Download (direct link):
that c = 0, 2c + 2(ln2)c = 0 so c = c = 0 and X = 1 is
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
substitution into 2 = 2 yields the equation
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"X x. The B.C. require that
c1+c4 = 0, c4 = 0, c1 + c2L + c3si^X L + c4co^X L = 0, and
c2 + a/X c3cosa/XL - a/Xc4si^VXL = 0. Thus c1 = c4 = 0 and for nontrivial solutions to exist X must satisfy the equation a/XLcosa/Xl - sin a/X L = 0. In this case c2 = (-a/X cosa/X L)(c3) and it follows that the eigenfunction ^1 is given by ^1(x) = sin(
x) - VX1 xcos^X1 L where X1 is the smallest positive solution of the equation a/X L = tanA/X l. A graphical or numerical estimate of X1
reveals that X1 = (4.4934)2/L .
25c. Assuming X ^ 0 the eigenvalue equation is
2(1-cosx) = xsinx, where x = a/X L. Graphing
f(x) = 2(1-cosx) and g(x) = xsinx we see that there is an
intersection for 6 < x < 7. Since both f(x) and g(x) are
zero for x = 2n, this then is the precise root and thus 22
X = (2n) /L . In addition, it appears there might be an
intersection for 0 < x < 1. Using a Taylor series representation for f(x) and g(x) about x = 0, however, shows there is no intersection for 0 < x < 1. Of course x = 0 is also an intersection, which yields X = 0, which gives the trivial solution and hence X = 0 is not an eigenvalue.
Section 11.3, Page 651
1. We must first find the eigenvalues and normalized
eigenfunctions of the associated homogeneous problem y" + Xy = 0, y(0) = 0, y(1) = 0. This problem has the solutions ? (x) = k sinnnx, for X = n2n2, n = 1,2,... .
T n n n
Choosing k so that 2 2dx = 1 we find k = 1/2. Hence n Jo n n v
the solution of the original nonhomogeneous problem is given by y = b^6n(x), where the coefficients b^ are
found from Eq.(12), b = c /(A -2) where c is given by
n n n n
c„ = ^ JO1 xsinnnxdx (Eq.9). [Note that the original
problem can be written as -y" = 2y + x and therefore comparison with Eq.(1) yields r(x) = 1 and f(x) = x]. Integrating the expression for cn by parts yields cn =
? I 2 (-1)n+1 __
V2 (-1)n+1/nn and thus y = X———-------- "\Z~2sinnrcx.
2. From Problem 1 of Section 11.2 we have
6 = -\/2sin[(2n-1) nx/2] for A = (2n-1) 2n 2/4 and from
Problem 7 of that section we have
cn = 4^2 (-1)n+1/(2n-1)2n2. Substituting these values into b = c / (A -2) and y = X b 6 yields the desired
n n n nT n
3. Referring to Problem 3 of Section 11.2 we have
y = b + / b(y2 cosnnx), where b = c / (A -2) for
O n ’ n n n
n = 0,1,2,... . The rest of the calculations follow those of Problem 1.
5. Note that the associated eigenvalue problem is the same as for Problem 1 and that |1-2x| = 1-2x for 0 < x < 1/2 while |1-2x| = 2x-1 for 1/2 < x < 1.
8. Writing the D.E. in the form Eq.(1), we have -y" = p.y + f(x),
so r(x) = 1. The associated eigenvalue problem is y" + Ay = 0, y'(0) = 0, y'(1) = 0 which has the eigenvalues An = n2n2, n = 0,1,2... and the normalized eigenfunctions 60 = 1, 6n(x) = V"2cosnnx, n = 1,2..., as found in Problem 3,
Section 11.2. Now, we assume y(x) = b0 + y bn6n(x) = X
bn6 n(x), Eq.(5), and thus bn = --------------, where cn = f(x)6 n(x)dx,
n = 0,1,2..., Eq.(9). Thus
cn ,— ^ cncosnnx
y(x) = I -6 n(x) = -co/M- + V 2 I -----------, where we
Xn-M n=1 ^n-M
have assumed M ^ Xn, n = 0,1,2... .
Since M = n2 is an eigenvalue of the corresponding homogeneous equation, Theorem 11.3.1 tells us that a solution will exist only if -(a+x) is orthogonal to the corresponding eigenfunction '\/~2sinTCx. Thus we require
J1(a+x)sinnxdx = 0, which yields a = -1/2. With a = -1/2,
we find that Y = (x-1/2)/n2 and yc = csinnx + dcosnx by
methods of Chapter 3. Setting y = yc + Y and choosing d
to satisfy the B.C. we obtain the desired family of solutions.
Note that in this case M = 4n2 and 62 = \pl sin2nx are the
eigenvalue and eigenfunction respectively of the corresponding homogeneous equation. However, there is no value of a for which -\J~2 J1(a+x)sin2nxdx = 0, and thus
there is no solution.
In this case a solution will exist only if -a is
orthogonal to i/~2cosrcx., that is if 11acosnxdx = 0.
Since this condition is valid for all a, a family of solutions exists.