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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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476
Chapter 9. Nonlinear Differential Equations and Stability
EXAMPLE
1
way of solving Eq. (14) to obtain the function H, so this approach is applicable only in special cases.
Find the trajectories of the system
dx/dt = y, dy/dt = x. (16)
In this case, Eq. (14) becomes
T = ^ (17)
dx y
This equation is separable since it can be written as
ydy = x dx,
and its solutions are given by
H(x, y) = y2 - x2 = c, (18)
where c is arbitrary. Therefore the trajectories of the system (16) are the hyperbolas
shown in Figure 9.2.4. The direction of motion on the trajectories can be inferred from
the fact that both dx/ dt and dy/ dt are positive in the first quadrant. The only critical point is the saddle point at the origin.
Another way to obtain the trajectories is to solve the system (16) by the methods of Section 7.5. We omit the details, but the result is
x = qe* + C2 e~t, y = cxel — C2 e—t.
Eliminating t between these two equations again leads to Eq. (18).
FIGURE 9.2.4 Trajectories of the system (16).
9.2 Autonomous Systems and Stability
477
EXAMPLE
2
Find the trajectories of the system From the equations
dx dy „ 2
— = 4 - 2y, — = 12 - 3x . dt J dt
(19)
4 - 2y = 0, 12 - 3x2 = 0
we find that the critical points of the system (19) are the points (-2, 2) and (2, 2). To determine the trajectories note that for this system Eq. (14) becomes
12 - ^ (20)
dy
dx
4 - 2y
Separating the variables in Eq. (20) and integrating, we find that solutions satisfy
H(x, y) = 4y - y2 - 12x + x3 =
c,
(21)
where c is an arbitrary constant. A computer plotting routine is helpful in displaying the level curves of H(x, y), some of which are shown in Figure 9.2.5. The direction of motion on the trajectories can be determined by drawing a direction field for the system (19), or by evaluating dx/dt and dy/dt at one or two selected points. From Figure 9.2.5 you can see that the critical point (2, 2) is a saddle point while the point (-2, 2) is a center. Observe that one trajectory leaves the saddle point (at t = — cx>), loops around the center, and returns to the saddle point (at t = +cx>).
PROBLEMS In each of Problems 1 through 4 sketch the trajectory corresponding to the solution satisfying
a the specified initial conditions, and indicate the direction of motion for increasing t.
1. dx/dt = -x, dy/dt = —2y; x(0) = 4, y(0) = 2
2. dx/dt = - x, dy/dt = 2y; x (0) = 4, y(0) = 2 and x (0) = 4, y(0) = 0
3. dx/dt = - y, dy/dt = x; x (0) = 4, y(0) = 0 and x (0) = 0, y(0) = 4
4. dx/dt = ay, dy/dt = —bx, a > 0, b > 0; x(0) = „Ja, y(0) = 0
478
Chapter 9. Nonlinear Differential Equations and Stability
For each of the systems in Problems 5 through 14:
(a) Find all the critical points (equilibrium solutions).
(b) Use a computer to draw a direction field and phase portrait for the system.
(c) From the plot(s) in part (b) determine whether each critical point is asymptotically stable,
stable, or unstable, and classify it as to type.
? 5. dx/dt = x — xy, dy/dt = y + 2xy
? 6. dx/dt = 1 + 2y, dy/dt = 1 — 3x2
? 7. dx/dt = x — x2 — xy, dy/dt = 2y — 1 y2 — 3xy
? 8. dx/dt =-(x — y)(1 — x — y), dy/dt = x(2 + y)
? 9. dx/dt = y(2 — x — y), dy/dt =—x — y — 2xy
? 10. dx/dt = (2 + x)(y — x), dy/dt = y(2 + x — x2)
? 11. dx/dt =—x + 2xy, dy/dt = y — x2 — y2
? 12. dx/dt = y, dy/dt = x — 1 x3 — 1 y
? 13. dx/dt = (2 + x)(y — x), dy/dt = (4 — x)(y + x)
? 14. The van der Pol equation: dx/dt = y, dy/dt = (1 — x2)y — x
In each of Problems 15 through 22:
(a) Find an equation of the form H(x, y) = c satisfied by the trajectories.
(b) Plot several level curves of the function H. These are trajectories of the given system. Indicate the direction of motion on each trajectory.
? 15. dx/dt = 2y, dy/dt = 8x ? 16. dx/dt = 2y, dy/dt =—8x
? 17. dx/dt = y, dy/dt = 2x + y ? 18. dx/dt =—x + y, dy/dt =—x — y
? 19. dx/dt =—x + y + x2, dy/dt = y — 2xy
? 20. dx/dt = 2x2y — 3x2 — 4y, dy/dt =—2xy2 + 6xy
? 21. Undamped pendulum: dx/dt = y, dy/dt =— sin x
? 22. Duffing’s equation: dx/dt = y, dy/dt =—x + (x3/6)
23. Given that x = 0(t), y = ft(t) is a solution of the autonomous system
dx/dt = F(x, y), dy/dt = G(x, y)
for a < t < ?, show that x = ฎ(t) = $(t — s), y = ^(t) = ft(t — s) is a solution for
a + s < t < ? + s for any real number s.
24. Prove that for the system
dx/dt = F(x, y), dy/dt = G(x, y)
there is at most one trajectory passing through a given point (x0, y0).
Hint: Let C0 be the trajectory generated by the solution x = 00 (t), y = ^0(t), with $0(t0) = x0, ^0(t0) = y and let C be the trajectory generated by the solution x = 01(t), y = ^ (t), with 0^(1 = x0, ^j(tj) = y0. Use the fact that the system is autonomous and also the existence and uniqueness theorem to show that C0 and C1 are the same.
25. Prove that if a trajectory starts at a noncritical point of the system
dx/dt = F(x, y), dy/dt = G(x, y),
then it cannot reach a critical point (x0, y0) in a finite length of time.
Hint: Assume the contrary; that is, assume that the solution x = 0(t), y = ft(t) satisfies
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