KREYSZIG E Advanced Engineering Mathematics (10th ed., Wiley 2011) Chapter 11 - Fourier analysis

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1 KREYSZIG E Advanced Engineering Mathematics (th ed., Wiley ) Chapter - Fourier analysis

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3 CHAPTER Fourier Analysis 474 This chapter on Fourier analysis covers three broad areas: Fourier series in Secs...4, more general orthonormal series called Sturm Liouville expansions in Secs..5 and.6 and Fourier integrals and transforms in Secs The central starting point of Fourier analysis is Fourier series. They are infinite series designed to represent general periodic functions in terms of simple ones, namely, cosines and sines. This trigonometric system is orthogonal, allowing the computation of the coefficients of the Fourier series by use of the well-known Euler formulas, as shown in Sec... Fourier series are very important to the engineer and physicist because they allow the solution of ODEs in connection with forced oscillations (Sec..3) and the approximation of periodic functions (Sec..4). Moreover, applications of Fourier analysis to PDEs are given in Chap.. Fourier series are, in a certain sense, more universal than the familiar Taylor series in calculus because many discontinuous periodic functions that come up in applications can be developed in Fourier series but do not have Taylor series expansions. The underlying idea of the Fourier series can be extended in two important ways. We can replace the trigonometric system by other families of orthogonal functions, e.g., Bessel functions and obtain the Sturm Liouville expansions. Note that related Secs..5 and.6 used to be part of Chap. 5 but, for greater readability and logical coherence, are now part of Chap.. The second expansion is applying Fourier series to nonperiodic phenomena and obtaining Fourier integrals and Fourier transforms. Both extensions have important applications to solving PDEs as will be shown in Chap.. In a digital age, the discrete Fourier transform plays an important role. Signals, such as voice or music, are sampled and analyzed for frequencies. An important algorithm, in this context, is the fast Fourier transform. This is discussed in Sec..9. Note that the two extensions of Fourier series are independent of each other and may be studied in the order suggested in this chapter or by studying Fourier integrals and transforms first and then Sturm Liouville expansions.. Fourier Series Prerequisite: Elementary integral calculus (needed for Fourier coefficients). Sections that may be omitted in a shorter course:.4.9. References and Answers to Problems: App. Part C, App.. Fourier series are infinite series that represent periodic functions in terms of cosines and sines. As such, Fourier series are of greatest importance to the engineer and applied mathematician. To define Fourier series, we first need some background material. A function f (x) is called a periodic function if f ( x) is defined for all real x, except

4 SEC.. Fourier Series 475 f(x) x p Fig. 58. Periodic function of period p possibly at some points, and if there is some positive number p, called a period of f (x), such that () f (x p) f (x) for all x. (The function f (x) tan x is a periodic function that is not defined for all real x but undefined for some points (more precisely, countably many points), that is x p>, 3p>, Á.) The graph of a periodic function has the characteristic that it can be obtained by periodic repetition of its graph in any interval of length p (Fig. 58). The smallest positive period is often called the fundamental period. (See Probs. 4.) Familiar periodic functions are the cosine, sine, tangent, and cotangent. Examples of functions that are not periodic are x, x, x 3, e x, cosh x, and ln x, to mention just a few. If f (x) has period p, it also has the period p because () implies f (x p) f ([x p] p) f (x p) f (x), etc.; thus for any integer n,, 3, Á, () f (x np) f (x) for all x. Furthermore if f (x) and g (x) have period p, then af (x) bg (x) with any constants a and b also has the period p. Our problem in the first few sections of this chapter will be the representation of various functions f (x) of period p in terms of the simple functions (3), cos x, sin x, cos x, sin x, Á, cos nx, sin nx, Á. All these functions have the period p. They form the so-called trigonometric system. Figure 59 shows the first few of them (except for the constant, which is periodic with any period). π ππ cos x π ππ cos x π ππ cos 3x π ππ π ππ π ππ sin x sin x sin 3x Fig. 59. Cosine and sine functions having the period p (the first few members of the trigonometric system (3), except for the constant )

5 476 CHAP. Fourier Analysis (4) The series to be obtained will be a trigonometric series, that is, a series of the form a a cos x b sin x a cos x b sin x Á a a n (a n cos nx b n sin nx). a, a, b, a, b, Á are constants, called the coefficients of the series. We see that each term has the period p. Hence if the coefficients are such that the series converges, its sum will be a function of period p. Expressions such as (4) will occur frequently in Fourier analysis. To compare the expression on the right with that on the left, simply write the terms in the summation. Convergence of one side implies convergence of the other and the sums will be the same. Now suppose that f (x) is a given function of period p and is such that it can be represented by a series (4), that is, (4) converges and, moreover, has the sum f (x). Then, using the equality sign, we write (5) f (x) a a (a n cos nx b n sin nx) n and call (5) the Fourier series of f (x). We shall prove that in this case the coefficients of (5) are the so-called Fourier coefficients of f (x), given by the Euler formulas () (6) (a) a p p a n p p p p (b) b n,, Á n p p f (x) sin nx dx. p f (x) dx f (x) cos nx dx n,, Á The name Fourier series is sometimes also used in the exceptional case that (5) with coefficients (6) does not converge or does not have the sum f (x) this may happen but is merely of theoretical interest. (For Euler see footnote 4 in Sec..5.) A Basic Example Before we derive the Euler formulas (6), let us consider how (5) and (6) are applied in this important basic example. Be fully alert, as the way we approach and solve this example will be the technique you will use for other functions. Note that the integration is a little bit different from what you are familiar with in calculus because of the n. Do not just routinely use your software but try to get a good understanding and make observations: How are continuous functions (cosines and sines) able to represent a given discontinuous function? How does the quality of the approximation increase if you take more and more terms of the series? Why are the approximating functions, called the

6 ` ` SEC.. Fourier Series 477 partial sums of the series, in this example always zero at and p? Why is the factor >n (obtained in the integration) important? EXAMPLE Periodic Rectangular Wave (Fig. 6) Find the Fourier coefficients of the periodic function f (x) in Fig. 6. The formula is (7) k if p x f (x) b k if x p and f (x p) f (x). Functions of this kind occur as external forces acting on mechanical systems, electromotive forces in electric circuits, etc. (The value of f (x) at a single point does not affect the integral; hence we can leave f (x) undefined at x and x p.) Solution. From (6.) we obtain a. This can also be seen without integration, since the area under the curve of f (x) between p and p (taken with a minus sign where is negative) is zero. From (6a) we obtain the coefficients a, a, Á f (x) of the cosine terms. Since f ( x) is given by two expressions, the integrals from p to p split into two integrals: a n p p f (x) cos nx dx p c p (k) cos nx dx k cos nx dx d p p p c k sin nx n p k sin nx n p ` d because sin nx at p,, and p for all n,, Á. We see that all these cosine coefficients are zero. That is, the Fourier series of (7) has no cosine terms, just sine terms, it is a Fourier sine series with coefficients b, b, Á obtained from (6b); b n p p f (x) sin nx dx p c p (k) sin nx dx k sin nx dx d p p p c k cos nx n p k cos nx n p ` d. Since cos (a) cos a and cos, this yields b n k k [cos cos (np) cos np cos ] ( cos np). np np Now, cos p, cos p, cos 3p, etc.; in general, for odd n, cos np b for even n, and thus for odd n, cos np b for even n. Hence the Fourier coefficients b n of our function are b 4k. p, b, b 3 4k 3p, b 4, b 5 4k 5p, Á Fig. 6. Given function f (x) (Periodic reactangular wave)

7 478 CHAP. Fourier Analysis Since the are zero, the Fourier series of f (x) is a n (8) 4k p (sin x 3 sin 3x 5 sin 5x Á ). The partial sums are S 4k sin x, p S 4k p asin x sin 3xb. 3 etc. Their graphs in Fig. 6 seem to indicate that the series is convergent and has the sum f (x), the given function. We notice that at x and x p, the points of discontinuity of f (x), all partial sums have the value zero, the arithmetic mean of the limits k and k of our function, at these points. This is typical. Furthermore, assuming that f (x) is the sum of the series and setting x p>, we have Thus f a p 4k b k p a 3 5 Á b Á p 4. This is a famous result obtained by Leibniz in 673 from geometric considerations. It illustrates that the values of various series with constant terms can be obtained by evaluating Fourier series at specific points. Fig. 6. First three partial sums of the corresponding Fourier series

8 SEC.. Fourier Series 479 Derivation of the Euler Formulas (6) The key to the Euler formulas (6) is the orthogonality of (3), a concept of basic importance, as follows. Here we generalize the concept of inner product (Sec. 9.3) to functions. THEOREM Orthogonality of the Trigonometric System (3) The trigonometric system (3) is orthogonal on the interval p x p (hence also on x p or any other interval of length p because of periodicity); that is, the integral of the product of any two functions in (3) over that interval is, so that for any integers n and m, (a) (9) (b) p cos nx cos mx dx (n m) p p sin nx sin mx dx (n m) p (c) p sin nx cos mx dx p (n m or n m). PROOF This follows simply by transforming the integrands trigonometrically from products into sums. In (9a) and (9b), by () in App. A3., Since m n (integer!), the integrals on the right are all. Similarly, in (9c), for all integer m and n (without exception; do you see why?) p p p p p p cos nx cos mx dx p cos (n m)x dx p sin nx sin mx dx p cos (n m)x dx p sin nx cos mx dx p sin (n m)x dx p Application of Theorem to the Fourier Series (5) We prove (6.). Integrating on both sides of (5) from p p to, we get We now assume that termwise integration is allowed. (We shall say in the proof of Theorem when this is true.) Then we obtain p p p p f (x) dx p c a a (a n cos nx b n sin nx) d dx. p n f (x) dx a p dx a p p p p n aa n p cos nx dx b n p sin nx dxb. p p p p cos (n m)x dx cos (n m)x dx. sin (n m)x dx. p

9 48 CHAP. Fourier Analysis The first term on the right equals pa. Integration shows that all the other integrals are. Hence division by p gives (6.). We prove (6a). Multiplying (5) on both sides by cos mx with any fixed positive integer m and integrating from p to p, we have () We now integrate term by term. Then on the right we obtain an integral of a cos mx, which is ; an integral of a n cos nx cos mx, which is a m p for n m and for n m by (9a); and an integral of b n sin nx cos mx, which is for all n and m by (9c). Hence the right side of () equals a m p. Division by p gives (6a) (with m instead of n). We finally prove (6b). Multiplying (5) on both sides by sin mx with any fixed positive integer m and integrating from p to p, we get () p p f (x) cos mx dx p c a a (a n cos nx b n sin nx) d cos mx dx. p n p p f (x) sin mx dx p c a a (a n cos nx b n sin nx) d sin mx dx. p n Integrating term by term, we obtain on the right an integral of a sin mx, which is ; an integral of a n cos nx sin mx, which is by (9c); and an integral of b n sin nx sin mx, which is b m p if n m and if n m, by (9b). This implies (6b) (with n denoted by m). This completes the proof of the Euler formulas (6) for the Fourier coefficients. Convergence and Sum of a Fourier Series The class of functions that can be represented by Fourier series is surprisingly large and general. Sufficient conditions valid in most applications are as follows. f(x) THEOREM f( ) f( + ) Fig. 6. Left- and right-hand limits ƒ( ), ƒ( ) _ of the function f (x) b x if x x> if x x Representation by a Fourier Series Let f (x) be periodic with period p and piecewise continuous (see Sec. 6.) in the interval p x p. Furthermore, let f (x) have a left-hand derivative and a righthand derivative at each point of that interval. Then the Fourier series (5) of f (x) [with coefficients (6)] converges. Its sum is f (x), except at points x where f (x) is discontinuous. There the sum of the series is the average of the left- and right-hand limits of f (x) at. x The left-hand limit of f (x) at x is defined as the limit of f (x) as x approaches x from the left and is commonly denoted by f (x ). Thus ƒ(x ) lim h* ƒ(x h) as h * through positive values. The right-hand limit is denoted by ƒ(x ) and ƒ(x ) lim h* ƒ(x h) as h * through positive values. The left- and right-hand derivatives of ƒ(x) at x are defined as the limits of f (x h) f (x ) h f (x h) f (x ) and, h respectively, as h * through positive values. Of course if ƒ(x) is continuous at x, the last term in both numerators is simply ƒ(x ).

10 SEC.. Fourier Series 48 PROOF We prove convergence, but only for a continuous function f (x) having continuous first and second derivatives. And we do not prove that the sum of the series is f (x) because these proofs are much more advanced; see, for instance, Ref. 3C4 listed in App.. Integrating (6a) by parts, we obtain a n p p p f (x) cos nx dx The first term on the right is zero. Another integration by parts gives a n f r(x) cos nx n p f (x) sin nx np p p The first term on the right is zero because of the periodicity and continuity of f r(x). Since f s is continuous in the interval of integration, we have ƒ f s(x) ƒ M p p p f s(x) cos nx dx. n p p np p p f r(x) sin nx dx. for an appropriate constant M. Furthermore, ƒ cos nx ƒ. It follows that ƒ a n ƒ n p p f s(x) cos nx dx p M dx M n p n. p Similarly, ƒ b n ƒ M>n for all n. Hence the absolute value of each term of the Fourier series of f (x) is at most equal to the corresponding term of the series p ƒ a ƒ M a 3 3 Áb which is convergent. Hence that Fourier series converges and the proof is complete. (Readers already familiar with uniform convergence will see that, by the Weierstrass test in Sec. 5.5, under our present assumptions the Fourier series converges uniformly, and our derivation of (6) by integrating term by term is then justified by Theorem 3 of Sec. 5.5.) EXAMPLE Convergence at a Jump as Indicated in Theorem The rectangular wave in Example has a jump at x. Its left-hand limit there is k and its right-hand limit is k (Fig. 6). Hence the average of these limits is. The Fourier series (8) of the wave does indeed converge to this value when x because then all its terms are. Similarly for the other jumps. This is in agreement with Theorem. Summary. A Fourier series of a given function f (x) of period p is a series of the form (5) with coefficients given by the Euler formulas (6). Theorem gives conditions that are sufficient for this series to converge and at each x to have the value f (x), except at discontinuities of f (x), where the series equals the arithmetic mean of the left-hand and right-hand limits of f (x) at that point.

11 48 CHAP. Fourier Analysis PROBLEM SET. 5 PERIOD, FUNDAMENTAL PERIOD The fundamental period is the smallest positive period. Find it for. cos x, sin x, cos x, sin x, cos px, sin px, cos px, sin px. 3. If f (x) and g (x) have period p, show that h (x) af (x) bg(x) (a, b, constant) has the period p. Thus all functions of period p form a vector space. 4. Change of scale. If f (x) has period p, show that f (ax), a, and f (x>b), b, are periodic functions of x of periods p>a and bp, respectively. Give examples. 5. Show that f const is periodic with any period but has no fundamental period. 6 GRAPHS OF p PERIODIC FUNCTIONS Sketch or graph f (x) which for p x p is given as follows x if p x 9. f (x) b p x if x p cos x if p x. f (x) b cos x if x p. Calculus review. Review integration techniques for integrals as they are likely to arise from the Euler formulas, for instance, definite integrals of x cos nx, x sin nx, e x cos nx, etc. FOURIER SERIES Find the Fourier series of the given function f (x), which is assumed to have the period p. Show the details of your work. Sketch or graph the partial sums up to that including cos 5x and sin 5x.. f (x) in Prob f (x) in Prob cos nx, sin pnx k sin nx, f (x) ƒ x ƒ f (x) ƒ sin x ƒ, f (x) sin ƒ x ƒ f (x) e ƒ x ƒ, f (x) ƒ e x ƒ f (x) x (p x p) f (x) x ( x p) π px cos k, π π π sin px k, cos pnx, k π π π π π π π π π π π π π π π π π π. CAS EXPERIMENT. Graphing. Write a program for graphing partial sums of the following series. Guess from the graph what f (x) the series may represent. Confirm or disprove your guess by using the Euler formulas. (a) (sin x 3 sin 3x 5 sin 5x Á ) ( sin x 4 sin 4x 6 sin 6x Á ) (b) 4 p acos x cos 3x 9 5 cos 5x Á b (c) 3 p 4(cos x 4 cos x 9 cos 3x 6 cos 4x Á ) 3. Discontinuities. Verify the last statement in Theorem for the discontinuities of f (x) in Prob.. 4. CAS EXPERIMENT. Orthogonality. Integrate and graph the integral of the product cos mx cos nx (with various integer m and n of your choice) from a to a as a function of a and conclude orthogonality of cos mx

12 SEC.. Arbitrary Period. Even and Odd Functions. Half-Range Expansions 483 and cos nx (m n) for a p from the graph. For what if f is continuous but f r df>dx is discontinuous, >n 3 m and n will you get orthogonality for a p>, p>3, if f and f r are continuous but f s is discontinuous, etc. p>4? Other a? Extend the experiment to cos mx sin nx Try to verify this for examples. Try to prove it by and sin mx sin nx. integrating the Euler formulas by parts. What is the 5. CAS EXPERIMENT. Order of Fourier Coefficients. practical significance of this? The order seems to be >n if f is discontinous, and >n. Arbitrary Period. Even and Odd Functions. Half-Range Expansions We now expand our initial basic discussion of Fourier series. Orientation. This section concerns three topics:. Transition from period p to any period L, for the function f, simply by a transformation of scale on the x-axis.. Simplifications. Only cosine terms if f is even ( Fourier cosine series ). Only sine terms if f is odd ( Fourier sine series ). 3. Expansion of f given for x L in two Fourier series, one having only cosine terms and the other only sine terms ( half-range expansions ). p. From Period to Any Period p L Clearly, periodic functions in applications may have any period, not just p as in the last section (chosen to have simple formulas). The notation p L for the period is practical because L will be a length of a violin string in Sec.., of a rod in heat conduction in Sec..5, and so on. The transition from period p to be period p L is effected by a suitable change of scale, as follows. Let f (x) have period p L. Then we can introduce a new variable v such that f (x), as a function of v, has period p. If we set () (a) x p p v, so that (b) v p p x p L x then v p corresponds to x L. This means that f, as a function of v, has period p and, therefore, a Fourier series of the form () f (x) f a L p vb a a n (a n cos nv b n sin nv) with coefficients obtained from (6) in the last section (3) a p p p f a L p vb dv, a n p p b n p p p p f a L p vb sin nv dv. f a L p vb cos nv dv,

13 484 CHAP. Fourier Analysis We could use these formulas directly, but the change to x simplifies calculations. Since (4) v p L x, we have dv p L dx and we integrate over x from L to L. Consequently, we obtain for a function f (x) of period L the Fourier series (5) f (x) a a n aa n cos np L x b n sin np L xb with the Fourier coefficients of f (x) given by the Euler formulas ( p>l in dx cancels >p in (3)) () a L L f (x) dx L (6) (a) (b) a n L L L b n L L L f (x) cos npx L dx f (x) sin npx L dx n,, Á n,, Á. Just as in Sec.., we continue to call (5) with any coefficients a trigonometric series. And we can integrate from to L or over any other interval of length p L. EXAMPLE Periodic Rectangular Wave Find the Fourier series of the function (Fig. 63) if x f (x) d k if x p L 4, L. Solution. From (6.) we obtain a k> (verify!). From (6a) we obtain a n Thus a n if n is even and if x f (x) cos npx dx k cos npx dx k np sin np. a n k>np if n, 5, 9, Á, a n k>np if n 3, 7,, Á. From (6b) we find that b n for n,, Á. Hence the Fourier series is a Fourier cosine series (that is, it has no sine terms) f (x) k k p acos p x 3 cos 3p x 5 cos 5p x Áb.

14 SEC.. Arbitrary Period. Even and Odd Functions. Half-Range Expansions 485 f(x) k f(x) k k x x Fig. 63. Example Fig. 64. Example EXAMPLE Periodic Rectangular Wave. Change of Scale Find the Fourier series of the function (Fig. 64) k if x f (x) c k if x p L 4, L. Solution. Since L, we have in (3) v px> and obtain from (8) in Sec.. with v instead of x, that is, the present Fourier series g(v) 4k p asin v 3 sin 3v 5 sin 5v Á b Confirm this by using (6) and integrating. f (x) 4k p asin p x 3 sin 3p x 5 sin 5p x Á b. EXAMPLE 3 Half-Wave Rectifier A sinusoidal voltage E sin vt, where t is time, is passed through a half-wave rectifier that clips the negative portion of the wave (Fig. 65). Find the Fourier series of the resulting periodic function if L t, u(t) c E sin vt if t L Solution. Since u when L t, we obtain from (6.), with t instead of x, and from (6a), by using formula () in App. A3. with x vt and y nvt, a n v p p>v E sin vt cos nvt dt ve p p>v [sin ( n) vt sin ( n) vt] dt. a v p p>v E sin vt dt E p If n, the integral on the right is zero, and if n, 3, Á, we readily obtain p L p v, L p v. a n ve p c cos ( n) vt cos ( n) vt p>v ( n) v ( n) v d E ( n)p acos p n If n is odd, this is equal to zero, and for even n we have cos ( n)p b. n a n E p a n n b E (n )(n )p (n, 4, Á ).

15 486 CHAP. Fourier Analysis In a similar fashion we find from (6b) that b E> and b n for n, 3, Á. Consequently, u(t) E p E sin vt E p a cos vt # 3 3 # 5 cos 4vt Á b. u(t) π/ ω π/ ω Fig. 65. Half-wave rectifier t Fig. 66. Even function x. Simplifications: Even and Odd Functions If f (x) is an even function, that is, f (x) f (x) (see Fig. 66), its Fourier series (5) reduces to a Fourier cosine series (5*) f (x) a a a n cos np ( f even) L x n with coefficients (note: integration from to L only!) x (6*) a L L f (x) dx, a n L L f (x) cos npx L dx, n,, Á. Fig. 67. Odd function If f (x) is an odd function, that is, f (x) f (x) (see Fig. 67), its Fourier series (5) reduces to a Fourier sine series (5**) f (x) a b n sin np ( f odd) L x with coefficients n (6**) b n L L f (x) sin npx L dx. These formulas follow from (5) and (6) by remembering from calculus that the definite integral gives the net area ( area above the axis minus area below the axis) under the curve of a function between the limits of integration. This implies (7) (a) L L g (x) dx g (x) dx L for even g (b) L L h (x) dx for odd h Formula (7b) implies the reduction to the cosine series (even f makes f (x) sin (npx>l) odd since sin is odd) and to the sine series (odd f makes f (x) cos (npx>l) odd since cos is even). Similarly, (7a) reduces the integrals in (6*) and (6**) to integrals from to L. These reductions are obvious from the graphs of an even and an odd function. (Give a formal proof.)

16 SEC.. Arbitrary Period. Even and Odd Functions. Half-Range Expansions 487 Summary Even Function of Period. If f is even and L p, then f (x) a a a n cos nx with coefficients a p p f (x) dx, a n p p f (x) cos nx dx, n Odd Function of Period p. If f is odd and L p, then n,, Á with coefficients f (x) a n b n sin nx b n p p f (x) sin nx dx, n,, Á. EXAMPLE 4 Fourier Cosine and Sine Series The rectangular wave in Example is even. Hence it follows without calculation that its Fourier series is a Fourier cosine series, the b n are all zero. Similarly, it follows that the Fourier series of the odd function in Example is a Fourier sine series. In Example 3 you can see that the Fourier cosine series represents u(t) E>p E sin vt. Can you prove that this is an even function? Further simplifications result from the following property, whose very simple proof is left to the student. THEOREM Sum and Scalar Multiple The Fourier coefficients of a sum f f are the sums of the corresponding Fourier coefficients of f and f. The Fourier coefficients of cf are c times the corresponding Fourier coefficients of f. EXAMPLE 5 Sawtooth Wave Find the Fourier series of the function (Fig. 68) f (x) x p if p x p and f (x p) f (x). f (x) Fig. 68. π π The function f(x). Sawtooth wave x

17 488 CHAP. Fourier Analysis y 5 S S S 3 S y π Fig. 69. Partial sums S, S, S 3, S in Example 5 π x Solution. We have f f f, where f x and f p. The Fourier coefficients of f are zero, except for the first one (the constant term), which is p. Hence, by Theorem, the Fourier coefficients are those of, except for, which is p. Since is odd, a for n,, Á a n, b n n, and f a Integrating by parts, we obtain b n p f b n p p f (x) sin nx dx p p x sin nx dx. c x cos nx n p n p Hence b, b, b 3 3, b 4 4, Á, and the Fourier series of f (x) is cos nx dx d cos np. n f (x) p asin x sin x 3 sin 3x Á b. (Fig. 69) 3. Half-Range Expansions Half-range expansions are Fourier series. The idea is simple and useful. Figure 7 explains it. We want to represent f (x) in Fig. 7. by a Fourier series, where f (x) may be the shape of a distorted violin string or the temperature in a metal bar of length L, for example. (Corresponding problems will be discussed in Chap..) Now comes the idea. We could extend f (x) as a function of period L and develop the extended function into a Fourier series. But this series would, in general, contain both cosine and sine terms. We can do better and get simpler series. Indeed, for our given f we can calculate Fourier coefficients from (6*) or from (6**). And we have a choice and can take what seems more practical. If we use (6*), we get (5*). This is the even periodic extension f of f in Fig. 7a. If we choose (6**) instead, we get (5**), the odd periodic extension f of f in Fig. 7b. Both extensions have period L. This motivates the name half-range expansions: f is given (and of physical interest) only on half the range, that is, on half the interval of periodicity of length L. Let us illustrate these ideas with an example that we shall also need in Chap..

18 SEC.. Arbitrary Period. Even and Odd Functions. Half-Range Expansions 489 f(x) L x () The given function f(x) f (x) L L x (a) f(x) continued as an even periodic function of period L f (x) L L x k EXAMPLE 6 (b) f(x) continued as an odd periodic function of period L Fig. 7. Even and odd extensions of period L Triangle and Its Half-Range Expansions Find the two half-range expansions of the function (Fig. 7) L/ L Fig. 7. The given function in Example 6 x Solution. (a) Even periodic extension. From (6*) we obtain a L c k L L> x dx k L L (L x) dx d k, k L x if x L f(x) e k L (L x) if L x L. L> a n L c k L L> x cos np L L x dx k L L L> (L x) cos np L x dx d. We consider a n. For the first integral we obtain by integration by parts L> x cos np L Lx x dx np sin np L> L x L np L> sin np L x dx Similarly, for the second integral we obtain L np sin np L n p acos np b. L (L x) cos np L> L L x dx np (L x) sin np L L x L> a L np al L b sin np b L np L sin np L x dx L> L n p acos np cos np b.

19 49 CHAP. Fourier Analysis We insert these two results into the formula for. The sine terms cancel and so does a factor. This gives a n L Thus, a n 4k n p a cos np cos np b. a 6k>( p ), a 6 6k>(6 p ), a 6k>( p ), Á and a n if n, 6,, 4, Á. Hence the first half-range expansion of f (x) is (Fig. 7a) f (x) k 6k p a cos p L x 6 cos 6p L x Áb. This Fourier cosine series represents the even periodic extension of the given function f (x), of period L. (b) Odd periodic extension. Similarly, from ( 6** ) we obtain (5) b n 8k n p sin np. Hence the other half-range expansion of f (x) is (Fig. 7b) f (x) 8k p a sin p x sin 3p L 3 L x 5 sin 5p L x Á b. The series represents the odd periodic extension of f (x), of period L. Basic applications of these results will be shown in Secs..3 and.5. L L x (a) Even extension L L x (b) Odd extension Fig. 7. Periodic extensions of f(x) in Example 6 PROBLEM SET. 7 EVEN AND ODD FUNCTIONS Are the following functions even or odd or neither even nor odd?. e x, e ƒ x ƒ, x 3 cos nx, x tan px, sinh x cosh x. sin x, sin (x ), ln x, x>(x ), x cot x 3. Sums and products of even functions 4. Sums and products of odd functions 5. Absolute values of odd functions 6. Product of an odd times an even function 7. Find all functions that are both even and odd. 8 7 FOURIER SERIES FOR PERIOD p = L Is the given function even or odd or neither even nor odd? Find its Fourier series. Show details of your work. 8.

20 SEC.. Arbitrary Period. Even and Odd Functions. Half-Range Expansions (b) Apply the program to Probs. 8, graphing the first few partial sums of each of the four series on common axes. Choose the first five or more partial sums until they approximate the given function reasonably well. Compare and comment.. 4. Obtain the Fourier series in Prob. 8 from that in Prob f (x) x ( x ), p f (x) x >4 ( x ), p HALF-RANGE EXPANSIONS Find (a) the Fourier cosine series, (b) the Fourier sine series. Sketch f (x) and its two periodic extensions. Show the details f (x) cos px ( x ), p π 5. π 4 6. π π π f (x) x ƒ x ƒ ( x ), p 6. π π 7. π π 8. Rectifier. Find the Fourier series of the function obtained by passing the voltage v(t) V cos pt through a half-wave rectifier that clips the negative half-waves. 9. Trigonometric Identities. Show that the familiar identities cos 3 x 3 and sin 3 x 3 sin x 4 cos x 4 cos 3x 4 4 sin 3x can be interpreted as Fourier series expansions. Develop cos 4 x.. Numeric Values. Using Prob., show that 9 6 Á 6 p. 4. CAS PROJECT. Fourier Series of L-Periodic Functions. (a) Write a program for obtaining partial sums of a Fourier series (5) π L π L π 9. f (x) sin x ( x p) 3. Obtain the solution to Prob. 6 from that of Prob. 7.

21 49 CHAP. Fourier Analysis.3 Forced Oscillations Fourier series have important applications for both ODEs and PDEs. In this section we shall focus on ODEs and cover similar applications for PDEs in Chap.. All these applications will show our indebtedness to Euler s and Fourier s ingenious idea of splitting up periodic functions into the simplest ones possible. From Sec..8 we know that forced oscillations of a body of mass m on a spring of modulus k are governed by the ODE () mys cyr ky r (t) where y y (t) is the displacement from rest, c the damping constant, k the spring constant (spring modulus), and r (t) the external force depending on time t. Figure 74 shows the model and Fig. 75 its electrical analog, an RLC-circuit governed by (*) LIs RIr I Er (t) C (Sec..9). We consider (). If r (t) is a sine or cosine function and if there is damping (c ), then the steady-state solution is a harmonic oscillation with frequency equal to that of r (t). However, if r (t) is not a pure sine or cosine function but is any other periodic function, then the steady-state solution will be a superposition of harmonic oscillations with frequencies equal to that of r(t) and integer multiples of these frequencies. And if one of these frequencies is close to the (practical) resonant frequency of the vibrating system (see Sec..8), then the corresponding oscillation may be the dominant part of the response of the system to the external force. This is what the use of Fourier series will show us. Of course, this is quite surprising to an observer unfamiliar with Fourier series, which are highly important in the study of vibrating systems and resonance. Let us discuss the entire situation in terms of a typical example. C Spring R L External force r(t) Mass m Dashpot E(t) Fig. 74. Vibrating system under consideration Fig. 75. Electrical analog of the system in Fig. 74 (RLC-circuit) EXAMPLE Forced Oscillations under a Nonsinusoidal Periodic Driving Force In (), let m (g), c.5 (g>sec), and k 5 (g>sec ), so that () becomes () ys.5yr 5y r (t)

22 SEC..3 Forced Oscillations 493 π r(t) π/ π π/ Fig. 76. Force in Example π t where r (t) is measured in g cm>sec. Let (Fig. 76) t p if p t, r (t) e r (t p) r (t). t p if t p, Find the steady-state solution y(t). Solution. We represent r (t) by a Fourier series, finding (3) r (t) 4. p acos t 3 cos 3t 5 cos 5t Á b Then we consider the ODE (4) ys.5yr 5y 4 n p cos nt (n, 3, Á ) whose right side is a single term of the series (3). From Sec..8 we know that the steady-state solution of (4) is of the form y n (t) (5) y n A n cos nt B n sin nt. By substituting this into (4) we find that (6) A n 4(5 n ), B where n n., pd n npd n D n (5 n ) (.5n). Since the ODE () is linear, we may expect the steady-state solution to be (7) y y y 3 y 5 Á where y n is given by (5) and (6). In fact, this follows readily by substituting (7) into () and using the Fourier series of r (t), provided that termwise differentiation of (7) is permissible. (Readers already familiar with the notion of uniform convergence [Sec. 5.5] may prove that (7) may be differentiated term by term.) From (6) we find that the amplitude of (5) is (a factor D n cancels out) Values of the first few amplitudes are C n A n B n 4 n pd n. C.53 C 3.88 C 5.37 C 7. C 9.3. Figure 77 shows the input (multiplied by.) and the output. For n 5 the quantity D n is very small, the denominator of C 5 is small, and C 5 is so large that y 5 is the dominating term in (7). Hence the output is almost a harmonic oscillation of five times the frequency of the driving force, a little distorted due to the term y, whose amplitude is about 5% of that of y 5. You could make the situation still more extreme by decreasing the damping constant c. Try it.

23 494 CHAP. Fourier Analysis y.3. Output. 3 3 t. Input. Fig. 77. Input and steady-state output in Example PROBLEM SET.3. Coefficients C n. Derive the formula for C n from A n and B n.. Change of spring and damping. In Example, what happens to the amplitudes C n if we take a stiffer spring, say, of k 49? If we increase the damping? 3. Phase shift. Explain the role of the B n s. What happens if we let c :? 4. Differentiation of input. In Example, what happens if we replace r (t) with its derivative, the rectangular wave? What is the ratio of the new C n to the old ones? 5. Sign of coefficients. Some of the A n in Example are positive, some negative. All B n are positive. Is this physically understandable? 6 GENERAL SOLUTION Find a general solution of the ODE ys v y r (t) with r (t) as given. Show the details of your work. 6. r (t) sin at sin bt, v a, b 7. r (t) sin t, v.5,.9,.,.5, 8. Rectifier. r (t) p/4 ƒ cos t ƒ if p and r (t p) r (t), ƒ v ƒ,, 4, Á t p 9. What kind of solution is excluded in Prob. 8 by ƒ v ƒ,, 4, Á?. Rectifier. r (t) p/4 ƒ sin t ƒ if and r (t p) r (t), ƒ v ƒ,, 4, Á t p if p t. r (t) b ƒ v ƒ, 3, 5, Á if t p,. CAS Program. Write a program for solving the ODE just considered and for jointly graphing input and output of an initial value problem involving that ODE. Apply the program to Probs. 7 and with initial values of your choice. 3 6 STEADY-STATE DAMPED OSCILLATIONS Find the steady-state oscillations of ys cyr y r (t) with c and r (t) as given. Note that the spring constant is k. Show the details. In Probs. 4 6 sketch r (t). N 3. r (t) a (a n cos nt b n sin nt) n if p t 4. r (t) b and r(t p) r(t) if t p 5. r (t) t (p t ) if p t p and r (t p) r (t) 6. r (t) t if p> t p> e and r(t p) r(t) p t if p> t 3p> 7 9 RLC-CIRCUIT Find the steady-state current I (t) in the RLC-circuit in Fig. 75, where R, L H, C F and with E (t) V as follows and periodic with period p. Graph or sketch the first four partial sums. Note that the coefficients of the solution decrease rapidly. Hint. Remember that the ODE contains Er(t), not E (t), cf. Sec E (t) b 5t if p t 5t if t p

24 SEC..4 Approximation by Trigonometric Polynomials 495 (t t ) if p t 8. E (t) b (t t ) if t p 9. E (t) t (p t ) (p t p). CAS EXPERIMENT. Maximum Output Term. Graph and discuss outputs of ys cyr ky r (t) with r (t) as in Example for various c and k with emphasis on the maximum and its ratio to the second largest ƒ C n ƒ. C n.4 Approximation by Trigonometric Polynomials Fourier series play a prominent role not only in differential equations but also in approximation theory, an area that is concerned with approximating functions by other functions usually simpler functions. Here is how Fourier series come into the picture. Let f (x) be a function on the interval p x p that can be represented on this interval by a Fourier series. Then the Nth partial sum of the Fourier series () N f (x) a a (a n cos nx b n sin nx) n is an approximation of the given f (x). In () we choose an arbitrary N and keep it fixed. Then we ask whether () is the best approximation of f by a trigonometric polynomial of the same degree N, that is, by a function of the form N () F (x) A a (A n cos nx B n sin nx) (N fixed). n Here, best means that the error of the approximation is as small as possible. Of course we must first define what we mean by the error of such an approximation. We could choose the maximum of ƒ f (x) F (x) ƒ. But in connection with Fourier series it is better to choose a definition of error that measures the goodness of agreement between f and F on the whole interval p x p. This is preferable since the sum f of a Fourier series may have jumps: F in Fig. 78 is a good overall approximation of f, but the maximum of ƒ f (x) F (x) ƒ (more precisely, the supremum) is large. We choose (3) E p p ( f F) dx. f F x x Fig. 78. Error of approximation

25 496 CHAP. Fourier Analysis This is called the square error of F relative to the function f on the interval p x p. Clearly, E. N being fixed, we want to determine the coefficients in () such that E is minimum. Since ( f F) f ff F, we have (4) We square (), insert it into the last integral in (4), and evaluate the occurring integrals. This gives integrals of cos nx and sin nx (n ), which equal p, and integrals of cos nx, sin nx, and (cos nx)(sin mx), which are zero (just as in Sec..). Thus We now insert () into the integral of ff in (4). This gives integrals of f cos nx as well as f sin nx, just as in Euler s formulas, Sec.., for a n and b n (each multiplied by A n or ). Hence B n With these expressions, (4) becomes (5) p f F dx p(a a A a Á A N a N B b Á B N b N ). p E p E p p p f dx p p p p N F dx p c A a (A n cos nx B n sin nx) d dx p n p(a A Á A N B Á B N ). f dx p c A a a N p c A a N n f F dx p (A n B n ) d. We now take A n a n and B n b n in (). Then in (5) the second line cancels half of the integral-free expression in the first line. Hence for this choice of the coefficients of F the square error, call it E*, is n p F dx. (A n a n B n b n ) d (6) p N E* f dx p c a a (a n b n ) d. p n We finally subtract (6) from (5). Then the integrals drop out and we get terms A and similar terms (B n b n ) n A n a n a n (A n an) : N E E* p e (A a ) a [(A n a n ) (B n b n ) ] f. n Since the sum of squares of real numbers on the right cannot be negative, E E*, thus E E*, and E E* if and only if A a, Á, B N b N. This proves the following fundamental minimum property of the partial sums of Fourier series.

26 SEC..4 Approximation by Trigonometric Polynomials 497 THEOREM Minimum Square Error The square error of F in () (with fixed N) relative to f on the interval p x p is minimum if and only if the coefficients of F in () are the Fourier coefficients of f. This minimum value E* is given by (6). From (6) we see that E* cannot increase as N increases, but may decrease. Hence with increasing N the partial sums of the Fourier series of f yield better and better approximations to f, considered from the viewpoint of the square error. Since E* and (6) holds for every N, we obtain from (6) the important Bessel s inequality (7) a a (a n b n ) p p n p f (x) dx for the Fourier coefficients of any function f for which integral on the right exists. (For F. W. Bessel see Sec. 5.5.) It can be shown (see [C] in App. ) that for such a function f, Parseval s theorem holds; that is, formula (7) holds with the equality sign, so that it becomes Parseval s identity 3 (8) a a (a n b n ) p p n p f (x) dx. EXAMPLE Minimum Square Error for the Sawtooth Wave Compute the minimum square error E* of F (x) with N,, Á,,, Á, and relative to f (x) x p (p x p) on the interval p x p. Solution. F (x) p (sin x by Example 3 in sin x 3 sin 3x Á ()N sin Nx) N Sec..3. From this and (6), π π π π x Fig. 79. F with N in Example Numeric values are: p N E* (x p) dx p ap 4 a p n n b. N E* N E* N E* N E* MARC ANTOINE PARSEVAL ( ), French mathematician. A physical interpretation of the identity follows in the next section.

27 498 CHAP. Fourier Analysis F S, S, S 3 are shown in Fig. 69 in Sec.., and F S is shown in Fig. 79. Although ƒ f (x) F (x) ƒ is large at p (how large?), where f is discontinuous, F approximates f quite well on the whole interval, except near p, where waves remain owing to the Gibbs phenomenon, which we shall discuss in the next section. Can you think of functions f for which E* decreases more quickly with increasing N? PROBLEM SET.4. CAS Problem. Do the numeric and graphic work in Example in the text. 5 MINIMUM SQUARE ERROR Find the trigonometric polynomial F (x) of the form () for which the square error with respect to the given f (x) on the interval p x p is minimum. Compute the minimum value for N,, Á, 5 (or also for larger values if you have a CAS) f (x) x (p x p) f (x) ƒ x ƒ (p x p) f (x) x (p x p) if p x 5. f (x) b if x p 6. Why are the square errors in Prob. 5 substantially larger than in Prob. 3? 7. f (x) x 3 (p x p) 8. f (x) ƒ sin x ƒ (p x p), full-wave rectifier 9. Monotonicity. Show that the minimum square error (6) is a monotone decreasing function of N. How can you use this in practice?. CAS EXPERIMENT. Size and Decrease of E*. Compare the size of the minimum square error E* for functions of your choice. Find experimentally the factors on which the decrease of E* with N depends. For each function considered find the smallest N such that E*.. 5 PARSEVALS S IDENTITY Using (8), prove that the series has the indicated sum. Compute the first few partial sums to see that the convergence is rapid Á p Use Example in Sec Á p Use Prob. 4 in Sec Á p Use Prob. 7 in Sec p cos 4 x dx 3p 4 p 5. p cos 6 x dx 5p 8 p.5 Sturm Liouville Problems. Orthogonal Functions The idea of the Fourier series was to represent general periodic functions in terms of cosines and sines. The latter formed a trigonometric system. This trigonometric system has the desirable property of orthogonality which allows us to compute the coefficient of the Fourier series by the Euler formulas. The question then arises, can this approach be generalized? That is, can we replace the trigonometric system of Sec.. by other orthogonal systems (sets of other orthogonal functions)? The answer is yes and will lead to generalized Fourier series, including the Fourier Legendre series and the Fourier Bessel series in Sec..6. To prepare for this generalization, we first have to introduce the concept of a Sturm Liouville Problem. (The motivation for this approach will become clear as you read on.) Consider a second-order ODE of the form

28 SEC..5 Sturm Liouville Problems. Orthogonal Functions 499 () [ p (x)yr]r [ q (x) lr (x)]y on some interval a x b, satisfying conditions of the form () (a) k y k yr at x a l y l yr at x b (b). Here l is a parameter, and k, k, l, l are given real constants. Furthermore, at least one of each constant in each condition () must be different from zero. (We will see in Example that, if p(x) r(x) and q(x), then sin lx and cos lx satisfy () and constants can be found to satisfy ().) Equation () is known as a Sturm Liouville equation. 4 Together with conditions (a), (b) it is know as the Sturm Liouville problem. It is an example of a boundary value problem. A boundary value problem consists of an ODE and given boundary conditions referring to the two boundary points (endpoints) x a and x b of a given interval a x b. The goal is to solve these type of problems. To do so, we have to consider Eigenvalues, Eigenfunctions Clearly, y is a solution the trivial solution of the problem (), () for any l because () is homogeneous and () has zeros on the right. This is of no interest. We want to find eigenfunctions y (x), that is, solutions of () satisfying () without being identically zero. We call a number l for which an eigenfunction exists an eigenvalue of the Sturm Liouville problem (), (). Many important ODEs in engineering can be written as Sturm Liouville equations. The following example serves as a case in point. EXAMPLE Trigonometric Functions as Eigenfunctions. Vibrating String Find the eigenvalues and eigenfunctions of the Sturm Liouville problem (3) ys ly, y (), y(p). This problem arises, for instance, if an elastic string (a violin string, for example) is stretched a little and fixed at its ends x and x p and then allowed to vibrate. Then y (x) is the space function of the deflection u (x, t) of the string, assumed in the form u (x, t) y (x)w(t), where t is time. (This model will be discussed in great detail in Secs,..4.) Solution. From () nad () we see that p, q, r in (), and a, b p, k l, k in (). For negative l a general solution of the ODE in (3) is y (x) c e x c e x l. From the boundary conditions we obtain c c, so that y, which is not an eigenfunction. For l the situation is similar. For positive l a general solution is y(x) A cos x B sin x. 4 JACQUES CHARLES FRANÇOIS STURM (83 855) was born and studied in Switzerland and then moved to Paris, where he later became the successor of Poisson in the chair of mechanics at the Sorbonne (the University of Paris). JOSEPH LIOUVILLE (89 88), French mathematician and professor in Paris, contributed to various fields in mathematics and is particularly known by his important work in complex analysis (Liouville s theorem; Sec. 4.4), special functions, differential geometry, and number theory.

29 5 CHAP. Fourier Analysis From the first boundary condition we obtain y () A. The second boundary condition then yields y (p) B sin p, thus,,, Á. For we have y. For l, 4, 9, 6, Á, taking B, we obtain y (x) sin x ( l,, Á ). Hence the eigenvalues of the problem are, where and corresponding eigenfunctions are, where, Á l,, Á, y(x) sin x. Note that the solution to this problem is precisely the trigonometric system of the Fourier series considered earlier. It can be shown that, under rather general conditions on the functions p, q, r in (), the Sturm Liouville problem (), () has infinitely many eigenvalues. The corresponding rather complicated theory can be found in Ref. [All] listed in App.. Furthermore, if p, q, r, and pr in () are real-valued and continuous on the interval a x b and r is positive throughout that interval (or negative throughout that interval), then all the eigenvalues of the Sturm Liouville problem (), () are real. (Proof in App. 4.) This is what the engineer would expect since eigenvalues are often related to frequencies, energies, or other physical quantities that must be real. The most remarkable and important property of eigenfunctions of Sturm Liouville problems is their orthogonality, which will be crucial in series developments in terms of eigenfunctions, as we shall see in the next section. This suggests that we should next consider orthogonal functions. Orthogonal Functions Functions y (x), y (x), Á defined on some interval a x b are called orthogonal on this interval with respect to the weight function r (x) if for all m and all n different from m, b (4) (y m, y n ) r (x) y m (x) y n (x) dx (m n). a (y m, y n ) is a standard notation for this integral. The norm y m of is defined by y m (5) y m (y m, y m ) b r (x)y m (x) dx. G a Note that this is the square root of the integral in (4) with. The functions y, y, Á n m are called orthonormal on a x b if they are orthogonal on this interval and all have norm. Then we can write (4), (5) jointly by using the Kronecker symbol 5, namely, d mn b if m n ( y m, y n ) r (x) y m (x) y n (x) dx d mn e if m n. a 5 LEOPOLD KRONECKER (83 89). German mathematician at Berlin University, who made important contributions to algebra, group theory, and number theory.

30 SEC..5 Sturm Liouville Problems. Orthogonal Functions 5 If r (x), we more briefly call the functions orthogonal instead of orthogonal with respect to r (x) ; similarly for orthognormality. Then b (y m, y n ) y m (x) y n (x) dx (m n), a y m (y m, y n ) b y m (x) dx. G a The next example serves as an illustration of the material on orthogonal functions just discussed. EXAMPLE Orthogonal Functions. Orthonormal Functions. Notation The functions y m (x) sin mx, m,, Á form an orthogonal set on the interval p x p, because for m n we obtain by integration [see () in App. A3.] (y m, y n ) p sin mx sin nx dx p cos (m n)x dx p p p p cos (m n)x dx, (m n). The norm y m (y m, y m ) equals p because p y m ( y m, y m ) sin mx dx p Hence the corresponding orthonormal set, obtained by division by the norm, is p (m,, Á ) sin x p, sin x p, sin 3x p, Á. Theorem shows that for any Sturm Liouville problem, the eigenfunctions associated with these problems are orthogonal. This means, in practice, if we can formulate a problem as a Sturm Liouville problem, then by this theorem we are guaranteed orthogonality. THEOREM Orthogonality of Eigenfunctions of Sturm Liouville Problems pr Suppose that the functions p, q, r, and in the Sturm Liouville equation () are real-valued and continuous and r (x) on the interval a x b. Let y m (x) and y n (x) be eigenfunctions of the Sturm Liouville problem (), () that correspond to different eigenvalues l m and l n, respectively. Then y m, y n are orthogonal on that interval with respect to the weight function r, that is, (6) b (y m, y n ) r (x)y m (x)y n (x) dx a (m n). If p (a), then (a) can be dropped from the problem. If p(b), then (b) can be dropped. [It is then required that y and yr remain bounded at such a point, and the problem is called singular, as opposed to a regular problem in which () is used.] If p(a) p(b), then () can be replaced by the periodic boundary conditions (7) y(a) y(b), yr(a) yr(b). The boundary value problem consisting of the Sturm Liouville equation () and the periodic boundary conditions (7) is called a periodic Sturm Liouville problem.