Math 412-501Theory of Partial Differential EquationsLecture 4-6:Review for the final exam.Math 412-501 Fall 2006Sample problems for the final examAny problem may be altered or replaced by a different one!Some possibly useful information• Parseval’s equality for the complex form of the Fourier serieson (−π, π):f (x) =∞Xn=−∞cneinx=⇒Zπ−π|f (x)|2dx = 2π∞Xn=−∞|cn|2.• Fourier sine and cosine transforms of the second derivative:S[f′′](ω) =2πf (0) ω − ω2S[f ](ω),C [f′′](ω) = −2πf′(0) − ω2C [f ](ω).• Laplace’s operator in polar coordinates r , θ:∇2u =∂2u∂r2+1r∂u∂r+1r2∂2u∂θ2.• Any nonzero solution of a regular Sturm-Liouville equation(pφ′)′+ qφ + λσφ = 0 (a < x < b)satisfies the Rayleigh quotient relationλ =−pφφ′ba+Zbap(φ′)2− qφ2dxZbaφ2σ dx.• Some table integrals:Zx2eiaxdx =x2ia+2xa2−2ia3eiax+ C , a 6= 0;Z∞−∞e−αx2ei βxdx =rπαe−β2/(4α), α > 0, β ∈ R;Z∞−∞e−α|x |ei βxdx =2αα2+ β2, α > 0, β ∈ R.Problem 1 Let f (x) = x2.(i) Find the Fourier series (complex form) of f (x) on the interval(−π, π).(ii) Rewrite the Fourier series of f (x) in the real form.(iii) Sketch the function to which the Fourier series converges.(iv) Use Parseval’s equality to evaluateP∞n=1n−4.Problem 1 Let f (x) = x2.(i) Find the Fourier series (complex form) of f (x) on the interval(−π, π).The required series is∞Xn=−∞cneinx, wherecn=12πZπ−πf (x)e−inxdx.In particular,c0=12πZπ−πf (x) dx =12πZπ−πx2dx =12πx33π−π=12π2π33=π23.If n 6= 0 then we have to integrate by parts twice.To save time, we could instead use the table integralZx2eiaxdx =x2ia+2xa2−2ia3eiax+ C , a 6= 0.According to this integral,cn=12πZπ−πx2e−inxdx =12π−x2in+2xn2+2in3e−inxπ−π=12π2π(e−inπ+ einπ)n2=2(−1)nn2.Thusx2∼π23+X−∞<n<∞n6=02(−1)nn2einx.(ii) Rewrite the Fourier series of f (x) in the real form.π23+X−∞<n<∞n6=02(−1)nn2einx=π23+∞Xn=12(−1)nn2(einx+ e−inx)=π23+∞Xn=14(−1)nn2cos nx.Thusx2∼π23+∞Xn=14(−1)nn2cos nx.(iii) Sketch the function to which the Fourier series converges.The series converges to the 2π-periodic function that coincideswith f (x) for −π ≤ x ≤ π.The sum is continuous and piecewise smooth hence theconvergence is uniform.The derivative of the sum has jump discontinuities at pointsπ + 2kπ, k ∈ Z.The graph is a scalloped curve.(iv) Use Parseval’s equality to evaluateP∞n=1n−4.In our case, Parseval’s equality can be written ashf , f i =∞Xn=−∞|hf , φni|2hφn, φni,where φn(x) = einxandhg, hi =Zπ−πg(x)h(x) dx.Since cn=hf ,φnihφn,φniand hφn, φni = 2π for all n ∈ Z, it can bereduced to an equivalent formZπ−π|f (x)|2dx = 2π∞Xn=−∞|cn|2.NowZπ−π|f (x)|2dx =Zπ−πx4dx =x55π−π=2π55,∞Xn=−∞|cn|2=π49+ 2∞Xn=14n4.Therefore12π2π55=π49+ 2∞Xn=14n4.It follows that∞Xn=11n4=18π45−π49=π490.Problem 2 Solve Laplace’s equation in a disk,∇2u = 0 (0 ≤ r < a), u(a, θ ) = f (θ).Laplace’s operator in polar coordinates r , θ:∇2u =∂2u∂r2+1r∂u∂r+1r2∂2u∂θ2.We search for the solution of the boundary value problem as asuperposition of solutions u(r , θ) = h(r )φ(θ) wit h separatedvariables.Solutions with separated variables satisfy periodic boundaryconditionsu(r, −π) = u(r, π),∂u∂θ(r, −π) =∂u∂θ(r, π)and the singular boundary condition|u(0, θ)| < ∞.Separation of variables provides the following solutions:u0= 1, un(r, θ) = rncos nθ, ˜un(r, θ) = rnsin nθ, n = 1, 2, . . .A superposition of these solutions is a seriesu(r, θ) = α0+X∞n=1rn(αncos nθ + βnsin nθ),where α0, α1, . . . and β1, β2, . . . are constants. Substituting theseries into the boundary condition u(a, θ) = f (θ), we getf (θ) = α0+X∞n=1an(αncos nθ + βnsin nθ).The boundary condition is satisfied if the right-hand side coincideswith the Fourier seriesA0+X∞n=1(Ancos nθ + Bnsin nθ)of the function f (θ) on (−π, π).Henceα0= A0, αn= a−nAn, βn= a−nBn, n = 1, 2, . . .andu(r, θ) = A0+X∞n=1ran(Ancos nθ + Bnsin nθ),whereA0=12πZπ−πf (θ) dθ, An=1πZπ−πf (θ) cos nθ dθ,Bn=1πZπ−πf (θ) sin nθ dθ, n = 1, 2, . . .Bonus Problem 7 Find a Green function implementing thesolution of Problem 2.The solution of Problem 2:u(r, θ) = A0+∞Xn=1ran(Ancos nθ + Bnsin nθ),whereA0=12πZπ−πf (θ0) dθ0, An=1πZπ−πf (θ0) cos nθ0dθ0,Bn=1πZπ−πf (θ0) sin nθ0dθ0, n = 1, 2, . . .It can be rewritten asu(r, θ) =Zπ−πG (r , θ; θ0) f (θ0) dθ0,whereG (r , θ; θ0) =12π+1π∞Xn=1ran(cos nθ cos nθ0+ sin nθ si n nθ0).This is the desired Green function. The expression can besimplified:G (r , θ; θ0) =12π+1π∞Xn=1ran(cos nθ cos nθ0+ sin nθ si n nθ0)=12π+1π∞Xn=1rancos n(θ − θ0)=12π+1π∞Xn=1ran·ein(θ−θ0)+ e−in(θ−θ0)2=12π∞Xn=0ra−1ei (θ−θ0)n+12π∞Xn=1ra−1e−i (θ−θ0)nG (r , θ; θ0) =12π 11 − ra−1ei (θ−θ0)+ra−1e−i (θ−θ0)1 − ra−1e−i (θ−θ0)!=12π aa − rei (θ−θ0)+re−i (θ−θ0)a − re−i (θ−θ0)!=12πa2− r2(a − rei (θ−θ0))(a − re−i (θ−θ0))=12πa2− r2a2− 2ar cos(θ −θ0) + r2.(Poisson’s formula)Problem 3 Find Green’s function for the boundary value problemd2udx2− u = f (x) (0 < x < 1), u′(0) = u′(1) = 0.The Green function G (x, x0) should satisfy∂2G∂x2− G = δ(x −x0),∂G∂x(0, x0) =∂G∂x(1, x0) = 0.It follows thatG (x, x0) =(aex+ be−xfor x < x0,cex+ de−xfor x > x0,where constants a, b, c, d may depend on x0. Then∂G∂x(x, x0) =(aex− be−xfor x < x0,cex− de−xfor x > x0.The boundary conditions imply that a = b and ce = de−1.Gluing conditions at x = x0are continuity of the function andjump discontinuity of the first derivative:G (x, x0)x =x0−= G (x, x0)x =x0+,∂G∂xx =x0+−∂G∂xx =x0−= 1.The two conditions imply thataex0+be−x0= cex0+de−x0, cex0−de−x0−(aex0−be−x0) = 1.Now we have 4 equations to determine 4 quantities a, b, c, d.Solution:c =ex0+ e−x02(1 − e2), a =ex0+ e2−x02(1 − e2),d =ex0+ e−x02(e−2− 1), b =ex0+ e2−x02(1 − e2).Finally,G (x, x0) =(ex0+ e2−x0)(ex+ e−x)2(1 − e2)for x < x0,(ex0+ e−x0)(ex+ e2−x)2(1 − e2)for x > x0.Observe that G (x, x0)
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