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Homework Help: Please help with Legendre problem.

  1. Jan 21, 2010 #1
    1. The problem statement, all variables and given/known data

    Determine the IVP has bounded solution:
    Legendre equation:
    [tex](1-x^2)y''-2xy'+6y=0[/tex] ; y(0)=0, y'(0)=1

    2. Relevant equations

    [tex]P_2 (x)=\frac{1}{2}[3x^2 -1][/tex]

    [tex]Q_2 (x)=P_2 (x)\int \frac{dx}{[P_2 (x)]^2 (1-x^2)}[/tex]

    [tex]y'(x)=c_1 P'_2 (x) + c_2 {P'_2(x)\int \frac{dx}{[P_2 (x)]^2 (1-x^2)}+c_2 P_2 (x)\frac{1}{[P_2 (x)]^2 (1-x^2)}}[/tex]


    3. The attempt at a solution

    n(n+1)=6 give n=2

    General solution is [tex]y(x)=c_1 P_n (x)+c_2 Q_n (x)[/tex]

    [tex]P_2 (x)=\frac{1}{2}[3x^2 -1]\Rightarrow P_2 (0)=1[/tex]

    [tex]y(0)=0 \Rightarrow c_1 = 0 \Rightarrow y(x)=c_2 Q_2 (x)[/tex] only.

    [tex]P'_2 (x)=6x [/tex] therefore

    [tex]y'(x)=c_2 {P'_2(x)\int \frac{dx}{[P_2 (x)]^2 (1-x^2)}+P_2 (x)\frac{1}{[P_2 (x)]^2 (1-x^2)}}[/tex]

    [tex]y'(x)=c_2 [6x\int\frac{dx}{(3x^2 -1)^2 (1-x^2)}+(3x^2 -1)\frac{1}{(3x^2 -1)^2 (1-x^2)}][/tex]

    x=0, the first part of y'(x)=0 because anything multiply by 6x equal zero. Therefore:

    For x=0, [tex]y'(x)=c_2 [\frac{1}{(3x^2 -1)(1-x^2)}][/tex]

    [tex] y'(0)=-c_2[/tex]


    Therefor there is a bounded solution for the problem.

    The answer from the book was [tex]Q_2 [/tex] is unbounded on (-1,1) therefore the Legendre differential equation has no bounded solution from the given boundary condition.
    What did I do wrong?
     
    Last edited: Jan 22, 2010
  2. jcsd
  3. Jan 21, 2010 #2
    Anyone please?
     
  4. Jan 21, 2010 #3

    vela

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    The book meant bounded over the interval when it asked for a bounded solution.
     
  5. Jan 21, 2010 #4

    vela

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    What's so special about x=0?
     
  6. Jan 21, 2010 #5
    Because the boundary condition given y(0)=0 and y'(0)=1

    I claim [tex]Q_2 (0)[/tex] is bounded. There is a bounded solution at x=0.

    I am still not quite sure about the bound or unbound of Q really means.
     
    Last edited: Jan 22, 2010
  7. Jan 21, 2010 #6

    vela

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    The solution to a differential equation is a function, not the function evaluated at some point.
     
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