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Eigenvalue problem

  1. Feb 28, 2005 #1
    I have this eigenvalue problem:
    [itex] \frac{\mbox{d}^2y}{\mbox{d}x^2}+\left(1-\lambda\right)\frac{\mbox{d}y}{\mbox{d}x}-\lambda y = 0 \ , \ x\in[0,1], \ \lambda\in\mathbb{R}
    [/itex]
    [itex] y(0)=0 [/itex]
    [itex] \frac{\mbox{d}y}{\mbox{d}x}(1)=0 [/itex]
    Then, I have to show that there exists only one eigenvalue [itex] \lambda [/itex], and find this eigenvalue and write the corresponding eigenfunctions.

    Thus far, I have solved the ODE's characteristic equation
    [itex] r^2+(1-\lambda)r-\lambda=0 [/itex].
    This gives me two solutions
    [itex] r=-1 [/itex] and [itex] r=\lambda [/itex].
    Thus the solution to the ODE is
    [itex]
    y=c_1\exp(-x)+c_2\exp(\lambda x) \ , \ x\in\mathbb{R} \ , \ c_1, \ c_2\in\mathbb{R}
    [/itex].

    Can I now conclude that because we only have real solutions to the characteristic equation, only one eigenvalue exists?

    Secondly, I am not completely sure how to find the sought eigenvalue. I know how to find the eigenvalue when I have a solution that involves [itex] \sin [/itex] and [itex] \cos [/itex], but here I am not sure how to do it. Could anyone give me a hint?

    Thirdly, when I have the eigenvalue there should not be any problems in writing the corresponding eigenfunctions, or is it?

    I would appreciate any help. I am not looking for a solution to my homework problem, but any hints to the problem mentioned above are welcome.
     
    Last edited: Feb 28, 2005
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  3. Feb 28, 2005 #2

    dextercioby

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    You miss text from the problem.In order for that to be a genuine Cauchy problem,you need to specify 2 initial conditions.Besides,you didn't use the one which u have posted...:wink:

    Daniel.
     
  4. Feb 28, 2005 #3

    HallsofIvy

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    An "eigenvalue" is a value of &lamba; such that the problem has a non-trivial solution. Have you determined what functions satisfy your other conditions?
    (You have y(0)= 0 but then is see only [itex]\frac{dy}{dx}[/itex]. Did you mean [itex]\frac{dy}{dx}(0)= 0[/itex] or [itex]\frac{dy}{dx}(1)= 0[/itex]?)

    The problem is a lot easier if it is [itex]\frac{dy}{dx}(0)= 0[/itex]!

    Put x= 0 into the formulas for y and dy/dx and you get two equations to solve for c1 and c2. For what values of &lamda; can you NOT solve for specific value of c1 and c2?
     
  5. Feb 28, 2005 #4

    dextercioby

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    He corrected his typo and now he can follow your advice for finding "lambda"...:approve:

    Daniel.
     
  6. Mar 2, 2005 #5
    Has anyone tried to calculate the eigenvalue? My suggestion is [itex] \lambda=-1 [/itex]. Can anyone either confirm or refute this?
     
  7. Mar 2, 2005 #6

    dextercioby

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    What is the system of equations for C_{1} & C_{2}...?

    Daniel.
     
  8. Mar 2, 2005 #7
    When I use the initial conditions, I get this system of equations:

    [tex] c_1+c_2=0 [/tex]
    [tex] -c_1\exp(-1)+c_2\lambda\exp(\lambda)=0 [/tex]

    I then guess that [itex] \lambda=-1 [/itex], and find out that with this eigenvalue you cannot solve the system for any particular value of [itex] c_1 [/itex] and [itex] c_2 [/itex].
     
    Last edited: Mar 2, 2005
  9. Mar 2, 2005 #8

    dextercioby

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    Yes,indeed,lambda=-1 makes the 2 equations identical.

    Now find the eigenfunctions corresponding to [itex]\lambda=-1[/itex]

    Daniel.
     
  10. Mar 2, 2005 #9

    HallsofIvy

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    Deleted- DexterCioby beat me to it!
     
    Last edited: Mar 2, 2005
  11. Mar 2, 2005 #10

    dextercioby

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    Halls,it has lambda =-1...:wink:

    I "guess" -1 can be called elementary,huh...?

    Daniel.
     
  12. Mar 2, 2005 #11
    Well, the eigenfunctions corresponding to [itex] \lambda=-1 [/itex] must then be

    [tex] y=c_1\exp(-x)+c_2\exp(-x)=(c_1+c_2)\exp(-x)=c\exp(-x),~~c_1,\,c_2,\,c\in\mathbb{R} [/tex]
     
  13. Mar 2, 2005 #12

    dextercioby

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    Yes.[itex] y(x)=C\exp(-x),C\in\mathbb{R}[/itex] is the awaited sollution.

    Daniel.
     
  14. Mar 3, 2005 #13
    After I have talked to the teacher, I have realised that the eigenfunction written in post #11 is wrong. The reason is that when [itex]\lambda=-1[/itex], the characteristic equation has a double root [itex]r=-1[/itex]. Thus, the solution is [itex]y=c_1\exp(-x)+c_2x\exp(-x)[/itex], and NOT [itex]y=c\exp(-x)[/itex] as written earlier. Do you agree dextercioby?
     
    Last edited: Mar 3, 2005
  15. Mar 4, 2005 #14

    saltydog

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    I'm confused: To meet the initial condition y(0)=0, [itex]c_1[/itex] has to be zero. But if that's the case, then any value of [itex]c_2[/itex] meets the derivative at the boundary condition specified above and thus we loose uniqueness.
     
  16. Mar 5, 2005 #15

    HallsofIvy

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    Yes, that's the whole point of "eigenvalue". If λ is not an eigenvalue, then the equation Ax= &lamda;x has a unique solution: x= 0. If λ is an eigenvalue, then there exist an infinite number of solutions: the set of eigenvectors corresponding to a given eigenvalue is is a subspace of the original vectdor space.
     
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