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Use differential equations to estimate the maximum population given census data

by RoRi

From the previous exercise we have the following population growth law:

    \[ x = \frac{M}{1+e^{-a(t-t_1)}}. \]

Census data is taken at equally spaced times t_1, t_2, t_3 and the population at these times is given by x_1, x_2, x_3, respectively. Prove that this data is sufficient to determine the value of the constant M and that

    \[ M = x_2 \frac{x_3(x_2 - x_1) -x_1(x_3 - x_2)}{x_2^2 - x_1x_3}. \]

Proof. First, we use x(t_1) = x_1 to solve for t_0:

    \begin{align*} x_1 = \frac{M}{1+e^{-\alpha(t_1-t_0)}} && \implies && e^{\alpha(t_1 - t_0)} &= \frac{x_1}{M-x_1} \\[9pt] && \implies && t_0 &= t_1 - \left( \frac{1}{\alpha} \right) \log \left( \frac{x_1}{M-x_1} \right). \end{align*}

Then, substituting into x(t_2) = x_2:

    \begin{align*} x_2 = \frac{M}{1+e^{-\alpha(t_2-t_0)}} && \implies && M &= x_2 + x_2 e^{-\alpha \left( t_2 - t_1 + \frac{1}{\alpha} \log \left( \frac{x_1}{M-x_1} \right) \right)} \\[9pt] && \implies && M &= x_2 + x_2 e^{-\alpha(t_2-t_1)} \cdot \frac{M-x_1}{x_1} \\[9pt] && \implies && \frac{Mx_1 - x_1 x_2}{M-x_1} &= x_2 e^{-\alpha(t_2 - t_1)}. \end{align*}

Since t_1, t_2, t_3 are equally spaced by assumption, we have t_2 - t_1 = t_3 - t_2 = 2 (t_3 - t_1). Therefore,

    \[ e^{-\alpha(t_3-t_1)} = \left( \frac{x_1 (M-x_1)}{x_2 (M-x_2)} \right)^2. \]

Finally, using x(t_3) = x_3, we have,

    \begin{align*} &&x_3 &= \frac{M}{1 + e^{-\alpha(t_3 - t_0)}} \\[9pt] \implies && M &= x_3 + x_3 e^{-\alpha \left( t_3 - t_1 + \frac{1}{\alpha} \log \left( \frac{x_1}{M-x_1} \right) \right)} \\[9pt] \implies && M &= x_3 + x_3 e^{-\alpha (t_3 -t_1)} \cdot \left( \frac{M-x_1}{x_1} \right) \\[9pt] \implies && M &= x_3 + x_3 \left( \frac{x_1 (M-x_2)}{x_2 (M-x_1)} \right)^2 \cdot \left( \frac{M-x_1}{x_1} \right) \\[9pt] \implies && M&= x_3 + \frac{x_1 x_3 (M-x_2)^2}{x_2^2 (M-x_1)} \\[9pt] \implies && M (M-x_1) &= \left( \frac{1}{x_2^2} \right) \left( Mx_3 x_2^2 - x_1 x_3 x_2^2 + M^2 x_1 x_3 - 2M x_1 x_2 x_3 + x_1 x_3 x_2^2 \right) \\[9pt] \implies && Mx_2^2 - x_1 x_2^2 &= x_3 x_2^2 + Mx_1 x_3 - 2x_1 x_2 x_3 \\[9pt] \implies && M(x_2^2 - x_1 x_3) &= x_2 (x_3 x_2 - x_1 x_3 + x_1 x_2 - x_1 x_3) \\[9pt] \implies && M &= x_2 \cdot \frac{x_3 (x_2 - x_1) - x_1 (x_3 - x_2)}{x_2^2 - x_1 x_3}. \qquad \blacksquare \end{align*}

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