DGT equilibrium density profile prediction in the NVT ensemble

Each value names the invariance the profile is assumed to have: planar is translation in two directions, spherical rotation about a point, prismatic translation in one direction, and axial rotation about an axis. What is not invariant is resolved.

Planar interface in a cubic domain \(V=\,]0,Z[^3\): profiles vary only in the \(z\)-direction.

Spherical interface in a ball domain \(V=\{\boldsymbol{x}\in\mathbb{R}^3\,\big\vert\, \lvert\boldsymbol{x}\rvert \lt R\}\): profiles vary only in the radial direction \(r\).

\(V\) must be large enough so that bulk phases are adjacent to the domain boundary \(\partial V\).

The two symmetries that resolve two coordinates are developer-only: prismatic over a square cross-section \(]0,Z[^2\), and axial over a meridional section \((0,R)\times(0,Z)\) with \(R=Z\) and \(V=\pi R^2 Z\). Both cost \(S^2\) cells rather than \(S\), so the number of cells below means cells per axis there.

The macroscopic configuration the run starts from. It selects which stationary state the solve converges to, since more than one can exist at the same state — a droplet on the axis and a cylinder spanning the domain, for instance.

Stripe keeps the profile one-dimensional. Every other name is the body the configuration has in space: cylinder (prismatic) shows as a circle in the section, ball is a free droplet on the axis, cap sits on the wall \(z=0\), and column spans both walls. The last three need the axial symmetry, the cylinder the prismatic one, and all four exactly two coexisting phases; otherwise the run falls back to the stripe and says so.

Impermeable: no-flux boundary condition at the 1D computational domain boundary.

Periodic: opposite 1D boundaries are connected periodically.

Spherical symmetry supports impermeable only.

Equidistant decomposition of \([0,Z]\); higher values improve spatial resolution at higher cost.

On a two-dimensional symmetry this is the number of cells per axis, so the cost grows as \(S^2\): switching to one of those symmetries lowers the value if it is above what such a run can afford.

Equilibrium: iterate until the stationary NVT system is solved directly (machine-precision chemical-potential residual); the tolerance \( \varepsilon_\mu \) is not used.

Tolerance: iterate the pseudo-time evolution until \( R_\mu \le \varepsilon_\mu \), then attempt the stationary solve once.