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Lost on eigenvalues

  1. Nov 20, 2004 #1
    I'm having trouble getting started on this problem... I just really don't understand what to do.


    Solve
    [tex]X'+2X'+(\lambda-\alpha)X=0, 0<x<1[/tex]
    [tex]X(0)=0[/tex]
    [tex]X'(1)=0[/tex]

    a. Is [tex]\lambda=1+\alpha[/tex] an eigenvalue? What is the corresponding eigenfunction?
    b. Find the equation that the other eigenvalues satisfy.


    I appreciate any help you can give me!

    Thanks,
    Chris
     
    Last edited: Nov 20, 2004
  2. jcsd
  3. Nov 20, 2004 #2
    are u sure that the edo is correct?

    you have there a linear second order d.o with constant coeficients and initial values, so your solution will be some linear comb. of exp[rt] (you have to calculate r of course).

    i dont really understand why would you end with an eigenvalue problem in this way but maybe im not well informed.
     
  4. Nov 21, 2004 #3
    Yeah, everything on there is correct. I'm not sure what you mean though...


    nm... figured it out.
     
    Last edited: Nov 22, 2004
  5. Nov 22, 2004 #4
    [tex]X''+2X'+(\lambda-\alpha)X=0[/tex]

    let [itex]X=e^{rt}[/itex] implies

    [tex]r^2+2r+(\lambda-\alpha)=0[/tex]

    so

    [tex]r=-1\pm\sqrt{1-(\lambda-\alpha)}[/tex]

    [tex]X(t)=Ae^{(-1+\sqrt{1-(\lambda-\alpha)})t}+Be^{(-1-\sqrt{1-(\lambda-\alpha)})t}[/tex]

    [tex]X(0)=A+B[/tex]

    so [itex]A=-B[/tex]

    [tex]X'(1)=A[r_{+}e^{r+}-r_{-}e^{r_{-}}]=0[/tex]

    [itex]A=0[/itex] would lead to the trivial solution, so

    [tex]r_{+}e^{\sqrt{1-(\lambda-\alpha)}}=r_{-}e^{-\sqrt{1-(\lambda-\alpha)}}[/tex]

    [itex]\lambda=1+\alpha[/itex] is clearly an eigenvalue

    and [itex]\lambda_{m}[/itex] satisfy the equation

    [tex]\frac{-1+\sqrt{1-(\lambda_{m}-\alpha)}}{-1-\sqrt{1-(\lambda_{m}-\alpha)}}e^{2\sqrt{1-(\lambda_{m}-\alpha)}}=1[/tex]
     
    Last edited: Nov 22, 2004
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