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Math Help - Bessel equation

  1. #1
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    Bessel equation

    Hi there. I'm working with the Bessel equation, and I have this problem. It says:
    a) Given the equation
    \frac{d^2y}{dt^2}+\frac{1}{t}\frac{dy}{dt}+4t^2y(t  )=0
    Use the substitution x=t^2 to find the general solution

    b) Find the particular solution that verifies y(0)=5
    c) Does any solution accomplish y'(0)=2? Justify.

    Well, so what I did is:
    x=t^2 \rightarrow t=\sqrt{x}

    \frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}=\frac{dy}  {dx}2t

    \frac{d^2y}{dt^2}=\frac{d^2y}{dx^2}4t^2+2\frac{dy}  {dx}=4x\frac{d^2y}{dx^2}+2\frac{dy}{dx}

    Then \frac{d^2y}{dt^2}+\frac{1}{t}\frac{dy}{dt}+4t^2y(t  )=\frac{d^2y}{dx^2}+\frac{1}{x}\frac{dy}{dx}+y(x)=  0

    Now I'm not pretty sure what I should do to solve this.

    Multiplying by x^2 I get the Bessel equation of order zero.
    \frac{d^2y}{dx^2}x^2+x\frac{dy}{dx}+x^2y(x)=0

    So the solution would be: y(x)=C_1J_0+C_2Y_0
    The first thing that is not clear to me is about Y_{\nu}=\frac{\cos \nu\pi J_{\nu}-J_{-\nu}}{\sin \nu\pi}, with \nu=0 I've got: Y_{0}=\frac{\cos 0\pi J_0-J_{-0}}{\sin 0\pi} which gives 0/0. I know from the books that its well defined for \nu=0, so I just accepted this fact. But now, in incise b) I have to evaluate y(0), which is not defined.
    Last edited by Ulysses; June 18th 2011 at 06:35 PM.
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  2. #2
    Member Ruun's Avatar
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    Re: Bessel equation

    I don't know how derive it, maybe using L'H˘pital's rule as \nu \to 0 but they are given by the last expression in this web Bessel Function of the Second Kind -- from Wolfram MathWorld
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  3. #3
    MHF Contributor
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    Re: Bessel equation

    Quote Originally Posted by Ulysses View Post
    Hi there. I'm working with the Bessel equation, and I have this problem. It says:
    a) Given the equation
    \frac{d^2y}{dt^2}+\frac{1}{t}\frac{dy}{dt}+4t^2y(t  )=0
    Use the substitution x=t^2 to find the general solution

    b) Find the particular solution that verifies y(0)=5
    c) Does any solution accomplish y'(0)=2? Justify.

    Well, so what I did is:
    x=t^2 \rightarrow t=\sqrt{x}

    \frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}=\frac{dy}  {dx}2t

    \frac{d^2y}{dt^2}=\frac{d^2y}{dx^2}4t^2+2\frac{dy}  {dx}=4x\frac{d^2y}{dx^2}+2\frac{dy}{dx}

    Then \frac{d^2y}{dt^2}+\frac{1}{t}\frac{dy}{dt}+4t^2y(t  )=\frac{d^2y}{dx^2}+\frac{1}{x}\frac{dy}{dx}+y(x)=  0

    Now I'm not pretty sure what I should do to solve this.

    Multiplying by x^2 I get the Bessel equation of order zero.
    \frac{d^2y}{dx^2}x^2+x\frac{dy}{dx}+x^2y(x)=0

    So the solution would be: y(x)=C_1J_0+C_2Y_0
    The first thing that is not clear to me is about Y_{\nu}=\frac{\cos \nu\pi J_{\nu}-J_{-\nu}}{\sin \nu\pi}, with \nu=0 I've got: Y_{0}=\frac{\cos 0\pi J_0-J_{-0}}{\sin 0\pi} which gives 0/0. I know from the books that its well defined for \nu=0, so I just accepted this fact. But now, in incise b) I have to evaluate y(0), which is not defined.
    Since you want y(0) and Y_0(0) is undefined (it \to -\infty as x \to 0) you will need to choos  c_2 = 0. Then pick c_1 as to make y(0) = 5.
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