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Math Help - Sturm-Liouville problem

  1. #1
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    Sturm-Liouville problem

    Hi i'm having a bit of trouble working out how to do this question. As i've not done Sturm-Liouville theory before and finding it hard to work it out.
    The question is
    y''-2y'+(1+lamda)y=0 with homogeneous bc on x=0,1
    Write in S-L form
    Write the orthagonality condition.

    Now i'm having trouble working out how this is done could anybody be kind enough to go through the steps needed to work this out?
    Thanks very much
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  2. #2
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    Quote Originally Posted by math_lete View Post
    Hi i'm having a bit of trouble working out how to do this question. As i've not done Sturm-Liouville theory before and finding it hard to work it out.
    The question is
    y''-2y'+(1+lamda)y=0 with homogeneous bc on x=0,1
    Write in S-L form
    Write the orthagonality condition.

    Now i'm having trouble working out how this is done could anybody be kind enough to go through the steps needed to work this out?
    Thanks very much
    The "Sturm-Liouville form" is \frac{d}{dx}\left(p(x)\frac{dy}{dx}\right)+ (q(x)+ \lambda)y= 0.

    You need to find p(x) so that \frac{d}{dx}\left(p(x)\frac{dy}{dx}\right)= p(x)\frac{d^2y}{dx^2}+ \frac{dp}{dx}\frac{dy}{dx}= p(x)\frac{d^2y}{dx^2}- 2p(x)\frac{dy}{dx}.

    In other words, you need to find p(x) so that \frac{dp}{dx}= -2p, then multiply the equation by that p. That's essentially the same as finding an "integrating factor".
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  3. #3
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    Quote Originally Posted by math_lete View Post
    Hi i'm having a bit of trouble working out how to do this question. As i've not done Sturm-Liouville theory before and finding it hard to work it out.
    The question is
    y''-2y'+(1+lamda)y=0 with homogeneous bc on x=0,1
    Write in S-L form
    Write the orthagonality condition.

    Now i'm having trouble working out how this is done could anybody be kind enough to go through the steps needed to work this out?
    Thanks very much
    So we solve the associated equation

    r^2-2r+(1+\lambda)=0

    Now if we complete the square we get

    (r-1)^2+\lambda \iff (r-1)^2=-\lambda \iff r=1 \pm\sqrt{-\lambda}

    Now we have 3 cases: \lambda > 0 \lambda = 0 \lambda < 0

    If \lambda = 0

    we get y=c_1+c_2x Now if you impose the boundry conditions you get the trivial solution y=0

    If you have \lambda < 0 we get

    y=c_1e^{-(1+\sqrt{-\lambda})x}+c_2e^{-(1-\sqrt{\lambda})x}

    now if x=0 we get

    0=c_1+c_2 \iff c_1=-c_2

    Now if x=1 we get

    0=c_1 e^{-(1+\sqrt{-\lambda})}+c_2 e^{-(1-\sqrt{-\lambda}) }

    0=-c_2 e^{-(1+\sqrt{-\lambda})}+c_2 e^{-(1-\sqrt{-\lambda}) }

    0=c_2 (-e^{-(1+\sqrt{-\lambda})}+e^{-(1-\sqrt{-\lambda})} )

    Since the exponential is a 1-1 they are not equal so c_2=0

    and we only get the trivial solution

    So finally we get to the last case

    \lambda > 0 we get

    y=c_1e^{-x}\cos(x\sqrt{-\lambda})+c_2e^{-x}\sin(x\sqrt{-\lambda})

    So when x=0 we get

    0=c_1 so the equation reduces to

    y=c_2e^{-x}\sin(x\sqrt{-\lambda})

    When x=1 we get


    0=c_2e^{-1}\sin(\sqrt{-\lambda}) \iff 0=\sin(\sqrt{-\lambda})

    Now we take the inverse sine of both sides to get

    \pi \cdot k =\sqrt{-\lambda} \mbox{ for }k \in \mathbb{Z}

    This gives

    \pi^2 k^2=-\lambda \mbox{ for } k \in \mathbb{Z}^+

    So your orthogonal basis is

    y_k=c_k\sin(\pi k x)
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