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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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is u = c1 + c2x. The B.C. require that c1 = 0,
c2(1-2L) = 0 so nontrivial solutions are possible only if L = 1/2. In this case the eigenfuction is 0(x) = xe-2x.
If X > 0, the general solution of the D.E. u" + 9Xu = 0 is u = c1sin3^/X x + c2cos^/X x. The B.C. require that c2 = 0, c1(^V"X cos3a/"X l - 2sin3^/~X L) = 0. In order to have nontrivial solutions X must satisfy the equation V~X = (2/3)tan^V"X L. A graphical analysis reveals that there is an infinite number of solutions to this eigenvalue equation. Thus the eigenfunctions are ^) = e"2xsin3y/Xn x where the eigenvalues Xn satisfy \/ = (2/3)tan3\/~Xn L.
20. This is an Euler equation whose characteristic equation
has roots r1 = X and r2 = 1. If X = 1 the general
solution of the D.E. is y = c x + c xlnx and the B.C.
12
238
Section 11.1
require that c1 = c2 = 0 and thus X = 1 is not an
eigenvalue. If X 1, y = c1x + c2xX is the general
solution and the B.C. require that c1 + c2 = 0 and
2c + c 2X _ (c + Xc 2X_1) = 0. Thus nontrivial solutions
1 2 1 2
exist if and only if X = 2(1_2_X). The graphs of
f(X) = X and g(X) = 2(1_2_X) intersect only at X = 1
(which has already been discussed) and X = 0. Thus the
only eigenvalue is X = 0 with corresponding eigenfunction
^) = x _ 1 (since c1 = _c2).
22a.For positive X, the general solution of the D.E. is y = c^inVX x + c2cos^/X x. The B.C. require that
VX c1 + ac2 = 0, c^inVX + c2cosVX = 0. Nontrivial
solutions exist if and only if VX cos VX _ asinVX = 0.
If a = 0 this equation is satisfied by the sequence Xn = [(2n_1)n/2]2, n = 1,2,... . If a 0, X must
satisfy the equation VX/a = tanVX . A plot of the graphs of f(VX) = VX/a and g(VX) = tanVX reveals that there is an infinite sequence of postive eigenvalues for a < 0 and a > 0.
22b. By procedures shown previously, the cases X < 0 and
X = 0, when a < 1, lead to only the trivial solution and thus by part a all real eigenvalues are positive. For 0 < a < 1, the graphs of f(VX) and g( VX ) (see part a) intersect once on 0 < VX < n/2. As a approaches 1 from below, the slope of f(VX) decreases and thus the intersection point approaches zero.
22c. If X = 0, then y(x) = c1 + c2x and the B.C. yield
ac1 + c2 = 0 and c1 + c2 = 0, which have a non_zero
solution if and only if a = 1.
22d. Let _X = 2, then y(x) = c1coshx + c2sinhx and thus the
B.C. yield ac1 + ^ = 0 and (cosh^^ + (sinh^c2 = 0, which
have non_zero solutions if and only if tanh = /a. For a>1, the straight line y = /a intersects the curve y = tnh in one point, which increases as a increases. Thus X = _2 decreases as a increases.
23. Using the D.E. m and following the hint yields:
liml ^ndx + Xml ndx = 0. Integrating the first term by
00
Section 11.2
239
parts yields: nl0 - f^r^ndx = -Xm| Lm ndx. Upon utilizing
00
he B.C. the first term on the left vanishes and thus
J4>mdx = X m I Lmndx. Similarly, the D.E. for yields 00
I mndx = XnI mdx and thus (Xn - Xm) I ^ndx = 0. If
0 0 0
Xn Xm the desired result follows.
24b. The general solution of the D.E. is
y = c1sinp,x + c2cosp.x + c3sinhp,x + c4coshp,x where X = 4.
The B.C. require that c2 + c4 = 0, -c2 + c4 = 0, c1sinp,L + c2cosp.L + c3sinhp,L + c4coshp,L = 0, and c1cosp.L - c2sinp,L + c3coshp,L + c4sinhp,L = 0. The first two
equations yield c2 = c4 = 0, and the last two have nontrivial solutions if and only if sinpLcoshpL - cospLsinhpL = 0. In this case the third equation yields c3 = -c1sinp,L/sinhp,L and thus the desired eigenfunctions are obtained. The quantity can be approximated by finding the intersetion of f(x) = tanx and g(x) = tanhx, where x = . The first intersection is at x = 3.9266, which gives X1 = 237.72/L4 and the second intersection is at x = 7.0686, which gives X2 = 2,496.5/L4.
Section 11.2, Page 639
1. We have y(x) = c1co^V"X x + c2sinV"X x and thus y(0) = 0
yields c1 = 0 and y'(1) = 0 yields "\pk c2cosV~X = 0. X = 0 gives y(x) = 0, so is not an eigenvalue. Otherwise X = (2n-1)2n2/4 and the eigenfuctions are sin[(2n-1)nx/2], n = 1,2,... . Thus, by Eq.(20), we must choose kn so that
J1{k sin[(2n-1)nx/2]}2dx = 1, since the weight function
0n
r(x) = 1 (by comparing the D.E. to Eq.(1)). Evaluating the integral yields ^/2 = 1 and thus kn = \p2 and the desired normalized eigenfuctions are obtained.
3. Note here that 0(x) = 1 satisifes Eq.(20) and hence it
is already normalized.
5. From Problem 17 of Section 11.1 we have exsinnnx,
n = 1,2,... as the eigenfunctions and thus k must be chosen
so that I1r(x)k2e2xsin2nnxdx = 1. To determine r(x), we must
0n
write the D.E. in the form of Eq.(1). That is, we multiply
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