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

ISBN 0-471-31999-6

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We note that f(x) is a straight line with a slope of one in any interval. Thus f(x) has the form x+b in any interval for the correct value of b. Since f(x+2) = f(x), we may set x = -1/2 to obtain f(3/2) = f(-1/2). Noting that 3/2 is

on the interval 1 < x < 2[f(3/2) = 3/2 + b] and that -1/2

is on the interval -1 < x < 0[f(-1/2) = -1/2 + 1], we

conclude that 3/2 + b = -1/2 + 1, or b = -1 for the

interval 1 < x < 2. For the interval 8 < x < 9 we have

f(x+8) = f(x+6) = ... = f(x) by successive applications of the periodicity condition. Thus for x = 1/2 we have f(17/2) = f(1/2) or 17/2 + b = 1/2 so b = -8 on the interval 8 < x < 9.

1,

By

Section 10.2

205

In Problems 13 through 18 it is often necessary to use integration by parts to evaluate the coefficients, although all the details will not be shown here.

13a. The function represents a sawtooth wave. It is periodic with period 2L.

13b. The Fourier series is of the form

f(x) = a0/2 + (amcosmnx/L + bmsinmnx/L ), where the

m=1

coefficients are computed from Eqs. (13) and (14). Substituting for f(x) in these equations yields

a0 = (l/L)IL (-x)dx = 0 and am = (l/L)IL (-x)cos(mnx/L)dx = 0, J-L J-L

m = 1,2... (these can be shown by direct integration, or using the fact that Jag(x)dx = 0 when g(x) is an odd

function). Finally, bm = (l/L)JL (-x)sin(mnx/L)dx

-L

L

(x/mn)cos(mnx/L)

- (l/mn)I L cos(mnx/L)dx

-L J-L

L

= 2L(-1) m/mn

= (2Lcosmn)/mn - (L/m2n 2)sin(mnx/L)

-L

Substituting these terms in the above Fourier series for f(x) yields the desired answer.

15a. See the next page.

15b. In this case f(x) is periodic of period 2n and thus

L = n in Eqs. (9), (13,) and (14). The constant a0 is

found to be a0 = (1/n)I°xdx = -n/2 since f(x) is zero on

J-n

the interval [0,n]. Likewise

an = (1/n)I°xcosnxdx = [1 - (-1)n]/n2n, using integration

J-n

by parts and recalling that cosnn = (-1)n. Thus an = 0 for n even and an = 2/n2n for n odd, which may be written as a2n-1 = 2/(2n-1) 2n since 2n-1 is always an odd number.

In a similar fashion bn = (1/n) 10xsinnxdx = (-1) n+1/n and

thus the desired solution is obtained. Notice that in this case both cosine and sine terms appear in the Fourier series for the given f(x).

206

Section 10.2

15a.

21a.

21b. a

1 f2—dx = — 212 2 12

n = 1 f 2-

12 nnx

4 a0 2

—, so — = — and

3 2 3

2x

-cos dx

2 -2 2 2

2

1 2x nnx 8x nnx

— [ sin + cos---------------------------

22 n2n 2

16 nnx

sin ]

n3n 3

2

-2

4 nn 2 n2n2 2

= (8/n2n2)cos(nn) = (-1) n8/n2n2 where the second line for an is found by integration by parts or a computer algebra system. Similarly,

bn = 1 -

2-

2 x2

sin— 2 -2 2 2

nnx 2 nnx

dx = 0, since x2sin is an odd

28

function. Thus f(x) = — + —

2

(-1) n nnx

--------cos--------

n=1

n

2 8 m ( -1 ) n

21c. As in Eq. (27), we have sm(x) = — + —^^ -----------------------------2—

nnx

n=1

2

3

x

2

3

2

cos

2

21d. Observing the graphs we see that the Fourier series

converges quite rapidly, except, at x = -2 and x = 2, where there is a sharp "point" in the periodic function.

25.

Section 10.3

207

27a. First we have Ja+Tg(x)dx = Jag(s)ds by letting x = s + T

in the left integral. Now, if 0 < a < T, then from elementary calculus we know that

J a+Tg(x)dx = J Tg(x)dx + J a+Tg(x)dx = J Tg(x)dx + J ag(x)dx Ja Ja JT Ja J0

using the equality derived above. This last sum is

I Tg(x)dx and thus we have the desired result.

0

Section 10.3, Page 562

2a. Substituting for f(x) in Eqs.(2) and (3) with L = n

yields a0 = (1/n) i^xdx = n/2;

0

am = (1/nn ^xcosmxdx = (cosmn - 1)/nm2 = 0 for m even and

0

= -2/nm2 for m odd; and

bm = (1/n) Pxsinmxdx = -(ncosmn)/mn = (-1) m+1/m,

0

m = 1,2... . Substituting these values into Eq.(1) with

L = n yields the desired solution.

2b. The function to which the series converges is indicated in the figure and is periodic with period 2n. Note that

h-

-371

the Fourier series converges to n/2 at x = -n, n, etc., even though the function is defined to be zero there. This value (n/2) represents the mean value of the left and right hand limits at those points. In (-n, 0), f(x) = 0 and f'(x) = 0 so both f and f' are continuous and have finite limits as x ^ -n from the right and as x ^ 0 from the left. In (0, n), f(x) = x and f'(x) = 1 and again both f and f' are continuous and have limits as x ^ 0 from the right and as x ^ n from the left. Since f and f' are piecewise continuous on [—n, n] the conditions of the Fourier theorem are satisfied.

4a. Substituting for f(x) in Eqs.(2) and (3), with L = 1 yields a0 = J1 (1-x2)dx = 4/3;

208

Section 10.3

4b.

7a.

7b.

7c.

12a.

= (1-x2)cosnnxdx = (2/nn)| 1xsinnnxdx

2 2 1 f1

= (-2/n2n2)[xcosnnx - cosnnxdx]

-1 J-1

= 4(-1) n+1/n2n2 ; and bn = l-1! (1-x2)sinnnxdx = 0. Substituting these values into Eq.(1) gives the desired series.

The function to which the series converges is shown in the figure and is periodic of fundamental period 2. In [-1,1] f(x) and f'(x) = -2x are both continuous and have finite limits as the endpoints of the interval are approached from within the interval.

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