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 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
Previous << 1 .. 377 378 379 380 381 382 < 383 > 384 385 386 387 388 389 .. 486 >> Next

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:
Previous << 1 .. 377 378 379 380 381 382 < 383 > 384 385 386 387 388 389 .. 486 >> Next