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

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 .. 479 480 481 482 483 484 < 485 > 486 487 488 489 490 491 .. 609 >> Next

moderating effect on the population fluctuations, keeping
the trajectory close to the center.
13. The critical points of the system are the solutions of the algebraic equations x(a - ox - ay) = 0, and y(-c + gx) = 0. the critical points are x = 0, y = 0;
x = a/o, y = 0; and x = c/g, y = a/a - co/ag = oA/a
where A = a/o - c/g > 0.
To study the critical point (0,0) we discard the nonlinear terms in the system of D.E. to obtain the corresponding linear system dx/dt = ax, dy/dt = -cy. The characteristic equation is r2 - (a+c)r - ac = 0 so ri = a, r2 = -c. Thus the critical point (0,0) is an unstable saddle point.
To study the critical point (a/o,0) we let x = (a/o) + u, y = 0 + v and substitute in the D.E. to obtain the almost linear system du/dt = -au - (aa/o)v - ou2 - auv, dv/dt = gAv + guv. The corresponding linear system is du/dt = -au - (aa/o)v, dv/dt = gAv. The characteristic equation is r2 + (a - gA)r - agA = 0 so r1 = -a, r2 = gA.
194
Section 9.6
Thus the critical point (a/o,0) is an unstable saddle point.
To study the critical point (c/g, oA/a) we let x = (c/g) + u, y = (oA/a) + v and substitute in the D.E. to obtain the almost linear system
du 2
— = -(co/g)u - (ac/g)v - ou2 - auv dt
dv
— = (oAg/a)u + guv dt
The corresponding linear system is du/dt = -(co/g)u - (ac/g)v, dv/dt = (oAg/a)u. The characteristic equation is r2 + (co/g)r + coA = 0, so r1,r2 = [-(co/g) ħ V (co/g) 2 - 4coA ] /2. Thus, depending
on the sign of the discriminant we have that (c/g, oA/a) is either an asymptotically stable spiral point or an asymptotically stable node. Thus for nonzero initial data (x,y) ^ (c/g, oA/a) as t ^ •.
Section 9.6, Page 519
1. Assuming that V(x,y) = ax2 + cy2 we find Vx(x,y) = 2ax,
Vy = 2cy and thus Eq.(7) yields V(x,y) = 2ax(-x3 + xy2) +
2cy(-2x2y - y3) = -[2ax4 + 2(2c-a)x2y2 + 2cy4]. If we choose a and c to be any positive real numbers with
2c > a, then V is a negative definite. Also, V is positive definite by Theorem 9.6.4. Thus by Theorem 9.6.1 the origin is an asymptotically stable critical point.
3. Assuming the same form for V(x,y) as in Problem 1, we
have
V(x,y) = 2ax(-x3 + 2y3) + 2cy(-2xy2) = -2ax4 + 4(a-c)xy3.
If we choose a = c > 0, then V(x,y) = -2ax4 < 0 in any
neighborhood containing the origin and thus V is negative semidefinite and V is positive definite. Theorem 9.6.1 then concludes that the origin is a stable critical point. Note that the origin may still be asymptotically stable, however, the V(x,y) used here is not sufficient to prove that.
6a. The correct system is dx/dt = y and dy/dt = -g(x). Since g(0) = 0, we conclude that (0,0) is a critical point.
Section 9.6
195
6b.
7b.
7c.
7d.
From the given conditions, the graph of g must be positive for 0 < x < k and negative for -k < x < 0. Thus
if 0 < x < k then xg(s)ds > 0,
0
x0
if -k < x < 0 then g(s)ds = - g(s)ds > 0.
0x
Since V(0,0) = 0 it follows that V(x,y) = y2/2 + xg(s)ds
0
is positive definite for -k < x < k, -• < y < •. Next,
• dx dy
we have V(x,y) = Vx + Vy— = g(x)y + y[-g(x)] = 0.
dt dt
Since V(x,y) is never positive, we may conclude that it is negative semidefinite and hence by Theorem 9.6.1 (0,0) is at least a stable critical point.
V is positive definite by Theorem 9.6.4. Since
Vx(x,y) = 2x, Vy(x,y) = 2y, we obtain
V(x,y) = 2xy - 2y2 - 2ysinx = 2y[-y + (x - sinx)]. If
x < 0, then V(x,y) < 0 for all y > 0. If x > 0, choose y
so that 0 < y < x - sinx. Then V(x,y) > 0. Hence V is not a Liapunov function.
Since V(0,0) = 0, 1 - cosx > 0 for 0 < |x| < 2p and y2 >
0 for y Ï 0, it follows that V(x,y) is positive definite
in a neighborhood of the origin. Next Vx(x,y) = sinx,
Vy(x,y) = y, so
V(x,y) = (sinx)(y) + y(-y - sinx) = -y2. Hence V is
negative semidefinite and (0,0) is a stable critical point by Theorem 9.6.1.
V(x,y) = (x+y)2/2 + x2 + y2/2 = 3x2/2 + xy + y2 is
positive definite by Theorem 9.6.4. Next
Vx(x,y) = 3x + y, Vy(x,y) = x + 2y so
V(x,y) = (3x+y)y - (x+2y)(y+sinx)
= 2xy - y2 - (x+2y)sinx
= 2xy - y2 - (x+2y)(x - ax3/6)
= -x2 - y2 + a(x+2y)x3/6
= -r2 + ar4 (cos9 + 2sin9)(cos39)/6 < -r2 + r4/2
= -r2(1-r2/2). Thus V is negative definite for
r < V2. From Theorem 9.6.1 it follows that the origin is an asymptotically stable critical point.
196
Section 9.6
8. Let x = u and y = du/dt to obtain the system dx/dt = y
and dy/dt = -c(x)y - g(x). Now consider
V(x,y) = y2/2 + xg(s)ds, which yields J0
V = g(x)y + y[-c(x)y - g(x)] = -y2c(x).
10b. Since Vx(x,y) = 2Ax + By, Vy(x,y) = Bx + 2Cy, we have
Previous << 1 .. 479 480 481 482 483 484 < 485 > 486 487 488 489 490 491 .. 609 >> Next