides, this relation about λ (λ → 1 − 0 as

ε → −0) is equivalent to the fact that the Lyapunov exponent converges to zero value

as ε→ −0.

Let us define the stretching rate σ′(Xn,a′n) ≡ ‖J−1(T−1

ε,∆τXn)a′

n‖2‖a′

n‖2= ‖J−1(Xn−1)a′

n‖2‖a′

n‖2which can be rewritten as

σ′(Xn,a′n) =

(An−1 + 1)a′21 −Bn−1a′1a

′2 + (Cn−1 + 1)a′22

a′21 + a′22

= 1 + An−1a′21

a′21 + a′22−Bn−1

a′1a′2

a′21 + a′22+ Cn−1

a′22a′21 + a′22

= 1 + An−1 cos2 φn − Bn−1 sin φn cos φn + Cn−1 sin

2 φn

where φn is defined such that cos2 φn =a′21 (n)

a′21 (n)+a′22 (n), sin2 φn =

a′22 (n)

a′21 (n)+a′22 (n)and

sin φn cos φn =a′1(n)a

′

2(n)

a′21 (n)+a′22 (n), − π < φ ≤ π. Then, if a′

n ∈ ll−(Xn), then

G(φn) ≡ An−1 cos2 φn − Bn−1 sin φn cos φn + Cn−1 sin

2 φn > 0, ε < 0,2

π(∆τ)2< ε (103)

and sincea′1(n)

a′2(n)< 0, the variable φ is defined on (−π

2, 0) ∪ (π

2, π). There are two cases

for Xn. The minimum value of σ′ for each case is calculated as follows.

The derivative of G(φ) is rewritten in

G′(φn) = (Cn−1 −An−1) sin 2φn − Bn−1 cos 2φn

=[(Cn−1 − An−1)

2 +B2n−1

] [Pn−1 sin 2φn −Qn−1 cos 2φn

]

=[(Cn−1 − An−1)

2 +B2n−1

] [sin(2φn − ψn−1

)]

where Pn−1, Qn−1 and ψ are defined by

Pn−1 ≡ Cn−1 − An−1

(Cn−1 −An−1)2 +B2n−1

, (104)

Qn−1 ≡ Bn−1

(Cn−1 −An−1)2 +B2n−1

, (105)

tan ψn−1 =Qn−1

Pn−1, ψn−1 ∈

(−π2,π

2

). (106)

(i) Case of Pn−1 ≥ 0 ⇔ Cn−1 ≥ An−1,

It holds that

tan ψn−1 =Qn−1

Pn−1> 0 ⇒ 0 < ψn−1 <

π

2. (107)

Proof of Irreversibility for Systems with Time Reversal Symmetry 23

The variable ψn−1 depends on T−1ε,∆τXn = Xn−1 not on a′

n. In this case, G(φn) increases

monotonously for −π2< φn < −π

2+ ψn−1

2and for π

2< φn <

π2+ ψn−1

2and decreases

monotonously for −π2+ ψn−1

2< φn < 0 and for π

2+ ψn−1

2< φn < π. Then the candidates

for the minimal value of G(φn) for −π2< φn < 0 and π

2< φn < π are

G(0) = G(π) = An, (108)

G(−π2

)= G

(π2

)= Cn. (109)

In this case, Cn ≥ An > 0. Then, the minimal value of σ′(Xn,a′n) by varying the vector

a′n is obtained as 1 + An. With this minimal value and An > 0, one has

σ′(Xn,a′n) ≥ 1 + An ≥

{1− πε(∆τ)2

}2+ (∆τ)2

{1− πε(∆τ)2

2

}2

> 1. (110)

(ii) Case of Pn−1 < 0 ⇔ Cn−1 < An−1,

It holds that

tan ψn−1 =Qn−1

Pn−1< 0 ⇒ −π

2< ψn−1 < 0. (111)

In this case, G(φn) increases monotonously for ψn−1

2< φn < 0 and π + ψn−1

2< φn < π

and decreases monotonously for −π2< φn <

ψn−1

2and π

2< φn < π + ψn−1

2. Then the

candidates for the minimal value of G(φn) for −π2< φn < 0 and π

2< φn < π are

G(π + ψn−1

2

)= F

(ψn−1

2

)

= An−1 cos2 ψn−1

2−Bn−1 sin

ψn−1

2cos ψn−1

2+ Cn−1 sin

2 ψn−1

2.

From An−1 > Cn−1 > 0, it holds that

G(π + ψn−1

2

)> Cn−1 − Bn−1 sin

ψn−1

2cos ψn−1

2> Cn−1,

∵ sin ψn−1 < 0, Bn−1 > 0.(112)

Then, the minimal value of σ′(Xn,a′n) by varying the vector a′

n is obtained as 1+Cn−1.

With this minimal value and Cn−1 > 0, one has

σ′(Xn,a′n) ≥ 1 + Cn−1 ≥ {2πε(∆τ)}2 +

{1− πε(∆τ)2

}2> 1 (113)

Thus from (110) and (113), by choosing

K = 1,

1λ

= min

√{1− πε(∆τ)2}2 + (∆τ)2

{1− πε(∆τ)2

2

}2

,√{2πε(∆τ)}2 + {1− πε(∆τ)2}2

, (114)

one sees that condition (32) is satisfied.

Now, we state the second main theorem in this paper.

Theorem B. Suppose that the condition ε < 0 or 2π(∆τ)2

< ε is satisfied. The map Tε,∆τon M ′ has the same ergodic properties as an Anosov diffeomorphism.

Proof of Irreversibility for Systems with Time Reversal Symmetry 24

Proof. From Lemma 25 and Lemma 26, the map Tε,∆τ satisfies the conditions (I), (II)

and (III) in Definition 1 where M ′ is not conpact. It is obvious that the map Tε,∆τ is a

diffeomorphism on M ′. The domain of Tε,∆τ is not compact. However, the map Tε,∆τ is

topologically semi-conjugate with Tε,∆τ through the transformation H . That is for any

orbit {(p1(n), q1(n))}, there exists at least one corresponding orbit {(sn, tn)}. Therefore,the dynamical system (M ′, Tε,∆τ , µ) has the same ergodic properties as (M, Tε,∆τ , µ).

Similar to the case of Tε,∆τ , the corollary provided below holds.

Corollary 27. Suppose that the condition ε < 0, 2π(∆τ)2

< ε is satisfied. Then,

the dynamical system, (M ′, Tε,∆τ , µ), is a K-system with the Lebesgue measure µ. In

addition, (M ′, Tε,∆τ , µ) has a mixing property.

According to Corollaries 18 and 27, the dynamical systems defined as the triplets

(M, Tε,∆τ , µ) and (M ′, Tε,∆τ , µ) have the mixing property where µ corresponds to the

Lebesgue measure. In general, the mixing property yields a relaxation process. We

adapted this general scenario to our case.

According to Gallavotti [2, 8], Tε,∆τ or Tε,∆τ , which has Anosov properties, has a

unique probability distribution (SRB distribution) on phase spaceM orM ′, respectively.

The Lebesgue measure µ is an invariant measure for Tε,∆τ on M and Tε,∆τ on M ′.

In addition, it is obvious that for every measurable partition ζ subordinate to an

unstable manifold, the conditional measures of the Lebesgue measure {µζx} are absolutelycontinuous with respect to the Lebesgue measure. Then, the uniform distribution

derived from the Lebesgue measure is the SRB distribution [30].

Convergence to the uniform distribution is proven for (M, Tε,∆τ , µ) and

(M ′, Tε,∆τ , µ), as described above for any initial density function defined except for a

zero volume set [8] when the condition (21) is satisfied. In particular, the map Tε,∆τ has

time-reversal symmetry. Thus, for the map Tε,∆τ , which satisfies the condition of time-

reversal symmetry and measure preserving, convergence to the uniform distribution was

proven.

In addition to this, correlation functions decay exponentially [2, 8, 9, 31, 32]. In

other words, in the dynamical system (M, Tε,∆τ , µ) in the regime, relaxation occurs,

such that

ρ(s, t) ≃ η(s)η(t), (115)

where ρ represents a density function onM and η represents the one-dimensional uniform

distribution defined on IδN,∆τ. In the dynamical system (M ′, Tε,∆τ , µ), for any orbit

{(p1(n), q1(n))}n∈Z of Tε,∆τ , there exists at least one corresponding orbit {(sn, tn)}n∈Zof Tε,∆τ as the map Tε,∆τ is topologically semi-conjugate with Tε,∆τ . In addition, from

Theorem B, any initial density function relaxes to the uniform distribution.

6. Lyapunov exponent

Let us discuss the Lyapunov exponent of Tε,∆τ . In the following, we consider the case in

which N is large enough so that one can identify IδN,∆τas(−1

2, 12

]and the case in which

Proof of Irreversibility for Systems with Time Reversal Symmetry 25

the condition (21) is satisfied. As the Lebesgue measure is the SRB measure, one can

obtain from the Pesin identity the Kolmogorov–Sinai (KS) entropy hKS(Tε,∆τ) (entropy

per unit time) [33] for almost all initial points as

hKS(Tε,∆τ) =

∫ ∫

M

log∣∣∣det

(DTε,∆τ |Eu

)∣∣∣µ(dx), (116)

where hKS(Tε,∆τ) is positive because Tε,∆τ is an Anosov diffeomorphism.

According to Kac [6] and Chrikov [34], it cannot be said that the relaxation always

occurs monotonically even though the system has the mixing property. However, there

are cases in which by assuming appropriate conditions, the KS entropy is consistent

with the time derivative of the entropy [10, 11, 12].

In calculating hKS(Tε,∆τ), one needs the eigenvalue corresponding to the unstable

direction, expressed by γ(t), t ∈ IδN,∆τwith

γ(t) =

{γ−(t), when ε > 2

π(∆τ)2,

γ+(t), when ε < 0,(117)

γ±(t) = 1− πε(∆τ)2

cos2(πt)±

√(1− πε(∆τ)2

cos2(πt)

)2

− 1, (118)

where hKS(Tε,∆τ) is equivalent to the positive Lyapunov exponent λ(ε,∆τ) owing to the

Pesin identity. To obtain an explicit formula of the Lyapunov exponent, we substituted

(118) into det(DTε,∆τ |Eu

).

(i) In the case of ε > 2π(∆τ)2

, the Lyapunov exponent is denoted as

λ(ε,∆τ) =

∫ 12

− 12

log

∣∣∣∣∣∣1− πε(∆τ)2

cos2(πt)−

√(1− πε(∆τ)2

cos2(πt)

)2

− 1

∣∣∣∣∣∣dt

=

∫ 12

− 12

log

∣∣∣∣cos2(πt)− πε(∆τ)2 −

√(cos2(πt)− πε(∆τ)2)2 − cos4(πt)

∣∣∣∣ dt

−∫ 1

2

− 12

log∣∣cos2(πt)

∣∣ dt

=

∫ 12

− 12

log

∣∣∣∣πε(∆τ)2 − cos2(πt) +

√(cos2(πt)− πε(∆τ)2)2 − cos4(πt)

∣∣∣∣ dt

−∫ 1

2

− 12

log∣∣cos2(πt)

∣∣ dt

=

∫ 12

− 12

log∣∣cos2(πt)− πε(∆τ)2

∣∣ dt−∫ 1

2

− 12

log∣∣cos2(πt)

∣∣ dt

+

∫ 12

− 12

log

1 +

√

1−(1− πε(∆τ)2

cos2(πt)

)−2 dt

= log[2πε(∆τ)2 − 1 + 2

√πε(∆τ)2 {πε(∆τ)2 − 1}

]

+

∫ 12

− 12

log

1 +

√

1−(1− πε(∆τ)2

cos2(πt)

)−2 dt

(119)

Proof of Irreversibility for Systems with Time Reversal Symmetry 26

(ii) In the case of ε < 0, the Lyapunov exponent is denoted as

λ(ε,∆τ) =

∫ 12

− 12

log

∣∣∣∣∣∣1− πε(∆τ)2

cos2(πt)+

√(1− πε(∆τ)2

cos2(πt)

)2

− 1

∣∣∣∣∣∣dt

=

∫ 12

− 12

log

∣∣∣∣cos2(πt)− πε(∆τ)2 +

√(cos2(πt)− πε(∆τ)2)2 − cos4(πt)

∣∣∣∣ dt

−∫ 1

2

− 12

log∣∣cos2(πt)

∣∣ dt

=

∫ 12

− 12

log∣∣cos2(πt)− πε(∆τ)2

∣∣ dt−∫ 1

2

− 12

log∣∣cos2(πt)

∣∣ dt

+

∫ 12

− 12

log

1 +

√

1−(1− πε(∆τ)2

cos2(πt)

)−2 dt

= log[1− 2πε(∆τ)2 + 2

√πε(∆τ)2 {πε(∆τ)2 − 1}

]

+

∫ 12

− 12

log

1 +

√

1−(1− πε(∆τ)2

cos2(πt)

)−2 dt

(120)

From Equation (120), in the condition such as |ε| ≪ 1, it holds that

log[1− 2πε(∆τ)2 + 2

√πε(∆τ)2 {πε(∆τ)2 − 1}

]

= O(−2πε(∆τ)2 + 2

√πε(∆τ)2 {πε(∆τ)2 − 1}

)

= O(|ε| 12 ).(121)

In terms of

∫ 12

− 12

log

1 +

√

1−(1− πε(∆τ)2

cos2(πt)

)−2 dt, it holds that

log

(1 +

√1−

(1− πε(∆τ)2

cos2(πt)

)−2),

= O

(√1−

(1− πε(∆τ)2

cos2(πt)

)−2)

= O(|ε| 12 ).

(122)

Thus, from Equations (121) and (122), when 0 < −ε ≪ 1, it holds that

λ(ε,∆τ) = O(|ε| 12 ). (123)

Therefore, the critical exponent of the Lyapunov exponent is 12as ε → −0. The value

of the critical exponent is the same as those in intermittent chaos [35, 36, 37]. Then,

there is the possibility that we can apply intermittency route to chaos to our maps.

As it is not necessary that such expressions (119) and (120) are simple enough,

we made the following assumption so that an explicit expression could be analytically

obtained.

(Assumption): One can obtain the equivalent value of (116) if one substitutes

〈cos2(πt)〉 = 1/2 into cos2(πt) in Equations (119) and (120).

Proof of Irreversibility for Systems with Time Reversal Symmetry 27

Based on this assumption, Equations (119) and (120) can be integrated as,

λ(ε,∆τ) ≃

log

[2πε(∆τ)2 − 1 + 4

√πε(∆τ)2 {πε(∆τ)2 − 1}+ 4πε(∆τ)2{πε(∆τ)2−1}

2πε(∆τ)2−1

], ε > 2

π(∆τ)2 ,

log

[1− 2πε(∆τ)2 + 4

√πε(∆τ)2 {πε(∆τ)2 − 1}+ 4πε(∆τ)2{πε(∆τ)2−1}

1−2πε(∆τ)2

], ε < 0.

(124)

Figure 3 shows the comparison between the numerical result and analytical formula

of the Lyapunov exponent. It is clear that the numerical result is consistent with the

analytical formula.

0

1

2

3

4

5

6

-10 -5 0 5 10

Lyap

unov

exp

onen

t

ε

Figure 3: Numerical calculation of Lyapunov exponent (red circle) and analytical formula

(lines) in the case of ∆τ = 1. The solid line corresponds to λ(ε,∆τ) for ε < 0 and

the broken line corresponds to ε > 2π. The iteration number is 104. The initial

deviation vector and point are given as follows: vs =1√2, vt =

1√2and

(s, t) = (√2−12 ,

√3−12 ). The vertical line corresponds to ε = 2

π.

7. Conclusion

On a microscopic scale, it has been proven that the map Tε,∆τ which has time-reversal

symmetry without scatterers, has Anosov properties for ε < 0 or 2π(∆τ)2

< ε namely,

the dynamical system derived from the Hamiltonian without scatterers has the mixing

property (ergodic).

Further, in the sense of distribution functions, we also have proven that the KS

entropy is positive for almost all initial points and that the Lebesgue measure is a unique

SRB measure in the parameter ranges. Thus, this indicates that a density function

converges to a uniform distribution in the sense of mixing and that the time derivative

of the Boltzmann entropy is positive after the system has been mixed sufficiently.

Proof of Irreversibility for Systems with Time Reversal Symmetry 28

This result has assured that dynamical systems can be handled probabilistically or

statistically using the uniform distribution.

In summary, the convergence of initial density functions to the uniform distribution

and positivity of the entropy rate were proven rigorously in our model derived from

the Hamiltonian system with time-reversal symmetry for the parameter ranges ε < 0

or 2π(∆τ)2

< ε. From these results, it can be claimed that the convergence of initial

density functions to the uniform distribution and positivity of the entropy rate are

compatible with microscopic time reversibility. Thus, from our analysis, a “macroscopic

unidirection” can be generated using the internally chaotic mechanism in Hamiltonian

dynamical systems with time-reversal symmetry even though the systems do not have

external artifacts such as walls, collisions, and scatterers.

Acknowledgments

One of the authors, Ken-ichi Okubo, thanks Dr. Yoshiyuki Y. Yamaguchi for his valuable

advice and informative discussions. Ken-ichi Okubo acknowledges the support of Grant-

in-Aid for JSPS Research Fellow Grant Number JP17J07694.

Appendix A. Exactness and Time-reversal symmetry

In this section, we will prove that the exactness is not compatible with time-reversal

symmetry as follows. When a map S is exact, from definition [29], it holds that

limn→∞

µ(SnA) = 1, ∀A ∈ A , µ(A) > 0 (A.1)

where µ is the invariant measure and A is the σ-algebra. If we assume the map S

has time-reversal symmetry in addition to exactness, for any set A ∈ A such that

0 < µ(A) < 1 it holds that

limn→∞

µ(SnA) = limn→∞

µ(R−1 ◦ S−n ◦RA). (A.2)

Since the value of measure is invariant if we apply the operator R or R−1, we have that

limn→∞

µ(SnA) = limn→∞

µ(S−nA). (A.3)

According to the property of measure preservation µ(S−1A) = µ(A), it holds that

µ(S−nA) = µ(A), ∀n ∈ N. Then it holds that

1 = limn→∞

µ(SnA) = limn→∞

µ(S−nA) = µ(A) < 1. (A.4)

This is a contradiction. Therefore, exactness is not compatible with time-reversal

symmetry.

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