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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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For m > 0, Eq. (13) yields
f0 1 r 2
1 ₯ m nx 1 fz mn x
am =t (—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
Qrn 2
2 . m n x
-------x sin-------
m n 2
2 \2 mnx
— cos _T“
m n ) 2
2
2
2 m n x ( 2
----x sin —— + ----------- I cos ,
m n 2 \ m n J 2
2
2
2
— I --- J + I --- J cos mn + I ---- I cos mn — I
\mn / \mn / \mο/ \mn
4
m = 1, 2,...
— -r(cos mn — 1),
(m ζ)2
_\ — 8/(mn]2, m odd,
0, m even.
Finally, from Eq. (14) it follows in a similar way that
bm = 0, m = 1, 2,.
(18)
(19)
By substituting the coefficients from Eqs. (17), (18), and (19) in the series (16) we obtain the Fourier series for f:
f (x) = 1 —
1
8
8
cos
3n x
2 32 2
“ cos(mnx/2)
5n x
+
?
m
2
1
m = 1,3,5,...
cos(2n — 1)n x/2
2
n1
(2n - 1)
2
(20)
0
2
0
2
2
Let
f(x) =
0, —3 < x < —1,
1, —1 < x < 1,
0, 1 < x < 3
(21)
and suppose that f (x + 6) = f (x); see Figure 10.2.3. Find the coefficients in the Fourier series for f.
10.2 Fourier Series
553
EXAMPLE
3
I
y
1
1 1 1 1 1 1 1 1 ,
-1 1 3 5 7 t
3
-
5
-
7
-
FIGURE 10.2.3 Graph of f (x) in Example 2.
Since f has period 6, it follows that L = 3 in this problem. Consequently, the Fourier series for f has the form
Li A
f (x) = f + ^ V
n=1
TO
?(¦
nnx nnX
a cos —--------+ h sin

(22)
3 n 3
where the coefficients an and bn are given by Eqs. (13) and (14) with L = 3. We have
1 f3 1 f1
η³.3 f (X) dX = η³.1
dx = -. =3
Similarly,
nnx 1 . nnx
cos-------dx = — sin-------------
3 nn 3
2 . nn
= — sin—, n = 1, 2,...
-1
and
b
nn x 1 nnx
sin--------dx =-----------cos---------
3 nn 3
nn
= 0,
3
n = 1, 2,...
(23)
(24)
(25)
1
Thus the Fourier series for f is ³ 00 >¦>
1 2 nn nnx
f (x) = - + > — sin — cos ——
3 nn 3 3
n1
cos(2n x/3) cos(4n x/3) cos(5n x/3)
cos (nx/3) -+-------------------------------—-—- -—- -----+
(26)
ao =
1
an =
1
Consider again the function in Example 1 and its Fourier series (20). Investigate the speed with which the series converges. In particular, determine how many terms are needed so that the error is no greater than 0.01 for all x.
The mth partial sum in this series,
, 4 , 8 m cos(2n — 1)nx/2 ^
s (x) = 1---------r > ------------------φ-----, (27)
m n2 n=1 (2n — 1)2
can be used to approximate the function f. The coefficients diminish as (2n — 1)—2, so the series converges fairly rapidly. This is borne out by Figure 10.2.4, where the partial sums for m = 1 and m = 2 are plotted. To investigate the convergence in more detail
554
Chapter 10. Partial Differential Equations and Fourier Series
we can consider the error em (x) = f (x) — sm (x). Figure 10.2.5 shows a plot of |e6(x) | versus x for 0 < x < 2. Observe that |e6(x)| is greatest at the points x = 0 and x = 2 where the graph of f (x) has corners. It is more difficult for the series to approximate the function near these points, resulting in a larger error there for a given n. Similar graphs are obtained for other values of m .
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