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Math Help - Sturm Liouville problem

  1. #1
    ux0
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    Sturm Liouville problem

    Show that \pi^2 is an eigenvalue of the Sturm- Liouville problem

    X'' + \lambda X=0

    X(0) - X'(0)=0

    \pi^2X(1/2)+X'(1/2)=0

    and find a corresponding eigenfunction.



    To start i posed that the solution of this problem will be in the form

    <br />
X(x)= A \sin{x\sqrt{\lambda}}+B\cos{x\sqrt{\lambda}}

    X'(x)=A \sqrt{\lambda} \cos{x\sqrt{\lambda}}-B \sqrt{\lambda}\sin{x \sqrt{\lambda}}

    using the end point conditions i get

    X(0)-X'(0)=A \sin{0\sqrt{\lambda}}+B\cos{0\sqrt{\lambda}}- A \sqrt{\lambda} \cos{0\sqrt{\lambda}}+B \sqrt{\lambda}\sin{0 \sqrt{\lambda}}=0

    this clearly gives us...

    B-A\lambda = 0

    clearly If A=B=0 this would give us a dummy answer so i toss it, and say that first this equation to be true,

    B=\sqrt{\lambda}

    A = 1

    Where i am stuck is when i use the second endpoint condition i get...


    <br />
\pi^2A\sin{\frac{\sqrt{\lambda}}{2}}+\pi^2B\cos{\f  rac{\sqrt{\lambda}}{2}}+A\lambda\cos{\frac{\sqrt{\  lambda}}{2}}-B\lambda\sin{\frac{\sqrt{\lambda}}{2}}=0

    with substitution + algebra putting the sin on one side and the cos on the other we get...


    <br />
(B-\pi^2A)\sin{\frac{\sqrt{\lambda}}{2}}=(\pi^2B+A)\c  os{\frac{\sqrt{\lambda}}{2}}


    divide both sides by cos.....


    <br />
\frac{\sin{\frac{\sqrt{\lambda}}{2}}}{\cos{\frac{\  sqrt{\lambda}}{2}}}=\frac{(\pi^2B+A)}{(B-\pi^2A)}


    substitute,

    <br />
 b= \sqrt{\lambda}

    <br />
 A = 1


    implies



    \frac{\sin{\frac{\pi}{2}}}{\cos{\frac{\pi}{2}}}=\f  rac{\pi^3 +1}<br />
{\pi - \pi^2}



    SO how do i use this to prove \pi^2 is an eigenvalue, then how do i find my function.... please help
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  2. #2
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    Quote Originally Posted by ux0 View Post
    Show that \pi^2 is an eigenvalue of the Sturm- Liouville problem

    X'' + \lambda X=0

    X(0) - X'(0)=0

    \pi^2X(1/2)+X'(1/2)=0

    and find a corresponding eigenfunction.
    To verify that \pi^2 is an eigenvalue, you just have to put \lambda = \pi^2 in the equation, and show that it then has a nonzero solution. So let X(x) = A\sin\pi x + B\cos\pi x and see what the initial conditions X(0) - X'(0)=0 and \pi^2X(1/2)+X'(1/2)=0 then tell you about A and B.
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  3. #3
    ux0
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    once again, doing it with this method, it tells me that

    A=1

    B=\sqrt{\lambda}=\pi

    But in either case

    initial condition a)

    <br />
B-A\pi=0

    initial condition b)

    <br />
A\pi^2-B\pi=0


    they both equal zero, does this mean that  \pi^2 is an eigenvalue of the Sturm-L-Problem?

    if so... how do i find the eigen function?
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  4. #4
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    Quote Originally Posted by ux0 View Post
    ... how do i find the eigen function?
    You have already found it! It is the function A\sin\pi x + B\cos\pi x with A=1 and B=π, namely f(x) = \sin\pi x + \pi\cos\pi x.
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