EGFP +
8.3. Extracción de RNA total de Líneas Celulares y de Tejidos
Stability
A first-order autonomous equation is an equation of the special form
x= f (x). (9.1)
Notice that the independent variable, which we have usually been denoting by t, does not appear explicitly on the right-hand side of equation (9.1). This is the defin-ing feature of an autonomous equation.
In Section 2.3, we derived the differential equation
v = −g − kv|v|/m, (9.2)
which models the velocityv of an object near the surface of the earth with air re-sistance proportional to the square of the velocity. Here g is the acceleration due to gravity, m is the mass of the object, and k is a proportionality constant. This is an autonomous equation, since the independent variable t does not appear on the right-hand side. Other examples of autonomous equations are
x = sin(x), y= y2+ 1, and z= ez. The equations
x= sin(tx), y= y2+ t, z= tz, and y= xy
are not autonomous, since the independent variable appears on the right-hand side of each equation.
Autonomous equations occur very frequently in applications. A differential equation model of any physical system that is evolving without external forces will almost always be autonomous. It is usually the external forces that give rise to terms that depend explicitly on time.
In this section, we will describe ways to discover the qualitative behavior of solutions to autonomous equations, without actually finding the solutions. Although autonomous equations are in principle solvable, finding the solutions explicitly may be difficult and the results may be so complicated that the formula does not reveal the behavior of the solutions. In contrast, qualitative methods are so easy that it will be useful to study the solutions qualitatively in addition to finding the solutions explicitly, when that is possible. In some cases, it might be sufficient to do the qualitative analysis without finding exact solutions.
In particular, we will learn how to discover the limiting behavior of solutions as the independent variable goes to±∞. If an autonomous equation is modeling some scientific phenomenon, the limiting behavior tells us the ultimate outcome. In addition our qualitative methods will enable us to sketch the graphs of all solutions to the autonomous equation.
The direction field and solutions
Since the function f(t, x) on the right-hand side of (9.1) does not depend on t, the slopes of the direction lines have the same feature. This is illustrated by the direction field for equation (9.2) shown in Figure 1. The slopes do not change as we move from right to left in this figure.
Because of this fact, we would expect the same behavior for the solution curves.
We would expect that one solution curve translated to the left or right would be another solution curve. We can see this analytically. Suppose that y(t) is a solution to the autonomous equation y = f (y). Consider the function y1(t) = y(t + C), where C is a constant. Since
y1(t) = y(t + C) = f (y(t + C)) = f (y1(t)),
2.9 Autonomous Equations and Stability 93
t
mg k
v
Figure 1. The direction field for equation (9.2).
we see that y1is also a solution. This means that we get different solution curves by translating one curve left and right. See Figure 2, which displays several solutions to equation (9.4).
t v
mg
k
Figure 2. Several solutions to equation (9.4).
Equilibrium points and solutions
The starting point to the qualitative analysis of an autonomous equation is the dis-covery of some easily found particular solutions. If f(x0) = 0, then the constant function x(t) = x0satisfies
x(t) = 0 = f (x0) = f (x(t)).
Hence this constant function is a particular solution to (9.1). We will call a point x0 such that f(x0) = 0 an equilibrium point. The constant function x(t) = x0 is called an equilibrium solution.
E x a m p l e 9 . 3 Find the equilibrium points and equilibrium solutions for the equation
v = −g − kv|v|/m, (9.4)
which models the velocity of an object subject to gravity and air resistance.
The right-hand side is the function
f(v) = −g − kv|v|/m =
−g − kv2/m for v ≥ 0,
−g + kv2/m for v < 0.
From these equations, we see that the graph of f consists of half of a parabola curved down forv ≥ 0, and half of a parabola curved up for v ≤ 0 as shown in the graph of f in Figure 3. Hence, f is a decreasing function, and can have only one zero. Since f(v) < 0 for v > 0, the zero of f must occur when v < 0, for which
f(v) g kvv/m
v mg
k
Figure 3. The graph of the right-hand side of equation (9.4).
f(v) = −g + kv2/m. Therefore, the equilibrium point is v = −
mg/k, and the corresponding equilibrium solution is
v1(t) = − mg/k.
The graph of this solution is shown in Figure 4. ●
Nonequilibrium solutions
Now suppose thatv(t) is a solution to (9.4) that satisfies v(t0) > −√
mg/k for some
t
mg k
v
Figure 4. The equilibrium solution for equation (9.4).
t0. Notice that f(v) = −g−kv|v|/m and its derivative d f/dv = −2k|v|/m are both continuous for allv. Thus equation (9.4) satisfies the hypotheses of the uniqueness theorem. Therefore, the graphs ofv and the equilibrium solution v1(t) = −√
mg/k cannot cross. Consequently we must have
v(t) > v1(t) = −
mg/k for all t. (9.5)
Since f(v) is decreasing and f (−√
mg/k) = 0, when v(t) satisfies (9.5), we have f(v(t)) < 0. Hence,
v(t) = −g − kv|v|/m < 0 for all t.
Because it has a negative derivative,v(t) is a monotone decreasing function. Since the decreasing functionv(t) is bounded below by v(t) > −√
mg/k for all t, we know thatv(t) approaches a limit as t → ∞. It can be shown that this limit must be−√
mg/k. A similar train of thought shows that v(t) → ∞ as t → −∞.
Notice that without solving the initial value problem we have learned three things about the solutionv(t) that has an initial value v(t0) > −√
mg/k:
1. v(t) is monotone decreasing 2. v(t) → −√
mg/k as t → ∞ 3. v(t) → ∞ as t → −∞
2.9 Autonomous Equations and Stability 95 Thus the solution curves have the appearance shown in Figure 2. We cannot say how fast v(t) → −√
mg/k as t → ∞, or v(t) → ∞ as t → −∞. For this reason, we have not included any tick marks along the t-axis in Figure 2. The same reasoning shows that ifv(0) = v0 < −√
mg/k, then v(t) is increasing to −√ mg/k as t → ∞, and tends to −∞ as t → −∞.
We have shown that as t increases, the velocity always tends to vterm= −
mg/k.
We reached the same result at the end of Section 2.3. Because of this fact, we called vtermthe terminal velocity. However, it is interesting to compare the amount of work involved in the two different methods used. Qualitative analysis is almost always easier when we want to discover the limiting behavior of solutions.
The phase line
An autonomous equation y = f (y) can be a mathematical model for a variety of
Vterm v
f(v) g kvv/m
Figure 5. The phase line for equation (9.4).
phenomena. Regardless of the actual application, it is useful to think of y as mea-suring the distance from 0 along a number line, a y-axis. With this interpretation, the equation y= f (y) describes the dynamics involved in the motion of y(t) along the line, which is called the phase line.
For example, consider the graph of the right-hand side f(v) = −g − kv|v|/m of equation (9.4) in Figure 5. The v-axis will be the phase line. We mark the equilibrium pointvterm = −√
mg/k with a solid point. To the left of vterm, f(v) >
0, so a solutionv(t) is increasing, and we indicate this on the phase line by a blue arrow pointing to the right. Similarly, to the rightvterm,v(t) is decreasing, indicated by a blue arrow pointing to the left. A solution on either side of the equilibrium point must approach the equilibrium point as t → ∞. Notice that the information on the phase line encapsulates everything we know about the solutions. In addition, notice that all of the information on the phase line was easily obtained from the graph of the f(v), the right-hand side of the differential equation.
Constructing and using the phase line
The analysis carried out above for equation (9.4) can be done for any autonomous equation. Let’s illustrate this with another example.
E x a m p l e 9 . 6 Discover the behavior as t→ ∞ of all solutions to the differential equation x= f (x) = (x2− 1)(x − 2). (9.7) We will use the phase line to do the analysis, and we start by graphing the right-hand side, f(x) = (x2− 1)(x − 2), as shown in Figure 6. The graph can be easily plotted on a computer or a calculator. However, we do not need great precision in this graph. We need to know the zeros precisely, but in the intervals limited by the zeros we only need to know the sign of f . Our phase line will be the x-axis in Figure 6. We will consider x to be the position of a point on this line with its motion modeled by equation (9.7).
Since we can factor the right-hand side of (9.7) as f(x) = (x −1)(x +1)(x −2), the zeros of f are x1 = −1, x2 = 1, and x3 = 2. These are the equilibrium points, and we plot them on the x-axis in Figure 6, either with a dot or a small circle.
Corresponding to these, we have three equilibrium solutions:
x(t) = −1, x(t) = 1, and x(t) = 2.
t f (x) (x2 1)(x 2)
5
5
1 1 2
Figure 6. The phase line for the equation x= (x2− 1)(x − 2).
Since these are constant functions, the position of the point on the phase line mod-eled by them is also constant.
The equilibrium solutions are plotted in blue in Figure 7. We easily show that the uniqueness theorem applies to equation (9.7). Therefore, the solution curves of nonequilibrium solutions cannot cross those of the equilibrium solutions. There are four intervals limited by the equilibrium points. If a solution x starts in one of these intervals, then x(t) is in that interval forever. Thus the motion of the solution point along the phase line is limited by the equilibrium points.
3 x
0 2 4
t
3
Figure 7. Solutions of the equation x= (x2− 1)(x − 2).
In each of the four intervals the right-hand side f(x) = (x2− 1)(x − 2) has a constant sign:
f(x) < 0 in (−∞, −1), f(x) > 0 in (−1, 1), f(x) < 0 in (1, 2), f(x) > 0 in (2, ∞).
2.9 Autonomous Equations and Stability 97 This information can be obtained from the graph of f in Figure 6. If you do not have a graph handy, you can check the sign of f at one point in each of the intervals.
If x(t) is a nonequilibrium solution to equation (9.7), and x(t) is in one of the intervals(−1, 1) and (2, ∞) where f (x) > 0, then x(t) = f (x(t)) > 0. Thus x is increasing. This means that x is moving to the right along the phase line. We indicate this on the phase line in Figure 6 by arrows pointing to the right in these intervals. Similarly, in the intervals(−∞, −1) and (1, 2) where f (x) < 0, x is moving to the left, and we have arrows pointing to left.
Because of its monotone behavior, a nonequilibrium solution in one of the inter-vals(−1, 1) and (1, 2) must move in the direction of the arrow, and must approach the equilibrium point x2 = 1 as t → ∞. By similar reasoning, an equilibrium so-lution in(2, ∞) must approach ∞, and one in (−∞, −1) must approach −∞. On the other hand, as t → −∞, the motion is in the direction opposite to the arrows, and approaches the equilibrium point in that direction. With this information we can sketch examples of the nonequilibrium solutions in each interval, as shown in
Figure 7. ●
Notice that we transferred all of the phase line information from the x-axis in Figure 6 to the x-axis in Figure 7 to assist us in sketching the solutions. Thus we have two examples of phase lines for equation (9.7). It should be emphasized that a phase line can be constructed on any x-axis. Another example is given in Figure 8.
x 3
1 2
3 1
Figure 8. The phase line for the equation x= (x2− 1)(x − 2).
Stability
Some equilibrium points, like x2 = 1 in Example 9.6, have the property that solu-tion curves which start near them approach the equilibrium point as t → ∞. These are called asymptotically stable equilibrium points. We have marked asymptotically stable equilibrium points with solid points on our phase lines. There are also equi-librium points, like x1 = −1 and x3 = 2 in Example 9.6, where some solutions move away. These are called unstable.6 Unstable equilibrium points are marked by open circles. If we focus our attention on the phase line near an equilibrium point, then we see that it is an asymptotically stable equilibrium point if and only if both adjacent arrows point toward the point. In fact, since each arrow can have only two directions, there are a total of four possibilities, only one of which represents an asymptotically stable equilibrium point.
These possibilities are shown in Figure 9, together with an indication of what the graph of f looks like near the associated equilibrium point. Notice that only Figure 9(b) depicts an asymptotically stable equilibrium point. In Figures 9(c) and (d), one arrow points toward the point and the other points away. Although these points might be called semistable, we will not stress that terminology. They are unstable equilibrium points.
Examining the possibilities, we see that an equilibrium point x0 for x = f (x) is asymptotically stable if and only if f is decreasing at x0. We can use this fact
6Consider the equation y= 0. For this equation, every point is an equilibrium point, and every solution is a constant function. These solutions do not move nearer to the equilibrium points, nor do they move away. The property of “not moving away” is described by saying that the equilibrium points are stable.
In one dimension, the equation y= 0 provides essentially the only example of stable equilibrium points that are not asymptotically stable. In higher dimensions, the concept of stability is more interesting.
(x0, 0)
(x0, 0)
(a) (b)
(c) (d)
(x0, 0)
(x0, 0)
Figure 9. Possible configurations of equilibrium points.
to derive a first derivative test for stability. In the figures in this section, we have systematically indicated asymptotically stable equilibrium points with solid points, and unstable equilibrium points with open circles.
THEOREM 9.8 Suppose that x0 is an equilibrium point for the differential equation x = f (x), where f is a differentiable function.
1. If f(x0) < 0, then f is decreasing at x0and x0is asymptotically stable.
2. If f(x0) > 0, then f is increasing at x0and x0 is unstable.
3. If f(x0) = 0, no conclusion can be drawn. ■ E x a m p l e 9 . 9 Classify the equilibrium points for the equation
x= (x2− 1)(x − 2) from Example 9.6.
We saw in Example 9.6 that the equilibrium points are−1, 1, and 2. We can analyze these by looking at the phase lines in Figures 6, 7, or 8, and noticing that the solutions starting near−1 or near 2 are driven away from these values. Hence these are unstable points. On the other hand, the solutions starting near 1, either above or below, are drawn toward 1 as t → ∞. Thus 1 is an asymptotically stable equilibrium point.
We could have also classified these equilibrium points by looking at the graph of the right-hand side in Figure 6. The right-hand side f(x) = (x2− 1)(x − 2) is decreasing when it passes through point 1, but increasing as it passes through the other two. Hence point 1 is asymptotically stable and the others are unstable.
Finally, a third way is to use Theorem 9.8. We compute that f(x) = 3x2−4x − 1. At the equilibrium points, we have f(−1) = 6, f(1) = −2, and f(2) = 3.
Thus−1 and 2 are unstable and 1 is asymptotically stable. ●
2.9 Autonomous Equations and Stability 99
Summary of the method
We now have a method for analyzing the solutions of the autonomous equation x= f (x).
Let’s summarize the procedure.
1. Graph the right-hand side f(x) and add the phase line information to the x-axis. Find, mark, and classify the equilibrium points where f(x) = 0. In each of the intervals limited by the equilibrium points, find the sign of f and draw an arrow to the right if f is positive and to the left if f is negative.
2. Create a t x-plane, transfer the phase line information to the x-axis,