Let $X$ be a smooth projective surface over $\mathbb{C}$, $D\subset X$ is an effective divisor. $(X,D)$ is a log CalabiYau pair if $K_X+D$ is a principal divisor. The complement $M=X\setminus D$ is a log CalabiYau surface if the curve $D$ has at worst nodal singularities. The pair $(X,D)$ is said to have maximal boundary if $D$ is actually nodal. As a consequence of Bertini's theorem, one can always find a smoothing $\widetilde{D}$ of $D$ in the same linear system so that $(X,\widetilde{D})$ becomes a log CalabiYau pair without maximal boundary. However, I'm interested in whether the converse is true, namely whether starting from a log CalabiYau pair $(X,\widetilde{D})$ such that $\widetilde{D}$ is not maximal, one can always find a degeneration $D$ of $\widetilde{D}$ so that $(X,D)$ is a log CalabiYau pair with maximal boundary?

1$\begingroup$ Why you can always find a smoothing? The linear system is not necessarily base point free. $\endgroup$ – Chen Jiang Jul 7 '19 at 2:24
Firstly, I do not think that every maximal boundary has a smoothing. For example, if the linear system $K_X$ as fixed part, then very likely it contains no smooth element.
On your question of the converse part, we can not expect to find a maximal boundary. For the easiest example, just consider $X=E\times \mathbb{P}^1$, where $E$ is an elliptic curve. Then every element in the linear system of $K_X$ Is just two copies of $E$, which is smooth. So there is mo nodal boundary.

$\begingroup$ ...or an abelian surface (with $D=0$). Maybe the question becomes more interesting if $X$ is rationally connected (or Fano?). $\endgroup$ – Piotr Achinger Jul 7 '19 at 10:21