Math Phys Anal Geom (2013) 16:1–17 DOI 10.1007/s11040-012-9118-6
Uniqueness of Gibbs Measure for Models with Uncountable Set of Spin Values on a Cayley Tree Yu. Kh. Eshkabilov · F. H. Haydarov · U. A. Rozikov
Received: 12 May 2011 / Accepted: 20 July 2012 / Published online: 11 August 2012 © Springer Science+Business Media B.V. 2012
Abstract We consider models with nearest-neighbor interactions and with the set [0, 1] of spin values, on a Cayley tree of order k 1. It is known that the ‘splitting Gibbs measures’ of the model can be described by solutions of a nonlinear integral equation. For arbitrary k 2 we find a sufficient condition under which the integral equation has unique solution, hence under the condition the corresponding model has unique splitting Gibbs measure. Keywords Cayley tree · Configuration · Gibbs measures · Uniqueness Mathematics Subject Classifications (2010) Primary 82B05 · 82B20; Secondary 60K35
1 Introduction In this paper we consider models (Hamiltonians) with a nearest neighbor interaction and uncountably many spin values on a Cayley tree. One of the central problems in the theory of Gibbs measures is to describe infinite-volume (or limiting) Gibbs measures corresponding to a given Hamiltonian. The existence of such measures for a wide class of Hamiltonians was established in the ground-breaking work of Dobrushin (see, e.g. [23]).
Yu. Kh. Eshkabilov · F. H. Haydarov National University of Uzbekistan, Tashkent, Uzbekistan e-mail:
[email protected] U. A. Rozikov (B) Institute of Mathematics, Tashkent, Uzbekistan e-mail:
[email protected]
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However, a complete analysis of the set of limiting Gibbs measures for a specific Hamiltonian is often a difficult problem. There are several papers devoted to models on Cayley trees, see for example [1–8, 10, 12, 16, 19–21, 24, 25, 27]. All these works are devoted to models with a finite set of spin values. These models have the following common property: The existence of finitely many translation-invariant and uncountable numbers of non-translation-invariant extreme Gibbs measures. Also for several models (see, for example, [7, 10, 20, 21]) it was proved that there exist three periodic Gibbs measures (which are invariant with respect to normal subgroups of finite index of the group representation of Cayley tree) and there are uncountably many non-periodic Gibbs measures. In [9] the Potts model with a countable set of spin values on a Cayley tree is considered and it was showed that the set of translation-invariant splitting Gibbs measures of the model contains at most one point, independently of the parameters of the Potts model with a countable set of spin values on Cayley tree. This is a crucial difference from the models with a finite set of spin values, since the last ones may have more than one translation-invariant Gibbs measures. How ‘rich’ is the set of translation-invariant Gibbs measures for models with an uncountable number of spin values? In [22] models with nearest-neighbor interactions and with the set [0, 1] of spin values, on a Cayley tree of order k 1 are considered and we reduced the problem of describing the ‘splitting Gibbs measures’ of the model to the description of the solutions of some nonlinear integral equation. For k = 1 we showed that the integral equation has a unique solution. In case k 2 some models (with the set [0, 1] of spin values) which have a unique splitting Gibbs measure are constructed. Note that paper [5] deals with spin systems on Zd that take values in a compact interval and interact via a summable nonnegative quadratic pair interaction. It is known that such systems admit a unique Gibbs measure [15]. The authors of [5] gave a simple proof of this fact using the attractiveness of the associated Gibbs sampler. In [18] the authors obtain various models on trees with a broken rotation symmetry, implying the existence of infinitely many translation-invariant extreme Gibbs measures. In [4] we have constructed several models (with uncountable spin values) which have more than one translation-invariant splitting Gibbs measures. In this paper we continue this investigation and give a sufficient condition on Hamiltonians for spin systems with uncountably many spin values under which the model has a unique translation-invariant splitting Gibbs measure.
2 Preliminaries A Cayley tree G k = (V, L) of order k 1 is an infinite homogeneous tree (see [1]), i.e., a graph without cycles, with exactly k + 1 edges incident to each vertices. Here V is the set of vertices and L that of edges (arcs).
On Gibbs Measure of a Models
3
Consider models where the spin takes values in the set [0, 1], and is assigned to the vertices of the tree. For A ⊂ V a configuration σ A on A is an arbitrary function σ A : A → [0, 1]. Denote A = [0, 1] A the set of all configurations on A. A configuration σ on V is then defined as a function x ∈ V → σ (x) ∈ [0, 1]; the set of all configurations is [0, 1]V . The (formal) Hamiltonian of the model is: H(σ ) = −J
ξσ (x)σ (y) ,
(2.1)
x,y∈L
where J ∈ R \ {0} and ξ : (u, v) ∈ [0, 1]2 → ξuv ∈ R is a given bounded, measurable function. As usually, x, y stands for nearest neighbor vertices. Let λ be the Lebesgue measure on [0, 1]. On the set of all configurations on A the a priori measure λ A is introduced as the |A|fold product of the measure λ. Here and further on |A| denotes the cardinality of A. We consider a standard sigma-algebra B of subsets of = [0, 1]V generated by the measurable cylinder subsets. A probability measure μ on (, B ) is called a Gibbs measure (with Hamiltonian H) if it satisfies the DLR equation, namely for any n = 1, 2, . . . and σn ∈ Vn : μ Vn where νω| W
n+1
σ ∈ : σ Vn = σn =
Vn μ(dω)νω| W
n+1
(σn ),
is the conditional Gibbs density
Vn νω| W
n+1
(σn ) =
1 Z n ω
exp −β H σn || ωW , n+1 Wn+1
and β = T1 , T > 0 is temperature. Here and below, Wl stands for a ‘sphere’ and Vl for a ‘ball’ on the tree, of radius l = 1, 2, . . ., centered at a fixed vertex x0 (an origin): Wl = {x ∈ V : d(x, x0 ) = l},
Vl = {x ∈ V : d(x, x0 ) l};
and Ln = {x, y ∈ L : x, y ∈ Vn }; distance d(x, y), x, y ∈ V, is the length of (i.e. the number of edges in) the in Vn shortest path connecting x with y. Vn is the set of configurations (and Wn that in Wn ; see below). Furthermore, σ Vn and ωWn+1 denote the restrictions of configurations σ, ω ∈ to Vn and Wn+1 , respectively. Next,
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σn : x ∈ Vn → σn (x) is a configuration in Vn and H σn || ωWn+1 is defined as the sum H (σn ) + U σn , ωWn+1 where H (σn ) = −J ξσn (x)σn (y) ,
U σn , ωWn+1 = −J
x,y∈Ln
ξσn (x)ω(y) .
x,y: x∈Vn ,y∈Wn+1
Finally, Z n ωWn+1 stands for the partition function in Vn , with the boundary condition ωWn+1 : exp −β H σn || ωWn+1 λVn (d σn ). Z n ω Wn+1 = Vn
Due to the nearest-neighbor character of the interaction, the Gibbs measure possesses a natural Markov property: for given a configuration ωn on Wn , random configurations in Vn−1 (i.e., ‘inside’ Wn ) and in V \ Vn+1 (i.e., ‘outside’ Wn ) are conditionally independent. We use a standard definition of a translation-invariant measure (see, e.g., [23]). The main object of study in this paper are translation-invariant Gibbs measures for the model (2.1) on Cayley tree. In [22] this problem of description of such measures was reduced to the description of the solutions of a nonlinear integral equation. For finite and countable sets of spin values this argument is well known (see, e.g. [2–9, 19, 24, 25, 27]). Write x < y if the path from x0 to y goes through x. Call vertex y a direct successor of x if y > x and x, y are nearest neighbors. Denote by S(x) the set of direct successors of x. Observe that any vertex x = x0 has k direct successors and x0 has k + 1. Let h : x ∈ V → hx = (ht,x , t ∈ [0, 1]) ∈ R[0,1] be mapping of x ∈ V \ {x0 } with |ht,x | < C where C is a constant which does not depend on t. Given n = 1, 2, . . ., consider the probability distribution μ(n) on Vn defined by ⎞ ⎛ μ(n) (σn ) = Z n−1 exp ⎝−β H(σn ) + (2.2) hσ (x),x ⎠ , x∈Wn
Here, as before, σn : x ∈ Vn → σ (x) and Z n is the corresponding partition function: ⎛ ⎞ exp ⎝−β H( σn ) + h σ (x),x ⎠ λVn (d σn ). (2.3) Zn = Vn
x∈Wn
The probability distributions μ(n) are compatible if for any n 1 and σn−1 ∈ Vn−1 : μ(n) (σn−1 ∨ ωn )λWn (d(ωn )) = μ(n−1) (σn−1 ). (2.4) Wn
On Gibbs Measure of a Models
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Here σn−1 ∨ ωn ∈ Vn is the concatenation of σn−1 and ωn . In this case there a unique measure μ on V such that, for any n and σn ∈ Vn , exists = μ(n) (σn ). μ σ = σn Vn
Definition 2.1 The measure μ is called a splitting Gibbs measure corresponding to Hamiltonian (2.1) and function x → hx , x = x0 . The following statement describes conditions on hx guaranteeing compatibility of the corresponding distributions μ(n) (σn ). Proposition 2.2 [22] The probability distributions μ(n) (σn ), n = 1, 2, . . ., in (2.2) are compatible if f for any x ∈ V \ {x0 } the following equation holds: 1 exp(Jβξtu ) f (u, y)du 0 . (2.5) f (t, x) = 1 y∈S(x) 0 exp(Jβξ0u ) f (u, y)du Here, and below f (t, x) = exp(ht,x − h0,x ), t ∈ [0, 1] and du = λ(du) is the Lebesgue measure. From Proposition 2.2 it follows that for any h = {hx ∈ R[0,1] , x ∈ V} satisfying (2.5) there exists a unique Gibbs measure μ and vice versa. However, the analysis of solutions to (2.5) is not easy. This difficulty depends on the given function ξ . In the next sections we will give a condition on such functions under which the corresponding integral equation has a unique solution.
3 Uniqueness of Translational—Invariant Solution of (2.5) In this section we consider ξtu to be a continuous function and we are going to obtain a condition on ξtu under which the (2.5) has a unique solution in the class of translational-invariant functions f (t, x), i.e f (t, x) = f (t), for any x ∈ V. For such functions (2.5) can be written as 1 f (t) =
0 1 0
K(t, u) f (u)du
k
K(0, u) f (u)du
,
where K(t, u) = exp(Jβξtu ) > 0, f (t) > 0, t, u ∈ [0, 1]. We put C+ [0, 1] = { f ∈ C[0, 1] : f (x) 0}. We are interested in positive continuous solutions to (3.1), i.e. such that f ∈ C0+ [0, 1] = { f ∈ C[0, 1] : f (x) 0} \ {θ ≡ 0}. Note that (3.1) is not linear for any k 1.
(3.1)
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Define the linear operator W : C[0, 1] → C[0, 1] by 1 (W f )(t) = K(t, u) f (u)du,
(3.2)
0
and defined the linear functional ω : C[0, 1] → R by 1 K(0, u) f (u)du. ω( f ) ≡ (W f )(0) =
(3.3)
0
Then (3.1) can be written as f (t) = (Ak f )(t) = ((B f )(t))k ,
(3.4)
where (B f )(t) =
(W f )(t) , (W f )(0)
f ∈ C0+ [0, 1], k 1.
(3.5)
3.1 Existence of Solutions to the Nonlinear Equation (3.4) In [22] for k = 1 we have proved that the (3.4) has unique solution for arbitrary K(·, ·) ∈ C+ [0, 1]2 and f (·) ∈ C+ [0, 1]. But for k 2 the uniqueness is not proved yet. Denote m k + Fk = f ∈ C [0, 1] : f (t) , k ∈ N, M0 where m = min K(t, u), t,u∈[0,1]
M0 = max K(0, u). u∈[0,1]
It is easy to see that Fk is a closed and convex subset of C[0, 1]. Moreover this set is invariant with respect to operator Ak , i.e. Ak (Fk ) ⊂ Fk . Proposition 3.2 The operator Ak is continuous on Fk for any k 2. Proof For arbitrary C > 0 we denote F0 = f ∈ C+ [0, 1] : f (t) C, ∀t ∈ [0, 1] . It is obvious that the operator A1 is continuous on the set F0 (see Lemma 2 in [22]). Let f ∈ Fk be an arbitrary element and { fn } ⊂ Fk such that limn→∞ fn = f . Since the operator A1 is continuous we have limn→∞ A1 fn = A1 f . Consequently, there exists C1 > 0 such that A1 fn C1 for n ∈ N. Moreover we have M , t ∈ [0, 1], (A1 f )(t) C2 = m0 where M = max K(t, u), t,u∈[0,1]
m0 = min K(0, u). u∈[0,1]
On Gibbs Measure of a Models
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We have Ak fn − Ak f = (B fn )k − (B f )k = qk,n (t)(A1 fn − A1 f ),
(3.6)
where k−1 (A1 fn )k− j−1 (t)(A1 f ) j(t) > 0, qk,n (t) =
t ∈ [0, 1].
j=0
Consequently, qk,n (t) C =
k−1 (C1 )k− j−1 (C2 ) j,
t ∈ [0, 1].
j=0
Hence
Ak fn − Ak f C A1 fn − A1 f ,
n ∈ N.
Since A1 is a continuous from the last inequality it follows that Ak is continuous on Fk . Denote
Fk0
=
+
f ∈ C [0, 1] :
m M0
k
f (t)
M m0
k .
Proposition 3.3 Let k 2. If f ∈ C0+ [0, 1] is a solution of the equation Ak f = f , then f ∈ Fk0 .
Proof Straightforward. Proposition 3.4 Let k 2. The set Ak (Fk0 ) is relatively compact in C[0, 1].
Proof By Arzelá-Ascoli’s theorem (see [26], ch.III, Section 3) it suffices to prove that the set of functions Ak (Fk0 ) is equi-continuous and there exists γ > 0 such that h(t) γ , ∀t ∈ [0, 1] and ∀h ∈ Ak (Fk0 ). Let h ∈ Ak (Fk0 ) be an arbitrary function, then we have 0 < h(t)
M m0
k ,
and there exists a function f ∈ Fk0 such that h = Ak f .
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Now we shall prove that the set Ak (Fk0 ) is equi-continuous. For arbitrary t, t ∈ [0, 1] we have (h = Ak f )
|h(t) − h(t )| = |(A1 f )k (t) − (A1 f )k (t )| =
k−1 (A1 f )k− j−1 (t)(A1 f ) j(t )|(A1 f )(t) − (A1 f )(t )| j=0
1 M k−1 1 k |K(t, u) − K(t , u)| f (u)du m0 ω( f ) 0 2k−1 1 M 1 k |K(t, u) − K(t , u)|du, m0 ω( f ) 0 where ω( f ) is defined in (3.3). We have ω( f ) m0 ·
m M0
k ,
f ∈ Fk0 .
Consequently, k |h(t) − h(t )| m0
M0 m
k
M m0
2k−1
1
|K(t, u) − K(t , u)|du.
0
Since the kernel K(t, u) is uniformly continuous on [0, 1]2 , we conclude that Ak (Fk0 ) also is equi-continuous. By Propositions 3.2–3.4 and Schauder’s theorem (see [17], p. 20) one gets the following Theorem 3.5 The equation Ak f = f has at least one solution in C0+ [0, 1] and the set of all solutions of the equation is a subset in Fk0 . 3.2 Hammerstein’s Nonlinear Equation For every k ∈ N we consider an integral operator Hk acting in C+ [0, 1] as follows: 1 (Hk f )(t) = K(t, u) f k (u)du. 0
If k 2 then the operator Hk is a nonlinear operator which is called Hammerstein’s operator of order k. Moreover the linear operator equation H1 f = f has a unique positive solution f in C[0, 1] (see [13], p. 80). For a nonlinear homogeneous operator A it is known that if there is one positive eigenfunction of the operator A then the number of positive eigenfunctions is uncountable (see [13], p. 186).
On Gibbs Measure of a Models
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Denote M0 = f ∈ C+ [0, 1] : f (0) = 1 . Lemma 3.6 The equation Ak f = f,
k 2,
(3.7)
has a strongly positive solution if f the equation Hk f = λ f,
k 2,
(3.8)
has a strongly positive solution in M0 . Proof Necessity. Let f0 ∈ C0+ [0, 1] be a solution of the (3.7). We have (W f0 )(t) = ω( f0 ) k f0 (t). From this equality we get (Hk h)(t) = λ0 h(t), k
where h(t) = f0 (t) and λ0 = ω( f0 ) > 0. It is easy to see that h ∈ M0 and h(t) is an eigenfunction of the operator Hk , corresponding the positive eigenvalue λ0 . Suf f iciency. Let k 2 and h ∈ M0 be an eigenfunction of the operator Hk . Then there is a number λ0 > 0 such that Hk h = λ0 h. From h(0) = 1 we get λ0 = (Hk h)(0) = ω(hk ). Then h(t) =
Hk h . ω(hk )
From this equality we get Ak f0 = f0 with f0 = hk ∈ C0+ [0, 1]. This completes the proof. Theorem 3.7 If k 2 then every number λ > 0 is an eigenvalue of Hammerstein’s operator Hk . Proof By Theorem 3.5 and Lemma 3.6 there exist λ0 > 0 and f0 ∈ M0 such that Hk f0 = λ0 f0 . Take λ ∈ (0, +∞), λ = λ0 . Define function h0 (t) ∈ C0+ [0, 1] by λ k−1 f0 (t), t ∈ [0, 1]. h0 (t) = λ0
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Then
Hk h0 = Hk
k−1
λ f0 λ0
= λh0 .
This completes the proof. Denote
K=
Pk =
f ∈ C+ [0, 1] : M · min f (t) m · max f (t) , t∈[0,1]
m · f ∈ C[0, 1] : M
1 M
1 k−1
t∈[0,1]
M · f (t) m
1 m
1 k−1
, k 2.
Proposition 3.8 Let k 2. (a) The following holds Hk (C+ [0, 1]) ⊂ K. (b) If a function f0 ∈ C0+ [0, 1] is a solution of the equation Hk f = f,
(3.9)
then f0 ∈ Pk . Proof (a) Let h ∈ Hk (C+ [0, 1]) be an arbitrary function. Then there exists a function f ∈ C+ [0, 1] such that h = Hk f . Since h is continuous on [0, 1], there are t1 , t2 ∈ [0, 1] such that hmin = min h(t) = h(t1 ) = (Hk f )(t1 ), t∈[0,1]
hmax = max h(t) = h(t2 ) = (Hk f )(t2 ). t∈[0,1]
Hence
1
hmin m
1
f k (u)du m
0
0
K(t2 , u) k m f (u)du = hmax , M M
i.e. h ∈ K. (b) Let f ∈ C0+ [0, 1] be a solution of the (3.9). Then we have f M f k . Consequently,
f
1 M
1 k−1
.
On Gibbs Measure of a Models
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By the property (a) we have f (t) fmin = min f (t) t∈[0,1]
m
f . M
Then we obtain f (t)
m M
Also we have
1 M 1
f (t) = (Hk f )(t) m 0
1 k−1
.
k f k (u)du mfmin .
k Then fmin mfmin , i.e.
fmin
1 m
1 k−1
.
Hence be the property (a) we get f (t) fmax
M M fmin m m
1 m
1 k−1
.
Thus we have f ∈ Pk . 3.3 The Uniqueness of Fixed Point of the Operators Ak and Hk
Now we shall give a condition on the interaction (Hamiltonian) under which the equations Ak f = f and Hk f = f have a unique solution in C0+ [0, 1]. Lemma 3.9 Assume function f ∈ C[0, 1] changes its sign on [0, 1]. Then for every a ∈ R the following inequality holds
fa
1
f , n+1
n ∈ N,
where fa = fa (t) = f (t) − a, t ∈ [0, 1]. Proof By conditions of lemma there are t1 , t2 ∈ [0, 1] such that fmin = f (t1 ) < 0,
fmax = f (t2 ) > 0.
In case a = 0 the proof is obvious. We assume a > 0 (a) Let | fmin | fmax . Then f = | fmin | = | f (t1 )|. Hence
fa = max{| f (t1 ) − a|, | f (t2 ) − a|} = | f (t1 ) − a| > | f (t1 )| = f
1
f , n+1
n ∈ N.
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(b) Let | fmin | < fmax and 12 f a. Then f = fmax = f (t2 ) and f − a a > 0. Consequently,
fa = max{| f (t1 ) − a|, | f (t2 ) − a|} | f (t2 ) − a| 1 1
f
f , 2 n+1
= f − a
n ∈ N.
(c) Let | fmin | < fmax and 12 f < a. Then f = f (t2 ) and
fa = max{| f (t1 ) − a|, | f (t2 ) − a|} | f (t1 ) − a| > a >
1
f , n+1
1
f 2
n ∈ N.
Thus for a > 0 the proof is completed. For a < 0 we put ga (t) = g(t) − a with g(t) = − f (t) and a = −a > 0. Then
fa = ga
1 1
g =
f , n+1 n+1
n ∈ N.
This completes the proof. Theorem 3.10 Let k 2. If the kernel K(t, u) satisf ies the condition k M m k 1 − < , m M k
(3.10)
then the operator Hk has a unique f ixed point in C0+ [0, 1]. Proof By Theorem 3.7 the equation Hk f = f has at least one solution. Assume that there are two solutions f1 ∈ C0+ [0, 1] and f2 ∈ C0+ [0, 1], i.e Hk fi = fi , i = 1, 2. Denote f (t) = f1 (t) − f2 (t). Then by Theorem 46.6 of [14] the function f (t) changes its sign on [0, 1]. From Lemma 3.9 we get 1 1 k max f (t) − (γ1 + γ2 ) f (s)ds f , t∈[0,1] 2 2 0 where γ1 =
m k M
,
γ2 =
M m
k .
By a mean value Theorem we have 1 f (t) = K(t, u)kξ k−1 (u) f (u)du, 0 +
here ξ ∈ C [0, 1] and min{ f1 (t), f2 (t)} ξ(t) max{ f1 (t), f2 (t)},
t ∈ [0, 1].
On Gibbs Measure of a Models
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By Proposition 3.8 we have ξ ∈ Pk , i.e. m M
1 M
1 k−1
ξ(t)
M m
1 m
1 k−1
,
t ∈ [0, 1].
Hence γ1 K(t, u)ξ k−1 (u) γ2 , Therefore
Then
t, u ∈ [0, 1].
k · K(t, u)ξ k−1 (u) − γ1 + γ2 γ2 − γ1 . 2 2 1 k (γ2 − γ1 ) f . f (t) − k (γ1 + γ2 ) f (u)du 2 2 0
(3.11)
Assume the kernel K(t, u) satisfies the condition (3.10). Then k(γ2 − γ1 ) < 1 and the inequality (3.11) contradicts to Lemma 3.9. This completes the proof. Theorem 3.11 Let k 2. If the kernel K(t, u) satisf ies the condition (3.10), then for every λ > 0 the equation Hk f = λ f has unique solution in C0+ [0, 1]. Proof Clearly the equation Hk f = λ f is equivalent to the following equation 1 Kλ (t, u) f k (u)du = f (t), (3.12) 0
where Kλ (t, u) = λ1 K(t, u). The kernel Kλ (t, u) satisfies the condition (3.10) ˜ = M . Consequently, by Theorem 3.10 it follows that the ˜ = mλ and M with m λ (3.12) has a unique solution in C0+ [0, 1]. Theorem 3.12 Let k 2. If the kernel K(t, u) satisf ies the condition (3.10), then the equation Ak f = f has a unique solution in C0+ [0, 1]. C+ [0, 1], f1 = f2 , i.e. Ak fi = Proof Assume there are two solutions f1 , f2 ∈ fi , i = 1, 2. By Lemma 3.6 the functions hi (t) = k fi (t), t ∈ [0, 1] are solutions of equation Hk hi = λi hi ,
i = 1, 2,
where λi = ω( fi ) > 0 and hi ∈ M0 . On the other hand Theorem 3.11 implies that λ1 = λ2 . Let h0 (t) ∈ C+ [0, 1] be a fixed point of the operator Hk . Then by Theorems 3.7 and 3.11 we get hi = k−1 λi h0 (t), i = 1, 2.
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Consequently, f1 (t) = γ k, f2 (t)
with γ =
k−1
λ1 . λ2
Using this equality we obtain f1 (t) = (Ak f1 )(t) = Ak (γ k f2 ) = Ak f2 (t) = f2 (t).
This completes the proof. Consider the following Hamiltonian1 H(σ ) = −J
ξσ (x)σ (y) = −
x,y∈L
ln K(σ (x), σ (y)),
(3.13)
x,y∈L
where J ∈ R \ {0} and K(t, u) satisfies the condition (3.10). Then as a corollary of Proposition 2.2 and Theorem 3.12 we get the following Theorem 3.13 Let k 2. If the function K(t, u) of the Hamiltonian (3.13) satisf ies the condition (3.10), then the model (3.13) has a unique translational invariant Gibbs measure. Example It is easy to see that the condition (3.10) is satisfied iff M < ηk = m
k
1+
√ 4k2 + 1 , 2k
k 2.
(3.14)
Consider the following function K(t, u) =
n m
cijti u j + a,
cij 0,
a > 0.
(3.15)
i=1 j=1
For this function we have m = a, M = obvious
m n i=1
j=1 cij
+ a. The following is
m n (a) If a1 i=1 j=1 cij ηk − 1 then for function (3.15) the condition (3.10) is satisfied. m n (b) If a1 i=1 j=1 cij > ηk − 1 then for function (3.15) the condition (3.10) is not satisfied.
1 See
Section 4 for the physical interpretation of the condition (3.10) and the Hamiltonian.
On Gibbs Measure of a Models
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4 Physical Interpretation and Discussions Here we shall give some physical interpretation of the condition (3.10), i.e. (3.14) and the Hamiltonian.2 Note that our interaction functions are defined as logarithms of certain kernels on which we put appropriate conditions. It is known that the spin systems on trees have produced the first and most tractable examples of certain qualitative phenomena. The function ξxy is interpreted as energy function, which is as usual nonconstant, symmetric and continuous. In [18] the authors studied several models on general infinite trees, including the classical Heisenberg and Potts models. Our model (2.1) (in particular (3.13)) naturally generalizes known models (for example rotor [18], spherical [28] and many other models) with nearest-neighbor interactions which have uncountable set of spin values. Our condition (3.10) can be reformulated with respect to temperature, or with respect to interaction parameter J, or with respect to energy function ξxy . Indeed Condition on the temperature We have M maxt,u∈[0,1] K(t, u) maxt,u∈[0,1] e Jβξtu = = m mint,u∈[0,1] K(t, u) mint,u∈[0,1] e Jβξtu exp (Jβ E) , if J > 0, = = exp(|J|β E) < ηk , exp (−Jβ E) , if J < 0
(4.1)
where E = maxt,u∈[0,1] ξtu − mint,u∈[0,1] ξtu . This gives T = β −1 > i.e. if the temperature is greater than invariant Gibbs measure.
|J|E ln ηk
|J|E , ln ηk then there exists unique translation
Condition on the interaction parameter J Fix T > 0 and find J from our condition: T T T |J| < ln ηk , i.e., − ln ηk < J < ln ηk . E E E Remark 4.1 In this paper by a translation-invariant Gibbs measure we mean a Gibbs measure which corresponds (by Proposition 2.2) to a solution f (t, x) of (2.5) with f (t, x) = f (t), for any x ∈ V, i.e. f (t, x) does not depend on vertices of the Cayley tree. In this class of functions the (2.5) is reduced to the (3.1). Thus our problem is reduced to find a sufficient condition for the uniqueness of the solutions to (3.1). Note that the (3.1) is an analogue of the (12.21) of the
2 This
section is added corresponding to a suggestion of reviewer.
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book [11] and in the terminology of this book the translation-invariant Gibbs measure (discussed in this paper) corresponds to a completely homogenous Markov chain. Acknowledgements Rozikov thanks the TWAS Research Grant: 09-009 RG/Maths/As-I; UNESCO FR: 3240230333. He also thanks IHES, Bures-sur-Yvette, France for support his visit to IHES and the Department of Algebra, University of Santiago de Compostela, Spain, for providing financial support to his visit to the Department. We thank the referee for careful reading of the manuscript; in particular, for picking an error in the earlier version of the (4.1) and for a number of suggestions which have improved the paper.
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