$$\mathcal{K}(\mathbf x, t) = \mathcal{K}_{L}(\mathbf x)(1-S(t)) 
+ \mathcal{K}_{H}(\mathbf x) S(t)$$ where a sigmoid function $0 = S(t_L) <  S(t) < S(t_H) = 1$ will smoothly interpolate between the two time frames. There are many choices available, such as that used in [@eriksson2012]. For this study, we used the following variant.

Start with the classical sigmoid $$S(z) = \left( 1 + e^{-z} \right)^{-1}
\label{eq:classicalsigmoid}$$ which is zero at $z = -\infty$ and unity at $z=+\infty$. The $-\infty
< z < +\infty$ interval is not what we want, but with a small change the following $t \mapsto z$ transformation permits many variants: $$\begin{aligned}
z & = \frac{2 \Delta T (t-t_{L}) -\Delta T^2}{((t-t_{L})(\Delta T - 
(t-t_{L})))^\nu}.
\label{eq:ourinterpolator}\end{aligned}$$ where $\Delta T = t_{H} - t_{L}$. Notice that $S(z(t_{L})) = 0$ and $S(z(t_{H})) = 1$. The exponent $\nu$ in (\[eq:ourinterpolator\]) gives some freedom in choosing a particular form for $z$ for almost any $\nu > 0$. If $\nu < 1/2$, ${\rm d}^{2} S/{\rm d}t^{2}$ has more than two sign changes, so $\nu \geq 1/2$ is preferable. By the choice $\nu=1/2$, the interpolant is nearly a straight line: see Figure \[fig:sigmoids\]. However, its turn-up at $t=t_{L}$ and turn-down at $t=t_{H}$ numerically resemble very quick derivative changes. 