ImplicitCondensation not immediate anymore

[milankl]

fixes #445

Explicit immediate condensation of humidity $q_i$ at time step $i$ given its saturation $q^\star$ calculated from temperature $T_i$ would be ($q > q^\star$ else $0$)

$$ \begin{aligned} q_{i+1} - q_i &= q^\star(T_i) - q_i \\ T_{i+1} - T_i &= -\frac{L_v}{c_p}( q^\star(T_i) - q_i ) \end{aligned} $$

with latent heat release of that condensation in the second equation. However, treating this explicitly poses the problem that because the saturation humidity is calculated from the current temperature $T_i$, which is increased due to the latent heat release, the humidity after this time step will be undersaturated. Ideally, one would want to condense towards the new saturation humidity $q^\star(T_{i+1})$ so that condensation draws the humidity back down to 100% not below it.
Taylor expansion at $i$ with $\delta T = T_{i+1} - T_i$ (and $\delta q$ similarly) to first order yields

$$q_{i+1} - q_i = q^\star(T_{i+1}) - q_i = q^\star(T_i) + (T_{i+1} - T_i) \frac{\partial q^\star}{\partial T} (T_i) + O(\delta T^2) - q_i$$

Now we make a linear approximation to the derivative, drop the $O(\delta T^2)$ term. Now inserting the (explicit) latent heat release

$$\delta q = q^\star(T_i) + -\frac{L_v}{c_p} \delta q \frac{\partial q^\star}{\partial T} (T_i) - q_i$$

And solving for $\delta q$ yields

$$ \left[ 1 + \frac{L_v}{c_p} \frac{\partial q^\star}{\partial T} (T^i) \right] \delta q = q^\star(T_i) - q_i $$

meaning that the implicit immediate condensation can be formulated as

$$ \begin{aligned} q_{i+1} - q_i &= \frac{q^\star(T_i) - q_i}{1 + \frac{L_v}{c_p} \frac{\partial q^\star}{\partial T}(T_i)} \\ T_{i+1} - T_i &= -\frac{L_v}{c_p}( q_{i+1} - q_i ) \end{aligned} $$

With Euler forward time stepping this is great, but with our leapfrog timestepping + RAW filter this is very dispersive (see #445) although the implicit formulation is already much better. We therefore introduce a timestep $\Delta t_c$ which makes the implicit condensation not immediate anymore but over several time steps $\Delta t$ of the leapfrogging.

$$ \begin{aligned} \frac{q_{i+1} - q_i}{\Delta t} &= \frac{q^\star(T_i) - q_i}{ \Delta t_c \left( 1 + \frac{L_v}{c_p} \frac{\partial q^\star}{\partial T}(T_i) \right)} \\ \frac{T_{i+1} - T_i}{\Delta t} &= -\frac{L_v}{c_p}( \frac{q_{i+1} - q_i}{\Delta t} ) \end{aligned} $$

For $\Delta t = \Delta t_c$ we have an immediate condensation, for $n = \frac{\Delta t_c}{\Delta t}$ condensation takes place over $n$ time steps.