Let $\frac{\partial^2}{\partial x^2}: L^2([0,1]) \to L^2([0,1])$ be the one-dimensional Laplacian, considered as an unbounded, densley defined operator with domain $\mathcal D(\frac{\partial^2}{\partial x^2}) = C^\infty([0,1])$ the smooth funtions without boundary conditions.
Question 1: Can one explicitly describe the domain $\mathcal D((\frac{\partial^2}{\partial x^2})^*)$ of the adjoint operator ?
By means of partial integration, one easily checks that
a) $B := \{ f \in C^2([0,1]): f(0) = f'(0) = f(1) = f'(1) = 0 \} \subseteq \mathcal D((\frac{\partial^2}{\partial x^2})^*)$, so that $(\frac{\partial^2}{\partial x^2})^*$ is densely defined (in particular, $\frac{\partial^2}{\partial x^2}$ is closable).
b) $(\frac{\partial^2}{\partial x^2})^*|_{B} = \frac{\partial^2}{\partial x^2}|_{B}$.
c) The restriction $\frac{\partial^2}{\partial x^2}|_{B}$ is symmetric and densley defined, hence closable.
By c), we may consider the minimal closure $\overline{\frac{\partial^2}{\partial x^2}|_{B}}$. Now since $(\frac{\partial^2}{\partial x^2})^*$ is a closed extension of $(\frac{\partial^2}{\partial x^2})^*|_{B} = \frac{\partial^2}{\partial x^2}|_{B}$ by b), we must have $\overline{\frac{\partial^2}{\partial x^2}|_{B}} \subset (\frac{\partial^2}{\partial x^2})^*$.
Question 2: Do we have the equality of closed operators $(\frac{\partial^2}{\partial x^2})^* = \overline{\frac{\partial^2}{\partial x^2}|_{B}}$. If not, what is their difference ?