Math 412-501Theory of Partial Differential EquationsLecture 4-1: Green’s functions.Dirac delta function.Heat equation on the infinite intervalInitial value problem:∂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.G (x, ˜x, t) =1√4πkte−(x −˜x )24ktG (x, ˜x, t) is:• defined for t > 0,• positive,• infinitely smooth.Z∞−∞G (x, ˜x, t) dx =Z∞−∞1√4πkte−(x −˜x )24ktdx=Z∞−∞1√4πkte−x24ktdx =Z∞−∞1√πe−y2dy = 1∂G (x, ˜x, t)∂x=1√4πkte−(x −˜x )24kt·−2(x − ˜x)4kt= −x − ˜x2ktG (x, ˜x, t);∂2G (x, ˜x, t)∂x2=(x − ˜x)24k2t2−12ktG (x, ˜x, t);∂G (x, ˜x, t)∂t= −12t−3/2·1√4πke−(x −˜x )24kt+1√4πkte−(x −˜x )24kt·(x − ˜x)24kt2=(x − ˜x)24kt2−12tG (x, ˜x, t).Lemma For any ˜x ∈ R the functionu(x, t) = G(x, ˜x, t) is a solution of the heatequation for t > 0.Besides, limt→0G (x, ˜x, t) =(∞ if x = ˜x,0 if x 6= ˜x.We say that G (x, ˜x, t) is the solution of the initialvalue problem∂u∂t= k∂2u∂x2(−∞ < x < ∞),u(x, 0) = δ(x − ˜x),where δ(x) is the Dirac delta function.Dirac delta functionδ(x) is a function on R such that• δ(x) = 0 for all x 6= 0,• δ(0) = ∞,•R∞−∞δ(x) dx = 1.For any continuous function f and any x0∈ R,Z∞−∞f (x)δ(x − x0) dx = f (x0).δ(x) is a generalized function (or distribution).That is, δ is a linear functional on a space of testfunctions f such that δ[f ] = f (0).Delta sequence/familyRegular function g can be regarded as a generalizedfunctionf 7→Z∞−∞f (x)g (x) dx.A delta sequence is a sequence of (sharplypeaked) functions g1, g2, . . . such thatlimn→∞Z∞−∞f (x)gn(x) dx = f (0)for any test function f (e.g., infinitely smooth andrapidly decaying). That is, gn→ δ as n → ∞ (asgeneralized functions).A delta fa mily is a family of functions hε,0 < ε ≤ ε0, such that limε→0hε= δ.hε(x) =1√πεe−x2/ε, ε > 0.Initial value problem:∂u∂t= k∂2u∂x2(−∞ < x < ∞),u(x, 0) = δ(x − ˜x).The initial condition means that gt(x) = u(x + ˜x, t)ought to be a delta family. Indeed, ifu(x, t) = G(x, ˜x, t), thengt(x) =1√4πkte−x2/(4kt).Now suppose thatu(x, t) =Z∞−∞G (x, ˜x, t) f (˜x) d˜x,where f is a test function (smooth and rapidlydecaying).Then u(x, t) is infinitely smooth for t > 0 andsolves the heat equation. Besides,limt→0u(x, t) = f (x).Actually, it is sufficient that f be continuous andbounded.Heat equation o n a semi-infinite intervalInitial-boundary value problem:∂u∂t= k∂2u∂x2(0 < x < ∞),∂u∂x(0, t) = 0,u(x, 0) = f (x).Solution: u(x, t) =Z∞0G1(x, ˜x, t) f (˜x) d˜x,where G1(x, ˜x, t) =1√4πkte−(x −˜x )24kt+ e−(x +˜x )24kt.Clearly, G1(x, ˜x, t) = G (x, ˜x, t) + G (−x, ˜x, t).Since G (x, ˜x, t) is a solution of the heat equationfor t > 0, so are G (−x, ˜x, t) and G1(x, ˜x, t).By definition, G1(x, ˜x, t) is even as a function of x.Therefore∂G1∂x(0, ˜x, t) = 0.Let f1denote the even extension of the function f toR, i.e., f1(x) = f1(−x) = f (x) for all x ≥ 0. Thenu(x, t) =Z∞0G1(x, ˜x, t)f (˜x) d ˜x =Z∞−∞G (x, ˜x, t)f1(˜x) d ˜x.It follows that u(x, t) is indeed the solution of theinitial-boundary value problem. Again, it is sufficientthat f be continuous and bounded.Heat equation o n a finite intervalInitial-boundary value problem:∂u∂t= k∂2u∂x2(0 < x < L),u(0, t) = u(L, t) = 0,u(x, 0) = f (x) (0 < x < L).Solution: u(x, t) =X∞n=1bne−λnktsinnπxL,where λn= (nπ/L)2,bn=2LZL0f (˜x) sinnπ˜xLd ˜x.For t > 0 we obtainu(x, t) =2L∞Xn=1e−λnktsinnπxLZL0f (˜x) sinnπ˜xLd ˜x=ZL0G2(x, ˜x, t)f (˜x) d ˜x,whereG2(x, ˜x, t)