MATH 220: DUHAMEL’S PRINCIPLEAlthough we have solved only the homogeneous heat equation on Rn, the samemethod employed there also solves the inhomogeneous PDE. As an application ofthese methods, let’s solve the heat equation on (0, ∞)t× Rnx:(1) ut− k∆u = f, u(0, x) = φ(x),with f ∈ C∞([0, ∞)t; S(Rnx)), φ ∈ S(Rn) (say) given. Namely, taking the partialFourier transform in x, and writing Fxu(t, ξ) = ˆu(t, ξ), gives∂ ˆu∂t(t, ξ) + k|ξ|2ˆu(t, ξ) =ˆf(t, ξ), ˆu(0, ξ) = (Fφ)(ξ).We again solve the ODE for each fixed ξ. To do so, we multiply through by ek|ξ|2t:ek|ξ|2t∂ ˆu∂t(t, ξ) + ek|ξ|2tk|ξ|2ˆu(t, ξ) = ek|ξ|2tˆf(t, ξ),and realize that the left hand side isddtek|ξ|2tˆu(t, ξ).Thus, integrating from t = 0 and using the fundamental theorem of calculus yieldsek|ξ|2tˆu(t, ξ) − ˆu(0, ξ) =Zt0ek|ξ|2sˆf(s, ξ) ds,soˆu(t, ξ) = e−k|ξ|2t(Fφ)(ξ) +Zt0e−k|ξ|2(t−s)ˆf(s, ξ) ds.Finally,u(t, x) = F−1ξe−k|ξ|2t(Fφ)(ξ) +Zt0e−k|ξ|2(t−s)ˆf(s, ξ) ds= F−1ξe−k|ξ|2t(Fφ)(ξ)+Zt0F−1e−k|ξ|2(t−s)ˆf(s, ξ)ds= (4πkt)−n/2ZRne−|x−y|2/4ktφ(y) dy+Zt0(4πk(t − s))−n/2ZRne−|x−y|2/4k(t−s)f(s, y) dy ds.Note that the parts of the solution formula c orresponding to the initial conditionand the forcing term are very similar. In fact, let the solution oper ator of thehomogeneous PDEut− k∆u = 0, u(0, x) = φ(x),be denoted by S(t), so(S(t)φ)(x) = u(t, x).Then(4πk(t − s))−n/2ZRne−|x−y|2/4k(t−s)f(s, y) dy = (S(t − s)fs)(x),12where we let fs(x) = f (s, x) be the restric tion of f to the time s slice. Thus, thesolution formula for the inhomogeneous PDE, (1), takes the form(2) u(t, x) = (S(t)φ)(x) +Zt0(S(t − s)fs)(x) ds.The fact that we can write the so lution of the inho mogeneous PDE in terms ofthe solution of the Cauchy problem for the homogeneous PDE is called Duhamel’sprinciple.Physically one may think of (2) as follows. The expression S(t − s)fsis thesolution of the heat equation at time t with initial condition f (s, x) imposed attime s. Thus, we think of the forcing as a superpositio n (namely, integral) ofinitial conditions given at times s between 0 (when the actual initial conditionis imposed) and time t (when the solution is evaluated). Conversely, one couldsay that the initial condition amounts to a delta-distributional forcing, φ(x)δ0(t),provided that f vanished for time ≤ 0 and we impose u(T, x) = 0 for some T < 0(say, T = −1): the delta distribution should be considered as being paired againstthe characteristic function [− T, t], giving S(t)φ as a result (at least if t > 0).In fact, this calculation did not depend significantly on particular features of theLaplacian on Rn. That is, suppose that we have another differential operator L onRn, and consider the PDE(3) ∂tu − Lu = f, u(0, x) = φ(x).Suppose that we can solve the homogeneous problem, i.e. that∂tu − Lu = 0, u(0, x) = φ(x),and let S(t) denote the solution operator, so that (S(t)φ)(x) = u(t, x), hence(S(0)φ)(x) = φ(x),as the initial condition holds, and(∂t− L)S(t)φ = 0,since the PDE holds. We claim that under these assumptions the solution of (3) isgiven by Duhamel’s formula,u(t, x) = (S(t)φ)(x) +Zt0(S(t − s)fs)(x) ds,just as above. To see this, first note that the initial condition is certainly satisfied,for the integral vanishes if t = 0. Next, as S(t) is the solution operator fo r thehomogeneous PDE, ∂t− L applied to the first term of Duhamel’s formula vanishes.Thus,(∂tu − Lu)(t, x) = (∂t− L)Zt0(S(t − s)fs)(x) ds= (S(0)ft)(x) +Zt0(∂t− L)(S(t − s)fs)(x) ds= f(t, x),where the S(0)ftterm arose by differentiating the upper limit of the integral, whilethe other term by differentiating under the integral sign, and we used that S(0)ft=ftand(∂t− L)(S(t − s)fs) =(∂τ− L)S(τ)fs|τ =t−s= 0.3We ca n also deal with inhomogeneous boundary conditions. For instance, wecan solveut− k∆u = f, xn≥ 0,u(x′, 0, t) = h(x′, t) (DBC),u(x, 0) = φ(x) (IC),(4)where f and φ are given functions for xn≥ 0, and h is a given function of (x′, t). Tosolve this, we reduce it to a PDE with homogeneous boundary co nditions. Thus,let F be any function on Rnx× [0, ∞)tsuch that F satisfies the boundary condition,i.e. F (x′, 0, t) = h(x′, t). For instance, we may take F (x′, 0, t) = h(x′, t). If u solves(4) then v = u − F solvesvt− k∆v = f − (Ft− k∆F ), xn≥ 0,v(x′, 0, t) = 0 (DBC),v(x, 0) = φ(x) − F (x, 0) (IC),(5)i.e. a PDE with homogeneous boundary condition, but different initial conditionand forcing term. Here we usedvt− k∆v = ut− Ft− k∆u + k∆F = f − (Ft− k∆F ).Conversely, if we solve (5 ), then u = v +F will solve (4). But we know how to solve(5): either use an appropriate (odd) extension of the data to Rnand solve the PDEthere, or use Duhamel’s principle and the solution of the homogeneous PDE (withvanishing right hand side) on the half space; see your homework. Thus, (4) can besolved as well.Similarly, for inhomogeneous Neumann boundary conditions,ut− k∆u = f, xn≥ 0,(∂xnu)(x′, 0, t) = h(x′, t) (DBC),u(x, 0) = φ(x) (IC),(6)we take some F such that F satisfies the boundary condition, i.e. (∂xnF )(x′, 0, t) =h(x′, t). For instance, we can takeF (x′, xn, t) = xnh(x′, t).Proceeding as above, it suffices to solvevt− k∆v = f − (Ft− k∆F ), xn≥ 0,(∂xnv)(x′, 0, t) = 0 (NBC),v(x, 0) = φ(x) − F (x, 0) (IC),(7)and let u = v + F . Since (7) has homo geneous boundary conditions, it c an besolved as above (either take even extensions, or use Duhamel’s principle on thehalf-space).One can also derive a version of Duhamel’s principle for the wave equation, orindeed equations of the form(8) (∂2t− L)u = f, u(x, 0) = φ(x), ut(x, 0) = ψ(x).Let S(t) be the solution operator corresponding to f = 0, φ = 0, i.e. for functionsψ, u(t, x) = (S(t)ψ)(x) solves(∂2t− L)u = 0, u(x, 0) = 0, ut(x, 0) = ψ(x),4so(∂2t− L)(S(t)ψ) = 0, S(0)ψ(x) = 0, (∂tS(t)ψ)(x)|t=0= ψ(x),Then the solution for (8) is(9) u(x, t) = ∂t(S(t)φ)(x) + (S(t)ψ)(x) +Zt0(S(t − s)fs)(x) ds.One checks easily that this indeed solves (8). First, the last term vanishes whent = 0 (since the integral is from 0 to 0 then) and its t-derivative at t = 0 isS(0)f0+Z00(∂tS(t − s)fs)(x) ds = 0,so it does not contribute to the initial conditions. We have S(0)ψ = 0 and∂t(S(t)ψ)|t=0= ψ.