# Elementary Differential Equations and Boundary Value Problems - Boyce W.E.

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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 ô (x) = ä/ 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-Ô „> = 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 ³—

f(x) = ----- / ------- V 2 sin[(2n-1)nx/2], 0 < x < 1,

Ï2 ^ (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 ô (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 ô for y in the D.E., multiplying both sides by ô, and integrating form 0 to 1 yields

Õ²1ãÔ 2dx = 11{-[p(x) ô']'ô + q(x^ 2}dx. Integrating the

00

first term on the right side once by parts, we obtain

1

0

Section 11.2

241

Õ²ãô 2dx = _p(1) ô'(1) ô (1) + p(0) ô'(0) ô (0) + Ð^ô' 2+qô2)dx.

00

If a2 Ô 0, b2 Ô 0, then ô'(1) = _Ü1ô(1)/Ü2 and

ô'(0) = _alô(0)/a2 and the result follows. If a2 = 0,

then ô(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 ôÛ = 1.

21c. If either q(x) Ô 0 or a1 Ô 0 or b1 Ô 0, there is at least one

positive term on the right and thus X must be positive.

23a. Using ôÛ = 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 ô(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 Ô 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 Ô 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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