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Thread: 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
    $\displaystyle \frac{d^2y}{dt^2}+\frac{1}{t}\frac{dy}{dt}+4t^2y(t )=0$
    Use the substitution $\displaystyle x=t^2$ to find the general solution

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

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

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

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

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

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

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

    So the solution would be: $\displaystyle y(x)=C_1J_0+C_2Y_0$
    The first thing that is not clear to me is about $\displaystyle Y_{\nu}=\frac{\cos \nu\pi J_{\nu}-J_{-\nu}}{\sin \nu\pi}$, with $\displaystyle \nu=0$ I've got: $\displaystyle Y_{0}=\frac{\cos 0\pi J_0-J_{-0}}{\sin 0\pi}$ which gives 0/0. I know from the books that its well defined for $\displaystyle \nu=0$, 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; Jun 18th 2011 at 05:35 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 $\displaystyle \nu \to 0$ 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
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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
    $\displaystyle \frac{d^2y}{dt^2}+\frac{1}{t}\frac{dy}{dt}+4t^2y(t )=0$
    Use the substitution $\displaystyle x=t^2$ to find the general solution

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

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

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

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

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

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

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

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