Fixed points of Lyapunov integral operators and Gibbs measures

In this paper we shall consider the connections between Lyapunov integral operators and Gibbs measures for four competing interactions of models with uncountable (i.e. $[0,1]$) set of spin values on a Cayley tree. And we shall prove the existence of fixed points of Lyapunov integral operators and give a condition of uniqueness of fixed points.


Preliminaries
A Cayley tree Γ k = (V, L) of order k ∈ N is an infinite homogeneous tree, 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).Two vertices x and y are called nearest neighbors if there exists an edge l ∈ L connecting them.We will use the notation l = x, y .The distance d(x, y), x, y ∈ V on the Cayley tree is defined by the formula d(x, y) = min{d| x = x 0 , x 1 , ..., x d−1 , x d = y ∈ V such that the pairs < x 0 , x 1 >, ..., < x d−1 , x d > are neighboring vertices}.Let x 0 ∈ V be a fixed and we set The set of the direct successors of x is denoted by S(x), i.e.

S(x)
We observe that for any vertex x = x 0 , x has k direct successors and x 0 has k + 1.The vertices x and y are called second neighbor which is denoted by > x, y <, if there exist a vertex z ∈ V such that x, z and y, z are nearest neighbors.We will consider only second neighbors > x, y <, for which there exist n such that x, y ∈ W n .Three vertices x, y and z are called a triple of neighbors and they are denoted by < x, y, z >, if < x, y >, < y, z > are nearest neighbors and x, z ∈ W n , y ∈ W n−1 , for some n ∈ N. Now we consider models with four competing interactions where the spin takes values in the set [0, 1].For some set A ⊂ V an arbitrary function σ A : A → [0, 1] is called a configuration and the set of all configurations on A we denote by are given bounded, measurable functions.Then we consider the model with four competing interactions on the Cayley tree which is defined by following Hamiltonian where the sum in the first term ranges all triples of neighbors, the second sum ranges all second neighbors, the third sum ranges all nearest neighbors and J, J where x 0 is a root of Cayley tree and C is a constant which does not depend on t.For some n ∈ N, σ n : x ∈ V n → σ(x) and Z n is the corresponding partition function we consider the probability distribution µ (n) on Ω Vn defined by Z n = ... where For n ∈ N we say that the probability distributions µ (n) are compatible if µ (n) satisfies the following condition: By Kolmogorov'sigma extension theorem there exists a unique measure µ on Ω V such that, for any n and σ n ∈ Ω Vn , µ ({σ| Vn = σ n }) = µ (n) (σ n ).The measure µ is called splitting Gibbs measure corresponding to Hamiltonian (1.1) and function x → h x , x = x 0 (see [1], [4], [5]).Denote and The following statement describes conditions on h x guaranteeing compatibility of the corresponding distributions µ (n) (σ n ).Proposition 1.1.[6] The measure µ (n) (σ n ), n = 1, 2, . . .satisfies the consistency condition (1.4) iff for any x ∈ V \ {x 0 } the following equation holds: where S(x) = {y, z}, < y, x, z > is a ternary neighbor.

Existence of fixed point of the operator L
Now we prove that there exist at least one fixed point of Lyapunov integral equation, namely there is a splitting Gibbs measure corresponding to Hamiltonian (1.1).Proposition 2.1.Let J 3 = J = α = 0 and J 1 = 0. Then (1.6) is equivalent to where Proof.For J 3 = J = α = 0 and Then (1.6) can be written as 2) is equivalent to (2.1).
Now we consider the case J 3 = 0, J = J 1 = α = 0 for the model (1.1) in the class of translational-invariant functions f (t, x) i.e f (t, x) = f (t), for any x ∈ V .For such functions equation (1.1) can be written as where We shall find positive continuous solutions to (2.3) i.e. such that Define a nonlinear operator H on the cone of positive continuous functions on [0, 1] : .
We'll study the existence of positive fixed points for the nonlinear operator H (i.e., solutions of the equation ( 2 We define the Lyapunov integral operator L on C[0, 1] by the equality (see [2]) Lemma 2.2.The equation Hf = f has a nontrivial positive solution iff the Lyapunov equation Lg = g has a nontrivial positive solution.
Proof.At first we shall prove that the equation has a positive solution iff the Lyapunov equation has a positive solution in M 0 for some λ > 0. Let λ 0 be a positive eigenvalue of the Lyapunov operator L. Then there exists . Then Lh 0 = λh 0 , i.e., the number λ is an eigenvalue of Lyapunov operator L corresponding the eigenfunction h 0 (t).It's easy to check that if the number λ 0 > 0 is an eigenvalue of the operator L, then an arbitrary positive number is eigenvalue of the operator L. Now we shall prove the lemma.Let equation (11) holds then the function 1 λ g(t) be a fixed point of the operator L. Analogously, since H is non-linear operator we can correspond to the fixed point if there exist any eigenvector.1], where m = min K(t, s, u) > 0.
Put Γ = {f : f = r, f ∈ C[0, 1]}.We define the set Γ + by Then we obtain inf Then by Schauder's theorem (see [3], p.20) there exists a number λ 0 > 0 and a function Denote by N f ix.p (H) and N f ix.p (L) are the set of positive numbers of nontrivial positive fixed points of the operators N f ix.p (H) and N f ix.p (L), respectively.By Lemma 2.2 and Proposition 2.3 we can conclude that:

The uniqueness of fixed point of the operator L
In this section we shall give a condition of the uniqueness of fixed point of the operator L. Theorem 3.1.Let the kernel K(t, u, v) satisfies the condition Then the operator L has the unique fixed point in Let s ∈ L(C + [0, 1]) be an arbitrary function.Then there exists a function h ∈ C + [0, 1] such that s = Lh.Since s is continuous on [0, 1], there exists t 1 , t 2 ∈ [0, 1] such that Consequently we get Since g is a fixed point of the operator L we have g Now we show that L has the unique fixed point.By Theorem Lg = g has at least one solution.Assume that there are two solutions g 1 ∈ C + 0 [0, 1] and g 2 ∈ C + 0 [0, 1], i.e Lg i = g i , i = 1, 2. Put ξ(t) = g 1 (t) − g 2 (t).Then ξ(t) changes its sign on [0, 1] and we get Since g i (t) ∈ G, i = 1, 2 we get ω Ω 2 ≤ η(u, v) ≤ Ω ω 2 , (u, v) ∈ [0, 1] 2 .Hence

Proposition 2 . 4 .
a)The equation(2.4) has at least one solution inC + 0 [0, 1].b) The equality N f ix.p (H) = N f ix.p (L) is hold.From Proposition 1.1 and Proposition 2.4 we get the following theorem.Theorem 2.5.The set of splitting Gibbs measures corresponding to Hamiltonian (1.1) is nonempty.