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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
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A = -(r1+2a), B = 2ar! + a2 + b2 and C = -r^a2+b2).
Hence, if AB = C, we have
Section 9.8
201
3c.
4.
5a.
V = 5b.
-(r1+2a)(2ar1+a2+b2) = -r1(a2+b2) or
-2a [r2 + 2ar1 + (a2+b2)] = 0 or -2a[(r1+a) 2+b2] = 0.
Since the square bracket term is positive, we conclude that if AB = C, then a = 0. That is, the conjugate complex roots are pure imaginary. Note that the converse is also true. That is, if the conjugate complex roots are pure imaginary then AB = C.
Comparing Eq.(9) to that of part b, we have A = 41/3,
B = 8(r+10)/3 and C = 160(r-1)/3. Thus AB = C yields r = 470/19.
We have V = 2x[o(-x+y)] + 2oy[rx-y-xz] + 2oz[-bz+xy]
= -2Ox2 + 2oxy + 2orxy - 2oy2 - 2obz2 = 2s{-[x2-(r+i)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 + sy2 + s(z-2r)2 = c > 0 yields dv
= 2rx[s(-x+y)] + 2sy(rx-y-xz) + 2s(z-2r)(-bz+xy). Thus
dt
-2s[rx2+y2 + b(z2 - 2rz)] = -2s [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/s c/s
M = max^br , r\fh , r), then the ellipsoid (i) is
x2 y2 (z-r)2
contained inside the sphere Si: + + ----------------- = 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 S
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 + c2sin^/2x is the general solution of the D.E. Thus y'(x) = - V"2c1si^V"2x + "\/2c2cos V2x and hence y'(0) = /2c2 = 1, which gives c2 = 1/^/2. Now,
y'(rc) = -^2cisi^^2 n + cosV"2 n = 0 then yields
cos^/2
c1 = ------- --- = co^ 2 / 2 . Thus the desired solution is
y2 siny2
y = (cot\/2cos\/2x + sin^/2x)/^/2 .
3. We have y(x) = c1cosx + c2sinx as the general solution and hence y(0) = c1 = 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) = c1cos2x + 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^/^ x + c2cosh^/|7 x where -X = . 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^/X x. The B.C. require that c2 = 0 and c1co^V"X = 0. The second condition is satisfied for X 0 and c1 0 if \[Xn = (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... .
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