Math 412-501Theory of Partial Differential EquationsLecture 4-4:Green’s function for the wave equation.Green’s function for the heat equationGreen’s function G (x, t; x0, t0) for the heat equationon the infinite interval satisfies∂G∂t= k∂2G∂x2+ δ(x − x0) δ(t − t0)subject to the causality principle:G (x, t; x0, t0) = 0 for t < t0.For t > t0we have thatG (x, t; x0, t0) =1p4πk(t − t0)e−(x −x0)24k(t−t0).General nonhomogeneous problemInitial value problem:∂u∂t= k∂2u∂x2+ Q(x, t) (−∞ < x < ∞, t > 0),u(x, 0) = f (x).Solution: u(x, t) ==Z∞0Z∞−∞G (x, t; x0, t0) Q(x0, t0) dx0dt0+Z∞−∞G (x, t; x0, 0) f (x0) dx0.General nonhomogeneous problemInitial value problem:∂u∂t= k∂2u∂x2+ Q(x, t) (−∞ < x < ∞, t > 0),u(x, 0) = f (x).Solution: u(x, t) ==Zt0Z∞−∞1p4πk(t − t0)e−(x −x0)24k(t−t0)Q(x0, t0) dx0dt0+Z∞−∞1√4πkte−(x −x0)24ktf (x0) dx0.Green’s function for the wave equationGreen’s function G (x, t; x0, t0) for the infiniteinterval describes vibrations of an infinite stringcaused by an instant unit force which is applied attime t0to the point x0.Formally, G solves the equation∂2G∂t2= c2∂2G∂x2+ δ(x −x0) δ(t − t0)subject to the conditionG (x, t; x0, t0) = 0 for t < t0.(causality principle)Apply the Fourier transform (relative to x) to bothsides of the equation:Fx∂2G∂t2= c2Fx∂2G∂x2+ Fx[δ(x −x0)] δ(t − t0).LetbG (ω, t; x0, t0) denote the Fourier transform of Grelative to x:bG (ω, t; x0, t0) =12πZ∞−∞G (x, t; x0, t0)e−i ωxdx.Fx∂2G∂t2=∂2bG∂t2, Fx∂2G∂x2= (iω)2bG = −ω2bG ,Fx[δ(x −x0)](ω) =12πe−i ωx0.=⇒∂2bG∂t2= −c2ω2bG +e−i ωx02πδ(t − t0).Besides,bG (ω, t; x0, t0) = 0 for t < t0.It follows thatbG (ω, t; x0, t0) =(0 for t < t0,aeicωt+ be−icωtfor t > t0,where a = a(ω, x0, t0), b = b(ω, x0, t0);∂bG∂tt=t0+−∂bG∂tt=t0−=e−i ωx02π;bGt=t0−=bGt=t0+.bGt=t0+= aeicωt0+ be−icωt0= 0,∂bG∂tt=t0+= icω · aeicωt0− icω · be−icωt0=e−i ωx02π.Then a =e−i ωx04πicωe−icωt0, b = −e−i ωx04πicωeicωt0.HencebG =e−i ωx04πicωeicω(t−t0)− e−icω(t−t0)=e−i ωx02csin(cω(t − t0))πωif t > t0.G (x, t; x0, t0) =(12cif |x −x0| < c(t − t0),0 if |x −x0| > c(t − t0).G (x, t; x0, t0) as a function of xNonhomogeneous problemsInitial value problem #1:∂2u∂t2= c2∂2u∂x2+ Q(x, t) (−∞ < x < ∞, t > 0),u(x, 0) = 0,∂u∂t(x, 0) = 0.Solution: u(x, t) =Z∞0Z∞−∞G (x, t; x0, t0) Q(x0, t0) dx0dt0=12cZZDx ,tQ(x0, t0) dx0dt0,where Dx ,t= {(x0, t0) : 0 < t0< t − c−1|x −x0|}.Domain of influence Dx ,tNonhomogeneous problemsInitial value problem #2:∂2u∂t2= c2∂2u∂x2(−∞ < x < ∞, t > 0),u(x, 0) = 0,∂u∂t(x, 0) = g(x).Solution: u(x, t) ==Z∞−∞G (x, t; x0, 0) g(x0) dx0=12cZx +ctx −ctg(x0) dx0.Nonhomogeneous problemsInitial value problem #3:∂2u∂t2= c2∂2u∂x2(−∞ < x < ∞, t > 0),u(x, 0) = f (x),∂u∂t(x, 0) = 0.Solution: u(x, t) =Z∞−∞G1(x, t; x0, 0) f (x0) dx0,where G1(x, t; x0, t0) is the solution of the equation∂2G1∂t2= c2∂2G1∂x2+ δ(x −x0) δ′(t − t0)subject to the causality principle.Since∂2G∂t2= c2∂2G∂x2+ δ(x −x0) δ(t − t0),it follows that G1= −∂G∂t0.Let H denote the Heaviside function: H(z) = 0 forz < 0 and H(z) = 1 for z > 0. ThenG (x, t; x0, t0) =12cHx −x0+ c(t − t0)−− Hx −x0− c(t − t0),∂G∂t0(x, t; x0, t0) = −12δx −x0+ c(t − t0)−−12δx −x0− c(t − t0).Nonhomogeneous problemsInitial value problem #3:∂2u∂t2= c2∂2u∂x2(−∞ < x < ∞, t > 0),u(x, 0) = f (x),∂u∂t(x, 0) = 0.Solution: u(x, t) == −Z∞−∞∂G∂t0(x, t; x0, 0) f (x0) dx0=f (x + ct) + f (x −ct)2.General nonhomogeneous problemInitial value problem:∂2u∂t2= c2∂2u∂x2+ Q(x, t) (−∞ < x < ∞, t > 0),u(x, 0) = f (x),∂u∂t(x, 0) = g(x).Solution: u(x, t) ==Z∞0Z∞−∞G (x, t; x0, t0) Q(x0, t0) dx0dt0−Z∞−∞∂G∂t0(x, t; x0, 0) f (x0) dx0+Z∞−∞G (x, t; x0, 0) g(x0) dx0.General nonhomogeneous problemInitial value problem:∂2u∂t2= c2∂2u∂x2+ Q(x, t) (−∞ < x < ∞, t > 0),u(x, 0) = f (x),∂u∂t(x, 0) = g(x).Solution: u(x, t) ==12cZZDx ,tQ(x0, t0) dx0dt0+f (x + ct) + f (x −ct)2+12cZx +ctx −ctg(x0)