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Then the second boundary condition leads to c 1 μ

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Then the second boundary condition leads toc1μcosh(μ) +μ2c1sinh(μ) = 0or (sincec1andμare both not zero),cosh(μ) +μsinh(μ) = 0.But, without any lose in generality, we may always assume thatμ >0, andsocosh(μ) +μsinh(μ)>1showing that it can never be zero.Thereforec=μ2>0leads to nosolutions. Next we assume thatc= 0. Then we getϕ00(x) = 0withϕ(0) = 0andϕ0(1) = 0which leads toϕ(x) = 0. Finally we assume thatc=λ2<0. This leadstoϕ00(x) +λϕ(x) = 0withϕ(0) = 0andϕ0(1)λ2ϕ(1) = 0.This problem was the subject of Problem #2 above and has the eigenfunc-tionsϕn(x) = sin(λnx)withcos(λn) =λnsin(λn).
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b.) Continuing with the separation of variables method, then we haveγ0n(t) +λ2nγn(t) = 0so thatγn(t) =eλ2ntforn= 1,2,3, . . .. Then we see thatu(x, t) =Xn=1aneλ2ntsin(λnx).The initial conditionu(x,0)then requires thatu(x,0) =f(x) =Xn=1ansin(λnx) =Xn=1anϕn(x).To solve foran, let us define an inner product byhF(x), G(x)i=Z10F(x)G(x)dxand computehf(x),ϕm(x)i=Xn=1hϕn(x),ϕm(x)ian.This is a system of infinity equation in the infinite number of unknowns{a1, a2, a3, . . . , an, . . .}.The fact thatϕn(x)andϕm(x)are non-orthogonal makes the remain un-coupled and therefore much more dicult to solve. Instead of solving thissystem, we may get around this problem by converting the set of functionsϕn(x) = sin(λnx)into an orthogonal setψn(x). Toward this end we use theGram-Schmidtprocessand setψ1(x) =ϕ1(x) = sin(λ1x)18
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 sucient number of these, then we may expandu(x, t)asu(x, t) =Xn=1bneλ2ntψn(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
––––––––––––––––––––––––––––––––––––Solution to Problem #6First we solve forup(x)so that it satisfiesxu00p(x)u0p(x) = 0,andup(0) = 1andup(1) = 0. Solving the differential equation forup(x), we getdu0pu0p=dxxorln(u0p) = ln(x) +c1which leads tou0p(x) =Ax, which then results in the general solutionup(x) =AZxdx+B=12Ax2+B.

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Term
Fall
Professor
car
Tags
Sin, Boundary value problem, Sturm Liouville theory, lim n

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