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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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y(-L) - y(L) = 0, y(-L) - y(L) = 0. (45)
This is not a Sturm-Liouville problem because the boundary conditions are not separated. The boundary conditions (45) are called periodic boundary conditions since they require that y and y assume the same values at x = L as at x =L. Nevertheless, it is straightforward to show that the problem (44), (45) is self-adjoint. A simple calculation establishes that k0 = 0 is an eigenvalue and that the corresponding eigenfunction is 0(x) = 1. Further, there are additional eigenvalues k1 = (n/L)2, k2 = (2/L )2,... ,kn = (nn/L )2,.... To each of these nonzero eigenvalues there correspond two linearly independent eigenfunctions; for example, corresponding to kn are the two eigenfunctions (x) = cos(nnx/L) and (x) = sin(nnx/L). This illustrates that the eigenvalues may not be simple when the boundary conditions are not separated. Further, if we seek to expand a given function f of period 2 L in a series of eigenfunctions of the problem (44), (45), we obtain the series
a0 / nn x nn x \
f (x) = IT + S \an cos_T + bn sin_r),
n= 1
which is just the Fourier series for f.
11.2 Sturm-Liouville Boundary Value Problems
639
PROBLEMS
We will not give further consideration to problems that have nonseparated boundary conditions, nor will we deal with problems of higher than second order, except in a few problems. There is, however, one other kind of generalization that we do wish to discuss. That is the case in which the coefficients p, q, and r in Eq. (1) do not quite satisfy the rather strict continuity and positivity requirements laid down at the beginning of this section. Such problems are called singular Sturm-Liouville problems, and are the subject of Section 11.4.
In each of Problems 1 through 5 determine the normalized eigenfunctions of the given problem.
1. + X7 = 0, 7(0) = 0, 0
)
2. + X7 = 0, 7(0) = 0, 7(1) = 0
3. + X7 = 0, 7(0) = 0, 7(1) = 0
4. + X7 = ^ 7(0) = 0, 7(1) + 7(1) = 0; see Section 11.1, Problem 8.
5. - 27 + (1 + X) 7 = 0, 7(0) = 0, 7(1) = 0; see Section 11.1, Problem 17
In each of Problems 6 through 9 find the eigenfunction expansion ^ an (x) of the given
n=1
function, using the normalized eigenfunctions of Problem 1.
6. f (x) = 1, 8. f(x) =
0 < x < 1
0 < x < 2 2 < x < i
7. f (x) = x, 0 < x < 1
9. f(x) =
2x,
1,
0 < x < 2 2 < x <1
In each of Problems 10 through 13 find the eigenfunction expansion ^ an (x) of the given
n=1 n n
function, using the normalized eigenfunctions of Problem 4.
10. f(x) = 1, 0 < x < 1
12. f (x) = 1 - x, 0 < x < 1
11.
13.
f(x) = 1x,
f(x) =
0 < x < 1
0 < x < 2 2 < x < 1
In each of Problems 14 through 18 determine whether the given boundary value problem is self-adjoint.
7' + + 2 y = 0, 7(0) = 0, 7(1) = 0
(1 + x2 )Z + 2x7 + 7 = 0, /(0) = 0, 7(1) + 2/(1) = 0
+ 7 = 7(0) - 7(1) = 0, 7(0) - 7(1) = 0
(1 + x2) + 2x7 + 7 = X(1 + x2)7, 7(0) - 7(1) = 0, 7(0) + 27(1) = 0
+ k7 = 0, 7(0) = 0, 7(n) + 7(n) = 0
14.
15.
16.
17.
18.
19. Show that if the functions u and v satisfy Eqs. (2), and either a2 = 0 or b2 = 0, or both, then
P(x)[U(x)v(x) - u(x)v'(x)]
= 0.
20.
21.
In this problem we outline a proof of the first part of Theorem 11.2.3: that the eigenvalues of the Sturm-Liouville problem (1), (2) are simple. For a given X suppose that 1 and 2 are two linearly independent eigenfunctions. Compute the Wronskian W(1, 2)(x) and use the boundary conditions (2) to showthat W (1, 2)(0) = 0. Then use Theorems 3.3.2 and 3.3.3 to conclude that 1 and 2 cannot be linearly independent as assumed.
Consider the Sturm-Liouville problem
- [P(x)/]' + q(x)7 = Xr(x)7, a17(0) + a2/(0) = 0, b17(1) + b2 7(1) = 0,
0
640
Chapter 11. Boundary Value Problems and Sturm-Liouville Theory
where p, q, and r satisfy the conditions stated in the text.
(a) Show that if k is an eigenvalue and a corresponding eigenfunction, then
f 1 f 1 b a
k 2 dx = (^'2 + q2) dx + -1 p(1^2(1)-------1 p(0)2(0),
0 0 b2 a2
provided that a2 = 0 and b2 = 0. How must this result be modified if a2 = 0 or b2 = 0?
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