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Thread: 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,
    $\displaystyle y''+\lambda y=0$
    $\displaystyle hy(0)+y'(0) = 0, y(1) = 0$
    Show that if $\displaystyle \lambda = 0$ is an eigenvalue, then $\displaystyle h$ must have a specific value and find this value. If $\displaystyle h \geq 1$, 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: $\displaystyle -(-y')' = \lambda (-1)y$, and taking that $\displaystyle \lambda = 0$, means the HLDE becomes $\displaystyle y'' = 0$. The general solution to this should be of the form $\displaystyle y(x)= Ax + B$. Using the boundary conditions,
    $\displaystyle hy(0)+y'(0) = hB + A = 0 \Rightarrow h = -\frac{A}{B}$
    $\displaystyle y(1) = A + B = 0 \Rightarrow A = -B$
    Therefore, $\displaystyle h = -\frac{(-B)}{B} = 1$
    Does this look correct?

    The next part of the question I get stuck on. The general solution for $\displaystyle -(-y')' = \lambda (-1)y$ is $\displaystyle y(x)= Ccos (\sqrt{\lambda}x) + Dsin(\sqrt{\lambda}x) $.
    Using the boundary conditions
    $\displaystyle hy(0)+y'(0) = h(Ccos (\sqrt{\lambda}0) + Dsin(\sqrt{\lambda}0)) + (-C\sqrt{\lambda}sin (\sqrt{\lambda}0) + D\sqrt{\lambda}cos(\sqrt{\lambda}0))$ $\displaystyle = h*C +\sqrt{\lambda}D = 0$
    $\displaystyle 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,
    $\displaystyle y''+\lambda y=0$
    $\displaystyle hy(0)+y'(0) = 0, y(1) = 0$
    Show that if $\displaystyle \lambda = 0$ is an eigenvalue, then $\displaystyle h$ must have a specific value and find this value. If $\displaystyle h \geq 1$, 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: $\displaystyle -(-y')' = \lambda (-1)y$, and taking that $\displaystyle \lambda = 0$, means the HLDE becomes $\displaystyle y'' = 0$. The general solution to this should be of the form $\displaystyle y(x)= Ax + B$. Using the boundary conditions,
    $\displaystyle hy(0)+y'(0) = hB + A = 0 \Rightarrow h = -\frac{A}{B}$
    $\displaystyle y(1) = A + B = 0 \Rightarrow A = -B$
    Therefore, $\displaystyle h = -\frac{(-B)}{B} = 1$
    Does this look correct? Yes.

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