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Elementary differential equations 7th edition - Boyce W.E

Boyce W.E Elementary differential equations 7th edition - Wiley publishing , 2001. - 1310 p.
ISBN 0-471-31999-6
Download (direct link): elementarydifferentialequat2001.pdf
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A comparison of Figures 2.5.1 and 2.5.4 reveals that solutions of the nonlinear equation (6) are strikingly different from those of the linear equation (1), at least for large values of t. Regardless of the value of K, that is, no matter how small the nonlinear term in Eq. (6), solutions of that equation approach a finite value as t ^ro, whereas solutions of Eq. (1) grow (exponentially) without bound as t ^ro. Thus even a tiny nonlinear term in the differential equation has a decisive effect on the solution for large t.
In many situations it is sufficient to have the qualitative information about the solution y = $(t) of Eq. (7) that is shown in Figure 2.5.4. This information was obtained entirely from the graph of f( y) versus y, and without solving the differential equation (7). However, if we wish to have a more detailed description of logistic growth—for example, if we wish to know the value of the population at some particular time—then we must solve Eq. (7) subject to the initial condition (3). Provided that y = 0 and y = K, we can write Eq. (7) in the form
dy
= rdt.
(1 - y/ K) y
2.5 Autonomous Equations and Population Dynamics
79
Using a partial fraction expansion on the left side, we have
1 1/K \
- + ? jj? I dy = rdt. y 1 — y/K)
Then by integrating both sides we obtain
ln |y| — ln
= rt + c, (9)
where c is an arbitrary constant of integration to be determined from the initial condition y(0) = y0. We have already noted that if 0 < y0 < K, then y remains in this interval for all time. Thus in this case we can remove the absolute value bars in Eq. (9), and by taking the exponential of both sides, we find that
y = Cert, (10)
1 — ( y/ K )
where C = ec. To satisfy the initial condition y(0) = y0 we must choose C = y0/[1 — (y0/K)]. Using this value for C in Eq. (10) and solving for y, we obtain
y = . ( K0 K ) —rt . (11)
y0 + (K — y0)e
We have derived the solution (11) under the assumption that 0 < y0 < K Ify0 > K, then the details of dealing with Eq. (9) are only slightly different, and we leave it to you to show that Eq. (11) is also valid in this case. Finally, note that Eq. (11) also contains the equilibrium solutions y = 01 (t) = 0 and y = &2(t) = K corresponding to the initial conditions y0 = 0 and y0 = K, respectively.
All the qualitative conclusions that we reached earlier by geometric reasoning can be confirmed by examining the solution (11). In particular, if y0 = 0, then Eq. (11) requires that y(t) = 0 for all t. If y0 > 0, and if we let t ^<x> in Eq. (11), then we obtain
lim y(t) = y0K/y0 = K.
Thus for each y0 > 0 the solution approaches the equilibrium solution y = 02(t) = K asymptotically (in fact, exponentially) as t ^?. Therefore we say that the constant solution 02(t) = K is an asymptotically stable solution of Eq. (7), or that the point y = K is an asymptotically stable equilibrium or critical point. This means that after a long time the population is close to the saturation level K regardless of the initial population size, as long as it is positive.
On the other hand, the situation for the equilibrium solution y = 01 ( t) = 0 is quite different. Even solutions that start very near zero grow as t increases and, as we have seen, approach K as t ^?. We say that ^1(t) = 0 is an unstable equilibrium solution or that y = 0 is an unstable equilibrium or critical point. This means that the only way to guarantee that the solution remains near zero is to make sure its initial value is exactly equal to zero.
80
Chapter 2. First Order Differential Equations
EXAMPLE
1
The logistic model has been applied to the natural growth of the halibut population in certain areas of the Pacific Ocean.6 Let y, measured in kilograms, be the total mass, or biomass, of the halibut population at time t. The parameters in the logistic equation are estimated to have the values r = 0.71/year and K = 80.5 x 106 kg. If the initial biomass is y0 = 0.25 K, find the biomass 2 years later. Also find the time t for which y(T) = 0.75K.
It is convenient to scale the solution (11) to the carrying capacity K; thus we write Eq. (11) in the form
y = y0/K
K (J0/K) + [1 - (J0/K)]e
Using the data given in the problem, we find that y(2) 0.25
(12)
K 0.25 + 0.75e-142 Consequently, y(2) = 46.7 x 106 kg.
= 0.5797.
FIGURE 2.5.5 y/ K versus t for population model of halibut in the Pacific Ocean.
To find t we can first solve Eq. (12) for t. We obtain
(VK)[1 - (y/k)] (y/K)[1 - (y,/K)]
e =
hence
^ (y0/K)[1 - (y/K)] t= ln — r (y/K)[1 - (y0/ K)]
(13)
Using the given values of r and y0/K and setting y/K = 0.75, we find that
1
t = —
ln
(0.25)(0.25)
1
0.71 (0.75)(0.75) 0.71
ln9 = 3.095 years.
A good source of information on the population dynamics and economics involved in making efficient use of a renewable resource, with particular emphasis on fisheries, is the book by Clark (see the references at the end of this chapter). The parameter values used here are given on page 53 of this book and were obtained as a result of a study by M. S. Mohring.
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