and then computeψ2(x)using,ψ2(x)=ϕ2(x)−hϕ2(x),ψ1(x)ihψ1(x),ψ1(x)iψ1(x)=sin(λ2x)−R10sin(λ2x) sin(λ1x)dxR10sin(λ1x) sin(λ1x)dxsin(λ1x)=sin(λ2x)−−sin(λ2) sin(λ1)12λ21sin2(λ1)sin(λ1x)and soψ2(x) = sin(λ2x) +2 sin(λ2)λ21sin(λ1)sin(λ1x).To computeψ3(x), we useψ3(x) =ϕ3(x)−hϕ3(x),ψ1(x)ihψ1(x),ψ1(x)iψ1(x)−hϕ3(x),ψ2(x)ihψ2(x),ψ2(x)iψ2(x).and so on. Once we have a suﬃcient number of these, then we may expandu(x, t)asu(x, t) =∞Xn=1bne−λ2ntψn(x).The initial conditionu(x,0)then requires thatu(x,0) =f(x) =∞Xn=1bnψn(x).To solve forbn, we then usebn=hf(x),ψn(x)ihψn(x),ψn(x)i.Of course,finding a general simple expression forψn(x)may not be easy.So either way,fitting the initial condition is not easy when the original setof eigenfunctions is not orthogonal.––––––––––––––––––––––––––––––––––––19