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PDE's Find the values of lambda (eigenvalues)

  1. Apr 6, 2013 #1
    the problem stays to find the values of Lambda for which the given problem has nontrivial solutions.
    Also to determine the corresponding nontrivial eigenfunctions.
    y''-2y'+[itex]\lambda[/itex]y=0

    0<x<[itex]\pi[/itex], y(0)=0, y([itex]\pi[/itex])=0

    [itex]r^{2}[/itex]-2r=-[itex]\lambda[/itex]
    r=1±i[itex]\sqrt{\lambda+1}[/itex]
    y=[itex]e^{x}[/itex]([itex]c_{1}[/itex]cos([itex]\sqrt{\lambda+1}[/itex]x)+[itex]c_{2}[/itex]sin([itex]\sqrt{\lambda+1}[/itex]x)

    for lambda i got [itex]\lambda_{n}[/itex]=[itex]n^{2}[/itex]-1

    and for the function i got
    [itex]y_{n}[/itex]=[itex]c_{n}[/itex][itex]e^{\pi}[/itex]sin(nx)
    is this anywhere close to being right?
     
  2. jcsd
  3. Apr 6, 2013 #2

    LCKurtz

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    Put the ##\lambda## back on the other side and check that quadratic formula again.

    Yes, it's close. Once you fix that ##\lambda## thing I think you will find a ##\lambda -1## under the radical. You have assumed the quantity under the radical is less than zero. To be complete you need to eliminate the cases where it is greater than or equal to zero.
     
  4. Apr 6, 2013 #3
    well what i did was put the lambda on the right side and then completed the square. and I am evaluating the case when λ>0, this is why I have i[itex]\sqrt{\lambda+1}[/itex]
     
  5. Apr 6, 2013 #4
    Ahh ok I think you assumed i was doing it for λ<0
     
  6. Apr 6, 2013 #5

    LCKurtz

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    Like I said, check it. It should be ##\sqrt{1-\lambda}##

    That isn't what I assumed or what you should do. The relevant cases are ##\lambda>1,\ \lambda < 1,\ \lambda = 1##.
     
  7. Apr 6, 2013 #6
    ahh ok got a bit confused there
     
  8. Apr 6, 2013 #7
    Thank you
     
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