Math 412-501Theory of Partial Differential EquationsLecture 3-10: Applications of Fouriertransforms (continued).Fourier transformGiven a function h : R → C, the functionˆh(ω) = F[h](ω) =12πZ∞−∞h(x)e−iωxdx, ω ∈ Ris called the Fourier transform of h.Given a function H : R → C, the functionˇH(x) = F−1[H](x) =Z∞−∞H(ω)eiωxdω, x ∈ Ris called the inverse Fourier transform of H.Initial value problem for the heat equation:∂u∂t= k∂2u∂x2(−∞ < x < ∞),u(x, 0) = f (x).Solution: u(x, t) =Z∞−∞G (x, ˜x, t) f (˜x) d ˜x,where G (x, ˜x, t) =1√4πkte−(x−˜x)24kt.The solution is in the integral operator form. Thefunction G is called the kernel of the operator.Also, G (x, ˜x, t) is called Green’s function of theproblem.Wave equation on an infinite intervalInitial value problem:∂2u∂t2= c2∂2u∂x2(−∞ < x < ∞),u(x, 0) = f (x),∂u∂t(x, 0) = g(x).We assume that f , g are smo oth and rapidlydecaying as x → ∞. We search for a solution withthe same properties.Apply the Fourier transform (relative to x) to bothsides of the equation:F∂2u∂t2= c2F∂2u∂x2.Let U = F[u]. That is,U(ω, t) = F[u(·, t)](ω) =12πZ∞−∞u(x, t)e−iωxdx.Then F∂2u∂t2=∂2U∂t2, F∂2u∂x2= (iω)2U(ω, t).Hence∂2U∂t2= c2(iω)2U(ω, t) = −c2ω2U(ω, t).General solution: U(ω, t) = a cos cωt + b sin cωt(ω 6= 0), where a = a(ω), b = b(ω).Apply the Fourier transform to the initial conditions:U(ω, 0) =ˆf (ω),∂U∂t(ω, 0) = ˆg(ω).Therefore U(ω, t) =ˆf (ω) cos cωt + ˆg(ω)sin cωtcω.We know that\χ[−a,a](ω) =sin aωπω, a > 0.Hence F−1sin cωtcω =πcχ[−ct,ct], t > 0. ThenU(ω, t) =12(eicωtˆf (ω) + e−icωtˆf (ω)+πcˆg(ω) \χ[−ct,ct](ω).By the shift theorem and the convolution theorem,u(x, t) =12f (x + ct) + f (x − ct)+12cg ∗ χ[−ct,ct](x).g ∗ χ[−ct,ct](x) =Z∞−∞g(˜x)χ[−ct,ct](x − ˜x) d ˜x=Zx+ctx−ctg(˜x) d ˜x.Initial value problem:∂2u∂t2= c2∂2u∂x2(−∞ < x < ∞),u(x, 0) = f (x),∂u∂t(x, 0) = g (x).Solution:u(x, t) =12f (x + ct) + f (x − ct)+12cZx+ctx−ctg(˜x) d ˜x.Sine and cosine transforms of derivativesSine transform: S[f ](ω) =2πZ∞0f (x) sin ωx dxCosine transform: C[f ](ω) =2πZ∞0f (x) cos ωx dxAssume that f and f′are continuous and absolutelyintegrable on [0, ∞). Then f (x) → 0 as x → ∞.HenceS[f′](ω) =2πZ∞0f′(x) sin ωx dx=2πf (x) sin ωx∞x=0−2πZ∞0f (x)(sin ωx)′dx= −ω C [f ](ω).Likewise, C [f′](ω) =2πZ∞0f′(x) cos ωx dx=2πf (x) cos ωx∞x=0−2πZ∞0f (x)(cos ωx)′dx= −2πf (0) + ω S[f ](ω).S[f′](ω) = −ω C [f ](ω)C [f′](ω) = −2πf (0) + ω S[f ](ω)Now assume that f , f′, f′′are continuous andabsolutely integrable on [0, ∞). By the above,S[f′′](ω) = −ω C [f′](ω) =2πf (0)ω − ω2S[f ](ω),C [f′′](ω) = −2πf′(0) + ω S[f′](ω) = −2πf′(0) − ω2C [f ](ω).S[f′′](ω) =2πf (0)ω − ω2S[f ](ω)C [f′′](ω) = −2πf′(0) − ω2C [f ](ω)Fourier 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 dxProposition Suppose thatR∞−∞|f (x)|dx < ∞.(i) If f is even, f (−x) = f (x), then F[f ] is alsoeven; moreover, C [f ](ω) = 2F[f ](ω) for all ω > 0.(ii) If f is odd, f (−x) = −f (x), then F[f ] is alsoodd; moreover, S[f ](ω) = 2i F[f ](ω) for all ω > 0.Heat equation on a semi-infinite intervalInitial-boundary value problem:∂u∂t= k∂2u∂x2(0 < x < ∞),∂u∂x(0, t) = 0,u(x, 0) = f (x).We search for a solution which is smooth and rapidlydecaying as x → ∞. Apply the cosine transform(relative to x) to both sides of the equation:C∂u∂t= k C∂2u∂x2.Let U(ω, t) = C [u](ω) =2πZ∞0u(x, t) cos ωx dx.Then C∂u∂t=∂U∂t,C∂2u∂x2= −ω2U(ω, t)−2π∂u∂x(0, t) = −ω2U(ω, t).Hence∂U∂t= −kω2U(ω, t).General solution: U(ω, t) = ce−ω2kt, where c = c(ω).Initial condition u(x, 0) = f (x) implies thatU(ω, 0) = C [f ](ω).Therefore U(ω, t) = C [f ](ω) e−ω2kt.Solution: u(x, t) =Z∞0c(ω)e−ω2ktcos ωx dω,where c(ω) =2πZ∞0f (˜x) cos ω˜x d ˜x.The same solution can be obtained by separation ofvariables. The solution can be rewritten in theintegral operator form:u(x, t) =Z∞0G (x, ˜x, t)f (˜x) d ˜x,where G (x, ˜x, t) =2πZ∞0e−ω2ktcos ωx cos ω˜x dω.Green’s function G (x, ˜x, t) ==1πZ∞0e−ω2ktcos (x − ˜x)ω + cos (x + ˜x)ωdωWe know that1πZ∞0e−αω2cos ωy dω =1√4παe−y24α, α > 0.It follows thatG (x, ˜x, t) =1√4πkte−(x−˜x)24kt+ e−(x+˜x)24kt.Laplace’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