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# Elementary differential equations 7th edition - Boyce W.E

Boyce W.E Elementary differential equations 7th edition - Wiley publishing , 2001. - 1310 p.
ISBN 0-471-31999-6
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13. Find the solution of the initial value problem
y" + a2y = sin nt, y(0) = 0, y (0) = 0,
where n is a positive integer and a2 = n2. What happens if a2 = n2?
14. Find the formal solution of the initial value problem
TO
y" + a2y = "Y2 bn sin nt, y(0) = 0, y'(0) = 0,
n=1
where a > 0 is not equal to a positive integer. How is the solution altered if a = m, where m is a positive integer?
15. Find the formal solution of the initial value problem
y" + a2 y = f (t), y(0) = 0, y'(0) = 0,
where f is periodic with period 2n and
f (t) =
1, 0 < t < n;
0, t = 0, n, In;
 1, n < t < In.
See Problem 1.
16. Find the formal solution of the initial value problem
y" + a>2y = f (t), y(0) = 1, y'(0) = 0,
where f is periodic with period 2 and
f (t) = ) 1  t  0 - t < 1;
f () { 1 + t, 1 - t < 2.
See Problem 8.
17. Assuming that
TO
a / nnx nnx\
f (x) = -j + X, (ancos ~l~ + bn sm~) > (i)
n=1
show formally that
1 f L a2
-I [f(x)]2dx = y+?a+bj).
L JL 2 n=1
564
Chapter 10. Partial Differential Equations and Fourier Series
This relation between a function f and its Fourier coefficients is known as Parsevals (1755-1836) equation. Parsevals equation is very important in the theory of Fourier series and is discussed further in Section 11.6.
Hint: Multiply Eq. (i)by f (x), integrate from  L to L, and use the Euler-Fourier formulas.
18. This problem indicates a proof of convergence of a Fourier series under conditions more restrictive than those in Theorem 10.3.1.
(a) If f and f' are piecewise continuous on  L < x < L, and if f is periodic with period 2L, show that na and nb are bounded as n -> oo.
5 n n
Hint: Use integration by parts.
(b) If f is continuous on  L < x < L and periodic with period 2L, and if f' and f" are piecewise continuous on  L < x < L, show that n2an and n2bn are bounded as n ^ <x. Use this fact to show that the Fourier series for f converges at each point in  L < x < L. Why must f be continuous on the closed interval?
Hint: Again, use integration by parts.
Acceleration of Convergence. In the next problem we show how it is sometimes possible to improve the speed of convergence of a Fourier, or other infinite, series.
19. Suppose that we wish to calculate values of the function g, where
^ (2n  1)
g(x) = > -------------------;r sin(2n  1)nx. (i)
1 + (2n  1)2
It is possible to show that this series converges, albeit rather slowly. However, observe that for large n the terms in the series (i) are approximately equal to [sin(2n  1)nx]/(2n  1) and that the latter terms are similar to those in the example in the text, Eq. (6).
(a) Show that
TO
[sin(2n  1)nx]/(2n  1) = (n/2)[f (x)  i], (ii)
n=1
where f is the square wave in the example with L = 1.
(b) Subtract Eq. (ii) from Eq. (i) and show that
n , sin(2n  1)n x
g(x) = -[f (x)  2]  ? ------------------------------±-2-----------------------------. (iii)
2 n=1 (2n  1)[1 + (2n  1) ]
The series (iii) converges much faster than the series (i) and thus provides a better way to calculate values of g(x).
10.4 Even and Odd Functions
Before looking at further examples of Fourier series it is useful to distinguish two classes of functions for which the Euler-Fourier formulas can be simplified. These are even and odd functions, which are characterized geometrically by the property of symmetry with respect to the y-axis and the origin, respectively (see Figure 10.4.1).
Analytically, f is an even function if its domain contains the point x whenever it contains the point x, and if
f (x) = f (x )
(1)
10.4 Even and Odd Functions
565
FIGURE 10.4.1 (a) An even function. (b) An odd function.
for each x in the domain of f. Similarly, f is an odd function if its domain contains x whenever it contains x, and if
f (-x) = -f (x )
(2)
for each x in the domain of f. Examples of even functions are 1, x , cos nx, |x|, and x2n. The functions x, x3, sin nx, and x2n+1 are examples of odd functions. Note that according to Eq. (2), f (0) must be zero if f is an odd function whose domain contains the origin. Most functions are neither even nor odd, for instance, ex. Only one function, f identically zero, is both even and odd.
Elementary properties of even and odd functions include the following:
1. The sum (difference) and product (quotient) of two even functions are even.
2. The sum (difference) of two odd functions is odd; the product (quotient) of two odd functions is even.
3. The sum (difference) of an odd function and an even function is neither even nor odd; the product (quotient) of two such functions is odd.5
The proofs of all these assertions are simple and follow directly from the definitions. For example, if both f1 and f2 are odd, and if g(x) = f1(x) + f2(x), then
g(-x) = f 1 ( x) + f2( x) = -f1(x) - f2(x)
= -[f1(x) + f2(x)] = -g(x), (3)
so f1 + f2 is an odd function also. Similarly, if h(x) = f1(x) f2(x), then
h(-x) = f1(-x )f2(-x) = [- f1(x)][- f2(x)] = f1(x )f2(x) = h(x), (4)
so that f1 f2 is even.
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