in black and white
Main menu
Share a book About us Home
Biology Business Chemistry Computers Culture Economics Fiction Games Guide History Management Mathematical Medicine Mental Fitnes Physics Psychology Scince Sport Technics

Elementary Differential Equations and Boundary Value Problems - Boyce W.E.

Boyce W.E. Elementary Differential Equations and Boundary Value Problems - John Wiley & Sons, 2001. - 1310 p.
Download (direct link): elementarydifferentialequations2001.pdf
Previous << 1 .. 43 44 45 46 47 48 < 49 > 50 51 52 53 54 55 .. 609 >> Next

FIGURE 2.5.4 Logistic growth: y versus t for dy/dt = r(1 — y/K)y.
Chapter 2. First Order Differential Equations
arrow in Figure 2.5.4. Further, note that if y is near zero or K, then the slope f (y) is near zero, so the solution curves are relatively flat. They become steeper as the value of y leaves the neighborhood of zero or K. These observations mean that the graphs of solutions of Eq. (7) must have the general shape shown in Figure 2.5.4, regardless of the values of r and K.
To carry the investigation one step further, we can determine the concavity of the solution curves and the location of inflection points by finding d2y/dt2. From the differential equation (1) we obtain (using the chain rule)
d2 y dy
dy = f (y) dt = f'(y) f (J)- (8)
The graph of y versus t is concave up when y" > 0, that is, when f and f have the same sign. Similarly, it is concave down when y" < 0, which occurs when f and f have opposite signs. The signs of f and f can be easily identified from the graph of f (y) versus y. Inflection points may occur when f'(y) = 0.
In the case ofEq. (7) solutions are concave up for 0 < y < K /2 where f is positive and increasing (see Figure 2.5.3), so that both f and f are positive. Solutions are also concave up for y > K where f is negative and decreasing (both f and f are negative). For K/2 < y < K solutions are concave down since here f is positive and decreasing, so f is positive but f is negative. There is an inflection point whenever the graph of y versus t crosses the line y = K/2. The graphs in Figure 2.5.4 exhibit these properties.
Finally, recall that Theorem 2.4.2, the fundamental existence and uniqueness theorem, guarantees that two different solutions never pass through the same point. Hence, while solutions approach the equilibrium solution y = K as t ^<x>, they do not attain this value at any finite time. Since K is the upper bound that is approached, but not exceeded, by growing populations starting below this value, it is natural to refer to K as the saturation level, or the environmental carrying capacity, for the given species.
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 ^<x>, whereas solutions ofEq. (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 = ô(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
= rdt.
(1 — y/ K) y
2.5 Autonomous Equations and Population Dynamics
Using a partial fraction expansion on the left side, we have
1 1/K \
- + i------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 = Ceft, (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 — Ó0)º
We have derived the solution (11) under the assumption that 0 < Ó0 < K.If Ó0 > 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 = ô1 (t) = 0 and y = ô2(³) = 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 ^ro in Eq. (11), then we obtain
Previous << 1 .. 43 44 45 46 47 48 < 49 > 50 51 52 53 54 55 .. 609 >> Next