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Laplace Transform of Bessel Diff Eq

  1. Mar 25, 2010 #1
    Hello PF, maybe you can help with this one!

    I need to show that the Laplace transform of J0(at) is (s^2 + a^2)^-1/2.

    My prof told me to start with the form:
    x2y'' + x y' + (x2 + p2)y = 0, where p = 0 ITC.

    What have I got so far?....

    Doing the Laplace transform on both sides, where Y stands for L[y]:
    x2(s2Y - sy(0) - y'(0)) + x(sY - y(0)) + x2Y = 0,

    ...then 3 lines of algebra...

    Y = y(0)* (1 + xs)/(xs2 + s + x) + y'(0)* x/(xs2 + s + x).

    Do I convolve these two terms on the RHS now? Do I factor the denominator first? Or do I use some recurrence relns for Bessel functions, which is what y(x) is?

    Your help would be much appreciated!!
  2. jcsd
  3. Apr 9, 2010 #2
    Something is not right here. I think.

    [tex]L\{x^2y' ' \}=\frac{d^2}{ds^2} (s^2Y(s)- sy(0) - y'(0))[/tex]

    http://en.wikipedia.org/wiki/Laplace_transform" [Broken]
    Last edited by a moderator: May 4, 2017
  4. Apr 9, 2010 #3
    I think it is easier to start with the integral representation

    [tex]J_{0}(x) =\frac{1}{2\pi}\int_{0}^{2\pi}\exp\left[i x\cos\left(\right\theta)\right]d\theta[/tex]

    You can then directly computethe Laplace transorm by multiplying this by exp(-sx) and then integrate over x from zero to infinity after reversing the order of integration.

    It also not difficult to apply the inverse Laplace transform formula

    [tex]f(x) = \frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\hat{f}(z)\exp(zx)dz[/tex]

    to the formula (s^2 + a^2)^-1/2 and then show that you get the Bessel function.
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