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Bessel equation

  1. Jun 18, 2011 #1
    Hi there. I'm working with the Bessel equation, and I have this problem. It says:
    a) Given the equation
    [tex]\frac{d^2y}{dt^2}+\frac{1}{t}\frac{dy}{dt}+4t^2y(t)=0[/tex]
    Use the substitution [tex]x=t^2[/tex] to find the general solution

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

    Well, so what I did is:
    [tex]x=t^2 \rightarrow t=\sqrt{x}[/tex]
    [tex]\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}=\frac{dy}{dx}2t[/tex]
    [tex]\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}[/tex]

    Then [tex]\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[/tex]

    Now I'm not pretty sure what I should do to solve this. I thought of using Frobenius, would that be right?
     
  2. jcsd
  3. Jun 18, 2011 #2

    LCKurtz

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    Have you solved the Bessel equation

    x2y'' + xy' +(x2-n2)y =0

    in class, with its solutions Jn(x) and Yn(x)? If you multilply your equation by x2 it looks like that with n = 0.

    If so, that would save you some work. Otherwise, yes, multiply it by x2 and do a series solution.
     
  4. Jun 18, 2011 #3
    Yes, you're right. Thanks.
     
  5. Jun 18, 2011 #4
    I have that one solution would be [tex]J_{\nu},\nu=0[/tex] but I'm not to sure about the other linear independent solution, cause if I use [tex]Y_{\nu}=\frac{\cos \nu\pi J_{\nu}-J_{-\nu}}{\sin \nu\pi}[/tex] I got that [tex]Y_{0}=\frac{\cos 0\pi J_0-J_{-0}}{\sin 0\pi}[/tex] which is not defined, right?

    I'm sorry, is the first time I'm working with this, probably I'm committing a really stupid mistake.

    It still no clear how is that [tex]Y_0[/tex] is well defined, but anyway, I've accepted that it is, and tried to go on. But now the problem asks me to evaluate my solution, which is: [tex]y(x)=C_1J_0+C_2Y_0[/tex] in zero, which is: [tex]y(0)=C_1J_0(0)+C_2Y_0(0)[/tex] to verify the condition, and the thing is that the function [tex]Y_{\nu}[/tex] goes to -infinity in x=0, right? how should I proceed?
     
    Last edited: Jun 18, 2011
  6. Jun 18, 2011 #5

    LCKurtz

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    Your boundary condition that y(0) be finite tells you that C2 must be zero because, as you have noted, Y0(x) blows up at x = 0.

    Don't forget that x = t2 when you back substitute. I think if you look at the series for J0 you will see how to pick C1 to satisfy the first boundary condition. Then use the series to answer the second question.
     
  7. Jun 18, 2011 #6
    Thanks :)
     
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