following discussion we consider the case when there are several solutions. The main results are summarize in Proposition \[pro3.4ter\]. We assume the validity of the $\varphi$-expansion.

Suppose first that the orbit of $1$ is not periodic and that the orbit of $0$ is periodic, with minimal period $p:=\min\{k\colon
T^k(0)=0\}>1$. Hence $0<\gamma<1$ and $0<\alpha<1$. Let $\ud{u}$ be a solution of equations and suppose furthermore that $\ud{w}$ is a $\varphi$-expansion of $1$ such that $$\forall n\colon\; \ud{u}\preceq\sigma^n\ud{u}\preceq
\ud{w}\quad\text{with}\quad \ophi^\ab(\ud{w})=1\;,\;
\ophi^\ab(\sigma\ud{w})\leq\gamma\,.$$ By Lemma \[lemelementary\] we conclude that $$u_j=u_j^\ab\quad\text{and}\quad
T_\ab^{j+1}(0)=\ophi^\ab(\sigma^{j+1}\ud{u})\,,\quad
j=1,\ldots,p-2\,.$$ Since $T^p(0)=0$, $T^{p-1}(0)\in \{a_1,\ldots,a_{k-1}\}$ and the equation $$T_\ab^{p-1}(0)=\ophi^\ab\big(u_{p-1}+\ophi^\ab(\sigma^{p}\ud{u})\big)$$ has two solutions. Either $u_{p-1}=u_{p-1}^\ab$ and $\ophi^\ab(\sigma^{p}\ud{u})=0$ or $u_{p-1}=u_{p-1}^\ab-1$ and $\ophi^\ab(\sigma^{p}\ud{u})=1$. Let $\ud{a}$ be the prefix of $\ud{u}^\ab$ of length $p$ and $\ud{a}^\prime$ the word of length $p$ obtained by changing the last letter of $\ud{a}$ into[^7] $u_{p-1}^\ab-1$. We have $\ud{a}^\prime<\ud{a}$. If $u_{p-1}=u_{p-1}^\ab$, then we can again determine uniquely the next $p-1$ letters $u_i$. The condition $\ud{u}\leq\sigma^k\ud{u}$ for $k=p$ implies that we have $u_{2p-1}=u_{p-1}^\ab$ so that, by iteration, we get the solution $\ud{u}=\ud{u}^\ab$ for the equations . If $u_{p-1}=u_{p-1}^\ab-1$, then $$1=\ophi^\ab(\sigma^{p}\ud{u})=\ophi^\ab\big(u_p+\ophi^\ab(\sigma^{p+1}\ud{u})\big)\,.$$ When $\ophi^\ab(\sigma^{p}\ud{u})=1$, by our hypothesis on $\ud{u}$ we also have $\ophi^\ab(\sigma^{p+1}\ud{u})=\gamma$. By Proposition \[pro3.4bis\] the equations $$\ophi^\ab(\sigma^{p}\ud{u})=1\quad\text{and}\quad
\ophi^\ab(\sigma^{p+1}\ud{u})=\gamma$$ have a unique solution, since we 