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r n n r n 1
X2 = -59.6795, X = -(2n+1)2n2/4 and 6 = sinp x - p cosp x.
n n rnrnrn
If X > 0, the general solution of the D.E. is
y(x) = c^osh^/X x + c2sinh^/X x. The B.C. respectively
require that c1 + \[X c2 = 0 and c^osh^/X + c2sinh^/X = 0 and
thus X must satisfy tanh^/X = j/X in order to have nontrivial solutions. The only solution of this equation is X = 0 and thus there are no positive eigenvalues.
11a. From Eq.(i) the coefficient of y' is pQ and from Eq.(ii) the
coefficient of y' is (pP)'. Thus (pP)' = pQ, which gives
11b. Eq.(iii) is both linear and separable. Using the latter
approach we have dp/p = [(Q-P')/P]dx and thus
lnP. Taking the exponential of
both sides yields Eq.(iv). The choice of x0 simply alters the constant of integration, which is immaterial here.
13. Since P(x) = x2 and Q(x) = x, we find that
p(x) = (1/x2)exp[^x' (s/s2)ds] = k/x, where k is an
arbitrary constant which may be set equal to 1. It follows that Bessel's equation takes the form (xy')' + (x2-u2/x)y = 0.
18a.Assuming y = s(x)u, we have y' = s'u + su' and
y" = s"u + 2s'u' + su" and thus the D.E. becomes su" + (2s'+4s)u' + [s" + 4s' + (4 + 9X)s]u = 0. Setting 2s' + 4s = 0 we find s(x) = e-2x and the D.E. becomes u" + 9Xu = 0. The B.C. y(0) = 0 yields s(0)u(0) = 0, or u(0) = 0 since s(0) ^ 0. The B.C. at L
is y'(L) = s' (L)u(L) + s(L)u'(L) = e-2L (-2u(L) + u'(L)) = 0
and thus u'(L) - 2u(L) = 0. Thus the B.V.P. satisfied by
u(x) is u" + 9Xu = 0, u(0) = 0, u'(L) - 2u(L) = 0.
If X < 0, the general solution of the D.E. u" + 9Xu = 0
is u = c1sinh3p.x + c2cosh3px where -X = p.2. The B.C.
require that c2 = 0, c1(3pcosh3pL - 2sinh3pL) = 0. In order to have nontrivial solutions p must satisfy the equation 3p/2 = tanh3pL. A graphical analysis reveals that for L < 1/2 this equation has no solutions for p ^ 0 so there are no negative eigenvalues for L < 1/2. If L > 1/2 there is one solution and hence one negative eigenvalue with eigenfuction ^-1(x) = e-2xsinh3px.
If X = 0, the general solution of the D.E. u" + 9Xu = 0
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 cos^V"X L - 2sin3\/X L) = 0. In order to have nontrivial solutions X must satisfy the equation ypk = (2/3)tan3^/"X L. A graphical analysis reveals that there is an infinite number of solutions to this eigenvalue equation. Thus the eigenfunctions are ^n(x) = e-2xsin3y/Xn x where the eigenvalues Xn satisfy \[K = (2/3)tan^^/"Xn L.
20. This is an Euler equation whose characteristic equation
has roots r! = X and r2 = 1. If X = 1 the general
solution of the D.E. is y = cx + c xlnx and the B.C.
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 12 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) = x - 1 (since c1 = -c2).
22a.For positive X, the general solution of the D.E. is y = c^in^/X x + c2cos^/X x. The B.C. require that
\[X c1 + ac2 = 0, c^in^/X + c2cos^X = 0. Nontrivial
solutions exist if and only if ypk cos a/X - asin^/X = 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 y/X/a = tan^X . A plot of the graphs of f(^/X) = "\[X /a and g(VX) = tan^X 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(^/X) and g(^/X ) (see part a) intersect once on 0 < < n/2. As a approaches 1 from
below, the slope of f(V"X) 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 = p2, then y(x) = c1coshp,x + c2sinhpx and thus the
B.C. yield ac1 + pc2 = 0 and (coshp)c1 + (sinhp)c2 = 0, which
have non-zero solutions if and only if tanhp = p/a. For a>1, the straight line y = p/a intersects the curve y = tanhp in one point, which increases as a increases. Thus X = -p2 decreases as a increases.
23. Using the D.E. ^m and following the hint yields:
liml LK6ndx + Xml l6m6ndx = 0. Integrating the first term by 00
parts yields: nlo - [L^m^ndx = -Xm|ndx. Upon utilizing
he B.C. the first term on the left vanishes and thus
JL6>mdx = Xml L6m6ndx. Similarly, the D.E. for 6n yields