assume that the orbit of $1$ is not periodic. The solution is $\sigma^{p}\ud{u}=\ud{v}^\ab$, so that $\ud{u}=\ud{a}^\prime\ud{v}^\ab\prec\ud{u}^\ab$ is also a solution of . In that case there is no other solution for . The borderline case $\alpha=1$ corresponds to the periodic orbit of the fixed point $0$, $\ud{u}^{1,\beta}=(1)^\infty$. Notice that $\ophi^{1,\beta}(\sigma\ud{u}^{1,\beta})\not=1$. We can also consider $\ophi^{1,\beta}$-expansions of $0$ with $u_0=0$ and $\ophi^{1,\beta}(\sigma\ud{u})=1$. Our hypothesis on $\ud{u}$ imply that $\ophi^{1,\beta}(\sigma^2\ud{u})=\gamma$. Hence, $\ud{u}=0\ud{v}^{1,\beta}=\ud{a}^\prime\ud{v}^\ab\prec\ud{u}^\ab$ is a solution of and a $\ophi^{1,\beta}$-expansion of $0$.

We can treat similarly the case when $\ud{u}^\ab$ is not periodic, but $\ud{v}^\ab$ is periodic. When both $\ud{u}^\ab$ and $\ud{v}^\ab$ are periodic we have more solutions, but the discussion is similar. Assume that $\ud{u}^\ab$ has (minimal) period $p>1$ and $\ud{v}^\ab$ has (minimal) period $q>1$. Define $\ud{a}$, $\ud{a}^\prime$ as before, $\ud{b}$ as the prefix of length $q$ of $\ud{v}^\ab$, and $\ud{b}^\prime$ as the word of length $q$ obtained by changing the last letter of $\ud{b}$ into $v_{q-1}^\ab+1$. When $0<\alpha<1$ and $0<\gamma<1$, one shows as above that the elements $\ud{u}\not=\ud{u}^\ab$ and $\ud{v}\not=\ud{v}^\ab$ which are $\oph^\ab$-expansions of $0$ and $1$ are of the form $$\ud{u}=\ud{a}^\prime\ud{b}^{n_1}\ud{b}^\prime\ud{a}^{n_2}\cdots\,,\,n_i\geq
0 \quad\text{and}\quad
\ud{v}=\ud{b}^\prime\ud{a}^{m_1}\ud{a}^\prime\ud{b}^{m_2}\cdots\,,\,m_i\geq
0 \,.$$ The integers $n_i$ and $m_i$ must be such that is verified. The largest solution 