Suppose $A$ is a finite alphabet. $x \notin A$. Let's call a word equation over $A$ a pair of words $(w, u) \in (A \cup {x})^* \times (A \cup {x})^*$. Let's call a word $\alpha \in A^*$ a solution of an equation $(w, u)$ iff $w[x := \alpha] = u[x := \alpha]$. Let's define the elementary equational language described by $(w, u)$ as $Eq(w, u)$ - the set of all solutions of $(w, u)$. Are all elementary equational languages regular?
On one hand, I managed to prove the finiteness of a large class of elementary equational languages.
Let's call equation $(u, w)$ balanced iff the numbers of occurrences of $x$ in $u$ and $w$ are the same, and unbalanced otherwise.
If $(u, w)$ is unbalanced, then $|Eq(u, w)| \leq 1$
Proof:
If $w[x := \alpha] = u[x := \alpha]$ then $|w[x := \alpha]| = |u[x := \alpha]|$. Thus $|\alpha|$ is uniquely defined by corresponding linear equation. Thus $Eq(u, w)$ consists of words of the same length.
Now suppose $L$ is some finite language, all words of which are of length $n$. Suppose $L := Eq(u, w)$, $(u, w)$ is unbalanced and the minimal number of $x$ among $u$ and $w$ is $m$. Then we can prove by double induction by $n$ and $m$ that $|L| \leq 1$.
Base for $n$: If $n = 0$ there is only one word of that length.
Base for $m$: If $m = 0$, then $\alpha$ is uniquely defined (if it exists) as a corresponding subword of $w$.
Double step: Suppose $n > 0$, $m > 0$ and for all lesser $n$ and $m$ the statement is true. Then $w = axb$, $u = cxd$ where $a, c \in ^*$, $b, d \in (A \cup x)^*$. Then either the equation has no solutions or (without the loss of generality) $a = ct$ for some $t \in A^*$ and our equation is equivalent to $(txb, xd)$. If $t = \Lambda$, then this equation can be simplified to $(b, d)$, which is also unbalanced and in which the word with the least number of $x$ has $m - 1$ of them. If $|t| \geq 1$, then $L = tK$, where $K = Eq(tx(b[x := tx]), x(d[x := tx]))$, and the length of words in $K$ is less than $n - |t| < n$.
Thus the statement of the proposition follows by mathematical induction.
Q.E.D.
On the other hand, we can get numerous infinite examples using balanced equations. For example, if $a$ is a primitive word in $A^*$, then $Eq(xa, ax) = \{a\}^*$. However, this family of examples is also regular...
Yes, the set of solutions form a regular language, as shown in [1, 2]. From the abstract:
[1] Laine, Markku; Plandowski, Wojciech. Word equations with one unknown. Developments in language theory, 348-359, LNCS 5583, Springer, Berlin, 2009.
[2] Laine, Markku; Plandowski, Wojciech. Word equations with one unknown. Internat. J. Found. Comput. Sci. 22 (2011), no. 2, 345--375.