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Solution of Parametric Bessel's equation.

  1. Dec 26, 2009 #1
    Problem to solve: [tex]x^{2}y''+xy'+(\lambda^{2}x^{2}-\frac{1}{4})=0[/tex] subject to [tex]y(0)=finite,y(\pi)=0[/tex]
    Let [tex]t=\lambda x[/tex] which give [tex]t^{2}\frac{d^{2}y}{dt^{2}}+t\frac{dy}{dt}+(t^{2}-\frac{1}{2})y=0[/tex]

    With that first linear independent solution is [tex]J_{\frac{1}{2}}[\sqrt{\frac{2}{\pi\lambda x}}sin(\lambda x)][/tex] and second independent solution is [tex]J_{-\frac{1}{2}}[\sqrt{\frac{2}{\pi\lambda x}}sin(\lambda x)][/tex]

    Since these two are linear independent solution, I should be able to write the general solution of y:

    [tex]y=C_{1}J_{\frac{1}{2}}[\sqrt{\frac{2}{\pi\lambda x}}sin(\lambda x)] + C_{2}J_{-\frac{1}{2}}[\sqrt{\frac{2}{\pi\lambda x}}sin(\lambda x)][/tex]

    But the book claimed the solution of y:[tex]y=C_{1}J_{\frac{1}{2}}[\sqrt{\frac{2}{\pi\lambda x}}sin(\lambda x)] + C_{2}Y_{-\frac{1}{2}}[\sqrt{\frac{2}{\pi\lambda x}}sin(\lambda x)][/tex]

    Question is why do I have to use Y in the second solution instead of both J to solve the boundary value problem? I always learn that if you get the two independent solution, that would be the answers!!!
     
    Last edited: Dec 27, 2009
  2. jcsd
  3. Dec 27, 2009 #2
    I forgot to mention, I understand [tex]J_{-p}[/tex] is independent only if p is not an integer, you have to use Bessel function of second kind......Y as the second linearly independent solution. In that case, you are stuck with Y.

    But in here, you don't need to. I guess my ultimate question is why people even use Y when p is not an integer to make their life more complicated!!!
     
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