# Several models have been proposed to explain the diversification of life during geological periods.

Exponential models

Several models have been proposed to explain the diversification of life during geological periods. According to Benton (1997), The diversification of marine families in the past 600 million years (Myr) appears to have followed two or three logistic curves, with equilibrium levels that lasted for up to 200 Myr. In contrast, continental organisms clearly show an exponential pattern of diversification, and although it is not clear whether the empirical diversification patterns are real or are artifacts of a poor fossil record, the latter explanation seems unlikely. In this problem, we will investigate three models fordiversification. They are analogous to models for populationgrowth, however, the quantities involved have a differentinterpretation. We denote by N(t) the diversification function,which counts the number of taxa as a function of time, and by rthe intrinsic rate of diversification.
(a) (Exponential Model) This model is described by $$\displaystyle{\frac{{{d}{N}}}{{{\left.{d}{t}\right.}}}}={r}_{{{e}}}{N}\ {\left({8.86}\right)}.$$ Solve (8.86) with the initial condition N(0) at time 0, and show that $$\displaystyle{r}_{{{e}}}$$ can be estimated from $$\displaystyle{r}_{{{e}}}={\frac{{{1}}}{{{t}}}}\ {\ln{\ }}{\left[{\frac{{{N}{\left({t}\right)}}}{{{N}{\left({0}\right)}}}}\right]}\ {\left({8.87}\right)}$$
(b) (Logistic Growth) This model is described by $$\displaystyle{\frac{{{d}{N}}}{{{\left.{d}{t}\right.}}}}={r}_{{{l}}}{N}\ {\left({1}\ -\ {\frac{{{N}}}{{{K}}}}\right)}\ {\left({8.88}\right)}$$ where K is the equilibrium value. Solve (8.88) with the initial condition N(0) at time 0, and show that $$\displaystyle{r}_{{{l}}}$$ can be estimated from $$\displaystyle{r}_{{{l}}}={\frac{{{1}}}{{{t}}}}\ {\ln{\ }}{\left[{\frac{{{K}\ -\ {N}{\left({0}\right)}}}{{{N}{\left({0}\right)}}}}\right]}\ +\ {\frac{{{1}}}{{{t}}}}\ {\ln{\ }}{\left[{\frac{{{N}{\left({t}\right)}}}{{{K}\ -\ {N}{\left({t}\right)}}}}\right]}\ {\left({8.89}\right)}$$ for $$\displaystyle{N}{\left({t}\right)}\ {<}\ {K}.$$
(c) Assume that $$\displaystyle{N}{\left({0}\right)}={1}$$ and $$\displaystyle{N}{\left({10}\right)}={1000}.$$ Estimate $$\displaystyle{r}_{{{e}}}$$ and $$\displaystyle{r}_{{{l}}}$$ for both $$\displaystyle{K}={1001}$$ and $$\displaystyle{K}={10000}.$$
(d) Use your answer in (c) to explain the following quote from Stanley (1979): There must be a general tendency for calculated values of $$\displaystyle{\left[{r}\right]}$$ to represent underestimates of exponential rates,because some radiation will have followed distinctly sigmoid paths during the interval evaluated.
(e) Explain why the exponential model is a good approximation to the logistic model when $$\displaystyle\frac{{N}}{{K}}$$ is small compared with 1.

2021-02-12

Step 1
Given: $$\displaystyle{N}{\left({t}\right)}$$ denotes the diversification function, which counts the number of taxa and r denotes the intrinsic rate of diversification.
Step 2
Calculation:
a) The exponential model is described as:
$$\displaystyle{\frac{{{d}{N}}}{{{\left.{d}{t}\right.}}}}={r}_{{{e}}}{N}$$
$$\displaystyle\Rightarrow\ {\frac{{{d}{N}}}{{{N}}}}={r}_{{{e}}}{\left.{d}{t}\right.}$$
Integrate both sides
$$\displaystyle\Rightarrow\ {\int_{{{N}{\left({0}\right)}}}^{{{N}{\left({t}\right)}}}}\ {\frac{{{d}{N}}}{{{N}}}}={r}_{{{e}}}\ {\int_{{{0}}}^{{{t}}}}\ {\left.{d}{t}\right.}$$
$$\Rightarrow\ \ln\ |N|_{N(0)}^{N(t)}=r_{e}t\mid_{0}^{t}$$
$$\displaystyle\Rightarrow\ {\ln{\ }}{\left[{\frac{{{N}{\left({t}\right)}}}{{{N}{\left({0}\right)}}}}\right]}={r}_{{{e}}}{t}$$
$$\Rightarrow\ \begin{array}{|c|}\hline r_{e}=\frac{1}{t}\ \ln\ \left[\frac{N(t)}{N(0)}\right]\\ \hline \end{array}$$
Step 3
b) The logistic model is described as:
$$\displaystyle{\frac{{{d}{N}}}{{{\left.{d}{t}\right.}}}}={r}_{{{l}}}{N}\ {\left({1}\ -\ {\frac{{{N}}}{{{K}}}}\right)}$$
$$\displaystyle\Rightarrow\ {\frac{{{d}{N}}}{{{\left.{d}{t}\right.}}}}={\frac{{{r}_{{{l}}}}}{{{K}}}}\ {\left({N}\right)}{\left({K}\ -\ {N}\right)}$$
$$\displaystyle\Rightarrow\ {\left[{\frac{{{K}}}{{{N}{\left({K}\ -\ {N}\right)}}}}\right]}{d}{N}={r}_{{{l}}}{\left.{d}{t}\right.}$$
$$\displaystyle\Rightarrow\ {d}{N}\ {\left[{\frac{{{1}}}{{{N}}}}\ +\ {\frac{{{1}}}{{{K}\ -\ {N}}}}\right]}={r}_{{{l}}}{\left.{d}{t}\right.}$$
$$\displaystyle\Rightarrow\ {\frac{{{d}{N}}}{{{N}}}}\ +\ {\frac{{{d}{N}}}{{{K}\ -\ {N}}}}={r}_{{{l}}}{\left.{d}{t}\right.}$$
Integrate both sides
$$\displaystyle{\int_{{{N}{\left({0}\right)}}}^{{{N}{\left({t}\right)}}}}\ {\frac{{{d}{N}}}{{{N}}}}\ +\ {\int_{{{N}{\left({0}\right)}}}^{{{N}{\left({t}\right)}}}}\ {\frac{{{d}{N}}}{{{k}\ -\ {N}}}}={r}_{{{l}}}\ {\int_{{{0}}}^{{{t}}}}\ {\left.{d}{t}\right.}$$
$$\Rightarrow\ln|N|\mid_{N(0)}^{N(t)}\ -\ \ln\ |K\ -\ N|\mid_{N(0)}^{N(t)}=r_{l}\mid_{0}^{t}\ dt$$
$$\displaystyle\Rightarrow\ {\ln{\ }}{\left({\frac{{{N}{\left({t}\right)}}}{{{N}{\left({0}\right)}}}}\right)}\ -\ {\ln{\ }}{\left({\frac{{{K}\ -\ {N}{\left({t}\right)}}}{{{K}\ -\ {N}{\left({0}\right)}}}}\right)}={r}_{{{l}}}{t}$$
$$\displaystyle\Rightarrow\ {\left[{\ln{{\left({N}{\left({t}\right)}\right)}}}\ -\ {\ln{{\left({K}\ -\ {N}{\left({t}\right)}\right)}}}\right]}\ +\ {\left[{\ln{{\left({N}{\left({0}\right)}\right)}}}\ +\ {\ln{{\left({K}\ -\ {N}{\left({0}\right)}\right)}}}\right]}={r}_{{{l}}}{t}$$
$$\displaystyle\Rightarrow\ {\ln{{\left[{\frac{{{N}{\left({t}\right)}}}{{{K}\ -\ {N}{\left({t}\right)}}}}\right]}}}\ +\ {\ln{\ }}{\left[{\frac{{{K}\ -\ {N}{\left({0}\right)}}}{{{N}{\left({0}\right)}}}}\right]}={r}_{{{l}}}{t}$$
$$\Rightarrow\ \begin{array}{|c|}\hline r_{l}=\frac{1}{t}\ \ln\ \left[\frac{N(t)}{K\ -\ N(t)}\right]\ +\ \frac{1}{t}\ \ln\ \left[\frac{K\ -\ N(0)}{N(0)}\right] \\ \hline \end{array}$$
Step 4
c) For exponential model:
$$\displaystyle{r}_{{{e}}}={\frac{{{1}}}{{{T}}}}\ {\ln{\ }}{\left[{\frac{{{N}{\left({t}\right)}}}{{{N}{\left({0}\right)}}}}\right]}$$
As $$\displaystyle{N}{\left({10}\right)}={1000}$$ and $$\displaystyle{N}{\left({0}\right)}={1}:$$
$$\displaystyle\Rightarrow\ {r}_{{{e}}}={\frac{{{1}}}{{{10}}}}\ {\ln{\ }}{\left[{\frac{{{N}{\left({10}\right)}}}{{{N}{\left({0}\right)}}}}\right]}$$
$$\displaystyle\Rightarrow\ {r}_{{{e}}}=\ {\frac{{{1}}}{{{10}}}}\ {\ln{\ }}{\left[{\frac{{{1000}}}{{{1}}}}\right]}$$
$$\Rightarrow\ \begin{array}{|c|}\hline r_{e}=0.69 \\ \hline \end{array}$$
For logistic model:
$$\displaystyle{r}_{{{l}}}={\frac{{{1}}}{{{t}}}}\ {\ln{\ }}{\left[{\frac{{{K}\ -\ {N}{\left({0}\right)}}}{{{N}{\left({0}\right)}}}}\right]}\ +\ {\frac{{{1}}}{{{t}}}}\ {\ln{\ }}{\left[{\frac{{{N}{\left({t}\right)}}}{{{K}\ -\ {N}{\left({t}\right)}}}}\right]}$$
As $$\displaystyle{N}{\left({10}\right)}={1000},\ {N}{\left({0}\right)}={1}$$ and $$\displaystyle{K}={1001}:$$
$$\displaystyle\Rightarrow\ {r}_{{{l}}}={\frac{{{1}}}{{{10}}}}\ {\left[{\ln{\ }}{\left({\frac{{{1001}\ -\ }}{{{1}}}}\right)}\ +\ {\ln{\ }}{\left({\frac{{{1000}}}{{{1001}\ -\ {1000}}}}\right)}\right]}$$
$$\displaystyle\Rightarrow\ {r}_{{{l}}}={\frac{{{1}}}{{{10}}}}\ {\left[{\ln{\ }}{\left({1000}\right)}\ +\ {\ln{{\left({1000}\right)}}}\right]}$$
$$\displaystyle\Rightarrow\ {r}_{{{l}}}={\frac{{{2}}}{{{10}}}}\ {\ln{{\left({1000}\right)}}}$$
$$\Rightarrow\ \begin{array}{|c|}\hline r_{l}=1.381 \\ \hline \end{array}$$
In case 2, $$\displaystyle{N}{\left({10}\right)}={1000},\ {N}{\left({0}\right)}={1}$$ and $$\displaystyle{K}={10},{000}:$$
$$\displaystyle\Rightarrow\ {r}_{{{l}}}={\frac{{{1}}}{{{10}}}}{\left[{\ln{\ }}{\left({\frac{{{10000}\ -\ {1}}}{{{1}}}}\right)}\ +\ {\ln{\ }}{\left({\frac{{{1000}}}{{{10000}\ -\ {100}}}}\right)}\right]}$$
$$\displaystyle\Rightarrow\ {r}_{{{l}}}={\frac{{{1}}}{{{10}}}}{\left[{\ln{{\left({9999}\right)}}}\ +\ {\ln{\ }}{\left({\frac{{{1000}}}{{{9000}}}}\right)}\right]}$$
$$\displaystyle\Rightarrow\ {r}_{{{l}}}={\frac{{{1}}}{{{10}}}}{\left[{\ln{\ }}{\left({9999}\right)}\ +\ {\ln{\ }}{\left({\frac{{{1}}}{{{9}}}}\right)}\right]}$$
$$\displaystyle\Rightarrow\ {r}_{{{l}}}={\frac{{{1}}}{{{10}}}}{\left[{\ln{\ }}{\left({9999}\ \cdot\ {\frac{{{1}}}{{{9}}}}\right)}\right]}$$
$$\displaystyle\Rightarrow\ {r}_{{{l}}}={\frac{{{1}}}{{{10}}}}{\ln{{\left({1111}\right)}}}$$
$$\Rightarrow\ \begin{array}{|c|}\hline r_{l}=0.7 \\ \hline \end{array}$$
Step 5
d) From part (c), it is observed that the calculated rate r in exponential model is less than the estimated rate in logistic model for both values of $$\displaystyle{K}{\left({K}={10001}\right.}$$ and $$\displaystyle{K}={10000}{)}.$$
The above observation justifies the following quote:
"There must be a general tendency for calculated values of $$\displaystyle{\left[{r}\right]}$$ to represent underestimates for exponential rates because some radiation will have followed distincly sigmoid paths during the interval evaluated."
e) From part (b):
$$\displaystyle{r}_{{{e}}}={\frac{{{1}}}{{{t}}}}{\left[{\ln{\ }}{\left[{\frac{{{K}\ -\ {N}{\left({0}\right)}}}{{{N}{\left({0}\right)}}}}\right]}\ +\ {\ln{\ }}{\left[{\frac{{{N}{\left({t}\right)}}}{{{K}\ -\ {N}{\left({t}\right)}}}}\right]}\right]}$$
$$\displaystyle\Rightarrow\ {r}_{{{e}}}={\frac{{{1}}}{{{t}}}}\ {\ln{\ }}{\left[{\frac{{{N}{\left({t}\right)}}}{{{N}{\left({0}\right)}}}}\ \cdot\ {\frac{{{K}\ -\ {N}{\left({0}\right)}}}{{{K}\ -\ {N}{\left({t}\right)}}}}\right]}$$
$$\displaystyle\Rightarrow\ {r}_{{{e}}}={\frac{{{1}}}{{{t}}}}\ {\ln{\ }}{\left[{\frac{{{N}{\left({t}\right)}}}{{{N}{\left({0}\right)}}}}\ \cdot\ {\left({\frac{{{1}\ -\ {\frac{{{N}{\left({0}\right)}}}{{{K}}}}}}{{{1}\ -\ {\frac{{{N}{\left({t}\right)}}}{{{K}}}}}}}\right)}\right]}$$
As $$\displaystyle{\frac{{{N}}}{{{K}}}}\ {\left\langle\ {1}\right.}$$
$$\displaystyle\Rightarrow\ {r}_{{{e}}}={\frac{{{1}}}{{{t}}}}{\left[{\ln{\ }}{\left({\frac{{{N}{\left({t}\right)}}}{{{N}{\left({0}\right)}}}}\ \cdot\ {1}\right)}\right]}$$
$$\Rightarrow\ \begin{array}{|c|}\hline r_{e}=\frac{1}{t}\ \ln\ \left[\frac{N(t)}{N(0)}\right]\\ \hline \end{array}$$
Hence, the exponential model is a good approximation to logistic model when
$$\displaystyle{\frac{{{N}}}{{{K}}}}\ {\left\langle\ {1}.\right.}$$