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The simplest model of population growth is dy/dt = ky, according to which populations grow exponentially. This may be true over short periods of time,

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Presentation on theme: "The simplest model of population growth is dy/dt = ky, according to which populations grow exponentially. This may be true over short periods of time,"— Presentation transcript:

1 The simplest model of population growth is dy/dt = ky, according to which populations grow exponentially. This may be true over short periods of time, but it is clear that no population can increase without limit. Therefore, population biologists use a variety of other differential equations that take into account environmental limitations to growth such as food scarcity and competition between species. One widely used model is based on the logistic differential equation: Here k > 0 is the growth constant, and A > 0 is a constant called the carrying capacity. Our next slide shows a typical S-shaped solution of a logistic differential equation. As in the previous section, we also denote dy/dt by

2 Solutions of the logistic equation with y 0 < 0 are not relevant to populations because a population cannot be negative (see Exercise 18).Exercise 18 Next Slide

3 The slope field shows clearly that there are three families of solutions, depending on the initial value y 0 = y (0). Slope field for If y 0 > A, then y (t) is decreasing and approaches A as t → ∞. If 0 < y 0 < A, then y (t) is increasing and approaches A as t → ∞. If y 0 < 0, then y (t) is decreasing and

4 has constant solutions: y = 0 and y = A. They correspond to the roots of ky(1 − y/A) = 0, and they satisfy because when y is a constant. Constant solutions are called equilibrium or steady-state solutions. The equilibrium solution y = A is a stable equilibrium because every solution with initial value y 0 close to A approaches the equilibrium y = A as t → ∞. By contrast, y = 0 is an unstable equilibrium because every nonequilibrium solution with initial value y 0 near y = 0 either increases to A or decreases to −∞. Graph

5 Having described the solutions qualitatively, let us now find the nonequilibrium solutions explicitly using separation of variables. Assuming that y 0 and y A, we have

6 For t = 0, we have a useful relation between C and the initial value y 0 = y (0): As C 0, we may divide by Ce kt to obtain the general nonequilibrium solution:

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8 Deer Population A deer population grows logistically with growth constant k = 0.4 year −1 in a forest with a carrying capacity of 1000 deer. (a) Find the deer population P(t) if the initial population is P 0 = 100. (b) How long does it take for the deer population to reach 500? `

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