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

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1 Instructor s Solutions Mnul PARTIA DIFFERENTIA EQUATIONS with FOURIER SERIES nd BOUNDARY VAUE PROBEMS Second Edition NAKHÉ H. ASMAR University of Missouri

2 Contents Prefce Errt v vi A Preview of Applictions nd Techniques. Wht Is Prtil Differentil Eqution?. Solving nd Interpreting Prtil Differentil Eqution 4 Fourier Series 3. Periodic Functions 3. Fourier Series.3 Fourier Series of Functions with Arbitrry Periods 35.4 Hlf-Rnge Epnsions: The Coe nd Sine Series 5.5 Men Squre Approimtion nd Prsevl s Identity 58.6 Comple Form of Fourier Series 63.7 Forced Oscilltions 73 Supplement on Convergence.9 Uniform Convergence nd Fourier Series 79. Dirichlet Test nd Convergence of Fourier Series 8 3 Prtil Differentil Equtions in Rectngulr Coordintes 8 3. Prtil Differentil Equtions in Physics nd Engineering Solution of the One Dimensionl Wve Eqution: The Method of Seprtion of Vribles D Alembert s Method The One Dimensionl Het Eqution Het Conduction in Brs: Vrying the Boundry Conditions The Two Dimensionl Wve nd Het Equtions plce s Eqution in Rectngulr Coordintes Poisson s Eqution: The Method of Eigenfunction Epnsions Neumnn nd Robin Conditions 5 4 Prtil Differentil Equtions in Polr nd Cylindricl Coordintes The plcin in Vrious Coordinte Systems 55

3 Contents iii 4. Vibrtions of Circulr Membrne: Symmetric Cse Vibrtions of Circulr Membrne: Generl Cse plce s Eqution in Circulr Regions plce s Eqution in Cylinder The Helmholtz nd Poisson Equtions 97 Supplement on Bessel Functions 4.7 Bessel s Eqution nd Bessel Functions Bessel Series Epnsions Integrl Formuls nd Asymptotics for Bessel Functions 8 5 Prtil Differentil Equtions in Sphericl Coordintes 3 5. Preview of Problems nd Methods 3 5. Dirichlet Problems with Symmetry Sphericl Hrmonics nd the Generl Dirichlet Problem The Helmholtz Eqution with Applictions to the Poisson, Het, nd Wve Equtions 4 Supplement on egendre Functions 5.5 egendre s Differentil Eqution egendre Polynomils nd egendre Series Epnsions 5 6 Sturm iouville Theory with Engineering Applictions Orthogonl Functions Sturm iouville Theory The Hnging Chin Fourth Order Sturm iouville Theory The Bihrmonic Opertor Vibrtions of Circulr Pltes 69

4 iv Contents 7 The Fourier Trnsform nd Its Applictions 7 7. The Fourier Integrl Representtion 7 7. The Fourier Trnsform The Fourier Trnsform Method The Het Eqution nd Guss s Kernel A Dirichlet Problem nd the Poisson Integrl Formul The Fourier Coe nd Sine Trnsforms Problems Involving Semi-Infinite Intervls Generlized Functions The Nonhomogeneous Het Eqution Duhmel s Principle 37

5 Prefce This mnul contins solutions with notes nd comments to problems from the tetbook Prtil Differentil Equtions with Fourier Series nd Boundry Vlue Problems Second Edition Most solutions re supplied with complete detils nd cn be used to supplement emples from the tet. There re lso mny figures nd numericl computions on Mthemtic tht cn be very useful for clssroom presenttion. Certin problems re followed by discussions tht im to generlize the problem under considertion. I hope tht these notes will sevre their intended purpose: To check the level of difficulty of ssigned homework problems; To verify n nswer or nontrivil computtion; nd to provide worked solutions to students or grders. As of now, only problems from Chpters 7 re included. Solutions to problems from the remining chpters will be posted on my website nkhle s I complete them nd will be included in future versions of this mnul. I would like to thnk users of the first edition of my book for their vluble comments. Any comments, corrections, or suggestions from Instructors would be gretly pprecited. My e-mil ddress is nkhle@mth.missouri.edu Nkhlé H. Asmr Deprtment of Mthemtics University of Missouri Columbi, Missouri 65

6 Errt The following mistkes ppered in the first printing of the second edition. Corrections in the tet nd figures p. 4, Eercise #3 is better done fter Section 4.4. p. 68, Eercise #8(b), n should be even. p.387, Eercise#, use y = I () not y = J (). p. 45 Figures 5 nd 6: Relbel the ticks on the -is s π, π/, π/, π, insted of π, π, π, π. p. 467, l. ( 3): Chnge reference () to (). p. 477 l. : (t) (, t). p. 477 l. 9: Chnge intervl to tringle p. 487, l.: Chnge is the equl to is equl p. 655, l.3: Chnge ln ln( + y ) to ln( + y ). Corrections to Answers of Odd Eercises Section 7., # 7: Chnge i to i. Section 7.8, # 3: f() = 3 for <<3 not <<. Section 7.8, # 35: Section 7.8, # 37: i π π (e iw ) w w 3 3 j= j (jw) 3 Section 7.8, # 5: π [δ δ ]. Section 7.8, # 57: The given nswer is the derivtive of the rel nswer, which should be ( ( +) ( ) ( ) ( ) ( ) ) U U +( +) U U + U U 3 +( +4) U3 U 4 π Section 7.8, # 59: The given nswer is the derivtive of the rel nswer, which should( be π ( +3) ( ) ( ) ( ) U 3 U +( +5) U U +( +4) U U +( +4) ( ) ( ) ( ) ) U U +( +5) U U +( +3) U U 3 Section 7., # 9: [ t ( + t)+ cos( + t) cos( t)]. Any suggestion or correction would be gretly pprecited. Plese send them to my e-mil ddress nkhle@mth.missouri.edu Nkhlé H. Asmr Deprtment of Mthemtics University of Missouri Columbi, Missouri 65

7 Section. Wht Is Prtil Differentil Eqution? Solutions to Eercises.. If u nd u re solutions of (), then t + = nd t + =. Since tking derivtives is liner opertion, we hve t (c u + c u )+ (c u + c u ) = c t + c t + c + c = {( }}{ = c t + ) {( }} ){ +c t ++ =, showing tht c u + c u is solution of ().. () Generl solution of (): u(, t) =f( t). On the -is (t = ): u(, ) = e = f( ) = f(). So u(, t) =f( t) =( t)e ( t). 3. () Generl solution of (): u(, t) =f( t). On the t-is ( = ): u(, t)= t = f( t) =f( t). Hence f(t) = t nd so u(, t) =f( t) = ( t) =t. 4. et α = + bt, β = c + dt, then = α α + β β = α + c β = α t α t + β β t = b α + d β. So () becomes ( + b) +(c + d) α β =. et =,b=,c=,d=. Then = α α =, which implies tht u is function of β only. Hence u = f(β) =f( t). 5. et α = + bt, β = c + dt, then Reclling the eqution, we obtin = α α + β β = α + c β = α t α t + β β t = b α + d β. t = (b ) +(d c) α β =. et =,b=,c=,d=. Then = u = f(β) u(, t) =f( + t), α where f is n rbitrry differentible function (of one vrible). 6. The solution is very similr to Eercise 5. et α = + bt, β = c + dt, then = α α + β β = α + c β = α t α t + β β t = b α + d β. =

8 Chpter A Preview of Applictions nd Techniques Reclling the eqution, we obtin t +3 = (b +3) +(d +3c) α β =. et =,b=,c=,d= 3. Then the eqution becomes = u = f(β) u(, t) =f(c + dt) =f( 3t), α where f is n rbitrry differentible function (of one vrible). 7. et α = + bt, β = c + dt, then = α α + β β = α + c β = α t α t + β β t = b α + d β. The eqution becomes (b ) +(d c) α β =. et =,b=,c=, d =. Then β = β =. Solving this ordinry differentil eqution in β, we get u = β + f(α) oru(, t) = + f( +t). 8. et α = c + d t, β = c + d t, then The eqution becomes = c α + c β = d t α + d β. (d + bc ) α +(d + bc ) β = u. et c =, d = b, c =,d = ( ). Then β = u u = f(α)eβ. Hence u(, t) =f( bt)e t. 9. () The generl solution in Eercise 5 is u(, t) =f( + t). When t =, we get u(, ) = f() =/( + ). Thus u(, t) =f( + t) = ( + t) +. (c) As t increses, the wve f() = + moves to the left. - - Figure for Eercise 9(b). 3. () The directionl derivtive is zero long the vector (, b). (b) Put the eqution in the form + b y =( ). The chrcteristic curves re obtined by solving dy d = b y = b + C y b = C.

9 Section. Wht Is Prtil Differentil Eqution? 3 et φ(, y) =y b. The chrcteristic curves re the level curves of φ. (c) The solution of + b y = is of the form u(, y) =f(φ(, y)) = f(y b ), where f is differentible function of one vrible.. The chrcteristic curves re obtined by solving dy d = y = C y 3 3 = C. et φ(, y) =y 3 3. The chrcteristic curves re the level curves of φ. The solution of is of the form u(, y) =f(φ(, y)) = f(y 3 3 ), where f is differentible function of one vrible.. We follow the method of chrcteristic curves. et s find the chrcteristic curves. For, + y y =; dy d = y y = C, where we hve used Theorem, Appendi A., to solve the lst differentil eqution. Hence the chrcteristic curves re y = C nd the solution of the prtil differentil eqution is u(, y) =f ( y ). To verify the solution, we use the chin rule nd get u = y f ( ) y nd uy = f ( y ). Thus u + yu y =, s desired. 3. To find the chrcteristic curves, solve dy d =. Hence y = cos + C or y + cos = C. Thus the solution of the prtil differentil eqution is u(, y) = f (y + cos ). To verify the solution, we use the chin rule nd get u = f (y + cos ) nd u y = f (y + cos ). Thus u + u y =, s desired. 4. Put the eqution in the form + e y =. The chrcteristic curves re obtined by solving dy d = e y = e + C y + e = C. et φ(, y) = y + e. The chrcteristic curves re the level curves of φ. The solution is of the form u(, y) =f(φ(, y)) = f(y + e ), where f is differentible function of one vrible.

10 4 Chpter A Preview of Applictions nd Techniques Eercises.. We hve So t ( ) = t t u t ( ) v nd ( ) v = t = v t nd v t = u. ( ). Assuming tht v t = v t, it follows tht u t = u, which is the one dimensionl wve eqution with c =. A similr rgument shows tht v is solution of the one dimensionl wve eqution.. () For the wve eqution in u, the pproprite initil conditions re u(, ) = f(), s given, nd u t (, ) = v (, ) = h (). (b) For the wve eqution in v, the pproprite initil conditions re v(, ) = h(), s given, nd v t (, ) = u (, ) = f (). 3. u = F ( + ct)+g ( + ct), u tt = c F ( + ct)+c G( ct). So u tt = c u, which is the wve eqution. 4. () Ug the chin rule in two dimensions: Similrly = α u = α + β β = α + β ( α + ) β = u α + u β α + u α β + u β = u α + u β α + u β. = α t α t + β β t = c α c β u = ( c t t α c ) β = c u α c u β α c u α β + c u β = c u α c u β α + c u β. Substituting into the wve eqution, it follows tht c u α + u β α + c u β = c u α u β α + c u β u α β =. (b) The lst eqution sys tht β is constnt in α. So β = g(β) where g is n rbitrry differentible function. (c) Integrting the eqution in (b) with respect to β, we find tht u = G(β)+F (α), where G is n ntiderivtive of g nd F is function of α only. (d) Thus u(, t) =F ( + ct)+g( ct), which is the solution in Eercise () We hve u(, t) =F ( + ct) +G( ct). To determine F nd G, we use the initil dt: u(, ) = + F ()+G() = + ; ()

11 Section. Solving nd Interpreting Prtil Differentil Eqution 5 t (, ) = cf () cg () = F () =G () F () =G()+C, () where C is n rbitrry constnt. Plugging this into (), we find G()+C = G() = [ ] + + C ; nd from () Hence F () = u(, t) =F ( + ct)+g( ct) = [ ] + + C. [ ] +( + ct) + +( ct). 6. We hve u(, t) = F ( + ct) + G( ct). To determine F nd G, we use the initil dt: u(, ) = e F ()+G() =e ; () t (, ) = cf () cg () = F () =G () F () =G()+C, () where C is n rbitrry constnt. Plugging this into (), we find G()+C = e G() = [ ] e C ; nd from () F () = [ ] e + C. Hence u(, t) =F ( + ct)+g( ct) = [ e (+ct) + e ( ct)]. 7. We hve u(, t) = F ( + ct) + G( ct). To determine F nd G, we use the initil dt: u(, ) = F ()+G() = ; () t (, ) = cf () cg () = e e cf () cg() = e d = e + C F () G() = e + C, () c where we rewrote C/c s C to denote the rbitrry constnt. Adding () nd (), we find nd from () F () = e c + C F () = c G() = c [ ] e + C. [ ] e + C ; Hence u(, t) =F ( + ct)+g( ct) = c [ e (+ct) e ( ct)].

12 6 Chpter A Preview of Applictions nd Techniques 8. We hve u(, t) =F ( + ct)+g( ct). To determine F nd G, we use the initil dt: u(, ) = F ()+G() = F () = G() F () = G (); () (, ) = t ( + ) cf () cg () = cf () = F () = where C is n rbitrry constnt. From (), ( + ) ( + ) F () = c( + ) d = 4c( + ) + C, G() = F () = 4c( + ) C; where the C here is the sme s the C in the definition of F (). So u(, t) =F ( + ct)+g( ct) = 4c [ ( + ( + ct)) + ( + ( ct)), (from ()) c( + ) 9. As the hint suggests, we consider two seprte problems: The problem in Eercise 5 nd the one in Eercise 7. et u (, t) denote the solution in Eercise 5 nd u (, t) the solution in Eercise 7. It is strightforwrd to verify tht u = u + u is the desired solution. Indeed, becuse of the linerity of derivtives, we hve u tt =(u ) tt +(u ) tt = c (u ) + c (u ), becuse u nd u re solutions of the wve eqution. But c (u ) + c (u ) = c (u + u ) = u nd so u tt = c u, showing tht u is solution of the wve eqution. Now u(, ) = u (, )+u (, ) = /(+ )+, becuse u (, ) = /(+ ) nd u (, ) =. Similrly, u t (, ) = e ; thus u is the desired solution. The eplicit formul for u is u(, t) = [ +( + ct) + +( ct) ]. ] + [ e (+ct) e ( ct)]. c. Resoning s in the previous eercise, we find the solution to be u = u + u, wehre u is the solution in Eercise 6 nd u is the solution in Eercise 8. Thus, u(, t) = [ e (+ct) + e ( ct)] + [ ] 4c (+( + ct)) + ( + ( ct)).. We hve so V I V = C t = ( I ) t RI = I t R I V GV t = C [ I t + G V t V t = C To check tht V verifies (), we strt with the right side C V t = C C V +(RC + G) t + RGV [ I t + G V t = I t R I G C = I t R I = V, ] +(RC + G) C =) {[ }}{ C V t + I ] + GV ; [ I + GV ]. ] ; [ ] I + GV + RGV

13 Section. Solving nd Interpreting Prtil Differentil Eqution 7 which shows tht V stisfies (). To show tht I stisfies (), you cn proceed s we did for V or you cn note tht the equtions tht relte I nd V re interchnged if we interchnge nd C, nd R nd G. However, () remins unchnged if we interchnge nd C, nd R nd G. SoI stisfies () if nd only if V stisfies ().. The function being grphed is the solution () with c = =: u(, t) = π cos πt. In the second frme, t =/4, nd so u(, t) =π cos π/4 = π. The mimum of this function (for <<πis ttined t =/nd is equl to, which is vlue greter thn /. 3. The function being grphed is u(, t) = π cos πt π cos πt + 3π cos 3πt. 3 In frmes, 4, 6, nd 8, t = m 4, where m =, 3, 5, nd 7. u(, t), we find Plugging this into u(, t) = π cos mπ 4 mπ π cos + 3mπ 3π cos 3 4. For m =, 3, 5, nd 7, the second term is, becuse cos mπ =. Hence t these times, we hve, for, m =, 3, 5, nd 7, u(, m 4 ) = π cos πt + 3π cos 3πt. 3 To sy tht the grph of this function is symmetric bout =/is equivlent to the ssertion tht, for <</, u(/+, m 4 )=u(/, m 4 ). Does this equlity hold? et s check: u(/+, m mπ ) = π( +/) cos mπ 3π( +/) cos 3 4 = cos π cos mπ 4 3mπ cos 3π cos 3 4, where we hve used the identities π( +/) = cos π nd π( +/) = cos 3π. Simillry, u(/, m mπ ) = π(/ ) cos mπ 3π(/ ) cos 3 4 = cos π cos mπ 4 3mπ cos 3π cos 3 4. So u(/+, m 4 )=u(/, m ), s epected Note tht the condition u(, t) = u(, t) holds in ll frmes. It sttes tht the ends of the string re held fied t ll time. It is the other condition on the first derivtive tht is specific to the frmes in question. The function being grphed is u(, t) = π cos πt π cos πt + 3π cos 3πt. 3 As we just stted, the equlity u(, t)=u(, t) holds for ll t. Now u(, t) =π cos π cos πt π cos π cos πt + π cos 3π cos 3πt. To sy tht u(, t) = nd u(, t) mens tht the slope of the grph (s function of ) is zero t = nd =. This is little difficult to see in the

14 8 Chpter A Preview of Applictions nd Techniques frmes, 4, 6, nd 8 in Figure 8, but follows by plugging t =/4, 3/4, 5/4 nd7/4 nd = or in the derivtive. For emple, when = nd t =/4, we obtin: u(, /4) = π cos π/4 π cos π/+πcos 3π/4 =π cos π/4+πcos 3π/4 =. 5. Since the initil velocity is, from (), we hve u(, t) = n= b n nπ cnπt cos. The initil condition u(, ) = f() = π + 4 3π implies tht n= b n nπ π =. The eqution is stisfied with the choice b =,b =, nd ll other b n s re zero. This yields the solution u(, t) = π cπt cos. Note tht the condition u t (, ) = is lso stisfied. 6. Since the initil velocity is, from (), we hve u(, t) = n= b n nπ cnπt cos. The initil condition u(, ) = f() = n= b n nπ = π + 4 3π π + 3π 4. implies tht Clerly, this eqution is stisfied with the choice b =, b 3 = 4, nd ll other b n s re zero. This yields the solution u(, t) = π cπt cos + 3π 3cπt cos 4. Note tht the condition u t (, ) = is lso stisfied. 7. Sme resoning s in the previous eercise, we find the solution u(, t) = π cπt cos + 3π 4 cos 3cπt + 7π 5 7cπt cos. 8. Since the initil displcement is, we use the functions following (): u(, t) = n= b n nπ cnπt. The initil condition u(, ) = is clerly stisfied. To stisfy the second intil

15 Section. Solving nd Interpreting Prtil Differentil Eqution 9 condition, we proceed s follows: Thus g() = π = u(, ) t = t = = = n= n= ( n= n= b n nπ b n nπ b n t cnπt ( cnπt ) b n nπ cnπ cos cnπt nπ cnπ. π = b cnπ n n= n= nπ cnπ. This equlity holds if we tke b cπ =orb = cπ, nd ll other b n =. Thus u(, t) = π cπt cπ. This solution stisfies both initil conditions. 9. Resoning s in the previous eercise, we strt with the solution u(, t) = b n nπ cnπt. The initil condition u(, ) = is clerly stisfied. To stisfy the second intil condition, we must hve [ ( 3π 4 )] 6π = b n nπ cnπt t n= t= cnπ nπ = b n. Thus nd ll other b 8 n re. Thus u(, t) = n= 4 = 3cπ b 3 b 3 = cπ ; = 6cπ b 6 b 6 = 6cπ ; cπ 3π. Write the initil condition s u(, ) = 4 5 or 6 nd you will get u(, t) = 4 t= ) 3cπt 6π 6cπ 4π t= 6cπt. 4π, then proceed s in Eercises 4cπt cos.. () We hve to show tht u(,t) is constnt for ll t>. With c = =, we hve u(, t) = π cos πt u(/, t) = π cos πt = for ll t>.

16 Chpter A Preview of Applictions nd Techniques (b) One wy for =/3not to move is to hve u(, t) = 3π cos 3πt. This is the solution tht corresponds to the initil condition u(, ) = 3π nd (, ) =. For this solution, we lso hve tht =/3does not move for t ll t.. () Resoning s in Eercise 7, we find the solution to be u(, t) = π cos πt + 4π cos 4πt. 4 (b) We used Mthemtic to plot the shpe of the string t times t =tot =by increments of.. The string returns to some of its previous shpes. For emple, when t =. nd when t =.9, the string hs the sme shpe. u _, t_ Sin Pi Cos Pit 4 Sin 4Pi Cos 4Pit tt Tble u, t, t,,,. ; Plot Evlute tt,,, Cos t Sin 4 Cos 4 t Sin The point =/ does not move. This is cler: If we put =/ in the solution, we obtin u(/, t) = for ll t. 3. The solution is u(, t) = π cos πt. The motions of the points = /4, /3, nd 3/4 re illustrted by the following grphs. Note tht the point =/ does not move, so the grph tht describes its motion is identiclly. u _, t_ Sin Pi Cos Pit Plot u 4, t,u 3, t,u 3 4, t, t,, Pi = /4.5 = / = 3/4

17 Section. Solving nd Interpreting Prtil Differentil Eqution In ech cse, we hve coe wve, nmely u(,t) = π cos πt, scled by fctor π. 4. The solution in Eercise is u(, t) = π cos πt + 4π cos 4πt. 4 The motions of the points = /4, /3, /, nd 3/4 re illustrted by the following grphs. As in the previous eercise, the point =/does not move, so the grph tht describes its motion is identiclly. u _, t_ Sin Pi Cos Pit 4 Sin 4Pi Cos 4Pit Plot u 4, t,u 3, t,u 3 4, t, t,, Pi Cos t Sin 4 Cos 4 t Sin Unlike the previous eercises, here the motion of point is not lwys coe wve. In fct, it is the sum of two scled coe wves: u(,t)= π cos πt + 4 4π cos 4πt. 5. The solution () is u(, t) = π Its initil conditions t time t = 3 c re u(, 3 ( π πc ) = c cos 3 ) c nd cos πct. = π cos 3π =; 3 (, t c )= πc π ( πc 3 ) = πc π c 3π = πc π. 6. We hve u(, t +(3)/(c)) = π πc 3 cos (t + c ) = π ( πct cos + 3π ) = π πct, where we used the indentity cos( + 3π ) =. Thus u(, t +(3)/(c)) is equl to the solution given by (7). Cll the ltter solution v(, t). We know tht v(, t) represents the motion of string tht strts with initil shpe f() = nd intil velocity g() = πc π. Now, ppeling to Eercise 5, we hve tht t time t =(3)/(c), the shpe of the solution u is u(, (3)/(c)) = nd its velocity is

18 Chpter A Preview of Applictions nd Techniques 3 πc π t (, c )=. Thus the subsequent motion of the string, u, t time t + t is identicl to the motion of string, v, strting t time t =, whenever v hs initil shpe u(, t ) nd the initil velocity t (, t ). 7. () The eqution is equivlent to r t κ r = u. The solution of this eqution follows from Eercise 8, Section., by tking = r nd b = κ r. Thus u(, t) =f( r + κ r t)e rt. Note tht this equivlent to u(, t) =f( κt)e rt, by replcing the function f() in the first formul by f( r). This is cceptble becuse f is rbitrry. (b) The number of prticles t time t is given by u(, t) d. We hve M = u(, ) d. But u(, ) = f(), so M = f() d. For t>, the number of prticles is u(, t) d = f( κt)e rt d = e rt f( κt) d = e rt f() d = Me rt, where, in evluting the integrl f( κt) d, we used the chnge of vribles κt, nd then used M = f() d.

19 Section. Periodic Functions 3 Solutions to Eercises.. () cos hs period π. (b) cos π hs period T = π π =. (c) cos 3 hs period T = π /3 =3π. (d) cos hs period π, cos hs period π, π, 3π,.. A common period of cos nd cos is π. So cos + cos hs period π.. () 7π hs period T = π 7π =/7. π (b) nπ hs period T = nπ = n. Since ny integer multiple of T is lso period, we see tht is lso period of nπ. (c) cos m hs period T = π m. Since ny integer multiple of T is lso period, we see tht π is lso period of cos m. (d) hs period π, cos hs period π; cos + so hs period π. (e) Write = cos 4. The function cos 4 hs period T = π 4 = π. So hs period π. 3. () The period is T =, so it suffices to describe f on n intervl of length. From the grph, we hve { if f() = <, if <. For ll other, we hve f( +)=f(). (b) f is continuous for ll k, where k is n integer. At the hlf-integers, = k+, ug the grph, we see tht lim h + f(h) = nd lim h f(h) =. At the integers, = k, from the grph, we see tht lim h + f(h) =nd lim h f(h) =. The function is piecewise continuous. (c) Since the function is piecewise constnt, we hve tht f () = t ll k, where k is n integer. It follows tht f (+) = nd f ( ) = (Despite the fct tht the derivtive does not eist t these points; the left nd right limits eist nd re equl.) 4. The period is T = 4, so it suffices to describe f on n intervl of length 4. From the grph, we hve { + if, f() = + if<<. For ll other, we hve f( +4)=f(). (b) The function is continuous t ll. (c) (c) The function is differentible for ll k, where k is n integer. Note tht f is lso 4-periodic. We hve { f if <, () = if <<. For ll other k, we hve f( +4) =f(). If =, ±4, ±8,..., we hve f (+) = nd f ( ) =. If = ±, ±6, ±,...,wehvef (+) = nd f ( ) =. 5. This is the specil cse p = π of Eercise 6(b). 6. () A common period is p. (b) The orthogonlity reltions re p cos mπ cos nπ d = ifm n, m, n =,,,...; p p p p p p p cos mπ p mπ p nπ p nπ p d = ifm n, m, n =,,...; d = for ll m =,,,..., n=,,... These formuls re estblished by ug vrious ddition formuls for the coe nd e. For emple, to prove the first one, if m n, then p cos mπ cos nπ d p = = [ p p p p [ (m + n)π cos + cos p p (m + n)π + m + n)π p ] (m n)π d p p (m n)π m n)π p ] p =. p

20 4 Chpter Fourier Series 7. Suppose tht Show tht f, f,..., f n,... re T -periodic functions. This mens tht f j ( + T )=f() for ll nd j =,,..., n. et s n () = f ()+ f ()+ + n f n (). Then s n ( + T ) = = f ( + T )+ f ( + T )+ + n f n ( + T ) = f ()+ f ()+ + n f n () =s n (); which mens tht s n is T -periodic. In generl, if s() = j= jf j () is series tht converges for ll, where ech f j is T -periodic, then nd so s() is T -periodic. s( + T )= j f j ( + T )= j f j () =s(); j= 8. () Since cos nd cos π, the eqution cos + cos π = holds if nd only if cos = nd cos π =. Now cos = implies tht =kπ, k nd integer, nd cos π = implies tht =m, m nd integer. So cos + cos π = implies tht m =kπ, which implies tht k = m = (becuse π is irrtionl). So the only solution is =. (b) Since f() = cos + cos π = tkes on the vlue only t =, it is not periodic. 9. () Suppose tht f nd g re T -periodic. Then f( + T ) g( + T )=f() g(), nd so f g is T periodic. Similrly, j= f( + T ) g( + T ) = f() g(), nd so f/g is T periodic. (b) Suppose tht f is T -periodic nd let h() =f(/). Then ( ) + T ( ) h( + T ) = f = f + T ( ) = f (becuse f is T -periodic) = h(). Thus h hs period T. Replcing by /, we find tht the function f() hs period T/. (c) Suppose tht f is T -periodic. Then g(f( + T )) = g(f()), nd so g(f()) is lso T -periodic.. () hs period π, so hs period π/ =π (by Eercise 9(b)). (b) cos hs period 4π nd hs period π (or ny integer multiple of it). So common period is 4π. Thus cos + 3 hd period 4π (by Eercise 7) (c) We cn write + = g(f()), where g() =/ nd f() = +. Since f is π-periodic, it follows tht + is π-periodic, by Eercise 9(c). (d) Sme s prt (c). Here we let f() =cosnd g() =e. Then e cos is π-periodic.. Ug Theorem, / π/ f() d = f() d = d=.. Ug Theorem, / f() d = f() d = π/ cos d=.

21 Section. Periodic Functions 5 3. / f() d = π/ 4. Ug Theorem, / π/ f() d = / f() d = d = π/. d = π et F () = f(t) dt. IfF is π-periodic, then F () =F ( +π). But F ( +π) = +π f(t) dt = f(t) dt + Since F () =F ( +π), we conclude tht Applying Theorem, we find tht +π +π f(t) dt = The bove steps re reversible. Tht is, +π f(t) dt =. f(t) dt = F ()+ f(t) dt =. +π f(t) dt. f(t) dt = +π f(t) dt = +π +π f(t) dt = f(t) dt + f(t) dt = f(t) dt F () =F ( +π); nd so F is π-periodic. 6. We hve F ( + T )= +T So F is T periodic if nd only if F ( + T )=F () f(t) dt = +T +T +T f(t) dt + +T f(t) dt + f(t) dt. f(t) dt = f(t) dt =, f(t) dt = f(t) dt where the lst ssertion follows from Theorem. 7. By Eercise 6, F is periodic, becuse f(t) dt = (this is cler from the grph of f). So it is enough to describe F on ny intervl of length. For <<, we hve F () = ( t) dt = t t =. For ll other, F (+) = F (). (b) The grph of F over the intervl [, ] consists of the rch of prbol looking down, with zeros t nd. Since F is -periodic, the grph is repeted over nd over.

22 6 Chpter Fourier Series 8. () We hve (b) (c) (n+)t nt +T (n+)t +T f() d = = = f() d = = = f() d = = = T T T nt nt nt f(s + nt ) ds f(s) ds f() d. f(s + T ) ds f(s) ds (n+)t f() d. (n+)t (n+)t nt (let = s + nt, d = ds) (becuse f is T -periodic) (let = s + T, d = ds) (becuse f is T -periodic) +T f() d + f() d + f() d = (n+)t nt T f() d f() d f() d (by (b)) (by ()). 9. () The plots re shown in the following figures. [ (b) et us show tht f() = p [ ] + p f( + p) = + p p p [ ] = p = f(). p p ] is p-periodic. [ ] = + p p p + = + p p ([ ] ) + p From the grphs it is cler tht f() = for ll <<p. To see this from the formul, use the fct tht [t]=if t<. So, if <p, we hve [ ] p <, so =, nd hence f() =. p. () Plot of the function f() = p [ ] +p for p =,, nd π. p

23 Section. Periodic Functions 7 p Plot pfloor p p,,, p p (b) [ ] [ ] ( +p)+p + p f( +p) = ( +p) p =( +p) p p p + ([ ] ) [ ] + p + p = ( +p) p + = p = f(). p p So f is p-periodic. For p <<p, we hve < +p p ] so f() = p =. [ +p p [ ] <, hence +p p =, nd. () With p =, the function f becomes f() = [ ] +, nd its grph is the first one in the group shown in Eercise. The function is -periodic nd is equl to on the intervl <<. By Eercise 9(c), the function g() =h(f() is - periodic for ny function h; in prticulr, tking h() =, we see tht g() =f() is -periodic. (b) g() = on the intervl <<, becuse f() = on tht intervl. (c) Here is grph of g() =f() = ( [ ]) +, for ll. Plot Floor ^,, 3, () As in Eercise, the function f() = [ ] + is -periodic nd is equl to on the intervl <<. So, by Eercise 9(c), the function [ ] g() = f() = + is -periodic nd is clerly equl to for ll <<. Its grph is tringulr wve s shown in (b).

24 8 Chpter Fourier Series Plot Abs Floor,, 3, (c) To obtin tringulr wve of rbitrry period p, we use the p-periodic function [ ] + p f() = p, p which is equl to on the intervl p <<p. Thus, [ ] g() = + p p p is p-periodic tringulr wve, which equl to in the intervl p <<p. The following grph illustrtes this function with p = π. p Pi Plot Abs p Floor p p,, Pi, Pi 3. () Since f( +p) =f(), it follows tht g(f( +p)) = g(f()) nd so g(f()) is p-periodic. For p <<p, f() = nd so g(f()) = g(). (b) The function e g(), with p =, is the -periodic etension of the function which equls e on the intervl <<. Its grph is shown in Figure, Section.6 (with = ). 4. et f,f,...,f n be the continuous components of f on the intervl [, T], s described prior to the eercise. Since ech f j is continuous on closed nd bounded intervl, it is bounded: Tht is, there eists M> such tht f j () M for ll in the domin of f j. et M denote the mimum vlue of M j for j =,,...,n. Then f() M for ll in [, T] nd so f is bounded.

25 Section. Periodic Functions 9 5. We hve F ( + h) F () = = +h +h f() d f() d f() d M h, where M is bound for f(), which eists by the previous eercise. (In deriving the lst inequlity, we used the following property of integrls: b f() d (b ) M, which is cler if you interpret the integrl s n re.) As h, M h nd so F (+h) F (), showing tht F (+h) F (), showing tht F is continuous t. (b) If f is continuous nd F () = f() d, the fundmentl theorem of clculus implies tht F () =f(). If f is only piecewise continuous nd is point of continuity of f, let ( j, j ) denote the subintervl on which f is continuous nd is in ( j, j ). Recll tht f = f j on tht subintervl, where f j is continuous component of f. For in ( j, j ), consider the functions F () = f() d nd G() = j f j () d. Note tht F () =G()+ j f() d = G() + c. Since f j is continuous on ( j, j ), the fundmentl theorem of clculus implies tht G () =f j () =f(). Hence F () =f(), ce F differs from G by constnt. 6. We hve F () = +T f() d f() d. By the previous eercise, F is sum of two continuous nd piecewise smooth functions. (The first term is trnslte of f() d by T, nd so it is continuous nd piecewise smooth.) Thus F is continuous nd piecewise smooth. Since ech term is differentible the points of continuity of f, we conclude tht F is lso differentible t the points of continuity of f. (b) By Eercise 5, we hve, t the points where f is continuous, F () =f( + T ) f() =, becuse f is periodic with period T. Thus F is piecewise constnt. (c) A piecewise constnt function tht is continuous is constnt (just tke left nd right limits t points of discontinuity.) So F is constnt. 7. () The function does not hve right or left limit s, nd so it is not piecewise continuous. (To be piecewise continuous, the left nd right limits must eist.) The reson is tht / tends to + s + nd so / oscilltes between + nd. Similrly, s, / oscilltes between + nd. See the grph. (b) The function f() = nd f() = is continuous t. The reson for this is tht / is bounded by, so, s, /, by the squeeze theorem. The function, however, is not piecewise smooth. To see this, let us compute its derivtive. For, f () = cos. As +, / +, nd so / oscilltes between + nd, while cos oscilltes between + nd. Consequently, f () hs no right limit t. Similrly, it fils to hve left limit t. Hence f is not piecewise smooth. (Recll tht to be piecewise smooth the left nd right limits of the derivtive hve to eist.) (c) The function f() = nd f() = is continuous t, s in cse (b).

26 Chpter Fourier Series Also, s in (b), the function is not piecewise smooth. To see this, let us compute its derivtive. For, f () = cos. As +,/ +, nd so /, while cos oscilltes between + nd. Hence, s +, cos oscilltes between + nd, nd so f () hs no right limit t. Similrly, it fils to hve left limit t. Hence f is not piecewise smooth. (d) The function f() = 3 nd f() = is continuous t, s in cse (b). It is lso smooth. We only need to check the derivtive t =. We hve f f(h) f() h 3 h () = lim = lim h h h h lim h h h =. For, we hve f () =3 cos. Since f () =f () s, we conclude tht f eists nd is continuous for ll. Plot Sin,,.3,.3, PlotPoints 5 Plot Sin,,.3,.3, PlotPoints ^ Sin,,.,. Plot ^3 Sin,,.,

27 Section. Fourier Series Solutions to Eercises.. The grph of the Fourier series is identicl to the grph of the function, ecept t the points of discontinuity where the Fourier series is equl to the verge of the function t these points, which is.. The grph of the Fourier series is identicl to the grph of the function, ecept t the points of discontinuity where the Fourier series is equl to the verge of the function t these points, which is in this cse. 3. The grph of the Fourier series is identicl to the grph of the function, ecept t the points of discontinuity where the Fourier series is equl to the verge of the function t these points, which is 3/4 in this cse. 4. Since the function is continuous nd piecewise smooth, it is equl to its Fourier series. 5. We compute the Fourier coefficients ug he Euler formuls. et us first note tht ce f() = is n even function on the intervl π <<π, the product f() n is n odd function. So b n = π π odd function {}}{ n d =, becuse the integrl of n odd function over symmetric intervl is. For the other

28 Chpter Fourier Series coefficients, we hve = π π = = π f() d = π π ( ) d + π π d d = d= π π π = π. In computing n (n ), we will need the formul cos d = cos() + () + C ( ), which cn be derived ug integrtion by prts. We hve, for n, Thus, the Fourier series is n = f() cos n d = cos n d π π π π = cos n d π = [ π n n + ] π cos n n = [ ( ) n π n ] n = [ ( ) n πn ] { if n is even = if n is odd. 4 πn π 4 π k= cos(k +). (k +) s n_, _ : Pi 4 Pi Sum k ^ Cos k, k,, n prtilsums Tble s n,, n,, 7 ; f _ Pi Floor Pi Pi g _ Abs f Plot g,, 3 Pi, 3 Pi Plot Evlute g, prtilsums,, Pi, Pi The function g() = nd its periodic etension Prtil sums of the Fourier series. Since we re summing over the odd integers, when n = 7, we re ctully summing the 5th prtil sum.

29 Section. Fourier Series 3 6. We compute the Fourier coefficients ug he Euler formuls. et us first note tht f() is n odd function on the intervl π <<π,so = π π f() d =, becuse the integrl of n odd function over symmetric intervl is. A similr rgument shows tht n = for ll n. This leves b n. We hve, for n, b n = π = π π f() n d π/( ) n d + π / = n d π = [ ] π n cos n π/ = [ ( ) n π n ] n = πn / [ nπ ] cos n d Thus the Fourier series is π n= cos nπ n n. (b) The new prt here is defining the periodic function. We willuse the sme construction s in the previous eercise. We will use the Which commnd to define function by severl formuls over the intervl - Pi to Pi. We then use the construction of Ercises nd of Section. to define the periodic etension. s n_, _ : Pi Sum Cos kpi k Sin k, k,, n prtilsums Tble s n,, n,,, ; f _ Which Pi,, Pi,, Pi,, Pi, g _ Pi Floor Pi Pi h _ f g Plot h,, 3 Pi, 3 Pi Plot Evlute h, prtilsums,, Pi, Pi - - The prtil sums of the Fourier series converge to the function t ll points of continuity. At the points of discontinuity, we observe Gibbs phenomenon. At the points of discontinuity, the Fourier series is converging to or ±/. depending on the loction of the point of discontinuity: For ll =kπ, the Fourier series converges to ; nd for ll =(k +)π/, the Fourier series converges to ( ) k /.

30 4 Chpter Fourier Series 7. f is even, so ll the b n s re zero. We hve = f() d = d= π π π π cos π = π. We will need the trigonometric identity For n, cos b = ( ( b) + ( + b) ). n = f() cos n d = cos n d π π π = ( ) ( n) + ( + n) d π = [ ] π n cos( n) π cos( + n) (if n ) ( + n) = [ π n ( ) n ( + n) ( )+n + n + ] +n { if n is odd = if n is even. If n =, we hve 4 π( n ) = π Thus, the Fourier series is : cos d= π = π 4 π k= () d =. cos k. (k) 8. You cn compute the Fourier series directly s we did in Eercise 7, or you cn use Eercise 7 nd note tht cos = ( + π ). This identity cn be verified by compring the grphs of the functions or s follows: So = = ( + π {}}{ ) = cos π {}}{ + cos π = cos. cos = ( + π ) = π 4 π = π 4 π k= k= ( (k) cos k( + π ) ) ( ) k cos k, (k) where we hve used cos (k( + π ) ) = cos k cos kπ k kπ =( ) k cos k. 9. Just some hints: () f is even, so ll the b n s re zero. () = π d = π 3.