Solution Manual for Applied Partial Differential Equations with Fourier Series and Boundary Value Problems, 5th Edition

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SOLUTIONSMANUALAPPLIEDPARTIALDIFFERENTIALEQUATION WITHFOURIERSERIESANDBOUNDARYVALUEPROBLEMSRichard HabermanSouthern Methodist University

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Chapter 1. Heat EquationSection 1.21.2.9 (d)Circular cross section means thatP= 2πr, A=πr2, and thusP/A= 2/r, whereris the radius.Alsoγ= 0.1.2.9 (e)u(x, t) =u(t) implies thatcρ dudt=−2hr u .The solution of this first-order linear differential equation with constant coefficients, which satisfies theinitial conditionu(0) =u0, isu(t) =u0exp[−2hcρr t].Section 1.31.3.2∂u/∂xis continuous ifK0(x0−) =K0(x0+), that is, if the conductivity is continuous.Section 1.41.4.1 (a)Equilibrium satisfies (1.4.14),d2u/dx2= 0, whose general solution is (1.4.17),u=c1+c2x. Theboundary conditionu(0) = 0 impliesc1= 0 andu(L) =Timpliesc2=T /Lso thatu=T x/L.1.4.1 (d)Equilibrium satisfies (1.4.14),d2u/dx2= 0, whose general solution (1.4.17),u=c1+c2x. Fromthe boundary conditions,u(0) =TyieldsT=c1anddu/dx(L) =αyieldsα=c2. Thusu=T+αx.1.4.1 (f) In equilibrium, (1.2.9) becomesd2u/dx2=−Q/K0=−x2, whose general solution (by integratingtwice) isu=−x4/12 +c1+c2x. The boundary conditionu(0) =Tyieldsc1=T, whiledu/dx(L) = 0yieldsc2=L3/3. Thusu=−x4/12 +L3x/3 +T.1.4.1 (h)Equilibrium satisfiesd2u/dx2= 0.One integration yieldsdu/dx=c2, the second integrationyields the general solutionu=c1+c2x.x= 0 :c2−(c1−T) = 0x=L:c2=αand thusc1=T+α.Therefore,u= (T+α) +αx=T+α(x+ 1).1.4.7 (a)For equilibrium:d2udx2=−1 impliesu=−x22+c1x+c2anddudx=−x+c1.From the boundary conditionsdudx(0) = 1 anddudx(L) =β, c1= 1 and−L+c1=βwhich is consistentonly ifβ+L= 1. Ifβ= 1−L, there is an equilibrium solution (u=−x22+x+c2). Ifβ6= 1−L,there isn’t an equilibrium solution.The difficulty is caused by the heat flow being specified at bothends and a source specified inside. An equilibrium will exist only if these three are in balance. Thisbalance can be mathematically verified from conservation of energy:ddt∫L0cρu dx=−dudx(0) +dudx(L) +∫L0Q0dx=−1 +β+L.Ifβ+L= 1, then the total thermal energy is constant and the initial energy = the final energy:∫L0f(x)dx=∫L0(−x22+x+c2)dx,which determinesc2.Ifβ+L6= 1, then the total thermal energy is always changing in time and an equilibrium is neverreached.1

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Section 1.51.5.9 (a)In equilibrium, (1.5.14) using (1.5.19) becomesddr(rdudr)= 0. Integrating once yieldsrdu/dr=c1and integrating a second time (after dividing byr) yieldsu=c1lnr+c2. An alternate general solutionisu=c1ln(r/r1) +c3.The boundary conditionu(r1) =T1yieldsc3=T1, whileu(r2) =T2yieldsc1= (T2−T1)/ln(r2/r1). Thus,u=1ln(r2/r1)[(T2−T1) lnr/r1+T1ln(r2/r1)].1.5.11 For equilibrium, the radial flow atr=a, 2πaβ, must equal the radial flow atr=b, 2πb. Thusβ=b/a.1.5.13 From exercise 1.5.12, in equilibriumddr(r2dudr)= 0. Integrating once yieldsr2du/dr=c1and integrat-ing a second time (after dividing byr2) yieldsu=−c1/r+c2. The boundary conditionsu(4) = 80andu(1) = 0 yields 80 =−c1/4 +c2and 0 =−c1+c2. Thusc1=c2= 320/3 oru=3203(1−1r).2

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Chapter 2. Method of Separation of VariablesSection 2.32.3.1 (a)u(r, t) =φ(r)h(t) yieldsφdhdt=khrddr(rdφdr). Dividing bykφhyields1khdhdt=1rφddr(rdφdr)=−λordhdt=−λkhand1rddr(rdφdr)=−λφ.2.3.1 (c)u(x, y) =φ(x)h(y) yieldshd2φdx2+φd2hdy2= 0.Dividing byφhyields1φd2φdx2=−1hd2hdy2=−λord2φdx2=−λφandd2hdy2=λh.2.3.1 (e)u(x, t) =φ(x)h(t) yieldsφ(x)dhdt=kh(t)d4φdx4. Dividing bykφh, yields1khdhdt=1φd4φdx4=λ.2.3.1 (f)u(x, t) =φ(x)h(t) yieldsφ(x)d2hdt2=c2h(t)d2φdx2. Dividing byc2φh, yields1c2hd2hdt2=1φd2φdx2=−λ.2.3.2 (b)λ= (nπ/L)2withL= 1 so thatλ=n2π2, n= 1,2, . . .2.3.2 (d)(i) Ifλ >0, φ=c1cos√λx+c2sin√λx. φ(0) = 0 impliesc1= 0, whiledφdx(L) = 0 impliesc2√λcos√λL= 0. Thus√λL=−π/2 +nπ(n= 1,2, . . .).(ii) Ifλ= 0, φ=c1+c2x.φ(0) = 0 impliesc1= 0 anddφ/dx(L) = 0 impliesc2= 0. Thereforeλ= 0is not an eigenvalue.(iii) Ifλ <0, letλ=−sandφ=c1cosh√sx+c2sinh√sx. φ(0) = 0 impliesc1= 0 anddφ/dx(L) = 0impliesc2√scosh√sL= 0. Thusc2= 0 and hence there are no eigenvalues withλ <0.2.3.2 (f) The simpliest method is to letx′=x−a. Thend2φ/dx′2+λφ= 0 withφ(0) = 0 andφ(b−a) = 0.Thus (from p. 46)L=b−aandλ= [nπ/(b−a)]2, n= 1,2, . . ..2.3.3 From (2.3.30),u(x, t) =∑∞n=1BnsinnπxLe−k(nπ/L)2t. The initial condition yields2 cos3πxL=∑∞n=1BnsinnπxL. From (2.3.35),Bn=2L∫L02 cos3πxLsinnπxLdx.2.3.4 (a)Total heat energy =∫L0cρuA dx=cρA∑∞n=1Bne−k(nπL)2t1−cosnπnπL, using (2.3.30) whereBnsatisfies (2.3.35).2.3.4 (b)heat flux to right =−K0∂u/∂xtotal heat flow to right =−K0A∂u/∂xheat flow out atx= 0 =K0A∂u∂x∣∣x=0heat flow out (x=L) =−K0A∂u∂x∣∣x=L2.3.4 (c)From conservation of thermal energy,ddt∫L0u dx=k∂u∂x∣∣∣L0=k∂u∂x(L)−k∂u∂x(0).Integrating fromt= 0 yields∫L0u(x, t)dx︸︷︷︸heat energyatt−∫L0u(x,0)dx︸︷︷︸initial heatenergy=k∫t0[∂u∂x(L)︸︷︷︸integral offlow in atx=L−∂u∂x(0)]dx︸︷︷︸integral offlow out atx=L.2.3.8 (a)The general solution ofkd2udx2=αu(α >0) isu(x) =acosh√αkx+bsinh√αkx.The boundaryconditionu(0) = 0 yieldsa= 0, whileu(L) = 0 yieldsb= 0. Thusu= 0.3

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2.3.8 (b)Separation of variables,u=φ(x)h(t) orφdhdt+αφh=khd2φdx2, yields two ordinary differentialequations (divide bykφh):1khdhdt+αk=1φd2φdx2=−λ.Applying the boundary conditions, yields theeigenvaluesλ= (nπ/L)2and corresponding eigenfunctionsφ= sinnπxL. The time-dependent part areexponentials,h=e−λkte−αt. Thus by superposition,u(x, t) =e−αt∑∞n=1bnsinnπxLe−k(nπ/L)2t, wherethe initial conditionsu(x,0) =f(x) =∑∞n=1bnsinnπxLyieldsbn=2L∫L0f(x) sinnπxLdx. Ast→ ∞,u→0, the only equilibrium solution.2.3.9 (a)Ifα <0, the general equilibrium solution isu(x) =acos√−αkx+bsin√−αkx.The boundaryconditionu(0) = 0 yieldsa= 0, whileu(L) = 0 yieldsbsin√−αkL= 0. Thus if√−αkL6=nπ, u= 0 isthe only equilibrium solution. However, if√−αkL=nπ, thenu=AsinnπxLis an equilibrium solution.2.3.9 (b)Solution obtained in 2.3.8 is correct. If−αk=(πL)2, u(x, t)→b1sinπxL, the equilibrium solution.If−αk<(πL)2, thenu→0 ast→ ∞. However, if−αk>(πL)2, u→ ∞(ifb16= 0). Note thatb1>0 iff(x)≥0. Other more unusual events can occur ifb1= 0. [Essentially, the other possible equilibriumsolutions are unstable.]Section 2.42.4.1 The solution is given by (2.4.19), where the coefficients satisfy (2.4.21) and hence (2.4.23-24).(a)A0=1L∫LL/21dx=12, An=2L∫LL/2cosnπxLdx=2L·LnπsinnπxL∣∣∣LL/2=−2nπsinnπ2(b) by inspectionA0= 6, A3= 4, others = 0.(c)A0=−2L∫L0sinπxLdx=2πcosπxL∣∣∣L0=2π(1−cosπ) = 4/π, An=−4L∫L0sinπxLcosnπxLdx(d) by inspectionA8=−3, others = 0.2.4.3 Letx′=x−π. Then the boundary value problem becomesd2φ/dx′2=−λφsubject toφ(−π) =φ(π)anddφ/dx′(−π) =dφ/dx′(π).Thus, the eigenvalues areλ= (nπ/L)2=n2π2, sinceL=π, n=0,1,2, ...with the corresponding eigenfunctions being both sinnπx′/L= sinn(x−π) = (−1)nsinnx=>sinnxand cosnπx′/L= cosn(x−π) = (−1)ncosnx=>cosnx.Section 2.52.5.1 (a)Separation of variables,u(x, y) =h(x)φ(y), implies that1hd2hdx2=−1φd2φdy2=−λ. Thusd2h/dx2=−λhsubject toh′(0) = 0 andh′(L) = 0.Thus as before,λ= (nπ/L)2, n= 0, l,2, . . .withh(x) =cosnπx/L. Furthermore,d2φdy2=λφ=(nπL)2φso thatn= 0 :φ=c1+c2y, whereφ(0) = 0 yieldsc1= 0n6= 0 :φ=c1coshnπyL+c2sinhnπyL, whereφ(0) = 0 yieldsc1= 0.The result of superposition isu(x, y) =A0y+∞∑n=1AncosnπxLsinhnπyL.The nonhomogeneous boundary condition yieldsf(x) =A0H+∞∑n=1AnsinhnπHLcosnπxL,so thatA0H= 1L∫L0f(x)dxandAnsinhnπHL= 2L∫L0f(x) cosnπxLdx.4

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2.5.1 (c)Separation of variables,u=h(x)φ(y), yields1hd2hdx2=−1φd2φdy2=λ.The boundary conditionsφ(0) = 0 andφ(H) = 0 yield an eigenvalue problem iny, whose solution isλ= (nπ/H)2withφ= sinnπy/H, n= 1,2,3, . . .The solution of thex-dependent equation ish(x) = coshnπx/Husingdh/dx(0) = 0. By superposition:u(x, y) =∞∑n=1AncoshnπxHsinnπyH.The nonhomogeneous boundary condition atx=Lyieldsg(y) =∑∞n=1AncoshnπLHsinnπyH, so thatAnis determined byAncoshnπLH=2H∫H0g(y) sinnπyHdy.2.5.1 (e)Separation of variables,u=φ(x)h(y), yields the eigenvaluesλ= (nπ/L)2and correspondingeigenfunctionsφ= sinnπx/L, n= 1,2,3, ...They-dependent differential equation,d2hdy2=(nπL)2h,satisfiesh(0)−dhdy(0) = 0.The general solutionh=c1coshnπyL+c2sinhnπyLobeysh(0) =c1,whiledhdy=nπL(c1sinhnπyL+c2coshnπyL)obeysdhdy(0) =c2nπL. Thus,c1=c2nπLand hencehn(y) =coshnπyL+LnπsinhnπyL.Superposition yieldsu(x, y) =∞∑n=1Anhn(y) sinnπx/L,whereAnis determined from the boundary condition,f(x) =∑∞n=1Anhn(H) sinnπx/L,and henceAnhn(H) = 2L∫L0f(x) sinnπx/L dx .2.5.2 (a)From physical reasoning (or exercise 1.5.8), the total heat flow across the boundary must equalzero in equilibrium (without sources, i.e. Laplace’s equation). Thus∫L0f(x)dx= 0 for a solution.2.5.3 In order foruto be bounded asr→ ∞, c1= 0 in (2.5.43) and ¯c2= 0 in (2.5.44). Thus,u(r, θ) =∞∑n=0Anr−ncosnθ+∞∑n=1Bnr−nsinnθ.(a) The boundary condition yieldsA0= ln 2, A3a−3= 4, otherAn= 0, Bn= 0.(b) The boundary conditions yield (2.5.46) witha−nreplacingan. Thus, the coefficients are determinedby (2.5.47) withanreplaced bya−n2.5.4By substituting (2.5.47) into (2.5.45) and interchanging the orders of summation and integrationu(r, θ) = 1π∫π−πf(¯θ)[12 +∞∑n=1(ra)n(cosnθcosn¯θ+ sinnθsinn¯θ)]d¯θ.Noting the trigonometric addition formula and cosz=Re[eiz], we obtainu(r, θ) = 1π∫π−πf(¯θ)[−12 +Re∞∑n=0(ra)nein(θ−¯θ)]d¯θ.Summing the geometric series enables the bracketed term to be replaced by−12 +Re11−raei(θ−¯θ)=−12 +1−racos(θ−¯θ)1 +r2a2−2racos(θ−¯θ) =12−12r2a21 +r2a2−2racos(θ−¯θ).5

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2.5.5 (a)The eigenvalue problem isd2φ/dθ2=−λφsubject todφ/dθ(0) = 0 andφ(π/2) = 0.It can beshown thatλ >0 so thatφ= cos√λθwhereφ(π/2) = 0 implies that cos√λπ/2 = 0 or√λπ/2 =−π/2 +nπ, n= 1,2,3, . . .The eigenvalues areλ= (2n−1)2.The radially dependent term satisfies(2.5.40), and hence the boundedness condition atr= 0 yieldsG(r) =r2n−1. Superposition yieldsu(r, θ) =∞∑n=1Anr2n−1cos(2n−1)θ.The nonhomogeneous boundary condition becomesf(θ) =∞∑n=1Ancos(2n−1)θorAn= 4π∫π/20f(θ) cos(2n−1)θ dθ.2.5.5 (c) The boundary conditions of (2.5.37) must be replaced byφ(0) = 0 andφ(π/2) = 0. Thus,L=π/2,so thatλ= (nπ/L)2= (2n)2andφ= sinnπθL= sin 2nθ.The radial part that remains bounded atr= 0 isG=r√λ=r2n. By superposition,u(r, θ) =∞∑n=1Anr2nsin 2nθ .To apply the nonhomogeneous boundary condition, we differentiate with respect tor:∂u∂r=∞∑n=1An(2n)r2n−1sin 2nθ .The bc atr= 1,f(θ) =∑∞n=12nAnsin 2nθ, determinesAn,2nAn=4π∫π/20f(θ) sin 2nθ dθ.2.5.6 (a)The boundary conditions of (2.5.37) must be replaced byφ(0) = 0 andφ(π) = 0. ThusL=π,so that the eigenvalues areλ= (nπ/L)2=n2and corresponding eigenfunctionsφ= sinnπθ/L=sinnθ, n= 1,2,3, ...The radial part which is bounded atr= 0 isG=r√λ=rn. Thus by superpositionu(r, θ) =∞∑n=1Anrnsinnθ .The bc atr=a,g(θ) =∑∞n=1Anansinnθ,determinesAn, Anan=2π∫π0g(θ) sinnθ dθ.2.5.7 (b)The boundary conditions of (2.5.37) must be replaced byφ′(0) = 0 andφ′(π/3) = 0.This willyield a cosine series withL=π/3, λ= (nπ/L)2= (3n)2andφ= cosnπθ/L= cos 3nθ, n= 0,1,2, . . ..The radial part which is bounded atr= 0 isG=r√λ=r3n.Thus by superpositionu(r, θ) =∞∑n=0Anr3ncos 3nθ .The boundary condition atr=a,g(θ) =∑∞n=0Ana3ncos 3nθ,determinesAn:A0=3π∫π/30g(θ)dθand (n6= 0)Ana3n=6π∫π/30g(θ) cos 3nθ dθ.2.5.8 (a) There is a full Fourier series inθ. It is easier (but equivalent) to choose radial solutions that satisfythe corresponding homogeneous boundary condition. Instead ofrnandr−n(1 and lnrforn= 0), wechooseφ1(r) such thatφ1(a) = 0 andφ2(r) such thatφ2(b) = 0 :φ1(r) ={ln(r/a)n= 0(ra)n−(ar)nn6= 0φ2(r) ={ln(r/b)n= 0(rb)n−(br)nn6= 0.6

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Then by superpositionu(r, θ) =∞∑n=0cosnθ[Anφ1(r) +Bnφ2(r)] +∞∑n=1sinnθ[Cnφ1(r) +Dnφ2(r)].The boundary conditions atr=aandr=b,f(θ) =∞∑n=0cosnθ[Anφ1(a) +Bnφ2(a)] +∞∑n=1sinnθ[Cnφ1(a) +Dnφ2(a)]g(θ) =∞∑n=0cosnθ[Anφ1(b) +Bnφ2(b)] +∞∑n=1sinnθ[Cnφ1(b) +Dnφ2(b)]easily determineAn, Bn, Cn, Dnsinceφ1(a) = 0 andφ2(b) = 0 :Dnφ2(a) =1π∫π−πf(θ) sinnθ dθ,etc.2.5.9 (a)The boundary conditions of (2.5.37) must be replaced byφ(0) = 0 andφ(π/2) = 0.This is asine series withL=π/2 so thatλ= (nπ/L)2= (2n)2and the eigenfunctions areφ= sinnπθ/L=sin 2nθ, n= 1,2,3, . . ..The radial part which is zero atr=aisG= (r/a)2n−(a/r)2n.Thus bysuperposition,u(r, θ) =∞∑n=1An[(ra)2n−(ar)2n]sin 2nθ.The nonhomogeneous boundary condition,f(θ) =∑∞n=1An[(ba)2n−(ab)2n]sin 2nθ,determinesAn:An[(ba)2n−(ab)2n]=4π∫π/20f(θ) sin 2nθ dθ.2.5.9 (b) The two homogeneous boundary conditions are inr, and henceφ(r) must be an eigenvalue problem.By separation of variables,u=φ(r)G(θ), d2G/dθ2=λGandr2d2φdr2+rdφdr+λφ= 0.The radial equationis equidimensional (see p.78) and solutions are in the formφ=rp.Thusp2=−λ(withλ >0) sothatp=±i√λ.r±i√λ=e±i√λlnr.Thus real solutions are cos(√λlnr) and sin(√λlnr). It is moreconvenient to use independent solutions which simplify atr=a,cos[√λln(r/a)] and sin[√λln(r/a)].Thus the general solution isφ=c1cos[√λln(r/a)] +c2sin[√λln(r/a)].The homogeneous conditionφ(a) = 0 yields 0 =c1, whileφ(b) = 0 implies sin[√λln(r/a)] = 0 . Thus√λln(b/a) =nπ, n= 1,2,3, ...and the corresponding eigenfunctions areφ= sin[nπln(r/a)ln(b/a)].Thesolution of theθ-equation satisfyingG(0) = 0 isG= sinh√λθ= sinhnπθln(b/a).Thus by superpositionu=∞∑n=1Ansinhnπθln(b/a) sin[nπln(r/a)ln(b/a)].The nonhomogeneous boundary condition,f(r) =∞∑n=1Ansinhnπ22 ln(b/a) sin[nπln(r/a)ln(b/a)],will determineAn. One method (for another, see exercise 5.3.9) is to letz= ln(r/a)/ln(b/a). Thena < r < b, lets 0< z <1. This is a sine series inz(withL= 1) and henceAnsinhnπ22 ln(b/a) = 2∫10f(r) sin[nπln(r/a)ln(b/a)]dz.Butdz=dr/rln(b/a). ThusAnsinhnπ22 ln(b/a) =2ln(b/a)∫10f(r) sin[nπln(r/a)ln(b/a)]dr/r.7

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Solution Manual for Applied Partial Differential Equations with Fourier Series and Boundary Value Problems, 5th Edition - Page 10 preview image

Chapter 3. Fourier SeriesSection 3.23.2.2 (a)x∼a0+∑∞n=1ancosnπx/L+∑∞n=1bnsinnπx/L.From (3.2.2),a0= 0 sincef(x) is odd,(n6= 0)an= 0 sincef(x) is odd, andbn=1L∫L−Lxsinnπx/L dx=2Lnπ(−1)n+1.3.2.2 (c)sinπx/L∼a0+∑∞n=1ancosnπx/L+∑∞n=1bnsinnπx/L. By inspection,b1= 1, all others = 0.3.2.2 (f)From (3.2.2),a0=12L∫L0dx= 1/2(n6= 0)an=1L∫L0cosnπx/L dx= 0bn=1L∫L0sinnπx/L dx=−1nπcosnπxL∣∣∣L0=1−cosnπnπThusbn= 2/nπ, nodd, butbn= 0,neven.3.3.2 (d)From (3.3.6),Bn=2L∫L/20sinnπx/L dx=−2nπcosnπx/L∣∣∣L/20=2nπ(1−cosnπ2).3.3.10f(−x) ={x2−x <0(orx >0)ex−x >0(orx <0).Thus.fe(x) =12[f(x) +f(−x)] =12{x2+exx <0x2+e−xx >0f0(x) =12[f(x)−f(−x)] =12{x2−exx <0e−x−x2x >03.3.13bn=2L∫L0f(x) sinnπx/L dx, given thatf(x) is even aroundL/2. Note (perhaps by graphing) thatsinnπx/Lis odd aroundL/2 forneven. Thusf(x) sinnπx/Lis odd aroundL/2 forneven, and hencebn= 0 forneven.Section 3.43.4.1 (a)∫ba=∫c−a+∫bc+. Thus∫bau dvdx dx=uv∣∣∣∣∣c−a+uv∣∣∣∣∣∣bc+−∫bav dudx dx=uv∣∣∣∣∣∣∣ba+uv∣∣∣∣∣∣∣∣c−c+−∫bav dudx dx.3.4.3 (a)We want to determine the sine coefficients ofdf /dx:dfdx=∑∞n=1bnsinnπxL,where the cosinecoefficients offare givenf=∞∑n=0ancosnπxL(n6= 0)an= 2L∫L0fcosnπxLdx.Here by integration by partsbn= 2L∫L0dfdxsinnπxLdx= 2L[fsinnπxL∣∣∣x0−0+fsinnπxL∣∣∣∣Lx0+−nπL∫L0fcosnπxLdx].Thusbn=2Lsinnπx0L(α−β)−nπLan.8

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Solution Manual for Applied Partial Differential Equations with Fourier Series and Boundary Value Problems, 5th Edition - Page 11 preview image

3.4.9∂u∂x=∑∞n=1(nπL)bncosnπxLsinceu(0) = 0 andu(L) = 0.∂2u∂x2∼ −∑∞n=1(nπL)2bnsinnπxL. Thus fromp. 119,∑∞n=1dbndtsinnπxL∼ −k∑∞n=1(nπL)2bnsinnπxL+q. Thusdbndt+k(nπL)2bn=2L∫L0qsinnπxLdx.3.4.12 The eigenfunctions of the related homogeneous problem are cosnπx/L, n= 0,1,2, . . ..Thusu∼∑∞n=0An(t) cosnπx/L, which can be differentiated (ifuis continuous) since it is a cosine series:∂u/∂x∼∑∞n=0An(−nπ/L) sinnπxL.This can be differentiated again (if∂u/∂xis continuous) onlybecause∂u/∂x(0) = 0 and∂u/∂x(L) = 0:∂2u/∂x2∼ −∑∞n=0An(nπ/L)2cosnπx/L. Thus from p.119∞∑n=0[dAndt+k(nπL)2An]cosnπxL=e−t+e−2tcos 3πxL.The right hand side is a simple cosine series (with only two non-zero terms). ThusdAndt+k(nπL)2An=e−tn= 0e−2tn= 30otherwise.The initial conditions areA0(0) =1L∫L0f(x)dxand (n6= 0)An(0) =2L∫L0f(x) cosnπxLdx.Thesolution of the differential equations aren6= 0,3An(t)=An(0)e−k(nπ/L)2tA0(t)=A0(0) + 1−e−t,obtained by integrationA3(t)=A3(0)e−k(nπ/L)2t+e−2t−e−k(nπ/L)2tk(3πL)2−2,obtained by using the method of undetermined coefficients (judicious guessing) for the particularsolution. This works ife−2tis not a homogeneous solution, i.e.,−26=−k(3π/L)2.Section 3.53.5.4 (a)Using 3.4.13 withf(x) = coshx(f(0) = 1, f(L) = coshL),sinhx∼1L(coshL−1) +∑∞n=1[nπLbn+2L((−1)ncoshL−1)]cosnπxL. Since this is a cosine series, itmay be differentiatedcoshx∼ −∞∑n=1[(nπL)2bn+ 2nπL2((−1)ncoshL−1)]sinnπxL.Thusbn=−(nπL)2bn−2nπL2[(−1)ncoshL−1] orbn=2nπL21−(−1)ncoshL1+(nπ/L)2.3.5.4 (b)Integrating yields sinhx=A0+∑∞n=1−Lnπbncosnπx/L, whereA0=1L∫L0sinhx dx=1L(coshL−1). Integrating again yields coshx−1 =A0x+∑∞n=1−(Lnπ)2bnsinnπx/L. Thus∞∑n=1bn[1 +(Lnπ)2]sinnπx/L= 1 +A0x.Using (3.3.8) and (3.3.l2)bn[1 +(Lnπ)2]=2nπ[1−(−1)n] +1L(coshL−1)2Lnπ(−1)n+1orbn=2nπ1−(−1)ncoshL1 +(Lnπ)2.3.5.7 Evaluate (3.5.6) atx=L/2:L28=L24−4L2π3(1−133+ 153−. . .)or 1−133+ 153−173=L2/84L2/π3=π3/32.9

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Solution Manual for Applied Partial Differential Equations with Fourier Series and Boundary Value Problems, 5th Edition - Page 12 preview image

Section 3.63.6.1The complex Fourier coefficient is defined by (3.6.7):cn=12L∫L−Lf(x)einπx/Ldx=12L4∫x0+4x0einπx/Ldx .Thuscn=12L4Linπ einπx/L∣∣∣∣x0+4x0=12inπ4einπx0/L(einπ4/L−1).Equivalently,cn=12inπ4einπ(x0+4/2)/L(einπ4/2L−e−inπ4/2L)orcn=1nπ4einπ(x0+4/2)/Lsin(nπ4/2L).10

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Solution Manual for Applied Partial Differential Equations with Fourier Series and Boundary Value Problems, 5th Edition - Page 13 preview image

Chapter 4. Vibrating Strings and MembranesSection 4.44.4.1 (a)Natural frequencies arec√λ, butλ= (nπ/L)2. Thus frequencies arenπc/L, n= 1,2,3, ...4.4.1 (b)Natural frequencies arec√λ.The boundary conditionφ(0) = 0 impliesφ=c1sin√λx, whiledφ/dx(H) = 0 yields√λH= (m−12)πwithm= 1,2,3. Thus the frequencies are (m−12)πc/Handthe eigenfunctions are sin(m−12)πx/H.4.4.2 (c)By separation of variables,u=φ(x)h(t),d2hdt2=−λhandT0d2φdx2+ (α+λρ0)φ= 0. Withφ(0) = 0andφ(L) = 0,(α+λρ0)/T0= (nπ/L)2, n= 1,2,3, . . .andφ= sinnπx/L. In generalh(t) involves alinear combination of sin√λtand cos√λt, but the homogeneous initial conditionu(x,0) = 0 impliesthere are no cosines. Thus by superpositionu(x, t) =∞∑n=1Ansin√λntsinnπx/L,where the frequencies of vibration are√λn=√(nπ/L)2T0−αρ0.The other initial condition,f(x) =∑∞n=1An√λnsinnπx/L,determinesAnAn√λn= 2L∫L0f(x) sinnπx/L dx.4.4.3 (b)By separation of variables,u=φ(x)h(t),ρ0h′′+βh′hT0=φ′′φ=−λ.The boundary conditionsφ(0) = 0andφ(L) = 0 yieldλ= (nπ/L)2withφ= sinnπx/L, n= 1,2,3, . . .. The time-dependent equationhas constant coefficients,ρ0h′′+βh′+(nπL)2T0h= 0,and hence can be solved by substitutionh=ert. This yields the quadratic equationρ0r2+βr+(nπL)2T0= 0,whose roots arer=−β±√β2−4ρ0T0(nπ/L)22ρ0.Sinceβ2<4ρ0T0(π/L)2, the discriminant is<0 for alln:r=−β2ρ0+iwn,wherewn=√T0ρ0(nπL)2−β24ρ20.Real solutions areh=e−βt/2ρ0(sinwnt,coswnt). Thus by superpositionu=e−βt/2ρ0∞∑n=1(ancoswnt+bnsinwnt) sinnπxL.The initial conditionu(x,0) =f(x) determinesan,an=2L∫L0f(x) sinnπxLdx,while∂u∂t(x,0) =g(x) isa little more complicated,g(x) =∑∞n=1bnwnsinnπxL−β2ρ0∞∑n=1ansinnπxL︸︷︷︸f(x),and thusbnwn=βan2ρ0+ 2L∫L0g(x) sinnπxLdx.11

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Solution Manual for Applied Partial Differential Equations with Fourier Series and Boundary Value Problems, 5th Edition - Page 14 preview image

Chapter 5. Sturm-Liouville Eigenvalue ProblemsSection 5.35.3.1 By separation of variables,u=φ(x)h(t),1hd2hdt2=1ρ0φ(x)[T0d2φdx2+αφ]=−λ.Thus,d2hdt2=−λhandT0d2φdx2+αφ+λρ0φ= 0.For the time-dependent equation (ifλ >0),h=c1cos√λt+c2sin√λt.The spatial equation is inSturm-Liouville form:ρ=T0constant,q=α(x), σ=ρ0(x).5.3.3Hd2φ/dx2=ddx(Hdφ/dx)−dH/dx dφ/dx. Thusddx(Hdφ/dx)+dφ/dx(Hα−dH/dx)+(λβ+γ)Hφ= 0.To be in standard Sturm-Liouville form,dHdx=αHorH=c1exp[∫xα(t)dt],letc1= 1.Thenρ(x) =H, q(x) =γH, andσ(x) =βH.5.3.4 (b)By separation of variables,u=φ(x)h(t),1hdhdt=1φ(kd2φdx2−v0dφdx)=−λ. The boundary valueproblem iskd2φdx2−v0dφdx+λφ= 0.φ=erximplieskr2−v0r+λ= 0 orr=v0±√v20−4λk2k=v0±i√4λk−v202k.To satisfy the boundary conditionsφ(0) = 0 andφ(L) = 0:√4λk−v202k=nπLandφ(x) =ev02kxsinnπxL.Thus, by superposition,u=∞∑n=1Anev02kxsinnπxLe−λnt=ev02kx∞∑n=1AnsinnπxLe−λnt.The initial value problem can be solvedf(x) =ev02kx∞∑n=1AnsinnπxLso thatAn= 2L∫L0f(x)e−v02kxsinnπxLdx.Note thatλ=v204k+k(nπL)2.5.3.9 (c) Since it is equidimensional,φ=xr, which impliesr(r−1)+r+λ= 0 orr2=−λ. Ifλ >0, r=±i√λ,wherex±i√λ=e±i√λlnx.Thus the general solution isφ=c1cos(√λlnx)+c2sin(√λlnx).Theboundary conditionφ(1) = 0 impliesc1= 0, and henceφ(b) = 0 yields sin(√λlnb)= 0 or√λlnb=nπ, n= 1,2, . . .. Thusλ= (nπ/lnb)2. Ifλ= 0 the differential equation becomesddx(xdφ/dx) = 0.Thus, the general solution isφ=c1+c2lnx. The conditionφ(1) = 0 yieldsc1= 0, and thenφ(b) = 0impliesc2= 0. Thusλ= 0 is not an eigenvalue.12

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Solution Manual for Applied Partial Differential Equations with Fourier Series and Boundary Value Problems, 5th Edition - Page 15 preview image

Section 5.45.4.2 The eigenfunctions satisfy (5.4.6) with the boundary conditions beingdφ/dx(0) = 0 anddφ/dx(L) = 0.From the Rayleigh quotient (5.4.16)λ≥0.Alsoλ= 0 only ifdφ/dx= 0 for allx.The boundaryconditions implyλ= 0 is an eigenvalue withφ= constant = 1 the corresponding eigenfunction. Thusby superposition using (5.4.5)u=∞∑n=1anφn(x)e−λnt(withλ1= 0, φ1= 1).The initial conditions,f(x) =∞∑n=1anφn(x),yieldan=∫L0f φncρ dx∫L0φ2ncρ dx ,since the eigenfunctions are orthogonal with weightcρ. Ast→ ∞,u→a1=∫L0f cρ dx∫L0cρ dx ,a weighted average of the initial temperature distribution. This can also be shown using conservationof total thermal energy.Since the ends are insulated, the final thermal energy (t→ ∞) equals theinitial thermal energy (t= 0).5.4.3 From (5.2.11) the eigenfunctions, denotedφn(r), are orthogonal with weightr. The time-dependentpart satisfies (5.2.10), and henceh(t) =e−λkt. By superpositionu=∞∑n=1anφn(r)e−λkt.The initial condition,f(r) =∞∑n=1anφn(r),determinesan,an=∫a0f(r)φn(r)r dr∫a0φ2n(r)r dr.5.4.6 By separation of variables,u=φ(x)h(t),T0d2φdx2+λρ0φ=0d2hdt2+λh=0.The boundary conditions are of the Sturm-Liouville type, and therefore the eigenfunctions denotedφn(x) are orthogonal with weightρ0(x). We call the eigenvaluesλn. Then the time-dependent problemhas solutions cos√λntand sin√λnt. Initially at rest means∂u/∂t(x,0) = 0, so that only cosines areneeded. Thus by superpositionu=∞∑n=1Anφn(x) cos√λnt.The initial position yieldsf(x) =∞∑n=1Anφn(x),13

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determiningAn,An=∫L0f(x)φn(x)ρ0(x)dx∫L0φ2nρ0dx.Section 5.55.5.1 (g)To be self-adjoint (ifpis constant),u(L)dVdx(L)−v(L)dudx(L) =u(0)dvdx(0)−v((0)dudx(0).The derivatives atx=Lmay be eliminated from the boundary conditions.After some algebra, weobtain(αδ−βγ)[u(0)dvdx(0)−v(0)dudx(0)]=u(0)dvdx(0)−v(0)dudx(0).Thus, it is self-adjoint only ifαδ−βγ= 1.5.5.9 Multiply the differential equation byφand integrate:∫10φ d4φdx4dx+λ∫10exφ2dx= 0.Note thatφ d4φdx4=ddx(φ d3φdx3)−dφdxd3φdx3=ddx(φ d3φdx3)−ddx(dφdxd2φdx2)+(d2φdx2)2.Thusφd3φdx3∣∣∣10−dφdxd2φdx2∣∣∣∣10+∫10(d2φdx2)2dx+λ∫10exφ2dx= 0.From the boundary conditions, the boundary contributions vanish. Thusλ=−∫10(d2φdx2)2dx∫10exφ2dx,so thatλ≤0.Section 5.65.6.1 (c)From (5.6.6) usingp= 1, q= 0, σ= 1 and using the boundary conditions,λ1≤u2T(1) +∫10(duTdx)2dx∫10u2Tdx.Any trial function should satisfy the boundary conditions (and be continuous with no zeroes).Ge-ometrically, a simple example is a parabola (uT(0) = 0, uT(1) = 1, duT/dx(1) =−1, from theboundary conditions.To obtain this parabola, we substituteuT=ax+bx2into the condition atx= 1 :a+ 2b+a+b= 0, a simple choice beinga= 3, b=−2 so thatuT= 3x−2x2anduT/dx= 3−4x. Thus,λ1≤1 +∫10(3−4x)2dx∫10(3x−2x2)2dx=1−112(3−4x)3∣∣∣10∫10(9x2−12x3+ 4x4)dx= 1−112(−1−27)3−3 +45= 40/124/5= 4 16.14

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