Students Solutions Manual PARTIAL DIFFERENTIAL EQUATIONS. with FOURIER SERIES and BOUNDARY VALUE PROBLEMS. NAKHLÉ H.ASMAR University of Missouri

Transcription

1 Students Solutions Manual PARTIAL DIFFERENTIAL EQUATIONS with FOURIER SERIES and BOUNDARY VALUE PROBLEMS Third Edition NAKHLÉ H.ASMAR University of Missouri

2 Contents A Preview of Applications and Techniques. What Is a Partial Differential Equation?. Solving and Interpreting a Partial Differential Equation 3 Fourier Series 9. Periodic Functions 9. Fourier Series 5.3 Fourier Series of Functions with Arbitrary Periods.4 Half-Range Expansions: The Cosine and Sine Series 9.5 Mean Square Approximation and Parseval s Identity 3.6 Complex Form of Fourier Series 36.7 Forced Oscillations 4 Supplement on Convergence.9 Uniform Convergence and Fourier Series 47. Dirichlet Test and Convergence of Fourier Series 48 3 Partial Differential Equations in Rectangular Coordinates Partial Differential Equations in Physics and Engineering Solution of the One Dimensional Wave Equation: The Method of Separation of Variables D Alembert s Method The One Dimensional Heat Equation Heat Conduction in Bars: Varying the Boundary Conditions The Two Dimensional Wave and Heat Equations Laplace s Equation in Rectangular Coordinates Poisson s Equation: The Method of Eigenfunction Expansions 9 3. Neumann and Robin Conditions 94

3 Contents iii 4 Partial Differential Equations in Polar and Cylindrical Coordinates The Laplacian in Various Coordinate Systems Vibrations of a Circular Membrane: Symmetric Case Vibrations of a Circular Membrane: General Case Laplace s Equation in Circular Regions Laplace s Equation in a Cylinder The Helmholtz and Poisson Equations 9 Supplement on Bessel Functions 4.7 Bessel s Equation and Bessel Functions Bessel Series Expansions Integral Formulas and Asymptotics for Bessel Functions 4 5 Partial Differential Equations in Spherical Coordinates 4 5. Preview of Problems and Methods 4 5. Dirichlet Problems with Symmetry Spherical Harmonics and the General Dirichlet Problem The Helmholtz Equation with Applications to the Poisson, Heat, and Wave Equations 53 Supplement on Legendre Functions 5.5 Legendre s Differential Equation Legendre Polynomials and Legendre Series Expansions 6 6 Sturm Liouville Theory with Engineering Applications Orthogonal Functions Sturm Liouville Theory The Hanging Chain Fourth Order Sturm Liouville Theory The Biharmonic Operator Vibrations of Circular Plates 78

4 iv Contents 7 The Fourier Transform and Its Applications The Fourier Integral Representation The Fourier Transform The Fourier Transform Method The Heat Equation and Gauss s Kernel 7.5 A Dirichlet Problem and the Poisson Integral Formula 7.6 The Fourier Cosine and Sine Transforms Problems Involving Semi-Infinite Intervals Generalized Functions 7.9 The Nonhomogeneous Heat Equation Duhamel s Principle 35 8 The Laplace and Hankel Transforms with Applications 38 A 8. The Laplace Transform Further Properties of the Laplace transform The Laplace Transform Method The Hankel Transform with Applications 6 Green s Functions and Conformal Mappings 68. Green s Theorem and Identities 68. Harmonic Functions and Green s Identities 7.3 Green s Functions 74.4 Green s Functions for the Disk and the Upper Half-Plane 76.5 Analytic Functions 77.6 Solving Dirichlet Problems with Conformal Mappings 86.7 Green s Functions and Conformal Mappings 96 Ordinary Differential Equations: Review of Concepts and Methods A. Linear Ordinary Differential Equations A98 A. Linear Ordinary Differential Equations with Constant Coefficients A38 A.3 Linear Ordinary Differential Equations with Nonconstant Coefficients A3 A.4 The Power Series Method, Part I A333 A.5 The Power Series Method, Part II A34 A.6 The Method of Frobenius A348 A98

5 Section. What Is a Partial Differential Equation? Solutions to Exercises.. If u and u are solutions of (, then u t + u x and u t + u x. Since taking derivatives is a linear operation, we have t (c u + c u + x (c u u + c u c t + c u t + c u x + c u x c {( }}{ u t + u x +c {( }} { u t + + u, x showing that c u + c u is a solution of (. 3. (a General solution of (: u(x, t f(x t. On the t-axis (x : u(, t t f( t f( t. Hence f(t t and so u(x, t f(x t (x t t x. 5. Let α ax + bt, β cx + dt, then Recalling the equation, we obtain u x u α α x + u β β x a u α + c u β u u α t α t + u β β t b u α + d u β. u t u (b a u + (d c u x α β. Let a, b, c, d. Then u u f(β u(x, t f(x + t, α where f is an arbitrary differentiable function (of one variable. 7. Let α ax + bt, β cx + dt, then The equation becomes Let a, b, c, d. Then u x u α α x + u β β x a u α + c u β u u α t α t + u β β t b u α + d u β. (b a u + (d c u α β. u β u β. Solving this ordinary differential equation in β, we get u β + f(α or u(x, t x + f(x + t. 9. (a The general solution in Exercise 5 is u(x, t f(x + t. When t, we get u(x, f(x /(x +. Thus u(x, t f(x + t (x + t +. (c As t increases, the wave f(x +x moves to the left.

6 Chapter A Preview of Applications and Techniques - - Figure for Exercise 9(b. 3. The characteristic curves are obtained by solving dy dx x y 3 x3 + C y 3 x3 C. Let φ(x, y y 3 x3. The characteristic curves are the level curves of φ. The solution of is of the form u(x, y f(φ(x, y f(y 3 x3, where f is a differentiable function of one variable. 3. To find the characteristic curves, solve dy sin x. Hence y cosx + dx C or y + cos x C. Thus the solution of the partial differential equation is u(x, y f (y + cosx. To verify the solution, we use the chain rule and get u x sin xf (y + cosx and u y f (y + cos x. Thus u x + sin xu y, as desired.

7 Section. Solving and Interpreting a Partial Differential Equation 3 Exercises.. We have So t ( u t t ( v x and x ( v t x u t v t x and v x t u x. ( u. x Assuming that v t x v x t, it follows that u u, which is the one dimensional t x wave equation with c. A similar argument shows that v is a solution of the one dimensional wave equation. 3. u xx F (x + ct + G (x + ct, u tt c F (x + ct + c G(x ct. So u tt c u xx, which is the wave equation. 5. (a We have u(x, t F(x + ct + G(x ct. To determine F and G, we use the initial data: u(x, + x F(x + G(x + x ; ( u t (x, cf (x cg (x F (x G (x F(x G(x + C, ( where C is an arbitrary constant. Plugging this into (, we find G(x + C G(x [ ] + x + x C ; and from ( Hence F(x u(x, t F(x + ct + G(x ct [ ] + x + C. [ ] + (x + ct +. + (x ct 7. We have u(x, t F(x + ct + G(x ct. To determine F and G, we use the initial data: u(x, F(x + G(x ; ( u t (x, xe x cf (x cg (x xe x cf(x cg(x xe x dx e x + C F(x G(x e x c + C, ( where we rewrote C/c as C to denote the arbitrary constant. Adding ( and (, we find F(x e x + C F(x [ ] e x + C ; c c and from ( G(x c [ ] e x + C. Hence u(x, t F(x + ct + G(x ct c [ e (x+ct e (x ct].

8 4 Chapter A Preview of Applications and Techniques 8. We 9. As the hint suggests, we consider two separate problems: The problem in Exercise 5 and the one in Exercise 7. Let u (x, t denote the solution in Exercise 5 and u (x, t the solution in Exercise 7. It is straightforward to verify that u u + u is the desired solution. Indeed, because of the linearity of derivatives, we have u tt (u tt + (u tt c (u xx + c (u xx, because u and u are solutions of the wave equation. But c (u xx + c (u xx c (u + u xx u xx and so u tt c u xx, showing that u is a solution of the wave equation. Now u(x, u (x, +u (x, /(+x +, because u (x, /(+x and u (x,. Similarly, u t (x, xe x ; thus u is the desired solution. The explicit formula for u is u(x, t. We have so [ + (x + ct + + (x ct ] + [ e (x+ct e (x ct]. c V x ( L I x t RI L I x t R I x ; I V C x t V GV t [ ] I C x + GV ; V t [ ] I C t x + G V. t To check that V verifies (, we start with the right side LC V t V + (RC + LG t + RGV LC [ I C t x + G V t L I t x R I x LG C L I t x R I x V x, ] + (RC + LG C {[ }}{ C V t + I ] x + GV [ ] I x + GV + RGV which shows that V satisfies (. To show that I satisfies (, you can proceed as we did for V or you can note that the equations that relate I and V are interchanged if we interchange L and C, and R and G. However, ( remains unchanged if we interchange L and C, and R and G. So I satisfies ( if and only if V satisfies (. 3. The function being graphed is u(x, t sin x cost sinx cost + sin 3x cos3t. 3 In frames, 4, 6, and 8, t m 4, where m, 3, 5, and 7. Plugging this into u(x, t, we find u(x, t sin x cos m 4 m sin x cos + 3m sin 3x cos 3 4. For m, 3, 5, and 7, the second term is, because cos m. Hence at these times, we have, for, m, 3, 5, and 7, u(x, m 4 sin x cost + sin 3x cos3t. 3

9 Section. Solving and Interpreting a Partial Differential Equation 5 To say that the graph of this function is symmetric about x / is equivalent to the assertion that, for < x < /, u(/ + x, m 4 u(/ x, m. Does this 4 equality hold? Let s check: u(/ + x, m 4 m sin (x + /cos 4 + 3m sin3(x + /cos 3 4 cos x cos m 4 3m cos3x cos 3 4, where we have used the identities sin (x + / cos x and sin (x + / cos 3x. Similalry, u(/ x, m 4 m sin (/ xcos 4 + 3m sin3(/ xcos 3 4 cos x cos m 4 3m cos3x cos 3 4. So u(/ + x, m 4 u(/ x, m 4, as expected. 5. Since the initial velocity is, from (, we have u(x, t n The initial condition u(x, f(x n b n sin nx L b n sin nx L cos cnt L. x sin L + 3x 4 sin L sin x L. implies that The equation is satisfied with the choice b, b, and all other b n s are zero. This yields the solution u(x, t sin x L cos ct L. Note that the condition u t (x, is also satisfied. 6. Since the initial velocity is, from (, we have u(x, t n The initial condition u(x, f(x n b n sin nx L cos cnt L. x sin L + 3x 4 sin L b n sin nx L x sin L + 3x sin 4 L. implies that Clearly, this equation is satisfied with the choice b, b 3 4, and all other b n s are zero. This yields the solution u(x, t sin x L ct cos L + 3x 3ct sin cos 4 L L. Note that the condition u t (x, is also satisfied. 7. Same reasoning as in the previous exercise, we find the solution u(x, t sin x L ct cos L + 3x 3ct sin cos 4 L L + 7x 7ct sin cos 5 L L. 9. Reasoning as in the previous exercise, we satrt with the solution u(x, t n b n sin nx L sin cnt L.

10 6 Chapter A Preview of Applications and Techniques The initial condition u(x, is clearly satisfied. To satisfy the second intial condition, we must have [ ( 3x sin 4 L ] 6x sin b n sin nx cnt sin L t L L n t cn L b n sin nx L. Thus n 4 3c L b 3 b 3 L c ; 6c L b 6 b 6 L 6c ; and all other b 8 n are. Thus u(x, t L c sin 3x L 3ct sin L L 6x 6ct sin sin 6c L L.. (a We have to show that u(, t is a constant for all t >. With c L, we have u(x, t sin x cost u(/, t sin cost for all t >. (b One way for x /3 not to move is to have u(x, t sin 3x cos3t. This is the solution that corresponds to the initial condition u(x, sin 3x and u t (x,. For this solution, we also have that x /3 does not move for all t.. (a Reasoning as in Exercise 7, we find the solution to be u(x, t sin x cost + sin 4x cos4t. 4 (b We used Mathematica to plot the shape of the string at times t to t by increments of.. The string returns to some of its previous shapes. For example, when t. and when t.9, the string has the same shape. u x_, t_ Sin Pix Cos Pit 4Sin 4Pix Cos 4Pit tt Table u x, t, t,,,. ; Plot Evaluate tt, x,, Cos t Sin x 4 Cos 4 t Sin 4 x The point x / does not move. This is clear: If we put x / in the solution, we obtain u(/, t for all t.

11 Section. Solving and Interpreting a Partial Differential Equation 7 3. The solution is u(x, t sin x cost. The motions of the points x /4, /3, and 3/4 are illustrated by the following graphs. Note that the point x / does not move, so the graph that describes its motion is identically. u x_, t_ Sin Pix Cos Pit Plot u 4, t, u 3, t, u 3 4, t, t,, Pi /4.5 / /4 In each case, we have a cosine wave, namely u(x, t sin x cost, scaled by a factor sin x. 4. The solution in Exercise is u(x, t sin x cost + sin 4x cos4t. 4 The motions of the points x /4, /3, /, and 3/4 are illustrated by the following graphs. As in the previous exercise, the point x / does not move, so the graph that describes its motion is identically. u x_, t_ Sin Pix Cos Pit 4Sin 4Pix Cos 4Pit Plot u 4, t, u 3, t, u 3 4, t, t,, Pi Cos t Sin x 4 Cos 4 t Sin 4 x Unlike the previous exercises, here the motion of a point is not always a cosine wave. In fact, it is the sum of two scaled cosine waves: u(x, t sin x cos t + 4 sin 4x cos 4t.

12 8 Chapter A Preview of Applications and Techniques 5. The solution ( is u(x, t sin x L Its initial conditions at time t 3L c are u(x, 3L ( x c sin c L cos L 3L c and cos ct L. sin x L cos 3 ; u 3L (x, t c c L sin x ( c L sin L 3L c x sin c L L sin 3 c L sin x L. 7. (a The equation is equivalent to r u t κ u r x u. The solution of this equation follows from Exercise 8, Section., by taking a r and b κ r. Thus u(x, t f( r x + κ r te rt. Note that this equivalent to u(x, t f(x κte rt, by replacing the function x f(x in the first formula by x f( rx. This is acceptable because f is arbitrary. (b The number of particles at time t is given by u(x, tdx. We have M u(x, dx. But u(x, f(x, so M f(xdx. For t >, the number of particles is u(x, tdx f(x κte rt dx e rt f(x κtdx e rt f(xdx Me rt, where, in evaluating the integral f(x κtdx, we used the change of variables x x κt, and then used M f(xdx.

13 Section. Periodic Functions 9 Solutions to Exercises.. (a cosx has period. (b cos x has period T. (c cos 3 x has period T /3 3. (d cos x has period, cosx has period,, 3,.. A common period of cos x and cos x is. So cos x + cos x has period. 3. (a The period is T, so it suffices to describe f on an interval of length. From the graph, we have { if f(x x <, if x <. For all other x, we have f(x + f(x. (b f is continuous for all x k, where k is an integer. At the half-integers, x k+, using the graph, we see that lim h x + f(h and lim h x f(h. At the integers, x k, from the graph, we see that lim h x + f(h and lim h x f(h. The function is piecewise continuous. (c Since the function is piecewise constant, we have that f (x at all x k, where k is an integer. It follows that f (x+ and f (x (Despite the fact that the derivative does not exist at these points; the left and right limits exist and are equal. 5. This is the special case p of Exercise 6(b. 7. Suppose that Show that f, f,..., f n,... are T-periodic functions. This means that f j (x + T f(x for all x and j,,..., n. Let s n (x a f (x + a f (x + + a n f n (x. Then s n (x + T a f (x + T + a f (x + T + + a n f n (x + T a f (x + a f (x + + a n f n (x s n (x; which means that s n is T-periodic. In general, if s(x j a jf j (x is a series that converges for all x, where each f j is T-periodic, then and so s(x is T-periodic. s(x + T a j f j (x + T a j f j (x s(x; j 9. (a Suppose that f and g are T-periodic. Then f(x+t g(x+t f(x g(x, and so f g is T periodic. Similarly, j f(x + T g(x + T f(x g(x, and so f/g is T periodic. (b Suppose that f is T-periodic and let h(x f(x/a. Then ( x + at ( x h(x + at f f a a + T ( x f (because f is T-periodic a h(x. Thus h has period at. Replacing a by /a, we find that the function f(ax has period T/a. (c Suppose that f is T-periodic. Then g(f(x + T g(f(x, and so g(f(x is also T-periodic.. Using Theorem, / / f(xdx f(xdx sin x dx.

14 Chapter Fourier Series 3. / f(xdx / 5. Let F(x x a F(x + x+ a / dx /. f(tdt. If F is -periodic, then F(x F(x +. But f(tdt x a f(tdt + Since F(x F(x +, we conclude that Applying Theorem, we find that x+ x x+ x f(tdt The above steps are reversible. That is, f(tdt and so F is -periodic. x+ x x a f(tdt f(tdt x a F(x F(x + ; x+ x f(tdt. f(tdt F(x + f(tdt. x+ x f(t dt. x+ x+ f(tdt + f(tdt f(t dt x a 7. By Exercise 6, F is periodic, because f(tdt (this is clear from the graph of f. So it is enough to describe F on any interval of length. For < x <, we have x F(x ( tdt t t x x x. For all other x, F(x+ F(x. (b The graph of F over the interval [, ] consists of the arch of a parabola looking down, with zeros at and. Since F is -periodic, the graph is repeated over and over. 9. (a The plots are shown in the following figures. [ (b Let us show that f(x x p f(x + p x p ] is p-periodic. [ ] x + p x + p p x + p p p [ ] x x p f(x. p [ ] x p + x + p p ([ ] x + p

15 Section. Periodic Functions From the graphs it is clear that f(x x for all < x < p. To see this from the formula, use the fact that [t] if t <. So, if x < p, we have x [ ] p <, so, and hence f(x x. x p. (a Plot of the function f(x x p [ ] x+p p for p,, and. p Plot x pfloor x p p, x,, p p (b [ ] [ ] (x + p + p x + p f(x + p (x + p p (x + p p p p + ([ ] [ ] x + p x + p (x + p p + x p f(x. p p [ ] So f is p-periodic. For p < x < p, we have < x+p p <, hence x+p p, and [ ] so f(x x p x+p p x.. (a With p, the function f becomes f(x x [ ] x+, and its graph is the first one in the group shown in Exercise. The function is -periodic and is equal to x on the interval < x <. By Exercise 9(c, the function g(x h(f(x is - periodic for any function h; in particular, taking h(x x, we see that g(x f(x is -periodic. (b g(x x on the interval < x <, because f(x x on that interval. (c Here is a graph of g(x f(x ( x [ ] x+, for all x. Plot x Floor x ^, x, 3, (a As in Exercise, the function f(x x [ ] x+ is -periodic and is equal to x on the interval < x <. So, by Exercise 9(c, the function [ ] g(x f(x x + x is -periodic and is clearly equal to x for all < x <. Its graph is a triangular wave as shown in (b.

16 Chapter Fourier Series Plot Abs x Floor x, x, 3, (c To obtain a triangular wave of arbitrary period p, we use the p-periodic function [ ] x + p f(x x p, p which is equal to x on the interval p < x < p. Thus, [ ] g(x x + p x p p is a p-periodic triangular wave, which equal to x in the interval p < x < p. The following graph illustrates this function with p. p Pi Plot Abs x pfloor x p p, x, Pi, Pi 3. (a Since f(x + p f(x, it follows that g(f(x + p g(f(x and so g(f(x is p-periodic. For p < x < p, f(x x and so g(f(x g(x. (b The function e g(x, with p, is the -periodic extension of the function which equals e x on the interval < x <. Its graph is shown in Figure, Section.6 (with a. 5. We have F(a + h F(a a a+h a a+h f(xdx f(x dx f(x dx M h, where M is a bound for f(x, which exists by the previous exercise. (In deriving

17 Section. Periodic Functions 3 the last inequality, we used the following property of integrals: b a f(x dx (b a M, which is clear if you interpret the integral as an area. As h, M h and so F(a+h F(a, showing that F(a+h F(a, showing that F is continuous at a. (b If f is continuous and F(a a f(xdx, the fundamental theorem of calculus implies that F (a f(a. If f is only piecewise continuous and a is a point of continuity of f, let (x j, x j denote the subinterval on which f is continuous and a is in (x j, x j. Recall that f f j on that subinterval, where f j is a continuous component of f. For a in (x j, x j, consider the functions F(a a f(xdx and G(a a x j f j (xdx. Note that F(a G(a + x j f(xdx G(a + c. Since f j is continuous on (x j, x j, the fundamental theorem of calculus implies that G (a f j (a f(a. Hence F (a f(a, since F differs from G by a constant. vspace7pt 7. (a The function sin x does not have a right or left limit as x, and so it is not piecewise continuous. (To be piecewise continuous, the left and right limits must exist. The reason is that /x tends to + as x + and so sin/x oscillates between + and. Similarly, as x, sin /x oscillates between + and. See the graph. (b The function f(x x sin x and f( is continuous at. The reason for this is that sin /x is bounded by, so, as x, x sin /x, by the squeeze theorem. The function, however, is not piecewise smooth. To see this, let us compute its derivative. For x, f (x sin x x cos x. As x +, /x +, and so sin /x oscillates between + and, while x cos x oscillates between + and. Consequently, f (x has no right limit at. Similarly, it fails to have a left limit at. Hence f is not piecewise smooth. (Recall that to be piecewise smooth the left and right limits of the derivative have to exist. (c The function f(x x sin x and f( is continuous at, as in case (b. Also, as in (b, the function is not piecewise smooth. To see this, let us compute its derivative. For x, f (x x sin x cos x. As x +, /x +, and so x sin /x, while cos oscillates between + x and. Hence, as x +, x sin x cos oscillates between + and, and so x f (x has no right limit at. Similarly, it fails to have a left limit at. Hence f is not piecewise smooth. (d The function f(x x 3 sin x and f( is continuous at, as in case (b. It is also smooth. We only need to check the derivative at x. We have For x, we have f f(h f( h 3 sin h ( lim lim lim h h h h h h sin h. f (x 3x sin x x cos x. Since f (x f ( as x, we conclude that f exists and is continuous for all x.

18 4 Chapter Fourier Series Plot Sin x, x,.3,.3, PlotPoints 5 Plot xsin x, x,.3,.3, PlotPoints x^sin x, x,.,. Plot x^3sin x, x,.,

19 Section. Fourier Series 5 Solutions to Exercises.. The graph of the Fourier series is identical to the graph of the function, except at the points of discontinuity where the Fourier series is equal to the average of the function at these points, which is. 3. The graph of the Fourier series is identical to the graph of the function, except at the points of discontinuity where the Fourier series is equal to the average of the function at these points, which is 3/4 in this case. 5. We compute the Fourier coefficients using he Euler formulas. Let us first note that since f(x x is an even function on the interval < x <, the product f(x sin nx is an odd function. So b n odd function {}}{ x sinnx dx, because the integral of an odd function over a symmetric interval is. For the other coefficients, we have a f(xdx x dx ( xdx + x dx x. x dx In computing a n (n, we will need the formula cos(a x x sin(a x x cosax dx a + + C (a, a which can be derived using integration by parts. We have, for n, a n f(xcos nx dx x cosnx dx x cosnx dx [ x n sinnx + ] cos nx n [ ( n ] [ ( n ] n n n { if n is even if n is odd. 4 n

20 6 Chapter Fourier Series Thus, the Fourier series is 4 k cos(k + x. (k + s n_, x_ : Pi 4 PiSum k ^Cos k x, k,, n partialsums Table s n, x, n,, 7 ; f x_ x PiFloor x Pi Pi g x_ Abs f x Plot g x, x, 3Pi, 3Pi Plot Evaluate g x, partialsums, x, Pi, Pi The function g(x x and its periodic extension Partial sums of the Fourier series. Since we are summing over the odd integers, when n 7, we are actually summing the 5th partial sum. 7. f is even, so all the b n s are zero. We have a f(xdx We will need the trigonometric identity For n, sin x dx cosx sin a cos b ( sin(a b + sin(a + b.. a n f(xcos nx dx sin x cosnx dx ( sin( nx + sin( + nx dx [ ] n cos( nx cos( + nx (if n ( + n [ n ( n ( + n ( +n + n + ] + n { if n is odd if n is even. If n, we have 4 ( n a sin x cosxdx sin(xdx.

21 Section. Fourier Series 7 Thus, the Fourier series is : sinx 4 9. Just some hints: ( f is even, so all the b n s are zero. ( (3 Establish the identity x cos(axdx using integration by parts. a x cos(a x a + k x dx 3. (k cos kx. ( + a x sin(a x a 3 + C (a,. We have f(x cosx and g(x + cosx. Both functions are given by their Fourier series. 3. You can compute directly as we did in Example, or you can use the result of Example as follows. Rename the function in Example g(x. By comparing graphs, note that f(x g(x +. Now using the Fourier series of g(x from Example, we get f(x n sin n( + x n 5. f is even, so all the b n s are zero. We have a f(xdx ( n+ sin nx. n n e x dx e x e. We will need the integral identity e ax cos(bxdx a ea x cos(b x a + b + b ea x sin(b x a + b + C (a + b, which can be established by using integration by parts; alternatively, see Exercise 7, Section.6. We have, for n, a n e x cosnx dx [ n n + e x sin nx ] n + e x cos nx [ e ( n + ] ( ( n e. (n + (n + Thus the Fourier series is e e + e n n + (e ( n cosnx. 7. Setting x in the Fourier series expansion in Exercise 9 and using the fact that the Fourier series converges for all x to f(x, we obtain f( n ( n n cos n n, where we have used cosn ( n. Simplifying, we find 6 n. n n

22 8 Chapter Fourier Series 9. (a Let f(x denote the function in Exercise and w(x the function in Example 5. Comparing these functions, we find that f(x w(x. Now using the Fourier series of w, we find [ ] f(x + sin(k + x k + + sin(k + x. k + k (b Let g(x denote the function in Exercise and f(x the function in (a. Comparing these functions, we find that g(x f(x. Now using the Fourier series of f, we find g(x 4 sin(k + x. k + k (c Let k(x denote the function in Figure 3, and let f(x be as in (a. Comparing these functions, we find that k(x f ( x +. Now using the Fourier series of f, we get k(x k k k sin(k + (x + k + ( ( {}}{ sin[(k + x] cos[(k + {}} k { ]+cos[(k + x] sin[(k + k + ] ( k cos[(k + x]. k + (d Let v(x denote the function in Exercise 3, and let k(x be as in (c. Comparing these functions, we find that v(x ( k(x +. Now using the Fourier series of k, we get v(x ( k cos[(k + x]. k +. (a Interpreting the integral as an area (see Exercise 6, we have k a 8. To compute a n, we first determine the equation of the function for < x <. From Figure 6, we see that f(x ( x if < x <. Hence, for n, k Also, a n {}}{{}}{ ( x cosnx dx / nx ( xsin n + / / [ ] n sin n n cosnx u v sin nx n / [ n sin n + ( n n n cos n dx ]. b n {}}{{}}{ ( x sin nx dx / nx ( xcos n / / [ n cos n + n sin n ]. u v cosnx n dx

23 Section. Fourier Series 9 Thus the Fourier series representation of f is f(x 8 + n { [ n sin n + ( n n cos n n [ + n cos n + n sin n ] } sin nx. ] cos nx (b Let g(x f( x. By performing a change of variables x x in the Fourier series of f, we obtain (see also Exercise 4 for related details Thus the Fourier series representation of f is g(x 8 + n { [ n n sin + ( n n n [ n cos n + n sin n n cos ] } sin nx. ] cosnx 3. This exercise is straightforward and follows from the fact that the integral is linear. 5. For (a and (b, see plots. (c We have s n (x n sin kx k. So s k n ( and s n ( for all n. Also, lim x + f(x, so the difference between s n(x and f(x is equal to /4 at x. But even we look near x, where the Fourier series converges to f(x, the difference s n (x f(x remains larger than a positive number, that is about.8 and does not get smaller no matter how large n. In the figure, we plot f(x s 5 (x. As you can see, this difference is everywhere on the interval (,, except near the points and, where this difference is approximately.8. The precise analysis of this phenomenon is done in the following exercise.

24 Chapter Fourier Series 7. The graph of the sawtooth function is symmetric with respect to the point / on the interval, ; that is, we have f(x f( x. The same is true for the partial sums of the Fourier series. So we expect an overshoots of the partial sums near of the same magnitude as the overshoots near. More precisely, since s N (x N n sin nx n, it follows that ( s N N N n sin ( n( N n N n sin ( n( N n N n sin(n N. n So, by Exercise 6(b, we have Similarly, ( lim s N ( lim N N s N N N sin x x dx. ( f ( s N N N ( + ( N + s N. N As N, we have ( + N and s ( N N I so ( lim f ( s N + I.7. N N N The overshoot occurs at N+ (N+ N+ (using the result from Exercise 6(f.

25 Section.3 Fourier Series of Functions with Arbitrary Periods Solutions to Exercises.3. (a and (b Since f is odd, all the a n s are zero and Thus the Fourier series is p b n sin n p p dx n cos n p [ ( n ] n { if n is even, if n is odd. 4 k 4 n (k + (k + sin x. At the points of discon- p tinuity, the Fourier series converges to the average value of the function. In this case, the average value is (as can be seen from the graph. 3. (a and (b The function is even so all the b n s are zero, a p p a [ ( ( x p ] dx a p (x 3p x3 p 3 a; and with the help of the integral formula from Exercise 9, Section., for n, a n a p a p 3 p ( x nx cos p p [x p nx cos (n p + 4a( n n. Thus the Fourier series is 3 a + 4a is continuous for all x. n dx a p 3 p3 (n ( n+ (n p 3 ( + (n cos( n p x cos nx p dx p x sin nx p ] p x. Note that the function 5. (a and (b The function is even. It is also continuous for all x. All the b n s are. Also, by computing the area between the graph of f and the x-axis, from x to x p, we see that a. Now, using integration by parts, we obtain a n p p 4c p ( c (x p/cos n p p {}}{ p n (x p/sin n p x p x x dx 4c p p n p 4c p p n cos n p x p 4c ( cosn x n { if n is even, 8c n Thus the Fourier series is if n is odd. f(x 8c k p ] cos [(k + p x (k +. v u {}}{{}}{ (x p/ cos n p x dx sin n p x dx 7. The function in this exercise is similar to the one in Example 3. Start with the Fourier series in Example 3, multiply it by /c, then change p p (this is not a

26 Chapter Fourier Series change of variables, we are merely changing the notation for the period from p to p and you will get the desired Fourier series f(x n ( nx sin p. n The function is odd and has discontinuities at x ±p + kp. At these points, the Fourier series converges to. 9. The function is even; so all the b n s are, a p p e cx dx p cp e cx e cp ; cp and with the help of the integral formula from Exercise 5, Section., for n, p a n e cx cos nx p p dx [ npe cx sin nx p n + p c p p ce cx cos nx ] p p pc [ ( n e cp]. n + p c Thus the Fourier series is pc ( e cp + cp n c p + (n ( e cp ( n cos( n p x.. We note that the function f(x x sin x ( < x < is the product of sinx with a familiar function, namely, the -periodic extension of x ( < x <. We can compute the Fourier coefficients of f(x directly or we can try to relate them to the Fourier coefficients of g(x x. In fact, we have the following useful fact. Suppose that g(x is an odd function and write its Fourier series representation as g(x b n sin nx, n where b n is the nth Fourier coefficient of g. Let f(x g(xsin x. Then f is even and its nth cosine Fourier coefficients, a n, are given by a b, a b, a n [b n+ b n ] (n. To prove this result, proceed as follows: f(x sin x b n sin nx n b n sin x sin nx n b n [ ] cos[(n + x] + cos[(n x]. n To write this series in a standard Fourier series form, we reindex the terms, as

27 Section.3 Fourier Series of Functions with Arbitrary Periods 3 follows: f(x n n n ( b n ( b n cos[(n + x] + cos nx + ( b n cos nx b + b cosx + n b n ( bn n ( bn+ cos[(n x] cos nx + b cos x + ( bn+ b n n cos nx. ( bn+ cosnx This proves the desired result. To use this result, we recall the Fourier series from Exercise : For < x <, g(x x ( n+ n n sin nx; so b, b / and b n ( n+ n for n. So, f(x x sin x n a n cosnx, where a, a /, and, for n, a n [ ( n+ n + ] ( n ( n+ n n. Thus, for < x <, x sin x cosx + n ( n+ n cos nx. The convergence of the Fourier series is illustrated in the figure. Note that the partial sums converge uniformly on the entire real line. This is a consequence of the fact that the function is piecewise smooth and continuous for all x. The following is the 8th partial sum. s n_, x_ Cos x Sum ^ k k^ Cos kx, k,, n ; Plot Evaluate xsin x, s 8, x, x, Pi, Pi.75 To plot the function over more than one period, we can use the Floor function to extend it outside the interval (,. In what follows, we plot the function and the 8th partial sum of its Fourier series. The two graphs are hard to distinguish from one another.

28 4 Chapter Fourier Series f x_ xsin x ; g x_ x PiFloor x Pi Pi ; h x_ f g x ; Plot Evaluate h x, s 8, x, x, Pi, Pi Take p in Exercise, call the function in Exercise f(x and the function in this exercise g(x. By comparing graphs, we see that g(x ( + f(x. Thus the Fourier series of g is ( + 4 sin(k + x (k + + k k sin(k + x. (k + f x_ Which x,, x,, x, s n_, x_ PiSum k Sin k Pix, k,, n ; Plot Evaluate f x, s, x, x,, Which x,, x,, x, The 4st partial sum of the Fourier series and the function on the interval (-,. 5. To match the function in Example, Section., take p a in Example of this section. Then the Fourier series becomes + 4 k cos(k + x, (k + which is the Fourier series of Example, Section..

29 Section.3 Fourier Series of Functions with Arbitrary Periods 5 7. (a Take x in the Fourier series of Exercise 4 and get p 3 4p n ( n n (b Take x p in the Fourier series of Exercise 4 and get p p 3 4p n n ( n n. ( n ( n n 6 n. Summing over the even and odd integers separately, we get 6 n n k But k (k 4 k k 4 6. So (k + + k (k. n 6 (k k k (k This is very similar to the proof of Theorem (i. If f(x n b n sin n p x, then, for all x, f( x b n sin( n p x b n sin n p x f(x, n and so f is odd. Conversely, suppose that f is odd. Then f(xcos n x is odd and, p from (, we have a n for all n. Use (5, (9, and the fact that f(xsin n p x is even to get the formulas for the coefficients in (ii. 3. From the graph, we have { x if < x <, f(x if < x <. So hence and f e (x f o (x f( x n { if < x <, + x if < x < ; f(x + f( x f(x f( x { x if < x <, x if < x <, { x if < x <, x if < x <, As expected, f(x f e (x+f o (x. Note that, f e (x x for < x <. Let g(x be the function in Example with p. Then f e (x g(x. So from Example with p, we obtain f e (x 4 k cos[(k + x]. (k + Note that f o (x x for < x <. The Fourier series of f o follows from Exercise 7 with p. Thus f o (x n sin(nx. n

30 6 Chapter Fourier Series Hence f(x 4 + (k + cos[(k + x] + k [ ( n cos(nx + sin(nx n n n n ]. sin(nx n Let s illustrate the convergence of the Fourier series. (This is one way to check that our answer is correct. Clear s f x_ Which x,, x, x, x,, x, s n_, x_ PiSum ^k Pik^ Cos kpix Sin kpix k, k,, n ; Plot Evaluate f x, s, x, x,, Since f is p-periodic and continuous, we have f( p f( p + p f(p. Now a p f (xdx p p f(x p (f(p f( p. p p p Integrating by parts, we get a n p p p f (xcos nx p dx {}}{{ f(xcos nx }}{ p p p +n p f(xsin nx p p p p dx n p b n. p b n Similarly, b n p p p f (xsin nx p dx {}}{{ f(xsin nx }}{ p p p n p f(xcos nx p p p p dx n p a n. p a n 7. The function in Exercise 5 is piecewise smooth and continuous, with a piecewise smooth derivative. We have { f c if < x < p, p (x c if p < x <. p

31 Section.3 Fourier Series of Functions with Arbitrary Periods 7 The Fourier series of f is obtained by differentiating term by term the Fourier series of f (by Exercise 6. So f (x 8c k (k + (k + sin p (k + x 8c p p k k + (k + sin x. p Now the function in Exercise is obtained by multiplying f (x by p. So to c obtain the Fourier series in Exercise, we multiply the Fourier series of f by p c and get 4 k k + (k + sin x. p 9. The function in Exercise 8 is piecewise smooth and continuous, with a piecewise smooth derivative. We have c if < x < d, f d (x if d < x < p, c d if d < x <. The Fourier series of f is obtained by differentiating term by term the Fourier series of f (by Exercise 6. Now the function in this exercise is obtained by multiplying f (x by d c. So the desired Fourier series is d c f (x d c cp d n cos dn p n ( n p sin n p x n cos dn p n sin n p x. 3. (a If lim n cos nx for some x, then any subsequence of (cos nx also converges to, in particular, lim n cos(nx. Furthermore, lim n cos nx. But cos nx +cos(nx, and taking the limit as n on both sides we get + or /, which is obviously a contradiction. Hence lim n cosnx holds for no x. (b If n cosnx converges for some x, then by the nth term test, we must have lim n cos nx. But this limit does not hold for any x; so the series does not converge for any x. 33. The function F(x is continuous and piecewise smooth with F (x f(x at all the points where f is continuous (see Exercise 5, Section.. So, by Exercise 6, if we differentiate the Fourier series of F, we get the Fourier series of f. Write F(x A + (A n cos np x + B n sin np x and f(x n n ( a n cos n p x + b n sin n p x. Note that the a term of the Fourier series of f is because by assumption p f(xdx. Differentiate the series for F and equate it to the series for f and get n ( n n A n sin p p x + n p B n cos n p x Equate the nth Fourier coefficients and get n A n n p b n A n p n b n; B n n p a n B n p n a n. (a n cos np x + b n sin np x.

32 8 Chapter Fourier Series This derives the nth Fourier coefficients of F for n. To get A, note that F( because of the definition of F(x x f(tdt. So F( A + A n A + p n b n; n and so A p n n b n. We thus obtained the Fourier series of F in terms of the Fourier coefficients of f; more precisely, F(x p n n n b n n + ( p n b n cos n p x + p n a n sin n p x. The point of this result is to tell you that, in order to derive the Fourier series of F, you can integrate the Fourier series of f term by term. Furthermore, the only assumption on f is that it is piecewise smooth and integrates to over one period (to guarantee the periodicity of F. Indeed, if you start with the Fourier series of f, f(t (a n cos np t + b n sin np t, n and integrate term by term, you get F(x x f(tdt x (a n cos n p t dt + b n n ( ( p a n n n p b n n + n x sin n p t dt sin n p t x ( dt + b n p cos n n p t x ( p n b n cos n p x + p n a n sin n p x n as derived earlier. See the following exercise for an illustration.,

33 Section.4 Half-Range Expansions: The Cosine and Sine Series 9 Solutions to Exercises.4. The even extension is the function that is identically. So the cosine Fourier series is just the constant. The odd extension yields the function in Exercise, Section.3, with p. So the sine series is 4 k sin((k + x. k + This is also obtained by evaluating the integral in (4, which gives b n sin(nxdx n cosnx n ( ( n. 3. The even extension is the function in Exercise 4, Section.3, with p. So the cosine Fourier series is 3 4 ( n+ n cosnx. n In evaluating the sine Fourier coefficients, we will use the formula (( + a x x cos(a x x sin(a x sin ax dx + + C (a, a 3 a which is obtained using integration by parts. For n, we have b n x sin(nxdx [ ( + (n x ] cos(n x x sin(n x (n 3 (n [ ( + (n ( n + ] (n 3 (n 3 [ ( n+ + ] n (n 3 (( n. Thus the sine series representation 5. (a Cosine series: [ ( n+ n a n p b a n a p + ] (n 3 (( n sin(nx. b a cos nx p dx p p n thus the even extension has the cosine series f e (x b a p + n n dx b a p ; ( sin nb na sin p p ( sin nb na sin p p ; cos nx p. Sine series: b n p b a sin nx p dx p p n ( cos na nb cos ; p p

34 3 Chapter Fourier Series thus the odd extension has the sine series f o (x ( cos na nb cos sin nx n p p p. n 7. The even extension is the function cosx. This is easily seen by plotting the graph. The cosine series is (Exercise 8, Section.: Sine series: b n 4 cosx 4 cosxsin nx dx n ( n (n cos(nx. [sin( + nx sin( nx] dx {}}{{}}{ ( + n ( + n [ + n + cos( cos( ] + n + n + n 8 n 4n ; thus the odd extension has the sine series f o (x 8 n 4n sin nx. 9. We have b n n x( xsin(nxdx. To evaluate this integral, we will use integration by parts to derive the following two formulas: for a, x cos(a x sin(a x x sin(axdx + + C, a a and So x sin(axdx x( xsin(axdx cos(a x a 3 x cos(a x x sin(a x + a a + C. cos(a x x cos(a x + x cos(a x sin(a x x sin(a x + + C. a 3 a a a a Applying the formula with a n, we get x( xsin(nxdx cos(n x x cos(n x (n 3 + x cos(n x sin(n x x sin(n x + n n (n (n (( n (n 3 { 4 (n 3 ( n n if n is odd, if n is even. + ( n n (( n (n 3

35 Section.4 Half-Range Expansions: The Cosine and Sine Series 3 Thus b n { 8 (n 3 if n is odd, if n is even, Hence the sine series in 8 3 k sin(k + x (k + 3. b k_ 8 Pi^3 k ^3; ss n_, x_ : Sum b k Sin k Pix, k,, n ; partialsineseries Table ss n, x, n,, 5 ; f x_ x x Plot Evaluate partialsineseries, f x, x,, Perfect!. The function is its own sine series. 3. We have sin x cosx sin x. This yields the desired -periodic sine series expansion. 5. We have b n e x e x sin nx dx + (n (sin nx n cos nx e ( n( n+ n + (n + + (n n ( + e( n+. + (n Hence the sine series is n n + (n ( + e( n+ sin nx.

36 3 Chapter Fourier Series 7. (b Sine series expansion: a h nx x sin a p b n p p dx + p a h [ x p nx cos a ap n p + p n + h (a pp h [ ap na cos pa n p + h a p a (x psin nx p dx cos nx ] p dx [ (x p ( p n cos(n x p p p + a a ] p na sin (n p + h [ p na (a pcos (a pp n p hp na[ sin (n p a ] a p hp na (n sin (p aa p. Hence, we obtain the given Fourier series. p na ] sin (n p p n cos n x ] dx p Solutions to Exercises.5. We have f(x The Fourier series representation is f(x 4 { if < x <, if < x < ; k The mean square error (from (5 is E N sin(k + x. k + f (xdx a N (a n + b n. n In this case, a n for all n, b k, b k+ 4 (k+, and f (xdx dx. So E (b Since b, it follows that E E. Finally, E (b + b We have f(x (x/ for < x <. Its Fourier series representation is f(x n ( n+ (n cosnx. Thus a 3, a n 4 ( n+ (n, and b n for all n. The mean square error

37 Section.5 Mean Square Approximation and Parseval s Identity 33 (from (5 is E N f (xdx a N (a n + b n n ( (x/ 4 dx ( x + x4 4 [ x x x5 5 4 ] N N n n n N n dx n 4 N n n 4 n 4 N n n 4 So E ; E E ; E 3 E We have E N f (xdx a N b n 8 n N (a n + b n n n odd N n. With the help of a calculator, we find that E 39.3 and E So take N (a Parseval s identity with p : f(x dx a + (a n + b n. n Applying this to the given Fourier series expansion, we obtain x 4 dx n n n For (b and (c see the solution to Exercise 7, Section.3. n 6 n. 9. We have f(x x x 3 for < x < and, for n, b n n 3 ( n+. By n

38 34 Chapter Fourier Series Parseval s identity n Simplifying, we find that ( n 3 ζ(6 n ( x x 3 ( 4 x x 4 + x 6 dx ( 4 3 x3 5 x5 + x7 7 ( n (8( 6 (5( For the given function, we have b n and a n n. By Parseval s identity, we have f (xdx n n 4 f (xdx n 5 ζ(4 n4 9, where we have used the table preceding Exercise 7 to compute ζ(4. 5. Let us write the terms of the function explicitly. We have f(x n Thus for the given function, we have By Parseval s identity, we have cosnx + cosx + cosx +. n b n for all n, a, a n for n. n f (xdx a n n n a n ( n n. To sum the last series, we use a geometric series: if r <, n r n r. n 4 n Hence and so n 4 n 4 4 3, f (xdx (

39 Section.5 Mean Square Approximation and Parseval s Identity For the given function, we have By Parseval s identity, we have a, a n 3 n, b n n f (xdx a + Using a geometric series, we find By Exercise 7(a, So + for n. (a n + b n n ( (3 n + n n n + n n n n n. n n 9 n 9 n n n f (xdx (

40 36 Chapter Fourier Series Solutions to Exercises.6. From Example, for a, ±i, ±i, ±3i,..., e ax sinha n ( n a in einx ( < x < ; consequently, e ax sinha n ( n a + in einx ( < x <, and so, for < x <, cosh ax eax + e ax sinha ( ( n a + in + e inx a in n a sinha ( n. n + a einx n 3. In this exercise, we will use the formulas cosh(iax cosax and sinh(iax isin ax, for all real a and x. (An alternative method is used in Exercise 4. To prove these formulas, write cosh(iax eiax + e iax cos ax, by Euler s identity. Similarly, sinh(iax eiax e iax isin ax. If a is not an integer, then ia, ±i, ±i, ±3i,..., and we may apply the result of Exercise to expand e iax in a Fourier series: cos(ax cosh(iax (iasinh(ia a sin(a n n ( n n a einx. ( n n + (ia einx 5. Use identities (; then cosx + sin 3x eix + e ix ie 3ix + e ix + e3ix e 3ix i + eix ie3ix. 7. You can use formulas (5 (8 to do this problem, or you can start with the

41 Section.6 Complex Form of Fourier Series 37 Fourier series in Exercise 3 and rewrite it as follows: cosax a sin(a a sin(a sin(a a sin(a a sin(a a 9. If m n then p p p If m n, then n n a sin(a a sin(a a sin(a ( n e inx n a ( n e inx n a n a sin(a ( n e inx ( n a a sin(a ( n n cos nx {}}{ e inx + e inx n a ( n cosnx n a. n e i m p x e i n p x dx p p p a a sin(a n n ( n e inx n a e i m p x m i e p x dx p dx. p p ( n e inx n a p p p e i m p x e i n p x dx p e i (m n p x dx p p i (m n p (m n ei p x p i (e i(m n e i(m n (m n i (cos[(m n] cos[ (m n]. (m n. The function in Example is piecewise smooth on the entire real line and continuous at x. By the Fourier series representation theorem, its Fourier series converges to the value of the function at x. Putting 4x4 in the Fourier series we thus get f( sinha n ( n a (a + in. + n The doubly infinite sum is to be computed by taking symmetric partial sums, as follows: But n ( n a + n(a + in lim N N n N a lim N N n N N n N ( n a + n(a + in ( n n a + n, ( n a + n + i lim because the summand is an odd function of n. So sinha ( n a sinha a + n(a + in n N N n N n ( n a + n, ( n n a + n.

42 38 Chapter Fourier Series which is equivalent to the desired identity. 3. (a At points of discontinuity, the Fourier series in Example converges to the average of the function. Consequently, at x the Fourier series converges to e a +e a cosh(a. Thus, plugging x into the Fourier series, we get cosh(a sinh(a n ( ( n {}}{ n a + n(a + in e in sinh(a The sum n in a +n is the limit of the symmetric partial sums Hence n in and so a +n cosh(a sinh(a i n N n N n a + n. a a + n coth(a a upon dividing both sides by sinh(a. Setting t a, we get coth t t n ( t + n n n n t t + (n, (a + in a + n. a + n, which is (b. Note that since a is not an integer, it follows that t is not of the form ki, where kis an integer. 5. (a This is straightforward. Start with the Fourier series in Exercise : For a, ±i, ±i, ±3i,..., and < x <, we have On the left side, we have cosh (axdx cosh ax a sinha n ( n n + a einx. cosh(ax + dx [ x + a sinh(ax] On the right side of Parseval s identity, we have Hence Simplifying, we get (a sinha n (n + a. [ + a sinh(a] (a sinha n n [ + a sinh(a]. (n + a. (n + a [ + (a sinha a sinh(a]. (b This part is similar to part (a. Start with the Fourier series of Exercise : For a, ±i, ±i, ±3i,..., and < x <, we have sinhax isinha n ( n n n + a einx.

43 Section.6 Complex Form of Fourier Series 39 On the left side, we have sinh (axdx cosh(ax dx [ x + a sinh(ax] On the right side of Parseval s identity, we have Hence Simplifying, we get sinh a n n (n + a. [ + a sinh(a] sinh (a n n [ + a sinh(a]. n (n + a. n (n + a [ + sinh (a a sinh(a]. 7. (a In this exercise, we let a and b denote real numbers such that a + b. Using the linearity of the integral of complex-valued functions, we have I + ii e ax cos bx dx + i e ax sin bx dx (e ax cos bx + ie ax sin bx dx {}}{ e ax (cos bx + isin bx dx e ax e ibx dx e x(a+ib dx e ibx a + ib ex(a+ib + C, where in the last step we used the formula e αx dx α eαx + C (with α a + ib, which is valid for all complex numbers α (see Exercise 9 for a proof. (b Using properties of the complex exponential function (Euler s identity and the fact that e z+w e z e w, we obtain I + ii a + ib ex(a+ib + C (a + ib (a + ib (a + ib eax e ibx + C a ib a + b eax( cos bx + isin bx + C e ax [( ( ] a cosbx + b sin bx + i b cosbx + a sinbx + C. a + b (c Equating real and imaginary parts in (b, we obtain and I I eax ( a cos bx + b sin bx a + b eax a + b ( b cosbx + a sin bx.

44 4 Chapter Fourier Series 9. The purpose of this exercise is to show you that the familiar formula from calculus for the integral of the exponential function, e ax dx a eax + C, holds for all nonzero complex numbers a. Note that this formula is equivalent to where the derivative of d dx eax means d dx eax ae ax, d dx eax d ( Re(e ax + iim (e ax d dx dx Re(eax + i d dx Im(eax. Write a α + iβ, where α and β are real numbers. Then e ax e x(α+iβ e αx e iβx e αx cos βx + ie αx sin βx. So d dx eax d dx eαx cos βx + i d dx eαx sin βx αe αx cosβx βe αx sin βx + i ( αe αx sinβx + βe αx cosβx e αx( α cosβx β sin βx + i ( α sin βx + β cosβx e αx (cos βx + isin βx(α + iβ ae αx e iβx ae αx+iβx ae iax, as claimed.. By Exercise 9, ( e it + e it dt i eit + i e it {}}{{}}{ i e i +i e i ( i + i. Of course, this result follows from the orthogonality relations of the complex exponential system (formula (, with p. 3. First note that cos t isin t e it eit. Hence cos t isin t dt e it dt ie it + C sint icos t + C. 5. First note that Hence + it it ( + it ( it( + it t + it + t t + t + i t + t. + it t it dt + t dt + i t + t dt ( + + t dt + i t + t dt t + tan t + iln( + t + C.

45 Section.7 Forced Oscillations 4 Solutions to Exercises.7. (a General solution of y + y + y. The characteristic equation is λ +λ+ or (λ+. It has one double characteristic root λ. Thus the general solution of the homogeneous equation y + y + y is y c e t + c te t. To find a particular solution of y + y + y 5 cost, we apply Theorem with µ, c, and k. The driving force is already given by its Fourier series: We have b n a n for all n, except a 5. So α n β n for all n, except α Aa A +B and β Ba A +B, where A 3 and B 4. Thus α and β 5 4, and hence a particular solution is y p 3 cost + 4 sin t. Adding the general solution of the homogeneous equation to the particular solution, we obtain the general solution of the differential equation y + y + y 5 cost y c e t + c te t 3 cost + 4 sin t. (b Since lim t c e t + c te t, it follows that the steady-state solution is y s 3 cost + 4 sin t. 3. (a General solution of 4y + 4y + 7y. The characteristic equation is 4λ + 4λ + 7. Its characteristic roots are λ ± 4 (4(7 4 ± 64 4 Thus the general solution of the homogeneous equation is y c e t/ cost + c e t/ sin t. ± i. It is easy to see that y /7 is a particular solution of 4y +4y +7y. (This also follows from Theorem. Hence the general solution is y c e t/ cost + c e t/ sint + 7. (b Since lim t c e t/ cos t + c e t/ sint, it follows that the steady-state solution is y s (a To find a particular solution (which is also the steady-state solution of y + 4y + 5y sin t sin t, we apply Theorem with µ, c 4, and k 5. The driving force is already given by its Fourier series: We have b n a n for all n, except b and b /. So α n β n for all n, except, possibly, α, α, β, and beta. We have A 4, A, B 4, and B 8. So α B b A + B α B b A + B β A b A + B β A b A + B 4 3 8, , 4 3 8, / 65 3.

46 4 Chapter Fourier Series Hence the steady-state solution is (b We have y p 8 cos t + 8 sin t cost sin t. 3 y p 8 cost + 8 sin t cost sin t, 3 (y p 8 sin t + 8 cost 8 65 sint cos t, 65 (y p 8 cost 6 sint 8 65 cost + sin t, ( 65 (y p + 4(y p + 5y p ( cost ( sin t + 3 ( sin t sin t ( sin t sin t, cos t which shows that y p is a solution of the nonhomogeneous differential equation. 9. (a Natural frequency of the spring is k ω µ (b The normal modes have the same frequency as the corresponding components of driving force, in the following sense. Write the driving force as a Fourier series F(t a + n f n(t (see (5. The normal mode, y n (t, is the steady-state response of the system to f n (t. The normal mode y n has the same frequency as f n. In our case, F is -periodic, and the frequencies of the normal modes are computed in Example. We have ω m+ m + (the n even, the normal mode is. Hence the frequencies of the first six nonzero normal modes are, 3, 5, 7, 9, and. The closest one to the natural frequency of the spring is ω 3 3. Hence, it is expected that y 3 will dominate the steady-state motion of the spring. 3. According to the result of Exercise, we have to compute y 3 (t and for this purpose, we apply Theorem. Recall that y 3 is the response to f sin 3t, the component of the Fourier series of F(t that corresponds to n 3. We have a 3, b 3 4 3, µ, c.5, k., A 3. 9., B 3 3(.5.5, So α 3 B 3b 3 A 3 + B 3 (.5(4/(3 (. + (.5.6 and β 3 A 3b 3 A 3 + B 3 y 3.6 cos3t +.4 sin3t. The amplitude of y 3 is (a In order to eliminate the 3rd normal mode, y 3, from the steady-state solution, we should cancel out the component of F that is causing it. That is, we 4 sin 3t 4 sin 3t must remove f 3 (t 3. Thus subtract 3 from the input function. The modified input function is 4 sin 3t F(t 3. Its Fourier series is he same as the one of F, without the 3rd component, f 3 (t. So the Fourier series of the modified input function is 4 sin t + 4 m sin(m + t. m +