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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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240
Section 11.2
the D.E. by r(x) to obtain ry" - 2ry' + ry + Xry = 0. Now choose r so that (ry')' = ry"-2ry', which yields r' = -2r or
-2x
r(x) = e . Thus the above integral becomes
J*1 2 2 / / x
k sin nnxdx = 1 and k = v 2 . Hence (x) = / 2 e sinnnx
0 n n n
are the normalized eigenfunctions.
7. Using Eq.(34) with r(x) = 1, we find that the
coefficients of the series (32) are determined by a - f- > = V2 f xsin[(2n-1)nx/2]dx
= (4^/"2/(2n-1) 2n 2)sin(2n-1)n/2. Thus Eq.(32) yields
4\p2 ^ (-1)n-1
f(x) = ----- / ------- V 2 sin[(2n-1)nx/2], 0 < x < 1,
2 ^ (2n-1)2
n=1
which agrees with the expansion using the approach developed in Problem 39 of Section 10.4.
10. In this case (x) = (v2 /a ) cosi/X x, where
T n * n V n
an = (1 + sin2^)1/2. Thus Eq.(34) yields
a = (V"2 /a ) 1co^/X xdx = yT2 sim/X /a i/X .
n * nj0 V n Vn^Vn
14. In this case L[y] = y" + y' + 2y is not of the form shown in Eq.(3) and thus the B.V.P. is not self adjoint.
17. In this case L[y] = [(1+x2)y']' + y and thus the D.E. has
the form shown in Eq.(3). However, the B.C. are not separated and thus we must determine by integration whether Eq.(8) is satisfied. Therefore, for u and v satisfying the B.C., integration by parts yields the following:
(L[u],v) = 11{[(1+x2)u']'+u}vdx = vu'(1+x2) -11{(1+x2)v'u'+uv}dx
J0 0 '0
= vu' (1+x2)
1
-uv' (1+x2)
0
+ 11{[(1+x2)v']'+v}udx
0
= (u,L[v])
since the integrated terms add to zero with the given B.C. Thus the B.V.P. is self-adjoint.
21a. Substituting for y in the D.E., multiplying both sides by , and integrating form 0 to 1 yields
ղ1 2dx = 11{-[p(x) ']' + q(x^ 2}dx. Integrating the
00
first term on the right side once by parts, we obtain
1
0
Section 11.2
241
ղ 2dx = _p(1) '(1) (1) + p(0) '(0) (0) + ^' 2+q2)dx.
00
If a2 0, b2 0, then '(1) = _1(1)/2 and
'(0) = _al(0)/a2 and the result follows. If a2 = 0,
then (0) = 0 and the boundary term at 0 will be missing.
A similar result is obtained if b2 = 0.
21b. From the text, p(x) > 0 and r(x) > 0 for 0 < x < 1 (following Eq.(4)). If q(x) > 0 and if b1/b2 and _a1/a2 are non_negative, then all terms in the final equation of part a are non_negative and thus X must be non_negative. Note that X could be zero if q(x) = 0, b1 = 0, a1 = 0 and = 1.
21c. If either q(x) 0 or a1 0 or b1 0, there is at least one
positive term on the right and thus X must be positive.
23a. Using = U(x) + iV(x) in Eq.(4) we have
L^] = L[U(x) + iV(x)] = Xr(x)[U(x)+iV(x)]. Using the linearity of L and the fact that X and r(x) are real we have L[U(x)] + iL[V(x)] = Xr(x)U(x) + iXr(x)V(x).
Equating the real and imaginary parts shows that both U and V satisfy Eq.(1). The B.C. Eq.(2) are also satisfied
by both U and V, using the same arguments, and thus both
U and V are eigenfunctions.
23b. By Theorem 11.2.3 each eigenvalue X has only one linearly independent eigenfuction. By part a we have U and V being eigenfunctions corresponding to X and thus U and V must be linearly dependent.
23c. From part b we have V(x) = cU(x) and thus (x) = U(x) + icuU(x) = (1+ic)U(x).
24. This is an Euler equation, so for y = xr we have
r2 _ (X+1)r + X = 0 or (r_1)(r_X) = 0. If X = 1, the general solution to the D.E. is y = c1x + c2xlnx. The B.C. require
that c = 0, 2c + 2(ln2)c = 0 so c = c = 0 and X = 1 is
112 12
not an eigenvalue. If X 1, the general solution to the
D.E. is y = cx + cxX. The B.C. require that c + c = 0 and
1 1 2 ^ 12
2c + 2Xc = 0. Nontrivial solutions exist if and only if 12
2X _ 2 = 0. If X is real, this equation has no solution (other than X = 1) and again y = 0 is the only solution to the boundary value problem. Suppose that X = a + bi with b 0. Then 2X = 2a+bl = 2a2bl = 2aexp(ibln2), which upon
X
substitution into 2 = 2 yields the equation
242
Section 11.3
ix
exp(ibln2) = 21 a. Since e = cosx + isinx, it follows that cos(bln2) = 21-a and sin(bln2) = 0, which yield a = 1 and b(ln2) = 2nn or b = 2nn/ln2, n = 1, 2,... . Thus the only
eigenvalues of the problem are Xn = 1 + i(2nn/ln2), n = 1, 2,... .
25b. For X < 0, there are no eigenfuctions. For X > 0 the general solution of the D.E. is
y = c1 + c2x + c3sinV"X x + c4co^V"Xx. The B.C. require that
c1+c4 = 0, c4 = 0, c1 + c2L + c3si^X L + c4co^X L = 0,
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