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Use differential equations to determine population under given growth assumptions

The simplest population growth law is given by

    \[ \frac{dx}{dt} = kx, \]

where x is the population and k is a constant dependent on the type of population in question. A more involved growth law in which the population is subject to a maximum constraint M, gives an equation for population growth of

    \[ \frac{dx}{dt} = kx(M-x), \]

where k is a constant, or possibly a function of time.

Express the population x as a function of t in each of these growth laws (with k and M both constant). Prove that the result in the second growth law can be expressed as:

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

where a is a constant and t_1 is the time at which x = \frac{M}{2}.

Proof. For the first growth law,

    \[ kx = \frac{dx}{dt} \quad \implies \quad x' - kx = 0. \]

This has solutions of the form

    \[ x = x_0 e^{k(t-t_0)} \]

where x = x_0 at time t_0.

For the second, more complicated, growth law we have

    \[ \frac{dx}{dt} = kx (M-x) \quad \implies \quad x' - kMx = -kx^2. \]

This is a Bernoulli equation (as seen in this exercise, Section 8.5 Exercise #13) so we know x^{-1} = v where v is the unique solution to

    \[ v' - P(t) v = - Q(t) \qquad \text{where} \quad P(t) = -kM, \quad Q(t) = -k. \]

Thus, we are looking for the unique solution of

    \[ v' + kMv = k \qquad \text{with} \qquad x(t_0) = x_0 \implies v(t_0) = \frac{1}{x_0}. \]

Using Theorem 8.3 (page 310 of Apostol) for the solutions of first-order linear differential equations, we have

    \begin{align*}  v &= \frac{1}{x_0} e^{-kM(t-t_0)} + e^{-kM(t-t_0)} \int_{t_0}^t ke^{km(u-t_0)} \, du \\[9pt]  &= \frac{1}{x_0} e^{-kM(t-t_0)} + e^{-kM(t-t_0)} \left( \frac{k}{e^{kMt_0}} \cdot \frac{1}{kM} \cdot \left( e^{kMt} - e^{kMt_0} \right) \right) \\[9pt]  &= \frac{1}{x_0} e^{-kM(t-t_0)} + \frac{1}{Me^{kMt}} \left( e^{kMt} - e^{kMt_0} \right) \\[9pt]  &= \frac{1}{x_0} e^{-kM(t-t_0)} + \frac{1}{M} - \frac{1}{M} \cdot e^{-kM(t-t_0)} \\[9pt]  &= \frac{1}{Mx_0} \left( e^{-kM(t-t_0)} (M- x_0) + x_0) \right). \end{align*}


    \[ x = Mx_0 \left( e^{-kM(t-t_0) (M-x_0) + x_0 \right)^{-1}. \]

So, if we have x(t_1) = \frac{M}{2}, then t_0 = t_1 and x_0 = \frac{M}{2}, so

    \[ x = \frac{M\left( \frac{M}{2} \right)}{e^{-\alpha(t-t_0)} \left(M - \frac{M}{2} \right) + \frac{M}{2}} = \frac{M}{e^{-\alpha(t-t_0)} + 1} \]

where \alpha = kM. \qquad \blacksquare


  1. tom says:

    Rori- wondering if you could explain how you got T0=T1 at the end. Is this just the obvious choice to remove the exponential? Thanks.

    • tom says:

      Okay think I got it now- T1 was just used for the initial condition ,a, and M/2 ,a constant value, is set as x(a)=b.

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