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

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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

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