part (
i
) is trivial since
f
(
h
) has a unique stable fixed point. Here we prove part (
ii
),
similar arguments also apply to part (
iii
). Let
θ
=
θ
c
.
In this case
f
0
(
α
1
)
<
1 and
f
0
(
α
2
) = 1
.
For any
x
0
we are aiming to prove the convergence of recurrent sequence
x
n
=
f
(
x
n
−
1
)
, n
≥
1 as
n
→ ∞
.
We consider the partition
(
−∞
,
+
∞
) = (
−∞
, α
2
)
∪ {
α
2
} ∪
(
α
2
, α
1
)
∪ {
α
1
} ∪
(
α
1
,
+
∞
)
.
Letting
x
∈
(
−∞
, α
2
)
,
then we have
x < f
(
x
)
< α
2
. Because the function
f
(
x
) is
increasing for
θ <
1, by iteration it follows that
f
n
−
1
(
x
)
< f
n
(
x
)
< α
2
,
which for
any
x
0
∈
(
−∞
, α
2
) we get
x
n
−
1
< x
n
< α
2
,
i.e.,
x
n
converges and its limit is a fixed
point of
f,
because
f
has unique fixed point
α
2
in (
−∞
, α
2
] we deduce that the limit
is
α
2
.
The rest of the proof runs as in the previous case.
Due to (
9
) the asymptotic behaviour of the vector
Y
n
(
w
) = (
Y
0
n
(
ω
)
, Y
1
n
(
ω
))
depends only on the vector
Y
1
(
ω
)
,
where
Y
l
1
(
ω
) =
J
(
c
l
(
ω
)
−
c
2
(
ω
))
, l
= 0
,
1
.
(17)
We denote the following sets:
D
=
{
ω
∈
Ω :
c
0
(
ω
) =
c
2
(
ω
)
}
,
D
+
=
{
ω
∈
D
:
J
(
c
1
(
ω
)
−
c
2
(
ω
))
> α
2
}
,
D
0
=
{
ω
∈
D
:
J
(
c
1
(
ω
)
−
c
2
(
ω
)) =
α
2
}
,
D
−
=
{
ω
∈
D
:
J
(
c
1
(
ω
)
−
c
2
(
ω
))
< α
2
}
.
Taking the coordinates of an initial state as in (
17
) then from Lemma
1
and
Lemma
4
we have
Theorem 1
(i) If
θ > θ
c
and
ω
∈
D
then
P
ω
=
µ
1
(
θ
)
.
(ii) If
θ
=
θ
c
then
P
ω
=
µ
2
(
θ
)
,
if
ω
∈
D
0
∪
D
−
,
µ
1
(
θ
)
,
if
ω
∈
D
+
.
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(iii) If
θ < θ
c
then
P
ω
=
µ
3
(
θ
)
,
if
ω
∈
D
−
,
µ
2
(
θ
)
,
if
ω
∈
D
0
,
µ
1
(
θ
)
,
if
ω
∈
D
+
.
3 Some examples of boundary conditions
In this section we construct several boundary configurations for the TISGMs
µ
i
, i
= 1
,
2
,
3
.
Case
µ
1
.
If
θ > θ
c
,
then
µ
1
is a unique measure. In this case we could take
any configuration which is
ω
∈
D
,
i.e.
c
0
(
ω
) =
c
2
(
ω
) (see Fig.
1
). If
θ
≤
θ
c
,
one
has to take
ω
∈
D
+
, i.e.,
J
(
c
1
(
ω
)
−
c
2
(
ω
))
> α
2
, c
0
(
ω
) =
c
2
(
ω
)
.
(18)
Recall
J
= ln
θ.
It is easy to see that
−
ln
θ > α
2
(
θ
)
(19)
is satisfied for all
θ
≤
θ
c
,
where
θ
c
≈
0
.
1414 (see Lemma
3
). Taking into
account (
19
), we could choose a configuration
ω
such that
c
0
(
ω
) =
c
2
(
ω
) = 2
and
c
1
(
ω
) = 1 (see, e.g., Fig.
1
) satisfying the system (
18
).
Remark 1
To obtain a boundary condition for the specific TISGM, we have to con-
struct configurations satisfying certain conditions, e.g., (
18
). Thus, one can deduce
that some TISGMs might possess an infinite set of boundary conditions.
Case
µ
2
.
The measure
µ
2
ceases to exist when
θ > θ
c
.
For this reason, we
should consider
θ
≤
θ
c
.
Let
θ
=
θ
c
.
In this case
ω
∈
D
0
∪
D
−
,
i.e.
J
(
c
1
(
ω
)
−
c
2
(
ω
))
≤
α
2
, c
0
(
ω
) =
c
2
(
ω
)
.
(20)
Note that the inequality
ln
θ
c
< α
2
(
θ
c
)
(21)
holds. If we take a configuration
ω
such that
c
0
(
ω
) =
c
2
(
ω
) = 1 and
c
1
(
ω
) = 2
then the system (
20
) is satisfied due to (
21
). Such a configuration is depicted
in Fig.
2
.
Let
θ < θ
c
.
Then by Theorem
1
we have the condition
ω
∈
D
0
,
i.e.
J
(
c
1
(
ω
)
−
c
2
(
ω
)) =
α
2
, c
0
(
ω
) =
c
2
(
ω
)
.
(22)
Note that
c
i
(
ω
)
∈ {
0
,
1
,
2
,
3
,
4
}
, i
= 0
,
1
,
2
.
Denote
n
0
=
c
1
(
ω
)
−
c
2
(
ω
)
∈
{−
4
,
−
3
,
−
2
,
−
1
,
0
,
1
,
2
}
.
A computer analysis shows that the equation
n
0
ln
θ
=
α
2
(
θ
)
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Fig. 1
(Geogebra) A boundary condition which corresponds to TISGMs
µ
1
such that
c
1
(
ω
) = 1 and
c
0
(
ω
) =
c
2
(
ω
) = 2
.
has no solution. In this case we have no information about which kind of
boundary condition is needed to obtain the TISGM
µ
2
.
Remark 2
For Potts model in [
2
] a set of configurations, satisfying a condition which
is similar to (
22
), is found for sufficiently large number of spin values. In our case we
consider the model with only three spin values, thus, arising uncertainty seems to be
relevant.
Case
µ
3
.
The measure
µ
3
exists when
θ
≤
θ
c
.
If
θ
=
θ
c
then
µ
3
coincides
with
µ
2
and it is examined in the previous case. If
θ < θ
c
we get the condition
ω
∈
D
−
,
i.e.,
J
(
c
1
(
ω
)
−
c
2
(
ω
))
< α
2
, c
0
(
ω
) =
c
2
(
ω
)
.
(23)
We consider the following inequality
ln
θ < α
2
(
θ
)
,
(24)
which holds when
θ < θ
c
.
Together with (
24
), the system (
23
) is satisfied, for
instance,
ω
is such that
c
0
(
ω
) =
c
2
(
ω
) = 1 and
c
1
(
ω
) = 2 (see Fig.
2
).
Remark 3
A configuration, depending on parameter
θ
, might be a boundary condi-
tion for different TISGMs. For example, the configuration which drawn in Fig.
2
is a
boundary condition for
µ
2
when
θ
=
θ
c
, at the same time, it is a boundary condition
for
µ
3
when
θ
≤
θ
c
.
Acknowledgments.
The authors are greatly indebted to Professor U.A.
Rozikov for suggesting the problem and for many stimulating conversations.
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The boundary condition problems for the three-state SOS model on the binary tree.
Fig. 2
(Geogebra) An example of a boundary condition for TISGM
µ
2
when
θ
=
θ
c
and
for TISGM
µ
3
when
θ
≤
θ
c
. Here we have
c
0
(
ω
) = 1
, c
1
(
ω
) = 2 and
c
2
(
ω
) = 1
.
4 Data Availability Statement
Not applicable
Conflicts of Interest.
The authors declare that they have no conflict of
interest
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M.M.,
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U.A.:
Boundary
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J.
Stat.
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167
,
1164-1179
(2017)
https://doi.org/10.1007/s10955-017-1771-5
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The boundary condition problems for the three-state SOS model on the binary tree.
11
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Document Outline - Introduction
- Translation-invariant limiting Gibbs measures
- Some examples of boundary conditions
- Data Availability Statement
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