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Math Help - Sturm-Liouville eigenvalue problem

  1. #1
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    Sturm-Liouville eigenvalue problem

    Hi,
    I got a question which I'm quite stuck on...
    Given the eigenvalue problem,
    y''+\lambda y=0
    hy(0)+y'(0) = 0, y(1) = 0
    Show that if \lambda = 0 is an eigenvalue, then h must have a specific value and find this value. If  h \geq 1, show that there is exactly one negative eigenvalue.

    I believe I've done the first part, as my working is as follows:
    Rearrange the HLDE into self adjoint form: -(-y')' = \lambda (-1)y, and taking that \lambda = 0, means the HLDE becomes y'' = 0. The general solution to this should be of the form y(x)= Ax + B. Using the boundary conditions,
    hy(0)+y'(0) = hB + A = 0 \Rightarrow h = -\frac{A}{B}
    y(1) = A + B = 0 \Rightarrow A = -B
    Therefore, h = -\frac{(-B)}{B} = 1
    Does this look correct?

    The next part of the question I get stuck on. The general solution for -(-y')' = \lambda (-1)y is y(x)= Ccos (\sqrt{\lambda}x) + Dsin(\sqrt{\lambda}x) .
    Using the boundary conditions
    hy(0)+y'(0) = h(Ccos (\sqrt{\lambda}0) + Dsin(\sqrt{\lambda}0)) + (-C\sqrt{\lambda}sin (\sqrt{\lambda}0) + D\sqrt{\lambda}cos(\sqrt{\lambda}0)) = h*C +\sqrt{\lambda}D = 0
    y(1) =Ccos (\sqrt{\lambda}) + Dsin(\sqrt{\lambda}) = 0
    From here, I not sure where to go... have I do the right thing?

    Thanks for your time.
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  2. #2
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    Quote Originally Posted by Silverflow View Post
    Hi,
    I got a question which I'm quite stuck on...
    Given the eigenvalue problem,
    y''+\lambda y=0
    hy(0)+y'(0) = 0, y(1) = 0
    Show that if \lambda = 0 is an eigenvalue, then h must have a specific value and find this value. If  h \geq 1, show that there is exactly one negative eigenvalue.

    I believe I've done the first part, as my working is as follows:
    Rearrange the HLDE into self adjoint form: -(-y')' = \lambda (-1)y, and taking that \lambda = 0, means the HLDE becomes y'' = 0. The general solution to this should be of the form y(x)= Ax + B. Using the boundary conditions,
    hy(0)+y'(0) = hB + A = 0 \Rightarrow h = -\frac{A}{B}
    y(1) = A + B = 0 \Rightarrow A = -B
    Therefore, h = -\frac{(-B)}{B} = 1
    Does this look correct? Yes.

    The next part of the question I get stuck on. The general solution for -(-y')' = \lambda (-1)y is y(x)= Ccos (\sqrt{\lambda}x) + Dsin(\sqrt{\lambda}x) .
    Using the boundary conditions
    hy(0)+y'(0) = h(Ccos (\sqrt{\lambda}0) + Dsin(\sqrt{\lambda}0)) + (-C\sqrt{\lambda}sin (\sqrt{\lambda}0) + D\sqrt{\lambda}cos(\sqrt{\lambda}0)) = h*C +\sqrt{\lambda}D = 0
    y(1) =Ccos (\sqrt{\lambda}) + Dsin(\sqrt{\lambda}) = 0
    You are looking for solutions with \lambda <0, so it's not a good idea to use functions like \cos(\sqrt{\lambda}x) and \sin(\sqrt{\lambda}x). Instead, notice that the equation y'' = (-\lambda) y has solutions y=e^{\pm\alpha x}, where \alpha = \sqrt{-\lambda}. If y = Ae^{\alpha x} + Be^{-\alpha x} then the boundary conditions lead to some equation like h = \frac{\alpha(e^{\alpha} + e^{-\alpha})}{e^{\alpha} - e^{-\alpha}}, and you need to show that for h\geqslant1 that has a unique solution for \alpha>0.
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  3. #3
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    That makes a lot more sense. Thank you very much!
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