FourierTrigonometricSeries
Chapter 2
1. Write y(t)= 3cos2t 4sin2t intheform y(t)= A cos(2π ft + φ) Wecandeterminetheconstantsbyexpandingthecosinefunction, y(t)= A cos(2π ft + φ)= A
Comparingthisto y(t)= 3cos2t 4sin2t,wesee2π f = 2and A cos φ = 3, A sin φ = 4.
Addingthesquaresoftheseequations, 25 = A2 cos2 φ + A2 sin2 φ = A2 ,
weobtain A = 5.Dividingthefirstequationintothesecond,tan φ = 4/3. So,wefind y(t)= 3cos2t 4sin2t = 5cos 2t + tan 1( 4 3 ) ≈ 5cos (2t + 0.927)
2. Determinetheperiodofthefollowingfunctions:
a. f (x)= cos x 3 T = 2π 1 3 = 6π
b. f (x)= sin2πx T = 2π 2π = 1.
c. f (x)= sin2πx 0.1cos3πx
Eachtermhasadifferentperiod:
Multiplesofeachgive nT = {1,2,3,4,5,6,...} nT = { 2 3 , 4 3 ,2, 8 3 , 10 3 ,4,...}.
Thesmallestcommonvalueistheperiodof f (x)= sin2πx 0.1cos3πx : T = 2.ThisisseeninFigure 2.1.
:Plotsof
Figure 2.2:Plotsof f (x)= 3sin πx 2 , f (x)= 2cos 3πx 4 ,and f (x)= 3sin πx 2 + 2cos 3πx 4
d. f (x)= | sin5πx|
Theperiodof f (x)= sin5πx is T = 2π 5π = 2 5 .However,thefrequencydoublesundertheabsolutevalue,so T = 1 5
e. f (x)= cot2πx.
Theperiodsofthetan x andcot x are T = π.So,forthisfunction wehave T = π 2π = 1 2
f. f (x)= cos2 x 2
Justliketheabsolutevalue,thefrequencyofthecosinefunction doubleswhenthefunctionissquared.So, T = π 1 2 = 2π.
g. f (x)= 3sin πx 2 + 2cos 3πx 4
Thisproblemissimilarto 2c.Eachtermhasadifferentperiod: T = 2π π 2 = 4and T = 2π 3π 4 = 8 3 .Multiplesofeachgive nT = {4,8,12,16,...} nT = { 8 3 , 16 3 ,8, 32 3 , 40 3 ,16,...}.
Thesmallestcommonvalueistheperiodof f (x)= 3sin πx 2 + 2cos 3πx 4 is T = 8.ThisisseeninFigure 2 2
(x) sin2πx 0.1cos3πx
3. Derivethecoefficients bn inEquation(5 24).
Thisderivationparallelsthatforthe an’s.WebeginwiththeFourier series
WemultiplythisFourierseriesbysin mx forsomepositiveinteger m and thenintegrate:
Integratingtermbyterm,therightsidebecomes
Wehaveshownthat 2
mxdx = 0,whichimpliesthatthefirstterm vanishes.Also,wehavethat
forintegers n and m
Westillneedtoevaluate 2π 0 sin nx sin mxdx whichwasnotdoneinthe book.Wecomputethisintegralbyusingtheproductidentityforsines.We havefor m = n that
For n = m,wehave
Now,wecanfinishthederivation.Wehaveshowntheorthogonalityof thesines,
Solvingfor bm,wehave
4. Let f (x) bedefinedfor x
.Parseval’sidentityisgivenby
AssumingtheFourierseriesof f (x) convergesuniformlyin ( L, L),prove Parseval’sidentitybymultiplyingtheFourierseriesrepresentationby f (x) andintegratingfrom x = L to x = L.[InSection 5.6.3,wewillencounter Plancherel’sFormulaforFouriertransforms,whichisacontinuousversion ofthisidentity.]
WebeginwiththeFourierseries
MultiplyingthisFourierseriesby f (x) andintegratingover x ∈ [ L, L],we obtain
RecalltheFouriercoefficientsaregivenby
ReplacingtheintegralsintheintegratedFourierserieswithFouriercoefficients,wehave
leadingtothesoughtresult,
5. Let f (x) bedefinedfor x ∈ [0, L].DerivetheParsevalidentities,similar toProblem 4,forthefollowingseries.
a. FortheFourierCosineseries,
MultiplytheFourierCosineseriesby
0to
b. FortheFourierSineseries,
MultiplytheFourierSineseriesby
andintegratefrom
Figure 2 3:Aplotoffirsttermsofthe Fourierseriesof f (x) inProblem 2.6.
6. Considerthesquarewavefunction
f (x)= 1,0 < x < π, 1, π < x < 2π
a. FindtheFourierseriesrepresentationofthisfunctionandplotthe first 50 terms.
Since f (x) isanoddfunctiononasymmetricinterval, an = 0, n = 0,1,....Weneedonlycomputethe bn’s.
TheFourierseriesisthen
)
Thefirst 50 terms (n = 50) areshowninFigure 2.3.
b. ApplyParseval’sidentityinProblem 5 totheresultinparta. Parseval’sidentitystates
Frompartawehave
Notingthat
Thus,wehave
c. Usetheresultofpartbtoshow π2 8 = ∞ ∑ n=1 1 (2n 1)2 . Multiplyingtheseriesinpartb,
by π2/16yieldsthisresult.
7. Forthefollowingsetsoffunctions:(i)showthateachisorthogonalon thegiveninterval,and(ii)determinethecorrespondingorthonormalset.
[Seepage 288.]
Orthogonalityandnormalizationcanbedoneusingsimplesubstitutions andrelatingtheintegralstothebasicorthogonalityrelationsbetweensines andcosines,
a. {sin2nx}, n = 1,2,3,...,0 ≤ x ≤ π
Let y = 2x, dy = 2 dx.Then,
Thesecanbenormalizedbyletting φn (x)=
So,wehave A = 2 π andtheorthonormalsetisgivenby { 2 π sin2nx}, n = 1,2,3,...,0 ≤ x ≤ π
b. {cos nπx}, n = 0,1,2,...,0 ≤ x ≤ 2.
Let y = πx, dy = π dx.Then, 2 0
Thesefunctionsarealreadyorthonormal.
c. {sin nπx L }, n = 1,2,3,..., x ∈ [ L, L].
Let y = πx/L, dy = π/Ldx.Then, L L sin nπx L
Thesecanbenormalizedbyletting
So,wehave A = 1 √L andtheorthonormalsetisgivenby { 1 √L sin nπx L }, n = 1,2,3,..., x ∈ [ L, L]
The a0’sarecomputedseparatelyfrom an ’swhendeterminingtheFouriercoefficients.
8. Consider f (x)= 4sin3 2x
a. Derivethetrigonometricidentitygivingsin3 θ intermsofsin θ and sin3θ usingDeMoivre’sFormula.
Wenotethat e3iθ =(eiθ )3.Writingbothsidesofthisequationin termsoftrigonometricfunctions,wehave e3iθ =(eiθ )3
cos3θ + i sin3θ =(cos θ + i sin θ)3 = cos3 θ + 3i
Equatingtherealandimaginarypartswehave cos3θ = cos3 θ 3cos θ sin2 θ, sin3θ = 3cos2 θ sin θ sin3 θ
Thesecondequationcanberearrangedtogettheresult. sin3θ = 3cos2 θ sin θ sin3 θ = 3(1 sin2 θ) sin θ sin3 θ = 3sin θ 4sin3 θ.
Therefore, sin3 θ = 3 4 sin θ 1 4 sin3θ.
b. FindtheFourierseriesof f (x)= 4sin3 2x on [0,2π] withoutcomputinganyintegrals.
Weneedonlylet θ = 2x inparta.Then, f (x)= 4sin3 2x = 3sin2x sin6x.
9. FindtheFourierseriesrepresentationsofthefollowing:
a. f (x)= x, x ∈ [0,2π]. WefirstcomputetheFouriercoefficients.Notethatwecompute a0 separatelyfrom an
TheFourierseriesisgivenby
AplotofthefirsttermsoftheseriesisgiveninFigure 2.4.
b. f (x)= x2 4 , |x| < π
(
) isanevenfunctionon
needthe an’s.
Figure 2 4:Aplotoffirsttermsofthe Fourierseriesof f (
Then,theFourierseriesisgivenas
Figure 2.5:Aplotoffirsttermsofthe Fourierseriesof f (x) inProblem 2 9b. c. f (
WefirstcomputetheFouriercoefficients.
Figure 2 6:Aplotoffirsttermsofthe Fourierseriesof f (x) inProblem 2 9
TheresultingFourierseriesis
AplotofthefirsttermsoftheseriesisgiveninFigure 2 8
WefirstcomputetheFouriercoefficients.
Figure 2 7:Aplotoffirsttermsofthe Fourierseriesof
TheresultingFourierseriesis
AplotofthefirsttermsoftheseriesisgiveninFigure
= π 2 an = 1 π 2π 0 f (x) cos nxdx = 1 π π 0 (π x) cos nxdx = 1 π 1 n (π x) sin nx 1 n2 cos nx π 0 = 1 cos nπ πn2 bn = 1 π 2π 0 f (x) sin nxdx = 1 π π 0 (π x) sin nxdx
= 1 n
TheresultingFourierseriesis
f (x) ∼ π 4 + ∞ ∑ n=1 1 cos nπ πn2 cos nx + 1 n sin nx
AplotofthefirsttermsoftheseriesisgiveninFigure ??.
10. FindtheFourierseriesrepresentationsofeachfunction f (x) ofperiod 2π.Foreachseries,plotthe Nthpartialsum, SN = a0 2 + N ∑ n=1 [an cos nx + bn sin nx] , for N = 5,10,50anddescribetheconvergence(Isitfast?Whatisitconvergingto?,etc.)[SomesimpleMaple,MATLAB,andPythoncodefor computingpartialsumsisshowninthenotes.]
a. f (x)= x, |x| < π.
Since f (x)= x isanoddfunctionon |x| < π,the an’svanishfor all n.So,wejustcomputethe bn’s. bn = 1 π π π x sin nxdx = 2 π π 0 x sin nxdx = 2 π 1 n
TheresultingFourierseriesis
f (x) ∼ 2 ∞ ∑ n=1 ( 1)n+1 sin nx n .
Figure 2 8:Aplotoffirsttermsofthe Fourierseriesof f (x) inProblem 2.9e.
Figure 2 9:Aplotoffirsttermsofthe Fourierseriesof f (x) inProblem 2 10a.
AsseeninFigure 2 9 theconvergenceisslowasthetermsdecay like 1 n .Thediscontinuitiesintheperiodicextensionalsoarean indication.
b. f (x)= |x|, |x| < π.
Since f (x)= |x| isanevenfunctionon |x| < π,the bn’svanishfor all n.Wecompute a0 and an. a0 = 1 π π π |x| dx = 2 π π 0 xdx = π an = 1 π π π |x| cos nxdx = 2 π π 0 x cos nxdx = 2 π 1 n x sin nx + 1 n2 cos nx π 0 = 2 πn2 (cos nπ 1)
TheresultingFourierseriesis f (x) ∼ π 2 + ∞ ∑ n=1 2(cos nπ 1) πn2 cos nx = π 2 4 π ∞ ∑ k=1 cos(2k 1)x (2k 1)2 .
Theconvergenceisfastasthetermsdecaylike 1 n2 .Therearenot discontinuitiesintheperiodicextension.SeeFigure 2.10.
c. f (x)= cos x, |x| < π. WhileonecancomputetheFouriercoefficientsbycarryingout integrations,itshouldbenoticedthatthisisatruncatedFourier seriesandnointegrationisneeded.Theresultis f (x)= cos x.
d. f (x)= 0, π < x < 0, 1,0 < x < π WeneedtocomputealloftheFouriercoefficients.
TheresultingFourierseriesis
Theconvergenceisslowasthetermsdecaylike 1 n .TherearediscontinuitiesintheperiodicextensionasseeninFigure 2.11.
11. FindtheFourierseriesrepresentationof f (x)= x onthegiveninterval. Plotthe Nthpartialsumsanddescribewhatyousee.
Figure 2 10:Aplotoffirsttermsofthe Fourierseriesof f (x) inProblem 2 10b.
Figure 2 11:Aplotoffirsttermsofthe
Fourierseriesof f (x) inProblem 2 10c. a. 0 < x < 2.
WecomputetheFouriercoefficients:
TheFourierseriesrepresentationisgivenby
Figure 2 12:Aplotoffirsttermsofthe
Fourierseriesof f (x) inProblem 2 11a.
AplotoftheFourierseriesisshowninFigure 2 12
b. 2 < x < 2.
Since f (x)= x isanoddfunctionon 2 < x < 2,weonlyneed the bn’s. bn = 2 4
TheFourierseriesis
AplotofthisFourierseriesisshowninFigure 2 13
c. 1 < x < 2. TheFouriercoefficientsarefoundas
Figure 2 13:Aplotoffirsttermsofthe Fourierseriesof f (
) inProblem 2 11b.
ThisgivestheFourierseries
AplotofthisFourierseriesisshowninFigure 2.14.
12. TheresultinProblem 9babovegivesaFourierseriesrepresentationof x2 4 .Bypickingtherightvaluefor x andalittlearrangementoftheseries, showthat[SeeExample
6.]
UsingtheresultsinProblem 5.12b,onehasthat
Figure 2.14:Aplotoffirsttermsofthe Fourierseriesof
inProblem 2 11c.
Letting x = π,wehave
Then,
Hint:Considerhowtheseriesinparta.canbeusedtodothis. Let
Notethat
Therefore,wehave
c. UsetheFourierseriesrepresentationresultinProblem 9e.toobtaintheseriesinpartb.
TheresultofProblem 9eis
Setting x = π,wehave
Therefore,
13. Sketch(byhand)thegraphsofeachofthefollowingfunctionsover fourperiods.Thensketchtheextensionseachofthefunctionsasbothan evenandoddperiodicfunction.DeterminethecorrespondingFouriersine andcosineseriesandverifytheconvergencetothedesiredfunctionusing Maple.
FortheseproblemswemakeuseoftheFouriercosineseries,
where
andtheFouriersineseries,
where
a. f (x)= x2,0 < x < 1.
ThegivenfunctionisshowninFigure 2 15.InFigure 2 16 thisfunctionisreflectedaboutthe y-axisandthenewfunctionisthenperiodicallyextendedtogivetheevenperiodicextension.InFigure 2 17 thefunctionisreflectedabouttheoriginandthenewfunction isthenperiodicallyextendedtogivetheoddperiodicextension. f (x) x 0 1
Figure 2 15:FunctiongiveninProblem 2 13a.
Figure 2 16:Sketchoftheevenperiodic extensionofthefunctiongiveninProblem 2 13a.
TheFouriercosineseriescoefficientsaregivenby
Figure 2 17:Sketchoftheoddperiodic extensionofthefunctiongiveninProblem 2 13a.
TheresultingFouriercosineseriesisgivenby
AplotofthisseriesrepresentationisshowninFigure 2.18. TheFouriersineseriescoefficientsaregivenby
TheresultingFouriersineseriesisgivenby
AplotofthisseriesrepresentationisshowninFigure 2 19
Figure 2 18:Aplotoffirsttermsofthe Fouriercosineseriesof f (x) inProblem
2 13a.
Figure 2 19:Aplotoffirsttermsofthe Fouriersineseriesof f (x) inProblem
2.13a.
b. f (x)= x(2 x),0 < x < 2. f (x) x 0 1 2
ThegivenfunctionisshowninFigure 2 20.InFigure 2 21 thisfunctionisreflectedaboutthe y-axisandthenewfunctionisthenperiodicallyextendedtogivetheevenperiodicextension.InFigure 2 22 thefunctionisreflectedabouttheoriginandthenewfunction isthenperiodicallyextendedtogivetheoddperiodicextension.
)
TheFouriercosineseriescoefficientsaregivenby
Figure 2 20:FunctiongiveninProblem 2 13b.
Figure 2 21:Sketchoftheevenperiodic extensionofthefunctiongiveninProblem 2 13b.
Figure 2 22:Sketchoftheoddperiodic extensionofthefunctiongiveninProblem 2 13b.
TheresultingFouriercosineseriesisgivenby
AplotofthisseriesrepresentationisshowninFigure 2 23
Figure 2 23:Aplotoffirsttermsofthe Fouriercosineseriesof f (x) inProblem 2 13b.
Figure 2 24:Aplotoffirsttermsofthe Fouriersineseriesof f (x) inProblem 2 13b.
TheFouriersineseriescoefficientsaregivenby
Figure 2 25:FunctiongiveninProblem 2.13c.
TheresultingFouriersineseriesisgivenby
Figure 2 26:Sketchoftheevenperiodic extensionofthefunctiongiveninProblem 2 13c.
AplotofthisseriesrepresentationisshowninFigure 2.24.
Figure 2 27:Sketchoftheoddperiodic extensionofthefunctiongiveninProblem 2 13c.
ThegivenfunctionisshowninFigure 2 25.InFigure 2 26 thisfunctionisreflectedaboutthe y-axisandthenewfunctionisthenperiodicallyextendedtogivetheevenperiodicextension.InFigure 2 27 thefunctionisreflectedabouttheoriginandthenewfunction isthenperiodicallyextendedtogivetheoddperiodicextension.
TheFouriercosineseriescoefficientsaregivenby
TheresultingFouriercosineseriesisgivenby
AplotofthisseriesrepresentationisshowninFigure 2 28
TheFouriersineseriescoefficientsaregivenby
TheresultingFouriersineseriesisgivenby
AplotofthisseriesrepresentationisshowninFigure 2 29
Figure 2 28:Aplotoffirsttermsofthe Fouriercosineseriesof f (x) inProblem 2 13c.
Figure 2 29:Aplotoffirsttermsofthe Fouriersineseriesof f (x) inProblem
2 13c.
(x)
Figure 2 30:FunctiongiveninProblem
2.13d.
Figure 2 31:Sketchoftheevenperiodic extensionofthefunctiongiveninProblem 2 13d.
ThegivenfunctionisshowninFigure 2 30.InFigure 2 31 thisfunctionisreflectedaboutthe y-axisandthenewfunctionisthenperiodicallyextendedtogivetheevenperiodicextension.InFigure 2 32 thefunctionisreflectedabouttheoriginandthenewfunction isthenperiodicallyextendedtogivetheoddperiodicextension.
Figure 2 32:Sketchoftheoddperiodic extensionofthefunctiongiveninProblem 2 13d.
TheFouriercosineseriescoefficientsaregivenby
TheresultingFouriercosineseriesisgivenby
AplotofthisseriesrepresentationisshowninFigure 2 33
TheFouriersineseriescoefficientsaregivenby
TheresultingFouriersineseriesisgivenby
AplotofthisseriesrepresentationisshowninFigure 2 34
. Considerthefunction
Showthat
TheFouriersineseriescoefficientsare
Figure 2 33:Aplotoffirsttermsofthe Fouriercosineseriesof f (x) inProblem 2 13d.
Figure 2 34:Aplotoffirsttermsofthe Fouriersineseriesof f (x) inProblem 2 13d.
ThisgivestheFouriersineseriesrepresentation f (x) ∼ 2 ∞ ∑ n=1 ( 1)n+1 sin nx n
b. Integratetheseriesinpartaandshowthat x
For f (x)= x,weconsidertheintegral
Integratingtheseriesaswell,wehave
Therefore,
Westillneedtoevaluate C.WecanusetheresultsinProblem 15 todothis.
Therefore,wehavethat
c. FindtheFouriercosineseriesof f (x)= x2 on [0, π] andcompare totheresultinpartb. WefirstdeterminetheFouriercosineseriescoefficients.
ThisgivestheFouriercosineseriesrepresentation
d. ApplyParseval’sidentityinProblem 5 totheseriesinparta.for f (x)= x on π < x < π.Thisgivesanothermeanstofindingthe value ζ (4),wheretheRiemannzetafunctionisdefinedby
FromtheParsevalidentity,wehave
Thisgives
15. Considerthefunction
a. FindtheFouriersineseriesrepresentationofthisfunctionandplot thefirst 50 terms.
TheFouriersineseriescoefficientsare
Figure 2 35:Aplotoffirsttermsofthe Fouriersineseriesof f (x) inProblem 2 15a.
ThisgivestheFouriersineseriesrepresentation
ThefirstfewtermsofthisseriesareshowninFigure 2 35
Figure 2 36:Aplotoffirsttermsofthe Fouriercosineseriesof f (x) inProblem 2 15b.
b. FindtheFouriercosineseriesrepresentationofthisfunctionand plotthefirst 50 terms.
TheFouriercosineseriescoefficientsare
ThisgivestheFouriercosineseriesrepresentation
ThefirstfewtermsofthisseriesareshowninFigure 2 36
c. ApplyParseval’sidentityinProblem 5 totheresultinpartb. Parseval’sidentitycanbeextendedtoFouriercosineseriesby slightlymodifyingthederivationinProblem 5.8.
WebeginwiththeParsevalidentity
Applyingthisresultto f (
)= x, x ∈ [0,2].Frompartb,wehave a0 = 2and
Then,
Thisgivesthesum
d. Usetheresultofpartc.tofindthesum
Theresultinpartcisnotquitethesumweseekastheterms involveonlytheoddterms.Wecanrearrangetheseriestomake useoftheresultinpartcandsolvefor
Therefore,wehaveshownthat
Figure 2 37:Plotoffirst 50 termsof(left) theFouriersineseriesof f (x)= x and (right)thederivativeofthesetermsfrom Problem 2 16
16. DifferentiatetheFouriersineseriesterm-by-terminthelastproblem. Showthattheresultisnotthederivativeof f (x)= x.. TheFouriersineseriesisgivenby
Asimpletermbytermdifferentiationgives
However,thisisadivergentseriesandcannotsumto f (x)= 1.
17. Considerthefunction f (x)= x sin x
a. FindtheFourierseriesrepresentationofthisfunctionif f (x) is definedon [0,2π] andplotthefirst 50 terms. WecomputetheFouriercoefficients:
TheFourierseriesrepresentationisgivenby
TheFourierseriesrepresentationisshowninFigure
.
b. FindtheFourierseriesrepresentationofthisfunctionif f (x) is definedon [ π, π] andplotthefirst 50 terms. WecomputetheFouriercoefficients:
Figure 2.38:Aplotoffirsttermsofthe Fourierseriesof f (x) inProblem 2
Figure 2 39:Aplotoffirsttermsofthe Fourierseriesof f (x) inProblem 2 17b.
TheFourierseriesrepresentationisgivenby
TheFourierseriesrepresentationisshowninFigure 2 39
18 Considerthefunction
a. FindtheFourierseriesrepresentationofthisfunctionandplotthe first 50 terms.
WecomputetheFouriercoefficients:
wehave
19. Thetemperature, u(x, t),ofaone-dimensionalrodoflength L satisfies theheatequation,
a. Showthatthegeneralsolution,
(x, t)=
satisfiestheone-dimensionalheatequationandtheboundaryconditions u(0, t)= 0and u(L, t)= 0.
Computing ut and uxx,wehave
Comparingthesederivatives,weseethat ut = kuxx.Notethat thevanishingofthefunctionattheintervalendpointsallowsthe differentiationofthesineseries.
Furthermore,wehave
u(0, t)= ∞
/L2 = 0, u(L, t)= ∞
/L2 = 0.
b. For k = 1and L = π,findthesolutionsatisfyingtheinitialcondition u(x,0)= sin x.Plotsixsolutionsonthesamesetofaxesfor t ∈ [0,1].
For k = 1and L = π,thegeneralsolutionis
u(x, t)= ∞ ∑ n=1 bn sin nxe n2t .
Usingtheinitialcondition, u(x,0)= sin x,wehave
sin x = ∞ ∑ n=1 bn sin nx.
TheFouriercoefficientsareeasilyfoundwithoutintegrationas b1 = 1and bn = 0, n > 1.Then,thesolutiontotheinitial-boundary valueproblemis
u(x, t)= sin xe t .
ThissolutionatsixtimesisshowninFigure 2.41.
c. For k = 1and L = 1,findthesolutionsatisfyingtheinitialcondition u(x,0)= x(1 x).Plotsixsolutionsonthesamesetofaxes for t ∈ [0,1]
For k = 1and L = 1,thegeneralsolutionis
Usingtheinitialcondition, u
),wehave
WeneedtodeterminetheFouriersinecoefficients:
Figure 2 41:Aplotofsolutionsforthe heatequationinProblem 2.19bfor t = 0,1/6,...,1.
Thesolutiontotheinitial-boundaryvalueproblemis
ThissolutionatsixtimesisshowninFigure 2.42.
20. Theheight, u(x, t),ofaone-dimensionalvibratingstringoflength L satisfiesthewaveequation,
Figure 2 42:Aplotofsolutionsforthe heatequationinProblem 2 19cfor t = 0,1/6,...,1.
a. Showthatthegeneralsolution,
satisfiestheone-dimensionalwaveequationandtheboundaryconditions u(0, t)= 0and u(L, t)= 0.
Inordertoverifythatthisisasolution,wecomputeafewderivatives:
Comparingderivatives,wehave utt = c2uxx.
Notethatthevanishingofthefunctionattheintervalendpoints allowsthedifferentiationofthesineseries. Wealsocanverifytheboundaryconditions: u(0, t)= ∞
n=1 An cos nπct L sin0 +Bn sin nπct L sin0 = 0. u(L, t)= ∞
n=1 An cos nπct L sin nπ +Bn sin nπct L sin nπ = 0.
b. For c = 1and L = 1,findthesolutionsatisfyingtheinitialconditions u(x,0)= x(1 x) and ut (x,0)= 0.Plotfivesolutionsfor t ∈ [0,1].
Forthisproblem,thegeneralsolutiontakestheform
Theinitialconditionsgive
Thesecondequationgives
Fromthepreviousproblemwehave
Therefore,thesolutiontotheinitial-boundaryvalueproblemis
ThissolutionatfivetimesisshowninFigure 2 43
c. For c = 1and L = 1,findthesolutionsatisfyingtheinitialcondition
and ut (x,0)= 0.Plotfivesolutionsfor t ∈ [0,0.5]
Forthisproblem,thegeneralsolutiontakestheform
Theinitialconditionsgive
Thesecondequationgives Bn = 0, n = 1,2,....
TheremainingFouriersinecoefficientscanbecomputedusing
Figure 2 43:Aplotofsolutionsforthe waveequationinProblem 2 20bfor t = 0,1/8,...,1/2.
Thus,
So,thesolutiontotheinitial-boundaryvalueproblemis
ThissolutionatfivetimesisshowninFigure 2.44.
21. Showthat
Figure 2 44:Aplotofsolutionsforthe waveequationinProblem 2 20cfor t = 0,1/8,...,1/2.
satisfiesthetwo-dimensionalwaveequation
andtheboundaryconditions,
Computingtheneededsecondparitalderivativesof
,wehave
Insertingthesederivativesintothethetwo-dimensionalwaveequation,we find
Thesolutioneasilysatisfiestheboundaryconditions.Forexample,
Theboundaryconditions, u(x,0, t)= 0, u(x, H, t)= 0,followinthesame way.
22 FindthedoubleFouriersineseriesrepresentationofthefollowing:
Afunction f (x, y) definedontherectangularregion [0, L] × [0, H] has adoubleFouriersineseriesrepresentation,
where
=
Thisrepresentationwillbeusedtoobtaintheseriesinthisproblem.
a. f (x, y)= sin πx sin2πy on [0,1] × [0,1].
Theseriesexpansionforthisproblemisgivenby
Itiseasytoseethattheisonlynonzeroterm,for n = 1and m = 2.
Thus, b12 = 1and bnm = 0,for n = 1and m = 2.
b. f (x, y)= x(2 x) sin y on [0,2] × [0, π].
Theseriesexpansionforthisproblemisgivenby
Here bnm = 0for m = 1.Weneedonlycomputethe bn1 terms. bn1 = 2
Thisgivestheseriesexpansion
c. f (x, y)= x2y3 on [0,1] × [0,1]
Theseriesexpansionforthisproblemisgivenby
TheFouriercoefficientscanbecomputeddirectly:
Eachintegralcanbecomputedseparately:
Thisgives
TheresultingFourierseriesis
23. DerivetheFouriercoefficientsinthedoubleFouriertrigonometricseries inEquation(2.124).
ThedoubleFouriertrigonometricseriesinEquation(2 124)isgivenby
Inordertoprovethis,oneneedstoconsidertheFourierbasisontherectangularregion [0, L] × [0, H],
φ00 = 1,
φn0 = cos
φ0n =
φ
Samplecomputationsarebelow,usingtheorthogonalityofthetrigonometricfunctions.
ThisgivestheFouriercoefficientsas
2.1: Plots of y(t) = A sin(2π ft) on [0, 5] for f = 2 Hz and f = 5 Hz.
Figure 2.2: Problems can occur while plotting. Here we plot the function y(t) = 2 sin 4 πt using N = 201, 200, 100, 101 points.
Figure 2.4: Plot of the functions y(t) = 2 sin(4 π t) and y(t) = 2 sin(4 πt + 7π/8) and their sum.
2.5: Plot of the function f (t) defined on [0, 2π] and its periodic extension.
2.6: Plot of discontinuous function in Example 2.3.
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