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Math Help - eigenvalues and eigenfunctions

  1. #1
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    eigenvalues and eigenfunctions

    Can someone help me with this?

    How do I find the eigenvalues and eigenfunctions for the Sturm-Liouville problem
    x^2y''+xy'+3y=uy;
    y(1)=0, y(2)=0

    Thanks in advance.
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  2. #2
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    Quote Originally Posted by PvtBillPilgrim View Post
    Can someone help me with this?

    How do I find the eigenvalues and eigenfunctions for the Sturm-Liouville problem
    x^2y''+xy'+3y=uy;
    y(1)=0, y(2)=0

    Thanks in advance.
    That equation is the same as x^2y"+ xy'+ (3-u)y= 0, an "Euler type" or "equipotential" equation. The change of variable t= ln(x) changes it to an equation with constant coefficients. You should be able to find the general solution. Obviously, y(x)= 0, for all x, satifies the equation as well as y(1)= y(2)= 0. The eigenvalues are the values of u for which there are other, non-trivial, solutions. And the eigenfunctions are the non-trivial solutions.
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    Quote Originally Posted by HallsofIvy View Post
    That equation is the same as x^2y"+ xy'+ (3-u)y= 0,(**) an "Euler type" or "equipotential" equation. The change of variable t= ln(x) changes it to an equation with constant coefficients. You should be able to find the general solution. Obviously, y(x)= 0, for all x, satifies the equation as well as y(1)= y(2)= 0. The eigenvalues are the values of u for which there are other, non-trivial, solutions. And the eigenfunctions are the non-trivial solutions.
    You might also want to try y = x^m in HallsofIvy equation (**) and look at the separate cases for m \;(> 0, = 0, < 0).
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    Quote Originally Posted by danny arrigo View Post
    You might also want to try y = x^m in HallsofIvy equation (**) and look at the separate cases for m \;(> 0, = 0, < 0).
    The problem with that is that in order that the boundary conditions to be satisfied your m must be complex! I think most people, who have studied d.e.s with constant coefficients will be more comfortable with e^{(a+bi)t}= e^{at}(cos(bt)+ i sin(bt)) than with x^{a+bi}.
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    Quote Originally Posted by HallsofIvy View Post
    The problem with that is that in order that the boundary conditions to be satisfied your m must be complex! I think most people, who have studied d.e.s with constant coefficients will be more comfortable with e^{(a+bi)t}= e^{at}(cos(bt)+ i sin(bt)) than with x^{a+bi}.
    In my experience, I find students have more of a problem transforming the DE under x = \ln t, than using y = x^m. Once the basic solutions are derived x^a \sin b \ln x and x^a \cos b \ln x, I find the students find these easier to use.
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    Hey I'm trying to answer the exact same problem for homework and am really confused..I get as far as;

     y = Ax^{ \sqrt{(3 - u)}} + Bx^{-\sqrt{(3 - u)}}

    But then I'm not sure what to do - I can see I'm going to get different results if u >3, <3, =3, but I don't know how to put that into an overall answer... at what point should I be using the boundary conditions?

    Also, the 2nd part of the question asks me to expand a hypothetical f(x) in terms of the eigenfunctions - I assume in the format  f(x) = \Sigma{(Ck)Yk} where Ck are the sturm-louiville coefficients. But how can I do this if the format of Yk changes depending on whether u is > or < 3? Confused... :-s
    Last edited by Squiggles; March 29th 2009 at 02:42 PM.
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    Quote Originally Posted by Squiggles View Post
    Hey I'm trying to answer the exact same problem for homework and am really confused..I get as far as;

     y = Ax^{ \sqrt{(3 - u)}} + Bx^{-\sqrt{(3 - u)}}

    But then I'm not sure what to do - I can see I'm going to get different results if u >3, <3, =3, but I don't know how to put that into an overall answer... at what point should I be using the boundary conditions?
    The best thing to do is consider the following three cases:

     u = 3 + \omega^2,\;\; u = 3,\;\; u = 3 - \omega^2. That corresponds to your three cases.

    Cases 1: u = 3 + \omega^2 then

    y = A x^{\omega} + B x^{-\omega}

    Now use your BC's.

    y(1) = A 1^{\omega} + B 1^{-\omega} = 0
    y(2) = A 2^{\omega} + B 2^{-\omega} = 0

    Two equations for the two unknowns A\;\; \text{and}\;\;B. The first says B = -A whereas the second
     A 2^{\omega} -A 2^{-\omega} = 0 so  A \left(2^{\omega} - 2^{-\omega}\right) = 0 which gives A=0 leading to the trival solution y = 0. Similarly for the second case. It's the third case that's interesting. You try now.
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  8. #8
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    Ah thank you, that makes sense! (I was trying to put all the possible solutions together before using the boundary conditions, d'oh! )
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