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

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

2 Contents Preface Errata v vi A Preview of Applications and Techniques. What Is a Partial Differential Equation?. Solving and Interpreting a Partial Differential Equation Fourier Series 4. Periodic Functions 4. Fourier Series 6.3 Fourier Series of Functions with Arbitrary Periods.4 Half-Range Expansions: The Cosine and Sine Series 4.5 Mean Square Approximation and Parseval s Identity 6.6 Complex Form of Fourier Series 8.7 Forced Oscillations Supplement on Convergence.9 Uniform Convergence and Fourier Series 7. Dirichlet Test and Convergence of Fourier Series 8 3 Partial Differential Equations in Rectangular Coordinates 9 3. 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 5 3. Neumann and Robin Conditions 5

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 65 Supplement on Bessel Functions 4.7 Bessel s Equation and Bessel Functions Bessel Series Expansions Integral Formulas and Asymptotics for Bessel Functions 79 5 Partial Differential Equations in Spherical Coordinates 8 5. Preview of Problems and Methods 8 5. Dirichlet Problems with Symmetry Spherical Harmonics and the General Dirichlet Problem The Helmholtz Equation with Applications to the Poisson, Heat, and Wave Equations 86 Supplement on Legendre Functions 5.5 Legendre s Differential Equation Legendre Polynomials and Legendre Series Expansions 9 6 Sturm Liouville Theory with Engineering Applications Orthogonal Functions Sturm Liouville Theory The Hanging Chain Fourth Order Sturm Liouville Theory 6.6 The Biharmonic Operator Vibrations of Circular Plates 4

4 iv Contents 7 The Fourier Transform and Its Applications 5 7. The Fourier Integral Representation 5 7. The Fourier Transform The Fourier Transform Method 7.4 The Heat Equation and Gauss s Kernel A Dirichlet Problem and the Poisson Integral Formula 7.6 The Fourier Cosine and Sine Transforms Problems Involving Semi-Infinite Intervals Generalized Functions The Nonhomogeneous Heat Equation Duhamel s Principle 34 8 The Laplace and Hankel Transforms with Applications 36 A 8. The Laplace Transform Further Properties of the Laplace transform The Laplace Transform Method The Hankel Transform with Applications 48 Green s Functions and Conformal Mappings 5. Green s Theorem and Identities 5. Harmonic Functions and Green s Identities 5.3 Green s Functions 53.4 Green s Functions for the Disk and the Upper Half-Plane 54.5 Analytic Functions 55.6 Solving Dirichlet Problems with Conformal Mappings 6.7 Green s Functions and Conformal Mappings 65 Ordinary Differential Equations: Review of Concepts and Methods A. Linear Ordinary Differential Equations A67 A. Linear Ordinary Differential Equations with Constant Coefficients A74 A.3 Linear Ordinary Differential Equations with Nonconstant Coefficients A8 A.4 The Power Series Method, Part I A87 A.5 The Power Series Method, Part II A9 A.6 The Method of Frobenius A97 A67

5 Preface This manual contains solutions with notes and comments to problems from the textbook Partial Differential Equations with Fourier Series and Boundary Value Problems Second Edition Most solutions are supplied with complete details and can be used to supplement examples from the text. Additional solutions will be posted on my website nakhle as I complete them and will be included in future versions of this manual. I would like to thank users of the first edition of my book for their valuable comments. Any comments, corrections, or suggestions would be greatly appreciated. My address is nakhle@math.missouri.edu Nakhlé H. Asmar Department of Mathematics University of Missouri Columbia, Missouri 65

6 Errata The following mistakes appeared in the first printing of the second edition (up-dates 4 March 5). Corrections in the text and figures p. 4, Exercise #3 is better done after Section 4.4. p. 68, Exercise #8(b), n should be even. p.387, Exercise#, use y I (x) not y J (x). p.4, line 7, the integrals should be from to. p. 45 Figures 5 and 6: Relabel the ticks on the x-axis as π, π/, π/, π, instead of π, π, π, π. p. 467, line ( 3): Change reference () to (). p. 477, line : (xt) (x, t). p. 477, line 9: Change interval to triangle p. 487, line : Change is the equal to is equal p. 655, line 3: Change ln ln(x + y ) to ln(x + y ). p. A38, the last two lines of Example should be: (a a )+(a a )x + m... m [(m +)a m+ + a m (m )]x m. Last page on inside back cover: Improper integrals, lines 3, the first integral should be from to and not from to. Corrections to Answers of Odd Exercises Section 7., # 7: Change i to i. Section 7.8, # 3: f(x) 3 for <x<3 not <x<. π (e Section 7.8, # 35: iw ) 3 w j j sin(jw). # 37: i 3, # 5: π w 3 π [δ δ ]. # 57: The given answer is the derivative of the real answer, which should be ( (x +) ( ) ( ) ( ) ( ) ) U U +( x +) U U + U U 3 +( x +4) U 3 U 4 π # 59: ( The given answer is the derivative of the real answer, which should be π (x +3) ( ) ( ) ( ) U 3 U +(x +5) U U +(x +4) U U +( x +4) ( ) ( ) ( ) ) U U +( x +5) U U +( x +3) U U 3 Section 7., # 9: [ t sin(x + t)+ cos(x + t) cos(x t)]. Appendix A., # 43: y c cos 3x + c sin 3x 8 x cos 3x + 6 n, n 3 # 49: y p... y p x(... # 67: y 8 ex + 3 e3x +( 8 x )e x. Appendix A.3, # 9:y c x + c [ x ln ( +x x ) ]. sin nx n(9 n ). # 5 ln(cos x) ln cos x. # 7 y c ( + x)+c e x x3 3 x. Appendix A.4, # 3 +4 n ( )n x n Appendix A.5, # 5 y 6x +3x x6 + Any suggestion or correction would be greatly appreciated. Please send them to my address nakhle@math.missouri.edu Nakhlé H. Asmar Department of Mathematics University of Missouri Columbia, Missouri 65

7 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 showing that c u + c u is a solution of (). 5. Let α ax + bt, β cx + dt, then Recalling the equation, we obtain {( }}{ u t + u x ) +c 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 u u (b a) +(d c) x α β. Let a,b,c,d. Then {( }} ){ u t ++ u, x u u f(β) u(x, t) f(x + t), α where f is an arbitrary differentiable function (of one variable). 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. - - Figure for Exercise 9(b) To find the characteristic curves, solve dy dx sin x. Hence y cos x + C or y + cos x C. Thus the solution of the partial differential equation is u(x, y) f (y + cos x). To verify the solution, we use the chain rule and get u x sin xf (y + cos x) and u y f (y + cos x). Thus u x + sin xu y, as desired.

8 Chapter A Preview of Applications and Techniques Exercises.. We have t ( ) u t t ( ) v x and x ( ) v t x ( ) u. x So u t v t x and v x t u x. Assuming that v t x v x t, it follows that u t u x, which is the one dimensional wave equation with c. A similar argument shows that v is a solution of the one dimensional 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, ) F (x)+g(x) +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 +x G(x) [ ] +x C ; and from () Hence F (x) u(x, t) F (x + ct)+g(x ct) [ ] +x + C. [ ] +(x + ct) + +(x ct). 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) [ +(x + ct) + +(x ct) 3. The function being graphed is ] + [ e (x+ct) e (x ct)]. c u(x, t) sin πx cos πt sin πx cos πt + sin 3πx cos 3πt. 3 In frames, 4, 6, and 8, t m, where m, 3, 5, and 7. Plugging this into 4 u(x, t), we find u(x, t) sin πx cos mπ 4 mπ sin πx cos + 3mπ sin 3πx cos 3 4.

9 Section. Solving and Interpreting a Partial Differential Equation 3 For m, 3, 5, and 7, the second term is, because cos mπ. Hence at these times, we have, for, m, 3, 5, and 7, u(x, m 4 ) sin πx cos πt + sin 3πx cos 3πt. 3 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 4 ). Does this equality hold? Let s check: u(/+x, m mπ ) sin π(x +/) cos mπ sin 3π(x +/) cos 3 4 cos πx cos mπ 4 3mπ cos 3πx cos 3 4, where we have used the identities sin π(x +/) cos πx and sin π(x +/) cos 3πx. Similalry, u(/ x, m mπ ) sin π(/ x) cos mπ sin 3π(/ x) cos 3 4 cos πx cos mπ 4 3mπ cos 3πx cos 3 4. So u(/+x, m 4 )u(/ x, m 4 ), as expected. 7. Same reasoning as in the previous exercise, we find the solution u(x, t) sin πx L cπt cos L + 3πx sin 4 L cos 3cπt L + 7πx sin 5 L 7cπt cos L.. (a) We have to show that u(,t) is a constant for all t>. With c L, we have u(x, t) sin πx cos πt u(/, t) sin π cos πt for all t>. (b) One way for x /3not to move is to have u(x, t) sin 3πx cos 3πt. This is the solution that corresponds to the initial condition u(x, ) sin 3πx and u(x, ). For this solution, we also have that x /3does not move for t all t. 5. The solution () is u(x, t) sin πx πct cos L L. Its initial conditions at time t 3L c are u(x, 3L ( πx πc ) sin c L cos L 3L ) sin πx c L cos 3π ; and u 3L (x, t c ) πc L sin πx ( πc L sin L 3L ) πc πx sin c L L sin 3π πc L sin πx L.

10 4 Chapter Fourier Series Solutions to Exercises.. (a) cos x has period π. (b) cos πx has period T π π. (c) cos 3 x has period T π /3 3π. (d) cos x has period π, cos x has period π, π, 3π,.. A common period of cos x and cos x is π. So cos x + cos x has period π. 5. This is the special case p π of Exercise 6(b). 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, 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. 3. π/ f(x) dx π/ π/ dx π/. 7. By Exercise 6, F is periodic, because f(t) dt (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) ( t) dt 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.. (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,

11 Section. Periodic Functions 5 5. We have F (a + h) F (a) a a+h a a+h f(x) dx f(x) dx f(x) dx M h, where M is a bound for f(x), which exists by the previous exercise. (In deriving the last inequality, we used the following property of integrals: b f(x) dx (b a) M, a 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(x) dx, 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. Fora in (x j,x j ), consider the functions F (a) a f(x) dx and G(a) a x j f j (x) dx. Note that F (a) G(a)+ x j f(x) dx 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.

12 6 Chapter Fourier Series 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. 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 π <x<π, the product f(x) sin nx is an odd function. So b n π π π odd function {}}{ x sinnx dx, because the integral of an odd function over a symmetric interval is. For the other coefficients, we have a π π π π π π f(x) dx π π π ( x) dx + π π π xdx π π x π. In computing a n (n ), we will need the formula x dx xdx x cos ax dx cos(ax) a + x sin(ax) a + C (a ), which can be derived using integration by parts. We have, for n, Thus, the Fourier series is π π a n f(x) cos nx dx x cos nx dx π π π π π x cos nx dx π [ x π n sin nx + ] π cos nx n [ ( ) n π n ] n [ ( ) n πn ] { if n is even if n is odd. 4 πn π 4 π k cos(k +)x. (k +)

13 Section. Fourier Series 7 s n_, x_ : Pi 4 Pi Sum k ^ Cos k x, k,, n partialsums Table s n, x, n,, 7 ; f x_ x Pi Floor x Pi Pi g x_ Abs f x Plot g x, x, 3 Pi, 3 Pi 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. 9. Just some hints: () f is even, so all the b n s are zero. () a π x dx π π 3. (3) Establish the identity x cos(ax) dx x cos(ax) ( +a x ) sin(ax) a + a 3 + C (a ), using integration by parts. 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 + π). Now using the Fourier series of g(x) from Example, we get f(x) n sin n(π + x) n ( ) n+ sin nx. n 7. Setting x π 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 π f(π) π 3 +4 n ( ) n n n cos nπ π 3 +4 n, where we have used cos nπ ( ) n. Simplifying, we find π 6 n. n n

14 8 Chapter Fourier Series. (a) Interpreting the integral as an area (see Exercise 6), we have a π π 8. To compute a n, we first determine the equation of the function for π <x<π. From Figure 6, we see that f(x) π (π x) if π <x<π. Hence, for n, Also, π u a n {}}{{}}{ (π x) cos nx dx π π/ π nx (π x)sin π π n + π sin nx dx π/ π π/ n [ ] π nπ π sin n π n cos nx π π/ [ π π n sin nπ + ( )n ] nπ cos. n n π b n {}}{{}}{ (π x) sin nx dx π π/ π nx (π x)cos π π n π π/ π [ π nπ π cos n + n sin nπ ]. Thus the Fourier series representation of f is f(x) 8 + π n u { [ π v v π/ cos nx n dx n sin nπ + ( )n n nπ cos n [ π + n cos nπ + n sin nπ ] } 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 + π n { [ π n nπ sin + ( )n n n [ π n cos nπ + nπ sin n nπ cos ] } sin nx. ] cos nx 5. For (a) and (b), see plots. (c) We have s n (x) n sin kx k. So s k n () and s n (π) for all n. Also, lim x + f(x) π, so the difference between s n(x) and f(x) is equal to π/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).

15 Section. Fourier Series 9 As you can see, this difference is everywhere on the interval (, π), except near the points and π, where this difference is approximately.8. The precise analysis of this phenomenon is done in the following exercise.

16 Chapter Fourier Series Solutions to Exercises.3. (a) and (b) Since f is odd, all the a n s are zero and b n p sin nπ p p dx nπ cos π nπ p [ ( ) n ] nπ { if n is even, if n is odd. Thus the Fourier series is 4 π k 4 nπ (k +) (k +)π 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. 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 nπ p p {}}{ p nπ (x p/) sin nπ p x p xdx 4c p x p nπ p 4c p p n π cos nπ p x p 4c x n ( cos nπ) π { if n is even, 8c n π Thus the Fourier series is if n is odd. f(x) 8c π k ] cos [(k +) πp x. (k +) v u p {}}{{}}{ (x p/) cos nπ p xdx sin nπ p xdx 9. The function is even; so all the b n s are, a p e cx dx p p cp e cx e cp ; cp and with the help of the integral formula from Exercise 5, Section., for n, a n p e cx cos nπx p p dx [ nπpe cx sin nπx p n π + p c p p ce cx cos nπx ] p p pc [ ( ) n e cp]. n π + p c Thus the Fourier series is pc ( e cp )+cp c p +(nπ) ( e cp ( ) n ) cos( nπ p x). n 3. 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)).

17 Section.3 Fourier Series of Functions with Arbitrary Periods Thus the Fourier series of g is ( ) + 4 sin(k +)πx π (k +) + sin(k +)πx. π (k +) k k f x_ Which x,, x,, x, s n_, x_ Pi Sum k Sin k Pi x, 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 (-, ). 7. (a) Take x in the Fourier series of Exercise 4 and get p 3 4p π ( ) n π n n (b) Take x p in the Fourier series of Exercise 4 and get p p 3 4p π n n ( ) n n. ( ) n ( ) n n π 6 n. Summing over the even and odd integers separately, we get π 6 n n k But k (k) 4 k π k 4 6.So (k +) + k (k). n π 6 (k +) + π 4 k. From the graph, we have k (k +) π 6 π 4 π 8. f(x) { x if <x<, +x if <x<. So hence { x if <x<, f( x) +x if <x<; f e (x) f(x)+f( x) { x if <x<, x if <x<,

18 Chapter Fourier Series and f o (x) f(x) f( x) { if <x<, if <x<. Note that, f e (x) x for <x<. The Fourier series of f is the sum of the Fourier series of f e and f o. From Example with p, f e (x) 4 π From Exercise with p, Hence f(x) + 4 π f o (x) 4 π k k k cos[(k +)πx]. (k +) sin[(k +)πx]. k + [ cos[(k +)πx] + π(k +) ] sin[(k +)πx]. k + 5. Since f is p-periodic and continuous, we have f( p) f( p + p) f(p). Now a p f (x) dx p p p f(x) p (f(p) f( p)). p p Integrating by parts, we get Similarly, a n p p p f (x) cos nπx p dx {}}{{ f(x) cos nπx }}{ p p p +nπ p f(x) sin nπx p p p p p dx nπ p b n. b n b n p p p f (x) sin nπx p dx {}}{ f(x) sin nπx p p p nπ p p nπ p a n. a n { }}{ p f(x) cos nπx p p dx p 9. The function in Exercise 8 is piecewise smooth and continuous, with a piecewise smooth derivative. We have c f d if <x<d, (x) if d< x <p, if d<x<. c d 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. So the desired Fourier series is c d c f (x) d c cp dπ n cos dnπ p n ( nπ p ) sin nπ p x π n cos dnπ p n sin nπ p x.

19 Section.3 Fourier Series of Functions with Arbitrary Periods 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 nπp x + B n sin nπp ) x n and ( f(x) a n cos nπ p x + b n sin nπ ) p x. n Note that the a term of the Fourier series of f is because by assumption p f(x) dx. Differentiate the series for F and equate it to the series for f and get n ( nπ nπ A n sin p p x + nπ p B n cos nπ ) p x n Equate the nth Fourier coefficients and get nπ A n p b n A n p nπ b n; nπ B n p a n B n p nπ a n. This derives the nth Fourier coefficients of F for n. F () because of the definition of F (x) x f(t) dt. So F () A + A n A + p nπ b n; n (a n cos nπp x + b n sin nπp x ). n To get A, note that and so A p n nπ b n. We thus obtained the Fourier series of F in terms of the Fourier coefficients of f; more precisely, F (x) p π b n n + ( p nπ b n cos nπ p x + p nπ a n sin nπ ) p x. n n 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 nπp t + b n sin nπp ) t, n and integrate term by term, you get x x F (x) f(t) dt (a n cos nπ p tdt+ b n n ( ( p ) a n sin nπ nπ p t x dt + b n n p b n π n + n n ( p nπ x ) cos nπ p t x ( p nπ b n cos nπ p x + p nπ a n sin nπ p x as derived earlier. See the following exercise for an illustration. sin nπ ) p tdt ) ),

20 4 Chapter Fourier Series 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 π k sin((k +)πx). k + This is also obtained by evaluating the integral in (4), which gives b n sin(nπx) dx nπ cos nπx nπ ( ( )n ). 9. We have b n x( x) sin(nπx) dx. To evaluate this integral, we will use integration by parts to derive the following two formulas: for a, x sin(ax) dx x cos(ax) a + sin(ax) a + C, and x sin(ax) dx cos(ax) a 3 x cos(ax) a + x sin(ax) a + C. So x( x) sin(ax) dx cos(ax) a 3 x cos(ax) a + x cos(ax) a + sin(ax) a x sin(ax) a + C. Applying the formula with a nπ, we get x( x) sin(nπx) dx cos(nπ x) x cos(nπ x) + x cos(nπ x) sin(nπ x) x sin(nπ x) + (nπ) 3 nπ nπ (nπ) (nπ) (( )n ) (nπ) 3 { 4 (nπ) 3 ( )n nπ if n is odd, if n is even. + ( )n nπ (( )n ) (nπ) 3 Thus Hence the sine series in b n 8 π 3 { 8 (nπ) 3 if n is odd, if n is even, k sin(k +)πx (k +) 3.

21 Section.4 Half-Range Expansions: The Cosine and Sine Series 5 b k_ 8 Pi^3 k ^3; ss n_, x_ : Sum b k Sin k Pi x, k,, n ; partialsineseries Table ss n, x, n,, 5 ; f x_ x x Plot Evaluate partialsineseries, f x, x,, Perfect! 3. We have sin πx cos πx sin πx. This yields the desired -periodic sine series expansion. 7. (b) Sine series expansion: b n p h ap h pa a h nπx x sin a p dx + p [ x p nπ + h (a p)p [ ap nπ cos nπx p p a a + p nπ [ (x p) ( p) nπ h a p a cos(nπx p ) p nπa ] sin p (x p) sin nπx p cos nπx p dx ] p + a a nπa cos p + p (nπ) + h [ p nπa (a p) cos (a p)p nπ p p nπa ] sin (nπ) p hp nπa[ sin (nπ) p a ] a p hp nπa (nπ) sin (p a)a p. Hence, we obtain the given Fourier series. dx p nπx ] cos dx nπ p

22 6 Chapter Fourier Series Solutions to Exercises.5. We have The Fourier series representation is { if <x<, f(x) if <x<; f(x) 4 π k The mean square error (from (5)) is E N sin(k +)πx. k + f (x) dx a N (a n + b n). n 4 In this case, a n for all n, b k,b k+ π(k+), and f (x) dx dx. So E (b ) 8 π.89. Since b, it follows that E E. Finally, 5. We have E (b + b 3) 8 π 8 9 π.99. E N f (x) dx a N b n 8 π 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 We have f(x) π x x 3 for π <x<πand, for n, b n n 3 ( ) n+.by Parseval s identity n Simplifying, we find that ( ) n 3 π ζ(6) n π π π ( π x x 3) ( π 4 x π x 4 + x 6) dx π ( ) π 4 π 3 x3 π 5 x5 + x7 π 7 ( π ) π6. n 6 (8)() (5)(44) π6 π6 945.

23 Section.5 Mean Square Approximation and Parseval s Identity 7 3. For the given function, we have b n and a n n. By Parseval s identity, we have π f (x) dx π π n 4 f π5 (x) dx π πζ(4) n4 9, π n π n where we have used the table preceding Exercise 7 to compute ζ(4). 7. For the given function, we have a,a n 3,b n n for n. n By Parseval s identity, we have π f (x) dx a + (a n + b π π n) n + ( (3 n ) + n ) n n Using a geometric series, we find By Exercise 7(a), So π π n + n 9 n 9 n n π 6. n 9 n n n n. f (x) dx π( π 6 )7π 8 + π3 6. n

24 8 Chapter Fourier Series Solutions to Exercises.6. From Example, for a, ±i, ±i, ±3i,..., e ax sinh πa π n ( ) n a in einx ( π <x<π); consequently, e ax and so, for π <x<π, sinh πa π n ( ) n a + in einx ( π <x<π), cosh ax eax + e ax sinh πa ( ( ) n π a + in + ) e inx a in n a sinh πa ( ) n π n + a einx. n. From Example, for a, ±i, ±i, ±3i,..., e ax sinh πa π n ( ) n a in einx ( π <x<π); consequently, e ax and so, for π <x<π, sinh πa π n ( ) n a + in einx ( π <x<π), sinh ax eax e ax sinh πa ( ( ) n π a in ) e inx a + in n i sinh πa ( ) n n π n + a einx. n 5. Use identities (); then cos x + sin 3x eix + e ix ie 3ix + e ix + e3ix e 3ix i + eix ie3ix. 9. If m n then p e i mπ p x e i nπ p x dx p p p p p e i mπ p x mπ i e p x dx p dx. p p

25 Section.6 Complex Form of Fourier Series 9 If m n, then p e i mπ p x e i nπ p x dx p p p p p e i (m n)π p x dx i (m n)π p (m n)π ei p x p i (e i(m n)π e i(m n)π) (m n)π i (cos[(m n)π] cos[ (m n)π]). (m n)π 3. (a) At points of discontinuity, the Fourier series in Example converges to the average of the function. Consequently, at x π the Fourier series converges to e aπ +e aπ cosh(aπ). Thus, plugging x π into the Fourier series, we get cosh(aπ) sinh(πa) π The sum n Hence n in a +n in a +n cosh(aπ) sinh(πa) π n ( ) n ( ) n {}}{ (a + in) e inπ a + n sinh(πa) π is the limit of the symmetric partial sums i and so n N n N n a + n. a a + n coth(aπ) a π upon dividing both sides by sinh(aπ). Setting t aπ, weget coth t t π n ( t π ) + n n n n t t +(πn), (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 kπi, where kis an integer. 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 + i sin 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

26 Chapter Fourier Series 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 + i sin bx ) + C e ax [( ) ( )] a cos bx + b sin bx + i b cos bx + a sin bx + C. a + b (c) Equating real and imaginary parts in (b), we obtain and. By Exercise 9, π I I eax ( ) a cos bx + b sin bx a + b eax a + b ( b cos bx + a sin bx ). ( e it +e it) dt i eit + π i e it {}}{{}}{ i e πi +i e πi ( i + i). Of course, this result follows from the orthogonality relations of the complex exponential system (formula (), with p π). 5. First note that +it it ( + it) ( it)( + it) t +it +t t +t + i t +t. Hence +it t it dt +t dt + i t +t dt ( + +t dt + i t +t dt t + tan t + i ln( + t )+C.

27 Section.7 Forced Oscillations 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 cos t, 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 Thus α 75 and β Ba, where A A 3 and B 4. +B 5 3 and β 5 4, and hence a particular solution is y p 3 cos t + 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 cos t y c e t + c te t 3 cos t + 4 sin t. (b) Since lim t c e t + c te t, it follows that the steady-state solution is y s 3 cos t + 4 sin t. 5. (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 Hence the steady-state solution is (b) We have 4 3 8, , 4 3 8, / y p 8 cos t + 8 sin t + 4 cos t sin t y p 8 cos t + 8 sin t + 4 cos t sin t, 65 3 (y p ) 8 sin t + 8 cos t 8 65 sin t cos t, 65 (y p ) 8 cos t 6 sin t cos t + sin t, ( 8 ) 65 ( 65 ) (y p ) +4(y p ) +5y p cos t + sin t ( ) sin t + 3 ( ) sin t sin t, 3 sin t + ( ) cos t which shows that y p is a solution of the nonhomogeneous differential equation.

28 Chapter Fourier Series 9. (a) Natural frequency of the spring is ω k 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 π-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 3 4 sin 3t, the 3π component of the Fourier series of F (t) that corresponds to n 3. We have a 3, b 3 4 3π, µ,c.5, k., A 3. 9., B 3 3(.5).5, α 3 B 3b 3 A 3 + B 3 (.5)(4)/(3π) (.) +(.5).6 and β 3 A 3b 3 A 3 + B 3.4. So y 3.6 cos 3t +.4 sin 3t. 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 must remove f 3 (t) 4sin3t 4sin3t 3π. Thus subtract 3π from the input function. The modified input function is F (t) 4sin3t 3π. 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 π sin t + 4 π m sin(m +)t. m + (b) The modified steady-state solution does not have the y 3 -component that we found in Exercise 3. We compute its normal modes by appealing to Theorem and using as an input function F (t) f 3 (t). The first nonzero mode is y ; the second nonzero normal mode is y 5. We compute them with the help of Mathematica. Let us first enter the parameters of the problem and compute α n and β n, using the definitions from Theorem. The input/output from Mathematica is the following

29 Section.7 Forced Oscillations 3 Clear a, mu, p, k, alph, bet, capa, capb, b, y mu ; c 5 ; k ; p Pi; a ; a n_ ; b n_ Pi n Cos npi ; alph a k; capa n_ k mu npi p ^ capb n_ cnpi p alph n_ capa n a n capb n b n capa n ^ capb n ^ bet n_ capa n b n capb n a n capa n ^ capb n ^ n n n 4 Cos n n n Cos n n n 4 n It appears that ( cos(nπ)) α n ( n 4 + ( ) n) π and β n ( ( n) ( cos(nπ)) n n 4 + ( ) n) π Note how these formulas yield when n is even. The first two nonzero modes of the modified solution are y (t) α cos t + β sin t.784 cos t +.43 sint and y 5 (t) α 5 cos 5t + β 5 sin 5t.8 cos 5t.698 sin5t. (c) In what follows, we use nonzero terms of the original steady-state solution and compare it with nonzero terms of the modified steady-state solution. The graph of the original steady-state solution looks like this: steadystate t_ Sum alph n Cos nt bet n Sin nt, n,, ; Plot Evaluate steadystate t, t,, 4 Pi

30 4 Chapter Fourier Series The modified steady-state is obtained by subtracting y 3 from the steady-state. Here is its graph. modifiedsteadystate t_ steadystate t alph 3 Cos 3t bet 3 Sin 3t ; Plot Evaluate modifiedsteadystate t, t,, 4 Pi In order to compare, we plot both functions on the same graph. Plot Evaluate steadystate t, modifiedsteadystate t, t,, 4 Pi It seems like we were able to reduce the amplitude of the steady-state solution by a factor of or 3 by removing the third normal mode. Can we do better? Let us analyze the amplitudes of the normal modes. These are equal to α n + β n.we have the following numerical values: amplitudes N Table Sqrt alph n ^ bet n ^, n,,.43,.,.4565,.,.69855,., ,.,.9979,.,.487,.,.668,.,.39489,.,.68454,.,.994,. It is clear from these values that y 3 has the largest amplitude (which is what we expect) but y also has a relatively large amplitude. So, by removing the first component of F, we remove y, and this may reduce the oscillations even further. Let s see the results. We will plot the steady-state solution y s, y s y 3, and y s y y 3.

31 Section.7 Forced Oscillations 5 modifiedfurther t_ modifiedsteadystate t alph Cos t bet Sin t ; Plot Evaluate modifiedfurther t, steadystate t, modifiedsteadystate t, t,, 4 Pi (a) The input function F (t) is already given by its Fourier series: F (t) cos t+sin 3t. Since the frequency of the component sin 3t of the input function is 3 and is equal to the natural frequency of the spring, resonance will occur (because there is no damping in the system). The general solution of y +9y cos t+sin 3t is y y h + y p, where y h is the general solution of y +9y andy p is a particular solution of the nonhomogeneous equation. We have y h c sin 3t + c cos 3t and, to find y p, we apply Exercise and get ( a y p cos t + b ) sin t + R(t), A A where a,b,a 9 5,a n,b n, and R(t) t cos 3t. 6 Hence y p 5 cos t t cos 3t 6 and so the general solution is y c sin 3t + c cos 3t + 5 cos t t cos 3t. 6 (b) To eliminate the resonance from the system we must remove the component of F that is causing resonance. Thus add to F (t) the function sin 3t. The modified input function becomes F modified (t) cos t. 5. The general solution is y c sin 3t + c cos 3t + 5 cos t t 6 cos 3t. Applying the initial condition y() we get c + 5 orc 5. Thus y c sin 3t 5 cos 3t + 5 cos t t cos 3t. 6 Applying the initial condition y (), we obtain y 3c cos 3t sin 3t 6 5 sin t 6 cos 3t + t sin 3t, y () 3c 6, y () c 8.

32 6 Chapter Fourier Series Thus y 8 sin 3t 5 cos 3t + 5 cos t t cos 3t. 6

33 Section.9 Uniform Convergence and Fourier Series 7 Solutions to Exercises.9. f n (x) sin nx asn. n n The sequence converges uniformly to for all real x, because n controls its size independently of x. 5. If x then f n () for all n. Ifx, then applying l Hospital s rule, we find lim f n n(x) x lim x lim. n n e nx n x e n The sequence does not converge uniformly on any interval that contains because f n ( n )e, which does not tend to. 9. cos kx k k M k for all x. Since M k < (p-series with p>), the series converges uniformly for all x. 7. ( )k x +k k M k for all x. Since M k < (p-series with p>), the series converges uniformly for all x.

34 8 Chapter Fourier Series Solutions to Exercises. 5. The cosine part converges uniformly for all x, by the Weierstrass M-test. The sine part converges for all x by Theorem (b). Hence the given series converges for all x. 9. (a) If lim k sin kx, then lim k sin kx lim ( k cos kx) lim k cos kx ( ). Also, if lim k sin kx, then lim k sin(k +)x). But sin(k +)x sin kx cos x + cos kx sin x, so lim k ( {}}{ sin kx cos x + cos kx sin x) lim k lim k cos kx sin x cos kx or sin x. By (*), cos kx does not tend to, so sin x, implying that x mπ. Consequently, if x mπ, then limk sin kx is not and the series k sin kx does not converge by the nth term test, which proves (b).

35 Section 3. Partial Differential Equations in Physics and Engineering 9 Solutions to Exercises 3.. u xx + u xy u is a second order, linear, and homogeneous partial differential equation. u x (,y) is linear and homogeneous. 5. u t u x +u xt u is second order and nonlinear because of the term u t u x. u(,t)+ u x (,t) is linear and homogeneous. 9. (a) Let u(x, y) e ax e by. Then So u x ae ax e by u y be ax e by u xx a e ax e by u yy b e ax e by u xy abe ax e by. Au xx +Bu xy + Cu yy + Du x + Eu y + Fu Aa e ax e by +Babe ax e by + Cb e ax e by +Dae ax e by + Ebe ax e by + Fe ax e by e ax e by( Aa +Bab + Cb + Da + Eb + F ) Aa +Bab + Cb + Da + Eb + F, because e ax e by for all x and y. (b) By (a), in order to solve u xx +u xy + u yy +u x +u y + u, we can try u(x, y) e ax e by, where a and b are solutions of a +ab + b +a +b +. But a +ab + b +a +b +(a + b +). So a + b +. Clearly, this equation admits infinitely many pairs of solutions (a, b). Here are four possible solutions of the partial differential equation: a,b u(x, y) e x e y a,b u(x, y) e y a /, b / u(x, y) e x/ e y/ a 3/, b/ u(x, y) e 3x/ e y/ 3. We follow the outlined solution in Exercise. We have A(u) ln(u), φ(x) e x, A(u(x(t)), t)) A(φ(x())) ln(e x() )x(). So the characteristic lines are x tx() + x() x() L(xt) x t +. ( ) x So u(x, t) f(l(x, t)) f. The condition u(x, ) e x implies that f(x) t+ e x and so u(x, t) e x t+. Check: u t e x t+ x (t+),u x e x t+ t+, u t + ln(u)u x e x t+ x (t +) + x t + e x t+ t +.

36 3 Chapter 3 Partial Differential Equations in Rectangular Coordinates 7. We have A(u) u, φ(x) x, A(u(x(t)), t)) A(φ(x())) x(). So the characteristic lines are x tx() + x() x()(t +) x. Solving for x(), we find and so Now So x() u(x, t) f x t +, ( ) x. t + u(x, ) f(x) x. x u(x, t) t +.

37 Section 3.3 Wave Equation, the Method of Separation of Variables 3 Solutions to Exercises 3.3. The solution is u(x, t) n sin nπx L ( b n cos c nπt L + b n sin c nπt ), L where b n are the Fourier sine coefficients of f and b n are times the Fourier coefficients of g. In this exercise, b n, since g,b.5; and b n for all n>, because f is already given by its Fourier sine series (period ). So u(x, t).5 sinπx cos t. 5. (a) The solution is u(x, t) L cnπ sin(nπx)(b n cos(4nπt)+b n sin(4nπt)), n where b n is the nth sine Fourier coefficient of f and b n is L/(cn) times the Fourier coefficient of g, where L and c 4. Since g, we have b n for all n. As for the Fourier coefficients of f, we can get them by using Exercise 7, Section.4, with p,h, and a /. We get Thus u(x, t) 8 π n 8 π k sin nπ n b n 8 nπ π sin n. sin(nπx) cos(4nπt) ( ) k sin((k +)πx) cos(4(k +)πt). (k +) (b) Here is the initial shape of the string. Note the new Mathematica command that we used to define piecewise a function. (Previously, we used the If command.) Clear f f x_ : x ; x f x_ : x ; x Plot f x, x,, Because the period of cos(4(k + )πt) is /, the motion is periodic in t with period /. This is illustrated by the following graphs. We use two different ways to plot the graphs: The first uses simple Mathematica commands; the second one is more involved and is intended to display the graphs in a convenient array.

38 3 Chapter 3 Partial Differential Equations in Rectangular Coordinates Clear partsum partsum x_, t_ : 8 Pi^ Sum Sin ^k k Pi x Cos 4 k Pi t k ^, k,, Plot Evaluate partsum x,, f x, x,, Here is the motion in an array. tt Table Plot Evaluate partsum x, t, x,,, PlotRange,,,, Ticks.5,,.5,.5,, DisplayFunction Identity, t,,, ; Show GraphicsArray Partition tt, The solution is u(x, t) sin(nπx)(b n cos(nπt)+b n sin(nπt)), n

39 Section 3.3 Wave Equation, the Method of Separation of Variables 33 where b π and all other b n. The Fourier coefficients of f are b n x( x) sin(nπx) dx. To evaluate this integral, we will use integration by parts to derive first the formula: for a, x sin(ax) dx x cos(ax) a + sin(ax) a + C, and x sin(ax) dx cos(ax) a 3 x cos(ax) a + x sin(ax) a + C; thus x( x) sin(ax) dx cos(ax) a 3 x cos(ax) a + x cos(ax) a Applying the formula with a nπ, weget x( x) sin(nπx) dx + sin(ax) a x sin(ax) a + C. cos(nπ x) x cos(nπ x) (nπ) 3 + x cos(nπ x) sin(nπ x) x sin(nπ x) + nπ nπ (nπ) (nπ) (( )n ) (nπ) 3 { 4 (nπ) 3 ( )n nπ if n is odd, if n is even. + ( )n nπ (( )n ) (nπ) 3 Thus b n { 8 (nπ) 3 if n is odd, if n is even, and so u(x, t) 8 π 3 3. To solve k sin((k +)πx) cos((k +)πt) (k +) 3 + sin(πx) sin(πt). π u t + u t u x, u(,t)u(π, t), u u(x, ) sin x, (x, ), t we follow the method of the previous exercise. We have c,k.5, L π, f(x) sin x, and g(x). Thus the real number Lk cπ.5 is not an

40 34 Chapter 3 Partial Differential Equations in Rectangular Coordinates integer and we have n> kl π for all n. So only Case III from the solution of Exercise needs to be considered. Thus u(x, t) e.5t sin nx ( a n cos λ n t + b n sin λ n t, n where Setting t, we obtain λ n (.5n). sin x a n sin nx. n Hence a and a n for all n>. Now since b n ka n λ n + λ n L L g(x) sin nπ x dx, n,,..., L it follows that b n for all n> and and the solution takes the form u(x, t) e.5t sin x ( cos λ t + b sin λ t ), where λ (.5).75 3 and b ka λ 3. So u(x, t) e.5t sin x ( cos( 3 t)+ 3 sin( 3 t)).

41 Section 3.4 D Alembert s Method 35 Solutions to Exercises 3.4. We will use (5), since g. The odd extension of period of f(x) sin πx is f (x) sin πx. So u(x, t) [ t sin(π(x + π )) + sin(π(x t π ))] [ ] sin(πx + t) + sin(πx t). 5. The solution is of the form u(x, t) [ f (x t)+f(x + t) ] + [ G(x + t) G(x t) ] [( f (x t) G(x t) ) + ( f (x + t)+g(x + t) )], where f is the odd extension of f and G is as in Example 3. In the second equality, we expressed u as the average of two traveling waves: one wave traveling to the right and one to the left. Note that the waves are not the same, because of the G term. We enter the formulas in Mathematica and illustrate the motion of the string. The difficult part in illustrating this example is to define periodic functions with Mathematica. This can be done by appealing to results from Section.. We start by defining the odd extensions of f and G (called big g) on the interval [-, ]. Clear f, bigg f x_ : x ; x f x_ : x ; x f x_ : x ; x bigg x_ x^ Plot f x, bigg x, x,, Here is a tricky Mathematica construction. (Review Section..)

42 36 Chapter 3 Partial Differential Equations in Rectangular Coordinates extend x_ : x Floor x periodicf x_ : f extend x periodicbigg x_ : bigg extend x Plot periodicf x, periodicbigg x, x, 3, Because f and G are -periodic, it follows immediately that f (x ± ct) and G(x ± ct) are /c-periodic in t. Since c, u is -periodic in t. The following is an array of snapshots of u. You can also illustrate the motion of the string using Mathematica (see the Mathematica notebooks). Note that in this array we have graphed the exact solution and not just an approximation using a Fourier series. This is a big advantage of the d Alembert s solution over the Fourier series solution.

43 Section 3.4 D Alembert s Method 37 u x_, t_ : periodicf x t periodicf x t periodicbigg x t periodicbigg x t tt Table Plot Evaluate u x, t, x,,, PlotRange,,,, Ticks.5,,.5,.5,, DisplayFunction Identity, t,,.3, 5 ; Show GraphicsArray Partition tt, 4 9. You can use Exercise, Section 3.3, which tells us that the time period of motion is T L c. So, in the case of Exercise, T π, and in the case of Exercise 5, T. You can also obtain these results directly by considering the formula for u(x, t). In the case of Exercise, u(x, t) [ sin(πx + t)+ sin(πx t) ] so u(x, t +π) ] [ sin(πx + tπ) + sin(πx tπ) u(x, t). In the case of Exercise 5, use the fact that f and G are both -periodic. 3. We have u(x, t) [f (x + ct)+f (x ct)] + c where f and g are odd and L-periodic. So x+ct x ct g (s) ds, x+ct+l u(x, t + L c ) [f (x + ct + L)+f (x ct L)] + g (s) ds. c x ct L Using the fact that f is odd, L-period, and satisfies f (L x) f (x) (this property is given for f but it extends to f ), we obtain f (x + ct + L) f (x + ct + L L) f (x + ct L) f (L x ct) f (L (x + ct)) f (x + ct). Similarly f (x ct L) f (L x + ct) f (L x + ct) f (L (x ct)) f (x ct).

44 38 Chapter 3 Partial Differential Equations in Rectangular Coordinates Also g (s + L) g ( s L) g ( s L +L) g (L s) g (s), by the given symmetry property of g. So, using a change of variables, we have x+ct+l g (s) ds x+ct g (s + L) ds x+ct g (s) ds. c x ct L c x ct c x ct Putting these identities together, it follows that u(x, t + L c ) u(x, t). 7. (a) To prove that G is even, see Exercise 4(a). That G is L-periodic follows from the fact that g is L-periodic and its integral over one period is, because it is odd (see Section., Exercise 5). Since G is an antiderivative of g, to obtain its Fourier series, we apply Exercise 33, Section 3.3, and get G(x) A L π n b n (g) n cos nπ L x, where b n (g) is the nth Fourier sine coefficient of g, b n (g) L L g(x) sin nπ L xdx and A L π n b n (g) n. In terms of b n, we have and so L b n (g) π n nπ G(x) L π n (b) From (a), it follows that n cb n L b n (g) n g(x) sin nπ L xdx cb n, L π n b n (g) n ( cos( nπ ) L x). cos nπ L x G(x + ct) G(x ct) n cb n cb n n [( cos( nπ ) ( L (x + ct)) cos( nπ )] L (x ct)) [ cos( nπ ] (x + ct)) cos(nπ L L (x ct))

45 Section 3.4 D Alembert s Method 39 (c) Continuing from (b) and using the notation in the text, we obtain c x+ct x ct g (s) ds [G(x + ct) G(x ct)] c b [ n cos( nπ ] (x + ct)) cos(nπ L L (x ct)) n b n sin( nπ L x) sin(nπ L ct) n n b n sin(nπ L x) sin(λ nt). (d) To derive d Alembert s solution from (8), Section 3.3, proceed as follows: u(x, t) n b n sin( nπ L x) cos(λ nt)+ b n sin( nπ L x) sin(λ nt) n ( f (x ct)+f (x + ct) ) + [G(x + ct) G(x ct)], c where in the last equality we used Exercise 6 and part (c).. Follow the labeling of Figure 8 in Section 3.4. Let P (x,t )bean arbitrary point in the region II. Form a characteristic parallelogram with vertices P, P, Q, Q, as shown in Figure 8 in Section 3.4. The vertices P and Q are on the characteristic line x +t and the vertex Q is on the boundary line x. From Proposition, we have u(p )u(q )+u(q ) u(p )u(q ) u(p ), because u(q ). We will find u(p ) and u(q ) by using the formula u(x, t) 4t + x x +8tx from Example 4, because P and Q are in the region I. The point Q is the intersection point of the characteristic lines x t x t and x +t. Adding the equations and then solving for x, we get x x + t. The second coordinate of Q is then t x +t. 4 The point Q is the intersection point of the characteristic line x +t x +t and x. Thus t x +t. The point P is the intersection point of the characteristic lines x+t and x t (x +t ). Solving for x and t, we find the coordinates of P to be x 3 x t and t +x +t. 4

46 4 Chapter 3 Partial Differential Equations in Rectangular Coordinates To simplify the notation, replace x and t by x and y in the coordinates of the points Q and P and let φ(x, t) 4t + x x +8tx. We have u(x, t) u(q ) u(p ) ( x + t φ, x +t ) 4 5 t 5x +tx, ( 3 x t φ, ) +x +t 4 where the last expression was derived after a few simplifications that we omit. It is interesting to note that the formula satisfies the wave equation and the boundary condition u(, t) for all t>. Its restriction to the line x +t (part of the boundary of region I) reduces to the formula for u(x, t) for (x, t) in region I. This is to be expected since u is continuous in (x, t).

47 Section 3.5 The One Dimensional Heat Equation 4 Solutions to Exercises 3.5. Multiply the solution in Example by We have where So b n u(x, t) 3 π u(x, t) k e (k+) t k + to obtain sin(k +)x. b n e (nπ)t sin(nπx), n [ x sin(nπx) dx x cos(nπx) + sin(nπx) ] nπ n π cos nπ nπ u(x, t) π ( )n+. nπ n ( ) n+ e (nπ)t sin(nπx). n 9. (a) The steady-state solution is a linear function that passes through the points (, ) and (, ). Thus, u(x) x. (b) The steady-state solution is a linear function that passes through the points (, ) and (, ). Thus, u(x). This is also obvious: If you keep both ends of the insulated bar at degrees, the steady-state temperature will be degrees. 3. We have u (x) 5 π x +. We use (3) and the formula from Exercise, and get (recall the Fourier coefficients of f from Exercise 3) u(x, t) 5 π x + [ 3 sin(n π + ) ( )] ( ) n π n e nt sin nx. nπ n 7. Fix t > and consider the solution at time t t : u(x, t ) b n sin nπ nt L xe λ. n We will show that this series converges uniformly for all x (not just x L) by appealing to the Weierstrass M-test. For this purpose, it suffices to establish the following two inequalities: (a) b n sin nπ L xe λ nt M n for all x; and (b) n M n <. To establish (a), note that b n L f(x) sin nπ L L xdx L f(x) sin nπ L L x dx (The absolute value of the integral is the integral of the absolute value.) L L f(x) dx A (because sin u for all u).