Math 412-501Theory of Partial Differential EquationsLecture 3-11: Review for Exam 3.Wave equation in polar coordinatesInitial-boundary value problem∂2u∂t2= c2∇2u in D,u|t=0= f ,∂u∂tt=0= g,∂u∂n∂D= 0,in a domain D = {(r, θ) : 0 < r < a, 0 < θ < π/2},a quarter-circle (given in polar coordinates).Initial conditions:u(r , θ, 0) = f (r , θ),∂u∂t(r, θ, 0) = g(r, θ).Normal derivative:∂u∂n(a, θ, t) =∂u∂r(a, θ, t),∂u∂n(r, π/2, t) = r−1∂u∂θ(r, π/2, t),∂u∂n(r, 0, t) = −r−1∂u∂θ(r, 0, t).Boundary conditions:∂u∂r(a, θ, t) = 0,∂u∂θ(r, 0, t) =∂u∂θ(r, π/2, t) = 0.Also, we will need the singular condition|u(0, θ, t)| < ∞.First we search for normal modes: solutionsu(r , θ, t) = f (r )h(θ)G (t) of the wave equation thatsatisfy the boundary conditions.Note that φ(r, θ) = f (r)h(θ) is going to be aneigenfunction of the Neumann Laplacian in D.Wave equation in polar coordinates:∂2u∂t2= c2∂2u∂r2+1r∂u∂r+1r2∂2u∂θ2.Substitute u(r , θ, t) = f (r )h(θ)G (t) into it:f (r)h(θ)G′′(t) = c2f′′(r)h(θ)G(t)+ r−1f′(r)φ(θ)G (t) + r−2f (r)h′′(θ)G (t).Divide both sides by c2f (r)h(θ)G(t):G′′(t)c2G (t)=f′′(r)f (r)+f′(r)r f (r)+h′′(θ)r2h(θ).It follows thatG′′(t)c2G (t)=f′′(r)f (r)+f′(r)r f (r)+h′′(θ)r2h(θ)= −λ = const.Hence G′′= −λc2G andf′′(r)h(θ) + r−1f′(r)h(θ) + r−2f (r)h′′(θ) = −λf (r)h(θ).The latter equation can be rewritten as∇2φ = −λφ, where φ(r, θ) = f (r)h(θ).Divide both sides by r−2f (r)h(θ):r2f′′(r)f (r)+r f′(r)f (r)+h′′(θ)h(θ)= −λr2.It follows thatr2f′′(r)f (r)+r f′(r)f (r)+ λr2= −h′′(θ)h(θ)= µ = const.Hence h′′= −µh andr2f′′(r) + r f (r ) + (λr2− µ)f (r) = 0.Boundary conditions∂u∂θ(r, 0, t) =∂u∂θ(r, π/2, t) = 0hold if h′(0) = h′(π/2) = 0.Boundary conditions∂u∂r(a, θ, t) = 0 and|u(0, θ, t)| < ∞ hold if f′(a) = 0 and |f (0)| < ∞.We obtain two eigenvalue problems:r2f′′+ rf′+ (λr2− µ)f = 0, f′(a) = 0, |f (0)| < ∞;h′′= −µh, h′(0) = h′(π/2) = 0.The second problem has eigenvalues µm= (2m)2,m = 0, 1, 2, . . . , and eigenfunctionshm(θ) = cos 2mθ. In particular, h0= 1.The first eigenvalue problem:r2f′′+ rf′+ (λr2− ν2)f = 0, |f (0)| < ∞, f′(a) = 0.Here ν =√µm= 2m. First assume that λ > 0.New coordinate z =√λ · r reduces the equationto Bessel’s equation of order ν:z2d2fdz2+ zdfdz+ (z2− ν2)f = 0.General solution: f (z) = c1Jν(z) + c2Yν(z), wherec1, c2are constants.Hence f (r) = c1Jν(√λ r) + c2Yν(√λ r).Boundary condition |f (0)| < ∞ holds if c2= 0.Nonzero solution exists if J′ν(√λ a) = 0.Now consider the case λ = 0. Herer2f′′+ rf′− ν2f = 0.General solution: f (r) =(c1rν+ c2r−νif ν > 0,c1+ c2log r if ν = 0,where c1, c2are constants.Boundary condition |f (0)| < ∞ holds if c2= 0.Nonzero solution exists only for ν = 0.Thus there are infinitely many eigenvalues λm,1, λm,2, . . . ,wherepλm,na = j′ν,nis the nth positive zero of J′ν(exception: j′0,1= 0).Corresponding eigenfunctions:fm,n(r) = Jν(pλm,nr) (note that f0,1= 1).Dependence on t: G′′= −λc2G=⇒ G (t) =c1cos(√λ ct) + c2sin(√λ ct), λ > 0c1+ c2t, λ = 0Normal modes:J2m(pλm,nr) · cos 2mθ ·(cos(pλm,nct)sin(pλm,nct))and t.The solution of the initial-boundary value problem isa superposition of normal modes:u(r , θ, t) = B0,1t ++∞Xm=0∞Xn=1Am,nJ2m(pλm,nr) cos 2mθ cos(pλm,nct)+∞Xm=0∞Xn=1Bm,nJ2m(pλm,nr) cos 2mθ sin(pλm,nct).Initial conditions u(r, θ, 0) = f (r , θ) and∂u∂t(r, θ, 0) = g(r, θ) imply thatu(r , θ, t) = b0,1t ++∞Xm=0∞Xn=1am,nJ2m(pλm,nr) cos 2mθ cos(pλm,nct)+∞Xm=0∞Xn=1bm,npλm,ncJ2m(pλm,nr) cos 2mθ sin(pλm,nct),wheref (r, θ) =∞Xm=0∞Xn=1am,nJ2m(pλm,nr) cos 2mθ,g(r, θ) =∞Xm=0∞Xn=1bm,nJ2m(pλm,nr) cos 2mθ .In particular, suppose that f (r, θ) = 0,g(r, θ) = h(r) cos 4θ .Then u(r , θ, t) ==∞Xn=1bnpλ2,ncJ4(pλ2,nr) cos 4θ sin(pλ2,nct),whereh(r) =∞Xn=1bnJ4(pλ2,nr)is the Fourier-Bessel series.Fourier transformsFourier transform: F[f ](ω) =12πZ∞−∞f (x)e−iωxdxSine transform: S[f ](ω) =2πZ∞0f (x) sin ωx dxCosine transform: C[f ](ω) =2πZ∞0f (x) cos ωx dxInverse Fourier transformsInverse Fourier transform:F−1[f ](ω) =Z∞−∞f (x)eiωxdxInverse sine transform:S−1[f ](ω) =Z∞0f (x) sin ωx dxInverse cosine transform:C−1[f ](ω) =Z∞0f (x) cos ωx dxLaplace’s equation in a half-planeBoundary value problem:∂2u∂x2+∂2u∂y2= 0 (−∞ < x < ∞, 0 < y < ∞),u(x, 0) = f (x).We assume that f is smooth and rapidly decaying atinfinity. We search for a solution with the sameproperties.Apply the Fourier transform Fx(relative to x) toboth sides of the equation:Fx∂2u∂x2+ Fx∂2u∂y2= 0.Let U(ω, y ) = Fx[u](ω) =12πZ∞−∞u(x, y)e−iωxdx.Then Fx∂2u∂y2=∂2U∂y2, Fx∂2u∂x2= (iω)2U(ω, y ).Hence∂2U∂y2= −(iω)2U(ω, y ) = ω2U(ω, y ).General solution: U(ω, y) = aeωy+ be−ωy(ω 6= 0),where a = a(ω), b = b(ω).Initial condition u(x, 0) = f (x) implies thatU(ω, 0) =ˆf (ω).Also, we have a boundary condition limy→∞U(ω, y ) = 0.Since U(ω, y ) → 0 as y → ∞, it follows thatU(ω, y ) =(b(ω)e−ωyif ω > 0,a(ω)eωyif ω < 0.Since U(ω, 0) =ˆf (ω), it follows thatU(ω, y ) =ˆf (ω)e−y|ω|.It turns out that F−1[e−α|ω|](x) =2αx2+ α2, α > 0.Hence U(ω, y ) =ˆf (ω)ˆg(ω, y ), whereg(x, y ) =2yx2+ y2.By the convolution theorem, u(x, y) = (2π)−1f ∗ g .Boundary value problem:∂2u∂x2+∂2u∂y2= 0 (−∞ < x < ∞, 0 < y < ∞),u(x, 0) = f (x).Solution:u(x, y) =12πZ∞−∞g(x − ˜x, y )f (˜x) d ˜x=1πZ∞−∞f (˜x)y(x − ˜x)2+ y2d ˜x.Properties of Fourier transformsLinearity and Shift Theorem(i) F[af + bg] = aF[f ] + bF[g] for all a, b ∈ C.(ii) If g(x) = f (x + α) then ˆg(ω) = eiαωˆf (ω).(iii) If h(x) = eiβxf (x) thenˆh(ω) =ˆf (ω − β).Convolution Theorem(i) F[f · g] = F[f ] ∗ F[g];(ii) F[f ∗ g] = 2π F[f ] · F[g].We know that bχ[−a,a](ω) =sin aωπω.Problem 1. Find the Fourier transform of χ[0,2a].Solution. Clearly, χ[0,2a](x) = χ[−a,a](x −a). Bythe shift theorem,bχ[0,2a](ω) = e−iaωbχ[−a,a](ω) = e−iaωsin aωπω.Problem 2. Computesin aωπω∗sin aωπω.Solution. By the convolution theorem,F−1sin aωπω∗sin
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