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Math Help - Bessel differencial equation

  1. #1
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    Smile Bessel differencial equation

    please please can someone hep me on this as i dont have a clue where to start. I know its a ,ong question but if someone coul do itthen would really give me a bse to understand the topic

    thanks loads
    Edgar

    Show that Jn(px) satisfies

    x^2y'' + xy' + (p^2x^2 - n^2)y = 0;

    and deduce

    [x d/dx Jn(px)] (Subscript)x + ((p^2)x -n^2/x) Jn(px) = 0:

    Show that the integral between l and 0 of xJn(px)Jn(qx) dx =

    l/(q^2 - p^2) [pJn(ql)J'n(pl) - qJn(pl)J'n(ql)] ;

    and, using l'Hopital's rule,

    the integral between l and 0 of
    xJn^2(px) dx =l^2/2 Jn'^2(pl) + 1 - (n^2/p^2l^2)Jn^2(pl)

    Deduce that, if l is such that Jn(pl), Jn(ql) are both zero (i.e. pl and ql are both
    roots of Jn(x)), then
    the integarl between l and 0 of
    xJn(px)Jn(qx) dx = 0; p doesnt equal q
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  2. #2
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    Quote Originally Posted by edgar davids View Post
    please please can someone hep me on this as i dont have a clue where to start. I know its a ,ong question but if someone coul do itthen would really give me a bse to understand the topic

    thanks loads
    Edgar

    Show that Jn(px) satisfies

    x^2y'' + xy' + (p^2x^2 - n^2)y = 0;
    What is "Jn(px)"?

    The equation,
    x^2y''+xy'+(p^2x^2-n^2)y=0
    Look like an analogue of the Bessel equation.
    I can show that there exists a solution satisfiying the Frobenius method by shows that these functions are Legendre polynomials of orders 1 and 2. If that is what you are asking.
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  3. #3
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    Jn(x) is a series solution to bessels equation
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  4. #4
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    Quote Originally Posted by edgar davids View Post
    Jn(x) is a series solution to bessels equation
    Okay.

    Given, the Bessel equation,
    x^2y''+xy'+(x^2-n^2)y=0
    The Bessel function of order n\,,
    J_n(x) satisfies the differencial equation.

    We need to show that,
    J_n(px) satisfies the differencial equation.
    x^2y''+xy'+(x^2p^2-n^2)y=0 (*)

    Since,
    x^2J''_n(x)+xJ'_n(x)+(x^2-n^2)J_n(x)=0
    We have that,
    (px)^2J''_n(px)+(px)J'_n(px)+((px)^2-n^2)J_n(px)=0
    \boxed{p^2x^2J''_n(px)+pxJ'_n(px)+(p^2x^2-n^2)J_n(px)=0}

    With the boxed equation we can show that y=J_n(px) satisfies the differencial equation (*).

    Note that,
    y=J_n(px)
    y'=pJ'_n(px)
    y''=p^2J''_n(px)

    Substitute that into (*) to get,
    p^2x^2J''_n(px)+pxJ'_n(px)+(x^2p^2-n^2)J_n(px)
    And this expression is equal to zero by the boxed equation.
    Hence J_n(px) satisfies (*).
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