# Elementary Differential Equations and Boundary Value Problems - Boyce W.E.

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It is noteworthy that a Fourier series may converge to a sum that is not differentiable, or even continuous, in spite of the fact that each term in the series (4) is continuous, and even differentiable infinitely many times. The example below is an illustration of this, as is Example 2 in Section 10.2.

EXAMPLE

1

Let

f (x) =

0, L < x < 0,

L, 0 < x < L,

(5)

and let f be defined outside this interval so that f (x + 2L) = f (x) for all x. We will temporarily leave open the definition of f at the points x = 0, ฑL, except that its value must be finite. Find the Fourier series for this function and determine where it converges.

y

L

1 1 1 III ,

-3L -2L - L L 2L 3L x

FIGURE 10.3.2 Square wave.

The equation y = f (x) has the graph shown in Figure 10.3.2, extended to infinity in both directions. It can be thought of as representing a square wave. The interval [L, L] can be partitioned to give the two open subintervals (-L, 0) and (0, L). In (0, L), f (x) = L and f(x) = 0. Clearly, both f and f' are continuous and furthermore have limits as x ^ 0 from the right and as x ^ L from the left. The situation in (-L, 0) is similar. Consequently, both f and f are piecewise continuous on [L, L), so f satisfies the conditions of Theorem 10.3.1. If the coefficients am and bm are computed from Eqs. (2) and (3), the convergence of the resulting Fourier series to f (x) is assured at all points where f is continuous. Note that the values of am and bm are the same regardless of the definition of f at its points of discontinuity. This is true because the value of an integral is unaffected by changing the value of the integrand at a finite number of points. From Eq. (2)

1 fL fL

~ I f (x) dx = I dx = L; L J-L J0

-L

L

1 mnx

f (x) cos- dx =

-L

= 0, m = 0

L

J0

mn x cos dx

Similarly, from Eq. (3),

bm =

1 fL m nx

lJ f (x) sm dx =

Lsi

J0

m n x sin dx

-----(1 cos mn)

mn

0, m even;

2L /m n, m odd.

a0 =

am =

10.3 The Fourier Convergence Theorem

561

Hence

L 2L ( nx 1 3nx 1 5nx

f (x) = - + sin + - sin - + - sin - + ฆ

2 n\ L 3 L 5 L

L 2L sin(m nx / L)

2 n ". ๒

๒ = 1,3,5,...

L 2L sin(2n 1)nx/L _

= T + (6)

2 n f 2n 1

n = 1

At the points x = 0, ฑnL, where the function f in the example is not continuous, all terms in the series after the first vanish and the sum is L /2. This is the mean value of the limits from the right and left, as it should be. Thus we might as well define f

at these points to have the value L/2. If we choose to define it otherwise, the series

still gives the value L /2 at these points, since none of the preceding calculations is altered in any detail; it simply does not converge to the function at those points unless f is defined to have this value. This illustrates the possibility that the Fourier series corresponding to a function may not converge to it at points of discontinuity unless the function is suitably defined at such points.

The manner in which the partial sums

L 2L / nx 1 . (2n 1)nx \

sn(x ) = 2 + V (sin T + ฆฆฆ + ฏร-าsin----------------------L----j n = ' 2

of the Fourier series (6) converge to f (x) is indicated in Figure 10.3.3, where L has been chosen to be one and the graph of s8 (x) is plotted. The figure suggests that at points where f is continuous the partial sums do approach f (x) as n increases. However, in the neighborhood of points of discontinuity, such as x = 0 and x = L, the partial sums do not converge smoothly to the mean value. Instead they tend to overshoot the mark at each end of the jump, as though they cannot quite accommodate themselves to the sharp turn required at this point. This behavior is typical of Fourier series at points of discontinuity, and is known as the Gibbs4 phenomenon.

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