# Extremality of Disordered Phase of \(\lambda\)-Model on Cayley Trees
## 3 Splitting Gibbs Measures
\tag{6}\]  
where \(A=\exp\{\beta\pi\},B=\exp\{\beta\overline{\xi}\},C_{k}=\exp\{\beta\overline{ \xi}_{k}\}\), \(k=1,2,3\). We recall that an SGM is called the _disordered phase_ of the \(\lambda\)-model if \(u_{1}=1,u_{2}=1\) is a solution of (6). This means that the corresponding SGM does not have external boundary conditions. **Remark 2**. : _We notice that, for symmetric models (interactions), the disordered Gibbs measure corresponds to the zero solution of the recurrent equations (Equation (4)). However, if the model is not symmetric, then the corresponding recurrent equation has no zero solution. In this case, the disordered measure is defined as a translation-invariant Gibbs measure that corresponds to the free boundary condition. Therefore, in some studies (see, for example, [29]), the disordered measure is called a free measure._  
Since \(u_{1}=1,u_{2}=1\) is a solution of (6), which yields  
\[B+C_{2}=A+C_{3},\ \ C_{1}=C_{3}. \tag{7}\]  
**Remark 3**. : _It is worth noting that (7) is satisfied for the Potts model. However, if one considers the SOS model (see Remark 1) the condition (7) is not satisfied, i.e., the model does not possess the disordered phase. Now, let us provide a model that is different from the Potts ones and for which the disordered phase exists._  
**Example 1**. : _Let \(\lambda(i,j)=J|i-j|+J_{1}(\delta_{2,i}+\delta_{2,j})\), where \(i,j\in\{1,2,3\}\) and \(J>0,J_{1}>0\). Since \(\lambda(1,1)=\lambda(3,3)\), then the first condition of (7) is reduced to_  
\[e^{2j\beta}+1=e^{(J+1)\beta}+e^{2j_{1}\beta}.\]  
_From this, we easily obtain_  
\[e^{j\beta}=\frac{1}{2}\left(e^{j_{1}\beta}+\sqrt{5e^{2j_{1}\beta}-4}\right). \tag{8}\]  
_Hence, if (8) is satisfied, then the considered model possesses the disordered phase._  
_Clearly, the model is totally different from the [PERSON] and SOS models._  
Hence, in the sequel, we assume that (7) is satisfied. Therefore, one infers that the line \(u_{1}=1\) in (6) is invariant, so the equation over this invariant line reduces to  
\[u_{2}=\left(\frac{C_{2}u_{2}+2B}{Bu_{2}+B+C_{2}}\right)^{k}. \tag{9}\]  
By denoting  
\[X=\frac{C_{2}u_{2}}{2B},\ \ \ a=\frac{1}{8B^{5}},\ \ \ b=\frac{C_{2}(B+C_{2})}{2B^{2}}, \tag{10}\]we rewrite (9) as follows:  
\[aX=\bigg{(}\frac{1+X}{b+X}\bigg{)}^{2}. \tag{11}\]  
One can see that \(X>0\), \(k\geq 1\), \(a>0\), and \(b>0\). Then, through the use of [4] (Proposition 10.7), one finds the following:  
**Lemma 1** ([24]). : __  
1. _If_ \(b\leq\bigg{(}\frac{k+1}{k-1}\bigg{)}^{2}\)_, then a solution to (_11_) is unique._
2. _If_ \(b>\bigg{(}\frac{k+1}{k-1}\bigg{)}^{2}\)_, then there are_ \(\zeta_{1}(b)\) _and_ \(\zeta_{2}(b)\)_, such that_ \(0<\zeta_{2}<\zeta_{2}\)_, and if_ \(\zeta_{1}<a<\zeta_{2}\)_, then (_11_) has three solutions._
3. _If_ \(a=\zeta_{1}\) _and_ \(a=\zeta_{2}\)_, then (_11_) has two solutions. The quantity_ \(\zeta_{1}\) _and_ \(\zeta_{2}\) _are determined from the formula_ \[\zeta_{i}(b)=\frac{1}{x_{i}}\bigg{(}\frac{1+x_{i}}{b+x_{i}}\bigg{)}^{k},\ i=1,2,\] (12) _where_ \(x_{1}\) _and_ \(x_{2}\) _are solutions to the equation_ \(z^{2}+(3-b)z+b=0\)_._  
The condition (i) of Lemma 1 reduces to  
\[\frac{C_{2}(B+C_{2})}{2B^{2}}<\bigg{(}\frac{k+1}{k-1}\bigg{)}^{2}\]  
which is equivalent to  
\[\bigg{(}\frac{C_{2}}{B}\bigg{)}^{2}+\frac{C_{2}}{B}-2\bigg{(}\frac{k+1}{k-1} \bigg{)}^{2}<0 \tag{13}\]  
by solving the last one, we obtain the following fact. **Theorem 2**. : _Assume that, for the \(\lambda\)-model given by (5), the condition (7) is satisfied. Then, the following statements hold:_  
1. _If_ \[\frac{C_{2}}{B}<\frac{1}{2}\bigg{(}\sqrt{1+8\bigg{(}\frac{k+1}{k-1}\bigg{)}^{ 2}}-1\bigg{)},\] (14) _then the disordered phase is unique;_
2. _If_ \[e^{2j\beta}+1=e^{(J+1)\beta}+e^{2j_{1}\beta}.\]0 (15) _then the phase transition occurs._