Time-stepping and the fractional step method
The time-integral of the momentum equation with the pressure decomposition from time step $n$ at $t = t_n$ to time step $n+1$ at $t_{n+1}$ is
\[ \begin{equation} \label{eq:momentum-time-integral} \boldsymbol{u}^{n+1} - \boldsymbol{u}^n = \int_{t_n}^{t_{n+1}} \Big [ - \boldsymbol{\nabla} \phi_{\rm{non}} - \boldsymbol{\nabla}_{h} \phi_{\rm{hyd}} - \left ( \boldsymbol{u} \boldsymbol{\cdot} \boldsymbol{\nabla} \right ) \boldsymbol{u} - \boldsymbol{f} \times \boldsymbol{u} + \boldsymbol{\nabla} \boldsymbol{\cdot} \boldsymbol{\tau} + \boldsymbol{F}_{\boldsymbol{u}} \Big ] \, \mathrm{d} t \, , \end{equation}\]
where the superscript $n$ and $n+1$ imply evaluation at $t_n$ and $t_{n+1}$, such that $\boldsymbol{u}^n \equiv \boldsymbol{u}(t=t_n)$. The crux of the fractional step method is to treat the pressure term $\boldsymbol{\nabla} \phi_{\rm{non}}$ implicitly using the approximation
\[\int_{t_n}^{t_{n+1}} \boldsymbol{\nabla} \phi_{\rm{non}} \, \mathrm{d} t \approx \Delta t \boldsymbol{\nabla} \phi_{\rm{non}}^{n+1} \, ,\]
while treating the rest of the terms on the right hand side of \eqref{eq:momentum-time-integral} explicitly. The implicit treatment of pressure ensures that the velocity field obtained at time step $n+1$ is divergence-free.
To effect such a fractional step method, we define an intermediate velocity field $\boldsymbol{u}^\star$ such that
\[ \begin{equation} \label{eq:intermediate-velocity-field} \boldsymbol{u}^\star - \boldsymbol{u}^n = \int_{t_n}^{t_{n+1}} \boldsymbol{G}_{\boldsymbol{u}} \, \mathrm{d} t \, , \end{equation}\]
where
\[\boldsymbol{G}_{\boldsymbol{u}} \equiv - \boldsymbol{\nabla}_h \phi_{\rm{hyd}} - \left ( \boldsymbol{u} \boldsymbol{\cdot} \boldsymbol{\nabla} \right ) \boldsymbol{u} - \boldsymbol{f} \times \boldsymbol{u} + \boldsymbol{\nabla} \boldsymbol{\cdot} \boldsymbol{\tau} + \boldsymbol{F}_{\boldsymbol{u}}\]
collects all terms on the right side of the time-integral of the momentum equation except the contribution of non-hydrostatic pressure $\boldsymbol{\nabla} \phi_n$. The integral on the right of the equation for $\boldsymbol{u}^\star$ may be approximated by a variety of explicit methods: for example, a forward Euler method uses
\[ \begin{equation} \int_{t_n}^{t_{n+1}} G \, \mathrm{d} t \approx \Delta t G^n \, , \label{eq:forward-euler} \end{equation}\]
for any time-dependent function $G(t)$, while a second-order Adams-Bashforth method uses the approximation
\[ \begin{equation} \label{eq:adams-bashforth} \int_{t_n}^{t_{n+1}} G \, \mathrm{d} t \approx \Delta t \left [ \left ( \tfrac{3}{2} + \chi \right ) G^n - \left ( \tfrac{1}{2} + \chi \right ) G^{n-1} \right ] \, , \end{equation}\]
where $\chi$ is a parameter. Ascher et al. (1995) claim that $\chi = \tfrac{1}{8}$ is optimal; $\chi=-\tfrac{1}{2}$ yields the forward Euler scheme.
Combining the equations for $\boldsymbol{u}^\star$ and the time integral of the momentum equation yields
\[ \begin{equation} \label{eq:fractional-step} \boldsymbol{u}^{n+1} - \boldsymbol{u}^\star = - \Delta t \boldsymbol{\nabla} \phi_{\rm{non}}^{n+1} \, . \end{equation}\]
Taking the divergence of fractional step equation and requiring that $\boldsymbol{\nabla} \boldsymbol{\cdot} \boldsymbol{u}^{n+1} = 0$ yields a Poisson equation for the potential $\phi_{\rm{non}}$ at time-step $n+1$:
\[ \nabla^2 \phi_{\rm{non}}^{n+1} = \frac{\boldsymbol{\nabla} \boldsymbol{\cdot} \boldsymbol{u}^{\star}}{\Delta t} \, .\]
With $\boldsymbol{u}^\star$ and $\phi_{\rm{non}}$, $\boldsymbol{u}^{n+1}$ is then computed via the fractional step equation.
Tracers are stepped forward explicitly via
\[ \begin{equation} \label{eq:tracer-timestep} c^{n+1} - c^n = \int_{t_n}^{t_{n+1}} G_c \, \mathrm{d} t \, , \end{equation}\]
where
\[ G_c \equiv - \boldsymbol{\nabla} \boldsymbol{\cdot} \left ( \boldsymbol{u} c \right ) - \boldsymbol{\nabla} \boldsymbol{\cdot} \boldsymbol{q}_c + F_c \, ,\]
and the same forward Euler or Adams-Bashforth scheme as for the explicit evaluation of the time-integral of $\boldsymbol{G}_u$ is used to evaluate the integral of $G_c$.