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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
Download (direct link): elementarydifferentialequat2001.pdf
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2 n \ m — n m + n J
-L
= 0,
550
Chapter 10. Partial Differential Equations and Fourier Series
as long as m + n and m — n are not zero. Since m and n are positive, m + n = 0. On the other hand, if m — n = 0, then m = n, and the integral must be evaluated in a different way. In this case
"L r L
/L . mnx . nnx fL / . mnx\2
sin sin dx = I I sin I dx
L L L -L L
=1L
1 f sin(2m n x / L)
- \x -
dx
L
-L
2 \ 2m n/L
= L .
This establishes Eq. (8); Eqs. (6) and (7) can be verified by similar computations.
The Euler-Fourier Formulas. Now let us suppose that a series of the form (1) converges, and let us call its sum f (x):
a0 / mnx mnx \
f (x) = f + E \m cos — + bm sin — ) • (9)
m = 1
The coefficients am and bm can be related to f (x) as a consequence of the orthogonality conditions (6), (7), and (8). First multiply Eq. (9) by cos(nn x / L), where n is a fixed positive integer (n > 0), and integrate with respect to x from — L to L. Assuming that the integration can be legitimately carried out term by term,2 we obtain
"L „ rL ^ r L
/L nnx a0 fL nnx fL mnx nnx
f (x) cos —-— dx = — I cos —-— dx + > al cos —-— cos —— dx
?L L 2 J—L L m=1 J—L L L
to f L
bm -LL si
m = 1 J L
L m n x nn x
+> .bml l sin ——cos ^ dx. (10)
L
Keeping in mind that n is fixed whereas m ranges over the positive integers, it follows from the orthogonality relations (6) and (7) that the only nonzero term on the right side of Eq. (10) is the one for which m = n in the first summation. Hence
rL nn x
f (x) cos dx = Lan, n = 1, 2,.... (11)
L L
To determine a0 we can integrate Eq. (9) from -L to L , obtaining
fL a0 fL -TO [L mnx -TO fL mnx
f (x) dx = — dx + > al cos - dx + > bl sin —— dx J—L 2 J—L m=1 J—L L m = 1 J—L L
= La0, (12)
since each integral involving a trigonometric function is zero. Thus
i/
f (x) cos dx, n = 0, 1, 2,.... (13)
2This is a nontrivial assumption, since not all convergent series with variable terms can be so integrated. For the
special case of Fourier series, however, term-by-term integration can always be justified.
an =
L
10.2 Fourier Series
551
By writing the constant term in Eq. (9) as a0/2, it is possible to compute all the an from Eq. (13). Otherwise, a separate formula would have to be used for a0.
A similar expression for bn may be obtained by multiplying Eq. (9) by sin(nnx/L), integrating termwise from —L to L, and using the orthogonality relations (7) and (8); thus
1 fL nn x
bn = — I f (x) sin dx, n = 1, 2, 3,.... (14)
L J—L L
Equations (13) and (14) are known as the Euler-Fourier formulas for the coefficients in a Fourier series. Hence, if the series (9) converges to f (x), and if the series can be integrated term by term, then the coefficients must be given by Eqs. (13) and (14).
Note that Eqs. (13) and (14) are explicit formulas for an and bn in terms of f, and that the determination of any particular coefficient is independent of all the other coefficients. Of course, the difficulty in evaluating the integrals in Eqs. (13) and (14) depends very much on the particular function f involved.
Note also that the formulas (13) and (14) depend only on the values of f (x) in the interval — L < x < L. Since each of the terms in the Fourier series (9) is periodic with period 2L, the series converges for all x whenever it converges in — L < x < L, and its sum is also a periodic function with period 2L. Hence f (x) is determined for all x by its values in the interval —L < x < L.
It is possible to show (see Problem 27) that if g is periodic with period T, then every integral of g over an interval of length T has the same value. If we apply this result to the Euler-Fourier formulas (13) and (14), it follows that the interval of integration, —L < x < L, can be replaced, if it is more convenient to do so, by any other interval of length 2L.
Assume that there is a Fourier series converging to the function f defined by
[—x, —2 < x < 0,
f(x) =• x, 0 < x < 2;
(15)
f (x + 4) = f (x ).
Determine the coefficients in this Fourier series.
This function represents a triangular wave (see Figure 10.2.2) and is periodic with period 4. Thus in this case L = 2 and the Fourier series has the form
w N ao ^ / m n x . mn x \
f (x ) = ~2 +^ \m cos^T + bm sin^T) ’ (16)
m=1
where the coefficients are computed from Eqs. (13) and (14) with L = 2. Substituting for f (x) in Eq. (13) with m = 0, we have
1 f0 1 f2
a0 = — I (-x) dx +— I x dx = 1 + 1 = 2. (17)
2 J-2 2 J0
For m > 0, Eq. (13) yields
f0 1 r 2
1 f m nx 1 f2 mn x
am =^ (-x) cos—-— dx + - I x cos—-— dx.
2 J-2 2 2 J0 2
552
Chapter 10. Partial Differential Equations and Fourier Series
EXAMPLE
2
These integrals can be evaluated through integration by parts, with the result that
1
am 2
2 . m n x
x sin-------------
m n 2
2 \2 mnx
— cos _T-
m n I 2
-2
2
m n x
2 m n x / 2
x sin —— + -------------- I cos ,
m n 2 \ m n ) 2
2
2
2
— |--- J + | -- J cos mn ^1 ---- I cos mn — I
\m^/ \m^/ \mn/ \mn
4
m = 1, 2,...
— -r(cos mn — 1),
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