Fourier Analysis Partial Differential Equations (PDEs)

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1 PART C Fourier Analysis. Partial Differential Equations (PDEs) CHAPTER CHAPTER Fourier Analysis Partial Differential Equations (PDEs) Chapter and Chapter are directly related to each other in that Fourier analysis has its most important applications in modeling and solving partial differential equations (PDEs) related to boundary and initial value problems of mechanics, heat flow, electrostatics, and other fields. However, the study of PDEs is a study in its own right. Indeed, PDEs are the subject of much ongoing research. Fourier analysis allows us to model periodic phenomena which appear frequently in engineering and elsewhere think of rotating parts of machines, alternating electric currents or the motion of planets. Related period functions may be complicated. Now, the ingeneous idea of Fourier analysis is to represent complicated functions in terms of simple periodic functions, namely cosines and sines. The representations will be infinite series called Fourier series. This idea can be generalized to more general series (see Sec..5) and to integral representations (see Sec..7). The discovery of Fourier series had a huge impetus on applied mathematics as well as on mathematics as a whole. Indeed, its influence on the concept of a function, on integration theory, on convergence theory, and other theories of mathematics has been substantial (see [GenRef7] in App. ). Chapter deals with the most important partial differential equations (PDEs) of physics and engineering, such as the wave equation, the heat equation, and the Laplace equation. These equations can model a vibrating string/membrane, temperatures on a bar, and electrostatic potentials, respectively. PDEs are very important in many areas of physics and engineering and have many more applications than ODEs. JEAN-BAPTISTE JOSEPH FOURIER (768 83), French physicist and mathematician, lived and taught in Paris, accompanied Napoléon in the Egyptian War, and was later made prefect of Grenoble. The beginnings on Fourier series can be found in works by Euler and by Daniel Bernoulli, but it was Fourier who employed them in a systematic and general manner in his main work, Théorie analytique de la chaleur (Analytic Theory of Heat, Paris, 8), in which he developed the theory of heat conduction (heat equation; see Sec..5), making these series a most important tool in applied mathematics. 473

2 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. 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 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

3 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] p) f (x p) f (x), etc.; thus for any integer n,, 3, Á f (x p), () 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 π cos x π π cos x π π cos 3x π π π π π π 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 )

4 476 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 p a n p p f (x) cos nx dx p (b) b n,, Á n p p f (x) sin nx dx. p f (x) 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

5 ` ` 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 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 Since cos (a) cos a and cos, this yields b n p p p c sin nx k n p c cos nx k n k k [cos cos (np) cos np cos ] ( cos np). np np Now, cos p, cos p, cos 3p, etc.; in general, p p p p p sin nx k ` n d cos nx k n p ` d. 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)

6 478 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

7 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 to p, 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

8 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 ) f (x h) f (x ) and, h 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 ).

9 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 p f s(x) cos nx dx. n 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 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 Á b 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.

10 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. cos nx, sin nx, cos pnx, k sin pnx k 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. f (x) b cos x if p x 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 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, 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 ) (b) (c) π π π π π π π π π π π π ( sin x 4 sin 4x 6 sin 6x Á ) 4 p acos x cos 3x 9 5 cos 5x Á b 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

11 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 ).. From Period p 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, b n p p p a n p p p f a L p vb sin nv dv. f a L p vb cos nv dv,

12 484 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 Á b.

13 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 p L p E sin vt if t L v, L p v. 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 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 ).

14 486 CHAP. Fourier Analysis In a similar fashion we find from (6b) that b and for n, 3, Á E> b n. 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 L g (x) dx g (x) dx 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.)

15 SEC.. Arbitrary Period. Even and Odd Functions. Half-Range Expansions 487 Summary Even Function of Period. If f is even and L p, then with coefficients a p p f (x) dx, f (x) a a a n cos nx n a n p p f (x) cos nx dx, 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

16 488 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 a are those of, except for, which is p. Since is odd, a for n,, Á 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 cos nx dx d cos np. n Hence b, b, b 3 3, b 4 4, Á, and the Fourier series of f (x) is 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..

17 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.

18 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 Á b. 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.

19 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.

20 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) (Sec..9). C 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)

21 SEC..3 Forced Oscillations 493 r(t) π/ π π t π/ Fig. 76. Force in Example where r (t) is measured in g cm>sec. Let (Fig. 76) r (t) e t p if p t, 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.

22 494 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). 3. N 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

23 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 p (3) E ( f F) dx. p f F x x Fig. 78. Error of approximation

24 496 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 p p E f dx f F dx F dx. p p N F dx p c A a (A n cos nx B n sin nx) d dx p n p f F dx p(a a A a Á A N a N B b Á B N b N ). p p N E f dx p c A a a p p p p(a A Á A N B Á B N ). p c A a N n (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 (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.

25 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.

26 498 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

27 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 (b) l y l yr at x 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.

28 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.

29 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 Suppose that the functions p, q, r, and pr 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.

30 5 CHAP. Fourier Analysis PROOF By assumption, and satisfy the Sturm Liouville equations y m y n ( py m r )r (q l m r)y m ( pyr n )r (q l n r)y n respectively. We multiply the first equation by y n, the second by y m, and add, (l m l n )ry m y n y m ( py n r )r y n ( pyr m )r [(pyr n ) y m [( pyr m ) y n ]r where the last equality can be readily verified by performing the indicated differentiation of the last expression in brackets. This expression is continuous on a x b since p and pr are continuous by assumption and y m, y n are solutions of (). Integrating over x from a to b, we thus obtain (8) b (l m l n ) b ry m y n dx [ p(yr n y m yr m y n )] a a (a b). The expression on the right equals the sum of the subsequent Lines and, (9) p(b)[yr n (b)y m (b) yr m (b)y n (b)] p (a)[ yr n (a)y m (a) yr m (a)y n (a)] (Line ) (Line ). Hence if (9) is zero, (8) with l m l n implies the orthogonality (6). Accordingly, we have to show that (9) is zero, using the boundary conditions () as needed. Case. Case. p (a) p (b). Clearly, (9) is zero, and () is not needed. p (a), p (b). Line of (9) is zero. Consider Line. From (a) we have k y n (a) k y n r(a), k y m (a) k yr m (a). Let k. We multiply the first equation by y m (a), the last by y n (a) and add, k [y n r (a)y m (a) yr m (a)y n (a)]. This is k times Line of (9), which thus is zero since k. If k, then k by assumption, and the argument of proof is similar. Case 3. p(a), p(b). Line of (9) is zero. From (b) it follows that Line of (9) is zero; this is similar to Case. Case 4. p(a), p(b). We use both (a) and (b) and proceed as in Cases and 3. Case 5. p(a) p(b). Then (9) becomes p(b)[ y n r (b)y m (b) y m r (b)y n (b) y n r (a)y m (a) y m r (a)y n (a)]. The expression in brackets [ Á ] is zero, either by () used as before, or more directly by (7). Hence in this case, (7) can be used instead of (), as claimed. This completes the proof of Theorem. EXAMPLE 3 Application of Theorem. Vibrating String The ODE in Example is a Sturm Liouville equation with. From Theorem it follows that the eigenfunctions y m sin mx (m,, Á p, q, and r ) are orthogonal on the interval x p.

31 SEC..5 Sturm Liouville Problems. Orthogonal Functions 53 Example 3 confirms, from this new perspective, that the trigonometric system underlying the Fourier series is orthogonal, as we knew from Sec... EXAMPLE 4 Application of Theorem. Orthogonlity of the Legendre Polynomials Legendre s equation ( x ) ys xyr n (n ) y may be written [( x ) yr]r ly l n (n ). Hence, this is a Sturm Liouville equation () with p x, q, and r. Since p () p (), we need no boundary conditions, but have a singular Sturm Liouville problem on the interval. We know that for n,, Á, hence l, #, # 3, Á x, the Legendre polynomials P n (x) are solutions of the problem. Hence these are the eigenfunctions. From Theorem it follows that they are orthogonal on that interval, that is, () P m (x)p n (x) dx (m n). What we have seen is that the trigonometric system, underlying the Fourier series, is a solution to a Sturm Liouville problem, as shown in Example, and that this trigonometric system is orthogonal, which we knew from Sec.. and confirmed in Example 3. PROBLEM SET.5. Proof of Theorem. Carry out the details in Cases 3 and 4. 6 ORTHOGONALITY. Normalization of eigenfunctions y m of (), () means that we multiply y m by a nonzero constant c m such that c m y m has norm. Show that z m cy m with any c is an eigenfunction for the eigenvalue corresponding to y m. 3. Change of x. Show that if the functions y y (x), Á (x), form an orthogonal set on an interval a x b (with r (x) ), then the functions y (ct k), y (ct k), Á, c, form an orthogonal set on the interval (a k)>c t (b k)>c. 4. Change of x. Using Prob. 3, derive the orthogonality of, cos px, sin px, cos px, sin px, Á on x (r(x) ) from that of, cos x, sin x, cos x, sin x, Á on p x p. 5. Legendre polynomials. Show that the functions P n (cos u), n,, Á, from an orthogonal set on the interval u p with respect to the weight function sin u. 6. Tranformation to Sturm Liouville form. Show that ys fyr (g lh) y takes the form () if you set p exp (f dx), q pg, r hp. Why would you do such a transformation? 7 5 STURM LIOUVILLE PROBLEMS Find the eigenvalues and eigenfunctions. Verify orthogonality. Start by writing the ODE in the form (), using Prob. 6. Show details of your work. 7. ys ly, y (), y () 8. ys ly, y (), y (L) 9. ys ly, y (), yr(l). ys ly, y () y (), yr() yr(). ( yr>x)r (l )y>x 3, y (), y (e p ). (Set x e t.). ys yr (l ) y, y (), y () 3. ys 8yr (l 6) y, y (), y (p) 4. TEAM PROJECT. Special Functions. Orthogonal polynomials play a great role in applications. For this reason, Legendre polynomials and various other orthogonal polynomials have been studied extensively; see Refs. [GenRef], [GenRef] in App.. Consider some of the most important ones as follows.

32 54 CHAP. Fourier Analysis (a) Chebyshev polynomials 6 of the first and second kind are defined by respectively, where n,, Á. Show that T, U, T n (x) cos (n arccos x) sin [(n ) arccos x] U n (x) x T (x) x, T 3 (x) 4x 3 3x, U (x) x, U 3 (x) 8x 3 4x. T (x) x. U (x) 4x, Show that the Chebyshev polynomials T n (x) are orthogonal on the interval x with respect to the weight function r (x) > x. (Hint. To evaluate the integral, set arccos x u.) Verify that T n (x), equation n,,, 3, satisfy the Chebyshev ( x )ys xyr n y. (b) Orthogonality on an infinite interval: Laguerre polynomials 7 are defined by L, and Show that L n (x) ex d n (x n e x ), n,, n! Á. dx n L n (x) x, L (x) x x >, L 3 (x) 3x 3x > x 3 >6. Prove that the Laguerre polynomials are orthogonal on the positive axis x with respect to the weight function r (x) e x. Hint. Since the highest power in L is x m, it suffices to show that e x x k m L n dx for k n. Do this by k integrations by parts..6 Orthogonal Series. Generalized Fourier Series Fourier series are made up of the trigonometric system (Sec..), which is orthogonal, and orthogonality was essential in obtaining the Euler formulas for the Fourier coefficients. Orthogonality will also give us coefficient formulas for the desired generalized Fourier series, including the Fourier Legendre series and the Fourier Bessel series. This generalization is as follows. Let y, y, y, Á be orthogonal with respect to a weight function r (x) on an interval a x b, and let f (x) be a function that can be represented by a convergent series () f (x) a m a m y m (x) a y (x) a y (x) Á. This is called an orthogonal series, orthogonal expansion, or generalized Fourier series. If the y m are the eigenfunctions of a Sturm Liouville problem, we call () an eigenfunction expansion. In () we use again m for summation since n will be used as a fixed order of Bessel functions. Given f (x), we have to determine the coefficients in (), called the Fourier constants of f (x) with respect to y, y, Á. Because of the orthogonality, this is simple. Similarly to Sec.., we multiply both sides of () by r (x)y n (x) (n fixed) and then integrate on 6 PAFNUTI CHEBYSHEV (8 894), Russian mathematician, is known for his work in approximation theory and the theory of numbers. Another transliteration of the name is TCHEBICHEF. 7 EDMOND LAGUERRE ( ), French mathematician, who did research work in geometry and in the theory of infinite series.

33 SEC..6 Orthogonal Series. Generalized Fourier Series 55 both sides from a to b. We assume that term-by-term integration is permissible. (This is justified, for instance, in the case of uniform convergence, as is shown in Sec. 5.5.) Then we obtain b b ( f, y n ) r fy n dx r a a a m y m b y n dx a a m b ry m y n dx a a m (y m, y n ). a a m m a m Because of the orthogonality all the integrals on the right are zero, except when m n. Hence the whole infinite series reduces to the single term a. Thus ( f, y n ) a n y n n (y n, y n ) a n y n. Assuming that all the functions y have nonzero norm, we can divide by y n n ; writing again m for n, to be in agreement with (), we get the desired formula for the Fourier constants () a m ( f, y m) y m y m b r (x) f (x)y m (x) dx a (n,, Á ). This formula generalizes the Euler formulas (6) in Sec.. as well as the principle of their derivation, namely, by orthogonality. EXAMPLE Fourier Legendre Series A Fourier Legendre series is an eigenfunction expansion f (x) a a m P m (x) a P a P (x) a P (x) Á a a x a ( 3 x ) Á m in terms of Legendre polynomials (Sec. 5.3). The latter are the eigenfunctions of the Sturm Liouville problem in Example 4 of Sec..5 on the interval x. We have r (x) for Legendre s equation, and () gives (3) m a m f (x)p m (x) dx, m,, Á because the norm is (4) P m G P m (x) dx B m (m,, Á ) as we state without proof. The proof of (4) is tricky; it uses Rodrigues s formula in Problem Set 5. and a reduction of the resulting integral to a quotient of gamma functions. For instance, let f (x) sin px. Then we obtain the coefficients, thus a 3, etc. a m x sin px dx 3 m (sin px)p m (x) dx p

34 56 CHAP. Fourier Analysis Hence the Fourier Legendre series of sin px is sin px.95493p (x).584p.p (x) Á 3 (x).99p 5 (x).664p 7 (x).68p 9 (x). The coefficient of P is about 3 # 7 3. The sum of the first three nonzero terms gives a curve that practically coincides with the sine curve. Can you see why the even-numbered coefficients are zero? Why a 3 is the absolutely biggest coefficient? EXAMPLE Fourier Bessel Series These series model vibrating membranes (Sec..9) and other physical systems of circular symmetry. We derive these series in three steps. Step. Bessel s equation as a Sturm Liouville equation. The Bessel function J n (x) with fixed integer n satisfies Bessel s equation (Sec. 5.5) x J ## n ( x ) xj # n ( x ) ( x n )J n ( x ) # where ## # J n dj and. We set. Then x and by the chain rule, J n dj n >d n >d x x kx x>k x (dj and J n J n s>k n >dx)/k. In the first two terms of Bessel s equation, and k drop out and we obtain J ## n d J n >d x k x J n s(kx) xj r n (kx) (k x n )J n (kx). Dividing by x and using (xj n r(kx))r xjs n (kx) J n r (kx) gives the Sturm Liouville equation n (5) [xj l k n r(kx)]r a x lxb J n(kx) with p (x) x, q (x) n >x, r (x) x, and parameter l k. Since p (), Theorem in Sec..5 implies orthogonality on an interval x R (R given, fixed) of those solutions J n (kx) that are zero at x R, that is, (6) J n (kr) (n fixed). Note that q (x) n >x is discontinuous at, but this does not affect the proof of Theorem. Step. Orthogonality. It can be shown (see Ref. [A3]) that has infinitely many zeros, say, x an, a n, Á J n (x ) (see Fig. in Sec. 5.4 for n and ). Hence we must have (7) kr a n,m thus This proves the following orthogonality property. k n,m a n,m >R (m,, Á ). THEOREM Orthogonality of Bessel Functions For each fixed nonnegative integer n the sequence of Bessel functions of the first kind J n (k n, x), J n (k n, x), Á with k n,m as in (7) forms an orthogonal set on the interval x R with respect to the weight function r (x) x, that is, R (8) xj n (k n,m x)j n (k n, j x) dx ( j m, n fixed). Hence we have obtained infinitely many orthogonal sets of Bessel functions, one for each of J, J, J, Á. Each set is orthogonal on an interval x R with a fixed positive R of our choice and with respect to the weight x. The orthogonal set for J n is J n (k n, x), J n (k n, x), J n (k n,3 x), Á, where n is fixed and k n,m is given by (7).

35 SEC..6 Orthogonal Series. Generalized Fourier Series 57 Step 3. Fourier Bessel series. The Fourier Bessel series corresponding to (n fixed) is (9) f (x) a a m J n (k n,m x) a J n (k n, x) a J n (k n, x) a 3 J n (k n,3 x) Á (n fixed). m The coefficients are (with a n,m k n,m R) J n () a m R x f (x) J n (k n,m x) dx, R J n(a n,m ) m,, Á because the square of the norm is () R J n (k n,m x) xj n (k n,m x) dx R J n(k n,m R) as we state without proof (which is tricky; see the discussion beginning on p. 576 of [A3]). EXAMPLE 3 Special Fourier Bessel Series For instance, let us consider f (x) x and take R and n in the series (9), simply writing l for a,m. Then k n,m a,m l.45, 5.5, 8.654,.79, etc. (use a CAS or Table A in App. 5). Next we calculate the coefficients by () a m a m x( x )J (lx) dx. J (l) This can be integrated by a CAS or by formulas as follows. First use [xj (lx)]r lxj (lx) from Theorem in Sec. 5.4 and then integration by parts, a m x( x )J (lx) dx J (l) J (l) c l ( x )xj (lx) ` l xj (lx)(x) dx d. The integral-free part is zero. The remaining integral can be evaluated by [x J (lx)]r lx J (lx) from Theorem in Sec This gives a m 4J (l) l J (l) (l a,m ). Numeric values can be obtained from a CAS (or from the table on p. 49 of Ref. [GenRef] in App., together with the formula J in Theorem of Sec. 5.4). This gives the eigenfunction expansion of x x J J in terms of Bessel functions J, that is, x.8j (.45x).398J (5.5x).455J (8.654x).J (.79x) Á. A graph would show that the curve of x and that of the sum of first three terms practically coincide. Mean Square Convergence. Completeness Ideas on approximation in the last section generalize from Fourier series to orthogonal series () that are made up of an orthonormal set that is complete, that is, consists of sufficiently many functions so that () can represent large classes of other functions (definition below). In this connection, convergence is convergence in the norm, also called mean-square convergence; that is, a sequence of functions is called convergent with the limit f if f k (*) lim f f ; k: k

36 58 CHAP. Fourier Analysis written out by (5) in Sec..5 (where we can drop the square root, as this does not affect the limit) () Accordingly, the series () converges and represents f if (3) where is the kth partial sum of (). (4) Note that the integral in (3) generalizes (3) in Sec..4. We now define completeness. An orthonormal set y, y, Á on an interval a x b is complete in a set of functions S defined on a x b if we can approximate every f belonging to S arbitrarily closely in the norm by a linear combination a, that is, technically, if for every we can find constants a, Á y Á a y a k y k P, a k (with k large enough) such that (5) f (a y Á a k y k ) P. Ref. [GenRef7] in App. uses the more modern term total for complete. We can now extend the ideas in Sec..4 that guided us from (3) in Sec..4 to Bessel s and Parseval s formulas (7) and (8) in that section. Performing the square in (3) and using (4), we first have (analog of (4) in Sec..4) b a s k lim k: b a lim k: b a b b b r (x)[s k (x) f (x)] dx rs k dx rfs k dx rf dx a b k r c a a The first integral on the right equals becausery m y l dx for m l, and ry. In the second sum on the right, the integral equals by () with y m m dx a m,. Hence the first term on the right cancels half of the second term, so that the right side reduces to (analog of (6) in Sec..4) This is nonnegative because in the previous formula the integrand on the left is nonnegative (recall that the weight r (x) is positive!) and so is the integral on the left. This proves the important Bessel s inequality (analog of (7) in Sec..4) k b (6) a a (k,, Á m f r (x) f (x) dx ), m r (x)[ f k (x) f (x)] dx. r (x)[s k (x) f (x)] dx s k (x) a k m a k m m a k a m y m d dx a m ga m a m y m (x). b a m rf dx. a a a a m b b rfy m dx rf dx. a a

37 SEC..6 Orthogonal Series. Generalized Fourier Series 59 Here we can let k :, because the left sides form a monotone increasing sequence that is bounded by the right side, so that we have convergence by the familiar Theorem in App. A.3.3 Hence (7) a m a m f. Furthermore, if y, y, Á is complete in a set of functions S, then (3) holds for every f belonging to S. By (3) this implies equality in (6) with k :. Hence in the case of completeness every f in S saisfies the so-called Parseval equality (analog of (8) in Sec..4) (8) a b a m f r (x) f (x) dx. m a As a consequence of (8) we prove that in the case of completeness there is no function orthogonal to every function of the orthonormal set, with the trivial exception of a function of zero norm: THEOREM Completeness Let y, y, Á be a complete orthonormal set on a x b in a set of functions S. Then if a function f belongs to S and is orthogonal to every y m, it must have norm zero. In particular, if f is continuous, then f must be identically zero. PROOF Since f is orthogonal to every y m, the left side of (8) must be zero. If f is continuous, then f implies f (x), as can be seen directly from (5) in Sec..5 with f instead of y m because r (x) by assumption. PROBLEM SET.6 7 FOURIER LEGENDRE SERIES Showing the details, develop x 5 9x 3 35x (x ) x 4, x, x, x 3, x 4 5. Prove that if f (x) is even (is odd, respectively), its Fourier Legendre series contains only P m (x) with even m (only P m (x) with odd m, respectively). Give examples. 6. What can you say about the coefficients of the Fourier Legendre series of f (x) if the Maclaurin series of contains only powers x 4m (m,,, Á f (x) )? 7. What happens to the Fourier Legendre series of a polynomial f (x) if you change a coefficient of f (x)? Experiment. Try to prove your answer. 8 3 CAS EXPERIMENT FOURIER LEGENDRE SERIES. Find and graph (on common axes) the partial sums up to S m whose graph practically coincides with that of f (x) within graphical accuracy. State m. On what does the size of m seem to depend? 8. f (x) sin px 9. f (x) sin px. f (x) e x. f (x) ( x ). f (x) J (a, x), a, the first positive zero of J (x) 3. f (x) J (a, x), a, the second positive zero of J (x)

38 5 CHAP. Fourier Analysis 4. TEAM PROJECT. Orthogonality on the Entire Real Axis. Hermite Polynomials. 8 These orthogonal polynomials are defined by He () and (9) REMARK. As is true for many special functions, the literature contains more than one notation, and one sometimes defines as Hermite polynomials the functions This differs from our definition, which is preferred in applications. (a) Small Values of n. Show that He (x) x, He 3 (x) x 3 3x, (b) Generating Function. A generating function of the Hermite polynomials is () He n (x) () n e x > dn H *, dx n (ex > ), H n *(x) () n e dn e x x dx. n He (x) x, He 4 (x) x 4 6x 3. e txt > a a n (x) t n n n,, Á. because He n (x) n! a n (x). Prove this. Hint: Use the formula for the coefficients of a Maclaurin series and note that tx t x (x t). (c) Derivative. Differentiating the generating function with respect to x, show that () He n r (x) nhe n (x). (d) Orthogonality on the x-axis needs a weight function that goes to zero sufficiently fast as x :, (Why?) Show that the Hermite polynomials are orthogonal on x with respect to the weight function r (x) e x >. Hint. Use integration by parts and (). (e) ODEs. Show that () Using this with n instead of n and (), show that y He n (x) satisfies the ODE (3) Show that w e x >4 y is a solution of Weber s equation (4) ws (n (n,, Á 4 x ) w ). 5. CAS EXPERIMENT. Fourier Bessel Series. Use Example and R, so that you get the series (5) Her n (x) xhe n (x) He n (x). ys xyr ny. f (x) a J (a, x) a J (a, x) a Á 3 J (a,3 x) With the zeros a, a,, Á from your CAS (see also Table A in App. 5). (a) Graph the terms J (a, x), Á, J (a, x) for x on common axes. (b) Write a program for calculating partial sums of (5). Find out for what f (x) your CAS can evaluate the integrals. Take two such f (x) and comment empirically on the speed of convergence by observing the decrease of the coefficients. (c) Take f (x) in (5) and evaluate the integrals for the coefficients analytically by (a), Sec. 5.4, with v. Graph the first few partial sums on common axes..7 Fourier Integral Fourier series are powerful tools for problems involving functions that are periodic or are of interest on a finite interval only. Sections. and.3 first illustrated this, and various further applications follow in Chap.. Since, of course, many problems involve functions that are nonperiodic and are of interest on the whole x-axis, we ask what can be done to extend the method of Fourier series to such functions. This idea will lead to Fourier integrals. In Example we start from a special function f L of period L and see what happens to its Fourier series if we let L :. Then we do the same for an arbitrary function f L of period L. This will motivate and suggest the main result of this section, which is an integral representation given in Theorem below. 8 CHARLES HERMITE (8 9), French mathematician, is known for his work in algebra and number theory. The great HENRI POINCARÉ (854 9) was one of his students.

39 SEC..7 Fourier Integral 5 EXAMPLE Rectangular Wave Consider the periodic rectangular wave f L (x) of period L given by f L (x) d if x The left part of Fig. 8 shows this function for L 4, 8, 6 as well as the nonperiodic function f(x), which we obtain from f L if we let L :, f (x) lim f if x L (x) e L: otherwise. We now explore what happens to the Fourier coefficients of f L as L increases. Since f L is even, b n for all n. For a n the Euler formulas (6), Sec.., give a L dx L, a n L cos npx if L x if x L. L dx L cos npx L dx L ƒ ƒ 4, 8, 6 ( sin w )>(Lw ) This sequence of Fourier coefficients is called the amplitude spectrum of f L because a n is the maximum amplitude of the wave a n cos (npx>l). Figure 8 shows this spectrum for the periods L 4, 8, 6. We see that for increasing L these amplitudes become more and more dense on the positive w n -axis, where w n np>l. Indeed, for L we have, 3, 7 amplitudes per half-wave of the function (dashed n n in the figure). Hence for L k we have k amplitudes per half-wave, so that these amplitudes will eventually be everywhere dense on the positive w n -axis (and will decrease to zero). The outcome of this example gives an intuitive impression of what about to expect if we turn from our special function to an arbitrary one, as we shall do next. sin (np>l) np>l. Waveform f L (x) Amplitude spectrum a n (w n ) n = w n = n π/l f L (x) n = 5 x w n L = 4 f L (x) _ n = 3 n = n = 7 n = 4 4 L = 8 x n = 6 n = 4 w n f L (x) _ 4 n = 4 n = 8 8 L = 6 x n = n = 8 w n f(x) x Fig. 8. Waveforms and amplitude spectra in Example

40 5 CHAP. Fourier Analysis From Fourier Series to Fourier Integral We now consider any periodic function f L (x) of period L that can be represented by a Fourier series f L (x) a a n (a n cos w n x b n sin w n x), w n np L and find out what happens if we let L :. Together with Example the present calculation will suggest that we should expect an integral (instead of a series) involving cos wx and sin wx with w no longer restricted to integer multiples w w n np>l of p>l but taking all values. We shall also see what form such an integral might have. If we insert a n and b n from the Euler formulas (6), Sec.., and denote the variable of integration by v, the Fourier series of f L (x) becomes We now set f L (x) L L f L (v) dv L a L n L L c cos w n x f L (v) cos w n v dv L sin w n x f L (v) sin w n v dv d. L (n )p w w n w n L np L p L. Then >L w>p, and we may write the Fourier series in the form () f L (x) L L f L (v) dv p a L L c (cos w n x) w f L (v) cos w n v dv n L L (sin w n x) w f L (v) sin w n v dv d. This representation is valid for any fixed L, arbitrarily large, but finite. We now let L : and assume that the resulting nonperiodic function f (x) lim f L (x) L: is absolutely integrable on the x-axis; that is, the following (finite!) limits exist: L () lim ƒ f (x) ƒ dx lim a: a b: b ƒ f (x) ƒ dx awritten ƒ f (x) ƒ dxb. Then >L :, and the value of the first term on the right side of () approaches zero. Also w p>l : and it seems plausible that the infinite series in () becomes an

41 SEC..7 Fourier Integral 53 integral from to, which represents f(x), namely, (3) f (x) p c cos wx f (v) cos wv dv sin wx f (v) sin wv dv d dw. If we introduce the notations (4) A (w) p f (v) cos wv dv, B (w) p f (v) sin wv dv we can write this in the form (5) f (x) [A(w) cos wx B (w) sin wx] dw. This is called a representation of f (x) by a Fourier integral. It is clear that our naive approach merely suggests the representation (5), but by no means establishes it; in fact, the limit of the series in () as w approaches zero is not the definition of the integral (3). Sufficient conditions for the validity of (5) are as follows. THEOREM Fourier Integral If f (x) is piecewise continuous (see Sec. 6.) in every finite interval and has a righthand derivative and a left-hand derivative at every point (see Sec.) and if the integral () exists, then f (x) can be represented by a Fourier integral (5) with A and B given by (4). At a point where f (x) is discontinuous the value of the Fourier integral equals the average of the left- and right-hand limits of f (x) at that point (see Sec..). (Proof in Ref. [C]; see App..) Applications of Fourier Integrals The main application of Fourier integrals is in solving ODEs and PDEs, as we shall see for PDEs in Sec..6. However, we can also use Fourier integrals in integration and in discussing functions defined by integrals, as the next example. EXAMPLE Single Pulse, Sine Integral. Dirichlet s Discontinuous Factor. Gibbs Phenomenon Find the Fourier integral representation of the function if ƒ x ƒ f (x) e if ƒ x ƒ (Fig. 8) f(x) Fig. 8. Example x

42 ` 54 CHAP. Fourier Analysis Solution. From (4) we obtain and (5) gives the answer A (w) p f (v) cos wv dv p cos wv dv B (w) p sin wv dv sin wv pw sin w pw (6) f (x) p cos wx sin w dw. w The average of the left- and right-hand limits of f (x) at x is equal to ( )>, that is,. Furthermore, from (6) and Theorem we obtain (multiply by p>) (7) cos wx sin w w p> if x, dw dp>4 if x, if x. We mention that this integral is called Dirichlet s discontinous factor. (For P. L. Dirichlet see Sec..8.) The case x is of particular interest. If x, then (7) gives (8*) sin w w dw p. We see that this integral is the limit of the so-called sine integral (8) Si(u) u sin w w dw as u :. The graphs of Si(u) and of the integrand are shown in Fig. 8. In the case of a Fourier series the graphs of the partial sums are approximation curves of the curve of the periodic function represented by the series. Similarly, in the case of the Fourier integral (5), approximations are obtained by replacing by numbers a. Hence the integral (9) p a cos wx sin w dw w approximates the right side in (6) and therefore f (x). Integrand y π.5 Si(u) 4π 3π π π π π 3π 4π u.5 π Fig. 8. Sine integral Si(u) and integrand

43 SEC..7 Fourier Integral 55 y y y a = 8 a = 6 a = 3 x x Fig. 83. The integral (9) for a 8, 6, and 3, illustrating the development of the Gibbs phenomenon x Figure 83 shows oscillations near the points of discontinuity of f (x). We might expect that these oscillations disappear as a approaches infinity. But this is not true; with increasing a, they are shifted closer to the points x. This unexpected behavior, which also occurs in connection with Fourier series (see Sec..), is known as the Gibbs phenomenon. We can explain it by representing (9) in terms of sine integrals as follows. Using () in App. A3., we have p a cos wx sin w w In the first integral on the right we set w wx t. Then dw>w dt>t, and w a corresponds to t (x ) a. In the last integral we set w wx t. Then dw>w dt>t, and w a corresponds to t (x ) a. Since sin (t) sin t, we thus obtain p a dw p a cos wx sin w dw w p (x) a From this and (8) we see that our integral (9) equals sin (w wx) w p Si(a[x ]) Si(a[x ]) p and the oscillations in Fig. 83 result from those in Fig. 8. The increase of a amounts to a transformation of the scale on the axis and causes the shift of the oscillations (the waves) toward the points of discontinuity and. Fourier Cosine Integral and Fourier Sine Integral Just as Fourier series simplify if a function is even or odd (see Sec..), so do Fourier integrals, and you can save work. Indeed, if f has a Fourier integral representation and is even, then B (w) in (4). This holds because the integrand of B (w) is odd. Then (5) reduces to a Fourier cosine integral sin t t dw p a dt p (x) a sin (w wx) dw. w sin t dt. t () f (x) A (w) cos wx dw where A (w) p f (v) cos wv dv. Note the change in A (w): for even f the integrand is even, hence the integral from to equals twice the integral from to, just as in (7a) of Sec... Similarly, if f has a Fourier integral representation and is odd, then A (w) in (4). This is true because the integrand of A (w) is odd. Then (5) becomes a Fourier sine integral () where B (w) p f (x) B (w) sin wx dw f (v) sin wv dv.

44 56 CHAP. Fourier Analysis Note the change of B (w) to an integral from to because B (w) is even (odd times odd is even). Earlier in this section we pointed out that the main application of the Fourier integral representation is in differential equations. However, these representations also help in evaluating integrals, as the following example shows for integrals from to. EXAMPLE 3 Fig. 84. f(x) in Example 3 Laplace Integrals We shall derive the Fourier cosine and Fourier sine integrals of f (x) e kx, where x and k (Fig. 84). The result will be used to evaluate the so-called Laplace integrals. Solution. (a) From () we have e kv cos wv dv. Now, by integration by parts, A (w) p ekv cos wv dv k w k w ekv a sin wv cos wvb. k If v, the expression on the right equals k>(k w ). If v approaches infinity, that expression approaches zero because of the exponential factor. Thus >p times the integral from to gives () k>p A (w) k w. By substituting this into the first integral in () we thus obtain the Fourier cosine integral representation f (x) e kx k p cos wx k w dw (x, k ). From this representation we see that (3) cos wx k w dw p k ekx (x, k ). (b) Similarly, from () we have e kv sin wv dv. By integration by parts, B (w) p ekv sin wv dv w k w ekv a k sin wv cos wvb. w This equals w>(k w ) if v, and approaches as v :. Thus (4) w>p B (w) k w. From (4) we thus obtain the Fourier sine integral representation From this we see that f (x) e kx p w sin wx k w dw. (5) w sin wx k w dw p ekx (x, k ). The integrals (3) and (5) are called the Laplace integrals.

45 SEC..7 Fourier Integral 57 PROBLEM SET.7 6 EVALUATION OF INTEGRALS Show that the integral represents the indicated function. Hint. Use (5), (), or (); the integral tells you which one, and its value tells you what function to consider. Show your work in detail. if x. cos xw w sin xw dx d p/ if x w pe x if x FOURIER COSINE INTEGRAL REPRESENTATIONS Represent f (x) as an integral (). 7. p sin pw sin xw sin x if x p dw b w if x p cos pw w cos pw p cos x if ƒ x ƒ p cos xw dw b w if ƒ x ƒ p sin w w cos w w w 3 sin xw w 4 4 dw pex cos x if x if x f(x) b if x sin xw dw b p if x p if x p px if x sin xw dw d 4p if x if x x if x 8. f (x) b if x 9. f (x) >( x ) [x. Hint. See (3).]. f (x) b a x if x a if x a sin x if x p. f (x) b if x p e x if x a. f (x) b if x a 3. CAS EXPERIMENT. Approximate Fourier Cosine Integrals. Graph the integrals in Prob. 7, 9, and as functions of x. Graph approximations obtained by replacing with finite upper limits of your choice. Compare the quality of the approximations. Write a short report on your empirical results and observations. 4. PROJECT. Properties of Fourier Integrals (a) Fourier cosine integral. Show that () implies (a) f (ax) a A a w a b cos xw dw (a ) (Scale change) (a) (a3) A as in () A* d A dw. (b) Solve Prob. 8 by applying (a3) to the result of Prob. 7. (c) Verify (a) for f (x) if x a and f (x) if x a. (d) Fourier sine integral. Find formulas for the Fourier sine integral similar to those in (a). 5. CAS EXPERIMENT. Sine Integral. Plot Si(u) for positive u. Does the sequence of the maximum and minimum values give the impression that it converges and has the limit p>? Investigate the Gibbs phenomenon graphically. 6 FOURIER SINE INTEGRAL REPRESENTATIONS Represent f(x) as an integral (). 6. if x 7. f (x) b if x cos x if x p 8. f (x) b if x p 9. xf (x) B * (w) sin xw dw, B* da dw, x f(x) A*(w) cos xw dw, x if x a f (x) b if x a f (x) b ex if x if x. f (x) b ex if x if x

46 58 CHAP. Fourier Analysis.8 Fourier Cosine and Sine Transforms An integral transform is a transformation in the form of an integral that produces from given functions new functions depending on a different variable. One is mainly interested in these transforms because they can be used as tools in solving ODEs, PDEs, and integral equations and can often be of help in handling and applying special functions. The Laplace transform of Chap. 6 serves as an example and is by far the most important integral transform in engineering. Next in order of importance are Fourier transforms. They can be obtained from the Fourier integral in Sec..7 in a straightforward way. In this section we derive two such transforms that are real, and in Sec..9 a complex one. Fourier Cosine Transform The Fourier cosine transform concerns even functions f (x). We obtain it from the Fourier cosine integral [() in Sec..7] f (x) A(w) cos wx dw, where A (w) p f (v) cos wv dv. Namely, we set A(w) >p fĉ(w), where c suggests cosine. Then, writing v x in the formula for A(w), we have (a) fĉ(w) B p f (x) cos wx dx and (b) f (x) B p fĉ (w) cos wx dw. Formula (a) gives from f (x) a new function fĉ(w), called the Fourier cosine transform of f (x). Formula (b) gives us back f (x) from fĉ(w), and we therefore call f (x) the inverse Fourier cosine transform of fĉ(w). The process of obtaining the transform fĉ from a given f is also called the Fourier cosine transform or the Fourier cosine transform method. Fourier Sine Transform Similarly, in (), Sec..7, we set B (w) >p fŝ(w), where s suggests sine. Then, writing v x, we have from (), Sec..7, the Fourier sine transform, of f (x) given by (a) fˆs(w) B p f(x) sin wx dx,

47 SEC..8 Fourier Cosine and Sine Transforms 59 and the inverse Fourier sine transform of fˆs (w), given by (b) f (x) B p fŝ (w) sin wx dw. The process of obtaining f s (w) from f (x) is also called the Fourier sine transform or the Fourier sine transform method. Other notations are f c ( f ) fĉ, f s ( f ) fŝ and and for the inverses of and, respectively. fc fs f c f s EXAMPLE Fourier Cosine and Fourier Sine Transforms Find the Fourier cosine and Fourier sine transforms of the function k a x Fig. 85. ƒ(x) in Example Solution. f (x) b k if x a if x a From the definitions (a) and (a) we obtain by integration (Fig. 85). fĉ (w) Bp k a cos wx dx B p aw k asin w b fŝ (w) Bp k a sin wx dx B p cos aw k a b. w This agrees with formulas in the first two tables in Sec.. (where k ). Note that for f (x) k const ( x ), these transforms do not exist. (Why?) EXAMPLE Fourier Cosine Transform of the Exponential Function Find f c (e x ). Solution. By integration by parts and recursion, f c (e x ) B p e x e x cos wx dx B p w (cos wx w sin wx) ` >p. w This agrees with formula 3 in Table I, Sec.., with a. See also the next example. What did we do to introduce the two integral transforms under consideration? Actually not much: We changed the notations A and B to get a symmetric distribution of the constant >p in the original formulas () and (). This redistribution is a standard convenience, but it is not essential. One could do without it. What have we gained? We show next that these transforms have operational properties that permit them to convert differentiations into algebraic operations (just as the Laplace transform does). This is the key to their application in solving differential equations.

48 5 CHAP. Fourier Analysis Linearity, Transforms of Derivatives If f (x) is absolutely integrable (see Sec..7) on the positive x-axis and piecewise continuous (see Sec. 6.) on every finite interval, then the Fourier cosine and sine transforms of f exist. Furthermore, if f and g have Fourier cosine and sine transforms, so does af bg for any constants a and b, and by (a) f c (af bg) B p a B p [af (x) bg (x)] cos wx dx f (x) cos wx dx b B p The right side is af c ( f ) bf c (g). Similarly for f s, by (). This shows that the Fourier cosine and sine transforms are linear operations, g (x) cos wx dx. (3) (a) (b) f c (af bg) af c ( f ) bf c (g), f s (af bg) af s ( f ) bf s (g). THEOREM Cosine and Sine Transforms of Derivatives Let f (x) be continuous and absolutely integrable on the x-axis, let f r(x) be piecewise continuous on every finite interval, and let f (x) : as x :. Then (4) (a) f c { f r(x)} w f s {f(x)} B p f (), (b) f s {f r(x)} wf c {f(x)}. PROOF This follows from the definitions and by using integration by parts, namely, and similarly, f c {f r(x)} B f s {f r(x)} B p B p c f (x) cos wx ` w f (x) sin wx dx d B p f () wf s{f (x)}; p f r(x) cos wx dx f r(x) sin wx dx B p c f (x) sin wx ` w f (x) cos wx dx d wf c {f(x)}.

49 SEC..8 Fourier Cosine and Sine Transforms 5 Formula (4a) with f r instead of f gives (when f r, f s satisfy the respective assumptions for f, f r in Theorem ) hence by (4b) f c {f s(x)} wf s {f r(x)} B p f r(); (5a) f c {f s(x)} w f c {f (x)} B p f r(). Similarly, (5b) f s {f s(x)} w f s {f (x)} B p wf (). A basic application of (5) to PDEs will be given in Sec..7. For the time being we show how (5) can be used for deriving transforms. EXAMPLE 3 An Application of the Operational Formula (5) Find the Fourier cosine transform f of f (x) e ax c (e ax ), where a. Solution. By differentiation, (e ax )s a e ax ; thus a f (x) f s(x). From this, (5a), and the linearity (3a), a f c ( f ) f c ( f s) w f c ( f ) B p f r() w f c ( f ) a B p. Hence (a w )f c ( f ) a>p. The answer is (see Table I, Sec..) f c (e ax ) B p a a a w b (a ). Tables of Fourier cosine and sine transforms are included in Sec...

50 5 CHAP. Fourier Analysis PROBLEM SET.8 8 FOURIER COSINE TRANSFORM. Find the cosine transform fĉ(w) of f (x) if x, f (x) if x, f (x) if x.. Find f in Prob. from the answer fĉ. 3. Find for f (x) x if x, f (x) if x. 4. Derive formula 3 in Table I of Sec.. by integration. 5. Find for f (x) x if x, f (x) if x. 6. Continuity assumptions. Find ĝ c (w) for g (x) if x, g (x) if x. Try to obtain from it fĉ(w) for f (x) in Prob. 5 by using (5a). 7. Existence? Does the Fourier cosine transform of x sin x ( x ) exist? Of x cos x? Give reasons. 8. Existence? Does the Fourier cosine transform of f (x) k const ( x ) exist? The Fourier sine transform? 9 5 FOURIER SINE TRANSFORM 9. Find f s (e ax ), a, by integration.. Obtain the answer to Prob. 9 from (5b).. Find f for f (x) x s (w) if x, f (x) if x.. Find f s (xe x > ) from (4b) and a suitable formula in Table I of Sec Find f s (e x ) from (4a) and formula 3 of Table I in Sec Gamma function. Using formulas and 4 in Table II of Sec.., prove ( ) p [(3) in App. A3.], a value needed for Bessel functions and other applications. 5. WRITING PROJECT. Finding Fourier Cosine and Sine Transforms. Write a short report on ways of obtaining these transforms, with illustrations by examples of your own..9 Fourier Transform. Discrete and Fast Fourier Transforms In Sec..8 we derived two real transforms. Now we want to derive a complex transform that is called the Fourier transform. It will be obtained from the complex Fourier integral, which will be discussed next. Complex Form of the Fourier Integral The (real) Fourier integral is [see (4), (5), Sec..7] where A(w) p f (v) cos wv dv, Substituting A and B into the integral for f, we have f (x) p f (x) [A(w) cos wx B(w) sin wx] dw B(w) p f (v) sin wv dv. f (v)[cos wv cos wx sin wv sin wx] dv dw.

51 SEC..9 Fourier Transform. Discrete and Fast Fourier Transforms 53 By the addition formula for the cosine [(6) in App. A3.] the expression in the brackets [ Á ] equals cos (wv wx) or, since the cosine is even, cos (wx wv). We thus obtain (*) The integral in brackets is an even function of w, call it F (w), because cos (wx wv) is an even function of w, the function f does not depend on w, and we integrate with respect to v (not w). Hence the integral of F (w) from w to is times the integral of F (w) from to. Thus (note the change of the integration limit!) () We claim that the integral of the form () with sin instead of cos is zero: () f (x) p B f (v) cos (wx wv)dvr dw. f (x) p p B f (v) sin (wx wv) dvr dw. B f (v) cos (wx wv) dvr dw. This is true since sin (wx wv) is an odd function of w, which makes the integral in brackets an odd function of w, call it G (w). Hence the integral of G (w) from to is zero, as claimed. We now take the integrand of () plus i ( ) times the integrand of () and use the Euler formula [() in Sec..] (3) e ix cos x i sin x. Taking wx wv instead of x in (3) and multiplying by f (v) gives f (v) cos (wx wv) if (v) sin (wx wv) f (v)e i(wxwv). Hence the result of adding () plus i times (), called the complex Fourier integral, is (4) f (x) p f (v)e iw(xv) dv dw (i ). To obtain the desired Fourier transform will take only a very short step from here. Fourier Transform and Its Inverse Writing the exponential function in (4) as a product of exponential functions, we have (5) f (x) B f (v)e iwv dvr e iwx dw. p p The expression in brackets is a function of w, is denoted by fˆ(w), and is called the Fourier transform of f ; writing v x, we have (6) fˆ(w) p f (x)e iwx dx.

52 54 CHAP. Fourier Analysis With this, (5) becomes (7) f (x) fˆ(w)e iwx dw p and is called the inverse Fourier transform of fˆ(w). Another notation for the Fourier transform is so that fˆ f( f ), f f ( fˆ). The process of obtaining the Fourier transform f( f ) fˆ from a given f is also called the Fourier transform or the Fourier transform method. Using concepts defined in Secs. 6. and.7 we now state (without proof) conditions that are sufficient for the existence of the Fourier transform. THEOREM Existence of the Fourier Transform If f (x) is absolutely integrable on the x-axis and piecewise continuous on every finite interval, then the Fourier transform fˆ(w) of f (x) given by (6) exists. EXAMPLE Fourier Transform Find the Fourier transform of f (x) if ƒ x ƒ and f (x) otherwise. Solution. Using (6) and integrating, we obtain fˆ(w) e iwx dx p p # e iwx iw ` iwp (eiw e iw ). As in (3) we have e iw cos w i sin w, e iw cos w i sin w, and by subtraction e iw e iw i sin w. Substituting this in the previous formula on the right, we see that i drops out and we obtain the answer p sin w fˆ(w) B w. EXAMPLE Fourier Transform Find the Fourier transform f (e ax ) of f (x) e ax if x and f (x) if x ; here a. Solution. From the definition (6) we obtain by integration f (e ax ) p e ax e iwx dx p This proves formula 5 of Table III in Sec... e (aiw)x (a iw) ` x p(a iw).

53 SEC..9 Fourier Transform. Discrete and Fast Fourier Transforms 55 Physical Interpretation: Spectrum The nature of the representation (7) of f (x) becomes clear if we think of it as a superposition of sinusoidal oscillations of all possible frequencies, called a spectral representation. This name is suggested by optics, where light is such a superposition of colors (frequencies). In (7), the spectral density fˆ(w) measures the intensity of f (x) in the frequency interval between w and w w ( w small, fixed). We claim that, in connection with vibrations, the integral ƒ fˆ(w) ƒ dw can be interpreted as the total energy of the physical system. Hence an integral of ƒ fˆ(w) ƒ from a to b gives the contribution of the frequencies w between a and b to the total energy. To make this plausible, we begin with a mechanical system giving a single frequency, namely, the harmonic oscillator (mass on a spring, Sec..4) mys ky. Here we denote time t by x. Multiplication by yr gives myrys kyry. By integration, mv ky E const where v yr is the velocity. The first term is the kinetic energy, the second the potential energy, and E the total energy of the system. Now a general solution is (use (3) in Sec..4 with t x) y a cos w x b sin w x c e iw x c e iw x, w k>m where c We write simply A c e iwx (a ib )>, c c (a ib )>., B c e iwx. Then y A B. By differentiation, v yr Ar Br iw (A B). Substitution of v and y on the left side of the equation for gives E mv ky m(iw ) (A B) k(a B). Here w as just stated; hence mw Also i k>m, k., so that E k[(a B) (A B) ] kab kc e iw x c e iw x kc c k ƒ c ƒ. Hence the energy is proportional to the square of the amplitude ƒ c ƒ. As the next step, if a more complicated system leads to a periodic solution y f (x) that can be represented by a Fourier series, then instead of the single energy term ƒ c ƒ we get a series of squares ƒ c n ƒ of Fourier coefficients c n given by (6), Sec..4. In this case we have a discrete spectrum (or point spectrum ) consisting of countably many isolated frequencies (infinitely many, in general), the corresponding ƒ c n ƒ being the contributions to the total energy. Finally, a system whose solution can be represented by an integral (7) leads to the above integral for the energy, as is plausible from the cases just discussed. E

54 56 CHAP. Fourier Analysis Linearity. Fourier Transform of Derivatives New transforms can be obtained from given ones by using THEOREM Linearity of the Fourier Transform The Fourier transform is a linear operation; that is, for any functions f (x) and g(x) whose Fourier transforms exist and any constants a and b, the Fourier transform of af bg exists, and (8) f(af bg) af ( f ) bf (g). PROOF This is true because integration is a linear operation, so that (6) gives f{af (x) bg (x)} p a p [af(x) bg (x)] e iwx dx f (x)e iwx dx b af{f (x)} bf{g (x)}. p g (x)e iwx dx In applying the Fourier transform to differential equations, the key property is that differentiation of functions corresponds to multiplication of transforms by iw: THEOREM 3 Fourier Transform of the Derivative of f (x) Let f (x) be continuous on the x-axis and f (x) : as ƒ x ƒ :. Furthermore, let f r(x) be absolutely integrable on the x-axis. Then (9) f {f r(x)} iwf {f (x)}. PROOF From the definition of the Fourier transform we have Integrating by parts, we obtain f{f r(x)} p f{f r(x)} p Bf (x)eiwx ` f r(x)e iwx dx. Since f (x) : as ƒ x ƒ :, the desired result follows, namely, (iw) f (x)e iwx dxr. f{f r(x)} iw f{f (x)}.

55 SEC..9 Fourier Transform. Discrete and Fast Fourier Transforms 57 Two successive applications of (9) give f ( f s) iwf ( f r) (iw) f ( f ). Since (iw) w, we have for the transform of the second derivative of f () f{f s(x)} w f{f (x)}. Similarly for higher derivatives. An application of () to differential equations will be given in Sec..6. For the time being we show how (9) can be used to derive transforms. EXAMPLE 3 Application of the Operational Formula (9) Find the Fourier transform of xe x from Table III, Sec.. Solution. We use (9). By formula 9 in Table III f (xe x ) f{ (e x )r} f{(ex )r} iwf(ex ) iw ew >4 iw ew >4. Convolution The convolution f * g of functions f and g is defined by () h (x) ( f * g) (x) f (p) g (x p) dp f (x p)g (p) dp. The purpose is the same as in the case of Laplace transforms (Sec. 6.5): taking the convolution of two functions and then taking the transform of the convolution is the same as multiplying the transforms of these functions (and multiplying them by p): THEOREM 4 Convolution Theorem Suppose that f (x) and g(x) are piecewise continuous, bounded, and absolutely integrable on the x-axis. Then () f ( f * g) p f ( f ) f (g).

56 58 CHAP. Fourier Analysis PROOF By the definition, f ( f * g) p f (p) g (x p) dp e iwx dx. An interchange of the order of integration gives f ( f * g) p f (p) g (x p) e iwx dx dp. Instead of x we now take x p q as a new variable of integration. Then x p q and f ( f * g) p f (p) g (q) e iw (pq) dq dp. This double integral can be written as a product of two integrals and gives the desired result f ( f * g) p f (p)e iwp dp g (q) e iwq dq [p f ( f )][p f (g)] p f ( f ) f (g). p By taking the inverse Fourier transform on both sides of (), writing fˆ f ( f ) and ĝ f (g) as before, and noting that p and > p in () and (7) cancel each other, we obtain (3) ( f * g) (x) fˆ(w)ĝ (w)e iwx dw, a formula that will help us in solving partial differential equations (Sec..6). Discrete Fourier Transform (DFT), Fast Fourier Transform (FFT) In using Fourier series, Fourier transforms, and trigonometric approximations (Sec..6) we have to assume that a function f (x), to be developed or transformed, is given on some interval, over which we integrate in the Euler formulas, etc. Now very often a function f (x) is given only in terms of values at finitely many points, and one is interested in extending Fourier analysis to this case. The main application of such a discrete Fourier analysis concerns large amounts of equally spaced data, as they occur in telecommunication, time series analysis, and various simulation problems. In these situations, dealing with sampled values rather than with functions, we can replace the Fourier transform by the so-called discrete Fourier transform (DFT) as follows.

57 SEC..9 Fourier Transform. Discrete and Fast Fourier Transforms 59 Let f (x) be periodic, for simplicity of period p. We assume that N measurements of f (x) are taken over the interval x p at regularly spaced points (4) x k pk N, k,, Á, N. We also say that f (x) is being sampled at these points. We now want to determine a complex trigonometric polynomial (5) N q (x) a c n e inx k n that interpolates f (x) at the nodes (4), that is, q (x k ) f (x k ), written out, with f k denoting f (x k ), (6) N f k f (x k ) q (x k ) a c n e inx k, n k,, Á, N. Hence we must determine the coefficients c, Á, c N such that (6) holds. We do this by an idea similar to that in Sec.. for deriving the Fourier coefficients by using the orthogonality of the trigonometric system. Instead of integrals we now take sums. Namely, we multiply (6) by e imx k (note the minus!) and sum over k from to N. Then we interchange the order of the two summations and insert x k from (4). This gives (7) N a f k e imx N k a k N a k n c n e i(nm)x k a N n N c n a e i (nm) pk>n. k Now e i (nm)pk>n [e i (nm)p>n ] k. We donote [ Á ] by r. For n m we have r e. The sum of these terms over k equals N, the number of these terms. For n m we have r and by the formula for a geometric sum [(6) in Sec. 5. with q r and n N ] N a r k r N r k because r N ; indeed, since k, m, and n are integers, r N e i(nm)pk cos pk(n m) i sin pk(n m). This shows that the right side of (7) equals c m N. Writing n for m and dividing by N, we thus obtain the desired coefficient formula (8*) c n N a N k f k e inx k f k f (x k ), n,, Á, N. Since computation of the c n (by the fast Fourier transform, below) involves successive halfing of the problem size N, it is practical to drop the factor >N from and define the c n

58 53 CHAP. Fourier Analysis discrete Fourier transform of the given signal to be the vector fˆ [ fˆ Á f [ f Á f N ] T fˆn ] with components (8) fˆn Nc n a N k f k e inx k, f k f (x k ), n, Á, N. This is the frequency spectrum of the signal. In vector notation, fˆ F N f, where the N N Fourier matrix F N [e nk ] has the entries [given in (8)] (9) e nk e inx k e pink>n w nk, w w N e pi>n, where n, k, Á, N. EXAMPLE 4 Discrete Fourier Transform (DFT). Sample of N 4 Values Let N 4 measurements (sample values) be given. Then w e pi>n e pi> iand thus w nk (i) nk. Let the sample values be, say f [ 4 9] T. Then by (8) and (9), w w w w 4 w w w w 3 i i 4 8i () fˆ F 4 f E U f E U E U E U. w w w 4 w w w 3 w 6 w 9 i i 9 4 8i From the first matrix in () it is easy to infer what F N looks like for arbitrary N, which in practice may be or more, for reasons given below. From the DFT (the frequency spectrum) fˆ F N f we can recreate the given signal fˆ F, as we shall now prove. Here F and its complex conjugate F N N f N N [wnk ] satisfy (a) F N F N F N F N NI where I is the N N unit matrix; hence has the inverse F N (b) F N N F N. PROOF We prove (). By the multiplication rule (row times column) the product matrix G N F N F N [g jk ] in (a) has the entries g jk Row j of F N times Column k of F N. That is, writing W w j w k, we prove that g jk (w j w k ) (w j w k ) Á (w j w k ) N W W Á if j k W N b N if j k.

59 SEC..9 Fourier Transform. Discrete and Fast Fourier Transforms 53 Indeed, when j k, then w k w k (ww) k (e pi>n e pi>n ) k k, so that the sum of these N terms equals N; these are the diagonal entries of G N. Also, when j k, then W and we have a geometric sum (whose value is given by (6) in Sec. 5. with q W and n N ) W W Á W N WN W because W N (w j w k ) N (e pi ) j (e pi ) k j # k. We have seen that fˆ is the frequency spectrum of the signal f (x). Thus the components fˆn of fˆ give a resolution of the p-periodic function f (x) into simple (complex) harmonics. Here one should use only n s that are much smaller than N>, to avoid aliasing. By this we mean the effect caused by sampling at too few (equally spaced) points, so that, for instance, in a motion picture, rotating wheels appear as rotating too slowly or even in the wrong sense. Hence in applications, N is usually large. But this poses a problem. Eq. (8) requires O (N) operations for any particular n, hence O (N ) operations for, say, all n N>. Thus, already for sample points the straightforward calculation would involve millions of operations. However, this difficulty can be overcome by the so-called fast Fourier transform (FFT), for which codes are readily available (e.g., in Maple). The FFT is a computational method for the DFT that needs only O (N) log N operations instead of O (N ). It makes the DFT a practical tool for large N. Here one chooses N p ( p integer) and uses the special form of the Fourier matrix to break down the given problem into smaller problems. For instance, when N, those operations are reduced by a factor >log. The breakdown produces two problems of size M N>. This breakdown is possible because for N M we have in (9) w N w M (e pi>n ) e 4pi>(M) e pi>(m) w M. The given vector f [ f Á fn ] T is split into two vectors with M components each, namely, f ev [ f f Á fn ] T containing the even components of f, and f od [ f f 3 Á fn ] T containing the odd components of f. For f ev and f od we determine the DFTs and fˆev [ f êv, f êv, Á f êv,n ] T F M f ev fˆod [ f ôd, f ôd,3 Á f ôd,n ] T F M f od involving the same M M matrix F M. From these vectors we obtain the components of the DFT of the given vector f by the formulas () (a) fˆn f êv,n w n N f ôd,n n, Á, M (b) fˆnm f êv,n w n N f ôd,n n, Á, M.

60 53 CHAP. Fourier Analysis For N p this breakdown can be repeated p times in order to finally arrive at N> problems of size each, so that the number of multiplications is reduced as indicated above. We show the reduction from N 4 to M N> and then prove (). EXAMPLE 5 Fast Fourier Transform (FFT). Sample of N 4 Values When N 4, then w w as in Example 4 and M N>, hence w w M e pi> e pi N i. Consequently, fˆev c fˆ fˆ fˆod c fˆ fˆ3 d F f ev c dcf d F f od c dcf f d c f f f f d d. f f 3 f 3 d c f f 3 From this and (a) we obtain Similarly, by (b), fˆ fˆev, w N fˆod, ( f f ) ( f f 3 ) f f f f 3 fˆ fˆev, w N fˆod, ( f f ) i( f f 3 ) f if f if 3. fˆ fˆev, w N fˆod, ( f f ) ( f f 3 ) f f f f 3 fˆ3 fˆev, w N fˆod, ( f f ) (i)( f f 3 ) f if f if 3. This agrees with Example 4, as can be seen by replacing,, 4, 9 with f, f, f, f 3. We prove (). From (8) and (9) we have for the components of the DFT N fˆn a w kn N f k. Splitting into two sums of M N> terms each gives M fˆn a w kn M N f k a w (k)n N f k. k k We now use w N w M and pull out from under the second sum, obtaining w N n k (3) M fˆn a w kn M n M f ev,k w N a w kn M f od,k. k k The two sums are f ev,n and f od,n, the components of the half-size transforms Ff ev and Ff od. Formula (a) is the same as (3). In (b) we have n M instead of n. This causes a sign changes in (3), namely before the second sum because w N n w M N e pim>n e pi> e pi. This gives the minus in (b) and completes the proof.

61 SEC..9 Fourier Transform. Discrete and Fast Fourier Transforms 533 PROBLEM SET.9. Review in complex. Show that >i i, e ix 7 USE OF TABLE III IN SEC... cos x i sin x, e ix e ix cos x, e ix e ix OTHER METHODS i sin x, e ikx cos kx i sin kx.. Find f ( f (x)) for f (x) xe x if x, f (x) if x, by (9) in the text and formula 5 in Table III FOURIER TRANSFORMS BY (with a ). Hint. Consider xe x and e x. INTEGRATION 3. Obtain f(e x > ) from Table III. Find the Fourier transform of f (x) (without using Table 4. In Table III obtain formula 7 from formula 8. III in Sec..). Show details. 5. In Table III obtain formula from formula. 6. TEAM PROJECT. Shifting (a) Show that if f (x). f (x) e eix if x has a Fourier transform, so does f (x a), and otherwise f{f (x a)} e iwa f{f (x)}. (b) Using (a), obtain formula in Table III, Sec.., if a x b 3. f (x) e from formula. otherwise (c) Shifting on the w-axis. Show that if fˆ (w) is the Fourier transform of f (x), then fˆ (w a) is the Fourier 4. f (x) e ekx if x (k ) transform of e iax f (x). if x (d) Using (c), obtain formula 7 in Table III from and formula 8 from. 5. f (x) e ex if a x a 7. What could give you the idea to solve Prob. by using otherwise the solution of Prob. 9 and formula (9) in the text? Would this work? 6. f (x) e ƒ x ƒ ( x ) 8 5 DISCRETE FOURIER TRANSFORM x if x a 7. f (x) e 8. Verify the calculations in Example 4 of the text. otherwise 9. Find the transform of a general signal f [ f f f 3 f 4 ] T of four values. 8. f (x) e xex if x. Find the inverse matrix in Example 4 of the text and otherwise use it to recover the given signal.. Find the transform (the frequency spectrum) of a ƒ x ƒ if x general signal of two values [ f 9. f (x) e f ] T. otherwise. Recreate the given signal in Prob. from the frequency spectrum obtained. x if x 3. Show that for a signal of eight sample values,. f (x) e otherwise w e i>4 ( i)>. Check by squaring. 4. Write the Fourier matrix F for a sample of eight values if x explicitly.. f (x) μ if x 5. CAS Problem. Calculate the inverse of the 8 8 Fourier matrix. Transform a general sample of eight otherwise values and transform it back to the given data.

62 534 CHAP. Fourier Analysis. Tables of Transforms Table I. Fourier Cosine Transforms See () in Sec..8. f (x) fˆc (w) f c ( f ) if x a e otherwise Bp sin aw w (a) x ((a) see App. A3..) Bp w a cos ap a ( a ) 3 eax (a ) Bp a a a w b 4 e x > e w > 5 e ax (a ) a ew >(4a) 6 x n e ax (a ) n! Bp (a w n Re (a iw)n ) Re Real part 7 e cos x if x a otherwise sin a( w) p c w sin a( w) d w 8 cos (ax ) (a ) cos aw a 4a p 4 b 9 sin (ax ) (a ) cos aw a 4a p 4 b sin ax p (a ) ( u(w a)) (See Sec. 6.3.) x B e x sin x x p arctan w J (ax) (a ) ( u(w a)) (See Secs. 5.5, 6.3.) Bp a w

63 SEC.. Tables of Transforms 535 Table II. Fourier Sine Transforms See (5) in Sec..8. f (x) fˆs (w) f s ( f ) if x a e otherwise Bp cos aw c w d > x > w 3 >x 3> w (a) 4 x ((a) see App. A3..) Bp w a sin ap a ( a ) 5 eax (a ) Bp a w a w b 6 e ax x (a ) Bp arctan w a 7 x n e ax (a ) n! Bp (a w n Im (a iw)n ) Im Imaginary part 8 xe x > we w > 9 xe ax (a ) w >4a ew 3> (a) sin x if x a e otherwise sin a( w) p c w sin a( w) d w cos ax p (a ) u (w a) (See Sec. 6.3.) x B arctan a (a ) p x sin aw w eaw

64 536 CHAP. Fourier Analysis Table III. Fourier Transforms See (6) in Sec..9. f (x) fˆ(w) f( f ) if b x b e otherwise Bp sin bw w if b x c e otherwise e ibw e icw iwp 3 4 x a (a ) x μ x b if x b if b x b otherwise B p e aƒwƒ a e ibw e ibw pw 5 e eax if x otherwise (a ) p(a iw) 6 e eax if b x c otherwise e (aiw)c e (aiw)b p(a iw) 7 e eiax if b x b otherwise Bp sin b(w a) w a 8 e eiax if b x c otherwise i e ib(aw) e ic(aw) p a w 9 e ax (a ) a ew >4a sin ax x (a ) B p if ƒ w ƒ a; if ƒ w ƒ a

65 Chapter Review Questions and Problems 537 CHAPTER REVIEW QUESTIONS AND PROBLEMS. What is a Fourier series? A Fourier cosine series? A half-range expansion? Answer from memory.. What are the Euler formulas? By what very important idea did we obtain them? 3. How did we proceed from p-periodic to generalperiodic functions? 4. Can a discontinuous function have a Fourier series? A Taylor series? Why are such functions of interest to the engineer? 5. What do you know about convergence of a Fourier series? About the Gibbs phenomenon? 6. The output of an ODE can oscillate several times as fast as the input. How come? 7. What is approximation by trigonometric polynomials? What is the minimum square error? 8. What is a Fourier integral? A Fourier sine integral? Give simple examples. 9. What is the Fourier transform? The discrete Fourier transform?. What are Sturm Liouville problems? By what idea are they related to Fourier series? FOURIER SERIES. In Probs., 3, 6, find the Fourier series of f (x) as given over one period and sketch f (x) and partial sums. In Probs., 4, 5, 7 9 give answers, with reasons. Show your work detail. if x. f (x) e if x. Why does the series in Prob. have no cosine terms? if x 3. f (x) e x if x 4. What function does the series of the cosine terms in Prob. 3 represent? The series of the sine terms? 5. What function do the series of the cosine terms and the series of the sine terms in the Fourier series of e x (5 x 5) represent? 6. f (x) ƒ x ƒ (p x p) 7. Find a Fourier series from which you can conclude that /3 /5 /7 Á p/4. 8. What function and series do you obtain in Prob. 6 by (termwise) differentiation? 9. Find the half-range expansions of f (x) x ( x ).. f (x) 3x (p x p) GENERAL SOLUTION Solve, ys v y r (t), where ƒ v ƒ,,, Á, r (t) is p-periodic and.. r (t) 3t (p t p) r (t) ƒ t ƒ (p t p) 3 5 MINIMUM SQUARE ERROR 3. Compute the minimum square error for f (x) x>p (p x p) and trigonometric polynomials of degree N, Á, How does the minimum square error change if you multiply f (x) by a constant k? 5. Same task as in Prob. 3, for f (x) ƒ x ƒ >p (p x p). Why is E* now much smaller (by a factor, approximately!)? 6 3 FOURIER INTEGRALS AND TRANSFORMS Sketch the given function and represent it as indicated. If you have a CAS, graph approximate curves obtained by replacing with finite limits; also look for Gibbs phenomena. 6. f (x) x if x and otherwise; by the Fourier sine transform 7. f (x) x if x and otherwise; by the Fourier integral 8. f (x) kx if a x b and otherwise; by the Fourier transform 9. f (x) x if x a and otherwise; by the Fourier cosine transform 3. f (x) e x if x and otherwise; by the Fourier transform

66 538 CHAP. Fourier Analysis SUMMARY OF CHAPTER Fourier Analysis. Partial Differential Equations (PDEs) Fourier series concern periodic functions f (x) of period p L, that is, by definition f (x p) f (x) for all x and some fixed p ; thus, f (x np) f (x) for any integer n. These series are of the form () f (x) a a aa n cos np (Sec..) L x b n sin np L xb with coefficients, called the Fourier coefficients of f (x), given by the Euler formulas (Sec..) () a, L L f (x) dx L n b n L L f (x) sin npx L dx where n,, Á. For period p we simply have (Sec..) L a n L L f (x) cos npx L dx L (*) f (x) a a n (a n cos nx b n sin nx) with the Fourier coefficients of f (x) (Sec..) a p p p f (x) dx, a n p p f (x) cos nx dx, b n p p f (x) sin nx dx. p Fourier series are fundamental in connection with periodic phenomena, particularly in models involving differential equations (Sec..3, Chap, ). If f (x) is even [ f (x) f (x)] or odd [ f (x) f (x)], they reduce to Fourier cosine or Fourier sine series, respectively (Sec..). If f (x) is given for x L only, it has two half-range expansions of period L, namely, a cosine and a sine series (Sec..). The set of cosine and sine functions in () is called the trigonometric system. Its most basic property is its orthogonality on an interval of length L; that is, for all integers m and n m we have p L cos mpx L L cos npx L dx, L sin mpx L sin npx dx L L and for all integers m and n, L cos mpx L sin npx dx. L L This orthogonality was crucial in deriving the Euler formulas ().

67 Summary of Chapter 539 Partial sums of Fourier series minimize the square error (Sec..4). Replacing the trigonometric system in () by other orthogonal systems first leads to Sturm Liouville problems (Sec..5), which are boundary value problems for ODEs. These problems are eigenvalue problems and as such involve a parameter l that is often related to frequencies and energies. The solutions to Sturm Liouville problems are called eigenfunctions. Similar considerations lead to other orthogonal series such as Fourier Legendre series and Fourier Bessel series classified as generalized Fourier series (Sec..6). Ideas and techniques of Fourier series extend to nonperiodic functions f (x) defined on the entire real line; this leads to the Fourier integral (3) f (x) [ A (w) cos wx B (w) sin wx] dw (Sec..7) where (4) A (w) p f (v) cos wv dv, B (w) p f (v) sin wv dv or, in complex form (Sec..9), (5) where (6) f (x) p fˆ (w)e iwx dw fˆ (w) p f (x)e iwx dx. (i ) Formula (6) transforms f (x) into its Fourier transform fˆ (w), and (5) is the inverse transform. Related to this are the Fourier cosine transform (Sec..8) (7) fˆc (w) B p and the Fourier sine transform (Sec..8) p (8) fˆs(w) f (x) sin wx dx. B f (x) cos wx dx The discrete Fourier transform (DFT) and a practical method of computing it, called the fast Fourier transform (FFT), are discussed in Sec..9.

68 CHAPTER Partial Differential Equations (PDEs) 54 A PDE is an equation that contains one or more partial derivatives of an unknown function that depends on at least two variables. Usually one of these deals with time t and the remaining with space (spatial variable(s)). The most important PDEs are the wave equations that can model the vibrating string (Secs..,.3,.4,.) and the vibrating membrane (Secs..8,.9,.), the heat equation for temperature in a bar or wire (Secs..5,.6), and the Laplace equation for electrostatic potentials (Secs..6,.,.). PDEs are very important in dynamics, elasticity, heat transfer, electromagnetic theory, and quantum mechanics. They have a much wider range of applications than ODEs, which can model only the simplest physical systems. Thus PDEs are subjects of many ongoing research and development projects. Realizing that modeling with PDEs is more involved than modeling with ODEs, we take a gradual, well-planned approach to modeling with PDEs. To do this we carefully derive the PDE that models the phenomena, such as the one-dimensional wave equation for a vibrating elastic string (say a violin string) in Sec.., and then solve the PDE in a separate section, that is, Sec..3. In a similar vein, we derive the heat equation in Sec..5 and then solve and generalize it in Sec..6. We derive these PDEs from physics and consider methods for solving initial and boundary value problems, that is, methods of obtaining solutions which satisfy the conditions required by the physical situations. In Secs..7 and. we show how PDEs can also be solved by Fourier and Laplace transform methods. COMMENT. Numerics for PDEs is explained in Secs..4.7, which, for greater teaching flexibility, is designed to be independent of the other sections on numerics in Part E. Prerequisites: Linear ODEs (Chap. ), Fourier series (Chap. ). Sections that may be omitted in a shorter course:.7,... References and Answers to Problems: App. Part C, App... Basic Concepts of PDEs A partial differential equation (PDE) is an equation involving one or more partial derivatives of an (unknown) function, call it u, that depends on two or more variables, often time t and one or several variables in space. The order of the highest derivative is called the order of the PDE. Just as was the case for ODEs, second-order PDEs will be the most important ones in applications.