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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
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9.6 Liapunovs Second Method
517
and Vis negative definite, then every solution of Eqs. (6) that starts at a point in DK approaches the origin as t approaches infinity.
In other words, Theorem 9.6.3 says that if x = (), y = ty(t) is the solution of Eqs. (6) for initial data lying in DK, then (x, y) approaches the critical point (0, 0) as t ^ to. Thus DK gives a region of asymptotic stability; of course, it may not be the entire basin of attraction. This theorem is proved by showing that (i) there are no periodic solutions of the system (6) in DK, and (ii) there are no other critical points in DK. It then follows that trajectories starting in DK cannot escape and, therefore, must tend to the origin as t tends to infinity.
Theorems 9.6.1 and 9.6.2 give sufficient conditions for stability and instability, respectively. However, these conditions are not necessary, nor does our failure to determine a suitable Liapunov function mean that there is not one. Unfortunately, there are no general methods for the construction of Liapunov functions; however, there has been extensive work on the construction of Liapunov functions for special classes of equations. An elementary algebraic result that is often useful in constructing positive definite or negative definite functions is stated without proof in the following theorem.
Theorem 9.6.4 The function
V(x, y) = ax2 + bxy + cy2 (14)
is positive definite if, and only if,
a > 0 and 4ac --- b2 > 0, (15)
and is negative definite if, and only if,
a < 0 and 4ac --- b2 > 0. (16)
The use of Theorem 9.6.4 is illustrated in the following example.
Show that the critical point (0, 0) of the autonomous system
dx/dt = x - xy2, dy/dt =-y - x2y (17)
is asymptotically stable.
We try to construct a Liapunov function of the form (14). Then Vx (x, y) = 2ax + by,
Vy(x, y) = bx + 2cy, so
V(x, y) = (2ax + by)(x xy2) + (bx + 2cy)(y x2y)
= [2a(x2 + x2 y2) + b(2xy + xy3 + x3 y) + 2c(y2 + x2 y2)].
If we choose b = 0, and a and c to be any positive numbers, then Vis negative definite
and V is positive definite by Theorem 9.6.4. Thus by Theorem 9.6.1 the origin is an asymptotically stable critical point.
518
Chapter 9. Nonlinear Differential Equations and Stability
EXAMPLE
4
Consider the system
dx/dt = x(1 - x - y),
(18)
dy/dt = y(0.75 - y - 0.5x).
In Example 1 of Section 9.4 we found that this systemmodels a certain pair of competing species, and that the critical point (0.5, 0.5) is asymptotically stable. Confirm this conclusion by finding a suitable Liapunov function.
It is helpful to transform the point (0.5, 0.5) to the origin. To this end let
x = 0.5 + u, y = 0.5 + v. (19)
Then, substituting for x and y in Eqs. (18), we obtain the new system
du/dt = 0.5u - 0.5v - u2 - uv,
2 (20)
dv/dt = 0.25u - 0.5v - 0.5uv - v .
To keep the calculations relatively simple, consider the function V(u, v) = u2 + v2 as a possible Liapunov function. This function is clearly positive definite, so we only need to determine whether there is a region containing the origin in the uv -plane where the derivative Vwith respect to the system (20) is negative definite. We compute V(u, v) and find that
V du dv
V(u, v) = V--------+ V
v ' udt v dt
= 2u(-0.5u - 0.5v - u2 - uv) + 2v(-0.25u - 0.5v - 0.5uv - v2),
or
V(u, v) = -[(u2 + 1.5uv + v2) + (2u3 + 2u2v + uv2 + 2v3)], (21)
where we have collected together the quadratic and cubic terms. We want to show that the expression in square brackets in Eq. (21) is positive definite, at least for u and v
sufficiently small. Observe that the quadratic terms can be written as
u2 + 1.5uv + v2 = 0.25 (u2 + v2) + 0.75(u + v)2, (22)
so these terms are positive definite. On the other hand, the cubic terms in Eq. (18) may be of either sign. Thus we must show that, in some neighborhood of u = 0, v = 0, the
cubic terms are smaller in magnitude than the quadratic terms; that is,
|2u3 + 2u2v + uv2 + 2v3| < 0.25(u2 + v2) + 0.75(u + v)2. (23)
To estimate the left side of Eq. (23) we introduce polar coordinates u = r cos , v = r sin . Then
|2u3 + 2u2v + uv2 + 2v3| = r3|2cos3 + 2cos2 sin + cos sin2 + 2 sin3 |
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