# Elementary differential equations 7th edition - Boyce W.E

ISBN 0-471-31999-6

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We have V = 2x[o(-x+y)] + 2oy[rx-y-xz] + 2oz[-bz+xy]

= -2Ox2 + 2oxy + 2orxy - 2oy2 - 2obz2 = 2o{-[x2-(r+1)xy+y2]-bz2}. For r < 1, the term in the square brackets remains positive for all

values of x and y, by Theorem 9.6.4, and thus V is negative definite. Thus, by the extension of Theorem 9.6.1 to three equations, we conclude that the origin is an asymptotically stable critical point.

V = rx2 + Oy2 + o(z-2r)2 = c > 0 yields dv

= 2rx[o(-x+y)] + 2oy(rx-y-xz) + 2O(z-2r)(-bz+xy). Thus

dt

-2o[rx2+y2 + b(z2 - 2rz)] = -2o[rx2 + y2 + b(z-r) 2 - br2].

From the proof of Theorem 9.6.1, we find that we need to

show that V, as found in part a, is always negative as it crosses V(x,y,z) = c. (Actually, we need to use the extension of Theorem 9.6.1 to three equations, but the proof is very similar using the vector calculus approach.) From part a we see that

V < 0 if rx2 + y2 + b(z-r)2 > br2, which holds if (x,y,z)

x2 y2 (z-r) 2

lies outside the ellipsoid -------- + —- + — = 1. (i)

br br2 r2

Thus we need to choose c such that V = c lies outside

Eq.(i). Writing V = c in the form of Eq.(i) we obtain

x2 y2 (z-2r) 2

the ellipsoid ------- + +------------------ = 1. (ii) Now let

c/r c/O c/O

M = max^br , r\fb , r), then the ellipsoid (i) is

x2 y2 (z-r)2

contained inside the sphere S1: — + — + ----------------- = 1.

M2 M2 M2

Let S2 be a sphere centered at (0,0,2r) with radius

202

Section 9.8

M+r:

y

(z-2r)

+ --------- = 1, then S1 is contained

(M+r2)

(M+r) 2 (M+r) 2

in S2. Thus, if we choose c, in Eq.(ii), such that c 2 c 2

— > (M+r) 2 and — > (M+r) 2, then V < 0 as the trajectory r o

crosses V(x,y,z) = c. Note that this is a sufficient condition and there may be many other "better" choices using different techniques.

8b. Several cases are shown. Results may vary, particularly for r = 24, due to the closeness of r to r @ 24.06.

2

x

203

CHAPTER 10

Section 10.1 Page 547

2. y(x) = c1co^V"2x + c2sinV"2x is the general solution of the D.E. Thus y'(x) = -V"2c1si^V"2x + "\/"2c2cosV"2x and hence y'(0) = /2 c2 = 1, which gives c2 = 1/^/2. Now,

y'(rc) = ^\[2c1sin\f2 n + cosV2n = 0 then yields

cosi/2 n I— I—

c! = ——--------------- = coty 2 rc/y 2 . Thus the desired solution is

y2 siny2 n

y = (cot\/2 ncos\/2x + sin^/2x)/^/2 .

3. We have y(x) = cicosx + c2sinx as the general solution and hence y(0) = c! = 0 and y(L) = c2sinL = 0. If sinL ^ 0, then c2 = 0 and y(x) = 0 is the only solution. If sinL = 0, then y(x) = c2sinx is a solution for arbitrary c2.

7. y(x) = cicos2x + c2sin2x is the solution of the related

1

homogeneous equation and yp(x) = —cosx is a particular

3

1

solution, yielding y(x) = c1cos2x + c2sin2x + —cosx as the

3

1

general solution of the D.E. Thus y(0) = c1 + — = 0 and

3

1

y(n) = c1 - — = 0 and hence there is no solution since there 3

is no value of c1 that will satisfy both boundary conditions.

11. If X < 0, the general solution of the D.E. is

y = c^inh^/l x + c2cos^V^ x where -X = p.. The two B.C.

require that c2 = 0 and c1 = 0 so X < 0 is not an eigenvalue.

If X = 0, the general solution of the D.E. is y = c1 + c2x. The B.C. require that c1 = 0, c2 = 0 so again X = 0 is not an eigenvalue. If X > 0, the general solution of the D.E. is y = c^in^/X x + c2cos^/Xx. The B.C. require that c2 = 0 and ‘sfX c1co^V"Xn = 0. The second condition is satisfied for X ^ 0 and c1 ^ 0 if = (2n-1)n/2, n = 1,2,... . Thus the

eigenvalues are Xn = (2n-1)2/4, n = 1,2,3... with the corresponding eigenfunctions yn(x) = sin[(2n-1)x/2], n = 1,2,3... .

15. For X < 0 there are no eigenvalues, as shown in Problem 11. For X = 0 we have y(x) = c1 + c2x, so y'(0) = c2 = 0 and

204

Section 10.2

y (n) = c2 = 0, and thus X = 0 is an eigenvalue, with y0(x) = 1 as the eigenfunction. For X > 0 we again have

y(x) = c^in^/X x + c2cos^/Xx, so y'(0) = \[X c1 = 0 and y'(L) = -c2\fX sin^/X L = 0. We know X > 0, in this case, so the eigenvalues are given by sin^X^L = 0 or \[X L = nn. Thus X n = (nn/L) 2 and yn(x) = cos(nnx/L) for n = 1,2,3... .

Section 10.2, Page 555

We look for values of T for which sinh2(x+T) = sinh2x for all x. Expanding the left side of this equation gives sinh2xcosh2T + cosh2xsinh2T = sinh2x, which will be satisfied for all x if we can choose T so that cosh2T = 1 and sinh2T = 0. The only value of T satisfying these two constraints is T = 0. Since T is not positive we conclude that the function sinh2x is not periodic.

We look for values of T for which tann(x+T) = tannx.

Expanding the left side gives

tann(x+T) = (tannx + tannT)/(1-tannxtannT) which is equal

to tannx only for tannT = 0. The only positive solutions of this last equation are T = 1,2,3... and hence tannx is periodic with fundamental period T = 1.

7.

To start, let n = 0, then f(x)

for n

I 0 -1 < x < 0

| 1 0 < x < 1

0 1 < x < 2 I 0 3 < x < 4

; and for n = 2, f(x) = \ .

[ 1 2 < x < 3 [ 1 4 < x < 5

continuing in this fashion, and drawing a graph, it can be

f(x) =

seen that T

2.

10.

The graph of f(x) is:

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